Home / Quantitative Aptitude Quiz- Speed, Distance & Time (Answers)

Quantitative Aptitude Quiz- Speed, Distance & Time (Answers)

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ANSWERS

1–(1)

If the length of train A be x meter, then length of train B = 2x meter.

When a train crosses a pole, it covers a distance equal to its own length.

∴ Required ratio = x/25 : 2x/75

= 1/25 × 75 : 2/75 × 75

= 3 : 2

 

2—(3)

Speed of tractor = Distance/ time= 575/23 = 25 kmph

∴ Speed of bus = 50 kmph

∴ Speed of car= 9/5 × 50 = 90 kmph

∴ Distance covered by car in 4 hours= 4 × 90 = 360 km

 

3–(4)

Average speed= Total distance/ Total time= (39 + 25)km/(45 + 35) minute

= 64/(80/60) kmph

= (64 × 60)/80 kmph= 48 kmph

 

4–(4)

Distance covered by aeroplane in 9 hours =Speed × Time= 9 × 756 = 6804 km

∴ Speed of helicopter= (2 × 6804)/48= 283.5 kmph

∴ Distance covered by helicopter in 18 hours= (283.5 × 18) km= 5303 km

 

5–(2)

Relative speed of train= (80 – 8) kmph= 72 kmph

= (72 × 5/18) m/sec.

= 20 m/sec.

∴Required time= Length of train/ Relative speed

= 360/20= 18 seconds

 

6–(1)

Rate downstream of boat = (9.5 + 2.5)kmph= 12 kmph

Rate upstream of boat= (9.5 – 2.5) kmph= 7 kmph

Distance between A and B= x km (let)

According to question,

x/7 + x/12 = 114/60

= (12x + 7x)/(12 × 7) = 114/5

= 19x/7 = 114/5

= 19x = 114/5 × 7

= x = (114 × 7)/(5 × 19)= 8.4 km.

 

7–(2)

Rate downstream= Distance/ Time = 16/2 = 8 kmph

Rate of upstream = 16/4 = 4 kmph

Speed of boat in still water= ½ (Rate downstream + Rate Upstream)

=1/2 (8 + 4) = 6 kmph

 

8–(3)

Let both cars meet each other after t hours from11 am.

Distance = Speed × Time

∴ 36 × t + 44 (t – 2) =592

=36t + 44t – 88 = 592

=80t= 592 +88 = 8.5 hours i.e.

= 7 : 30 pm

 

9–(5)

Speed of train= [Length of (train + platform)]/Time takento cross= 440/22 = 20 m/s

= 20 × 18/5 = 72 km/hr

 

10–(1)

The distance walked on the first day = 2 kms

The distance walked on subsequent days is half th distance walked on the previous days.

∴ Total distance walked= 2 + 1 + ½ + ¼ + ….

This is a Geometric series Whose First term, a = 2 common ratio, r = ½

Maximum total distance walked by the person in his lifttime menas the number of terms in the series would be infinite.

Hence, the series would be an infinite Geometric series.

Sum of an infinite Geometric series is given by

S = a/(1 – r)

= S = 2/(1-1/2)

= S = 2/(1/2

= s = 4 kms)

 

11– E

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