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# Time & Distance Quiz For Upcoming Exams Answers

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1.(4) [As nothing can be said about the speed]
Distance = 200 meter
time = 16 seconds
Speed = distance/ time = 200/16
= 12.5 m/sec.
= 12.5 m/sec.
= 12.5 × 18/5
= 45 km/hr.

2.(1) Let the employee travelled x kms by taxi.
∴ Distance covered by him by his own car = (90 – x) km.
According to the question.
x × 7 + (90 – x) km.
According to the question.
x × 7 + (90 – x) × 6 = 675
= 7x + 540 – 6x = 675
x = 675 – 540 = 135
∴ Required distance = 135 km.

3.(1) Let the speed of train C be x kmph.
Speed of train B relative to C
= (120 – x) kmph
= [(120 – x) × 5/18] m/sec
= (600 – 5x)/18 m/sec.
Distance covered
= 100 + 200 = 300m
∴ 300/(600 – 5x)/18 = 120
= 300 = 120(600 – 5x)/18
= 10 × 9 = 2 (600 – 5x)
= 90 = 1200 – 10x
= 10x = 1200 – 90
= x = 1110/10 = 111
Hence, the speed of train C is 111 kmph.

4.(4) When a train crosses a platform it covers a
distance equal to the sum of lengths of platform and the
train itself. If the length of train be x meters, then
= (x + 300)/38 m/sec. ……….. (i)
When the train crosses a signal post it covers its own
length.
∴ Speed of train
= x / 18 m/sec. ……….. (ii)
From equati0ns (i) and (ii)
= (x + 300)/38 = x/18
= 38x – 18x = 300 × 18
= 20x = 300 × 18
= x= (300 × 18)/20
= x = 270 meters
∴ Speed of train = 270/18
= 15 m/sec.
= 15 × 18/5 = 54 kmph

5.(2) Average speed = 2xy/(x + y)
(when the same distances are covered)
= (2 × 24 × 36)/(24 + 36) kmph
= (2 × 24 × 36)/60 = 28.8 kmph

6.(3) 2 kmph = (2 ×5)/18 meter/sec.
= 10/9 meter/sec.
Let the length of the train be x meter and its speed be y
meter/sec.
Then,
x/(y – 5/9) = 9
= 9y – 5 = x
∴ 9y – x = 5 ……………. (i)
and x(9y- 10) = 9x
= 10 (9y – 10) = 9x
= 90 y – 9x = 100 ……….. (ii)
By equation (i) × 10 – equation (ii)
we have
90y – 10x = 50
90y – 9x = 100
– + .
– x = – 50
= Length of train = 50 m

7.(2) Let the length of train B =x
∴ Length of train A = 3x/4 meter
∴ Required ratio = 3x/4 × 33 : x/55
= 5 : 4

8.(3) If the distance be x km, then
x/(3/2) + x/(9/2) = 6
= (2x/3 + 2x/9) = 6
(6x + 2x) /9 = 6
= 8x = 9 × 6
= x = 54/8 = 27/4 = 27/4 km

9.(4) C = k x2
When x = 16 kmph
C = Rs. 64
∴64 = k × 162 = k = ¼
∴ C = ¼ x2
Total expenditure per hour
= ¼ x2 + 400
When speed = 40 kmph, the expenditure will be
minimum.

10.(2) Total expenditure
= ¼ × 40 × 40 × 10 + 400 × 10
= Rs. 8000