# Quantitative Aptitude: Time and Distance Set 1

1. Vimal completes the first part of his journey at 30kmph and the next at 60kmph, covering the entire journey at an average speed of 40 kmph. What is the ratio of the distance that he covered at 30kmph to that he covered at 60kmph?
A) 2 : 1
B) 1 : 5
C) 3 : 1
D) 3 : 5
E) None of these
Option C
Solution:

Distance of first part of his journey = x
Distance of first part of his journey = y
x/30 + y/60 = (x + y)/40
6x + 2y = 3x + 3y
x/y = 3:1
2. Suresh takes 7 hours 40 minutes in walking to a certain place and riding back. If he walks on both ways he will lose one hour. The time he would take to ride both ways is?
A) 5 hours 20 minutes
B) 7 hours 40 minutes
C) 6 hours 20 minutes
D) 6 hours 40 minutes
E) None of these
Option D
Solution:

Walking + Riding = 7 hours 40 minutes
Walking + Walking = 8 hours 40 minutes
So Walking = 4 hours 20 minutes
Riding + Riding = 6 hours 40 minutes
3. Anil completed his journey in 10 hours. He travels first half of the journey at the rate of 22 kmph and second half at the rate of 18 kmph. Find the total journey in km.
A) 145 km
B) 198 km
C) 220 km
D) 180 km
E) None of these
Option B
Solution:

Total hours = 10 hours
Average Speed = 2xy/(x+y) = 2*22*18/(40) = 19.8 kmph
Total Journey in Km = 19.8 * 10 = 198 km
4. Two cars start from a place with a speed of 40 kmph at an interval of 10 minutes. What is the speed of a man coming from the opposite direction towards the place if he meets the cars at an interval of 8 minutes?
A) 10 kmph
B) 13 kmph
C) 14 kmph
D) 16 kmph
E) None of these
Option A
Solution:

Distance covered in 10 minutes at 60 kmph = distance covered in 8 minutes at (60+x) kmph.
40*10/60 = 8/60 * (40+x)
20 * 5 = 80 + 2x
x = 10 kmph
5. Waking 3/4 of his normal speed, Ravi was 18 minutes late in reaching his office. The usual time took to cover the distance between his home and office was:
A) 36 minutes
B) 24 minutes
C) 42 minutes
D) 54 minutes
E) None of these
Option D
Solution:

3/4 of speed = 4/3 of original time
4/3 of original time = original time + 18 minutes;
1/3rd of original time = 18 minutes;
Thus, original time = 18*3 = 54 minutes.
6. Mr. Ravi completes a certain journey by a car. If he covered 40% of the distance at the speed of 20kmph, 50% of the distance at 25 kmph and the remaining of the distance at 10 kmph, then what will be the speed?
A) 15 kmph
B) 20 kmph
C) 18 kmph
D) 14 kmph
E) None of these
Option B
Solution:

Assume Total distance = 100 km.
So speed = 100/[(40/20)+(50/25)+(10/10];
speed = 100/[(2)+(2)+(1)];
= 100/
= 20 kmph.
7. Ajay walked at 10 kmph for certain part of the journey and then he took an auto for the remaining part of the journey travelling at 30 kmph. If he took 10 hours for the entire journey, what part of journey did he traveled by auto if the average speed of the entire journey be 18 kmph?
A) 132 km
B) 145 km
C) 128 km
D) 120 km
E) None of these
Option D
Solution:

Total distance = 18*10 = 180
Journey traveled by auto = x hours
30 * x + (10-x )10 = 180
30x + 100 – 10x = 180
20x = 80
x = 4 hours
Distance traveled by auto = 4 * 30 = 120 km
8. Rani started walking to the station half a km from her home at 1 kmph to catch the train in time. After 6 minutes she realized that she had forgotten her purse at home and returned with increased, but constant speed to get it succeeded in catching the train. Find her latter speed in kmph.
A) 1.5 kmph
B) 1.2 kmph
C) 2.2 kmph
D) 3.5 kmph
E) None of these
Option A
Solution:

Distance covered in 6 minute = 6*(1000/60) = 100
She has to cover (500+100) meters in 24 minutes
Required speed = (600/1000)/(24/60) = 1.5kmph
9. Two Rabbits started running towards each other, one from A to B and another from B to A. They cross each other after 1.2 hours and the first Rabbit reaches B, 1 hour before the second rabbit reaches A. If the distance between A and B is 60 km, what is the speed of the slower rabbit?
A) 10 kmph
B) 15 kmph
C) 25 kmph
D) 18 kmph
E) 20 kmph
Option E
Solution:

Let speed of faster rabbit = x
& speed of slower rabbit = y
x + y = 60/1.2
x + y = 50 —- (A)
60/y – 60/x = 1 —- (B)
Solve equations, we get y = 150 and 20 bt y cannot be 120 as x+y = 50. So y = 20 kmph and x = 30 kmph
10. Walking at 3/2 of his normal speed Sehwag takes 40 minutes less than the usual time. What is the new time taken by Sehwag?
A) 6 hours
B) 5 hours
C) 4 hours
D) 8 hours
E) 2 hours