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Bunuel
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What should the length of a race be so that two drivers with speeds of 03 Nov 2025, 00:43
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55% (hard)

Question Stats:

63% (02:26) correct 37% (02:28) wrong based on 57 sessions

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What should the length of a race be so that two drivers with speeds of 22 m/s and 25 m/s reach the end point simultaneously even though the slower one had a head-start of 6 minutes (i.e. the slower one starts racing first and after 6 minutes the faster one starts)?

A. 44 km
B. 50 km
C. 55 km
D. 66 km
E. 75 km



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D
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arushi118
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What should the length of a race be so that two drivers with speeds of 03 Nov 2025, 02:38
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Distance travelled by the slower one in the first 6 minutes: 22*360m = 7920m

Then in the next t seconds:
25t = 22t + 7920
t = 2640s
Hence the total distance travelled by the faster one: (25*2640)/1000 = 66km
Hence Option D is correct.
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What should the length of a race be so that two drivers with speeds of 03 Nov 2025, 04:32
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Here is the quick solution of the question:-

speed of the slower = 22m/s
speed of the faster = 25m/s
head start = 6 min = 360 s

Now let race length = D

D/22 = D/25 + 360
D(1/22 - 1/25) = 360
D(3/550) = 360
D = 360 * 550/3 = 66,000 m = 66km

Final correct answer: Option (D)


Bunuel
What should the length of a race be so that two drivers with speeds of 22 m/s and 25 m/s reach the end point simultaneously even though the slower one had a head-start of 6 minutes (i.e. the slower one starts racing first and after 6 minutes the faster one starts)?

A. 44 km
B. 50 km
C. 55 km
D. 66 km
E. 75 km



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Dereno
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What should the length of a race be so that two drivers with speeds of 03 Nov 2025, 09:31
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Bunuel
What should the length of a race be so that two drivers with speeds of 22 m/s and 25 m/s reach the end point simultaneously even though the slower one had a head-start of 6 minutes (i.e. the slower one starts racing first and after 6 minutes the faster one starts)?

A. 44 km
B. 50 km
C. 55 km
D. 66 km
E. 75 km



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We need to find the length of the race = d

Speed of the two racers are 22 m/s and 25 m/s respectively.

The slower person ( 22 m/s ) is allowed a head start of 6 minutes = 6*60 seconds = 360 seconds.

1 second = 22 m

360 second = x ?

x = 360 * 22 = 7920 m.

We know that speed is proportional to distance.

The slower speed person travels a distance of ( d - 7920) m ,

While the Faster person travels a distance of d metres.


(22/25). = [ (d-7920) / d ]

22 d = 25 d. - 7920*25

3d = 7920*25

d = 25*2640

d = 66000 m

d = 66 Km.

Option D
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What should the length of a race be so that two drivers with speeds of 03 Nov 2025, 10:44
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Since they have to reach together, we can treat this as a catch-up problem.

Distance travelled by the slower one before the faster one starts = 22m/sec * 6 * 60 sec = 7920 m

Now Time to catch up = 7920/(25-23) sec

Distance travelled by the faster one in that duration = 7920 * 25 / (1000 * 3) = 66 km

Option D
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