Time and Work — SSC CGL Quantitative Aptitude Formulas and Traps

Work Rates

Workers together: add the one-day rates

Working together

\text{Rate} = \frac{1}{n} \qquad T = \frac{ab}{a + b} \qquad T = \frac{1}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}

$n$ days one worker needs alone
$a, b, c$ days each worker needs alone
$T$ days needed when they work together

Part of the job done in a given time

Work done and work left

W = \frac{t}{n} \qquad \text{Left} = 1 - W \qquad \text{Whole time} = \frac{d}{f}

$t$ days worked
$n$ days the worker needs for the whole job
$W$ fraction of the job done
$f$ fraction of the job finished in d days
$d$ days taken for that fraction

One worker's time from the team's time

Finding one worker

x = \frac{ab}{b - a} \qquad A + B + C = \frac{(A + B) + (B + C) + (C + A)}{2}

$a$ days the pair takes together
$b$ days the known worker takes alone
$x$ days the other worker takes alone
$A, B, C$ each worker's one-day rate

Sharing the pay by work done

Pay shared by work

\text{Share} = \frac{\text{work done}}{\text{whole job}} \times \text{total pay}

Common traps

Adding the days instead of the rates

Two workers taking 10 and 15 days do not take 25 days together, nor 12.5 days (the average). Together is always quicker than the faster worker alone: here

\frac{10 \times 15}{25} = 6

days.

Using product over sum for three workers

\frac{abc}{a + b + c}

is not the time for three workers. For 12, 20 and 30 days it gives about 116 days, while the true time is 6. For three or more workers, add the rates or use LCM units.

Marking the part done when the part left is asked

If a team does

\frac{5}{8}

of a job in the days given, the part left is

\frac{3}{8}

. SSC usually prints both fractions among the options, so read the last line of the question again before marking.

Scaling the days the wrong way

If

\frac{3}{4}

of a job takes 18 days, the whole takes

18 \div \frac{3}{4} = 24

days, not

18 \times \frac{3}{4} = 13.5

. The whole job always takes longer than any part of it.

Forgetting to halve the pair total

When the times of A and B, B and C, and C and A are given, the sum of the three pair rates is twice the rate of all three, because each person appears in two pairs. Not halving doubles the team's speed and halves its time.

Subtracting the times

If a pair takes 10 days and one of them 15 days, the other does not take 5 days. Subtract rates:

\frac{1}{10} - \frac{1}{15} = \frac{1}{30}

, so 30 days.

Sharing the pay in the ratio of days

Pay follows work, so it is shared in the ratio of the rates, which is the inverse of the days. When both work the whole time, a worker who needs 10 days earns more than one who needs 15.

Giving a helper an equal share

A helper who joins for all the days does not get an equal part of the money. His share is the work he did: 1 minus what the others did in those days.

Efficiency Ratios

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Efficiency ratio and time ratio

Efficiency and time

\frac{T_A}{T_B} = \frac{E_B}{E_A} \qquad \text{Work} = E \times T

$E_A, E_B$ efficiencies (work per day) of A and B
$T_A, T_B$ days each needs alone
$E, T$ one worker's efficiency and the days that worker needs

A faster worker who takes fewer days

Faster worker, fewer days

t = \frac{D}{k - 1} \qquad T_{\text{together}} = \frac{kt}{k + 1}

$k$ how many times as fast A is as B
$t$ days A takes alone
$D$ how many fewer days A takes than B

Common traps

Not flipping the ratio

If the efficiencies are

5 : 2

, the times are

2 : 5

, not

5 : 2

. The more efficient worker always takes less time.

Reading 60% more efficient as 60% less time

A worker 60% more efficient takes

\frac{100}{160} = \frac{5}{8}

of the time, a cut of 37.5%, not 60%.

Dividing the gap by k

If A is 3 times as fast as B and takes 40 days less, A's time is

\frac{40}{3 - 1} = 20

days, not

\frac{40}{3}

. The gap is

k - 1

of A's times.

Workers Joining and Leaving

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One worker starts and another finishes

Work in stretches

\sum (\text{rate} \times \text{days}) = 1 \qquad pA + qB = p(A + B) + (q - p)B

$A, B$ one-day rates of the two workers
$p, q$ days each of them works

A team that changes partway

Team changes partway

x(A + B) + dB = 1

$x$ days the full team works
$d$ days the smaller team works after the change
$A, B$ one-day rates

Leaving a few days before the end

Leaving k days before the end

\sum r_i (T - k_i) = 1

$T$ total days the job takes
$r_i$ one-day rate of worker i
$k_i$ days before the end that worker i leaves (0 if they stay)

Common traps

Taking the days for the rest as the whole time

If B finishes the remaining

\frac{3}{4}

of a job in 27 days, B's time for the whole job is

27 \div \frac{3}{4} = 36

days, not 27.

Giving the days after the change when the total is asked

"How many more days" wants only the second stretch. "In how many days was the work completed" wants both stretches added. SSC prints both numbers among the options.

Forgetting the old members in a joiner's rate

If a team finishes the rest at

\frac{1}{12}

of the job a day after C joins, C's rate is

\frac{1}{12}

minus the rates of those already working, not

\frac{1}{12}

itself.

Taking the leaving days off the leaver's own time

Leaving 4 days before the END means working

T - 4

of the job's

T

days. It does not turn the leaver's own time of, say, 24 days into 20.

Working on Alternate Days

Learn this subtopic in the notes

Common traps

Rounding a part day up to a whole day

After the last whole cycle, work day by day. If 3 units are left and it is the turn of a worker who does 4 units a day, the last day is

\frac{3}{4}

of a day.

Ignoring who starts

When the job ends inside a cycle, the starter matters. A 6-day worker and a 15-day worker on alternate days finish in

8\frac{2}{5}

days if the faster one starts and in 9 days if the slower one starts.

Taking one cycle too many

Count whole cycles only while work remains after them. A 30-unit job with a 7-unit cycle takes 4 whole cycles (28 units) and then single days, not 5 cycles (35 units).

Putting the helpers' day first

Helpers who join on every third day work on days 3, 6, 9 and so on, not on days 1, 4, 7. In a part block at the end they may never appear.

Reading a day off after every 4 days as a 4-day rhythm

A worker who takes a day off after every four days of work is off on days 5, 10, 15: the rhythm is 5 days long, 4 working and 1 off.

Man-Days and Man-Hours

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Workers times days is fixed for a job

Man-days rule

M_1 D_1 = M_2 D_2 \qquad \frac{M_1 D_1}{W_1} = \frac{M_2 D_2}{W_2}

$M_1, M_2$ numbers of workers
$D_1, D_2$ days taken
$W_1, W_2$ amounts of work (1 for the whole job)

Hours a day and part of a job

Man-hours rule

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

$M_1, M_2$ numbers of workers
$D_1, D_2$ days worked
$H_1, H_2$ hours worked each day
$W_1, W_2$ parts of the job done

Workers added or leaving partway

Changing workforce

\text{Days left} = \frac{n(D - t)}{n'} \qquad dn - \frac{kd(d - 1)}{2} = Dn

$n$ workers at the start
$D$ days the job would take them
$t$ days already worked
$n'$ workers after the change
$k$ workers who leave at the end of each day
$d$ days the job actually takes

Common traps

Working the proportion the wrong way

More workers means fewer days. If 6 men take 20 days, 10 men take

20 \times \frac{6}{10} = 12

days, not

20 \times \frac{10}{6}

.

Giving the total when "how many more" is asked

If 27 men are needed and 18 are already working, the answer to "how many more" is 9, and 27 is usually printed among the options too.

Ignoring a change in hours a day

When the hours a day change, man-days are not equal; man-hours are. 10 men for 12 days at 6 hours is 720 man-hours, the same labour as 9 men for 10 days at 8 hours.

Putting the part of the job on the wrong side

Divide each team's labour by the work that team did. The 30% goes under the labour that did the 30%, and the remaining 70% under the labour that does the rest.

Starting the leaving sequence one day early

If

k

men leave after each day, day 1 still has all

n

men and the last of

d

days has

n - (d - 1)k

. The amount taken away is

k(0 + 1 + \cdots + (d - 1))

, not

k(1 + 2 + \cdots + d)

.

Men, Women and Equivalent Workers

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Converting men into women

Equivalent workers

a \, d_1 \text{ man-days} = b \, d_2 \text{ woman-days} \qquad 1 \text{ man} = \frac{b \, d_2}{a \, d_1} \text{ women}

$a$ men in the first team
$b$ women in the second team
$d_1, d_2$ days each team takes

Two mixed teams doing the same job

Two mixed teams

d_1(x_1 m + y_1 w) = d_2(x_2 m + y_2 w)

$x_1, x_2$ men in each team
$y_1, y_2$ women in each team
$d_1, d_2$ days each team takes
$m, w$ one man's and one woman's daily work

Common traps

Averaging the two teams' times

Two teams that take 12 and 24 days alone take 8 days together, not 18. Teams combine like single workers: add the rates.

Setting men equal to women without the days

8 men in 10 days and 10 women in 12 days do not make 8 men equal to 10 women. Compare labour: 80 man-days

= 120

woman-days, so 1 man

= 1.5

women.

Converting the wrong way

If 1 man

= 1.5

women, then 4 men

= 6

women and 1 woman

= \frac{2}{3}

of a man. Multiply when going from men to women, and divide when going back.

Setting the teams equal when the days differ

If the two teams take different numbers of days, multiply each team by its days before setting them equal. Only when the days are the same can the teams be set equal directly.

Not checking with the second team

After finding

m

in terms of

w

, work out the job in woman-days from both teams. If the two totals differ, the ratio is wrong.

Pipes and Cisterns

Learn this subtopic in the notes

Inlets and outlets open together

Net rate of pipes

\text{Net} = \sum \frac{1}{a_i} - \sum \frac{1}{b_j} \qquad T = \frac{ab}{b - a}

$a_i$ hours each filling pipe needs alone
$b_j$ hours each emptying pipe needs to empty a full tank
$a, b$ the same two times for a single inlet and a single outlet
$T$ hours to fill the tank with both open

Finding a leak from the slower fill

Time for the leak to empty the tank

T_{\text{leak}} = \frac{ab}{b - a}

$a$ hours the pipe needs with no leak
$b$ hours it needs with the leak

One pipe so many times as fast as another

Pipes in a speed ratio

T_{\text{slow}} = (k + 1)T \qquad T_{\text{fast}} = \frac{(k + 1)T}{k}

$k$ how many times as fast the faster pipe is
$T$ time to fill with both pipes open

Common traps

Adding an outlet's rate

An emptying pipe works against the others, so its rate is taken away. Pipes of 12 and 15 hours with a 20-hour drain fill the tank in 10 hours; adding the drain gives a wrong 5 hours.

Answering in the wrong unit

If the options are in minutes, convert the hours:

\frac{9}{4}

hours is 135 minutes.

Taking the extra hours as the leak's time

If a leak makes a 10-hour fill take 2 hours longer, the leak does not empty the tank in 2 hours. It takes

\frac{10 \times 12}{12 - 10} = 60

hours.

Multiplying the joint time by k instead of k + 1

If one pipe is 3 times as fast and together they take 30 minutes, the slower pipe alone takes

4 \times 30 = 120

minutes, not

3 \times 30

. The two pipes together are worth

k + 1

slow pipes.

Keeping the wrong root

In

\frac{1}{x} + \frac{1}{x + d} = \frac{1}{T}

one root makes a time negative or shorter than the joint time. Only the root that keeps every time positive and longer than the joint time is the answer.

Running whole cycles with an outlet in them

With an inlet adding 3 units an hour and an outlet taking 2, an 18-unit tank fills in an inlet hour once the level reaches 15, after 31 hours. Dividing 18 by the cycle's net 1 unit gives a wrong 36 hours.

Time and Work formulas

Work Rates

Workers together: add the one-day rates

Part of the job done in a given time

One worker's time from the team's time

Sharing the pay by work done

Common traps

Efficiency Ratios

Efficiency ratio and time ratio

A faster worker who takes fewer days

Common traps

Workers Joining and Leaving

One worker starts and another finishes

A team that changes partway

Leaving a few days before the end

Common traps

Working on Alternate Days

Common traps

Man-Days and Man-Hours

Workers times days is fixed for a job

Hours a day and part of a job

Workers added or leaving partway

Common traps

Men, Women and Equivalent Workers

Converting men into women

Two mixed teams doing the same job

Common traps

Pipes and Cisterns

Inlets and outlets open together

Finding a leak from the slower fill

One pipe so many times as fast as another

Common traps

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