Practice Test: Time & Work- 2 - UPSC MCQ

# Practice Test: Time & Work- 2 - UPSC MCQ

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## 15 Questions MCQ Test CSAT Preparation - Practice Test: Time & Work- 2

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Practice Test: Time & Work- 2 - Question 1

### Mayank can do 50% more work than Shishu in the same time. Shishu alone can do a piece of work in 30 hours. Shishu starts working and he had already worked for 12 hours when Mayank joins him. How many hours should Shishu and Mayank work together to complete the remaining work?

Detailed Solution for Practice Test: Time & Work- 2 - Question 1

Practice Test: Time & Work- 2 - Question 2

### A can do a piece of work in 90 days, B in 40 days and C in 12 days. They work for a day each in turn, i.e., first day A does it alone, second day B does it alone and 3rd day C does it alone. After that the cycle is repeated till the work is finished. They get Rs 240 for this job. If the wages are divided in proportion to the work each had done. Find the amount A will get?

Detailed Solution for Practice Test: Time & Work- 2 - Question 2

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Practice Test: Time & Work- 2 - Question 3

### Read the passage below and solve the questions based on it. The tank at a water supply station is filled with water by several pumps. At first three pumps of Ihe same capacity are turned on: 2.5 hours later, two more pumps (both the same) of a different capacity are set into operation. After 1 hour, the additional pumps were set into operation; the tank was almost filled to its capacity (15 m3 were still lacking): in another hour the tank was full. One of the two additional pumps could have filled the tank in 40 hours Q. What is the volume of the tank?

Detailed Solution for Practice Test: Time & Work- 2 - Question 3

Let us assume that, first three pumps fills the tank in x hours .

so,

→ Efficiency of each pump = (1/x) m³ / hour .

then,

→ Efficiency of three pump = (3/x) m³ / hour .

now,

→ First three pumps works for = 2.5h + 1h + 1h = 4.5 hours.

so,

→ Water filled by 3 pumps in 4.5 hours = 4.5 * (3/x) = (13.5/x) m³ .

now, given that,

→ Time taken by additional pump to fill the tank = 40 hours.

so,

→ Efficiency of 2 additional tanks = 2 * (1/40) = (1/20) m³ / h .

and,

→ Additional pumps work for = 1 + 1 = 2 hours.

so,

→ Water filled by additional pumps in 2 hours = 2 * (1/20) = (1/10) m³ .

therefore,

→ (13.5/x) + (1/10) = 1

→ (13.5/x) = 1 - (1/10)

→ (13.5/x) = (9/10)

→ x = 135/9 = 15 hours.

since given that, in last 1 hour they filled 15 m³ .

hence,

→ 3 * (1/15) + (1/20) = 15 m³

→ (1/5) + (1/20) = 15

→ (4 + 1)/20 = 15

→ (5/20) = 15

→ (1/4) = 15

→ 1 = 60  (Ans.) (Option A)

Practice Test: Time & Work- 2 - Question 4

Read the passage below and solve the questions based on it.

There are three taps A, B and C and an outlet pipe D. A, B and C can fill the tank in the Panikam locality in 10, 20 and 25 h respectively. The outlet pipe can empty the same tank in 100 h. There are 2,000 houses in the locality. The tank has a capacity of 50,000 litres

Q.

If all the taps and the outlet pipe are opened simultaneously, how much water is thrown into the tank every hour?

Detailed Solution for Practice Test: Time & Work- 2 - Question 4

Total time required to fill the tank: 1/10 + 1/20 + 1/25 - 1/100 = 18/100
Time required to fill the tank in 1hr = capacity/total time
= 50000*18/100
= 9000litre

Practice Test: Time & Work- 2 - Question 5

There are three taps A, B and C in a tank. They can fill the tank in 25 hrs, 20 hrs and 10 hrs respectively. At first all of them are opened simultaneously. Then after 1 hrs, tap C is closed and tap A and B are kept running. After the 4th hour, tap B is also closed. The remaining work is done by tap A alone.
Find the percentage of work done by tap A itself?

Detailed Solution for Practice Test: Time & Work- 2 - Question 5

Tap A can fill the tank in 25 hrs.

Tap B cab fill the tank in 20 hrs.

Tap C can fill the tank in 10 hrs.

Total volume of the tank filled by 3 pipes = LCM of (25, 20, 10) = 100 units

⇒ Pipe A can fill 4 units of water in 1 hr.

⇒ Pipe B can fill 5 units of water in 1 hr.

⇒ Pipe C can fill 10 units of water in 1 hr.

⇒ Pipe (A + B + C) can fill 19 units of water in 1 hr.

According to question,

Pipe (A + B + C) are opened for 1 hr, then tap C is closed.

⇒ 19 units of water are filled in the tank.

After 4th hr, tap B is also closed

⇒ for 3 hrs, pipe A and B are open

⇒ (4 + 5) × 3 = 27 units of water are filled.

⇒ 100 - (19 + 27) = 54 units of water are filled by tap A alone.

Total units of water filled by tap A is 4 × 1 + 4 × 3 + 54 = 4 + 12 + 54 = 70 units

∴ Percentage of work done by tap A = 70%

Practice Test: Time & Work- 2 - Question 6

Anup can dig a well in 10 days. but particularly in difficult time the work is such that due to fatigue every subsequent day efficiency of a worker falls by 10%.If Anup is given a task of digging one such well in the difficult time, then in how many days will he finish the work?

Detailed Solution for Practice Test: Time & Work- 2 - Question 6

Let us assume that digging one well = 40 unit

Efficiency of Anup = 40/10 = 4 unit/day

Now efficiency is falling by 10%

So, this is case of GP

⇒ 4,18/5,81/25,…… where common ratio = 9/10

Now, sum of infinite GP = a / (1 – r)

⇒ 4/ (1 – 9/10) = 40

Hence, it is clear that Anup will takes infinite time, so he will never finish the work.

Practice Test: Time & Work- 2 - Question 7

A and B together can complete a work in 3 days. They start together but after 2 days, B left the work. If the work is completed after two more days, B alone could do the work in

Detailed Solution for Practice Test: Time & Work- 2 - Question 7

(A+B)'s one day's work = 1/3 part
(A+B) works 2 days together = 2/3 part
Remaining work = 1−2/3 = 1/3 part

1/3 part of work is completed by A in two days
Hence, one day's work of A = 1/6
Then, one day's work of B = 1/3−1/6 = 1/6
So, B alone can complete the whole work in 6 days.

Practice Test: Time & Work- 2 - Question 8

Refer to the data below and answer the questions that follow.

Anoop was writing the reading comprehension sections in Lhe DOG entrance examinations, There were four passages of exactly equal length in terms of number of words and die four passages had 5, 8, 8 and 6 questions following each of them respectively. It is known that Anoop can answer exactly 12 questions in the time he takes to read any one of the four passages. Assume that his rate of reading and answering questions remains the same throughout the section.

Q.

By what per cent should Anoop increase his reading speed if he has to cut down on his total time spent on the section by 20%? Assume that the time spent on answering the questions is constant and as given in the directions.

Detailed Solution for Practice Test: Time & Work- 2 - Question 8

To solve this problem, let's first find out the total time Anoop takes to read all four passages and answer all the questions.

Let the time he takes to read one passage be T. Since there are four passages, he takes 4T time to read all the passages. It is given that he can answer 12 questions in the time he takes to read one passage. So, the time he takes to answer one question is T/12.

There are a total of 5+8+8+6 = 27 questions. The time he takes to answer all the questions is 27 * (T/12) = 27T/12 = 9T/4.

Now, the total time spent on the section is the sum of the time spent on reading all the passages and answering all the questions: 4T + 9T/4 = 25T/4.

To cut down on his total time spent on the section by 20%, the new total time should be 80% of the original time, which is 0.8 * (25T/4) = 5T.

Since the time spent on answering the questions remains constant, the time spent on reading should reduce to 5T - 9T/4 = 11T/4. The new time he takes to read one passage is (11T/4) / 4 = 11T/16.

Now, let's find out the percentage increase in reading speed. The original time to read one passage is T, and the new time is 11T/16. Since speed is inversely proportional to time, the new speed will be 16/11 times the original speed.

The percentage increase in speed is [(16/11 - 1) * 100] = [(5/11) * 100] = 45.45%.

So, Anoop should increase his reading speed by 45.45% to cut down on his total time spent on the section by 20%.

Practice Test: Time & Work- 2 - Question 9

A can work twice as fast as B. A and C together can work thrice as fast as B. If A, B and C complete a job in 30 days working together, in how many days can each of them complete the work?

Detailed Solution for Practice Test: Time & Work- 2 - Question 9

A, B and C complete a job in 30 days working together,

⇒ 1/A + 1/B + 1/C = 1/30

A can work twice as fast as B,

⇒ 1/B = 1/2A

A and C together can work thrice as fast as B,

⇒ 1/B = 1/3(1/A + 1/C)

Solving,

⇒ 3/B = 1/A + 1/C

⇒ 3/B = 1/30 – 1/B

⇒ 1/B = 120

Then,

⇒ 1/A = 1/60

⇒ 1/C = 1/120

∴ A, B and C can complete work in 60, 120 and 120 days respectively.

Practice Test: Time & Work- 2 - Question 10

In what time would a cistern be filled by three pipes of diameter of 1 cm, 2 cm and 3 cm if the largest pipe alone can fill the cistern in 49 minutes, the amount of water flowing through each pipe being proportional to the square of its diameter?

Detailed Solution for Practice Test: Time & Work- 2 - Question 10

Since the amount of water flowing through each pipe is proportional to square of its diameter so if efficiency of longest pipe (3 cm) = 1/49

Then efficiency of pipe (2 cm) = 4/(49 x 9)

and efficiency of pipe (1 cm) = 1/ (49 x 9)

Now let cistern is filled by all three pipes in x minutes.

Practice Test: Time & Work- 2 - Question 11

Sagar is 20% more efficient than Pranesh. If Pranesh can complete a piece of work in 30 days, then in how many days can both Sagar and Pranesh complete the work?

Detailed Solution for Practice Test: Time & Work- 2 - Question 11

Practice Test: Time & Work- 2 - Question 12

A factory manufactures dyes. It has 12 men and two machines which can be operated by all of its men. It takes 4 hours to manufacture one dye on the machine with the operator. The machines can work continuously without a break. Without the machine each of the men can manufacture a dye in 8 hours. The policy is such that the production is maximized and the men are ready to work in three shifts of 8 hours per day. What will be the average cost incurred per dye if 1 man hour costs Rs 20 and 1 machine hour costs Rs 15?

Detailed Solution for Practice Test: Time & Work- 2 - Question 12

To maximize the production, four men will work in each shift. 2 men will work with machines and 2 men work alone.
Total cost incurred in one hour

Practice Test: Time & Work- 2 - Question 13

480 persons working 10 hours per day complete one-fourth of a work in 10 days. How many additional persons are to be employed in order to complete the remaining work in 20 days, working 8 hours per day?

Detailed Solution for Practice Test: Time & Work- 2 - Question 13

Practice Test: Time & Work- 2 - Question 14

Anil, Bhuvan and Chandan take 10,20 and 25 days to complete a job. What is the minimum time required to finish the job if not more than 2 of them work together on a single day and no two consecutive days have the same pair of people working?

Detailed Solution for Practice Test: Time & Work- 2 - Question 14

Assume total amount of work = 100 units.
A does = 10 units/day, B does = 5 units/day, and C does = 4 units/day
Possible Pairs:
A+B = 15 units/day, A+C = 14 units/day, B+C = 9 units/day
To minimize time, we will use the first two pairs.
So, 15 +14 +15 + 14 + 15 +14 + 15 = 102 units.
So, little less than 7 days are required

Hence, the correct option is 'B'.

Practice Test: Time & Work- 2 - Question 15

Anuj can do a piece of work in a certain number of days. To do the same piece of work, Bhanu takes thrice the number of days as Anuj takes whereas Chandu takes thrice as many days as Bhanu does and Dodo takes thrice as many days as Chandu does. Now, they are paired and two groups are formed. The first pair takes one-third the time taken by the second pair to complete the work, which is the first pair?

Detailed Solution for Practice Test: Time & Work- 2 - Question 15

Let Anuj do the work in x days.

So, Bhanu = 3x days, Chandu = 9x days and Dodo = 27x days

Now, using options,

Anuj + Chandu = 1/x + l/9x = 10/9x,

So, they take 9x/10 days

Bhanu + Dodo = l/3x + 1/27x = 10/27x, So, they take 27x/10 days

Now, 1/3 of 27x/ 10 = 9x /10.

So, Anuj and Chandu is first pair.

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