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Fred and Sam are standing 45 miles apart and they start walking in a straight line toward each other at...

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Fred and Sam are standing 45 miles apart and they start walking in a straight line toward each other at...

by AAPL » Thu Dec 17, 2020 4:44 pm

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Fred and Sam are standing 45 miles apart and they start walking in a straight line toward each other at the same time. If Fred walks at a constant speed of 4 miles per hour and Sam walks at a constant speed of 5 miles per hour, how many miles has Sam walked when they meet?

A. 5
B. 9
C. 25
D. 30
E. 45

OA C
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Re: Fred and Sam are standing 45 miles apart and they start walking in a straight line toward each other at...

by Scott@TargetTestPrep » Wed Dec 23, 2020 4:00 pm
AAPL wrote: ↑
Thu Dec 17, 2020 4:44 pm
 Princeton Review

Fred and Sam are standing 45 miles apart and they start walking in a straight line toward each other at the same time. If Fred walks at a constant speed of 4 miles per hour and Sam walks at a constant speed of 5 miles per hour, how many miles has Sam walked when they meet?

A. 5
B. 9
C. 25
D. 30
E. 45

OA C
Solution:

Let t denote the time (in hours) until they meet. We can create the equation:

4t + 5t = 45

9t = 45

t = 5 hours

Thus, Sam has walked 5 x 5 = 25 miles when they meet.

Answer: C

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Re: Fred and Sam are standing 45 miles apart and they start walking in a straight line toward each other at...

by deloitte247 » Sun Dec 27, 2020 9:10 am
$$Total\ relative\ dis\tan ce\ apart=45miles$$
$$Fred's\ speed=4\ miles\ per\ hour$$
$$Sam's\ speed=5\ miles\ per\ hour$$
$$Speed=\frac{dis\tan ce}{timetaken}$$
$$Time\ taken=\frac{dis\tan ce}{speed}$$
$$\sin ce\ both\ Fred\ and\ Sam\ are\ moving\ in\ opposite\ direction;$$
$$their\ relative\ dis\tan ce\ will\ reduce\ at\ their\ combined\ speed=(4+5)mph=9mph$$
$$Time\ taken\ for\ both\ to\ meet=\frac{45}{9}=5hours$$
$$And\ in\ 5\ hours\ at\ Sam's\ speed\ of\ 5miles\ per\ hour;$$
$$dis\tan ce\ covered\ by\ Sam$$
$$=25miles$$
$$Answer=C$$
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