Work rates — SSC CGL Quant Notes

Concepts and rules

Treat the whole job as 1 unit. If a person finishes it alone in T days, the rate is 1/T of the job per day, and in t days the person does t/T of the job.

Rates add, times do not. If A alone takes a days and B alone takes b days, together they take ab/(a+b) days; for three people the time is abc/(ab+bc+ca) days.

Efficiency means rate. For the same job, time is inversely proportional to efficiency: if A is 50% more efficient than B, the rate ratio A:B is 3:2, so the time ratio A:B is 2:3.

If A alone takes a days and A and B together take c days, then B alone takes ac/(a−c) days, because 1/c − 1/a = (a−c)/ac.

Pair rule: if A+B, B+C and C+A finish the job in x, y and z days, then the rate of A+B+C = 1/2 × (1/x + 1/y + 1/z), because each person is counted twice in the three pairs.

Man-days rule: M1 × D1 × H1 / W1 = M2 × D2 × H2 / W2 (men, days, hours per day, amount of work). It holds only when every worker has the same efficiency; otherwise convert first, for example 1 man = 2 boys.

Wages are shared in the ratio of work done (rate × days worked). If everyone works the same number of days, this is simply the ratio of their rates.

Method and shortcuts

Step 1: Take total work = LCM of the given times. Each person's rate = total work ÷ his time, in whole units per day. This removes fractions from the working.

Step 2: Write the timeline in phases (who works, and for how many days). Work in a phase = combined rate of those present × number of days.

Step 3: Total work − work already done = remaining work; remaining work ÷ rate of whoever finishes it = days needed.

Leaving shortcut: if A and B start together and A leaves x days before completion, let the total time be T; then A works T − x days and B works T days, so A × (T − x) + B × T = total work.

Alternate days: find the work done in one cycle (one day of each person), take as many full cycles as fit, then let the next person in turn finish the leftover. The last cycle is often partial, so never simply divide total work by the work per cycle.

If a time is a fraction such as 15/2 days, the rate is 2/15 of the job per day; take total work as a multiple of 15 and the units stay whole.

Example 1: Two workers together

Question: A can finish a job in 12 days and B in 18 days. In how many days will they finish it working together?

Solution: Total work = LCM(12, 18) = 36 units. A's rate = 36 ÷ 12 = 3 units/day and B's rate = 36 ÷ 18 = 2 units/day. Together = 5 units/day, so time = 36 ÷ 5 = 36/5 days.

Check with the formula: ab/(a+b) = 12 × 18/30 = 216/30 = 36/5 days.

Answer: 36/5 days = 7 1/5 days.

Example 2: Three workers together

Question: A, B and C can do a job alone in 10, 15 and 30 days respectively. How long will all three take together?

Solution: Total work = LCM(10, 15, 30) = 30 units. Rates: A = 3, B = 2, C = 1 unit/day. Together = 6 units/day, so time = 30 ÷ 6 = 5 days.

Answer: 5 days.

Example 3: Pair rates

Question: A and B together can do a job in 12 days, B and C in 15 days, and C and A in 20 days. In how many days can all three together finish it, and in how many days can C alone?

Solution: Total work = LCM(12, 15, 20) = 60 units. A+B = 5, B+C = 4, C+A = 3 units/day. Adding, 2(A+B+C) = 12, so A+B+C = 6 units/day and all three take 60 ÷ 6 = 10 days.

C = (A+B+C) − (A+B) = 6 − 5 = 1 unit/day, so C alone takes 60 ÷ 1 = 60 days.

Answer: 10 days together; C alone 60 days.

Example 4: Man-days with hours

Question: 12 men working 8 hours a day finish a job in 15 days. In how many days will 18 men working 10 hours a day finish a job twice as large? (a) 12 (b) 15 (c) 16 (d) 20

Solution: Use M1 × D1 × H1 / W1 = M2 × D2 × H2 / W2 with W1 = 1 and W2 = 2: 12 × 15 × 8 / 1 = 18 × D2 × 10 / 2.

So 1440 = 90 × D2, giving D2 = 16 days.

Answer: (c) 16

Example 5: Workers leaving midway

Question: A, B and C can do a job alone in 20, 30 and 60 days respectively. All three start together; A leaves after 4 days, and B leaves 3 days before the job is completed. In how many days is the job completed? (a) 16 (b) 18 (c) 20 (d) 22

Solution: Total work = LCM(20, 30, 60) = 60 units. Rates: A = 3, B = 2, C = 1 unit/day. Let the job take T days.

A works 4 days, B works T − 3 days and C works all T days, so 3 × 4 + 2 × (T − 3) + 1 × T = 60.

12 + 2T − 6 + T = 60 ⇒ 3T = 54 ⇒ T = 18 days. Check: A did 12, B did 2 × 15 = 30 and C did 18 units; total 60.

Answer: (b) 18

Example 6: Efficiency chain and wages

Question: A is twice as efficient as B, and B is 50% more efficient than C. Working together, A, B and C finish a job in 12 days and are paid ₹8,800. In how many days would C alone finish the job, and what is C's share? (a) 66 days, ₹1,600 (b) 72 days, ₹1,600 (c) 66 days, ₹2,400 (d) 60 days, ₹1,800

Solution: Take C's rate = 2 units/day. Then B = 1.5 × 2 = 3 and A = 2 × 3 = 6 units/day. Together = 11 units/day.

Total work = 11 × 12 = 132 units, so C alone needs 132 ÷ 2 = 66 days.

All three worked the same 12 days, so wages go in the ratio of rates 6 : 3 : 2. C's share = 8,800 × 2/11 = ₹1,600.

Answer: (a) 66 days, ₹1,600

Common mistakes

Adding days instead of rates: A in 12 days and B in 18 days do not take 30 days, or 15 days, together. Add the rates 1/12 + 1/18 = 5/36 and invert to get 36/5 days.

Reading 50% more efficient as 50% more time. More efficiency means less time: A takes 2/3 of B's time, which is 33 1/3% less time, not 50% less.

In leaving problems, giving the full T days to the worker who left. A worker who leaves x days before the end works only T − x days; one who leaves after k days works only k days.

Sharing wages in the ratio of days worked alone. Share = rate × days, so a faster worker earns more for the same days.

Using M1D1 = M2D2 when men and boys (or men and women) are mixed. Convert everyone into one unit of efficiency first.

Practice questions

Q1. A alone can do a job in 15 days, and A and B together can do it in 6 days. In how many days can B alone do it? (a) 9 (b) 10 (c) 12 (d) 8

Q2. 2 men and 3 boys can finish a job in 10 days, while 3 men and 2 boys can finish it in 8 days. In how many days can 2 men and 1 boy finish it? (a) 12.5 (b) 12 (c) 15 (d) 10

Q3. A can do a job in 12 days and B in 18 days. They work on alternate days, A starting first. In how many days is the job finished? (a) 14 1/2 (b) 15 (c) 14 (d) 14 1/3

Answers

Q1. (b) B's rate = 1/6 − 1/15 = (5 − 2)/30 = 1/10 of the job per day, so B alone takes 10 days.

Q2. (a) Let 1 man do m and 1 boy do b of the job per day: 2m + 3b = 1/10 and 3m + 2b = 1/8. Adding, m + b = 9/200; subtracting, m − b = 1/40 = 5/200. So m = 7/200, b = 1/100 and 2m + b = 16/200 = 2/25, giving 25/2 = 12.5 days.

Q3. (d) Total = 36 units; A = 3 and B = 2 units/day, so 2 days give 5 units. 7 cycles = 14 days give 35 units; the remaining 1 unit is done by A on day 15 in 1/3 day. Total = 14 1/3 days.

Quick revision

Total work = LCM of times; rate = work ÷ time; add rates, never days.

Two workers: ab/(a+b). B alone from A and A+B: ac/(a−c). Three pairs: add the pair rates and halve.

The efficiency ratio is the inverse of the time ratio; wages follow work done.

M1D1H1/W1 = M2D2H2/W2 only for equally efficient workers.

Leaving and alternate-day problems: write the phases, and stop the moment the work is complete.

Concepts and rules

An inlet that fills a tank alone in a hours adds 1/a of the tank per hour. An outlet or leak that empties a full tank in b hours removes 1/b per hour, so its rate is taken as negative.

Net rate = sum of inlet rates − sum of outlet rates. If the net rate is positive the tank fills; if it is negative a full tank empties; if it is zero the level does not change.

Two inlets taking a and b hours fill the tank together in ab/(a+b) hours. An inlet taking a hours with an outlet taking b hours (b > a) fills the tank in ab/(b − a) hours.

Leak rule: if a pipe normally fills a tank in a hours but takes b hours because of a leak, the leak alone empties the full tank in ab/(b − a) hours.

Only the empty part needs filling: if the tank is already 1/3 full, time = (2/3) ÷ net rate. In the same way, only the filled part needs emptying.

Capacity in litres = flow in litres per minute × minutes. If a pipe fills a tank in a minutes, its flow is capacity/a litres per minute.

The tank stops filling the moment it is full: an outlet that would act after that moment does not count. This decides the answer in alternate-pipe problems.

Method and shortcuts

Step 1: Take capacity = LCM of all the times, in units. Give each inlet a positive and each outlet a negative rate in units per hour.

Step 2: Split the time into phases whenever a pipe is opened or closed, and find the content of the tank at the end of each phase.

Step 3: Remaining capacity ÷ net rate in the last phase = time still needed.

Shortcut for one inlet (a hours) and one outlet (b hours): time = ab/(b − a). The same expression gives the leak's emptying time in the leak rule.

Alternate pipes: find the net gain per cycle, but first find the content at which the filling pipe can finish within its own turn. Count full cycles only up to that point.

Example 1: Two inlets

Question: Pipe A fills a tank in 10 hours and pipe B in 15 hours. How long will both take together?

Solution: Capacity = LCM(10, 15) = 30 units. A = 3 and B = 2 units/hour, together 5 units/hour, so time = 30 ÷ 5 = 6 hours. Formula: 10 × 15/25 = 6.

Answer: 6 hours.

Example 2: Inlet with an outlet

Question: Pipe A fills a tank in 8 hours and pipe B empties the full tank in 12 hours. If both are opened when the tank is empty, when will it be full?

Solution: Capacity = LCM(8, 12) = 24 units. A = +3, B = −2, net = +1 unit/hour, so time = 24 hours. Formula: 8 × 12/(12 − 8) = 24.

Answer: 24 hours.

Example 3: Leak rule

Question: A pipe can fill a tank in 6 hours, but because of a leak at the bottom it takes 8 hours. In how many hours can the leak alone empty the full tank?

Solution: Capacity = LCM(6, 8) = 24 units. Pipe = 4 units/hour; pipe with leak = 3 units/hour, so the leak removes 1 unit/hour and empties 24 units in 24 hours. Formula: 6 × 8/(8 − 6) = 24.

Answer: 24 hours.

Example 4: Three pipes at once

Question: Pipes A and B can fill a tank in 12 hours and 15 hours, and pipe C can empty it in 20 hours. If all three are opened together when the tank is empty, in how many hours will it be full? (a) 8 (b) 9 (c) 12 (d) 10

Solution: Capacity = LCM(12, 15, 20) = 60 units. A = +5, B = +4, C = −3, so net = 6 units/hour.

Time = 60 ÷ 6 = 10 hours.

Answer: (d) 10

Example 5: Pipe opened later

Question: Pipes A and B fill a tank in 20 minutes and 30 minutes, and pipe C empties it in 15 minutes. A and B are opened together in an empty tank, and C is opened 6 minutes later. How many more minutes after C is opened will the tank take to fill? (a) 24 (b) 36 (c) 30 (d) 20

Solution: Capacity = LCM(20, 30, 15) = 60 units. A = +3, B = +2, C = −4 units/minute.

Phase 1 (A and B for 6 minutes): 5 × 6 = 30 units filled. Phase 2 (all three): net = 3 + 2 − 4 = 1 unit/minute.

The remaining 60 − 30 = 30 units at 1 unit/minute take 30 more minutes (36 minutes in all, which is the trap option).

Answer: (c) 30

Example 6: Alternate hours

Question: Pipe A fills a tank in 4 hours and pipe B empties the full tank in 6 hours. Starting with an empty tank, they are opened alternately for 1 hour each, A first. After how many hours will the tank be full for the first time? (a) 19 (b) 24 (c) 20 (d) 18

Solution: Capacity = LCM(4, 6) = 12 units. A = +3, B = −2, so each 2-hour cycle gains 1 unit.

A can finish within its own hour once the content is at least 12 − 3 = 9 units. After 9 cycles (18 hours) the content is 9 units.

In hour 19, A adds 3 units and the tank is full, so B never runs again. Dividing 12 by 1 to get 24 hours is the trap.

Answer: (a) 19

Common mistakes

Adding the rate of an emptying pipe. An outlet's rate is negative; subtract it from the inlets.

In alternate-pipe problems, dividing capacity by the gain per cycle. The tank can fill during a filling turn before the cycle ends.

Confusing the leak rule: ab/(b − a) is the time the leak takes to empty the full tank, not the filling time with the leak (that is b).

Ignoring water already in the tank. If the tank is 2/5 full, only 3/5 has to be filled.

Mixing minutes and hours, or giving the total time when the question asks for the time after a pipe was opened or closed.

Practice questions

Q1. Pipe A fills a tank in 18 minutes and pipe B in 36 minutes. How many minutes will both take together? (a) 12 (b) 15 (c) 10 (d) 27

Q2. A tank is 2/5 full. Pipe A can fill the whole tank in 10 hours and pipe B can empty the full tank in 15 hours. If both are opened, in how many hours will the tank be full? (a) 12 (b) 15 (c) 18 (d) 30

Q3. A tap can fill an empty tank in 20 minutes and a leak can empty the full tank in 30 minutes. Both are open for 10 minutes and then the leak is plugged. How many more minutes will the tap take to fill the tank? (a) 15 (b) 16 2/3 (c) 18 (d) 20

Answers

Q1. (a) Capacity = 36 units; A = 2 and B = 1 unit/minute, so 36 ÷ 3 = 12 minutes.

Q2. (c) Net rate = 1/10 − 1/15 = 1/30 per hour; the empty part is 3/5, so time = 3/5 × 30 = 18 hours.

Q3. (b) Capacity = 60 units; tap = 3, leak = −2. In 10 minutes the content is 10 units; the remaining 50 units at 3 units/minute take 50/3 = 16 2/3 minutes.

Quick revision

Capacity = LCM of times; inlet rates positive, outlet and leak rates negative.

Two inlets: ab/(a+b). Inlet a with outlet b: ab/(b − a). Leak from a and b: ab/(b − a).

Fill only the empty part; split the time into phases whenever a pipe changes.

Alternate pipes: stop counting cycles once the next filling turn can complete the tank.