MP Board Class 6th Maths Solutions Chapter 12 Ratio and Proportion Ex 12.3
Question 1.
If the cost of 7 m of cloth is Rs. 294, find the cost of 5 m of cloth.
Solution:
Cost of 7 m of cloth = Rs. 294
∴ Cost of 1 m of cloth = Rs. \(\frac{294}{7}\) = Rs. 42
∴ Cost of 5 m of cloth = Rs. 42 × 5 = Rs. 210
Thus, the cost of 5 m of cloth is Rs. 210.
Question 2.
Ekta earns Rs. 1500 in 10 days. How much will she earn in 30 days?
Solution:
Earning of 10 days = Rs. 1500
∴ Earning of 1 day = Rs. \(\frac{1500}{10}\) = Rs.150
∴ Earning of 30 days = Rs. 150 × 30 = Rs. 4500
Thus, the earning of 30 days is Rs. 4,500.
Question 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.
Solution:
Rain fall in 3 days = 276 mm 276
∴ Rain fall in 1 day = \(\frac{276}{3}\) mm = 92 mm
∴ Rain fall in 7 days = 92 × 7 mm = 644 mm
Thus, the rain fall in one full week is 644 mm.
Question 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?
Solution:
(a) Cost of 5 kg of wheat = Rs. 30.50
∴ Cost of 1 kg of wheat = Rs. \(\frac{30.50}{5}\) = Rs. \(\frac{3050}{500}\)
= Rs. 6.10
∴ Cost of 8 kg of wheat = Rs. 6.10 × 8 = Rs. 48.80
(b) From Rs. 30.50, quantity of wheat can be purchased = 5 kg
∴ From Re. 1, quantity of wheat can be purchased = \(\frac{5}{30.50}\) kg
∴ From Rs. 61, quantity of wheat can be purchased = \(\frac{5}{30.50}\) × 61 kg
= \(\frac{5}{3050}\) × 6100 kg = 10 kg
Question 5.
The temperature dropped 15 degree 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?
Solution:
Temperature dropped in last 30 days = 15 degrees
∴ Temperature dropped in 1 day
= \(\frac{15}{30}\) degree = \(\frac{1}{2}\) degree
∴ Temperature will drop in next 10 days
= \(\frac{1}{2}\) × 10 degrees = 5 degrees
Thus, 5 degree Celsius temperature will drop in next 10 days.
Question 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?
Solution:
Rent paid for 3 months = Rs. 7500
∴ Rent paid for 1 month = Rs. \(\frac{7500}{3}\)
= Rs. 2500
∴ Rent paid for 12 months = Rs. 2500 × 12
= Rs. 30,000
Thus, the total rent of one year is Rs. 30,000.
Question 7.
Cost of 4 dozens bananas is Rs. 60. How many bananas can be purchased for Rs. 12.50?
Solution:
Cost of 4 dozens bananas = Rs. 60
i.e., cost of 48 bananas = Rs. 60 [∵ 4 dozens = 4 × 12 = 48]
From Rs. 60, number of bananas can be purchased = 48
∴ From Re. 1, number of bananas can be purchased = \(\frac{48}{60}=\frac{4}{5}\)
∴ FromRs. 12.50, number of bananas can be purchased
= \(\frac{4}{5}\) × 12.50 = \(\frac{4}{5} \times \frac{1250}{100}=\frac{250}{25}\) = 10
Thus, 10 bananas can be purchased for Rs. 12.50.
Question 8.
The weight of 72 books is 9 kg. What is the weight of 40 such books?
Solution:
The weight of 72 books = 9 kg
∴ The weight of 1 book = \(\frac{9}{72}\) kg = \(\frac{1}{8}\) kg
∴ The weight of 40 books = \(\frac{1}{8}\) × 40 kg = 5 kg
Thus, the weight of 40 books is 5 kg.
Question 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?
Solution:
For covering 594 km, required diesel = 108 litres
∴ For covering 1 km, required diesel
= \(\frac{108}{594}\) litres = \(\frac{2}{11}\) litres
∴ For covering 1650 km, required diesel
= \(\frac{2}{11}\) × 1650 litres = 300 litres
Thus, 300 litres of diesel will be required by the truck to cover a distance of 1650 km.
Question 10.
Raju purchases 10 pens for Rs. 150 and Manish buys 7 pens for Rs. 84. Can you say who got the pens cheaper?
Solution:
Cost of 10 pens for Raju = Rs. 150
∴ Cost of 1 pen for Raju = Rs. \(\frac{150}{10}\) = Rs. 15
Cost of 7 pens for Manish = Rs. 84
∴ Cost of 1 pen for Manish = Rs. \(\frac{84}{7}\) = Rs. 12
Thus, Manish got the pens cheaper.
Question 11.
Anish made 42 runs in 6 overs and Anup made 63 runs in 7 overs. Who made more runs per over?
Solution;
Runs made by Anish in 6 overs = 42
∴ Runs made by Anish in 1 over = \(\frac{42}{6}\) = 7
Runs made by Anup in 7 overs = 63
∴ Runs made by Anup in 1 over = \(\frac{63}{7}\) = 9
Thus, Anup made more runs per over.