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# Time and Distance Questions

Home > Quantitative Aptitude > Time and Distance > General Questions
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Two people A and B starts running at 10:00 AM towards north at a speed of 40Kmph and 60Kmph respectively. After 2 hours(12:00 noon ), A turned left and B turned right. They stopped at 2:00 PM.Find the distance between A and B at 2:00 PM.

A120

B190

C230

D200

Explanation:

We should only focus on the distances travelled by them when A & B start running in opposite direction i.e after 12 noon. Both run for two hours till 2pm
So distance travelled by A= 40*2=80km And distance travelled by B = 60*2=120km
So separation between them is 120+80=200km

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The time taken to travel from city X to city Y is 12 hrs. If a person starts at 6 A.M from city X and another person leaves the city Y at 8 A.M. When they will meet. (Speed of Both person is same.)

A01:00 PM

B07:00 PM

C08:00 PM

DNone of these

Explanation:

Let's assume Person "A" travel from City X to City Y at 6:00 AM.

and Person "B" travel from City Y to City X at 08:00 AM.

If the distance between two cities (City X and City Y) = 12 KM

Person A travel distance of 2 hrs = 2 KM.

Distance remaining = 10 Km to be traveled by both person

The time required to travel 10 Km by both person = 5 hrs.

So, They will meet after 5 hrs. when Y start. i.e 1:00 PM

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A man leaves office daily at 7:00 PM. A driver with car comes from his home to pick him from office and bring back home. One day he gets free at 5:30 PM and instead of waiting for driver he starts walking towards home. In the way he meets the car and returns home on car. He reaches home 20 minutes earlier than usual. In how much time does the man reach home usually?

A1 hr 20 min

B1 hr 40 min

C2hrs 20 min

D2 hrs 40 min

Explanation:

Since the car has met the person 20 minutes beforehand, it has actually saved 10 mins of the journey (to and fro)

since the man has started 1.30 hrs before and the car has met him 10 mins before the actual time he takes to reach daily is 1hr and 20 mins

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A car is 15 minutes late by running at a speed of 4/5th of its actual speed. The actual time taken by the car to cover that distance is

A30 minutes

B1 hour

C2 hours

D3 hours

Explanation:

Let the distance be d. Then time taken at x speed is d/x ( This d/x is the value which we have to find.
Now the new speed is 4/5x. So, time taken will be= 5d/4x
Acc. to the question= 5d/4x-d/x=1/4
Solving we get d/x = 1hr

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A boy standing idle sounds a whistle to his friend at a distance of 1200 m moving away from him in a speeding car at 108 kmph. Find the duration (seconds) after which his friend is going to hear him. (Speedof sound = 330m/sec).

A3.6

B4

C40

DNone of these

Explanation:

The logic behind this question is to get that extra distance which the car will cover while the sound approaches him while traveling 1200m.
Time taken by sound to cover 1200m =1200/330= 40/11 sec= 3.63sec (i)
In this time of 40/11sec the car will cover an extra distance of 30m/sec*40/11sec= 109.09m
So extra time taken by sound to cover this distance will be= 109.09/330= 0.33sec (ii)
Adding both the times we get 3.96 approx = 4 sec.

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A man takes 5 hours 45 min in walking to a certain place and riding back. He would have gained 2 hours by riding both ways. The time he would take to walk both ways is

A11 hrs

B8 hrs 5 min

C7 hrs 45 min

D9 hrs 20 min

Explanation:

Let his walking speed be x km/hr and riding speed be ykm/hr
First case: d/x +d/y = 23/4
Second case: 2d/y= 15/4 (ii)
We have to calculate 2d/x
Putting the value of d/y from eq.(ii) in eq (i) we get d/x as 31/8
So, 2d/x will be equal to 62/8= 7hrs 45 minutes.

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A person traveled from his house to office at 30 kmph, then he was late to his office by 5 minutes. If he increases his speed by 10 kmph he would be early by 15 minutes to his office. What should be his speed so that he reaches his office on time ?

A36 kmph

B32 kmph

C34 kmph

D35 kmph

Explanation:

Let the distance between house and office be x km.
=> (x/30) – (x/40) = 20/40;
=> x/120 = (1/3)x = 40 km
Travelling at 40 kmph, he reaches office in 1 hour i.e. 15 minutes early. So, required speed = 40/ 5/4 = 40 * 4/5 = 160/5 = 32 kmph

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A person travels from Chennai to Pondicherry in cycle at 7.5 Kmph. Another person travels the same distance in train at a speed of 30 Kmph and reached 30 mins earlier. Find the distance.

A5 Km

B10 Km

C15 Km

D20 Km

Explanation:

Let the distance be x kms.
then A/Q,
x/(7.5) - x/(30) =30/60 => x= 5 Km.

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Two cars starts from A and B and travel towards each other at the speed of 50 kmph and 60 kmph respectively. At the time of their meeting the second car has travelled 120 km more than the first, the distance between A and B is:

A600 km

B3120 km

C1320 km

D720 km

Explanation:

Given both cars travelled for the same time. the time they met , let it be t.
as speed= dist./time

Time = Dist/Speed [Formula]

Let the distance travelled by first = x
Therefore the distance travelled by the second is (120 + x)

x/50 = x+120/60 -------- from [Formula] As time taken by both is same

By solving above eq. x=600

So, Total distance between A & B = 600+(120+600)
= 1320

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A bus leaves Mumbai at 3 pm. It travels for 1.5 hours at 60km/hr and then halts for 30 minutes. It then travels at an average speed of 50km/hr for the remaining duration to reach Pune at 6pm. What is the distance between Mumbai and Pune?

A100 km

B110 km

C120 km

D140 km

ENone of these

Explanation:

bus covered a distance in (1.5 hour and 30 min. halt) 2 hour is = 1.5*60 = 90 km
Remaing time to reach pune is only 1 hour then bus travels with the speed of 50km/h
Now reamin. distance = 1*50 = 50 km
Now total 50+90 = 140 km

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