Problem on Time Speed and Distance
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Some Useful Tricks On Time and Distance
If a man travels different distances $d_1,d_2,d_3$,... and so on in different time $t_1,t_2,t_3$ respectively then,
Average speed = $\text"total travelled distance"/\text"total time taken in travelling distance"$
= ${d_1 + d_2 + d_3 +...}/{t_1 + t_2 + t_3 +...}$
If a bus travels from A to B with the speed x km/h and returns from B to A with the speed y km/h,then the average speed will be $({2xy}/{x + y})$
If a man travels different distances $d_1,d_2,d_3$, and so on with different speeds $s_1,s_2,s_3$, respectively then,
Average speed = $({d_1 + d_2 + d_3 + ...})/{d_1/S_1 + d_2/S_2 + d_3/S_3 + ...}$
If a man travels at the speed of $s_1$, he reaches his destination $t_1$ late while he reaches $t_2$ before when he travels at $s_2$ speed, then the distance between the two places is
D = ${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$
Some Problems on Time, Speed & Distance
Question-1)
A person crosses a 600 m long street in 5 minutes. What is his speed in km per hour?
A) 3.6 km/hr
B) 7.2 km/hr
C) 8.4 km/hr
D) 10 km/hr
Ans- B
Solution:
Speed = ${600}/ {5} × 60$ m/sec. = 2 m/sec.
= 2 x ${18}/{5}$km/hr =7.2 km/hr
Question-2)
Negesh walks from his home to Bus station. If he walks at the speed of 9 kmph, he reaches the bus station 3 minutes before the arrival of the bus. However, if he walks at the speed of 6 kmph, he misses the bus by 2 minutes. Find the distance covered by him to reach the bus station.
A) 3.6 km
B) 1.5 km
C) 5.5 km
D) 12 km
Ans- B
Solution:
$d/6 - d/9$ = ${3+2}/60$
${3d-2d}/28 = 1/12$
Therefore, d = $18/12 = 1.5$ km
Question-3)
A man riding his bicycle covers 150 metres in 25 seconds. What is his speed in km per hour ?.
A) 21.6 km/hr
B) 15.5 km/hr
C) 25.5 km/hr
D) 12 km/hr
Ans- A
Solution:
Speed = $150/25$ = 6 m/sec
= $6 × 18/5 = 108/5$ = 21.6 km/hr
Question-4)
Two men start together to walk a certain distance, one at 4 km/h and another at 3 km/h. The former arrives half an hour before the latter. Find the distance?
A) 5 kms
B) 6 kms
C) 7 kms
D) 7.5 kms
Ans- B
Solution:
Here $S_1 = 4, t_1 = x, S_2 = 3, t_2 = x + 1/2$
$S_1t_1 = S_2t_2$
$4 × x = 3(x + 1/2)$
$4x - 3x = 3/2 x = 3/2$
Distance= $4 × 3/2$ = 6 kms
Question-5)
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) 5 kms
B) 7 kms
C) 8 kms
D) 8.5 kms
Ans- A
Solution:
If a man travels at the speed of $S_1$, he reaches his destination $t_1$ late while he reaches $t_2$ before when he travels at $S_2$ speed,
then the distance between the two places is D = ${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$
Here, $S_1 = 4, t_1 = 5, S_2 = 5, t_2$= 10
Distance =${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$
=${(4 × 5)(5 + 10)}/{5 - 4}$
=20 × $15/60$ = 5 kms
Test instructions :
- Total number of questions : 20
- Each question carries 1 mark
- Negative marks of Each Questions: 0.25
- Skiping questions No marks will deduct.
- Time allotted : 30 minutes