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Arithmetic II

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Time & Work
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CAT 2025 Lesson : Time & Work - Worker-Day Method

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2.1 Completion of work on schedule

2.1.1 Some workers had to leave

In the following two examples, two methods have been outlined. Product constancy is the recommended and the quickest method, if you are sound in the logic to be applied. Otherwise, we suggest you use the standard worker-day approach.

Example 3

1212 men can build a wall in 2020 days. 55 days after they started working, 33 of them had to leave. If no other worker leaves or joins this group, what is the expected delay (in days) in construction?

Solution

1212 men worked for 55 days and then 33 men left. Let's define overall work as 12×20=24012 \times 20 = 240 man-days.

Work completed in 55 days =12×5=60= 12 \times 5 = 60 man-days

Work Remaining =240−60=180= 240 - 60 = 180 man-days

As remaining work is completed by 99 workers in xx days,
9×x=1809 \times x = 180
⇒ x=20 x = 20

Total time taken =20+5=25= 20 + 5 = 25 days

This is 5\bm{5} days more than the initial time taken of 2020 days.

Alternatively (Product Constancy)

After 1212 men worked for 55 days, 1515 days of work is left. However, 33 of the 1212 men left.

As number of workers decreased by 14\dfrac{1}{4}, the time taken would have increased by 14−1\dfrac{1}{4 - 1}.

Work Delay =13×15== \dfrac{1}{3} \times 15 = 5\bm{5} days

Answer: 55 days


Example 4

88 women take 1515 hours to complete a piece of work. 99 hours after the 88 women started working, some of them had to leave. If it took a total of 1717 hours to complete the work, how many women left at the end of 99 hours?

Solution

Total work =8×15=120= 8 \times 15 = 120 women-hours (w−h)(w-h)

Work completed in 99 hours =8×9=72w= 8 \times 9 = 72w − h- \ h

Remaining work (120−72=48w−h)(120 - 72 = 48 w-h) is completed in 88 more hours.

Number of women left =488=6= \dfrac{48}{8} = 6 women

Number of women who left =8−6== 8 - 6 = 2\bm{2} women

Alternatively (Product Constancy)

After 88 women worked for 99 hours, instead of taking 66 more hours, it took 88 more hours to finish the work.

As time increased by 13\dfrac{1}{3}, women would have decreased by 13+1=14\dfrac{1}{3+1} = \dfrac{1}{4}.

Number of women who left =14×8== \dfrac{1}{4} \times 8 = 2\bold2 women

Alternatively

88 women work for the first 99 hours in both scenarios. So, this can be ignored.

88 women work for the remaining 66 hours in a normal scenario. When a few women left, the remaining xx women took 88 hours to complete the work.

As the work done is the same in both the scenarios,
8×6=x×88 \times 6 = x \times 8
⇒ x=6 x = 6

∴\therefore Number of women who left =8−6== 8 - 6 = 2\bm{2} women

Answer: 22 women


2.1.2 Wrong assessment of time taken to complete

As stated in 2.1.12.1.1, apply the product constancy method, only if you're sound in this logic.

Example 5

Kunal takes a contract to construct a warehouse in 6060 days and deploys 4040 of his men for this job. At the end of 2020 days, if only 25%25 \% of the construction is complete, how many more men does he need to deploy on this project, in order to complete the construction on schedule?

Solution

Work done by 4040 men in 2020 days =20×40=800= 20 \times 40 = 800 man-days

If this is 25%25 \% of the total work, then total work =800×4=3200= 800 \times 4 = 3200 man-days

Work remaining after 2020 days =3200−800=2400= 3200 - 800 = 2400 man-days

Time left to complete the work on schedule =60−20=40= 60 - 20 = 40 days

Number of men required to complete the work on schedule =240040=60= \dfrac{2400}{40}= 60 men

Extra men required =60−40=20= 60 - 40 = 20 men

Alternatively (Product Constancy)

25%25 \% of the work is completed by 4040 men in 2020 days. So, with 4040 men, the remaining 75%75 \% of the work will be completed in 6060 more days.

However, the remaining work is to be completed in 4040 more days.

So, if time taken decreases by 13\dfrac{1}{3}, then men required will increase by 13−1=12\dfrac{1}{3 - 1} = \dfrac{1}{2}

Extra men required =12×40=20= \dfrac{1}{2} \times 40 = 20 men

Answer: 2020 men


2.1.3 Several variables influencing work

In these questions, work can be expressed as area, volume, etc. The same can be expressed as worker-hours or worker-days as well. These two works will be directly proportional to each other. We can then use variation to solve these questions.

Example 6

8080 workers can tile a rectangular floor of dimensions 80m×72m80\text{m} \times 72\text{m} in 4040 hours. How many hours will it take 5050 workers to tile a floor of dimensions 90m×60m90 \text{m} \times 60 \text{m}?

Solution

It takes 80×4080 \times 40 worker-hours to tile an area of 80×7280 \times 72 m2\text{m}^{2}.

Let tt hours be the time taken by 5050 workers. It takes 50×t50 \times t worker-hours to tile an area of 90×6090 \times 60 m2\text{m}^{2}.

Here, there is a direct relationship between area, which is the work to be completed, and worker-hours.

80×4050×t=80×7290×60\dfrac{80 \times 40}{50 \times t} = \dfrac{80 \times 72}{90 \times 60}



⇒ 85×40t=89×65\dfrac{8}{5} \times \dfrac{40}{t} = \dfrac{8}{9} \times \dfrac{6}{5}

⇒ t=t = 60\bm{60} hours

Answer: 6060 hours


For additional variables in a 33-dimensional figure and hours per day worked, the approach remains the same.

Example 7

5050 workers take 88 days to build 66 walls that are each 40m40 \text{m}, 50m50 \text{m} and 60m60 \text{m} in length, breadth and height respectively, working 66 hours on each day. If 6060 workers work 99 hours a day for 66 days, then how many wall of dimensions 54m×30m×40m54 \text{m} \times 30 \text{m} \times 40 \text{m} can they build?

Solution

This question has more variables than those in example 66. However, the approach remains the same. Let nn be the number of walls built in the second case.



50×8×660×6×9=6×40×50×60n×54×30×40\dfrac{50 \times 8 \times 6}{60 \times 6 \times 9} = \dfrac{6 \times 40 \times 50 \times 60}{n \times 54 \times 30 \times 40}

⇒ n=15n = 15 walls

Answer: 1515 walls

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