Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 11 Ratio and Proportion Ex 11.3 Textbook Exercise Questions and Answers.
PSEB Solutions for Class 6 Maths Chapter 11 Ratio and Proportion Ex 11.3
1. The cost of 1 kg apples is ₹ 45. What is the cost of 7 kg apples?
Solution:
Cost of 1 kg apples = ₹ 45
Cost of 7 kg apples = ₹ 45 × 7
= ₹ 315
2. A car travels 224 km in 7 litres of petrol. How much distance will it cover in 1 litre?
Solution:
Distance covered in 7 litres = 224 km
Distance covered in 1 litres = \(\frac {224}{7}\)
= 32 km
3. A pipe can fill 10 water tanks in 12 hours. How much time will it take to fill 15 such water tanks?
Solution:
Time taken to fill 10 water tanks = 12 hours
Time taken to fill 1 water tank = \(\frac {12}{10}\) hours
Time taken to fill 15 water tanks = \(\frac {12}{10}\) × 15 hours
= 18 hours
4. The cost of 18 m cloth is ₹ 810. What is the cost of 25 m cloth?
Solution:
Cost of 18 m cloth = ₹ 810
Cost of 1 m cloth = ₹ \(\frac {810}{18}\)
Cost of 25 m cloth = ₹ \(\frac {810}{18}\) × 25
= ₹ 1125
5. The weight of 24 books is 6 kg. What is the weight of 36 such books?
Solution:
Weight of 24 books = 6 kg
Weight of 1 book = \(\frac {6}{24}\) kg
Weight of 36 books = \(\frac {6}{24}\) × 36 kg
= 9 kg
Aliter:
By cross product, we have
24 × x = 6 × 36
x = \(\frac{6 \times 36}{24}\)
⇒ x = 9
Hence, weight of 36 books is 9 kg
6. ‘A’ runs 28 km in 5 hours. How many kilometres does it run in 9 hours?
Solution:
A runs in 5 hours = 28 km
A runs in 1 hour = \(\frac {28}{5}\) km
A runs in 9 hours= \(\frac {28}{5}\) × 9 km
= \(\frac {252}{5}\) km
50.4 km
7. A 12 m high pole casts a shadow of 30 m. Find the height of the pole that casts a shadow of 45 m.
Solution:
If shadow cast is 30 m, then height of Pole = 12 m
If shadow cast is 1 m, then height of Pole = \(\frac {12}{30}\) m
If shadow cast is 45 m, then height of Pole
= \(\frac {12}{30}\) × 45 m
= 18 m
8. A man earns ₹ 11200 in 7 months.
Question (i)
How much will he earn in 18 months?
Solution:
A man earns in 7 months = ₹ 11200
A man earns in 1 month = ₹ \(\frac {11200}{7}\)
A man earns in 18 months = ₹ \(\frac {11200}{7}\) × 18
= ₹ 28800
Question (ii)
In how many months will he earn ₹ 40,000?
Solution:
A man earns ₹ 1600 = 1 month
A man earns ₹ 40000 = \(\frac {1}{1600}\) × 40000
= 25 months
9. If the cost of a dozen soaps is ₹ 153.60. What will be the cost of 16 such soaps?
Solution:
Cost of 12 soaps = ₹ 153.60
(1 dozen =12 pieces)
Cost of 1 soap = ₹ \(\frac {153.60}{12}\)
Cost of 16 soap = ₹ \(\frac {153.60}{12}\) × 16
= ₹ 204.80
10. Cost of 105 envelops is ₹ 35. How many envelops can be purchased for ₹ 10?
Solution:
Number of envelops purchased for ₹ 35 = 105
Number of envelops purchased for ₹ 1 = \(\frac {105}{35}\)
Number of envelops purchased for ₹ 10 = \(\frac {105}{35}\) × 10
= 30
11. A bus travels 90 km in 2\(\frac {1}{2}\) hours.
Question (i)
How much time is required to cover 54 km with the same speed?
Solution:
Time required to cover 90 km
= 2\(\frac {1}{2}\) hours
= \(\frac {5}{2}\) hours
Time required to cover 1 km
= \(\frac{5}{2} \times \frac{1}{90}\) hours
Time required to cover 54 km
= \frac{5}{2} \times \frac{1}{90} × 54 hours
= \(\frac {3}{2}\) hours
= 1\(\frac {1}{2}\) hours
Question (ii)
Find the distance covered in 4 hours with the same speed?
Solution:
Distance covered in \(\frac {5}{2}\) hours
= 90 km
Distance covered 1 hour
= 90 × \(\frac {2}{5}\) km
Distance covered 4 hours
= 4 × 90 × \(\frac {2}{5}\) km
= 144 km
12. Anshul made 57 runs in 6 overs. In how many overs he made 95 runs with same strike rate?
Solution:
Number of overs to make 57 runs = 6 overs
Number of overs to make 1 run = \(\frac {6}{57}\) over
Number of overs to make 95 runs = \(\frac {6}{57}\) × 95 overs
= 10 overs
13. Cost of 5 kg rice is ₹ 32.50.
Question (i)
What will be the cost of 14 kg such rice?
Solution:
Cost of 5 kg rice = ₹ 32.50
Cost of 1 kg rice = ₹ \(\frac {32.50}{5}\)
Cost of 14 kg rice = ₹ \(\frac {32.50}{5}\) × 14
= ₹ 91
Question (ii)
What quantity of rice can be purchased in ₹ 162.50?
Solution:
Qunatity of rice for ₹ 32.50 = 5 kg
Qunatity of rice for ₹ 1 = \(\frac {5}{32.50}\) kg
Qunatity of rice for ₹ 162.50 = \(\frac {5}{32.50}\) × 162.50
= 25 kg
14. If a cow grazes 21 sq. m of a field in 6 days. How much area will it graze in 27 days?
Solution:
Field grazed in 6 days = 21 sq. m
Field grazed in 1 day = \(\frac {21}{6}\) sq. m
Field grazed in 27 days = \(\frac {21}{6}\) × 27 sq. m
= 94.5 sq. m