A plane traveled k miles in its first 96 minutes of flight..

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A plane traveled k miles in its first 96 minutes of flight time. If it completed the remaining 300 miles of the trip in t minutes, what was its average speed, in miles per hour, for the entire trip?

$$A.\ 60\frac{\left(k+300\right)}{96+t}$$
$$B.\ kt+96\frac{\left(300\right)}{96t}$$
$$C.\ \frac{k+300}{60\left(96+t\right)}$$
$$D.\ \frac{5k}{8}+\frac{60\left(300\right)}{t}$$
$$E.\ \frac{5k}{8}+5t$$

The OA is A.

The total distance will be,
$$T_{dist}=k+300\ miles$$
And the total time will be,
$$T_{time}=\frac{96+t\ }{60}hours$$
Then the total average speed will be,
$$T_{avg\ speed}=\frac{T_{dist}}{T_{time}}=\frac{k+300}{\frac{96+t}{60}}=60\frac{\left(k+300\right)}{96+t}miles\ per\ hour$$
Now, my question is, is there another way to solve this PS question? Experts, can you help me, please? Thanks!

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by EconomistGMATTutor » Fri Feb 09, 2018 8:19 am
AAPL wrote:A plane traveled k miles in its first 96 minutes of flight time. If it completed the remaining 300 miles of the trip in t minutes, what was its average speed, in miles per hour, for the entire trip?

$$A.\ 60\frac{\left(k+300\right)}{96+t}$$
$$B.\ kt+96\frac{\left(300\right)}{96t}$$
$$C.\ \frac{k+300}{60\left(96+t\right)}$$
$$D.\ \frac{5k}{8}+\frac{60\left(300\right)}{t}$$
$$E.\ \frac{5k}{8}+5t$$

The OA is A.

The total distance will be,
$$T_{dist}=k+300\ miles$$
And the total time will be,
$$T_{time}=\frac{96+t\ }{60}hours$$
Then the total average speed will be,
$$T_{avg\ speed}=\frac{T_{dist}}{T_{time}}=\frac{k+300}{\frac{96+t}{60}}=60\frac{\left(k+300\right)}{96+t}miles\ per\ hour$$
Now, my question is, is there another way to solve this PS question? Experts, can you help me, please? Thanks!
Hello AAPL.

The way you solve it is perfect.

Another way you could solve it is making the conversion from minutes to hours from the beginning. That is to say, $$96\ \text{minutes}\ =\ \frac{96}{60}\text{hours}$$ and $$t\ \text{minutes}\ =\ \frac{t}{60}\text{hours}.$$ Hence, $$T_{dist}=k+300\ \ \ \ \ and\ \ \ \ \ \ \ \ T_{time}\ =\ \frac{96}{60}+\frac{t}{60}=\frac{96+t}{60}.$$ Therefore, $$T_{avg\ speed}=\frac{T_{dist}}{T_{time}}\ =\ \frac{300+k}{\frac{96+t}{60}}=60\ \frac{\left(300+k\right)}{96+t}$$ So, the correct answer is the option A.

I hope this answer also can help you.

I'm available if you'd like a follow-up.

Regards.
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by Scott@TargetTestPrep » Mon Jul 01, 2019 4:19 pm
AAPL wrote:A plane traveled k miles in its first 96 minutes of flight time. If it completed the remaining 300 miles of the trip in t minutes, what was its average speed, in miles per hour, for the entire trip?

$$A.\ 60\frac{\left(k+300\right)}{96+t}$$
$$B.\ kt+96\frac{\left(300\right)}{96t}$$
$$C.\ \frac{k+300}{60\left(96+t\right)}$$
$$D.\ \frac{5k}{8}+\frac{60\left(300\right)}{t}$$
$$E.\ \frac{5k}{8}+5t$$
We are given that a plane traveled k miles in 96 minutes. Since we need the average speed in miles per hour we can convert 96 minutes to hours.

96 minutes = 96/60 = 8/5 hours

We are also given that the plane completed the remaining 300 miles in t minutes. We must also convert t minutes to hours.

t minutes = t/60 hours

Now we can calculate the average speed for the entire trip, using the formula for average speed.

Average speed = total distance/total time

Average speed = (k + 300)/(8/5 + t/60)

Average speed = (k + 300)/(96/60+t/60)

Average speed = (k + 300)/((96 + t)/60)

Average speed = 60(k + 300)/(96 + t)

Answer: A

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