Men-Days, Alternate Days and Work and Wages

Man-Days: Measuring a Job in Worker-Days

When a group of equally good workers does a job, the size of the job is the number of workers × the number of days. This product is the man-days (or worker-days) in the job.

"14 men can lay a road in 15 days." The road is 14 × 15 = 210 man-days of work. That one number answers every follow-up question:

• 21 men take 210 ÷ 21 = 10 days.
• One man alone would take 210 days.
• To finish in 6 days you need 210 ÷ 6 = 35 men.

$$M_1 D_1 = M_2 D_2$$

Men and days move opposite ways. More men means fewer days, in the same ratio. If the men go up by half (14 to 21), the days drop to two-thirds (15 to 10). This is inverse proportion.

Hours a day. If the hours a day change too, count man-hours instead. "14 men working 8 hours a day lay the road in 15 days." The road is 14 × 15 × 8 = 1,680 man-hours.

• 20 men working 12 hours a day: 1,680 ÷ (20 × 12) = 7 days.
• 28 men who must finish in 10 days: 1,680 ÷ (28 × 10) = 6 hours a day.

$$M_1 D_1 H_1 = M_2 D_2 H_2$$

Different people, different hours. "A finishes a job in 10 days working 8 hours a day, and B in 12 days working 10 hours a day. How many days will they take together, working 6 hours a day?" Count each person's hours: A needs 80 hours and B 120 hours. LCM 240: A does 3 units an hour and B 2, so together 5 units an hour and 48 hours in all. At 6 hours a day that is 8 days.

Man-days and the units method are the same idea. In the LCM method each person got their own units a day. Here every worker is 1 unit a day, so the job size is just men × days (× hours). When the workers are not equally good (men and women, or a faster team), each kind gets its own units a day; that comes later in this section.

Finding the answer in the right unit. Questions may ask for days, men, or hours a day. Write the fixed total first (1,680 man-hours), then divide by whatever two things are known. That one habit handles all three forms.

A group that is only part of the way through. If 14 men have already worked 5 of the 15 days, the work left is 14 × 10 = 140 man-days, whatever happens next. This "work left in man-days" idea drives every leaving and joining question later in this section.

Before you multiply, decide whether the answer should be bigger or smaller. If 14 men take 15 days and your answer for 21 men is 22.5 days, you have used direct proportion where it should be inverse.

The Chain Rule: Men, Days, Hours and the Amount of Work

When the amount of work changes as well, it goes under the line:

$$\frac{M_1 D_1 H_1}{W_1} = \frac{M_2 D_2 H_2}{W_2}$$

The left side is the man-hours spent on each unit of work. It stays the same as long as the workers are equally good.

Finding the men. "21 men dig a 56 m trench in 8 days, working 6 hours a day. How many men are needed to dig an 80 m trench in 10 days, working 8 hours a day?"

$$\frac{21 \times 8 \times 6}{56} = \frac{M_2 \times 10 \times 8}{80} \Rightarrow M_2 = \frac{21 \times 8 \times 6 \times 80}{56 \times 10 \times 8} = 18$$

Finding the days. "10 men print 4,000 pages in 6 days, working 8 hours a day. In how many days will 12 men print 6,000 pages, working 10 hours a day?"

$$D_2 = \frac{10 \times 6 \times 8 \times 6000}{4000 \times 12 \times 10} = 6 \text{ days}$$

The reasoning way. Start from the quantity you want and change it one factor at a time, asking "more or less?" each time.

For the trench: 21 × $\frac{80}{56}$ (more work) × $\frac{8}{10}$ (more days) × $\frac{6}{8}$ (longer hours) = 18 men.

Work measured by size. For walls, fields or roads, the work is the length, the area or the volume. "8 men build a wall 30 m long and 2 m high in 10 days. How many days will 12 men take to build a wall 45 m long and 4 m high?" The walls are 60 and 180 square metres, so

$$D_2 = \frac{8 \times 10 \times 180}{60 \times 12} = 20 \text{ days}$$

Workers who are better or worse. Add an efficiency factor E: $\frac{M_1 D_1 H_1 E_1}{W_1} = \frac{M_2 D_2 H_2 E_2}{W_2}$. If the trench diggers in the second job were each 1.5 times as good as the first lot, each counts as 1.5 men, and only 18 ÷ 1.5 = 12 men would be needed.

Finishing Early, and Men Leaving or Joining Midway

More men to finish early. "28 men can do a job in 30 days. How many more men are needed to finish it in 21 days?" The job is 840 man-days. In 21 days you need 840 ÷ 21 = 40 men, so 12 more. The trap is 40, which is the total, not the extra.

Men leaving midway. "60 men start a job that they can finish in 28 days. After 10 days, 12 men leave. How long does the job take in all?"

• Work left = what 60 men would do in the other 18 days = 60 × 18 = 1,080 man-days.
• Only 48 men are left, so they need 1,080 ÷ 48 = 22.5 days.
• Total = 10 + 22.5 = 32.5 days, which is 4.5 days late.

Men joining midway. Same job, but after 10 days 12 more men join. The 1,080 man-days are done by 72 men in 1,080 ÷ 72 = 15 days. Total = 25 days, so the job ends 3 days early.

The key step in both: find the work left in man-days, using the original team and the original plan, then divide by the new team.

Behind schedule. "A contractor hires 90 men to finish a job in 50 days. After 20 days only $\frac{1}{3}$ of the job is done. How many more men must he hire to finish on time?"

• 90 × 20 = 1,800 man-days did $\frac{1}{3}$ of the job, so the other $\frac{2}{3}$ needs 3,600 man-days.
• 30 days are left, so he needs 3,600 ÷ 30 = 120 men, that is 30 more.

Don't use the planned 90 × 50 = 4,500 man-days here. The job turned out bigger (5,400 man-days) than planned, and the work already done tells you its real size.

Ahead of schedule. The same steps tell you how many men can be released. If more than the planned share is done, the remaining man-days are fewer, and the extra men can go.

[Tier 2] Men dropping out every day. "A group of men planned to finish a job in 10 days. But 4 men dropped out at the end of every day, and the job was finished at the end of the 15th day. How many men were there at the start?" With N men at the start, the days had N, N − 4, N − 8, and so on down to N − 56 men. Their total is $15N - 4(1 + 2 + \cdots + 14) = 15N - 420$, and this must equal the planned 10N man-days. So 5N = 420 and N = 84.

[SSC Tip] Check whether the question wants the total number of men or the extra men, and the total time or the time after the change. SSC usually puts both versions in the options.

Men, Women and Children: Turning Everyone into One Unit

When a job has men, women, boys or children, they don't work at the same speed. Turn everyone into one kind of worker, or give each kind its own units a day, and then work in man-days as before.

"2 men or 5 women can do a job in 15 days." This means 2 men alone take 15 days, and 5 women alone also take 15 days. So 2 men do as much as 5 women: 2M = 5W.

The neat way is units. Give each man 5 units a day and each woman 2 units a day, so that 2 men and 5 women both do 10. The job is 10 × 15 = 150 units.

• "How long will 4 men and 5 women take?" 4 × 5 + 5 × 2 = 30 units a day, so 150 ÷ 30 = 5 days.
• "How many women must join 4 men to finish in 3 days?" In 3 days you need 50 units a day. 4 men give 20, so the women must give 30 more: 15 women.

The same in woman-days: the job is 5 × 15 = 75 woman-days, and 4 men are worth 10 women, so 4 men and 5 women count as 15 women: 75 ÷ 15 = 5 days.

A chain of equals. "16 men can finish a job in 9 days. If 2 men = 3 women = 6 children in work, how long will 4 men, 6 women and 12 children take?" Turn everyone into men: 6 women = 4 men and 12 children = 4 men, so the group is 4 + 4 + 4 = 12 men. The job is 16 × 9 = 144 man-days, so 144 ÷ 12 = 12 days.

Efficiency from individual times. "A man can do a job in 10 days and a woman in 15 days." A man and a woman work in the ratio 15 : 10 = 3 : 2 (flip the times, as in the Time and Work: Rates, Efficiency and Working Together section). So 2 men = 3 women.

Reading the wording.

• "2 men or 5 women" means each group alone; it does not mean 2 men and 5 women together.
• "A man does twice as much work as a woman" means 1 man = 2 women.
• "A woman takes twice as long as a man" also means 1 man = 2 women.
• "A man is 50% more efficient than a woman" means 2 men = 3 women.

After converting, check the direction: 4 men are worth 10 women, which must be more than 4 women, because a man does more. If your group shrinks when you turn men into women, the exchange rate is upside down.

Two Groups, Two Equations, and How Many Are Needed

When the exchange rate is not given. "6 men and 3 women can finish a job in 6 days, while 1 man and 4 women take 18 days. How long would 2 men and 1 woman take?"

Both groups do the same job, so (daily work) × (days) is the same:

$$6(6M + 3W) = 18(M + 4W) \Rightarrow 36M + 18W = 18M + 72W \Rightarrow 18M = 54W \Rightarrow M = 3W$$

So 1 man = 3 women. Give a woman 1 unit a day and a man 3.

• The job is 6 × (6 × 3 + 3 × 1) = 6 × 21 = 126 units. Check: 18 × (3 + 4) = 126.
• 2 men and 1 woman do 7 units a day, so they take 126 ÷ 7 = 18 days.
• One woman alone would take 126 days, and one man alone 42 days.

How many of one kind. "With the same job, how many men must work with 6 women to finish it in 7 days?" In 7 days you need 18 units a day. 6 women give 6, so the men must give 12, which is 4 men.

Individual times for each kind. "A man can do a job in 6 days, a woman in 10 days and a boy in 15 days. How many boys must help 3 men and 1 woman to finish the job in 1 day?" LCM 30: a man does 5 units a day, a woman 3 and a boy 2. All 30 units must be done in the one day. 3 men and 1 woman give 15 + 3 = 18 units, so the boys must give 12, which is 6 boys.

Men and boys, women and children. The same three steps work for any two kinds of worker:

1. Write the two groups as the same job: group 1's daily work × days = group 2's daily work × days.
2. Find the exchange rate (1 man = k boys).
3. Switch to units a day and answer what is asked.

When only one group's time is given. "3 men and 4 boys take 12 days. A man is twice as good as a boy. How long for 1 man and 3 boys?" Give a boy 1 unit and a man 2. The job is 12 × (6 + 4) = 120 units, and 1 man and 3 boys do 5 units a day, taking 120 ÷ 5 = 24 days.

Never subtract one group's days from the other's; the equation always comes from both groups doing the same job. And if "how many boys" comes out as 5.4 boys, go back and check, because SSC answers in whole workers.

Alternate Days: Two People Taking Turns

"A and B work on alternate days" means A works on day 1, B on day 2, A on day 3, and so on (or B first, if the question says so). Work in 2-day cycles.

When the cycles fit exactly. A takes 15 days and B 30 days. LCM 30: A does 2 units a day and B 1, so each 2-day cycle does 3 units. 10 cycles do all 30 units, so the job takes 20 days, whoever starts.

When there is a part left. A takes 10 days and B 16 days. LCM 80: A does 8 units a day and B 5, so each cycle does 13.

• 6 full cycles (12 days) do 78 units, and 2 units are left.
• If A started, day 13 is A's turn: 2 units at 8 a day take $\frac{1}{4}$ of a day. Total $12\frac{1}{4}$ days.
• If B started, day 13 is B's turn: 2 units at 5 a day take $\frac{2}{5}$ of a day. Total $12\frac{2}{5}$ days.

When the next person cannot finish it alone. A takes 20 days and B 24 days. LCM 120: A does 6 units a day and B 5, so a cycle does 11.

• 10 cycles (20 days) do 110 units, and 10 are left.
• A first: on day 21 A does 6, leaving 4. On day 22 B does those 4 in $\frac{4}{5}$ of a day. Total $21\frac{4}{5}$ days.
• B first: on day 21 B does 5, leaving 5. On day 22 A does them in $\frac{5}{6}$ of a day. Total $21\frac{5}{6}$ days.

The method. Divide the job by one cycle's work and take the whole number of cycles. Then go one day at a time, checking whose turn it is. Never just double "job ÷ cycle": $\frac{120}{11} \times 2 = 21\frac{9}{11}$ days is the wrong option in the last case, because it treats the two people as sharing every day at an average speed, when on the last day only one of them works.

The last day is a part day. When the answer is $12\frac{1}{4}$ days, the job ends a quarter of the way into the 13th day. If the question asks "on which day is the work completed?", the answer is the 13th day.

Comparing the two starts. When the cycles do not fit exactly, the job ends sooner if the faster person starts, because the faster one gets the extra day. When the cycles fit exactly, the start makes no difference.

[SSC Tip] After the full cycles, write the remaining units and the next person's name before doing anything else. Most wrong answers come from giving the last day to the wrong person, or from letting one person finish more than they can do in a day.

A Helper Every Third Day, Turns of Three, and Shifts

A helper who comes every third day. "A can do a job in 15 days and B in 45 days. A works every day, and B helps him on every third day (days 3, 6, 9 and so on). How long does the job take?" LCM 45: A does 3 units a day and B 1. Each 3-day block does 3 + 3 + (3 + 1) = 10 units.

• 4 blocks (12 days) do 40 units, and 5 are left.
• Day 13 is A alone: 3 units, leaving 2. Day 14 is A alone again: $\frac{2}{3}$ of a day.
• Total $13\frac{2}{3}$ days. B's next turn (day 15) never comes.

Alone one day, together the next. "A works every day and B joins him on every second day." Use 2-day blocks of A's two days plus B's one day, and follow the same steps.

Three people in turn. "A, B and C can do a job in 8, 10 and 20 days. They work one at a time: A on day 1, B on day 2, C on day 3, then A again." LCM 40: A does 5 units a day, B 4 and C 2, so a 3-day cycle does 11 units.

• 3 cycles (9 days) do 33 units, and 7 are left.
• Day 10 is A's: 5 units, leaving 2. Day 11 is B's: $\frac{2}{4} = \frac{1}{2}$ of a day.
• Total $10\frac{1}{2}$ days.

Shifts of a few hours. "A can finish a job in 9 hours and B in 12 hours. They work in turns of 2 hours each, A first." LCM 36: A does 4 units an hour and B 3. A's 2-hour shift does 8 units and B's does 6, so each 4-hour round does 14.

• 2 rounds (8 hours) do 28 units, and 8 are left.
• A's next shift does exactly 8 units in 2 hours.
• Total 8 + 2 = 10 hours. If B had taken the first shift, the last 8 units would need B's full shift (6) and half an hour of A's, so $10\frac{1}{2}$ hours.

Blocks of several days. "A works for 3 days, then B for 2 days, and so on" works the same way: find one block's work, take full blocks, and finish day by day.

Write one block as a pattern (A, A, A + B) before you add anything up. The usual wrong options give the helper a day he did not work, or divide the job by the average work per day.

Negative Work, and Speed That Changes

One builds, one breaks. "A can build a wall in 15 days, and B can knock it down in 24 days. If both work at the same time, how long will the wall take?" LCM 120: A adds 8 units a day and B removes 5, so the wall grows by 3 units a day: 120 ÷ 3 = 40 days. With fractions: $\frac{1}{15} - \frac{1}{24} = \frac{8 - 5}{120} = \frac{1}{40}$. This is the same idea as a pipe filling a tank while a leak empties it, in the Pipes and Cisterns section.

Two builders and a breaker. A and B can build a wall in 12 and 24 days, and C can break it down in 16 days. LCM 48: 4 + 2 − 3 = 3 units a day, so all three together need 16 days.

Builder and breaker on alternate days. "A builds a wall in 5 days and B knocks it down in 10 days. They work on alternate days, A first." LCM 10: A adds 2 units and B removes 1, so a 2-day cycle adds only 1 unit. The trap answer is 10 cycles = 20 days. Watch the last day: the wall is complete the moment A's day takes it to 10, before B gets a turn.

• After 8 cycles (16 days) the wall stands at 8 units.
• On day 17, A adds 2 and the wall reaches 10.
• Answer: 17 days.

So stop the cycles as soon as the builder's next day can finish the job, here at 8 + 2 = 10.

Speed that drops. "A can finish a job in 24 days. After 6 days he falls ill and works at $\frac{3}{4}$ of his speed. How long does the job take?" Take the job as 96 units, with A doing 4 a day. His 6 days give 24 units; the other 72 units at 3 a day take 24 days. Total 30 days.

Speed that rises with practice. "B can do a job in 30 days at his starting speed. After 10 days his speed goes up by 25%." In 10 days B does $\frac{1}{3}$ of the job. His new speed is $\frac{5}{4} \times \frac{1}{30} = \frac{1}{24}$ of the job a day, so the other $\frac{2}{3}$ takes 16 days. Total 26 days.

[Tier 2] A little more each day. "A worker does 2 units on day 1, 4 on day 2, 6 on day 3, and so on. The job is 110 units." After n days he has done 2 + 4 + … + 2n = n(n + 1) units, and n(n + 1) = 110 gives n = 10 days.

Work and Wages

The money for a job is shared in the ratio of the work each person did, not the days they worked.

Working together for the whole job. When everyone works all the days, the work done is in the ratio of their efficiencies, which is the inverse ratio of their times. "A can do a job in 27 days and B in 36 days. They do it together and get ₹7,000." LCM 108: A does 4 units a day and B 3, so the money is split 4 : 3. A gets ₹4,000 and B ₹3,000.

A helper's share. "A and B can do a job in 30 and 40 days. With C's help they finish it in 12 days and get ₹6,000. What is C's share?" LCM 120: A does 4 units a day and B 3. In 12 days A does 48 units and B 36, so C did 120 − 84 = 36 units. C's share = $\frac{36}{120}$ × 6,000 = ₹1,800. A gets ₹2,400 and B ₹1,800.

Quick way: C's share is simply the fraction of the job C did, which is 1 − (the part done by A and B).

People working different numbers of days. "A can do a job in 20 days and B in 36 days. A works for 5 days and leaves, and B finishes the rest. They get ₹4,800." A did $\frac{5}{20} = \frac{1}{4}$ of the job, so A gets ₹1,200 and B ₹3,600. B worked 27 days (the other $\frac{3}{4}$ of 36 days). Sharing by days, 5 : 27, would give A only ₹750, and that is the trap option.

Daily wages. "2 men and 3 women together earn ₹2,900 a day, while 3 men and 2 women earn ₹3,100 a day." Adding: 5 men and 5 women earn ₹6,000, so 1 man and 1 woman earn ₹1,200. Subtracting: a man earns ₹200 more than a woman. So a man earns ₹700 a day and a woman ₹500. A job that needs 5 men and 4 women for 8 days costs 8 × (3,500 + 2,000) = ₹44,000.

Comparing costs. When different teams can do a job, find each team's total bill: workers × days × daily wage. "A job needs 60 man-days or 90 woman-days. Men are paid ₹700 a day and women ₹500." Men cost 60 × 700 = ₹42,000 and women cost 90 × 500 = ₹45,000, so the men are cheaper by ₹3,000 even though each man is paid more.

Solved Examples

Question 1

24 workers can build a boundary wall in 15 days. In how many days can 18 workers build it? (a) $11\frac{1}{4}$ days (b) 21 days (c) 20 days (d) 18 days

Answer. The wall is 24 × 15 = 360 worker-days of work. 18 workers need 360 ÷ 18 = (c) 20 days. Fewer workers must take longer, so $11\frac{1}{4}$ days (15 × $\frac{18}{24}$, direct proportion) cannot be right.

Question 2

20 men working 9 hours a day can finish a job in 14 days. In how many days will 28 men working 6 hours a day finish it? (a) 15 days (b) 10 days (c) 12 days (d) 18 days

Answer. The job is 20 × 14 × 9 = 2,520 man-hours. 28 men working 6 hours a day put in 168 man-hours a day, so they need 2,520 ÷ 168 = (a) 15 days.

Question 3

20 men can dig a trench 120 m long in 9 days, working 8 hours a day. How many men are needed to dig a trench 150 m long in 10 days, working 6 hours a day? (a) 24 men (b) 25 men (c) 36 men (d) 30 men

Answer. Use $\frac{M_1 D_1 H_1}{W_1} = \frac{M_2 D_2 H_2}{W_2}$: $M_2 = \frac{20 \times 9 \times 8 \times 150}{120 \times 10 \times 6} = \frac{216000}{7200}$ = (d) 30 men. By reasoning: 20 × $\frac{150}{120}$ (more work, more men) × $\frac{9}{10}$ (more days, fewer men) × $\frac{8}{6}$ (shorter hours, more men) = 30.

Question 4

45 men can complete a job in 32 days. How many more men are needed to complete it in 24 days? (a) 60 (b) 15 (c) 12 (d) 20

Answer. The job is 45 × 32 = 1,440 man-days. In 24 days it needs 1,440 ÷ 24 = 60 men, so the extra men needed are 60 − 45 = (b) 15.

Question 5

A contractor took up a job to be finished in 54 days and put 80 men on it. After 18 days, only $\frac{1}{4}$ of the job was done. How many more men must he employ to finish the job on time? (a) 120 (b) 60 (c) 40 (d) 20

Answer. 80 men in 18 days put in 1,440 man-days, and that did $\frac{1}{4}$ of the job. So the other $\frac{3}{4}$ needs 3 × 1,440 = 4,320 man-days. 36 days are left, so he needs 4,320 ÷ 36 = 120 men, which means (c) 40 more. Using the original plan (80 × 54 man-days) would wrongly say no extra men are needed, because the job turned out bigger than planned.

Question 6

36 men can finish a job in 25 days. They work for 10 days, and then 9 of them leave. How many days in all does the job take? (a) 30 days (b) 20 days (c) $33\frac{1}{3}$ days (d) 35 days

Answer. The work left is what 36 men would do in the remaining 15 days: 36 × 15 = 540 man-days. Now 27 men are left, so they need 540 ÷ 27 = 20 days. Total = 10 + 20 = (a) 30 days.

Question 7

30 men can finish a job in 40 days. After 16 days, 10 more men join them. How many days earlier than planned is the job finished? (a) 10 days (b) 18 days (c) 4 days (d) 6 days

Answer. The work left after 16 days is 30 × 24 = 720 man-days. 40 men do it in 720 ÷ 40 = 18 days, so the job ends on day 16 + 18 = 34, which is (d) 6 days early. 18 days is the time after the new men joined, not the time saved.

Question 8

6 men or 8 women can finish a job in 20 days. In how many days will 9 men and 4 women finish it? (a) 12 days (b) 10 days (c) 15 days (d) 8 days

Answer. 6 men = 8 women, so 3 men = 4 women. Give a man 4 units a day and a woman 3. The job is 6 × 4 × 20 = 480 units. 9 men and 4 women do 36 + 12 = 48 units a day, so they need 480 ÷ 48 = (b) 10 days. (In man-days: the job is 6 × 20 = 120, and 4 women are worth 3 men, so the group is 12 men and 120 ÷ 12 = 10.) Check: the group is worth 12 men, twice the 6 men who need 20 days, so half the time is right.

Question 9

7 men and 2 women can finish a job in 5 days, while 3 men and 4 women take 8 days. In how many days can 1 man alone finish the job? (a) 80 days (b) 20 days (c) 40 days (d) 35 days

Answer. Same job, so 5(7M + 2W) = 8(3M + 4W), which gives 35M + 10W = 24M + 32W, so 11M = 22W and 1 man = 2 women. Give a woman 1 unit a day and a man 2. The job is 5 × (14 + 2) = 80 units (check: 8 × (6 + 4) = 80). One man alone takes 80 ÷ 2 = (c) 40 days.

Question 10

A man can finish a job in 10 days, a woman in 20 days and a boy in 40 days. How many boys must help 2 men and 4 women so that the job is finished in 2 days? (a) 4 (b) 2 (c) 5 (d) 8

Answer. LCM 40: a man does 4 units a day, a woman 2 and a boy 1. To finish in 2 days the team must do 20 units a day. 2 men and 4 women give 8 + 8 = 16 units, so the boys must give 4 units: (a) 4 boys. Check: 2 × (8 + 8 + 4) = 40.

Question 11

9 men can finish a job in 16 days. In work, 1 man = 2 women = 3 children. In how many days will 2 men, 6 women and 9 children finish the job? (a) 16 days (b) 8 days (c) 24 days (d) 18 days

Answer. Turn everyone into men: 6 women = 3 men and 9 children = 3 men, so the group is 2 + 3 + 3 = 8 men. The job is 9 × 16 = 144 man-days, so 8 men take 144 ÷ 8 = (d) 18 days.

Question 12

A can finish a job in 16 days and B in 28 days. They work on alternate days, A starting on the first day. In how many days is the job finished? (a) $20\frac{1}{2}$ days (b) $20\frac{2}{7}$ days (c) $20\frac{4}{11}$ days (d) 21 days

Answer. LCM 112: A does 7 units a day and B 4, so each 2-day cycle does 11. 10 cycles (20 days) do 110 units, leaving 2. Day 21 is A's turn, and 2 units at 7 a day take $\frac{2}{7}$ of a day. Total = (b) $20\frac{2}{7}$ days.

Question 13

A can do a job in 8 days and B in 10 days. They work on alternate days, and B works on the first day. In how many days is the job finished? (a) $8\frac{4}{5}$ days (b) $8\frac{8}{9}$ days (c) 9 days (d) $9\frac{1}{2}$ days

Answer. LCM 40: A does 5 units a day and B 4, so a 2-day cycle does 9. 4 cycles (8 days) do 36 units, leaving 4. Day 9 is B's turn (B started), and B does exactly 4 units a day, so the job ends at (c) 9 days.

Question 14

A can do a job in 10 days and B in 25 days. A works every day, and B helps him only on every third day. In how many days is the job finished? (a) $8\frac{6}{7}$ days (b) $8\frac{14}{17}$ days (c) 9 days (d) $7\frac{1}{7}$ days

Answer. LCM 50: A does 5 units a day and B 2. A 3-day block does 5 + 5 + (5 + 2) = 17 units. 2 blocks (6 days) do 34 units, leaving 16. Day 7 (A alone): 5, leaving 11. Day 8 (A alone): 5, leaving 6. Day 9 is a helper day, so A and B do 7 units a day, and 6 units take $\frac{6}{7}$ of it. Total = (a) $8\frac{6}{7}$ days.

Question 15

A, B and C can do a job in 5, 12 and 15 days respectively. They work one at a time: A on the first day, B on the second, C on the third, then A again, and so on. In how many days is the job finished? (a) $8\frac{4}{7}$ days (b) 9 days (c) $2\frac{6}{7}$ days (d) $8\frac{1}{4}$ days

Answer. LCM 60: A, B and C do 12, 5 and 4 units a day, so a 3-day cycle does 21. 2 cycles (6 days) do 42 units, leaving 18. Day 7 (A): 12, leaving 6. Day 8 (B): 5, leaving 1. Day 9 is C's turn, and 1 unit takes C $\frac{1}{4}$ of a day. Total = (d) $8\frac{1}{4}$ days.

Question 16

A can build a wall in 8 days and B can knock the same wall down in 12 days. If A builds while B breaks it down at the same time, in how many days will the wall be built? (a) $4\frac{4}{5}$ days (b) 24 days (c) 20 days (d) 4 days

Answer. LCM 24: A adds 3 units a day and B removes 2, so the wall grows by 1 unit a day and takes 24 ÷ 1 = (b) 24 days. By fractions: $\frac{1}{8} - \frac{1}{12} = \frac{3 - 2}{24} = \frac{1}{24}$.

Question 17

A and B can build a wall in 9 days and 36 days respectively, and C can knock it down in 12 days. If all three work at the same time, in how many days will the wall be built? (a) $4\frac{1}{2}$ days (b) 12 days (c) 18 days (d) $7\frac{1}{5}$ days

Answer. LCM 36: A adds 4 units a day, B adds 1 and C removes 3. Net = 4 + 1 − 3 = 2 units a day, so the wall takes 36 ÷ 2 = (c) 18 days.

Question 18

A can build a wall in 6 days and B can knock it down in 9 days. They work on alternate days, A starting first. In how many days will the wall be completed? (a) 31 days (b) 36 days (c) 33 days (d) 18 days

Answer. LCM 18: A adds 3 units on his days and B removes 2 on his, so each 2-day cycle adds 1 unit. After 15 cycles (30 days) the wall stands at 15 units. On day 31, A adds 3 and the wall reaches 18, so it is finished at (a) 31 days before B can break any of it. 36 days, 18 full cycles, misses that A finishes on his own day.

Question 19

A can finish a job in 20 days. After working for 8 days, his efficiency drops by 25%. In how many days in all is the job finished? (a) 23 days (b) $26\frac{2}{3}$ days (c) 25 days (d) 24 days

Answer. In 8 days A does $\frac{8}{20} = \frac{2}{5}$ of the job, so $\frac{3}{5}$ is left. At his normal speed this would take 12 days, but his speed is now $\frac{3}{4}$ of normal, so it takes 12 × $\frac{4}{3}$ = 16 days. Total = 8 + 16 = (d) 24 days. A 25% slower speed means $\frac{4}{3}$ times the time, not 1.25 times, so 23 days is wrong.

Question 20

A can do a job in 15 days and B in 10 days. They do it together and are paid ₹5,000. What is B's share? (a) ₹2,000 (b) ₹3,000 (c) ₹2,500 (d) ₹3,200

Answer. LCM 30: A does 2 units a day and B 3, and both work every day, so the money is shared 2 : 3. B gets $\frac{3}{5}$ × 5,000 = (b) ₹3,000. Sharing in the ratio of the times, 15 : 10, would give B only ₹2,000 and pay the slower worker more.

Question 21

A can do a job in 10 days and B in 40 days. With the help of C they finish it in 5 days, and the three are paid ₹5,600 in all. What is C's share? (a) ₹2,800 (b) ₹700 (c) ₹2,100 (d) ₹3,360

Answer. LCM 40: A does 4 units a day and B 1. In 5 days A does 20 units and B 5, so C did 40 − 25 = 15 units. C's share = $\frac{15}{40}$ × 5,600 = (c) ₹2,100.

Question 22

3 men and 2 women together earn ₹3,400 a day, while 2 men and 3 women earn ₹3,100 a day. How much will 4 men and 5 women earn in 6 days? (a) ₹5,700 (b) ₹36,000 (c) ₹33,000 (d) ₹34,200

Answer. Adding the two days: 5 men and 5 women earn ₹6,500, so 1 man and 1 woman earn ₹1,300. Subtracting: a man earns ₹300 more than a woman. So a man earns ₹800 a day and a woman ₹500. 4 men and 5 women earn 3,200 + 2,500 = ₹5,700 a day, and in 6 days (d) ₹34,200.