# Quantitative Aptitude DI - Sep 22 2020

Here we are providing new series of Quantitative Aptitude Questions for upcoming exams, so the aspirants can practice it on a daily basis.

Direction (1 - 5): Study the following data carefully and answer the questions:

There are total five trains A, B, C, D and E, first pie chart given below shows the per cent distribution of length (meters) of five trains respectively while the second pie chart given below shows the distribution (degree) of speed (km/h) of those five trains respectively.

Sum of first chart is 1250 meters while the sum of second pie chart is 180 km/h.

1) Train P can cross a pole in 72 seconds, then what is the time taken by train A to cross train P when both the trains are travelling in opposite direction if time taken by train A to cross train P when both the trains are travelling in same direction is 79.2 seconds?

a) 27 seconds

b) 48 seconds

c) 42 seconds

d) 36 seconds

e) None of these

2) Train B can cross a platform in 72 seconds, then what is the time taken by another train P of length equal to that of platform to cross train B if both the trains are travelling in same direction and speed of train P is 63 km/h?

a) 90 seconds

b) 60 seconds

c) 75 seconds

d) 120 seconds

e) None of these

3) If a car can cross train C in 27 seconds when travelling in same direction, then what is the total distance covered by the car in 2.5 hours if it travels with speed 20% less than its original speed and also the length of car is negligible with respect to that of train and car is faster than train?

a) 100 km

b) 90 km

c) 80 km

d) 120 km

e) None of these

4) Train D can cross a platform of certain length in 36 seconds. If a person boarded in the train and he started running inside the train with speed 10 km/h and in the opposite direction as that of train, then what is the time taken by the person to cross that platform?

a) 24 seconds

b) 36 seconds

c) 27 seconds

d) 18 seconds

e) None of these

5) Train C can cross a person travelling in opposite direction as that of train in 13.5 seconds, then what is the time taken by train E to cross that persons when he is travelling in opposite direction to that of train with speed one – third of his original speed?

a) 36 seconds

b) 24 seconds

c) 48 seconds

d) 42 seconds

e) None of these

Directions (1 - 5):

Length of train A = 20% of 1250 = 250 meters

Speed of train A = 180 * (80/360) = 40 km/h

Length of train B = 24% of 1250 = 300 meters

Speed of train B = 180 * (70/360) = 35 km/h

Length of train C = 12% of 1250 = 150 meters

Speed of train C = 180 * (50/360) = 25 km/h

Length of train D = 16% of 1250 = 200 meters

Speed of train D = 180 * (100/360) = 50 km/h

Length of train E = 28% of 1250 = 350 meters

Speed of train E = 180 * (60/360) = 30 km/h

Let the length and speed of train P is ‘x’ meters and ‘y’ km/h respectively.

According to the question:

x = y * 72 * (5/18)

x = 20y…………. (1)

(x + 250) = (40 – y) * 79.2 * (5/18)

From equation (1):

[20y + 250] = 22(40 – y)

20y + 250 = 880 – 22y

42y = 630

y = 15 km/h

x = 300 meters

Let time taken by train A to cross train P when both the trains are travelling in opposite direction = t

(250 + 300) = (40 + 15) * t * (5/18)

550 = 55 * t * (5/18)

t = 36 seconds

Let the length of platform and train P is ‘x’ meters.

According to the question:

(300 + x) = 35 * 72 * (5/18)

300 + x = 700

x = 400 meters

Now,

(300 + 400) = (63 – 35) * t * (5/18)

700 = 28 * t * (5/18)

t = 90 seconds

Let the speed of car = ‘x’ km/h

According to the question:

150 = (x – 25) * 27 * (5/18)

(x – 25) = 20

x = 45 km/h

Reduced speed of car = 80% of 45 = 36 km/h

Let the length of platform = ‘x’ meters

According to the question:

(200 + x) = 50 * 36 * (5/18)

(200 + x) = 500

x = 300 meters

Persons is running inside the train with speed 10 km/h and in opposite direction as that of train, then effective speed of person inside the train

= 50 – 10 = 40 km/h

Time taken by person to cross the train = (300/40) * (18/5) = 27 seconds

Let the speed of person = ‘x’ km/h

According to the question:

150 = (25 + x) * 13.5 * (5/18)

25 + x = 40

x = 15 km/h

Reduced speed of persons = x/3 = 5 km/h

Now,

350 = (30 + 5) * t * (5/18)

t = 36 seconds