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# Time and Distance Questions

Home > Quantitative Aptitude > Time and Distance > General Questions
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A car departs at 11:10am. It travels at 110km/hr until 2:30 then it reduces its speed by 5%for the remainder of trip. It reaches destination at 7:30. How much kilometer did he travel?

A889.16 Kms

B879 Kms

C798 Kms.

DNone of these

Explanation:

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A cycled from P to Q at 10 kmph and returned at the rate of 9 kmph. B cycled both ways at 12 kmph. In the whole journey B took 10 minutes less than A. Find the distance between P and Q.

A4 Km.

B4.5 Km.

C3 Km.

D3.75 Km.

Explanation:

Lets assume distance between P and Q = d km.
Time taken by A in both side = d/10 + d/9 = 19d/90 hrs.
Time taken by B in both side = 2d/12 = d/6 hrs.
B tool 10 min. or 1/6 hrs. less than A.
So, 19d/90 - d/6 = 1/6
=> (19d-15d)/90 = 1/6
=> 4d/90 = 1/6
=> d = 15/4 km = 3.75 km.

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A man traveled from the village to the post-office at the rate of 25 kmph and walked back at the rate of 4 kmph. If the whole journey took 5 hours 48 minutes, find the distance of the post-office from the village.

A12 Km.

B32 Km

C22 Km.

D20 Km.

Explanation:

Average speed = (2*a*b)/(a + b) here a = 25 b = 4
Average speed = 2*25 * 4/(25 + 4) = 200/29 kmph.
Distance covered in 5 hours 48 minutes = Speed * time
Distance = (200/29)*(29/5) = 40 kms
Distance covered in 5 hours 48 minutes = 40 kms
Distance of the post office from the village = (40/2) =20 km.

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A car starts from rest and accelerates uniform upto speed 90 km/hrs in 5 sec. Total distance covered by car in this time will be

A62.5 m

B60 m

C54.5 m

DNone of these

Explanation:

Speed of car at last instant of given interval
= 90*5/ 18 m/s = 25 m/s
so average speed = 0+25 / 2
Therefore distance = 5 * 25/2 = 125/2 = 62.5 m

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How many rotations per minute must a wheel of diameter 1.5m make to achieve a speed of 45 kmph?

A155

B159

C125

DNone of these

Explanation:

For 45kmph, distance to cover in one hour = 45 * 1000 = 45000 m.
Distance to be covered per minute (d) = 45000 / 60 = 750 m/min
Distance covered in 1 rotation of wheel =Pi*diameter=Pi*1.5 m.
So rotation /minute Required =750/(Pi*1.5)=750/[(22/7)*1.5]=159.09 ~ 159

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Juan is a gold medalist in athletics. In the month of May, if Juan takes 11 seconds to run "y" yards. How many seconds will it take him to run "x" yards at the same rate?

A11y/x

B11x/y sec.

Cxy/11

D11/xy

Explanation:

Juan takes 11 sec. to run "y" yards.
So, to cover 1 yard he need 11/y sec.
So to cover x yard he need 11x/y sec.

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Two vehicles A and B leaves from city Y to X. A overtakes B at 10:30 am and reaches city X at 12:00 pm. It waits for 2 hrs. and return to city Y. On its way it meets B at 3:00 pm and reaches city Y at 5:00 pm. B reaches city X, waits for 1hr and returns to city Y. After how many hours will B reach city Y from the time A overtook him for the first time?

A50 hrs

B49.5 hrs

C41.5 hrs

D37.5 hrs

Explanation:

Vehicle A overtaken B at 10.30 am and reached X at 12 pm.Â  It started at 2 pm and met B at 3 pm at Q. It means, Vehicle A took one hour to cover distance 'n', So it should be at Q at 11 pm.Â  It is clear that Vehicle A takes 0.5 hour to cover distance 'm'.Â

Now vehicle B travelled from 10.30 am to 3 pm to meet A. So it took 4.5 hours to cover m.Â  So Speeds ratio = 4.5 : 0.5 = 9 : 1.

Now Vehicle A took a total of 1.5 + 3 = 4.5 hours to travel fro P to Y. So It must take 4.5 * 9 + 1 = 41.5 hours

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

A600 kms

B729 kms

C1320kms

D1230 kms

Explanation:

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The five tyres of a car (four road tyres and one spare) were used equally in a journey of 40,000 kms. The number of kms of use of each tyre was

A40000

B10000

C32000

D8000

Explanation:

Total kilometers travelled by 4 tyre = 40000 * 4 = 1,60,000.
This has to be share by 5 tyres.
So each tyre capacity = 1,60,000 / 5 = 32, 000.
You have a doubt, after we travel 32,000 km, we are left with 4 worn tyes and one new tyre.
But If the tyres are rotated properly after each 8000 km,

All the tyres are equally used.

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Jake and paul start together to walk 10kms jake walks 1.5kmph faster than paul and arrives at his destination 1.5 hours earlier than paul jakes speed walking is

A6kmph

B2kmph

C4 kmph

D8 kmph

Explanation:

Let speed of jake = x kmph.
So, speed of paul = ( x - 1.5 ) kmph.
Time difference =1.5hours
Total Distance = 10km
10/(x-1.5) -10/x = 1.5
By solving the above eq.
x = 4 kmph

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