Practice Test: Time & Work- 1


10 Questions MCQ Test IBPS Clerk Prelims - Study Material, Mock Tests | Practice Test: Time & Work- 1


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QUESTION: 1

Walter White, who lives in the countryside, caught a train for home earlier than usual yesterday. His wife normally drives to the station to meet him. But yesterday he set out on foot from the station to meet his wife on the way. He reached home 12 minutes earlier than he would have done had he waited at the station for his wife. The car travels at a uniform speed, which is 5 times Walter White’s speed on foot. Walter White reached home at exactly 6 O’clock. At what time would he have reached home if his wife, forewarned of his plan, had met him at the station?

Solution:

The wife drives for 12 minutes less than her driving on normal days. Thus, she would have saved 6 minutes each way.

Hence, Walter White would have walked for 30 minutes (since his speed is 1/5th of the car’s speed). In effect, Walter White spends 24 minutes extra on the walking (rather than if he had travelled the same distance by car).

Thus, if Walter White had got the car at the station only, he would have saved 24 minutes more and reached at 5:36. 

QUESTION: 2

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 slower horse = (t + 5) hr,
total time taken by faster horse = t hr.
Therefore, speed of the slower horse = d/(t + 5) km/hr
speed of the faster horse = d/t 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. Therefore,
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 = 3t2+15t
3t2−5t−50 = 0
3t2+10t−15t−50 = 0
t(3t+10) − 5(3t+10) = 0
(3t+10)(t−5) = 0
t = 5
(ignoring -ve value) 
i.e., total time taken by slower horse 
= 5+5 = 10hr

QUESTION: 3

A pedestrian and a cyclist start simultaneously towards each other from Hyderabad and Uppal which are 40 km apart and meet 2 hours after the start. Then they resumed their trips and the cyclist arrived at Hyderabad 7 hours 30 minutes earlier than the pedestrian arrived at Uppal. Which of these could be the speed of the pedestrian?

Solution:

The relative speed is 20 kmph. Also, the pedestrian should take 7:30 hours more than the cyclist.

Using option (a) the speeds of the two people are 4km/hr and 16 km/hr respectively.

At this speed, the respective times would be 10 hrs and 2:30 hours, giving the required answer. 

QUESTION: 4

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.

Also, their relative speed is 12 kmph and the distance between the two would get to 0 in 40 / 12 = 3.33 hours.

Distance covered = 28 X 3.33 = 93.33 km.

QUESTION: 5

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:

Let the time at which they meet be t minutes past 10.

So, distance run by Charlie + distance run by Alan = 10 km

=> (60 + t)5 / 60 [t + 60 because he would have traveled for 9 am to 10 am and t minutes more before meeting Alan] +

(15 + t) * 10 / 60 [15 + t because he would have traveled from 9:45 to 10:00 and t minutes more] = 10

=> 300 + 5t + 150 + 10t = 600 => t = 10

So, they meet at 10.10 am

QUESTION: 6

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
Per minute -> 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: 7

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, we can infer that 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.

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

QUESTION: 8

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 kmph, his original speed would have been 24 kmph.

The time difference can be seen to be 6 minutes in this case:
60 / 24 – 60 / 25 = 0.1 hrs = 6 mins. Thus, this is the correct answer. 

QUESTION: 9

Two joggers left Chennai for Tambaram simultaneously. The first jogger stopped 42 min later when he was 1 km short of Tambaram and the other one stopped 52 min later when he was 2 km short of Tambaram. If the first jogger jogged as many kilometres as the second and the second as many kilometres as the first, the first one would need 17 min less than the second. Find the distance between Chennai and Tambaram.

Solution:

Solve using options. The first option you would check for (given the values in the questions) would be option (b).

This would give that the first jogger would run at 3 min per km, while the second jogger would run at 4 min per km.

In the new condition, the first jogger would jog for 13 km while the second jogger would jog for 14 km and their respective times would be 39 mins and 56 minutes.

This is consistent with the condition in the question which talks about a difference of 17 minutes in their respective times.

QUESTION: 10

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 h 30 minutes while it goes from P to R in 9 h. How long would it take to go from R to P?

Solution:

Since P to R is double the distance of P to Q, it is evident that the time is taken for 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 4hours (Since 9 + 4 = 13).

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