Unitary Method — Exercise 8(B)

The cost of $\dfrac{3}{5}$ kg of ghee is ₹ 96, find the cost of : (i) one kg ghee. (ii) $\dfrac{5}{8}$ kg ghee.

Answer

Given:

Cost of $\dfrac{3}{5}$ kg ghee = ₹ 96

(i) Cost of 1 kg ghee = $96 \div \dfrac{3}{5} = 96 \times \dfrac{5}{3} = \dfrac{480}{3}$ = ₹ 160

Hence, the cost of 1 kg ghee is ₹ 160.

(ii) Cost of $\dfrac{5}{8}$ kg ghee = $160 \times \dfrac{5}{8} = \dfrac{800}{8}$ = ₹ 100

Hence, the cost of $\dfrac{5}{8}$ kg ghee is ₹ 100.

$3\dfrac{1}{2}$ m of cloth costs ₹ 168, find the cost of $4\dfrac{1}{3}$ m of the same cloth.

Answer

Given:

Cost of $3\dfrac{1}{2} \text { m, i.e.} \dfrac{7}{2}$ m of cloth = ₹ 168

Cost of 1 m of cloth = $168 \div \dfrac{7}{2} = 168 \times \dfrac{2}{7} = \dfrac{336}{7}$ = ₹ 48

$4\dfrac{1}{3} \text { m }= \dfrac{13}{3}$ m

Cost of $\dfrac{13}{3}$ m of cloth = $\dfrac{13}{3} \times 48 = 13 \times 16$ = ₹ 208

Hence, the cost of $4\dfrac{1}{3}$ m of cloth is ₹ 208.

A wrist-watch loses 10 sec in every 8 hours. In how much time will it lose 15 sec?

Answer

Given:

The watch loses 10 sec in 8 hours.

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

Time to lose 1 sec = $\dfrac{8}{10}$ hour

Time to lose 15 sec = $15 \times \dfrac{8}{10} = \dfrac{120}{10}$ = 12 hours

Hence, the watch will lose 15 sec in 12 hours.

In 2 days and 20 hours a watch gains 20 sec. Find, how much time the watch will take to gain 35 sec.

Answer

Given:

2 days 20 hours = (2 × 24) + 20 = 68 hours

The watch gains 20 sec in 68 hours.

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

Time to gain 1 sec = $\dfrac{68}{20}$ hour

Time to gain 35 sec = $35 \times \dfrac{68}{20} = \dfrac{2380}{20}$ = 119 hours

Converting into days: 119 hours = 96 hours + 23 hours = 4 days 23 hours

Hence, the watch will take 4 days 23 hours to gain 35 sec.

50 men mow 32 hectares of land in 3 days. How many days will 15 men take to mow it?

Answer

Given:

50 men mow the land in 3 days.

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

1 man will mow the land in 50 × 3 days.

15 men will mow the land in $\dfrac{50 \times 3}{15} = \dfrac{150}{15}$ = 10 days.

Hence, 15 men will take 10 days to mow the land.

The wages of 10 workers for a six days week are ₹ 1,200. What are the one day wages:

(i) of one worker?

(ii) of 4 workers?

Answer

Given:

Wages of 10 workers for 6 days = ₹ 1,200

Total number of one-day works = 10 workers × 6 days = 60

(i) One day wage of 1 worker = $\dfrac{1200}{60}$ = ₹ 20

Hence, the one day wage of one worker is ₹ 20.

(ii) One day wage of 4 workers = 4 × 20 = ₹ 80

Hence, the one day wage of 4 workers is ₹ 80.

If 32 apples weigh 2 kg 800 g, how many apples will there be in a box, containing 35 kg of apples?

Answer

Given:

Weight of 32 apples = 2 kg 800 g = 2800 g

Weight of apples in the box = 35 kg = 35000 g

Weight of 1 apple = $\dfrac{2800}{32}$ = 87.5 g

Number of apples in the box = $\dfrac{35000}{87.5}$ = 400

Hence, there will be 400 apples in the box.

A truck uses 20 litres of diesel for 240 km. How many litres will be needed for 1200 km?

Answer

Given:

Diesel used for 240 km = 20 litres

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

Diesel used for 1 km = $\dfrac{20}{240} = \dfrac{1}{12}$ litre

Diesel used for 1200 km = $1200 \times \dfrac{1}{12} = \dfrac{1200}{12}$ = 100 litres

Hence, 100 litres of diesel will be needed for 1200 km.

A garrison of 1200 men has provisions for 15 days. How long will the provisions last if the garrison be increased by 600 men?

Answer

Given:

1200 men have provisions for 15 days.

Total number of men now = 1200 + 600 = 1800

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

For 1 man, the provisions will last for 1200 × 15 days.

For 1800 men, the provisions will last for $\dfrac{1200 \times 15}{1800} = \dfrac{18000}{1800}$ = 10 days.

Hence, the provisions will last for 10 days.

A camp has provisions for 60 pupils for 18 days. In how many days, the same provisions will finish off if the strength of the camp is increased to 72 pupils?

Answer

Given:

60 pupils have provisions for 18 days.

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

For 1 pupil, the provisions will last for 60 × 18 days.

For 72 pupils, the provisions will last for $\dfrac{60 \times 18}{72} = \dfrac{1080}{72}$ = 15 days.

Hence, the provisions will finish off in 15 days.

With the uniform speed of 27 km per hour, a vehicle runs for 1 hour 40 minutes. Find how much time will it take to cover the same distance with the speed of 30 km per hour.

Answer

Given:

Speed = 27 km/h, Time = 1 hour 40 minutes = $1\dfrac{40}{60} \text { h } = \dfrac{5}{3}$ h

Distance = Speed × Time = $27 \times \dfrac{5}{3} = \dfrac{135}{3}$ = 45 km

Now, with speed = 30 km/h to cover the same 45 km:

Time = $\dfrac{\text{Distance}}{\text{Speed}} = \dfrac{45}{30} = \dfrac{3}{2}$ hours = 1 hour 30 minutes

Hence, the vehicle will take 1 hour 30 minutes at the speed of 30 km per hour.

A fort has provisions for 300 men for 70 days. For how many days will the same provisions be sufficient for 250 men?

[Assume that each man, in both the cases, consumes the same amount of provisions].

Answer

Given:

300 men have provisions for 70 days.

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

For 1 man, the provisions will last for 300 × 70 days.

For 250 men, the provisions will last for $\dfrac{300 \times 70}{250} = \dfrac{21000}{250}$ = 84 days.

Hence, the provisions will be sufficient for 84 days.