NCERT Solutions Class 6 Mathematics Chapter 12 Ratio and Proportion

Ratio and Proportion

Exercise: 12.1

Q.1) There are 20 girls and 15 boys in a class.
(a) What is the ratio of number of girls to the number of boys?
(b) What is the ratio of number of girls to the total number of students in the class?
Sol.1) a) Number of girls: number of boys = 20: 15 = 20/15 = 4/3 = 4: 3
b) Number of girls = 20
Number of boys = 15
Total number of students in the class = 35
Number of girls: total number of students in the class = 20: 35 = 20/35 = 4: 7

Q.2) Out of 30 students in a class, 6 like football, 12 like cricket and remaining like tennis.
Find the ratio of
(a) Number of students liking football to number of students liking tennis.
(b) Number of students liking cricket to total number of students.
Sol.2) Given
Total number of students in the class = 30
Number of students who like football = 6
Number of students who like cricket = 12
So, Number of students who like tennis = Total number of students in the class – (Sum of
students who like football and cricket) =30 – (6 + 12) = 12
a) Number of students who like football: number of students who like tennis = 6: 12 = 6/12 = 1: 2
b) Number of students who like cricket: total number of students = 12: 30 = 12/30 = 2: 5

Q.3) See the figure and find the ratio of
(a) Number of triangles to the number of circles inside the rectangle.
(b) Number of squares to all the figures inside the rectangle.
(c) Number of circles to all the figures inside the rectangle.

Sol.3) (a) The ratio of triangles to the number of circles inside the rectangle = 3/2 = 3 ∶ 2
(b)The ratio of Number of squares to all the figures inside the rectangle. = 2/7 = 2: 7
(c)The ratio of Number of circles to all the figures inside the rectangle. = 2/7 = 2: 7

Q.4) Distances travelled by Hamid and Akhtar in an hour are 9 km and 12 km. Find the ratio of speed of Hamid to the speed of Akhtar.
Sol.4) We know that, 𝑆𝑝𝑒𝑒𝑑 = 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒/𝑇𝑖𝑚𝑒
Distance travelled by Hamid in 1 ℎ𝑜𝑢𝑟 = 9 𝑘𝑚
Speed of Hamid = distance/time = 9 𝑘𝑚/ℎ𝑟
Distance travelled by Akhtar in 1 ℎ𝑜𝑢𝑟 = 12 𝑘𝑚
Speed of Hamid = 12 𝑘𝑚/ℎ𝑟
Ratio of speed of Hamid : Akhtar = 9 ∶ 12 = 3 ∶ 4

Q .5) Fill in the following blanks:

Q.7) Find the ratio of the following:
(a) 30 minutes to 1.5 hours
(b) 40 cm to 1.5 m
(c) 55 paisa to Re 1
(d) 500 ml to 2 litres
Sol.7) The two quantities are not in the same unit. So, We can only find ratio here by converting them in same units.
It is always easy to convert into smaller of unit
(a) 30 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 𝑡𝑜 1.5 ℎ𝑜𝑢𝑟
1.5 ℎ𝑜𝑢𝑟𝑠 = 1.5 × 60 = 90 𝑚𝑖𝑛𝑢𝑡𝑒𝑠                     [∵ 1 hour = 60 minutes]
Now, ratio of 30 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 𝑡𝑜 1.5 ℎ𝑜𝑢𝑟 = 30 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 ∶ 1.5 ℎ𝑜𝑢𝑟
⇒ 30 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 ∶ 90 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 = 1 ∶ 3

(b) 40 𝑐𝑚 𝑡𝑜 1.5 𝑚
1.5 𝑚 = 1.5 × 100 𝑐𝑚 = 150 𝑐𝑚                   [∵ 1 m = 100 cm]
Now, ratio of 40 𝑐𝑚 𝑡𝑜 1.5 𝑚 = 40 𝑐𝑚 ∶ 1.5 𝑚
⇒ 40 𝑐𝑚 ∶ 150 𝑐𝑚 = 4 ∶ 15

(c) 55 𝑝𝑎𝑖𝑠𝑎 𝑡𝑜 𝑅𝑒. 1
𝑅𝑒. 1 = 100 𝑝𝑎𝑖𝑠𝑎
Now, ratio of 55 𝑝𝑎𝑖𝑠𝑎 𝑡𝑜 𝑅𝑒. 1 = 55 𝑝𝑎𝑖𝑠𝑎 ∶ 100 𝑝𝑎𝑖𝑠𝑎
⇒ 11 ∶ 20

(d) 500 𝑚𝑙 𝑡𝑜 2 𝑙𝑖𝑡𝑟𝑒𝑠
2 𝑙𝑖𝑡𝑒𝑟𝑠 = 2 × 1000 𝑚𝑙 = 2000 𝑚𝑙                 [∵ 1 litre = 1000 ml]
Now, ratio of 500 𝑚𝑙 𝑡𝑜 2 𝑙𝑖𝑡𝑟𝑒𝑠 = 500 𝑚𝑙 ∶ 2 𝑙𝑖𝑡𝑟𝑒𝑠
⇒ 500 𝑚𝑙 ∶ 2000 𝑚𝑙 = 1 ∶ 4

Q.8) In a year, Seema earns 𝑅𝑠 1,50,000 and saves 𝑅𝑠 50,000. Find the ratio of
(a) Money that Seema earns to the money she saves.
(b) Money that she saves to the money she spends.
Sol.8) Amount earned by Seema = 𝑅𝑠 1,50,000
Amount saved = 𝑅𝑠. 50,000
Amount spent = 1,50,000 – 50,000 = 1,00,000
a) Money Seema earns : money she saves =1,50,000 : 50,000
= 150000/50000 = 15/5 = 3/1 = 3: 1
b) Money saved : money spent = 50,000 : 1,00,000
= 50000/100000 = 5/10 = 1/2 = 1: 2

Q.9) There are 102 teachers in a school of 3300 students. Find the ratio of the number of teachers to the number of students.
Sol.9) Number of teachers in the school = 102
Number of students in the school = 3300
Number of teachers : number of students = 102 ∶ 3300
= 102/3300 = 102 ÷ 3/3300 ÷ 3 
= 34/1100 = 17/550 = 17: 550

Q.10) In a college, out of 4320 students, 2300 are girls. Find the ratio of
(a) Number of girls to the total number of students.
(b) Number of boys to the number of girls.
(c) Number of boys to the total number of students.
Sol.10) Total number of students in the school = 4320
Number of girls = 2300
(a) The ratio of Number of girls to the total number of students = 2300/4320 = 115/216 = 115: 216
(b) The Numbers of boys= 4320 − 2300 = 2020
The ratio of Number of boys to the number of girls = 2020/2300 = 101/115 = 101: 115
(c) The ratio of Number of boys to the total number of students = 2020/4320 = 101/216 = 101: 216

Q.11) Out of 1800 students in a school, 750 opted basketball, 800 opted cricket and remaining opted table tennis. If a student can opt only one game, find the ratio of
(a) Number of students who opted basketball to the number of students who opted table tennis.
(b) Number of students who opted cricket to the number of students opting basketball.
(c) Number of students who opted basketball to the total number of students.
Sol.11) (a) The number of students who opted table tennis = 1800 − 750 − 800 = 250
The ratio of Number of students who opted basketball to the number of students who opted table tennis = 750/250 = 3/1 = 3: 1
(b) The ratio of Number of students who opted cricket to the number of students opting basketball. = 800/750 = 16/15 = 16 ∶ 15
(c) The ratio of Number of students who opted basketball to the total number of students = 750/1800 = 5/12 = 5: 12

Q.12) Cost of a dozen pens is Rs.180 and cost of 8 ball pens is Rs.56. Find the ratio of the cost of a pen to the cost of a ball pen.
Sol.12) The Cost of a dozen pens is 𝑅𝑠. 180 and
∴ cost of 1 pen is = 180/12 = 𝑅𝑠 15
The cost of 8 ball pens is = 𝑅𝑠. 56
∴ cost of 1 ball pen is = 56/8 = 𝑅𝑠. 7
The ratio of cost of a pen to the cost of a ball pen = 15/7 = 15 ∶ 7

Q.13) Consider the statement : Ratio of breadth and length of a hall is 2 : 5. Complete the following table that shows some possible breadths and lengths of the hall

Q.14) Divide 20 pens between Sheela and Sangeeta in the ratio of 3: 2.
Sol.14) The two parts are 3 and 2. Sum of the parts is 5.
So, Sheela gets 3 parts and Sangeeta gets 2 parts out of every 5 parts.
                                        OR
Sheela gets 3/5 of the total pens and Sangeeta gets 2/5 of the total pens
Number of pens Sheela gets = (3/5) × 20 = 12
Number of pens Sangeeta gets = (2/5) × 20 = 8
So, Sheela gets 12 pens and Sangeeta gets 8 pens.

Q.15) Mother wants to divide Rs.36 between her daughters Shreya and Bhoomika in the ratio of their ages. If age of Shreya is 15 years and age of Bhoomika is 12 years, find how much Shreya and Bhoomika will get.
Sol.15) Ratio of the age of Shreya to that of Bhoomika = 5 ∶ 4
Thus, Rs. 36 divide between Shreya and Bhoomika in the ratio of 5 ∶ 4.
Shreya gets = 5/9
𝑜𝑓 𝑅𝑠. 36 = 𝑅𝑠. 20
Bhoomika gets = 4/9
𝑜𝑓 𝑅𝑠. 36 = 𝑅𝑠. 16
Shreya gets Rs. 20 and Bhoomika gets Rs. 16

Q.16) Present age of father is 42 years and that of his son is 14 years. Find the ratio of
(a) Present age of father to the present age of son.
(b) Age of the father to the age of son, when son was 12 years old.
(c) Age of father after 10 years to the age of son after 10 years.
(d) Age of father to the age of son when father was 30 years old.
Sol.16) Given
Present age of father = 42 𝑦𝑒𝑎𝑟𝑠
Present age of his son = 14 𝑦𝑒𝑎𝑟𝑠
a) Age of father: age of son = 42: 14 = 42/14 
                                                          = 3/1 = 3: 1

b) When the son was 12 = (14 – 2) years old
Then father would have been 40 = (42 – 2) years old.
Age of father: Age of son = 40: 12 = 10: 3

c) Age of father after 10 𝑦𝑒𝑎𝑟𝑠 = Present age + 10 = 42 + 10 = 52 𝑦𝑒𝑎𝑟𝑠
Age of son after 10 𝑦𝑒𝑎𝑟𝑠 = 14 + 10 = 24 𝑦𝑒𝑎𝑟𝑠
Age of father: Age of son = 52: 24 = 13: 6

d) When was the father 30 years old?
Since, 42 – 30 = 12, we know that 12 years back, the father was 30 years old.
So Age of son 12 years back was = present age – 12 = 14 − 2 = 2 𝑦𝑒𝑎𝑟
Ratio of father: Age of son = 30: 2 = 15: 1

Exercise 12.2

Q.1) Determine if the following are in proportion.
(a) 15, 45, 40, 120      (b) 33, 121, 9,96    (c) 24, 28, 36, 48
(d) 32, 48, 70, 210      (e) 4, 6, 8, 12          (f) 33, 44, 75, 100
Sol.1) If two ratios are equal, we say that they are in proportion.
a) Ratio of 15 and 45 = 15/45 = 3/9 
                                            = 1/3 = 1: 3
Ratio of 40 and 120 = 40/120 = 4/12 
                                            = 1/3 = 1: 3
Since 15 ∶ 45 and 40 ∶ 120 are equal, we say that they are in proportion.
15 ∶ 45 ∶ : 40 ∶ 120

b) Ratio of 33 and 121 = 33/121 = 3/11 = 3: 11
Ratio of 9 and 96 = 9/96 = 3/32 = 3: 32
Since 33 ∶ 121 and 9 ∶ 96 are unequal, we say that they are not in proportion.

c) Ratio of 24 and 28 = 24/28 = 12/14 
                                            = 6/7 = 6: 7
Ratio of 36 and 48 = 36/48 = 3/4 = 3: 4
Since 24 ∶ 28 and 36 ∶ 48 are unequal, we say that they are not in proportion.

d) Ratio of 32 and 48 = 32/48 = 16/24 
                               = 8/12 = 2/3 = 2: 3
Ratio of 70 and 210 = 70/210 = 7/21 
                              = 1/3 = 1: 3
Since 32 ∶ 48 and 70 ∶ 210 are unequal, we say that they are not in proportion.

e) Ratio of 4 and 6 = 4/6 = 2/3 = 2: 3
Ratio of 8 and 12 = 8/12 = 2/3 = 2: 3
Since 4 ∶ 6 and 8 ∶ 12 are equal, we say that they are in proportion.
4 ∶ 6 ∶ : 8 ∶ 12

f) Ratio of 33 and 44 = 33/44 = 3/4 = 3: 4
Ratio of 75 and 100 =
75/100 = 15/20 = 3/4 = 3: 4
Since 33 ∶ 44 and 75 ∶ 100 are equal, we say that they are in proportion.
33 ∶ 44 ∶ : 75 ∶ 100

Q.2) Write True (T) or False (F) against each of the following statements:
(a) 16 : 24 :: 20 : 30 (b) 21: 6 :: 35 : 10 (c) 12 : 18 :: 28 : 12
(d) 8 : 9 :: 24 : 27 (e) 5.2 : 3.9 :: 3 : 4 (f) 0.9 : 0.36 :: 10 : 4
Sol.2) (a) 16 : 25 : : 20 : 30
16: 25 ⇒ 4/6 = 2/3 = 2: 3 
20/30 ⇒ 2/3 = 2: 3
Hence, it is True.

(b) 21 : 6 : : 35 : 10
21: 6 ⇒ 7/2 = 7: 2
35: 10 ⇒ 7/2 = 7: 2
Hence, it is True.

(c) 12 : 18 : : 28 : 12
12: 18 ⇒ 6/9 = 2/3
28: 22 ⇒ 14/16 = 7/3 = 7: 3
The ratios are unequal. Hence False.

(d) 8 : 9 : : 24 : 27
8: 9 ⇒ 8: 9
24: 27 ⇒ 8/9 = 8: 9
Hence, it is True.

(e) 5.2 : 3.9 : : 3 : 4
5.2 ∶ 3.9 ⇒ 52/39 = 4/3 = 4: 3
3: 4 ⇒ 3: 4
Hence, it is False.

(f) 0.9 : 0.36 : : 10 : 4
0.9 ∶ 0.36 ⇒ 90/36 = 10/4 = 5/2 = 5: 2
10: 4 ⇒ 5/2 = 5: 2
Hence the ratios are equal it is True.

Q.3) Are the following statements true?
(a) 40 persons: 200 persons = Rs 15: Rs 75
(b) 7.5 litres: 15 litres = 5 kg: 10 kg
(c) 99 kg: 45 kg = Rs 44: Rs 20
(d) 32 m: 64 m = 6 sec: 12 sec
(e) 45 km: 60 km = 12 hours: 15 hours

Q.4) Determine if the following ratios form a proportion. Also, write the middle terms and extreme terms where the ratios form a proportion.
(a) 25 cm: 1 m and Rs 40: Rs 160 (b) 39 litres: 65 litres and 6 bottles: 10 bottles
(c) 2 kg: 80 kg and 25 g: 625 g (d) 200 mL: 2.5 litre and Rs 4: Rs 50
Sol.4) (a) 25 𝑐𝑚 ∶ 1 𝑚
= 25 𝑐𝑚 ∶ (1 𝑥 100) 𝑐𝑚 = 25 𝑐𝑚 ∶ 100 𝑐𝑚 = 1 ∶ 4
𝑅𝑠. 40 ∶ 𝑅𝑠. 160 = 1 ∶ 4
Since the ratios are equal, therefore these are in proportion.
Middle terms = 1 𝑚, 𝑅𝑠. 40 and Extreme terms = 25 𝑐𝑚, 𝑅𝑠. 160
(b) 39 𝑙𝑖𝑡𝑒𝑟𝑠 ∶ 65 𝑙𝑖𝑡𝑒𝑟𝑠 =
6 𝑏𝑜𝑡𝑡𝑙𝑒𝑠 ∶ 10 𝑏𝑜𝑡𝑡𝑙𝑒𝑠 = 3 ∶ 5
Since the ratios are equal, therefore these are in proportion.
Middle terms = 65 𝑙𝑖𝑡𝑒𝑟𝑠, 6 𝑏𝑜𝑡𝑡𝑙𝑒𝑠 and Extreme terms = 39 𝑙𝑖𝑡𝑒𝑟𝑠, 10 𝑏𝑜𝑡𝑡𝑙𝑒𝑠
(c) 2 𝑘𝑔 ∶ 80 𝑘𝑔 = 1 ∶ 40
25 𝑔 ∶ 625 𝑔 = 1 ∶ 25
Since the ratios are not equal, therefore these are not in proportion.
(d) 200 𝑚𝑙 ∶ 2.5 𝑙𝑖𝑡𝑒𝑟𝑠
= 200 𝑚𝑙 ∶ (25 𝑥 1000) 𝑙𝑖𝑡𝑒𝑟𝑠 = 200 𝑚𝑙 ∶ 2500 𝑚𝑙 = 2 ∶ 25
𝑅𝑠. 4 ∶ 𝑅𝑠. 50 = 2 ∶ 25
Since the ratios are equal, therefore these are in proportion.
Middle terms = 2.5 𝑙𝑖𝑡𝑒𝑟𝑠, 𝑅𝑠. 4 and Extreme terms = 200 𝑚𝑙, 𝑅𝑠. 50

Exercise 12.3

Q.1) If the cost of 7 𝑚 of cloth is 𝑅𝑠. 294, find the cost of 5 𝑚 of cloth.
Sol.1) Cost of 7 𝑚 of cloth = 𝑅𝑠. 294
∴ Cost of 1 𝑚 of cloth = 𝑅𝑠. 42
∴ Cost of 5 𝑚 of cloth = 42 × 5 = 𝑅𝑠. 210
Thus, the cost of 5 𝑚 of cloth is 𝑅𝑠. 210.

Q.2) Ekta earns Rs 1500 in 10 days. How much will she earn in 30 days?
Sol.2) Amount Ekta earns in 10 days = 𝑅𝑠. 1500
So, amount earned in 1 day = 𝑅𝑠 1500 ÷ 10 = 𝑅𝑠. 150
Amount she will earn in 30 days = 𝑅𝑠. 150 × 30 = 𝑅𝑠. 4500

Q.3) If it has rained 276 𝑚𝑚 in the last 3 days, how many 𝑐𝑚 of rain will fall in one full week (7 days)? Assume that the rain continues to fall at the same rate.
Sol.3) Measure of rain in 3 𝑑𝑎𝑦𝑠 = 276 𝑚𝑚 = 27.6 𝑐𝑚
So, measure of rain in 1 𝑑𝑎𝑦 = 27.6 ÷ 3 = 9.2 𝑐𝑚
We know that one week has 7 days.
So, measure of rain in 7 𝑑𝑎𝑦𝑠 = 9.2 × 7 = 64.4 𝑐𝑚

Q.4) Cost of 5 kg of wheat is 𝑅𝑠. 30.50.
(a) What will be the cost of 8 𝑘𝑔 of wheat?
(b) What quantity of wheat can be purchased in 𝑅𝑠. 61?
Sol.4) Cost of 5 𝑘𝑔 of wheat = 𝑅𝑠. 30.50
Cost of 1 𝑘𝑔 = 𝑅𝑠. 30.50 ÷ 5 = 𝑅𝑠. 6.10
a) Cost of 8 kg of wheat = 8 × 𝑅𝑠. 6.10 = 𝑅𝑠. 48.80
b) Amount of wheat purchased for 𝑅𝑠. 6.10 = 1 𝑘𝑔
Amount of wheat purchased for 𝑅𝑠. 61 = 61 ÷ 6.10 = 6100 ÷ 610 = 10 𝑘𝑔

Q.5) The temperature dropped 15 degrees Celsius in the last 30 days. If the rate of temperature drop remains the same, how many degrees will the temperature drop in the next ten days?
Sol.5) Temperature drop in 30 days = 15°𝐶
Temperature drop in 1 day = 15 ÷ 30 = (1/2)°𝐶
Temperature drop in 10 days = (1/2)°𝐶 × 10 = 5°𝐶
Thus, 5 degree Celsius temperature dropped in 10 days.

Q.6) Shaina pays Rs.7500 as rent for 3 months. How much does she has to pay for a whole year, if the rent per month remains same?
Sol.6) Rent paid by Shaina for 3 months = 𝑅𝑠 7500
Rent paid by Shaina per month = 7500 ÷ 3 = 𝑅𝑠 2500
We know that in a year, there are 12 𝑚𝑜𝑛𝑡ℎ𝑠.
So, the rent paid by Shaina for a year = 𝑅𝑠. 2500 × 12 = 𝑅𝑠 30,000

Q.7) Cost of 4 dozen bananas is Rs.60. How many bananas can be purchased for Rs.12.50?
Sol.7) We know that
1 dozen = 12
So, 4 dozen = 48
The cost of 48 bananas = 𝑅𝑠. 60
Cost of 1 banana= 60 ÷ 48 = 5 ÷ 4 = 𝑅𝑠 1.25
For 𝑅𝑠. 1.25, number of bananas that can be purchased = 1
For 𝑅𝑠. 12.50, number of bananas that can be purchased = 12.50 × 1 ÷ 1.25
= 12.50 ÷ 1.25 = 1250 ÷ 125 = 10
So, 10 bananas can be purchased.

Q.8) The weight of 72 books is 9 𝑘𝑔. What is the weight of 40 such books?
Sol.8) Given, Weight of 72 books = 9 𝑘𝑔
Weight of 1 𝑏𝑜𝑜𝑘 = 9 ÷ 72 = 1/8 𝑘𝑔
Weight of 40 𝑏𝑜𝑜𝑘𝑠 = 40 × 1/8 = 5 𝑘𝑔
∴ the weight of 40 books is 5 kg.

Q.9) A truck requires 108 litres of diesel for covering a distance of 594 km. How much diesel will be required by the truck to cover a distance of 1650 km?
Sol.9) Amount of Diesel required by truck for covering a distance for 594 𝑘𝑚 = 108𝑙
Amount of Diesel required by truck for covering a distance for 1 𝑘𝑚 = 108 ÷ 594 = (2/11)𝑙
So Amount of diesel required by truck for covering a distance for 1650 𝑘𝑚 = 1650 × (2/11) = 300 𝑙𝑖𝑡𝑟𝑒

Q.10) Raju purchases 10 pens for 𝑅𝑠. 150 and Manish buys 7 pens for 𝑅𝑠 84. Can you say who got the pens cheaper?
Sol.10) Amount paid by Raju for 10 𝑝𝑒𝑛𝑠 = 𝑅𝑠. 150
Amount paid for 1 𝑝𝑒𝑛 = 150 ÷ 10 = 𝑅𝑠 15
Amount paid by Manish for 7 𝑝𝑒𝑛𝑠 = 𝑅𝑠. 84
Amount paid for 1 𝑝𝑒𝑛 = 84 ÷ 7 = 𝑅𝑠 12
Raju paid 𝑅𝑠. 15 for 1 pen but Manish paid 𝑅𝑠. 12
So, Manish got the pens cheaper.

Q.11) Anish made 42 runs in 6 overs and Anup made 63 runs in 7 overs. Who made more runs per over?
Sol.11) Runs made by Anish in 6 𝑜𝑣𝑒𝑟𝑠 = 42
Runs made in 1 𝑜𝑣𝑒𝑟 = 42 ÷ 6 = 7
Runs made by Anup in 7 𝑜𝑣𝑒𝑟𝑠 = 63
Runs made in 1 𝑜𝑣𝑒𝑟 = 63 ÷ 7 = 9
In one over, Anup made 9 runs but Anish made 7 runs. So, Anup made more runs in an over than Anish.