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at given speed the time taken ; 100/s
and at speed +5kmph ; 100/s+5
so ∆ of time ; 100/s- 100/(s+5) = 1
solve for s
100* (s+5 -s) = s^2+5s
s^2+5s-500= 0
(s+25)(s-20)
s= -25 , 20
IMO A: 20 kmph


Bunuel
A jeep travels a distance of 100 km at a uniform speed. If the speed of the jeep is 5 kmph more, then it takes 1 hour less to cover the same distance. The original speed of the jeep is

A. 20 kmph
B. 25 kmph
C. 30 kmph
D. 50 kmph
E. 60 kmph


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Bunuel
A jeep travels a distance of 100 km at a uniform speed. If the speed of the jeep is 5 kmph more, then it takes 1 hour less to cover the same distance. The original speed of the jeep is

A. 20 kmph
B. 25 kmph
C. 30 kmph
D. 50 kmph
E. 60 kmph

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Given: A jeep travels a distance of 100 km at a uniform speed. If the speed of the jeep is 5 kmph more, then it takes 1 hour less to cover the same distance.

Asked: The original speed of the jeep is ?

Let the original speed of the jeep be s km/hr and original time taken to cover 100 km be t hrs

100 = st = (s+5)(t-1) = st +5t -s - 5
5t -s - 5 = 0
s = 5t - 5

100 = (5t-5) t
t(t-1) = 20
t = 5 hrs

s = 5*4 = 20 km/hr

IMO A­
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Suppose, the original speed is X.

We know, Time = Distance/ Speed

So, with original speed, Time = 100/ X

And, with additional speed, Time = 100/ (X+5)

According to the question,

100/X - 100/ (X+5) = 1

DO NOT SOLVE FOR X

Plug in the answer choices. For X = 20, Left Hand Side = Right Hand Side.

Option: A
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I just took 20 km/hr and plugged it in, andgotthe difference between uniform speed and new speed as 1 hour.
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