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

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Source: Veritas Prep

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

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by himalaya savalia » Sun Dec 23, 2018 5:07 am
Average speed is calculated as
$$Average\ speed\ =\frac{Total\ Dis\tan ce}{Total\ Time}$$
In this case, total distance = (k + 300) km
and total time = (96 + t) min
But we are asked speed in km / hr
$$So\ total\ time=\frac{96\ +\ t}{60}\ hr$$

$$So,\ Average\ speed\ =\ \frac{60\ \left(\ 300\ +\ k\right)}{\left(96\ +\ t\right)}$$

I think there is a typo in Ans option A.

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by [email protected] » Sun Dec 23, 2018 10:45 am
Hi All,

We're told that a plane traveled K miles in its first 96 minutes of flight time and it completed the remaining 300 miles of the trip in T minutes. We're asked for the average speed of the plane, in miles per hour, for the entire trip. This question can be solved by TESTing VALUES.

IF... K = 300 and T = 96...
then the plane traveled 300+300 = 600 miles in 96+96 = 192 minutes

We're asked for the average speed in miles/HOUR though, so we have to go the extra step of converting minutes to hours. This can be done in a couple of different ways: either by dividing 192 by 60 OR by multiplying the fraction (600/192) by 60. If you take the second approach, then you can actually avoid the physical math and just look for the answer that matches the 'look' of the fraction: 60(600/192). There's only one answer that matches...

Final Answer: A

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by fskilnik@GMATH » Sun Dec 23, 2018 5:11 pm
BTGmoderatorLU wrote:Source: Veritas Prep

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?
$$A.\ \frac{60\left(k+300\right)}{\left(96+t\right)} \,\,\,\,B.\ \frac{\left(kt+96\left(300\right)\right)}{96t} \,\,\,\,C.\ \frac{\left(k+300\right)}{60\left(96+t\right)}\,\,\,\,D.\ \frac{5k}{8}+\frac{60\left(300\right)}{t}\,\,\,\,E.\ \frac{5k}{8}+5t$$
Let´s use UNITS CONTROL, one of the most powerful tools of our method!

$$? = {{\left( {k + 300} \right)\,\,{\rm{miles}}} \over {\left( {96 + t} \right)\,\,\min }}\,\,\,\left( {{{60\,\,\min } \over {1\,\,{\rm{h}}}}} \right)\,\,\,\,\, = \,\,\,\,\,{{60\left( {k + 300} \right)} \over {96 + t}}\,\,\,{\rm{mph}}$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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by Scott@TargetTestPrep » Fri Feb 15, 2019 5:22 pm
BTGmoderatorLU wrote:Source: Veritas Prep

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