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18 Mar 2024, 01:12
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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
A. 25
B. 30
C. 50
D. 100
E. 150
18 Mar 2024, 01:12
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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
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
03 Oct 2024, 05:06
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WHY DO WE SUBTRACT 2/5D IN THE FIRST PART OF THE EQUATION
