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Martha takes a road trip from point A to point B. She drives x percent of the distance at 60 miles per hour...

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Martha takes a road trip from point A to point B. She drives x percent of the distance at 60 miles per hour...

by BTGmoderatorLU » Mon Oct 26, 2020 9:53 am

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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
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Re: Martha takes a road trip from point A to point B. She drives x percent of the distance at 60 miles per hour...

by swerve » Tue Oct 27, 2020 4:49 am
BTGmoderatorLU wrote: ↑
Mon Oct 26, 2020 9:53 am
 Source: Manhattan Prep

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
Let total distance \(= D\).

\(d_1 = \dfrac{xD}{100} \cdot v_1 = 60 t_1 = \dfrac{d_1}{v_1} = \dfrac{xD}{6000}\)

\(d_2 = \dfrac{(1-x)D}{100} \cdot v_2 = 50 t_2 = \dfrac{d_2}{v_2} = \dfrac{(1-x)D}{5000}\)

For the total trip:

\(V = \dfrac{D}{T} = \dfrac{d_1+d_2}{t_1+t_2}\)

Note that \(d_1+d_2 = D\)

\(V=\dfrac{D}{t_1+t_2} = \dfrac{D}{\dfrac{xD}{6000}+\dfrac{(1-x)D}{5000}}\)

For this, it's easy to see that the numerator in the reduced form will be the least common multiple of 5000 and 6000, which is 30,000.

Therefore, E
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Re: Martha takes a road trip from point A to point B. She drives x percent of the distance at 60 miles per hour...

by Scott@TargetTestPrep » Sat Oct 31, 2020 1:11 pm
BTGmoderatorLU wrote: ↑
Mon Oct 26, 2020 9:53 am
 Source: Manhattan Prep

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
Solution:
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

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