A cyclist, after covering 2/5th of the race, noticed that to reach the

Question

A cyclist, after covering 2/5th of the race, noticed that to reach the midpoint of the race, she needs to cover 15 kilometers less than she had already covered. How long was the race in kilometers ?

A. 25
B. 30
C. 50
D. 100
E. 150

trammtrann · Answer

A is the distance of the race -> A/2 - 2/5A = 2/5A-15 => A=50

SarthakSharma8 · Answer

Let Distance be 'D'
So it has to cover 15 kilometers less than 2D/5 to reach the mid point.

so the Distance between the 2D/5 mark to mid point (D/2) will be the same as 15 kilometers less then 2D/5
Hence,

2D/5 - 15 = D/2 - 2D/5
4D/5 - D/2 = 15
3D/10 = 15
3D = 150
D = 50 ---- Answer

Bunuel · Answer

Official Solution:

A cyclist, after covering  \frac{2}{5}^{th}  of the race, noticed that to reach the midpoint of the race, she needs to cover 15 kilometers less than she had already covered. How long was the race in kilometers?

A. 25
B. 30
C. 50
D. 100
E. 150

Let's assume the race is  d  kilometers long. Then half of the race would be  \frac{d}{2}  kilometers long. We are told that to reach the midpoint of the race, so to cover additional  \frac{d}{2} - \frac{2}{5}d  kilometers, she needs to cover 15 kilometers less than she had already covered, so she needs to cover  \frac{2}{5}d -15 . Hence, we'd have:
 
 \frac{d}{2} - \frac{2}{5}d = \frac{2}{5}d-15 
 
Multiplying by 10 to get rid of the fractions yields:
 
 5d - 4d = 4d-150

3d = 150

d = 50 
 
Answer: C

Vishalp05 · Answer

Assume x= 50x (avoiding fractions)

2/5th of 50x= 20x and she needs to cover 20x-15 to reach the midpoint. Hence, the remaining distance would be 25x.

20x + (20x-15) + 25x= 50x
20x+ 20x -15 + 25x= 50x
65x- 50x= 15
15x= 15
x=1

Hence, Total distance= 50x= 50(1)= 50

A cyclist, after covering 2/5th of the race, noticed that to reach the

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GMAT Problem Solving (PS) Questions

A cyclist, after covering 2/5th of the race, noticed that to reach the

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