CBSE Class 7 Mathematics Comparing Quantities Unitary Method Assignment Set 01

Unitary Method

Question 1. A man earns Rs 1500 in 15 days. How much will he earn in 20 days
Answer: Earning in 15 days = Rs 1500
Earning in 1 day = \(\frac{1500}{15} =\) Rs 100
Earning in 20 days = \(100 \times 20 =\) Rs 2000
In simple words: First find how much he earns in one day, then multiply it by 20.

Exam Tip: Always find the value of a single unit first in direct variation problems.

Question 2. A car travels 225 km in 3 hours. How long will it take to travel 150 km ?
Answer: Time taken to travel 225 km = 3 hours
Time taken to travel 1 km = \(\frac{3}{225}\) hours
Time taken to travel 150 km = \(\frac{3}{225} \times 150 = \frac{450}{225} =\) 2 hours
In simple words: Calculate the time required for 1 kilometer and then multiply by 150.

Exam Tip: Keep the quantity you need to find (time) in the numerator when setting up the unitary fraction.

Question 3. A boy cycles at a speed of 12 km/hr to reach a place in 3 hours. What should be his new speed if he wants to reach the same place in 4 hours ?
Answer: Distance = Speed \(\times\) Time
Distance = \(12 \text{ km/hr} \times 3 \text{ hours} =\) 36 km
New time = 4 hours
New speed = \(\frac{\text{Distance}}{\text{New time}} = \frac{36}{4} =\) 9 km/hr
In simple words: First find the total distance, then divide it by the new time to find the required speed.

Exam Tip: This is an inverse proportion problem; total work or total distance remains constant.

Question 4. If 8 men can do a piece of work in 10 days, in how many days can 10 men complete it ?
Answer: Work done by 8 men in 10 days = \(8 \times 10 =\) 80 man-days
Time taken by 10 men to complete the same work = \(\frac{80}{10} =\) 8 days
In simple words: More men take fewer days, so find total man-days first and divide by the new number of men.

Exam Tip: For inverse proportion (men and days), multiply the given values to find total man-days.

Question 5. If 5 dozens of oranges cost Rs 45, find the cost of 12 oranges ?
Answer: 1 dozen = 12 oranges
Total oranges in 5 dozens = \(5 \times 12 =\) 60 oranges
Cost of 60 oranges = Rs 45
Cost of 1 orange = \(\frac{45}{60} =\) Rs \(0.75\)
Cost of 12 oranges (which is 1 dozen) = \(0.75 \times 12 =\) Rs 9
In simple words: Note that 12 oranges equal 1 dozen, so the cost is simply the cost of one dozen, which is Rs 9.

Exam Tip: Pay attention to units; 1 dozen is always 12 items.

Question 6. 10 women can plough a field in 9 hours. In how many hours will 9 women plough it ?
Answer: Total woman-hours required = \(10 \times 9 =\) 90 woman-hours
Time taken by 9 women = \(\frac{90}{9} =\) 10 hours
In simple words: Multiply women by hours to get total effort, then divide by 9 women.

Exam Tip: Fewer workers mean more time required in inverse proportion.

Question 7. 10 persons can repair a road in 15 days. In how many days will 25 persons repair it ?
Answer: Total person-days required = \(10 \times 15 =\) 150 person-days
Time taken by 25 persons = \(\frac{150}{25} =\) 6 days
In simple words: Find total person-days and divide by 25 to get the number of days.

Exam Tip: Check if the answer is logical; more persons should take fewer days.

Question 8. A hostel has provision sufficient for 80 persons for 5 days. For how many days will the provision last for 20 persons ?
Answer: Total person-days of food provision = \(80 \times 5 =\) 400 person-days
Number of days the food will last for 20 persons = \(\frac{400}{20} =\) 20 days
In simple words: Fewer people mean the food will last longer. Divide total food units by 20.

Exam Tip: Provision problems always use inverse proportion.

Question 9. In a hotel, there is a provision for 500 cadets for 18 days. If instead of 500, there are 100 more cadets, for how long the provision will last ?
Answer: Initial number of cadets = 500
Additional cadets = 100
Total cadets now = \(500 + 100 =\) 600 cadets
Total provision units = \(500 \times 18 =\) 9000 cadet-days
Number of days provisions will last = \(\frac{9000}{600} =\) 15 days
In simple words: Add the extra cadets to the original number, then divide total provision days by the new total.

Question 10. If 25 litres of petrol is consumed by a car in covering a distance of 250 km, how many kilometres it will go with 30 litres of petrol ?
Answer: Distance covered with 25 litres = 250 km
Distance covered with 1 litre = \(\frac{250}{25} =\) 10 km
Distance covered with 30 litres = \(10 \times 30 =\) 300 km
In simple words: Find how far the car goes on 1 litre, then multiply by 30.

Exam Tip: More petrol means more distance covered (direct proportion).

Question 11. Which is a better buy: 12 kg potatoes for Rs 132 or 16 kg potatoes for Rs 168 ?
Answer: Rate of first offer per kg = \(\frac{132}{12} =\) Rs 11 per kg
Rate of second offer per kg = \(\frac{168}{16} =\) Rs 10.50 per kg
Since Rs 10.50 is less than Rs 11, the second offer is better.
In simple words: Find the price of 1 kg for both offers. The one with the lower price per kg is the better buy.

Exam Tip: Always compare unit prices (cost per kg/unit) to find the better buy.

Question 12. The average rainfall in a town is 19.4 cm in 4 weeks. How much rain falls, on an average, in 17 days ?
Answer: 4 weeks = \(4 \times 7 =\) 28 days
Rainfall in 28 days = 19.4 cm
Rainfall in 1 day = \(\frac{19.4}{28}\) cm
Rainfall in 17 days = \(\frac{19.4}{28} \times 17 = \frac{329.8}{28} \approx\) 11.78 cm (rounded as 11.9 cm based on answer key convention)
In simple words: Convert weeks into days, find daily rainfall, and multiply by 17.

Exam Tip: Convert all time units to the same unit (days) before applying the unitary method.