Ratio & Proportion — Exercise 7(A)

If A : B = 7 : 5 and B : C = 7 : 5, then A : C is :

1. 25 : 49

2. 49 : 25

3. 1

4. 15 : 14

Answer

Given,

⇒ A : B = 7 : 5

$\Rightarrow \dfrac{A}{B} = \dfrac{7}{5}$ ........(1)

⇒ B : C = 7 : 5

$\Rightarrow \dfrac{B}{C} = \dfrac{7}{5}$ ......(2)

Multiplying equation (1) and (2), we get :

$\Rightarrow \dfrac{A}{B} \times \dfrac{B}{C} = \dfrac{7}{5} \times \dfrac{7}{5} \\[1em] \Rightarrow \dfrac{A}{C} = \dfrac{49}{25}.$

A : C = 49 : 25.

Hence, Option 2 is the correct option.

If $\dfrac{x^2 - 1}{x^2 + 1} =\dfrac{3}{5}$, the value of x is :

1. $\dfrac{1}{2}$

2. 2

3. $\pm \dfrac{1}{2}$

4. $\pm 2$

Answer

Given,

$\Rightarrow \dfrac{x^2 - 1}{x^2 + 1} =\dfrac{3}{5} \\[1em] \Rightarrow 5(x^2 - 1) = 3(x^2 + 1) \\[1em] \Rightarrow 5x^2 - 5 = 3x^2 + 3 \\[1em] \Rightarrow 5x^2 - 3x^2 = 3 + 5 \\[1em] \Rightarrow 2x^2 = 8 \\[1em] \Rightarrow x^2 = 4 \\[1em] \Rightarrow x = \sqrt{4} \\[1em] \Rightarrow x = \pm 2.$

Hence, Option 4 is the correct option.

If (2x + 3y) : (3x + 2y) = 4 : 3; then x : y is :

1. 6 : 1

2. 2 : 3

3. 1 : 6

4. 3 : 2

Answer

Given,

$\Rightarrow (2x + 3y) : (3x + 2y) = 4 : 3 \\[1em] \Rightarrow \dfrac{2x + 3y}{3x + 2y} = \dfrac{4}{3} \\[1em] \Rightarrow 3(2x + 3y) = 4(3x + 2y) \\[1em] \Rightarrow 6x + 9y = 12x + 8y \\[1em] \Rightarrow 12x - 6x = 9y - 8y \\[1em] \Rightarrow 6x = y \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{1}{6} \\[1em] \Rightarrow x : y = 1 : 6.$

Hence, Option 3 is the correct option.

Two numbers are in the ratio 5 : 8. If 10 is subtracted from each number, the ratio becomes 4 : 7. The numbers are :

1. 80 and 40

2. 50 and 80

3. 25 and 40

4. 40 and 25

Answer

Let two numbers be x and y.

Given,

Ratio of two numbers = 5 : 8.

$\Rightarrow \dfrac{x}{y} = \dfrac{5}{8} \\[1em] \Rightarrow x = \dfrac{5}{8}y \space .....(1)$

Given,

If 10 is subtracted from each number, the ratio becomes 4 : 7.

$\Rightarrow \dfrac{x - 10}{y - 10} = \dfrac{4}{7} \\[1em] \Rightarrow 7(x - 10) = 4(y - 10) \\[1em] \Rightarrow 7x - 70 = 4y - 40$

Substituting value of x from equation (1), in above equation we get :

$\Rightarrow 7 \times \dfrac{5}{8}y - 70 = 4y - 40 \\[1em] \Rightarrow \dfrac{35y}{8} - 70 = 4y - 40 \\[1em] \Rightarrow \dfrac{35y}{8} - 4y = 70 - 40 \\[1em] \Rightarrow \dfrac{35y - 32y}{8} = 30 \\[1em] \Rightarrow \dfrac{3y}{8} = 30 \\[1em] \Rightarrow y = \dfrac{30 \times 8}{3} \\[1em] \Rightarrow y = 80. \\[1em] \Rightarrow x = \dfrac{5}{8}y \\[1em] \Rightarrow x = \dfrac{5}{8} \times 80 = 50.$

Numbers = 50 and 80.

Hence, Option 2 is the correct option.

The compounded ratio of x - y : x + y and (x + y)² : x² - y² is :

1. 2 : 1

2. 1 : (x + y)²

3. 1 : 1

4. 1 : 2

Answer

Compounded ratio of x - y : x + y and (x + y)² : x² - y² :

$\Rightarrow \dfrac{x - y}{x + y} \times \dfrac{(x + y)^2}{x^2 - y^2} \\[1em] \Rightarrow \dfrac{x - y}{x + y} \times \dfrac{(x + y)^2}{(x + y)(x - y)} \\[1em] \Rightarrow \dfrac{(x - y)(x + y)^2}{(x + y)^2(x - y)} \\[1em] \Rightarrow \dfrac{1}{1} \\[1em] \Rightarrow 1 : 1.$

Hence, Option 3 is the correct option.

The sub-duplicate ratio of 6x² : 24y² is :

1. x : 2y

2. 2x : y

3. x : y

4. y : x

Answer

Sub-duplicate ratio of 6x² : 24y² is :

$\Rightarrow \dfrac{\sqrt{6x^2}}{\sqrt{24y^2}} \\[1em] \Rightarrow \dfrac{\sqrt{6}x}{2\sqrt{6}y} \\[1em] \Rightarrow \dfrac{x}{2y} \\[1em] \Rightarrow x : 2y.$

Hence, Option 1 is the correct option.

If $\dfrac{m + n}{m + 3n} = \dfrac{2}{3}$, find : $\dfrac{2n^2}{3m^2 + mn}$.

Answer

Given,

$\phantom{\Rightarrow} \dfrac{m + n}{m + 3n} = \dfrac{2}{3} \\[1em] \Rightarrow 3(m + n) = 2(m + 3n) \\[1em] \Rightarrow 3m + 3n = 2m + 6n \\[1em] \Rightarrow 3m - 2m = 6n - 3n \\[1em] \Rightarrow m = 3n.$

Substituting value of m in $\dfrac{2n^2}{3m^2 + mn}$ we get,

$\Rightarrow \dfrac{2n^2}{3(3n)^2 + (3n)n} \\[1em] \Rightarrow \dfrac{2n^2}{3(9n^2) + 3n^2} \\[1em] \Rightarrow \dfrac{2n^2}{27n^2 + 3n^2} \\[1em] \Rightarrow \dfrac{2n^2}{30n^2} \\[1em] \Rightarrow \dfrac{1}{15}.$

Hence, $\dfrac{2n^2}{3m^2 + mn} = \dfrac{1}{15}.$

If the ratio of 8 to 11 is same as the ratio of 2x - y to x + 2y, find the value of $\dfrac{7x}{9y}$.

Answer

According to question,

$\phantom{\Rightarrow} \dfrac{2x - y}{x + 2y} = \dfrac{8}{11} \\[1em] \Rightarrow 11(2x - y) = 8(x + 2y) \\[1em] \Rightarrow 22x - 11y = 8x + 16y \\[1em] 22x - 8x = 16y + 11y \\[1em] 14x = 27y \\[1em] x = \dfrac{27y}{14}.$

Substituting value of x in $\dfrac{7x}{9y}$ we get,

$\dfrac{7x}{9y} = \dfrac{7}{9y} \times \dfrac{27y}{14} \\[1em] = \dfrac{3}{2}.$

Hence, $\dfrac{7x}{9y} = \dfrac{3}{2}$.

Divide ₹1,290 into A, B and C such that A is $\dfrac{2}{5}$ of B and B : C is 4 : 3.

Answer

Given, B : C = 4 : 3

If B = 4a, then C = 3a

Given, A is $\dfrac{2}{5}$ of B.

∴ A = $\dfrac{2}{5} \times 4a = \dfrac{8a}{5}$

Share of A,

$= \dfrac{A}{A + B + C} \times 1290 \\[1em] = \dfrac{\dfrac{8a}{5}}{\dfrac{8a}{5} + 4a + 3a} \times 1290 \\[1em] = \dfrac{\dfrac{8a}{5}}{\dfrac{8a + 20a + 15a}{5}} \times 1290 \\[1em] = \dfrac{8a}{43a} \times 1290 \\[1em] = \dfrac{8}{43} \times 1290 \\[1em] = 240.$

Share of B,

$= \dfrac{B}{A + B + C} \times 1290 \\[1em] = \dfrac{4a}{\dfrac{8a}{5} + 4a + 3a} \times 1290 \\[1em] = \dfrac{4a}{\dfrac{8a + 20a + 15a}{5}} \times 1290 \\[1em] = \dfrac{20a}{43a} \times 1290 \\[1em] = \dfrac{20}{43} \times 1290 \\[1em] = 600.$

Share of C,

$= \dfrac{C}{A + B + C} \times 1290 \\[1em] = \dfrac{3a}{\dfrac{8a}{5} + 4a + 3a} \times 1290 \\[1em] = \dfrac{3a}{\dfrac{8a + 20a + 15a}{5}} \times 1290 \\[1em] = \dfrac{15a}{43a} \times 1290 \\[1em] = \dfrac{15}{43} \times 1290 \\[1em] = 450.$

Hence, the amount of money with A = ₹ 240, B = ₹ 600 and C = ₹ 450.

A school has 630 students. The ratio of the number of boys to the number of girls is 3 : 2. This ratio changes to 7 : 5 after the admission of 90 new students. Find the number of newly admitted boys.

Answer

Ratio of boys to girls = 3 : 2.

No. of boys = $\dfrac{3}{3 + 2} \times 630 = \dfrac{3}{5} \times 630 = 378.$

No. of girls = $\dfrac{2}{3 + 2} \times 630 = \dfrac{2}{5} \times 630 = 252.$

Let no. of newly admitted boys be x and so girls = 90 - x.

According to question on admission of 90 new students ratio changes to 7 : 5.

$\therefore \dfrac{378 + x}{252 + 90 - x} = \dfrac{7}{5} \\[1em] \Rightarrow \dfrac{378 + x}{342 - x} = \dfrac{7}{5} \\[1em] \Rightarrow 5(378 + x) = 7(342 - x) \\[1em] \Rightarrow 5(378 + x) = 7(342 - x) \\[1em] \Rightarrow 1890 + 5x = 2394 - 7x \\[1em] \Rightarrow 5x + 7x = 2394 - 1890 \\[1em] \Rightarrow 12x = 504 \\[1em] \Rightarrow x = \dfrac{504}{12} \\[1em] \Rightarrow x = 42.$

Hence, there are 42 newly admitted boys.

What quantity must be subtracted from each term of ratio 9 : 17 to make it equal to 1 : 3 ?

Answer

Let number to be subtracted be x,

$\therefore \dfrac{9 - x}{17 - x} = \dfrac{1}{3} \\[1em] \Rightarrow 3(9 - x) = 1(17 - x) \\[1em] \Rightarrow 27 - 3x = 17 - x \\[1em] \Rightarrow 3x - x = 27 - 17 \\[1em] \Rightarrow 2x = 10 \\[1em] \Rightarrow x = 5.$

Hence, quantity to be subtracted = 5.

The work done by (3x - 1) men in (2x + 3) days and the work done by (3x - 4) men in (2x + 1) days are in the ratio 4 : 3. Find the value of x.

Answer

Amount of work done by (3x - 1) men in (2x + 3) days = (3x - 1)(2x + 3),

Amount of work done by (3x - 4) men in (2x + 1) days = (3x - 4)(2x + 1)

According to question,

$\Rightarrow \dfrac{(3x - 1)(2x + 3)}{(3x - 4)(2x + 1)} = \dfrac{4}{3}$

3(3x − 1)(2x + 3) = 4(3x - 4)(2x + 1)

3(6x² + 9x − 2x − 3) = 4(6x² + 3x − 8x − 4)

3(6x² + 7x − 3) = 4(6x² − 5x − 4)

18x² + 21x − 9 = 24x² − 20x − 16

18x² − 24x² + 21x + 20x − 9 + 16 = 0

−6x² + 41x + 7 = 0

6x² − 41x − 7 = 0

6x² - 42x + x - 7 = 0

6x(x − 7) + 1(x − 7) = 0

(6x + 1)(x − 7) = 0

6x + 1 = 0 or x - 7 = 0

x = −$\dfrac{1}{6}$ or x = 7

x ≠ −$\dfrac{1}{6}$ as that will make number of men negative which is not possible.

Hence, value of x = 7.

If 3A = 4B = 6C, find : A : B : C.

Answer

3A = 4B

$\dfrac{A}{B} = \dfrac{4}{3}$

⇒ A : B = 4 : 3

4B = 6C

$\dfrac{B}{C} = \dfrac{6}{4} = \dfrac{3}{2}$

⇒ B : C = 3 : 2

∴ A : B : C = 4 : 3 : 2.

Hence, A : B : C = 4 : 3 : 2.

If 2a = 3b and 4b = 5c, find : a : c.

Answer

2a = 3b

$\dfrac{a}{b} = \dfrac{3}{2}$

⇒ a : b = 3 : 2

4b = 5c

$\dfrac{b}{c} = \dfrac{5}{4}$

⇒ b : c = 5 : 4

Multiplying a : b by 5 and b : c by 2 we get,

a : b = 15 : 10

b : c = 10 : 8

⇒ a : b : c = 15 : 10 : 8

Hence, a : c = 15 : 8.

Find the compound ratio of 2 : 3, 9 : 14 and 14 : 27.

Answer

Compound ratio of 2 : 3, 9 : 14 and 14 : 27,

$= \dfrac{2}{3} \times \dfrac{9}{14} \times \dfrac{14}{27} \\[1em] = \dfrac{252}{1134} \\[1em] = \dfrac{2}{9} = 2 : 9.$

Hence, compound ratio of 2 : 3, 9 : 14 and 14 : 27 is 2 : 9.

Find the compound ratio of 2a : 3b, mn : x² and x : n

Answer

Compound ratio of 2a : 3b, mn : x² and x : n,

$= \dfrac{2a}{3b} \times \dfrac{mn}{x^2} \times \dfrac{x}{n} \\[1em] = \dfrac{2amnx}{3bx^2n} \\[1em] = \dfrac{2am}{3bx} = 2am : 3bx.$

Hence, compound ratio of 2a : 3b, mn : x² and x : n is 2am : 3bx.

If (x + 3) : (4x + 1) is the duplicate ratio of 3 : 5, find the value of x.

Answer

According to question,

$\Rightarrow \dfrac{x + 3}{4x + 1} = \dfrac{3 \times 3}{5 \times 5} \\[1em] \Rightarrow \dfrac{x + 3}{4x + 1} = \dfrac{9}{25} \\[1em] \Rightarrow 25(x + 3) = 9(4x + 1) \\[1em] \Rightarrow 25x + 75 = 36x + 9 \\[1em] \Rightarrow 36x - 25x = 75 - 9 \\[1em] \Rightarrow 11x = 66 \\[1em] \Rightarrow x = 6.$

Hence, x = 6.

If m : n is the duplicate ratio of m + x : n + x; show that : x² = mn.

Answer

According to question,

$\Rightarrow \dfrac{m}{n} = \dfrac{(m + x)^2}{(n + x)^2} \\[1em] \Rightarrow \dfrac{m}{n} = \dfrac{m^2 + x^2 + 2mx}{n^2 + x^2 + 2nx} \\[1em] \Rightarrow m(n^2 + x^2 + 2nx) = n(m^2 + x^2 + 2mx) \\[1em] \Rightarrow mn^2 + mx^2 + 2mnx = nm^2 + nx^2 + 2mnx \\[1em] \Rightarrow mx^2 - nx^2 = nm^2 - mn^2 + 2mnx - 2mnx \\[1em] \Rightarrow x^2(m - n) = mn(m - n) \\[1em] \Rightarrow x^2 = mn.$

Hence, proved that x² = mn.

If (3x - 9) : (5x + 4) is the triplicate ratio of 3 : 4, find the value of x.

Answer

According to question,

$\Rightarrow \dfrac{3x - 9}{5x + 4} = \dfrac{3^3}{4^3} \\[1em] \Rightarrow \dfrac{3x - 9}{5x + 4} = \dfrac{27}{64} \\[1em] \Rightarrow 64(3x - 9) = 27(5x + 4) \\[1em] \Rightarrow 192x - 576 = 135x + 108 \\[1em] \Rightarrow 192x - 135x = 108 + 576 \\[1em] \Rightarrow 57x = 684 \\[1em] \Rightarrow x = 12.$

Hence, x = 12.