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Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions (Including Operations on Algebraic Expressions)

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions (Including Operations on Algebraic Expressions)

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ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 11 Algebraic Expressions (Including Operations on Algebraic Expressions). You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Algebraic Expressions Exercise 11A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Separate the constants and variables from the following :

Solution:
Clearly constants are : -7, √5, 8 – 5

Question 2.
Write the number of terms in each of the following polynomials.
(i) 5x² + 3 x ax
(ii) ax ÷ 4 – 7
(iii) ax – by + y x z
(iv) 23 + a x b ÷ 2.
Solution:

Question 3.
Separate monomials, binomials, trinomials and polynomials from the following algebraic expressions :

Solution:

Question 4.
Write the degree of each polynomial given below :

Solution:
(i) degree = 2 (Polynomial is xy + 7z)
(ii) degree = 3 (Polynomial is x² – 6x³ + 8 y)
(iii) degree = 8 (Polynomial is y – 6y² + 5y⁸)
(iv) degree = 3 (Polynomial is xyz – 3)
(v) degree = 4 (Polynomial is xy + yz² – xz³)
(vi) degree = 12 (Polynomial is x⁵y⁷ – 8x³y⁸ + 10x⁴, y⁴z⁴)

Question 5.
Write the coefficient of :
(i) ab in 7abx ,
(ii) 7a in 7abx ;
(iii) 5x² in 5x² – 5x ;
(iv) 8 in a² – 8ax + a ;
(v) 4xy in x² – 4xy + y².
Solution:
(i) The coefficient of ab in 7abx = 7x
(ii) The coefficient of 7a in 7abx = bx
(iii) The coefficient of 5x² in 5x² – 5x = 1
(iv) The coefficient of 8 in a² – 8ax + a = – ax
(v) The coefficient of 4xy in x² – 4xy + y² = -1

Question 6.
In \(\frac { 5 }{ 7 }\) xy²z³, write the coefficient of

Solution:

Question 7.
In each polynomial, given below, separate the like terms :

Solution:

Algebraic Expressions Exercise 11B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Evaluate :

Solution:

Question 2.
Add :

Solution:

Question 3.
Find the total savings of a boy who saves ₹ (4x – 6y) ; ₹ (6x + 2y) ; ₹ (4y – x) and ₹ (y – 2x) for four consecutive weeks.
Solution:

Question 4.
Subtract :

Solution:

Question 5.
(i) Take away – 3x³ + 4x² – 5x+ 6 from 3x³ – 4x² + 5x – 6
(ii) Take m² + m + 4 from -m² + 3m + 6 and the result from m² + m + 1.
Solution:

Question 6.
Subtract the sum of 5y² + y – 3 and y² – 3y + 7 from 6y² + y – 2.
Solution:

Question 7.
What must be added to x⁴ – x³ + x² + x + 3 to obtain x⁴ + x² – 1 ?
Solution:

Question 8.
(i) How much more than 2x² + 4xy + 2y² is 5x² + 10xy – y² ?
(ii) How much less 2a² + 1 is than 3a² – 6 ?
Solution:

Question 9.
If x = 6a + 86 + 9c ; y = 2b – 3a – 6c and z = c – b + 3a ; find
(i) x + y + z
(ii) x – y + z
(iii) 2x – y – 3z
(iv) 3y – 2z – 5x
Solution:

Question 10.
The sides of a triangle are x² – 3xy + 8, 4x² + 5xy – 3 and 6 – 3x² + 4xy. Find its perimeter.
Solution:

Question 11.
The perimeter of a triangle is 8y² – 9y + 4 and its two sides are 3y² – 5y and 4y² + 12. Find its third side.
Solution:

Question 12.
The two adjacent sides of a rectangle are 2x² – 5xy + 3z² and 4xy – x² – z². Find its perimeter.
Solution:

Question 13.
What must be subtracted from 19x⁴ + 2x³ + 30x – 37 to get 8x⁴ + 22x³ – 7x – 60 ?
Solution:

Question 14.
How much smaller is 15x – 18y + 19z than 22x – 20y – 13z + 26 ?
Solution:
The required result is
(22x – 20y – 13z + 26) – (15x – 18y + 19z)
= 22x – 20y – 13z + 26 – 15x + 18y – 19z
= 7x – 2y – 32z + 26

Question 15.
How much bigger is 15x²y² – 18xy² – 10x²y than -5x² + 6x²y – 7xy ?
Solution:

Algebraic Expressions Exercise 11C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Multiply :

Solution:

Question 2.
Multiply :

Solution:

Question 3.
Simplify :
(i) (7x – 8) (3x + 2)
(ii) (px – q) (px + q)
(iii) (5a + 5b – c) (2b – 3c)
(iv) (4x – 5y) (5x – 4y)
(v) (3y + 4z) (3y – 4z) + (2y + 7z) (y + z)
Solution:

Question 4.
The adjacent sides of a rectangle are x² – 4xy + 7y² and x³ – 5xy². Find its area.
Solution:

Question 5.
The base and the altitude of a triangle are (3x – 4y) and (6x + 5y) respectively. Find its area.
Solution:
Reqd. Area = \(\frac { 1 }{ 2 }\) (base) x (altitude)

Question 6.
Multiply -4xy³ and 6x²y and verify your result for x = 2 and y= 1.
Solution:

Question 7.
Find the value of (3x³) x (-5xy²) x (2x²yz³) for x = 1, y = 2 and z = 3.
Solution:

Question 8.
Evaluate (3x⁴y²) (2x²y³) for x = 1 and y = 2.
Solution:

Question 9.
Evaluate (x⁵) x (3x²) x (-2x) for x = 1.
Solution:

Question 10.
If x = 2 and y = 1; find the value of (-4x²y³) x (-5x²y⁵).
Solution:

Question 11.
Evaluate:
(i) (3x – 2)(x + 5) for x = 2.
(ii) (2x – 5y)(2x + 3y) for x = 2 and y = 3.
(iii) xz (x² + y²) for x = 2, y = 1 and z= 1.
Solution:

Question 12.
Evaluate:
(i) x(x – 5) + 2 for x = 1.
(ii) xy²(x – 5y) + 1 for x = 2 and y = 1.
(iii) 2x(3x – 5) – 5(x – 2) – 18 for x = 2.
Solution:

Question 13.
Multiply and then verify :
-3x²y² and (x – 2y) for x = 1 and y = 2.
Solution:

Question 14.
Multiply:
(i) 2x² – 4x + 5 by x² + 3x – 7
(ii) (ab – 1)(3 – 2ab)
Solution:

Question 15.
Simplify : (5 – x)(6 – 5x)(2 -x).
Solution:

Algebraic Expressions Exercise 11D – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Divide :

Solution:

Question 2.
Find the quotient and the remainder (if any) when :

Solution:

Question 3.
The area of a rectangle is x³ – 8x² + 7 and one of its sides is x – 1. Find the length of the adjacent side.
Solution:

Question 4.
The product of two numbers-is 16x⁴ – 1. If one number is 2x – 1, find the other.
Solution:

Question 5.
Divide x⁶ – y⁶ by the product of x² + xy + y² and x – y.
Solution:

Simplification
(Using removal of brackets)
The signs for different types of brackets are :

1. ____ ; Vinculum or bar brackets,
2. ( ); Parenthesis or small brackets,
3. { }; Curly brackets or middle brackets,
4. [ ]; Square brackets or big brackets.
   In a combined operation, the brackets must be removed in the same order as written above:

Algebraic Expressions Exercise 11E – Selina Concise Mathematics Class 8 ICSE Solutions

Simplify :
Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution:

Question 11.

Solution:

Question 12.

Solution:

Question 13.

Solution:

Question 14.

Solution:

Question 15.

Solution:

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers)

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 2 Exponents (Powers)

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ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 2 Exponents (Powers). You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Exponents Exercise 2A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Evaluate:

Solution:

Question 2.
If 1125 = 3^m x 5ⁿ; find m and n.
Solution:

Question 3.
Find x, if 9 × 3^x = (27)^2x-3
Solution:

Exponents Exercise 2B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Compute:


Solution:

Question 2.
Simplify:

Solution:

Question 3.
Evaluate:

Solution:

Question 4.
Simplify:

Solution:

Question 5.
Simplify:
(x^a+b)^a-b.(x^b+c)^b-c.(x^c+a)^c-a
Solution:

Question 6.
Simplify:

Solution:


Question 7.
(i) (a^-2)^-2. (ab)^-3
(ii) (xⁿy^-m)⁴ × (x³y^-2)^-n

Solution:

Question 8.
Show that:

Solution:

Question 9.
Evaluate:

Solution:

Question 10.
Evaluate:

Solution:

Question 11.
(m+n)^-1 (m^-1 + n^-1) = (mn)^-1
Solution:

Question 12.
Prove that:

Solution:

Question 13.
Find the values of n, when:

Solution:

Question 14.
Simplify:

Solution:

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes (Including Polygons)

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes (Including Polygons)

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ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 16 Understanding Shapes (Including Polygons). You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Understanding Shapes Exercise 16A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
State which of the following are polygons :

If the given figure is a polygon, name it as convex or concave.
Solution:
Only Fig. (ii), (iii) and (v) are polygons.
Fig. (ii) and (iii) are concave polygons while
Fig. (v) is convex.

Question 2.
Calculate the sum of angles of a polygon with :
(i) 10 sides
(ii) 12 sides
(iii) 20 sides
(iv) 25 sides
Solution:
(i) No. of sides n = 10
sum of angles of polygon = (n – 2) x 180°
= (10 – 2) x 180° = 1440°
(ii) no. of sides n = 12
sum of angles = (n – 2) x 180°
= (12 – 2) x 180° = 10 x 180° = 1800°
(iii) n = 20
Sum of angles of Polygon = (n – 2) x 180°
= (20 – 2) x 180° = 3240°
(iv) n = 25
Sum of angles of polygon = (n – 2) x 180°
= (25 – 2) x 180° = 4140°

Question 3.
Find the number of sides in a polygon if the sum of its interior angles is :
(i) 900°
(ii) 1620°
(iii) 16 right-angles
(iv) 32 right-angles.
Solution:
(i) Let no. of sides = n
Sum of angles of polygon = 900°
(n – 2) x 180° = 900°
n – 2 = \(\frac { 900 }{ 180 }\)
n – 2 = 5
n = 5 + 2
n = 7
(ii) Let no. of sides = n
Sum of angles of polygon = 1620°
(n – 2) x 180° = 1620°
n – 2 = \(\frac { 1620 }{ 180 }\)
n – 2 = 9
n = 9 + 2
n = 11
(iii) Let no. of sides = n
Sum of angles of polygon = 16 right angles = 16 x 90 = 1440°
(n – 2) x 180° = 1440°
n – 2 = \(\frac { 1440 }{ 180 }\)
n – 2 = 8
n = 8 + 2
n = 10
(iv) Let no. of sides = n
Sum of angles of polygon = 32 right angles = 32 x 90 = 2880°
(n – 2) x 180° = 2880
n – 2 = \(\frac { 2880 }{ 180 }\)
n – 2 = 16
n = 16 + 2
n = 18

Question 4.
Is it possible to have a polygon ; whose sum of interior angles is :
(i) 870°
(ii) 2340°
(iii) 7 right-angles
(iv) 4500°
Solution:
(i) Let no. of sides = n
Sum of angles = 870°
(n – 2) x 180° = 870°
n – 2 = \(\frac { 870 }{ 180 }\)
n – 2 = \(\frac { 29 }{ 6 }\)
n = \(\frac { 29 }{ 6 }\) + 2
n = \(\frac { 41 }{ 6 }\)
Which is not a whole number.
Hence it is not possible to have a polygon, the sum of whose interior angles is 870°
(ii) Let no. of sides = n
Sum of angles = 2340°
(n – 2) x 180° = 2340°
n – 2 = \(\frac { 2340 }{ 180 }\)
n – 2 = 13
n = 13 + 2 = 15
Which is a whole number.
Hence it is possible to have a polygon, the sum of whose interior angles is 2340°.
(iii) Let no. of sides = n
Sum of angles = 7 right angles = 7 x 90 = 630°
(n – 2) x 180° = 630°
n – 2 = \(\frac { 630 }{ 180 }\)
n – 2 = \(\frac { 7 }{ 2 }\)
n = \(\frac { 7 }{ 2 }\) + 2
n = \(\frac { 11 }{ 2 }\)
Which is not a whole number. Hence it is not possible to have a polygon, the sum of whose interior angles is 7 right-angles.
(iv) Let no. of sides = n
(n – 2) x 180° = 4500°
n – 2 = \(\frac { 4500 }{ 180 }\)
n – 2 = 25
n = 25 + 2
n = 27
Which is a whole number.
Hence it is possible to have a polygon, the sum of whose interior angles is 4500°.

Question 5.
(i) If all the angles of a hexagon are equal ; find the measure of each angle.
(ii) If all the angles of a 14-sided figure are equal ; find the measure of each angle.
Solution:
(i) No. of sides of hexagon, n = 6
Let each angle be = x°
Sum of angles = 6x°
(n – 2) x 180° = Sum of angles
(6 – 2) x 180° = 6x°
4 x 180 = 6x

Question 6.
Find the sum of exterior angles obtained on producing, in order, the sides of a polygon with :
(i) 7 sides
(ii) 10 sides
(iii) 250 sides.
Solution:
(i) No. of sides n = 7
Sum of interior & exterior angles at one vertex = 180°

Question 7.
The sides of a hexagon are produced in order. If the measures of exterior angles so obtained are (6x – 1)°, (10x + 2)°, (8x + 2)° (9x – 3)°, (5x + 4)° and (12x + 6)° ; find each exterior angle.
Solution:
Sum of exterior angles of hexagon formed by producing sides of order = 360°

i.e. 41° ; 72°, 58° ; 60° ; 39° and 90°

Question 8.
The interior angles of a pentagon are in the ratio 4 : 5 : 6 : 7 : 5. Find each angle of the pentagon.
Solution:
Let the interior angles of the pentagon be 4x, 5x, 6x, 7x, 5x.
Their sum = 4x + 5x + 6x + 7x + 5x = 21x

Question 9.
Two angles of a hexagon are 120° and 160°. If the remaining four angles are equal, find each equal angle.
Solution:
Two angles of a hexagon are 120°, 160°
Let remaining four angles be x, x, x and x.
Their sum = 4x + 280°
But sum of all the interior angles of a hexagon

Question 10.
The figure, given below, shows a pentagon ABCDE with sides AB and ED parallel to each other, and ∠B : ∠C : ∠D = 5 : 6 : 7.

(i) Using formula, find the sum of interior angles of the pentagon.
(ii) Write the value of ∠A + ∠E
(iii) Find angles B, C and D.
Solution:
(i) Sum of interior angles of the pentagon

Question 11.
Two angles of a polygon are right angles and the remaining are 120° each. Find the number of sides in it.
Solution:
Let number of sides = n

n = \(\frac { 300 }{ 60 }\)
n = 5

Question 12.
In a hexagon ABCDEF, side AB is parallel to side FE and ∠B : ∠C : ∠D : ∠E = 6 : 4 : 2 : 3. Find ∠B and ∠D.
Solution:

Question 13.
the angles of a hexagon are x + 10°, 2x + 20°, 2x – 20°, 3x – 50°, x + 40° and x + 20°. Find x.
Solution:

Question 14.
In a pentagon, two angles are 40° and 60°, and the rest are in the ratio 1 : 3 : 7. Find the biggest angle of the pentagon.
Solution:
In a pentagon, two angles are 40° and 60° Sum of remaining 3 angles = 3 x 180°
= 540° – 40° – 60° = 540° – 100° = 440°
Ratio in these 3 angles =1 : 3 : 7
Sum of ratios =1 + 3 + 7 = 11
Biggest angle = \(\frac { 440\times 7 }{ 11 }\) = 280°

Understanding Shapes Exercise 16B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Fill in the blanks :
In case of regular polygon, with :

Solution:

Question 2.
Find the number of sides in a regular polygon, if its each interior angle is :
(i) 160°
(ii) 135°
(iii) \(1\frac { 1 }{ 5 }\) of a right-angle
Solution:

Question 3.
Find the number of sides in a regular polygon, if its each exterior angle is :
(i) \(\frac { 1 }{ 3 }\) of a right angle
(ii) two-fifth of a right-angle.
Solution:
(i) Each exterior angle = \(\frac { 1 }{ 3 }\) of a right angle
= \(\frac { 1 }{ 3 }\) x 90
= 30°
Let number of sides = n

Question 4.
Is it possible to have a regular polygon whose each interior angle is :
(i) 170°
(ii) 138°
Solution:
(i) No. of sides = n
each interior angle = 170°


Which is not a whole number.
Hence it is not possible to have a regular polygon having interior angle of 138°.

Question 5.
Is it possible to have a regular polygon whose each exterior angle is :
(i) 80°
(ii) 40% of a right angle.
Solution:
(i) Let no. of sides = n each exterior angle = 80°

Which is not a whole number.
Hence it is not possible to have a regular polygon whose each exterior angle is of 80°.
(ii) Let number of sides = n
Each exterior angle = 40% of a right angle

Which is a whole number.
Hence it is possible to have a regular polygon whose each exterior angle is 40% of a right angle.

Question 6.
Find the number of sides in a regular polygon, if its interior angle is equal to its exterior angle.
Solution:
Let each exterior angle or interior angle be = x°

Question 7.
The exterior angle of a regular polygon is one-third of its interior angle. Find the number of sides in the polygon.
Solution:
Let interior angle = x°
Exterior angle = \(\frac { 1 }{ 3 }\) x°

Question 8.
The measure of each interior angle of a regular polygon is five times the measure of its exterior angle. Find :
(i) measure of each interior angle ;
(ii) measure of each exterior angle and
(iii) number of sides in the polygon.
Solution:
Let exterior angle = x°
Interior angle = 5x°
x + 5x = 180°
6x = 180°
x = 30°
Each exterior angle = 30°
Each interior angle = 5 x 30° = 150°
Let no. of sides = n

Question 9.
The ratio between the interior angle and the exterior angle of a regular polygon is 2 : 1. Find :
(i) each exterior angle of the polygon ;
(ii) number of sides in the polygon
Solution:
Interior angle : exterior angle = 2 : 1
Let interior angle = 2x° & exterior angle = x°

Question 10.
The ratio between the exterior angle and the interior angle of a regular polygon is 1 : 4. Find the number of sides in the polygon.
Solution:
Let exterior angle = x° & interior angle = 4x°

Question 11.
The sum of interior angles of a regular polygon is twice the sum of its exterior angles. Find the number of sides of the polygon.
Solution:
Let number of sides = n
Sum of exterior angles = 360°
Sum of interior angles = 360° x 2 = 720°
Sum of interior angles = (n – 2) x 180°
720° = (n – 2) x 180°
n – 2 = \(\frac { 720 }{ 180 }\)
n – 2 = 4
n = 4 + 2
n = 6

Question 12.
AB, BC and CD are three consecutive sides of a regular polygon. If angle BAC = 20° ; find :
(i) its each interior angle,
(ii) its each exterior angle
(iii) the number of sides in the polygon.
Solution:

Question 13.
Two alternate sides of a regular polygon, when produced, meet at the right angle. Calculate the number of sides in the polygon.
Solution:

Question 14.
In a regular pentagon ABCDE, draw a diagonal BE and then find the measure of:
(i) ∠BAE
(ii) ∠ABE
(iii) ∠BED
Solution:
(i) Since number of sides in the pentagon = 5
Each exterior angle = \(\frac { 360 }{ 5 }\) = 72°
∠BAE = 180° – 72°= 108°

Question 15.
The difference between the exterior angles of two regular polygons, having the sides equal to (n – 1) and (n + 1) is 9°. Find the value of n.
Solution:
We know that sum of exterior angles of a polynomial is 360°
(i) If sides of a regular polygon = n – 1

Question 16.
If the difference between the exterior angle of a n sided regular polygon and an (n + 1) sided regular polygon is 12°, find the value of n.
Solution:
We know that sum of exterior angles of a polygon = 360°
Each exterior angle of a regular polygon of 360°

Question 17.
The ratio between the number of sides of two regular polygons is 3 : 4 and the ratio between the sum of their interior angles is 2 : 3. Find the number of sides in each polygon.
Solution:
Ratio of sides of two regular polygons = 3 : 4
Let sides of first polygon = 3n
and sides of second polygon = 4n
Sum of interior angles of first polygon

Question 18.
Three of the exterior angles of a hexagon are 40°, 51 ° and 86°. If each of the remaining exterior angles is x°, find the value of x.
Solution:
Sum of exterior angles of a hexagon = 4 x 90° = 360°
Three angles are 40°, 51° and 86°
Sum of three angle = 40° + 51° + 86° = 177°
Sum of other three angles = 360° – 177° = 183°
Each angle is x°
3x = 183°
x = \(\frac { 183 }{ 3 }\)
Hence x = 61

Question 19.
Calculate the number of sides of a regular polygon, if:
(i) its interior angle is five times its exterior angle.
(ii) the ratio between its exterior angle and interior angle is 2 : 7.
(iii) its exterior angle exceeds its interior angle by 60°.
Solution:
Let number of sides of a regular polygon = n
(i) Let exterior angle = x
Then interior angle = 5x
x + 5x = 180°
=> 6x = 180°

Question 20.
The sum of interior angles of a regular polygon is thrice the sum of its exterior angles. Find the number of sides in the polygon.
Solution:
Sum of interior angles = 3 x Sum of exterior angles
Let exterior angle = x
The interior angle = 3x
x + 3x=180°
=> 4x = 180°
=> x = \(\frac { 180 }{ 4 }\)
=> x = 45°
Number of sides = \(\frac { 360 }{ 45 }\) = 8

Understanding Shapes Exercise 16C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Two angles of a quadrilateral are 89° and 113°. If the other two angles are equal; find the equal angles.
Solution:
Let the other angle = x°
According to given,
89° + 113° + x° + x° = 360°
2x° = 360° – 202°
2x° = 158°
x° = \(\frac { 158 }{ 2 }\)
other two angles = 79° each

Question 2.
Two angles of a quadrilateral are 68° and 76°. If the other two angles are in the ratio 5 : 7; find the measure of each of them.
Solution:
Two angles are 68° and 76°
Let other two angles be 5x and 7x
68° + 76°+ 5x + 7x = 360°
12x + 144° = 360°
12x = 360° – 144°
12x = 216°
x = 18°
angles are 5x and 7x
i.e. 5 x 18° and 7 x 18° i.e. 90° and 126°

Question 3.
Angles of a quadrilateral are (4x)°, 5(x+2)°, (7x – 20)° and 6(x+3)°. Find :
(i) the value of x.
(ii) each angle of the quadrilateral.
Solution:
Angles of quadrilateral are,

Question 4.
Use the information given in the following figure to find :
(i) x
(ii) ∠B and ∠C

Solution:

Question 5.
In quadrilateral ABCD, side AB is parallel to side DC. If ∠A : ∠D = 1 : 2 and ∠C : ∠B = 4 : 5
(i) Calculate each angle of the quadrilateral.
(ii) Assign a special name to quadrilateral ABCD
Solution:

Question 6.
From the following figure find ;
(i) x
(ii) ∠ABC
(iii) ∠ACD
Solution:

(i) In Quadrilateral ABCD,
x + 4x + 3x + 4x + 48° = 360°
12x = 360° – 48°
12x = 312

Question 7.
Given : In quadrilateral ABCD ; ∠C = 64°, ∠D = ∠C – 8° ; ∠A = 5(a+2)° and ∠B = 2(2a+7)°.
Calculate ∠A.
Solution:
∠C = 64° (Given)
∠D = ∠C – 8° = 64°- 8° = 56°
∠A = 5(a+2)°
∠B = 2(2a+7)°
Now ∠A + ∠B + ∠C + ∠D = 360°
5(a+2)° + 2(2a+7)° + 64° + 56° = 360°
5a + 10 + 4a + 14° + 64° + 56° = 360°
9a + 144° = 360°
9a = 360° – 144°
9a = 216°
a = 24°
∠A = 5 (a + 2) = 5(24+2) = 130°

Question 8.
In the given figure : ∠b = 2a + 15 and ∠c = 3a + 5; find the values of b and c.

Solution:
Stun of angles of quadrilateral = 360°
70° + a + 2a + 15 + 3a + 5 = 360°
6a + 90° = 360°
6a = 270°
a = 45°
b = 2a + 15 = 2 x 45 + 15 = 105°
c = 3a + 5 = 3 x 45 + 5 = 140°
Hence ∠b and ∠c are 105° and 140°

Question 9.
Three angles of a quadrilateral are equal. If the fourth angle is 69°; find the measure of equal angles.
Solution:
Let each equal angle be x°
x + x + x + 69° = 360°

3x = 360°- 69
3x = 291
x = 97°
Each, equal angle = 97°

Question 10.
In quadrilateral PQRS, ∠P : ∠Q : ∠R : ∠S = 3 : 4 : 6 : 7.
Calculate each angle of the quadrilateral and then prove that PQ and SR are parallel to each other
(i) Is PS also parallel to QR ?
(ii) Assign a special name to quadrilateral PQRS.
Solution:

Question 11.
Use the informations given in the following figure to find the value of x.

Solution:
Take A, B, C, D as the vertices of Quadrilateral and BA is produced to E (say).
Since ∠EAD = 70°
∠DAB = 180° – 70°= 110°
[EAB is a straight line and AD stands on it ∠EAD+ ∠DAB = 180°]
110° + 80° + 56° + 3x – 6° = 360°
[sum of interior angles of a quadrilateral = 360°]
3x = 360° – 110° – 80° – 56° + 6°
3x = 360° – 240° = 120°
x = 40°

Question 12.
The following figure shows a quadrilateral in which sides AB and DC are parallel. If ∠A : ∠D = 4 : 5, ∠B = (3x – 15)° and ∠C = (4x + 20)°, find each angle of the quadrilateral ABCD.

Solution:
Let ∠A = 4x
∠D = 5x
Since ∠A + ∠D = 180° [AB||DC]
4x + 5x = 180°
=> 9x = 180°
=> x = 20°
∠A = 4 (20) = 80°,
∠D = 5 (20) = 100°
Again ∠B + ∠C = 180° [ AB||DC]
3x – 15° + 4x + 20° = 180°
7x = 180° – 5°
=> 7x = 175°
=> x = 25°
∠B = 75° – 15° = 60°
and ∠C = 4 (25) + 20 = 100°+ 20°= 120°

Question 13.
Use the following figure to find the value of x

Solution:
The sum of exterior angles of a quadrilateral

=> y + 80° + 60° + 90° = 360°
=> y + 230° = 360°
=> y = 360° – 230° = 130°
At vertex A,
∠y + ∠x = 180° (Linear pair)
x = 180° – 130°
=> x = 50°

Question 14.
ABCDE is a regular pentagon. The bisector of angle A of the pentagon meets the side CD in point M. Show that ∠AMC = 90°.
Solution:

Given : ABCDE is a regular pentagon.
The bisector ∠A of the pentagon meets the side CD at point M.
To prove : ∠AMC = 90°
Proof: We know that, the measure of each interior angle of a regular pentagon is 108°.
∠BAM = \(\frac { 1 }{ 2 }\) x 108° = 54°
Since, we know that the sum of a quadrilateral is 360°
In quadrilateral ABCM, we have
∠BAM + ∠ABC + ∠BCM + ∠AMC = 360°
54° + 108° + 108° + ∠AMC = 360°
∠AMC = 360° – 270°
∠AMC = 90°

Question 15.
In a quadrilateral ABCD, AO and BO are bisectors of angle A and angle B respectively. Show that:
∠AOB = \(\frac { 1 }{ 2 }\) (∠C + ∠D)
Solution:
Given : AO and BO are the bisectors of ∠A and ∠B respectively.
∠1 = ∠4 and ∠3 = ∠5 ……..(i)

To prove : ∠AOB = \(\frac { 1 }{ 2 }\) (∠C + ∠D)
Proof: In quadrilateral ABCD
∠A + ∠B + ∠C + ∠D = 360°
\(\frac { 1 }{ 2 }\) (∠A + ∠B + ∠C + ∠D) = 180° …………(ii)
Now in ∆AOB
∠1 + ∠2 + ∠3 = 180° ………(iii)
Equating equation (ii) and equation (iii), we get
∠1 + ∠2 + ∠3 = ∠A + ∠B + \(\frac { 1 }{ 2 }\) (∠C + ∠D)
∠1 + ∠2 + ∠3 = ∠1 + ∠3 + \(\frac { 1 }{ 2 }\) (∠C + ∠D)
∠2 = \(\frac { 1 }{ 2 }\) (∠C + ∠D)
∠AOB = \(\frac { 1 }{ 2 }\) (∠C + ∠D)
Hence proved.

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 9 Simple and Compound Interest

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ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 9 Simple and Compound Interest. You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Simple and Compound Interest Exercise 9A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Find the interest and the amount on:
(i) ₹ 750 in 3 years 4 months at 10% per annum.
(ii) ₹ 5,000 at 8% per year from 23rd December 2011 to 29th July 2012.
(iii) ₹ 2,600 in 2 years 3 months at 1% per month.
(iv) ₹ 4,000 in 1\(\frac { 1 }{ 3 }\) years at 2 paise per rupee per month.
Solution:

Question 2.
Rohit borrowed Rs. 24,000 at 7.5 percent per year. How much money will he pay at the end of 4th years to clear his debt ?
Solution:

Question 3.
The interest on a certain sum of money is Rs. 1,480 in 2 years and at 10 per cent per year. Find the sum of money.
Solution:
Let P = Rs. x
Time (T) = 2 years
Rate (R) = 0%

Question 4.
On what principal will the simple interest be Rs. 7,008 in 6 years 3 months at 5% per year ?
Solution:

Question 5.
Find the principal which will amount to Rs. 4,000 in 4 years at 6.25% Per annum.
Solution:

Question 6.
(i) At what rate per cent per annum will Rs. 630 produce an interest of Rs. 126 in 4 years ?
(ii) At what rate per cent per year will a sum double itself in 6\(\frac { 1 }{ 4 }\) years ?
Solution:

Question 7.
(i) In how many years will Rs.950 produce Rs.399 as simple interest at 7% ?
(ii) Find the time in which Rs.1200 will amount to Rs.1536 at 3.5% per year.
Solution:

Question 8.
The simple interest on a certain sum of money is \(\frac { 3 }{ 8 }\) of the sum in 6\(\frac { 1 }{ 4 }\) years. Find the rate percent charged.
Solution:

Question 9.
What sum of money borrowed on 24th May will amount to Rs. 10210.20 on 17th October of the same year at 5 percent per annum simple interest.
Solution:
A = Rs. 10210.20
R = 5% P.A.
T=May + June + July + August + Sept.+ Oct.

Question 10.
In what time will the interest on a certain sum of money at 6% be \(\frac { 5 }{ 8 }\) of itself ?
Solution:

Question 11.
Ashok lent out Rs.7000 at 6% and Rs.9500 at 5%. Find his total income from the interest in 3 years.
Solution:

Question 12.
Raj borrows Rs.8,000; out of which Rs. 4500 at 5% and remainder at 6%. Find the total interest paid by him in 4 years.
Solution:

Question 13.
Mohan lends Rs.4800 to John for 4\(\frac { 1 }{ 2 }\) years and Rs.2500 to Shy am for 6 years and receives a total sum of Rs.2196 as interest. Find the rate percent per annum, it being the same in both the cases.
Solution:

Question 14.
John lent Rs. 2550 to Mohan at 7.5 per cent per annum. If Mohan discharges the debt after 8 months by giving an old black and white television and Rs. 1422.50; find the price of the television.
Solution:

Simple and Compound Interest Exercise 9B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
The interest on a certain sum of money is 0.24 times of itself in 3 years. Find the rate of interest.
Solution:

Question 2.
If ₹ 3,750 amount to ₹ 4,620 in 3 years at simple interest. Find:
(i) the rate of interest
(ii) the amount of Rs. 7,500 in 5\(\frac { 1 }{ 2 }\) years at the same rate of interest
Solution:

Question 3.
A sum of money, lent out at simple interst, doubles itself in 8 years. Find :
(i) the rate of interest
(ii) in how many years will the sum become triple (three times) of itself at the same rate per cent ?
Solution:

Question 4.
Rupees 4000 amount to Rs.5000 in 8 years ; in what time will Rs.2100 amount to Rs.2800 at the same rate ?
Solution:

Question 5.
What sum of money lent at 6.5% per annum will produce the same interest in 4 years as Rs.7500 produce in 6 years at 5% per annum ?
Solution:

Question 6.
A certain sum amounts to Rs.3825 in 4 years and to Rs.4050 in 6 years. Find the rate percent and the sum.
Solution:

Question 7.
At what rafepercent of simple interest will the interest on Rs.3750 be one-fifth of itself in 4 years ? To what will it amount in 15 years ?
Solution:

Question 8.
On what date will ₹ 1950 lent on 5th January, 2011 amount to ₹ 2125.50 at 5 percent per annum simple interest?
Solution:

Question 9.
If the interest on Rs.2400 be more than the interest on Rs.2000 by Rs.60 in 3 years at the same rate percent ; find the rate.
Solution:

Question 10.
Divide Rs. 15,600 into two parts such that the interest on one at 5 percent for 5 years may be equal to that on the other at 4\(\frac { 1 }{ 2 }\) per cent for 6 years.
Solution:

Simple and Compound Interest Exercise 9C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
A sum of Rs. 8,000 is invested for 2 years at 10% per annum compound interest. Calculate:
(i) interest for the first year.
(ii) principal for the second year.
(iii) interest for the second year.
(iv) final amount at the end of second year
(v) compound interest earned in 2 years.
Solution:
(i) Here Principal (P) = Rs. 8,000
Rate of interest = 10%

Question 2.
A man borrowed Rs. 20,000 for 2 years at 8% per year compound interest. Calculate :
(i) the interest of the first year.
(ii) the interest of the second year.
(iii) the final amount at the end of second year.
(iv) the compound interest of two years.
Solution:

Question 3.
Calculate the amount and the compound interest on Rs. 12,000 in 2 years and at 10% per year.
Solution:

Question 4.
Calculate the amount and the compound interest on Rs. 10,000 in 3 years at 8% per annum.
Solution:

Question 5.
Calculate the compound interest on Rs. 5,000 in 2 years ; if the rates of interest for successive years be 10% and 12% respectively.
Solution:

Question 6.
Calculate the compound interest on Rs. 15,000 in 3 years ; if the rates of interest for successive years be 6%, 8% and 10% respectively.
Solution:

Question 7.
Mohan borrowed Rs. 16,000 for 3 years at 5% per annum compound interest. Calculate the amount that Mohan will pay at the end of 3 years.
Solution:

Question 8.
Rekha borrowed Rs. 40,000 for 3 years at 10% per annum compound interest. Calculate the interest paid by her for the second year.
Solution:

Question 9.
Calculate the compound interest for the second year on Rs. 15000 invested for 5 years at 6% per annum.
Solution:

= 159 x 6 = Rs. 954

Question 10.
A man invests Rs. 9600 at 10% per annum compound interest for 3 years. Calculate :
(i) the interest for the first year.
(ii) the amount at the end of the first year.
(iii) the interest for the second year.
(iv) the interest for the third year.
Solution:

Question 11.
A person invests Rs. 5,000 for two years at a certain rate of interest compounded annually. At the end of one year, this sum amounts to Rs. 5,600. Calculate :
(i) the rate of interest per year.
(ii) the amount at the end of the second year.
Solution:

Question 12.
Calculate the difference between the compound interest and the simple interest on ₹ 7,500 in two years and at 8% per annum.
Solution:

Question 13.
Calculate the difference between the compound interest and the simple interest on ₹ 8,000 in three years and at 10% per annum.
Solution:

Question 14.
Rohit borrowed ₹ 40,000 for 2 years at 10% per annum C.I. and Manish borrowed the same sum for the same time at 10.5% per annum simple interest. Which of these two gets less interest and by how much?
Solution:

Question 15.
Mr. Sharma borrowed ₹ 24,000 at 13% p.a. simple interest and an equal sum at 12% p.a. compound interest. Find the total interest earned by Mr. Sharma in 2 years.
Solution:

Question 16.
Peter borrows ₹ 12,000 for 2 years at 10% p.a. compound interest. He repays ₹ 8,000 at the end of first year. Find:
(i) the amount at the end of first year, before making the repayment.
(ii) the amount at the end of first year, after making the repayment.
(iii) the principal for the second year.
(iv) the amount to be paid at the end of second year, to clear the account.
Solution:

Question 17.
Gautam takes a loan of ₹ 16,000 for 2 years at 15% p.a. compound interest. He repays ₹ 9,000 at the end of first year. How mucH must he pay at the end of second year to clear the debt?
Solution:
Loan taken (P) = ₹ 16000
Rate (R) = 15% p.a.
Time (T) = 2 years

Question 18.
A certain sum of money, invested for 5 years at 8% p.a. simple interest, earns an interest of ₹ 12,000. Find:
(i) the sum of money.
(ii) the compound interest earned by this money in two years and at 10% p.a. compound interest.
Solution:

Question 19.
Find the amount and the C.I. on ₹ 12,000 at 10% per annum compounded half-yearly.
Solution:

Question 20.
Find the amount and the C.I. on ₹ 8,000 in 1\(\frac { 1 }{ 2 }\) years at 20% per year compounded half- yearly.
Solution:

Question 21.
Find the amount and the compound interest on ₹ 24,000 for 2 years at 10% per annum compounded yearly.
Solution:

Question 22.
Find the amount and the compound interest on ₹ 16,000 for 3 years at 5% per annum compounded annually.
Solution:

Question 23.
Find the amount and the compound interest on ₹ 20,000 for 1\(\frac { 1 }{ 2 }\) years at 10% per annum compounded half-yearly.
Solution:

Question 24.
Find the amount and the compound interest on ₹ 32,000 for 1 year at 20% per annum compounded half-yearly.
Solution:

Question 25.
Find the amount and the compound interest on ₹ 4,000 in 2 years, if the rate of interest for first year is 10% and for the second year is 15%.
Solution:

Question 26.
Find the amount and the compound interest on ₹ 10,000 in 3 years, if the rates of interest for the successive years are 10%, 15% and 20% respectively.
Solution:

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots (Including use of tables for natural numbers)

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 4 Cubes and Cube-Roots (Including use of tables for natural numbers)

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ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 4 Cubes and Cube-Roots (Including use of tables for natural numbers). You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Cubes and Cube-Roots Exercise 4A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Find the cube of :
(i) 7
(ii) 11
(iii) 16
(iv) 23
(v) 31
(vi) 42
(vii) 54
Solution:
(i) (7)³ = 7 x 7 x 7 = 343
(ii) (11)³ =11 x 11 x 11 = 1331
(iii) (16)³ = 16 x 16 x 16 = 4096
(iv) (23)³ = 23 x 23 x 23 = 12167
(v) (31)³ = 31 x 31 x 31 = 29791
(vi) (42)³ = 42 x 42 x 42 = 74088
(vii) (54)³ = 54 x 54 x 54 = 157464

Question 2.
Find which of the following are perfect cubes :
(i) 243
(ii) 588
(iii) 1331
(iv) 24000
(v) 1728
(vi) 1938
Solution:



1938 = 2 x 3 x 17 x 19
1938 is not a perfect cube.

Question 3.
Find the cubes of :
(i) 2.1
(ii) 0.4
(iii) 1.6
(iv) 2.5
(v) 0.12
(vi) 0.02
(vii) 0.8
Solution:

Question 4.
Find the cubes of :
(i) \(\frac { 3 }{ 7 }\)
(ii) \(\frac { 8 }{ 9 }\)
(iii) \(\frac { 10 }{ 13 }\)
(iv) \(1\frac { 2 }{ 7 }\)
(v) \(2\frac { 1 }{ 2 }\)
Solution:

Question 5.
Find the cubes of :
(i) -3
(ii) -7
(iii) -12
(iv) -18
(v) -25
(vi) -30
(vii) -50
Solution:

Question 6.
Which of the following are cubes of:
(i) an even number
(ii) an odd number
216, 729, 3375, 8000, 125, 343, 4096 and 9261.
Solution:



(i) Cubes of an even number are 216, 8000, 4096.
(ii) Cubes of an odd number are 729, 3375, 125, 343, 9261.

Question 7.
Find the least number by which 1323 must be multiplied so that the product is a perfect cube.
Solution:
The prime factor of 1323 are =3 x 3 x 3 x 7 x 7
= (3 x 3 x 3) x 7 x 7
Clearly, 1323 must be multiplied by 7.

Question 8.
Find the smallest number by which 8768 must be divided so that the quotient is a perfect cube.
Solution:

Question 9.
Find the smallest number by which 27783 be multiplied to get a perfect square number.
Solution:

Question 10.
With what least number must 8640 be divided so that the quotient is a perfect cube?
Solution:

Question 11.
Which is the smallest number that must be multiplied to 77175 to make it a perfect cube?
Solution:

Cubes and Cube-Roots Exercise 4B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Find the cube-roots of :
(i) 64
(ii) 343
(iii) 729
(iv) 1728
(v) 9261
(vi) 4096
(vii) 8000
(viii) 3375
Solution:

Question 2.
Find the cube-roots of :
(i) \(\frac { 27 }{ 64 }\)
(ii) \(\frac { 125 }{ 216 }\)
(iii) \(\frac { 343 }{ 512 }\)
(iv) 64 x 729
(v) 64 x 27
(vi) 729 x 8000
(vii) 3375 x 512
Solution:

Question 3.
Find the cube-roots of :
(i) -216
(ii) -512
(iii) -1331
(iv) \(\frac { -27 }{ 125 }\)
(v) \(\frac { -64 }{ 343 }\)
(vi) \(\frac { -512 }{ 343 }\)
(vii) -2197
(viii) -5832
(ix) -2744000
Solution:

Question 4.
Find the cube-roots of :
(i) 2.744
(ii) 9.261
(iii) 0.000027
(iv) -0.512
(v) -15.625
(vi) -125 x 1000
Solution:

Question 5.
Find the smallest number by which 26244 may be divided so that the quotient is a perfect cube.
Solution:
The prime factors of 26244 are

Clearly, 26244 must be divided by
3 x 3 x 2 x 2 = 36

Question 6.
What is the least number by which 30375 should be multiplied to get a perfect cube?
Solution:
The prime factors of 30375 are

Question 7.
Find the cube-roots of :
(i) 700 x 2 x 49 x 5
(ii) -216 x 1728
(iii) -64 x -125
(iv) \(\frac { -27 }{ 343 }\)
(v) \(\frac { 729 }{ -1331 }\)
(vi) 250.047
(vii) -175616
Solution:

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Board – Indian Certificate of Secondary Education (ICSE), www.cisce.org
Class – Class 10
Subject – Maths
Year of Examination – 2020, 2019, 2018, 2017.

ICSE Class 10 Maths Solved Question Papers Last Ten Years

• ICSE 2019 Maths Question Paper Solved
• ICSE 2018 Maths Question Paper Solved
• ICSE 2017 Maths Question Paper Solved
• ICSE 2016 Maths Question Paper Solved
• ICSE 2015 Maths Question Paper Solved
• ICSE 2014 Maths Question Paper Solved
• ICSE 2013 Maths Question Paper Solved
• ICSE 2012 Maths Question Paper Solved
• ICSE 2011 Maths Question Paper Solved
• ICSE 2010 Maths Question Paper Solved
• ICSE 2009 Maths Question Paper Solved
• ICSE 2008 Maths Question Paper Solved
• ICSE 2007 Maths Question Paper Solved

We hope the ICSE Class 10 Maths Previous Years Question Papers Solved Last 10 Years Pdf Free Download with Solutions will help you. If you have any query regarding ICSE Class 10 Maths Solved Question Papers Last Ten Years with Answers, drop a comment below and we will get back to you at the earliest.

APlusTopper.com provides ICSE Class 10 Geography Previous Year Board Question Papers Solved Pdf Free Download with Solutions, Answers and Marking Scheme. Here we have given ICSE Class 10 Geography Solved Question Papers Last Ten Years. Students can view or download the ICSE Board 10th Geography Previous Year Question Papers with Solutions for their upcoming examination.

These ICSE Class 10 Geography Previous 10 Years Board Question Papers with Answers are useful to understand the pattern of questions asked in the board exam. Know about the important concepts to be prepared for ICSE Class 10 Board Exam and Score More marks.

Board – Indian Certificate of Secondary Education (ICSE), www.cisce.org
Class – Class 10
Subject – Geography
Year of Examination – 2020, 2019, 2018, 2017.

ICSE Class 10 Geography Solved Question Papers Last Ten Years

• ICSE 2019 Geography Question Paper Solved
• ICSE 2018 Geography Question Paper Solved
• ICSE 2017 Geography Question Paper Solved
• ICSE 2016 Geography Question Paper Solved
• ICSE 2015 Geography Question Paper Solved
• ICSE 2014 Geography Question Paper Solved
• ICSE 2013 Geography Question Paper Solved
• ICSE 2012 Geography Question Paper Solved
• ICSE 2011 Geography Question Paper Solved
• ICSE 2010 Geography Question Paper Solved
• ICSE 2009 Geography Question Paper Solved
• ICSE 2008 Geography Question Paper Solved
• ICSE 2007 Geography Question Paper Solved

We hope the ICSE Class 10 Geography Previous Years Question Papers Solved Last 10 Years Pdf Free Download with Solutions will help you. If you have any query regarding ICSE Class 10 Geography Solved Question Papers Last Ten Years with Answers, drop a comment below and we will get back to you at the earliest.

ICSE Previous Year Question Papers Class 10

APlusTopper.com provides ICSE Class 10 Biology Previous Year Board Question Papers Solved Pdf Free Download with Solutions, Answers and Marking Scheme. Here we have given ICSE Class 10 Biology Solved Question Papers Last Ten Years. Students can view or download the ICSE Board 10th Biology Previous Year Question Papers with Solutions for their upcoming examination.

These ICSE Class 10 Biology Previous 10 Years Board Question Papers with Answers are useful to understand the pattern of questions asked in the board exam. Know about the important concepts to be prepared for ICSE Class 10 Board Exam and Score More marks.

Board – Indian Certificate of Secondary Education (ICSE), www.cisce.org
Class – Class 10
Subject – Biology
Year of Examination – 2020, 2019, 2018, 2017.

ICSE Class 10 Biology Solved Question Papers Last Ten Years

ICSE Biology Class 10 Solved Question Papers

• ICSE 2019 Biology Question Paper Solved
• ICSE 2018 Biology Question Paper Solved
• ICSE 2017 Biology Question Paper Solved
• ICSE 2016 Biology Question Paper Solved
• ICSE 2015 Biology Question Paper Solved
• ICSE 2014 Biology Question Paper Solved
• ICSE 2013 Biology Question Paper Solved
• ICSE 2012 Biology Question Paper Solved
• ICSE 2011 Biology Question Paper Solved
• ICSE 2010 Biology Question Paper Solved
• ICSE 2009 Biology Question Paper Solved
• ICSE 2008 Biology Question Paper Solved
• ICSE 2007 Biology Question Paper Solved

We hope the ICSE Class 10 Biology Previous Years Question Papers Solved Last 10 Years Pdf Free Download with Solutions will help you. If you have any query regarding ICSE Class 10 Biology Solved Question Papers Last Ten Years with Answers, drop a comment below and we will get back to you at the earliest.

ICSE Previous Year Question Papers Class 10