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# A plane traveled k miles in its first 96 minutes of flight time. If it

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Re: A plane traveled k miles in its first 96 minutes of flight time. If it [#permalink]
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arslanama 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 ( k + 300 )/96 + t
(B) kt + 96 ( 300 ) / 96t
(C) k + 300 / 60 ( 96 + y )
(D) (5k / 8) + (60(300) / t)
(E) (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) = (k + 300)/(96/60+t/60)

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

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

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Re: A plane traveled k miles in its first 96 minutes of flight time. If it [#permalink]
arslanama 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 ( k + 300 )/(96 + t)
(B) kt + 96 ( 300 ) / 96t
(C) k + 300 / 60 ( 96 + y )
(D) (5k / 8) + (60(300) / t)
(E) (5k / 8) + 5t

Simple application of UNITS CONTROL, one of the most powerful tools of our method!

$$? = \frac{{\left[ {{\text{total}}\,\,\# \,\,{\text{miles}}} \right]}}{{\left[ {{\text{total}}\,\,\# \,\,\min } \right]}}\,\,\frac{{{\text{miles}}}}{{\min }}\,\,\,\left( {\frac{{60\,\,\min }}{{1\,\,\,{\text{h}}}}\,\,\begin{array}{*{20}{c}}\\ \nearrow \\ \\ \nearrow \\ \end{array}} \right)\,\,\, = \,\,60 \cdot \left( {\frac{{\left[ {{\text{total}}\,\,\# \,\,{\text{miles}}} \right]}}{{\left[ {{\text{total}}\,\,\# \,\,\min } \right]}}} \right)\,\,\,{\text{mph}}\,\,\,$$

(Arrows indicate licit converter.)

$$\left[ {{\text{total}}\,\,\# \,\,{\text{miles}}} \right] = k + 300$$

$$\left[ {{\text{total}}\,\,\# \,\,\min } \right]\,\,{\text{ = }}\,\,96 + t$$

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

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

Regards,
Fabio.
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Re: A plane traveled k miles in its first 96 minutes of flight time. If it [#permalink]
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...

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Re: A plane traveled k miles in its first 96 minutes of flight time. If it [#permalink]
arslanama 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) $$\frac{60(k + 300)}{(96 + t)}$$
(B) $$\frac{(kt + 96(300))}{96t}$$
(C) $$\frac{(k + 300)}{60(96 + t)}$$
(D) $$\frac{5k}{8} + \frac{60(300)}{t}$$
(E) $$\frac{5k}{8} + 5t$$

Total Distance $$= k + 300$$

Total Time (in minutes) $$= 96 + t$$

Total Time (in hours) $$= \frac{96}{60} + \frac{t}{60} = \frac{96 + t}{60}$$

Average Speed $$= \frac{Distance}{Time}$$

Average Speed $$= \frac{(k +300)}{\frac{(96+t)}{60}}$$ $$= \frac{60 (k+300)}{(96+t)}$$

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Re: A plane traveled k miles in its first 96 minutes of flight time. If it [#permalink]
arslanama 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) $$\frac{60(k + 300)}{(96 + t)}$$
(B) $$\frac{(kt + 96(300))}{96t}$$
(C) $$\frac{(k + 300)}{60(96 + t)}$$
(D) $$\frac{5k}{8} + \frac{60(300)}{t}$$
(E) $$\frac{5k}{8} + 5t$$

KeyWord =miles per hour

60 mins 1 hr
1 min = 1/60 hr
96 mins = 96/60 hr = 8/5 hr

8/5 hr, k miles

1 min = 1/60 hr
t mins = t/60 hr

Average speed = 300 + k / [ 8/5 + t/60]

$$\frac{60(k + 300)}{(96 + t)}$$

A
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Re: A plane traveled k miles in its first 96 minutes of flight time. If it [#permalink]
total distance= K+300 miles
total time= 96+t minutes or (96+t)/60 hrs
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Re: A plane traveled k miles in its first 96 minutes of flight time. If it [#permalink]
Bunuel Why can't we apply the formula 2uv/(u+v) over here? u,v are the speeds in the 2 intervals.
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Re: A plane traveled k miles in its first 96 minutes of flight time. If it [#permalink]
Ruchi007 wrote:
Bunuel Why can't we apply the formula 2uv/(u+v) over here? u,v are the speeds in the 2 intervals.

You can get the average speed with that formula when the two intervals are equal, which is not the case in this question.

If you assume k = 300 miles, with that formula you'd get:

$$\frac{2*\frac{300 }{(\frac{96}{60})}*\frac{300}{(\frac{t}{60})} }{\frac{300 }{(\frac{96}{60})}+ \frac{300}{(\frac{t}{60})} }=\frac{36000}{t + 96}$$.

And if you substitute k = 300 ins option A you'd get the same: $$\frac{60(300 + 300)}{96 + t}=\frac{36000}{t + 96}$$

Hope it's clear.
Re: A plane traveled k miles in its first 96 minutes of flight time. If it [#permalink]
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