RD Sharma 2014 Solutions for Class 7 Math Chapter 9 Ratio And Proportion are provided here with simple step-by-step explanations. These solutions for Ratio And Proportion are extremely popular among class 7 students for Math Ratio And Proportion Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma 2014 Book of class 7 Math Chapter 9 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma 2014 Solutions. All RD Sharma 2014 Solutions for class 7 Math are prepared by experts and are 100% accurate.

Page No 9.10:

Question 1:

Which ratio is larger in the following pairs?
(i) 3 : 4 or 9 : 16
(ii) 15 : 16 or 24 : 25
(iii) 4 : 7 or 5 : 8
(iv) 9 : 20 or 8 : 13
(v) 1 : 2 or 13 : 27

Answer:

(i) Writing the ratios as fractions, we have
        3 : 4 = 34 and 9 : 16 = 916
     Now, LCM of 4 and 16 = 16.
     Making the denominator of each fraction = 16, we have
         34 = 3 × 44 × 4 = 1216 and the other fraction = 916
         Of 1216 and 916,  clearly 1216>916.
          Therefore, 34>916.
(ii) Writing the ratios as fractions, we have
        15 : 16 = 1516 and 24 : 25 = 2425
     Now, LCM of 16 and 25 = 400.
     Making the denominator of each fraction = 400, we have
         151615 × 2516 × 25= 375400 and the other fraction = 24 × 1625 × 16= 384400
        Clearly, 384 > 375. So, 384400 > 375400.
          Therefore, 2425>1516.
(iii) Writing the ratios as fractions, we have
        4 : 7 = 47 and 5 : 8 = 58
     Now, LCM of 7 and 8 = 56.
     Making the denominator of each fraction = 56, we have
            4× 87 × 8= 3256 and the other fraction = 5 × 78 × 7= 3556
        Clearly, 36 > 32. So, 3556 > 3256.
          Therefore, 58>47.
(iv) Writing the ratios as fractions, we have
       9 : 20 = 920 and 8 : 13 = 813
     Now, LCM of 20 and 13 = 260.
     Making the denominator of each fraction = 260, we have
          9× 1320 × 13= 117260 and the other fraction = 8 × 2013 × 20= 160260
        Clearly, 160 > 117. So, 160260 > 117260.
          Therefore, 813>920.
(v) Writing the ratios as fractions, we have
       1 : 2 = 12 and 13 : 27 = 1327
     Now, LCM of 2 and 27 = 54.
     Making the denominator of each fraction = 54, we have
          1× 272 × 27= 2754 and the other fraction = 13 × 227 × 2= 2654
        Clearly, 27 > 26. So, 2754 > 2654.
          Therefore, 12>1327.

Page No 9.10:

Question 2:

Give two equivalent ratios of 6 : 8.

Answer:

We have
              68 = 6÷28÷2 = 34
     Therefore, 3 : 4 is an equivalent ratio of 6 : 8.
               68 = 6×28×2 = 1216
 Hence, 3 : 4 and 12 : 16 are equivalent ratios of 6 : 8.

Page No 9.10:

Question 3:

Fill in the following blanks: 1220=   5=9   

Answer:

1220=  5= 9
 Let 1220= x 5= 9y.
Then, 1220= x 5 ⇒ 12 × 5 = 20x  ⇒ x = 12 × 520 = 3.
Also, 1220= 9 y ⇒ 12y = 20 × 9 ⇒ y = 20 × 912 = 15.
Therefore, 1220= 3 5= 915.



Page No 9.13:

Question 1:

Find which of the following are in proportion?
(i) 33, 44, 66, 88
(ii) 46, 69, 69, 46
(iii) 72, 84, 186, 217

Answer:

(i) We have
                 Product of extremes = 33 × 88 = 2904
                 Product of means  = 44 × 66 = 2904
   Therefore, the product of the extremes is equal to the product of the means.
    Hence, 33, 44, 66, 88 are in proportion.
(ii) We have
                 Product of extremes = 46 × 46 = 2116
                 Product of means  = 69 × 69 = 4761
   Therefore, the product of the extremes is not equal to the product of the means.
    Hence, 46, 69, 69, 46 are not in proportion.
(iii) We have
                 Product of extremes = 72 × 217 = 15624
                 Product of means  = 84 × 186 = 15624
   Therefore, the product of the extremes is equal to the product of the means.
    Hence, 72, 84, 186, 217 are in proportion.

Page No 9.13:

Question 2:

Find x in the following proportions:
(i) 16 : 18 = x : 96
(ii) x : 92 = 87 : 116

Answer:

(i) 16 : 18 = x : 96
    ⇒ 16, 18, x, and 96 are in proportion.
    ⇒ Product of extremes = Product of means
    ⇒  16 × 96 = 18 × x
    ⇒ x = 16 × 9618 = 2563
(ii) x : 92 = 87 : 116
      ⇒ x, 92, 87, and 116 are in proportion.
      ⇒ Product of extremes = Product of means
      ⇒  x ×  116 = 87 ×  92
      ⇒ x = 87 × 92116 = 69



Page No 9.14:

Question 3:

The ratio of the income to the expenditure of a family is 7 : 6. Find the savings if the income is Rs 1400.

Answer:

The ratio of the income of a family to its expenditure = 7 : 6.
Let us assume that the income and expenditure of the family are '7x' and '6x', respectively.
But the income = Rs. 1400.
Therefore, 7x = 1400
                  x = 14007 = 200
The expenditure = 6x = 6 × 200 = Rs. 1200.
Now, savings = Income - expenditure = Rs. (1400 - 1200) = Rs. 200.

Page No 9.14:

Question 4:

The scale of a map is 1 : 4000000. What is the actual distance between the two towns if they are 5 cm apart on the map?

Answer:

The scale of the map = 1 : 4000000.
This means that 1 unit of distance on the map is equal to 4000000 units of the actual distance.
So, let us assume that the actual distance between the towns = 'x' cm.
Now, it is given that
   1 : 4000000 = 5 : x
Hence, 1, 4000000, 5 and x are in proportion.
Therefore, product of extremes = product of means
        =  1 × x = 5 × 4000000
       = x = 5 ×40000001 = 20000000 cm
 Since 1 km = 1000 m =1000×1 m =1000×100 cm = 100000 cm (1 m =100 cm),
 x = 20000000100000 = 200 km
                  

Page No 9.14:

Question 5:

The ratio of income of a person to his savings is 10 : 1. If his savings of one year are Rs 6000, what is his income per month?

Answer:

Savings in one year = Rs. 6000
So, savings per month = 600012 = Rs. 500.
Let the income per month be Rs 'x'.
Then, x : 500 = 10 : 1.
So, x, 500, 10 and 1 are in proportion.
Product of extremes = Product of means
 x × 1 = 10 × 500
 x = 10 × 5001 = Rs. 5000

Page No 9.14:

Question 6:

An electric pole casts a shadow of length 20 metres at a time when a tree 6 metres high casts a shadow of length 8 metres. Find the height of the pole.

Answer:

Length of the shadow of the electric pole = 20 m
Length of the shadow of the tree = 8 m
Height of the tree = 6 m
Now, let us assume that the height of the pole is 'x' m.
Height of the electric pole : length of the shadow of the electric pole = Height of the tree : length of the shadow of the tree
 x : 20 = 6 : 8
Thus, x, 20, 6 and 8 are in proportion.
Product of extremes = Product of means
 = x × 8 = 20 × 6
 = x = 20 × 68 = 15 m



Page No 9.6:

Question 1:

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

Answer:

It is given that
                        x : y = 3 : 5 ⇒ xy = 35
                                Now, 3x + 4y : 8x + 5y
                                        = 3x + 4y8x + 5y
                                        = 3x + 4yy8x + 5yy                        {dividing the numerator and the denominator by 'y'}
                                        = 3xy + 48xy + 5 = 335 + 4835 + 5 = 95+ 4245 + 5
                                        = 9 + 20524 + 255 = 295495 = 2949

Page No 9.6:

Question 2:

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

Answer:

It is given that
x : y = 8 : 9 ⇒ xy  =   89
Now, 7x - 4y : 3x + 2y
= 7x - 4y3x + 2y
= 7x - 4yy3x+ 2yy       {dividing the numerator and the denominator by 'y'}
= 7xy - 43xy +2 = 789 - 4389 +2 = 569 - 4249 + 2
= 56 - 36924 + 189 = 2042 = 1021
Hence, 7x - 4y : 3x + 2y = 10 : 21.

Page No 9.6:

Question 3:

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

Answer:

Let the two numbers be 'x' and 'y' such that x : y = 6 : 13 ⇒ xy = 613 .         
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×y = 312 × k
                                           ⇒  6k × 13k  =  312 × k      
                                           ⇒  k = 3126×13 = 4
                                           ⇒  k = 4                                 
                                           Thus, x = 6k = 6 ×4  = 24, and y = 13 × 4 = 52.
                                     

Page No 9.6:

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:

Let the two numbers in ratio be x and y such that
                                                                 x : y = 3 : 5
                                                                  = xy  = 35 ⇒ x = 3y5.     ------- (1)
Now, 8 is added to each number, which means
                                                                  = x + 8y + 8 = 23       
                                                                  =  3y5 + 8y+ 823  ------ From (1)
                                                                  = 3y + 405y + 8 = 23
             On cross-multiplying, we get     ⇒ 3(3y + 40) = 2 ×5(y + 8)
                                                             ⇒ 9y + 120 = 10y + 80
                                                             ⇒ 120 - 80 = 10y - 9y
                                                             ⇒ y = 40
                                                           x = 3y5 = 3 × 405  = 24
                          So, the numbers are 24 and 40.

Page No 9.6:

Question 5:

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

Answer:

Let the numbers that must be added to the ratio 7 : 13 be 'x'.
So, 7 + x13 + x= 23
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.
                                        

Page No 9.6:

Question 6:

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

Answer:

We have
Sum of the terms of the ratio = 2 +3 + 5 = 10.
Sum of the numbers = 800.
Therefore, first number = 210× 800
                                       = 160
        or, Second number = 310× 800
                                       = 240
         or,  Third number = 510× 800
                                     = 400

Page No 9.6:

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:

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, 5x-  187x - 18 = 813
13(5x - 18) =  8(7x - 18)
65x - 234 = 56x - 144
65x - 56x = 234 - 144
9x = 90
x = 909 = 10
So, their ages are 5x = 5×10 = 50 years and 7x = 7 × 10 = 70 years.                                                       

Page No 9.6:

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:

Let the two numbers be 'x' and 'y'.
Given that x : y = 7 : 11 = xy = 711 = x = 7y11     ------- (1)
    Now, 7 is added to each of the numbers, which means that
                                  x + 7y + 7= 23
                                7y11 + 7y + 7 = 23
                                7y + 7711y + 7 = 23
                                  3 (7y + 77) = 2 × 11 (y + 7)
                                  21y + 231 = 22y + 154
                                  22y - 21y = 231 - 154
                                  Therefore, y = 77, and x = 7y11 = 7 × 7711= 49.
         Thus, the two numbers are 49 and 77.

Page No 9.6:

Question 9:

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

Answer:

We have
Sum of the terms of the ratio = 2 + 7 = 9.
Sum of the numbers = 810.
Therefore, first number = 29× 810  = 180
Second number = 79× 810  = 630    

Page No 9.6:

Question 10:

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

Answer:

We have
             Sum of the terms of the ratio = 2 + 3 = 5
              Therefore, Ravish's share = Rs 25×1350 = Rs 540
                 Sikha's share = Rs 35×1350 = Rs 810

Page No 9.6:

Question 11:

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

Answer:

We have
                   Sum of the terms of the ratio = 2 +3 +5 = 10
                    Therefore, P's share =Rs 210× 2000 = Rs 400
                                    Q's share = Rs 310× 2000 = Rs 600
                                     R's share = Rs 510× 2000 = Rs 1000

Page No 9.6:

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:

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 = 55011 = 50
Hence, the number of boys = 7x = 7× 50 = 350, and the number of girls = 4x = 4 × 50 = 200.

Page No 9.6:

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:

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 5002 = Rs 250
            Thus, the income = 7x = 7 × 250 = Rs 1750.
     Now, expenditure = Income - savings = Rs 1750 - Rs 500 = Rs 1250.
 Thus, the income = Rs 1750, and the expenditure = Rs 1250.

Page No 9.6:

Question 14:

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

Answer:

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 = 366 = 6
 Thus, the sides of the triangle = x = 6 cm; 2x = 2×6 = 12 cm; 3x = 3 ×6 = 18 cm.
    So, the sides of the triangle = 6 cm, 12 cm and 18 cm.



Page No 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:

We have
Sum of the terms of the ratio = 2 + 3 = 5, and the total sum = Rs 5500
Therefore, Raman's share = 25×5500 = Rs 2200
Aman's share = 35×5500 = Rs 3300

Page No 9.7:

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:

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 = 11.79 = 1.3
 Hence, the weight of zinc in the alloy = 7x = 7×1.3 = 9.1 kg.

Page No 9.7:

Question 17:

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

Answer:

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 = 408 = 5
Therefore, the antecedent = 7x = 7×5 = 35.

Page No 9.7:

Question 18:

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

Answer:

We have
            Sum of the ratio of the terms = 2 +7 = 9
           Therefore, first part = Rs. 29×351 = Rs. 78
         Similarly, second part = Rs. 79×351 = Rs. 273

Page No 9.7:

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:

We have
 Cost of 1 score of pencils = Rs. 16
 Since 1 score = 20 items,
 Cost of one pencil = Rs. 1620 = Rs. 0.8
 Cost of 1 dozen ball pens = Rs. 8.40
 Since 1 dozen =12 items,
 Cost of one ball pen = Rs. 8.4012 = Rs. 0.7
 So, price of pencil : price of ball pen = 0.8 : 0.7 = 0.80.7 = 87
       Price of pencil : price of ball pen = 8 : 7

Page No 9.7:

Question 20:

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

Answer:

We have
One out of every six student fails, which means that 16th of the total students fail in the class.
And total number of students in the class = 42.
Therefore, the number of students who fail = 16×  42 = 7.
So, the number of students who pass = (Total students -the number of students who fail) = 42 - 7 = 35.



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