Bunuel wrote:
Rachel drove the 120 miles from A to B at a constant speed. What was this speed?
(1) If she had driven 50% faster, her new time would have by 2/3 of her original time.
(2) If she drove 20 mph faster, she would have arrived an hour sooner.
Kudos for a correct solution.
MAGOOSH OFFICIAL SOLUTION:Rachel drove the 120 miles from A to B at a constant speed. What was this speed?
Let’s say the original variables are D = 120, R, and T, and the original case is 120 = RT.
Statement #1: If she had driven 50% faster, her new time would have by 2/3 of her original time.
To increase by 50%, we will multiply by the
multiplier 1.5. This means that the new speed, R2, is R2 = 1.50*R = (3/2)*R. The new time is T2 = (2/3)*T. Well, we know, that 120 = (R2)*(T2) — the new speed and time must have the small product as the original speed and time. 120 = (R2)*(T2) = [(3/2)*R]*[(2/3)*T] = (3/2)*(2/3)*RT = RT. This information leads in a big logical circle right back to the original equation 120 = RT. It gives us no new information at all. Therefore, this statement, by itself, does not provide any insight into the answer to the prompt question. This statement is insufficient.
Statement #2: If she drove 20 mph faster, she would have arrived an hour sooner.
Now, we know R2 = R + 20 and T2 = T – 1. Again, we know that 120 = (R2)*(T2), so
120 = (R + 20)*(T – 1) = RT – R + 20T – 20
120 – RT = 0 = 20T – R – 20
That is one equation with two unknowns, and 120 = RT is a second equation with two unknowns. We have
two clearly different equations for the two unknowns, so even without solving, this is sufficient to determine a unique value for the variables R & T.
Answer = B.
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