# Time speed and distance Aptitude questions

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## Some Useful Tricks On Time and Distance

Trick-1

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 = $\text"total travelled distance"/\text"total time taken in travelling distance"$

= ${d_1 + d_2 + d_3 +...}/{t_1 + t_2 + t_3 +...}$

Trick-2

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 $({2xy}/{x + y})$

Trick-3

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 = $({d_1 + d_2 + d_3 + ...})/{d_1/S_1 + d_2/S_2 + d_3/S_3 + ...}$

Trick-4

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 = ${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$

## Time speed and distance Aptitude questions & Answer

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

### 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