Category: OP Malhotra Solutions

Parents can use Class 10 ICSE Maths Solutions S Chand Chapter 15 Three Dimensional Solids Ex 15(d) to provide additional support to their children.

S Chand Class 10 ICSE Maths Solutions Chapter 15 Three Dimensional Solids Ex 15(d)

Question 1.
Find the volume and curved surface of the sphere if \(\left(\pi=\frac{22}{7}\right)\)
(i) radius = 1 cm
(ii) radius = 14 cm
(iii) radius = 28 cm
(iv) radius = 5\(\frac { 1 }{ 4 }\) cm
(v) diameter = 14 cm
(vi) diameter = 35 cm
Solution:
(i) Radius of sphere (r) = 1 cm
∴ volume = \(\frac { 4 }{ 3 }\)πr³ = \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 1 × 1 × 1 cm³ = \(\frac { 88 }{ 21 }\)cm³
and curved surface area = 4πr²
= 4 × \(\frac { 22 }{ 7 }\) × 1 × 1 = \(\frac { 88 }{ 7 }\) cm²

(ii) Radius (r) = 14 cm
∴ volume = \(\frac { 4 }{ 3 }\)πr³
= \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 14 × 14 × 14 cm³
= \(\frac { 34496 }{ 3 }\) = 11498\(\frac { 2 }{ 3 }\) cm³
curved surface area = 4πr²
= 4 × \(\frac { 22 }{ 7 }\) × 14 × 14 cm² = 2464 cm²

(iii) Radius (r) = 28 cm
volume = \(\frac { 4 }{ 3 }\)πr³ = \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 28 × 28 × 28 cm³
= \(\frac { 275968 }{ 3 }\) = 91989\(\frac { 1 }{ 3 }\) cm³
curved surface area = 4πr²
= 4 × \(\frac { 22 }{ 7 }\) × 28 × 28 cm³ = 9856 cm²

(iv) Radius (r) = 5\(\frac { 1 }{ 4 }\) cm = \(\frac { 21 }{ 4 }\) cm
∴ volume = \(\frac { 4 }{ 3 }\)πr³
= \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × \(\frac { 21 }{ 4 }\) × \(\frac { 21 }{ 4 }\) × \(\frac { 21 }{ 4 }\) cm³ = \(\frac { 4851 }{ 8 }\) = 606\(\frac { 3 }{ 8 }\) cm³
curved surface area = 4πr²
= 4 × \(\frac { 22 }{ 7 }\) × \(\frac { 21 }{ 4 }\) × \(\frac { 21 }{ 4 }\) cm²
= \(\frac { 693 }{ 2 }\) = 346\(\frac { 1 }{ 2 }\) cm²

(v) Diameter = 14 cm
∴ Radius (r) = \(\frac { 14 }{ 2 }\) = 7 cm
∴ volume = \(\frac { 4 }{ 3 }\)πr³ = \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 7 × 7 × 7 cm³
= \(\frac { 4312 }{ 3 }\) = 1437\(\frac { 1 }{ 3 }\) cm³ = 1437.3 cm³
curved surface area = 4πr²
= 4 × \(\frac { 22 }{ 7 }\) × 7 × 7 cm² = 616 cm²

(vi) Diameter = 35 cm
∴ Radius (r) = \(\frac { 35 }{ 2 }\) cm
∴ volume = \(\frac { 4 }{ 3 }\) πr³
= \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × \(\frac { 35 }{ 2 }\) × \(\frac { 35 }{ 2 }\) × \(\frac { 35 }{ 2 }\) cm³
= \(\frac { 67375 }{ 3 }\) = 22458\(\frac { 1 }{ 3 }\) cm³
and curved surface are = 4πr²
=4 × \(\frac { 22 }{ 7 }\) × × \(\frac { 35 }{ 2 }\) × × \(\frac { 35 }{ 2 }\) = 3850 cm²

Question 2.
Find the radius of the sphere whose surface area is equal to its volume.
Solution:
Let r be the radius of a sphere, then
Surface area = 4πr²
and volume = \(\frac { 4 }{ 3 }\)πr³
∵ Surface area = volune
∴ 4πr² = \(\frac { 4 }{ 3 }\)πr³
1 = \(\frac { 1 }{ 3 }\)r ⇒ r = 3
∴ Radius = 3 units

Question 3.
Find the diameter of the sphere whose surface area is equal to the area of a circle of diameter 2.8 cm.
Solution:
Let r be the radius of sphere = r
∴ Surface area = 4πr²
Diameter of a circle = 2.8 cm
Radius = 2.8 = 1.4 cm
∴ Area of the circle = πr² = π × 1.4 × 1.4 = 1.96 π
∵ The surface area of sphere = area of the circle
∴ 4πr² = 1.96π
r² = \(\frac{1.96 \pi}{4 \pi}\) = 0.49 = (0.7)²
∴ r = 0.7 cm
∴ Diameter of sphere = 2r = 2 × 0.7 = 1.4 cm

Question 4.
Find the radius of the sphere whose volume is \(\frac { 32 }{ 3 }\)π cm³.
Solution:
Volume of sphere = \(\frac { 32 }{ 3 }\)π cm³
Let r be the radius of the sphere, then
\(\frac { 4 }{ 3 }\)πr³ = \(\frac { 32 }{ 3 }\)π
r³ = \(\frac{32 \pi \times 3}{3 \times 4 \times \pi}\) = 8 = (2)³
∴ r = 2 cm
∴ Radius of the sphere = 2 cm

Question 5.
The volume of a sphere is 4.851 cm³. Find its surface area.
Solution:
Let r be the radius of the sphere
∴ Volume = \(\frac { 4 }{ 3 }\)πr³ = 4.851
∴ \(\frac { 4 }{ 3 }\)πr³ = 4.851
\(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\)r³ = 4.851 ⇒ r³ = \(\frac{4.851 \times 3 \times 7}{4 \times 22}\)
⇒ r³ = 1.157625 = (1.05)³
∴ r = 1.05 cm
Now surface area = 4πr²
= \(\frac{4 \times 22}{7}\) × (1.05)² cm² = 13.86 cm²

Question 6.
Find the volume of a hollow sphere whose outer diameter is 10 cm and the thickness of the material of which it is made is 1 cm.
Solution:
Outer diameter of hollow sphere = 10 cm
∴ Outer radius (R) = \(\frac { 10 }{ 2 }\) = 5 cm
Thickness of material = 1 cm
∴ Inner radius (r) = 5 – 1 = 4 cm
∴ Volume of hollow sphere = \(\frac { 4 }{ 3 }\)π [R³ – r³]
= \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) [(5)³ – (4)³]
= \(\frac { 88 }{ 21 }\) [125 – 64] = \(\frac { 88 }{ 21 }\) × 61 cm³
= 255.619 = 255.6 cm³

Question 7.
The surface areas of two spheres are in the ratio 4 : 9. Find the ratio of their volumes.
Solution:
Let r₁ and r₂ be the radii of two spheres
Then surface area of the first sphere = 4πr₁²
and surface area of the second = 4πr₂²
∴ 4πr₁² : 4πr₂² = 4 : 9
⇒ r₁² : r₂² = 4 : 9
⇒ \(\frac{r_1^2}{r_2^2}\) = \(\frac { 4 }{ 9 }\) = \(\left(\frac{r_1}{r_2}\right)^2\) = \(\left(\frac{2}{3}\right)^2\)
∴ \(\frac{r_1}{r_2}\) = \(\frac { 2 }{ 3 }\)
Now volume of the first square = \(\frac { 4 }{ 3 }\)πr₁³
and volume of the second sphere = \(\frac { 4 }{ 3 }\)πr₂³

∴ Ratio in the volumes of two spheres will be 8 : 27

Question 8.
The volumes of two spheres are in the ratio 64 : 27. Find their radii if the sum 4 of their radii is 21 cm.
Solution:
Let r₁ and r₂ be the radii of two spheres
Then volume of first sphere = \(\frac { 4 }{ 3 }\)πr₁³
and volume of second sphere = \(\frac { 4 }{ 3 }\)πr₂³

∴ Their radii are 12 cm and 9 cm

Question 9.
Find the cost of painting a hemispherical dome of diameter 10 m at the rate of Rs. 1.40 per m².
Solution:
Diameter of hemisphere dome = 10 m
∴ Radius (r) = \(\frac { 10 }{ 2 }\) = 5 m
Now curved surface area = 2πr
= 2 × \(\frac { 22 }{ 7 }\) × 5 × 5 = \(\frac { 1100 }{ 7 }\) m²
Rate of painting = Rs. 1.40 per m²
∴ Total cost of painting = \(\frac { 1100 }{ 7 }\) × 1.40
= \(\frac{1100 \times 140}{7 \times 100}\) = Rs. 220

Question 10.
A certain spherical ball of diameter 4 cm weighs 8 kg. Find the weight of a spherical ball of the same material whose inner and outer diameters are 8 cm and 10 cm respectively.
Solution:
Diameter of a ball = 4 cm
∴ Radius (r) = \(\frac { 4 }{ 2 }\) = 2 cm
and volume = \(\frac { 4 }{ 3 }\)πr³
= \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 2 × 2 × 2 cm³ = \(\frac { 704 }{ 21 }\)cm³
Weight of ball = 8 kg
∴ weight of 1 cm³ = \(\frac{8 \times 21}{704}\) kg
= \(\frac{8 \times 21 \times 1000}{704}\) gm
= \(\frac{21 \times 1000}{88}\) = \(\frac{21 \times 125}{11}\) gm = \(\frac{2625}{11}\) gm
Now diameters of spherical ball are 8 cm and 10 cm
∴ Their radii will be 4 cm and 5 cm
i.e., R = 5 cm and r = 4 cm
∴ Volume of spherical ball
= \(\frac { 4 }{ 3 }\)π[R³ – r³] = \(\frac { 4 }{ 3 }\) × \(\frac{22}{7}\) [(5)³ – (4)³]
= \(\frac { 88 }{ 21 }\) [125 – 64]cm³ = \(\frac { 88 }{ 21 }\) × 61 cm³
∴ weight of the ball = \(\frac{88 \times 61}{21}\) × \(\frac{2625}{11}\)
= 61000 gm = \(\frac{61000}{1000}\) kg = 61 kg

Question 11.
(i) Prove that the surface area of a sphere of diameter d is πd² and the volume is \(\frac { 1 }{ 6 }\)πd³.
(ii) The volumes and diameters of a cone and sphere are equal. Prove that the height of the cone is twice the diameter of the sphere.
Solution:
(i) Diameter of a sphere = d
∴ Radius (r) = \(\frac { d }{ 2 }\)
∴ Surface area = 4πr² = 4π × \(\left(\frac{d}{2}\right)^2\)
= 4π × \(\frac{d^2}{4}\) = πd²
and volume = \(\frac { 4 }{ 2 }\)πr³ = \(\frac { 4 }{ 3 }\)π \(\left(\frac{d}{2}\right)^3\)
= \(\frac { 4 }{ 3 }\)π × \(\frac{d^3}{8}\) = \(\frac { 1 }{ 6 }\)πd³

(ii) Let diameter of cone and sphere be 2r
∴ Radius of each one = r
Now volume of cone = \(\frac { 1 }{ 3 }\)πr²h
and volume of sphere = \(\frac { 4 }{ 3 }\)πr³
∵ Their volume are equal
∴ \(\frac { 1 }{ 3 }\)πr²h = \(\frac { 4 }{ 3 }\)πr³
⇒ h = 4 r = 2(2r) = 2 × diameter
Hence height of cone = twice of the diameter of sphere
Hence proved.

Question 12.
The external diameter of a hollow metal sphere is 12 cm, and its thickness is 2 cm. Find the radius of a solid sphere made of the same material and having the same weight as the hollow sphere.
Solution:
External diameter of a hollow sphere = 12 cm
∴ Radius (r) = \(\frac { 12 }{ 2 }\) = 6 cm
Thickness of the metal used = 2 cm
∴ Internal radius = 6 – 2 = 4 cm
∴ Volume of the metal = \(\frac { 4 }{ 3 }\)π [R³ – r³]
= \(\frac { 4 }{ 3 }\)π [(6)³ – (4)³] cm³
= \(\frac { 4 }{ 3 }\)π (216 – 64) cm³
= \(\frac { 4 }{ 3 }\)π (512) cm³
= \(\frac { 608 }{ 3 }\) π cm³

Now the volume of solid sphere = \(\frac { 608 }{ 3 }\) π cm³
Let radius of the sphere = x
Then \(\frac { 4 }{ 3 }\) π x³ = \(\frac { 608 }{ 3 }\) π
x³ = \(\frac{608 \pi \times 3}{3 \times 4 \pi}\) = 152
∴ x = \(\sqrt[3]{152}\) = \((152)^{\frac{1}{3}}\) = 5.3368 = 5.34
∴ Radius of solid sphere = 5.34 cm

Question 13.
A hollow copper ball has an external diameter of 12 cm, and a thickness of 0.1 cm. Find
(i) the outer surface area of the ball;
(ii) the weight of the ball if 1 cm³ of copper weighs 8.88 g. (Take π to be 3.14).
Solution:
External diameter of a hollow spherical ball = 12 cm

∴ Outer radius (R) = \(\frac { 12 }{ 6 }\) = 6 cm
Thickness of the metal used = 0.1 cm
∴ Internal radius (r) = 6 – 0.1 = 5.9 cm

(i) Outer surface area = 4πR²
= 4 × 3.14 × (6)² cm²
= 4 × 3.14 × 36 = 452.16 cm²

(ii) Volume of metal used = \(\frac { 4 }{ 3 }\)π[R³ – r³]
= \(\frac { 4 }{ 3 }\) × (3.14) × [(6)³ – (5.9)³] cm³
= \(\frac { 4 }{ 3 }\) × (3.14) × [216 – 205.379] cm³
= \(\frac { 4 }{ 3 }\) × (3.14) × (10.621) cm²
= 44.467 cm³
weight of 1 cm³ = 8.88 cm
∴ Total weight of the ball = 44.467 × 8.88 gm = 394.87 gm

Question 14.
Marbles of diameter 1.4 cm, are dropped into a beaker containing some water and are fully submerged. The diameter of the beaker is 7 cm. Find how many marbles have been dropped in it if the water rises by 5.6 cm. \(\left(\pi=\frac{22}{7}\right)\)
Solution:
Diameter of each marble = 1.4 cm
∴ Radius of each marble (r) = \(\frac { 1.4 }{ 2 }\) = 0.7 cm
Volume of one marble = \(\frac { 4 }{ 3 }\)πr³
= \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × (0.7)³ = \(\frac { 88 }{ 21 }\) × 0.343 cm³
= 1.437333 cm³
Diameter of beaker = 7 cm
∴ Radius of the base of beaker (r₁) = \(\frac { 7 }{ 2 }\) cm
Height of water level = 5.6 cm
∴ Volume of water = πr²h
= \(\frac { 22 }{ 7 }\) × \(\frac { 7 }{ 2 }\) × \(\frac { 7 }{ 2 }\) × 5.6 cm³ = 215.6 cm³

Question 15.
A cylindrical tub of radius 5 cm and length 9.8 cm is full of water. A solid in the form of a right circular cone mounted on a hemisphere is immersed into the tub. If the radius of the hemisphere is 3.5 cm and height of the cone outside the hemisphere is 5 cm, find the volume of water left in the tub. (Take π = 22/7)
Solution:
Radius of tub (r) = 5 cm and height (length) (h) = 9.8 cm

∴ Volume of water filled in it = πr²h
= \(\frac { 22 }{ 7 }\) × 5 × 5 × 9.8 cm³ = 770 cm³
In the solid,
Radius of the hemisphere (r₁) = 3.5 cm and height of cone (h₁) = 5 cm
∴ Volume of solid = \(\frac { 2 }{ 3 }\)πr₁³ + \(\frac { 1 }{ 3 }\)πr₁²h₁
= \(\frac { 2 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × (3.5)³ + \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × (3.5)² × 5
= \(\frac { 2 }{ 3 }\) × \(\frac { 22 }{ 7 }\) (3.5)² [2 (3.5) + 5]
= \(\frac { 22 }{ 21 }\) × 12.25 × [7 + 5] cm³
= \(\frac { 22 }{ 21 }\) × 12.35 × 12 cm³ = 154 cm³
∴ Water left in the tub = (770 – 154) cm3 = 616 cm3

Question 16.
A toy is in the form of a cone mounted on a hemisphere of radius 3.5 cm. The total height of the toy is 15.5 cm. Find the total surface area.
Solution:
Total height of the toy = 15.5 cm
Radius of hemispherical part = 3.5 cm
∴ Height of conical part (h)=15.5 – 3.5 = 12 cm

∴ Slant height (l) = \(\sqrt{r^2+h^2}\) = \(\sqrt{(3.5)^2+(12)^2}\) = \(\sqrt{12.25+144}\) = \(\sqrt{156.25}\) = 12.5 cm
Now total surface area of the toy
= πrl + 2πr² = πr(l + 2r)
= \(\frac { 22 }{ 7 }\) × 3.5 (12.5 + 2 × 3.5) cm²
= 11 (12.5 + 7) = 11 × 19.5 cm²
= 214.5 cm²

Question 17.
The largest sphere is carved out of a cube of side 7 cm. Find the volume of the sphere.
Solution:
Side of cube = 7 cm
∴ Diameter of the largest sphere which is carved out of it = 7 cm

∴ Radius of the sphere = \(\frac { 7 }{ 2 }\) cm
Now volume of sphere = \(\frac { 4 }{ 3 }\)πr³
= \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × \(\left(\frac{7}{2}\right)^3\) cm³
= \(\frac { 88 }{ 21 }\) × \(\frac { 343 }{ 8 }\) = \(\frac { 549 }{ 3 }\) = 179.666
= 179.7 cm³

Question 18.
A cylindrical bucket, whose base has a radius of 15 cm, is filled with water up to a height of 20 cm. A heavy iron spherical ball of a radius 10 cm is dropped to submerge completely in water in the bucket. Find the increase in the level of water.
Solution:
Radius of the cylindrical bucket = 15 cm and height of water level = 20 cm

∴ Volume of water in the bucket
= πr²h = \(\frac { 22 }{ 7 }\) × 15 × 15 × 20 cm³
= \(\frac { 99000 }{ 7 }\) cm³
Radius of spherical ball = 10 cm
∴ Volume of ball = \(\frac { 4 }{ 3 }\)πr³
= \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 10 × 10 × 10 cm³ = \(\frac { 88000 }{ 21 }\) cm³
Let the level of water arise upto h cm in the bucket
∴ Volume = πr²h
\(\frac { 88000 }{ 21 }\) = \(\frac { 22 }{ 7 }\) × 15 × 15h
⇒ \(\frac { 88000 }{ 21 }\) = \(\frac { 22 }{ 7 }\) × 225h
⇒ h = \(\frac { 88000 }{ 21 }\) × \(\frac{7}{22 \times 225}\)
= \(\frac { 160 }{ 27 }\) = 5\(\frac { 25 }{ 27 }\) cm
∴ Increase in water level = 5\(\frac { 25 }{ 27 }\) cm

Question 19.
The radius of the base of a cone and the radius of a sphere are the same, each being 8 cm. Given that the volumes of these two solids are also the same, calculate the slant height of the cone, correct to one place of decimal.
Solution:
Radius of the base of a cone (r) = 8 cm
and radius of the sphere (r) = 8 cm
Volume of cone = volume of sphere Let h be in height of the cone,
∴ \(\frac { 4 }{ 3 }\)πr³ = \(\frac { 1 }{ 3 }\)πr²h
h = \(\frac{4 \pi r^3 \times 3}{3 \times \pi r^2}\) = 4r = 4 × 8 = 32 cm
∴ Slant height (l) of cone = \(\sqrt{r^2+h^2}\) = \(\sqrt{(8)^2+(32)^2}\) = \(\sqrt{64+1024}\) cm = \(\sqrt{1088}\) cm = 32.98 cm = 33 cm

Question 20.
A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top which is open is 5 cm. It is filled with water up to the rim. When lead shots, each of which is a sphere of radius 0.5 cm, are dropped into the vessel, one- fourth of the water flows out. Find the number of lead shots dropped in the vessel. \(\left(\text { Take } \pi \text { to be } \frac{22}{7}\right)\)
Solution:
Radius of the conical vessal (r) = 5 cm
and height (h) = 8 cm
∴ Volume of water filled in it = \(\frac { 1 }{ 3 }\)πr²h
= \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 5 × 5 × 8 cm³
= \(\frac { 4400 }{ 21 }\) cm³

∴ volume of lead shots = \(\frac { 1 }{ 4 }\) of \(\frac { 4400 }{ 21 }\) = \(\frac { 1100 }{ 21 }\) cm³
Radius of each lead shot (r₁) = 0.5 cm
∴Volume of one lead shot = \(\frac { 4 }{ 3 }\)πr₁³
= \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 0.5 × 0.5 × 0.5
= \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × \(\frac { 1 }{ 2 }\) × \(\frac { 1 }{ 2 }\) × \(\frac { 1 }{ 2 }\) cm³ = \(\frac { 11 }{ 21 }\)cm³
∴ Number of lead shots = \(\frac { 1100 }{ 21 }\) ÷ \(\frac { 11 }{ 21 }\) = \(\frac { 1100 }{ 21 }\) × \(\frac { 21 }{ 11 }\) = 100

Question 21.
A buoy is made in the form of a hemisphere surmounted by a right cone whose circular base conicides with the plane surface of the hemisphere. The radius of the base of the cone is 3.5 metres and its volume is \(\frac { 2 }{ 3 }\) of the hemisphere. Calculate the height of the cone and the surface of the buoy correct to 2 places of decimals. \(\left(\text { Take } \pi=3 \frac{1}{7}\right)\)
Solution:
Radius of cone = 3.5 m
Volume of cone = \(\frac { 2 }{ 3 }\) of the volume of hemisphere

Let h be the height of the cone = h
volume of hemisphere = \(\frac { 2 }{ 3 }\)πr³
= \(\frac { 2 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × (3.5)³ m³
= \(\frac { 2 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × \(\frac { 7 }{ 2 }\) × \(\frac { 7 }{ 2 }\) × \(\frac { 7 }{ 2 }\) m³ = \(\frac { 539 }{ 6 }\) m³
∴ volume of cone = \(\frac { 2 }{ 3 }\) of volume of hemisphere
= \(\frac { 2 }{ 3 }\) × \(\frac { 539 }{ 6 }\) = \(\frac { 539 }{ 9 }\) m³
But volume of cone = \(\frac { 1 }{ 3 }\) πr²h
∴ \(\frac { 1 }{ 3 }\) πr²h = \(\frac { 539 }{ 9 }\)
⇒ \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × \(\frac { 7 }{ 2 }\) × \(\frac { 7 }{ 2 }\)h = \(\frac { 539 }{ 9 }\)
⇒ \(\frac { 77 }{ 6 }\)h = \(\frac { 539 }{ 9 }\)
⇒ h = \(\frac { 539 }{ 9 }\) × \(\frac { 6 }{ 77 }\)
⇒ h = \(\frac{7 \times 2}{3}\) = \(\frac { 14 }{ 3 }\) m = 4\(\frac { 2 }{ 3 }\) m
∴ Height of cone = 4\(\frac { 2 }{ 3 }\) m

(ii) slant height (l) = \(\sqrt{r^2+h^2}\)
= \(\sqrt{\left(\frac{7}{2}\right)^2+\left(\frac{14}{3}\right)^2}\) = \(\sqrt{\frac{49}{4}+\frac{196}{9}}\)
= \(\sqrt{\frac{441+784}{36}}\) = \(\sqrt{\frac{1225}{36}}\) = \(\frac{35}{6}\) m
∴ Surface area of the buoy = πrl + 2πr²
= \(\frac { 22 }{ 7 }\) × \(\frac { 7 }{ 2 }\) × \(\frac { 35 }{ 6 }\) + 2 × \(\frac { 22 }{ 7 }\) × \(\frac { 7 }{ 2 }\) × \(\frac { 7 }{ 2 }\) = \(\frac { 385 }{ 6 }\) + 77 = \(\frac{385+462}{6}\)
= \(\frac { 847 }{ 6 }\) = 141\(\frac { 1 }{ 6 }\) m²

Question 22.
A cylinder, whose height is equal to its diameter, has the same volume as a sphere of radius 4 cm. Calculate the radius of the base of the cylinder, correct to 1 decimal place.
Solution:
Height of a cylinder = its diameter
⇒ h = 2r
and volume of cylinder = volume of sphere Radius of sphere = 4 cm
∴ Volume of sphere = \(\frac { 4 }{ 3 }\)πr³
= \(\frac { 4 }{ 3 }\)π × 4 × 4 × 4 cm³ = \(\frac { 256 }{ 3 }\)π cm³
∴ Volume of cylinder = \(\frac{256 \pi}{3}\) cm³
But volume of cylinder = πr²h
= πr² × 2r = 2πr³
∴ 2πr³ = \(\frac { 256 }{ 3 }\)π
r³ = \(\frac{256 \pi}{3 \times 2 \pi}\) = \(\frac { 128 }{ 3 }\)
∴ r = \(\sqrt[3]{\frac{128}{3}}\) = \(\sqrt[3]{42.667}\) = 3.49 cm
∴ Radius = 3.49 cm = 3.5 cm

Interactive OP Malhotra Class 10 ICSE Solutions Chapter 10 Reflection Ex 10(b) engage students in active learning and exploration.

S Chand Class 10 ICSE Maths Solutions Chapter 10 Reflection Ex 10(b)

Question 1.
State the coordinates of
(a) Points (i) (5, 7), (ii) (3, -4), (iii) (-8, 9), (iv) (-1, -2) under reflection in the x-axis.
(b) Points (0 (2, 5), (ii) (1, -4), (iii) (-7, -13), (iv) (-6, 8) under reflection in’ the y-axis.
Solution:
(a) Points (i) (5, 7), (ii) (3, -4), (iii) (-8, 9), (iv) (-1, -2) under reflection in the x-axis will be (i) (5, -7), (ii) (3, 4), (iii) (-8, -9) and (iv) (-1, 2)

(b) Points (i) (2, 5), (ii) (1, -4), (iii) (-7, – 13), (iv) (-6, 8) under reflection in the y-axis will be (i) (-2, 5), (ii) (-1, -4), (iii) (7, -13) and (iv) (6, 8)

Question 2.
Parallelogram ABCD has vertices A (-3, 2), B (5, 2), C (7, -1), D (-1, -1). Determine the image A’, B’, C’, D’ of A, B, C, D respectively, under reflection in the x-axis.
Solution:
The vertices of a parallelogram ABCD are A (-3, 2) B (5, 2), C (7, -1), D (-1, -1)
The image of these points in the x-axis will be A’ (-3, -2), B’ (5, -2), C’ (7, 1) and D’ (-1, 1)

Question 3.
(i) Plot each of the given points on graph paper, reflect them in the x-axis and then reflect the points obtained in the y-axis. Write down the co-ordinates of the points obtained.
(a) (3, 4)
(b) (-3, 2)
(c) (5, 0) (d) (-3, -3)
(ii) What happens if you reflect first in the y-axis and then in the x-axis ?
Solution:
(a) Reflection of the point A (3, 4) in x-axis will be (3, -4) and then in y-axis, will be (-3, -4).

(b) Similarly the refection of B (-3, 2) in x-axis will be (-3, 2) and then in y-axis will be (3, 2).

(c) Reflection of C (5, 0) in x-axis will be (5,0) and then in y-axis will be (-5, 0).

(d) Reflection of D (-3, -3) in x-axis will be (-3, 3) and then in y-axis will be (3, 3)
Now (a) Reflection of A (3, 4) in y-axis will be (-3, 4) and then in x-axis, it will be (-3, -4) x’ which is same as in above (a).
Similarly we see that the reflection of each point first in y-axis and then in x-axis, the- result will same as in above.
So, we get the same points.

Question 4.
A point P (a, b) is reflected in the y-axis to
P’ (-3, 5). Write down the values of a and b.
P” is the image of P, when reflected in the x-axis. Write down the coordinates of P”. Find the coordinates of P'”, when P is reflected in the line, parallel to the x-axis, such that its equation is y = -3.
Solution:
A point P (a, b) is reflected in y-axis, the image is P’ (- a, b) but the reflection P’ is given (- 3, 5)
∴ Comparing, we get – a = – 3 ⇒ a = 3
and b = 5, then coordinates of P will be (3, 5)
∵ P” is the image of P when reflected in x-aixs the coordinate of P” will be (3, – 5)
Again P”‘ is the image of P when reflected in the line parallel to x-axis such that its equation isy
= – 3
Whose coordinates will be 2b – y where b = – 3 and y = 5
∴ 2b – y = 2 (- 3) – 5 = – 6 – 5 = – 11
and abscissa, x = 3
∴ Coordinates will be (3, -11)

Question 5.
Use graph paper for this question.
(i) Plot the points P (2, -4). Use 1 cm = 1 unit on both the axes.
(ii) P’ is the image of P when reflected in AB, which is parallel to the x-axis and is at a distance 1 on the positive side of y-axis. (Le. the line y = 1).
(iii) P” is the image of P’ when reflected in the line LM which is parallel to the y-axis is at a distance 1 on the negative side of the x-axis. (i.e., the line x = – 1).
Solution:
(i) Point P (2, -4) is given.

(ii) The image of P’ when reflected to a line AB which is parallel to x-axis at a distance of 1 unit i.e. y= 1
∴ Co-ordinates of P’ will be (2, 2b – y) or [2, 2 x 1 – (- 4)]
or (2, 2 + 4), or (2, 6)

(iii) The co-ordinates of P”, which is the image of P’ when reflected in the line LM which is parallel to y-axis at a distance of 1 unit on the left side i.e. x = – 1

∴ Co-ordinates of P” will be (2a – x, y) or [2 x (-1) -2, 6], [(-2 -2), 6] or (-4, 6)

Question 6.
A point P is reflected in the jr-axis. Co-ordinates of its image are (8, -6).
(i) Find the co-ordinates of P
(ii) Find the co-ordinates of the image of P under reflection in the .y-axis. (ICSE)
Solution:
(i) Let P’ be the image of point P (x, y), whose co-ordinates are (8, -6) i.e. P’ (8, -6) when reflected in x-axis.
Co-ordinates of P will be (8, 6)
Let P” is the image of P when reflected iny-axis the co-ordinates of P” will be (-8, 6)

Question 7.
The image of the point (1, 5) when reflected in a line LM is (9, 5). Write down the equation of the line LM.
Solution:
Let P’ be the image of point P (1, 5) when reflected a line LM, is (9, 5)
Here we see the co-ordinates of y are same in P and P’
∴ The line LM is parallel to y-axis (x = 0)
i.e. equation of the line will be = 2a – x
⇒ 9 = 2a – 1 ⇒ 2a = 9 + 1 = 10
⇒ a = 5
Equation of the line parallel to y-axis will be x = a ⇒ x = 5

Question 8.
Use graph paper to plot the triangle ABC where A is (1, 2), B is (3, 4) and C is (6,1). On the same graph paper plot.
(i) the image A, B, C of the triangle ABC under reflection in the origin O (0, 0);
(ii) the image of triangle ABC under reflection in the y-axis followed by a reflection in the x-axis.
Compare your results for (i) and (ii) above and make a statement connecting the two results.
Solution:
Plot the points A (1, 2), B (3, 4) and C (6, 1) on the graph and join them to form a ∆ABC.
(i) The image of A, B and C under reflection in the origin are A’, B’ and C’ respectively and co-ordinate of A’, B’ and C’ will be A’ (-1, -2), B’ (-3, -4) and C’ (-6, -1) and are plotted on the graph as shown.

(ii) Under reflection of AABC in y-axis, the images of A, B and C will be A” (-1, 2), B” (-3,4) and C”(-6, 1)
And again under reflection it in the x-axis, we get the image of the ∆A”B”C” as ∆A’B’C’ which is the same as in (i).

(iii) So, we can say the results are same as in (i) and in (ii).

Question 9.
On the graph paper, taking 1 cm = 1 unit, plot the triangle ABC whose vertices are at the points A (3, 1), B (5, 0) and C (7, 4). On the same diagram, draw the image of the AABC under reflection in the line x = 2. Mark I, the point invariant under this reflection.
Solution:
Plot the vertices of ∆ABC whose vertices are A (3, 1), B (5, 0) and C (7, 4) .
Draw a line LM parallel to y-axis at a distance of 2 units on positive side which is x = 2 on reflection of AABC in the line x = 2, the image of A, B and C will be A’, B’ and C’ whose co-ordinates will be A’ (1, 1), B’ (-1, 0) and C’ (-3, 4) respectively.
Point is marked on the graph as the point in variant under this reflection whose co-ordinate are (2, 0).

Question 10.
B, C have co-ordinates (3, 2) and (0, 3). Find
(i) the image B’ of B under the reflection in the x-axis;
(ii) the image C’ of C under reflection in the line BB’;
(iii) Calculate the length of B’C’.
Solution:
Plot the points B (3, 2) and C (0, 3) on the graph.
(i) The image B’ of B under reflection in the x-axis will be B’ (3, -2).
(ii) Join BB’.
(iii) The image C’ (0, 3) of C under reflection in the line BB’ is C’ (6, 3).

(iv) Join B’C’. Draw perpendicular from C’ and B’ meeting each other at D.
Now C’D = 5 units and B’D = 3 units
∴ In right angled AC’B’D,
(B’C’)² = (B’D)² + (C’D)² (Pythagoras Theorem)
= (5)² + (3)² = 25 + 9 = 34
B’C’ = \(\sqrt{34}\) units

Question 11.
The image of a point P under reflection in the x-axis is (5, -2). Write down the co-ordinates of P.
Solution:
Let co-ordinates of P be (x, y), then
the image of P under reflection in the x-axis will be P’ (x, -y)
But P’ is given (5, -2)
Comparing, we get
x = 5, – y = – 2 ⇒ y = 2
∴Co-ordinates of P are (5, 2)

Question 12.
Draw a square whose vertices are (3, 3), (5, 3), (5, 5) and (3, 5). Reflect the square in the y-axis and then reflect the image in the origin. What single transformation would give the same result ?
Solution:
Plot the points A (3, 3), B (5, 3), C (5, 5), D (3, 5) and join AB, BC, CD and DA forming a square. Draw the image of A, B, C and D under reflection in, y-axis. The co-ordinates of images are A’ (-3, 3), B’ (-5, 3), C (-5, 5) and D’ (-3, 5)
Again reflect the points A’, B’, C’ and D’ in origin
The images are A” (3, -3), B” (5, -3), C” (5, -5) and D” (3, -5) respectively.
We see that reflection A”B”C”D” is the image of ABCD under reflection in x-axis.

Question 13.
The triangle ABC, where A (1, 2), B (4, 8), C (6, 8), is reflected in the x-axis to triangle A’B’C’. Triangle A’B’C’ is then reflected in the origin to triangle A”B”C”. Write down the coordinates of A”, B”, C”. Write down a single transformation that maps ABC onto A”B”C”.
Solution:
The vertices of a AABC are A (1,2), B (4, 8) and C (6, 8) and are reflected in the x-axis to AA’B’C’.
Then the co-ordinates of A’, B’ and C’ will be (1, -2), (4, -8) and (6, -8).
The ∆A’B’C’ is again reflected in the origin to AA”B”C”. Then the co-ordinates of A”, B” and C” will be (-1, 2), (-4, 8) and (-6, 8) We see that the single transformation of A”, B” and C” are the images of ∆ABC when reflected in y-axis.

Question 14.
The point P (a, b) is first reflected in the origin, and then reflected in the y-axis to P’. If P’ has coordinates (3, -4), evaluate a, b.
Solution:
Point P (a, b) is reflected under reflection in the origin
∴ Co-ordinates of the image of P will be P, (-a, -b)
Again P, (-a, -b) is reflected under reflection in the j-axis is to P’ whose co-ordinates will be (a, -b)
But co-ordinates of P’ are (3, – 4)
Then comparing, we get
a = 3, – b = – 4 ⇒ b = 4
∴ a = 3, b = 4

Question 15.
The point P (-3, -2) on reflection in the x-axis is mapped as P’. Then P’ on reflection in the origin is mapped as P”. Find the coordinates of P’ and P”.
Solution:
P’ is the image of point P (-3, -2) in the x-axis
∴ Co-ordinates of P’ will be (-3, 2)
P’ is again reflected under reflection in the origin to P”
Then the co-ordinates of P” will be (3, -2)

Question 16.
Point A (5, -1) on reflection in the x-axis is mapped as A’. Also A on reflection in the 7-axis is mapped as A”. Write the co-ordinates of A’ and A”. Also calculate the distance AA”.
Solution:
Point A (5, -1) is reflected in the x-axis to A’
Then co-ordinates of A’ will be (5, 1)
Point A (5, -1) is reflected in the 7-axis to A”
The co-ordinates of A” will be (- 5, – 1)
Length of AA” = \(\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}\)
= \(\sqrt{(5-(-5))^2+(-1-(-1))^2}\)
= \(\sqrt{(10)^2+(0)^2}=\sqrt{100+0}\)
= \(\sqrt{100}\)
= 10 units

Question 17.
Point A (2, – 4) is reflected in origin as A’. Point B (- 3, 2) is reflected in x-axis as B’. Write the coordinates of A’ and B’. Calculate the distance A’B’. Give your answer correct to 1 decimal place. (Do not consult tables)
Solution:
Point A (2, – 4) is reflected in the origin to A’
Then co-ordinates of A’ will be (-2, 4)
Point B (-3, 2) is reflected in x-axis to B’
Then co-ordinates of B’ will be (-3, -2)
Now length of A’B’
= \(\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}\)
= \(\sqrt{[-2-(-3)]^2+[4-(-2)]^2}\)
= \(\sqrt{(-2+3)^2+(4+2)^2}\)
= \(\sqrt{(1)^2+(6)^2}=\sqrt{1+36}\)
= \(\sqrt{37}\) = 6.1 units

Question 18.
Point A (4, -1) is reflected as A’ in the y- axis. Point B on reflection in the x-axis is mapped as B’ (-2, 5).
Write the coordinates of A’ and B.
Write the coordinates of the middle point of the line segment A’B.
Solution:
Point A (4, -1) is reflected in y-axis as A’
∴ Co-ordinates of A’ will be (-4, -1)
Point B is reflected in x-axis as B’ (-2, 5)
Let co-ordinates of B’ be (x, y) then B’ will be (x, -y)
Comparing, we get
x = – 2, – y = 5 or x = – 2, y = – 5
∴ Co-ordinates of B are (- 2, – 5)
Let P (x₁, y₂) be the middle point of line segment A’B
∴ Co-ordinates of P will be
\(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)
or \(\left(\frac{-4-2}{2}, \frac{-1-5}{2}\right)\)
or \(\left(\frac{-6}{2}, \frac{-6}{2}\right)\) or (- 3, – 3)

Question 19.
Point A (1, -5) is mapped as A’ on reflection in the x-axis. Point B (3, 2) is mapped as B’ on reflection in the origin. Write the coordinates of A’ and B’. Calculate AB’.
Solution:
Point A(1, – 5) is mapped as A’ on reflection in the x-axis
∴ Co-ordinates of A’ will be (1, 5)
Point B (3, 2) is mapped as B’ on reflection in origin
∴ Co-ordinates of B’ will be (-3, -2)
Length of AB’ = \(\sqrt{[1-(-3)]^2+[-5-(-2)]^2}\)
= \(\sqrt{(1+3)^2+(-5+2)^2}\)
= \(\sqrt{16+9}\)
= \(\sqrt{25}\) = 5

Question 20.
Attempt this question on the graph paper.
(i) Plot A (3, 2) and B (5, 4) on the graph paper. Take 2 cm = 1 unit on both axes.
(ii) Reflect A and B in the x-axis to A’ and B’. Plot these on the same graph paper.
(iii) Write down
(a) the geometrical name of the figure ABBA’,
(b) the axis of symmetry of ABB’A,
(c) the measure of angle ABB’,
(d) the image of A” of A, when A is reflection in the origin,
(e) the single transformation that maps A’ onto A”.
Solution:
(i) Plot the point A (3, 2) and B (5, 4) on the graph.

(ii) A’ and B’ are the reflection under reflection in x-axis
∴ Co-ordinates of A’ (3, -2) and of B (5, -4)

(iii) (a) Joining AB, BA’, A’B’ and B’A, the figure ABB’A’ is formed.
The figure so formed is of an isosceles trapezium.
(b) The axis of symmetry of the figure trapezium is the x-axis.
(c) The measure of ∠ABB’ is 45°.
(d) The A” is the image of A when reflected in origin.
The co-ordinates of A” will be (-3, -2).
(e) The single transformation that maps A’ onto A” is the reflection of A’ in y-axis.

Self Evaluation And Revision
(LATEST ICSE QUESTIONS)

Question 1.
Points (3, 0) and (-1, 0) are invariant points under reflection L₁ ; points (0, -3) and (0, 1) are invariant points under reflection in line L₂.
(i) Name or write equations for the lines L₁ and L₂.
(ii) Write down the images of P (3, 4) and Q (-5, -2) on reflection in L₁. Name the images as P’ and Q’ respectively.
(iii) Write down the images of P and Q on reflection in L₂. Name the images as P” and Q” respectively.
(iv) State or describe a single transformation that maps P’ onto P”.
Solution:
(i) ∵ Points (3,0) and (-1,0) are invariant points under reflection L₁
Then the equation of L₁ will be x-axis is or y = 0
∵ Points (0, -3) and (0, 1) are invariant points under reflection L₂
Then the equaiton ofL, will bey-axis or x = 0

(ii) The images of point P (3, 4) and Q (-5, -2) on reflection is L, or in x-axis is will be P’ (3, -4) and Q’ (-5, 2)

(iii) The image of P and Q on reflection is L₂ i.e. y-axis will be P” (-3, 4) and Q” (5, -2)

(iv) Single transformation of P and Q that maps P’ onto P” is reflection in origin.

Question 2.
(i) Point P (a, b) is reflected in the x-axis to P’ (5, -2). Write down the values of a and b.
(ii) P” is the image of P when reflected in the y-axis. Write down the coordinate of P”.
(iii) Name a single transformation that maps P’ to P”.
Solution:
(i) The reflection of point P (a, b) in x-axis will be P’ (a, -b)
But P’ is (5, -2)
∴ Comparing, we get a = 5, – b = – 2 or b = 2
∴ Co-ordinates of P are (5, 2)

(ii) P” is the image of P when reflected in y-axis
∴ Co-ordinates of P” will be (-5, 2)

(iii) Single transformation that maps P’ to P” is in origin

Question 3.
A point P (a, b) is reflected in the x-axis to P’ (2, -3). Write down the value of a and b. P” is the image of P, when reflected in the y-axis. Write down the coordinates of P”. Find the coordinates of P'” when P is reflected in the line parallel to the y- axis, such that x = 4.
Solution:
(i) A point P (a, b) is reflected in x-axis to P’ (a, -b) but P’ is given (2, -3)
∴ Comparing, we get
a = 2, – b = – 3 or b = 3
∴ a = 2, b = 3
∴ Co-ordinates of P will be (2, 3)

(ii) P” is the image of P when reflected in y-axis
∴ Co-ordinates of P” will be (-2, 3)

(iii) When P is reflected in a line parallel to y-axis
such that x = 4, to P”
∴ Co-ordinates of P” will be (2a – x, y)
or (2 x 4 – 2, 3) or (8 – 2, 3) or (6, 3)

Question 4.
Use graph paper for this question.
(i) Plot the points A (3, 5) and B (-2, -4). Use 1 cm – 1 unit on both axes.
(ii) A’ is the image of A when reflected in the x-axis. Write down the coordinates of A’ and plot it on the graph paper.
(iii) B’ is the image of B when reflected in the y-axis, followed by reflection in the origin. Write down the coordinates of B’ and plot it on the graph paper.
(iv) Write down the geometrical name of AA’ BB’.
(v) Name two invariant points under reflection in the x-axis.
Solution:
(i) Plot the points A (3, 5) and B (-2, -4) on the graph.
(ii) ∵ A’ is the image of A (3, 5) when reflected in x-axis
∴ Co-ordinates of A’ will be (3, -5) which has been plotted.

(iii) ∵ B’ is the image of B (-2, -4) in the y-axis
∴ Co-ordinates of B’ will be (2, -4) which has been plotted.

(iv) By joining AA’ BB’, the figure so formed is of an isosceles trapezium.

(v) Points which are invariant under reflection in x-axis are C (3, 0) and D (-2, 0).

Question 5.
Write down the coordinates of the image of the point (3, -2) when :
(i) reflected in the x-axis;
(ii) reflected in the y-axis;
(iii) reflected in the x-axis followed by the reflection in the y-axis;
(iv) reflected in the origin;
(v) reflected in the line y = 5.
Solution:
Let point A (3, -2) be given.
(i) The image of A (3, -2) when reflected in x- axis will be A’ and its co-ordinates will be (3, 2).

(ii) The image of A (3, -2) when reflected in y- axis will be A” and its co-ordinates will be (-3, -2).

(iii) The image of A (3, -2) when reflected in x- axis followed by the reflection in y-axis be A'” whose co-ordinates will be (-3, 2).

(iv) The image of point A (3, -2) when reflected in origin will be (-3, 2).

(v) The image of point A (3, -2) when reflected in a line y = 5 The co-ordinates will be (x, 2y-y)
or [3, 2 x 5 – (-2)] or [3, (10+ 2)] or (3, 12).

Question 6.
Use graph paper for this question.
The points P (5, 3) was reflected in the origin to get the image P’.
(a) Write down the coordinates of P’.
(b) If M is the foot of the perpendicular from P to the x-axis, find the coordinates of M.
(c) If N is the foot of the perpendicular from P’ to the x-axis. Find the coordinates of N.
(d) Name the figure PMP’N.
(e) Find the area of the figure PMP’N.
Solution:
(a) The point P (5,3) has been plotted on the graph.
P’ is the image of P (5, 3) under reflected in origin
∴ Co-ordinates of P’ will be (-5, -3)

(b) PM is perpendicular on x-axis
Co-ordinates of M are (5, 0)

(c) P’N is perpendicular on y-axis
∴ Co-ordinates of N will be (-5, 0) x1

(d) By joining PM, MP’, P’N and NP, the figure PMP’N so found is a parallelogram whose opposite sides of are parallel and equal.

(e) Area of parallelogram PMP’N’ = 2 x area of ∆PMN
= 2 x \(\frac { 1 }{ 2 }\) x MN x PM = MN x PM
= 10 x 3 = 30 sq. units

Question 7.
The point P (3, 4) is reflected to P’ in the x-axis, and O’ is the image of O (the origin) when reflected in the line PP’. Using graph paper, give
(i) The coordinates of P’ and O’.
(ii) The lengths of the segments PP’ and OO’.
(iii) The perimeter of the quadrilateral POP’O’.
(iv) The geometrical name of the figure POP’O’.
Solution:
(i) P’ is the image of point P (3, 4) under reflection in x-axis, then the co-ordinates of P’ will be (3, – 4)
O’ is the image of origin O (0, 0) when reflected in the line PP’.
Then the co-ordinates of O’ will be (6, 0)

(ii) Length of PP = 4 + 4 = 8
and length of OO’ = 3 + 3 = 6

(iii) Join OP, PO’, O’P’ and P’O forming a rhombus
Area of figure OPO’P’ = \(\frac { 1 }{ 2 }\) PP’ x OO’
= \(\frac { 1 }{ 2 }\) x 8 x 6 = 24 sq. units.

Question 8.
Use a graph paper for this question. (Take 10 small division = 1 unit on both axes.) Plot the points P (3, 2) and Q (-3, -2). From P and Q draw perpendiculars PM and QN on the x-axis.
(a) Name the image of P on reflection in the origin.
(b) Assign the special name to the geometrical figure PMQN and find its area.
(c) Write the coordinates of the point to which M is mapped on reflection in (i) x- axis; (ii) y-axis, (iii) origin.
Solution:
Plot points P (3, 2) and Q (-3, -2) on the graph.
Draw PM and QN perpendicular on x-axis Co-ordinates of M are (3, 0) and of N are (-3, 0)
(a) The image of P on reflection in origin is Q (-3, -2)

(b) The figure so formed by joining PN and QM is a parallelogram
Area = 2 x area of ∆PMN = 2 x \(\frac { 1 }{ 2 }\) x MN x PM
= MN x PM = 6 x 2 = 12 sq. units

(c) (i) The image of M when reflected in x-axis is M itself.
(ii) The image of M when reflected in y-axis is N(-3, 0).
(iii) The image of M when reflected in origin is N (-3, 0).

Question 9.
Use a graph paper for this question.
A (1,1), B (5,1), C (4, 2) and D (2, 2) are the vertices of a quadrilateral.
Name the quadrilateral ABCD. A, B, C and D are reflected in the origin onto A’, B’, C’ and D’ respectively. Locate A’, B’, C’ and D’ on the graph sheet and write their co-ordinates. Are D, A, A’ and D’ collinear?
Solution:
Plot points A (1, 1), B, (5, 1) C (4, 2) and D (2, 2) on the graph and join then to form a quadrilateral ABCD, which is an isosceles trapezium.
The images of A, B, C, D are A’, B’, C’, and D’ when reflected in origin.
∴ The co-ordinates of A’ are (-1, -1), of B’ are (-5, -1), of C are (-4, -2) and of D’ are (-2, -2). The figure A’B ’C’D’ is the image of the figure ABCD in origin.
We see that D, A, A’ and D’ are collinear when joined.

Question 10.
Use a graph paper for this question (take 10 small division = 1 unit on both axes). P and Q have co-ordinates (0, 5) and (-2, 4).
(i) P is invariant when reflected in an axis. Name the axis.
(ii) Find the image of Q on reflection in the axis found in (i).
(iii) (0, k) on reflection in the origin is invariant. Write the value of k.
(iv) Write the co-ordinates of the image of Q obtained by reflecting it in the origin followed by reflection in the x-axis.
Solution:

(i) Plot points P (0, 5) is invariant on y-axis and Q (-2, 4) on the graph.

(ii) Q’ is the image of Q on reflection in the line y-axis (x = 0)
The co-ordinates of Q’ are (2, 4)

(iii) ∵ (0, k) is invariant in the origin (0, 0)
∴ k = 0

(iv) The image of Q on reflection in the origin followed by the reflection in x-axis is Q”
The co-ordinates of Q” will be (2, 4)

Question 11.
Use graph paper for this question.
The points A (2, 3), B (4, 5) and C (7, 2) are the vertices of AABC.
(i) Write down the co-ordinates of A’, B’, C’, if ∆A’B’C’ is the image of ∆ABC, when reflected in the origin.
(ii) Write down the co-ordinates of A”, B”, C” if ∆A”B”C” is the image of ∆A’B’C’, when reflected in the x-axis.
(iii) Mention the special name of the quadrilateral BC C”B” and find its area.
Solution:
Plot the points A (2, 3), B (4, 5) and C (7, 2) and join them to form the AABC
(i) A’, B’, C’ are the images of A, B, C respectively
Therefore ∆A’B’C’ is the image of ∆ABC when reflected in origin
∴ Co-ordinates of A’ are are (-2, -3), of B’ are (-4, -5) and of C’ are (-7, -2).

(ii) A”, B” and C” are the images of A, B, C under reflection in x-axis
∴ Co-ordinates of A” (2, -3), of B” (4, -5) and C” (7, -2)
Thus ∆A”B”C” is the image of AABC reflected in x-axis.

(iii) By joining C, C” and B, B”, the figure so formed is an isosceles trapezium BCC”B”.

Question 12.
Use a graph paper for this question.
(i) The point P (2, -4) is reflected about the line x = 0 to get the image Q. Find the co-ordinates of Q.
(ii) Point Q is reflected about the line y = 0 to get the image R. Find the co-ordinates of R.
(iii) Name the figure PQR.
(iv) Find the area of the figure PQR.
Solution:

(i) Plot point P (2, – 4) on the graph,
Q is the image of P under reflection iny-axis i.e. x = 0
∴ Co-ordinates of Q will be (-2, -4)

(ii) Point Q (-2, -4) is reflected about the line = 0, i.e. x-axis we get the image R of Q, whose co-ordinates are (-2, 4).

(iii) Join PQ, QR and RP, the figure so formed PQR is a right angled triangle.

(iv) Area of ∆PQR = \(\frac { 1 }{ 2 }\) x base x height
= \(\frac { 1 }{ 2 }\) PQ x QR
= \(\frac { 1 }{ 2 }\) x 4 x 8 = 16 sq. units

Question 13.
Use graph paper to answer this question:
(i) Plot the points A (4, 6) and B (1, 2).
(ii) A’ is the image of A when reflected in ac-axis.
(iii) B’ is the image of B when B is reflected in the line AA’.
(iv) Give the geometrical name for the figure AB A’B’.
Solution:
(i), (ii), (iii)
(iv) Figure ABA’B’ is a kite.

Question 14.
Use a graph paper for this question. Y’
A (0, 3), B (3, -2) and O (0, 0) are the vertices of a ∆ABO.
(i) Plot the triangle on the graph paper taking 2 cm = 1 unit on both the axes.
(ii) Plot D the reflection of B in the y-axis, and write its coordinates.
(iii) Give the geometrical name of the figure ABOD.
(iv) Write the equation of the line of symmetry of the figure ABOD.
Solution:
(i)

(ii) Coordinates of D are (-3, -2)
(iii) Arrow head.
(iv) Equation of line of symmetry of figure ∆BOD is x = 0 i.e., y-axis.

Question 15.
Use a graph paper to answer the following questions. (Take 1 cm = 1 unit on both axes)
(i) Plot A (4, 4), B (4, -6) and C (8, 0) the vertices of a triangle AABC.
(ii) Reflect ABC on the y-axis and name it as A’B’C’.
(iii) Write the coordinates of the images A’, B’ and C’.
(iv) Give a geometrical name for the figure AA’C’B’BC.
(v) Identify the line of symmetry of AA’C’ B’BC.
Solution:

(iii) Coordinates of A’ (-4, 4), B’ (-4, -6) and C’ (-8, 0)
(iv) Hexagon
(v) YY’ is line of symmetry

Question 16.
Using graph paper and taking 1 cm = 1 unit along both xr-axis and y-axis :
(i) Plot the points A (-4, 4) and B (2, 2)
(ii) Reflect A and B in the origin to get the images A’ and B’ respectively.
(iii) Write down the co-ordinates of A’ and B’.
(iv) Give the geometrical name for the figure ABA’B’.
(v) Draw and name its lines of symmetry.
Solution:
(i) The points are plotted in the graph.

(ii) The points A and B have been reflected in the origin to get the image A’ and B’ respectively.

(iii) Coordinates of A’ and B’ are (4, -4) and (-2, -2) respectively.

(iv) In the figure, AB = BA’ = A’B’ = B’A. i.e., ABA’B’ is a rhombus.

(v) As we know a rhombus has two lines of symmetry namely its two diagonals. So, these lines of symmetry have been drawn.

Question 17.
Using a graph paper, plot the points A (6, 4) and B (0, 4).
(i) Reflect A and B in the origin to get the images A’ and B’.
(ii) Write the co-ordinates of A’ and B’.
(iii) State the geometrical name for the figure ABA’B’.
(iv) Find its perimeter.
Solution:

Question 18.
Use a graph paper to answer the following questions. (Take 2 cm = 1 unit on both axis)
(i) Plot the points A (-4, 2) and B (4, 2)
(ii) A’ is the image of A when reflected in the y-axis. Plot it on the graph paper and write the coordinates of A’.
(iii) B’ is the image of B when reflected in the line AA’. Write the coordinates of B’.
(iv) Write the geometric name of the figure ABA’B’.
(v) Name a line of symmetry of the figure formed.
Solution:

Question 19.
Use a graph paper for this question taking 1 cm = 1 unit along both the x-axis and j-axis years:
(i) Plot the points A (0, 5), B (2, 5), C (5, 2), D (5, -2), E (2, -5) and F (0, -5).
(ii) Reflect the points B, C D and E on the j-axis and name them respectively as B’, C’, D’ and E’.
(iii) Write the coordinates of B’, C’, D’ and E’.
(iv) Name the figure formed by BCDEE’D’C’B’.
(v) Name a line of symmetry for the figure formed.
Solution:
(i)

(ii) Reflection of points on the y-axis will result in the change of the x-coordinate.
(iii) Points will be B’ (-2, 5), C’ (-5, 2), D’ (-5, -2), E’ (-2, -5).
(iv) The figure BCDEE’D’C’B’ is a octagon.
(v) The lines of symmetry are x-axis, y-axis, Y = X and Y = -X, total 8.

Question 20.
Use graph paper for this question.
(Take 2 cm = 1 unit along both x and y axis.
Plot the points O (0, 0), A (-4, 4), B (-3, 0) and C (0, -3)
(i) Reflect the points A and B on they-axis and name them A’ and B’ respectively. Write down their coordinates.
(ii) Name the figure OABCB’A’.
(iii) State the line of symmetry of the figure.
Solution:
(i)

(ii) The figure OABCB’A’ is an irregular hexagon.
(iii) They-axis is the line of symmetry of figure OABCB’A’.

Utilizing OP Malhotra Class 10 ICSE Solutions Chapter 10 Reflection Ex 10(a) as a study aid can enhance exam preparation.

S Chand Class 10 ICSE Maths Solutions Chapter 10 Reflection Ex 10(a)

Question 1.
Copy the figure and show the images which result from reflection in the line m.

Solution:
The images have been dawn in reflection to the line m.

Question 2.
Reflect the circle with arrow in the line m in the figure.

Solution:
A circle with centre O’ is drawn as image of the given circle with centre O in the reflection to the line m.

Question 3.
In the figure, mark accurately.
(i) P₁ is the image of P under reflection in line AB.
(ii) P₂ is the image of P under reflection in line BC.
(iii) P₃ is the image of P under reflection in line CA.
What do you observe about the points P₁, P₂ and P₃.

Solution:
P₁ is the image of P under reflection in line AB.
P₂ is the image of P under reflection in line BC.
P₃ is the image’ of P under reflection in the line CA.
We see that P₁, P₃ and P₂ are in the same line. Which is parallel to CA drawn from B.

Question 4.
Answer true or false :
Under reflection, lines which are parallel to the mirror, have image, which are also parallel to the mirror line.
Solution:
True. The image of a line parallel to the mirror when reflected, is also parallel to the mirror.

The availability of step-by-step OP Malhotra Class 10 ICSE Solutions Chapter 9 Arithmetic and Geometric Progression Ex 9(d) can make challenging problems more manageable.

S Chand Class 10 ICSE Maths Solutions Chapter 9 Arithmetic and Geometric Progression Ex 9(d)

Question 1.
Find the sum of the terms of the indicated geometric sequence.
(i) 3, -6, 12,… to 6 terms
(ii) -2, -6, -18,… to 7 terms
(iii) \(\frac { 1 }{ 9 }\), \(\frac { 1 }{ 3 }\), 1, … to 5 terms 1
(iv) 2, 1, \(\frac { 1 }{ 2 }\), … to 6 terms
(v) 1, y, … to 10 terms
(vi) 0.15, 0.015, 0.0015, … 20 terms
Solution:
Find the sum of
(i) 3, – 6, 12, … to 6 terms
Here, a = 3, r = \(\frac { -6 }{ 3 }\) = – 2, n = 6
S₆ = \(\frac{a\left(1-r^n\right)}{1-r}\) (∵ r < 1)
= \(\frac{3\left[1-(-2)^6\right]}{1-(-2)}=\frac{3(1-64)}{1+2}\)
= \(\frac { 3(-63) }{ 3 }\) = – 63

(ii) -2, -6, -18,… to 7 terms
Here, a = – 2, r = \(\frac { -6 }{ -2 }\) = 3, n = 7
∴ S₇ = \(\frac{a\left(r^n-1\right)}{r-1}\) (∵ r > 1)
= \(\frac{-2\left(3^7-1\right)}{3-1}=\frac{-2(2187-1)}{2}\)
= – (2186) = – 2186

(iii) \(\frac { 1 }{ 9 }\), \(\frac { 1 }{ 3 }\), 1, … to 5 terms
Here, a = \(\frac { 1 }{ 9 }\) r = \(\frac { 1 }{ 3 }\) ÷ \(\frac { 1 }{ 9 }\) = \(\frac { 1 }{ 3 }\) x \(\frac { 9 }{ 1 }\) = 3,
S₅ = \(\frac{a\left(r^n-1\right)}{r-1}=\frac{\frac{1}{9}\left(3^5-1\right)}{3-1}\)
= \(\frac{1}{9 \times 2}(243-1)=\frac{1}{18} \times 242\)
= \(\frac { 121 }{ 9 }\) = 13\(\frac { 4 }{ 9 }\)

(iv) 2, 1, \(\frac { 1 }{ 2 }\), … to 6 terms
Here, a = 2, r = 1 ÷ 2 = \(\frac { 1 }{ 2 }\), n = 6

(v) 1, \(\frac { 2 }{ 3 }\), \(\frac { 4 }{ 9 }\), … to 10 terms

(vi) 0.15, 0.015, 0.0015, … 20 terms
Here, a = 0.15, r = \(\frac { 0.015 }{ 0.15 }\) = \(\frac { 1 }{ 10 }\) = 0.1, n = 20
∴ S₂₀ = \(\frac{a\left(1-r^n\right)}{1-r}=\frac{0.15\left[1-(0.1)^{20}\right]}{1-0.1}\)
= \(\frac{0.15}{0.9}\left[1+(0.1)^{20}\right]=\frac{15}{90}\left[1+(0.1)^{20}\right]\)
= \(\frac { 1 }{ 6 }\)[1 + (0.1)²⁰]

Question 2.
Find the sum of the following series
(i) 12 + 6 + 3 + 1.5 + … to 10 terms
(ii) 6 – 3 + 1\(\frac { 1 }{ 2 }\) – \(\frac { 3 }{ 4 }\) + … to 15 terms
(iii) 2 + 6 + 18 + 54 + … to 12 terms
(iv) 6 + 12 + 24 + … + 1536
Solution:
(i) 12 + 6 + 3 + 1.5 + … to 10 terms
Here, a = 12, r = \(\frac { 6 }{ 12 }\) = \(\frac { 1 }{ 2 }\), n = 10

(ii) 6 – 3 + 1\(\frac { 1 }{ 2 }\) – \(\frac { 3 }{ 4 }\)+… to 15 terms
Here, a = 6, r = \(\frac { -3 }{ 6 }\) = \(\frac { -1 }{ 2 }\), n = 15

(iii) 2 + 6 + 18 + 54 + … to 12 terms 6
Here, a = 2, r = \(\frac { 6 }{ 2 }\) = 3, n = 12
S_n = \(\frac{a\left(r^n-1\right)}{r-1}\) (∵ r > 1)
S₁₂ = \(\frac{2\left(3^{12}-1\right)}{3-1}=\frac{2\left(3^{12}-1\right)}{2}\) = 3¹² – 1

(iv) 6 + 12 + 24 + … + 1536 12
Here, a = 6, r = \(\frac { 12 }{ 6 }\) = 2, l = 1536
T_n = l = a(rⁿ – 1)
1536 = 6(2^n-1)

Question 3.
Find the sum to n terms of the following series.
(i) 12 + 6 + 3 + 1 \(\frac { 1 }{ 2 }\) + …
(ii) 20 – 10 + 5 – 2\(\frac { 1 }{ 2 }\) + …
(iii) 9 – 3, +1 – \(\frac { 1 }{ 3 }\) + …
(iv) \(\sqrt{3}\) + 3 + 3\(\sqrt{3}\) + 9 + …
(v) 0.9 + 0.09 + 0.009 + 0.0009 + ….
Solution:
(i) 12 + 6 + 3 + 1 \(\frac { 1 }{ 2 }\) + … n terms

(ii) 20 – 10 + 5 – 2\(\frac { 1 }{ 2 }\) + … n terms
Here, a = 20, r = \(\frac { -10 }{ 20 }\) = \(\frac { -1 }{ 2 }\)

(iii) 9 – 3, + 1 – \(\frac { 1 }{ 3 }\)+… n terms

(iv) \(\sqrt{3}\) + 3 + 3\(\sqrt{3}\) + 9 + … n terms
Here, a = \(\sqrt{3}\), r = \(\frac{3}{\sqrt{3}}\) = \(\sqrt{3}\)

(v) 0.9 + 0.09 + 0.009 + 0.0009 + ….
Here, a = 0.9, r = \(\frac { 0.09 }{ 0.9 }\) = \(\frac { 1 }{ 10 }\) = 0.1
S_n = \(\frac{a\left(1-r^n\right)}{1-r}=\frac{0.9\left[1-(0.1)^n\right]}{1-0.1}\)
= \(\frac { 0.9 }{ 0.9 }\)[1- (0.1)ⁿ]
= 1 – (0.1)ⁿ

Question 4.
Use the formula S_n = \(\frac{a\left(r^n-1\right)}{r-1}\) to find:
(i) a₁ and S_n if a_n = 1000, r = 10 and n = 7
(ii) n and S if a_n = 5, a₁ = 320, r = 2
(iii) a₉ and a₁ if n = 9, r = 2, s_n = 1022
Solution:
S_n = \(\frac{a\left(r^{n^{\prime}}-1\right)}{r-1}\)
(i) a_n = 1000, r = 10 and n = 7, a₁ and S_n
a_n = ar^n-1 ⇒ a x 10^7-1 = 1000

(ii) a₁ = 5, a_n = 320, r = 2, n and S_n
a_n = ar^n-1 ⇒ 320 = 5 × 2^2n-1
2^n-1 = \(\frac { 320 }{ 5 }\) = 64 = 2⁶
∴ n – 1 = 6 ⇒ n = 6 + 1 = 7
and S_n = \(\frac{a\left(r^n-1\right)}{r-1}=\frac{5\left(2^7-1\right)}{2-1}\)
= \(\frac{5(128-1)}{1}\) = 5 × 127 = 635

(iii) n = 9, r = 2, S_n = 1022, a₉ and a₁

Question 5.
If {a_n} is a G.S. (i.e., geometric sequence) and a₁ = 4, r = 5, find a₆ and S₆.
Solution:
{a_n} is in G.S.
a₁ = 4, r = 5, find a₆ and S₆,
S₆ = ar^n-1 = 4 x (5)^6-1 = 4 x 5⁵
= 4 x 5 x 5 x 5 x 5 x 5 = 12500
S₆ = \(\frac { 1 }{ 2 }\)
= 5⁶ – 1 = 15625 – 1 = 15624

Question 6.
How many terms of the G.P. 3, \(\frac { 3 }{ 2 }\), \(\frac { 3 }{ 4 }\),…are needed to give the sum
Solution:
G.P. is
3, \(\frac { 3 }{ 2 }\), \(\frac { 3 }{ 4 }\), …. sum = \(\frac { 3069 }{ 512 }\)

Question 7.
The sum of some terms of a G.P. is 315 whose first term and the common ratio are 5 and 2 respectively. Find the last term and the number of terms.
Solution:
Sum of some terms = 315
Ratio in first term (a₁) and common ratio (r) = 5:2
Let number of terms be n
and first term (a₁) be 5 and r = 2
Now, S_n = \(\frac{a\left(r^n-1\right)}{r-1}\)
315 = \(
63 = 2^{n – 1} ⇒ 2ⁿ = 63 + 1 = 64 = 2⁶
∴ n = 6
and T₆ or a₆ = a^{n -1} = 5(2)^6-1
= 5 x 2⁵ = 5 x 32 = 160

Question 8.
Given a GP. with a = 729 and 7th term = 64, determine S₇.
Solution:
In a G.P.
a = 729, T₇ = 64, S₇ T₇ = ar^n-1
729.r^7-1 = 64 ⇒ 729.r⁶ = 64

Question 9.
Find the sum of the series 2 + 6 + 18 + … + 4374.
Solution:
G.P. is
2 + 6 + 18 + … + 4374 6
Here, a = 2, r = [latex]\frac { 6 }{ 2 }\) = 3, and l = 4374
Now, a₇ = l = ar^n-1
⇒ 4374 = 2 x 3^n-1 ⇒ 3^n-1 = \(\frac { 4374 }{ 2 }\)
⇒ 3^n-1 = 2187 = 3⁷
Comparing, we get
n – 1 = 7 ⇒ n = 7 + 1 = 8
∴ n = 8
Now, S₈ = \(\frac{a\left(r^n-1\right)}{r-1}=\frac{2\left(3^8-1\right)}{3-1}\)
= \(\frac{2(6561-1)}{2}\) = 6560

Question 10.
How many terms of the sequence \(\sqrt{3}\), 3, 3\(\sqrt{3}\), … must be taken to make the sum 39 + 13\(\sqrt{3}\)?
Solution:
G.P. is
\(\sqrt{3}\), 3, 3\(\sqrt{3}\), ….. and S_n = 39 + 13\(\sqrt{3}\)
Here, a = \(\sqrt{3}\), r = \(\sqrt{3}\)
S_n = \(\frac{a\left(r^n-1\right)}{r-1}\) (∵ r > 1)
39 + 13 \(\sqrt{3}=\frac{\sqrt{3}\left[(\sqrt{3})^n-1\right]}{\sqrt{3}-1}\)
⇒ \((\sqrt{3})^n-1=\frac{(\sqrt{3}-1)(39+13 \sqrt{3})}{\sqrt{3}}\)
= (13\(\sqrt{3}\) + (\(\sqrt{3}\) – 1)
= (13\(\sqrt{3}\) + 1) (\(\sqrt{3}\) – 1)
= 13(\(\sqrt{3}\) – 1) = 13 x 2 = 26
∴ (\(\sqrt{3}\))ⁿ = 26 + 1 = 27 = 3³ = (\(\sqrt{3}\))⁶
Comparing, we get n = 6
Number of terms = 6
Comparing, we get
n = 6
∴ Number of terms = 6

Students often turn to OP Malhotra Class 10 ICSE Solutions Chapter 9 Arithmetic and Geometric Progression Ex 9(c) to clarify doubts and improve problem-solving skills.

S Chand Class 10 ICSE Maths Solutions Chapter 9 Arithmetic and Geometric Progression Ex 9(c)

Question 1.
Determine whether the following sequences are geometric progressions or not? If yes then find the common ratio.
(i) 27, 9, 3, 1, …
(ii) -1, 2, 4, 8, …
(iii) 2, \(\frac { 1 }{ 2 }\), \(\frac { 1 }{ 8 }\), \(\frac { 1 }{ 32 }\), …
(iv) -12, -6, 0, 6, …
Solution:
(i) 27, 9, 3, 1, …
Here, a = 27, r = \(\frac { 9 }{ 27 }\) = \(\frac { 1 }{ 3 }\), \(\frac { 3 }{ 9 }\) = \(\frac { 1 }{ 3 }\), ….
∴ It is a G.P. and r = \(\frac { 1 }{ 3 }\)

(ii) -1, 2, 4, 8, …
Here, a = – 1, r = \(\frac { 2 }{ -1 }\) = – 2, \(\frac { 4 }{ 2 }\) = 2, \(\frac { 4 }{ 2 }\) = 2, \(\frac { 8 }{ 4 }\) = 2
∵ differs
∴ It is not a G.P.

(iii) 2, \(\frac { 1 }{ 2 }\), \(\frac { 1 }{ 8 }\), \(\frac { 1 }{ 32 }\), …
Here, a = 2
r = \(\frac{1}{2} \div 2,=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}\)
\(\frac{1}{8} \div \frac{1}{2}=\frac{1}{8} \times \frac{2}{1}=\frac{1}{4}\)
\(\frac{1}{32} \div \frac{1}{8}=\frac{1}{32} \times \frac{8}{1}=\frac{1}{4}\) …
∴ It is G.P. and r = \(\frac { 1 }{ 4 }\)

(iv) – 12, – 6, 0, 6, …
a = – 12, r = \(\frac { -6 }{ -12 }\) = \(\frac { 1 }{ 2 }\)
= \(\frac { 0 }{ -6 }\) = 0
∴ It is not G.P.

Question 2.
Write the next three terms in each of the GPs given below:
(i) 2, 6, …
(ii) \(\frac { 1 }{ 16 }\), – \(\frac { 1 }{ 8 }\), …
(iii) 0.3, 0.06, …
Solution:
Next 3 terms of G.P.
(i) 2, 6, … (r = \(\frac { 6 }{ 2 }\) = 3)
2, 6, 18, 54, 162

(ii) \(\frac { 1 }{ 16 }\), – \(\frac { 1 }{ 8 }\), …. (r = \(-\frac{1}{8} \div \frac{1}{16}=-\frac{1}{8} \times \frac{16}{1}\) = – 2)
\(\frac{1}{16},-\frac{1}{8}, \frac{1}{4},-\frac{1}{2}\), 1

(iii) 0.3, 0.06, …
r = \(\frac { 0.06 }{ 0.3 }\) = \(\frac { 0.06 }{ 0.30 }\) = \(\frac { 1 }{ 5 }\) = -.2
0.3, 0.03, 0.012, 0.0024, 0.00048

Question 3.
Find the:
(i) 6th term of the G.P. 2, 10, 50 …
(ii) 11th term of the GP. 4, 12, 36 …
Solution:
(i) 6th term of the G.P. 2, 10, 50 …
Here, a = 2, r = \(\frac { 10 }{ 2 }\) = 5
∴ T₆ = ar^n-1 = 2 x 5^{6 – 1}
= 2 x 55 = 2 x 3125
= 6250

(ii) 11th term of the G.P. 4, 12, 36 …
Here, a = 4, r = \(\frac { 12 }{ 4 }\) = 3
T₁₁ = ar^n-1 = 4 x (3)^11-1 = 4 x 3¹⁰
= 4 x 243 x 243
= 4 x 59049
= 236196

Question 4.
Write the first five terms of the G.P. where nth term is given as:
(i) 4.3^n-1
(ii) \(\frac{5^{n-1}}{2^{n+1}}\)
Solution:
(i) T_n = 4.3^n-1.
∴ T₁ = 4.3¹ = 4.3° = 4 x l = 4
T₂= 4.3² = 4.3¹ = 4 x 3 = 12
T₃ = 4.33³ = 4.3² = 4 x 9 = 36
T₄ = 4.3^4-1 = 4.3³ = 4 x 27 = 108
T₅ = 4.3^5-1 = 4.3⁴ = 4 x 81 = 324
∴ Terms are 4, 12, 36, 108, 324

(ii) T_n = \(\frac{5^{n-1}}{2^{n+1}}\)

∴ 5 terms are , \(\frac{1}{4}, \frac{5}{8}, \frac{25}{16}, \frac{125}{32}, \frac{625}{64}\)

Question 5.
Write down the nth term of each of the following GPs whose first two terms are given as follows. Also find the term stated besides each G.P.
(i) 12, -36, … sixth term
(ii) 3, – \(\frac { 1 }{ 3 }\), …, 8th term
(iii) b²c³, b³c², …, 5th term
Solution:
(i) 12, -36, … sixth term -36
Here, a = 12, r = \(\frac { -36 }{ 12 }\) = – 3
∴ T₆ = ar^n-1 = 12 x (- 3)^6-1
= 12(- 3)⁵ = 12 x (- 243) = – 2916

(ii) 3, – \(\frac { 1 }{ 3 }\), …, 8th term
Here, a = 3, r = – \(\frac { 1 }{ 3 }\) ÷ 3 = – \(\frac { 1 }{ 3 }\) x \(\frac { 1 }{ 3 }\) = – \(\frac { 1 }{ 9 }\)
∴ T₈ = ar^n-1 = 3(-\(\frac { 1 }{ 9 }\))^8-1 = 3(-\(\frac { 1 }{ 9 }\))⁷

(iii) b²c³, b³c², …, 5th term

Question 6.
Which term of the G.P. 27, -18,12, -8, … is \(\frac { 1024 }{ 2187 }\).
Solution:
G.P. is 27,-18, 12, -8, … is \(\frac { 1 }{ 2 }\)
Here, a = 27, r = \(\frac { -18 }{ 27 }\) = \(\frac { -2 }{ 3 }\)
Let \(\frac { 1027 }{ 2187 }\) be the nth term, then

Comparing, we get n – 1 = 10 ⇒ n = 10 + 1 = 11
It is 11th term.

Question 7.
Write the GP. whose 4th term is 54 and the 7th term is 1458.
Solution:
In a G.P.
T₄ = 54, T₇ = 1458
Let a be the first term and r be the common ratio.
∴ T₄ = ar^n-1 = ar³ = 54 … (i)
and T₇ = ar^7-1 = ar⁶ = 1458 … (ii)
Dividing, we get
\(\frac{a r^6}{a r^3}=\frac{1458}{54}\) ⇒ r³ = 27 = (3)³
∴ r = 3
Now, T₄ = ar^4-1
54 = a(3)³
∴ r = 3
54 = 27a
∴ a = \(\frac { 54 }{ 27 }\) = 2
∴ a = 2, r = 3
∴ G.P. will be 2, 6, 18, 54, …

Question 8.
The last term of the G.P. : 3, 3\(\sqrt{3}\) , 9,… is 2187. How many terms in all there in the GP.?
Solution:
In a G.P. 3, 3\(\sqrt{3}\), 9, …
Last term (l) = 2187
Here, a – 3, r = \(\frac{3 \sqrt{3}}{3}\) = \(\sqrt{3}\)

Comparing, n – 1 = 12
B = 12 + 1 = 13
∴ It is 13th term.

Question 9.
Find the value of x + y + z if 1, x, y, z, 16 are in GP.
Solution:
1, x, y, z, 16 are in G.P.
∴ a (first term) = 1
Common ratio (r) = \(\frac { x }{ 1 }\)
T_s = 16
T₅ = ar^{(n – 1)}
= ar^(5-1)
16 = ar⁴
16 = 1(r)⁴
16 = (2)⁴
r⁴ = (2)⁴
By comparing,
r = 2
i.e. the common ratio = 2
Now, r = \(\frac { x }{ 1 }\) = \(\frac { 2 }{ 1 }\)
i.e. x = 2 … (i)
Also, common ratio
\(\frac { 16 }{ z }\) = 2
z = \(\frac { 16 }{ 2 }\)
z = 8 … (ii)
Also, common ratio
\(\frac { z }{ y }\) = 2
\(\frac { 8 }{ y }\) = 2
y = \(\frac { 8 }{ 2 }\)
y = 4 … (iii)
As per condition,
From (i), (ii) and (iii)
⇒ x + y + z = 2 + 8 + 4 = 14

Question 10.
The third term of a G.P. is 18 and its seventh term is 3\(\frac { 5 }{ 9 }\). Find the tenth term of the G.P.
Solution:
In a G.P.
T₃ = 18, T₇ = 3\(\frac { 5 }{ 9 }\) = \(\frac { 32 }{ 9 }\)
Let a be the first term and r be the common ratio, then
T_n = ar^n-1
T₃ = ar^3-1 = ar² = 18 … (i)
T₇ = ar^7-1 = ar⁶ = \(\frac { 50 }{ 9 }\) … (ii)
Dividing, we get

Question 11.
The 5th, 8th and 11th terms of a G.P. are P, Q and S respectively. Show that Q² = PS.
Solution:
In a G.P.
T₅ = P, T₈ = Q, T₁₁ = S
To prove : Q² = PS
Let a be the first term and r be the common ratio
∴ T₅ = ar^5-1 = ar⁴ = P
T₈ = ar^8-1 = ar⁷ = Q
T₁₁ = ar^11-1= ar¹⁰ = S
Q² = (ar⁷)² = a²r¹⁴
and P x S = ar⁴ x ar¹⁰ = a²r⁴⁺¹⁰ = a²r¹⁴
Hence, Q² = P x 5

Students can track their progress and improvement through regular use of Class 10 ICSE Maths Solutions S Chand Chapter 15 Three Dimensional Solids Ex 15(c)

S Chand Class 10 ICSE Maths Solutions Chapter 15 Three Dimensional Solids Ex 15(c)

Question 1.
Complete the following table. Measurements of the cone are in centimetres. Do not substitute the value of π.

Solution:
(i) Base radius (r) = 3 cm
Height (h) = 4 cm
∴ Slant height (l) = \(\sqrt{r^2+h^2}\) = \(\sqrt{(3)^2+(4)^2}\) = \(\sqrt{9+16}\) = \(\sqrt{25}\) = 5 cm
Curved surface area = πrl
= π × 3 × 5 = 15π cm²
Area of the base = πr²
= π × 3 × 3 = 9π cm²
Total surface area = 15π + 9π = 24π cm²

volume = \(\frac { 1 }{ 3 }\)πr²h = \(\frac { 1 }{ 3 }\)π (3)² × 4
= \(\frac { 1 }{ 3 }\) × π × 9 × 4 cm³ = 12π cm³

(ii) Base radius (r) = 20 cm
slant height (l) = 25
∴ Height (h) = \(\sqrt{l^2-r^2}\) = \(\sqrt{(25)^2-(20)^2}\) = \(\sqrt{625-400}\) = \(\sqrt{225}\) = 15 cm
Curved surface area = πrl = π × 20 × 25 cm² = 500π cm²
Area of the base = πr² = π × (20)²cm² = 400π cm²
Total surface area = πrl + πr² = 500π + 400π = 900π cm²
Volume = \(\frac { 1 }{ 3 }\)πr²h = \(\frac { 1 }{ 3 }\)π (20)² × 15 cm³
= \(\frac { 1 }{ 3 }\)π × 400 × 15 = 2000π cm³

(iii) Base radius = ?
Height (h) = 18
Slant height (l) = 30
∴ Radius (r) = \(\sqrt{l^2-h^2}\) = \(\sqrt{(30)^2-(18)^2}\) = \(\sqrt{900-324}\) = \(\sqrt{576}\) = 24 cm
Curved surface area = πrl = π × 24 × 30 cm² = 720π cm²
Area of the base = πr² = π (24)² = 576π cm²
Total surface area = 720π + 576π = 1296π cm²
Volume = \(\frac { 1 }{ 3 }\)πr²h = \(\frac { 1 }{ 3 }\)π (24)² × 18 cm³
= \(\frac { 1 }{ 3 }\)π × 576 × 18 = 3456π cm³

(iv) Base radius (r) = 27
Height (h) = 36
∴ Slant height (l) = \(\sqrt{r^2+h^2}\)
= \(\sqrt{(27)^2+(36)^2}\) = \(\sqrt{729+1296}\)
= \(\sqrt{2025}\) = 45 cm
Curved surface area = πrl = π × 27 × 45 = 1215π cm²
Area of base = πr² = π(27)² = π × 729 = 729 π cm²
Total surface area = 1215π + 729π = 1944π cm²
volume = \(\frac { 1 }{ 3 }\)πr²h = \(\frac { 1 }{ 3 }\)π × (27)² × 36 cm³
= \(\frac { 1 }{ 3 }\)π × 729 × 36 = 8748π cm³

(v) Base radius (r) = 5 cm
Curved surface area πrl = 65π
Slant height (l) = \(\frac{65 \pi}{5 \pi}\) = 13 cm
and height (h) = \(\sqrt{l^2-r^2}\) = \(\sqrt{(13)^2-(5)^2}\) = \(\sqrt{169-25}\) = \(\sqrt{144}\) = 12 cm
Area of base = πr² = π(5)² = 25π cm²
Total surface area = 65π + 25π = 90π cm²
Volume = \(\frac { 1 }{ 3 }\)πr²h = \(\frac { 1 }{ 3 }\)π (5)² × 12 cm³
= \(\frac { 1 }{ 3 }\)π × 25 × 12 = 100π cm³

(vi) Base radius (r) = 35 cm
Total surface area =πrl + πr² = 4410π
⇒ π × 35 l + π × (35)² = 4410π
⇒ π(35 l + 1225) = 4410π
⇒ 35 l + 1225 = 4410
⇒ 35 l = 4410 – 1225 = 3185
l = \(\frac { 3185 }{ 35 }\) = 91
∴ Slant height = 91 cm
Height (h) = \(\sqrt{l^2-r^2}\) = \(\sqrt{91^2-35^2}\) = \(\sqrt{8281-1225}\) = \(\sqrt{7056}\) = 84 cm
Curved surface = πrl = π × 35 × 91 = 3185π
Area of base = πr² = π(35)² = 1225π
Total surface area = 4410π
volume = \(\frac { 1 }{ 3 }\)πr²h = \(\frac { 1 }{ 3 }\)π (35)² × 84 cm³
= \(\frac { 1 }{ 3 }\)π × 1225 × 84 = 34300π cm³

Question 2.
Find the volume of the cone, given :
(i) Height 8 m; area of base 156 m².
(ii) Slant height 17 cm, radius 8 cm.
(iii) Height 8 cm, slant length 10 cm.
(iv) Height 5 cm, perimeter of base 8 cm.
Solution:
(i) Area of base (πr²) = 156 m²
Height (h) = 8 m
∴ Volume = \(\frac { 1 }{ 3 }\)πr²h = Area of base × \(\frac { 1 }{ 3 }\)h
= \(\frac{156 \times 8}{3}\) = \(\frac{1248}{3}\) m³ = 416 m³

(ii) Slant height (l) = 17 cm
Radius (r) = 8 cm
Height (h) = \(\sqrt{l^2-r^2}\) = \(\sqrt{(17)^2-(8)^2}\) = \(\sqrt{289-64}\) = \(\sqrt{225}\) = 15 cm
∴ Volume = \(\frac { 1 }{ 3 }\)πr²h = \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × (8)² × 15 cm³
= \(\frac { 22 }{ 21 }\) × 64 × 15 = 1005.71 cm³

(iii) Height (h) = 8 cm
and slant height (l) = 10 cm
∴ Radius (r) = \(\sqrt{l^2-h^2}\) = \(\sqrt{(10)^2-(8)^2}\) = \(\sqrt{100-64}\) = \(\sqrt{36}\) = 6 cm
∴ volume = \(\frac { 1 }{ 3 }\)πr²h = \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × (6)² × 8 cm³
= \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 36 × 8 = \(\frac { 2112 }{ 7 }\) = 301.71 cm³

(iv) Height (h) = 5 cm
Perimeter of base (2πr) = 8 cm
Radius (r) = \(\frac{8}{2 \pi}\) = \(\frac{4}{\pi}\) = 4 × \(\frac { 7 }{ 22 }\) = \(\frac { 14 }{ 11 }\) cm
∴ volume = \(\frac { 1 }{ 3 }\)πr²h = \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × \(\left(\frac{14}{11}\right)^2\) × 5 cm³
= \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × \(\frac { 196 }{ 121 }\) × 5 = \(\frac { 280 }{ 33 }\) cm³
= 8.4848 = 8.49 cm³

Question 3.
Find the area of curved surface of the circular cone, given :
(i) Height 8 m, slant height 10 m.
(ii) Perimeter of the base 88 cm, slant height 2 dm.
(iii) Area of the base 154 cm², height 24 cm.
Solution:
(i) Height (h) = 8 m
Slant height (l) = 10 m
∴ Radius (r) = \(\sqrt{l^2-h^2}\) = \(\sqrt{(10)^2-(8)^2}\) = \(\sqrt{100-64}\) = \(\sqrt{36}\) = 6 m
∴ Curved surface area = πrl = \(\frac { 22 }{ 7 }\) × 6 × 10 m²
= \(\frac { 1320 }{ 7 }\) = 188.57 m² = 188.6 m²

(ii) Perimeter of the base (2πr) = 88 cm
Radius (r) = \(\frac{88}{2 \pi}\) = \(\frac{88 \times 7}{2 \times 22}\) = 14 cm
Slant height (l) = 2 dm = 20 cm
∴ Height (h) = \(\sqrt{l^2-r^2}\) = \(\sqrt{(20)^2-(14)^2}\) = \(\sqrt{400-196}\) = \(\sqrt{204}\) cm
∴ Curved surface area = πrl
= \(\frac { 22 }{ 7 }\) × 14 × 20 cm² = 880 cm²

(iii) Area of the base \(\left(\pi r^2\right)\) = 154 cm
Height (h) = 24 cm
Now πr² = 154 ⇒ r² = \(\frac{154}{\pi}\)
⇒ r² = \(\sqrt{r^2+h^2}\) = \(\sqrt{(7)^2+(24)^2}\) = \(\sqrt{49+576}\) = \(\sqrt{625}\) = 25 cm
∴ Curved surface area = πrl
= \(\frac { 22 }{ 7 }\) × 7 × 25 cm² = 550 cm²

Question 4.
Find the height of the cone whose base- radius is 5 cm and volume 50π cm³.
Solution:
Radius (r) = 5 cm
volume \(\left(\frac{1}{3} \pi r^2 h\right)\) = 50π cm³
⇒ \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × (5)²h = 50π ⇒ \(\frac { 22 }{ 21 }\) × 25h = 50π
h = \(\frac{50 . \pi \times 21}{22 \times 25}\) = \(\frac{50 \times 22 \times 21}{22 \times 7 \times 25}\) = 6 cm
∴ Height = 6 cm

Question 5.
The curved surface (area) of a right circular cone of radius 11.3 cm is 710 cm². what is the slant height of the cone? \(\left(\text { Take } \pi=\frac{355}{113}\right)\)
Solution:
Radius of a cone (r) = 11.3 cm
Curved surface area = 710 cm²
∴ πrl = 710
\(\frac { 355 }{ 113 }\) × 11.3 × l = 710
⇒ l = \(\frac{710 \times 113 \times 10}{355 \times 113}\) = 20 cm
∴ Slant height = 20 cm

Question 6.
If the radius of the base of a circular cone is halved, keeping the height same, what is the ratio of the volume of the reduced cone to that of the original cone ?
Solution:
Let in first cone, radius = r and height = h
∴ volume = \(\frac { 1 }{ 3 }\)πr²h
In second case,
The radius = \(\frac { r }{ 2 }\)
and height = h
∴ volume = \(\frac { 1 }{ 3 }\)π \(\left(\frac{r}{2}\right)^2\) h = \(\frac{\frac{1}{3} \pi r^2 h}{4}\) = \(\frac{1}{12}\) πr²h
∴ Ratio \(\frac { 1 }{ 12 }\)πr²h : \(\frac { 1 }{ 3 }\)πr²h
⇒ \(\frac { 1 }{ 12 }\) : \(\frac { 1 }{ 3 }\) : \(\frac { 1 }{ 4 }\) : 1 ⇒ 1 : 4

Question 7.
A conical tent requires 264 m² of canvas. If the slant height is 12 m, find the vertical height.
Solution:
Area of canvas = 264 m²
and slant height (l) = 12 m
∴πrl = 264
\(\frac { 22 }{ 7 }\) × r × 12 = 264
r = \(\frac{264 \times 7}{22 \times 12}\) = 7 cm
∴ vertical heiht (h) = \(\sqrt{l^2-r^2}\) = \(\sqrt{(12)^2-(7)^2}\) = \(\sqrt{144-49}\) = \(\sqrt{95}\) cm
= 9.746 = 9.75 cm

Question 8.
The vertical height of a right circular cone is twice its diameter and its volume is 36π cm³. Find the height.
Solution:
Let height of the cone = h
∴ Diameter = \(\frac { h }{ 2 }\)
∴ Radius (r) = \(\frac { h }{ 4 }\)
Volume = 36π cm³
⇒ \(\frac { 1 }{ 3 }\) πr²h = 36π ⇒ \(\frac { 1 }{ 3 }\) r²h = 36
r²h = 36 × 3
\(\left(\frac{h}{4}\right)^2\) × h = 108 ⇒ \(\frac{h^2}{16}\) × h = 108 (∵ h = 2d or 4r)
⇒ h³ = 108 × 16 = 1728 = (12)³
∴ h = 12 cm
∴ Height = 12 cm

Question 9.
The radius and height of a cone are in the ratio 3 : 4. If its volume is 301.44 cm³, what is its radius? What is its slant height ? (Take π = 3.14)
Solution:
Ratio in radius and height of a cone = 3 : 4
Let radius (r) =3x
Then height (h) = 4 x
Volume = \(\frac { 1 }{ 3 }\)πr²h
⇒ 301.44 =\(\frac { 1 }{ 3 }\) × 3.14 × (3x)² × 4x
⇒ \(\frac{301.44 \times 3}{3.14}\) = 36x³
⇒ x³ = \(\frac{301.44 \times 3}{3.14 \times 36}\) = 8 = (2)³
∴ x = 2
∴ Radius (r) = 3x = 3 × 2 = 6 cm
and height (h) = 4x = 4 × 2 = 8 cm
∴ Slant height (l) = \(\sqrt{r^2+h^2}\) = \(\sqrt{(6)^2+(8)^2}\) = \(\sqrt{36+64}\) = \(\sqrt{100}\) = 10 cm

Question 10.
The radius and slant height of a cone are in the ratio of 4 : 7. If its curved surface area is 792 cm², find its radius. (Use π = 22/7)
Solution:
Ratio in radius and slant height = 4 : 7
Let radius (r) =4 x
The slant height (l) = 7 x
Curved surface area = πrl
⇒ 792 = \(\frac { 22 }{ 7 }\) × 4x × 7x
⇒ 792 = 88x² ⇒ x² = \(\frac { 792 }{ 88 }\) = 9 = (3)²
∴ x = 3
∴ Radius = 4x = 4 × 3 = 12 cm

Question 11.
The base radii of two right circular cones of the same height are in the ratio 3 : 5. Find the ratio of their volumes.
Solution:
Ratio between the radii of two cones r₁ : r₂ = 3 : 5
Let r₁ = 3 x and r₂ = 5 x and height in each case =h
∴ Volume of first cone =\(\frac { 1 }{ 3 }\)πr₁²h = \(\frac { 1 }{ 3 }\)π (3x)²h
= \(\frac { 1 }{ 3 }\)π × 9x²h = 3πx²h
and volume of second cone
= \(\frac { 1 }{ 3 }\)πr₂h = \(\frac { 1 }{ 3 }\)π(5x)₂h = \(\frac { 1 }{ 3 }\)π × 25x²h = \(\frac { 25 }{ 3 }\)πr₂h
∴ Ratio between their volumes
= 3πr₂h : \(\frac { 25 }{ 3 }\)πr₂h = 9πr₂h : 25πr₂h = 9 : 25

Question 12.
The circumference of the base of a 10 m high conical tent is 44 metres. Calculate the length of canvas used in making the tent if width of canvas is 2 cm. (Use π = 22/7)
Solution:
Height of tent (h) = 10 m
Circumference of the base = 44 m
i.e. 2nr = 44 ⇒ \(\frac{2 \times 22}{7}\)r = 44
⇒ r = \(\frac{44 \times 7}{2 \times 22}\) = 7 m
Slant height (l) = \(\sqrt{r^2+h^2}\) = \(\sqrt{(7)^2+(10)^2}\) = \(\sqrt{49+100}\) = \(\sqrt{149}\) m
∴ Curved surface area = πrl
= \(\frac { 22 }{ 7 }\) × 7 × 12.2 = 268.4 m²
Width of canvas = 2 m = 12.2 m
∴ Length = 268.4 ÷ 2 = 134.2 m
∴ Length of canvas = 134.2 m

Question 13.
How many metres of cloth 5 m wide will be required to make to conical tent, the radius of whose base is 7 m and whose height is 24 m ? (Take π = 22/7)
Solution:
Radius of the base of the tent (r) = 7 m and height (h) = 24 m
∴ Slant height (l) = \(\sqrt{r^2+h^2}\) = \(\sqrt{(7)^2+(24)^2}\) = \(\sqrt{49+576}\) = \(\sqrt{625}\) = 25
∴ Curved surface area = πrl
= \(\frac { 22 }{ 7 }\) × 7 × 25 m² = 550 m²
width of the cunvass = 5 m
∴ Length of convas required = \(\frac { 550 }{ 5 }\) = 110 m

Question 14.
A conical tent of capacity 1232 m² stands on a circular base of area 154 m². Find in m² the area of the canvas.
Solution:
Capacity of the tent = 1232 m² and area of base of the tent = 154 m²
Let r be the radius h be the height of the tent, then
πr² = 154 ⇒ \(\frac { 22 }{ 7 }\) r² = 154
⇒ r² = \(\frac{154 \times 7}{22}\) = 49 = (7)²
∴ r = 7 m
∴ volume = 1232
∴ \(\frac { 1 }{ 3 }\) πr²h = 1232
⇒ \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) (7)²h = 1232
⇒ \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 49h = 1232
⇒ h = \(\frac{1232 \times 3 \times 7}{22 \times 49}\) = 24 m
Now slant height (l) = \(\sqrt{r^2+h^2}\) = \(\) = \(\) = \(\) = 25 m
∴ Curved surface area = πrl
= \(\frac { 22 }{ 7 }\) × 7 × 25 = 550 m²
∴ Area of convas = 550 m²

Question 15.
A conical tent is to accommodate 11 persons, each person must have 4 m² of space on the ground and 20 m³ of air to breathe. Find the height of the cone.
Solution:
In the tent, accommodation is available for = 11 persons
Space required for each person on the ground = 4 m²
∴ Area of the base of the tent =11 × 4 = 44 m²
Air is required for each person = 20 m³
∴ Volume of the air of the tent = 20 × 11 = 220 m³
Let r be the radius and h be the height of the tent
∴ πtr² = 44
⇒ \(\frac { 22 }{ 7 }\) r² = 44 ⇒ r² = \(\frac{44 \times 7}{22}\) = 14
and \(\frac { 1 }{ 3 }\) πr²h = 220
⇒ \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 14h = 220 (∵ r² = 14)
⇒ h = \(\frac{220 \times 3 \times 7}{22 \times 14}\) = 15
∴ Height of the conical tent = 15 m

Question 16.
Find out whether the following statement is true or false :
The volume of a cone is one-half of the volume of the cylinder of the same radius and height.
Solution:
It is false as the volume of a cone is one third (\(\frac { 1 }{ 3 }\)) of the cylinder of the same radius and height.

Question 17.
The volume of a cone is the same as that of a cylinder whose height is 9 cm and diameter 40 cm. Find the radius of the base of cone if its height is 108 cm. \(\left(\text { Take } \pi \text { to be } \frac{22}{7}\right)\)
Solution:
Diameter of cylinder = 40 cm
∴ Radius (r₁) = \(\frac { 40 }{ 2 }\) = 20 cm
Height (h₁) = 9 cm
∴ Volume = πr₁²h₁ = π(20)² × 9 cm³ = 3600π cm³
Height (h₂) = 108 cm
Let r₂ be the radius, then
\(\frac { 1 }{ 3 }\) πr₂²h = 3600π
\(\frac { 1 }{ 3 }\) πr₂² × 108 = 3600π
r₂² = \(\frac{3600 \pi \times 3}{108 \pi}\) = 100 = (10)²
∴ r₂ = 10 cm
∴ Radius of the base of the cone = 10 cm

Question 18.
A canvas tent is in the shape of a cylinder surmounted by a conical roof. The common diameter of the cone and cylinder is 14 m. The height of the cylinderical part is 8 m and the height of the conical roof is 4 m. Find the area of the canvas used to make the tent. Give your answer in m² correct to one decimal place. \(\left(\text { Take } \pi \text { to be } \frac{22}{7}\right)\)
Solution:
Diameter of the tent = 14 m
∴ Radius (r) = \(\frac { 14 }{ 2 }\) = 7 m

Height cf cylinderical part (h₁) = 8 m
and height of conical part (h₂) = 4 m
∴ Slant height (l) = \(\sqrt{r^2+h_2^2}\) = \(\sqrt{(7)^2+(4)^2}\) = \(\sqrt{49+16}\) m = \(\sqrt{65}\) m = 8.06 m
Now area of convas = curved surface area of cylindrical part + curved surface area of conical part = 2πrh + πrl
= 2 × \(\frac { 22 }{ 7 }\) × 7 × 8 + \(\frac { 22 }{ 7 }\) × 7 × 8.06 m²
= 352 + 177.3 = 529.3 m²

Question 19.
A circus tent is cylindrical to a height 3 m and conical above it. If its diameter is 105 m and slant height of the cone is 53 m, calculate the total area of the canvas
required. \(\left(\text { Use } \pi=\frac{22}{7}\right)\)
Solution:
Diameter of the tent = 105 m

∴ Radius (r) = \(\frac { 105 }{ 2 }\) m
Height of the cylinderical part (h) = 3 m
and slant height of the conical part (l) = 53 m
Total area of the convas used to make it = 2 πrh + πrl
= 2 × \(\frac { 22 }{ 7 }\) × \(\frac { 105 }{ 2 }\) × 3 + \(\frac { 22 }{ 7 }\) × \(\frac { 105 }{ 2 }\) × 53
= 990 + 8745 = 9735 m²

Question 20.
Find the volume of the largest circular cone that can be cut out of a cube whose edge is 9 cm.

Solution:
Edge of cube = 9 cm
∴ Diameter of the largest cone = 9 cm (equal to edge of cube)
and radius (r) = \(\frac { 9 }{ 2 }\) cm
Height (h) = 9 cm (equal to edge of cube)
∴ Volume of the largest cone = \(\frac { 1 }{ 3 }\) πr²h
= \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × \(\frac { 9 }{ 2 }\) × \(\frac { 9 }{ 2 }\) × 9 cm³
= 190.9285 = 190.93 cm³

Question 21.
A tent is of the shape of a right circular cylinder upto a height of 3 metres and then becomes a right circular cone with a maximum height of 13.5 metres above the ground. Calculate the cost of painting the inner side of the tent at the rate of Rs. 2 per square metre, if the radius of the base is 14 metres.
Solution:
Total height of the tent = 13.5 m
Radius of the base of the tent (r) = 14 m
Height of the cylindrical part (h₁) = 3 m and height of the conical part (h₂)
= 13.5 – 3.0 = 10.5 m
Now slant height (l) = \(\sqrt{r^2+h_2^2}\) = \(\sqrt{(14)^2+(10.5)^2}\) = \(\sqrt{196+110.25}\) = \(\sqrt{306.25}\) = 17.5 m

Now curved surface area of the tent = πrl + 2πrh₁ = πr (l + 2h₁)
= \(\frac { 22 }{ 7 }\) × 14 (17.5 + 2 × 3) m²
= 44(23.5) = 1034 m²
Rate of painting the inner side = Rs. 2 per m²
∴ Tota cost = 1034 × 2 = Rs. 2068

Question 22.
A girl fills a cylindrical bucket 32 cm in height and 18 cm in radius with sand. She empties the bucket on the ground and makes a conical heap of the sand. If the height of the conical heap is 24 cm, find (i) the radius and (ii) the slant height of the heap. Give your answer correct to one place of decimal.
Solution:
Radius of the bucket (r) = 18 cm
Height (h) = 32 cm

∴ Volume of sand in it = πr²h
= π × 18 × 18 × 32 cm³ = 10368π cm³
∴ Volume of conical heap of the sand = 10368π cm³
Height of heap = 24 cm
∴ \(\frac { 1 }{ 3 }\)πr²h = 10368π
\(\frac { 1 }{ 3 }\)πr² × 24 = 10368π
r² = \(\frac{10368 \pi \times 3}{\pi \times 24}\) = 1296 = (36)²
∴ Radius of cone = 36 cm and slant heigth (l) = \(\sqrt{r^2+h^2}\)
= \(\sqrt{(36)^2+(24)^2}\)
= \(\sqrt{1296+576}\) cm
= \(\sqrt{1872}\) cm = 43.266 cm
= 43.3 cm

Question 23.
From a solid cylinder whose height is 8 cm and radius is 6 cm, a conical cavity of height 8 cm and of base radius 6 cm is hollowed out. Find the volume of the remaining solid correct to 4 places of decimals. (π = 3.1416)
Solution:
Radius of a cylinder (r) = 6 cm and height (h) = 8 cm

∴ Volume of the cylinder = πr2²h
= 3.1416 × (6)² × 8 cm³
= 3.1416 × 36 × 8 cm³ = 904.7808 cm³
Radius of cone (r) = 6 cm and height (h) = 8 cm
∴ Volume of cone carved = \(\frac { 1 }{ 3 }\)πr²h
= \(\frac { 1 }{ 3 }\) × 3.1416 × 36 × 8 cm²
= 301.5936 cm³
∴ Volume of the remaining solid
= 904.7808 – 301.5936 = 603.1872 cm³

Question 24.
A metallic cylinder has radius 3 cm and height 5 cm. It is made of a metal A. To reduce its weight, a conical hole is drilled in the cylinder as shown in the figure and it is completely filled with a lighter metal B. The conical hole has a radius of \(\frac { 3 }{ 2 }\) cm and its depth is \(\frac { 8 }{ 9 }\) cm. Calculate the ratio of the volume of the metal A to the volume of the metal B in the solid.

Solution:
Radius of cylinder (R) = 3 cm
and height (H) = 5 cm
Volume of whole cylindrical metal = πR²H
= π(3)² × 5 cm³ = 45π cm³
Radius of conical portion (r) = \(\frac { 3 }{ 2 }\) cm and height (h) = \(\frac { 8 }{ 9 }\)

∴ volume = \(\frac { 1 }{ 3 }\)πr2²h = \(\frac { 1 }{ 3 }\)π \(\left(\frac{3}{2}\right)^2\) × \(\frac { 8 }{ 9 }\)cm³
= \(\frac { 1 }{ 3 }\)π × \(\frac { 9 }{ 4 }\) × \(\frac { 8 }{ 9 }\) = \(\frac { 2 }{ 3 }\)π cm³
Metal in the remaining portion
= 45π – \(\frac { 2 }{ 3 }\)π = \(\frac{(135-2) \pi}{3}\) = \(\frac { 133 }{ 3 }\)π cm
∴ Ratio = \(\frac { 133 }{ 3 }\)π : \(\frac { 2 }{ 3 }\)π = 133 : 2

Question 25.
An open cylindrical vessel of internal diameter 7 cm and height 8 cm stands on a horizontal table. Inside this is placed a solid metallic right circular cone, the diameter of whose base is \(\frac { 7 }{ 2 }\) cm and height 8 cm. Find the volume of water required to fill the vessel. If the cone is replaced by another cone, whose height is 1\(\frac { 3 }{ 4 }\) cm and the radius of whose base is 2 cm, find the drop in the water level.
Solution:
Diameter of the cylindrical vessel = 7 cm
∴ Radius (r₁) = \(\frac { 7 }{ 2 }\) cm
Height (h₁) = 8 cm

In First case,
Radius of cone (r₂) = \(\frac { 7 }{ 4 }\) cm
and heigth (h₂) = 8 cm
∴ Volume of cylinder = πr₁₂h₁
= \(\frac { 22 }{ 7 }\) × \(\left(\frac{7}{2}\right)^2\) × 8 cm
= \(\frac { 22 }{ 7 }\) × \(\frac { 7 }{ 2 }\) × \(\frac { 7 }{ 2 }\) × 8 = 308 cm³
and volume of cone = \(\frac { 1 }{ 3 }\)πr₂²h₂
= \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × \(\frac { 7 }{ 4 }\) × \(\frac { 7 }{ 4 }\) × 8 = \(\frac { 77 }{ 3 }\) cm³
∴ volume of water to be filled in the vessel = \(\frac { 308 }{ 1 }\) – \(\frac { 77 }{ 3 }\) = \(\frac { 924 – 77 }{ 3 }\)
= \(\frac { 847 }{ 3 }\) = 282\(\frac { 1 }{ 3 }\) cm³
In second case, when the cone is replaced by another one of the height = 1\(\frac { 3 }{ 4 }\) cm and radius = 2 cm
∴ volume = \(\frac { 1 }{ 3 }\) πr₃h₃
= \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × (2)² × \(\frac { 7 }{ 4 }\) cm³
= \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 4 × \(\frac { 7 }{ 4 }\) = \(\frac { 22 }{ 3 }\) cm³
∴ Change in volume of the cones
= \(\frac { 77 }{ 3 }\) – \(\frac { 22 }{ 3 }\) = \(\frac { 55 }{ 3 }\) cm³
Let the drop in water level be x cm, then
\(\frac { 22 }{ 7 }\) × \(\left(\frac{7}{2}\right)^2\) × x = \(\frac { 55 }{ 3 }\)
⇒ \(\frac { 22 }{ 7 }\) × \(\frac { 49 }{ 4 }\) x = \(\frac { 55 }{ 3 }\)
⇒ \(\frac { 77 }{ 2 }\)x = \(\frac { 55 }{ 3 }\)
⇒ x = \(\frac { 55 }{ 3 }\) × \(\frac { 2 }{ 77 }\) = \(\frac { 10 }{ 21 }\) cm
∴ Fall in water level = \(\frac { 10 }{ 21 }\) cm

Accessing OP Malhotra Class 10 ICSE Solutions Chapter 9 Arithmetic and Geometric Progression Ex 9(b) can be a valuable tool for students seeking extra practice.

S Chand Class 10 ICSE Maths Solutions Chapter 9 Arithmetic and Geometric Progression Ex 9(b)

Question 1.
Find the sum of the:
(i) first 15 terms of the A.P: 2, 5, 8, 11, …
(ii) first 50 terms of the A.P: -27, -23, -19, …
(iii) first 17 terms of the A.P: = \(\frac{1}{5}, \frac{-3}{10}, \frac{-4}{5}\) …
(iv) first 24 terms of A.P: 0.6, 1.7, 2.8, …
Solution:
(i) First 15 terms of the A.P: 2, 5, 8, 11, …
Here, a = 2, d= 5 – 2 = 3, n = 15
S_n = \(\frac{n}{2}[2 a+(n-1) d]\)
S₁₅ = \(\frac{15}{2}[2 \times 2+(15-1) \times 3]\)
= \(\frac{15}{2}[4+42]\)
= \(\frac{15}{2} \times 46\) = 345

(ii) First 50 terms of the A.P: – 27, – 23, – 19, …
Here, a = – 27, d = – 23 – (- 27)
= – 23 + 27 = 4
n = 50
∴ S_n = \(\frac{n}{2}[2 a+(n-1) d]\)
= \(\frac{50}{2}[2 \times(-27)+(50-1) \times 4]\)
S₅₀ = 25[-54 + 49 x 4] = 25[- 54 + 196] = 25 x 142 = 3550

(iii) First 17 terms of the A.P: = \(\frac{1}{5}, \frac{-3}{10}, \frac{-4}{5}\) …
Here, a = \(\frac {1}{5}\), d = \(\frac { -3 }{ 10 }\) – \(\frac { 1 }{ 5 }\)
\(\frac{-3-2}{10}=\frac{-5}{10}=\frac{-1}{2}\)
and n = 1
S_n = \(\frac { n }{ 2 }\)[2a + (n – 1)d]
S₁₇ = \(\frac{17}{2}\left[2 \times \frac{1}{5}+(17-1)\left(\frac{-1}{2}\right)\right]\)
= \(\frac{17}{2}\left[\frac{2}{5}+16\left(\frac{-1}{2}\right)\right]\)
= \(\frac{17}{2}\left[\frac{2}{5}-8\right]\)
= \(\frac{17}{2}\left(\frac{2-40}{5}\right)=\frac{17}{2} \times \frac{-38}{5}\)
= \(\frac { -323 }{ 5 }\) = – 64\(\frac { 3 }{ 5 }\)

(iv) First 24 terms of A.P: 0.6, 1.7, 2.8, …
Here, a = 0.6, d = 1.7 – 0.6 = 1.1, n = 24
∴ S_n = \(\frac{n}{2}[2 a+(n-1) d]\)
S₂₄ = \(\frac { 24 }{ 2 }\)[2 x 0.6 + (24 – 1) (1.1)]
= 12[1.2 + 23 x 1.1]
= 12[1.2 + 25.3]
= 12 x 26.5
= 318.0

Question 2.
Find the sum:
(i) 34 + 32 + 30 + … + 2
(ii) 7 + 9 \(\frac { 1 }{ 2 }\) + 12 + … + 67.
Solution:
(i) 34 + 32 + 30 + … + 2
Here, a = 34, d = 32 – 34 = – 2, l = 2
a_n = a + (n – 1)d
⇒ 2 = 34 + (n – 1) x (- 2)
⇒ 2 – 34 = – 2(n – 1)
⇒ \(\frac { -32 }{ -2 }\) = n – 1
⇒ n – 1 = 16
n = 16 + 1 = 17
∴ S_n = \(\frac{n}{2}(a+l)=\frac{17}{2}(34+2)\)
= \(\frac { 17 }{ 2 }\) x 36 = 306

(ii) 7 + 9 \(\frac { 1 }{ 2 }\) + 12 + … + 67
Here, a = 7, d = 9\(\frac { 1 }{ 2 }\) – 7 = 2\(\frac { 1 }{ 2 }\) = \(\frac { 5 }{ 2 }\), l = 67
l = (a_n) = a + (n – 1 )d
⇒ 67 = 7 + (n – 1)\(\frac { 1 }{ 2 }\)
67 – 7 = \(\frac { 5 }{ 2 }\)(n – 1) ⇒\(\frac { 60×2 }{ 5 }\) n = 1
⇒ n – 1 = 24 ⇒ n = 24 + 1 = 25
∴ S₂₅ = \(\frac { n }{ 2 }\)[a + l] = \(\frac { 25 }{ 2 }\)[7 + 67]
= \(\frac { 25 }{ 2 }\) x 74 = 925

Question 3.
The common difference of an A.P. is – 2. Find its sum, if its first term is 100 and the last term is – 10.
Solution:
In an AP
d = – 2, a = 100, l = – 10
l = (a_n) = a + (n – 1 )d
– 10 = 100 + (n – 1)(- 2)
(n – 1)(- 2) = – 10 – 100 = – 110
n – 1 = \(\frac { -110 }{ – 2 }\) = 55
⇒ n = 55 + 1 = 56
∴ S₅₆ = \(\frac { n }{ 2 }\)[a + l] = \(\frac { 56 }{ 2 }\) [100 – 10]
= 28 x 90 = 2520

Question 4.
How many terms of the A.P. 54, 51, 48, … should be taken so that their sum is 513?
Solution:
AP is 54, 51, 48, … and S_n = 513
Here, a = 54, d = 51 – 54 = – 3
S_n = \(\frac { n }{ 2 }\)[2a + (n – 1)d]
513 = \(\frac { n }{ 2 }\)[2 x 54 + (n – 1) x (- 3)]
⇒ 513 x 2 = n[108 – 3n + 3]
⇒ 1026 = 108n – 3n² + 3n
⇒ 3n² – 111n + 1026 = 0
⇒ n² – 37n + 342 = 0
⇒ n² – 18n – 19n + 342 = 0

⇒ n(n – 18) – 19(n – 18) = 0
⇒ (n – 18)(n – 19) = 0
Either n – 18 = 0, then n = 18
or n – 19 = 0, then n = 19
∴ Number of terms = 18 or 19

Question 5.
If the sum of first 9 terms of an A.P. is 72 and the common difference is 5, find the first term and the 10th term of the A.P.
Solution:
S₉ = 72, d = 5
Let a be the first term,
n = 9
∴ S₉ = \(\frac { n }{ 2 }\)[2a + (n – 1)d]
72 = \(\frac { 9 }{ 2 }\)[2a + (9 – 1) x 5]
\(\frac { 72×2 }{ 2 }\) = 2a + 40
⇒ 16 = 2a+ 40
2a =16-40 = – 24
a = \(\frac { -24 }{ 2 }\) = -12
∴ a = – 12
a₁₀ = a + (n – 1 )d
= – 12 + (10 – 1) x 5
= – 12 + 45
= – 33

Question 6.
The sum of the first six terms of an A.P. is 42. The ratio of its 10th term to its 30th term is 1 : 3. Calculate the first and thirteenth term of the AP.
Solution:
In an AP,
Sum of first 6 terms = 42
a₁₀ : a₃₀ = 1 : 3
Let a be the first term and d be the common difference, then
S₆ = \(\frac { 6 }{ 2 }\) [2a + (K – 1)d]
42 = 3(2a + 5 d) = 6 a + 15 d
⇒ 6a + 15d = 42 … (i)
a₁₀ : a₃₀ = 1 : 3
\(\frac{a+(10-1) d}{a+(30-1) d}=\frac{1}{3} \Rightarrow \frac{a+9 d}{a+29 d}=\frac{1}{3}\)
3a + 27d = a + 29 d
3a – a = 29d – 27d
⇒ 2a = 2d
⇒ a = d
From (i)
6a + 15a = 42 ⇒ 21a = 42
⇒ a = \(\frac { 42 }{ 2 }\) = 2
∴ a = 2, d = 2
∴ First term = 2
a₁₃ = a + (n – 1)d = 2 + (13 – 1) x 2
= 2 + 12 x 2 = 2 + 24 = 26

Question 7.
The 13th term of an A.P. is four times its 3rd term. If its fifth term is 16, then find the sum of its first ten terms.
Solution:
In an AP,
a₁₃ = 4 x a₃
a₅ = 16
Let a be the first term and d be the common difference.
∴ a₅ = a + (n – 1)d = a + (5 – 1)d = a + 4d
∴ a + 4d = 16 … (i)
Similarly,
a₁₃ = a + 12d and a₃ = a + 2d
∴ a + 12d = 4 x (a + 2d)
a + 12d = 4a + 8d
12d – 8d = 4a – a ⇒ 3a = 4d
a = \(\frac { 4 }{ 3 }\)d
From (i)
\(\frac { 4 }{ 3 }\) + 4d = 16 ⇒ \(\frac { 16 }{ 3 }\)d = 16
⇒ d = \(\frac { 16×3 }{ 16 }\) = 3
∴ d = 3
and a = \(\frac { 4 }{ 3 }\) d = \(\frac { 4 }{ 3 }\) x 3 = 4
a – 4, d = 3
Now, S₁₀ = \(\frac { n }{ 2 }\)[2a + (n – 1)d]
= \(\frac { 10 }{ 2 }\)[2 x 4+ (10 – 1) x 3] = 5(8 + 27)
= 5 x 35 = 175

Question 8.
Find the 60th term of the AP 8,16,12,… if it has a total of 60 terms. Hence find the sum of its last 10 terms.
Solution:
AP is 8, 10, 12, …
Here a = 8, d = 10 – 8 = 2, n = 60
∴ T_n = a +(n – 1)d
⇒ 60 = 8 + (n – 1) x 2
⇒ (n – 1) x 2 = 60 – 8 = 52
n – 1 = \(\frac { 52 }{ 2 }\) = 26
n = 26 + 1 = 27
T₆₀ = a + (60 – 1)d = 8 + 59 x 2
= 8 + 61 = 69
Sum of last 10 terms = S₆₀ – S₅₀
= \(\frac { 60 }{ 2 }\) (2a + 59d) – \(\frac { 50 }{ 2 }\)(2a + 49d)
= 30(2a + 59d) – 25(2a + 49d)
= 60a + 1770d – 50a – 1225d
= 10a + 545d = 10 x 8 + 545 x 2
= 80 + 1090 = 1170

Question 9.
In an A.P. if the 12th term is -13 and the sum of its first four terms is 24, find the sum of its first 10 terms.
Solution:
In an AP,
a₁₂ or T₁₂ = – 13
S₂₄ = 24
Let a be the first term and d be the common difference, then
a₁₂ = a + (n – 1 )d
⇒ – 13 = a + (12 – 1)d
⇒ a + 11d = -13 ⇒ a = – 13 – 11d
S₄ = \(\frac { n }{ 2 }\)(2a + (n – 1)d]
24 = \(\frac { 4 }{ 2 }\)[2a + 3d] = 2[2 x (- 13 – 11d) + 3d]
\(\frac { 24 }{ 2 }\)= – 26 – 22d + 3d ⇒ 12 = – 26 – 19d
⇒ 12 + 26 = – 19d ⇒ – 19d = 38
d = \(\frac { 38 }{ -19 }\) = – 2
and a = – 13 – 11d = – 13 + 11 x 2
= – 13 + 22 = 9
∴ a = 9, d = – 2
Now, S₁₀ = \(\frac { n }{ 2 }\)(2a + (n – 1)d]
= \(\frac { 10 }{ 2 }\)[2 x 9 + (10 – 1)(- 2)]
= 5[18 + 9(-2)] = 5(18 – 18)
= 5 x 0 = 0

Question 10.
Find the sum of the natural numbers between 101 and 999 which are divisible by both 2 and 5.
Solution:
Natural numbers between 101 and 999 which are divisible by 2 and 5 both are 110, 120, 130, …, 990
Here a = 110 and d= 10, l = 990
Now, l = a_n = a + (n – 1 )d
990 = 110 + (n – 1) x 10
⇒ 990 – 110 = 10(n – 1) ⇒ 10(n – 1) = 880
⇒ n – 1 = \(\frac { 880 }{ 10 }\) ⇒ n – 1 = 88
∴ n = 88 + 1 = 89
Now, S₈₉ = \(\frac { n }{ 2 }\)[2a + (n – 1)d]
= \(\frac { 89 }{ 2 }\) [2 x 110 + (89 – 1) x 10]
= \(\frac { 89 }{ 2 }\) [220 + 88 x 10] = \(\frac { 88 }{ 2 }\) [220 + 880]
= \(\frac { 89 }{ 2 }\) x 1100 = 48950

Question 11.
Find the sum of ail two digits numbers greatest than 50 which when divided by 7, leave a remainder of 4.
Solution:
2-digit number greater than 50 which are divisible by 7, leaves a remainder of 4 are: 53, 60, 67, 74, 81, 88, 95
Here a = 53, d = 7, l = 95
∴ a_n(l) = a + (n – 1)d
95 = 53 + (n – 1) x 7 ⇒ 95 – 53 = 7(n – 1)
⇒ 7(n – 1) = 42 ⇒ n – 1 = \(\frac { 42 }{ 7 }\) = 6
∴ n = 6 + 1 = 7
Now, S_n = \(\frac { n }{ 2 }\)[2a + (n – 1 )d]
= \(\frac{7}{2}[2 \times 53+(7-1) \times 7]\)
= \(\frac { 7 }{ 2 }\)[106 + 6 × 7]
= \(\frac { 1 }{ 2 }\)

Question 12.
Find the sum of the integers between 100 and 200 that are (i) divisible by 9 (ii) not divisible by 9.
Solution:
Integers between 100 and 200
(i) Which are divisible by 9 are
108, 117, 126, …, 198
Here, a = 108, d = 9, l = 198
∴ a_n(l) = a + (n – 1 )d
⇒ 198 = 108 + (n – 1) x 9
⇒ 198 – 108 = 9(n – 1)
⇒ 90 = 9 (n – 1)
⇒ n – 1 = \(\frac { 90 }{ 9 }\) = 10
∴ n = 10 + 1 = 11
Now, S₁₁ = \(\frac { n }{ 2 }\) [2a + (n – 1)d]
= \(\frac { 11 }{ 2 }\)[2 x 108 + (11 – 1) x 9]
= \(\frac { 11 }{ 2 }\)[216 + 10 x 9] = \(\frac { 11 }{ 2 }\)[216 + 90]
= \(\frac { 11 }{ 2 }\)[306] = 1683

(ii) Now, sum of integers from 101 to 199
Here, a = 101, d = 1, l = 199, n = 99
S_n = \(\frac { n }{ 2 }\) [2a + (n – 1)d]
= \(\frac { 99 }{ 2 }\)[2 x 101 +(99 – 1) x 1]
= \(\frac { 99 }{ 2 }\) [202 + 98] = \(\frac { 99 }{ 2 }\) x 300 = 14850
∴ Sum of integers which are not divisible by 9 = 14850 – 1683 = 13167

Question 13.
Find the middle term of the sequence formed by all the numbers between 9 and 95 which leave a remainder 1 when divided by 3. Also find the sum of numbers on both the sides of the middle term separately.
Solution:
Numbers between 9 and 95 when divided
by 3, leaves remainder 1
10, 13, 16, 19, …, 94
Here, a = 10, d = 3, l = 94
a_n(l) = a + (n – 1)d
94 = 10 + (n – 1) x 3 ⇒ 94 – 10 = 3 (n – 1)
⇒ 84 = 3 (n – 1) ⇒ n – 1 = \(\frac { 84 }{ 3 }\) = 28
∴ n = 28 + 1 = 29
Now middle term = \(\frac { 29+1 }{ 2 }\)th = 15th term Sum of first 14 terms
= \(\frac { n }{ 2 }\)[2 a + (n – 1 )d]
= \(\frac { n }{ 2 }\)[2a + (14 – 1)d]
= \(\frac { 14 }{ 2 }\)[2 x 10 + 13 x 3]
= 7[20 + 39] = 59 x 2 = 413
Middle term = 10 + 15 x 3 = 10 + 45 = 55
After middle term, number of terms = 14
Whose first term (a) = 55, d = 3 and n = 14 Similarly,
S₁₄ = \(\frac { 19 }{ 2 }\)[2 x 55 + (14 – 1) x 3]
= \(\frac { 19 }{ 2 }\)[110 + 13 x 13] = 7[110 + 39]
= 7 x 149 = 1043

Question 14.
If Sn denotes the sum of first n terms of an A.P., prove that S₁₂ = 3(S₈ – S₄).
Solution:
S_n = Sum of first n terms of an A.P.
S_n = \(\frac { n }{ 2 }\)[2a + (n – 1)d]
To prove : S₁₂ = 3(S₈ – S₄)
RHS = 3[S₈ – S₄]
= 3[ \(\frac { 8 }{ 2 }\)[2a + (8 – 1)d] – \(\frac { 4 }{ 2 }\)[2a + (4 – 1)d] ]
= 3[ \(\frac { 8 }{ 2 }\) (2a + 7J) – \(\frac { 4 }{ 2 }\)(2a + 3d) ]
= 3 [4(2 a + 7d) – 2(2 a + 3d)]
= 3[8a + 28 d – 4a – 6d]
= 3[4a + 22d] = 12a + 66d
= 6[2 a+ 11 d]
= \(\frac { 12 }{ 2 }\) [2a + (12 – 1)d] = S₁₂ = LHS

Question 15.
Show that the sum of first n even natural numbers is equal to (1 + \(\frac { 1 }{ n }\)) times the sum of the first n odd natural numbers.
Solution:
Sum of first n even natural number = (1 + \(\frac { 1 }{ n }\))
(Sum of first n odd natural numbers) Sum of first n even natural numbers (2, 4, 6, …, 2n)
= \(\frac { n }{ 2 }\) [2a + (n – 1 )d]
= \(\frac { n }{ 2 }\) [2 x 2 + (n- 1) x 2]
= \(\frac { n }{ 2 }\) [4 + 2n – 2]
= \(\frac { n }{ 2 }\) [2 + 2n]
= n( 1 + n)
and sum of n odd natural numbers (1, 3, 5, …., (2n – 1)
= \(\frac { n }{ 2 }\) [2x l + (n – 1)2]
= \(\frac { n }{ 2 }\) [2 + 2n – 2]
= \(\frac { n }{ 2 }\)[2n] = n²
Now, n² x (1 + \(\frac { 1 }{ n }\)) = n²(\(\frac { n+1 }{ n }\))
= n( 1 + n)

Question 16.
The sum of n terms of an A.P. whose first term is 5 and common difference is 36 is equal to the sum of In terms of another A.P. whose first term is 36 and common difference is 5. Find n.
Solution:
First term (a₁) = 5
Common difference (d₁) = 36
S_n = Sum of 2n terms of another AP whose
first term (a₂) = 36 and common difference
(d₂) = 5
In first AP
S_n = \(\frac { n }{ 2 }\)[2a + (n- 1 )d]
= \(\frac { n }{ 2 }\)[2 x 5 + (n – 1) x 36]
= \(\frac { n }{ 2 }\) [10 + 36n – 36] = \(\frac { n }{ 2 }\)[36n – 26]
= n(18n – 13)
In second AP,
S_2n= \(\frac { n }{ 2 }\) [2 x 36 + (2n – 1)5]
= n[72 + 10n – 5]
= n[67 + 10n]
∵ s_n = s_2n
n(18n – 13) = n(67 + 10n)
⇒ 18n – 10n = 67 + 13 ⇒ 8n = 80
n = \(\frac { 80 }{ 8 }\) = 10
∴ n = 10

Question 17.
If the sum of first ten terms of an A.P. is 4 times the sum of the first five terms, then find the ratio of the first term of the common difference.
Solution:
Sum of first 10 terms of an AP = 4 x Sum of first 5 terms
Let a be the first term and d be the common difference, then
S_n = \(\frac { n }{ 2 }\)[2 a + (n – 1 )d]
S₁₀ = \(\frac { 10 }{ 2 }\) [2a + (10 – 1)d = 5(2a + 9d)
and S₅ = \(\frac { 5 }{ 2 }\) [2a + (5 – 1 )d] = \(\frac { 5 }{ 2 }\) [2a + 4d]
According to the condition,
5(2 a + 9d) = 4 x \(\frac { 5 }{ 2 }\) (2a + 4d)
10a + 45d = 20a + 40d
45d – 40d = 20a – 10a
⇒ 10a = 5d ⇒ \(\frac { a }{ d }\) = \(\frac { 5 }{ 10 }\) = \(\frac { 1 }{ 2 }\)
∴ Ratio in first term to the common difference = 1 : 2

Question 18.
In a nursery, 37 plants have been arranged in the first row, 35 in the second, 33 in the third and so on. If there are 5 plants in the last row, how many plants are there in the nursery.
Solution:
In the first row, plants are = 37
In second row = 35
In third row = 33
and in the last row = 5
Here, a = 37, d = 35 – 37 = – 2
a_n = a + (n – 1)d
5 = 37 + (n – 1)(- 2)
5 – 37 = – 2(n – 1)
⇒ – 32 = – 2(n – 1)
n – 1 = \(\frac { -32 }{ -2 }\) = 16
∴ n = 16 + 1 = 17
∴ Number of rows = 17
Now, total plants (S_n) = \(\frac { n }{ 2 }\) [2a + (n – 1 )d]
= \(\frac { 17 }{ 2 }\)[2 x 37 + (17 – 1)(- 2)]
= \(\frac { 17 }{ 2 }\) [74 + 16 x (- 2)]
= \(\frac { 17 }{ 2 }\)[74 – 32] = \(\frac { 17 }{ 2 }\) x 42 = 357
∴ Total plants = 357

Question 19.
200 logs are stacked in the following manner : 20 legs in the bottom row, 19 in the next, above it, 18 in the row above it and so on. In how many rows are the 200 logs placed and how many logs are there in the top row?
Solution:
Total logs = 200
In the bottom row, number of logs = 20
and next row above it = 19
Next row above it = 18
and so on
So, AP is 20, 19, 18, 17, …
and S_n = 200
Let the top row be the nth row
∴ a_n = a + (n – 1 )d
and S_n = 200
∴ s_n = \(\frac { n }{ 2 }\)[2 a + (n – 1 )d]
200 = \(\frac { n }{ 2 }\) [2 x 20 + (n – 1)(- 1)]
⇒ 400 = n(40 – n + 1)
⇒ 400 = 41n – n²
⇒ n² – 41n + 400 = 0
⇒ n² – 25n – 16n + 400 = 0
⇒ n(n – 25) – 16(n – 25) = 0
⇒ (n – 25) (n – 16) = 0
Either n – 25 = 0, then n = 25
or n – 16 = 0, then n = 16
But n = 25 is not possible
As number of logs in the first row = 20
and d = – 1
∴ Number by rows = 16
and number of log in 16th row
= a + (n – 1)d = 20 + (16 – 1) (- 1)
= 20 + 15(- 1) = 20 – 15 = 5 logs

Question 20.
A sum of ₹ 1890 is to be used to give seven cash prizes. If each prize is ₹ 50 less than the preceding prize, find the value of each prize?
Solution:
Total amount = ₹ 1890
and number of cash prizes = 7
and each prize is 750 less than the proceeding
prize.
Let first prize = ₹ a
Then second prize a – 50
Third prize = a – 100 and so on
Here, first term = a and d = – 50, S_n = 1890 n = 7
s_n = \(\frac { n }{ 2 }\) [2a + (n – 1)d]
1890 = \(\frac { n }{ 2 }\)[2a + (7 – 1)(- 50)]
\(\frac { 2×1890 }{ 7 }\) = 2a – 300
540 + 300 = 2a ⇒ 2a = 840
a = \(\frac { 840 }{ 2 }\) = 420
Prizes are ₹ 420, ₹ 370, ₹ 320, ₹ 270, ₹ 220, ₹ 170 and ₹ 120

Practicing OP Malhotra Class 10 ICSE Solutions Chapter 9 Arithmetic and Geometric Progression Ex 9(a) is the ultimate need for students who intend to score good marks in examinations.

S Chand Class 10 ICSE Maths Solutions Chapter 9 Arithmetic and Geometric Progression Ex 9(a)

Question 1.
Which of the following situations does the list of numbers involved make an arithmetic progression and why?
(i) the amount of air present in a balloon when \(\frac { 1 }{ 5 }\)th of the air remaining in the balloon is removed by a vacuum pump at a time.
(ii) the cost of digging a well after every metre of digging, if it costs ₹ 125 for the first metre and increases by ₹ 65 for every subsequent metre.
Solution:
(i) Yes.

(ii) Yes, it is an arithmetic progression as cost of digging in the beginning is ₹ 125 for the first metre and then increasing ₹ 65 for every subsequent metre.

Question 2.
Write the first five terms of an AP. in which
(i) a = 10, d = 7
(ii) a = 55, d = – 4
(iii) a = – 1, d = \(\frac { 1 }{ 2 }\)
(iv) a = – 3.25, d = – 0.25
Solution:
(i) a = 10, d = 7
∴ 10, 17, 24, 31, 38

(ii) a = 55, d = -4
∴AP = 55, 51, 47, 43, 39

(iii) a = – 1, d = \(\frac { 1 }{ 2 }\)
∴ AP = – 1, – \(\frac { 1 }{ 2 }\), 0, \(\frac { 1 }{ 2 }\), 1

(iv) a = – 3.25, d = – 0.25
∴ AP = – 3.25, – 3.50, – 3.75, – 4.00, – 4.25

Question 3.
Choose the correct answer in the following:
(i) the list of numbers -12, -9, -6, -3, 0,3 is
(a) not an AP
(b) an AP with d = – 3
(c) an AP with d = 0
(d) an AP with d- 3

(ii) In an AP if a = 3, d = 0 and n = 7, then a_n will be
(a) 4
(b) 1
(c) 3
(d) 2
Solution:
(i) List of numbers : -12, -9, -6, -3, 0, 3
Here a = -12, d = -9 – (-12) = – 9 + 12 = 3
a = – 12, d = 3

(ii) In an AP, a = 3, d = 0, n = 7, then
a_n = a + (n – 1 )d = 3 + (7 – 1) + 0
= 3 + 0 = 3

Question 4.
What is the:
(i) 25th term of the A.P.: – 5, – \(\frac { 5 }{ 2 }\), 0, \(\frac { 5 }{ 2 }\), …
(ii) 30th term of the A.P.: 27, 22, 17, …
(iii) 10th term of the A.P.: -0.1, -0.2, -0.3,…
Solution:
(i) AP = – 5, – \(\frac { 5 }{ 2 }\), 0, \(\frac { 5 }{ 2 }\), …
Here, a = -5, d = – \(\frac { 5 }{ 2 }\) – (- 5)
∴ a₂₅ = a + (n – 1 )d = – 5 + (25 – 1) x \(\frac { 5 }{ 2 }\)
= – 5 + 24 x \(\frac { 5 }{ 2 }\) = – 5 + 60 = 55

(ii) AP is 27, 22, 17, …
Here, a = 27, d= (22 – 27) = – 5
∴ a₃₀ = a + (n – 1 )d
= 27 + (30 – 1) x (- 5) = 27 + 29(- 5)
= 27 – 145 = – 118

(iii) AP = -0.1, -0.2, -0.3, …
Here, a = -0.1
d = – 0.2 – (- 0.1) = – 0.2+ 0.1 = – 0.1
∴ a₁₀ = a + (n – 1)d = – 0.1 +(10 – 1) (- 0.1)
= – 0.1 + 9 x (- 0.1)
= – 0.1 – 0.9 = – 1

Question 5.
If k, 2k – 1 and 2k + 1 are the three consecutive terms of an AP, then find the value of k.
Solution:
k, 2k – 1 and 2k + 1 are the three consecutive terms of an AP, then
a = k, d = 2k – 1 – k = k – 1 … (i)
and d = 2k + 1 – (2k – 1) = 2 k + 1 – 2k + 1 = 2 …(ii)
From (i) and (ii),
∴ k – 1 = 2 ⇒ k = 2 + 1 = 3
∴ k = 3

Question 6.
What is the common difference of the
A.P. \(\frac{1}{3 q}, \frac{1-6 q}{3 q}, \frac{1-12 q}{3 q}\).
Solution:
AP is \(\frac{1}{3 q}, \frac{1-6 q}{3 q}, \frac{1-12 q}{3 q}\)
d = \(\frac{1-6 q}{3 q}-\frac{1}{3 q}=\frac{1-6 q-1}{3 q}=\frac{-6 q}{3 q}\)
= – 2
∴ Common difference = – 2

Question 7.
Which term of the A.P., 5, 13, 21, … is 181?
Solution:
Which term of AP 5, 13, 21, … is 181
Let it be nth term, then
In AP a = 5, d = 13 – 5 = 8
a_n = a + (n – 1 )d
181 =5 + (n- 1) x 8 ⇒ 181 = 5 + 8n – 8
⇒ 8n = 181 – 5 + 8 = 184 ⇒ n = \(\frac { 1 }{ 2 }\) = 23
∴ 181 is 23rd term.

Question 8.
Which term of the A.P. 3, 10, 17,… will be 84 more than its 13th term?
Solution:
AP is 3, 10, 17,…
Here, a = 3, d = 10 – 3 = 7
Let nth term is 84 more than its 13th term
a₁₃ = a + (n – 1)d = 3 + (13 – 1) x 3
= 3 + 12 x 7 = 87
a_n = 3 + (n – 1) x 7 = 3 + 7n – 7 = 7n – 4
Now, 7n – 4 = 84 + 87 = 171
7n = 171 + 4 = 175
n = \(\frac { 175 }{ 7 }\)
∴ 25th is the required term.

Question 9.
Find the AP if the 6th term of the A.P. is 19 and the 16th term is 15 more than the 11th term.
Solution:
In an AP
6th term = 19 and
16th term = 15 + 11th term
Let a be the first term and d be the common difference
∴ a₆ = a + (n – 1 )d ⇒ 19 = a + 5d … (i)
⇒ a + 5d = 19
Similarly,
a₁₁ = a + 10d and a₁₆ = a + 15d
Now, a + 15d = a + 10d + 15
15d – 10d = 15 ⇒ 5d = 15
⇒ d = \(\frac { 15 }{ 5 }\) = 3
From (i)
a + 5 x 3 = 19 ⇒ a + 15 = 19
⇒ a = 19 – 15 = 4
or = 4, d = 3
∴ AP is 4, 7, 10, 13, 16, …

Question 10.
The sum of the 2nd and 7th terms of an A.P. is 30. If its 15th term is 1 less than twice its 8th term, find the AP.
Solution:
2nd term + 7th term = 30
15th term = 2 x 8th term – 1
Let a be the first term and d be the common
difference, then
a₂ = a + (n – 1 )d = a + (2 – 1 )d = a + d
Similarly, a₇ = a + (7 – 1)d ⇒ a + 6d
a₁₅ = a + (15 – 1)d ⇒ a + 14d and
a₈ = a + (8 – 1 )d ⇒ a + 7d
Now, a₂ + a₇ = 30 ⇒ a + d + a + 6d = 30
⇒ 2a + 7d = 30 … (i)
and a + 14d = 2 x (a + 7d) – 1
a + 14d = 2a + 14d – 1
2a – a – 1 ⇒ a = 1
From (i), 2 x 1 + 7d= 30 ⇒ 2 + 7d = 30
⇒ 7d = 30 – 2 = 28
⇒ d = \(\frac { 28 }{ 7 }\) ⇒ d = 4
∴ d = 4, a = 1
∴ AP = 1, 5, 9, 13, 17, …

Question 11.
The fourth term of an A.P. is 11. The sum of the fifth and seventh terms of the A.P. is 34, find its common difference.
Solution:
In an AP
4th term (a₄) = 11
5th term + 7th term = 34
Let a be the first term and d be the common
difference, then
a₄ – a + (n – 1 )d = a + (4 – 1)d
⇒ a + 3d = 11 … (1)
Similarly,
a₇ = a + (7 – 1 )d ⇒ a + 6d
We know that, a₄ + a₇ = 34
∴ a + Ad + a + 6d = 34
⇒ 2a + 10d = 34
a + 5d = 17 … (ii)
Subtracting (ii) from (i),
= (a + 3d) – (a + 5d) = 11 – 17
= a + 3d – a – 5d = – 6
– 2d = – 6
2d = 6 ⇒ d = \(\frac { 6 }{ 2 }\) = 3
∴ Common difference = 3

Question 12.
Find the middle term of the A.P. 213,205, 197, …, 37.
Solution:
AP = 213, 205, 197, …, 37
Let 37 be the nth term
Now, a = 213 and d = 205 – 213 = – 8
∴ a_n = a + (n – 1 )d
⇒ 37 = 213 + (n – 1) (- 8)
⇒ 37 = 213 – 8n + 8
8n = 213 + 8 – 37 = 221 – 37 = 184
n = \(\frac { 184 }{ 8 }\) = 23
∴ There are 23 term in the AP 23 + 1
∴ Middle term = \(\frac { 23+1 }{ 2 }\) = 12th term
Nw, a₁₂ = a + 11 d
= 213 + 11 (- 8) = 213 – 88 = 125

Question 13.
If the 3rd and 9th terms of an A.P. are 4 and – 8 respectively, then which term of this A.P. is zero?
Solution:
In an AP
3rd term = 4
9th term = – 8
Which term is 0
Let a be the first term and d be the common difference, then
a₃ = a + (n – 1 )d
⇒ 4 = a + (3 – 1)d
⇒ a + 2d = 4 … (i)
Similarly, a + ad = – 8 … (ii)
Subtracting (i) from (ii),
⇒ (a + 8d) – (a + 2d) = – 8 – 4
6d = – 12 ⇒ d = \(\frac { -12 }{ 6 }\) = – 2
and a + 2(- 2) = 4 ⇒ a – 4 = 4
⇒ a = 4 + 4 = 8
∴ a = 8, d = – 2
Let a_n be equal to 0, then
a + (n – 1 )d = 0
⇒ 8 + (n – 1) (- 2) = 0
⇒ + 8 – 2n + 2 = 0
⇒ 10 – 2n = 0
⇒ 2n = 10
n = \(\frac { 10 }{ 2 }\) = 5
∴ 5th term is zero.

Question 14.
The 8th term of an A.P. is zero. Prove that its 38th term is three times its 18th term.
Solution:
In an AP
8th term (a₈) = 0
To prove that 38th term = 3 x 18th term
Let a be the first term and d be the common difference, then
a₈ = a + (n – 1)d
⇒ a + (8 – 1)d = 0
⇒ a + 7d = 0
⇒ a = – 7d … (i)
Similarly,
a₁₈ = a + 17d
a₃₈ = a +31d
Now, a + 17d = – 7d + 17d = 10d
and a + 37d = – 7d + 37d = 30d
∴ 30d = 3 x 10d = 30d
Hence, a₃₈ = 3 x a₁₈

Question 15.
Which term of the A.P. 120, 116, 112,… is its first negative term?
Solution:
AP is 120, 116, 112,…
Which first term of this AP will be negative
Here, a = 120, d = 116 – 120 = – 4
Let nth term of the given AP be the negative term a_n or T_n < 0
⇒ a + (n – 1)d < 0
⇒ 120 + (n – 1) (- 4) < 0
⇒ 120 – 4n + 4 < 0
⇒ 124 – 4n < 0 ⇒ 4n > 124
⇒ n > \(\frac { 124 }{ 4 }\) = 31
∴ n = 32 be the first negative term.
⇒ 32nd term is the first negative term

Question 16.
How many 3-digit natural numbers are divisible by 7?
Solution:
3-digit terms are 100, 101, 102, …, 999
and terms which are divisible by 7 will be 105, 112, 119, …, 994
Here, a = 105, d = 7
Last term (a_n) = 994
∴ a_n = a + (n – 1 )d
994 = 105 + (n – 1) x 7
= 105 + 7K – 7 = 7n + 98
∴ 7n = 994 – 98 = 896
n = \(\frac { 896 }{ 7 }\) = 128
∴ Number of 3-digit term divisible by 7 = 128

Question 17.
Two APs have the same common difference. The first term of one of these is – 1 and that of the other is – 8. What is the difference between their 4th terms?
Solution:
Let d be the common difference of the two AP series ax and a2 are the first term of the two AP’s respectively.
In first AP, a₁ = 1, a₂ = – 8
Difference between their 4th terms = (a₁ + 3d) – (a₂ + 3d) = – 1 – (- 8)
a₁ + 3d – a₂ – 3d = – 1 + 8 ⇒ a₁ – a₂ = 7
or a₂ + 3d – a₁ – 3d = – 8 – (- 1)
a₂ – a₁ = – 8 + 1 = – 7
∴ Difference = 7 or – 7

Question 18.
If an AP, ratio of the 4th and 9th terms is 1 : 3, find the ratio of 12 term and 5th term?
Solution:
In an AP
Ratio between 4th term and 9th term = 1 : 3
Let a be the first term and d be the common difference, then
4th term (a₄) = a + (n – 1 )d= a + (4 – 1 )d = a + 3d
Similarly 9th term = a + 8d
Now,
∴ \(\frac{a+3 d}{a+8 d}=\frac{1}{3}\)
3a + 9d = a + 8d
⇒ 3a – a = 8d – 9d
⇒ 2a = – d
⇒ d = – 2a
Now, a₁₂ = a + (12 – 1 )d = a + 11d
and a₅ = a + (5 – 1 )d = a + 4d
∴ \(\frac{a+11 d}{a+4 d}=\frac{a+(-2 a) \times 11}{a+4(-2 a)}\)
\(\frac{a-22 a}{a-8 a}=\frac{-21 a}{-7 a}\)
= \(\frac { 3 }{ 1 }\) = 3 : 1

Question 19.
Find the 7th term from the end of the
A.P.7,10, 13 184.
Solution:
7th term from the end of AP
7, 10, 13, …, 184
Here, a = 7, d = 10 – 7 = 3,
T_n(l) = 184
T_n = a + (n – 1)d
⇒ 7 +(n – 1) x 3
⇒ 7 + 3n – 3 = 4 + 3n
∴ 4 + 3M = 184
⇒ 3n = 184 – 4 = 180
n = \(\frac { 180 }{ 3 }\) = 60
7th term from the last will be 60 – (7 – 1) = 60 – 6 = 54th from the beginning.
∴ T₅₄ = a + 53d
= 7 + 53 x 3 = 7 + 159 = 166

Question 20.
The four angles of a quadrilateral form an A.P. If the sum of the first three angles is twice the fourth angle, then find all the angles.
Solution:
Four angles of a quadrilateral are in AP
Sum of the first 3 angles = 2 x 4th angle
Adding 4th angle to both sides
Sum of 4 angles = 2 x 4th angle + 4th angle = 3 x 4th angle
But sum of 4 angles of a quadrilateral = 360
∴ 3 x 4th angle = 360°
⇒ 4th angle = \(\frac { 360° }{ 3 }\) = 120°
⇒ 4th angle = 120°
Let a be the first term (angle) and d be the common difference.
∴ a + 3d= 120° … (i)
and a + a + d+a + 2d = 360° – 120° = 240
3a + 3d = 240 … (ii)
Subtracting 2a = 120 ⇒ a = \(\frac { 120° }{ 2 }\) = 60°
But a + 3d = 120
⇒ 60 + 3d = 120 ⇒ 3d = 120 – 60 = 60
d = \(\frac { 60° }{ 3 }\) = 20°
∴ Angles are 60°, 80°, 100°, 120°

Question 21.
The sum of three numbers in A.P. is – 3 and their product is 8. Find the numbers.
Solution:
Sum of 3 numbers in AP = – 3
and product = 8
Let three numbers in AP be
a – d, a, a + d
∴ a – d + a + a + d = – 3
⇒ 3a = – 3 ⇒ a = \(\frac { -3 }{ 3 }\) = – 1
and (a – d)a(a + d) = 8
⇒ (a² – d²)a = 8
[(- 1 )² – d²)](-1) = 8
1 – d² = – 8
d² = 1 + 8
d² = 9
d₂ = (±3)²
∴ d = + 3
∴ Numbers be – 1 – 3, – 1, – 1 + 3
= – 4, – 1, 2
or – 1 + 3, – 1, – 1 – 3
= 2, – 1, – 4

Question 22.
Ram prasad saved ₹ 10 in the first week of the year and then increased his weekly savings by ₹ 2.75. If in the nth week, his savings become ₹ 59.50, find n.
Solution:
Ram Prasad’s savings in first week = ₹ 10
Then his weekly saving is increasing = ₹ 2.75
∴ a = ₹ 10 and d = ₹ 2.75
nth week saving = ₹ 59.50
a_nn = a + {n – 1 )d
59.50 = 10 + (n – 1)(2.75)
59.50 – 10.00 = (n – 1)(2.75)
49.50 = (n – 1)(2.75)
∴ n – 1 = \(\frac { 49.50 }{ 2.75 }\)
∴ n = 18 + 1 = 19

Question 23.
For an A.P. show that T_p + T_p+2q = 2T_p+q.
Solution:
For an AP
T_p + T_p+2q = 2T_p+q.
Let a be the first term and d be the common difference, then
T_p = a + (p – 1)d,
T_p+2q = a + (p + 2q – 1)d
T_p+q = a + (p + q – 1)d
Now, LHS = T_p + T_p+2q = a + (p – 1)d + a + (p + 2q – 1 )d
= a + dp – d + a + pd + 2qd – d
= 2a + 2dp + 2qd – 2d
= 2(a + dp + qd – d)
= 2[a + (p + q – 1)d]
= 2 x T_p+q
= RHS

Well-structured OP Malhotra Class 10 Solutions Chapter 8 Matrices Exercise 8(c) facilitate a deeper understanding of mathematical principles.

S Chand Class 10 ICSE Maths Solutions Chapter 8 Matrices Exercise 8(c)

Question 1.
If A = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\), B = \(\left[\begin{array}{ll}
6 & 1 \\
1 & 1
\end{array}\right]\), C = \(\left[\begin{array}{cc}
-2 & -3 \\
0 & 1
\end{array}\right]\), find each of the following:
(a) AB, BA, AC, CA, BC, CB
(b) A², B², C²
(c) A(BC), (AB)C, B(CA), (BC)A.
Solution:

Question 2.
Answer true or false :
(i) \(\frac { 1 }{ 2 }\) is the identity matrix for addition of 2 x 2 matrices.
(ii) If A, B and C are any 2 x 2 matrices, then
A (B – C) = A. B – A. C
Solution:
(i) False : (∵ For addition it is false)
(ii) True.

Question 3.
If \(\left[\begin{array}{ll}
a & 3 \\
1 & 2
\end{array}\right]\left[\begin{array}{c}
2 \\
-1
\end{array}\right]=\left[\begin{array}{l}
5 \\
0
\end{array}\right]\), find the value of a.
Solution:
\(\left[\begin{array}{ll}
a & 3 \\
1 & 2
\end{array}\right]\left[\begin{array}{c}
2 \\
-1
\end{array}\right]=\left[\begin{array}{l}
5 \\
0
\end{array}\right]\)
⇒ \(\left[\begin{array}{l}
a \times 2+3 \times(-1) \\
1 \times 2+2 \times(-1)
\end{array}\right]=\left[\begin{array}{l}
5 \\
0
\end{array}\right]\)
⇒ \(\left[\begin{array}{c}
2 a-3 \\
2-2
\end{array}\right]=\left[\begin{array}{l}
5 \\
0
\end{array}\right] \Rightarrow\left[\begin{array}{c}
2 a-3 \\
0
\end{array}\right]=\left[\begin{array}{l}
5 \\
0
\end{array}\right]\)
Comparing the corresponding terms;
2a – 3 = 5 ⇒ 2a = 5 + 3 ⇒ 2a = 8
⇒ a = \(\frac { 8 }{ 2 }\)
⇒ a = 4
∴ a = 4

Question 4.
If A = \(\left[\begin{array}{ll}
1 & 4 \\
2 & 1
\end{array}\right]\), B = \(\left[\begin{array}{cc}
-3 & 2 \\
4 & 0
\end{array}\right]\), C = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 2
\end{array}\right]\), simplify A² + BC.
Solution:

Question 5.
Find x and y if \(\left[\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}\right]\left[\begin{array}{l}
3 \\
4
\end{array}\right]=\left[\begin{array}{l}
x \\
y
\end{array}\right]\).
Solution:
\(\left[\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}\right]\left[\begin{array}{l}
3 \\
4
\end{array}\right]=\left[\begin{array}{l}
x \\
y
\end{array}\right]\)
⇒ \(\left[\begin{array}{c}
1 \times 3+0 \times 4 \\
0 \times 3+(-1) \times 4
\end{array}\right]=\left[\begin{array}{l}
x \\
y
\end{array}\right] \Rightarrow\left[\begin{array}{l}
3+0 \\
0-4
\end{array}\right]=\left[\begin{array}{l}
x \\
y
\end{array}\right]\)
⇒ \(\left[\begin{array}{c}
3 \\
-4
\end{array}\right]=\left[\begin{array}{l}
x \\
y
\end{array}\right]\)
Comparing the corresponding elements,
x = 3, y = – 4

Question 6.
Let M x \(\left[\begin{array}{ll}
1 & 1 \\
0 & 2
\end{array}\right]\) = [1 2], where M is a matrix,
(i) State the order of the matrix M.
(ii) Find the matrix M.
Solution:
(i) The order of the matrix will be 1 x 2

(ii) Let M = [x y], then
[x y] x \(\left[\begin{array}{ll}
1 & 1 \\
0 & 2
\end{array}\right]\) = [1 2]
⇒ [x x 1 + x o x × 1 + y x 2] = [1 2]
[x + 0 x + 2y] = [1 2]
⇒ [x x + 2y] = [1 2]
Comparing the corresponding elements
x = 1
x + 2y = 2 ⇒ 1 +2y = 2 ⇒ 2y = 2 – 1 = 1
⇒ y = \(\frac { 1 }{ 2 }\)
∴ M = [x y] = \(\frac { 1 }{ 2 }\)

Question 7.
Let A = \(\left[\begin{array}{ll}
4 & -2 \\
6 & -3
\end{array}\right]\), B = \(\left[\begin{array}{cc}
0 & 2 \\
1 & -1
\end{array}\right]\), C = \(\left[\begin{array}{cc}
-2 & 3 \\
1 & -3
\end{array}\right]\), find
(i) A²
(ii) BC
(iii) A² – A + BC
Solution:

Question 8.
If P = \(\left[\begin{array}{cc}
1 & 2 \\
2 & -1
\end{array}\right]\), Q = \(\left[\begin{array}{ll}
1 & 0 \\
2 & 1
\end{array}\right]\), then compute
(i) P² – Q²
(ii) (P + Q) (P – Q)
Is (P + Q) (P – Q) = P² – Q² true for matrix algebra?
Solution:

We see that (P + Q) (P – Q) ≠ P² – Q² for matrix.

Question 9.
If \(\left[\begin{array}{ll}
3 & 4 \\
2 & 5
\end{array}\right]=\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\left[\begin{array}{rr}
1 & 0 \\
0 & -1
\end{array}\right]\) write down the values of a, b, c and d.
Solution:
\(\left[\begin{array}{ll}
3 & 4 \\
2 & 5
\end{array}\right]=\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
= \(\left[\begin{array}{ll}
a \times 1+b \times 0 & a \times 0+b \times 1 \\
c \times 1+d \times 0 & c \times 0+d \times 1
\end{array}\right]\)
= \(\left[\begin{array}{ll}
a+0 & 0+b \\
c+0 & 0+d
\end{array}\right]=\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\)
Comparing we get,
a = 3, b = 4, c = 2, d = 5

Question 10.
If A = \(\left[\begin{array}{cc}
2 & 0 \\
-3 & 1
\end{array}\right]\), and B = \(\left[\begin{array}{cc}
0 & 1 \\
-2 & 3
\end{array}\right]\), find (i) BA (ii) A².
Solution:
A = \(\left[\begin{array}{cc}
2 & 0 \\
-3 & 1
\end{array}\right]\), B = \(\left[\begin{array}{cc}
0 & 1 \\
-2 & 3
\end{array}\right]\)
(i) BA = \(\left[\begin{array}{cc}
0 & 1 \\
-2 & 3
\end{array}\right]\left[\begin{array}{cc}
2 & 0 \\
-3 & 1
\end{array}\right]\)
= \(\left[\begin{array}{cc}
0 \times 2+1 \times(-3) & 0 \times 0+1 \times 1 \\
-2 \times 2+3 \times(-3) & -2 \times 0+3 \times 1
\end{array}\right]\)
= \(\left[\begin{array}{cc}
0-3 & 0+1 \\
-4+(-9) & 0+3
\end{array}\right]=\left[\begin{array}{cc}
-3 & 1 \\
-13 & 3
\end{array}\right]\)

A² = A x A = \(\left[\begin{array}{cc}
2 & 0 \\
-3 & 1
\end{array}\right] \times\left[\begin{array}{cc}
2 & 0 \\
-3 & 1
\end{array}\right]\)
= \(\left[\begin{array}{cc}
2 \times 2+0 \times(-3) & 2 \times 0+0 \times 1 \\
-3 \times 2+1 \times(-3) & -3 \times 0+1 \times 1
\end{array}\right]\)
= \(\left[\begin{array}{cc}
4+0 & 0+0 \\
-6-3 & 0+1
\end{array}\right]=\left[\begin{array}{cc}
4 & 0 \\
-9 & 1
\end{array}\right]\)

Question 11.
If A = \(\frac { 1 }{ 2 }\) and B = \(\frac { 1 }{ 2 }\), find the value of x, given that A² = B.
Solution:

Question 12.
(i) Find the integers p and q such that [p q] \(\left[\begin{array}{l}
p \\
q
\end{array}\right]\) = [25].

(ii) \(\left[\begin{array}{ll}
1 & 3 \\
0 & 0
\end{array}\right]\left[\begin{array}{c}
2 \\
-1
\end{array}\right]=\left[\begin{array}{l}
p \\
q
\end{array}\right]\), findp.
Solution:
[p q] \(\left[\begin{array}{l}
p \\
q
\end{array}\right]\) = [25]
[p² + q²] = [25]
∴ 25 = sum of two squares which can be
(±0)² + (±5)²; (±5)² + (±0)²; (±3)² + (±4)². (±4)² + (+3)²
∴ p = 0, and q = ± 5
p = ± 5, q = o
P = ±3, q = ± 4
P = ± 4, q = ± 3

(ii) \(\left[\begin{array}{ll}
1 & 3 \\
0 & 0
\end{array}\right]\left[\begin{array}{c}
2 \\
-1
\end{array}\right]=\left[\begin{array}{l}
p \\
q
\end{array}\right]\)
⇒ \(\left[\begin{array}{l}
1 \times 2+3 \times(-1) \\
0 \times 2+0 \times(-1)
\end{array}\right]=\left[\begin{array}{l}
p \\
q
\end{array}\right]\)
⇒ \(\left[\begin{array}{l}
2-3 \\
0+0
\end{array}\right]=\left[\begin{array}{l}
p \\
q
\end{array}\right] \Rightarrow\left[\begin{array}{c}
-1 \\
0
\end{array}\right]=\left[\begin{array}{l}
p \\
q
\end{array}\right]\)
Comparing corresponding elements, P = – 1

Question 13.
If A and B are any two 2 x 2 matrices such that AB = B and B is not a zero matrix, what can you say about the matrix A?
Solution:
A and B are any two 2 x 2 matrices and AB = B
∵ B is not a zero matrix
∴ A matrix will be unit matrix or identity matric = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)

Question 14.
If A = \(\left[\begin{array}{cc}
3 & 2 \\
4 & -3
\end{array}\right]\), B = \(\left[\begin{array}{cc}
2 & -3 \\
-4 & 5
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
1 & -3 \\
-4 & 4
\end{array}\right]\), find (i) BC (ii) A² + A.
Solution:

Question 15.
(i) Find a and b if
\(\left[\begin{array}{ll}
a-b & b-4 \\
b+4 & a-2
\end{array}\right]\left[\begin{array}{ll}
2 & 0 \\
0 & 2
\end{array}\right]=\left[\begin{array}{cc}
-2 & -2 \\
14 & 0
\end{array}\right]\)

(ii) Given \(\left[\begin{array}{cc}
8 & -2 \\
1 & 4
\end{array}\right] X=\left[\begin{array}{l}
12 \\
10
\end{array}\right]\), , write down
(a) the order of matrix X, (b) the matrix X.
Solution:
(i) \(\left[\begin{array}{ll}
a-b & b-4 \\
b+4 & a-2
\end{array}\right]\left[\begin{array}{ll}
2 & 0 \\
0 & 2
\end{array}\right]=\left[\begin{array}{cc}
-2 & -2 \\
14 & 0
\end{array}\right]\)
⇒ \(\left[\begin{array}{ll}
(a-b) \times 2+(b-4) \times 0 & (a-b) \times 0+(b-4) \times 2 \\
(b+4) \times 2+(a-2) \times 0 & (b+4) \times 0+(a-2) \times 2
\end{array}\right]\)
= \(\left[\begin{array}{cc}
-2 & -2 \\
14 & 0
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
2 a-2 b+0 & 0+2 b-8 \\
2 b+8+0 & 0+2 a-4
\end{array}\right]=\left[\begin{array}{cc}
-2 & -2 \\
14 & 0
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
2 a-2 b & 2 b-8 \\
2 b+8 & 2 a-4
\end{array}\right]=\left[\begin{array}{cc}
-2 & -2 \\
14 & 0
\end{array}\right]\)
Comparing the corresponding elements
2b – 8 = – 2 ⇒ 2b = 8 – 2 = 6 ⇒ b = \(\frac { 6 }{ 2 }\) = 3
2a – 4 = 0 ⇒ 2a = 4 ⇒ a = \(\frac { 4 }{ 2 }\) = 2
Hence a = 2, 6 = 3

(ii) \(\left[\begin{array}{cc}
8 & -2 \\
1 & 4
\end{array}\right] X=\left[\begin{array}{l}
12 \\
10
\end{array}\right]\)
(a) Here matrix X is of the order 2 x 1
(b) Let X = \(\left[\begin{array}{l}
a \\
b
\end{array}\right]\), then
\(\left[\begin{array}{cc}
8 & -2 \\
1 & 4
\end{array}\right]\left[\begin{array}{l}
a \\
b
\end{array}\right]=\left[\begin{array}{l}
12 \\
10
\end{array}\right]\)
⇒ \(\left[\begin{array}{c}
8 a-2 b \\
a+4 b
\end{array}\right]=\left[\begin{array}{c}
12 \\
10
\end{array}\right]\)
Comparing the corresponding elements
8a – 26 = 12 … (i)
a + 46 = 10 … (ii)
Multiply (i) by 2 and (ii) by 1

From (ii) 2 + 4b = 10 ⇒ 4b = 10 – 2 = 8
= \(\frac { 8 }{ 4 }\) = 2
∴ X = \(\left[\begin{array}{l}
2 \\
2
\end{array}\right]\)

Question 16.
If A = \(\left[\begin{array}{cc}
1 & -1 \\
2 & 3
\end{array}\right]\), show that A² – 4A + 5I = 0, where I is the unit matrix.
Solution:

Question 17.
If A = \(\left[\begin{array}{ll}
1 & 4 \\
1 & 0
\end{array}\right]\), B = \(\left[\begin{array}{cc}
2 & 1 \\
3 & -1
\end{array}\right]\) and C = \(\left[\begin{array}{ll}
2 & 3 \\
0 & 5
\end{array}\right]\), compute (AB) C and (CB) A.
Is (AB) C = (CB) A?
Solution:

Question 18.
(i) Write as a single matrix : \(\left[\begin{array}{cc}
6 & -2 \\
1 & 2
\end{array}\right]\) \(\left[\begin{array}{ll}
5 & 3 \\
2 & 4
\end{array}\right]\).
(ii) Find x and y if \(\left[\begin{array}{cc}
2 & 3 \\
-1 & 0
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{c}
7 \\
-2
\end{array}\right]\).
Solution:
(i) \(\left[\begin{array}{cc}
6 & -2 \\
1 & 2
\end{array}\right]\left[\begin{array}{ll}
5 & 3 \\
2 & 4
\end{array}\right]\)
= \(\left[\begin{array}{cc}
6 \times 5+(-2) \times 2 & 6 \times 3+(-2) \times 4 \\
1 \times 5+2 \times 2 & 1 \times 3+2 \times 4
\end{array}\right]\)
= \(\left[\begin{array}{cc}
30-4 & 18-8 \\
5+4 & 3+8
\end{array}\right]=\left[\begin{array}{cc}
26 & 10 \\
9 & 11
\end{array}\right]\)

(ii) \(\left[\begin{array}{cc}
2 & 3 \\
-1 & 0
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{c}
7 \\
-2
\end{array}\right]\)
⇒ \(\left[\begin{array}{c}
2 x+3 y \\
-x+0
\end{array}\right]=\left[\begin{array}{c}
7 \\
-2
\end{array}\right] \Rightarrow\left[\begin{array}{c}
2 x+3 y \\
-x
\end{array}\right]=\left[\begin{array}{c}
7 \\
-2
\end{array}\right]\)
Comparing the corresponding elements,
– x = – 2 ⇒ x = 2
2x + 3y = 7 ⇒ 2 x 2 + 3y = 7
⇒ 4 + 3y = 7 ⇒ 3y = 7 – 4 = 3
⇒ y = \(\frac { 3 }{ 3 }\) = 1
∴ x = 2, y = 1

Question 19.
If A = \(\left[\begin{array}{ll}
3 & 1 \\
2 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
1 & -2 \\
5 & 3
\end{array}\right]\), verify if (A – B)² = A² – 2AB + B².
Solution:

Question 20.
Given that A = \(\left[\begin{array}{ll}
3 & 0 \\
0 & 4
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
a & b \\
0 & c
\end{array}\right]\) and that AB = A + B, find the values of a, b and c.
Solution:
A = \(\left[\begin{array}{ll}
3 & 0 \\
0 & 4
\end{array}\right]\), B = \(\left[\begin{array}{ll}
a & b \\
0 & c
\end{array}\right]\)
AB = \(\left[\begin{array}{ll}
3 & 0 \\
0 & 4
\end{array}\right]\left[\begin{array}{ll}
a & b \\
0 & c
\end{array}\right]\)
= \(\left[\begin{array}{cc}
3 a+0 & 3 b+0 \\
0+0 & 0+4 c
\end{array}\right]=\left[\begin{array}{cc}
3 a & 3 b \\
0 & 4 c
\end{array}\right]\)
A + B = \(\left[\begin{array}{ll}
3 & 0 \\
0 & 4
\end{array}\right]+\left[\begin{array}{ll}
a & b \\
0 & c
\end{array}\right]\)
= \(\left[\begin{array}{ll}
3+a & 0+b \\
0+0 & 4+c
\end{array}\right]=\left[\begin{array}{cc}
3+a & b \\
0 & 4+c
\end{array}\right]\)
Now ∵ AB = A + B
∴ \(\left[\begin{array}{cc}
3 a & 3 b \\
0 & 4 c
\end{array}\right]=\left[\begin{array}{cc}
3+a & b \\
0 & 4+c
\end{array}\right]\)
Comparing the corresponding elements,
3a = 3 + a ⇒ 3a – a = 3 ⇒ 2a = 3 ⇒ a = \(\frac { 3 }{ 2 }\)
3b = b ⇒ 3b – b = 0 ⇒ 2b = 0 4c = 4 + c ⇒ 4c – c = 4 ⇒ 3c = 4
⇒ \(\frac { 4 }{ 3 }\)
Hence a = \(\frac { 3 }{ 2 }\), b = 0, c = \(\frac { 4 }{ 3 }\)

Question 21.
(i) Solve the matrix equation \(\left[\begin{array}{ll}
2 & 1 \\
5 & 0
\end{array}\right]\) – 3X = \(\left[\begin{array}{cc}
-7 & 4 \\
2 & 6
\end{array}\right]\).
Solution:

Question 22.
If A = \(\left[\begin{array}{ll}
9 & 1 \\
5 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
1 & 5 \\
7 & -11
\end{array}\right]\), find the matrix X such that 3A + 5B – 2X = 0.
Solution:

Question 23.
(i) Find x and y if, \(\left[\begin{array}{ll}
x & 3 x \\
y & 4 y
\end{array}\right]\left[\begin{array}{l}
2 \\
1
\end{array}\right]=\left[\begin{array}{c}
5 \\
12
\end{array}\right]\)
(ii) A = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right], B=\left[\begin{array}{ll}
2 & 1 \\
4 & 2
\end{array}\right], C=\left[\begin{array}{ll}
5 & 1 \\
7 & 4
\end{array}\right]\).
Compute A (B + C) and (B + C) A.
Solution:

Question 24.
(i) If \(\left[\begin{array}{cc}
x-2 & 5 \\
3 & 3
\end{array}\right]=\left[\begin{array}{ll}
4 & 2 \\
y & 5
\end{array}\right]\) + \(\left[\begin{array}{cc}
-4 & 3 \\
-1 & -2
\end{array}\right]\), find the values of x and y.
(ii) If A = \(\left[\begin{array}{ll}
1 & 2 \\
2 & 3
\end{array}\right]\), B = \(\left[\begin{array}{ll}
2 & 1 \\
3 & 2
\end{array}\right]\), C = \(\left[\begin{array}{ll}
1 & 3 \\
3 & 1
\end{array}\right]\). Find the matrix C (B – A).
Solution:

Question 25.
If A = \(\left[\begin{array}{ccc}
1 & -2 & 1 \\
2 & 1 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
2 & 1 \\
3 & 2 \\
1 & 1
\end{array}\right]\), write down the matrix AB. Would it be possible to find the product BA ? If so, compute it, and if not, give reasons.
Solution:

Hence it is possible.

Self Evaluation And Revision
(LATEST ICSE QUESTIONS)

Question 1.
Evaluate x, y if
Solution:
\(\left[\begin{array}{cc}
3 & -2 \\
-1 & 4
\end{array}\right]\left[\begin{array}{c}
2 x \\
1
\end{array}\right]=\left[\begin{array}{c}
6 x-2 \\
-2 x+4
\end{array}\right]\)
∴ \(\left[\begin{array}{cc}
3 & -2 \\
-1 & 4
\end{array}\right]\left[\begin{array}{c}
2 x \\
1
\end{array}\right]+2\left[\begin{array}{c}
-4 \\
5
\end{array}\right]=\left[\begin{array}{c}
6 x-2 \\
-2 x+4
\end{array}\right]\) + \(\left[\begin{array}{c}
-8 \\
10
\end{array}\right]=\left[\begin{array}{c}
8 \\
4 y
\end{array}\right]\)
⇒ \(\left[\begin{array}{c}
6 x-2-8 \\
-2 x+4+10
\end{array}\right]=\left[\begin{array}{c}
8 \\
4 y
\end{array}\right]\)
⇒ \(\left[\begin{array}{c}
6 x-10 \\
-2 x+14
\end{array}\right]=\left[\begin{array}{c}
8 \\
4 y
\end{array}\right]\)
Comparing the corresponding elements
6x – 10 = 8 ⇒ 6x = 8 + 10 = 18
⇒ x = \(\frac { 18 }{ 6 }\) = 3
and – 2x + 14 = 4y ⇒ – 2 x 3 + 14 = 4y
⇒ 4y = 14 – 6 = 8 ⇒ y = \(\frac { 8 }{ 4 }\) = 2
Hence x = 3, y = 2

Question 2.
Evaluate :

Solution:

We know that cos 60° = \(\frac { 1 }{ 2 }\), sin 30° = \(\frac { 1 }{ 2 }\)
tan 45° = 1, cos 0° = 1
cot 45° = 1 cosec 30° = \(\frac { 1 }{ sin 30° }\) = \(\frac { 2 }{ 1 }\),
sec 60° = \(\frac { 1 }{ cos 60° }\) = \(\frac { 2 }{ 1 }\), sin 90° = 1
∴ \(\left[\begin{array}{cc}
2 \times \frac{1}{2} & -2 \times \frac{1}{2} \\
-1 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right]\)
= \(\left[\begin{array}{cc}
1 & -1 \\
-1 & 1
\end{array}\right]\left[\begin{array}{cc}
1 & 2 \\
2 & 1
\end{array}\right]\)
= \(\left[\begin{array}{cc}
1 \times 1+(-1) \times 2 & 1 \times 2+(-1)(1) \\
-1 \times 1+1 \times 2 & -1 \times 2+1 \times 1
\end{array}\right]\)
= \(\left[\begin{array}{cc}
1-2 & 2-1 \\
-1+2 & -2+1
\end{array}\right]=\left[\begin{array}{cc}
-1 & 1 \\
1 & -1
\end{array}\right]\)

Question 3.
Find the value of x and y; if
\(\left[\begin{array}{ll}
1 & 2 \\
3 & 3
\end{array}\right]\left[\begin{array}{ll}
x & 0 \\
0 & y
\end{array}\right]=\left[\begin{array}{ll}
x & 0 \\
9 & 0
\end{array}\right]\)
Solution:
\(\left[\begin{array}{ll}
1 & 2 \\
3 & 3
\end{array}\right]\left[\begin{array}{ll}
x & 0 \\
0 & y
\end{array}\right]=\left[\begin{array}{ll}
x & 0 \\
9 & 0
\end{array}\right]\)
⇒ \(\left[\begin{array}{ll}
1 \times x+2 \times 0 & 1 \times 0+2 \times y \\
3 \times x+3 \times 0 & 3 \times 0+3 \times y
\end{array}\right]=\left[\begin{array}{ll}
x & 0 \\
9 & 0
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
0+x & 0+2 y \\
3 x+0 & 0+3 y
\end{array}\right]=\left[\begin{array}{cc}
x & 0 \\
9 & 0
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
x & 2 y \\
3 x & 3 y
\end{array}\right]=\left[\begin{array}{ll}
x & 0 \\
9 & 0
\end{array}\right]\)
Comparing the corresponding elements,
2y = 0 ⇒ y = 0, 3x = 9 ⇒ x = \(\frac { 9 }{ 3 }\) = 3
Hence x = 3, y = 0

Question 4.
Find the 2 x 2 matrix X which satisfies the equation.
\(\left[\begin{array}{ll}
3 & 7 \\
2 & 4
\end{array}\right]\left[\begin{array}{ll}
0 & 2 \\
5 & 3
\end{array}\right]+2 X=\left[\begin{array}{cc}
1 & -5 \\
-4 & 6
\end{array}\right]\)
Solution:

Comparing the corresponding elements,
35 + 2 a = 1 ⇒ 2a = 1 – 35 = -34
⇒ a = \(\frac { -34 }{ 2 }\) = -17
27 + 2b = – 5 ⇒ 2b = -5 – 27 = – 32
⇒ b = \(\frac { -32 }{ 2 }\) = – 16
20 + 2c = – 4 ⇒ 2c = – 4 – 20 = – 24
⇒ c = \(\frac { -24 }{ 2 }\) = – 12
16 + 2d = 6 ⇒ 2d = 6 – 16 = – 10
⇒ d = \(\frac { -10 }{ 2 }\) = – 5
⇒ x = \(\left[\begin{array}{cc}
-17 & -16 \\
-12 & -5
\end{array}\right]\)

Question 5.
Given A = \(\left[\begin{array}{ll}
1 & 1 \\
8 & 3
\end{array}\right]\), evaluate A² – 4A.
Solution:

Question 6.
Find x and y, if
\(\left[\begin{array}{cc}
-3 & 2 \\
0 & -5
\end{array}\right]\left[\begin{array}{l}
x \\
2
\end{array}\right]=\left[\begin{array}{c}
-5 \\
y
\end{array}\right]\)
Solution:
\(\left[\begin{array}{cc}
-3 & 2 \\
0 & -5
\end{array}\right]\left[\begin{array}{l}
x \\
2
\end{array}\right]=\left[\begin{array}{c}
-5 \\
y
\end{array}\right]\)
⇒ \(\left[\begin{array}{c}
-3 x+2 \times 2 \\
0 \times x+(-5) \times 2
\end{array}\right]=\left[\begin{array}{c}
-5 \\
y
\end{array}\right]\)
⇒ \(\left[\begin{array}{c}
-3 x+4 \\
0-10
\end{array}\right]=\left[\begin{array}{c}
-5 \\
y
\end{array}\right] \Rightarrow\left[\begin{array}{c}
-3 x+4 \\
-10
\end{array}\right]=\left[\begin{array}{c}
-5 \\
y
\end{array}\right]\)
Comparing the corresponding elements,
– 3x + 4 = – 5
⇒ – 3x = – 5 – 4
⇒ – 3x = – 9
⇒ x = \(\frac { -9 }{ -3 }\) = 3
y = – 10
Hence x = 3, y = – 10

Question 7.
Find x and y, if \(\left[\begin{array}{ll}
x & 3 x \\
y & 4 y
\end{array}\right]\left[\begin{array}{l}
2 \\
1
\end{array}\right]=\left[\begin{array}{c}
5 \\
12
\end{array}\right]\).
Solution:
\(\left[\begin{array}{ll}
x & 3 x \\
y & 4 y
\end{array}\right]\left[\begin{array}{l}
2 \\
1
\end{array}\right]=\left[\begin{array}{c}
5 \\
12
\end{array}\right]\)
⇒ \(\left[\begin{array}{l}
2 x+3 x \\
2 y+4 y
\end{array}\right]=\left[\begin{array}{c}
5 \\
12
\end{array}\right] \Rightarrow\left[\begin{array}{l}
5 x \\
6 y
\end{array}\right]=\left[\begin{array}{c}
5 \\
12
\end{array}\right]\)
Comparing the corresponding elements, 5
5x = 5 ⇒ x = \(\frac { 5 }{ 5 }\) = 1
6y = 12 ⇒ y = \(\frac { 12 }{ 6 }\) = 2
Hence x = 1, y = 2

Question 8.
Find x and y, if:
\(\left[\begin{array}{cc}
3 & -2 \\
-1 & 4
\end{array}\right]\left[\begin{array}{c}
2 x \\
1
\end{array}\right]+2\left[\begin{array}{c}
-4 \\
5
\end{array}\right]=4\left[\begin{array}{l}
2 \\
y
\end{array}\right]\)
Solution:
\(\left[\begin{array}{cc}
3 & -2 \\
-1 & 4
\end{array}\right]\left[\begin{array}{c}
2 x \\
1
\end{array}\right]+2\left[\begin{array}{c}
-4 \\
5
\end{array}\right]=4\left[\begin{array}{l}
2 \\
y
\end{array}\right]\)
\(\left[\begin{array}{c}
6 x-2 \\
-2 x+4
\end{array}\right]+\left[\begin{array}{c}
-8 \\
10
\end{array}\right]=\left[\begin{array}{c}
8 \\
4 y
\end{array}\right]\)
\(\left[\begin{array}{c}
6 x-2-8 \\
-2 x+4+10
\end{array}\right]=\left[\begin{array}{c}
8 \\
4 y
\end{array}\right]\)
Now, comparing the corresponding elements of two equal matrices
6x – 10 = 8
∴ 6x = 18
∴ x = \(\frac { 18 }{ 6 }\) = 3
and 4y = – 2x + 14(Using x = 3)
4y = – 2 x 3 + 14
⇒ 4y = 14 – 6 = 8
y = \(\frac { 8 }{ 4 }\)
Hence x = 3, y = 2

Question 9.
Given A = \(\left[\begin{array}{cc}
2 & -1 \\
2 & 0
\end{array}\right]\), B = \(\left[\begin{array}{cc}
-3 & 2 \\
4 & 0
\end{array}\right]\) and C = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 2
\end{array}\right]\), find the matrix X, such that A + X = 2B + C.
Solution:

Comparing the corresponding elements,
2 + a = – 5 ⇒ a = – 5 – 2 ⇒ a = – 7
– 1 + b = 4 ⇒ b = 4 + 1 ⇒ b = 5
2 + c = 8 ⇒ c = 8 – 2 ⇒ c = 6
d = 2
Hence matrix X = \(\left[\begin{array}{cc}
-7 & 5 \\
6 & 2
\end{array}\right]\)

Question 10.
Find the value of x given that
A = \(\left[\begin{array}{cc}
2 & 12 \\
0 & 1
\end{array}\right], B=\left[\begin{array}{ll}
4 & x \\
0 & 1
\end{array}\right]\) and A² = B.
Solution:
A = \(\left[\begin{array}{cc}
2 & 12 \\
0 & 1
\end{array}\right], B=\left[\begin{array}{ll}
4 & x \\
0 & 1
\end{array}\right]\)
A² = A x A = \(\left[\begin{array}{cc}
2 & 12 \\
0 & 1
\end{array}\right]\left[\begin{array}{cc}
2 & 12 \\
0 & 1
\end{array}\right]\)
= \(\left[\begin{array}{cc}
2 \times 2+12 \times 0 & 2 \times 12+12 \times 1 \\
0 \times 2+1 \times 0 & 0 \times 12+1 \times 1
\end{array}\right]\)
= \(\left[\begin{array}{cc}
4+0 & 24+12 \\
0+0 & 0+1
\end{array}\right]=\left[\begin{array}{cc}
4 & 36 \\
0 & 1
\end{array}\right]\)
But A² = B
∴ \(\left[\begin{array}{cc}
4 & 36 \\
0 & 1
\end{array}\right]=\left[\begin{array}{ll}
4 & x \\
0 & 1
\end{array}\right]\)
Comparing the corresponding elements x = 36

Question 11.
Let A = \(\left[\begin{array}{cc}
4 & -2 \\
6 & -3
\end{array}\right], B=\left[\begin{array}{cc}
0 & 2 \\
1 & -1
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
-2 & 3 \\
1 & -1
\end{array}\right]\). Find A² – A + BC.
Solution:

Question 12.
(a) If 2\(\left[\begin{array}{ll}
3 & 4 \\
5 & x
\end{array}\right]+\left[\begin{array}{ll}
1 & y \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
7 & 0 \\
10 & 5
\end{array}\right]\), find the values of x and y.
(b) Let A = Find A² + AB + B².
Solution:
(a) 2\(\left[\begin{array}{ll}
3 & 4 \\
5 & x
\end{array}\right]+\left[\begin{array}{ll}
1 & y \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
7 & 0 \\
10 & 5
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
6 & 8 \\
10 & 2 x
\end{array}\right]+\left[\begin{array}{ll}
1 & y \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
7 & 0 \\
10 & 5
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
6+1 & 8+y \\
10+0 & 2 x+1
\end{array}\right]=\left[\begin{array}{cc}
7 & 0 \\
10 & 5
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
7 & 8+y \\
10 & 2 x+1
\end{array}\right]=\left[\begin{array}{cc}
7 & 0 \\
10 & 5
\end{array}\right]\)
Comparing the corresponding elements, 2x + 1 = 5 ⇒ 2x = 5 – 1 = 4
⇒ x = \(\frac { 4 }{ 2 }\)
and 8 + y = 0 ⇒ y = – 8
Hence x = 2, y = – 8

(b) \(A=\left[\begin{array}{ll}
1 & 0 \\
2 & 1
\end{array}\right], B=\left[\begin{array}{cc}
2 & 3 \\
-1 & 0
\end{array}\right]\)

Question 13.
(a) If \(\left[\begin{array}{cc}
1 & 4 \\
-2 & 3
\end{array}\right]\) + 2M = 3\(\left[\begin{array}{cc}
3 & 2 \\
0 & -3
\end{array}\right]\), find the matrix M.
(b) Given, \(\mathbf{A}=\left[\begin{array}{ll}
p & 0 \\
0 & 2
\end{array}\right], \mathbf{B}=\left[\begin{array}{cc}
0 & -q \\
1 & 0
\end{array}\right]\), C = \(\left[\begin{array}{cc}
2 & -2 \\
2 & 2
\end{array}\right]\) and BA = C². Find the values of p and q.
Solution:
(a) \(\left[\begin{array}{cc}
1 & 4 \\
-2 & 3
\end{array}\right]+2 \mathrm{M}=3\left[\begin{array}{cc}
3 & 2 \\
0 & -3
\end{array}\right]\)
Let M = \(\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\), then 2M = \(\left[\begin{array}{ll}
2 a & 2 b \\
2 c & 2 d
\end{array}\right]\)
∴ \(\left[\begin{array}{cc}
1 & 4 \\
-2 & 3
\end{array}\right]+\left[\begin{array}{cc}
2 a & 2 b \\
2 c & 2 d
\end{array}\right]=\left[\begin{array}{cc}
9 & 6 \\
0 & -9
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
1+2 a & 4+2 b \\
-2+2 c & 3+2 d
\end{array}\right]=\left[\begin{array}{cc}
9 & 6 \\
0 & -9
\end{array}\right]\)
Comparing the corresponding elements
1 + 2a = 9 ⇒ 2a = 9 – 1 = 8 ⇒ a = \(\frac { 8 }{ 2 }\) = 4
4 + 2b = 6 ⇒ 2b = 6 – 4 = 2 ⇒ b = \(\frac { 2 }{ 2 }\) = 1
– 2 + 2c = 0 ⇒ 2c = 2 ⇒ c = \(\frac { 1 }{ 2 }\) = 1
3 + 2d = – 9 ⇒ 2d = – 9 – 3 ⇒ 2d = – 12
⇒ d = \(\frac { -12 }{ 2 }\) = – 6
∴ M = \(\left[\begin{array}{cc}
4 & 1 \\
1 & -6
\end{array}\right]\) = – 6

(b) A = \(\left[\begin{array}{ll}
p & 0 \\
0 & 2
\end{array}\right], \mathrm{B}=\left[\begin{array}{cc}
0 & -q \\
1 & 0
\end{array}\right]\), C = \(\left[\begin{array}{cc}
2 & -2 \\
2 & 2
\end{array}\right]\)

Comparing the corresponding elements,
p = 8, -2 q = – 8 ⇒ q = \(\frac { -8 }{ -2 }\) = 4
⇒ p = 8, q = 4

Question 14.
Find x and y, if \(\left[\begin{array}{cc}
2 x & x \\
y & 3 y
\end{array}\right]\left[\begin{array}{l}
3 \\
2
\end{array}\right]=\left[\begin{array}{c}
16 \\
9
\end{array}\right]\)
Solution:
Given : \(\left[\begin{array}{cc}
2 x & x \\
y & 3 y
\end{array}\right]\left[\begin{array}{l}
3 \\
2
\end{array}\right]=\left[\begin{array}{c}
16 \\
9
\end{array}\right]\)
= \(\left[\begin{array}{l}
2 x \times 3+2 \times x \\
3 \times y+3 y \times 2
\end{array}\right]=\left[\begin{array}{c}
16 \\
9
\end{array}\right]\)
∴ \(\left[\begin{array}{l}
6 x+2 x \\
3 y+6 y
\end{array}\right]=\left[\begin{array}{c}
16 \\
9
\end{array}\right]=\left[\begin{array}{l}
8 x \\
9 y
\end{array}\right]=\left[\begin{array}{c}
16 \\
9
\end{array}\right]\)
⇒ 8x =16 and 9y = 9
⇒ x = 2 and y = 1

Question 15.
Given A = \(\left[\begin{array}{cc}
3 & -2 \\
-1 & 4
\end{array}\right]\), B = \(\left[\begin{array}{l}
6 \\
1
\end{array}\right]\), C = \(\left[\begin{array}{c}
-4 \\
5
\end{array}\right]\) and D = \(\left[\begin{array}{l}
2 \\
2
\end{array}\right]\), find AB + 2C – 4D.
Solution:

Question 16.
Evaluate \(\left[\begin{array}{cc}
4 \sin 30^{\circ} & 2 \cos 60^{\circ} \\
\sin 90^{\circ} & 2 \cos 0^{\circ}
\end{array}\right]\left[\begin{array}{ll}
4 & 5 \\
5 & 4
\end{array}\right]\)
Solution:
\(\left[\begin{array}{cc}
4 \sin 30^{\circ} & 2 \cos 60^{\circ} \\
\sin 90^{\circ} & 2 \cos 0^{\circ}
\end{array}\right]\left[\begin{array}{ll}
4 & 5 \\
5 & 4
\end{array}\right]\)
= \(\left[\begin{array}{cc}
4 \times \frac{1}{2} & 2 \times \frac{1}{2} \\
1 & 2 \times 1
\end{array}\right]\left[\begin{array}{ll}
4 & 5 \\
5 & 4
\end{array}\right]\)
= \(\left[\begin{array}{ll}
2 & 1 \\
1 & 2
\end{array}\right]\left[\begin{array}{ll}
4 & 5 \\
5 & 4
\end{array}\right]\)
= \(\left[\begin{array}{cc}
8+5 & 10+4 \\
4+10 & 5+8
\end{array}\right]=\left[\begin{array}{cc}
13 & 14 \\
14 & 13
\end{array}\right]\)

Question 17.
If A = \(\left[\begin{array}{cc}
3 & 5 \\
4 & -2
\end{array}\right]\) and B = \(\left[\begin{array}{l}
2 \\
4
\end{array}\right]\), is the product AB possible? Give a reason. If yes, find AB.
Solution:
Yes, the product is possible because number of column in A = number of row in B i.e., (2 x 2). (2 x 1) = (2 x 1) is the order of matrix.
AB = \(\left[\begin{array}{cc}
3 & 5 \\
4 & -2
\end{array}\right]\left[\begin{array}{l}
2 \\
4
\end{array}\right]=\left[\begin{array}{c}
3 \times 2+5 \times 4 \\
4 \times 2+(-2) \times 4
\end{array}\right]\)
= \(\left[\begin{array}{c}
6+20 \\
8-8
\end{array}\right]=\left[\begin{array}{c}
26 \\
0
\end{array}\right]\)

Question 18.
If A = \(\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]\) and I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\), find A² – 5A + 7I.
Solution:

Question 19.
Given A = \(\left[\begin{array}{cc}
2 & -6 \\
2 & 0
\end{array}\right]\), B = \(\left[\begin{array}{cc}
-3 & 2 \\
4 & 0
\end{array}\right]\), C = \(\left[\begin{array}{ll}
4 & 0 \\
0 & 2
\end{array}\right]\)
Find the matrix X such that A + 2X = 2B + C.
Solution:

Question 20.
Let A = \(\left[\begin{array}{cc}
2 & 1 \\
0 & -2
\end{array}\right]\), B = \(\left[\begin{array}{cc}
4 & 1 \\
-3 & -2
\end{array}\right]\) and C = \(\left[\begin{array}{ll}
-3 & 2 \\
-1 & 4
\end{array}\right]\). Find A² + AC – 5B.
Solution:

Question 21.
If A = \(\left[\begin{array}{ll}
3 & x \\
0 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
9 & 16 \\
0 & -y
\end{array}\right]\), find x and y when A² = B.
Solution:
Given : A = \(\left[\begin{array}{ll}
3 & x \\
0 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
9 & 16 \\
0 & -y
\end{array}\right]\), find x and y when A² = B
Now, A² = A x A
= \(\left[\begin{array}{ll}
3 & x \\
0 & 1
\end{array}\right] \times\left[\begin{array}{ll}
3 & x \\
0 & 1
\end{array}\right]\)
= \(\left[\begin{array}{cc}
9 & 3 x+x \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
9 & 4 x \\
0 & 1
\end{array}\right]\)
We have A² = B
Two matrices are equal if each and every corresponding element is equal.
Thus, \(\left[\begin{array}{cc}
9 & 4 x \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
9 & 16 \\
0 & -y
\end{array}\right]\)
⇒ 4x = 16 and 1 = – y
⇒ x = 4 and y = – 1

Question 22.
Given A = \(\frac { 1 }{ 2 }\) and I = \(\frac { 1 }{ 2 }\) and A² = 9A + ml. Find m.
Solution:
A² = 9A + ml
⇒ A² – 9A = ml … (i)
A² = AA

Question 23.
Given matrix A = \(\left[\begin{array}{cc}
4 \sin 30^{\circ} & \cos 0^{\circ} \\
\cos 0^{\circ} & 4 \sin 30^{\circ}
\end{array}\right]\) and B = \(\left[\begin{array}{l}
4 \\
5
\end{array}\right]\)
If AX = B
(i) Write the order of matrix ‘X’.
(ii) Find the matrix ‘X’.
Solution:
A = \(\left[\begin{array}{cc}
4 \sin 30^{\circ} & \cos 0^{\circ} \\
\cos 0^{\circ} & 4 \sin 30^{\circ}
\end{array}\right]\) and B = \(\left[\begin{array}{l}
4 \\
5
\end{array}\right]\)
(i) Order of matrix A is 2 x 2
Order of matrix B is 2 x 1
∴ Order of matrix X is 2 x 1

(ii) Let the matrix X = \(\left[\begin{array}{l}
4 \\
5
\end{array}\right]\)
AX = B
⇒ \(\left[\begin{array}{cc}
4 \sin 30^{\circ} & \cos 0^{\circ} \\
\cos 0^{\circ} & 4 \sin 30^{\circ}
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
4 \\
5
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
4\left(\frac{1}{2}\right) & 1 \\
1 & 4\left(\frac{1}{2}\right)
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
4 \\
5
\end{array}\right]\)
⇒ \(\left[\begin{array}{ll}
2 & 1 \\
1 & 2
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
4 \\
5
\end{array}\right]\)
⇒ \(\left[\begin{array}{l}
2 x+y \\
x+2 y
\end{array}\right]=\left[\begin{array}{l}
4 \\
5
\end{array}\right]\)
⇒ 2x + y = 4 … (i)
and x + 2y – 5 … (ii)
Subtracting (ii) from (i), we get
⇒ 2x+ y – (x + 2y) = 4 – 5
⇒ 2x + y – x – 2y = 4 – 5
x – y = – 1 … (iii)
Adding (i) and (ii), we get
⇒ 2x + y + x + 2y = 4 + 5
⇒ 3x + 3y = 9
⇒ x + y = 3
Adding (iii) and (iv), we get
2x = 2 ⇒ x – 1
Substitute x in (iv), we get = 2
Hence, the matrix X = \(\left[\begin{array}{l}
1 \\
2
\end{array}\right]\)

Well-structured Class 10 ICSE Maths Solutions S Chand Chapter 15 Three Dimensional Solids Ex 15(b) facilitate a deeper understanding of mathematical principles.

S Chand Class 10 ICSE Maths Solutions Chapter 15 Three Dimensional Solids Ex 15(b)

Question 1.
Find the volume of the cylinders whose radii and heights are given below. Take π to be \(\frac { 22 }{ 7 }\).
(i) r = 7 cm, h = 8 cm
(ii) r = 7 cm, h = 12 cm
(iii) r = 14 cm, h = 16 cm
(iv) r = 21 cm, h = 40 cm
Solution:
(i) Radius of a cylinder (r) = 7 cm and height (h) = 8 cm
∴ Volume = πr²h = \(\frac { 22 }{ 7 }\) × 7 × 7 × 8 cm³ = 1232 cm³

(ii) Radius of cylinder (r) 7 cm and height (h) = 12 cm
∴ Volume = πr²h = \(\frac { 22 }{ 7 }\) × 7 × 7 × 12 cm³ = 1848 cm³

(iii) Radius of the cylinder (r)= 14 cm and height (h)= 16 cm
∴ Volume = πr²h = \(\frac { 22 }{ 7 }\) × 14 × 14 × 16 = 9856 cm³

(iv) Radius of the cylinder (r) = 21 cm and height (h) = 40 cm
∴ Volume = πr²h = \(\frac { 22 }{ 7 }\) × 21 × 21 × 40 cm³ = 55440 cm³

Question 2.
Find the diameter of the circular cylinders if:
(a) Volume is 44 cm³, height 3.5 cm;
(b) Volume 385 cm³, height 1 dm.
Solution:
(a) Volume of a cylinder = 44 cm³
Height (h) = 3.5 cm = \(\frac { 35 }{ 10 }\)cm
∴ Radius = \(\sqrt{\frac{\text { Volume }}{\pi h}}\)
= \(\sqrt{\frac{44 \times 7 \times 10}{22 \times 35}}\) = √4 = 2 cm
∴ Diameter = 2 × r = 2 × 2 = 4 cm

(b) Volume of the cylinder = 385 cm²
and height (h) = 1 dm = 10 cm
∴ Radius = \(\sqrt{\frac{\text { Volume }}{\pi h}}\)
= \(\sqrt{\frac{385 \times 7}{22 \times 10}}\) = \(\sqrt{\frac{49}{4}}\) = \(\frac{7}{2}\) cm
∴ Diameter = 2r = \(\frac{7}{2}\) × 2 = 7 cm

Question 3.
Find the height of the circular cylinders if:
(a) Volume is 66 cm³, radius 2 cm;
(b) Volume 4 litres, radius 5 cm.
Solution:
(a) Volume of a cylinder = 66 cm³ and radius (r) = 2 cm
∴ Height = \(\frac{\text { Volume }}{\pi r^2}\)
= \(\frac{66 \times 7}{22 \times 2 \times 2}\) = \(\frac{21}{4}\)cm = 5.25 cm

(b) Volume of a cylinder = 4 litres = 4000 cm³
Radius (r) = 5 cm
∴ Height = \(\frac{\text { Volume }}{\pi r^2}\)
= \(\frac{4000 \times 7}{22 \times 5 \times 5}\)cm = \(\frac{560}{11}\)cm = 50.9 cm

Question 4.
A wooden pole is 7 m high and 20 cm in diameter. Find its weight if the wood weighs 225 kg per m³.
Solution:
Diameter of a wooden pole = 20 cm
∴ Radius (r) = \(\frac{20}{2}\) = 10 cm
and height (h) = 7 m = 700 cm
∴ Volume = πr²h = \(\frac{22}{7}\) × 10 × 10 × 7 × 100 cm³
= 220000 cm³ = \(\frac{220000}{1000000}\) m³
Weight of 1 m³ = 225 kg
Total weight = \(\frac{220000}{1000000}\) × 225 kg
= \(\frac{198}{4}\) = 49\(\frac{1}{2}\)kg = 49.5 kg

Question 5.
Find the volume of metal in the hollow pipe of internal radius 3 cm, metal 1 cm thick and length 6 cm.
Solution:
Length of metal hollow pipe (h) = 6 cm
Internal radius (r) = 3 cm
Thickness of pipe = 1 cm
∴ Outer radius (R) = 3 + 1 = 4 cm
∴ Volume of the metal used = πh(R² – r²)
= \(\frac{22}{7}\) × 6 × (4² – 3²)
= \(\frac{22}{7}\) × 6 × 7 × 1 = 132 cm³

Question 6.
The sum of the radius of the base and the height of a solid cylinder is 37 cm. If the total surface area of the solid cylinder is 1628 cm², find the volume of the cylinder. (Take π = \(\frac{22}{7}\))
Solution:
Sum of the radius of the base and height of a solid cylinder = 37 cm
Total surface area = 1628 cm²
∴ 2πr(h + r) = 1628
⇒ 2 × \(\frac{22}{7}\) × r × 37 = 1628
r = \(\frac{1628 \times 7}{2 \times 22 \times 37}\) = 7 cm
∴ Height = 37 – 7 = 30 cm
Now volume = πr²h
= \(\frac{22}{7}\) × 7 × 7 × 30 cm³ = 4620 cm³

Question 7.
A cylindrical tank has capacity 6160 cu m. Find its depth if the diameter of its base is 28 m. Also, find the area of the inside curved surface of the tank. (Take π = \(\frac{22}{7}\) )
Solution:
Capacity of a cylindrical tank = 6160 cu m Diameter of its base = 28 m
∴ Radius (r) = \(\frac{28}{2}\) = 14 m
∴ Depth (h) = \(\frac{\text { Volume }}{\pi r^2}\)
= \(\frac{6160 \times 7}{22 \times 14 \times 14}\) = 10 m
Area of curved surface of its inner side = 2πrh
= 2 × \(\frac{22}{7}\) × 14 × 10 = 880 m²

Question 8.
The area of the curved surface of a cylinder is 4400 cm², and the circum-ference of its base is 110 cm. Find
(i) the height of the cylinder.
(ii) the volume of the cylinder.
\(\left(\text { Take } \pi \text { to be } \frac{22}{7}\right)\)
Solution:
Curved surface area of a cylinder = 4400cm²
Circumference its base = 110 cm
∴ 2πrh = 4400
and 2πr = 110 cm ⇒ r = \(\frac{110 \times 7}{2 \times 22}\) = \(\frac{35}{2}\)cm

(i) ∴ Height (h) = \(\frac{4400}{110}\) = 40 cm
(ii) Volume = πr²h = \(\frac{22}{7}\) × \(\frac{35}{2}\) × \(\frac{35}{2}\) × 40 = 38500 cm²

Question 9.
(i) How many cubic metres of earth must be dug out to make a well 20 metres deep and 2 metres in diameter? \(\left(\text { Take } \pi \text { to be } \frac{22}{7}\right)\)
(ii) If the inner curved surface of the well in part (i) above is to be plastered at the rate of ₹5 per m², find the cost of Plastering \(\left(\text { Take } \pi \text { to be } \frac{22}{7}\right)\)
Solution:
Depth of a well (h) = 20 m
Diameter = 2 m
∴ Radius (r) = \(\frac{2}{2}\) = 1 m

(i) ∴ Volume of earth dugout = πr²h
= \(\frac{22}{7}\) × 1 × 1 × 20 = \(\frac{440}{7}\) m³
= 66\(\frac{6}{7}\) m³

(ii) Area of inner curved surface = 2πrh
= 2 × \(\frac{22}{7}\) × 1 × 20 = \(\frac{880}{7}\) m²
Rate of Plastering the inner surface = ₹5 per m²
∴ Total cost = \(\frac{880}{7}\) × 5
= ₹ \(\frac{4400}{7}\) = ₹628.57

Question 10.
A cylinder has a diameter of 20 cm. The area of curved surface is 1000 cm² (sq cm). Find
(i) the height of the cylinder correct to one decimal place.
(ii) the volume of the cylinder correct to one decimal place. (Take n to be 3.14)
Solution:
Diameter of a cylinder = 20 cm
∴ Radius (r) = \(\frac{20}{2}\) = 10 cm
Curved surface area = 1000 cm²

(i) ∴ Height = \(\frac{\text { Curved surface area }}{\text { 2πr }}\)
= \(\frac{1000}{2 \times 3.14 \times 10}\) = \(\frac{100}{6.28}\)cm = 15.9 cm

(ii) Volume = πr²h
= 3.14 × 10 × 10 × 15.9 cm³
= 314 × 15.9 = 4992.6 cm³

Question 11.
An electric geyser is cylindrical in shape, having a diameter of 35 cm and height of 1.2 m. Neglecting the thickness of its walls, calculate
(i) its outer lateral surface area;
(ii) its capacity in liters \(\left(\text { Take } \pi \text { to be } \frac{22}{7}\right)\)
Solution:
Diameter of a geyser = 35 cm
∴ Radius (r) = \(\frac{35}{2}\)cm
and height (h) = 1.2 m = 120 cm

(i) ∴ Outer lateral surface area
= 2nrh = 2 × \(\frac{22}{7}\) × \(\frac{35}{2}\) × 120 cm²
= 13200 cm²

(ii) Capacity = πr²h
= \(\frac{22}{7}\) × \(\frac{35}{2}\) × \(\frac{35}{2}\) × 120 cm³ = 115500 cm³
= \(\frac{115500}{1000}\) litres = 115.5 litres

Question 12.
How many cylindrical glasses, diameter 8 cm, height 15 cm, can be filled from cylindrical vessel, diameter 30 cm, height 80 cm, full of milk. (There is no need to substitute for π).
Solution:
Diameter of a cylindrical vessel = 30 cm
∴ Radius (R) = \(\frac{30}{2}\) = 15 cm
and height (H) = 80 cm
∴ Volume = πR²H
= π(15)² × 80 = π × 225 × 80 cm³ Diameter of cylindrical glass = 8 cm
∴ Radius (r) = \(\frac{8}{2}\) = 4 cm
and height (h) = 15 cm
∴ Volume of glass = πr²h
= π(4)² × 15 = π × 4 × 4 × 15 cm³
= 240π cm³
∴ Number of glasses = \(\frac{\text { Volume of vessel }}{\text { volume of glass }}\) = \(\frac{225 \times 80 \pi}{240 \pi}\) = 75 glasses

Question 13.
A path 2 m wide surrounds a circular pond of diameter 40 m. How many cubic metre of gravel are required to gravel the path to a depth of 7 cm?
Solution:
Width of circular path = 2 m
Diameter of circular pond = 40 m
Inner radius (r) = \(\frac{40}{2}\) = 20 m
Outer radius (R) = 20 + 2 = 22 m
∴ Area of path = π[R² – r²]
= \(\frac{22}{7}\) [22² – 20²]
= \(\frac{22}{7}\) × 42 × 2 = 264 m²
Depth of path = 7 cm = \(\frac{7}{100}\) m
∴ Volume of gravel required = 264 × \(\frac{7}{100}\) m³
= \(\frac{1848}{100}\) = 18.48 m³

Question 14.
An iron block is in the form of a cylinder of diameter 0.5 m and length 3.5 m. This block is to be rolled into the form of a bar having square section of side 25 cm. Find the length of the bar.
Solution:
Diameter of iron cylindrical block = 0.5 m
∴ Radius (r) = \(\frac{0.5}{2}\) = 0.25 m
= 25 cm
Length = 3.5 m = 350 cm
Volume of block = πr²h
= \(\frac{22}{7}\) × 25 × 25 × 350 cm³
= 687500 cm³
∴ Volume of bar = \(\frac{687500}{25 \times 25}\) = 1100 cm = 11 m

Question 15.
A swimming pool 70 m long, 44 m wide, 3 m deep is fdled by water issuing from a pipe of diameter 14 cm, at 2 m per second. How many hours does it take to fill the bath?
Solution:
Length of swimming pool (l) = 70 m
Breadth (b) = 44 m
and depth (d) = 3 m
∴ Volume of water in it = 70 × 44 × 3 m³ = 9240 m³
Diameter of the pipe = 14 cm
∴ Radius (r) = \(\frac{14}{2}\) = 7 cm = 0.07 m
Length of water flow
= \(\frac{\text { Volume of water }}{\text { Area of mouth of pipe }}\)
= \(\frac{9240}{\pi r^2}\) = \(\frac{9240 \times 7}{22 \times 0.07 \times 0.07}\)m
= \(\frac{60}{0.0001}\) = 60 × 10000 m
Speed of water flow = 2 m per second
∴ Time taken = \(\frac{60 \times 10000}{2}\)seconds
= \(\frac{60 \times 10000}{2 \times 60 \times 60}\)hrs. = \(\frac{250}{3}\)hrs = 83\(\frac{1}{3}\)hours

Question 16.
What length of solid cylinder 2 cm in diameter must be taken to be cast into a hollow cylinder of external diameter 12 cm, 0.25 cm thick, 15 cm long?
Solution:
External diameter of a hollow cylinder = 12 cm
∴ Radius (R) = \(\frac{12}{2}\) = 6 cm
Thickness of sheet = 0.25 cm
∴ Inner radius (r) = 6 – 0.25 = 5.75 cm and length (h)= 15 cm
∴ Volume of metal used = πh(R² -r²)
= \(\frac{22}{7}\) × 15 × (6² – 5.75²) cm³
= \(\frac{330}{7}\) (11.75 × 0.25) cm³
= \(\frac{330}{7}\) × 11.75 × 0.25 cm³

Now volume of solid cylinder
= \(\frac{330}{7}\) × 11.75 × 0.25 cm³
Diameter of cylinder = 2 cm
∴ Radius = \(\frac{2}{2}\) = 1 cm
∴ Length of cylinder = \(\frac{\text { Volume }}{\pi r^2}\) = \(\frac{330 \times 11.75 \times 0.25}{7 \times \pi \times 1 \times 1}\)
= \(\frac{330 \times 1175 \times 25 \times 7}{7 \times 100 \times 100 \times 22}\) = \(\frac{705}{16}\) = 44.0625 cm

Question 17.
A cylindrical tube open at both ends is made of metal. The internal diameter of the tube is 11.2 cm, and its length is 21 cm. The metal every where is 0.4 cm thick. Calculate the volume of the metal to 1 place od decimal. \(\left(\pi=\frac{22}{7}\right)\)
Solution:
Internal diameter of the open tube = 11.2cm
∴ Radius (r) = \(\frac{11.2}{2}\) = 5.6 cm
Length (h) = 21 cm
Thickness of tube = 0.4 cm
∴ External radius (R) = 5.6 + 0.4 = 6.0 cm
∴ Volume = πh(R² – r²)
= \(\frac{22}{7}\) × 21 × (6² – 5.6²) cm³
= 306.24 cm³ = 306.2 cm³

Question 18.
Water flows along a pipe of radius 0.6 cm at 8 cm per second. This pipe is draining the water from a tank which holds 1000 litres of water when full. How long would it take to completely empty the tank?
Solution:
Water in the tank = 1000 l = 1 m³
∴ Volume of water = 1000 × 1000 m³ = 1000000 cm³
Radius of pipe = 0.6 cm
Area of mouth of pipe = πr²
= \(\frac { 22 }{ 7 }\) × 0.6 × 0.6 = \(\frac { 792 }{ 700 }\) cm²
∴ Length of flow of water in pipe = \(\frac{1000000 \times 7 \times 10 \times 10}{22 \times 6 \times 6}\)
= \(\frac{700000000}{792}\)cm
speed of water from 8 cm per sec.
∴ Time taken = \(\frac{700000000}{792 \times 8}\)sec.
= \(\frac{700000000}{792 \times 8 \times 60 \times 60}\)hr
= \(\frac{1104780}{60 \times 60}\)hr = 30.69 hours

Question 19.
A cylindrical bucket 28 cm in diameter; 72 cm high and full of water, is emptied into a rectangular tank 66 cm long, 28 cm wide. Find the height of the water-level in the tank. \(\left(\pi=\frac{22}{7}\right)\)
Solution:
Diameter of cylindrical bucket = 28 cm
∴ Radius (r) = \(\frac { 28 }{ 2 }\) = 14 cm
Height (h) = 72 cm
∴ Volume of water in it = πr²h
= \(\frac { 22 }{ 7 }\) × 14 × 14 × 72 cm³ = 44352 cm³
Water in the tank = 44352 cm³
Length of rectangular tank = 66 cm
Breadth = 28 cm
∴ Height = \(\frac{\text { Volume }}{l \times b}\) = \(\frac{44352}{66 \times 28}\) = 24 cm

Question 20.
There is some water in a cylindrical vessel of base diameter 14 cm. When an iron- cube is entirely immersed in it, the height of the water rises by 8\(\frac { 9 }{ 14 }\) cm. Find the length of the edge of the cube \(\left(\pi=\frac{22}{7}\right)\)
Solution:
Diameter of cylindrical vessel = 14 cm
∴ Radius (r) = \(\frac { 14 }{ 2 }\) = 7 cm
Height of water = 8\(\frac { 9 }{ 14 }\) = \(\frac { 121 }{ 14 }\) cm
∴ Volume of water = πr²h
= \(\frac { 22 }{ 7 }\) × 7 × 7 × \(\frac { 121 }{ 14 }\)cm³ = 1331 cm³
Now volume of cube = 1331 cm³

Question 21.
If the radius of the base of a right circular cylinder is halved, keeping the height same, what is the ratio of the volume of the reduced cylinder to that of the original one?
Solution:
Let radius of the base of a cylinder = r and height = h
∴ Volume πr²h
If radius is halved, then
Radius (r¹) = \(\frac { r }{ 2 }\)
Height = h
Volume = π(r¹)²h
∴ Volume = π\(\left(\frac{r}{2}\right)^2\) h = \(\frac{\pi r^2 h}{4}\)
∴ Ratio in the volume is reduced to the original cylinder = \(\frac{\pi r^2 h}{4}: \pi r^2 h\)
= 1 : 4

Question 22.
A rectangular sheet of paper 22 cm long and 12 cm broad can be formed into the curved surface of a cylinder in two ways. Find the difference between the volumes of the two cylinders which can be formed.
Solution:
Length of sheet = 22 cm
and breadth = 12 cm
If it is folded length wire, then Circumference of so formed cylinder = 22cm
∴ Radius (r) = \(\frac{C}{2 \pi}\) = \(\frac{22 \times 7}{2 \times 22}\) = \(\frac{7}{2}\) cm
and height (h) = 12 cm
Volume = πr²h = \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\) × 12 cm³ = 462 cm³
If it is folded breadth wire, then
Circumference = 12 cm
∴ Radius = \(\frac{12}{2 \pi}\) = \(\frac{12 \times 7}{2 \times 22}\) = \(\frac{2}{11}\)cm
and height (h) = 22 cm
∴ volume = πr²h = \(\frac{22}{7}\) × \(\frac{21}{11}\) × \(\frac{21}{11}\) × 22 = 252 cm³
Difference volumes = 462 – 252 = 210 cm³

Question 23.
A 20 m deep well with diameter 7 m is dug up and the earth from digging is spread evenly to form a platform 22 m × 14 m. Determine the height of the platform.
Solution:
Depth of a well = 20 m
Diameter = 7 m
∴Radius = \(\frac{7}{2}\) m
∴ Volume of earth so dugout = πr²h
= \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\) × 20 m³ = 770 m³
Now volume of the platform = 770 m³
Size of the platform = 27 m × 14 m
∴ Height = \(\frac{\text { Volume }}{l \times b}\) = \(\frac{770}{22 \times 14}\) = \(\frac { 5 }{ 2 }\)m = 2.5 m

Question 24.
The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 min in diameter. A full barrel of ink in the pen will be used up when writing 310 words on an average. How many words would use up a bottle of ink containing one fifth of a litre?
(Answer correct to the nearest 100 words) \(\left(\pi=\frac{22}{7}\right)\)
Solution:
Length of cylindrical barrel of a pen = 7 cm and diameter – 5 mm
∴ Radius = \(\frac{5}{2}\)mm = \(\frac{5}{20}\) = \(\frac{1}{4}\)cm
∴ volume of link in it = πr²h
= \(\frac{22}{7}\) × \(\frac{1}{4}\) × \(\frac{1}{4}\) × 7 = \(\frac{11}{8}\) cm³
Volume of link in the bottle = \(\frac{1}{5}\)l = \(\frac{1000}{5}\) = 200 cm³
∴ Total number of barrels = 200 ÷ \(\frac{11}{8}\)
= \(\frac{200 \times 8}{11}\) = \(\frac{1600}{11}\)
words written in one barrel = 310
Total number of words = \(\frac{1600}{11}\) × 310
= 45090 words
= 45100 words (approx)

Question 25.
A closed rectangular box 40 cm long, 30 cm wide and 25 cm deep, has the same volume as that of a cylindrical tin of radius 17.5 cm. Calculate the height of the cylindrical tin correct to 1 decimal place. (Take π to be 3.14)
Solution:
Length of a rectangular box (l) = 40 cm
Breadth (b) = 30 cm
and height (h) = 25 cm
Volume = lbh = 40 × 30 × 25 cm³ = 30000 cm³
Radius of cylindrical tin = 17.5 cm
Volume = 30000 cm³
Height = \(\frac{\text { Volume }}{\pi r^2}\) = \(\frac{30000}{3.14 \times 17.5 \times 17.5}\)cm = \(\frac{30000 \times 10 \times 10}{3.14 \times 175 \times 175}\) = 31.197 cm = 31.2 cm

Question 26.
Earth taken out on digging a circular tank of diameter 17.5 m is spread all around the tank uniformly to a width of 4 m, to form an embankment of height 2 m. Calculate the depth of the circular tank, correct to two decimal places.

Solution:
Diameter of a circular tank = 17.5 m
∴ Radius = \(\frac{17.5}{2}\) = 8.75 m
width of embankment = 4 m
∴ Outer radius = 8.75 + 4 = 12.75 m
and height (h) = 2 m
∴ Volume of the embankment
= πh(R² – r²) = \(\frac{22}{7}\) × 2 × (12.75² – 8.75²)
= \(\frac{44}{7}\)(21.5 × 4) m³
= \(\frac{44}{7}\) × 86 m³ = \(\frac{3784}{7}\) m³
∴ Volume of earth dugout of the tank = \(\frac{3784}{7}\) m³
∴ Height = \(\frac{\text { Volume }}{\pi r^2}\) = \(\frac{3784 \times 7}{7 \times 8.75 \times 8.75 \times 22}\)m
= 2.246 = 2.25 m

Question 27.
Water is flowing at the rate of 7 metres per second through a circular pipe whose internal diameter is 2 cm into a cylindrical tank, the radius of whose base is 40 cm. Determine the increase in the water level in \(\frac{1}{2}\) hour.
Solution:
Speed of waterflow = 7 m per sec.
Internal diameter of pipe = 2 cm
∴ Radius (r) = \(\frac{2}{2}\) = 1 cm
Radius of the base of the circular tank = 40 cm
Time = \(\frac{1}{2}\) hour
Waterflow in \(\frac{1}{2}\) hour (h) =\(\frac{7 \times 60 \times 60}{2}\)m
= 12600 m = 12600 × 100 = 1260000 cm
∴ Volume of water = πr²h
= \(\frac{22}{7}\) × 1 × 1 × 126000 cm³
= 3960000 cm³
Now volume of water in the tank = 3960000 cm³
Radius = 40 cm
∴ Height = \(\frac{\text { Volume }}{\pi r^2}\)
= \(\frac{3960000 \times 7}{22 \times 40 \times 40}\) = \(\frac{6300}{8}\) cm = 787.5 cm

Question 28.
Water flow through a cylindrical pipe of internal diameter 7 cm at 36 km/hr. Calculate time in minutes it would take to fill a cylindrical tank, the radius of whose base is 35 cm and height is 1 m.
Solution:
Diameter of circular end of pipe = 7 cm
Radius (r₁) of circular end of pipe = \(\frac{7}{2}\) cm
= 3.5 cm or 0.035 m
Area of cross section = π × r²
= π × (0.35)² m²
= 1225 × 10^-6 π m²
Speed of water = 36 km/h
= 36 × \(\frac{1000}{60}\) = 600 m
Volume of water that flows in 1 minute from pipe = 600 × 1225 × 10^-6 π m²
Volume of water that flows in t minutes from pipe = t × 735 × 10^-3 π m²
Radius (r₂) of cicular end of cylindrical tank = 35 cm = 0.35 m
Depth (h₂) of cylindrical tank = 1 m
Let the tank be filled completely in t minutes.
Volume of water filled in tank in t minutes is equal to the volume of water flowed in t minutes from the pipe.
Volume of water that flows in t times = Volume of water in tank
⇒ t × 735 × 10^-3 π m² = π × (r₂)² × h²
⇒ t = 0.35 × 0.35 × 1 × \(\frac{1000}{735}\)
⇒ t = 0.166 minutes

Question 29.
Find the number of coins, 1.5 cm in diameter and 0.2 cm thick to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.
Solution:
Diameter of a coin = 1.5 cm
∴ Radius (r) = \(\frac{1.5}{2}\) = 0.75 cm
and thickness = 0.2 cm
∴ Volume of one coin = πr²h
= π × 0.75 × 0.75 × 0.2 cm3 = 0.11257π cm³
Diameter of cylinder = 4.5 cm
∴ Radius = \(\frac{4.5}{2}\) = 2.25 cm
and height = 10 cm
∴ Volume = π × 2.25 × 2.25 × 10 cm³ = 50.625π cm³
∴ Number of coins = \(\frac{50.625 \pi}{0.1125 \pi}\) = 450

Question 30.
If 1 cubic cm of cast iron weighs 21 g, then find the weight of a cast iron pipe of length 1 m with a bore of 3 cm in which the thickness of the metal is 1 cm.
Solution:
Weight of 1 cm³ = 21 g
Length of pipe (h) = 1 m = 100 cm
Internal diameter = 3 cm
∴ Radius (r) = \(\frac{3}{2}\) = 1.5 cm
Thickness of metal = 1 cm
∴ External radius (R) = 1.5 + 1 = 2.5 cm
Now volume of iron = πh(R² – r²)
= \(\frac{22}{7}\) × 100(2.5² = 1.5²) cm³
= \(\frac{2200}{7}\)(4 × 1) cm³
= \(\frac{8800}{7}\) cm³
∴ Total weight of the metal = \(\frac{8800}{7}\) × 21 g
\(\frac{8800 \times 3}{1000}\) kg = 26.4 kg

Practicing Class 10 ICSE Maths Solutions S Chand Chapter 15 Three Dimensional Solids Ex 15(a) is the ultimate need for students who intend to score good marks in examinations.

S Chand Class 10 ICSE Maths Solutions Chapter 15 Three Dimensional Solids Ex 15(a)

Question 1.
Find the area of curved surface and total surface area of the cylinders whose heights and radii are given below:
(i) h = 12 cm, r = 7 cm
(ii) h = 10 cm, r = 7 cm
(iii) h = 5 cm, r = 21 cm
(iv) h = 20 cm, r = 14 cm
(v) h = 16 m, r = 10.5 m
Solution:
(i) Height of a cylinder (h) = 12 cm and radius (r) = 7 cm
∴ Curved surface area = 2πrh
= 2 × \(\frac { 22 }{ 7 }\) × 7 × 12 = 528 cm²
Total surface area = 2πr(h + r)
= 2 × \(\frac { 22 }{ 7 }\) × 7(12 + 7) cm²
= 44 × 19 = 836 cm²

(ii) Height of the cylinder (h) = 10 cm and radius (r) = 1 cm
∴ Curved surface area = 2πrh
= 2 × \(\frac { 22 }{ 7 }\) × 7 × 10 = 440 cm²
and total surface area = 2πr(h + r)
= 2 × \(\frac { 22 }{ 7 }\) × 7(10 + 7) cm²
= 44 × 17 = 748 cm²

(iii) Height of the cylinder (h) = 5 cm and radius (r) = 21 cm
Curved surface area = 2nrh
= 2 × \(\frac { 22 }{ 7 }\) × 21 × 5 = 660 cm²
and total surface area = 2πr(h + r)
= 2 × \(\frac { 22 }{ 7 }\) × 21(5 + 21) cm²
= 132 × 26 = 3432 cm²

(iv) Height of cylinder (h) = 20 cm and radius (r)= 14 cm
∴ Curved surface area = 2πrh
= 2 × \(\frac { 22 }{ 7 }\) × 14 × 20 cm² = 1760 cm²
Total surface area = 2πr(h + r)
= 2 × \(\frac { 22 }{ 7 }\) × 14 × (20 + 14) cm²
= 88 × 34 cm² = 2992 cm²

(v) Height of cylinder (l) = 16 m
and radius (r) = 10.5 m = \(\frac { 21 }{ 2 }\) m
∴ Curved surface area = 2πrh
= 2 × \(\frac { 22 }{ 7 }\) × \(\frac { 21 }{ 2 }\) × 16 m² = 1056 m²
and total surface area = 2πr(h + r)
= 2 × \(\frac { 22 }{ 7 }\) × \(\frac { 21 }{ 2 }\) \(\left(16+\frac{21}{2}\right)\) m²
= 66 × \(\frac { 53 }{ 2 }\) = 1749 m²

Question 2.
A cylindrical tank 7 m in diameter, contains water to a depth of 4 m. Find the total area of the wet surface. (π = 3.14)
Solution:
Diameter of the base of a tank = 7 m
∴ Radius (r) = \(\frac { 7 }{ 2 }\) m
Depth (h) = 4 m
∴ Total surface area of the wet surface
= 2πrh + πr² = πr(2h + r) m²
= 3.14 × \(\frac { 7 }{ 2 }\)(2 × 4 + 3.5) m²
= 3.14 × \(\frac { 7 }{ 2 }\) × 11.5 m²
= 126.385 m² = 126.39 m²

Question 3.
Find the whole surface of a hollow cylinder open at both ends, whose external diameter is 14 cm, thickness 2 cm and height 20 cm.
Solution:
External diameter of a hollow cylinder = 14 cm
Radius (R) = \(\frac { 14 }{ 2 }\) = 7 cm
Thickness of cylinder = 2 cm
∴ Inner radius (r) = 7 – 2 = 5 cm
Height (h) = 20 cm
Inner and external surface area
2πrh + 2πRh
= 2πh(r + R)
=2 × π × 20(7 + 5) cm²
=40π × 12 = 480π cm²
Area of two rings (upper and lower)
= 2 × π(R² – r²) = 2(7² – 5²)
= 2π × (49 – 25) = 2π × 24 = 48π cm²
∴ Total surface area – 480π × 48π
= 528π cm²

Question 4.
Find the radius of the cylinder if area of its curved surface is 110 cm2, and height 5 cm. \(\left(\text { Take } \pi=\frac{22}{7}\right)\)
Solution:
Area of curved surface of a cylinder = 110 cm²
and height (h) = 5 cm
Curved surface area = 2πrh
Then radius (r) = \(\frac{\text { Curved surface area }}{\text { 2πh }}\)

Question 5.
Find the height of the cylinder if area of its curved surface is 13.2 cm², and radius 6 cm.
Solution:
Area of curved surface of a cylinder = 13.2 cm²
Radius (r) = 6 cm
Curved surface area = 2πrh
∴ Height (h) = \(\frac{\text { Surface area }}{\text { 2πr }}\)

Question 6.
A garden roller is 75 cm in diameter and 105 cm in width. What area does it cover in 14 revolutions?
Solution:
Diameter of a garden roller = 75 cm
∴ Radius (r) = \(\frac { 75 }{ 2 }\) cm
and width (h) = 105 cm
∴ Curved surface area = 27πrh
= 2 × \(\frac { 22 }{ 7 }\) × \(\frac { 75 }{ 2 }\) × 105 cm² = 24750 cm²
Area covered in 14 revolutions
= 24750 × 14 cm² = 346500 cm²
= \(\frac{346500}{100 \times 100}\) = 34.65 m²

Question 7.
10 cylindrical pillars of a building have to be painted. If the diameter of each pillar is 50 cm and the height 4 m, what will be the cost of painting these at the
rate of 50 paise per m²? (Use π = 3.14)
Solution:
Diameter of each cylinder = 50 cm
∴ Radius (r) = \(\frac { 50 }{ 2 }\) = 25 cm = \(\frac { 25 }{ 100 }\) m
and height (h) = 4 m
∴ Curved surface area = 2πrh
= 2 × 3.14 × \(\frac { 25 }{ 100 }\) × 4 m² = 6.28 m²
Area of 10 cylinders = 6.28 × 1o m² = 62.8m²
Rate of painting = 50 paise per m²
Total cost 62.8 × \(\frac { 50 }{ 100 }\) = ₹ 31.40

Question 8.
The diameter of a roller 120 cm long is 84 cm. If it takes 500 complete revolutions to level a playground, determine the cost of levelling at the rate of 30 paise per m². \(\left(\text { Use } \pi=\frac{22}{7}\right)\)
Solution:
Diameter of a roller = 84 cm
∴ Radius (r) = \(\frac { 84 }{ 2 }\) = 42 cm
and length (h) 120 cm
Curved surface area = 2πrh
= 2 × \(\frac { 22 }{ 7 }\) × 42 × 120 cm² = 31680 cm²
Area of leveling playground by 500 revolutions = 31680 × 500 cm²
= \(\frac{31680 \times 500}{100 \times 100}\) = 1584 cm²
Rate of leveling the ground =30 paise per m²
∴ Total cost = ₹ \(\frac{1584 \times 30}{100}\) = ₹ 475.20

Question 9.
The curved surface of a cylinder is 1000 cm². A wire of diameter 5 mm is wound round it so as to cover it completely. Find the length of the wire.
Solution:
Curved surface area of a cylinder = 1000cm²
It is covered by a wire of 5 mm diameter
Let h be the length of the ‘wire, then
2πrh = 1000 cm²
Let h be the height of the cylinder, then
Number of coils = \(\frac{\text { h }}{\text { Diameter of wire }}\) = \(\frac{h}{0.5}\)
Now total length of wire = 2 × 2πRh
= 2 × 1000 cm = \(\frac{2 \times 1000}{100}\) = 20 m

Question 10.
An iron pipe 20 cm long has exterior diameter equal to 25 cm. If the thickness of the pipe is 1 cm, find the whole surface of the pipe.
Solution:
Length of iron pipe = 20 cm
Exterior diameter = 25 cm
∴ Radius (R) = \(\frac{25}{2}\) cm
Thickness of pipe = 1 cm
∴ Inner radius (r) = \(\frac{25}{2}\) – 1 = \(\frac{23}{2}\)cm
∴ whole surface area

Question 11.
50 circular plates, each of radius 7 cm and thickness 0.5 cm, are placed one above the other to form a right circular cylinder. Find Its total surface area.
Solution:
Radius of each plate (r) = 7 cm
and thickness = 0.5 cm
∴ Height (thickness) of 50 plates (h)
= 0.5 × 50 = 25 cm
∴ Curved surface area of so formed cylinder
= 2 × πrh = 2 × \(\frac{22}{7}\) × 7 × 25 cm² = 1100 cm²
Area of top and bottom = 2 × πr²
= 2 × \(\frac{22}{7}\) × 7² = 308 cm²
Total surface area 1100 cm² + 308 cm²
= 1408 cm²

Students can track their progress and improvement through regular use of OP Malhotra Class 10 Solutions Chapter 8 Matrices Exercise 8(b).

S Chand Class 10 ICSE Maths Solutions Chapter 8 Matrices Exercise 8(b)

Question 1.
Given matrix A = \(\left[\begin{array}{l}
3 \\
2
\end{array}\right]\), B = \(\left[\begin{array}{l}
-2 \\
-1
\end{array}\right]\), C = \(\left[\begin{array}{c}
1 \\
-3
\end{array}\right]\), find the matrix X in each of the following cases:
(i) X = A + B + C
(ii) X = A + B – C
(iii) X – A = B
(iv) A + X = B + C
(v) 2X = A – C.
Solution:

Question 2.
If B = \(\left[\begin{array}{cc}
2 & -3 \\
-4 & 5
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
1 & -3 \\
-4 & 4
\end{array}\right]\), find the matrix B – C.
Solution:
B = \(\left[\begin{array}{cc}
2 & -3 \\
-4 & 5
\end{array}\right]\), C = \(\left[\begin{array}{cc}
1 & -3 \\
-4 & 4
\end{array}\right]\)
B – C = \(\left[\begin{array}{cc}
2 & -3 \\
-4 & 5
\end{array}\right]-\left[\begin{array}{cc}
1 & -3 \\
-4 & 4
\end{array}\right]\)
= \(\left[\begin{array}{cc}
2-1 & -3+3 \\
-4+4 & 5-4
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)

Question 3.
If A \(\left[\begin{array}{cc}
2 & 0 \\
-3 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
0 & 1 \\
-2 & 3
\end{array}\right]\), find 2A – 3B.
Solution:

Question 4.
Given A = \(\left[\begin{array}{ll}
1 & 4 \\
2 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
-4 & -1 \\
-3 & -2
\end{array}\right]\)
(i) Find the matrix 2A + B;
(ii) Find a matrix C such that C + B = \(\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)
Solution:

Question 5.
If P = \(\left[\begin{array}{cc}
4 & 6 \\
2 & -8
\end{array}\right]\) and Q = \(\left[\begin{array}{cc}
2 & -3 \\
-1 & 1
\end{array}\right]\), find P + 2Q.
Solution:
P = \(\left[\begin{array}{cc}
4 & 6 \\
2 & -8
\end{array}\right]\), Q = \(\left[\begin{array}{cc}
2 & -3 \\
-1 & 1
\end{array}\right]\)
∴ P + 2Q = \(\left[\begin{array}{cc}
4 & 6 \\
2 & -8
\end{array}\right]+2\left[\begin{array}{cc}
2 & -3 \\
-1 & 1
\end{array}\right]\)
= \(\left[\begin{array}{cc}
4 & 6 \\
2 & -8
\end{array}\right]+\left[\begin{array}{cc}
4 & -6 \\
-2 & 2
\end{array}\right]\)
= \(\left[\begin{array}{cc}
4+4 & 6-6 \\
2-2 & -8+2
\end{array}\right]=\left[\begin{array}{cc}
8 & 0 \\
0 & -6
\end{array}\right]\)

Question 6.
If 2\(\left[\begin{array}{ll}
3 & 4 \\
5 & x
\end{array}\right]+\left[\begin{array}{ll}
1 & y \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
7 & 0 \\
10 & 5
\end{array}\right]\), find the values of x and y.
Solution:
2\(\left[\begin{array}{ll}
3 & 4 \\
5 & x
\end{array}\right]+\left[\begin{array}{ll}
1 & y \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
7 & 0 \\
10 & 5
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
6 & 8 \\
10 & 2 x
\end{array}\right]+\left[\begin{array}{ll}
1 & y \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
7 & 0 \\
10 & 5
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
6+1 & 8+y \\
10+0 & 2 x+1
\end{array}\right]=\left[\begin{array}{cc}
7 & 0 \\
10 & 5
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
7 & 8+y \\
10 & 2 x+1
\end{array}\right]=\left[\begin{array}{cc}
7 & 0 \\
10 & 5
\end{array}\right]\)
Comparing the corresponding elements,
2x + 1 = 5 ⇒ 2x = 5 – 1 = 4
⇒ x = \(\frac { 4 }{ 2 }\) = 2
and 8 +7 = 0 ⇒ y = – 8
∴ x = 2, y = – 8

Question 7.
If \(\left[\begin{array}{ll}
a & 3 \\
4 & 2
\end{array}\right]+\left[\begin{array}{cc}
2 & b \\
1 & -2
\end{array}\right]-\left[\begin{array}{cc}
1 & 1 \\
-2 & c
\end{array}\right]\) = \(\left[\begin{array}{ll}
5 & 0 \\
7 & 3
\end{array}\right]\), find the value of a, b and c.
Solution:
\(\left[\begin{array}{ll}
a & 3 \\
4 & 2
\end{array}\right]+\left[\begin{array}{cc}
2 & b \\
1 & -2
\end{array}\right]-\left[\begin{array}{cc}
1 & 1 \\
-2 & c
\end{array}\right]=\left[\begin{array}{ll}
5 & 0 \\
7 & 3
\end{array}\right]\)
⇒ \(\left[\begin{array}{ll}
a+2-1 & 3+b-1 \\
4+1+2 & 2-2-c
\end{array}\right]=\left[\begin{array}{ll}
5 & 0 \\
7 & 3
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
a+1 & b+2 \\
7 & -c
\end{array}\right]=\left[\begin{array}{ll}
5 & 0 \\
7 & 3
\end{array}\right]\)
Comparing the corresponding elements a + 1 = 5 ⇒ a = 5 – 1 = 4
b + 2 = 0 ⇒ b = – 2
– c = 3 ⇒ c = – 3
∴ a = 4, b = – 2, c = – 3

Question 8.
If \(\left[\begin{array}{cc}
2 & -1 \\
2 & 0
\end{array}\right]+2 A=\left[\begin{array}{cc}
-3 & 5 \\
4 & 3
\end{array}\right]\), find A.
Solution:

Question 9.
If 2\(\left[\begin{array}{ll}
3 & x \\
0 & 1
\end{array}\right]+3\left[\begin{array}{ll}
1 & 3 \\
y & 2
\end{array}\right]=\left[\begin{array}{cc}
z & -7 \\
15 & 8
\end{array}\right]\), find the values of x, y and z.
Solution:
2\(\left[\begin{array}{ll}
3 & x \\
0 & 1
\end{array}\right]+3\left[\begin{array}{ll}
1 & 3 \\
y & 2
\end{array}\right]=\left[\begin{array}{cc}
z & -7 \\
15 & 8
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
6 & 2 x \\
0 & 2
\end{array}\right]+\left[\begin{array}{cc}
3 & 9 \\
3 y & 6
\end{array}\right]=\left[\begin{array}{cc}
z & -7 \\
15 & 8
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
6+3 & 2 x+9 \\
0+3 y & 2+6
\end{array}\right]=\left[\begin{array}{cc}
z & -7 \\
15 & 8
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
9 & 2 x+9 \\
3 y & 8
\end{array}\right]=\left[\begin{array}{cc}
z & -7 \\
15 & 8
\end{array}\right]\)
Comparing the corresponding elements
2x + 9 = – 7 ⇒ 2x + – 7 – 9 ⇒ 2x = – 16
⇒ x = \(\frac { – 16 }{ 2 }\) = – 8
3y = 15 ⇒ y = \(\frac { 15 }{ 3 }\) = 5
z = 9
Hence x = – 8, y = 5, z = 9

Question 10.
Given that M = \(\left[\begin{array}{ll}
2 & 0 \\
1 & 2
\end{array}\right]\) and N = \(\left[\begin{array}{cc}
2 & 0 \\
-1 & 2
\end{array}\right]\), find M + 2N.
Solution:
M = \(\left[\begin{array}{ll}
2 & 0 \\
1 & 2
\end{array}\right]\) and N = \(\left[\begin{array}{cc}
2 & 0 \\
-1 & 2
\end{array}\right]\)
∴ M + 2N = \(\left[\begin{array}{ll}
2 & 0 \\
1 & 2
\end{array}\right]+2\left[\begin{array}{cc}
2 & 0 \\
-1 & 2
\end{array}\right]\)
= \(\left[\begin{array}{ll}
2 & 0 \\
1 & 2
\end{array}\right]+\left[\begin{array}{cc}
4 & 0 \\
-2 & 4
\end{array}\right]\)
= \(\left[\begin{array}{ll}
2+4 & 0+0 \\
1-2 & 2+4
\end{array}\right]=\left[\begin{array}{cc}
6 & 0 \\
-1 & 6
\end{array}\right]\)

Question 11.
(a) If A = \(\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right]\), B = \(\left[\begin{array}{ll}
1 & 3 \\
0 & 1
\end{array}\right]\), C = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\), find A – B + C.
(b) Is the following statement true or false? You do not have to give any reason or proof for your answer. “If A and B are two ‘2 x 2’ matrices such that A – B = A + B, then B is a zero matrix.
Solution:
(a) A = \(\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right]\), B = \(\left[\begin{array}{ll}
1 & 3 \\
0 & 1
\end{array}\right]\), C = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
∴ A – B + C = \(\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right]-\left[\begin{array}{ll}
1 & 3 \\
0 & 1
\end{array}\right]+\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
= \(\left[\begin{array}{cc}
3-1+1 & 0-3+0 \\
0-0+0 & 3-1+1
\end{array}\right]\)
= \(\left[\begin{array}{cc}
3 & -3 \\
0 & 3
\end{array}\right]\)

(b) ∵ A and B are ‘2 x 2’ matrices
A – B = A + B
⇒ – B = B
It is possible if B is a zero matrix.

Question 12.
If P = \(\left[\begin{array}{cc}
-3 & 1 \\
2 & 5
\end{array}\right]\), Q = \(\left[\begin{array}{cc}
1 & 6 \\
-4 & 0
\end{array}\right]\) and R = \(\left[\begin{array}{cc}
4 & -1 \\
2 & 3
\end{array}\right]\)
Find the value of :
(i) 2P + 3Q – R
(ii) 4p – 2Q + 3R
Solution:

Question 13.
Compute :

Solution:

Question 14.
If A = \(\left[\begin{array}{ll}
x & y \\
z & w
\end{array}\right]\), B = \(\left[\begin{array}{cc}
x & -y \\
-z & w
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
0 & y \\
2 z & 0
\end{array}\right]\)
Show that (A + B) + C = A + (B + C).
Solution:

Question 15.
If A = \(\left[\begin{array}{ll}
8 & 1 \\
5 & 2
\end{array}\right]\), B = \(\left[\begin{array}{cc}
-4 & 6 \\
-2 & 12
\end{array}\right]\), find the matrix C such that 2A + 3B + 4C is a null matrix.
Solution:

Question 16.
If P = \(\left[\begin{array}{ll}
6 & -2 \\
4 & -6
\end{array}\right]\), Q = \(\left[\begin{array}{ll}
5 & 3 \\
2 & 0
\end{array}\right]\), find X such that 3P – 2Q + 3X = 0.
Solution:

Question 17.
Find X and Y if X + Y = \(\left[\begin{array}{cc}
9 & 7 \\
2 & 11
\end{array}\right]\) and X – Y = \(\left[\begin{array}{cc}
-5 & -6 \\
4 & 3
\end{array}\right]\).
Solution:

Question 18.
Solve for A and B, when
2A + B = \(\left[\begin{array}{cc}
3 & -4 \\
2 & 7
\end{array}\right]\) and A – 2B = \(\left[\begin{array}{ll}
4 & 3 \\
1 & 1
\end{array}\right]\).
Solution: