MGMAT free CAT Test #28

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MGMAT free CAT Test #28

by tpz » Tue May 27, 2014 10:04 pm
Can someone please help me figure out this problem? It's #28 in the MGMAT CAT Test- the 1st free test. The explanation that MGMAT gives is not helping me at all.

Data Sufficiency:

What is the distance between Harry's home and his office?

(1) Harry's average speed on his commute to work this Monday was 30 miles per hour.

(2) If Harry's average speed on his commute to work this Monday had been twice as fast, his trip would have been 15 minutes shorter.

MGMAT says that per (2) alone, we know that Harry's actual commute has to be 30 minutes. How do we know this?

Any tips or easy answers would be greatly appreciated!
Source: — Data Sufficiency |

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by Brent@GMATPrepNow » Tue May 27, 2014 10:35 pm
tpz wrote: What is the distance between Harry's home and his office?

(1) Harry's average speed on his commute to work this Monday was 30 miles per hour.

(2) If Harry's average speed on his commute to work this Monday had been twice as fast, his trip would have been 15 minutes shorter.
Let d = distance between Harry's home and his office

Target question: What is the distance between Harry's home and his office?

Statement 1: Harry's average speed on his commute to work this Monday was 30 miles per hour.
To determine the distance (d), we need the travel time.
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: If Harry's average speed on his commute to work this Monday had been twice as fast, his trip would have been 15 minutes shorter.
Travel time = (distance)/(speed)
Let v = Monday's speed,

Start with a word equation:
(Monday's travel time) = (travel time at twice Monday's speed) + 15 minutes
Or..., (Monday's travel time) = (travel time at twice Monday's speed) + 0.25 hours
So, we can write d/v = d/2v + 0.25
Multiply both sides by 2v to get: 2d = d + 0.5v
Simplify: d = 0.5v
Divide both sides by v to get: d/v = 0.5
NOTE: distance/speed = time [in other words, time = d/v]
So, the Monday's travel time = 0.5 hours
So, statement 2 allows us to determine Monday's travel time, but we don't know Harry's speed.
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that Harry's speed was 30 mph
Statement 2 tells us that Harry drove for 0.5 hours
Distance = (speed)(time)
So, distance = (30)(0.5) = 15 miles
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer = C

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by GMATGuruNY » Wed May 28, 2014 11:22 am
What is the distance between Harry's home and his office?

(1) Harry's average speed on his commute to work this Monday was 30 miles per hour.

(2) If Harry's average speed on his commute to work this Monday had been twice as fast, his trip would have been 15 minutes shorter.
RATE and TIME are RECIPROCALS.
If Harry travels at 2 times his actual rate, the trip will take 1/2 his actual time.
If Harry travels at 3 times his actual rate, the trip will take 1/3 his actual time.
If Harry travels at 4 times his actual rate, the trip will take 1/4 his actual time.

Statement 2: If Harry's average speed on his commute to work this Monday had been twice as fast, his trip would have been 15 minutes shorter.
As noted above, if Harry travels at 2 times his actual rate, the trip will take 1/2 his actual time.
Here, taking 1/2 the actual time saves 15 minutes.
Thus, the 15 minutes in time savings must be equal to 1/2 the actual time:
15 = (1/2)t
t = 30 minutes.
No way to determine the distance.
INSUFFICIENT.

Statement 1: Harry's average speed on his commute to work this Monday was 30 miles per hour.
No way to determine the distance.
INSUFFICIENT.

Statements combined:
d = r*t = (30 miles per hour)(1/2 hour) = 15 miles.
SUFFICIENT.

The correct answer is C.
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by [email protected] » Wed May 28, 2014 5:47 pm
Hi tpz,

You can approach this question using "system" math, which is a category in Algebra that you'll see a few times on Test Day.

**As a warning, It's actually quite common for a DS question that "looks like" a "system" question to NOT be a system question, so you need to be sure to write everything down on the pad and do the necessary work to avoid making a silly mistake.**

We're asked for the DISTANCE between home and office. The information in the two Facts imply that we'll be using the Distance Formula, so I'll start there...

Distance = (Rate)(Time)

or

D = RT = ?

To start, we have 1 equation. To solve for Distance, we'll either need the exact value of RT or two additional unique equations that involve these variables.

Fact 1: Harry's speed = 30mph.

This gives us R. It's only one equation though, so it's not enough. Here's why:

Plugging this in, we get...

D = (30)(T) There's no way to get the exact distance with this information.
Fact 1 is INSUFFICIENT

Fact 2: If Harry's speed was twice as fast, his trip would have been 15 minutes shorter.

Since rate is in miles per HOUR, I'm going to translate 15 minutes into 1/4 hour. We can translate this equation as...

D = (2R)(T - 1/4)

Combined with the initial equation...

D = RT

We now have 2 equations, but 3 variables. No amount of algebra will get us the exact value of D.
Fact 2 is INSUFFICIENT

Combined, we have....
D = RT
D = (2R)(T - 15)
R = 30

We have three variables and three unique equations, so we CAN solve this problem and get the exact values for D, R and T.
Combined, SUFFICIENT.

Final Answer: C

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by ceilidh.erickson » Sat May 31, 2014 8:54 am
tpz wrote:
(2) If Harry's average speed on his commute to work this Monday had been twice as fast, his trip would have been 15 minutes shorter.

MGMAT says that per (2) alone, we know that Harry's actual commute has to be 30 minutes. How do we know this?
I find that it's a lot easier to think of this in terms of real-world logic than algebra.

If the speed is doubled, what happens to the time? It's halved - twice the speed means half the time. You could use R*T = D to get there, but isn't it more obvious intuitively?

If half the time is 15 min less, then the actual time must be 30 min.
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