sanju09 wrote:How long in minutes does it normally take Mr. Robinson to drive home from his office?
I. If Mr. Robinson drives 6 miles per hour more than his normal speed, it would take him 6 fewer minutes than normal to drive home from his office.
II. The number of minutes Mr. Robinson normally takes to drive home from his office is twice the number of miles he needs to drive home from his office.
Rate and time are RECIPROCALS.
2 times as fast implies 1/2 the time.
3 times as fast implies 1/3 the time.
Statement 1: If Mr. Robinson drives 6 miles per hour more than his normal speed, it would take him 6 fewer minutes than normal to drive home from his office.
Case 1: Normal speed = 6 miles per hour.
6 miles per hour faster = 6+6 = 12 miles per hour.
Since 12/6 = 2, he travels at twice the normal speed, implying that the trip takes 1/2 the normal time.
Since the time decreases by 1/2, and he arrives 6 minutes early, we get:
(1/2)t = 6
t = 12.
Case 2: Normal speed = 3 miles per hour.
6 miles per hour faster = 3+6 = 9 miles per hour.
Since 9/3 = 3, he travels at 3 times the normal speed, implying that the trip takes 1/3 the normal time.
Since the time decreases by 2/3, and he arrives 6 minutes early, we get:
(2/3)t = 6
t = 9.
Since different times are possible, INSUFFICIENT.
Statement 2: The number of minutes Mr. Robinson normally takes to drive home from his office is twice the number of miles he needs to drive home from his office.
Thus:
t = 2d
t = 2(rt)
r = 1/2.
Thus, r = 1/2 mile per minute = 30 miles per hour.
No way to determine the time.
INSUFFICIENT.
Statements combined:
Since r = 30 miles per hour, 6 miles per hour faster = 30+6 = 36 miles per hour.
Since 36/30 = 6/5, he travels at 6/5 the normal speed, implying that the trip takes 5/6 the normal time.
Since the time decreases by 1/6, and he arrives 6 minutes early, we get:
(1/6)t = 6
t = 36.
SUFFICIENT.
The correct answer is
C.
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