Suppose a train travels x miles in y hours...

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Suppose a train travels x miles in y hours...

by AAPL » Mon Oct 30, 2017 9:27 am
Suppose a train travels x miles in y hours and 15 minutes. Its average speed in miles per hour is

$$(A)\ \frac{(y+15)}{x}$$
$$(B)\ x(y−\frac{1}{4})$$
$$(C)\ \frac{x}{\left(y+\frac{1}{4}\right)}$$
$$(D)\ \frac{x}{y+15}$$
$$(E)\frac{(y+\frac{1}{4})}{x}$$

The OA is C.

Please, can any expert assist me with this PS question? I don't have it clear. Thanks.
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by [email protected] » Mon Oct 30, 2017 2:12 pm
Hi APL,

This question can be solved in a couple of different ways; since it makes a direct reference to the AVERAGE SPEED formula, you might find it easiest to just set up that formula and solve:

Total Distance = (Average Speed)(Total Time)

Given the information in the prompt, we have....

(X miles ) = (Average Speed)(Y + 1/4 hours)
Average Speed = (X)/(Y + 1/4)

Final Answer: C

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by Scott@TargetTestPrep » Mon Nov 11, 2019 1:24 pm
AAPL wrote:Suppose a train travels x miles in y hours and 15 minutes. Its average speed in miles per hour is

$$(A)\ \frac{(y+15)}{x}$$
$$(B)\ x(y−\frac{1}{4})$$
$$(C)\ \frac{x}{\left(y+\frac{1}{4}\right)}$$
$$(D)\ \frac{x}{y+15}$$
$$(E)\frac{(y+\frac{1}{4})}{x}$$

The OA is C.

Please, can any expert assist me with this PS question? I don't have it clear. Thanks.
The time, in hours, is y + 15/60 = y + ¼. So, the average rate is x/(y + 1/4) miles per hour.

Answer: C

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