Welcome to freeonlinetest online web portal, In this section you will learn about Time speed and distance Aptitude questions, shortcut tips, formulas, solved examples, quantitative aptitude MCQ Questions, quiz, solutions, pdf downloads.
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 = ${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$
Question-1)
In a race of 200 metres, B can give a start of 10 metres to A, and C can give a start of 20 metres to B. The start that C can give to A, in the same race, is
A) 19 metres
B) 29 metres
C) 39 metres
D) 45 metres
Ans- B
Solution:
According to the question,
Since, When B runs 200 m metres, A runs 190 metres
When B runs 180 metres, A runs
= $190/200 × 180$ = 171 metres
When C runs 200m, B runs 180 metres.
Hence, C will give a start to A by
= 200 - 171 = 29 metres
Question-2)
In a one-kilometre race A, B and C are the three participants. A can give B a start of 50 m. and C a start of 69 m. The start, which B can allow C is
A) 10 metres
B) 14 metres
C) 20 metres
D) 45 metres
Ans- C
Solution:
Let the time taken to complete the race by A,B, and C be x minutes.
Speed of A = $1000/x$,
B = ${1000 - 50}/x = 950/x$
C = ${1000 - 69}/x = 931/x$
Now, time taken to complete the race by
B = $1000/{950/x} = {1000 × x}/950$
and distance travelled by C in
${1000x}/950$ min = ${1000x}/950 × 931/x$ = 980 km.
B can allow C
= 1000 - 980 = 20 m
Question-3)
In a 100m race, Saroj defeats Manoj by 5 seconds. If the speed of Saroj is 18 Kmph, then the speed of Manoj is
A) 14.4 kmph
B) 20 kmph
C) 25.5 kmph
D) 30 kmph
Ans- A
Solution:
Time taken by Saroj
= $100/{18 × 5/18}$ = 20 seconds
Time taken by Manoj
= 20 + 5 = 25 seconds
Manoj's speed
= $100/25$ =4 m/sec = ${4 × 18}/5$ kmph
= 14.4 kmph.
Question-4)
Geeta goes to a multiplex at the speed of 3 km/hr to see a movie and reaches 5 minutes late. If she travels at the speed of 4 km/hr she reaches 5 minutes early. Then the distance of the multiplex from her starting point is
A) 1 km
B) 2 km
C) 2.5 km
D) 4 km
Ans- B
Distance between starting point and multiplex = x metre
Time = $\text"Distance"/ \text"Speed"$
According to the question,
$x/3 - x/4 = {5 + 5}/60$
${4x - 3x}/12 = 1/6$
$x/12 = 1/6$
$x = 12/6$ = 2 km.
Question-5)
A can give 40 metres start to B and 70 metres to C in a race of one kilometre How many metres start can B give to C in a race of one kilometre ?
A) 45 metre
B) 31$1/4$ metre
C) 25 metre
D) 23$1/4$ metre
Ans- B
According to the question,
When A covers 1000m, B covers
= 1000 - 40 = 960 m
and C covers =1000 - 70 = 930 m
When B covers 960m, C covers 930 m.
When B covers 1000m, C covers
= $930/960 × 1000$ = 968.75 metre
Hence, B gives C a start of
= 1000 - 968.75 = 31.25 metre= 31$1/4$ metre
Question-6)
A jeep is chasing a car which is 5km ahead. Their respective speed are 90 km/hr and 75 km/ hr. After how many minutes will the jeep catch the car ?
A) 12 min
B) 25 min
C) 40 min
D) 20 min
Ans- D
Relative speed
= 95 - 75 = 15 kmph
Required time
= $\text"Distance"/ \text"Relative speed"$
= $5/15$ hours = $5/15 × 60$ minutes
= 20 minutes
Question-7)
In a race of 1000 m, A can beat B by 100m. In a race of 400 m, B beats C by 40m. In a race of 500m. A will beat C by
A) 68 metres
B) 70 metres
C) 95 metres
D) 85 metres
Ans- C
When A runs 1000m, B runs 900m.
When A runs 500m, B runs 450 m.
Again, when B runs 400m, C runs 360 m.
When B runs 450m, C runs
= $360/400 × 450$ = 405 metres
Required distance
= 500 - 405 = 95 metres
Question-8)
A is twice as fast as B, and B is thrice as fast as C is. The journey covered by C in 1$1/2$ hours will be covered by A in
A) 12 min
B) 15 min
C) 40 min
D) 20 min
Ans- B
Time taken by C = t hours
Time taken by B = $t/3$ hours
and time taken by A = $t/6$ hours
Here, t = $3/2$ hours
∴ Required time taken by A
= $3/{2/6}$ hour = $1/4$ hour
= $(1/4 × 60)$ minutes = 15 minutes
Question-9)
In a race of one kilometre, A gives B a start of 100 metres and still wins by 20 seconds. But if A gives B a start of 25 seconds, B wins by 50 metres. The time taken by A to run one kilometre is
A) $300/29$ seconds
B) $700/29$ seconds
C) 500 seconds
D) $500/29$ seconds
Ans- D
Let A take x seconds in covering 1000m and b takes y seconds
According to the question,
x + 20 = $900/1000$ y
x + 20 = ${9y}/10$ ...(i)
and, $950/1000$ x + 25 = y ...(ii)
From equation (i),
${10x}/9 + 200/9 = y$
${10x}/9 + 200/9 = {950x}/1000 + 25$
${10x}/9 + 200/9 = {19x}/20 + 25$
${10x}/9 - {19x}/20 = 25 - 200/9$
${200x - 171x}/180 = {225 - 200}/9$
${29x}/180 = 25/9$
$x = 25/9 × 180/29 = 500/29$ seconds.
Question-10)
In a race of one kilometre, A gives B a start of 100 metres and still wins by 20 seconds. But if A gives B a start of 25 seconds, B wins by 50 metres. The time taken by A to run one kilometre is
A) 1 hours and 45 minutes
B) 2 hours
C) 2 hours and 45 minutes
D) 2 hours and 15 minutes
Ans- B
Relative speed
= 12 + 10 = 22 kmph
Distance covered
= 55 - 11 = 44 km
∴ Required time
= $(44/22)$ hours = 2 hours