To make the math easier, we can replace the given values with rounder numbers, as follows:
fskilnik@GMATH wrote:GMATH practice exercise (Quant Class 19)
If Rebeca drives to work at x mph she will be one hour late, but if she drives at y mph she will be one hour early. How far (in miles) does Rebeca drive to work?
(1) x and y differ by 10 miles per hour.
(2) y is 50% greater than x.
Since Rebeca arrives 1 hour late when traveling at the lower speed and 1 hour early when traveling at the higher speed, the time at the lower speed must be 2 hours greater than the time at the higher speed.
Statement 1:
Case 1: x = 10 mph and y = 20 mph
Since the rate ratio = 10:20 = 1:2, the time ratio = 2:1 = 4:2, implying 4 hours at the lower speed and 2 hours at the higher speed.
Since the trip takes 4 hours when traveling at the lower speed of 10 mph, the distance = 10*4 = 40 miles.
Case 2: x = 20 mph and y = 30 mph
Since the rate ratio = 20:30 = 2:3, the time ratio = 3:2 = 6:4, implying 6 hours at the lower speed and 4 hours at the higher speed.
Since the trip takes 6 hours when traveling at the lower speed of 20 mph, the distance = 20*6 = 120 miles.
Since the distance can be different values, INSUFFICIENT.
Statement 2:
Case 2 also satisfies Statement 2.
In Case 2, the distance = 120 miles.
Case 3: x = 2 mph and y = 3 mph
Since the rate ratio = 2:3, the time ratio = 3:2 = 6:4, implying 6 hours at the lower speed and 4 hours at the higher speed.
Since the trip takes 6 hours when traveling at the lower speed of 2 mph, the distance = 2*6 = 12 miles.
Since the distance can be different values, INSUFFICIENT.
Statements combined:
Only Case 2 satisfies both statements.
In Case 2, the distance = 120 miles.
SUFFICIENT.
The correct answer is
C.
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