mikemcgarry wrote:
Cars A and B are traveling from Town X to Town Y on the same route at constant speeds. Car A is initially behind Car B, and Car A’s speed is 1.25 times Car B’s speed. Car A passes Car B at 1:30 pm. At 3:15 pm, Car A reaches Town Y, and at that moment, Car B is still 35 miles away from Town Y. What is the speed of Car A?
(A) 60 mph
(B) 75 mph
(C) 80 mph
(D) 96 mph
(E) 100 mphFor a discussion of Motion questions on the GMAT, with five practice problems, including the OE of this particular question, see:
http://magoosh.com/gmat/2014/gmat-pract ... on-motion/Mike
Here's an alternative explanation! We can use backsolving, but before doing so lets set up the equations!
This formula is key: Rate*Time = Distance
Car A’s speed is 1.25 times Car B’s speed =
Rate(A) = 1.25Rate(B)Car A passes Car B at 1:30 pm. At 3:15 pm, Car A reaches Town Y =
The Total Time that Car A travels is 1 hour and 45 mins (or 7/4 hours)Rate(A) * \(\frac{7}{4} Hours\) = D
Rate(B) * \(\frac{7}{4} Hours\) = D - 35 miles
So when we choose the answer choices, we can plug those numbers into Rate(A) and Rate(B) to find out which speed would have a 35 mile difference!
Because the time is \(\frac{7}{4} Hours\) and
Rate(A) = 1.25Rate(B), it would be smart to pick an answer choice that is divisible by 4, so lets test out C = 80
If Rate(A) = 80 mph, then Rate(B) = 64 mph
80 mph * \(\frac{7}{4} Hours\) = 140 miles
64 mph * \(\frac{7}{4} Hours\) = 112 miles
Difference is only 28, so Rate(A) has to be LARGER than 80, so eliminate A,B,C
Now let's choose answer choice E, as it's also divisible by 4 too
If Rate(A) = 100 mph, then Rate(B) = 80 mph
100 mph * \(\frac{7}{4} Hours\) = 175 miles
80 mph * \(\frac{7}{4} Hours\) = 140 miles
The difference is exactly 35 miles, so
E has to be the Answer!