Courses

# Time And Distance - MCQ 1

## 20 Questions MCQ Test Quantitative Ability for SSC CHSL | Time And Distance - MCQ 1

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

### A car started from Indore to Bhopal at a certain speed. The Car missed an accident at 40Kms away from Indore, then the driver decided to reduce Car speed to 4/5 of the original speed. Due to this, he reached Bhopal by a late of 1hr 15min.Suppose if he missed an accident at 80Km away from Indore and from then he maintained 4/5 of original speed then he would reach Bhopal by a late of 1hour. Then what is the original speed of the Car?

Solution:

(x+40/s) + 5/4 = 40/s + 5x/4s — — — 1
(x+40/s) + 1 = 80/s + 5(x-40)/4s — — — 2
s = 40

QUESTION: 2

### Two places A and B are at a certain distance. Ramu started from A towards B at a speed of 40 kmph. After 2 hours Raju started from B towards A at a speed of 60 kmph. If they meet at a place C then ratio of ratio of time taken by Raju to Ramu to reach Place C is 2:3. Then what is the distance between A and B?

Solution:

40t 1 +60t 2 = d
T 1 /t 2 = 3/2
T 1 – t 2 = 2
Solving d = 480 Km

QUESTION: 3

### Ramu started from A towards B at a speed of 20Km/hr and Raju started from B towards A. They crossed each other after one hour. Raju reached his destination 5/6 hour earlier than Ramu reached his destination.Then what is the distance between A and B?

Solution:

1 = x/20 = D-x/s 2
5/6 = D(1/20 – 1/s 2)
D =50

QUESTION: 4

Two Cars started at same time, same place and towards same direction. First Car goes at uniform speed of 12Km/hr. Second Car goes at speed of 4 Km/hr in first hour and increases it speed by 1 Km/hr for every hour. Then what is the distance traveled by car B when the both the Cars meet for the first time?

Solution:

12*x = x/2(2*4+(x-1)*1)
X= 17
D = 17*12 =204

QUESTION: 5

A man traveled 100 km by Bike in 2 hours. He then traveled in Bus for 8 hrs and then Train in 9 hrs. Ratio of Speeds of Bus to Train is 4:5. If speed of train is 4/5 of Bike speed then the entire journey covered by him in Km is?

Solution:

Speed of train = 50
Bus = 32
Train = 40
Distance = 100+32*8+40*9 = 716

QUESTION: 6

Car A leaves Delhi at a certain time, after 5 hours Car B leaves Delhi in the same direction as of A. Speed of Car A is 20Km/hr and Speed of Car B is 40Km/hr. In how much time Car B will be 20Km ahead of Car A?

Solution:

20 =40*(t-5) – 20*t
T = 11
11-5 = 6

QUESTION: 7

Two places A and B are at a distance of 480Km. Sita started from A towards B at the speed of 40Kmph. After 2 hours Gita started from B towards A at speed of 60 Kmph. They meet at a Place C then what is the difference between the time taken by them to reach their destinations from Place C?

Solution:

t*40+ (t-2)*60 = 480
t=6
From Place C to A = C to B = 240
Then time taken by Sita = 6hours and GIta = 4 hours
Difference = 6-4 = 2 hours

QUESTION: 8

Two Cars started at same time, same place and towards same direction. First Car goes at a speed of 5km/hr in first and 7 Km/hr in second hour and repeats the cycle over the entire journey. Similarly second Car goes at 4km/hr in first hour and 9km/hr in second hour and repeats the cycle over the entire journey.Then these two Cars for the first time for after how many hours of journey?

Solution:

First car : 5,12,17
Second Car: 4,3,17

QUESTION: 9

A police saw a Thief at a distance of 2km. When Police started chasing him Thief also started running. If the ratio of Speeds of Police to Thief is 5:4. Then thief was caught at a certain distance then how many Kms did police run to catch the Thief?

Solution:

2+x/5 = x/4
X = 8
Police = 8+2 = 10

QUESTION: 10

If a Car runs at 45Km/hr it reaches its destination by 10 min late. If it runs at 60Km/hr it is late by 4min. Then what is the correct time for the journey?

Solution:

45(t+10) = 60(t+4)
T = 14

QUESTION: 11

The distance of the School and house of Suresh is 80km. One day he was late by 1 hour than the normal time to leave for the college, so he increased his speed by 4km/h and thus he reached to college at the normal time. What is the changed speed of Suresh?

Solution:

80/x – 80/(x+4) = 1
x(x+20) – 16(x+20) = 0
x = 16kmph
Increased speed = 20 kmph

QUESTION: 12

Anita goes to College at 20 km/h and reaches college 4 minutes late. Next time she goes at 25 km/h and reaches the college 2 minutes earlier than the scheduled time. What is the distance of her school?

Solution:

20*25/(25-20)*6/60=10.

QUESTION: 13

Two places R and S are 800 km apart from each other. Two persons start from R towards S at an interval of 2 hours. Whereas A leaves R for S before B. The speeds of A and B are 40 kmph and 60 kmph respectively. B overtakes A at M, which is on the way from R to S. What is the ratio of time taken by A and B to meet at M?

Solution:

Time taken by B to reach at M = 4h
Time taken by A to reach at M = 6h
Ratio = 6:4 = 3:2

QUESTION: 14

Two places R and S are 800 km apart from each other. Two persons start from R towards S at an interval of 2 hours. Whereas A leaves R for S before B. The speeds of A and B are 40 kmph and 60 kmph respectively. B overtakes A at M, which is on the way from R to S. What is the extra time taken by A to reach at S?

Solution:

Time taken by A to reach at Q = 800/40 = 20 hours
Time taken by B to reach at Q = 800/60 = 13 hours and 20 min
A takes 6hr 40 minutes extra time to reach at Q.

QUESTION: 15

Ajay covers certain distance with his own speed but when he reduces his speed by 10kmph his time duration for the journey increases by 40 hours while if he increases his speed by 5 kmph from his original speed he takes 10 hours less than the original time taken. Find the distance covered by him.

Solution:

x/(y – 10)  – x/y = 40
x = 4y(y-10) —(i)
x/y  – x/(y + 5) = 10
x = 2y(y + 5) — (ii) From (i) and (ii) ⇒ y = 25; x = 1500

QUESTION: 16

The driver of an ambulance sees a college bus 40 m ahead of him after 20 seconds, the college bus is 60 meter behind. If the speed of the ambulance is 30 km/h, what is the speed of the college bus?

Solution:

Relative Speed = (Total distance)/total time
= (60+40) /20 = 5 m/s = (5*18)/5 = 18 kmph
Relative Speed = (speed of ambulance – speed of College bus)
Speed of College bus = speed of ambulance – relative speed.
= 30-18 = 12 kmph.

QUESTION: 17

Two places R and S are 800 km apart from each other. Two persons start from R towards S at an interval of 2 hours. Whereas A leaves R for S before B. The speeds of A and B are 40 kmph and 60 kmph respectively. B overtakes A at M, which is on the way from R to S. What is the distance from R, where B overtakes A?

Solution:

Distance between R and M = 4 * 60 = 240

QUESTION: 18

Two rabbits start running towards each other, one from A to B and another from B to A. They cross each other after one hour and the first rabbit reaches B, 5/6 hour before the second rabbit reaches A. If the distance between A and B is 50 km. what is the speed of the slower rabbit?

Solution:

Let second rabbit takes x hr with speed s2
First rabbit takes x-5/6 hr with speed s1
Total distance = 50km
S1 = 50/(x-(5/6))
S2 = 50/x
As they cross each other in 1hr…
Total speed = s1 + s2
Now, T = D / S
50/(s1+s2) = 1
x = 5/2, 1/3
Put x = 5/2 in s2 –> 20km/hr

QUESTION: 19

Pranav walked at 5 kmph for certain part of the journey and then he took an auto for the remaining part of the journey travelling at 25 kmph. If he took 10 hours for the entire journey, what part of journey did he travelled by auto if the average speed of the entire journey be 17 kmph

Solution:

Total distance = 17*10 = 170

Let Journey travelled by auto in x hr

25 * x + (10-x ) 5 = 170

25 x + 50 – 5x = 170

x = 6

Required Distance = 6 * 25 = 150 km

QUESTION: 20

Aravind started for the station half a km from his home walking at 1 km/h to catch the train in time. After 3 minutes he realised that he had forgotten a document at home and returned with increased, but constant speed to get it succeded in catching the train. Find his latter speed in kmph?

Solution:

Distance covered in 3 minutes = 3*(1000/60) = 50

Now he has to cover (500+50)m in (30-3) minutes

Required speed = (550/1000)/(27/60) = 11/9 km/h