CTET Maths - Time and Work

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16. Everyday Sachin exercises for hours. He exercises by walking, jogging and running. His speeds while walking, jogging and running in the ratio 1 : 2: 4. Which of the following statements are true?
I. If he spends equal time on the three activities, then he runs twice the distance that he jogs.
II. If he runs at 8 km per hour and distance covered by running and jogging together is 12 km, then he walks a distance of 1 km in 1 hour.
III. If the time for which he walks, jogs and runs is in the ratio 2 : 2 : 1 and he covers a total distance of 10 km, then his speeds while walking, jogging and running are 2 km/hr, 4 km/hr and 8 km/hr respectively.

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Option : C
Explanation :
I. Ratio of speed = 1 : 2 : 4
If equal time is spent on each of these, then ratio of distance = 1 : 2 : 4
So, Sachin runs twice the distance that he jogs. Thus, I is true.
II. He runs at 8 km/hr and therefore walks at 2 km/hr. Hence II is not true.
III. Ratio of speed = 1 : 2 : 4
Ratio of time = 2: 2 : 1
Ratio of distance = 2 x l: 2 x 2 : 4 x 1
= 2: 4: 4
If Sachin covers a total distance of 10 km, then he

Thus, III is true.

17. A monkey climbing up a greased pole ascends 10 metres and slips down 2 metres in alternate minutes. If the pole is 64 metres high, how long will it take him to reach the top?

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Option : B
Explanation :
In 1 minute, monkey ascends 10 metres but he takes 1 minute to slip down 2 metres.
Thus, atthe end of 2 minutes, net ascending of the monkey is = 10 - 2 = 8 metres.
Thus, to have a net ascending of 8 metres, process of ascending and then slipping happens once.
So,to cover 64 metres, above process is repeated 64/8 or 8 times. It is clear that in 8 such happenings, the monkey will slip 7 times, because 8th time, he will ascend to the top.
Thus, in climbing 7 times and slipping 7 times, he covers (7 x 8) or 56 metres.
Time taken to cover 56 metres = (56 x 2)/8 = 14 minutes
Remaining distance = 64 - 56 = 8 metres
Time taken to ascend 8 metres = 8/10 = 4/5 min
Total time taken = 14 minutes + 4/5 min
= 14 mins. 48 sec.

18. A train traveling at 10 m/sec from A to B at 7 a.m. meets a train leaving B at 7:20 a.m. and coming to A at a speed 1/3 times faster than the first train. If the distance from A to B is 68 km. then, at what distance from A will the two trains meet?

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Option : B
Explanation :
If they meet t hrs after 7 a.m,

36t + 48t - 16 = 68
t = 1 hr.
⇒ They meet at a distance of
⇒ 10 x 60 x 60 = 36000 m i.e., 36 km from A.

19. Two trains A and B start from stations X and Y towards each other. B leaves station Y half an hour after train A leaves station X. Two hours after train A has started, the distance between trains A and B is 19/30 th of the distance between stations X and Y. How much time would it take each train (A and B) to cover the distance X to Y, if train A reaches half an hour later to its destination as compared to B?

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Option : C
Explanation :
Two equations are

where T₁ and T₂ are time taken by trains A and B to cover the whole distance
and
Solving equations (i) and (ii), we get
T₁ = 10 hrs. and T₂ = 9 hrs.

Refer to the data below and answer the questions that follow.
The variation in the speed of a car on a particular day at the respective times is shown in the table below:

s(km/hr)	0	40	50	85	10	10
t(hr)	11.00 am	11.30 am	1.00 pm	1.30 pm	3.30 pm	4.30 pm

20. What is the distance travelled by the car from 11 a.m. to 1 p.m.?

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Option : B
Explanation :
Graph of speed vs time is plotted as shown. Area under the graph and time axis gives the distance.

Required distance
= A(ΔOAP) + A(ΔBXA) + A(▢APQX)

= 7.5 km.