Time and Work: Rates, Efficiency and Working Together

Work as One Whole Job, and One Day's Work

In time and work questions the whole job counts as 1. A person's speed is the part of the job they finish in one day (or one hour).

• If Kabir can paint a fence in 9 days, he paints $\frac{1}{9}$ of it every day.
• In 4 days he paints $\frac{4}{9}$ of it, and $\frac{5}{9}$ is still left.
• Turned round: if someone does $\frac{2}{15}$ of a job each day, the whole job takes $\frac{15}{2} = 7\frac{1}{2}$ days.

$$\text{one day's work} = \frac{1}{\text{days for the whole job}} \qquad \text{work done} = \text{one day's work} \times \text{days}$$

Working together. When people work together, add their one day's work, never their days. Kabir paints the fence in 9 days and Ruchi in 18 days. In one day together they paint

$$\frac{1}{9} + \frac{1}{18} = \frac{2 + 1}{18} = \frac{3}{18} = \frac{1}{6}$$

of the fence, so together they need 6 days. For exactly two people this turns into one formula:

$$T = \frac{xy}{x + y} = \frac{9 \times 18}{9 + 18} = \frac{162}{27} = 6$$

The answer need not be a whole number. If A takes 8 days and B takes 12, one day together is $\frac{1}{8} + \frac{1}{12} = \frac{3 + 2}{24} = \frac{5}{24}$, so the job takes $\frac{24}{5} = 4\frac{4}{5}$ days. Options may write this as $4\frac{4}{5}$ days or as 4.8 days, so keep your answer exact until the end.

A quick check. Two people together are faster than the faster one alone, but at most twice as fast. So the time together is less than the faster person's time and at least half of it. With 9 and 18 days the answer has to sit between 4.5 and 9, and 6 does. Two people who each take x days finish together in $\frac{x}{2}$ days.

Hours work the same way. A machine that finishes a print run in 5 hours does $\frac{1}{5}$ of it every hour. "Day" in the formulas can be any unit of time, as long as everyone is measured in the same unit.

Reading the question.

• "A can finish the work in 12 days": A does $\frac{1}{12}$ of it a day.
• "A finishes $\frac{1}{4}$ of the work in 3 days": the whole job would take A 12 days, so again $\frac{1}{12}$ a day.
• "A and B together take 12 days": this is the speed of the pair, not of A or of B.

[SSC Tip] Adding the days (9 + 18 = 27) or averaging them (13.5) gives the two wrong options you will usually see. Both are more than the faster person's time, so the quick check above rules them out.

The Total-Work (LCM) Method

Fractions work, but they get heavy once someone leaves or joins. The quicker way is to give the job a size in units, using the LCM of the times. This is the method used through the rest of the chapter.

Steps.

1. Total work = LCM of everyone's time.
2. Each person's units a day = total work ÷ their time.
3. For people working together, add their units a day. Time = units to be done ÷ units a day.

A paints a hall in 22 days and B in 33 days. The LCM of 22 and 33 is 66, so call the job 66 units.

• A does 66 ÷ 22 = 3 units a day, and B does 66 ÷ 33 = 2 units a day.
• Together they do 5 units a day, so the hall takes 66 ÷ 5 = $13\frac{1}{5}$ days.

The fraction method gives the same thing: $\frac{1}{22} + \frac{1}{33} = \frac{3 + 2}{66} = \frac{5}{66}$, so the time is $\frac{66}{5}$ days. The units are simply the numerators of these fractions.

Why it pays off. Say A and B work together for 6 days and then B leaves. In units: 6 × 5 = 30 units done, 36 left, and A takes 36 ÷ 3 = 12 more days. With fractions: $6 \times \frac{5}{66} = \frac{5}{11}$ done, $\frac{6}{11}$ left, and $\frac{6}{11} \div \frac{1}{22} = 12$ days. Same answer, but every step in units is mental arithmetic.

For a time like $7\frac{1}{2}$ days, take a total that both times divide into exactly: 30 ÷ 7.5 = 4 and 30 ÷ 10 = 3.

When fractions are quicker.

• The only thing asked is "how long together" for two people. $\frac{xy}{x + y}$ is one line.
• The question hands you fractions of the job ("A finishes $\frac{3}{8}$ of it in 6 days…"). Either turn each fraction into a full time first (A: 6 × $\frac{8}{3}$ = 16 days) and then use units, or stay with fractions.
• The pair's time is known but one person's is not (A and B together in 12 days, then B alone…). Fractions of the pair's work are quicker there, as in the Part of the Work part below.

When efficiencies are given instead of times ("A works twice as fast as B"), skip the LCM and take the efficiencies themselves as units a day. That is in the Efficiency and Time part below.

Any common multiple works as the total, even the product of the times; the LCM just keeps the numbers small. Once you have units a day, stay in units to the end, because switching to fractions halfway is where mistakes creep in.

Finding One Person's Time from the Pair's Time

Two people. "A and B together finish a job in 12 days, and A alone takes 20 days. How long does B take alone?"

• Units: the LCM of 12 and 20 is 60. Together they do 5 units a day, and A does 3. So B does 5 − 3 = 2 units a day and needs 60 ÷ 2 = 30 days.
• Fractions: $\frac{1}{12} - \frac{1}{20} = \frac{5 - 3}{60} = \frac{2}{60} = \frac{1}{30}$.

As a formula, if the pair takes p days and A takes a days, B takes $\frac{ap}{a - p}$ days. Here $\frac{20 \times 12}{20 - 12} = \frac{240}{8} = 30$.

Three people. "A, B and C together finish in 7 days. A alone takes 14 days and B alone 42 days. How long does C take alone?" The LCM of 7, 14 and 42 is 42. All three do 6 units a day, A does 3 and B does 1, so C does 6 − 3 − 1 = 2 units a day and takes 21 days.

When one takes longer than the other by a known gap. "B takes 8 days more than A. Together they finish in 3 days. Find each one's time." If A takes x days,

$$\frac{1}{x} + \frac{1}{x + 8} = \frac{1}{3} \Rightarrow x^2 + 2x - 24 = 0 \Rightarrow x = 4$$

So A takes 4 days and B takes 12. In the exam, try the options instead of solving: 4 and 12 days give $\frac{1}{4} + \frac{1}{12} = \frac{4}{12} = \frac{1}{3}$, which fits. Any option pair that does not differ by 8 can be crossed out at once.

[Tier 2] Extra time over the pair. "Working alone, A takes 2 days more than A and B take together, and B takes 50 days more." Then the time together is

$$T = \sqrt{2 \times 50} = 10 \text{ days}$$

so A takes 12 days and B takes 60. Check: $\frac{1}{12} + \frac{1}{60} = \frac{5 + 1}{60} = \frac{1}{10}$. This works for any two people, because $T = \frac{xy}{x + y}$ can be rearranged to $(x - T)(y - T) = T^2$.

The person you find by subtraction always takes longer than the pair, so if B's time comes out shorter than the pair's time, or negative, you subtracted the wrong way round. Subtract units a day (6 − 3 − 1), never days (7 − 14 − 42).

Three People Together, and Pair Data

All three together. A, B and C can finish a job in 12, 16 and 24 days. LCM 48 gives 4, 3 and 2 units a day, so together they do 9 units a day and need 48 ÷ 9 = $5\frac{1}{3}$ days.

Fractions: $\frac{1}{12} + \frac{1}{16} + \frac{1}{24} = \frac{4 + 3 + 2}{48} = \frac{9}{48} = \frac{3}{16}$, so $\frac{16}{3}$ days. Same answer.

Pair data. "A and B together take 36 days, B and C take 45 days, and C and A take 60 days. How long do all three take together, and how long does each take alone?"

The LCM of 36, 45 and 60 is 180.

• A + B = 5 units a day, B + C = 4, C + A = 3.
• Adding the three pairs counts everyone twice: 2(A + B + C) = 12, so A + B + C = 6 units a day, and all three together take 180 ÷ 6 = 30 days.
• A = (A + B + C) − (B + C) = 6 − 4 = 2 units a day, so A alone takes 180 ÷ 2 = 90 days.
• B = 6 − 3 = 3 units a day, so 60 days. C = 6 − 5 = 1 unit a day, so 180 days.

Fractions: $\frac{1}{2}\left(\frac{1}{36} + \frac{1}{45} + \frac{1}{60}\right) = \frac{1}{2} \times \frac{5 + 4 + 3}{180} = \frac{1}{30}$, so 30 days.

Pairs plus the trio. "A and B take 9 days, B and C take 12 days, and all three take 6 days. How long do A and C take together?" LCM 36: A + B = 4, B + C = 3 and all three = 6 units a day. So C = 6 − 4 = 2, A = 6 − 3 = 3 and B = 1. A + C = 5 units a day, which needs 36 ÷ 5 = $7\frac{1}{5}$ days.

One person as good as two others. "A is as efficient as B and C together" means A's units a day equal B's plus C's. If B and C take 20 and 30 days (LCM 60: 3 and 2 units a day), A does 5 units a day and takes 12 days alone, and all three together do 10 units a day and finish in 6 days.

[SSC Tip] The usual slip in pair data is forgetting to halve: 12 units a day is two of everyone, not one. A quick check on your working: each person's units a day must be smaller than every pair total they belong to, and none can come out zero or negative.

Efficiency and Time

Efficiency is the work done in one day. For the same job, more efficiency means fewer days, in exactly the same proportion:

$$\text{time} \propto \frac{1}{\text{efficiency}} \qquad \frac{T_A}{T_B} = \frac{E_B}{E_A}$$

If A and B work in the ratio 3 : 2, their times are in the ratio 2 : 3. "A does half as much work as B" (in the same time) is the same as "B is twice as efficient as A", so A takes twice B's time.

Use the efficiencies as units a day. "A is twice as efficient as B. Together they finish a job in 14 days. How long does each take alone?" Give A 2 units a day and B 1. Together that is 3 units a day for 14 days, so the job is 42 units. A alone takes 42 ÷ 2 = 21 days and B alone 42 days.

"x% more efficient". "A is 40% more efficient than B" means A does 140 for every 100 that B does, so A : B = 7 : 5 and their times are 5 : 7. If B takes 28 days, A takes 28 × $\frac{5}{7}$ = 20 days. Together: the job is 5 × 28 = 140 units, done at 12 a day, which takes $\frac{140}{12} = 11\frac{2}{3}$ days.

More and less are not the same percentage. A is 40% more efficient than B, but B is only $\frac{2}{7}$ = $28\frac{4}{7}\%$ less efficient than A, because the base has changed. Time works the same way: B takes 40% more time than A, while A takes $28\frac{4}{7}\%$ less time than B.

"Takes x% more time". "A takes 25% more time than B" means the times are 5 : 4, so the efficiencies are 4 : 5. B is 25% more efficient than A, and A is 20% less efficient than B.

Comparing two people from their times. A finishes a job in 18 days and B in 24 days. The efficiencies are 24 : 18 = 4 : 3, so A is $\frac{1}{3}$ = $33\frac{1}{3}\%$ more efficient than B, and B is $\frac{1}{4}$ = 25% less efficient than A.

Efficiency with a time gap. "A is 3 times as efficient as B and finishes a job 30 days earlier than B." The times are 1 : 3, so the gap is 2 parts = 30 days and one part is 15. A takes 15 days and B 45 days. Together: B does 1 unit a day and A does 3, the job is 45 units, and 4 units a day finish it in $11\frac{1}{4}$ days.

[Tier 2] Less work in less time. "A does half as much work as B in two-thirds of the time." In the same time A would do $\frac{1}{2} \div \frac{2}{3} = \frac{3}{4}$ of what B does, so A : B = 3 : 4. If together they finish in 12 days, the job is 7 × 12 = 84 units: A alone needs 28 days and B alone 21 days.

Part of the Work Done, Then the Rest

From a fraction to the full time. If $\frac{3}{7}$ of a job takes 9 days, the whole job takes $9 \div \frac{3}{7} = 9 \times \frac{7}{3}$ = 21 days. Percentages work the same way: 35% of a job in 7 days means 7 ÷ 0.35 = 20 days for all of it.

One works first, another finishes. "A does $\frac{2}{5}$ of a job in 8 days. B then finishes the rest in 9 days. How long would they take working together?"

• A would take 8 × $\frac{5}{2}$ = 20 days for the whole job.
• B did $\frac{3}{5}$ in 9 days, so B would take 9 × $\frac{5}{3}$ = 15 days.
• Together: LCM 60 gives 3 + 4 = 7 units a day, so $\frac{60}{7} = 8\frac{4}{7}$ days.

Work left after some days. A takes 14 days and B 35 days (LCM 70: 5 and 2 units a day). They work together for 6 days: 42 units are done and 28 are left, so $\frac{28}{70} = \frac{2}{5}$ of the job is left.

A works for some days, then B finishes. Same A and B. A works alone for 6 days and does 30 units. B does the other 40 units at 2 a day, which takes 20 days.

Two different ways to finish the same job. "A job is finished if A works for 6 days and B for 10 days, or if A works for 9 days and B for 4 days. How long does each take alone?" Both ways finish the same job, so

$$6a + 10b = 9a + 4b \Rightarrow 6b = 3a \Rightarrow a = 2b$$

Take B = 1 unit a day and A = 2. The job is 6 × 2 + 10 × 1 = 22 units, so A alone needs 11 days and B alone 22 days. Check with the second way: 9 × 2 + 4 × 1 = 22.

A pair works, then one finishes. "A and B together can finish a job in 12 days. They work together for 4 days, then A leaves and B finishes the rest in 10 days. How long would A take alone?" In 4 days the pair does $\frac{4}{12} = \frac{1}{3}$ of the job. B does the other $\frac{2}{3}$ in 10 days, so B alone needs 15 days. Then A's one day's work is $\frac{1}{12} - \frac{1}{15} = \frac{5 - 4}{60} = \frac{1}{60}$, so A alone takes 60 days. Fractions are quicker here because B's time is not known at the start.

[SSC Tip] "$\frac{3}{7}$ of the job in 9 days" means the whole job takes longer than 9 days. Multiplying 9 by $\frac{3}{7}$ (and getting less than 4 days) is the trap; divide by the fraction.

When Someone Leaves or Joins Partway

The safe method for every version of this question: count how many days each person actually works, multiply by their units a day, and make the total equal to the whole job.

All the cases below use A = 32 days and B = 48 days. LCM 96: A does 3 units a day, B does 2, and together they do 5 (working together throughout, they would need $19\frac{1}{5}$ days).

1. Both start, and one leaves after some days. They work together for 6 days and do 30 units, so 66 units are left.

• If B leaves, A finishes in 66 ÷ 3 = 22 days. The job takes 6 + 22 = 28 days in all.
• If A leaves instead, B needs 66 ÷ 2 = 33 days, a total of 39 days.

2. One leaves a few days before the end. "B leaves 2 days before the work is finished." Now the total T is unknown. A works all T days and B works T − 2 days:

$$3T + 2(T - 2) = 96 \Rightarrow 5T = 100 \Rightarrow T = 20 \text{ days}$$

Another way to see it: in the last 2 days A alone does 6 units, so the first 90 units were done together at 5 a day, in 18 days. Total 18 + 2 = 20.

3. One joins later. "A starts alone and B joins after 4 days." A's 4 days give 12 units. The other 84 units are done together in 84 ÷ 5 = $16\frac{4}{5}$ days, so the job takes $20\frac{4}{5}$ days in all.

4. The total is known; find when someone left. "They start together, A leaves after some days, and the job is finished in 36 days in all." B works all 36 days and does 72 units. A did the other 24 units at 3 a day, so A left after 8 days.

5. Joining and leaving in one question. "A starts the work. B joins after 4 days, and A leaves 2 days before the work is finished." A works T − 2 days, and B works from day 5 to the end, which is T − 4 days:

$$3(T - 2) + 2(T - 4) = 96 \Rightarrow 5T = 110 \Rightarrow T = 22 \text{ days}$$

Check: A works 20 days (60 units) and B works 18 days (36 units), and 60 + 36 = 96.

A range check. Every answer above lies between $19\frac{1}{5}$ days (both working all the time) and 48 days (B alone), which helps you cross out options.

[SSC Tip] In "leaves 2 days before the end", only the person who leaves loses those 2 days. Taking 2 days off both people, or adding 2 days to the together time ($19\frac{1}{5}$ + 2), are the usual wrong options.

Three People: Leaving, Joining and Changing Teams

The day-count method carries straight over to three people. Use A = 18 days, B = 24 days and C = 72 days. LCM 72: A does 4 units a day, B does 3 and C does 1, so all three together do 8 and would finish in 9 days.

One leaves early, another leaves before the end. "All three start. A leaves after 2 days, and B leaves 4 days before the work is finished. How long does the work take?" A works 2 days, B works T − 4 days and C works all T days:

$$4 \times 2 + 3(T - 4) + 1 \times T = 72 \Rightarrow 4T = 76 \Rightarrow T = 19 \text{ days}$$

Check: A does 8 units, B works 15 days for 45 units, and C works 19 days for 19 units. 8 + 45 + 19 = 72.

People leave one after another. "All three start. After 4 days A leaves, and after 4 more days B leaves. C finishes the job. How long does it take in all?"

• First 4 days, all three: 32 units.
• Next 4 days, B and C: 16 units.
• Left: 72 − 48 = 24 units for C alone at 1 a day, which is 24 days.
• Total: 4 + 4 + 24 = 32 days.

Find when someone left. "All three start, A leaves after some days, and the job is finished in 12 days." B and C work all 12 days: 4 × 12 = 48 units. A did the other 24 units at 4 a day, so A left after 6 days.

Joining later. "A and B start, and C joins after 6 days. When is the work finished?" A and B do 7 units a day, so 42 units in 6 days. The other 30 units at 8 a day take $3\frac{3}{4}$ days. Total: $9\frac{3}{4}$ days.

The last, part day. When the answer is not a whole number, the job ends partway through the last day. In the joining case, after 9 full days 66 units are done, and the last 6 units take $\frac{6}{8} = \frac{3}{4}$ of day 10. So the job ends during the 10th day, at $9\frac{3}{4}$ days. If a question asks "on which day is the work finished", the answer is the 10th day.

A small table keeps this straight: who works, from which day to which day, and how many units a day. The usual wrong options let someone work after he has left, or leave out the person who stays till the end.

Jobs of a Different Size

So far the job has been "one job". SSC also asks about two walls, half a field, or $\frac{3}{5}$ of a job.

Same people, bigger or smaller job. The time grows in step with the size of the job. If A and B together build a wall in 6 days, then

• 2 such walls take 12 days, and $2\frac{1}{2}$ walls take 15 days;
• $\frac{3}{4}$ of a wall takes $4\frac{1}{2}$ days.

In units: if one job is 60 units, two jobs are 120 units at the same units a day.

A fraction of the job given for each person. "A can do $\frac{3}{5}$ of a job in 12 days, and B can do $\frac{3}{4}$ of it in 21 days. How long will they take to do 3 such jobs together?"

• A would take 12 × $\frac{5}{3}$ = 20 days for one job, and B would take 21 × $\frac{4}{3}$ = 28 days.
• LCM 140: A does 7 units a day and B 5, so together 12 units a day.
• Three jobs are 420 units, which take 420 ÷ 12 = 35 days.

Work counted in pieces. When the job is counted in things (rooms, pages, chairs), find each person's things per day. "A paints 2 rooms in 3 days and B paints 3 rooms in 6 days. How long will they take to paint 14 rooms together?" A paints $\frac{2}{3}$ of a room a day and B $\frac{1}{2}$. Together that is $\frac{2}{3} + \frac{1}{2} = \frac{7}{6}$ of a room a day, so 14 rooms take $14 \times \frac{6}{7}$ = 12 days.

Part of the job already done. "A, B and C together can finish a job in 9 days. Someone else has already done $\frac{1}{3}$ of it. How long will A, B and C take for the rest?" The rest is $\frac{2}{3}$ of the job, so $\frac{2}{3}$ × 9 = 6 days.

One person, a longer piece of the same work. "A builds a 20 m wall in 8 days. How long will he take for a 35 m wall of the same kind?" A builds 2.5 m a day, so 35 m takes 14 days. When the number of workers or the hours a day also change, use the chain rule in the Men-Days, Alternate Days and Work and Wages section.

Solved Examples

Question 1

A can finish a piece of work in 28 days and B can finish it in 84 days. In how many days can they finish it working together? (a) 21 days (b) 56 days (c) 112 days (d) 18 days

Answer. Take the work as the LCM of 28 and 84, which is 84 units. A does 3 units a day and B does 1, so together they do 4 units a day. Time = 84 ÷ 4 = (a) 21 days. By fractions: $\frac{1}{28} + \frac{1}{84} = \frac{3 + 1}{84} = \frac{1}{21}$. The average of the times, 56 days, is more than A's own 28 days, which cannot happen when B helps.

Question 2

A and B together can finish a job in 9 days. A alone can finish it in 15 days. In how many days can B alone finish it? (a) 6 days (b) 24 days (c) 22.5 days (d) 18 days

Answer. LCM of 9 and 15 is 45 units. Together they do 5 units a day and A does 3, so B does 2 units a day and needs 45 ÷ 2 = (c) 22.5 days. By fractions: $\frac{1}{9} - \frac{1}{15} = \frac{5 - 3}{45} = \frac{2}{45}$. B's time has to be longer than the pair's 9 days, and 22.5 is.

Question 3

A, B and C can do a piece of work in 12, 28 and 42 days respectively. In how many days can they finish it working together? (a) $27\frac{1}{3}$ days (b) 7 days (c) $8\frac{2}{5}$ days (d) 6 days

Answer. LCM of 12, 28 and 42 is 84 units. A, B and C do 7, 3 and 2 units a day, 12 units a day in all. Time = 84 ÷ 12 = (b) 7 days.

Question 4

A and B together can finish a job in 21 days, B and C together in 24 days, and C and A together in 56 days. In how many days can B alone finish it? (a) 28 days (b) 84 days (c) $11\frac{1}{5}$ days (d) $18\frac{2}{3}$ days

Answer. LCM of 21, 24 and 56 is 168 units. A + B = 8, B + C = 7 and C + A = 3 units a day. Adding them gives 2(A + B + C) = 18, so all three do 9 units a day. B = 9 − (C + A) = 9 − 3 = 6 units a day, so B alone needs 168 ÷ 6 = (a) 28 days. Not halving the 18 gives B = 15 units a day and $11\frac{1}{5}$ days.

Question 5

A, B and C together can finish a job in 8 days. A alone can finish it in 18 days and B alone in 36 days. In how many days can C alone finish it? (a) 12 days (b) 46 days (c) 10 days (d) 24 days

Answer. LCM of 8, 18 and 36 is 72 units. All three do 9 units a day, A does 4 and B does 2. So C does 9 − 4 − 2 = 3 units a day and needs 72 ÷ 3 = (d) 24 days.

Question 6

A is thrice as efficient as B. Working together, they finish a job in 21 days. In how many days can B alone finish it? (a) 28 days (b) 63 days (c) 84 days (d) 42 days

Answer. Give A 3 units a day and B 1. Together they do 4 units a day for 21 days, so the job is 84 units. B alone needs 84 ÷ 1 = (c) 84 days, and A alone needs 84 ÷ 3 = 28 days.

Question 7

A is 50% more efficient than B. Together they finish a job in 24 days. How many days will A alone take? (a) 60 days (b) 40 days (c) 36 days (d) 48 days

Answer. 50% more efficient means A : B = 150 : 100 = 3 : 2. Take 3 and 2 units a day. Together that is 5 units a day for 24 days, so the job is 120 units. A alone needs 120 ÷ 3 = (b) 40 days, and B alone 120 ÷ 2 = 60 days.

Question 8

A is twice as efficient as B and takes 12 days less than B to finish a job. In how many days can they finish it working together? (a) 12 days (b) 8 days (c) 18 days (d) 6 days

Answer. Twice as efficient means the times are 1 : 2. The gap of 1 part is 12 days, so A takes 12 days and B takes 24. LCM 24: A does 2 units a day and B 1, so together 3 units a day and 24 ÷ 3 = (b) 8 days.

Question 9

A can finish a job in 15 days and B can finish it in 24 days. By what per cent is A more efficient than B? (a) 37.5% (b) 9% (c) 62.5% (d) 60%

Answer. Efficiencies are in the inverse ratio of the times, so A : B = 24 : 15 = 8 : 5. A does 3 units more for every 5 that B does, so A is $\frac{3}{5}$ = (d) 60% more efficient. 37.5% is how much less efficient B is than A (3 out of 8), which uses A as the base.

Question 10

A takes 27 days less than B to finish a job. Working together, they finish it in 18 days. How many days does B alone take? (a) 54 days (b) 27 days (c) 36 days (d) 45 days

Answer. Try the options: if B takes 54 days, A takes 54 − 27 = 27 days, and $\frac{1}{27} + \frac{1}{54} = \frac{2 + 1}{54} = \frac{1}{18}$, which matches. So B takes (a) 54 days. (By algebra: $\frac{1}{x} + \frac{1}{x + 27} = \frac{1}{18}$ gives $x^2 - 9x - 486 = 0$, so x = 27 for A.)

Question 11

A can do $\frac{2}{3}$ of a job in 16 days and B can do $\frac{3}{5}$ of it in 18 days. In how many days can they finish the whole job working together? (a) 27 days (b) 34 days (c) $13\frac{1}{3}$ days (d) 12 days

Answer. A would take 16 ÷ $\frac{2}{3}$ = 24 days for the whole job, and B 18 ÷ $\frac{3}{5}$ = 30 days. LCM 120: A does 5 units a day and B 4, so together 9 units a day, and 120 ÷ 9 = $\frac{40}{3}$ = (c) $13\frac{1}{3}$ days.

Question 12

A can do a job in 40 days and B can do it in 32 days. A works alone for 10 days and then leaves. In how many days will B finish the rest? (a) 22 days (b) 24 days (c) 8 days (d) 30 days

Answer. In 10 days A does $\frac{10}{40} = \frac{1}{4}$ of the job, so $\frac{3}{4}$ is left. B needs $\frac{3}{4} \times 32$ = (b) 24 days. (In units: LCM 160, A does 4 a day and B 5; A's 10 days give 40 units, and B does the other 120 in 24 days.)

Question 13

A can finish a job in 18 days and B in 30 days. They work together for 5 days. What fraction of the job is still left? (a) $\frac{5}{9}$ (b) $\frac{4}{9}$ (c) $\frac{2}{3}$ (d) $\frac{7}{9}$

Answer. LCM 90: A does 5 units a day and B 3, so 8 units a day together. In 5 days they do 40 units, and 50 of the 90 are left. Fraction left = $\frac{50}{90}$ = (a) $\frac{5}{9}$.

Question 14

A job is finished if A works for 8 days and B for 6 days, or if A works for 4 days and B for 12 days. How many days will B alone take to finish it? (a) 12 days (b) 15 days (c) 20 days (d) 18 days

Answer. Both ways finish the same job, so 8a + 6b = 4a + 12b, which gives 4a = 6b and a : b = 3 : 2. Take A = 3 units a day and B = 2. The job is 8 × 3 + 6 × 2 = 36 units (check: 4 × 3 + 12 × 2 = 36). B alone needs 36 ÷ 2 = (d) 18 days, and A alone 12 days.

Question 15

A can do a job in 21 days and B in 42 days. They start working together, but A leaves after 7 days. In how many days in all is the job finished? (a) 21 days (b) $17\frac{1}{2}$ days (c) 28 days (d) 42 days

Answer. LCM 42: A does 2 units a day and B 1. In 7 days together they do 21 units, leaving 21. B alone does these in 21 days, so the job takes 7 + 21 = (c) 28 days. 21 days is only B's part after A left.

Question 16

A and B can finish a job in 25 days and 50 days respectively. They start together, but B leaves 4 days before the work is finished. In how many days is the work finished? (a) 18 days (b) $16\frac{2}{3}$ days (c) $20\frac{2}{3}$ days (d) 20 days

Answer. LCM 50: A does 2 units a day and B 1. In the last 4 days A works alone and does 8 units. The other 42 units were done together at 3 a day, in 14 days. Total = 14 + 4 = (a) 18 days. (As an equation: 2T + (T − 4) = 50 gives T = 18.)

Question 17

A can finish a job in 18 days and B in 45 days. A starts the work alone, and B joins him after 4 days. In how many days in all is the job finished? (a) 10 days (b) 14 days (c) $12\frac{6}{7}$ days (d) 16 days

Answer. LCM 90: A does 5 units a day and B 2. A's first 4 days give 20 units. The other 70 units are done together at 7 a day, in 10 days. Total = 4 + 10 = (b) 14 days.

Question 18

A and B can finish a job in 15 days and 25 days respectively. They start together. A leaves after some days, and B finishes the rest. The whole job takes 20 days. After how many days did A leave? (a) 5 days (b) 4 days (c) 3 days (d) 6 days

Answer. LCM 75: A does 5 units a day and B 3. B works all 20 days and does 60 units. A did the other 75 − 60 = 15 units at 5 a day, so A left after (c) 3 days. Check: 3 days together = 24 units, then B's 17 days = 51 units, and 24 + 51 = 75.

Question 19

A, B and C can finish a job in 12, 20 and 30 days respectively. All three start together. A leaves after 3 days, and B leaves 5 days before the work is finished. In how many days is the work finished? (a) 10 days (b) 15 days (c) 11 days (d) 12 days

Answer. LCM 60: A, B and C do 5, 3 and 2 units a day. A works 3 days, B works T − 5 days and C works all T days: 5 × 3 + 3(T − 5) + 2T = 60, so 5T = 60 and T = (d) 12 days. Check: A does 15 units, B works 7 days for 21 units, and C works 12 days for 24 units; 15 + 21 + 24 = 60.

Question 20

A and B together can finish a job in 15 days. They work together for 6 days, then B leaves and A finishes the rest in 30 days. In how many days can A alone finish the whole job? (a) 50 days (b) 36 days (c) 45 days (d) 75 days

Answer. In 6 days the pair does $\frac{6}{15} = \frac{2}{5}$ of the job. A does the other $\frac{3}{5}$ in 30 days, so A alone needs 30 ÷ $\frac{3}{5}$ = (a) 50 days. 36 days, 6 + 30, is the total time taken, not A's time alone.

Question 21

A can do $\frac{3}{8}$ of a job in 6 days and B can do $\frac{5}{6}$ of it in 40 days. Working together, how many days will they take to finish 2 such jobs? (a) 12 days (b) 24 days (c) 64 days (d) 6 days

Answer. A would take 6 ÷ $\frac{3}{8}$ = 16 days for one job, and B 40 ÷ $\frac{5}{6}$ = 48 days. LCM 48: A does 3 units a day and B 1, so 4 a day together. Two jobs are 96 units, which take 96 ÷ 4 = (b) 24 days. 6 days halves the time for two jobs instead of doubling it.

Question 22

[Tier 2] Working alone, A takes 18 days more than A and B take together, and B takes 32 days more than A and B take together. How many days does A alone take? (a) 24 days (b) 56 days (c) 25 days (d) 42 days

Answer. If the pair takes T days, then $T^2 = 18 \times 32 = 576$, so T = 24 days. A takes 24 + 18 = (d) 42 days and B takes 24 + 32 = 56 days. Check: $\frac{1}{42} + \frac{1}{56} = \frac{4 + 3}{168} = \frac{1}{24}$.