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Martha takes a road trip from point A to point B.

Expert replies
by M7MBA » Sat Dec 02, 2017 5:33 am
Martha takes a road trip from point A to point B. She drives x percent of the distance at 60 miles per hour and the remainder at 50 miles per hour. If Martha's average speed for the entire trip is represented as a fraction in its reduced form, in terms of x, which of the following is the numerator?

(A) 110
(B) 300
(C) 1,100
(D) 3,000
(E) 30,000

The OA is E.

What are the equations I should use on this PS question? Experts, can you help me?
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Source: — Problem Solving |

TTT

by OWN » Sat Dec 02, 2017 5:18 pm
Hi M7MBA,

Though I am not a "GMAT Expert," I can help you with this problem.

First, let's write down what we know:
X% of the distance was traveled at 60mph
The remaining distance was traveled at 50 mph.

We also know that the total trip must add up to 100%, so we can say
X% @ 60 mph
(100-X)% @ 50mph

Since the question asks for average speed, we need to use the average speed equation:
$$Average\ Speed\ =\ \frac{Total\ Dis\tan ce}{Total\ Time}$$

In this case, we do not have a total distance provided, but we do know that it does need to add up to 100%. So let's assume the distance is 100 miles (if you don't like assuming distances, you could alternatively express the total distance as 100% *D and the individual distances as X%*D and (100-X)%*D, but assuming 100 miles is better and more efficient):
So Total distance = 100 miles
Distance travelled at 60 mph = X miles
Distance travelled at 50 mph = 100-X miles

We are missing the time to calculate average speed, so we can go ahead and find that using the standard rate/speed equation:
$$Speed\ =\ \frac{dis\tan ce}{time}$$

Therefore, total time can be calculated using:
$$Total\ time\ =\frac{Dist_1}{Speed_1}+\frac{Dist_2}{Speed_2}\ =\ \frac{X\ miles}{60\ mph}+\frac{100-X\ miles}{50\ mph}$$



To simplify the fraction, use 300 as a common denominator and fire away:
$$Total\ time\ =\ \ \frac{5\left(X\ \right)}{300}+\frac{6\left(100-X\ \right)}{300}\ =\ \frac{5X\ -\ 6X\ +\ 600}{300}\ =\ \frac{600-X}{300}$$

Plug this into the average speed equation:

$$Average\ Speed=\ \frac{100}{\left(\frac{600-X}{300}\right)}\ =\ \frac{300\times100}{600-X}\ =\ \frac{30,\ 000}{600-X}$$

Which gives you answer choice E. In reality, you could have stopped once you came up with the common denominator and arrived at the same answer; however, because the question asks for the numerator in its reduced form, it is important to simplify the final equation as a whole in case the equation can further be reduced once you combine like terms, etc.
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by Scott@TargetTestPrep » Thu Oct 03, 2019 11:09 am
M7MBA wrote:Martha takes a road trip from point A to point B. She drives x percent of the distance at 60 miles per hour and the remainder at 50 miles per hour. If Martha's average speed for the entire trip is represented as a fraction in its reduced form, in terms of x, which of the following is the numerator?

(A) 110
(B) 300
(C) 1,100
(D) 3,000
(E) 30,000

The OA is E.

What are the equations I should use on this PS question? Experts, can you help me?
d/[(d*x/100)/60 + (d*(100 - x)/100)/50]

d/[d*x/6000 + d*(100 - x)/5000]

Multiplying by 30,000/30,000, we have:

30,000d/[5dx + 6d(100 - x)]

Divide by d/d, we have:

30,000/[5x + 600 - 6x]

30,000/[600 - x]

Answer: E

Scott Woodbury-Stewart
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[email protected]

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