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Joseph travels the first of the distance to his home on bicycle an
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30 Dec 2024, 01:26
30 Dec 2024, 01:26
Expert Reply
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Question Stats:
100% (01:41) correct
0% (00:00) wrong based on 5 sessions
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Joseph travels the first \(\frac{2}{3}\) of the distance to his home on bicycle and the remaining distance on foot. What is the ratio of the Joseph’s speed on the bicycle to his speed on foot if he takes three times as long on foot as on the bicycle?
A. 1:6
B. 1:3
C. 2:1
D. 3:1
E. 6:1
A. 1:6
B. 1:3
C. 2:1
D. 3:1
E. 6:1
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Joseph travels the first of the distance to his home on bicycle an
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08 Feb 2025, 11:18
08 Feb 2025, 11:18
Expert Reply
Let the speed at which Joseph traveled on bicycle and on foot be B \& F respectively.
Since Joseph travelled $\(\frac{2}{3}\)$ of the distance on bicycle and the rest of the distance on foot, so using the formula Time $\(=\frac{ Distance}{ Speed} \)$, the time taken by him for the respective travels is
$\(\frac{\frac{2d}{3}}{B}\)\(=\frac{ 2d}{3B}\)& \(\frac{\frac{d}{3}}{F}\)\(=\frac{d}{3F}\)$
(Assumed that the total distance travelled is ' d ')
So, we get $\(\frac{2 d}{3 B}: \frac{d}{3 F}=1: 3 \Rightarrow B: F=6: 1\)$
Hence the answer is (E).
Since Joseph travelled $\(\frac{2}{3}\)$ of the distance on bicycle and the rest of the distance on foot, so using the formula Time $\(=\frac{ Distance}{ Speed} \)$, the time taken by him for the respective travels is
$\(\frac{\frac{2d}{3}}{B}\)\(=\frac{ 2d}{3B}\)& \(\frac{\frac{d}{3}}{F}\)\(=\frac{d}{3F}\)$
(Assumed that the total distance travelled is ' d ')
So, we get $\(\frac{2 d}{3 B}: \frac{d}{3 F}=1: 3 \Rightarrow B: F=6: 1\)$
Hence the answer is (E).
gmatclubot
Joseph travels the first of the distance to his home on bicycle an [#permalink]
08 Feb 2025, 11:18
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