NCERT Solutions Class 6th Maths Chapter – 12 Ratio and Proportion All Examples

Solution –

Length of the rectangular field = 50 m
Breadth of the rectangular field = 15 m
The ratio of the length to the breadth is 50 : 15
The ratio can be written as 50/15 = 50 ÷ 5/15 ÷ 5 = 10/3 = 10 : 3
Thus, the required ratio is 10 : 3.

Example 2. Find the ratio of 90 cm to 1.5 m.

Solution –

The two quantities are not in the same units. Therefore, we have to convert them into same units.
1.5 m = 1.5 × 100 cm = 150 cm.

Therefore, the required ratio is 90 : 150.
= 90/150 = 90 × 30/150 × 30 = 3/5
Required ratio is 3 : 5.

Example 3. There are 45 persons working in an office. If the number of females is 25 and the remaining are males, find the ratio of-

(a) The number of females to number of males.
(b) The number of males to number of females.

Solution –

Number of females = 25
Total number of workers = 45
Number of males = 45 – 25 = 20

Therefore, the ratio of number of females to the number of males
= 25 : 20 = 5 : 4

And the ratio of number of males to the number of females
= 20 : 25 = 4 : 5.
(Notice that there is a difference between the two ratios 5 : 4 and 4 : 5)

Example 4. Give two equivalent ratios of 6 : 4.

Solution –

Ratio 6 : 4 = 6/4 = 6 × 2/4 × 2 = 12/8.
Therefore, 12 : 8 is an equivalent ratio of 6 : 4
Similarly, the ratio 6 : 4 = 6/4 = 6 × 2/4 × 2 = 3/2.
So, 3:2 is another equivalent ratio of 6 : 4.

Therefore, we can get equivalent ratios by multiplying or dividing the
numerator and denominator by the same number.

Write two more equivalent ratios of 6 : 4.

Example 5. Fill in the missing numbers-

Solution –

In order to get the first missing number, we consider the fact that 21 = 3 × 7. i.e. when we divide 21 by 7 we get 3. This indicates that to get the missing number of second ratio, 14 must also be divided by 7. When we divide, we have, 14 ÷ 7 = 2

Hence, the second ratio is 2/3.
Similarly, to get third ratio we multiply both terms of second ratio by 3. (Why? )
Hence, the third ratio is 6/9.

Example 6. Ratio of distance of the school from Mary’s home to the distance of the school from John’s home is 2 : 1.

(a) Who lives nearer to the school?
(b) Complete the following table which shows some possible distances that Mary and John could live from the school.

(c) If the ratio of distance of Mary’s home to the distance of Kalam’s home from school is 1 : 2, then who lives nearer to the school?

Solution –

(a) John lives nearer to the school (As the ratio is 2 : 1).
(b) 

(c) Since the ratio is 1 : 2, so Mary lives nearer to the school.

Example 7. Divide ₹ 60 in the ratio 1 : 2 between Kriti and Kiran.

Solution –

The two parts are 1 and 2.
Therefore, sum of the parts = 1 + 2 = 3.
This means if there are ₹ 3, Kriti will get ₹ 1 and Kiran will get ₹ 2. Or, we can say that Kriti gets 1 part and Kiran gets 2 parts out of every 3 parts.

Therefore, Kriti’s share = 1/3 × 60 = ₹ 20
And Kiran’s share = 2/3 × 60 = ₹ 40.

Example 8. Are the ratios 25g : 30g and 40 kg : 48 kg in proportion?

Solution –

25 g : 30 g = 25/30 = 5 : 6
40 kg : 48 kg = 40/48 = 5 : 6
So, 25 : 30 = 40 : 48.

Therefore, the ratios 25 g : 30 g and 40 kg : 48 kg are in proportion, i.e. 25 : 30 :: 40 : 48
The middle terms in this are 30, 40 and the extreme terms are 25, 48.

Example 9. Are 30, 40, 45 and 60 in proportion?

Solution –

Ratio of 30 to 40 = 30/40 = 3 : 4.
Ratio of 45 to 60 = 45/60 = 3 : 4.
Since, 30 : 40 = 45 : 60.
Therefore, 30, 40, 45, 60 are in proportion.

Example 10. Do the ratios 15 cm to 2 m and 10 sec to 3 minutes form a
proportion?

Solution –

Ratio of 15 cm to 2 m = 15 : 2 × 100 (1 m = 100 cm)
= 3 : 40

Ratio of 10 sec to 3 min = 10 : 3 × 60 (1 min = 60 sec)
= 1 : 18
Since, 3 : 40 ≠ 1 : 18, therefore, the given ratios do not form a proportion.

Example 11. If the cost of 6 cans of juice is ₹ 210, then what will be the cost of 4 cans of juice?

Solution –

Cost of 6 cans of juice = ₹ 210
Therefore, cost of one can of juice = 210/6 = ₹ 35
Therefore, cost of 4 cans of juice = ₹ 35 × 4 = ₹ 140.
Thus, cost of 4 cans of juice is ₹ 140.

Example 12. A motorbike travels 220 km in 5 litres of petrol. How much distance will it cover in 1.5 litres of petrol?

Solution –

In 5 litres of petrol, motorbike can travel 220 km.
Therefore, in 1 litre of petrol, motor bike travels = 220/5 km
Therefore, in 1.5 litres, motorbike travels = 220/5 × 1 5. km
= 220/5 × 15/10 km = 66 km.
Thus, the motorbike can travel 66 km in 1.5 litres of petrol.

Example 13. If the cost of a dozen soaps is ₹ 153.60, what will be the cost of 15 such soaps?

Solution –

We know that 1 dozen = 12
Since, cost of 12 soaps = ₹ 153.60

Therefore, cost of 1 soap = 153.60/12 = ₹ 12.80
Therefore, cost of 15 soaps = ₹ 12.80 × 15 = ₹ 192
Thus, cost of 15 soaps is ₹ 192.

Example 14. Cost of 105 envelopes is ₹ 350. How many envelopes can be
purchased for ₹ 100?

Solution –

In ₹ 350, the number of envelopes that can be purchased = 105
Therefore, in ₹ 1, number of envelopes that can be purchased = 105/350
Therefore, in ₹ 100, the number of envelopes that can be
purchased = 105/350 × 100 = 30
Thus, 30 envelopes can be purchased for ₹ 100.

Example 15. A car travels 90 km in 2 1/2 hours.

(a) How much time is required to cover 30 km with the same speed?
(b) Find the distance covered in 2 hours with the same speed.

Solution –

(a) In this case, time is unknown and distance is known. Therefore, we proceed as follows-
2 1/2 hours = 5/2 hours = 5/2 × 60 minutes = 150 minutes.
90 km is covered in 150 minutes
Therefore, 1 km can be covered in 150/90 minutes
Therefore, 30 km can be covered in 150/90 × 30 minutes i.e. 50 minutes Thus, 30 km can be covered in 50 minutes.

(b) In this case, distance is unknown and time is known. Therefore, we proceed as follows-
Distance covered in 2 1/2 hours (i.e. 5/2 hours) = 90 km
Therefore, distance covered in 1 hour = 90 ÷ 5/2 km = 90 × 2/5 = 36 km
Therefore, distance covered in 2 hours = 36 × 2 = 72 km.
Thus, in 2 hours, distance covered is 72 km.