Aptitude Time And DistancePage 5
33. | | An aeroplane covers a certain distance at a speed of 240 kmph in 5 hours. To cover the same distance in 5/3 hours, it must travel at a speed of: | | (a)300 kmph | | (b)360 kmph | | (c)600 kmph | | (d)720 kmph |
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Answer is: DDistance = (240 x 5) = 1200 km.
Speed = Distance/Time
Speed = 1200/(5/3) km/hr.
Required speed = 1200 x (3/5) km/hr = 720 km/hr.
34. | | A person travels from P to Q at a speed of 40 kmph and returns by increasing his speed by 50%. What is his average speed for both the trips ? | | (a)36 kmph | | (b)45 kmph | | (c)48 kmph | | (d)50 kmph |
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Answer is: CSpeed on return trip = 150% of 40 = 60 kmph
Average speed = (2x40x60)/(40+60) = 48 kmph
35. | | Mac travels from A to B a distance of 250 miles in 5 hours 30 minutes. He returns to A in 4 hours 30 minutes. His average speed is : | | (a)44 kmph | | (b)46 kmph | | (c)48 kmph | | (d)50 kmph |
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Answer is: DSpeed from A to B = (250 x 2/11) mph = 500/11 mph.
Speed from B to A = (250 x 2/9) mph = 500/9 mph.
Average speed = [2x(500/11) x (500/9)]/[500/11 + 500/9] mph
Average speed = 500000/(4500+5500) = 50 mph
36. | | A boy goes to his school from his house at a speed of 3 kmph and returns at a speed of 2 kmph. If he takes 5 hours in going and coming, the distance between his house and school is: | | (a)5 km | | (b)5.5 km | | (c)6 km | | (d)6.5 km |
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Answer is: CAverage speed = [(2 x 3 x 2)/(3 + 2)] kmph = 12/5 kmph
Distance travelled = (12/5) x 5 km = 12 km.
Distance between house and school = 12/2 km = 6 km
37. | | A man on tour travels first 160 km at 64 kmph and the next 160 km at 80 kmph. The average speed for the first 320 km of the tour is : | | (a)35.55 kmph | | (b)36 kmph | | (c)71.11 kmph | | (d)71 kmph |
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Answer is: CTotal time taken = (160/64 + 160/8) hours = 9/2 hours.
So, average speed = [320 x (2/9)] kmph = 71.11 kmph.
38. | | Walking 6/7th of his usual speed, a man is 12 minutes too late. The usual time taken by him to cover that distance is : | | (a)1 hour | | (b)1 hour 12 min. | | (c)1 hour 15 min | | (d)1 hour 20 min. |
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Answer is: BNew speed = 6/7 of usual speed.
New time = 7/6 of usual time.
(7/6 of usual time) - (usual time) = 1/5 hour.
Therefore, 1/6 of usual time = 1/5 hour
Usual time = 6/5 hours = 1 hour 12 min.
39. | | In covering a certain distance, the speeds of A and B are in the ratio 3 : 4. A takes 30 min. more than B to reach the destination. The time taken by A to reach the destination is : | | (a)1 hour | | (b)1 hour 30 min. | | (c)2 hours | | (d)2 hours 30 min. |
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Answer is: CRatio of speeds = 3 : 4
Ratio of times taken = 4 : 3.
Suppose A takes 4X hours and B takes 3X hours to reach the destination. Then, 4X - 3X = 30/60
X = 1/2.
Time taken by A = 4X hours = 4 x 1/2 = 2 hours.
40. | | In covering a distance of 30 km, Abhay takes 2 hours more than Sameer. If Abhay doubles his speed, then he would take 1 hour less than Sameer. Abhay's speed is: | | (a)5 kmph | | (b)6 kmph | | (c)6.5 kmph | | (d)7.5 kmph |
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Answer is: ALet Abhay's speed be X km/hr.
Then, 30/X - 30/2X = 3
6X = 30.
X = 5 kmph
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