RS Aggarwal 2019,2020 solution class 7 chapter 8 Ratio and Proportion Test Paper 8

Test Paper 8

page-131

Question 1:

Compare 4 : 5 and 7 : 9.

Answer 1:

The given fractions are $\frac{4}{5} \mathrm{and} \frac{7}{9}$.
LCM of 5 and 9 = 5 $\times$ 9 = 45

$$\begin{gathered} \mathrm{Now}, \mathrm{we} \mathrm{have}: \\ \frac{4\times 9}{5\times 9}=\frac{36}{45} \mathrm{and} \frac{7\times 5}{9\times 5}= \frac{35}{45} \\ \mathrm{Clearly}, \frac{35}{45}<\frac{36}{45} \\ \mathrm{Hence}, (7 : 9 ) <(4 : 5) \end{gathered}$$

Question 2:

Divide Rs 1100 among A, B and C in the ratio 2 : 3 : 5.

Answer 2:

The sum of ratio terms is 10.

Then, we have:

A's share = Rs $\frac{2}{10}\times 1100 = \mathrm{Rs} 220$

$\mathrm{B}\text{'}\mathrm{s} \mathrm{share} =\mathrm{Rs} \frac{3}{10}\times 1100 =\mathrm{Rs} 330$

$\mathrm{C}\text{'}\mathrm{s} \mathrm{share} =\mathrm{Rs} \frac{5}{10}\times 1100 = \mathrm{Rs} 550$

Question 3:

Show that the numbers 25, 36, 5, 6 are not in proportion.

Answer 3:

Product of the extremes = 25 $\times$ 6 = 150
Product of the means = 36 $\times$ 5 = 180

The product of the extremes is not equal to that of the means.

Hence, 25, 36, 5 and 6 are not in proportion.

Question 4:

If x,18,108 are in continued proportion, find the value of x.

Answer 4:

x : 18 :: 18 : 108

 $$\begin{gathered} \Rightarrow x \times 108 = 18\times 18 (\mathrm{Product} \mathrm{of} \mathrm{extremes} =\mathrm{Product} \mathrm{of} \mathrm{means}) \\ \Rightarrow 108x = 324 \\ \Rightarrow x =3 \\ \mathrm{Hence}, \mathrm{the} \mathrm{value} \mathrm{of} x \mathrm{is} 3. \end{gathered}$$

Question 5:

Two numbers are in the ratio 5 : 7. If the sum of these numbers is 84, find the numbers.

Answer 5:

Suppose that the numbers are 5x and 7x.
Then, 5x + 7x = 84
⇒ 12x = 84
⇒ x = 7

Hence, the numbers are (5 $\times$ 7 =) 35 and (7 $\times$ 7 =) 49.

Question 6:

The ages of A and B are in the ratio 4 : 3. Eight years ago, their ages were in the ratio 10 : 7. Find their present ages.

Answer 6:

Suppose that the present ages of A and B are 4x yrs and 3x yrs, respectively.
Eight years ago, age of A = (4x − 8) yrs
Eight years ago, age of B = (3x − 8) yrs
Then, (4x − 8) : (3x − 8) = 10 : 7

$$\begin{gathered} \Rightarrow \frac{4x-8}{3x - 8}=\frac{10}{7} \\ \Rightarrow 28x - 56 = 30x - 80 \\ \Rightarrow 2x = 24 \\ \Rightarrow x= 12 \\ \mathrm{Hence}, \mathrm{present} \mathrm{age} \mathrm{of} \mathrm{A} = 4\times 12 \mathrm{yrs} = 48 \mathrm{yrs} \\ \mathrm{and} \mathrm{present} \mathrm{age} \mathrm{of} \mathrm{B} = 3\times 12 \mathrm{yrs} = 36 \mathrm{yrs} \end{gathered}$$

Question 7:

If a car covers 54 km in an hour, how much distance will it cover in 40 minutes?

Answer 7:

Distance covered in 60 min = 54 km
Distance covered in 1 min = $\frac{54}{60} \mathrm{km}$

∴ Distance covered in 40 min = $\frac{54}{60}\times 40 = 36 \mathrm{km}$

Question 8:

Find the third proportional to 8 and 12.

Answer 8:

Suppose that the third proportional to 8 and 12 is x.
Then, 8 : 12 :: 12 : x

⇒ 8x = 144  (Product of extremes = Product of means)
⇒ x = 18

Hence, the third proportional is 18 .

Question 9:

If 40 men can finish a piece of work in 60 days, in how many days will 75 men finish the same work?

Answer 9:

40 men can finish the work in 60 days.
1 man can finish the work in 60 $\times$ 40 days.     [Less men, more days]
75 men can finish the work in $\frac{60\times 40}{75} = 32 \mathrm{days}$

Hence, 75 men will finish the same work in 32 days.

Question 10:

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

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

Answer 10:

(d)  6 : 4 : 3

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

Question 11:

Mark (✓) against the correct answer
$\mathrm{If}\frac{A}{2}=\frac{B}{3}=\frac{C}{4}$ then A : B : C = ?
(a) 2 : 3 : 4
(b) 4 : 3 : 2
(c) 3 : 2 : 4
(d) none of these

Answer 11:

(a)  2 : 3 : 4


$$\begin{gathered} A = \frac{2}{3}B \\ C = \frac{4}{3}B \\ \therefore A : B : C = \frac{2}{3}B : B : \frac{4}{3} B \\ = 2 : 3 : 4 \end{gathered}$$

Question 12:

Mark (✓) against the correct answer
If (x : y) = 3 : 4, then (7x + 3y) : (7x − 3y) = ?

(a) 7 : 3
(b) 5 : 2
(c) 11 : 3
(d) 14 : 9

Answer 12:

(c) 11 : 3

$$\begin{gathered} \mathrm{We} \mathrm{have} x = \frac{3}{4}y \\ \mathrm{Now}, \frac{7x + 3y}{7x - 3y} = \frac{7\times {\displaystyle \frac{3}{4}}y +3y}{7\times {\displaystyle \frac{3}{4}}y - 3y} = \frac{21y + 12y}{21y -12y} \\ =\frac{33y}{9y}=\frac{11}{3} \end{gathered}$$

Question 13:

Mark (✓) against the correct answer
What least number must 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 13:

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

$$\begin{gathered} \Rightarrow \frac{15 - x}{19 - x} =\frac{3}{4} \\ \Rightarrow 60 - 4x = 57 - 3x\Rightarrow x = 3 \end{gathered}$$

Question 14:

Mark (✓) against the correct answer
If Rs 840 is divided between A and B in the ratio 4 : 3, then B's share is

(a) Rs 480
(b) Rs 360
(c) Rs 320
(d) Rs 540

Answer 14:

(b) 360

Sum of the ratio terms = 4 + 3 = 7

∴ B's share = $\mathrm{Rs} 840 \times \frac{3}{7}$ = Rs 360

Question 15:

Mark (✓) against the correct answer
The ages of A and B are in the ratio 5 : 2. After 5 years, their ages will be in the ratio 15 : 7. The present age of A is

(a) 48 years
(b) 36 years
(c) 40 years
(d) 35 years

Answer 15:

(c) 40 years

Suppose that the present ages of A and B are 5x yrs and 2x yrs, respectively.
After 5 years, the ages of A and B will be (5x+5) yrs and (2x+5) yrs, respectively.

Then, (5x + 5) : (2x + 5) = 15 : 7

⇒ $\frac{5x+5}{2x +5}=\frac{15}{7}$

Cross multiplying, we get:

35x + 35 = 30x + 75
⇒ 5x = 40
⇒ x = 8

Hence, the present age of A is 5 $\times$ 8 = 40 yrs.

Question 16:

Mark (✓) against the correct answer
The boys and girls in a school are in the ratio 9 : 5. If the number of girls is 320, then the total strengh of the school is

(a) 840
(b) 896
(c) 920
(d) 576

Answer 16:

(b)  896

Suppose that the number of boys in the school is x.
Then, x : 320 = 9 : 5
⇒ 5x = 2880
⇒ x = 576

Hence, total strength of the school = 576 + 320 = 896

Question 17:

Fill in the blanks.

(i) If A : B = 2 : 3 and B : C = 4 : 5, then C : A = ...... .
(ii) If 16% of A = 20% of B, then A : B = ...... .
(iii) If $A=\frac{1}{3}B and B=\frac{1}{2}C,$ then A : B : C = ...... .
(iv) If A : B = 5 : 7 and B : C = 6 : 11, then A : B : C = ...... .

Answer 17:

(i) 15 : 8

$\frac{A}{C } = \frac{A}{B}\times \frac{B}{C} = \frac{2}{3}\times \frac{4}{5}= \frac{8}{15}$

∴ C : A=15 : 8

(ii) 5 : 4

$$\begin{gathered} \frac{16}{100}A = \frac{20}{100}B \\ \Rightarrow \frac{A}{B}= \frac{20}{16}= \frac{5}{4} \end{gathered}$$

(iii) 1 : 3 : 6

$A : B : C = \frac{1}{3} B : B : 2B = 1 : 3 : 6$

(iv)  30 : 42 : 77

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

Question 18:

Write 'T' for true and 'F' for false

(i) Mean proportional between 0.4 and 0.9 is 6.
(ii) The third proportional to 9 and 12 is 10.5.
(iii) If 8 : x :: 48 : 18, then x = 3.
(iv) If (3a + 5b) : (3a − 5b) = 5 : 1, then a : b = 5 : 2

Answer 18:

(i) F
Suppose that the mean proportional is x.
Then, 0.4 : x :: x : 0.9
$\Rightarrow 0.9 \times 0.4 =x \times x (\mathrm{Product} \mathrm{of} \mathrm{extremes} =\mathrm{Product} \mathrm{of} \mathrm{means})$
$$\begin{gathered} \Rightarrow x^2 = 0.36 \\ \Rightarrow x = 0.6 \end{gathered}$$


(ii)  F
Suppose that the third proportional is x.
Then, 9 : 12 :: 12 : x
⇒ 9x = 144                      (Product of extremes = Product of means)
⇒ x = 16

(iii)  T
8 : x :: 48 : 18
⇒ 144 = 48x               (Product of extremes = Product of means)        
⇒ x = 3

(iv) T

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

⇒ a : b = 5 : 2