vaibhav101 wrote:Mr. David usually starts at 9 am and reaches his offices just in time, driving at his regular speed. Last Wednesday, he started at 9:30 am and drove 25% faster than his regular speed. Did he reach the office in time.

1) last Monday, he started to his office 20minutes early, drove 20% slower than his regular speed and reached his office in time.
2) last Tuesday, he started to his office 10 minutes early, and reached the office 10 minutes early driving at his regular speed .

Regular day:
Let r = the regular rate and t = the regular time, implying that the distance to the office = rt.

Last Wednesday:
Since Mr. David drives 25% faster -- 5/4 of his regular speed -- the rate = (5/4)r.
Since Mr. David leaves 30 minutes late -- reducing the time by 30 minutes -- the time = t-30.
Thus, the distance traveled between 9:30am and the desired arrival time = (5/4)(r)(t-30).
For Mr. David to arrive at the office in time, the distance traveled by the desired arrival time must be equal to the distance to the office:
(5/4)(r)(t-30) = rt
(5/4)(t-30) = t
5(t-30) = 4t
5t - 150 = 4t
t = 150.
Question stem, rephrased:
Does t = 150?

Statement 1:
Since Mr. David drives 20% slower -- 4/5 of his regular speed -- the rate = (4/5)r.
Since Mr. David leaves 20 minutes early -- increasing the time by 30 minutes -- the time = t+20.
Thus, the distance traveled between 9:30am and the desired arrival time = (4/5)(r)(t+20).
Since Mr. David arrives at the office in time, the distance traveled by the desired arrival time is equal to the distance to the office:
(4/5)(r)(t+20) = rt
(4/5)(t+20) = t
4(t+20) = 5t
4t + 80 = 5t
80 = t.
Thus, the answer to the rephrased question stem is NO.
SUFFICIENT.

Statement 2:
Since he started 10 minutes early and reached his office 10 minutes early, Mr. David traveled at his usual speed last Tuesday.
No new information about last Wednesday.
INSUFFICIENT.

The correct answer is A.