RS Aggarwal 2019,2020 solution class 7 chapter 8 Ratio and Proportion Exercise 8C

Exercise 8C

Page-128

Question 1:

Mark (✓) against the correct answer
If a : b = 3 : 4 and b : c = 8 : 9, then a : c = ?

(a) 1 : 2
(b) 3 : 2
(c) 1 : 3
(d) 2 : 3

Answer 1:

The correct option is (d).


$$\begin{gathered} \frac{a}{c}= \frac{a}{b}\times \frac{b}{c} = \frac{3}{4}\times \frac{8}{9} \\ = \frac{2}{3} \end{gathered}$$

Hence, a : c = 2 : 3

Question 2:

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If A : B = 2 : 3 and B : C = 4 : 5, then C : A = ?

(a) 15 : 8
(b) 6 : 5
(c) 8 : 5
(d) 8 : 15

Answer 2:

(a) 15 : 8

$$\begin{gathered} \frac{A}{B}= \frac{2}{3} \\ \frac{B}{C}= \frac{4}{5} \\ \mathrm{Then}, \frac{A}{B}\times \frac{B}{C } = \frac{2}{3}\times \frac{4}{5}= \frac{8}{15} \\ \mathrm{Hence}, C : A = 15 : 8 \end{gathered}$$

Question 3:

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If 2A = 3B and 4B = 5C, then A : C = ?

(a) 4 : 3
(b) 8 : 15
(c) 3 : 4
(d) 15 : 8

Answer 3:

The correct option is (d).


$$\begin{gathered} A = \frac{3B}{2} \\ C = \frac{4B}{5} \\ \therefore A : C = \frac{A}{C} = \frac{{\displaystyle \frac{3B}{2}}}{{\displaystyle \frac{4B}{5}}}= \frac{15}{8} \end{gathered}$$
Hence, A : C = 15 : 8

Question 4:

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If 15% of A = 20% of B, then A : B = ?

(a) 3 : 4
(b) 4 : 3
(c) 17 : 16
(d) 16 : 17

Answer 4:

The correct option is (b).

$$\begin{gathered} \frac{15}{100}A =\frac{20}{100}B \\ \Rightarrow \frac{A}{B }=\frac{4}{3} \end{gathered}$$

Hence, A : B = 4 : 3

Question 5:

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If $A=\frac{1}{3}B and B=\frac{1}{2}C,$ then A : B : C = ?

(a) 1 : 3 : 6
(b) 2 : 3 : 6
(c) 3 : 2 : 6
(d) 3 : 1 : 2

Answer 5:

(a)  1 : 3 : 6

$$\begin{gathered} A = \frac{1}{3}B \\ C = 2B \\ \therefore A : B : C = \frac{1}{3}B : B : 2B = 1 : 3 : 6 \end{gathered}$$

Page-129

Question 6:

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If A : B = 5 : 7 and B : C = 6 : 11, then A : B : C = ?

(a) 30 : 42 : 55
(b) 30 : 42 : 77
(c) 35 : 49 : 66
(d) none of these

Answer 6:

(b)  30 : 42 : 77


$$\begin{gathered} \frac{A}{B} = \frac{5}{7} \\ \Rightarrow A = \frac{5B}{7}\frac{B}{C} = \frac{6}{11}\Rightarrow C =\frac{11B}{6} \\ \therefore A : B : C = \frac{5B}{ 7} : B : \frac{11B}{6} = 30 : 42 : 77 \end{gathered}$$

Question 7:

Mark (✓) against the correct answer
If 2A = 3B = 4C, then A : B : C = ?

(a) 2 : 3 : 4
(b) 4 : 3 : 2
(c) 6 : 4 : 3
(d) 3 : 4 : 6

Answer 7:

(c)  6 : 4 : 3

$$\begin{gathered} 2A=3B = 4C \\ \mathrm{Then}, A = \frac{3B}{2} \mathrm{and} C = \frac{3B}{4} \\ \therefore A : B : C = \frac{3B}{2} : B : \frac{3B}{4} = 6 : 4 : 3 \end{gathered}$$

Question 8:

Mark (✓) against the correct answer
$\mathrm{If} \frac{A}{3}=\frac{B}{4}=\frac{C}{5},$ then A : B : C = ?

(a) 3 : 4 : 5
(b) 4 : 3 : 5
(c) 5 : 4 : 3
(d) 20 : 15 : 12

Answer 8:

(a) 3 : 4 : 5

$$\begin{gathered} A =\frac{3B}{4} \\ C = \frac{5B}{4} \\ \therefore A : B : C = \frac{3B}{4 } : B : \frac{5B}{4} \end{gathered}$$

                 = 3 : 4 : 5

Question 9:

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$\mathrm{If}\frac{1}{\mathrm{x}}:\frac{1}{\mathrm{y}}:\frac{1}{\mathrm{z}}=2:3:5,$ then, x : y : z = ?

(a) 2 : 3 : 5
(b) 15 : 10 : 6
(c) 5 : 3 : 2
(d) 6 : 10 : 15

Answer 9:

(b)  15 : 10 : 6

$$\begin{gathered} \frac{1}{x }:\frac{1}{y}=2 : 3 \\ \mathrm{Then}, y : x = 2 : 3 \mathrm{and} y = \frac{2}{3}x \\ \frac{1}{y}:\frac{1}{z }= 3 : 5 \\ \mathrm{Then}, z : y =3 : 5 \mathrm{and} z = \frac{3}{5}y \\ \therefore x : y : z = x : \frac{2}{3}x : \frac{3}{5}y = x : \frac{2}{3}x: \frac{3}{5}\times \frac{2}{3}x \\ =x : \frac{2}{3}x : \frac{2}{5}x = 15 : 10 : 6 \end{gathered}$$

Question 10:

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If x : y = 3 : 4, then (7x + 3y) : (7x − 3y) = ?

(a) 4 : 3
(b) 5 : 2
(c) 11 : 3
(d) 37 : 39

Answer 10:

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

$$\begin{gathered} x = \frac{3y}{4} \\ \therefore \frac{7x + 3y}{7x - 3y} = \frac{7{\displaystyle \frac{3y}{4}}+3y}{7{\displaystyle \frac{3y}{4}} - 3y} \\ =\frac{21y + 12y}{21y - 12y} = \frac{33y}{9y}= \frac{11}{3} \end{gathered}$$

Hence, (7x + 3y) : (7x − 3y) = 11 : 3

The correct option is (c).

Question 11:

Mark (✓) against the correct answer
If (3a + 5b) : (3a − 5b) = 5 : 1, then a : b = ?

(a) 2 : 1
(b) 3 : 2
(c) 5 : 2
(d) 5 : 3

Answer 11:

(c) 5 : 2

$$\begin{gathered} \frac{3a + 5b}{3a - 5b}=\frac{5}{1} \\ 3a + 5b = 15a - 25b \\ 12a = 30b \\ \frac{a}{b}= \frac{30}{12}=\frac{5}{2} \end{gathered}$$

∴ a : b = 5 : 2

Question 12:

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If 7 : x :: 35 : 45, then x = ?

(a) 11
(b) 15
(c) 9
(d) 5

Answer 12:

(c)  9

$$\begin{gathered} 7 \times 45 = x \times 35 (\mathrm{Product} \mathrm{of} \mathrm{extremes} = \mathrm{Product} \mathrm{of} \mathrm{means}) \\ \Rightarrow 35x = 315 \\ \Rightarrow x = 9 \end{gathered}$$

Question 13:

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What number has to be added to each term of 3 : 5 to make the ratio 5 : 6?

(a) 6
(b) 7
(c) 12
(d) 11

Answer 13:

(b) 7

Suppose that x is the number that is to be added.

Then, (3 + x) : (5 + x) = 5 : 6

$$\begin{gathered} \Rightarrow \frac{3 +x}{5 + x}= \frac{5}{6} \\ \Rightarrow 18 + 6x = 25+ 5x \\ \Rightarrow x =7 \end{gathered}$$

Question 14:

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Two numbers are in the ratio 3 : 5. If each number is increased by 10, the ratio becomes 5 : 7. The sum of the numbers is

(a) 8
(b) 16
(c) 35
(d) 40

Answer 14:

(d) 40

Suppose that the numbers are x and y.

Then, x : y = 3 : 5 and (x + 10) : (y + 10) = 5 : 7

$$\begin{gathered} \frac{x}{y}=\frac{3}{5} \\ x = \frac{3y}{5} \\ =>\frac{x + 10}{y + 10}=\frac{5}{7}=>7x+70 = 5y + 50=>7\times \frac{3y}{5} + 70 =5y + 50=>5y -\frac{21y}{5} = 20=>\frac{4y}{5}= 20=>y = 25\mathrm{Therefore}, x =\frac{3 \times 25}{5}= 15 \end{gathered}$$

Hence, sum of numbers = 15 + 25 = 40

Question 15:

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What least number is to be subtracted from each term of the ratio 15 : 19 to make the ratio 3 : 4?

(a) 3
(b) 5
(c) 6
(d) 9

Answer 15:

(a)  3
Suppose that x is the number that is to be subtracted.
Then, (15 − x) : (19 − x) = 3 : 4

$$\begin{gathered} \Rightarrow \frac{15 - x}{19- x}=\frac{3}{4} \\ \mathrm{Cross} \mathrm{multiplying}, \mathrm{we} \mathrm{get}: \\ 60 - 4x = 57 - 3x \\ \Rightarrow x = 3 \end{gathered}$$

Question 16:

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If Rs 420 is divided between A and B in the ratio 3 : 4, then A's share is

(a) Rs 180
(b) Rs 240
(c) Rs 270
(d) Rs 210

Answer 16:

(a)  Rs 180

A's share = $\frac{3}{7} \times 420 = 180$

Question 17:

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The boys and girls in a school are in the ratio 8 : 5. If the number of girls is 160, what is the total strength of the school?

(a) 250
(b) 260
(c) 356
(d) 416

Answer 17:

(d) 416

Let x be the number of boys.
Then, 8 : 5 = x : 160

$$\begin{gathered} \Rightarrow \frac{8}{5}= \frac{x}{160 } \\ \Rightarrow x=\frac{8\times 160}{5}= 256 \\ \therefore \mathrm{Total} \mathrm{strength} \mathrm{of} \mathrm{the} \mathrm{school} =256 + 160 = 416 \end{gathered}$$

Question 18:

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Which one is greater out of (2 : 3) and (4 : 7)?

(a) 2 : 3
(b) 4 : 7
(c) both are equal

Answer 18:

(a) (2 :3)

LCM of 3 and 7 = $7\times 3=$21

$$\begin{gathered} \frac{2\times 7}{3\times 7} = \frac{14}{21} \mathrm{and} \frac{4\times 3}{7\times 3}= \frac{12}{21} \\ \mathrm{Clearly}, \frac{12}{21}<\frac{14}{21} \\ \mathrm{H}\mathrm{ence}, (4 : 7) < (2 : 3) \end{gathered}$$

Question 19:

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The third proportional to 9 and 12 is

(a) 10.5
(b) 8
(c) 16
(d) 21

Answer 19:

(c) 16

Suppose that the third proportional is x.
Then, 9 : 12 :: 12 : x

$$\begin{gathered} \Rightarrow 9 \times x = 12 \times 12 (\mathrm{Product} \mathrm{of} \mathrm{extremes}= \mathrm{Product} \mathrm{of} \mathrm{means}) \\ \Rightarrow 9x = 144 \\ \Rightarrow x = 16 \end{gathered}$$                               

Question 20:

Mark (✓) against the correct answer
The mean proportional between 9 and 16 is

(a) 12.5
(b) 12
(c) 5
(d) none of these

Answer 20:

(b) 12

Suppose that the mean proportional is x.

Then, 9 : x :: x : 16

$$\begin{gathered} 9 \times 16 = x \times x (\mathrm{Product} \mathrm{of} \mathrm{extremes} =\mathrm{Product} \mathrm{of} \mathrm{means}) \\ \Rightarrow x^2 = 144 \\ \Rightarrow x = 12 \end{gathered}$$

Question 21:

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The ages of A and B are in the ratio 3 : 8. Six years hence, their ages will be in the ratio 4 : 9. The present age of A is

(a) 18 years
(b) 15 years
(c) 12 years
(d) 21 years

Answer 21:

(a)  18 years

Suppose that the present ages of A and B are 3x yrs and 8x yrs, respectively.
After six years, the age of A will be (3x+6) yrs and that of B will be (8x+6) yrs.
Then, (3x +6) : (8x + 6) = 4 : 9

$$\begin{gathered} \Rightarrow \frac{3x + 6}{8x+6}= \frac{4}{9} \\ \Rightarrow 27x + 54 = 32x + 24 \\ \Rightarrow 5x = 30 \\ \Rightarrow x = 6 \\ \mathrm{Hence}, \mathrm{the} \mathrm{present} \mathrm{ages} \mathrm{of} \mathrm{A} \mathrm{and} \mathrm{B} \mathrm{are} 18 \mathrm{yrs} \mathrm{and} 48 \mathrm{yrs}, \mathrm{respectively}. \end{gathered}$$