PYQ Vault
Ebooks

AFCAT Numerical Ability · Formula sheet

Time and Work formulas

26 formulas and 35 common traps for AFCAT Numerical Ability Time and Work, grouped by subtopic.

Full notes with worked examples

Work as a Rate, and Efficiency

Learn this subtopic in the notes

Part of the work in some days

Whole job from a part

pq of the job in d days  ⇒  T=d×qp\tfrac pq \text{ of the job in } d \text{ days} \;\Rightarrow\; T = d \times \dfrac qp

Work done in stages, in percentages

A share of the remainder

done=k100×(work left)\text{done} = \tfrac{k}{100} \times (\text{work left})

Efficiency and time

Time and efficiency

TATB=EBEA\dfrac{T_A}{T_B} = \dfrac{E_B}{E_A}

Output per hour: machines and shifts

Output

output=rate×time\text{output} = \text{rate} \times \text{time}

Common traps

Scale the part the right way

If 35\tfrac 35 of a job takes 9 days, the whole job takes 9×53=159 \times \tfrac 53 = 15 days, not 9×359 \times \tfrac 35. The whole job is bigger than the part, so the days must go up.

'More days' is what is left

When the question asks how many more days, take away the days already worked. A 15-day job that is 9 days in needs 6 more days, not 15.

Of the remaining work

If X does 75% of a job and Y then does 40% of the remaining work, Y has done only 10% of the whole job. 15% is still left, not none.

More efficient is not the same percentage less time

25% more efficient cuts the time to 100125\tfrac{100}{125} of it, which is 20% less, not 25% less. 50% more efficient cuts the time by a third.

The ratio of times turns round

Efficiencies 2:32 : 3 mean times 3:23 : 2. Writing the times as 2:32 : 3 gives the slower worker the shorter time.

Efficiency scales the rate

A machine 25% less efficient than one making 80 units an hour makes 60 an hour. It needs 43\tfrac 43 of the time, not 25% more time.

Working Together: Adding Rates

Learn this subtopic in the notes

Adding rates

Two people together

T=aba+bT = \dfrac{ab}{a + b}

One person's time from the joint time

The other person's time

b=aTa−Tb = \dfrac{aT}{a - T}

Pairs working together

All three from the pairs

A+B+C=12[(A+B)+(B+C)+(C+A)]A + B + C = \tfrac 12\left[(A + B) + (B + C) + (C + A)\right]

Efficiency ratios with a joint time

One person from a ratio

TA=(a+b) TaT_A = \dfrac{(a + b)\,T}{a}

Work done in a given time, and each share

Work done together

W=d(1a+1b)W = d\left(\dfrac 1a + \dfrac 1b\right)

Common traps

Never add or average the days

A in 10 days and B in 15 days do not take 25 days together, or 12.5. They take 10×1525=6\tfrac{10 \times 15}{25} = 6 days, less than either alone.

Subtract rates, not days

Together 9 days and A alone 12 days does not make B 3 days. B's rate is 19−112=136\tfrac 19 - \tfrac 1{12} = \tfrac 1{36}, so B takes 36 days.

Halve the total

Adding the three pair rates counts everyone twice. For pairs taking 8, 12 and 8 days the sum is 13\tfrac 13, so all three together take 6 days, not 3.

Which way the ratio runs

'A takes twice as long as B' makes B the faster worker, with efficiencies A : B = 1 : 2. Giving A the 2 units answers for the wrong person.

Done, or left?

After people work for a while, read whether the question asks for the work done or the work left. With 23\tfrac 23 done, 13\tfrac 13 is left, and both are usually among the options.

Pay follows the work, not the days

Two people who work the same number of days at different speeds are not paid equally. Split the money in the ratio of the work each one did.

Someone Leaves or Takes Over

Learn this subtopic in the notes

Work in stages: one stops, another finishes

Stages

∑(rate×days)=1\sum (\text{rate} \times \text{days}) = 1

Leaving a few days before the end

Leaving k days early

t−ka+tb=1\dfrac{t - k}{a} + \dfrac{t}{b} = 1

Common traps

More days, or days in all?

If A works 5 days and B then needs 12 more, the job took 17 days. Check which of the two the question asks for before you answer.

Use the rate of whoever is left

After A leaves, the rest of the work goes at B's rate alone. Dividing it by the joint rate makes the last stage far too short.

The one who left worked fewer days

If the job takes tt days and A leaves 3 days before the end, A worked t−3t - 3 days. Writing t+3t + 3, or giving A the full tt, loses the mark.

Working on Alternate Days, and Other Work Patterns

Learn this subtopic in the notes

Two people on alternate days

One cycle

work per cycle=1a+1b\text{work per cycle} = \dfrac 1a + \dfrac 1b

Turns, helpers and rest days

Repeating blocks

blocks=workwork per block\text{blocks} = \dfrac{\text{work}}{\text{work per block}}

Common traps

Do not run the cycles past the job

If each cycle does 27\tfrac 27, three cycles do 67\tfrac 67 and a fourth would overshoot. Stop at three and finish day by day.

The starter works first in the last cycle

After the whole cycles, the next day belongs to whoever started. Giving it to the other person changes the answer whenever their rates differ.

Rest after the end does not count

Someone working 6 days a week who needs 16 working days finishes on day 18, not day 19: the rest days after the last working day are never taken.

A helper's day is one day per block

If helpers join every fourth day, a 4-day block has four days of the main worker but only one day of each helper.

Man-Days, Man-Hours and Groups of Workers

Learn this subtopic in the notes

Man-days: workers × days

Man-days

M1D1=M2D2M_1 D_1 = M_2 D_2

Man-hours: when the working day changes

Man-hours

M1H1D1=M2H2D2M_1 H_1 D_1 = M_2 H_2 D_2

When the amount of work changes

Work with output

M1H1D1W1=M2H2D2W2\dfrac{M_1 H_1 D_1}{W_1} = \dfrac{M_2 H_2 D_2}{W_2}

Converting men, women and children

Equivalent workers

p men=q women  ⇒  1 man=qp womenp \text{ men} = q \text{ women} \;\Rightarrow\; 1 \text{ man} = \tfrac qp \text{ women}

Two mixed teams: solve for one worker

Equal work

(p m+q w) d1=(r m+s w) d2(p\,m + q\,w)\, d_1 = (r\,m + s\,w)\, d_2

Common traps

More workers, fewer days

If 10 men take 12 days, 15 men take 8 days, not 18. Multiply men by days and divide; never scale the days up with the men.

Hours a day are part of the work

If the hours a day change and you count only men × days, the answer is out by the ratio of the hours. A team that halves its hours needs twice the days.

Output goes under the labour

More output needs more labour. Writing MHDW\tfrac{MHD}{W} keeps that straight: doubling the walls doubles the man-days instead of halving them.

'Or' is not 'and'

'4 men or 6 women in 20 days' means either group alone takes 20 days. 4 men and 6 women working together are two full teams and take 10 days.

Equal work, not equal daily work

Two teams finishing the same job in different times have equal TOTAL work. Setting their daily work equal is right only when their days are equal too.

When Workers Leave or Join Mid-Job

Learn this subtopic in the notes

Work left after the team changes

Days after a change

more days=MD−MdM′\text{more days} = \dfrac{M D - M d}{M'}

Extra days against the plan

Extra days

extra=L×RM−L\text{extra} = \dfrac{L \times R}{M - L}

Keeping to the deadline

Team needed

workers=work leftdays left\text{workers} = \dfrac{\text{work left}}{\text{days left}}

Hours, efficiency and team size change together

Standard workers

effective team=regular workers+p100×others\text{effective team} = \text{regular workers} + \tfrac{p}{100} \times \text{others}

Common traps

Take away the work already done

If 30 men need 16 days and 6 leave after 4 days, the 24 men share only the 360 man-days that are left: 15 more days. Dividing the whole 480 man-days by 24 counts the first 4 days twice.

Extra is not the same as more

When 40 workers are planned for 30 days and 8 leave after 10 days, the job needs 25 more days, which is 5 beyond the plan. A question asking for extra days wants 5; one asking for more days wants 25.

Joining saves days

When JJ workers join, the days saved are JRM+J\dfrac{J R}{M + J}: the team grows, so the sum goes below the line.

Count from the workers still there

If 10 workers are needed and 2 of the original 10 have left, hire 2. Answering 10 hires a whole new team on top of the one already working.

Efficiency changes the team, not the job

Five trainees at 60% efficiency count as three workers. Adding all five as full workers makes the rest of the job look shorter than it is.

Each stage at its own hours

When the hours a day change partway, the work done so far was at the old hours. Only the work left is done at the new hours.

Pipes and Cisterns

Learn this subtopic in the notes

Filling and emptying together

Net rate

1T=1a+1b−1c\dfrac 1T = \dfrac 1a + \dfrac 1b - \dfrac 1c

Finding a leak from the delay

Time for the leak

L=abb−aL = \dfrac{ab}{b - a}

Pipes opened or closed partway

Stages

∑(net rate×time)=1\sum (\text{net rate} \times \text{time}) = 1

Capacity, pipes in turn and mixtures

Capacity

capacity=litres per minuteshare of the tank per minute\text{capacity} = \dfrac{\text{litres per minute}}{\text{share of the tank per minute}}

Common traps

Take the outlet away

Taps filling in 20 and 30 hours take 12 hours together. With a drain that empties the tank in 15 hours also open, they take 60 hours, not less than 12.

A negative net rate never fills

If the emptying pipe is faster than the filling pipes together, an empty tank stays empty. Look for an option that says the tank is never filled.

The leak is slower than both times

A pump taking 4 hours, and 5 hours with a leak, does not mean the leak empties the tank in 1 hour, or in 9. Its rate is the difference of the two rates, so it takes 20 hours.

Leave out the closed pipe

After a pipe is shut, only the pipes still open fill the rest. Keeping the closed pipe's rate in the sum makes the answer too short.

The tank can fill partway through the cycles

With a filling pipe and an emptying pipe in turn, the level peaks at the end of each filling turn. Dividing the whole tank by the gain per cycle overshoots, because the last filling turn finishes the job before the emptying turn comes round.

More AFCAT Numerical Ability formula sheets