Mathematics Solutions Solutions for Class 6 Maths Chapter 11 - Ratio Proportion

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Page No 59:

Question 1:

In each example below, find the proportion of the first number to the second.
(1) 24, 56 (2) 63, 49 (3) 52, 65 (4) 84, 60 (5) 35, 65 (6) 121, 99

Answer:

(1) 24 : 56 = $\frac{24}{56} = \frac{24 \div 8}{56 \div 8} = \frac{3}{7}$ = 3 : 7   (HCF of 24 and 56 = 8)

(2) 63 : 49 = $\frac{63}{49} = \frac{63 \div 7}{49 \div 7} = \frac{9}{7}$ = 9 : 7   (HCF of 63 and 49 = 7)

(3) 52 : 65 = $\frac{52}{65} = \frac{52 \div 13}{65 \div 13} = \frac{4}{5}$ = 4 : 5   (HCF of 52 and 65 = 13)

(4) 84 : 60 = $\frac{84}{60} = \frac{84 \div 12}{60 \div 12} = \frac{7}{5}$ = 7 : 5   (HCF of 84 and 60 = 12)

(5) 35 : 65 = $\frac{35}{65} = \frac{35 \div 5}{65 \div 5} = \frac{7}{13}$ = 7 : 13 (HCF of 35 and 65 = 5)

(6) 121 : 99 = $\frac{121}{99} = \frac{121 \div 11}{99 \div 11} = \frac{11}{9}$ = 11 : 9 (HCF of 121 and 99 = 11)

Page No 59:

Question 2:

Find the ratio of the first quantity to the second.
(1) 25 beads, 40 beads (2) 40 rupees, 120 rupees (3) 15 minutes, 1 hour
(4) 30 litres, 24 litres (5) 99 kg, 44000 grams (6) 1 litre, 250 ml
(7) 60 paise, 1 rupee (8) 750 grams, $\frac{1}{2}$ kg (9) 125 cm, 1 metre

Answer:

(1) 25 beads : 40 beads = $\frac{25}{40} = \frac{25 \div 5}{40 \div 5} = \frac{5}{8}$ = 5 : 8 (HCF of 25 and 40 = 5)

(2) 40 rupees : 120 rupees = $\frac{40}{120} = \frac{40 \div 40}{120 \div 40} = \frac{1}{3}$ = 1 : 3 (HCF of 40 and 120 = 40)

(3) 15 minutes : 1 hour = 15 minutes : 60 minutes = $\frac{15}{60} = \frac{15 \div 15}{60 \div 15} = \frac{1}{4}$ = 1 : 4 (HCF of 15 and 60 = 4)

(4) 30 litres : 24 litres = $\frac{30}{24} = \frac{30 \div 6}{24 \div 6} = \frac{5}{4}$ = 5 : 4 (HCF of 30 and 24 = 6)

(5) 44000 g = $\frac{44000}{1000}$ = 44 kg (1 kg = 1000 g)

∴ 99 kg : 44000 g = 99 kg : 44 kg = $\frac{99}{44} = \frac{99 \div 11}{44 \div 11} = \frac{9}{4}$ = 9 : 4 (HCF of 99 and 44 = 11)

(6) 1 L : 250 mL = 1000 mL : 250 mL = $\frac{1000}{250} = \frac{1000 \div 250}{250 \div 250} = \frac{4}{1}$ = 4 : 1 (HCF of 1000 and 250 = 250)

(7) 60 paise : 1 rupee = 60 paise : 100 paise = $\frac{60}{100} = \frac{60 \div 20}{100 \div 20} = \frac{3}{5}$ = 3 : 5 (HCF of 60 and 100 = 20)

(8) $\frac{1}{2}$ kg = $\frac{1}{2}$ × 1000 g = 500 g (1 kg = 1000 g)

∴ 750 g : $\frac{1}{2}$ kg = 750 g : 500 g = $\frac{750}{500} = \frac{750 \div 250}{500 \div 250} = \frac{3}{2}$ = 3 : 2 (HCF of 750 and 500 = 250)

(9) 125 cm : 1 metre = 125 cm : 100 cm = $\frac{125}{100} = \frac{125 \div 25}{100 \div 25} = \frac{5}{4}$ = 5 : 4 (HCF of 125 and 100 = 25)

Page No 59:

Question 3:

Reema has 24 notebooks and 18 books. Find the ratio of notebooks to books.

Answer:

Number of notebooks = 24

Number of books = 18

∴ Ratio of notebooks to books = Number of notebooks : Number of books

= 24 : 18

= $\frac{24}{18}$

= $\frac{24 \div 6}{18 \div 6}$ (HCF of 24 and 18 = 6)

= $\frac{4}{3}$

= 4 : 3

Thus, the ratio of notebooks to books is $\frac{4}{3}$ or 4 : 3.

Page No 59:

Question 4:

30 cricket players and 20 kho-kho players are training on a field. What is the ratio of cricket players to the total number of players?

Answer:

Number of cricket players = 30

Number of kho-kho players = 20

∴ Total number of player = Number of cricket players + Number of kho-kho players = 30 + 20 = 50

Ratio of cricket players to the total number of players = Number of cricket players : Total number of player

= 30 : 50

= $\frac{30}{50}$

= $\frac{30 \div 10}{50 \div 10}$ (HCF of 30 and 50 = 10)

= $\frac{3}{5}$

= 3 : 5

Thus, the ratio of cricket players to the total number of players is $\frac{3}{5}$ or 3 : 5.

Page No 59:

Question 5:

Snehal has a red ribbon that is 80 cm long and a blue ribbon, 2.20 m long. What is the ratio of the length of the red ribbon to that of the blue ribbon?

Answer:

Length of red ribbon = 80 cm

Length of blue ribbon = 2.20 m = 2.20 × 100 = 220 cm

∴ Ratio of the length of the red ribbon to that of the blue ribbon = Length of red ribbon : Length of blue ribbon

= 80 cm : 2.20 m

= 80 cm : 220 cm

= $\frac{80}{220}$

= $\frac{80 \div 20}{220 \div 20}$ (HCF of 80 and 220 = 20)

= $\frac{4}{11}$

= 4 : 11

Thus, the ratio of the length of the red ribbon to that of the blue ribbon is $\frac{4}{11}$ or 4 : 11.

Page No 59:

Question 6:

Shubham’s age today is 12 years and his father’s is 42 years. Shubham’s mother is younger than his father by 6 years. Find the following ratios.
(1) Ratio of Shubham’s age today to his mother’s age today.
(2) Ratio of Shubham’s mother’s age today to his father’s age today
(3) The ratio of Shubham’s age to his mother’s age when Shubham was 10 years old.

Answer:

Shubham’s age today = 12 years

Shubham’s father age today = 42 years

∴ Shubham’s mother age today = Shubham’s father age today − 6 years = 42 years − 6 years = 36 years

(1)
Ratio of Shubham’s age today to his mother’s age today = Shubham’s age today : Shubham’s mother age today

= 12 years : 36 years

= $\frac{12}{36}$

= $\frac{12 \div 12}{36 \div 12}$ (HCF of 12 and 36 = 12)

= $\frac{1}{3}$

= 1 : 3

Thus, the ratio of Shubham’s age today to his mother’s age today is $\frac{1}{3}$ or 1 : 3.

(2)
Ratio of Shubham’s mother’s age today to his father’s age today = Shubham’s mother age today : Shubham’s father age today

= 36 years : 42 years

= $\frac{36}{42}$

= $\frac{36 \div 6}{42 \div 6}$ (HCF of 36 and 42 = 6)

= $\frac{6}{7}$

= 6 : 7

Thus, the ratio of Shubham’s mother’s age today to his father’s age today is $\frac{6}{7}$ or 6 : 7.

(3)
Difference between the present age of Shubham and when he was 10 year old = 12 years − 10 years = 2 years

∴ Shubham's mother age when Shubham was 10 years old = 36 years − 2 years = 34 years

Ratio of Shubham’s age to his mother’s age when Shubham was 10 years old

= 10 years : 34 years

= $\frac{10}{34}$

= $\frac{10 \div 2}{34 \div 2}$ (HCF of 10 and 34 = 2)

= $\frac{5}{17}$

= 5 : 17

Thus, the ratio of Shubham’s age to his mother’s age when Shubham was 10 years old is $\frac{5}{17}$ or 5 : 17.

Page No 60:

Question 1:

Solve the following.
If 20 metres of cloth cost â‚¹ 3600, find the cost of 16 m of cloth.

Answer:

Cost of 20 m of cloth = â‚¹ 3,600

∴ Cost of 1 m of cloth = â‚¹ $\frac{3600}{20}$ = â‚¹ 180

⇒ Cost of 16 m of cloth = â‚¹ 180 × 16 = â‚¹ 2,880

Thus, the cost of 16 m of cloth is â‚¹ 2,880.

Page No 60:

Question 2:

Solve the following.
Find the cost of 8 kg of rice, if the cost of 10 kg is â‚¹ 325.

Answer:

Cost of 10 kg of rice = â‚¹ 325

∴ Cost of 1 kg of rice = â‚¹ $\frac{325}{10}$ = â‚¹ 32.50

⇒ Cost of 8 kg of rice = â‚¹ 32.50 × 8 = â‚¹ 260

Thus, the cost of 8 kg of rice is â‚¹ 260.

Page No 60:

Question 3:

Solve the following .
If 14 chairs cost â‚¹ 5992, how much will have to be paid for 12 chairs?

Answer:

Cost of 14 chairs = â‚¹ 5,992

∴ Cost of 1 chair = â‚¹ $\frac{5992}{14}$ = â‚¹ 428

⇒ Cost of 12 chairs = â‚¹ 428 × 12 = â‚¹ 5,136

Thus, the amount of money to be paid for 12 chairs is â‚¹ 5,136.

Page No 60:

Question 4:

Solve the following .
The weight of 30 boxes is 6 kg. What is the weight of 1080 such boxes?

Answer:

Weight of 30 boxes = 6 kg

∴ Weight of 1 box = $\frac{6}{30} = \frac{1}{5}$ kg

⇒ Weight of 1080 boxes = $\frac{1}{5}$ × 1080 = 216 kg

Thus, the weight of 1080 such boxes is 216 kg.

Page No 60:

Question 5:

Solve the following .
A car travelling at a uniform speed covers a distance of 165 km in 3 hours. At that same speed,
(a) How long will it take to cover a distance of 330 km?
(b) How far will it travel in 8 hours?

Answer:

(a)
Time taken by car to cover 165 km = 3 hours

∴ Time taken by car to cover 1 km = $\frac{3}{165} = \frac{1}{55}$ hours

⇒ Time taken by car to cover 330 km = $\frac{1}{55} \times 330$ = 6 hours

Thus, the time taken by car to cover a distance of 330 km is 6 hours.

(b)
Distance covered by the car in 3 hours = 165 km

∴ Distance covered by the car in 1 hour = $\frac{165}{3}$ = 55 km

⇒ Distance covered by the car in 8 hours = 55 × 8 = 440 km

Thus, the distance travelled by the car in 8 hours is 440 km.

Page No 60:

Question 6:

Solve the following .
A tractor uses up 12 litres of diesel while ploughing 3 acres of land. How much diesel will be needed to plough 19 acres of land?

Answer:

Amount of diesel used by tractor to plough 3 acres of land = 12 litres

∴ Amount of diesel used by tractor to plough 1 acre of land = $\frac{12}{3}$ = 4 litres

⇒ Amount of diesel used by tractor to plough 19 acres of land = 4 × 19 = 76 litres

Thus, the amount of diesel needed to plough 19 acres of land is 76 litres.

Page No 60:

Question 7:

Solve the following .
At a sugar factory, 5376 kg of sugar can be obtained from 48 tonnes of sugarcane. If Savitatai has grown 50 tonnes of sugarcane, how much sugar will it yield?

Answer:

Amount of sugar obtained from 48 tonnes of sugarcane = 5376 kg

∴ Amount of sugar obtained from 1 ton of sugarcane = $\frac{5376}{48}$ = 112 kg

⇒ Amount of sugar obtained from 50 tonnes of sugarcane = 112 × 50 = 5600 kg

Thus, the amount of sugar yielded from 50 tonnes of sugarcane is 5600 kg.

Page No 60:

Question 8:

Solve the following .
In an orchard, there are 128 mango trees in 8 rows. If all the rows have an equal number of trees, how many trees would there be in 13 rows?

Answer:

Number of mongo trees in 8 rows = 128

∴ Number of mango trees in 1 row = $\frac{128}{8}$ = 16

⇒ Number of mango trees in 13 rows = 16 × 13 = 208

Thus, there are 208 mango trees in 13 rows in an orchard.

Page No 60:

Question 9:

Solve the following.
A pond in a field holds 120000 litres of water. It costs 18000 rupees to make such a pond. How many ponds will be required to store 480000 litres of water, and what would be the expense?

Answer:

Number of ponds required to store 120000 litres of water = 1

∴ Number of ponds required to store 1 litre of water = $\frac{1}{120000}$

⇒ Number of ponds required to store 480000 litres of water = $\frac{1}{120000}$ × 480000 = 4

Thus, 4 ponds required to store 480000 litres of water.

Now,

Cost of making 1 pond = â‚¹ 18,000

∴ Cost of making 4 ponds = â‚¹ 18,000 × 4 = â‚¹ 72,000

Thus, the expense to make 4 ponds to store 480000 litres of water is â‚¹ 72,000.

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