ML Aggarwal Solution Class 10 Chapter 7 Ratio and Proportion Test

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20 Q21 Q22 Q23

 Test

Question 1

Find the compound ratio of:

$(a+b)^{2}:(a-b)^{2}$

$\left(a^{2}-b^{2}\right):\left(a^{2}+b^{2}\right)$

$\left(a^{4}-b^{4}\right):(a+b)^{4}$

Sol :

$(a+b)^{2}:(a-b)^{2}$

$\left(a^{2}-b^{2}\right):\left(a^{2}+b^{2}\right)$

$\left(a^{4}-b^{4}\right):(a+b)^{4}$

$=\frac{(a+b)^{2}}{(a-b)^{2}} \times \frac{a^{2}-b^{2}}{a^{2}+b^{2}} \times \frac{a^{4}-b^{4}}{(a+b)^{4}}$

$=\frac{(a+b)^{2}}{(a-b)^{2}} \times \frac{(a+b)(a-b)}{a^{2}+b^{2}}\times \frac{\left(a^{2}+b^{2}\right)(a+b)(a-b)}{(a+b)^{4}}$

$=\frac{1}{1}$ = 1 : 1

Question 2

If (7 p + 3 q) : (3 p – 2 q) = 43 : 2 find p : q

Sol :

(7p + 3q) : (3p – 2q) = 43 : 2

$\Rightarrow \frac{7 p+3 q}{3 p-2 q}=\frac{43}{2}$

$\Rightarrow \quad 129 p-86 q=14 p+6 q$

$\Rightarrow \quad 129 p-14 p=6 q+86 q$

$\Rightarrow \quad 115 p=92 q$

$\Rightarrow \quad \frac{p}{q}=\frac{92}{115}=\frac{4}{5}$

∴p : q= 4 : 5 

Question 3

If a : b = 3 : 5, find (3a + 5b): (7a – 2b).

Sol :

a : b = 3 : 5

$\Rightarrow \frac{a}{b}=\frac{3}{5}$

⇒ 3a + 5n : 7a – 2b

Dividing each term by b

$3 \frac{a}{b}+5: 3 \frac{a}{b}-2$

$\Rightarrow 3 \times \frac{3}{5}+5: 7 \times \frac{3}{5}-2$

$\Rightarrow\left(\frac{9}{5}+5\right):\left(\frac{21}{5}-2\right)$

$\Rightarrow \frac{9+25}{5}: \frac{21-10}{5}$

$\Rightarrow \frac{34}{5}: \frac{11}{5}$

=34:11

Question 4

The ratio of the shorter sides of a right angled triangle is 5 : 12. If the perimeter of the triangle is 360 cm, find the length of the longest side.

Sol :

Let the two shorter sides of a right-angled triangle be 5x and 12x.

Third (longest side)

$=\sqrt{(5 x)^{2}+(12 x)^{2}}$

$=\sqrt{25 x^{2}+144 x^{2}}$

$=\sqrt{169 x^{2}}=13 x$

But 5x+12x+13x=360cm

⇒30x=360

$\Rightarrow x=\frac{360}{30}=12$

∴Length of the longest side=13x

=13×12cm=156cm

Question 5

The ratio of the pocket money saved by Lokesh and his sister is 5 : 6. If the sister saves Rs 30 more, how much more the brother should save in order to keep the ratio of their savings unchanged?

Sol :

Let the savings of Lokesh and his sister are 5x and 6x.

and the Lokesh should save Rs y more Now, according to the problem,

$\Rightarrow \frac{5 x+y}{6 x+30}=\frac{5}{6}$

⇒ 30x + 6y = 30x + 150

⇒ 6y=150

$\therefore y=\frac{150}{6}=25$

Hence, Lokesh should save 25 more

Question 6

In an examination, the number of those who passed and the number of those who failed were in the ratio of 3 : 1. Had 8 more appeared, and 6 less passed, the ratio of passed to failures would have been 2 : 1. Find the number of candidates who appeared.

Sol :

Let number of passed = 3 x

and failed = x

Total candidates appeared = 3x + x = 4x.

In second case

No. of candidates appeared = 4 x + 8

and No. of passed = 3 x – 6

and failed = 4x + 8 – 3x + 6 = x + 14

then ratio will be = 2 : 1

Now according to the condition

$\frac{3 x-6}{x+14}=\frac{2}{1}$

⇒ 3x-6=2x+28

⇒ 3x-2x=28+6

⇒ x=34

∴No. of candidates appeared

=4x=4×34=136

Question 7

What number must be added to each of the numbers 15, 17, 34 and 38 to make them proportional ?

Sol :

Let x be added to each number, then numbers will be

15 + x, 17 + x, 34 + x, and 38 + x.

Now according to the condition

$\frac{15+x}{17+x}=\frac{34+x}{38+x}$

$\Rightarrow(15+x)(38+x)=(34+x)(17+x)$

$\Rightarrow 570+53 x+x^{2}=578+51 x+x^{2}$

$\Rightarrow x^{2}+53 x-x^{2}-51 x=578-570$

$\Rightarrow 2 x=8 \Rightarrow x=4$

$\therefore 4$ is to be added.

Question 8

If (a + 2 b + c), (a – c) and (a – 2 b + c) are in continued proportion, prove that b is the mean proportional between a and c.

Sol :

(a + 2 b + c), (a – c) and (a – 2 b + c) are in continued proportion

$\Rightarrow \frac{a+2 b+c}{a-c}=\frac{a-c}{a-2 b+c}$

$\therefore \frac{a+2 b+c}{a-c}=\frac{a-c}{a-2 b+c}$

$\Rightarrow \quad(a+2 b+c)(a-2 b+c)=(a-c)^{2}$

$\Rightarrow \quad a^{2}-2 a b+a c+2 a b-4 b^{2}+2 b c+a c-2 b c+c^{2}=a^{2}-2 a c+c^{2}$

$\Rightarrow \quad a^{2}-2 a b+a c+2 a b-4 b^{2}+2 b c+a c-2 b c+c^{2}-a^{2}+2 a c-c^{2}=0$

$\Rightarrow \quad 4 a c-4 b^{2}=0 \Rightarrow a c-b^{2}=0$

$\Rightarrow \quad b^{2}=a c$

Here b is the mean proportional between a and c

Question 9

If 2, 6, p, 54 and q are in continued proportion, find the values of p and q.

Sol :

2, 6, p, 54 and q are in continued proportional then

$\Rightarrow \frac{2}{6}=\frac{6}{p}=\frac{p}{54}=\frac{54}{9}$

(i) $\because \frac{2}{6}=\frac{6}{p}$ then 

2p=36

$ \Rightarrow p=18$

(ii) $\frac{p}{54}=\frac{54}{q} \Rightarrow p q=54 \times 54$

$\Rightarrow 18 q=54 \times 54 $

$\Rightarrow q=\frac{54 \times 54}{18}=162$

Hence , p=18, q=162

Question 10

If a, b, c, d, e are in continued proportion, prove that: $a: e=a^{4}: b^{4}$

Sol :

a, b, c, d, e are in continued proportion

$\Rightarrow \frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{e}=k($ say $)$

$d=e k, c=e k^{2}, b=e k^{3}$ and $a=e k^{4}$

Now, $\mathrm{L.H.S.}=\frac{a}{e}=\frac{e k^{4}}{e}=k^{4}$

R.H.S $\frac{a^{4}}{b^{4}}=\frac{\left(e k^{4}\right)^{4}}{\left(e k^{3}\right)^{4}}=\frac{e^{4} k^{16}}{e^{4} k^{12}}$

$=k^{16-12}=k^{4}$

∴L.H.S=R.H.S

Question 11

Find two numbers whose mean proportional is 16 and the third proportional is 128.

So :

Let x and y be two numbers

Their mean proportion = 16

and third proportion = 128

$\therefore \sqrt{x y}=16 \Rightarrow x y=256$

$\Rightarrow x=\frac{256}{y}$...(i)

and $\frac{y^{2}}{x}=128 \Rightarrow x=\frac{y^{2}}{128}$...(ii)

From (i) and (ii)

$\frac{256}{y}=\frac{y^{2}}{128} $

$\Rightarrow y^{3}=256 \times 128=32768$

$\Rightarrow y^{3}=(32)^{3} \Rightarrow y=32$

$\therefore x=\frac{256}{y}=\frac{256}{32}=8$

$\therefore$ Numbers are 8,32

Question 12

If q is the mean proportional between p and r, prove that:

$p^{2}-3 q^{2}+r^{2}=q^{4}\left(\frac{1}{p^{2}}-\frac{3}{q^{2}}+\frac{1}{r^{2}}\right)$

Sol :

q is mean proportional between p and r

q² = pr

L.H.S. $=p^{2}-3 q^{2}+r^{2}=p^{2}-3 p r+r^{2}$

$\mathrm{R.H.S.}=q^{4}\left(\frac{1}{p^{2}}-\frac{3}{q^{2}}+\frac{1}{r^{2}}\right)$

$=\left(q^{2}\right)^{2}\left(\frac{1}{p^{2}}-\frac{3}{q^{2}}+\frac{1}{r^{2}}\right)=(p r)^{2}\left(\frac{1}{p^{2}}-\frac{3}{q^{2}}+\frac{1}{r^{2}}\right)$

$=(p r)^{2}\left(\frac{1}{p^{2}}-\frac{3}{p r}+\frac{1}{r^{2}}\right)$

$=p^{2} r^{2}\left(\frac{r^{2}-3 p r+p^{2}}{p^{2} r^{2}}\right)=r^{2}-3 p r+p^{2}$

∴L.H.S=R.H.S

Question 13

If $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}$ , prove that each ratio is 

(i) $\sqrt{\frac{3 a^{2}-5 c^{2}+7 e^{2}}{3 b^{2}-5 d^{2}+7 t^{2}}}$

(ii) $\left[\frac{2 a^{3}+5 c^{3}+7 e^{3}}{2 b^{3}+5 d^{3}+7 f^{3}}\right]^{\frac{1}{3}}$

Sol :

$\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=k(\mathrm{say})$

∴ a = k, c = dk, e = fk

(i) $\sqrt{\frac{3 a^{2}-5 c^{2}+7 e^{2}}{3 b^{2}-5 d^{2}+7 f^{2}}}$

$=\sqrt{\frac{3 b^{2} k^{2}-5 d^{2} k^{2}+7 f^{2} k^{2}}{3 b^{2}-5 d^{2}-7 f^{2}}}$

$=k \sqrt{\frac{3 b^{2}-5 d^{2}+7 f^{2}}{3 b^{2}-5 d^{2}+7 f^{2}}}=k$

hence proved

(ii) $\left[\frac{2 a^{3}+5 c^{3}+7 e^{3}}{2 b^{3}+5 d^{3}+7 f^{3}}\right]^{\frac{1}{3}}$

$=\left[\frac{2 b^{3} k^{3}+5 d^{3} k^{3}+7 f^{3} k^{3}}{2 b^{3}+5 d^{3}+7 f^{3}}\right]^{\frac{1}{3}}$

$=k\left[\frac{2 b^{3}+5 d^{3}+7 f^{3}}{2 b^{3}+5 d^{3}+7 f^{3}}\right]^{\frac{1}{3}}=k$

hence proved

Question 14

If $\frac{x}{a}=\frac{y}{b}=\frac{z}{c},$ prove that $\frac{3 x^{3}-5 y^{3}+4 z^{3}}{3 a^{3}-5 b^{3}+4 c^{3}}=\left(\frac{3 x-5 y+4 z}{3 a-5 b+4 c}\right)^{3}$

Sol :

$\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k$( say )

x=a k, y=b k, z=c k

L.H.S. $=\frac{3 x^{3}-5 y^{3}+4 z^{3}}{3 a^{3}-5 b^{3}+4 c^{3}}$

$=\frac{3 a^{3} k^{3}-5 b^{3} k^{3}+4 c^{3} k^{3}}{3 a^{3}-5 b^{3}+4 c^{3}}$

$=\frac{k^{3}\left(3 a^{3}-5 b^{3}+4 c^{3}\right)}{3 a^{3}-5 b^{3}+4 c^{3}}=k^{3}$

R.H.S. $=\left(\frac{3 x-5 y+4 z}{3 a-5 b+4 c}\right)^{3}$

$=\left(\frac{3 a k-5 b k+4 c k}{3 a-5 b+4 c}\right)^{3}=\left(\frac{k(3 a-5 b+4 c)}{(3 a-5 b+4 c)}\right)^{3}$

$=(k)^{3}=k^{3}$

∴L.H.S=R.H.S

Question 15

If x : a = y : b, prove that

$\frac{x^{4}+a^{4}}{x^{3}+a^{3}}+\frac{y^{4}+b^{4}}{y^{3}+b^{3}}=\frac{(x+y)^{4}+(a+b)^{4}}{(x+y)^{3}+(a+b)^{3}}$

Sol :

$\frac{x}{a}=\frac{y}{b}=k(s a y)$

x = ak, y = bk

L.H.S. $=\frac{x^{4}+a^{4}}{x^{3}+a^{3}}+\frac{y^{4}+b^{4}}{y^{3}+b^{3}}$

$=\frac{a^{4} k^{4}+a^{4}}{a^{3} k^{3}+a^{3}}+\frac{b^{4} k^{4}+b^{4}}{b^{3} k^{3}+b^{3}}$

$=\frac{a^{4}\left(k^{4}+1\right)}{a^{3}\left(k^{3}+1\right)}+\frac{b^{4}\left(k^{4}+1\right)}{b^{3}\left(k^{3}+1\right)}$

$=\frac{a\left(k^{4}+1\right)}{k^{3}+1}+\frac{b\left(k^{4}+1\right)}{k^{3}+1}$

$=\frac{a\left(k^{4}+1\right)+b\left(k^{4}+1\right)}{k^{3}+1}=\frac{\left(k^{4}+1\right)(a+b)}{k^{3}+1}$

R.H.S. $=\frac{(x+y)^{4}+(a+b)^{4}}{(x+y)^{3}+(a+b)^{3}}$

$=\frac{(a k+b k)^{4}+(a+b)^{4}}{(a k+b k)^{3}+(a+b)^{3}}$

$=\frac{k^{4}(a+b)^{4}+(a+b)^{4}}{k^{3}(a+b)^{3}(a+b)^{3}}=\frac{(a+b)^{4}\left(k^{4}+1\right)}{(a+b)^{3}\left(k^{3}+1\right)}$

$=\frac{(a+b)\left(k^{4}+1\right)}{k^{3}+1}=\frac{\left(k^{4}+1\right)(a+b)}{k^{3}+1}$

∴L.H.S=R.H.S

Hence proved

Question 16

If $\frac{x}{b+c-a}=\frac{y}{c+a-b}=\frac{z}{a+b-c}$ prove that each ratio’s equal to :

$\frac{x+y+z}{a+b+c}$

Sol :

$\frac{x}{b+c-a}=\frac{y}{c+a-b}=\frac{z}{a+b-c}=k($ say $)$

x = k(b + c – a),

y = k(c + a – b),

z = k(a + b – c)

$\frac{x+y+z}{a+b+c}$

$=\frac{k(b+c-a)+k(c+a-b)+k(a+b-c)}{a+b+c}$

$=\frac{k(b+c-a+c+a-b+a+b-c)}{a+b+c}$

$=\frac{k(a+b+c)}{a+b+c}=k$

Hence proved

Question 17

If a : b = 9 : 10, find the value of

(i) $\frac{5 a+3 b}{5 a-3 b}$

(ii) $\frac{2 \alpha^{2}-3 b^{2}}{2 a^{2}+3 b^{2}}$

Sol :

a : b = 9 : 10

$\Rightarrow \frac{a}{b}=\frac{9}{10}$

(i) $\frac{5 a+3 b}{5 a-3 b}=\frac{\frac{5 a}{b}+\frac{3 b}{b}}{\frac{5 a}{b}-\frac{3 b}{b}}=\frac{\frac{5 a}{b}+3}{\frac{5 a}{b}-3}$

(dividing by b)

$=\frac{5 \times \frac{9}{10}+3}{5 \times \frac{9}{10}-3}$

$=\frac{\frac{9}{2}+3}{\frac{9}{2}-3}$

$=\frac{\frac{15}{2}}{\frac{3}{2}}$

(Substituting the value of $\frac{a}{b}$)

$=\frac{15}{2} \times \frac{2}{3}=5$

(ii) \frac{2 a^{2}-3 b^{2}}{2 a^{2}+3 b^{2}}$

$=\frac{\frac{2 a^{2}}{b^{2}}-\frac{3 b^{2}}{b^{2}}}{\frac{2 a^{2}}{b^{2}}-\frac{3 b^{2}}{h^{2}}}

(Dividing by $b^{2}$)

=\frac{2\left(\frac{a}{b}\right)^{2}-3}{2\left(\frac{a}{b}\right)^{2}+3}$

$=\frac{2\left(\frac{9}{10}\right)^{2}-3}{2\left(\frac{9}{10}\right)^{2}+3}

$=\frac{2 \times \frac{81}{100}-3}{2 \times \frac{81}{100}+3}$

$=\frac{\frac{81}{50}-3}{\frac{81}{50}+3}$

$=\frac{\frac{81-150}{50}}{\frac{81+150}{50}}$

$=\frac{-69}{50} \times \frac{50}{231}=\frac{-69}{231}=\frac{-23}{77}$

Question 18

If (3x² + 2y²) : (3x² – 2y²) = 11 : 9, find the value of  $\frac{3 x^{4}+25 y^{4}}{3 x^{4}-25 y^{4}}$

Sol :

$\frac{3 x^{4}+25 y^{4}}{3 x^{4}-25 y^{4}}=\frac{11}{9}$

Applying componendo and dividendo

$\frac{3 x^{2}+2 y^{2}+3 x^{2}-2 y^{2}}{3 x^{2}+2 y^{2}-3 x^{2}+2 y^{2}}=\frac{11+9}{11-9}$

$\Rightarrow \frac{6 x^{2}}{4 y^{2}}=\frac{20}{2}$

$ \Rightarrow  \frac{3}{2} \frac{x^{2}}{y^{2}}=10$

$\Rightarrow \frac{x^{2}}{y^{2}}=10 \times \frac{2}{3}=\frac{20}{3}$

$\frac{3 x^{4}+25 y^{4}}{3 x^{4}-25 y^{4}}$

$=\frac{\frac{3 x^{4}}{y^{4}}+\frac{25 y^{4}}{y^{4}}}{\frac{3 x^{4}}{y^{4}}-\frac{25 y^{4}}{y^{4}}}$

$=\frac{3\left(\frac{x^{2}}{y^{2}}\right)^{2}+25}{3\left(\frac{x^{2}}{y^{2}}\right)^{2}-25}$

$=\frac{3 \times\left(\frac{20}{3}\right)^{2}+25}{3\left(\frac{20}{3}\right)^{2}-25}$

$=\frac{3 \times \frac{400}{9}+25}{3 \times \frac{400}{9}-25}$

$=\frac{\frac{400}{3}+\frac{25}{1}}{\frac{400}{3}-\frac{25}{1}}$

$=\frac{\frac{400+75}{3}}{\frac{400-75}{3}}$

$=\frac{475}{3} \times \frac{3}{325}=\frac{19}{13}$

Question 19

If $x=\frac{2 m a b}{a+b}$ , find the value of

$\frac{x+m a}{x-m a}+\frac{x+m b}{x-m b}$

Sol :

$x=\frac{2 m a b}{a+b}$

$\Rightarrow \frac{x}{m a}+\frac{2 b}{a+b}$

Applying componendo and dividendo

$\frac{x+m a}{x-m a}=\frac{2 b+a+b}{2 b-a-b}=\frac{3 b+a}{b-a}$...(i)

Again, $\frac{x}{m b}=\frac{2 a}{a+b}$

Applying componendo and dividendo,

$\frac{x+m b}{x-m b}=\frac{2 a+a+b}{2 a-a-b}=\frac{3 a+b}{a-b}$..(ii)

Adding (i) and (ii)

$\frac{x+m a}{x-m a} \cdot+\frac{x+m b}{x-m b}=\frac{3 b+a}{b-a}+\frac{3 a+b}{a-b}$

$=-\frac{3 b+a}{a-b}+\frac{3 a+b}{a-b}=\frac{-3 b-a+3 a+b}{a-b}$

$=\frac{2 a-2 b}{a-b}=\frac{2(a-b)}{a-b}=2$

Question 20

If $x=\frac{p a b}{a+b},$ prove that $\frac{x+p a}{x-p a}-\frac{x+p b}{x-p b}=\frac{2\left(a^{2}-b^{2}\right)}{a b}$

Sol :

$x=\frac{p a b}{a+b}$

$\Rightarrow \frac{x}{p a}+\frac{b}{a+b}$

Applying componendo and dividendo

$\frac{x+p a}{x-p a}=\frac{b+a+b}{b-a-b}=\frac{a+2 b}{-a}$..(i)

Again $\frac{x}{p b}=\frac{a}{a+b}$

Applying componendo and dividendo,

$\frac{x+p b}{x-p b}=\frac{a+a+b}{a-a-b}=\frac{2 a+b}{-b}$..(ii)

$\mathrm{L.H.S.}=\frac{x+p a}{x-p a}-\frac{x+p b}{x-p b}$

$=\frac{a+2 b}{-a}-\frac{2 a+b}{-b}$

$=\frac{a+2 b}{-a}+\frac{2 a+b}{b}$

$=\frac{a b+2 b^{2}-2 a^{2}-a b}{-a b}$

$=\frac{2 b^{2}-2 a^{2}}{-a b}$

$=\frac{-2 a^{2}+2 b^{2}}{-a b}$

$=\frac{-2\left(a^{2}-b^{2}\right)}{-a b}=\frac{2\left(a^{2}-b^{2}\right)}{a b}$

=R.H.S

Question 21

Find x from the equation $\frac{a+x+\sqrt{a^{2}-x^{2}}}{a+x-\sqrt{a^{2}-x^{2}}}=\frac{b}{x}$

Sol :

$\frac{a+x+\sqrt{a^{2}-x^{2}}}{a+x-\sqrt{a^{2}-x^{2}}}=\frac{b}{x}$

Applying componendo and dividendo,

$\frac{a+x+\sqrt{a^{2}-x^{2}}+a+x-\sqrt{a^{2}-x^{2}}}{a+x+\sqrt{a^{2}-x^{2}}-a-x+\sqrt{a^{2}-x^{2}}}$

$=\frac{b+x}{b-x} $

$\Rightarrow \frac{2(a+x)}{2 \sqrt{a^{2}-x^{2}}}=\frac{b+x}{b-x}$

$\Rightarrow \frac{a+x}{\sqrt{a^{2}-x^{2}}}=\frac{b+x}{b-x}$

Squaring both sides,

$\frac{(a+x)^{2}}{a^{2}-x^{2}}=\frac{(b+x)^{2}}{(b-x)^{2}}$

$\Rightarrow \frac{(a+x)^{2}}{(a+x)(a-x)}=\frac{(b+x)^{2}}{(b-x)^{2}}$

$\Rightarrow \frac{a+x}{a-x}=\frac{(b+x)^{2}}{(b-x)^{2}}$

Again applying componendo and dividendo,

$\frac{a+x+a-x}{a+x-a+x}=\frac{(b+x)^{2}+(b-x)^{2}}{(b+x)^{2}-(b-x)^{2}}$

$\Rightarrow \frac{2 a}{2 x}=\frac{2\left(b^{2}+x^{2}\right)}{4 b x}$

$ \Rightarrow \frac{a}{x}=\frac{b^{2}+x^{2}}{2 b x}$

$\Rightarrow \frac{2 a}{2 x}=\frac{2\left(b^{2}+x^{2}\right)}{4 b x}$

$ \Rightarrow \frac{a}{x}=\frac{b^{2}+x^{2}}{2 b x}$

$\Rightarrow x^{2}=2 a b-b^{2}$

$x=\sqrt{2 a b-b^{2}}$ 

Question 22

If $x=\frac{\sqrt[3]{a+1}+\sqrt[3]{a-1}}{\sqrt[3]{a+1}-\sqrt[3]{a-1}},$ prove that

$x^{3}-3 a x^{2}+3 x-a=0$

Sol :

$x=\frac{\sqrt[3]{a+1}+\sqrt[3]{a-1}}{\sqrt[3]{a+1}-\sqrt[3]{a-1}}$

Applying componendo and dividendo,

$\frac{x+1}{x-1}=\frac{\sqrt[3]{a+1}+\sqrt[3]{a-1}+\sqrt[3]{a+1}-\sqrt[3]{a-1}}{\sqrt[3]{a+1}+\sqrt[3]{a-1}-\sqrt[3]{a+1}+\sqrt[3]{a-1}}$

$\frac{x+1}{x-1}=\frac{2 \sqrt[3]{a+1}}{2 \sqrt[3]{a-1}}$

$ \Rightarrow \frac{x+1}{x-1}=\frac{\sqrt[3]{a+1}}{\sqrt[3]{a-1}}$

Cubing both sides

$\frac{(x+1)^{3}}{(x-1)^{3}}=\frac{a+1}{a-1}$

Again applying componendo and dividendo,

$\frac{(x+1)^{3}+(x-1)^{3}}{(x+1)^{3}-(x-1)^{3}}=\frac{a+1+a-1}{a+1-a+1}$

$\Rightarrow \frac{2\left(x^{3}+3 x\right)}{2\left(3 x^{2}+1\right)}=\frac{2 a}{2}$

$ \Rightarrow \frac{x^{3}+3 x}{3 x^{2}+1}=\frac{a}{1}$

$\Rightarrow x^{3}+3 x=3 a x^{2}+a $

$\Rightarrow x^{3}-3 a x^{2}+3 x-a=0$

Question 23

If $\frac{b y+c z}{b^{2}+c^{2}}=\frac{c z+a x}{c^{2}+a^{2}}=\frac{a x+b y}{a^{2}+b^{2}}$, prove that each of these ratio is equal to 

Sol :

$\frac{b y+c z}{b^{2}+c^{2}}=\frac{c z+\alpha x}{c^{2}+a^{2}}=\frac{a x+b y}{a^{2}+b^{2}}$

$=\frac{2(a x+b y+c z)}{2\left(a^{2}+b^{2}+c^{2}\right)}=\frac{a x+b y+c z}{a^{2}+b^{2}+c^{2}}$ (Adding)

Now $\frac{b y+c z}{b^{2}+c^{2}}=\frac{a x+b y+c z}{a^{2}+b^{2}+c^{2}}$

$\Rightarrow \frac{b y+c z}{a x+b y+c z}=\frac{b^{2}+c^{2}}{a^{2}+b^{2}+c^{2}}$ (By alternendo)

$\Rightarrow \frac{b y+c z-a x-b y-c z}{a x+b y+c z}$

$=\frac{b^{2}+c^{2}-a^{2}-b^{2}-c^{2}}{a^{2}+b^{2}+c^{2}}$

$\Rightarrow \frac{-a x}{a x+b y+c z}=\frac{-a^{2}}{a^{2}+b^{2}+c^{2}}$

$\Rightarrow \frac{x}{a x+b y+c z}=\frac{a}{a^{2}+b^{2}+c^{2}}$

$\Rightarrow \frac{x}{a}=\frac{a x+b y+c z}{a^{2}+b^{2}+c^{2}}$ ..(i)

Similarly we can prove that

$\frac{y}{b}=\frac{a x+b y+c z}{a^{2}+b^{2}+c^{2}}$..(ii)

and $\frac{z}{c}=\frac{a x+b y+c z}{a^{2}+b^{2}+c^{2}}$...(iii)

From (i), (ii) and (iii)

Hence $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$

ML Aggarwal Solution Class 10 Chapter 7 Ratio and Proportion MCQs

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10

 MCQs

Question 1

The ratio of 45 minutes to $5 \frac{3}{4}$  hours is

(a) 180:23

(b) 3:23

(c) 23:3

(d) 6:23

Solution:

ratio of 45 minutes to $5 \frac{3}{4}$ hours is

45 minutes to : $5 \frac{3}{4}$ hours

$=45: 5 \frac{3}{4} \times 60$

$=45: \frac{23}{4} \times 60$

$=45: 345$

=3: 23

Ans (b)

Question 2

The ratio of 4 litres to 900 mL is

(a) 4 : 9

(b) 40 : 9

(c) 9 : 40

(d) 20 : 9

Sol :

4l : 900 ml

= 4000 ml : 900 ml

= 4000 : 900

= 40 : 9 (b)

Question 3

When the number 210 is increased in the ratio 5 : 7, the the new number is

(a) 150

(b) 180

(c) 294

(d) 420

Sol :

210 is increased in the ratio 5 : 7, then

New increased number will be

$=210 \times \frac{7}{5}$

=294

Ans (c)

Question 4

Two numbers are in the ratio 7 : 9. If the sum of the numbers is 288, then the smaller number is

(a) 126

(b) 162

(c) 112

(d) 144

Sol :

Ratio in two number = 7 : 9

Sum of numbers = 288

Sum of ratios = 7 + 9

= 16

Smaller number $=\frac{288 \times 7}{16}$

= 126 

Ans (a)

Question 5

A ratio equivalent to the ratio $\frac{2}{3}: \frac{5}{7}$ is

(a) 4:6

(b) 5:7

(c) 15:14

(d) 14:15

Sol :

$\frac{2}{3}: \frac{5}{7}$

Multiply and divide $\frac{2}{3}$ by 7 and 

Multiply and divide $\frac{5}{7}$ by 3

$\therefore \frac{2}{3} \times \frac{7}{7}: \frac{5}{7} \times \frac{3}{3}$

$\Rightarrow \frac{14}{21}: \frac{15}{21} \Rightarrow 14: 15$

$\therefore \frac{2}{3}: \frac{5}{7}$ is equivalent 14: 15

Ans (d)

Question 6

The ratio of number of edges of a cube to the number of its faces is

(a) 2 : 1

(b) 1 : 2

(c) 3 : 8

(d) 8 : 3

Sol :

No. of edges of the cube = 12

No. of faces = 6

Ratio in edges a cube to the number of faces = 12 : 6

= 2 : 1

Ans  (a)

Question 7

If x, 12, 8 and 32 are in proportion, then the value of x is

(a) 6

(b) 4

(c) 3

(d) 2

Sol :

x, 12, 8, 32 are in proportion, then

x × 32 = 12 × 8 (∵ ad = bc)

$\Rightarrow x=\frac{12 \times 8}{32}=3$

x = 3

Ans (c)

Question 8

The fourth proportional to 3, 4, 5 is

(a) 6

(b) $\frac{20}{3}$

(c) $\frac{15}{4}$

(d) $\frac{12}{5}$

Sol :

The fourth proportion to 3, 4, 5 will be

$=\frac{4 \times 5}{3}$

$=\frac{20}{3}$

Ans (b)

Question 9

The third proportional to $6 \frac{1}{4}$ and 5 is

(a) 4

(b) $8 \frac{1}{2}$

(c) 3

(d) none of these

Sol :

The third proportional to $6 \frac{1}{4}$ and 5 is 

$\Rightarrow 6 \frac{1}{4}: 5: 5: x$

$\Rightarrow \frac{25}{4}: 5: 5: x$

$\Rightarrow x=\frac{5 \times 5}{25} \times 4$

$\Rightarrow 4(a)$

Question 10

The mean proportional between $\frac{1}{2}$ and 128 is 

(a) 64

(b) 32

(c) 16

(d) 8

Sol :

The mean proportional between $\frac{1}{2}$ and 128 is 

$=\sqrt{\frac{1}{2} \times 128}$

$=\sqrt{64}$

=8

Ans (d)

ML Aggarwal Solution Class 10 Chapter 7 Ratio and Proportion Exercise 7.3

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20

 Exercise 7.3

Question 1

If a : b : : c : d, prove that

(i) $\frac{2 a+5 b}{2 a-5 b}=\frac{2 c+5 d}{2 c-5 d}$

(ii) $\frac{5 a+11 b}{5 c+11 d}=\frac{5 a-11 b}{5 c-11 d}$

(iii) (2a + 3b)(2c – 3d) = (2a – 3b)(2c + 3d)

(iv) (la + mb) : (lc + mb) :: (la – mb) : (lc – mb)

Sol :

(i) a : b : : c : d

then $\frac{a}{b}=\frac{c}{d}$

$\Rightarrow \frac{2 a}{5 b}=\frac{2 c}{5 d}$ (multiply by $\left.\frac{2}{5}\right)$

Applying componendo and dividendo,

$\frac{2 a+5 b}{2 a-5 b}=\frac{2 c+5 d}{2 c-5 d}$

(ii) ∵ a: b:: c: d

$\therefore \frac{a}{b}=\frac{c}{d} \Rightarrow \frac{5 a}{11 b}=\frac{5 c}{11 d}\left(\right.$ Multiply by $\left.\frac{5}{11}\right)$

Applying componendo and dividendo,

$\frac{5 a+11 b}{5 a-11 b}=\frac{5 c+11 d}{5 c-11 d}$

$\Rightarrow \frac{5 a+11 b}{5 c+11 d}=\frac{5 a-11 b}{5 c-11 d}$ (Applying alternendo)

(iii) ∵ a: b:: c: d

$\therefore \frac{a}{b}=\frac{c}{d} \Rightarrow \frac{2 a}{3 b}=\frac{2 c}{3 d}\left(\right.$ Multiply by $\left.\frac{2}{3}\right)$

Applying componendo and dividendo,

$\frac{2 a+3 b}{2 a-3 b}=\frac{2 c+3 d}{2 c-3 d}$

⇒(2 a+3 b)(2 c-3 d)

=(2 a-3 b)(2 c+3 d) (By cross multiplication)

(iv) ∵ a: b:: c: d

$\therefore \frac{a}{b}=\frac{c}{d}$

$\Rightarrow \quad \frac{l a}{m b}=\frac{l c}{m d}\left(\right.$ Multiply by $\left.\frac{l}{m}\right)$

Applying componendo and dividendo,

$\frac{l a+m b}{l a-m b}=\frac{l c+m d}{l c-m d}$

$\Rightarrow \quad \frac{l a+m b}{l c+m d}=\frac{l a-m b}{l c-m d} \quad$ (By alternendo)

⇒(l a+m b):(l c+m d)::(l a-m b) :(l c-m d)

Question 2

(i) If $\frac{5 x+7 y}{5 u+7 y}=\frac{5 x-7 y}{5 u-7 y},$ Show that $\frac{x}{y}=\frac{u}{v}$

(ii) $\frac{8 a-5 b}{8 c-5 d}=\frac{8 a+5 b}{8 c+5 d},$ prove that $\frac{a}{b}=\frac{c}{d}$

Sol :

(i) $\frac{5 x+7 y}{5 x+7 y}=\frac{5 x-7 y}{5 x-7 y}$

Applying alternendo $\frac{5 x+7 y}{52+7 y}=\frac{5 x-7 y}{54-7}$

Applying componendo and dividendo

$\frac{5 x+7 y+5 x-7 y}{5 x+7 y-5 x+7 y}=\frac{5 u+7 v+5 u-7 v}{5 u+7 u-5 u+7 v}$

$\Rightarrow \frac{10 x}{14 y}=\frac{10 u}{14 v} \Rightarrow \frac{x}{y}=\frac{u}{v}$

Hence proved. ( Dividing by $\left.=\frac{10}{14}\right)$

(ii) $\frac{8 a-5 b}{8 c-5 d}=\frac{8 a+5 b}{8 c+5 d}$

$\Rightarrow \frac{8 a+5 b}{8 a-5 b}=\frac{8 c+5 d}{8 c-5 d} \quad$ (using alternendo)

Applying compounde and dividendo,

$\frac{8 a+5 b+8 a-5 b}{8 a+5 b-8 a+5 b}=\frac{8 c+5 d+8 c-5 c}{8 c+5 d-8 c+5 d}$

$\therefore \frac{16 a}{10 b}=\frac{16 c}{10 d}$

$\Rightarrow \frac{a}{b}=\frac{c}{d}$

(Dividing by $\left.\frac{16}{10}\right)$

Hence proved.

Question 3

If (4a + 5b) (4c – 5d) = (4a – 5d) (4c + 5d), prove that a, b, c, d are in proporton.

Sol :

(4a + 5b) (4c – 5d) = (4a – 5d) (4c + 5d)

$\Rightarrow \frac{4 a+5 b}{4 a-5 b}=\frac{4 c+5 d}{4 c-5 d}$

Applying componendo and dividendo

$\frac{4 a+5 b+4 a-5 b}{4 a+5 b-4 a+5 b}=\frac{4 c+5 d+4 c-5 d}{4 c+5 d-4 c+5 d}$

$\Rightarrow \frac{8 a}{10 b}=\frac{8 c}{10 d} \Rightarrow \frac{a}{b}=\frac{c}{d}$

Hence, a,b,c,d are in proportion

Question 4

If (pa + qb) : (pc + qd) :: (pa – qb) : (pc – qd) prove that a : b : : c : d

Sol :

(pa + qb) : (pc + qd) :: (pa – qb) : (pc – qd)

$\Rightarrow \frac{p \alpha+q b}{p c+q d}=\frac{p q-q b}{p c-q d}$

$\Rightarrow \frac{p a+q b}{p c-q d}=\frac{p q+q b}{p c-q d}$

Applying componendo and dividendo

$\Rightarrow \frac{p a+q b+p \alpha-q b}{p a+q b-p \alpha+q b}=\frac{p c+q d+p c-q d}{p c-q d-p c+q d}$

$\Rightarrow \frac{2 p a}{2 q b}=\frac{2 p c}{2 q d}$

$\Rightarrow \frac{a}{b}=\frac{c}{d}$ $\left(\right.$ Dividing by $\left.\frac{2 p}{2 q}\right)$

hence a : b:: c : d

Question 5

If (ma + nb): b :: (mc + nd) : d, prove that a, b, c, d are in proportion.

Sol :

(ma + nb): b :: (mc + nd) : d

$\Rightarrow \frac{m a+n b}{b}=\frac{m c+n d}{d}$

⇒ mad + nbd = mbc + nbd

⇒ mad = mbc

⇒ ad = bc

$\Rightarrow \frac{a}{b}=\frac{c}{d}$

Hence a : b :: c : d.

Question 6

If (11a² + 13b²) (11c² – 13d²) = (11a² – 13b²)(11c² + 13d²), prove that a : b :: c : d.

Sol :

(11a² + 13b²) (11c² – 13d²) = (11a² – 13b²)(11c² + 13d²)

$\Rightarrow \frac{11 a+13 b^{2}}{11 a^{2}-13 b^{2}}=\frac{11 c^{2}+13 d^{2}}{11 c^{2}-13 d^{2}}$

Applying componendo and dividendo

$\frac{11 a^{2}+13 b^{2}+11 a^{2}-13 b^{2}}{11 a^{2}+13 b^{2}-11 a^{2}+13 b^{2}}=\frac{11 c^{2}+13 d^{2}+11 c^{2}-13 d^{2}}{11 c^{2}+13 d^{2}-11 c^{2}+13 d^{2}}$

$\Rightarrow \frac{22 a^{2}}{26 b^{2}}=\frac{22 c^{2}}{26 d^{2}}$

$\Rightarrow \frac{a^{2}}{b^{2}}=\frac{c^{2}}{d^{2}}$ $\left(\right.$ Dividing by $\left.\frac{22}{26}\right)$

$\Rightarrow \frac{a}{b}=\frac{c}{d}$

hence a:b::c:d

Question 7

If (a + 3b + 2c + 6d) (a – 3b – 2c + 6d) = (a + 3b – 2c – 6d) (a – 3b + 2c – 6d), prove that a : b :: c : d.

Sol :

$\frac{a+3 b+2 c+6 d}{a-3 b+2 c-6 d}=\frac{a+3 b-2 c-6 d}{a-3 b-2 c+6 d}$

$\Rightarrow \frac{a+3 b+2 c+6 d}{a+3 b-2 c-6 d}=\frac{a-3 b+2 c-6 d}{a-3 b-2 c+6 d}$ (by altenendo)

Applying componendo and dividendo

$\frac{a+3 b+2 c+6 d+a+3 b-2 c-6 d}{a+3 b+2 c+6 d-a-3 b+2 c+6 d}$

$=\frac{a-3 b+2 c-6 d+a-3 b-2 c+6 d}{a-3 b+2 c-6 d-a+3 b+2 c-6 d}$

$\Rightarrow \frac{2(a+3 b)}{2(2 c+6 d)}=\frac{2(a-3 b)}{2(2 c-6 d)}$ (Dividing by 2)

$\Rightarrow \frac{a+3 b}{2 c+6 d}=\frac{a-3 b}{2 c-6 d}$ (by alternendo)

Again applying componendo and dividendo)

$\frac{a+3 b+a-3 b}{a+3 b-a+3 b}=\frac{2 c+6 d+2 c-6 d}{2 c+6 d-2 c+6 d}$

$\Rightarrow \frac{2 a}{6 b}=\frac{4 c}{12 d}=\frac{2 c}{6 d}$

$\Rightarrow \frac{a}{b}=\frac{c}{d}$  $\left[\right.$ Dividing by $\left.\frac{2}{6}\right]$

Question 8

If $x=\frac{2 a b}{a+b}$ find the value of $\frac{x+a}{x-a}+\frac{x+b}{x-b}$

Sol :

$x=\frac{2 a b}{a+b}$

$\Rightarrow \frac{x}{a}=\frac{2 b}{a+b}$

Applying componendo and dividendo,

$\frac{x+a}{x-a}=\frac{2 b+a+b}{2 b-a-b}=\frac{3 b+a}{b-a}$...(i)

Again $\frac{x}{b}=\frac{2 a}{a+b}$

Applying componendo and dividendo,

$\frac{x+b}{x-b}=\frac{2 a+a+b}{2 a-a-b}=\frac{3 a+b}{a-b}$..(ii)

Adding (i) and (ii)

$\frac{x+a}{x-a}+\frac{x+b}{x-b}=\frac{3 b+a}{b-a}+\frac{3 a+b}{a-b}$

$=-\frac{a+3 b}{a-b}+\frac{3 a+b}{a-b}$

$=\frac{-a-3 b+3 a+b}{a-b}$

$=\frac{2 a-2 b}{a-b}=\frac{2(a-b)}{a-b}=2$

Question 9

If $x=\frac{8 a b}{a+b}$ find the value of $\frac{x+4 a}{x-4 a}+\frac{x+4 b}{x-4 b}$

Sol :

$x=\frac{8 a b}{a+b}$

$\Rightarrow \frac{x}{4 a}=\frac{2 b}{a+b}$

Applying componendo and dividendo,

$\frac{x+4 a}{x-4 a}=\frac{2 b+a+b}{2 b-a-b}=\frac{3 b+a}{b-a} \quad \ldots(i)$

$\operatorname{Again} \frac{x}{4 b}=\frac{2 a}{a+b}$

Applying compondndo and dividendo,

$\frac{x+4 b}{x-4 b}=\frac{2 a+a+b}{2 a-a-b}=\frac{3 a+b}{a-b}$..(ii)

Adding (i) and (ii)

$\frac{x+4 a}{x-4 a}+\frac{x+4 b}{x-4 b}=\frac{3 b+a}{b-a}+\frac{3 a+b}{a-b}$

$=-\frac{a+3 b}{a-b}+\frac{3 a+b}{a-b}$

$=\frac{-a-3 b+3 a+b}{a-b}=\frac{2 a-2 b}{a-b}$

$=\frac{2(a-b)}{a-b}=2$

Question 10

If $x=\frac{4 \sqrt{6}}{\sqrt{2}+\sqrt{3}}$ find the value of $\frac{x+2 \sqrt{2}}{x-2 \sqrt{2}}+\frac{x+2 \sqrt{3}}{x-2 \sqrt{3}}$

Sol :

$x=\frac{4 \sqrt{6}}{\sqrt{2}+\sqrt{3}}$

$\Rightarrow \frac{4 \sqrt{2} \times \sqrt{3}}{\sqrt{2}+\sqrt{3}}$

$\frac{x}{2 \sqrt{2}}=\frac{2 \sqrt{3}}{\sqrt{2}+\sqrt{3}}$

Applying componendo and dividendo,

$\frac{x+2 \sqrt{2}}{x-2 \sqrt{2}}=\frac{2 \sqrt{3}+\sqrt{2}+\sqrt{3}}{2 \sqrt{3}-\sqrt{2}-\sqrt{3}}$

$=\frac{3 \sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$..(i)

Again $\frac{x}{2 \sqrt{3}}=\frac{2 \sqrt{2}}{\sqrt{2}+\sqrt{3}}$

Applying componendo and dividendo,

$\frac{x+2 \sqrt{3}}{x-2 \sqrt{3}}=\frac{2 \sqrt{2}+\sqrt{2}+\sqrt{3}}{2 \sqrt{2}-\sqrt{2}-\sqrt{3}}$

$=\frac{3 \sqrt{2}+\sqrt{3}}{\sqrt{2}-\sqrt{3}}$..(ii)

Adding (i) and (ii)

$\frac{x+2 \sqrt{2}}{x-2 \sqrt{2}}+\frac{x+2 \sqrt{3}}{x-2 \sqrt{3}}$

$=\frac{3 \sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\frac{3 \sqrt{2}+\sqrt{3}}{\sqrt{2}-\sqrt{3}}$

$=\frac{3 \sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}-\frac{3 \sqrt{2}+\sqrt{3}}{\sqrt{3}-\sqrt{2}}$

$=\frac{3 \sqrt{3}+\sqrt{2}-3 \sqrt{2}-\sqrt{3}}{\sqrt{3}-\sqrt{2}}$

$=\frac{2 \sqrt{3}-2 \sqrt{2}}{\sqrt{3}-\sqrt{2}}=\frac{2(\sqrt{3}-\sqrt{2})}{\sqrt{3}-\sqrt{2}}=2$

Question 11

Solve $x: \frac{\sqrt{36 x+1}+6 \sqrt{x}}{\sqrt{36 x+1}-6 \sqrt{x}}=9$

Sol :

$\frac{\sqrt{36 x+1}+6 \sqrt{x}}{\sqrt{36 x+1}-6 \sqrt{x}}=\frac{9}{1}$

Applying componendo and dividendo,

$\frac{\sqrt{36 x+1}+6 \sqrt{x}+\sqrt{36 x+1}-6 \sqrt{x}}{\sqrt{36 x+1}+6 \sqrt{x}-\sqrt{36 x-1}+6 \sqrt{x}}$

$=\frac{9+1}{9-1}$

$ \Rightarrow \frac{2 \sqrt{36 x+1}}{12 \sqrt{x}}=\frac{10}{8}$

$\Rightarrow \frac{\sqrt{36 x+1}}{6 \sqrt{x}}=\frac{5}{4}$ (Squaring both sides)

$\frac{36 x+1}{36 x}=\frac{25}{16}$

$\Rightarrow  36 x \times 25=16(36 x+1)$

$\Rightarrow 900 x=576 x+16$

$ \Rightarrow 900 x-576 x=16$

$\Rightarrow 324 x=16$

$\therefore x=\frac{16}{324}=\frac{4}{81}$

Question 12

Find x from the following equations :

(i) $\frac{\sqrt{2-x}+\sqrt{2+x}}{\sqrt{2-x}-\sqrt{2+x}}=3$

(ii) $\frac{\sqrt{x+4}+\sqrt{x-10}}{\sqrt{x+4}-\sqrt{x-10}}=\frac{5}{2}$

(iii) $\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}=\frac{a}{b}$

(iv) $\frac{\sqrt{12 x+1}+\sqrt{2 x-3}}{\sqrt{12 x+1}-\sqrt{2 x-3}}=\frac{3}{2}$

(v) $\frac{3 x+\sqrt{9 x^{2}-5}}{3 x-\sqrt{9 x^{2}-5}}=5$

(vi) $\frac{\sqrt{a+x}+\sqrt{a-x}}{\sqrt{a+x}-\sqrt{a-x}}=\frac{c}{d}$

Sol :

(i) $\frac{\sqrt{2-x}+\sqrt{2+x}}{\sqrt{2-x}-\sqrt{2+x}}=3$

Applying componendo and dividendo,

$\frac{\sqrt{2-x}+\sqrt{2+x}+\sqrt{2-x}-\sqrt{2+x}}{\sqrt{2-x}+\sqrt{2+x}-\sqrt{2-x}+\sqrt{2+x}}=\frac{3+1}{3-1}$

$\Rightarrow \frac{2 \sqrt{2-x}}{2 \sqrt{2+x}}=\frac{4}{2} \Rightarrow \frac{\sqrt{2-x}}{\sqrt{2+x}}=\frac{2}{1}$

Squaring both sides

$\frac{2-x}{2+x}=\frac{4}{1} \Rightarrow 8+4 x=2-x$

$4 x+x=2-8 \Rightarrow 5 x=-6$

$\therefore x=\frac{-6}{5}$ Ans.

(ii) $\frac{\sqrt{x+4}+\sqrt{x-10}}{\sqrt{x+4}-\sqrt{x-10}}=\frac{5}{2}$

Applying componendo and dividendo,

$\frac{\sqrt{x+4}+\sqrt{x-10}+\sqrt{x+4}-\sqrt{x-10}}{\sqrt{x+4}+\sqrt{x-10}-\sqrt{x+4}+\sqrt{x-10}}=\frac{5+2}{5-2}$

$\Rightarrow \quad \frac{2 \sqrt{x+4}}{2 \sqrt{x-10}}=\frac{7}{3} \Rightarrow \frac{\sqrt{x+4}}{\sqrt{x-10}}=\frac{7}{3}$

Squaring both sides,

$\frac{x+4}{x-10}=\frac{49}{9}$

$ \Rightarrow 49 x-490=9 x+36$

$\Rightarrow 49 x-9 x=36+490 \Rightarrow \quad 40 x=526$

$\therefore x=\frac{526}{40}=\frac{263}{20}$

(iii) $\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}=\frac{a}{b}$

Applying componendo and dividendo,

$\frac{\sqrt{1+x}+\sqrt{1-x}+\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}-\sqrt{1+x}+\sqrt{1-x}}=\frac{a+b}{a-b}$

$\Rightarrow \frac{2 \sqrt{1+x}}{2 \sqrt{1-x}}=\frac{a+b}{a-b} \Rightarrow \frac{\sqrt{1+x}}{\sqrt{1-x}}=\frac{a+b}{a-b}$

Squaring both sides, $\frac{1+x}{1-x}=\frac{(a+b)^{2}}{(a-b)^{2}}$

Again applying componendo and dividendo,

$\frac{1+x+1-x}{1+x-1+x}=\frac{(a+b)^{2}+(a-b)^{2}}{(a+b)^{2}-(a-b)^{2}}$

$\Rightarrow \frac{2}{2 x}=\frac{2\left(a^{2}+b^{2}\right)}{4 a b}$

$ \Rightarrow \frac{1}{x}=\frac{a^{2}+b^{2}}{2 a b}$

$\therefore x=\frac{2 a b}{a^{2}+b^{2}}$

(iv) $\frac{\sqrt{12 x+1}+\sqrt{2 x-3}}{\sqrt{12 x+1}-\sqrt{2 x-3}}=\frac{3}{2}$

Applying componendo and dividendo,

$\frac{\sqrt{12 x+1}+\sqrt{2 x-3}+\sqrt{12 x+1}-\sqrt{2 x-3}}{\sqrt{12 x+1}+\sqrt{2 x-3}-\sqrt{12 x+1}+\sqrt{2 x-3}}$

$=\frac{3+2}{3-2}$

$\Rightarrow \frac{2 \sqrt{12 x+1}}{2 \sqrt{2 x-3}}=\frac{5}{1}$

$\Rightarrow \frac{\sqrt{12 x+1}}{\sqrt{2 x-3}}=\frac{5}{1}$

Squaring both sides, $\frac{12 x+1}{2 x-3}=\frac{25}{1}$

$\Rightarrow \quad 50 x-75=12 x+1$

$\Rightarrow \quad 50 x-12 x=1+75$

$\Rightarrow \quad 38 x=76 \Rightarrow x=\frac{76}{38}=2$

∴ x=2

(v) $\frac{3 x+\sqrt{9 x^{2}-5}}{3 x-\sqrt{9 x^{2}-5}}=\frac{5}{1}$

Applying componendo and dividendo.

$\frac{3 x+\sqrt{9 x^{2}-5}+3 x-\sqrt{9 x^{2}-5}}{3 x+\sqrt{9 x^{2}-5}-3 x+\sqrt{9 x^{2}-5}}=\frac{5+1}{5-1}$

$\frac{6 x}{2 \sqrt{9 x^{2}-5}}=\frac{6}{4} \Rightarrow \frac{3 x}{\sqrt{9 x^{2}-5}}=\frac{3}{2}$

Squaring both sides

$\frac{9 x^{2}}{9 x^{2}-5}=\frac{9}{4} \Rightarrow 81 x^{2}-45=36 x^{2}$

$\Rightarrow \quad 81 x^{2}-36 x^{2}=45 \Rightarrow 45 x^{2}=45$

$\Rightarrow \quad x^{2}=1 \Rightarrow x=\pm 1$

$\therefore x=1,-1$

CHECK 

(i) When x=1, then in the given equation

$\frac{3 \times 1+\sqrt{9 \times 1-5}}{3 \times 1-\sqrt{9 \times 1-5}}=\frac{3+\sqrt{4}}{3-\sqrt{4}}=\frac{3+2}{3-2}=\frac{5}{1}$

which is given

∴x=1

(ii) When x=-1, then

$3(-1)+\sqrt{9(-1)^{2}-5}$

$3(-1)-\sqrt{9(-1)^{2}-5}$

$=\frac{-3+\sqrt{9-5}}{-3-\sqrt{9-5}}=\frac{-3+\sqrt{4}}{-3-\sqrt{4}}$

$=\frac{-3+2}{-3-2}=\frac{-1}{-5}=\frac{1}{5} \neq \frac{5}{1}$

$\therefore x=-1$ is not its solution.

Hence x=1

(vi) $\frac{\sqrt{a+x}+\sqrt{a-x}}{\sqrt{a+x}-\sqrt{a-x}}=\frac{c}{d}$

Applying componendo and dividendo,

$\frac{\sqrt{a+x}+\sqrt{a-x}+\sqrt{a+x}-\sqrt{a-x}}{\sqrt{a+x}+\sqrt{a-x}-\sqrt{a+x}+\sqrt{a-x}}=\frac{c+d}{c-d}$

$\Rightarrow \frac{2 \sqrt{a+x}}{2 \sqrt{a-x}}=\frac{c+d}{c-d} \Rightarrow \frac{\sqrt{a+x}}{\sqrt{a-x}}=\frac{c+d}{c-d}$

Squaring both sides $\frac{a+x}{a-x}=\frac{(c+d)^{2}}{(c-d)^{2}}$

Again applying componendo and dividendo

$\frac{a+x+a-x}{a+x-a+x}=\frac{(c+d)^{2}+(c-d)^{2}}{(c+d)^{2}-(c-d)^{2}}$

$\Rightarrow \frac{2 a}{2 x}=\frac{2\left(c^{2}+d^{2}\right)}{4 c d} \Rightarrow \frac{a}{x}=\frac{c^{2}+d^{2}}{2 c d}$

$\Rightarrow  x\left(c^{2}+d^{2}\right)=2 a c d$

$\Rightarrow  x=\frac{2 a c d}{c^{2}+d^{2}}$

Question 13

Solve $\frac{1+x+x^{2}}{1-x+x^{2}}=\frac{62(1+x)}{63(1-x)}$

Sol :

$\frac{1+x+x^{2}}{1-x+x^{2}}=\frac{62(1+x)}{63(1-x)}$

$\Rightarrow \frac{(1-x)\left(1+x+x^{2}\right)}{(1+x)\left(1-x+x^{2}\right)}=\frac{62}{63}$

$\Rightarrow \frac{(1+x)\left(1-x+x^{2}\right)}{(1-x)\left(1+x+x^{2}\right)}=\frac{63}{62}$

$\Rightarrow \quad \frac{1+x^{3}}{1-x^{3}}=\frac{63}{62}$

Applying componendo and dividendo,

$\frac{1+x^{3}+1-x^{3}}{1+x^{3}-1+x^{3}}=\frac{63+62}{63-62}$

$\Rightarrow \frac{2}{2 x^{3}}=\frac{125}{1}$

$ \Rightarrow  \frac{1}{x^{3}}=\frac{125}{1}$

$\Rightarrow x^{3}=\frac{1}{125}=\left(\frac{1}{5}\right)^{3}$

$\therefore x=\frac{1}{5}$ Ans.

Question 14

Solve for $x: 16\left(\frac{a-x}{a+x}\right)^{3}=\frac{a+x}{a-x}$

Sol :

$x : 16\left(\frac{a-x}{a+x}\right)^{3}=\frac{a+x}{a-x}$

$\Rightarrow\left(\frac{a+x}{a-x}\right) \times\left(\frac{a+x}{a-x}\right)^{3}=16$

$\Rightarrow\left(\frac{a+x}{a-x}\right)^{4}=16=(\pm 2)^{4}$

$ \Rightarrow \frac{a+x}{a-x}=\pm 2$

When $\frac{a+x}{a-x}=\frac{2}{1}$

Applying componendo and dividendo,

$\frac{a+x+a-x}{a+x-a+x}=\frac{-2+1}{-2-1}$

$ \Rightarrow=\frac{2 a}{2 x}=\frac{-1}{-3}$

$\Rightarrow \frac{a}{x}=\frac{1}{3}$

$ \Rightarrow x=3 a$

Hence $x=\frac{a}{3}, 3 a$ 

Question 15

If $x=\frac{\sqrt{a+x}+\sqrt{a-1}}{\sqrt{a+1}-\sqrt{a-1}}$ , using properties of proportion , show that x² – 2ax + 1 = 0

Sol :

We have $x=\frac{\sqrt{a+x}+\sqrt{a-1}}{\sqrt{a+1}-\sqrt{a-1}}$

$\Rightarrow \frac{x+1}{x-1}=\frac{2 \sqrt{a+1}}{2 \sqrt{a-1}}$

(Applying componendo and dividendo)

$\Rightarrow \frac{(x+1)^{2}}{(x-1)^{2}}=\frac{a+1}{a-1}$

$\Rightarrow \frac{(x+1)^{2}+(x-1)^{2}}{(x+1)^{2}-(x-1)^{2}}=\frac{2 a}{2}$

(Again applying componendo and dividendo)

$\Rightarrow \frac{x^{2}+1+2 x+x^{2}+1-2 x}{x^{2}+1+2 x-x^{2}-1+2 x}=a$

$\Rightarrow \frac{2 x^{2}+2}{4 x}=a \Rightarrow \frac{2\left(x^{2}+1\right)}{4 x}=a$

$\Rightarrow \frac{\left(x^{2}+1\right)}{2 x}=a \Rightarrow 2 a x=x^{2}+1$

$\Rightarrow x^{2}-2 a x+1=0 \quad$

Question 16

Given : $x=\frac{\sqrt{\alpha^{2}+b^{2}}+\sqrt{\alpha^{2}-b^{2}}}{\sqrt{\alpha^{2}+b^{2}}-\sqrt{\alpha^{2}-b^{2}}}$  Use componendo and dividendo to prove that 

Sol :

If $\frac{x}{1}=\frac{\sqrt{\alpha^{2}+b^{2}}+\sqrt{\alpha^{2}-b^{2}}}{\sqrt{\alpha^{2}+b^{2}}-\sqrt{\alpha^{2}-b^{2}}}$

Applying componendo and dividendo both sides

$\frac{x+1}{x-1}=\frac{\sqrt{a^{2}+b^{2}}+\sqrt{a^{2}-b^{2}}+\sqrt{a^{2}+b^{2}}-\sqrt{a^{2}-b^{2}}}{\sqrt{a^{2}+b^{2}}+\sqrt{a^{2}-b^{2}}-\sqrt{a^{2}+b^{2}}+\sqrt{a^{2}-b^{2}}}$

$\Rightarrow \frac{x+1}{x-1}=\frac{2 \sqrt{a^{2}+b^{2}}}{2 \sqrt{a^{2}+b^{2}}}$

$ \Rightarrow \frac{x+1}{x-1}=\frac{\sqrt{a^{2}+b^{2}}}{\sqrt{a^{2}+b^{2}}}$

Squaring, both sides we have

$\Rightarrow \frac{(x+1)^{2}}{(x-1)^{2}}=\frac{a^{2}+b^{2}}{a^{2}-b^{2}} $

$\Rightarrow \frac{x^{2}+1+2 x}{x^{2}+1-2 x}=\frac{a^{2}+b^{2}}{a^{2}-b^{2}}$

Applying componendo and dividendo both sides

$\Rightarrow \frac{x^{2}+1+2 x+x^{2}+1-2 x}{x^{2}+1+2 x-x^{2}-1+2 x}=\frac{a^{2}+b^{2}+a^{2}-b^{2}}{a^{2}+b^{2}-a^{2}+b^{2}}$

$\frac{2 x^{2}+2}{4 x}=\frac{2 a^{2}}{2 b^{2}}$

$ \Rightarrow \frac{x^{2}+1}{2 x}=\frac{a^{2}}{b^{2}} \Rightarrow b^{2}=\frac{2 a^{2} x}{x^{2}+1}$

Question 17

Given that $\frac{a^{3}+3 a b^{2}}{b^{3}+3 a^{2} b}=\frac{63}{62}$. Using componendo and dividendo find a: b. (2009)

Sol :

Given that $\frac{a^{3}+3 a b^{2}}{b^{3}+3 a^{2} b}=\frac{63}{62}$

By componendo and dividendo 

$\frac{a^{3}+3 a b^{2}+b^{3}+3 a^{2} b}{a^{3}+3 a b^{2}-b^{3}-3 a^{2} b}=\frac{63+62}{63-62}=\frac{125}{1}$

$\Rightarrow \frac{(a+b)^{3}}{(a-b)^{3}}=\left(\frac{5}{1}\right)^{3}$

$ \Rightarrow \frac{a+b}{a-b}=5 \Rightarrow a+b=5 a-5 b$

⇒5a-a-5b-b=0

⇒4a-6b=0

⇒4a=60

$\Rightarrow \frac{a}{b}=\frac{6}{4}$

$ \Rightarrow \frac{a}{b}=\frac{3}{2}$

a : b = 3 : 2

Question 18

Give $\frac{x^{3}+12 x}{6 x^{2}+8}=\frac{y^{3}+27 y}{9 y^{2}+27}$ Using componendo and dividendo find x : y.

Sol :

Given : 

$\frac{x^{3}+12 x}{6 x^{2}+8}=\frac{y^{3}+27 y}{9 y^{2}+27}$

Using componendo-dividendo, we have

$\frac{x^{3}+12 x+6 x^{2}+8}{x^{3}+12 x-6 x^{2}-8}=\frac{y^{3}+27 y+9 y^{2}+27}{y^{3}+27 y-9 y^{2}-27}$

$\Rightarrow \frac{(x+2)^{3}}{(x-2)^{3}}=\frac{(y+3)^{3}}{(9 y-3)^{3}}$

$\Rightarrow\left(\frac{x+2}{x-2}\right)^{3}=\left(\frac{y+3}{y-3}\right)^{3}$

$\Rightarrow \frac{x+2}{x-2}=\frac{y+3}{y-3}$

Again using componendo-dividendo, we get

$\frac{x+2+x-2}{x+2-x+2}=\frac{y+3+y-3}{y+3-y+3}$

$\Rightarrow \frac{2 x}{4}=\frac{2 y}{3}$

$\Rightarrow \frac{x}{2}=\frac{y}{3}$

$\Rightarrow \frac{x}{y}=\frac{2}{3}$

Thus the required ratio is x: y=2: 3

Question 19

Using the properties of proportion, solve the following equation for x; given

$\frac{x^{3}+3 x}{3 x^{2}+1}=\frac{341}{91}$

Sol :

$\frac{x^{3}+3 x}{3 x^{2}+1}=\frac{341}{91}$

Applying componendo and dividendo

$\frac{x^{3}+3 x+3 x^{2}+1}{x^{3}+3 x-3 x^{2}-1}=\frac{341+91}{341-91}$

$\Rightarrow \frac{x^{3}+3 x^{2}+3 x+1}{x^{3}-3 x^{2}+3 x-1}=\frac{432}{250}=\frac{216}{125}$

$\Rightarrow \frac{(x+1)^{3}}{(x-1)^{3}}=\frac{216}{125}=\left(\frac{6}{5}\right)^{3}$

$\therefore \frac{x+1}{x-1}=\frac{6}{5}$

⇒6 x-6=5 x+5

x=11

Question 20

If $\frac{x+y}{a x+b y}=\frac{y+z}{a y+b z}=\frac{z+x}{a z+b x}$ , prove that each of these ratio is equal to $\frac{2}{a+b}$ unless x + y + z = 0

Sol :

$\frac{x+y}{a x+b y}=\frac{y+z}{a y+b z}=\frac{z+x}{a z+b x}$

$=\frac{x+y+y+z+z+x}{a x+b y+a y+b z+\alpha z+b x}$

$=\frac{2(x+y+z)}{x(a+b)+y(a+b)+z(a+b)}$

$=\frac{2(x+y+z)}{(a+b)(x+y+z)}=\frac{2}{a+b}$ if $x+y+z \neq 0$

ML Aggarwal Solution Class 10 Chapter 7 Ratio and Proportion Exercise 7.2

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20 Q21 Q22 Q23

 Exercise 7.2

Question 1

Find the value of x in the following proportions :

(i) 10 : 35 = x : 42

(ii) 3 : x = 24 : 2

(iii) 2.5 : 1.5 = x : 3

(iv) x : 50 :: 3 : 2

Sol :

(i) 10 : 35 = x : 42

⇒ 35×x = 10 × 42

$\therefore \quad x=\frac{10 \times 42}{35}=2 \times 6=12$

(ii) $3: x=24: 2$

$\Rightarrow x \times 24=3 \times 2$

$\therefore x=\frac{3 \times 2}{24}=\frac{1}{4}$

(iii) $2 \cdot 5: 1 \cdot 5=x: 3$

$\Rightarrow 1 \cdot 5 \times x=2 \cdot 5 \times 3$

$x=\frac{2 \cdot 5 \times 3}{1 \cdot 5}=5 \cdot 0$

(iv) $x: 50:: 3: 2$

$\Rightarrow x \times 2=50 \times 3$

$x=\frac{50 \times 3}{2}=75$

Question 2

Find the fourth proportional to

(i) 3, 12, 15

(ii) $\frac{1}{3}, \frac{1}{4}, \frac{1}{5}$

(iii) 1.5, 2.5, 4.5

(iv) 9.6 kg, 7.2 kg, 28.8 kg

Sol :

(i) Let fourth proportional to

3, 12, 15 be x.

then 3 : 12 :: 15 : x

$\Rightarrow 3 \times x=12 \times 15$

$x=\frac{12 \times 15}{3}=60$

(ii) Let fourth proportional to $\frac{1}{3}, \frac{1}{4}, \frac{1}{5}$ be x

then $\frac{1}{3}: \frac{1}{4}:: \frac{1}{5}: x$

$\Rightarrow \frac{1}{3} \times x=\frac{1}{4} \times \frac{1}{5}$

$\Rightarrow x=\frac{1}{4} \times \frac{1}{5} \times \frac{3}{1}=\frac{3}{20}$

(iii) Let fourth proportional to

$1 \cdot 5,2 \cdot 5,4 \cdot 5$ be $x$

then $1 \cdot 5: 2 \cdot 5:: 4 \cdot 5: x$

$\therefore 1 \cdot 5 \times x=2 \cdot 5 \times 4 \cdot 5$

$x=\frac{2 \cdot 5 \times 4 \cdot 5}{1 \cdot 5}=7 \cdot 5$

(iv) Let fourth proportional to 9.6kg, 7.2 kg, 28.8 kg be x

then 9.6 : 7.2 :: 28.8 : x

$\Rightarrow \quad 9 \cdot 6 \times x=7 \cdot 2 \times 28 \cdot 8$

$x=\frac{7 \cdot 2 \times 28 \cdot 8}{9 \cdot 6}=21 \cdot 6$

Question 3

Find the third proportional to

(i) 5, 10

(ii) 0.24, 0.6

(iii) Rs. 3, Rs. 12

(iv) $5 \frac{1}{4}$ and 7

Sol :

(i) Let x be the third proportional to 5, 10,

then 5 : 10 :: 10 : x

$\therefore 5 \times x=10 \times 10 \Rightarrow x=\frac{10 \times 10}{5}=20$

∴Third proportional =20

(ii) Let x be the third proportional to 0.24, 0.6

then 0.24 : 0.6 :: 0.6 : x

$\therefore 0 \cdot 24 \times x=0 \cdot 6 \times 0 \cdot 6$

$x=\frac{0 \cdot 6 \times 0 \cdot 6}{0 \cdot 24}=1 \cdot 5$

∴ Third proportional =1.5

(iii) Let x be the third proportional to Rs. 3 and Rs. 12 then Rs. 3: Rs. 12: Rs. 12: x

$\therefore x=\frac{12 \times 12}{3}=48$

$\therefore$ Third proportional Rs 48

(iv) Let x be the third proportional to $5 \frac{1}{4}$ and 7

then $5 \frac{1}{4}: 7:: 7: x$

$\Rightarrow \frac{21}{4}: 7:: 7: x$

$\therefore \frac{21}{4} \times x=7 \times 7$

$x=\frac{7 \times 7 \times 4}{21}$

$=\frac{28}{3}=9 \frac{1}{3}$

∴ Third proportional $=9 \frac{1}{3}$ Ans.

Question 4

Find the mean proportion of:

(i) 5 and 80

(ii) $\frac{1}{12}$ and $\frac{1}{75}$

(iii) 8.1 and 2.5

(iv) (a – b) and (a³ – a²b), a> b

Solution:

(i) Let x be the mean proportion of 5 and 80 ,

then 5 : x : : x : 80

x² = 5 x 80

⇒ x = √5×80=√400= 20

x = 20

Hence mean proportion=20

(ii) Let x be the mean proportion of $\frac{1}{12}$ and $\frac{1}{75}$

then $\frac{1}{12}: x:: x: \frac{1}{75}$

$\therefore x^{2}=\frac{1}{12} \times \frac{1}{75}=\frac{1}{900}$

$\therefore x=\sqrt{\frac{1}{900}}=\frac{1}{30}$

Hence the mean proportion $=\frac{1}{30}$

(iii) Let the x be the mean proportion of 8.1 and 2.5 

∴ 8.1: x:: x: 2.5

$\therefore \quad x^{2}=8 \cdot 1 \times 2 \cdot 5$

$\therefore \quad x=\sqrt{8 \cdot 1 \times 2 \cdot 5}=\sqrt{20 \cdot 25}=4 \cdot 5$

Hence mean proportion =4.5

(iv) Let x be the mean proportion to $(a-b)$ and $\left(a^{3}-a^{2} b\right), a>b$

then $(a-b): x:: x:\left(a^{3}-a^{2} b\right)$

$x^{2}=(a-b)\left(a^{3}-a^{2} b\right)$

$=(a-b) a^{2}(a-b)=a^{2}(a-b)^{2}$

∴ x=a(a-b)

Hence the mean proportion=a(a-b)

Question 5

If a, 12, 16 and b are in continued proportion find a and b.

Sol :

∵ a, 12, 16, b are in continued proportion, then

$\frac{a}{12}=\frac{12}{16}=\frac{16}{b}$

$ \Rightarrow \frac{a}{12}=\frac{12}{16} \Rightarrow 16 a=144$

$\Rightarrow a=\frac{144}{16}=$

and $\frac{12}{16}=\frac{16}{b} \Rightarrow 12 b=16 \times 16=256$

$b=\frac{256}{12}=\frac{64}{3}=21 \frac{1}{3}$

Hence $a=9, b=\frac{64}{3}$ or $21 \frac{1}{3}$

Question 6

What number must be added to each of the numbers 5, 11, 19 and 37 so that they are in proportion ? (2009)

Sol :

Let x be added to 5, 11, 19 and 37 to make them in proportion.

5 + x : 11 + x : : 19 + x : 37 + x

⇒ (5+x)(37+x)=(11+x)(19+x)

$\Rightarrow 185+5 x+37 x+x^{2}=209+11 x+19 x+x^{2}$

$\Rightarrow 185+42 x+x^{2}=209+30 x+x^{2}$

$\Rightarrow 42 x-30 x+x^{2}-x^{2}=209-185$

⇒ 12 x=24

⇒ x=2

∴ Least number to be added =2

Question 7

What number should be subtracted from each of the numbers 23, 30, 57 and 78 so that the remainders are in proportion ? (2004)

Sol :

Let x be subtracted from each term, then

23 – x, 30 – x, 57 – x and 78 – x are proportional

23 – x : 30 – x : : 57 – x : 78 – x

$\Rightarrow  \frac{23-x}{30-x}=\frac{57-x}{78-x}$

⇒ (23-x)(78-x)=(30-x)(57-x)

$\Rightarrow 1794-23 x-78 x+x^{2}$

$=1710-30 x-57 x+x^{2}$

$\Rightarrow x^{2}-101 x+1794=x^{2}-87 x+1710$

$\Rightarrow  x^{2}-101 x+1794-x^{2}+87 x-1710=0$

⇒-14 x+84=0

⇒ 14 x=84

$\therefore x=\frac{84}{14}=6$

Hence 6 is to be subtracted Ans.

Question 8

If 2x – 1, 5x – 6, 6x + 2 and 15x – 9 are in proportion, find the value of x.

Sol :

∵ 2x – 1, 5x – 6, 6x + 2 and 15x – 9 are in proportion.

then (2x – 1) (15x – 9) = (5x – 6) (6x + 2)

$\Rightarrow 30 x^{2}-18 x-15 x+9 \Rightarrow 30 x^{2}+10 x-36 x-12$

$\Rightarrow 30 x^{2}-33 x+9=30 x^{2}-26 x-12$

$\Rightarrow 30 x^{2}-33 x-30 x^{2}+26 x=-12-9$

$\Rightarrow-7x=-21$

$\therefore \quad x=\frac{-21}{-7}=3$

Hence x=3

Question 9

If 2x – 1, 5x – 6, 6x + 2 and 15x – 9 are in proportion, find the value of x.

Sol :

∵ 2x – 1, 5x – 6, 6x + 2 and 15x – 9 are in proportion.

then (2x – 1) (15x – 9) = (5x – 6) (6x + 2)

$\Rightarrow 30 x^{2}-18 x-15 x+9 \Rightarrow 30 x^{2}+10 x-36 x-12$

$\Rightarrow 30 x^{2}-33 x+9=30 x^{2}-26 x-12$

$\Rightarrow 30 x^{2}-33 x-30 x^{2}+26 x=-12-9$

⇒-7 x=-21

$\therefore x=\frac{-21}{-7}=3$

Hence x=3

Question 10

What number must be added to each of the numbers 16, 26 and 40 so that the resulting numbers may be in continued proportion?

Sol :

Let x be added to each number then

16 + x, 26 + x and 40 + x

are in continued proportion.

$\Rightarrow \frac{16+x}{26+x}=\frac{26+x}{40+x}$

Cross Multiplying,

(16+x)(40+x)=(26+x)(26+x)

$\Rightarrow 640+16 x+40 x+x^{2}=676+26 x+26 x+x^{2}$

$\Rightarrow 640+56 x+x^{2}=676+52 x+x^{2}$

$\Rightarrow 56 x+x^{2}-52 x-x^{2}=676-640$

⇒4x=36

$\Rightarrow x=\frac{36}{4}=9$

∴9 is to be added.

Question 11

Find two numbers such that the mean proportional between them is 28 and the third proportional to them is 224.

Sol :

Let the two numbers are a and b.

∵ 28 is the mean proportional

∵ a : 28 : : 28 : b

$\therefore a b=(28)^{2}=784$

$ \Rightarrow a=\frac{784}{b} \quad \ldots(i)$

∵ 224 is the third proportional

∵ a: b:: b: 224

$\Rightarrow \quad b^{2}=224 a$

Substituting the value of a in (ii)

$b^{2}=224 \times \frac{784}{b} \Rightarrow b^{3}=224 \times 784$

$\Rightarrow b^{3}=175616=(56)^{3}$

∴ b=56

Now substituting the value of b in (i) 

$a=\frac{784}{56}=14$

Hence numbers are 14,56

Question 12

If b is the mean proportional between a and c, prove that a, c, a² + b², and b² + c² are proportional.

Sol :

∵ b is the mean proportional between a and c, then,

b² = a × c ⇒ b² = ac …(i)

Now $a, c, a^{2}+b^{2}$ and $b^{2}+c^{2}$ are in proportion

if $\frac{a}{c}=\frac{a^{2}+b^{2}}{b^{2}+c^{2}}$

if $a\left(b^{2}+c^{2}\right)=c\left(a^{2}+b^{2}\right)$

if $a\left(a c+c^{2}\right)=c\left(a^{2}+a c\right)[$ from $(i)]$ if $\quad a c(a+c)=a^{2} c+a c^{2}$

if ac(a+c)=ac(a+c) which is true. Hence proved.

Question 13

If b is the mean proportional between a and c, prove that (ab + bc) is the mean proportional between (a² + b²) and (b² + c²).

Sol :

b is the mean proportional between a and c then

b² = ac …(i)

Now if (ab + bc) is the mean proportional

between $\left(a^{2}+b^{2}\right)$ and $\left(b^{2}+c^{2}\right),$ then $(a b+b c)^{2}=\left(a^{2}+b^{2}\right)\left(b^{2}+c^{2}\right)$

Now L.H.S. $=(a b+b c)^{2}=a^{2} b^{2}+b^{2} c^{2}+2 a b^{2} c$

$=a^{2}(a c)+a c(c)^{2}+2 a$ ac. $c \quad[$ from $(i)]$

$=a^{3} c+a c^{3}+2 a^{2} c^{2}$

$=a c\left(a^{2}+c^{2}+2 a c\right)=a c(a+c)^{2}$

$\mathrm{R.H.S.}=\left(a^{2}+b^{2}\right)\left(b^{2}+c^{2}\right)$

$=\left(a^{2}+a c\right)\left(a c+c^{2}\right) \quad[$ from $(i)]$

$=a(a+c) c(a+c)=a c(a+c)^{2}$

$\therefore L.H.S=R.H.S$

Question 14

If y is mean proportional between x and z, prove that

xyz (x + y + z)³ = (xy + yz + zx)³.

Sol :

∵ y is the mean proportional between

x and z, then

y² = xz …(i)

L.H.S. $=x y z(x+y+z)^{3}$

$=x z, y(x+y+z)^{3}$

$=y^{2} y(x+y+z)^{3}$ [from (i)]

$=y^{3}(x+y+z)^{3}$

$=[y(x+y+z)]^{3}$

$=\left[x y+y^{2}+y z\right]^{3}$

$=(x y+y z+z x)^{3}$ (from i)

=R.H.S

Question 15

If a + c = mb and  $\frac{1}{b}+\frac{1}{d}=\frac{m}{c}$ , prove that a, b, c and d are in proportion.

Sol :

a + c = mb and $\frac{1}{b}+\frac{1}{d}=\frac{m}{c}$

a + c = mb

$\frac{a}{b}+\frac{c}{d}=m$ (Dividing by b)...(i)

and $\frac{1}{b}+\frac{1}{d}=\frac{m}{c}$

$\frac{c}{b}+\frac{c}{d}=m$ (Multiplying by c)...(ii)

From (i) and (ii)

$\frac{a}{b}+\frac{c}{b}=\frac{c}{b}+\frac{c}{d} \Rightarrow \frac{a}{b}=\frac{c}{d}$

Hence, a, b, c and d are proportional.

Question 16

If $\frac{x}{a}=\frac{y}{b}=\frac{z}{c},$ prove that

(i) $\frac{x^{3}}{a^{2}}+\frac{y^{3}}{b^{2}}+\frac{z^{3}}{c^{2}}=\frac{(x+y+z)^{3}}{(a+b+c)^{2}}$

(ii) $\left[\frac{a^{2} x^{2}+b^{2} y^{2}+c^{2} z^{2}}{a^{3} x+b^{3} y+c^{3} z}\right]^{3}=\frac{x y z}{a b c}$

(iii) $\frac{a x-b y}{(a+b)(x-y)}+\frac{b y-c z}{(b+c)(y-z)}+\frac{c z-a x}{(c+a)(z-x)}=3$

Sol :

$\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$

$\therefore x=a k, y=b k, z=c k$

(i) L.H.S. $=\frac{x^{3}}{a^{2}}+\frac{y^{3}}{b^{2}}+\frac{z^{3}}{c^{2}}$

$=\frac{a^{3} k^{3}}{a^{2}}+\frac{b^{3} k^{3}}{b^{2}}+\frac{c^{3} k^{3}}{c^{2}}$

$=a k^{3}+b k^{3}+c k^{3}=k^{3}(a+b+c)$

R.H.S. $=\frac{(x+y+z)^{3}}{(a+b+c)^{2}}$

$=\frac{(a k+b k+c k)^{3}}{(a+b+c)^{2}}=\frac{k^{3}(a+b+c)^{3}}{(a+b+c)^{2}}$

$=k^{3}(a+b+c)$

Hence L.H.S. = R.H.S.

(ii) L.H.S $=\left[\frac{a^{2} x^{2}+b^{2} y^{2}+c^{2} z^{2}}{a^{3} x+b^{3} y+c^{3} z}\right]^{3}$

$=\left[\frac{a^{2} \cdot a^{2} k^{2}+b^{2} \cdot b^{2} k^{2}+c^{2} \cdot c^{2} k^{2}}{a^{3} \cdot a \cdot k+b^{3} \cdot b k+c^{3} \cdot c k}\right]^{3}$

$=\left[\frac{a^{4} k^{2}+b^{4} k^{2}+c^{4} k^{2}}{a^{4} k+b^{4} k+c^{4} k}\right]^{3}$

$=\left[\frac{k^{2}\left(a^{4}+b^{4}+c^{4}\right)}{k\left(a^{4}+b^{4}+c^{4}\right)}\right]^{3}=k^{3}$

$\mathrm{R.H.S}=\frac{x y z}{a b c}=\frac{a k \cdot b k \cdot c k}{a b c}=k^{3}$

∴L.H.S=R.H.S

(iii) L.H.S $\frac{a x-b y}{(a+b)(x-y)}+\frac{b y-c z}{(b+c)(y-z)}+\frac{c z-a x}{(c+a)(z-x)}$

$=\frac{a \cdot a k-b \cdot b k}{(a+b)(a k-b k)}+\frac{b \cdot b k-c \cdot c k}{(b+c)(b k-c k)}+\frac{c . c k-a . a k}{(c+a)(c k-a k)}$

$=\frac{a^{2} k-b^{2} k}{(a+b) k(a-b)}+\frac{b^{2} k-c^{2} k}{(b+c) k(b-c)}+\frac{c^{2} k-a^{2} k}{(c+a) k(c-a)}$

$=\frac{k\left(a^{2}-b^{2}\right)}{k\left(a^{2}-b^{2}\right)}+\frac{k\left(b^{2}-c^{2}\right)}{k\left(b^{2}-c^{2}\right)}+\frac{k\left(c^{2}-a^{2}\right)}{k\left(c^{2}-a^{2}\right)}$

=1+1+1=3=R.H.S

Question 17

If $\frac{a}{b}=\frac{c}{d}=\frac{e}{t}$ prove that

(i) (b² + d² + f²) (a² + c² + e²) = (ab + cd + ef)²

(ii) $\frac{\left(a^{3}+c^{3}\right)^{2}}{\left(b^{3}+d^{3}\right)^{2}}=\frac{e^{6}}{f^{6}}$

(iii) $\frac{a^{2}}{b^{2}}+\frac{c^{2}}{d^{2}}+\frac{e^{2}}{f^{2}}=\frac{a c}{b d}+\frac{c e}{d f}+\frac{a e}{d f}$

(iv) $b d f\left[\frac{a+b}{b}+\frac{c+d}{d}+\frac{c+f}{f}\right]^{3}$=27(a+b)(c+d)(e+f)

Sol :

$\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=k(\mathrm{say})$

∴ a = bk, c = dk, e =fk

(i) L.H.S. $=\left(b^{2}+d^{2}+f^{2}\right)\left(a^{2}+c^{2}+e^{2}\right)$

$=\left(b^{2}+d^{2}+f^{2}\right)\left(b^{2} k^{2}+d^{2} k^{2}+f^{2} k^{2}\right)$

$=\left(b^{2}+d^{2}+f^{2}\right) k^{2}\left(b^{2}+d^{2}+f^{2}\right)$

$=k^{2}\left(b^{2}+d^{2}+f^{2}\right)$

$\mathrm{R} \cdot \mathrm{H} \cdot \mathrm{S}=(a b+c d+e f)^{2}$

$=(b . k b .+d k . d+f k . f)^{2}$

$=\left(k b^{2}+k d^{2}+k f^{2}\right)=k^{2}\left(b^{2}+d^{2}+f^{2}\right)^{2}$

$\therefore \mathrm{L} \cdot \mathrm{H.S}=\mathrm{R} \cdot \mathrm{H.S}$

(ii) $\frac{\left(a^{3}+c^{3}\right)^{2}}{\left(b^{3}+d^{3}\right)^{2}}=\frac{\left(b^{3} k^{3}+d^{3} k^{3}\right)^{2}}{\left(b^{3}+d^{3}\right)^{2}}$

$=\frac{\left[k^{3}\left(b^{3}+d^{3}\right)\right]^{2}}{\left(b^{3}+a^{3}\right)^{2}}=\frac{k^{6}\left(b^{3}+d^{3}\right)^{2}}{\left(b^{3}+d^{3}\right)^{2}}=k^{6}$

$\mathrm{R} . \mathrm{H.S}=\frac{e^{6}}{f^{6}}=\frac{f^{6} k^{6}}{f^{6}}=k^{6}$

$\therefore \mathrm{L} \cdot \mathrm{H.S}=\mathrm{R} . \mathrm{H.S}$

(iii) $\mathrm{L.H.S}=\frac{a^{2}}{b^{2}}+\frac{c^{2}}{d^{2}}+\frac{e^{2}}{f^{2}}=\frac{b^{2} k^{2}}{b^{2}}$

$+\frac{d^{2} k^{2}}{d^{2}}+\frac{f^{2} k^{2}}{f^{2}}=k^{2}+k^{2}+k^{2}=3 k^{2}$

$\mathrm{R} \cdot \mathrm{H.S}=\frac{a c}{b d}+\frac{c e}{d f}+\frac{a e}{b f}$

$=\frac{b k \cdot d k}{b \cdot d}+\frac{d k \cdot f k}{d \cdot f}+\frac{b k \cdot f k}{b \cdot f}$

$=k^{2}+k^{2}+k^{2}=3 k^{2}$

$\therefore \mathrm{L} \cdot \mathrm{H.S}=\mathrm{R} . \mathrm{H.S}$

(iv) L.H.S $=b d f\left[\frac{a+b}{b}+\frac{c+d}{d}+\frac{e+f}{f}\right]^{3}$

$=b d f\left[\frac{b k+b}{b}+\frac{d k+d}{d}+\frac{f k+f}{f}\right]^{3}$

$=b d f\left[\frac{b(k+1)}{b}+\frac{d(k+1)}{d}+\frac{f(k+1)}{f}\right]$

$=b d f(k+1+k+1+k+1)^{3}$

$=b d f(3 k+3)^{3}=27 b d f(k+1)^{3}$

R.H.S $=27(a+b)(c+d)(e+f)$

$=27(b k+b)(d k+d)(f k+f)$

$=27 b(k+1) d(k+1) f(k+1)$

$=27 b d f(k+1)^{3}$

$\therefore \mathrm{L} \cdot \mathrm{H.S}=\mathrm{R} . \mathrm{H.S}$

Question 18

If ax = by = cz; prove that

$\frac{x^{2}}{y z}+\frac{y^{2}}{z x}+\frac{z^{2}}{x y}=\frac{b c}{a^{2}}+\frac{c a}{b^{2}}+\frac{a b}{c^{2}}$

Sol :

Let ax = by = cz = k

$\therefore  x=\frac{k}{a}, y=\frac{k}{b}, z=\frac{k}{c}$

L.H.S. $=\frac{x^{2}}{y z}+\frac{y^{2}}{z x}+\frac{z^{2}}{x y}$

$=\frac{\frac{k^{2}}{a^{2}}}{\frac{k}{b} \cdot \frac{k}{c}}+\frac{\frac{k^{2}}{b^{2}}}{\frac{k}{c} \cdot \frac{k}{a}}+\frac{\frac{k^{2}}{c^{2}}}{\frac{k}{a} \cdot \frac{k}{b}}$

$=\frac{\frac{k^{2}}{a^{2}}}{\frac{k^{2}}{b c}}+\frac{\frac{k^{2}}{b^{2}}}{\frac{k^{2}}{c a}}+\frac{\frac{k^{2}}{c^{2}}}{\frac{k^{2}}{a b}}$

$=\frac{k^{2}}{a^{2}} \times \frac{b c}{k^{2}}+\frac{k^{2}}{b^{2}} \times \frac{c a}{k^{2}}+\frac{k^{2}}{c^{2}} \times \frac{a b}{k^{2}}$

$=\frac{b c}{a^{2}}+\frac{c a}{b^{2}}+\frac{a b}{c^{2}}=\mathrm{R} \cdot \mathrm{H.S}$

Question 19

If a, b, c and d are in proportion, prove that:

(i) (5a + 7b) (2c – 3d) = (5c + 7d) (2a – 3b)

(ii) (ma + nb) : b = (mc + nd) : d

(iii) (a4 + c4) : (b4 + d4) = a2 c2 : b2 d2.

(iii) $\left(a^{4}+c^{4}\right):\left(b^{4}+d^{4}\right)=a^{2} c^{2}: b^{2} d^{2}$

(iv) $\frac{a^{2}+a b}{c^{2}+c d}=\frac{b^{2}-2 a b}{d^{2}-2 c d}$

(v) $\frac{(a+c)^{3}}{(b+d)^{3}}=\frac{a(a-c)^{2}}{b(b-d)^{2}}$

(vi) $\frac{a^{2}+a b+b^{2}}{a^{2}-a b+b^{2}}=\frac{c^{2}+c d+d^{2}}{c^{2}-c d+d^{2}}$

(vii) $\frac{a^{2}+b^{2}}{c^{2}+d^{2}}=\frac{a b+a d-b c}{b c+c d-a d}$

(viii) abcd $\left[\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{d^{2}}\right.=a^{2}+b^{2}+c^{2}+d^{2}$

Sol :

∵ a, b, c, d are in proportion

$\frac{a}{b}=\frac{c}{d}=k($ say $)$

a=bk, c=dk

(i) L.H.S. $=(5 a+7 b)(2 c-3 d)$

$=(5 \cdot b k+7 b)(2 d k-3 d)$

$=k(5 b+7 b) k(2 d-3 d)$

$=k^{2}(12 b) \times(-d)=-12 b d k^{2}$

$\mathrm{R} . \mathrm{H.S.}=(5 c+7 d)(2 a-3 b)$

$=(5 d k+7 d)(2 k \cdot b-3 b)$

$=k(5 d+7 d) k(2 b-3 b)$

$=k^{2}(12 d)(-b)=-12 k^{2} b d=-12 b d k^{2}$

$\therefore \mathrm{L} \cdot \mathrm{H.S}=\mathrm{R} \cdot \mathrm{H.S}$

(ii) $(m a+n b): b=(m c+n d)^{*}: d$

$\Rightarrow \frac{m a+n b}{b}=\frac{m c+n d}{d}$

L.H.S. $=\frac{m b k+n b}{b}=\frac{b(m k+n)}{b}$

=m k+n

$\mathrm{R} . \mathrm{H.S.}=\frac{m c+n d}{d}=\frac{m d k+n d}{d}$

$=\frac{d(m k+n)}{d}=m k+n$

$\therefore \mathrm{L} \cdot \mathrm{H.S}=\mathrm{R} . \mathrm{H.S}$

(iii) $\left(a^{4}+c^{4}\right):\left(b^{4}+d^{4}\right)=a^{2} c^{2}: b^{2} d^{2}$

$\frac{a^{4}+c^{4}}{b^{4}+d^{4}}=\frac{a^{2} c^{2}}{b^{2} d^{2}}$

L. H.S. $=\frac{a^{4}+c^{4}}{b^{4}+d^{4}}=\frac{b^{4} k^{4}+d^{4} k^{4}}{b^{4}+d^{4}}$

$=\frac{k^{4}\left(b^{4}+d^{4}\right)}{\left(b^{4}+d^{4}\right)}=k^{4}$

R.H.S. $=\frac{a^{2} c^{2}}{b^{2} d^{2}}=\frac{k^{2} b^{2} \cdot k^{2} d^{2}}{b^{2} \cdot d^{2}}=k^{4}$

Hence L.H.S.=R.H.S.

(iv) $\quad$ L.H.S $=\frac{a^{2}+a b}{c^{2}+c d}=\frac{k^{2} b^{2}+b k \cdot b}{k^{2} d^{2}+d k \cdot d}$

$=\frac{k b^{2}(k+1)}{d^{2} k(k+1)}=\frac{b^{2}}{d^{2}}$

$\mathrm{R.H.S}=\frac{b^{2}-2 a b}{d^{2}-2 c d}=\frac{b^{2}-2 \cdot b k b}{d^{2}-2 d k d}$

$=\frac{b^{2}(1-2 k)}{d^{2}(1-2 k)}=\frac{b^{2}}{d^{2}}$

$\therefore \mathrm{L} \cdot \mathrm{H.S}=\mathrm{R} \cdot \mathrm{H.S}$

(v) L.H.S. $=\frac{(a+c)^{3}}{(b+d)^{3}}=\frac{(b k+d k)^{3}}{(b+d)^{3}}$

$=\frac{k^{3}(b+d)^{3}}{(b+d)^{3}}=k^{3}$

R.H.S. $=\frac{a(a-c)^{2}}{b(b-d)^{2}}=\frac{b k(b k-d k)^{2}}{b(b-d)^{2}}$

$=\frac{b k \cdot k^{2}(b-d)^{2}}{b(b-d)^{2}}=k^{3}$

$\therefore \mathrm{L} . \mathrm{H.S}=\mathrm{R} . \mathrm{H.S}$

$(v i) \quad L . H . S=\frac{a^{2}+a b+b^{2}}{a^{2}-a b+b^{2}}$

$=\frac{b^{2} k^{2}+b k \cdot b+b^{2}}{b^{2} k^{2}-b k \cdot b+b^{2}}$

$=\frac{b^{2}\left(k^{2}+k+1\right)}{b^{2}\left(k^{2}-k+1\right)}=\frac{k^{2}+k+1}{k^{2}-k+1}$

$\mathrm{R} \cdot \mathrm{H.S}=\frac{c^{2}+c d+d^{2}}{c^{2}-c d+d^{2}}$

$=\frac{d^{2} k^{2}+d k d+d^{2}}{d^{2} k^{2}-d k \cdot d+d^{2}}$

$=\frac{d^{2}\left(k^{2}+k+1\right)}{d^{2}\left(k^{2}-k+1\right)}=\frac{k^{2}+k+1}{k^{2}-k+1}$

$\therefore \mathrm{L} \cdot \mathrm{H.S}=\mathrm{R} \cdot \mathrm{H.S}$

$(vi) L . H . S=\frac{a^{2}+a b+b^{2}}{a^{2}-a b+b^{2}}$

$=\frac{b^{2} k^{2}+b k \cdot b+b^{2}}{b^{2} k^{2}-b k \cdot b+b^{2}}$

$=\frac{b^{2}\left(k^{2}+k+1\right)}{b^{2}\left(k^{2}-k+1\right)}=\frac{k^{2}+k+1}{k^{2}-k+1}$

$\mathrm{R} \cdot \mathrm{H.S}=\frac{c^{2}+c d+d^{2}}{c^{2}-c d+d^{2}}$

$=\frac{d^{2} k^{2}+d k d+d^{2}}{d^{2} k^{2}-d k \cdot d+d^{2}}$

$=\frac{d^{2}\left(k^{2}+k+1\right)}{d^{2}\left(k^{2}-k+1\right)}=\frac{k^{2}+k+1}{k^{2}-k+1}$

$\therefore \mathrm{L} \cdot \mathrm{H.S}=\mathrm{R} \cdot \mathrm{H.S}$

(viii) $\mathrm{L.H.S.}=a b c d\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{d^{2}}\right)$

(viii) L.H.S. $=a b c d\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{d^{2}}\right)$

$=b k$.b. dk.d $\left[\frac{1}{b^{2} k^{2}}+\frac{1}{b^{2}}+\frac{1}{d^{2} k^{2}}+\frac{1}{d^{2}}\right]$

$=k^{2} b^{2} d^{2}\left[\frac{d^{2}+d^{2} k^{2}+b^{2}+b^{2} k^{2}}{b^{2} d^{2} k^{2}}\right]$

$=d^{2}\left(1+k^{2}\right)+b^{2}\left(1+k^{2}\right)=\left(1+k^{2}\right)\left(b^{2}+d^{2}\right)$

R.H.S $=a^{2}+b^{2}+c^{2}+d^{2}$

$=b^{2} k^{2}+b^{2}+d^{2} k^{2}+d^{2}$

$=b^{2}\left(k^{2}+1\right)+d^{2}\left(k^{2}+1\right)=\left(k^{2}+1\right)\left(b^{2}+d^{2}\right)$

$\therefore$ L.H.S = R.H.S.

Question 20

If x, y, z are in continued proportion, prove that: $\frac{(x+y)^{2}}{(y+z)^{2}}=\frac{x}{z}$. (2010)

Sol :

x, y, z are in continued proportion

Let $\frac{x}{y}=\frac{y}{z}=k$

Then y=k z

$x=y k=k z \times k=k^{2} z$

Now L.H.S. $=\frac{(x+y)^{2}}{(y+z)^{2}}$

$=\frac{\left(k^{2} z+k z\right)^{2}}{(k z+z)^{2}}=\frac{\{k z(k+1)\}^{2}}{\{z(k+1)\}^{2}}$

$=\frac{k^{2} z^{2}(k+1)^{2}}{z^{2}(k+1)^{2}}=k^{2}$

$\mathrm{R.H.S.}=\frac{x}{z}=\frac{k^{2} z}{z}=k^{3}$

∴L.H.S=R.H.S

Question 21

If a, b, c are in continued proportion, prove that:

$\frac{p \alpha^{2}+q a b+r b^{2}}{p b^{2}+q b c+r c^{2}}=\frac{\alpha}{c}$

Sol :

Given a, b, c are in continued proportion

$\frac{p \alpha^{2}+q \alpha b+r b^{2}}{p b^{2}+q b c+r c^{2}}=\frac{a}{c}$

Let $\frac{a}{b}=\frac{b}{c}=k$

⇒a=bk and b=ck...(i)

⇒a=(ck)k$=c k^{2}$  [using (i)]

b=ck

$\mathrm{L.H.S.}=\frac{a}{c}=\frac{c k^{2}}{c}=k^{2}$

R.H.S. $=\frac{p\left(c k^{2}\right)^{2}+q\left(c k^{2}\right) c k+r(c k)^{2}}{p(c k)^{2}+q(c k) c+r c^{2}}$

$=\frac{p c^{2} k^{4}+q c^{2} k^{3}+r c^{2} k^{2}}{p c^{2} k^{2}+q c^{2} k+r c^{2}}$

$=\frac{c^{2} k^{2}}{c^{2}}\left[\frac{p k^{2}+q k+r}{p k^{2}+q k+r}\right]=k^{2}$...(iii)

From (ii) and (iii) L.H.S=R.H.S

∴b=ck, a=bk=ckk$=c k^{2}$

(i) L.H.S

$=\frac{a+b}{b+c}=\frac{c k^{2}+c k}{c k+c}=\frac{c k(k+1)}{c(k+1)}=k$

$\mathrm{R} . \mathrm{H.S}=\frac{a^{2}(b-c)}{b^{2}(a-b)}$

$=\frac{\left(c k^{2}\right)^{2}(c k-c)}{(c k)^{2}\left(c k^{2}-c k\right)}$

$=\frac{c^{2} k^{4} c(k-1)}{c^{2} k^{2} c k(k-1)}$

Question 22

If a, b, c are in continued proportion, prove that:

(i) $\frac{a+b}{b+c}=\frac{a^{2}(b-c)}{b^{2}(a-b)}$

(ii) $\frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}=\frac{a}{b^{2} c^{2}}+\frac{b}{c^{2} a^{2}}+\frac{c}{a^{2} b^{2}}$

(iii) $a: c=\left(a^{2}+b^{2}\right):\left(b^{2}+c^{2}\right)$

(iv) $a^{2} b^{2} c^{2}\left(a^{-4}+b^{-4}+c^{-4}\right)=b^{-2}\left(a^{4}+b^{4}+c^{4}\right)$

(v) $a b c(a+b+c)^{3}=(a b+b c+c a)^{3}$

(vi) $(a+b+c)(a-b+c)=a^{2}+b^{2}+c^{2}$

Sol :

As a, b, c, are in continued proportion

Let $\frac{a}{b}=\frac{b}{c}=k$

$=\frac{c^{3} k^{4}(k-1)}{c^{3} k^{3}(k-1)}=k$

∴L.H.S=R.H.S

(ii) L.H.S. $=\frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}$

$=\frac{1}{\left(c k^{2}\right)^{3}}+\frac{1}{(c k)^{3}}+\frac{1}{c^{3}}$

$=\frac{1}{c^{3} k^{6}}+\frac{1}{c^{3} k^{3}}+\frac{1}{c^{3}}$

$=\frac{1}{c^{3}}\left[\frac{1}{k^{6}}+\frac{1}{k^{3}}+\frac{1}{1}\right]$

R.H.S. $=\frac{a}{b^{2} c^{2}}+\frac{b}{c^{2} a^{2}}+\frac{c}{a^{2} b^{2}}$

$=\frac{c k^{2}}{(c k)^{2} c^{2}}+\frac{c k}{c^{2}\left(c k^{2}\right)^{2}}+\frac{c}{\left(c k^{2}\right)^{2}(c k)^{2}}$

$=\frac{c k^{2}}{c^{4} k^{2}}+\frac{c k}{c^{4} k^{4}}+\frac{c}{c^{4} k^{6}}$

$=\frac{1}{c^{3}}+\frac{1}{c^{3} k^{3}}+\frac{1}{c^{3} k^{6}}$

$=\frac{1}{c^{3}}\left[1+\frac{1}{k^{3}}+\frac{1}{k^{6}}\right]$

$=\frac{1}{c^{3}}\left[\frac{1}{k^{6}}+\frac{1}{k^{3}}+1\right]$

∴ L.H.S = R.H.S.

(iii) $a: c=\left(a^{2}+b^{2}\right):\left(b^{2}+c^{2}\right)$

$\Rightarrow \frac{a}{c}=\frac{a^{2}+b^{2}}{b^{2}+c^{2}}$

L.H.S. $\frac{a}{c}=\frac{c k^{2}}{c}=k^{2}$

R.H.S. $=\frac{\left(c k^{2}\right)^{2}+(c k)^{2}}{(c k)^{2}+c^{2}}$

$=\frac{c^{2} k^{4}+c^{2} k^{2}}{c^{2} k^{2}+c^{2}}$

$=\frac{c^{2} k^{2}\left(k^{2}+1\right)}{c^{2}\left(k^{2}+1\right)}=k^{2}$

∴ L.H.S = R.H.S.

(iv) L.H.S. $=a^{2} b^{2} c^{2}\left(a^{-4}+b^{-4}+c^{-4}\right)$

$=a^{2} b^{2} c^{2}\left[\frac{1}{a^{4}}+\frac{1}{b^{4}}+\frac{1}{c^{4}}\right]$

$=\frac{a^{2} b^{2} c^{2}}{a^{4}}+\frac{a^{2} b^{2} c^{2}}{b^{4}}+\frac{a^{2} b^{2} c^{2}}{c^{4}}$

$=\frac{b^{2} c^{2}}{a^{2}}+\frac{c^{2} a^{2}}{b^{2}}+\frac{a^{2} b^{2}}{c^{2}}$

$=\frac{(c k)^{2} \cdot c^{2}}{\left(c k^{2}\right)^{2}}+\frac{c^{2}\left(c k^{2}\right)^{2}}{(c k)^{2}}+\frac{\left(c k^{2}\right)^{2}(c k)^{2}}{c^{2}}$

$=\frac{c^{2} k^{2} \cdot c^{2}}{c^{2} k^{4}}+\frac{c^{2} \cdot c^{2} k^{4}}{c^{2} k^{2}}+\frac{c^{2} k^{4} \cdot c^{2} k^{2}}{c^{2}}$

$=\frac{c^{2}}{k^{2}}+\frac{c^{2} k^{2}}{1}+\frac{c^{2} k^{6}}{1}$

$=c^{2}\left[\frac{1}{k^{2}}+k^{2}+k^{6}\right]$

$=\frac{c^{2}}{k^{2}}\left[1+k^{4}+k^{8}\right]$

R.H.S. $=b^{-2}\left[a^{4}+b^{4}+c^{4}\right]$

$=\frac{1}{b^{2}}\left[a^{4}+b^{4}+c^{4}\right]$

$=\frac{1}{(c k)^{2}}\left[\left(c k^{2}\right)^{4}+(c k)^{4}+c^{4}\right]$

$=\frac{1}{c^{2} k^{2}}\left[c^{4} k^{8}+c^{4} k^{4}+c^{4}\right]$

$=\frac{c^{4}}{c^{2} k^{2}}\left[k^{8}+k^{4}+1\right]$

$=\frac{c^{2}}{k^{2}}\left[1+k^{4}+k^{8}\right]$

∴ L.H.S = R.H.S.

(v) L.H.S. $=a b c(a+b+c)^{3}$

$=c k^{2}$.ck.c $\left[c k^{2}+c k+c\right]^{3}$

$=c^{3} k^{3}\left\{c\left(k^{2}+k+1\right)\right]^{3}$

$=c^{3} k^{3} \cdot c^{3} \cdot\left(k^{2}+k+1\right)^{3}=c^{6} k^{3}\left(k^{2}+k+1\right)^{3}$

$\mathrm{R} . \mathrm{H.S.}=(a b+b c+c a)^{3}$

$=\left(c k^{2} \cdot c k+c k \cdot c+c \cdot c k^{2}\right)^{3}$

$=\left(c^{2} k^{3}+c^{2} k+c^{2} k^{2}\right)^{3}=\left(c^{2} k^{3}+c^{2} k^{2}+c^{2} k\right)^{3}$

$=\left[c^{2} k\left(k^{2}+k+1\right)\right]^{3}=c^{6} k^{3}\left(k^{2}+k+1\right)^{3}$

∴ L.H.S = R.H.S.

(vi) L.H.S. $=(a+b+c)(a-b+c)$

$=\left(c k^{2}+c k+c\right)\left(c k^{2}-c k+c\right)$

$=c\left(k^{2}+k+1\right) c\left(k^{2}-k+1\right)$

$=c^{2}\left(k^{2}+k+1\right)\left(k^{2}-k+1\right)$

$=c^{2}\left(k^{4}+k^{2}+1\right)$

R.H.S. $=a^{2}+b^{2}+c^{2}$

$=\left(c k^{2}\right)^{2}+(c k)^{2}+(c)^{2}$

$=c^{2} k^{4}+c^{2} k^{2}+c^{2}=c^{2}\left(k^{4}+k^{2}+1\right)$

∴ L.H.S = R.H.S.

Question 23

If a, b, c, d are in continued proportion, prove that:

(i) $\frac{a^{3}+b^{3}+c^{3}}{b^{3}+c^{3}+3^{3}}=\frac{a}{d}$

(ii) $\left(a^{2}-b^{2}\right)\left(c^{2}-d^{2}\right)=\left(b^{2}-c^{2}\right)^{2}$

(iii) $(a+d)(b+c)-(a+c)(b+d)=(b-c)^{2}$

(iv) $a: d=$ triplicate ratio of $(a-b):(b-c)$

(v) $\left(\frac{a-b}{c}+\frac{a-c}{b}\right)^{2}-\left(\frac{d-b}{c}+\frac{d-c}{b}\right)^{2}$.

$=(a-d)^{2}\left(\frac{1}{c^{2}}-\frac{1}{b^{2}}\right)$

Sol :

a, b, c, d are in continued proportion

$\therefore \frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k(s a y)$

$\therefore c=d k, b=c k=d k . k=d k^{2}$

$a=b k=d k^{2} \cdot k=d k^{3}$

(i) L.H.S. $=\frac{a^{3}+b^{3}+c^{3}}{b^{3}+c^{3}+d^{3}}$

$=\frac{\left(d k^{3}\right)^{3}+\left(d k^{2}\right)^{3}+(d k)^{3}}{\left(d k^{2}\right)^{3}+(d k)^{3}+d^{3}}$

$=\frac{d^{3} k^{9}+d^{3} k^{6}+d^{3} k^{3}}{d^{3} k^{6}+d^{3} k^{3}+d^{3}}$

$=\frac{d^{3} k^{3}\left(k^{6}+k^{3}+1\right)}{d^{3}\left(k^{6}+k^{3}+1\right)}=k^{3}$

$\mathrm{R.H.S.}=\frac{a}{d}=\frac{d k^{3}}{d}=k^{3}$

∴ L.H.S = R.H.S.

(ii) L.H.S. $=\left(a^{2}-b^{2}\right)\left(c^{2}-d^{2}\right)$

$=\left[\left(d k^{3}\right)^{2}-\left(d k^{2}\right)^{2}\right]\left[(d k)^{2}-d^{2}\right]$

$=\left(d^{2} k^{6}-d^{2} k^{4}\right)\left(d^{2} k^{2}-d^{2}\right)$

$=d^{2} k^{4}\left(k^{2}-1\right) d^{2}\left(k^{2}-1\right)=d^{4} k^{4}\left(k^{2}-1\right)^{2}$

$\mathrm{R} \cdot \mathrm{H} . \mathrm{S} .=\left(b^{2}-c^{2}\right)^{2}=\left[\left(d k^{2}\right)^{2}-(d k)^{2}\right]^{2}$

$=\left[d^{2} k^{4}-d^{2} k^{2}\right]^{2}=\left[d^{2} k^{2}\left(k^{2}-1\right)\right]^{2}$

$=d^{4} k^{4}\left(k^{2}-1\right)^{2}$

∴ L.H.S = R.H.S.

(iii) L.H.S. $=(a+d)(b+c)-(a+c)(b+d)$

$=\left(d k^{3}+d\right)\left(d k^{2}+d k\right)-\left(d k^{3}+d k\right)\left(d k^{2}+d\right)$

$=d\left(k^{3}+1\right) d k(k+1)-d k\left(k^{2}+1\right) d\left(k^{2}+1\right)$

$=d^{2} k(k+1)\left(k^{3}+1\right)-d^{2} k\left(k^{2}+1\right)\left(k^{2}+1\right)$

$=d^{2} k\left[k^{4}+k^{3}+k+1-k^{4}-2 k^{2}-1\right]$

$=d^{2} k\left[k^{3}-2 k^{2}+k\right]=d^{2} k^{2}\left[k^{2}-2 k+1\right]$

$=d^{2} k^{2}(k-1)^{2}$

$\mathrm{R} . \mathrm{H.S.}=(b-c)^{2}=\left(d k^{2}-d k\right)^{2}=d^{2} k^{2}(k-1)^{2}$

∴ L.H.S = R.H.S.

(iv) a : d= triplicate ratio of (a-b) : (b-c)

$=(a-b)^{3}:(b-c)^{3}$

L.H.S. $=a: d=\frac{a}{d}=\frac{d k^{3}}{d}=k^{3}$

$\mathrm{R} . \mathrm{H.S.}=\frac{(a-b)^{3}}{(b-c)^{3}}$

$=\frac{\left(d k^{3}-d k^{2}\right)^{3}}{\left(d k^{2}-d k\right)^{3}}=\frac{d^{3} k^{6}(k-1)^{3}}{d^{3} k^{3}(k-1)^{3}}=k^{3}$

∴ L.H.S = R.H.S.

(v) L.H.S=

$\left(\frac{a-b}{c}+\frac{a-c}{b}\right)^{2}-\left(\frac{d-b}{c}+\frac{d-c}{b}\right)^{2}$

$=\left(\frac{d k^{3}-d k^{2}}{d k}+\frac{d k^{3}-d k}{d k^{2}}\right)^{2}-\cdot\left(\frac{d-d k^{2}}{d k}+\frac{d-d k}{d k^{2}}\right)^{2}$

$=\left(\frac{d k^{2}(k-1)}{d k}+\frac{d k\left(k^{2}-1\right)}{d k^{2}}\right)^{2}-\left(\frac{d\left(1-k^{2}\right)}{d k}+\frac{d(1-k)}{d k^{2}}\right)^{2}$

$=\left(k(k-1)+\frac{k^{2}-1}{k}\right)^{2}-\left(\frac{1-k^{2}}{k}+\frac{1-k}{k^{2}}\right)^{2}$

$=\left(\frac{k^{2}(k-1)+\left(k^{2}-1\right)}{k}\right)^{2}-\left(\frac{k\left(1-k^{2}\right)+1-k}{k^{2}}\right)^{2}$

$=\left(\frac{k^{3}-k^{2}+k^{2}-1}{k}\right)^{2}-\left(\frac{k-k^{3}+1-k}{k^{2}}\right)^{2}$

$=\frac{\left(k^{3}-1\right)^{2}}{k^{2}}-\frac{\left(-k^{3}+1\right)^{2}}{k^{4}}$

$=\frac{\left(k^{3}-1\right)^{2}}{k^{2}}-\frac{\left(1-k^{3}\right)^{2}}{k^{4}}$

$=\left(\frac{\left(k^{3}-1\right)}{k^{2}}\right)^{2}\left(1-\frac{1}{k^{2}}\right)=\frac{\left(k^{3}-1\right)^{2}\left(k^{2}-1\right)}{k^{4}}$

$=\frac{\left(k^{3}-1\right)^{2}\left(k^{2}-1\right)}{k^{4}}$

R.H.S. $=(a-d)^{2}\left(\frac{1}{c^{2}}-\frac{1}{b^{2}}\right)$

$=\left(d k^{3}-d\right)^{2}\left(\frac{1}{d^{2} k^{2}}-\frac{1}{d^{2} k^{4}}\right)$

$=d^{2}\left(k^{3}-1\right)^{2}\left(\frac{k^{2}-1}{d^{2} k^{4}}\right)=\frac{\left(k^{3}-1\right)^{2}\left(k^{2}-1\right)}{k^{4}}$

∴ L.H.S = R.H.S.

ML Aggarwal Solution Class 10 Chapter 7 Ratio and Proportion Exercise 7.1

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20 Q21

 Exercise 7.1

Question 1

An alloy consists of $27 \frac{1}{2}$ kg of copper and $2 \frac{3}{4}$ kg of tin. Find the ratio by weight of tin to the alloy

Sol :

Copper $=27 \frac{1}{2} \mathrm{~kg}=\frac{55}{2} \mathrm{~kg}$

Tin $=2 \frac{3}{4} \mathrm{~kg}=\frac{11}{4} \mathrm{~kg}$

Total alloy$=\frac{55}{2}+\frac{11}{4}=\frac{110+11}{4}=\frac{121}{4}$ kg

Now Ratio between tin and alloy$=\frac{11}{4} \mathrm{~kg}: \frac{121}{4} \mathrm{~kg}$

=11:121

=1:11

Question 2

Find the compounded ratio of:

(i) 2 : 3 and 4 : 9

(ii) 4 : 5, 5 : 7 and 9 : 11

(iii) (a – b) : (a + b), $(a+b)^{2}:\left(a^{2}+b^{2}\right)$ and $\left(a^{4}-b^{4}\right):\left(a^{2}-b^{2}\right)^{2}$

Sol :

(i) 2 : 3 and 4 : 9

Compound ratio $=\frac{2}{3} \times \frac{4}{9}$

$=\frac{8}{27}$ or 8: 27

(ii) 4 : 5, 5 : 7 and 9 : 11

Compound ratio $=\frac{4}{5} \times \frac{5}{7} \times \frac{9}{11}=\frac{36}{77}$

or 36: 77

(iii) $(a-b):(a+b),(a+b)^{2}:\left(a^{2}+b^{2}\right)$

and $\left(a^{4}-b^{4}\right):\left(a^{2}-b^{4}\right)^{2}$

Compound ratio

$=\frac{a-b}{a+b} \times \frac{(a+b)^{2}}{a^{2}+b^{2}} \times \frac{a^{4}-b^{4}}{\left(a^{2}-b^{2}\right)^{2}}$

$=\frac{a-b}{a+b} \times \frac{(a+b)(a+b)}{a^{2}+b^{2}}\times \frac{\left(a^{2}+b^{2}\right)(a+b)(a-b)}{(a+b)^{2}(a-b)^{2}}$

$=\frac{1}{1}$ or 1: 1

Question 3

Find the duplicate ratio of

(i) 2 : 3

(ii) √5 : 7

(iii) 5a : 6b

Sol :

(i) Duplicate ratio of 2 : 3 $=(2)^{2}:(3)^{2}=4 \cdot 9$

(ii) Duplicate ratio of √5 : 7 $=(\sqrt{5})^{2}:(7)^{2}=5: 49$

(iii) Duplicate ratio of 5a : 6b $=(5 a)^{2}:(6 b)^{2}=25 a^{2}: 36 b^{2}$

Question 4

Find the triplicate ratio of

(i) 3 : 4

(ii) $\frac{1}{2}: \frac{1}{3}$

(iii) $1^{3}: 2^{3}$

Sol :

(i) Triplicate ratio of 3 : 4

$=(3)^{3}:(4)^{3}$

=27 : 64

(ii) Triplicate ratio of $\frac{1}{2}: \frac{1}{3}=\left(\frac{1}{2}\right)^{3}:\left(\frac{1}{3}\right)^{3}$

$=\frac{1}{8}: \frac{1}{27}$

=27: 8

(iii) Triplicate ratio of $1^{3}: 2^{3}=\left(1^{3}\right)^{3}:\left(2^{3}\right)^{3}$

$=(1)^{3}:(8)^{3}$

=1 : 512

Question 5

Find the sub-duplicate ratio of

(i) 9 : 16

(ii) $\frac{1}{4}: \frac{1}{9}$

(iii) $9 a^{2}: 49 b^{2}$

Sol :

(i) Sub-duplicate ratio of 9 : 16

= √9 : √16

= 3 : 4

$=\sqrt{9}: \sqrt{16}=3: 4$

(ii) Sub-duplicate ratio of $\frac{1}{4}: \frac{1}{9}=\sqrt{\frac{1}{4}}: \sqrt{\frac{1}{9}}$

$=\frac{1}{2}: \frac{1}{3}=3: 2$

(iii) Sub-duplicate ratio of $9 a^{2}: 49 b^{2}$

$=\sqrt{9 a^{2}}: \sqrt{49 b^{2}}$

=3 a: 7 b

Question 6

Find the sub-triplicate ratio of

(i) 1 : 216

(ii) $\frac{1}{8}: \frac{1}{125}$

(iii) $27 a^{3}: 64 b^{3}$

Sol :

(i) Sub-triplicate ratio of 1 : 216

$=\sqrt[3]{1}: \sqrt[3]{216}$

$=\left(1^{3}\right)^{\frac{1}{3}}:\left(6^{3}\right)^{\frac{1}{3}}=1: 6$

(ii) Sub-triplicate ratio of $\frac{1}{8}: \frac{1}{125}$

$=\left(\frac{1}{8}\right)^{\frac{1}{3}}:\left(\frac{1}{125}\right)^{\frac{1}{3}}$

$=\left[\left(\frac{1}{2}\right)^{3}\right]^{\frac{1}{3}}:\left[\left(\frac{1}{5}\right)^{3}\right]^{\frac{1}{3}}$

$=\frac{1}{2}: \frac{1}{5}$

=5: 2

(iii) Sub-triplicate ratio of $27 a^{3}: 64 b^{3}$

$=\left[(3 a)^{3}\right]^{\frac{1}{3}}:\left[(4 b)^{3}\right]^{\frac{1}{3}}$

=3a: 4b

Question 7

Find the reciprocal ratio of

(i) 4 : 7

(ii) $3^{2}: 4^{2}$

(iii) $\frac{1}{9}: 2$

Sol :

(i) Reciprocal ratio of 4 : 7 = 7 : 4

(ii) Reciprocal ratio of $3^{2}: 4^{2}=4^{2}: 3^{2}$

=16: 9

(iii) Reciprocal ratio of $\frac{1}{9}: 2=2: \frac{1}{9}=18: 1$

Question 8

Arrange the following ratios in ascending order of magnitude:

2 : 3, 17 : 21, 11 : 14 and 5 : 7

Sol :

Writing the given ratios in fraction

$\frac{2}{3}, \frac{17}{21}, \frac{11}{14}, \frac{5}{7}$

L.C.M of 3,21,14,7=42

Converting the given ratio as equivalent

$\frac{2}{3}=\frac{2 \times 14}{3 \times 14}=\frac{28}{42}$

$\frac{17}{21}=\frac{17 \times 2}{21 \times 2}=\frac{34}{42}$

$\frac{11}{14}=\frac{11 \times 3}{14 \times 3}=\frac{33}{42} $

$ \frac{5}{7}=\frac{5 \times 6}{7 \times 6}=\frac{30}{42}$

From above, writing in ascending order;

$\frac{28}{42}, \frac{30}{42}, \frac{33}{42}, \frac{34}{42}$ or 

$\frac{2}{3}, \frac{5}{7}, \frac{11}{14}, \frac{17}{21}$

or 2:3 ; 5:7; 11:14 and 17:21

Question 9

(i) If A : B = 2 : 3, B : C = 4 : 5 and C : D = 6 : 7, find A : D

(ii) If x : y = 2 : 3, and y : z = 4 : 7, find x : y : z

Sol :

(i)

Let A : B = 2 : 3, B : C = 4 : 5 and C : D = 6 : 7

$\frac{A}{B}=\frac{2}{3}, \frac{B}{C}=\frac{4}{5}, \frac{C}{D}=\frac{6}{7}$

Multiplying $\frac{A}{B} \times \frac{B}{C} \times \frac{C}{D}=\frac{2}{3} \times \frac{4}{5} \times \frac{6}{7}$

$\therefore \frac{\mathrm{A}}{\mathrm{D}}=\frac{16}{35}$

$ \Rightarrow \mathrm{A}: \mathrm{D}=16: 35 \mathrm{Ans}$

(ii) L.C.M of y's terms 3 and 4=12

Making equals of y as 12

$\frac{x}{y}=\frac{2}{3}=\frac{2 \times 4}{3 \times 4}=\frac{8}{12}$ or 8: 12

$\frac{y}{z}=\frac{4}{7} \times \frac{3}{3}=\frac{12}{21}$ or 12: 21

Then x : y : z=8:12:21

Question 10

(i) If $A: B=$\frac{1}{4}: \frac{1}{5}$ and B : C$\frac{1}{7}: \frac{1}{6}$, find A : B : C

(ii) If 3A = 4B = 6C, find A : B : C

Sol :

$A: B=\frac{1}{4} \times \frac{5}{1}=\frac{5}{4}$

$B: C=\frac{1}{7} \times \frac{6}{1}=\frac{6}{7}$

LCM of B's terms 4 and 6=12

 Making terms of B's; as 12

$\frac{A}{B}=\frac{5 \times 3}{4 \times 3}=\frac{15}{12}=15: 12$

$\frac{B}{C}=\frac{6 \times 2}{7 \times 2}=\frac{12}{14}=12: 14$

∴ A: B: C=15: 12: 14

(ii) 3A=4B

$\Rightarrow \frac{A}{B}=\frac{4}{3}$

or A : B= 4 : 3

and 4B=6C

$\Rightarrow \quad \frac{B}{C}=\frac{6}{4}=\frac{3}{2}$ 

or B: C=3: 2

∴A : B : C=4 : 3 : 2

Question 11

(i) If $\frac{3 x+5 y}{3 x-5 y}=\frac{7}{3}$, Find x : y

(ii) If a : b=3 : 11, find (15a-3b) : (9a+5b).a

Sol :

(i) $\frac{3 x+5 y}{3 x-5 y}=\frac{7}{3}$

⇒ 9x + 15y = 21x – 35y [By cross multiplication]

⇒ 21x – 9x = 15y + 35y

⇒12 x=50y 

$\Rightarrow \frac{x}{y}=\frac{50}{12}=\frac{25}{6}$

Hence, x : y=25 : 6

(ii) a: b=3: 11 or $\frac{a}{b}=\frac{3}{11}$

Now $\frac{15 a-3 b}{9 a+5 b}=\frac{\frac{15 a}{b}-\frac{3 b}{b}}{\frac{9 a}{b}+\frac{5 b}{b}}$

[Dividing by b]

$=\frac{\frac{15 a}{b}-3}{\frac{9 a}{b}+5}=\frac{15 \times \frac{3}{11}-3}{9 \times \frac{3}{11}+5}$

$\left(\right. \text{ Substituting the value of  }\left.\frac{a}{b}\right)$

$=\frac{\frac{45}{11}-3}{\frac{27}{11}+5}$

$=\frac{\frac{45-33}{11}}{\frac{27+55}{11}}$

$=\frac{\frac{12}{11}}{\frac{82}{11}}$

$=\frac{12}{11} \times \frac{11}{82}$

$=\frac{12}{82}=\frac{6}{41}$

∴(15a-3b) : (9a+5b)

=6:41

Question 12

(i) If (4x² + xy) : (3xy – y²) = 12 : 5, find (x + 2y) : (2x + y).

(ii) If y (3x – y) : x (4x + y) = 5 : 12. Find (x² + y²) : (x + y)².

Sol :

(4x² + xy) : (3xy – y²) = 12 : 5

$\Rightarrow \frac{4 x^{2}+\infty y}{3 x y-y^{2}}=\frac{12}{5}$

⇒ 20x² + 5xy = 36xy – 12y²

$\Rightarrow 20 x^{2}+5 x y-36 x y+12 y^{2}=0$

$\Rightarrow \quad 20 x^{2}-31 x y+12 y^{2}=0$

$\Rightarrow \quad \frac{20 x^{2}}{y^{2}}-\frac{31 x y}{y^{2}}+\frac{12 y^{2}}{y^{2}}=0$ (Dividing by $y^{2}$ )

$\Rightarrow 20\left(\frac{x}{y}\right)^{2}-31\left(\frac{x}{y}\right)+12=0$

$\Rightarrow 20\left(\frac{x}{y}\right)^{2}-15\left(\frac{x}{y}\right)-16\left(\frac{x}{y}\right)+12=0$

$\Rightarrow 5\left(\frac{x}{y}\right)\left[4\left(\frac{x}{y}\right)-3\right]-4\left[4\left(\frac{x}{y}\right)-3\right]=0$

$\Rightarrow\left[4\left(\frac{x}{y}\right)-3\right]\left[5\left(\frac{x}{y}\right)-4\right]=0$

Either $4\left(\frac{x}{y}\right)-3=0,$ then $4\left(\frac{x}{y}\right)=3$

$\Rightarrow \frac{x}{y}=\frac{3}{4}$

or $5\left(\frac{x}{y}\right)-4=0$

then $5\left(\frac{x}{y}\right)=4$

$\Rightarrow \quad \frac{x}{y}=\frac{4}{5}$

Now $\frac{x+2 y}{2 x+y}=\frac{\frac{x}{y}+2}{2 \frac{x}{y}+1}$ (Dividing by y)

(a) When $\frac{x}{y}=\frac{3}{4},$ then

$=\frac{\frac{x}{y}+2}{2 \frac{x}{y}+1}$

$=\frac{\frac{3}{4}+2}{2 \times \frac{3}{4}+1}$

$=\frac{\frac{11}{4}}{\frac{3}{2}+1}$

$=\frac{\frac{11}{4}}{\frac{5}{2}}$

$=\frac{11}{4} \times \frac{2}{5}=\frac{11}{10}$

∴(x+2y) : (2x+y)=11 : 10

(b) When $\frac{x}{y}=\frac{4}{5},$ then

$\frac{x+2 y}{2 x+y}$

$=\frac{\frac{x}{y}+2}{2 \frac{x}{y}+1}$

$=\frac{\frac{4}{5}+2}{2 \times \frac{4}{5}+1}$

$=\frac{\frac{14}{5}}{\frac{8}{5}+1}$

[Dividing by y]

$=\frac{\frac{14}{5}}{\frac{13}{5}}=\frac{14}{5} \times \frac{5}{13}=\frac{14}{13}$

Hence $\frac{x+2 y}{2 x+y}=\frac{11}{10}$ or $\frac{14}{13}$

$\therefore \quad \frac{x+2 y}{2 x+y}$

=11: 10 or 14: 13

(ii) If y(3 x-y): x(4 x+y)=5: 12

Find $\left(x^{2}+y^{2}\right):(x+y)^{2}$

$\frac{3 x y-y^{2}}{4 x^{2}+x y}=\frac{5}{12}$

$\Rightarrow 36 x y-12 y^{2}=20 x^{2}+5 x y$

$\Rightarrow 20 x^{2}+5 x y-36 x y+12 y^{2}=0$

$\Rightarrow 20 x^{2}-31 x y+12 y^{2}=0$

$\Rightarrow 20 \frac{x^{2}}{y^{2}}-31 \frac{x y}{y^{2}}+\frac{12 y^{2}}{y^{2}}=0$

[Dividing by $y^{2}$]

$\Rightarrow 20\left(\frac{x^{2}}{y^{2}}\right)-31\left(\frac{x y}{y^{2}}\right)+12=0$

$\Rightarrow 20\left(\frac{x}{y}\right)^{2}-15\left(\frac{x}{y}\right)-16\left(\frac{x}{y}\right)+12=0$

$\Rightarrow 5\left(\frac{x}{y}\right)\left[4\left(\frac{x}{y}\right)-3\right]-4\left[4\left(\frac{x}{y}\right)-3\right]=0$

$\Rightarrow\left[4\left(\frac{x}{4}\right)-3\right]\left[5\left(\frac{x}{y}\right)-4\right]=0$

Either $\left[4\left(\frac{x}{y}\right)-3\right]=0$

then $4\left(\frac{x}{y}\right)=3 \Rightarrow \frac{x}{y}=\frac{3}{4}$

or $\left[5\left(\frac{x}{y}\right)-4\right]=0$

then $5\left(\frac{x}{y}\right)=4 \Rightarrow \frac{x}{y}=\frac{4}{5}$

(a) When $\frac{x}{y}=\frac{3}{4}$

then $\left(x^{2}+y^{2}\right):(x+y)^{2}$

$=\frac{x^{2}+y^{2}}{(x+y)^{2}}=\frac{\frac{x^{2}}{y^{2}}+\frac{y^{2}}{y^{2}}}{\frac{1}{y^{2}}(x+y)^{2}}$

(Dividing by $y^{2}$)

$=\frac{\frac{x^{2}}{y^{2}}+1}{\left(\frac{x}{y}+1\right)^{2}}$

$=\frac{\left(\frac{3}{4}\right)^{2}+1}{\left(\frac{3}{4}+1\right)^{2}}=\frac{\frac{9}{16}+1}{\left(\frac{7}{4}\right)^{2}}$

$=\frac{\frac{25}{16}}{\frac{49}{16}}=\frac{25}{16} \times \frac{16}{49}=\frac{25}{49}$

$\because\left(x^{2}+y^{2}\right):(x+y)^{2}=25: 49$

(b) $\because\left(x^{2}+y^{2}\right):(x+y)^{2}=25: 49$

$\frac{x^{2}+y^{2}}{(x+y)^{2}}$

$=\frac{\frac{x^{2}}{y^{2}}+1}{\left(\frac{x}{y}+1\right)^{2}}$

$=\frac{\left(\frac{x}{y}\right)^{2}+1}{\left(\frac{x}{y}+1\right)^{2}}$

$=\frac{\left(\frac{4}{5}\right)^{2}+1}{\left(\frac{4}{5}+1\right)^{2}}$

$=\frac{\frac{16}{25}+1}{\left(\frac{9}{5}\right)^{2}}$

$=\frac{\frac{41}{25}}{\frac{81}{25}}$

$=\frac{41}{25} \times \frac{25}{81}=\frac{41}{81}$

$\because\left(x^{2}+y^{2}\right):(x+y)^{2}=41: 81$

Question 13

(i) If (x – 9) : (3x + 6) is the duplicate ratio of 4 : 9, find the value of x.

(ii) If (3x + 1) : (5x + 3) is the triplicate ratio of 3 : 4, find the value of x.

(iii) If (x + 2y) : (2x – y) is equal to the duplicate ratio of 3 : 2, find x : y.

Sol :

(i) $\frac{x-9}{3 x+6}=\left(\frac{4}{9}\right)^{2}$

$\Rightarrow \frac{x-9}{3 x+6}=\frac{16}{81}$

$\Rightarrow 81 x-729=48 x+96$

$\Rightarrow 81 x-48 x=96+729$

$\Rightarrow 33 x=825 $

$\Rightarrow x=\frac{825}{33}=25$

(ii) If (3 x+1):(5 x+3) is the triplicate ratio of 3: 4

then $\frac{3 x+1}{5 x+3}=\frac{(3)^{3}}{(4)^{3}}=\frac{27}{64}$

$\Rightarrow 64(3 x+1)=27(5 x+3)$

$\Rightarrow 192 x+64=135 x+81$

$\Rightarrow 192 x-135 x=81-64$

$\Rightarrow 57 x=17 $

$\Rightarrow x=\frac{17}{57}$

Hence $x=\frac{17}{57}$

(iii) If (x+2 y):(2 x-y) is the duplicate ratio of 3: 2

then $\frac{x+2 y}{2x-y}=\frac{(3)^{2}}{(2)^{2}}=\frac{9}{4}$

$\Rightarrow 9(2 x-y)=4(x+2 y)$

$\Rightarrow 18 x-9 y=4 x+8 y$

$\Rightarrow 18 x-4 x=8 y+9 y$

$\Rightarrow 14 x-17 y $

$ \Rightarrow \quad \frac{x}{y}=\frac{17}{14}$

$\therefore x: y=17: 14$

Question 14

(i) Find two numbers in the ratio of 8 : 7 such that when each is decreased by $12 \frac{1}{2}$ , they are in the ratio 11 : 9.

(ii) The income of a man is increased in the ratio of 10 : 11. If the increase in his income is Rs 600 per month, find his new income.

Sol :

(i) The ratio is 8 : 7

Let the numbers be 8x and 7x,

According to condition,

$\frac{8 x-\frac{25}{2}}{7 x-\frac{25}{2}}=\frac{11}{9} $

$\Rightarrow \frac{\frac{16 x-25}{2}}{\frac{14 x-25}{2}}=\frac{11}{9}$

$\Rightarrow \frac{(16 x-25) \times 2}{2(14 x-25)}=\frac{11}{9}$

$\Rightarrow \frac{16 x-25}{14 x-25}=\frac{11}{9}$

$\Rightarrow 154 x-275=144 x-225$

$\Rightarrow 154 x-144 x=275-225$

$\Rightarrow 10 x=50$

$\therefore x=\frac{50}{10}=5$

∴ Numbers are 8x=8×5=40

and 7x=7×5=35

(ii) Let the present income=10x

then increased income=11x

∴increased per month=11x-10x=x

∴x=600

Now his new income=11x=11×600

=6600

Question 15

(i) A woman reduces her weight in the ratio 7 : 5. What does her weight become if originally it was 91 kg.

(ii) A school collected Rs 2100 for charity. It was decided to divide the money between an orphanage and a blind school in the ratio of 3 : 4. How much money did each receive?

Sol :

(i) Ratio between the original weight and reduced weight = 7 : 5

Let original weight = 7x

then reduced weight = 5x

If original weight = 91 kg.

then weight $=\frac{91 \times 5 x}{7 x}=65$ kg

(ii) Total amount to be distributed=2100

Ratio between orphanage and a blind school

=3 : 4

Sum of ratio=3+4=7

∴Orphanage school's share

$=₹ 2100 \times \frac{3}{7}$

=900

Blind school's share

$=₹ 2100 \times \frac{4}{7}=₹ 1200$

Question 16

(i) The sides of a triangle are in the ratio 7 : 5 : 3 and its perimeter is 30 cm. Find the lengths of sides.

(ii) If the angles of a triangle are in the ratio 2 : 3 : 4, find the angles.

Sol :

(i) Perimeter of a triangle = 30 cm.

Ratio among sides = 7 : 5 : 3

Sum of ratios 7 + 5 + 3 = 15

Length of first side $=30 \times \frac{7}{15}=14 \mathrm{~cm}$

Length of second side $=30 \times \frac{5}{15}=10 \mathrm{~cm}$

Length of third side $=30 \times \frac{3}{15}=6 \mathrm{~cm}$

∴ Sides are 14 cm, 10 cm, 6 cm

(ii) Sum of angles of a triangle=180°

Ratio among angles=2 : 3 : 4

Sum of ratios=2+3+4=9

∴First angle $=180^{\circ} \times \frac{2}{9}=40^{\circ}$

Second angle $=180^{\circ} \times \frac{3}{9}=60^{\circ}$

Third angle $=180^{\circ} \times \frac{4}{9}=80^{\circ}$

∴Angles are 40°, 60° and 80°

Question 17

Three numbers are in the ratio $\frac{1}{2}: \frac{1}{3}: \frac{1}{4}$ If the sum of their squares is 244, find the numbers.

Sol :

The ratio of three numbers $\frac{1}{2}: \frac{1}{3}: \frac{1}{4}$

$=\frac{6: 4: 3}{12}$

=.6 : 4 : 3

Let first number 6x, second 4x and third 3x

∴ According to the condition

$(6 x)^{2}+(4 x)^{2}+(3 x)^{2}=244$

$\Rightarrow 36 x^{2}+16 x^{2}+9 x^{2}=244$

⇒$61x^{2}=244$

$\Rightarrow x^{2}=\frac{244}{61}=4=(2)^{2}$

∴ x=2

∴ first number =6x=6×2=12

second number =4 x=4×2=8

and third number =3x=3×2=6

Question 18

(i) A certain sum was divided among A, B and C in the ratio 7 : 5 : 4. If B got Rs 500 more than C, find the total sum divided.

(ii) In a business, A invests Rs 50000 for 6 months, B Rs 60000 for 4 months and C, Rs 80000 for 5 months. If they together earn Rs 18800 find the share of each.

Sol :

(i) Ratio between A, B and C = 7 : 5 : 4

Let A’s share = 7x

B’s share = 5x

and C’s share = 4x

Total sum = 7x + 5x + 4x = 16x

Now, according to the condition

5x-4x=500

x=500

∴Total sum=16x=16×500=8000

(ii) A's 6 months investment =₹ 50000

∴ A's 1 month investment

=₹ 50000 × 6=₹ 300000

B's 4 month's investment =₹ 60000

∴ B's 1 month investment

=60000×4=₹ 240000

C's 5 months investment =₹ 80000

∴ C's 1 month investment

=₹ 80000×5=₹ 400000

∴ Ratio between their investments

=300000: 240000: 400000

=30: 24: 40

Sum of ratios =30+24+40=94 

Total earnings=₹ 18800

∴ A's share $=\frac{30}{94} \times 18800=₹ 6000$

B's share $=\frac{24}{94} \times 18800=₹ 4800$

C's share $=\frac{40}{94} \times 18800=₹ 8000$

Question 19

(i) In a mixture of 45 litres, the ratio of milk to water is 13 : 2. How much water must be added to this mixture to make the ratio of milk to water as 3 : 1 ?

(ii) The ratio of the number of boys to the number of girls in a school of 560 pupils is 5 : 3. If 10 new boys are admitted, find how many new girls may be admitted so that the ratio of the number of boys to the number of girls may change to 3 : 2.

Sol :

(i) Mixture of milk and water = 45 litres

Ratio of milk and water =13 : 2

Sum of ratio = 13 + 2 = 15

∴ Quantity of milk $=\frac{45 \times 13}{15}=39$ litres

and quantity of water $=45 \times \frac{2}{15}=6$ litres

Let x litre of water be added, then water =(6+x) litres

Now new ratio=3 : 1

∴39 : (6+x)=3 : 1

$\frac{39}{6+x}=\frac{3}{1}$

39=18+3x

3x=39-18=21

$\therefore x=\frac{21}{3}=7$ litres.

∴7 litres of water is to be added

(ii) ratio between boys and girls= 5 : 3

No. of pupils=560

Sum of ratios=5+3=8

∴No. of boys $=\frac{5}{8} \times 560=350$

and no. of girls $=\frac{3}{8} \times 560=210$

No. of new boys admitted =10

∴ Total number of boys =350+10=360

Let the no. of girls admitted =x

∴ Total number of girls =210+x

Now according to the condition,

360: 210+x=3: 2 

$\Rightarrow \frac{360}{210+x}=\frac{3}{2}$

⇒ 630+3 x=720

⇒ 3 x=720-630=90

$\therefore x=\frac{90}{3}=30$

∴No of girls to be admitted =30

Question 20

(i) The monthly pocket money of Ravi and Sanjeev are in the ratio 5 : 7. Their expenditures are in the ratio 3 : 5. If each saves Rs 80 every month, find their monthly pocket money.

(ii) In class X of a school, the ratio of the number of boys to that of the girls is 4 : 3. If there were 20 more boys and 12 less girls, then the ratio would have been 2 : 1, How many students were there in the class?

Sol :

(i) Let the monthly pocket money of Ravi and Sanjeev be 5x and 7x respectively.

Also, let their expenditure be 3y and 5y respectively.

So, 5 x-3 y=80..(i)

and 7 x-5 y=800..(ii)

Multiplying (i) by 7 and (ii) by 5 and subtracting, we get

$\begin{array}{r}
35 x-21 y=560 \\
35 x-25 y=400 \\-\phantom{35 x}+\phantom{25 y}-\phantom{400 } \\
\hline 4 y=160
\end{array}$

From (i), 5 x=80+3×40=200

⇒ x=40

So, monthly pocket money of Ravi

=5×40=200

(ii) Let the number of students in the class =x

Ratio of boys and girls =4 : 3

∴No. of boys $=\frac{4 x}{7}$

and no. of girls $=\frac{3 x}{7}$

According to the problem,

$\left(\frac{4 x}{7}+20\right):\left(\frac{3 x}{7}-12\right)=2: 1$

$\frac{4 x+140}{7}: \frac{3 x-84}{7}: 2: 1$

$\Rightarrow \frac{4 x+140}{7} \times \frac{7}{3 x-84}=\frac{2}{1}$

$\Rightarrow \frac{4 x+140}{3 x-84}=\frac{2}{1}$

$\Rightarrow 6 x-168=4 x+140$

$\Rightarrow 6 x-4 x=140+168$

$\Rightarrow 2 x=308$

$ \Rightarrow   x=\frac{308}{2}=154$

Hence, number of students =154

Question 21

In an examination, the ratio of passes to failures was 4 : 1. If 30 less had appeared and 20 less passed, the ratio of passes to failures would have been 5 : 1. How many students appeared for the examination

Sol :

Let the number of passes = 4x

and number of failures = x

The total number of students appeared = 4x + x = 5x

In the second case, the number of students appeared = 5x – 30

and number of passes = 4x – 20

∴No. of failures=(5x-30)-(4x-20)

=5x-30-4x+20

=x-10

According to the condition

$\frac{4 x-20}{x-10}=\frac{5}{1}$

$\Rightarrow 5 x-50=4 x-20$

$\Rightarrow 5 x-4 x=-20+50$

$\Rightarrow x=30$

∴Number of students appeared=5x

=5×30=150