CAT Questions on circular motion

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CAT Questions on circular motion:

From Time, Speed and Distance (TSD) topic questions will be asked on linear or circular motion. In Circular motion questions like clocks and circular tracks are very popular. You can learn circular tracks shortcuts and can apply them for the following circular motion and circular track problems. You can also read about some important time, speed and distance CAT level problems. In circular tracks “number of distinct meeting points, meeting for the 1st time” etc were commonly asked. So we have provided some circular motion CAT problems for practice with solutions.

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Question 1:

Two persons A and B are racing along a circular track. The speed of A is twice the speed of B. The length of the circular track is 800 m and the length of the race is 7200 m. After the start of the race, A meets B for the first time at the end of the third minute. If A and B start the race from the same point, find the time taken by A to finish the race.

a) 19 minutes
b) 9.5 minutes
c) 27 minutes
d) 13.5 minutes

Question 2:

Amar and Rajesh are running on a circular track. Their 8th point of meeting is same as their 20th meeting point. If they are running in the opposite direction and the ratio of their speeds is x:1 (x is a natural number), which of the following cannot be the value of x?

a) 5
b) 2
c) 4
d) 11
e) 3

Question 3:

What is the minor angle made by the minute hand and hour hand when the time is 3:48 PM?

a) 176
b) 174
c) 172
d) 170

Question 4:

Prakash and Deepika start running simultaneously from point A on a circular track in the same direction. The ratio of their speeds is 2:7. How many times are they diametrically opposite to each other before they meet again at point A?

a) 7
b) 6
c) 5
d) 4
e) 0

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Solutions: (1 to 4)

1) Answer (d)

Let the speed of A be ‘2s’ m/s. Speed of B = ‘s’ m/s
Relative speed between A and B = 2s – s = s m/s
Time taken for A to meet B = length of the track / relative speed between A and B = 800 / s = 3 minutes
=> 800 / s = 180
=> s = 800/180 m/s
So, time taken by A to finish the race = 7200 / 2*(800/180) = 7200 * 180 / 2*800 = 9 * 90 seconds = 810 seconds = 13.5 minutes

2) Answer (c)

If 8th meeting point is same as 20th meeting point, there can be a maximum of 12 distinct meeting points. Also, any factor of 12 can also be the number of distinct meeting points of Amar and Rajesh.
So, number of distinct meeting point will be 12/(x+1). Thus, x+1 has to be a factor of 12.
So, x can take values of 1,2,3,5,11.

3) Answer (b)

Each minute is equal to 360/60 = 6°
48 minutes => 48 * 6 = 288°
Hour hand covered 3.8 hours.
Each hour = 360/12 = 30°
3.8 hours => 3.8 * 30 = 114°
Hence, the minor angle made = 288°
– 114°
= 174°

4) Answer (c)

The ratio of their speeds is 2:7 which is in its lowest form i.e. numerator and denominator are co-prime. Thus, the number of distinct points where they meet on the circular track is |2-7|=5. The following figure represents the distinct points-

They will meet for the first time at point B. Between this, Deepika must have completed one round and the distance between A and B while Prakash must have travelled only the distance between A and B. Thus, before meeting for the first time, they must be diametrically opposite to each other exactly once. Thus, before meeting at point A again, they must have been diametrically opposite 5 times.

Question 5:

Two persons, A and B, start simultaneously from starting point on a circular track and drive around the track in the same direction, with speeds of 29 km/hr and 19 km/hr respectively. Every time A overtakes B (anywhere on the track), both of them decrease their respective speeds by 1 km/hr. If the length of the track is 1 km, how many times do they meet at the starting point before B comes to rest?

Answer: 3

Solution:
Initial difference between the speeds of A and B = 10km/hr
The difference in the distance is 10 km in 1 hour
It would be 1km in 1/10 hours.

Distance covered by A and B in 0.1 hours = 2.9km and 1.9km.
They will meet for the first time 0.1 km away from starting point (10).
Now speeds of A and B = 28 and 18
They will meet at 0.8 km from point 10 which means point 8.
Then they will meet at 0.7 km from point 8, that is point 5.
They will meet at 0.6 km from point 5, that is point 1 which is starting point. At this instance speed of A and B = 26 and 16.
This process will continue and at when the speed of B = 5km/hr and 1km/hr they will meet.
Total = 3 points.

Question 6:

Ramesh and Suresh decide to run on a circular track. They run in the same direction and the point at which they meet for 7th time is same as the point at which they meet for the 25th time. If the ratio of the speeds of Ramesh and Suresh is k:1, where k is a natural number, then for how many values of k is the given situation possible?

Answer: 6

Solution:

Ramesh and Suresh are running in the same direction and the ratio of their speed is k:1, so they will meet at k-1 distinct points. If L is the length of the track, then these points would be $$\frac{L}{K-1}$$, $$\frac{2L}{K-1}$$, $$\frac{3L}{K-1}$$…..L. They meet at each of these points once before they meet at the same point twice. Since 7th meeting point is same as 25th, the number of distinct meeting points is 25-7 = 18. So for all the values of k for which k-1 is a factor of 18, their 7th and 25th meeting point would be same. Factors of 18 are 1,2,3,6,9,18, so possible values of k are 2,3,4,7,10,19. So there are 6 values of k which are possible.

They will meet for the first time at point B. Between this, Deepika must have completed one round and the distance between A and B while Prakash must have travelled only the distance between A and B. Thus, before meeting for the first time, they must be diametrically opposite to each other exactly once. Thus, before meeting at point A again, they must have been diametrically opposite 5 times.