Unitary Method — Exercise 8(A)

Weight of 8 identical articles is 4.8 kg. What is the weight of 11 such articles?

Answer

Given:

Weight of 8 articles = 4.8 kg

Weight of 1 article = $\dfrac{4.8}{8}$ = 0.6 kg

Weight of 11 articles = 11 × 0.6 = 6.6 kg

Hence, the weight of 11 articles is 6.6 kg.

6 books weigh 1.260 kg. How many books will weigh 3.150 kg?

Answer

Given:

Weight of 6 books = 1.260 kg

Weight of 1 book = $\dfrac{1.260}{6}$ = 0.21 kg

Number of books weighing 3.150 kg = $\dfrac{3.150}{0.21}$ = 15 books

Hence, 15 books will weigh 3.150 kg.

8 men complete a work in 6 hours. In how many hours will 12 men complete the same work?

Answer

Given:

8 men complete the work in 6 hours.

1 man will complete the work in 8 × 6 = 48 hours.

12 men will complete the work in $\dfrac{48}{12}$ = 4 hours.

Hence, 12 men will complete the work in 4 hours.

If a 25 cm long candle burns for 45 minutes, how long will another candle of the same material and same thickness but 5 cm longer than the previous one burn?

Answer

Given:

A 25 cm long candle burns for 45 minutes.

Length of the other candle = 25 + 5 = 30 cm

A 1 cm long candle burns for $\dfrac{45}{25}$ = 1.8 minutes.

A 30 cm long candle burns for 30 × 1.8 = 54 minutes.

Hence, the 30 cm long candle will burn for 54 minutes.

A typist takes 80 minutes to type 24 pages. How long will he take to type 87 pages?

Answer

Given:

Time taken to type 24 pages = 80 minutes

Time taken to type 1 page = $\dfrac{80}{24} = \dfrac{10}{3}$ minutes

Time taken to type 87 pages = $87 \times \dfrac{10}{3} = \dfrac{870}{3}$ = 290 minutes

Hence, the typist will take 290 minutes to type 87 pages.

₹750 support a person for 15 days. For how many days will ₹2,500 support the same person?

Answer

Given:

₹ 750 supports a person for 15 days.

₹ 1 supports a person for $\dfrac{15}{750}$ day.

₹ 2,500 will support the person for $2500 \times \dfrac{15}{750} = \dfrac{37500}{750}$ = 50 days.

Hence, ₹ 2,500 will support a person for 50 days.

400 men have provisions for 23 weeks. They are joined by 60 men. How long will the provisions last?

Answer

Given:

400 men have provisions for 23 weeks.

This is a case of inverse variation (more men, provisions last for fewer weeks).

For 1 man, the provisions will last for 23 × 400 weeks.

Total number of men now = 400 + 60 = 460

For 460 men, the provisions will last for $\dfrac{23 \times 400}{460} = \dfrac{9200}{460}$ = 20 weeks.

Hence, the provisions will last for 20 weeks.

200 men have provisions for 30 days. If 50 men have left, for how many days the same provisions would last for the remaining men?

Answer

Given:

200 men have provisions for 30 days.

Number of remaining men = 200 − 50 = 150

This is a case of inverse variation (fewer men, provisions last for more days).

For 1 man, the provisions will last for 200 × 30 days.

For 150 men, the provisions will last for $\dfrac{200 \times 30}{150} = \dfrac{6000}{150}$ = 40 days.

Hence, the provisions will last for 40 days for the remaining men.

8 men can finish a certain amount of provisions in 40 days. If 2 more men join them, find for how many days will the same amount of provisions be sufficient.

Answer

Given:

8 men can finish the provisions in 40 days.

Total number of men now = 8 + 2 = 10

This is a case of inverse variation (more men, provisions last for fewer days).

For 1 man, the provisions will last for 8 × 40 days.

For 10 men, the provisions will last for $\dfrac{8 \times 40}{10} = \dfrac{320}{10}$ = 32 days.

Hence, the provisions will be sufficient for 32 days.

If the interest on ₹ 200 be ₹ 25 in a certain time, what will be the interest on ₹ 750 for the same time?

Answer

Given:

Interest on ₹ 200 = ₹ 25

This is a case of direct variation (more principal, more interest).

Interest on ₹ 1 = ₹ $\dfrac{25}{200}$

Interest on ₹ 750 = $750 \times \dfrac{25}{200} = \dfrac{18750}{200}$ = ₹ 93.75

Hence, the interest on ₹ 750 will be ₹ 93.75.

If 3 dozen eggs cost ₹ 90, find the cost of 3 scores of eggs. [1 score = 20]

Answer

Given:

3 dozen eggs = 3 × 12 = 36 eggs

Cost of 36 eggs = ₹ 90

1 score = 20

3 scores of eggs = 3 × 20 = 60 eggs

This is a case of direct variation (more eggs, more cost).

Cost of 1 egg = $\dfrac{90}{36}$ = ₹ 2.50

Cost of 60 eggs = 60 × 2.50 = ₹ 150

Hence, the cost of 3 scores of eggs is ₹ 150.

If the fare for 48 km is ₹ 288, what will be the fare for 36 km?

Answer

Given:

Fare for 48 km = ₹ 288

This is a case of direct variation (more distance, more fare).

Fare for 1 km = $\dfrac{288}{48}$ = ₹ 6

Fare for 36 km = 36 × 6 = ₹ 216

Hence, the fare for 36 km is ₹ 216.

What will be the cost of 3.20 kg of an item, if 3 kg of it costs ₹ 360?

Answer

Given:

Cost of 3 kg = ₹ 360

This is a case of direct variation (more weight, more cost).

Cost of 1 kg = $\dfrac{360}{3}$ = ₹ 120

Cost of 3.20 kg = 3.20 × 120 = ₹ 384

Hence, the cost of 3.20 kg of the item is ₹ 384.

If 9 lines of a print, in a column of a book, contain 36 words, how many words will a column of 51 lines contain?

Answer

Given:

9 lines contain 36 words.

This is a case of direct variation (more lines, more words).

1 line contains $\dfrac{36}{9}$ = 4 words.

51 lines contain 51 × 4 = 204 words.

Hence, a column of 51 lines will contain 204 words.

125 students have food sufficient for 18 days. If 25 more students join them, how long will the food last now?

What assumption have you made to come to your answer?

Answer

Given:

125 students have food for 18 days.

Total number of students now = 125 + 25 = 150

This is a case of inverse variation (more students, food lasts for fewer days).

For 1 student, the food will last for 125 × 18 days.

For 150 students, the food will last for $\dfrac{125 \times 18}{150} = \dfrac{2250}{150}$ = 15 days.

Hence, the food will last for 15 days.

Assumption: Each student consumes the same amount of food every day.

A carpenter prepares a new chair in 3 days, working 8 hours a day.

At least how many hours per day must he work in order to make the same chair in 4 days?

Answer

Given:

Working 8 hours a day, the chair is prepared in 3 days.

Total hours needed to make one chair = 3 × 8 = 24 hours

This is a case of inverse variation (more days, fewer hours per day).

Hours per day required to finish in 4 days = $\dfrac{24}{4}$ = 6 hours

Hence, he must work at least 6 hours per day.

A man earns ₹ 5,800 in 10 days. How much will he earn in the month of February of a leap year?

Answer

Given:

Earning in 10 days = ₹ 5,800

In a leap year, February has 29 days.

This is a case of direct variation (more days, more earning).

Earning in 1 day = $\dfrac{5800}{10}$ = ₹ 580

Earning in 29 days = 29 × 580 = ₹ 16,820

Hence, he will earn ₹ 16,820 in the month of February of a leap year.

A machine makes 500 rubber balls in 30 minutes. How many rubber balls will it make in $3\dfrac{1}{2}$ hours?

Answer

Given:

500 rubber balls are made in 30 minutes.

$3\dfrac{1}{2} \text { hours } = \dfrac{7}{2} \times 60$ = 210 minutes

This is a case of direct variation (more time, more balls).

Number of balls made in 1 minute = $\dfrac{500}{30} = \dfrac{50}{3}$

Number of balls made in 210 minutes = $210 \times \dfrac{50}{3} = \dfrac{10500}{3}$ = 3500

Hence, the machine will make 3500 rubber balls in $3\dfrac{1}{2}$ hours.

In a school's hostel mess, 20 children consume a certain quantity of ration in 6 days. However, 5 children did not return to the hostel after holidays. How long will the same amount of ration last now?

Answer

Given:

20 children consume the ration in 6 days.

Number of children now = 20 − 5 = 15

This is a case of inverse variation (fewer children, ration lasts for more days).

For 1 child, the ration will last for 20 × 6 days.

For 15 children, the ration will last for $\dfrac{20 \times 6}{15} = \dfrac{120}{15}$ = 8 days.

Hence, the ration will last for 8 days.

120 identical articles can be prepared by 25 men in 10 days. Find how many articles of the same kind can be prepared by 35 men in the same time?

Answer

Given:

25 men prepare 120 articles (in 10 days).

Since the time is the same in both cases, this is a case of direct variation (more men, more articles).

Number of articles prepared by 1 man = $\dfrac{120}{25} = \dfrac{24}{5}$

Number of articles prepared by 35 men = $35 \times \dfrac{24}{5} = \dfrac{840}{5}$ = 168

Hence, 35 men can prepare 168 articles in the same time.