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OG 13 #132

This topic has 2 expert replies and 1 member reply
kheba Junior | Next Rank: 30 Posts Default Avatar
Joined
10 Aug 2008
Posted:
22 messages

OG 13 #132

Post Sat Aug 18, 2012 5:36 pm
Joanna bought only $0.15 and $0.29 stamps. How many $0.15 stamps did she buy?

1. She bought $4.40 worth of stamps
2. She bought an equal number of $0.15 stamps and $0.29 stamps.

OA A

I'd like to see what are the efficient ways of solving it. I took more than 2 minutes for this problem.
Thanks

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SmartAssJun Junior | Next Rank: 30 Posts Default Avatar
Joined
21 Aug 2012
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Post Wed Aug 22, 2012 9:02 pm
Stuart Kovinsky wrote:
Hi!

Normally for this type of question we can rely on our old friend, "number of equations vs number of unknowns". However, there are some cases, such as this one, in which you have more info that you originally thought and you don't actually need as many equations as you may think.

The key to solving this question is noting that stamps are indivisible - in other words, Joanna has to by an integer number of each denomination. Always be on the lookout for objects that are indivisible!

Let's call the two types of stamps c (for cheap) and e (for expensive). From (1), we know that:

15c + 29e = 440

Now, without knowing that c and e have to be integers, we'd quickly say that (1) is insufficient, since we have 2 variables and only 1 equation. However, since c and e have to be integers, it's possible that there's only one pair of values that satisfy the equation. So, we need a quick way to determine if there's one unique solution.

A great way to solve quickly is to pay attention to the units digit of the sum. Since 440 ends in a 0, and since 15c will end in either 5 or 0 every time, 29e must also end in 5 or 0. Focusing on values of e that are multiples of 5 should make the process fairly quick.

e=5... 29e = 145. 440-145=295. Is 295 divisible by 15? NO (since 15 goes into 300)
e=10... 29e = 290. 440-290=150. Is 150 divisible by 15? YES
e=15... 20e = 435. 440-435=5. Is 5 divisible by 15? NO

So the only possible value for e is 10 (which means there's also a unique value for c). (1) is sufficient alone... choose (A)!

(I skipped over looking at (2) - we can pick any equal values for c and e to make (2) true.)

kheba wrote:
Joanna bought only $0.15 and $0.29 stamps. How many $0.15 stamps did she buy?

1. She bought $4.40 worth of stamps
2. She bought an equal number of $0.15 stamps and $0.29 stamps.

OA A

I'd like to see what are the efficient ways of solving it. I took more than 2 minutes for this problem.
Thanks
Same way I did it!

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Post Sun Aug 19, 2012 7:09 am
This question illustrates a common trap on the GMAT.

In high school we learned that, if we're given 1 equation with 2 variables, we cannot find the value of either variable. However, if we restrict the variables to positive integers within a certain range of values, then there are times when we can find the value of a variable if we're given 1 equation with 2 variables.

This common GMAT trap (along with other common traps) is addressed in the following free videos:
http://www.gmatprepnow.com/module/gmat-data-sufficiency?id=1105
http://www.gmatprepnow.com/module/gmat-data-sufficiency?id=1106
http://www.gmatprepnow.com/module/gmat-data-sufficiency?id=1107

Cheers,
Brent

_________________
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SmartAssJun Junior | Next Rank: 30 Posts Default Avatar
Joined
21 Aug 2012
Posted:
28 messages
Upvotes:
2
Post Wed Aug 22, 2012 9:02 pm
Stuart Kovinsky wrote:
Hi!

Normally for this type of question we can rely on our old friend, "number of equations vs number of unknowns". However, there are some cases, such as this one, in which you have more info that you originally thought and you don't actually need as many equations as you may think.

The key to solving this question is noting that stamps are indivisible - in other words, Joanna has to by an integer number of each denomination. Always be on the lookout for objects that are indivisible!

Let's call the two types of stamps c (for cheap) and e (for expensive). From (1), we know that:

15c + 29e = 440

Now, without knowing that c and e have to be integers, we'd quickly say that (1) is insufficient, since we have 2 variables and only 1 equation. However, since c and e have to be integers, it's possible that there's only one pair of values that satisfy the equation. So, we need a quick way to determine if there's one unique solution.

A great way to solve quickly is to pay attention to the units digit of the sum. Since 440 ends in a 0, and since 15c will end in either 5 or 0 every time, 29e must also end in 5 or 0. Focusing on values of e that are multiples of 5 should make the process fairly quick.

e=5... 29e = 145. 440-145=295. Is 295 divisible by 15? NO (since 15 goes into 300)
e=10... 29e = 290. 440-290=150. Is 150 divisible by 15? YES
e=15... 20e = 435. 440-435=5. Is 5 divisible by 15? NO

So the only possible value for e is 10 (which means there's also a unique value for c). (1) is sufficient alone... choose (A)!

(I skipped over looking at (2) - we can pick any equal values for c and e to make (2) true.)

kheba wrote:
Joanna bought only $0.15 and $0.29 stamps. How many $0.15 stamps did she buy?

1. She bought $4.40 worth of stamps
2. She bought an equal number of $0.15 stamps and $0.29 stamps.

OA A

I'd like to see what are the efficient ways of solving it. I took more than 2 minutes for this problem.
Thanks
Same way I did it!

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