Question: Mr. Alex usually starts at 9:00am and reaches his office just in time, driving at his regular speed. Last Wednesday, he started at 9:30am and drove 25% faster than his usual speed. Did he reach the office in time?

(1) Last Monday, he started to his office 20 minutes early, drove 20% slower than his regular speed, and reached his office just 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.

1. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
2. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
3. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
4. EACH statement ALONE is sufficient.
5. Statements (1) and (2) TOGETHER are not sufficient.

Correct Answer: A

Solution and Explanation:
Approach Solution 1:
The problem statement states that:

Given:

• Mr. Alex starts at 9:00 am and arrives at his office just in time.
• He drives at his regular speed.
• Last Wednesday, he started at 9:30 am and drove 25% faster than his usual speed.

Asked:

• To find out if he reached the office in time.

Statement 1 alone: Last Monday, he started to his office 20 minutes early, drove 20% slower than his regular speed, and reached his office just in time.

Let the speed of the cat be x and the time be t.
Since the distance traveled by car is equal, then we can say,
x*t = .8x(t+1/3)
On solving, it gives
t=4/3 or 80 minutes
Now as per the question, since again distance traveled is equal
1.25x* t1= 4/3*x

t1= 64 minutes.
Therefore if he began at 9:30 am, he did not reach on time. Since he had to reach by 10:20.

Hence the statement is sufficient.

Statement 2 alone: Last Tuesday, he started to his office 10 minutes early, and reached the office 10 minutes early driving at his regular speed.
It does not give us an equation.

Hence the statement is not sufficient.

Approach Solution 2:
Let’s assume d = distance between his office and home
s = regular speed
t = regular time taken to reach office.
d= s∗t
Now, it is given that on one day, Alex begins at 9.30 (30 mins late) for the office with a 25% speed higher than usual speed.
d′ =1.25s∗ t′

The problem statement asked if Alex travels the same distance on Wednesday, as he does usually:
i.e. is s∗t=1.25s ∗ t′ ?

Statement 1: Last Monday, he started to his office 20 minutes early, drove 20% slower than his regular speed, and reached his office just in time.

This statement implies that:
s∗t=0.8s ∗ (t+20)
By solving for t, we will get,
t = 80 mins, the usual time he takes to reach the office.

Applying this, we get:
d=s∗80 (the time he reaches the office usually is 10.20)....(i)
Since on Wednesday, he leaves the office at 9.30, to reach his office in time, time taken by him should be 80-30 = 50 mins,
so d′ = 1.25s ∗ 50= 62.5s ....(ii)
From (i) and (ii) we can infer that Alex does not reach the office on time.

Therefore, Statement 1 is sufficient.

Statement 2: Last Tuesday, he started to his office 10 minutes early, and reached the office 10 minutes early driving at his regular speed.
This statement provides no additional information than already given.

Therefore, Statement 2 is not sufficient.

Approach Solution 3:

The problem statement states that:

Given:

• Mr. Alex starts at 9:00 am and arrives at his office just in time.
• He drives at his regular speed.
• Last Wednesday, he started at 9:30am and drove 25% faster than his usual speed.

Asked:

To find out if he reached the office in time.

Therefore, whether he arrived on time or not relies on what time he takes usually to reach the office.

Say, when he begins at 9, he usually reaches the office at 9:35 am.
Now, if he starts at 9:30, raising his speed by 25% will not be enough to reach the office on time. Therefore he will not arrive office on time.

On the other hand, if he usually arrives at 11:30 when he starts at 9:00. By raising his speed by 25% will be enough to reach the office on time when he starts at 9:30. In this circumstance, he will arrive at the office on time.
Usual speed:Increased speed = 4:5
Time taken usually:Time taken now = 5:4 (since the same distance has to be covered)
The difference of 1 in time is .5 hrs. So the usual time taken must be .5*5 = 2.5 hrs
Therefore, when he begins at 9:00, he reaches at 11:30.
Last Wednesday, time taken = 0.5*4 = 2 hrs.
Since he begins at 9:30, he must have arrived at 11:30.

Therefore, whether he reached the office on time or not relies on the time he usually reaches the office. Hence we need the statements:

Statement 1: Last Monday, he started to his office 20 minutes early, drove 20% slower than his regular speed, and reached his office just in time.

Usual Speed:Last Monday's Speed = 5:4
Usual time taken: Time taken last Monday = 4:5
The difference of 1 account for 20 mins i.e. (1/3)hr.
Therefore, usual time taken = 4*(1/3) = 80 mins
Hence, when he beings at 9, he arrives at 10:20.
Thus, on Wednesday, he did not reach on time.

Therefore, the statement alone is sufficient.

Statement 2: Last Tuesday, he started to his office 10 minutes early, and reached the office 10 minutes early driving at his regular speed.

It does not provide any new information.
If he travels at his usual speed, he will need the same time as always.
Therefore, if he starts 10 min early, he will reach 10 mins early.

Thus the statement alone is not sufficient.

“Mr. Alex usually starts at 9:00am and reaches his office just”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book "Manhattan Review GMAT Quantitative Question Bank [6th Edition]". The GMAT Quant section constitutes a sum of 31 questions. GMAT Data Sufficiency questions come up with a problem statement followed by two factual statements. GMAT data sufficiency includes a total of 15 questions which are two-fifths of the total 31 GMAT quant questions.

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