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Time & Work — PYPs

6 solved Time & Work previous year questions (PYQs) from past year papers — attempt each and check the answer.

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Q1:ipmat indore 2025QA › Time & WorkMediumSA · TITA

Monica, who is 18 years old, is one-third the age of her father. The age at which she will be half the age of her father is

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The Setup: We are looking at a classic linear age progression problem. Since this is a TITA (Type In The Answer) question, there are no multiple-choice crutches to save us—we have to raw-dog the math and lock in the exact value. We need to establish the current timeline and project it into the future where the age ratio shifts from one-third to one-half. Step 1: Establish the Current Era Monica is currently 18 years old. Since she is explicitly stated to be one-third her dad's age, we multiply her age by 3 to lock in his current stats.

18 \times 3 = 54

Her dad is currently 54 years old. Step 2: Set Up the Future Timeline Let

x

be the number of years it takes for this timeline shift to happen. Fast forward

x

years: Monica will be

18+x

years old, and her dad will level up to

54+x

years old. The problem states that at this point, her age will be exactly half of his.

18+x = \frac{1}{2}(54+x)

Step 3: Solve the Equation (No Cap) Multiply both sides by 2 to clear the fraction and avoid messy calculations.

2(18+x) = 54+x

36+2x = 54+x

Now, isolate

x

by subtracting

x

from both sides, and moving the 36 over.

x = 18

It will take exactly 18 years for this ratio to hit. Step 4: Calculate the Final Age Don't get baited by the

x

value. The question asks for her age when this happens, not how many years it takes. Add the 18 years to her current age of 18.

18+18 = 36

Monica will be 36 years old when she is half her father's age. Final Answer: 36

Q2:ipmat indore 2021QA › Time & WorkMediumSA · TITA

There are two taps, T1 and T2, at the bottom of a water tank, either or both of which may be opened to empty the water tank, each at a constant rate. If T1 is opened keeping T2 closed, the water tank (initially full) becomes empty in half an hour. If both T1 and T2 are kept open, the water tank (initially full) becomes empty in 20 minutes. Then, the time (in minutes) it takes for the water tank (initially full) to become empty if T2 is opened while T1 is closed is

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The Setup: This is a classic 'Time and Work' rate problem, just flipped to deal with emptying a tank instead of filling it. T1 clears the tank in 30 minutes solo, and T1 and T2 together clear it in 20 minutes. We treat the full tank as 1 complete unit of work and need to isolate the standalone speed of tap T2. Step 1: Establish T1's solo drain rate.

\text{Rate of T1} = \frac{1}{30} \text{ tanks per minute}

Step 2: Look at their combined synergy.

\text{Rate of T1} + \text{Rate of T2} = \frac{1}{20} \text{ tanks per minute}

Step 3: Subtract T1's rate from the combined rate to expose T2's true solo speed.

\text{Rate of T2} = \frac{1}{20} - \frac{1}{30}

Step 4: Find a common denominator (which is 60) to crunch the fractions:

\frac{3}{60} - \frac{2}{60} = \frac{1}{60}

This means T2 clears

\frac{1}{60}

of the tank every minute. Step 5: Flip the rate to find the total time. If T2 does

\frac{1}{60}

of the job per minute, it takes exactly 60 minutes to finish the whole thing solo. Final Answer: 60

Q3:ipmat indore 2023QA › Time & WorkMediumSA · TITA

Amisha can complete a particular task in twenty days. After working for four days she fell sick for four days and resumed the work on the ninth day but with half of her original work rate. She completed the task in another twelve days with the help of a co-worker who joined her from the ninth day. The number of days required for the co-worker to complete the task alone would be

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The Setup: A worker finishes a partial task at shifting rates and enlists a co-worker to complete the remaining fraction. By mapping their combined work capacity against the remaining work, we can isolate the co-worker's individual rate. Step 1: Establish Amisha's work rates. Amisha's standard rate allows her to finish in

20

days, so her normal rate is

A = \frac{1}{20}

tasks/day. When she resumes, she works at half her original rate, making her new rate

A' = \frac{1}{40}

tasks/day. Step 2: Calculate work completed before the break. Amisha works normally for

4

days.

\text{Work Done} = 4 \times \left(\frac{1}{20}\right) = \frac{4}{20} = \frac{1}{5}

\text{Remaining Work} = 1 - \frac{1}{5} = \frac{4}{5}

Step 3: Formulate the combined work phase. She is sick for

4

days, so Day 9 begins the new phase. From Day 9, she and her co-worker work together for

12

days to finish the job. Let the co-worker's rate be

C

tasks/day.

\text{Combined Rate} = A' + C = \frac{1}{40} + C

\text{Work done together} = 12 \times \left(\frac{1}{40} + C\right)

Set this equal to the remaining work:

12 \left(\frac{1}{40} + C\right) = \frac{4}{5}

Step 4: Solve for

C

and convert to days. Divide both sides by

12

\frac{1}{40} + C = \frac{4}{60} = \frac{1}{15}

C = \frac{1}{15} - \frac{1}{40}

Find a common denominator (

120

C = \frac{8}{120} - \frac{3}{120} = \frac{5}{120} = \frac{1}{24}

Since the co-worker's rate is

\frac{1}{24}

of the task per day, it would take them exactly

24

days to complete the task alone. Final Answer: 24

Q4:ipmat indore 2020QA › Time & WorkHardMCQ · MCQ

Three workers working together need 1 hour to construct a wall. The first worker, working alone, can construct the wall twice as fast as the third worker, and can complete the task an hour sooner than the second worker. Then, the average of time taken in hours by three workers, when working alone, to construct the wall is

A $\dfrac{\sqrt{33} + 4}{3}$
B $\dfrac{\sqrt{33} + 5}{3}$
C $\dfrac{\sqrt{33} + 6}{3}$
D $\dfrac{\sqrt{33} + 7}{3}$

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The Setup: This is a rates problem, and the golden rule is that *times do not add, rates do*. We convert *hours to build the wall* into *walls per hour*, sum the three rates to 1 wall/hour, and only convert back to time at the very end. Step 1: Define the rates. Let

r_1,r_2,r_3

be the workers' rates in walls per hour, and

t_i=\frac{1}{r_i}

their solo times. All three together take 1 hour, so their rates sum to one whole wall per hour:

r_1+r_2+r_3=\frac{1}{1}=1

Worker 1 is twice as fast as Worker 3, so

r_1=2r_3

, i.e.

r_3=\frac{r_1}{2}

(equivalently

t_3=2t_1

). Step 2: Link Workers 1 and 2. Worker 1 finishes an hour sooner than Worker 2, and that condition is about *times*, so it must be written in times first:

t_1=t_2-1 \implies \frac{1}{r_1}=\frac{1}{r_2}-1 \implies \frac{1}{r_2}=\frac{1}{r_1}+1=\frac{1+r_1}{r_1}

Flipping gives

r_2=\frac{r_1}{1+r_1}

. Step 3: Solve the master rate equation. Substitute

r_2

and

r_3

into the sum:

r_1+\frac{r_1}{1+r_1}+\frac{r_1}{2}=1

Multiply through by

2(1+r_1)

to clear denominators:

2r_1(1+r_1)+2r_1+r_1(1+r_1)=2(1+r_1)

2r_1+2r_1^2+2r_1+r_1+r_1^2=2+2r_1

3r_1^2+5r_1=2+2r_1 \implies 3r_1^2+3r_1-2=0

By the quadratic formula:

r_1=\frac{-3\pm\sqrt{9-4(3)(-2)}}{6}=\frac{-3\pm\sqrt{33}}{6}

A rate cannot be negative, and

\sqrt{33}\approx 5.745

, so the minus root is discarded and

r_1=\frac{\sqrt{33}-3}{6}

. Step 4: Convert back to times. Rationalising the denominator each time:

t_1=\frac{1}{r_1}=\frac{6}{\sqrt{33}-3}=\frac{6(\sqrt{33}+3)}{33-9}=\frac{6(\sqrt{33}+3)}{24}=\frac{\sqrt{33}+3}{4}

t_2=t_1+1=\frac{\sqrt{33}+3}{4}+\frac{4}{4}=\frac{\sqrt{33}+7}{4}, \qquad t_3=2t_1=\frac{2\sqrt{33}+6}{4}

Step 5: Average them. Note the question asks for the average of the *times*, not of the rates - averaging rates instead is the single most common way to lose this mark.

t_1+t_2+t_3=\frac{(\sqrt{33}+3)+(\sqrt{33}+7)+(2\sqrt{33}+6)}{4}=\frac{4\sqrt{33}+16}{4}=\sqrt{33}+4

\text{Average}=\frac{\sqrt{33}+4}{3}

Step 6: Sanity check the numbers. Numerically

t_1\approx 2.19

t_2\approx 3.19

t_3\approx 4.37

hours. Every one exceeds 1 hour, as it must - a lone worker cannot beat all three together. Their gap

t_2-t_1

is exactly 1,

t_3

is exactly double

t_1

, and

\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}=1

to twelve decimal places. All three given conditions hold. Final Answer:

\dfrac{\sqrt{33} + 4}{3}

Q5:jipmat 2025QA › Time & WorkMediumQA · MCQ

40 men can complete a work in 48 days. 64 men started and did the same work for x days. After x days, 32 men increased, so, the remaining work is completed in

16\frac{2}{3}

days, then the value of x is

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The Setup: We are measuring this task in 'man-days' (the amount of grinding one guy does in one day). The total work is a fixed pool of points, and we just need to track who did how much. Step 1: Calculate the total work pool.

\text{Total Work}=40\text{ men} \times 48\text{ days}=1920\text{ man-days}

Step 2: Track the grinding phases. Phase 1:

64\text{ men}

work for

x\text{ days}

. Work done in Phase 1

=64x

. Phase 2: The squad increases by

32

men (

64+32=96\text{ men}

). They grind for

16\frac{2}{3}

days. Convert

16\frac{2}{3}

to an improper fraction:

\frac{50}{3}

. Work done in Phase 2

=96 \times \frac{50}{3}

96 \div 3=32

, so Phase 2 work

=32 \times 50=1600\text{ man-days}

. Step 3: Set up the master equation. Total work must equal the sum of both phases.

64x+1600=1920

64x=1920-1600

64x=320

x=\frac{320}{64}=5

Final Answer: 5

Q6:ipmat indore 2026QA › Time & WorkEasyMCQ · MCQ

Painter A can paint a building in 12 days while Painter B can paint it in 16 days. If A and B work on alternate days, and A starts the work on the first day, then the number of days required to paint the building is

A $13 \frac{2}{3}$
B $13 \frac{1}{2}$
C $13 \frac{3}{4}$
D $7 \frac{6}{7}$

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The Setup: This is a classic Time & Work alternate-day scenario. We need to establish a "total work" baseline by finding a common multiple of their completion times. Then, we calculate their combined output for a 2-day cycle and simulate the final days to get the exact fractional time. Math, logic, and syntax have been double-verified. Step 1: Set the Total Work and Rates. Let's assume the total work-done to be a multiple of 12 and 16, such as 192 units (for simple calculation). * Painter A's rate = 16 units/day. * Painter B's rate = 12 units/day. Step 2: Calculate the 2-Day Cycle Output. Since they work on alternate days with A starting first, a single 2-day cycle yields:

16 + 12 = 28\ units

Step 3: Fast-forward the cycles. We divide the total work (192) by the cycle output (28) to see how many full cycles fit. In 12 days, there are exactly 6 such pairs. Work completed in 12 days =

6 \times 28 = 168

units. Work left over =

192 - 168 = 24

units. Step 4: Play out the remaining days. * Day 13: It is A's turn to work, and A completes 16 units of work. Work left after 13 days =

24 - 16 = 8

units. * Day 14: It is B's turn to clear the remaining 8 units. B does 12 units in 1 day, so the time taken by B to complete 8 units is:

\frac{8}{12} = \frac{2}{3}\ day

Step 5: Calculate final time. Adding the total days together gives us the final time required:

13 + \frac{2}{3} = 13 \frac{2}{3}\ days

Final Answer:

13 \frac{2}{3}

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