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08 Nov 2023, 02:30
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C
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Difficulty:
15%
(low)
Question Stats:
80% (01:43) correct
20%
(01:38)
wrong
based on 118
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Starting 20 miles apart, two runners, A and B, begin running directly toward each other. What is the sum of the distances they travel in the last 15 minutes before they meet?
(1) Runner A runs at a constant speed that is one mile per hour slower than Runner B's constant speed.
(2) Runner B runs one-and-a-half miles in the last 15 minutes before they meet.
(1) Runner A runs at a constant speed that is one mile per hour slower than Runner B's constant speed.
(2) Runner B runs one-and-a-half miles in the last 15 minutes before they meet.
09 Nov 2023, 05:46
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Bunuel
The distance travelled by them in \(15\) minutes would be given by \(s_a*15 +s_b*15 \)
Where \(s_a \) and \(s_b\) are the speeds of runner \(a\) and \(b\) respectively.
(1) Runner A runs at a constant speed that is one mile per hour slower than Runner B's constant speed.
\((s_b-1 )*15 +s_b*15\)
We do not know \(s_b\)
INSUFF.
(2) Runner B runs one-and-a-half miles in the last 15 minutes before they meet.
Hence \(s_b = \frac{1.5}{15}\) miles per min. or \(6\) miles per hour
Now we know speed of \(b\), however we still do not know the speed of \(a\)
INSUFF.
1+2
From (2) we know speed of \(b\) and using this info with (1) we can find the speed of \(a \).
Thus we can find distance travelled in the last \(15\) minutes :
\(5*\frac{1}{4} + 6 *\frac{1}{4} = \frac{11}{4} \) miles.
SUFF.
Ans C
Hope it helps.
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