Quantitative Aptitude- Speed & Distance (Answer)

ANSWER

1.(4)

[As nothing can be said about the speed]

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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 ymeter/sec.

Then,

x/(y – 5/9) == 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

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