Time and Distance Questions for SSC CHSL Set-2 PDF
Download SSC CHSL Time and Distance questions with answers Set-2 PDF based on previous papers very useful for SSC CHSL Exams. Top-15 Very Important Questions for SSC Exam
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Question 1: I walk a certain distance and ride back taking a total time of 37 minutes. I could walk both ways in 55 minutes. How long would it take me to ride both ways ?
a) 30 minutes
b) 19 minutes
c) 37 minutes
d) 20 minutes
Question 2: You arrive at your school 5 minutes late if you walk with a speed of 4 km/h, but you arrive 10 minutes before the scheduled time if you walk with a speed of 5 km/h. The distance of your school from your house (in km) is
a) 4
b) 5
c) 10
d) 2
Question 3: The distance between the points (0, 0) and the intersecting point of the graphs of x= 3 and y = 4 is
a) 4 unit
b) 3 unit
c) 2 unit
d) 5 unit
Question 4: A student goes to school at the rate of 2.5 km/h and reaches 6 minutes late. If he travels at the speed of 3 km/h. he is 10 minutes early. The distance (in km) between the school and his house is
a) 5
b) 4
c) 3
d) 1
Question 5: A thief is noticed by a policeman from a distance of 200 m. The thief starts running and the policeman chases him. The thief and the policeman run at the rate of 10 km and 11 km per hour respectively. The distance (in metres) between them after 6 minutes is
a) 190
b) 200
c) 100
d) 150
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Question 6: A person can row a distance of one km upstream in ten minutes and downstream in four minutes. What is the speed of the stream ?
a) 4.5 km/h
b) 4 km/h
c) 9 km/h
d) 5.6 km/h
Question 7: Two trains starting at the same time from two stations 200 km apart and going in opposite directions cross each other at a distance of 110 km from one of the stations. The ratio of their speeds is
a) 11 : 20
b) 20 : 9
c) 9 : 20
d) 11 : 9
Question 8: By walking at 3/4 of his usual speed, a man reaches his office 20 minutes later than usual. His usual time is
a) 30 min.
b) 75 min.
c) 90 min.
d) 60 min.
Question 9: A can finish a work in 24 days, B in 9 days and C in 12 days. B and C start the work but they are forced to leave after 3 days. The remaining work was done by A in
a) 10 days
b) 10 1/2 days
c) 5 days
d) 6 days
Question 10: Mandar works 3 times as fast as Samarth. If Samarth can complete a job alone in 28 days, then in how many days can they together finish the job?
a) 4 days
b) 5 days
c) 8 days
d) 7 days
Question 11: Two cars travel from city A to city B at a speed of 36 and 54 km/hr respectively. If one car takes 3.5 hours lesser time than the other car for the journey, then the distance between City A and City B is
a) 454 km
b) 567 km
c) 302 km
d) 378 km
Question 12: Two cars travel from city A to city B at a speed of 30 and 48 km/hr respectively. If one car takes 3 hours lesser time than the other car for the journey, then the distance between City A and City B is
a) 288 km
b) 240 km
c) 360 km
d) 192 km
Question 13: Two cars travel from city A to city B at a speed of 36 and 48 km/hr respectively. If one car takes 3 hours lesser time than the other car for the journey, then the distance between City A and City B is
a) 518 km
b) 432 km
c) 648 km
d) 346 km
Question 14: Two cars travel from city A to city B at a speed of 42 and 60 km/hr respectively. If one car takes 2 hours lesser time than the other car for the journey, then the distance between City A and City B is:
a) 336 km
b) 280 km
c) 420 km
d) 224 km
Question 15: Manjeet can do a work in 18 hours. If he is joined by Jaya who is 100% more efficient, in what time will they together finish the work?
a) 6 hours
b) 3 hours
c) 12 hours
d) 24 hours
Answers & Solutions:
1) Answer (B)
time to taken walk single way = $\frac{55}{2}$ = 27.5 m
time taken to ride 1 way = 37-27.5 = 9.5m
time taken to ride both ways = 9.5*2 = 19 m
2) Answer (B)
Here the distance between home and school will remain same , let the distance be D
Let the original time taken be T hours
as in both cases the distance remain constant, we can use
$\frac{S1}{S2}$ = $\frac{T2}{T1}$
T1 = T + $\frac{1}{12}$ hr , S1 = 4 km/hr
T2 = T – $\frac{1}{6}$ hr , S2 = 5 km/hr
So, $\frac{4}{5}$ = $\frac{(T-\frac{1}{6})}{(T+\frac{1}{12})}$
T = $\frac{7}{6}$ hour
Distance = Speed x Time
Distance of School = 4 x $\frac{15}{12}$ = 5 km
3) Answer (D)
The lines x=3 and y=4 will intersect at the point (3,4)
$x_1 = 0$
$y_1 = 0$
$x_2 = 3$
$y_2 = 4$
Using distance formula
Distance = $\sqrt{(x_2 – x_1 )^2 + (y_2 – y_1 )^2}$
= $\sqrt{(3 – 0 )^2 + (4 – 0 )^2}$
=$\sqrt{(9 +16)}$
=$\sqrt{25} = 5$
Hence option D is the correct answer.
4) Answer (B)
When speed is 2.5 km/hour time = t + (6/60) hours
When speed is 3 km/hour time is t – (10/60) hours
Distance is same.Hence the product of speed and time must be same.
2.5 (t + (6/60)) = 3(t – (10/60)
2.5t + (15/60) = 3t – (30/60)
0.5t=(15/60)+(30/60)
0.5t = 45/60 hours
t = 3/2 hours
Distance = 3 ( t – (10/60) )
=3t – (30/60)
=3 (3/2) – (30/60) = (9/2)-(1/2) = (8/2)=4 kms
Hence Option B is the correct answer.
5) Answer (C)
When two bodies move in the same direction their relative speed is the difference of their speeds.
The relative speed between the police and the thief is 1 km per hour.
1km per hour = 1 km in 60 minutes.
Therefore 100 m in 6 minutes.This is the distance between them after 6 minutes.
Hence Option C is the correct answer.
6) Answer (A)
Let the speed of the person be $x$ km/h and speed of stream be $y$ km/h
=> Speed when going upstream = $(x – y)$ km/h
and speed while going downstream = $(x + y)$ km/h
=> $\frac{1}{x – y} = \frac{10}{60}$
=> $x – y = 6$
Also, $\frac{1}{x + y} = \frac{4}{60}$
=> $x + y = 15$
Subtracting the two equations, we get :
$2y = 9$
=> Speed of stream = $y = 4.5$ km/h
7) Answer (D)
Distance between two trains = 200 km
Since one of the trains covered 110 km
=> Other train covered = 200 – 110 = 90 km
Also, $speed \propto distance$
$\therefore$ Ratio of speed = $\frac{110}{90}$
= 11 : 9
8) Answer (D)
Let the original speed of man be $4x$ and he reaches his office in $t$ min.
New speed = $3x$ and new time taken = $(t + 20)$ min
$\because speed \propto \frac{1}{time}$
=> $\frac{4x}{3x} = \frac{t + 20}{t}$
=> $4t = 3t + 60$
=> $t = 60$ min
9) Answer (A)
Let the total work be 72 units
Rate at which A finishes the work alone = $\frac{72}{24}$ = 3 units/day
Rate at which B finishes the work alone = $\frac{72}{9}$ = 8 units/day
Rate at which C finishes the work alone = $\frac{72}{12}$ = 6 units/day
Now, B & C work together for 3 days = (8 + 6) * 3 = 42 units
Work left = 72 – 42 = 30 units
Remaining work is done by A in = $\frac{30}{3}$ = 10 days
10) Answer (D)
Let total work to be done = 28 units
Samarth’s efficiency = $\frac{28}{28} = 1$ unit/day
Mandar works 3 times as fast as Samarth
=> Mandar’s efficiency = $3 \times 1 = 3$ units/day
Thus, (Samarth + Mandar)’s 1 day’s work = 1 + 3 = 4 units/day
$\therefore$ Time taken by them to finish the work together = $\frac{28}{4} = 7$ days
=> Ans – (D)
11) Answer (D)
Let the distance between City A and City B = $d$ km
Speed of first car = 36 km/hr and speed of second car = 54 km/hr
Let time taken by first car = $t$ hrs and time taken by second car = $(t – 3.5)$ hrs
Using, speed = distance/time for first car :
=> $\frac{d}{t} = 36$
=> $d = 36t$ ————–(i)
For second car, => $\frac{d}{t – 3.5} = 54$
Substituting value of $d$ from equation (i), we get :
=> $36t = 54t – 189$
=> $54t – 36t = 18t = 189$
=> $t = \frac{189}{18} = \frac{21}{2}$ hrs
From equation (i), => $d = 36 \times \frac{21}{2} = 378$ km
=> Ans – (D)
12) Answer (B)
Let the distance between City A and City B = $d$ km
Speed of first car = 30 km/hr and speed of second car = 48 km/hr
Let time taken by first car = $t$ hrs and time taken by second car = $(t – 3)$ hrs
Using, speed = distance/time for first car :
=> $\frac{d}{t} = 30$
=> $d = 30t$ ————–(i)
For second car, => $\frac{d}{t – 3} = 48$
Substituting value of $d$ from equation (i), we get :
=> $30t = 48t – 144$
=> $48t – 30t = 18t = 144$
=> $t = \frac{144}{18} = 8$ hrs
From equation (i), => $d = 30 \times 8 = 240$ km
=> Ans – (B)
13) Answer (B)
Let the distance between City A and City B = $d$ km
Speed of first car = 36 km/hr and speed of second car = 48 km/hr
Let time taken by first car = $t$ hrs and time taken by second car = $(t – 3)$ hrs
Using, speed = distance/time for first car :
=> $\frac{d}{t} = 36$
=> $d = 36t$ ————–(i)
For second car, => $\frac{d}{t – 3} = 48$
Substituting value of $d$ from equation (i), we get :
=> $36t = 48t – 144$
=> $48t – 36t = 12t = 144$
=> $t = \frac{144}{12} = 12$ hrs
From equation (i), => $d = 36 \times 12 = 432$ km
=> Ans – (B)
14) Answer (B)
Let the distance between City A and City B = $d$ km
Speed of first car = 42 km/hr and speed of second car = 60 km/hr
Let time taken by first car = $t$ hrs and time taken by second car = $(t – 2)$ hrs
Using, speed = distance/time for first car :
=> $\frac{d}{t} = 42$
=> $d = 42t$ ————–(i)
For second car, => $\frac{d}{t – 2} = 60$
Substituting value of $d$ from equation (i), we get :
=> $42t = 60t – 120$
=> $60t – 42t = 18t = 120$
=> $t = \frac{120}{18} = \frac{20}{3}$ hrs
From equation (i), => $d = 42 \times \frac{20}{3} = 280$ km
=> Ans – (B)
15) Answer (A)
Let total work to be done = 18 units
Manjeet’s efficiency = $\frac{18}{18} = 1$ unit/hr
Jaya is 100% more efficient, => Jaya’s efficiency = $1 + \frac{100}{100} \times 1 = 2$ units/hr
(Manjeet + Jaya)’s 1 day’s work together = 1 + 2 = 3 units/hr
$\therefore$ Time taken by Manjeet and Jaya together to finish the work = $\frac{18}{3} = 6$ hours
=> Ans – (A)
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