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22 Jul 2024, 01:30
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
70% (02:45) correct
30%
(02:54)
wrong
based on 361
sessions
History
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Time
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Not Attempted Yet
Josh and Mike live 13 miles apart. Yesterday Josh started to ride his bicycle toward Mike’s house. A little later Mike started to ride his bicycle toward Josh’s house. When they met, Josh had ridden for twice the length of time as Mike and at four-fifths of Mike’s rate. How many miles had Mike ridden when they met?
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8
22 Jul 2024, 06:58
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Given - Josh and Mike live 13 miles apart. Yesterday Josh started to ride his bicycle toward Mike’s house. A little later Mike started to ride his bicycle toward Josh’s house. When they met, Josh had ridden for twice the length of time as Mike and at four-fifths of Mike’s rate.
To find - How many miles had Mike ridden when they met?
J House -----------------------<13>------------------------M House
A little later Mike started to ride his bicycle toward Josh’s house. When they met, Josh had ridden for twice the length of time as Mike and at four-fifths of Mike’s rate.
\(Tj = 2*Tm\)
\(Sj = \frac{4}{5} * Sm\)
and, \(Dj + Dm = 13\)
\(Distance = Speed * Time\)
Dj = Sj*Tj and Dm = Sm*Tm
Now Dj in terms of Sm and Tm
\(Dj = \frac{4}{5} * Sm * 2Tm\)
\(Dj + Dm = \frac{4}{5} * Sm * 2Tm\) + \(Sm*Tm\) = \(13\)
on soloving,
Sm*Tm = 5
Which is Dm = 5 miles.
Answer is B.
To find - How many miles had Mike ridden when they met?
J House -----------------------<13>------------------------M House
A little later Mike started to ride his bicycle toward Josh’s house. When they met, Josh had ridden for twice the length of time as Mike and at four-fifths of Mike’s rate.
\(Tj = 2*Tm\)
\(Sj = \frac{4}{5} * Sm\)
and, \(Dj + Dm = 13\)
\(Distance = Speed * Time\)
Dj = Sj*Tj and Dm = Sm*Tm
Now Dj in terms of Sm and Tm
\(Dj = \frac{4}{5} * Sm * 2Tm\)
\(Dj + Dm = \frac{4}{5} * Sm * 2Tm\) + \(Sm*Tm\) = \(13\)
on soloving,
Sm*Tm = 5
Which is Dm = 5 miles.
Answer is B.
02 Sep 2025, 04:26
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Here's a simpler approach:
Let distance travelled when they meet -> by \(J = x; M = 13-x\)
speed of \(J = 0.8M\)
\(\frac{X}{0.8M} = \frac{(2)(13-x)}{M}\) (time ridden by J is double of M's)
\(x = 8\)
Therefore, \(M = 5\)
Let distance travelled when they meet -> by \(J = x; M = 13-x\)
speed of \(J = 0.8M\)
\(\frac{X}{0.8M} = \frac{(2)(13-x)}{M}\) (time ridden by J is double of M's)
\(x = 8\)
Therefore, \(M = 5\)
