Continuous practice using OP Malhotra Class 10 Solutions Chapter 6 Ratio and Proportion Ex 6(a) can lead to a stronger grasp of mathematical concepts.
S Chand Class 10 ICSE Maths Solutions Chapter 6 Ratio and Proportion Ex 6(a)
Question 1.
Find the value of x in each case :
(i) 8 : 14 : : x : 28
(ii) x : 9 : : 5 : 3
Solution:
(i) 8 : 14 : : x : 28
⇒ 14 × x = 8 x 28 ⇒ x = \(\frac{8 \times 28}{14}\) = 16 (∵ ad = bc)
(ii) x : 9 :: 5 : 3
⇒ x × 3 = 9 x 5 ⇒ x = \(\frac{9 \times 5}{3}\) = 16 (∵ ad = bc)
(iii) 12 : x :: 4 : 15
⇒ x × 4 = 12 x 15 ⇒ x = \(\frac{12 \times 15}{4}\) = 45 (∵ ad = bc)
Question 2.
Find the fourth proportional to :
(i) 25, 15, 40
(ii) 3a² b², a³, b³
(iii) a² – 5a + 6, a² a – 6, a² – 9
Solution:
We know that if a : b : : c : d, then d = \(\frac { b×c }{ a }\)
(i) 25, 15, 40
∴ Fourth proportional = \(\frac { bc }{ a }\) = \(\frac{15 \times 40}{25}\) = 24
(ii) 3a² b², a3, b3
Fourth proportional = \(\frac { bc }{ a }\)
= \(\frac{a^3 \times b^3}{3 a^2 b^2}=\frac{1}{3} a b=\frac{a b}{3}\)
(iii) a² – 5a + 6, a² + a – 6, a² – 9
a² – 5a + 6 = a² – 3a – 2a + 6
= a (a – 3) – 2 (a – 3)
= (a – 3) (a – 2)
a² + a – 6 = a² + 3a – 2a – 6
= a (a + 3) – 2 (a + 3)
= (a + 3) (a – 2)
a² – 9 = (a)² – (3)² = (a + 3) (a – 3)
Now fourth proportional = \(\frac { bc }{ a }\)
= \(\frac{(a+3)(a-2)(a+3)(a-3)}{(a-3)(a-2)}\)
= (a + 3) (g + 3) = (a + 3)²
Question 3.
Find the third proportional to :
(i) 16 and 36
(ii) \(\frac { x }{ y }\) + \(\frac { y }{ x }\) and
(iii) a² – b², a + b
Solution:
If a, b :: b : c, then
Third proportional (c) = \(\frac { b² }{ a }\)
(i) 16 and 36
Third proportional \(\frac{b^2}{a}=\frac{36 \times 36}{16}\) = 81
(ii) \(\frac { x }{ y }\) + \(\frac { y }{ x }\) and \(\frac { x }{ y }\)
∴ Third proportional = \(\frac{\frac{x}{y} \times \frac{x}{y}}{\frac{x}{y}+\frac{y}{x}}=\frac{\frac{x^2}{y^2}}{\frac{x^2+y^2}{x y}}\)
= \(\frac{x^2}{y^2} \times \frac{x y}{x^2+y^2}=\frac{x^3}{y\left(x^2+y^2\right)}\)
(iii) a² – b², (g + b)
Third proportional = \(\frac{(a+b)(a+b)}{a^2-b^2}\)
= \(\frac{(a+b)(a+b)}{(a+b)(a-b)}=\frac{a+b}{a-b}\)
Question 4.
Find the mean proportional between the following:
(i) 5 and 80
(ii) 360a4 and 250a² b²
(iii) (x – y) and (x³ – x²y)
Solution:
We know that mean proportional (b) = \(\sqrt{ac}\)
(i) 5, 80
Mean proportional = \(\sqrt{5×80}\) = \(\sqrt{400}\) = 20
(ii) 360a4 and 250a² b²
Mean proportional = \(\sqrt{360 a^4 \times 250 a^2 b^2}\)
= \(\sqrt{360 \times 250 \times a^{4+2} \times b^2}\)
= \(\sqrt{90000 \times a^6 \times b^2}\)
= 300 x a³ x b
= 300 a³ b
(iii) (x – y) and (x³ – x²y)
Mean proportioanl = \(\sqrt{(x-y)\left(x^3-x^2 y\right)}\)
= \(\sqrt{(x-y) x^2(x-y)}\)
= \(\sqrt{x^2(x-y)^2}\) = x(x – y)
Question 5.
(i) If x, 16, 48, y are in continued proportion, find the value of x and y.
(ii) If x, 9, y, 16 are in continued proportion, then find the value of x and y.
Solution:
(i) ∵ x, 16, 48, y are in continued proportion,
then \(\frac{x}{16}=\frac{16}{48}=\frac{48}{y}\)
(ii) ∵ x, 9, y, 16 are in continued proportion,
Question 6.
What number must be added to 3, 5, 7, 10 each in order to get four numbers in proportion?
Solution:
Let x be added to each number,
then 3 +x, 5 + x, 7 + x and 10 + x are in proportion
⇒ \(\frac{3+x}{5+x}=\frac{7+x}{10+x}\)
⇒ (3 + x) (10 + x) = (7 + x) (5 + x)
⇒ 30 + 3x + 10x + x² = 35 + 7x + 5x + x²
⇒ 30 + 13x + x² = 35 + 12x + x²
⇒ x² + 13x – x² – 12x = 35 – 30
⇒ x = 5
∴ 5 must be added
Question 7.
What number must be subtracted from each of the numbers 28, 53, 19, 35 so that they are in proportion?
Solution:
Let x be subtracted from each number, then 28 – x, 53 – x, 19 – x, 35 – x are in proportion,
∴ \(\frac{28-x}{53-x}=\frac{19-x}{35-x}\)
⇒ (28 – x) (35 – x) = (19 – x) (53 – x)
⇒ 980 – 28x – 35 x + x² = 1007 – 19x – 53x + x²
⇒ 980 – 63x + x² = 1007 – 72x + x²
⇒ – 63x + x² + 72x – x² = 1007 – 98027
⇒ 9x = 27 ⇒ x = \(\frac { 27 }{ 9 }\) = 3
∴ 3 must be added.
Question 8.
(i) Find two numbers such that the mean proportional between them is 14 and the third proportional to them is 112.
(ii) Find two numbers such that the mean proportional between them is 18 and the third proportional to them is 144.
Solution:
(i) Let a and b be the two numbers whose mean
proportional = 14 and third proportional = 112
∴ Numbers are 7, 28
(ii) Let a and b be the numbers whose mean proportional = 18 and third proportional = 144
∴ Numbers are 9, 36
Question 9.
If p + r = 2q and \(\frac { 1 }{ q }\) + \(\frac { 1 }{ s }\) = \(\frac { 2 }{ r }\), then prove that p : q = r : s.
Solution:
p + r = 2q … (i)
\(\frac { 1 }{ q }\) + \(\frac { 1 }{ s }\) = \(\frac { 2 }{ r }\) .. (ii)
⇒ r (q + s) = (p + r) s {from (i)}
⇒ qr + rs = ps + rs
⇒ qr = ps
⇒ \(\frac { p }{ q }\) = \(\frac { r }{ s }\)
⇒ p : q = r : s
Hence proved.
Question 10.
If b is the mean proportional between a and c, prove that a, c
and c, prove that a, c, a² + b² and b² + c² are proportional.
Solution:
∵ b is the mean proportional between a and c
∴ b² = ac … (i)
Now a, c, a² + b² and b² + c² are proportional
If \(\frac{a}{c}=\frac{a^2+b^2}{b^2+c^2}\)
= \(\frac{a^2+a c}{a c+c^2}=\frac{a(a+c)}{c(a+c)}=\frac{a}{c}\)
∴ Which is true.
Hence proved.
Question 11.
If x + 7 is the mean proportional between (x + 3) and (x + 12), find the value of x.
Solution:
x + 1 is the mean proportional between
(x + 3) and (x + 12)
∴ (x + 7)² = (x + 3) (x + 12)
⇒ x² + 14x + 49 = x² + 12x + 3x + 36
⇒ x² + 14x – 12x – 3x – x² = 36 – 49
⇒ – x = – 13
⇒ x = 13
∴ x = 13
Question 12.
If \(\frac{a^2+c^2}{a b+c d}=\frac{a b+c d}{b^2+d^2}\), Prove that \(\frac { a }{ b }\) = \(\frac { c }{ d }\).
Solution:
∵ \(\frac{a^2+c^2}{a b+c d}=\frac{a b+c d}{b^2+d^2}\)
⇒ (a² + c²) (b² + d²) = (ab + cd)²
⇒ a²b² + a²d² + b²c² + c²d² (By cross multiplication)
= a²b² + a²d² + 2abcd
⇒ b²c² + a²d² – 2abcd = 0
= (be – ad)² = 0
⇒ \(\frac { a }{ b }\) = \(\frac { c }{ d }\)