# Quantitative Aptitude: Time, Speed and Distance Set 3

1. The distance between two towns A and B is 545 km. A train starts from town A at 8 A.M. and travels towards town B at 80 km/hr. Another train starts from town B at 9 : 30 A.M. and travels towards town A at 90 km/hr. At what time will they meet each other?
A) 11:30 AM
B) 12:30 PM
C) 12:00 Noon
D) 1:00 PM
E) 11:00 AM
Option C
Solution:

With 80 km/hr, distance travelled in 1 n half hours (9:30AM – 8AM) is 3/2 * 80 = 120 Km
Now second train also starts, and at this time distance between both trains is (545-120) = 425 km
Relative speed = 80+90 = 170 km/hr (when travelling in opposite direction, add speed)
So time when they meet = 425/170 = 2.5 hrs
So after 9:30 AM they meet after 2.5 hrs, so 12 AM
2. A bus can travel 560 km in 8 hours. The ratio of speed to train to that of car is 13 : 8. If the speed of bus is 7/8 of the speed of car, find in how much time train can cover 520 km distance.
A) 3 hours
B) 4 hours
C) 6 hours
D) 5 hours
E) 2 hours
Option B
Solution:
Speed of bus = 560/8 = 70 km/hr
So speed of car = 8/7 * 70 = 80 km/hr
So speed of train = 130 km/hr
So time taken by train to cover 520 km = 520/130 = 4 hours
3. A person has to travel from point A to point B in car in a scheduled time at uniform speed. Due to some problem in car engine, the speed of car has to be decreased by 1/5th of the original speed after covering 30 km. With this speed he reaches point B 45 minutes late than the scheduled time. Had the engine be malfunctioned after 48 km, he would have reached late by only 36 minutes. Find the distance between points A and B.
A) 120 km
B) 80 km
C) 100 km
D) 150 km
E) 70 km
Option A
Solution:
Let total distance be d km, speed = u, and time = t hours
So case 1:
30 km with speed u, (d-30) with speed 1 – 1/5 = 4/5 of u
If he would have travelled (d-30) by speed u, then time = (d-30)/u
But now time is = (d-30)/(4u/5) = 5(d-30)/4u
And difference in timings is 45 minutes = 3/4 hour
So  5(d-30)/4u  – (d-30)/u  = 3/4
Solve (d-30)/u = 3
case 2:
48 km with speed u, (d-48) with speed 1 – 1/5 = 4/5 of u
If he would have travelled (d-48) by speed u, then time = (d-48)/u
But now time is = (d-48)/(4u/5) = 5(d-48)/4u
And difference in timings is 36 minutes = 3/5 hour
So  5(d-48)/4u  – (d-48)/u  = 3/5
Solve (d-48)/4u = 3/5
Divide both equations, d = 120 km
4. Towns A and B are 225 km apart. Two cars P and Q travel from towards each other from towns A and B respectively and meet after 3 hours. If the speed of P be 1/2 of its original speed and Q be 2/3 of its original speed, they would have met after 5 hours. Find the speed of the faster car.
A) 50 km/hr
B) 40 km/hr
C) 45 km/hr
D) 30 km/hr
E) 60 km/hr
Option C
Solution:
Let speeds be x km/hr and y km/hr
So 225/(x+y) = 3
And 225/(x/2 + 2y/3) = 5
Solve, x = 30, y = 45
5. From point A, Priya and Bhavna start cycling towards point B which is 60 km away from A. The speed of Priya is 10 km/hr more than the speed of Bhavna. After reaching point B, Priya returns towards point A and meets Bhavna 12 km away from point B. Find the speed of Bhavna.
A) 40 km/hr
B) 15 km/hr
C) 30 km/hr
D) 20 km/hr
E) 45 km/hr
Option D
Solution:
Speed of Bhavna = x km/hr, of priya = (x+10) km/hr
Distance covered by Priya = 60+12 = 72 km
And by Bhavna = 60-12 = 48 km
So
72/(x+10) = 48/x
Solve, x = 20
6. A train crosses 2 men running in the same direction at speeds 5 km/hr and 8 km/hr in 12 seconds and 15 seconds respectively. Find the speed of the train.
A) 30 km/hr
B) 24 km/hr
C) 25 km/hr
D) 35 km/hr
E) 20 km/hr
Option E
Solution:
Let the speed of the train is s km/hr and its length is a m.
So
a/[(s-5)*(5/18)] = 12; [In same direction relative speed is obtained by subtracting. Also changing km/hr to m/s]
Solve 3a = 10s – 50 . . . . . . . . (i)
And also
a/[(s-8)*(5/18)] = 15;
6a = 25s-200 . . . . . . . . . .. . . (ii)
Solve (i) and (ii)
s = 20 km/hr
7. A train which is travelling at 80 km/hr meets another train travelling in same direction and then leaves it 150 m behind in next 20 seconds. Find the speed of the second train.
A) 72 km/hr
B) 53 km/hr
C) 64 km/hr
D) 59 km/hr
E) 65 km/hr
Option B
Solution:

Let speed of the 2nd train is s m/sec.
80 km/hr = (80*5)/18 = 200/9 m/sec.
Trains are travelling in same direction. So
(200/9) – s = 150/20
Solve, s = 265/18 m/sec = 265/18 * 18/5 = 53 km/hr
8. In a 500 m race C can beat B by 30 m, and in a 400 m race B can beat C by 20 m. Then in 200 m race A will beat C by how much distance (in m)?
A) 58.2 m
B) 68.4 m
C) 63.5 m
D) 72.8 m
E) 55.2 m
Option B
Solution:

When A runs 500 m, B runs 470 m
So when A runs 200 m, B runs 470/500 * 200 = 188 m
When B runs 400 m, C runs 280 m
So when B runs 188 m, C runs, 280/400 * 188 = 131.6 m
So A will beat C by (200-131.6) = 68.4 m
9. 2 towns A and B are 300 km apart. 2 trains start travelling from town A towards town B such that the second train leaves 8 hours late than the first one. They both arrive at town B simultaneously. If the speed of the faster train is 10 km/hr more than the speed of the slower train, find the time taken by the slower train to complete the journey.
A) 25 hours
B) 22 hours
C) 14 hours
D) 18 hours
E) Cannot be determined
Option E
Solution:

Let speed of the slower train is x km/hr, then speed of faster is  (x+10) kmph.
Let faster train takes t hours to cover the distance 300 km, then slower one takes (t+8) hours.
Distance is same. So
x/(x+10) = t/(t+8)
Solve, 4x = 5t
10. A man leaves from point A at 4 AM and reaches point B at 6 AM. Another man leaves from point B at 5 AM and reaches point A at 8 AM. Find the time when they meet.
A) 6:20 AM
B) 6:15 AM
C) 5:45 AM
D) 5:36 AM
E) 5:30 AM
Option D
Solution:

Use formula:
4 AM + (6-4)*(8-4)/[(6-4)+(8-5)]
gives 4 AM + 8/5
8/5 hours = 1 3/5 hours = 1 3/5*60 = 1 hour 36 minutes
So 4 AM + 1 hour 36 minutes = 5:36 AM

## 7 Thoughts to “Quantitative Aptitude: Time, Speed and Distance Set 3”

1. Eng M AwesomE

check ques no 4
Let speeds be x km/hr and y km/hr
So 225/(x+y) = 3
And 225/(x/2 + 2y/3) = 5
Solve, x = 75, y = 45

1. Oliver Queen

x =30
y=45

2. Shubhra

Yes 30

2. jaga

Q10 KA FORMULA KYA HAI?

1. Shubhra

Train A starts at x and reaches at y
Train B starts at a and reaches at b

Formula = x + (y-x)(b-a)/[(y-x)+(b-a)

3. Ayushi Srivastava

que 3 plz koi smjha do

4. AmRITa @ bank po

Q1.3 approches
1-d-120/80=d/90
2-d/80-(545-d)/90=1.5