Aptitude Time And DistancePage 2
9. | | A bus left P for Q and travelled at an average speed of 40 kmph and reached Q one hour later than the scheduled time, Had the bus travelled at an average speed of 60 kmph, it would have reached Q one hour earlier than the scheduled time. What should be the average speed of the bus to reach Q on time? | | (a)48 kmph | | (b)50 kmph | | (c)52 kmph | | (d)54 kmph |
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Answer is: ALet the distance between P and Q be d km, and the required time be t hours.
(d/40) - 1 = (d/60) + 1 = t =» (d/40) - (d/60) = 2
=» d(60 - 40) = 4800 =» d = 240 km.
So, t = (240/40) - 1 = 5 hours
Therefore , required speed = 240/5 = 48 kmph.
10. | | A beats B by 10 meters in a 100 meters race. If B beats C by 20 meters in the same race, then A beats C by how many meters, in that race? | | (a)32 | | (b)30 | | (c)28 | | (d)26 |
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Answer is: CRatio of speeds of A and B is 100 : 90 = 10 : 9
Ratio of speeds of B and C is 100 : 80 = 10 : 8
Ratio of speeds of A, B and C is 100 : 90 : 72
So, A : C = 100 : 72
Hence, A beats C by 28 meters.
11. | | Two persons start simultaneously from A and B towards B and A respectively. The first person reached B in 10 minutes and the second person reached A in 15 minutes from the time of their start. In how many minutes from their start do they meet each other? | | (a)5 | | (b)8 | | (c)7 | | (d)6 |
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Answer is: DLet the distance between A and B is 'd' km.
To meet, they have to cover together a distance, 'd' km.
In one minute the total distance travelled by them = d/10 + d/15 = d/6
Hence, in 6 minutes they will meet each other.
12. | | Vijay left P for Q at 10:00 am. At the same time Ajay left Q for P. After their meeting at a point on the way, Vijay took 24 minutes to reach Q and Ajay took 54 minutes to reach P. At what time did they meet? | | (a)10:37 am | | (b)10:36 am | | (c)10:39 am | | (d)10:38 am |
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Answer is: BLet Vijay and Ajay take t minutes to meet at a point A.(note point A is between P and Q anywhere)
=» in such cases , t = √(t₠x t₂) → (1)
Here, tâ‚ = 24
tâ‚‚ = 54
Put the value in (1) t = √(24X54) = √1296 = 36.
Hence, they meet at 10:36 am
13. | | Had a man maintained his average speed at (4/3)rd of the original average speed, then the time taken by him to cover the same distance would have been? | | (a)(4/3)rd of the original time | | (b)(3/4)th of the original time | | (c)Neither (A) nor (B) | | (d)cannot be determined |
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Answer is: BIf speed 's' becomes (m/n)s, then time t becomes (n/m)t, provided the distance is same.
Here, m/n = 4/3
Hence, n/m = 3/4.
So, t becomes (3/4)t.
14. | | About to leave for home from his office, a person notices a problem with his car and travels home at a speed which is less than his usual speed and takes 20% more of usual time. What is the average speed of the car in this journey? | | (a)(6/5)th of the original speed | | (b)(5/6)th of the original speed | | (c)Neither (A) nor (B) | | (d)cannot be determined |
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Answer is: BTime taken is 20% more than the usual time.
i.e. If t is the time, then (6/5)t is the actual time taken.
i. e. t has become (6/5)t
Hence, speed will become (5/6)s, where s is actual speed.
15. | | In a 1000 m race A beats B by 100 m. What is the ratio of speeds of A and B? | | (a) 9 : 10 | | (b)10 : 9 | | (c) Neither (A) nor (B) | | (d) cannot be determined |
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Answer is: BRatio of speeds of A and B
= 1000 : (1000-100)
= 10 : 9
16. | | In a 1000 m race A beats B by 10 sec. What is the ratio of speeds of A and B? | | (a)99 : 100 | | (b)100 : 99 | | (c)Neither (A) nor (B) | | (d) cannot be determined |
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Answer is: DAs we don't know either the speed of A or the time taken by A, we cannot find the ratio of their speeds.
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