Unitary Method

Given:

Weight of 8 articles = 4.8 kg

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

Weight of 11 articles = 11 × 0.6 = 6.6 kg

Hence, the weight of 11 articles is 6.6 kg.

Given:

Weight of 6 books = 1.260 kg

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

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

Hence, 15 books will weigh 3.150 kg.

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 $\frac{48}{12}$ = 4 hours.

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

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 $\frac{45}{25}$ = 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.

Given:

Time taken to type 24 pages = 80 minutes

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

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

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

Given:

₹ 750 supports a person for 15 days.

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

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

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

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 $\frac{23 \times 400}{460} = \frac{9200}{460}$ = 20 weeks.

Hence, the provisions will last for 20 weeks.

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 $\frac{200 \times 30}{150} = \frac{6000}{150}$ = 40 days.

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

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 $\frac{8 \times 40}{10} = \frac{320}{10}$ = 32 days.

Hence, the provisions will be sufficient for 32 days.

Given:

Interest on ₹ 200 = ₹ 25

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

Interest on ₹ 1 = ₹ $\frac{25}{200}$

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

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

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 = $\frac{90}{36}$ = ₹ 2.50

Cost of 60 eggs = 60 × 2.50 = ₹ 150

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

Given:

Fare for 48 km = ₹ 288

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

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

Fare for 36 km = 36 × 6 = ₹ 216

Hence, the fare for 36 km is ₹ 216.

Given:

Cost of 3 kg = ₹ 360

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

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

Cost of 3.20 kg = 3.20 × 120 = ₹ 384

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

Given:

9 lines contain 36 words.

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

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

51 lines contain 51 × 4 = 204 words.

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

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 $\frac{125 \times 18}{150} = \frac{2250}{150}$ = 15 days.

Hence, the food will last for 15 days.

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

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 = $\frac{24}{4}$ = 6 hours

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

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 = $\frac{5800}{10}$ = ₹ 580

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

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

Given:

500 rubber balls are made in 30 minutes.

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

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

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

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

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

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 $\frac{20 \times 6}{15} = \frac{120}{15}$ = 8 days.

Hence, the ration will last for 8 days.

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 = $\frac{120}{25} = \frac{24}{5}$

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

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

Given:

Cost of $\frac{3}{5}$ kg ghee = ₹ 96

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

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

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

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

Given:

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

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

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

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

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

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 = $\frac{8}{10}$ hour

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

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

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 = $\frac{68}{20}$ hour

Time to gain 35 sec = $35 \times \frac{68}{20} = \frac{2380}{20}$ = 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.

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 $\frac{50 \times 3}{15} = \frac{150}{15}$ = 10 days.

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

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 = $\frac{1200}{60}$ = ₹ 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.

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 = $\frac{2800}{32}$ = 87.5 g

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

Hence, there will be 400 apples in the box.

Given:

Diesel used for 240 km = 20 litres

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

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

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

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

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 $\frac{1200 \times 15}{1800} = \frac{18000}{1800}$ = 10 days.

Hence, the provisions will last for 10 days.

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 $\frac{60 \times 18}{72} = \frac{1080}{72}$ = 15 days.

Hence, the provisions will finish off in 15 days.

Given:

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

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

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

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

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

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 $\frac{300 \times 70}{250} = \frac{21000}{250}$ = 84 days.

Hence, the provisions will be sufficient for 84 days.

Given:

A can do the work in 6 days ⇒ A's 1 day's work = $\frac{1}{6}$

B can do the work in 8 days ⇒ B's 1 day's work = $\frac{1}{8}$

(A + B)'s 1 day's work

$$\begin{aligned}= \frac{1}{6} + \frac{1}{8}\\[1em] \text { L.C.M. of 6 and 8 = 24 }\\[1em] = \frac{4}{24} + \frac{3}{24}\\[1em] = \frac{4 + 3}{24}\\[1em] = \frac{7}{24}\end{aligned}$$

Time taken by both together = $\frac{24}{7} = 3\frac{3}{7}$ days

Hence, A and B together will complete the work in $3\frac{3}{7}$ days.

Given:

(A + B)'s 1 day's work = $\frac{1}{10}$

B's 1 day's work = $\frac{1}{15}$

A's 1 day's work

$$\begin{aligned}= \frac{1}{10} - \frac{1}{15}\\[1em] \text { L.C.M. of 10 and 15 = 30 }\\[1em] = \frac{3}{30} - \frac{2}{30}\\[1em] = \frac{3 - 2}{30}\\[1em] = \frac{1}{30}\end{aligned}$$

Time taken by A alone = 30 days

Hence, A alone will take 30 days to do the work.

Given:

A's 1 day's work = $\frac{1}{4}$

B's 1 day's work = $\frac{1}{5}$

(i) (A + B)'s 1 day's work

$$\begin{aligned}= \frac{1}{4} + \frac{1}{5}\\[1em] \text { L.C.M. of 4 and 5 = 20 }\\[1em] = \frac{5}{20} + \frac{4}{20}\\[1em] = \frac{5 + 4}{20}\\[1em] = \frac{9}{20}\end{aligned}$$

Hence, in one day they can do $\frac{9}{20}$ of the work.

(ii) Work done in 2 days

$$\begin{aligned}= 2 \times \frac{9}{20}\\[1em] = \frac{18}{20}\\[1em] = \frac{9}{10}\end{aligned}$$

Hence, in 2 days they can do $\frac{9}{10}$ of the work.

Work left after 2 days

$$\begin{aligned}= 1 - \frac{9}{10}\\[1em] = \frac{10 - 9}{10}\\[1em] = \frac{1}{10}\end{aligned}$$

Hence, $\frac{1}{10}$ of the work will be left.

Given:

A's 1 hour's work = $\frac{1}{6}$

B's 1 hour's work = $\frac{1}{9}$

(i) Work done by A in 1 hour = $\frac{1}{6}$

Work done by B in 2 hours = $2 \times \frac{1}{9} = \frac{2}{9}$

Total work done in these 3 hours

$$\begin{aligned}= \frac{1}{6} + \frac{2}{9}\\[1em] \text { L.C.M. of 6 and 9 = 18 }\\[1em] = \frac{3}{18} + \frac{4}{18}\\[1em] = \frac{3 + 4}{18}\\[1em] = \frac{7}{18}\end{aligned}$$

Hence, $\frac{7}{18}$ of the work is done in these 3 hours.

(ii) Work still left

$$\begin{aligned}= 1 - \frac{7}{18}\\[1em] = \frac{18 - 7}{18}\\[1em] = \frac{11}{18}\end{aligned}$$

Hence, $\frac{11}{18}$ of the work is still left.

Given:

A's 1 day's work = $\frac{1}{12}$

B's 1 day's work = $\frac{1}{15}$

C's 1 day's work = $\frac{1}{20}$

(A + B + C)'s 1 day's work

$$\begin{aligned}= \frac{1}{12} + \frac{1}{15} + \frac{1}{20}\\[1em] \text { L.C.M. of 12, 15 and 20 = 60 }\\[1em] = \frac{5}{60} + \frac{4}{60} + \frac{3}{60}\\[1em] = \frac{5 + 4 + 3}{60}\\[1em] = \frac{12}{60}\\[1em] = \frac{1}{5}\end{aligned}$$

Time taken by all together = 5 days

Hence, A, B and C together will take 5 days to complete the work.

Given:

First tap fills in 10 hours ⇒ 1 hour's work = $\frac{1}{10}$

Second tap fills in 8 hours ⇒ 1 hour's work = $\frac{1}{8}$

Third tap empties in 15 hours ⇒ 1 hour's work = $-\frac{1}{15}$ (emptying)

Work done in 1 hour when all are opened together:

$$\begin{aligned}= \frac{1}{10} + \frac{1}{8} - \frac{1}{15}\\[1em] \text { L.C.M. of 10, 8 and 15 = 120 }\\[1em] = \frac{12}{120} + \frac{15}{120} - \frac{8}{120}\\[1em] = \frac{12 + 15 - 8}{120}\\[1em] = \frac{19}{120}\end{aligned}$$

Time taken to fill the cistern = $\frac{120}{19} = 6\frac{6}{19}$ hours

Hence, the cistern will be filled in $6\frac{6}{19}$ hours.

Given:

Mohit's 1 day's work = $\frac{1}{50}$

Anuj's 1 day's work = $\frac{1}{40}$

(i) Work done by Mohit in 20 days

$$\begin{aligned}= 20 \times \frac{1}{50}\\[1em] = \frac{20}{50}\\[1em] = \frac{2}{5}\end{aligned}$$

Hence, Mohit does $\frac{2}{5}$ of the work in 20 days.

(ii) Work left

$$\begin{aligned}= 1 - \frac{2}{5}\\[1em] = \frac{5 - 2}{5}\\[1em] = \frac{3}{5}\end{aligned}$$

Hence, $\frac{3}{5}$ of the work is left.

(iii) Time taken by Anuj to do $\frac{3}{5}$ of the work

$$\begin{aligned}= \frac{3}{5} \div \frac{1}{40}\\[1em] = \frac{3}{5} \times 40\\[1em] = 24 \text{ days}\end{aligned}$$

Hence, Anuj will take 24 days to complete the remaining work.

Given:

Joseph's 1 hour's work = $\frac{1}{20}$

Peter's 1 hour's work = $\frac{1}{25}$

(Joseph + Peter)'s 1 hour's work

$$\begin{aligned}= \frac{1}{20} + \frac{1}{25}\\[1em] \text { L.C.M. of 20 and 25 = 100 }\\[1em] = \frac{5}{100} + \frac{4}{100}\\[1em] = \frac{5 + 4}{100}\\[1em] = \frac{9}{100}\end{aligned}$$

The total work done by Joseph and Peter in 1 hour = $\frac{9}{100}$.

(i) Work done by both in 4 hours

$$\begin{aligned}= 4 \times \frac{9}{100}\\[1em] = \frac{36}{100}\\[1em] = \frac{9}{25}\end{aligned}$$

Hence, both together do $\frac{9}{25}$ of the work in 4 hours.

(ii) Work left

$$\begin{aligned}= 1 - \frac{9}{25}\\[1em] = \frac{25 - 9}{25}\\[1em] = \frac{16}{25}\end{aligned}$$

Hence, $\frac{16}{25}$ of the work is left.

(iii) Time taken by Peter to do $\frac{16}{25}$ of the work

$$\begin{aligned}= \frac{16}{25} \div \frac{1}{25}\\[1em] = \frac{16}{25} \times 25\\[1em] = 16 \text{ hours}\end{aligned}$$

Hence, Peter will take 16 hours to complete the remaining work.

Given:

A completes $\frac{1}{3}$ of the work in 10 hours.

B completes $\frac{2}{5}$ of the work in 12 hours.

(i) A's 1 hour's work

$$\begin{aligned}= \frac{1}{3} \div 10\\[1em] = \frac{1}{3} \times \frac{1}{10}\\[1em] = \frac{1}{30}\end{aligned}$$

Hence, A can do $\frac{1}{30}$ of the work in 1 hour.

(ii) B's 1 hour's work

$$\begin{aligned}= \frac{2}{5} \div 12\\[1em] = \frac{2}{5} \times \frac{1}{12}\\[1em] = \frac{2}{60}\\[1em] = \frac{1}{30}\end{aligned}$$

Hence, B can do $\frac{1}{30}$ of the work in 1 hour.

(iii) (A + B)'s 1 hour's work

$$\begin{aligned}= \frac{1}{30} + \frac{1}{30}\\[1em] = \frac{1 + 1}{30}\\[1em] = \frac{2}{30}\\[1em] = \frac{1}{15}\end{aligned}$$

Time taken to complete the work together = 15 hours

Hence, working together they will complete the work in 15 hours.

Given:

Shaheed's 1 day's work = $\frac{1}{3}$

Shaif's 1 day's work = $\frac{1}{4}$

(Shaheed + Shaif)'s 1 day's work

$$\begin{aligned}= \frac{1}{3} + \frac{1}{4}\\[1em] \text { L.C.M. of 3 and 4 = 12 }\\[1em] = \frac{4}{12} + \frac{3}{12}\\[1em] = \frac{4 + 3}{12}\\[1em] = \frac{7}{12}\end{aligned}$$

The total work done by Shaheed and Shaif in 1 day = $\frac{7}{12}$

(i) Time taken to prepare 1 chair = $\frac{12}{7} = 1\frac{5}{7}$ days

Hence, working together they will prepare one chair in $1\frac{5}{7}$ days.

(ii) Time taken to prepare 14 chairs

$$\begin{aligned}= 14 \times \frac{12}{7}\\[1em] = 2 \times 12\\[1em] = 24 \text{ days}\end{aligned}$$

Hence, they will prepare 14 chairs in 24 days.

Given:

(A + B + C)'s 1 day's work = $\frac{1}{4}$

A's 1 day's work = $\frac{1}{8}$

B's 1 day's work = $\frac{1}{12}$

C's 1 day's work

$$\begin{aligned}= \frac{1}{4} - \frac{1}{8} - \frac{1}{12}\\[1em] \text { L.C.M. of 4, 8 and 12 = 24 }\\[1em] = \frac{6}{24} - \frac{3}{24} - \frac{2}{24}\\[1em] = \frac{6 - 3 - 2}{24}\\[1em] = \frac{1}{24}\end{aligned}$$

Time taken by C alone = 24 days

Hence, C will take 24 days to finish the work.

This is a case of direct variation (more students, more computers).

Computers required for 1 student = $\frac{8}{10}$

Computers required for 40 students = $40 \times \frac{8}{10} = \frac{320}{10}$ = 32

Hence, option 1 is the correct option.

Cost of 6 pens = ₹ 90

Cost of 1 pen = $\frac{90}{6}$ = ₹ 15

Cost of 15 pens = 15 × 15 = ₹ 225

Hence, option 1 is the correct option.

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

1 man will complete the work in 6 × 90 days.

15 men will complete the work in $\frac{6 \times 90}{15} = \frac{540}{15}$ = 36 days.

Hence, option 2 is the correct option.

Speed = 24 km per hour

Time = 20 minutes = $\frac{20}{60}$ hours (60 minutes = 1 hour)

Distance = Speed × Time = $24 \times \frac{20}{60}$ = 8 km

Now, 15 minutes $= \frac{15}{60} \text { h }= \frac{1}{4}$ h:

Speed = $\frac{\text{Distance}}{\text{Time}} = \frac{8}{\frac{1}{4}} = 8 \times 4$ = 32 km per hour

Hence, option 4 is the correct option.

Total work = 20 × 28 = 560 worker-days

Half of the work = $\frac{560}{2}$ = 280 worker-days

Number of days for 8 workers = $\frac{280}{8}$ = 35 days

Hence, option 1 is the correct option.

A's 1 day's work = $\frac{1}{6}$

B's 1 day's work = $\frac{1}{12}$

(A + B)'s 1 day's work

$$\begin{aligned}= \frac{1}{6} + \frac{1}{12} \\[1em] \text { L.C.M. of 6 and 12 = 12 }\\[1em] = \frac{2}{12} + \frac{1}{12} \\[1em] = \frac{3}{12} \\[1em] = \frac{1}{4}\end{aligned}$$

Time taken together = 4 days

Hence, option 2 is the correct option.

Filling pipe's 1 hour's work = $\frac{1}{6}$

Emptying pipe's 1 hour's work = $\frac{1}{9}$

Work done in 1 hour when both pipes are open

$$\begin{aligned}= \frac{1}{6} - \frac{1}{9} \\[1em] \text { L.C.M. of 6 and 9 = 18 }\\[1em] = \frac{3}{18} - \frac{2}{18} \\[1em] = \frac{3 - 2}{18} \\[1em] = \frac{1}{18}\end{aligned}$$

Time taken to fill the tank = 18 hours

Hence, option 4 is the correct option.

A : B = $\frac{1}{2} : \frac{1}{3}$

Multiply both terms by 6

= $\frac{1}{2} \times 6:\frac{1}{3} \times 6$ = 3 : 2

A : B = 3 : 2

Let A = 3x, B = 2x

Given, A - B = 50

⇒ 3x − 2x = 50

⇒ x = 50

B = 2x = 2 × 50 = 100

Hence, option 3 is the correct option.

A : B : C = 5 : 2 : 3

Let A = 5x, B = 2x, C = 3x

Given, A + B − C = 200

⇒ 5x + 2x − 3x = 200

⇒ 4x = 200

⇒ x = $\frac{200}{4}$ = 50

B = 2x = 2 × 50 = 100

Hence, option 3 is the correct option.

In a proportion, product of extremes = product of means:

⇒ 5 × 6 = 30x × 15

⇒ 30 = 450x

⇒ x = $\frac{30}{450} = \frac{1}{15}$

Hence, option 2 is the correct option.

If a man completes a work in n days, then his one day's work = $\frac{1}{n}$, i.e. the n^th part of the work. So, Statement 1 is true.

The number of days and the one day's work are reciprocals of each other. If a woman's one day's work is n, then the time taken to complete the whole work = $\frac{1}{n}$ days, not n days. So, Statement 2 is false.

Hence, option 3 is the correct option.

Given:

56 men can do the work in 42 days.

Let 48 men complete the same work in x days.

Since less men take more days to complete the same work, the number of men and the number of days are in inverse variation.

⇒ 56 × 42 = 48 × x

$$\begin{aligned}\Rightarrow x = \frac{56 \times 42}{48}\\[1em] = \frac{2352}{48}\\[1em] = 49 \text{ days }\end{aligned}$$

Since 48 men will take 49 days and not 48 days,

Thus, Assertion (A) is false.

We know that,

Less men at work means more days are taken to complete the same work, and more men at work means less days are taken to complete the same work.

Thus, Reason (R) is true.

A is false and R is true.

Hence, option 2 is the correct option.

Given:

6 pipes can fill the tank in 64 minutes.

Let x pipes fill the same tank in 96 minutes.

Since more pipes take less time to fill the same tank, the number of pipes and the time taken are in inverse variation.

⇒ 6 × 64 = x × 96

$$\begin{aligned}\Rightarrow x = \frac{6 \times 64}{96}\\[1em] = \frac{384}{96}\\[1em] = 4 \text{ pipes }\end{aligned}$$

Since 4 pipes are required to fill the tank in 96 minutes,

Thus, Assertion (A) is true.

We know that,

When an increase in one quantity causes a corresponding decrease in the other quantity, and vice versa, such a variation is called an inverse variation.

Thus, Reason (R) is true.

Both A and R are true.

Hence, option 3 is the correct option.

Given:

A and B together can do the work in 20 days.

⇒ (A + B)'s 1 day's work = $\frac{1}{20}$

B alone can finish the work in 60 days.

⇒ B's 1 day's work = $\frac{1}{60}$

A's 1 day's work = (A + B)'s 1 day's work − B's 1 day's work

$$\begin{aligned}= \frac{1}{20} - \frac{1}{60}\\[1em] = \frac{3 - 1}{60}\\[1em] = \frac{2}{60}\\[1em] = \frac{1}{30}\end{aligned}$$

Since A's 1 day's work is $\frac{1}{30}$ and not $\frac{1}{40}$,

Thus, Assertion (A) is false.

We know that,

If a person completes a work in n days, then his 1 day's work = $\frac{1}{n}$, i.e. the number of days required to complete a work and the one day's work are reciprocal to each other.

Thus, Reason (R) is true.

A is false and R is true.

Hence, option 2 is the correct option.

Given:

X can do the work in 3 days.

⇒ X's 1 day's work = $\frac{1}{3}$

Y can do the work in 6 days.

⇒ Y's 1 day's work = $\frac{1}{6}$

(X + Y)'s 1 day's work

$$\begin{aligned}= \frac{1}{3} + \frac{1}{6}\\[1em] = \frac{2 + 1}{6}\\[1em] = \frac{3}{6}\\[1em] = \frac{1}{2}\end{aligned}$$

Time taken by X and Y together to complete the work

$$\begin{aligned}= \frac{\text{Work to be done}}{\text{Work done in 1 day}}\\[1em] = \frac{1}{\frac{1}{2}}\\[1em] = 2 \text{ days }\end{aligned}$$

Since X and Y together complete the work in 2 days,

Thus, Assertion (A) is true.

We know that,

Time required to do a certain work = $\frac{\text{Work to be done}}{\text{Work done in unit time}}$.

Thus, Reason (R) is true.

Both A and R are true.

Hence, option 3 is the correct option.