Peer review of OP Malhotra Class 10 Solutions Chapter 6 Ratio and Proportion Ex 6(b) can encourage collaborative learning.
S Chand Class 10 ICSE Maths Solutions Chapter 6 Ratio and Proportion Ex 6(b)
Question 1.
If \(\frac {x}{y}\) = \(\frac {p}{q}\), prove that \(\frac{5 x+7 y}{5 x-7 y}\) = \(\frac{5 p+7 q}{5 p-7 q}\).
Solution:
\(\frac {x}{y}\) = \(\frac {p}{q}\)
\(\frac {5x}{7y}\) = \(\frac {5p}{7q}\) (Multipling both sides bye \(\frac {5}{7}\)
Applying componendo and dividendo,
\(\frac{5 x+7 y}{5 x-7 y}=\frac{5 p+7 q}{5 p-7 q}\)
Hence proved.
Question 2.
If (4a + 9b): (4a – 9b) = (4c + 9d): (4c – 9d), Show that a : b : : c : d.
Solution:
(4a + 9b) : (4a-9b) = (4c + 9d): (4c – 9d)
⇒ \(\frac{4 a+9 b}{4 a-9 b}=\frac{4 c+9 d}{4 c-9 d}\)
Applying componendo and dividendo,
\(\frac{4 a+9 b+4 a-9 b}{4 a+9 b-4 a+9 b}=\frac{4 c+9 d+4 c-9 d}{4 c+9 d-4 c+9 d}\)
⇒ \(\frac{8 a}{18 b}=\frac{8 c}{18 d}\)
⇒ \(\frac{a}{b}=\frac{c}{d}\) (Dividing by \(\frac { 8 }{ 18 }\))
∴ a : b :: c : d
Hence proved.
Question 3.
If (5a + 116) : (5c -11 d) : : (5a – 116) (5c + 11d), prove that \(\frac { a }{ b }\) = \(\frac { c }{ d }\).
Solution:
(5a + 116) : (5c -11 d) : : (5a – 116) (5c + 11d)
⇒ \(\frac {x}{y}\)
Applying componendo and dividendo,
\(\frac{5 a+11 b+5 a-11 b}{5 a+11 b-5 a+11 b}=\frac{5 c+11 d+5 c-11 d}{5 c+11 d-5 c+11 d}\)
⇒ \(\frac{10 a}{22 b}=\frac{10 c}{22 d}\)
⇒ \(\frac {a}{b}\) = \(\frac {c}{d}\) (Dividing by \(\frac {10}{22}\))
Hence proved.
Question 4.
Show that a, b, c, d are in proportion if
(i) ma² + nb²: mc² + nd² :: ma² – nb² : mc² – nd².
(ii) (a + b + c + d)(a – b – c + d) = (a + b – c – d) (a – b + c – d).
Solution:
(i) ma² + nb²: mc² + nd² :: ma² – nb² : mc² – nd².
Taking square root of
⇒ \(\frac {a}{b}\) = \(\frac {c}{d}\) ⇒ a : b : : c : d
∴ a, b, c and d are in proportion.
(ii) (a + b + c + d)(a – b – c + d) = (a + b – c – d) (a – b + c – d)
⇒ \(\frac{a+b+c+d}{a+b-c-d}=\frac{a-b+c-d}{a-b-c+d}\)
Applying componendo and dividendo,
∴ a, b, c and d are in proportion.
Question 5.
If a : b = c : d and e : f = g : h, prove that ae + bf : ae – bf = eg + dh : eg – dh.
Solution:
a: b = c : dand e : f = g : h
∴ \(\frac { a }{ b }\) = \(\frac {c}{d}\) and \(\frac {e}{f}\) = \(\frac {g}{h}\) (Multiplying each other)
\(\frac { ae }{ bf }\) = \(\frac {cg}{dh}\)
Now applying componendo and dividendo
\(\frac{a e+b f}{a e-b f}=\frac{c g+d h}{c g-d h}\)
∴ (ae + bf) : (ae – bf) = (cg + dh) : (cg – dh)
Hence proved.
Question 6.
If x = \(\frac { 10pq }{ p+q }\), find the value of \(\frac{x+5 p}{x-5 p}\) + \(\frac{x+5 q}{x-5 q}\).
Solution:
x = \(\frac{10 p q}{p+q} \Rightarrow \frac{x}{5 p}=\frac{2 q}{p+q}\)
Applying componendo and dividendo,
Question 7.
If x = \(\frac{6 p q}{p+q}\), find the value of \(\frac{x+3 p}{x-3 p}\) + \(\frac{x+3 q}{x-3 q}\)
Solution:
x = \(\frac{6 p q}{p+q} \Rightarrow \frac{x}{3 p}=\frac{2 q}{p+q}\)
Applying componendo and dividendo,
\(\frac{x+3 p}{x-3 p}=\frac{2 q+p+q}{2 q-p-q}=\frac{3 q+p}{q-p}\) … (i)
Again x = \(\frac{6 p q}{p+q} \Rightarrow \frac{x}{3 q}=\frac{2 p}{p+q}\)
Applying componendo and dividendo,
\(\frac{x+3 q}{x-3 q}=\frac{2 p+p+q}{2 p-p-q}=\frac{3 p+q}{p-q}\) … (ii)
Adding (i) and (ii)
\(\frac{x+3 p}{x-3 p}+\frac{x+3 q}{x-3 q}=\frac{3 q+p}{q-p}+\frac{3 p+q}{p-q}\)
= \(\frac{-(3 q+p)}{p-q}+\frac{3 p+q}{p-q}\)
= \(\frac{-3 q-p+3 p+q}{p-q}\)
= \(\frac{2 p-2 q}{p-q}=\frac{2(p-q)}{p-q}\) = 2
Question 8.
Solve for x, using the properties of proportion.
(i) \(\frac{3 x+\sqrt{9 x^2-5}}{3 x-\sqrt{9 x^2-5}}\) = 5
(ii) \(\frac{\sqrt{x+2}+\sqrt{x-3}}{\sqrt{x+2}-\sqrt{x-3}}\) = 5
(iii) \(\frac{x^3+3 x}{3 x^2+1}=\frac{341}{91}\)
(iv) \(\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}=\frac{4 x-1}{2}\)
Solution:
(i) \(\frac{3 x+\sqrt{9 x^2-5}}{3 x-\sqrt{9 x^2-5}}\) = 5
Applying componendo and dividendo,
\(\frac{3 x+\sqrt{9 x^2-5}+3 x-\sqrt{9 x^2-5}}{3 x+\sqrt{9 x^2-5}-3 x+\sqrt{9 x^2-5}}=\frac{5+1}{5-1}\)
⇒ \(\frac{6 x}{2 \sqrt{9 x^2-5}}=\frac{6}{4} \Rightarrow \frac{3 x}{\sqrt{9 x^2-5}}\) = \(\frac { 3 }{ 2 }\)
⇒ \(3 \sqrt{\left(9 x^2-5\right)}=6 x \Rightarrow \sqrt{9 x^2-5}\) = 2x
(Dividing by 3)
Squaring both sides,
9x² – 5 = 4x² ⇒ 9x² – 4x² = 5
5x² = 5 ⇒ x² = \(\frac { 5 }{ 5 }\) = 1
∴ x = 1
(ii) \(\frac{\sqrt{x+2}+\sqrt{x-3}}{\sqrt{x+2}-\sqrt{x-3}}=\frac{5}{1}\)
Appling componendo and dividendo,
\(\frac{\sqrt{x+2}+\sqrt{x-3}+\sqrt{x+2}-\sqrt{x-3}}{\sqrt{x+2}+\sqrt{x-3}-\sqrt{x+2}+\sqrt{x-3}}=\frac{5+1}{5-1}\)
⇒ \(\frac{2 \sqrt{x+2}}{2 \sqrt{x-3}}=\frac{6}{4} \Rightarrow \frac{\sqrt{x+2}}{\sqrt{x-3}}=\frac{3}{2}\)
Squaring both sides,
⇒ \(\frac { x+2 }{ x-3 }\) = \(\frac { 9 }{ 4 }\) ⇒ 9x – 27 = 4x + 8
⇒ 9x – 4x = 8 + 27 ⇒ 5x = 35
⇒ x = \(\frac { 35 }{ 5 }\) = 7
∴ x = 7
(iii) \(\frac{x^3+3 x}{3 x^2+1}=\frac{341}{91}\)
Appling componendo and dividendo,
\(\frac{x^3+3 x+3 x^2+1}{x^3+3 x-3 x^2-1}=\frac{341+91}{341-91}\)
⇒ \(\frac{x^3+3 x^2+3 x+1}{x^3-3 x^2+3 x-1}=\frac{432}{250}=\frac{216}{125}\)
⇒ \(\frac{(x+1)^3}{(x-1)^3}=\frac{(6)^3}{(5)^3}\)
Taking cube root,
\(\frac { x+1 }{ x-1 }\) = \(\frac { 6 }{ 5 }\)
⇒ 6x – 5x = 5 + 6 ⇒ x = 11
∴ x = 11x
(iv) \(\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}=\frac{4 x-1}{2}\)
Applying componendo and dividendo,
Question 9.
Solve for x : 16\(\left[\frac{a-x}{a+x}\right]^3=\frac{a+x}{a-x}\).
Solution:
Question 10.
Solve for x : 16\(\frac{1+x+x^2}{1-x+x^2}=\frac{171(1+x)}{172(1-x)}\), x ≠ 1, – 1.
Solution:
Taking cube root of,
x = \(\frac { 1 }{ 7 }\)
Question 11.
Solve for x = \(\frac{\sqrt{a^2+b^2}+\sqrt{a^2-b^2}}{\sqrt{a^2+b^2}-\sqrt{a^2-b^2}}\) show that b²x² – 2a²x + b² = 0.
Solution:
Question 12.
If y = \(\frac{\sqrt{a+3 b}+\sqrt{a-3 b}}{\sqrt{a+3 b}-\sqrt{a-3 b}}\) show that 3by² – 2ay + 3b = 0.
Solution:
Question 13.
If y = \(\frac{a^3+3 a b^2}{3 a^2 b+b^3}=\frac{x^3+3 x y^2}{3 x^2 y+y^3}\) show that \(\frac { x }{ a }\) = \(\frac { y }{ b }\)
Solution:
\(\frac{a^3+3 a b^2}{3 a^2 b+b^3}=\frac{x^3+3 x y^2}{3 x^2 y+y^3}\)
Applying componendo and dividendo,
