Practice Test: Time & Work- 1


10 Questions MCQ Test UPSC Prelims Paper 2 CSAT - Quant, Verbal & Decision Making | Practice Test: Time & Work- 1


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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?

Solution:

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

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?

Solution:

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.

QUESTION: 3

Chetan is thrice as efficient as Mamta and together they can finish a piece of work in 60 days. Mamta will take how many days to finish this work alone?

Solution:
  • Chetan is thrice as efficient as Mamta.
  • Let, Mamta takes 3x days and Chetan takes x days to complete the work.
  • ∴ 1/x + 1/3x = 1/60 ⇒ x = 80.
  • ∴ Mamta will take 80 × 3 = 240 days to complete the 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.

Solution:

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. 

QUESTION: 5

Leonard, a tourist, leaves Ellora on a bicycle. Having travelled for 1.5 h at 16 km/h, he makes a stop for 1.5 h and then pedals on with the same speed. Four hours after Leonard started, his friend and local guide Howard leaves Ellora on a motorcycle and rides with a speed of 28 km/h in the same direction as Leonard had gone. What distance will they cover before Howard overtakes Leonard?

Solution:

When Howard starts, Leonard would have covered 40 km.
∵ Their relative speed is 12 km/h
∴ The distance between the two would get to 0 in 40 / 12 = 3.33 hours
⇒  Distance covered = 28 * 3.33 = 93.33 km

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?

Solution:
  • 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.
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).

Solution:

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.

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?

Solution:

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

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?

Solution:
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.

Solution:

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.

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