A driver estimated a trips distance and constant speed.

Question

A driver estimated a trip’s distance and constant speed. The actual distance was 10% greater than estimated, and the actual speed was 10% lower than estimated.

Was the actual trip time more than 30 minutes longer than estimated?

(1) The estimated trip time was two hours. 
(2) The estimated trip distance was 120 kilometres.

MarkGMATMentor · Answer

Like many Data Sufficiency questions, this question benefits from doing some work with the prompt before testing the statements.
 
Let's use the distance / rate equation, distance = rate * time.

Call the estimated distance d and estimated rate r, so estimated time is  \frac{d}{r} .

Then the actual distance is  \frac{11}{10}d  and the actual rate is  \frac{9}{10}r . The actual time is  \frac{\frac{11}{10}d}{\frac{9}{10}r} = \frac{11d}{9r} .

The question can be restated as:
Is  \frac{11d}{9r}>\frac{d}{r }+ \frac{1}{2} ?

Or, more simply:
Is  \frac{2d}{9r}>\frac{1}{2} ?

Statement 1: 
This tells us estimated time, which is equal to  \frac{d}{r} . You can substitute 2 hours for  \frac{d}{r}  and solve. Though it's not hard to do so, no need to actually calculate.  Sufficient.

Statement 2: 
This gives us estimated distance, or d, but no way to calculate rate or time. Think of extremes. If the rate were extremely fast (1,000,000 km per hour, say), the answer to the question would be NO. If the rate were extremely slow (say 1 km per hour), the answer would be YES.  Not sufficient.

The answer is A.

It’s easy to get lost here if you aren’t careful. Make sure you deduce what you can from the question prompt, including restating the question when you can, before testing the statements.

dhart · Answer

estimated disted d, estimated rate r , estimated d/r , actual time 1.1d/.9r or 11d/9r-> 
is 11d/9r > d/r+1/2

stmnt 1 d/r was 2 hours
11d/9r >d/r +1/2
2d/9r > 1/2 where d/2 =2 suff

stmt 2 rate value is not avlb ns A

ohhno · Answer

Let estimated distance = DD, estimated speed = SS.
Estimated time:
T=DST=\frac{D}{S}
Actual distance = 1.1D1.1D
Actual speed = 0.9S0.9S
Therefore actual time:
Ta=1.1D0.9S=119TT_a=\frac{1.1D}{0.9S} =\frac{11}{9}T
So the increase in time is:
Ta−T=(119−1)T=29TT_a-T=\left(\frac{11}{9}-1ight)T =\frac{2}{9}T
We need to know whether this increase is  more than 30 minutes .
Statement (1)
Estimated time = 2 hours = 120 minutes.
Increase:
29(120)=26.67 minutes\frac{2}{9}(120)=26.67	ext{ minutes}
That's  not more than 30 minutes .
So  Statement 1 is sufficient  because we can answer definitively:  No .
Statement (2)
Estimated distance = 120 km.
We still don't know the estimated speed, so we don't know the estimated time.
For example:
 
 At 60 km/h → time = 2 hours → increase = 26.67 min 
 At 48 km/h → time = 2.5 hours → increase = 33.33 min  
So the answer could be  No or Yes .
Statement 2 is  not sufficient .
✅ Answer:  (A) 
 Statement 1 alone is sufficient, but statement 2 alone is not.

A driver estimated a trips distance and constant speed.

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GMAT Data Sufficiency (DS) Questions

A driver estimated a trips distance and constant speed.

Prep Toolkit

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