RD Sharma solution class 7 chapter 9 Ratio and Proportion Exercise 9.1

Exercise 9.1

Page-9.6

Question 1:

If x : y = 3 : 5, find the ratio 3x + 4y : 8x + 5y.

Answer 1:

It is given that
                        x : y = 3 : 5 ⇒ $\frac{x}{y} = \frac{3}{5}$
                                Now, 3x + 4y : 8x + 5y
                                        = $\frac{3x + 4y}{8x + 5y}$
                                        = $\frac{{\displaystyle \frac{3x + 4y}{y}}}{{\displaystyle \frac{8x + 5y}{y}}}$                        {dividing the numerator and the denominator by 'y'}
                                        = $\frac{3\left({\displaystyle \frac{x}{y}}\right) + 4}{8\left({\displaystyle \frac{x}{y}}\right) + 5}$ = $\frac{3\left({\displaystyle \frac{3}{5}}\right) + 4}{8\left({\displaystyle \frac{3}{5}}\right) + 5}$ = $\frac{{\displaystyle \frac{9}{5}}+ 4}{{\displaystyle \frac{24}{5}} + 5}$
                                        = $\frac{{\displaystyle \frac{9 + 20}{5}}}{{\displaystyle \frac{24 + 25}{5}}}$ = $\frac{{\displaystyle \frac{29}{5}}}{{\displaystyle \frac{49}{5}}}$ = $\frac{29}{49}$

Question 2:

If x : y = 8 : 9, find the ratio (7x − 4y) : 3x + 2y.

Answer 2:

It is given that
x : y = 8 : 9 ⇒ $\frac{x}{y}$  =   $\frac{8}{9}$
Now, 7x $-$ 4y : 3x + 2y
= $\frac{7x - 4y}{3x + 2y}$
= $\frac{{\displaystyle \frac{7x - 4y}{y}}}{{\displaystyle \frac{3x+ 2y}{y}}}$       {dividing the numerator and the denominator by 'y'}
= $\frac{7\left({\displaystyle \frac{x}{y}}\right) - 4}{3\left({\displaystyle \frac{x}{y}}\right) +2}$ = $\frac{7\left({\displaystyle \frac{8}{9}}\right) - 4}{3\left({\displaystyle \frac{8}{9}}\right) +2}$ = $\frac{{\displaystyle \frac{56}{9}} - 4}{{\displaystyle \frac{24}{9}} + 2}$
= $\frac{{\displaystyle \frac{56 - 36}{9}}}{{\displaystyle \frac{24 + 18}{9}}}$ = $\frac{20}{42}$ = $\frac{10}{21}$
Hence, 7x $-$ 4y : 3x + 2y = 10 : 21.

Question 3:

If two numbers are in the ratio 6 : 13 and their l.c.m. is 312, find the numbers.

Answer 3:

Let the two numbers be 'x' and 'y' such that x : y = 6 : 13 ⇒ $\frac{x}{y} = \frac{6}{13} .$         
We can assume that the HCF of 'x' and 'y' is a number 'k'.
So, x = 6k, and y = 13k.
Now, the product of any two numbers 'x' and 'y' is always equal to the product of their LCM and HCF
                                           ⇒  x$\times$y = 312 $\times$ k
                                           ⇒  6k $\times$ 13k  =  312 $\times$ k      
                                           ⇒  k = $\frac{312}{6\times 13} = 4$
                                           ⇒  k = 4                                 
                                           Thus, x = 6k = 6 $\times$4  = 24, and y = 13 $\times$ 4 = 52.
                                     

Question 4:

Two numbers are in the ratio 3 : 5. If 8 is added to each number, the ratio becomes 2 : 3. Find the numbers.

Answer 4:

Let the two numbers in ratio be x and y such that
                                                                 x : y = 3 : 5
                                                                  = $\frac{x}{y} = \frac{3}{5}$ ⇒ x = $\frac{3y}{5}$.     ------- (1)
Now, 8 is added to each number, which means
                                                                  = $\frac{x + 8}{y + 8}$ = $\frac{2}{3}$       
                                                                  =  $\frac{{\displaystyle \frac{3y}{5}} + 8}{y+ 8}$ = $\frac{2}{3}$  ------ From (1)
                                                                  = $\frac{{\displaystyle \frac{3y + 40}{5}}}{y + 8} = \frac{2}{3}$
             On cross-multiplying, we get     ⇒ 3(3y + 40) = 2 $\times$5(y + 8)
                                                             ⇒ 9y + 120 = 10y + 80
                                                             ⇒ 120 $-$ 80 = 10y $-$ 9y
                                                             ⇒ y = 40
                                                           x = $\frac{3y}{5}$ = $\frac{3 \times 40}{5}$  = 24
                          So, the numbers are 24 and 40.

Question 5:

What should be added to each term of the ratio 7 : 13 so that the ratio becomes 2 : 3

Answer 5:

Let the numbers that must be added to the ratio 7 : 13 be 'x'.
So, $\frac{7 + x}{13 + x}= \frac{2}{3}$
After cross-multiplication, we get
3(7 + x) = 2(13 + x)
21 + 3x = 26 + 2x
3x - 2x = 26 - 21
x = 5
Thus, 5 must be added to each term to make the ratio = 2 : 3.
                                        

Question 6:

Three numbers are in the ratio 2 : 3 : 5 and the sum of these numbers is 800. Find the numbers.

Answer 6:

We have
Sum of the terms of the ratio = 2 +3 + 5 = 10.
Sum of the numbers = 800.
Therefore, first number = $\left(\frac{2}{10}\times 800\right)$
                                       = 160
        or, Second number = $\left(\frac{3}{10}\times 800\right)$
                                       = 240
         or,  Third number = $\left(\frac{5}{10}\times 800\right)$
                                     = 400

Question 7:

The ages of two persons are in the ratio 5 : 7. Eighteen years ago their ages were in the ratio 8 : 13. Find their present ages.

Answer 7:

Let the present ages of the two persons be '5x' and '7x'  years.
Ratio of their present ages = 5 : 7.
Eighteen years ago, their ages were (5x $-$ 18) and (7x $-$ 18), respectively.
But eighteen years ago the ratio of their ages was 8 : 13.
So, $\frac{5x- 18}{7x - 18}$ = $\frac{8}{13}$
13(5x $-$ 18) =  8(7x $-$ 18)
65x $-$ 234 = 56x $-$ 144
65x $-$ 56x = 234 $-$ 144
9x = 90
x = $\frac{90}{9}$ = 10
So, their ages are 5x = 5$\times$10 = 50 years and 7x = 7 $\times$ 10 = 70 years.                                                       

Question 8:

Two numbers are in the ratio 7 : 11. If 7 is added to each of the numbers, the ratio becomes 2 : 3. Find the numbers.

Answer 8:

Let the two numbers be 'x' and 'y'.
Given that x : y = 7 : 11 = $\frac{x}{y} = \frac{7}{11}$ = x = $\frac{7y}{11}$     ------- (1)
    Now, 7 is added to each of the numbers, which means that
                                  $\frac{x + 7}{y + 7}= \frac{2}{3}$
                                $\frac{{\displaystyle \frac{7y}{11}} + 7}{y + 7}$ = $\frac{2}{3}$
                                $\frac{{\displaystyle \frac{7y + 77}{11}}}{y + 7}$ = $\frac{2}{3}$
                                  3 (7y + 77) = 2 $\times$ 11 (y + 7)
                                  21y + 231 = 22y + 154
                                  22y $-$ 21y = 231 $-$ 154
                                  Therefore, y = 77, and x = $\frac{7y}{11}$ = $\frac{7 \times 77}{11}$= 49.
         Thus, the two numbers are 49 and 77.

Question 9:

Two numbers are in the ratio 2 : 7. If the sum of the numbers is 810, find the numbers.

Answer 9:

We have
Sum of the terms of the ratio = 2 + 7 = 9.
Sum of the numbers = 810.
Therefore, first number = $\frac{2}{9}\times 810$  = 180
Second number = $\frac{7}{9}\times 810$  = 630    

Question 10:

Divide Rs 1350 between Ravish and Shikha in the ratio 2 : 3.

Answer 10:

We have
             Sum of the terms of the ratio = 2 + 3 = 5
              Therefore, Ravish's share = Rs $\left(\frac{2}{5}\times 1350\right)$ = Rs 540
                 Sikha's share = Rs $\left(\frac{3}{5}\times 1350\right)$ = Rs 810

Question 11:

Divide Rs 2000 among P, Q, R in the ratio 2 : 3 : 5.

Answer 11:

We have
                   Sum of the terms of the ratio = 2 +3 +5 = 10
                    Therefore, P's share =Rs $\left(\frac{2}{10}\times 2000\right)$ = Rs 400
                                    Q's share = Rs $\left(\frac{3}{10}\times 2000\right)$ = Rs 600
                                     R's share = Rs $\left(\frac{5}{10}\times 2000\right)$ = Rs 1000

Question 12:

The boys and the girls in a school are in the ratio 7 : 4. If total strength of the school be 550, find the number of boys and girls.

Answer 12:

We have the ratio boys : girls = 7 : 4.
So, let there be 7x boys and 4x girls. It is given that there are a total of 550 students in the school.
 Therefore, 7x + 4x = 550
                 11x = 550
                    x = $\frac{550}{11}$ = 50
Hence, the number of boys = 7x = 7$\times$ 50 = 350, and the number of girls = 4x = 4 $\times$ 50 = 200.

Question 13:

The ratio of monthly income to the savings of a family is 7 : 2. If the savings be of Rs 500, find the income and expenditure.

Answer 13:

We have the ratio of income : savings = 7 : 2.
 So, let the income be 7x and the savings be 2x. It is given that the savings are Rs 500.
   Therefore, 2x = 500
                    x = Rs $\frac{500}{2}$ = Rs 250
            Thus, the income = 7x = 7 $\times$ 250 = Rs 1750.
     Now, expenditure = Income $-$ savings = Rs 1750 $-$ Rs 500 = Rs 1250.
 Thus, the income = Rs 1750, and the expenditure = Rs 1250.

Question 14:

The sides of a triangle are in the ratio 1 : 2 : 3. If the perimeter is 36 cm, find its sides.

Answer 14:

We have the ratio of the sides of the triangle = 1 : 2 : 3.
 Now, let the sides of the triangle be x, 2x and 3x, respectively.
 Therefore, the perimeter = x + 2x + 3x = 36
                                    ⇒ 6x = 36
                                    ⇒ x = $\frac{36}{6}$ = 6
 Thus, the sides of the triangle = x = 6 cm; 2x = 2$\times$6 = 12 cm; 3x = 3 $\times$6 = 18 cm.
    So, the sides of the triangle = 6 cm, 12 cm and 18 cm.

Page-9.7

Question 15:

A sum of Rs 5500 is to be divided between Raman and Aman in the ratio 2 : 3. How much will each get?

Answer 15:

We have
Sum of the terms of the ratio = 2 + 3 = 5, and the total sum = Rs 5500
Therefore, Raman's share = $\left(\frac{2}{5}\times 5500\right)$ = Rs 2200
Aman's share = $\left(\frac{3}{5}\times 5500\right)$ = Rs 3300

Question 16:

The ratio of zinc and copper in an alloy is 7 : 9. If the weight of the copper in the alloy is 11.7 kg, find the weight of the zinc in the alloy.

Answer 16:

We have
Weight of zinc : weight of copper = 7 : 9
So, let the weight of zinc in the alloy be '7x' kg and the weight of copper in the alloy be '9x' kg.
But the weight of copper in the alloy is given to be 11.7 kg.
Therefore, 9x = 11.7
                  x = $\frac{11.7}{9}$ = 1.3
 Hence, the weight of zinc in the alloy = 7x = 7$\times$1.3 = 9.1 kg.

Question 17:

In the ratio 7 : 8, if the consequent is 40, what is the antecedent?

Answer 17:

In a ratio a : b, 'a' is known as the antecedent and 'b' is known as the consequent.
In the given ratio, let the antecedent be 7x and the consequent be 8x, respectively,
But consequent = 8x = 40
                              x = $\frac{40}{8}$ = 5
Therefore, the antecedent = 7x = 7$\times$5 = 35.

Question 18:

Divide Rs 351 into two parts such that one may be to the other as 2 : 7.

Answer 18:

We have
            Sum of the ratio of the terms = 2 +7 = 9
           Therefore, first part = Rs. $\left(\frac{2}{9}\times 351\right) =$Rs. 78
         Similarly, second part = Rs. $\left(\frac{7}{9}\times 351\right) =$Rs. 273

Question 19:

Find the ratio of the price of pencil to that of ball pen, if pencils cost Rs 16 per score and ball pens cost Rs 8.40 per dozen.

Answer 19:

We have
 Cost of 1 score of pencils = Rs. 16
 Since 1 score = 20 items,
 Cost of one pencil = Rs. $\left({\displaystyle \frac{16}{20}}\right)$ = Rs. 0.8
 Cost of 1 dozen ball pens = Rs. 8.40
 Since 1 dozen =12 items,
 Cost of one ball pen = Rs. $\left({\displaystyle \frac{8.40}{12}}\right)$ = Rs. 0.7
 So, price of pencil : price of ball pen = 0.8 : 0.7 = $\frac{0.8}{0.7}$ = $\frac{8}{7}$
       Price of pencil : price of ball pen = 8 : 7

Question 20:

In a class, one out of every six students fails. If there are 42 students in the class, how many pass?

Answer 20:

We have
One out of every six student fails, which means that $\frac{1}{6}$th of the total students fail in the class.
And total number of students in the class = 42.
Therefore, the number of students who fail = $\left(\frac{1}{6}\times 42\right)$ = 7.
So, the number of students who pass = (Total students $-$the number of students who fail) = 42 $-$ 7 = 35.