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take Mr. Robinson to drive home

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take Mr. Robinson to drive home

by sanju09 » Thu Feb 09, 2012 4:17 am
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.



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by Anurag@Gurome » Thu Feb 09, 2012 11:03 pm
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.
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Let normally Mr. Robinson drives at x mph or x/60 miles per minute and also assume that it takes t minutes when he drives at his normal speed.
Distance = speed * time = (x/60) * t miles

(1) (x/60 + 6)(t - 6) = xt/60
6t - x/10 - 36 = 0; NOT sufficient.

(2) t = 2 * (x/60) * t implies x = 30 miles per minute
But we cannot find t from here, as we do not know the distance traveled; NOT sufficient.

Combining (1) and (2), 6t = x/10 + 36
6t = 30/10 + 36; SUFFICIENT.

The correct answer is C.
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by sana.noor » Fri Jun 14, 2013 3:32 am
Hi anurag! cant we pick any smart number for distance let suppose 30 miles to solve the question with statement 1..?
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by GMATGuruNY » Sat Jun 15, 2013 3:54 am
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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