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Practice Test: Time & Work - UPSC MCQ


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10 Questions MCQ Test CSAT Preparation - Practice Test: Time & Work

Practice Test: Time & Work for UPSC 2024 is part of CSAT Preparation preparation. The Practice Test: Time & Work questions and answers have been prepared according to the UPSC exam syllabus.The Practice Test: Time & Work MCQs are made for UPSC 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Practice Test: Time & Work below.
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Practice Test: Time & Work - Question 1

In a stream, Q lies in between P and R such that it is equidistant from both P and R. A boat can go from P to Q and back in 6 hours 30 minutes while it goes from P to R in 9 hours. How long would it take to go from R to P?

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

Since P to R is double the distance of P to Q,
Therefore, it is evident that the time taken from P to R and back would be double the time taken from P to Q and back (i.e. double of 6.5 hours = 13 hours).

Since going from P to R takes 9 hours, coming back from R to P would take 4 hours i.e. 139 = 4

So Option A is correct

Practice Test: Time & Work - Question 2

If A and B together can complete a piece of work in 15 days and B alone in 20 days, in how many days can A alone complete the work?

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

A and B complete a work in = 15 days
⇒ One day's work of (A + B) = 1/ 15

B complete the work in = 20 days
⇒ One day's work of B = 1/20

⇒ A's one day's work = 1/15 − 1/20 = (4−3)/6 = 1/60

Thus, A can complete the work in = 60 days.

So Option A is correct

Practice Test: Time & Work - Question 3

A and B together can do a piece of work in 50 days. If A is 40% less efficient than B, in how many days can A working alone complete 60% of the work?

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

Given:

A and B together can do a piece of work in 50 days.

A is 40% less efficient than B

Concept used:

Total work = Efficiency of the workers × time taken by them

Calculation:

Let the efficiency of B be 5a

So, efficiency of A = 5a × 60%

⇒ 3a

So, total efficiency of them = 8a

Total work = 8a × 50

⇒ 400a

Now,

60% of the work = 400a × 60%

⇒ 240a

Now,

Required time = 240a/3a

⇒ 80 days

∴ A can complete 60% of the work working alone in 80 days.

Practice Test: Time & Work - Question 4

Sheldon had to cover a distance of 60 km. However, he started 6 minutes later than his scheduled time and raced at a speed 1 km/h higher than his originally planned speed and reached the finish at the time he would reach it if he began to race strictly at the appointed time and raced with the assumed speed. Find the speed at which he travelled during the journey described.

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

Solve this question through options.
⇒  For instance, if he travelled at 25 km/h, his original speed would have been 24 km/h.
⇒ The time difference can be seen to be 6 minutes in this case = 60 / 24 – 60 / 25 = 0.1 hrs = 6 mins

Thus, 25 km/h is the correct answer. 

So Option A is correct

Practice Test: Time & Work - Question 5

Two sprinters run the same race of100 m One runs at a speed of 10 m/s and the other runs at 8 m/s. By what time will the first sprinter beat the other sprinter?

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

Correct option is C
Time taken by first sprinter 
= 100/10 = 10sec
Time taken by second sprinter 
= 100/80 = 12.5sec
Difference = 12.5 - 10 = 2.5 sec

Practice Test: Time & Work - Question 6

X can do a piece of work in 20 days. He worked at it for 5 days and then Y finished it in 15 days. In how many days can X and Y together finish the work?

Detailed Solution for Practice Test: Time & Work - Question 6
  • X’s five day work = 5/20 = 1/4. Remaining work = 1 – 1/4 = 3/4.
  • This work was done by Y in 15 days. Y does 3/4th of the work in 15 days, he will finish the work in 15 × 4/3 = 20 days.  
  • X & Y together would take 1/20 + 1/20 = 2/20 = 1/10 i.e. 10 days to complete the work.

So Option C is correct

Practice Test: Time & Work - Question 7

A dog sees a cat. It estimates that the cat is 25 leaps away. The cat sees the dog and starts running with the dog in hot pursuit. If in every minute, the dog makes 5 leaps and the cat makes 6 leaps and one leap of the dog is equal to 2 leaps of the cat. Find the time in which the cat is caught by the dog (assume an open field with no trees).

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

Initial distance = 25 dog leaps
Per-minute dog makes 5 dog leaps and cat makes 6 cat leaps = 3 dog leaps
⇒  Relative speed = 2 dog leaps / minutes
⇒  An initial distance of 25 dog leaps would get covered in 12.5 minutes.

So Option D is correct

Practice Test: Time & Work - Question 8

Two cars started simultaneously towards each other and met each other 3 h 20 min later. How much time will it take the slower car to cover the whole distance if the first arrived at the place of departure of the second 5 hours later than the second arrived at the point of departure of the first?

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

Let distance between the two places = d km
Let total time taken by faster horse = t hr
⇒ Total time taken by slower horse = (t + 5) hr,

Therefore,
speed of the faster horse = d/t km/hr
speed of the slower horse = d/(t + 5) km/hr 
The two horses meet each other in 3 hour 20 min i.e. in 3(1/3) hr = 10/3 hr
In this time, total distance travelled by both the horses together is d. 

d/(t+5) * 10/3 + d/t * 10/3 = d
⇒ 10/(3(t+5)) + 10/3t = 1
⇒ 10t + 10(t+5) = 3t(t+5)
⇒ 20t + 50 = 3t+ 15t
⇒ 3t− 5t − 50 = 0
⇒ 3t+ 10t − 15t − 50 = 0
⇒ t(3t + 10) − 5(3t + 10) = 0
⇒ (3t + 10)(t − 5) = 0
t = 5 (ignoring -ve value) 

Thus, Total time taken by slower horse = 5 + 5 = 10 hr

So Option B is correct

Practice Test: Time & Work - Question 9

Charlie and Alan run a race between points A and B, 5 km apart. Charlie starts at 9 a.m. from A at a speed of 5 km/hr, reaches B, and returns to A at the same speed. Alan starts at 9:45 a.m. from A at a speed of 10 km/hr, reaches B and comes back to A at the same speed. At what time do Charlie and Alan first meet each other?

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

To solve this problem, we can first calculate the time it takes each person to run the full distance between points A and B. Charlie runs at a speed of 5 km/hr, so it takes him 5 km / (5 km/hr) = 1 hour to run the distance between A and B. Alan runs at a speed of 10 km/hr, so it takes him 5 km / (10 km/hr) = 0.5 hours to run the distance between A and B.

Then, we can calculate the time at which each person reaches point B. Charlie starts at 9 a.m. and it takes him 1 hour to reach B, so he reaches B at 9 a.m. + 1 hour = 10 a.m. Alan starts at 9:45 a.m. and it takes him 0.5 hours to reach B, so he reaches B at 9:45 a.m. + 0.5 hours = 10:15 a.m.

Now Charlie is returning to point A and Alan is going towards B, so their relative speed is 10km/h. So after 1/4 * 10 minutes, the distance between them is 2.5km.
Therefore, Charlie and Alan will first meet each other at a time of 10:10 a.m.  The answer is 3: 10:10 a.m.

Practice Test: Time & Work - Question 10

Two rabbits start simultaneously from two rabbit holes towards each other. The first rabbit covers 8% of the distance between the two rabbit holes in 3 hours, The second rabbit covered 7 / 120 of the distance in 2 hours 30 minutes. Find the speed (feet / h) of the second rabbit if the first rabbit travelled 800 feet to the meeting points.

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

Since the second rabbit covers 7/120 of the distance in 2 hours 30 minutes
⇒ it covers 8.4 / 120 = 7% of the distance in 3 hours.

Thus, in 3 hours both rabbits together cover 15% of the distance which means 5% per hour so they will meet in 20 hours.

The ratio of speeds = 8 : 7.
⇒ the second rabbit would cover 700 ft to the meeting point in 20 hours and its speed would be 35 feet/hr.

So Option is correct

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