0
6309

# Time and Distance Questions for SSC-CHSL PDF

Download SSC CHSL Time and Distance Questions with answers  PDF based on previous papers very useful for SSC CHSL Exams. Top-15 Very Important Questions for SSC Exams.

Question 1: Buses start from a bus terminal with a speed of 20 km/hr at intervals of 10 minutes. What is the speed of a man coming from the opposite direction towards the bus terminal if he meets the buses at intervals of 8 minutes ?

a) 3 km/hr

b) 4 km/hr

c) 5 km/hr

d) 7 km/hr

Question 2: By walking at 3/4 of his usual speed, a man reaches his office 20 minutes later than his usual time. The usual time taken by him to reach his office is

a) 75 minutes

b) 60 minutes

c) 40 minutes

d) 30 minutes

Question 3: In a 100m race, Kamal defeats Bimal by 5 seconds. If the speed of Kamal is 18 Kmph, then the speed of Bimal is

a) 15.4 kmph

b) 14.5,kmph

c) 14.4 kmph

d) 14 kmph

Question 4: A train, 240 in long crosses a man walking along the line in opposite direction at the rat4 of 3 kmph in 10 seconds. The speed of the train is

a) 63 kmph

b) 75 km ph

c) 83.4 kmph

d) 86.4 kmph

Question 5: The diameter of a wheel is 98 cm. The number of revolutions in which it will have to cover a distance of 1540 m is

a) 500

b) 600

c) 700

d) 800

SSC CHSL Study Material (FREE Tests)

Question 6: Walking at 3/4th of his usual speed , aman is $1\frac{1}{2}$ hours late. his usual time to cover the same distance, in hours, is?

a) $4\frac{1}{2}$

b) $4$

c) $5\frac{1}{2}$

d) $5$

Question 7: Four runners started running simultaneously from a point on a circular track. They took 200 seconds, 300 seconds, 360 seconds and 450 seconds to complete the round. After how much time to they meet at the starting point for the first time?

a) 1800 seconds

b) 3600 seconds

c) 2400 seconds

d) 4800 seconds

Question 8: Walking at 6/7th of this usual speed a man is 25 minutes too late. His usual time to cover this distance is

a) 2 hours 30 minutes

b) 2 hours 15 minutes

c) 2 hours 25 minutes

d) 2 hours 10 minutes

Question 9: Walking at 5 km/hr a student reaches his school from his house 15 minutes early and walking at 3 km/hr he is late by 9 minutes. What is the distance between his school and his house?

a) 5 km

b) 8 km

c) 3 km

d) 2 km

Question 10: With average speed of 40 km/hour, a train reaches its destination in time. If it goes with an average speed of 35 km hour, it is late by 15 minutes. The total journey is

a) 30 km

b) 40 km

c) 70 km

d) 80 km

Question 11: A rail road curve is to be laid out on a circle. What radius should be used if the track is to change direction by 25$^{\circ}$ in a distance of 40 metres?

a) 91.64 metres

b) 90.46 metres

c) 89.64 metres

d) 93.64 metres

Question 12: Two cars are moving with speeds $v1$ and $v2$ towards a crossing along two roads. If their distance from the crossing be 40 m and 50 m at an instant of time, then they do not collide, if their speeds are such that

a) $v_1:v_2 \neq 4:5$

b) $v_1:v_2 \neq 5:4$

c) $v_1:v_2=16:25$

d) $v_1:v_2 = 25:16$

Question 13: A man rides at the rate of 18 km/hr, but stops for 6 mins. to change horses at the end of every 7th km. The time that he will take to cover a distance of 90 km is

a) 6 hrs.

b) 6 hrs 12 min.

c) 6 hrs 18 min.

d) 6 hrs 24 min.

Question 14: A farmer travelled a distance of 61 km in 9 hrs. He travelled partly on foot at the rate of 4 km/hr and partly on bicycle at the rate of 9 km/hr. The distance travelled on foot is

a) 14 km

b) 171cm

c) 16 km

d) 15 km

Question 15: Raj and Prem walk in opposite directions at the rate of 3 km and 2 km per hour respectively. How far will they be from each other after 2 hours ?

a) 10 km

b) 8 km

c) 6 km

d) 2 km

Distance between buses will $20\times\frac{10}{60}$ = $\frac{10}{3}$ km.
Now man is travelling this distance in 8 min. with the relative speed of (20+$x$) (let’s assume speed of man is $x$ km/hr )
hence (20+$x$) = $\frac{\frac{10}{3}}{\frac{8}{60}}$
$x$= 5

As distance is constant and we know s = \frac{d}{t} (where s is distance and t is time)
hence st = constant
or $s_{1}t_{1}$= $s_{2}t_{2}$
or $s_{1}t_{1}$=$\frac{3}{4}s_{1}(t_{1}+ 20)$
hence $t_{1}$ = 60

As we know distance is constant for vimal and kamal and that is equal to 100 m.
hence $V_{bimal}\times t_{bimal}$ = 100
or $V_{bimal}\times (t_{kamal} + 5)$ = 100
where $t_{kamal}$ will be $\frac{Distance}{V_{kamal}}$ i.e. $\frac{100 m}{18 kmph}$
so $V_{bimal}\times (\frac{100 m}{18 kmph} + 5)$ = 100
After  solving above equation we will get $V_{bimal}$ = 4 $\frac{m}{s}$ or 14.4 kmph

As man and train are coming to each other
and we assume speed of train = $x$
hence its relative speed will be ($x$ + 3)
relative distance travelled = 240m = .24 km
time taken = 10 sec. = $\frac{10}{3600}$ hr

so ($x$ + 3) = $\frac{.24}{\frac{10}{3600}}$

solving the eq. we will $x$ equals to 83.4 kmph

In 1 revolution wheel will complete a distance of $2\pi r = 2 \times \pi \times \frac{98}{2}$ = 308 cm.
Hence to cover 1540 m. , revolutions will be = $\frac{1540\times100}{308} = 500$

As distance is constant
Hence $v_1 \times t_1 = \frac{3v_1}{4} \times (t_1 + \frac{3}{2})$ (Where is $v_1$ is speed, t_1 is time taken to travel)
On solving above equation, we will get $t_1 = \frac{9}{2}$

Meeting at the first time will be L.C.M. of time taken by individuals to complete
i.e. L.C.M. of 200,300,360 and 450 will be equal to 1800 sec.

Let the initial speed and time be s,t  respectively,
then speed and time in the next case are 6s/7 and (t+25)
As distance = speed * time, and distance travelled in both cases is the same,
(6s/7)*(t+25) = s*t
Solving the  above equation results in  t=150min

Let the time and distance be t mins and d respectively,
In the first case:
Total time taken = (t – 15) mins = (t-15)/60 hrs.
Distance travelled  =  5*(t-15)/60
In the second case:
Total time taken = (t + 9) mins = (t+9)/60 hrs.
Distance travelled  = 3*(t+9)/60

So, 5*(t-15)/60 = 3*(t+9)/60,
Solving the above equation we get, t=51
So, d=3*(51+9)/60
=3 KMs

Let the time and distance be t mins and d km respectively,
If it goes with an average speed of 40 km/hour, a train reaches its destination in time.
So, distance = (40*t)/60
If it goes with an average speed of 35 km hour, it is late by 15 minutes.
So, distance = 35*(t+15)/60
In both the cases, distance is same,
So,  40*t = 35*(t+15)
Solving the above equation gives t = 105
and d = (40*105)/60 = 70

With 25 degree angle length of arc will be ($r \times$ ($\theta$ in radian) = 40m (where $r$ is radius of arc and $\theta$ will be angle made by it)
Now solving above equation, we will get $r$ = 91.64 m.

they will collide if time to reach the crossing is same.

it is the case when

time taken by 1st car = time taken by 2nd car.

40/v1 = 50/v2

v1/v2 = 4/5

but condition not to collide is v1/v2 $\neq$ 4/5.

so the answer is option A.

Speed of man = 18 km/h

Total distance = 90 km

As he stops after 7th km, => $90=(12\times7)+6$

=> He stops 12 times in the journey.

Total stoppage time = $12\times6=72$ min

Real time = $\frac{90}{18}=5$ hours

$\therefore$ Required time taken = 5hr + 72 min

= 6 hrs 12 min

=> Ans – (B)

Let the distance travelled on foot be $x km$

=> Distance travelled on bicycle = $(61-x) km$

Time taken to travel on foot = $\frac{x}{4}$ hrs

Time taken to travel on bicycle = $\frac{61-x}{9}$ hrs

=> Total time = 9 = $\frac{x}{4} + \frac{61-x}{9}$

=> $9x + 244 – 4x = 324$

=> $5x = 324-244$

=> $x = \frac{80}{5} = 16$ km

Speed, $s$ = 3+2 = 5 kmph