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22 Jul 2018, 23:07
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E
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Difficulty:
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76% (01:30) correct
24%
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wrong
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John runs from his home to his school at an average speed of 6 miles per hour, and then walks home along the same route at an average speed of 3 miles per hour. If his whole journey took one hour, how many miles is his home from his school?
A. 9
B. 6
C. 4
D. 3
E. 2
A. 9
B. 6
C. 4
D. 3
E. 2
Bunuel
let t=time running
1-t=time walking
d=distance
6t+3(1-t)=2d→
t=2d/3-1
if t<1, d must be 2
E
23 Jul 2018, 07:17
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Bunuel
Distance
Equal distances. Each leg's distance = D. Each leg's time is expressed in terms of Distance/rate
Leg 1, at \(r=6\), TIME: \(\frac{D}{r}=\frac{D}{6}\)
Leg 2, at \(r=3\), TIME: \(\frac{D}{3}\)
(Leg 1 time + Leg 2 time) = 1 hour
\((\frac{D}{6}+\frac{D}{3})=1\)
\(\frac{3D+6D}{18}=1\)
\(9D=18\)
\(D=2\) miles
Answer E
Time, \(t\), to find distance
Total time: 1 hour
Let \(t=\) Leg 1 time at 6 mph
\((1 - t)=\) Leg 2 time at 3 mph
Leg 1: \((r*t)=(6* t)=6t=D\)
Leg 2: \(3*(1-t)=3(1-t)=D\)
Distances are equal, so
\(6t=3(1-t)\)
\(6t=3-3t\)
\(9t=3\)
\(t=\frac{1}{3}\)hr
Use Leg 1 to find D
\(D =(r*t) = (6mph*\frac{1}{3}hr)\)
\(D=2\) miles
Answer E
