Shortcuts & Speed Methods

The LCM Units Method

Time and work questions are fraction questions in disguise, and in Tier 1 you have about 36 seconds for each. Almost every method below turns those fractions into whole-number units that you can add and subtract in your head. Each one ends with where it goes wrong, and in this chapter that usually means taking a rate for a time, giving a helper a day he did not work, or counting an outlet as an inlet.

Total work = LCM of the times

When to use it: two or more people (or pipes) with whole-number times, where you would otherwise add fractions like $\frac{1}{15} + \frac{1}{35}$.

Fast way: call the whole job the LCM of the times. Each person's rate is then a whole number of units a day (units ÷ own time). Add or subtract the rates, and divide the units left by the rate.

For example: A can do a job in 15 days and B in 35 days. LCM = 105 units, so A does 7 units a day and B does 3. Together they do 10 a day and finish in 105 ÷ 10 = $10\frac{1}{2}$ days. If A works alone for 3 days and leaves, he has done 21 units, and B does the other 84 at 3 a day in 28 days.

Three people are no harder. A takes 7 days, B 14 and C 28: LCM = 28 units, rates 4, 2 and 1, so all three do 7 a day and finish in 4 days.

Times that are not whole numbers

If A takes $4\frac{1}{2}$ days and B takes 6, pick a total both go into: 18 units. A does 18 ÷ 4.5 = 4 a day and B does 3, so together they take $\frac{18}{7}$ = $2\frac{4}{7}$ days.

When fractions come first

If the question already talks in fractions ('A does $\frac{2}{5}$ of the job in 6 days'), turn that into a full-job time first: A takes 6 × $\frac{5}{2}$ = 15 days. Then use units as usual. For a bigger job, scale the units: three such jobs for A and B above are 3 × 105 = 315 units, which take 315 ÷ 10 = $31\frac{1}{2}$ days.

Where it fails:

• The rate is not the answer. 10 units a day does not mean 10 days. The time is units ÷ rate.
• Any common multiple works. 210 units give the same $10\frac{1}{2}$ days; the LCM just keeps the numbers small. Using the product 15 × 35 = 525 is slow, not wrong.
• When the question gives efficiencies, not times ('A is twice as fast as B'), take the efficiencies as the units a day (A 2, B 1) and get the total from rate × time.
• Units belong to one job. If the second job is twice the size, double the total before you divide.

Working Together: Two People, Three People and Pairs

Two people: product over sum

When to use it: two times are given and only the together time is asked.

Fast way:

$$T = \frac{xy}{x + y}$$

For example: 20 days and 80 days. T = $\frac{20 \times 80}{100}$ = 16 days.

One person from the pair

If the pair takes T days and A alone takes x days, then B alone takes $\frac{xT}{x - T}$ days. Together 16 days and A alone 20 days give B = $\frac{20 \times 16}{4}$ = 80 days.

Three people

$$T = \frac{xyz}{xy + yz + zx}$$

For example: 20, 36 and 45 days. T = $\frac{32400}{720 + 1620 + 900} = \frac{32400}{3240}$ = 10 days. Units are quicker: 180 units, rates 9 + 5 + 4 = 18 a day, 10 days.

Pair data: add and halve

When to use it: the times for A + B, B + C and C + A are given.

Fast way: take the LCM of the three pair times and find each pair's units a day. Add the three pair rates and halve: that is A + B + C. Each person alone = all three − the pair he is not in.

For example: A and B take 5 days, B and C 6 days, C and A 10 days. LCM = 30 units, and the pair rates 6, 5 and 3 add to 14. So all three do 7 a day and finish in $4\frac{2}{7}$ days. A alone does 7 − 5 = 2 a day (15 days), B does 7 − 3 = 4 ($7\frac{1}{2}$ days) and C does 7 − 6 = 1 (30 days).

Where it fails:

• Adding the times (20 + 80 = 100 days) or averaging them (50 days). Two people together are always faster than the faster one alone.
• Forgetting to halve. The pair rates count each person twice; without halving, the example gives 30 ÷ 14 = $2\frac{1}{7}$ days.
• Taking away the wrong pair. A alone = all three − (B + C), the pair without A.
• $\frac{xT}{x - T}$ needs x bigger than T. If one person alone is quicker than the pair, the numbers in your working are wrong.

Extra time over the pair. If A alone takes a days more than A and B together, and B alone takes b days more, the together time is $\sqrt{ab}$. Extra times of 16 and 9 days give a together time of 12 days, so A takes 28 and B takes 21. Check: $\frac{28 \times 21}{49}$ = 12.

Efficiency and Time

When to use it: 'A is 20% more efficient than B', 'A is twice as good a workman', 'A does in 6 days what B does in 9', or a ratio of efficiencies.

Fast way: for the same job, time goes the opposite way to efficiency. An efficiency ratio of a : b means a time ratio of b : a. For 'x% more efficient' multiply the time by $\frac{100}{100 + x}$; for 'x% less efficient' multiply by $\frac{100}{100 - x}$.

For example: B takes 36 days and A is 20% more efficient. A takes 36 × $\frac{100}{120}$ = 30 days. Someone 25% less efficient than a 15-day worker takes 15 × $\frac{100}{75}$ = 20 days.

Efficiency with a together time

Use the efficiencies as units a day. A is twice as good as B, and together they take 26 days. A does 2 a day and B 1, so the job is 3 × 26 = 78 units. A alone takes 39 days and B alone 78 days.

Efficiency with a gap in days

A is 50% more efficient than B and takes 17 days less. Efficiency 3 : 2 means time 2 : 3. The gap is 1 part = 17 days, so A takes 34 days and B 51 days.

Same work, different times

'A does in 6 days what B does in 9 days' gives time 2 : 3, so efficiency 3 : 2. 'A does half as much as B in three-quarters of the time' gives A's rate as $\frac{1}{2} \div \frac{3}{4} = \frac{2}{3}$ of B's, so A : B = 2 : 3.

Where it fails:

• 20% more efficient does not mean 20% less time. 36 days become 30, which is only $16\frac{2}{3}\%$ less. 36 × 0.8 = 28.8 days is the usual trap.
• Flipping the wrong ratio. Efficiency 3 : 2 gives time 2 : 3. The better worker takes fewer days.
• 'B is 25% less efficient than A' is not the same as 'A is 25% more efficient than B'. Here A : B = 4 : 3, so A is $33\frac{1}{3}\%$ more efficient.

Turn the words into a ratio before anything else: 20% more is 6 : 5, 50% more is 3 : 2, twice as good is 2 : 1, 25% less is 3 : 4. Flip it for time, then one part does the rest.

Joining Late, Leaving Early and the Work Left

For the first two parts, A takes 21 days and B takes 28: 84 units, A does 4 a day and B does 3.

Joins after n days, or leaves after n days

When to use it: one person starts alone and the other joins later, or both start and one leaves after some days.

Fast way: go forward in time. Take the units done in the first stretch off the total and divide what is left by the new rate.

For example:

• A starts and B joins after 7 days. A does 28 units, the other 56 take 56 ÷ 7 = 8 days, so 15 days in all.
• Both start and A leaves after 6 days. They do 42 units, B does the other 42 in 14 days, so 20 days in all.

Leaves n days before the end: work backwards

When to use it: 'B leaves 14 days before the work is finished' or 'A was away for the last 4 days'.

Fast way: let the whole job take T days. A works all T days and B works T − n. So aT + b(T − n) = total, which gives

$$T = \frac{\text{total} + bn}{a + b}$$

In words: pretend B never left. That adds bn units to the job, and now both work the whole time.

For example: B leaves 14 days before the end. T = (84 + 3 × 14) ÷ 7 = 18 days. Check: A works 18 days (72 units) and B works 4 days (12 units).

The work-left method

When to use it: one person works some days, another finishes, and someone's full time is asked.

Fast way: turn each stretch into a fraction of the job. The finisher's share is days worked ÷ his own time, and the rest belongs to the first person.

For example: A works 5 days and leaves; B, who takes 21 days alone, finishes in 14 days. B did $\frac{14}{21} = \frac{2}{3}$, so A's 5 days did $\frac{1}{3}$ and A alone takes 5 × 3 = 15 days.

Where it fails:

• 'Leaves after 6 days' counts from the start; 'leaves 6 days before the end' counts from the finish. They are easy to mix up, so check which one the question says.
• Adding n to the together time. Together they take 12 days, but 12 + 14 = 26 days is not the answer when B leaves 14 days early.
• In the work-left method, the finisher did the fraction that was left, not the fraction you found first.

[SSC Tip] When several people leave at different times, give each person his own number of working days and write one line: sum of (rate × days) = total units. It handles any mix of joining and leaving.

Men-Days and the Chain Rule

When to use it: groups of workers, hours a day, and jobs of different sizes.

Fast way: the work a group does is men × days × hours a day. So for two groups

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

Put the unknown alone and cancel before you multiply.

For example: 16 men working 5 hours a day lay 360 m of pipeline in 9 days. How many days will 24 men working 7 hours a day take to lay 504 m? D = $\frac{16 \times 9 \times 5 \times 504}{360 \times 24 \times 7}$ = 6 days. Another way to see it: each metre needs 2 man-hours, so 504 m need 1,008 man-hours, and the new group gives 168 man-hours a day.

More men to finish early

40 men can finish a job in 30 days. After 12 days, how many more men are needed to finish 6 days early? The job is 1,200 man-days and 480 are done. The other 720 must be done in 30 − 12 − 6 = 12 days, which needs 60 men, so 20 more men.

Some men leave midway

48 men can finish a job in 35 days. After 15 days, 8 men leave. The job is 1,680 man-days, 720 are done, and the 40 men left do the other 960 in 24 more days, 39 days in all.

Where it fails:

• Putting the whole job on the second side when part of it is done. Only the work left goes there. In the 'more men' example, 1,200 man-days in 24 days gives 50 men and '10 more', which is the trap.
• Days left = total days − days gone − days early. Write all three numbers before you divide.
• Mixed units. W must be in the same unit on both sides (metres with metres), and H is hours a day, not total hours.
• More men means fewer days. If your answer grows when the men grow, you have flipped a ratio.

Men, Women and Children in One Unit

When to use it: a job done by mixed groups (men, women, children, boys), and the time for another mix or the number of one kind is asked.

Fast way: turn everyone into one unit, picking the one that keeps the numbers whole. Then it is a man-days question.

For example (ratio given): in work, 1 man = 2 women = 4 children. A job takes 6 men 10 days, which is 60 man-days. 4 men, 4 women and 8 children are worth 4 + 2 + 2 = 8 men, so they take 60 ÷ 8 = $7\frac{1}{2}$ days.

Another one (two groups, same job): 2 men and 3 women finish a job in 16 days; 3 men and 2 women finish it in 14 days. The work is the same, so

16(2m + 3w) = 14(3m + 2w), which gives 32m + 48w = 42m + 28w, so 20w = 10m.

1 man = 2 women. The job is 16 × 7 = 112 woman-days, so 7 men (14 women's worth) take 8 days.

How many of one kind?

The same 112 woman-day job must be done in 4 days by 5 men and some women. 4 days need 28 women's worth of work each day. 5 men give 10, so 18 women are needed.

'Or' lists

'2 men or 3 women or 4 boys can do it in 12 days' means 2 men = 3 women = 4 boys. So 1 man = 1.5 women = 2 boys, and the job is 24 man-days.

Where it fails:

• Counting heads. 5 men and 5 women are not 10 workers of one kind.
• Flipping the ratio. 4 men = 6 women means 1 man = $\frac{3}{2}$ women, not $\frac{2}{3}$.
• Writing the equation from days instead of work. Equal jobs mean (group strength) × days is equal.
• Using the work ratio for money. If wages are given in another ratio, use that one for wages only.

It is easiest to count in the slowest worker when the others are whole multiples of him. If 1 man = 2 women = 4 children, count in children and call a man 4 and a woman 2. If 2 men = 3 women, count in pairs: every 2 men become 3 women.

Alternate Days and Turns

When to use it: A and B work on alternate days, or people take turns, or a helper comes every third day.

Fast way:

1. One cycle is one day for each person, so it does a + b units in 2 days.
2. Run as many full cycles as you can without finishing the job.
3. Hand the leftover to whoever's turn it is, then to the next person if needed.

For example: A takes 8 days and B takes 12. 24 units, A does 3 a day and B 2, so a cycle does 5 units in 2 days. 4 cycles (8 days) do 20 units and leave 4.

• A starts: day 9 is A's, 3 units, 1 left. Day 10 is B's, and 1 unit is half of his day. $9\frac{1}{2}$ days.
• B starts: day 9 is B's, 2 units, 2 left. Day 10 is A's, and 2 of his 3 units take $\frac{2}{3}$ of a day. $9\frac{2}{3}$ days.

The half-day finish

When the last bit is half a day's work for the person on duty, the answer ends in $\frac{1}{2}$, as with A starting above. Options like $9\frac{1}{2}$, $9\frac{3}{5}$ and $9\frac{2}{3}$ often sit side by side, so do the last two days by hand.

A helper every third day

If A works every day and B joins him on every third day, one cycle is 3 days and does 3a + b units. Run whole cycles, then do the leftover from day 1 of the next cycle, when A is alone.

Where it fails:

• Dividing the total by the cycle rate and doubling: 24 ÷ 5 × 2 = $9\frac{3}{5}$ days. The last day belongs to one person, not to an average of the two.
• Giving the leftover to the wrong person. After a whole number of cycles, the person who started is on duty again.
• If the cycles finish the job exactly, the last day is the second person's and the answer is the same whoever starts. With 6 days and 12 days (12 units, 3 per cycle) it is 8 days either way.

As a check, the answer is always close to twice the together time. Here the together time is $4\frac{4}{5}$ days, so the answer must be near $9\frac{3}{5}$; only the last day or two need care.

Negative Work: Builders and Breakers

When to use it: one person builds and another breaks (a wall and a demolisher, a digger and a filler), or one pipe fills while another empties.

Fast way: give the breaker a minus sign. When both work at the same time, the net rate is build − break.

For example: A builds a wall in 10 days and B can knock it down in 25 days. 50 units: A +5 a day, B −2. Working together, the wall grows 3 units a day and is finished in 50 ÷ 3 = $16\frac{2}{3}$ days.

Alternate days: the last-day overshoot

Now A and B work on alternate days, A first. Each 2-day cycle adds only 3 units, so it looks like 50 ÷ 3 × 2 = $33\frac{1}{3}$ days. That is the trap. The wall is finished the moment one of A's days takes it to 50, and B never gets to knock that day's work down.

So stop the cycles early. After k cycles the wall stands at 3k units, and A's next day adds 5. We need $3k + 5 \ge 50$, so k = 15. After 30 days the wall is 45 units high, and on day 31 A adds the last 5. 31 days.

$$\text{days} = 2 \times \frac{50 - 5}{3} + 1 = 31$$

The rule: (total − builder's one day) ÷ net gain per cycle gives the full cycles; then add the builder's last day. If that division is not whole, round the cycles up, and the builder needs only part of his last day.

Who starts matters

If B starts on an empty site, his first day does nothing, because he cannot knock down a wall that is not there. The job then takes 32 days.

Where it fails:

• Using 2 × (total ÷ net) for alternate days. Here that gives $33\frac{1}{3}$ days, the classic wrong option.
• If the breaker is faster than the builder, the net rate is below zero and the wall is never finished.
• Mixing up 'together' and 'alternate'. Together the answer is total ÷ net; alternately it is about twice that, less the overshoot.

Work and Wages

When to use it: a payment for a job is shared among the people who did it.

Fast way: wages go in the ratio of work done, and work done = rate × days worked. With units this is quick: each share = that person's units ÷ total units.

For example (everyone works throughout): A takes 10 days and B 15. They work together and are paid ₹4,500. In 30 units, A does 3 a day and B does 2, so the split is 3 : 2: A gets ₹2,700 and B ₹1,800. The faster worker earns more; splitting in the ratio of times, 10 : 15, gets it backwards.

Another one (a helper): with C's help, A and B finish the same job in 5 days. All three do 30 ÷ 5 = 6 units a day, so C does 6 − 3 − 2 = 1. The shares go 3 : 2 : 1, and out of ₹6,000 C gets ₹1,000.

One more (different days): A works 4 days (12 units) and leaves; B does the other 18 units in 9 days. Shares go 12 : 18 = 2 : 3, not 4 : 9 (the days) and not 3 : 2 (the rates).

Daily wages

When the daily rate is fixed, total wages go with man-days. 12 men earn ₹28,800 in 6 days, which is ₹400 a man-day. 15 men in 8 days give 120 man-days, so they earn ₹48,000.

If a man's daily wage is twice a woman's, convert to one unit first, exactly as you do for work.

Where it fails:

• Sharing by days worked, or by the times taken alone. Both put money with the wrong person.
• The helper's share is the work he did: the together rate with him minus the rate without him.
• When work and wages come in different ratios (a man works like 2 women but is paid like 3), use the work ratio for time and the wage ratio for money.

Pipes and Cisterns as Signed Work

When to use it: any pipe question.

Fast way: a tank is a job. Inlets are + and outlets or leaks are −. Take the tank as the LCM of the times, add the signed rates, and divide.

For example: A fills a tank in 20 hours, B fills it in 60 hours and C empties it in 40 hours. 120 units: A +6, B +2, C −3. With all three open the tank gains 5 units an hour and fills in 24 hours.

The leak formula

A pipe fills a tank in x hours, but with a leak it takes y hours (y is more than x). The leak alone empties the full tank in

$$\frac{xy}{y - x}$$

For example: 10 hours without the leak and 12 hours with it. The leak alone takes $\frac{120}{2}$ = 60 hours. Check: $\frac{1}{10} - \frac{1}{60} = \frac{1}{12}$.

Capacity in litres from a rate difference

Find the gap in units, match it with the litres given, and scale up.

For example: a leak empties a full tank in 12 hours. An inlet pouring 3 litres a minute is opened, and now the full tank empties in 20 hours. Take 60 units. The leak takes out 5 an hour and the net loss is 3, so the inlet puts in 2 units an hour. The inlet gives 3 × 60 = 180 litres an hour, so 1 unit = 90 litres and the tank holds 5,400 litres.

Where it fails:

• Swapping x and y in the leak formula. y − x must be positive, because the tank with the leak always takes longer.
• If the outlets beat the inlets, the net rate is below zero and the tank never fills.
• Mixing minutes and hours. 3 litres a minute is 180 litres an hour.
• A leak's time is for a full tank. A leak that empties a full tank in 6 hours empties a half-full one in 3.

[SSC Tip] In the leak formula both numbers are filling times. If the question says 'it takes 2 hours longer because of the leak', add first: y = x + 2.

Pipes Opened and Closed at Different Times

For the first three examples, A fills a tank in 15 hours and B in 60: 60 units, A +4 an hour and B +1.

Closed after n hours, or opened later

Fast way: go forward in time, stretch by stretch.

For example:

• Both open, B closed after 4 hours: 20 units are in, and A adds the other 40 in 10 hours. 14 hours in all.
• A opened, B opened 3 hours later: A puts in 12 units, and the other 48 take 48 ÷ 5 = 9.6 hours. 12 hours 36 minutes in all.

Closed n hours before the tank is full: work backwards

This is the 'leaves before the end' idea. The pipe that stays open runs all T hours, and the other runs T − n.

For example: both open, A closed 5 hours before the tank is full. 4(T − 5) + T = 60, so 5T = 80 and T = 16 hours. Check: A runs 11 hours (44 units) and B runs 16 hours (16 units).

Inlet and outlet in turns

Fast way: stop the cycles one inlet hour early. After k cycles the level is k × (net gain); the tank fills during the next inlet hour once that level + one inlet hour reaches the top.

For example: A fills a tank in 5 hours and B empties it in 6 hours. They are opened alternately for an hour each, A first. 30 units: A +6, B −5, so each 2-hour cycle adds 1. We need $k + 6 \ge 30$, so k = 24: after 48 hours there are 24 units, and in the 49th hour A adds the last 6. 49 hours, not 60.

Two inlets in turns are just alternate days. Pipes taking 5 and 7 hours (35 units, 7 + 5 per cycle) give 24 units in 2 cycles. The 5th hour, A's, adds 7, and B needs $\frac{4}{5}$ of the 6th hour for the last 4: 5 hours 48 minutes.

Where it fails:

• 30 ÷ 1 × 2 = 60 hours ignores that the tank fills during an inlet hour, before the outlet gets its turn.
• An outlet opened first on an empty tank does nothing in its hour, but that hour still counts in the time.
• 'Closed after 3 hours' and 'closed 3 hours before it is full' are different questions. Read which end the n hours are counted from.

With the inlet going first, the answer is an even number of hours plus part or all of one inlet hour, so a round even number like 60 hours is almost certainly wrong.

Option Elimination

When to use it: before the full working, or when you are short of time.

Fast way: run quick checks on the four options.

• Together beats the fastest. Two people together take less time than the faster one alone and more than half of his time. With 18 and 21 days, the answer lies between 9 and 18 days.
• The answer fits the LCM. The time is (LCM) ÷ (units a day), so its denominator must divide the units a day. If the rates add to 7, look for a 7 in the denominator.
• Leaks slow things down. With a leak the tank takes longer, and a leak's own time must be longer than the filling time (or the tank would never fill).
• Alternate days are about twice the together time.
• The faster worker earns more when both work the same days.

For example: A can do a job in 18 days and B in 21 days. Together they take (a) $19\frac{1}{2}$ days (b) $9\frac{9}{13}$ days (c) 39 days (d) 9 days.

(a) and (c) are more than 18, so they are out. (d) is half of 18, which would need B to be as fast as A, so it is out too. That leaves (b), and the LCM check agrees: 126 units, 7 + 6 = 13 a day, $\frac{126}{13}$ days.

Another one: a pipe fills a tank in 15 hours, but with a leak it takes 20 hours. The leak alone empties the full tank in (a) 5 hours (b) $8\frac{4}{7}$ hours (c) 35 hours (d) 60 hours.

The leak must be slower than the pipe, so (a) and (b) are out. (c) just adds the times. The formula gives $\frac{15 \times 20}{5}$ = 60, so the answer is (d). ((b) is the together time of 15 and 20, a common slip.)

Where it fails:

• With builders and breakers, the together time is longer than the builder's own time, so the 'together beats the fastest' check flips.
• When two options both pass, do the working.

[SSC Tip] Trap options come from a short list: the sum or average of the times, the units a day given as days, 2 × total ÷ net for alternate days, the leak formula with the times swapped, and wages split by days. If your answer is one of these, go over that one step again before you mark it.

Solved Examples

Every answer below uses units or a one-line formula, and the time check shows roughly how long it should take you.

Question 1

A can finish a job in 26 days and B in 39 days. They work together for 6 days, and then A leaves. In how many more days will B finish the job?

Answer. Take 78 units: A does 3 a day and B does 2. In 6 days together they do 30 units, so 48 are left. B does those at 2 a day in 24 days.

Time check: about 15 seconds.

Question 2

A and B can do a job in 27 days, B and C in 36 days, and C and A in 54 days. How long will A take alone?

Answer. Take 108 units. The pair rates are 4, 3 and 2, which add to 9, so all three together do 4.5 a day. A alone does 4.5 − 3 (the B and C rate) = 1.5 a day, so A takes 108 ÷ 1.5 = 72 days. Skipping the halving step gives 9 − 3 = 6 a day and 18 days.

Time check: about 20 seconds.

Question 3

A is 20% more efficient than B, and together they finish a job in 30 days. How long would A take alone?

Answer. A : B = 120 : 100 = 6 : 5. The job is 11 × 30 = 330 units, so A alone takes 330 ÷ 6 = 55 days (and B takes 66).

Time check: about 12 seconds.

Question 4

A and B can do a job in 24 days and 36 days. They start together, but A leaves 6 days before the job is finished. How long does the job take in all?

Answer. Take 72 units: A does 3 a day, B does 2. B works all T days and A works T − 6. So 3(T − 6) + 2T = 72, which gives 5T = 90 and T = 18 days. Check: A works 12 days (36 units) and B works 18 days (36 units).

Time check: about 20 seconds.

Question 5

A can do a job in 45 days. He works alone for 15 days, then B joins him and they finish the rest in 12 days. How long would B take alone?

Answer. A's 15 days are $\frac{1}{3}$ of the job, so $\frac{2}{3}$ is left. The pair does $\frac{2}{3}$ in 12 days, so they would take 18 days for the whole job. Then B = $\frac{45 \times 18}{45 - 18}$ = 30 days.

Time check: about 20 seconds.

Question 6

18 men working 7 hours a day lay 630 m of cable in 10 days. How many men working 8 hours a day are needed to lay 960 m in 12 days?

Answer. $\frac{18 \times 10 \times 7}{630} = \frac{M \times 12 \times 8}{960}$. The left side is 2 and the right side is $\frac{M}{10}$, so M = 20 men.

Time check: about 20 seconds.

Question 7

A contractor puts 50 men on a road that is to be built in 60 days. After 20 days only $\frac{1}{4}$ of the road is built. How many more men must he hire to finish on time?

Answer. 50 × 20 = 1,000 man-days built $\frac{1}{4}$ of the road, so the whole road needs 4,000. The other 3,000 must be done in 40 days, which takes 75 men. He needs 25 more men.

Time check: about 20 seconds.

Question 8

4 men or 6 women can do a job in 15 days. How long will 2 men and 5 women take?

Answer. 4 men = 6 women, so 1 man = 1.5 women. The job is 6 × 15 = 90 woman-days. 2 men and 5 women are worth 3 + 5 = 8 women, so they take 90 ÷ 8 = $11\frac{1}{4}$ days.

Time check: about 15 seconds.

Question 9

A can do a job in 16 days and B in 20 days. They work on alternate days, A first. In how many days is the job done?

Answer. Take 80 units: A does 5 a day and B 4, so a 2-day cycle does 9. 8 cycles (16 days) do 72 units, leaving 8. Day 17 is A's: 5 units, 3 left. Day 18 is B's, and 3 of his 4 units take $\frac{3}{4}$ of the day. $17\frac{3}{4}$ days.

Time check: about 25 seconds.

Question 10

Ravi can build a wall in 4 days and Mohan can pull it down in 10 days. They work on alternate days, Ravi first. On which day is the wall finished?

Answer. Take 20 units: Ravi +5 a day, Mohan −2, so each 2-day cycle adds 3. Stop the cycles early: $3k + 5 \ge 20$ gives k = 5. After 10 days the wall is 15 units high, and on day 11 Ravi adds the last 5. Day 11. Working out 20 ÷ 3 × 2 = $13\frac{1}{3}$ days misses that Ravi finishes on his own day.

Time check: about 20 seconds.

Question 11

A can do a job in 18 days and B in 45 days. A works for 6 days and leaves, and B finishes the rest. They are paid ₹8,400 for the job. What is B's share?

Answer. A's 6 days are $\frac{6}{18} = \frac{1}{3}$ of the job, so B did $\frac{2}{3}$ (it took him 30 days). B gets $\frac{2}{3}$ of 8,400 = ₹5,600. Splitting by the rates, 5 : 2, would give B only ₹2,400, but B worked far more days.

Time check: about 15 seconds.

Question 12

Two pipes can fill a tank in 28 and 70 minutes. Because of a leak at the bottom, together they take 24 minutes. How long would the leak alone take to empty the full tank?

Answer. Without the leak, together they take $\frac{28 \times 70}{98}$ = 20 minutes. With it, 24 minutes. Leak formula: $\frac{20 \times 24}{24 - 20}$ = 120 minutes, which is 2 hours.

Time check: about 15 seconds.

Question 13

Pipe A fills a tank in 5 hours and pipe B empties it in 15 hours. Starting with an empty tank, they are opened alternately for an hour each, A first. When will the tank be full?

Answer. Take 15 units: A +3, B −1, so each 2-hour cycle adds 2. We need $2k + 3 \ge 15$, so k = 6: after 12 hours there are 12 units, and in the 13th hour A adds the last 3. 13 hours.

Time check: about 20 seconds.

Question 14

A pipe can fill a tank in 10 hours. A drain at the bottom lets out 8 litres a minute, and with both open the tank fills in 15 hours. Find the capacity of the tank.

Answer. Take 30 units. The pipe puts in 3 an hour and the net is 2, so the drain takes out 1 unit an hour. That unit is 8 × 60 = 480 litres, so the tank holds 30 × 480 = 14,400 litres.

Time check: about 20 seconds.

Question 15

Pipes A and B can fill a tank in 20 and 36 minutes. Both are opened together. After how many minutes should B be closed so that the tank is full in 15 minutes?

Answer. Take 180 units: A does 9 a minute, B does 5. A runs all 15 minutes and puts in 135 units. B must give the other 45 at 5 a minute, so B is closed after 9 minutes.

Time check: about 15 seconds.