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# Time And Distance - MCQ 3

## 20 Questions MCQ Test Quantitative Ability for SSC CHSL | Time And Distance - MCQ 3

Description
This mock test of Time And Distance - MCQ 3 for Quant helps you for every Quant entrance exam. This contains 20 Multiple Choice Questions for Quant Time And Distance - MCQ 3 (mcq) to study with solutions a complete question bank. The solved questions answers in this Time And Distance - MCQ 3 quiz give you a good mix of easy questions and tough questions. Quant students definitely take this Time And Distance - MCQ 3 exercise for a better result in the exam. You can find other Time And Distance - MCQ 3 extra questions, long questions & short questions for Quant on EduRev as well by searching above.
QUESTION: 1

### Two places are A and B are 200 kms from each other. A train leaves from A for B at the same time another train leaves B for A. The two trains meet at the end of 8 hours. If the train travelling from A to B travels 8km/hr faster than the other. Find the speed of the faster train?

Solution:

Speed of trains  = s + 8 and s km/hr.
At some P distance from A both train will meet. So,
8 = P/(s +8) ; 8 = (200 – P)/s
Solve both equation, we get s = 8.5 so faster train speed 16.5 km/hr
SHORTCUT:
200/8 = 25
A+B = 25
A- B = 8 (given)
A= 16.5 km/hr

QUESTION: 2

### A truck driving on a highway passed a man walking at the rate of 15km/hr in the same direction. He could see the truck for 2 minutes and up to 500 meters. Find the speed of the truck?

Solution:

x-15=(500/120) *18/5
x=30

QUESTION: 3

### A train leaves Chennai for Bangalore at 2:15 p.m. and travels at the rate of 50 kmph. Another train leaves Bangalore for Chennai at 1:35 p.m. and travels at the rate of 60 kmph. If the distance between Bangalore and Chennai is 590 km at what distance from Chennai will the two trains meet?

Solution:

Total Distance = 590 km
Distance in 40 min = 60*40/60=40km
Remaining=550 km
Time = 550/110=5hr
Reqd Distance = 50*5=250 km

QUESTION: 4

A bus was travelling from Mumbai to Pune was delayed by 16 minutes and made up for the delay on a section of 80 km travelling with a speed 10 km per hour higher than its normal speed. Find the original speed of the bus?

Solution:

80/x – 80/(x+10) = 16/60= 4/15
300/x -300/x+10 =1
X2 + 10x – 3000 = 0
X = -60, +50
So 50km/hr

QUESTION: 5

Two cities A and B are at a distance of 120 km from each other. Two persons P and Q start from First city at a speed of 20km/hr and 10km/hr respectively. P reached the second city B and returns back and meets Q at Y. Find the distance between A and Y.

Solution:

P = 120/20 = 6hrs
In 6hrs Q travels 60km
120-60 = 60km
60/(20+10) = 2hrs
Point Y → 60 + 10*2hrs = 80

QUESTION: 6

Vijay takes 4 hr in walking at certain place and return back. While it takes 3 hrs in walking at certain place and riding back. Find the time Vijay will take to ride both sides

Solution:

W + W = 4. W =2
W+ R = 3. R =1
So to ride both direction, it will take 1+1 = 2 hrs.

QUESTION: 7

Rakesh traveled 2000 kilometer by air which covered 3/5 of the total journey. He travels 1/4 of the trip by car and the remaining trip by train. Find the distance traveled by train.

Solution:

3=2000
5=10000/3
Distance by train=3×10000/3×4=2500
Another method :
Total 20
12====2000
20====2000*20/12
5=====2000*5/12
3=====2000*5*3/12=2500

QUESTION: 8

Rahul has to travel from one point to another point in a certain time. Travelling at a speed of 6kmph he reaches 40m late and travelling at a speed of 8kmph he reaches 12 m earlier.What is the distance between this two points ?

Solution:

Explanation :
t + 40/60 = d/6
t – 12/60 = d/8
By the solving these two equations we get.
d = 20.8 km ~ 21 km
Another method :
6………………..4.           (40m late)
………24…………
8…………………..3.            (12 m ear)
60 = 52
1 = 52/60 = 13/15
So distance = 24 *13/15
= 8*13/5 = 104/5 = 20.8km

QUESTION: 9

Vivek travelled a distance of 50km in 7hrs. He travelled the distance partly on foot at 5kmph and partly on bicycle at 8kmph. What is the distance that he travelled on foot ?

Solution:

x/5 + 50-x/8 =7
8x+250-5x = 7*40
3x+250 = 280
3x = 30
X = 30/3 = 10km

QUESTION: 10

Mani drove at the speed of 45 kmph. From home to a resort. Returning over the same route, he got stuck in traffic and took an hour longer, also he could drive only at the speed of 40 kmph. How many kilometers did he drive each way ?

Solution:

x/40 – x/45 = 1
9x-8x/360 = 1
x/360 = 1
x = 360km

QUESTION: 11

Rohit starts cycling along the boundaries of the squares. He starts from a point A and after 90 minutes he reached to point C diagonally opposite to A. If he is travelling with 20km/hr, then find the area of square field.

Solution:

D = 20*3/2 = 30 km. So side of square is 15km, so area – 225km2

QUESTION: 12

A truck covers a distance of 376 km at a certain speed in 8 hours. How much time would a car take at an average speed which is 18 kmph more than that of the speed of the truck to cover a distance which is 14 km more than that travelled by the truck ?

Solution:

Speed of the truck = Distance/time = 376/8 = 47 kmph
Now, speed of car = (speed of truck + 18) kmph = (47 + 18) = 65 kmph
Distance travelled by car = 376 + 14 = 390 km
Time taken by car = Distance/Speed = 390/65 = 6 hours.

QUESTION: 13

Two cars start together in the same direction from the same place. The first goes with uniform speed of 10 kmph. The second goes at a speed of 8 kmph in the first hour and increases its speed by 1/2 kmph each succeeding hours. After how many hours will the second car overtake the first, if both cars go non stop?

Solution:

The second car overtake the first car in x hours
Distance covered by the first car in x hours = Distance covered by the second car in x hours
10x = x/2[2a + (x-1)d] 10x = x/2[2*8 + (x-1)1/2] x = 40 -31 = 9

QUESTION: 14

A car runs at the speed of 50 km per hour when not serviced and runs at 60 kmph when serviced. After servicing the car covers a certain distance in 6 hours. How much time will the car take to cover the same distance when not serviced?

Solution:

Time = 60*6 / 50 = 7 hours 12 mins

QUESTION: 15

Two cars namely A and B start simultaneously from a certain place at the speed of 40 kmph and 55 kmph, respectively.The car B reaches the destination 2 hours earlier than A. How much time is taken by the cars from  the starting point and destination?

Solution:

Let the time taken by car A to reach destination is T hours
So, the time taken by car B to reach destination is (T – 2) hours.
S1T1 = S2T2
⇒ 40(T) = 55 (T – 2)
⇒ 40T = 55T -110
⇒ 15T = 110
T = 7 hours 20 minutes

QUESTION: 16

A thief is spotted by a policeman from a distance of 200 metre. When the policeman starts chasing , the thief also starts running. If the speed of the thief be 16kmph and that of policeman be 20kmph, how far the thief will have run before he is overtaken?

Solution:

d = 200 m, a = 16kmph = 40/9 m/s, b = 20kmph = 50/9 m/s
Required Distance D = d*(a/b-a)b= 200*(40/9/10/9) = 800m

QUESTION: 17

A bus travels at the rate of 54 kmph without stoppages and it travels at 45 kmph with stoppages. How many minutes does the bus stop on an average per hour?

Solution:

Due to stoppages, the bus can cover 9 km less per hour[54 -45 = 9] Time taken to cover 9 km =(9/54) x 60 = 10 minutes.

QUESTION: 18

The ratio between the rate of speed of travelling of A and B is 2:3 and therefore A takes 20 minutes more than time taken by B to reach a particular destination. If A had walked at double the speed, how long would he have taken to cover the distance?

Solution:

Let B and A takes T minutes and (T + 20) minutes respectively.
Speed Inversely proportional to time, So time taken by A and B is
(T + 20) : T = 1/2 : 1/3 = 3 : 2
⇒ (T + 20)/T = 3/2
⇒ 2T + 40 = 3T
T = 40
A takes (T + 20) = (40 + 20) = 60 min. If A had walked at double the speed then the time taken by A is 30 minutes.

QUESTION: 19

Anu normally takes 4 hours more than the time taken by Sachin to walk D km. If Anu doubles her speed, she can make it in 2 hours less than that of Sachin. How much time does Sachin require for walking D km?

Solution:

Let Sachin takes x hours to walk D km.
Then, Anu takes (x + 4) hours to walk D km.
With double of the speed, Anu will take (x + 4)/2 hours.
x – (x + 4)/2 = 2
⇒ 2x – (x + 4) = 4
⇒ 2x – x – 4 = 4
x = 4 + 4 = 8 hours

QUESTION: 20

Sohail covers a distance by walking for 6 hours. While returning, his speed decreases by 2kmph and he takes 9 hours to cover the same distance. What was his speed while returning?

Solution:

The speed of Sohail in return journey = x
6(x + 2) = 9x
⇒ 6x + 12 = 9x
⇒ 9x – 6x = 12
x = 4kmph