Mary bought a number of pencils. If each pencil was either a

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Mary bought a number of pencils. If each pencil was either a 24-cent pencil or a 30-cent pencil, how many 24-cent pencils did Mary buy?

1) The total price of the pencils Mary bought was 102 cents
2) Mary bought a total of 4 pencils

Source : Math Revolution
Official Answer : A

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by [email protected] » Sun Apr 02, 2017 9:08 am
Hi ziyuenlau,

Since the numbers involved in this prompt are relatively small, you can use 'brute force' to get to the correct answer without too much trouble.

We're told that there are two types of pencils: those that cost 24 cents and those that cost 30 cents. We're asked for the number of 24-cent pencils that Mary bought.

1) The total price of the pencils Mary bought was 102 cents

Notice that 102 cents is a relatively small total and there must be at least a few 24-cent pencils in that group (since the unit's digit of 102 is "2" and there's no way to get that "2" when taking a multiple of 30).

Increments of 24: 24, 48, 72, 96, 120

The only option that fits this information and that has a unit's digit of "2" is 72.... so there MUST be three 24-cent pencils and one 30-cent pencil.
Fact 1 is SUFFICIENT

2) Mary bought a total of 4 pencils

With this Fact, the answer can be 0, 1, 2, 3 or 4.
Fact 2 is INSUFFICIENT

Final Answer: A

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by DavidG@VeritasPrep » Sun Apr 02, 2017 11:08 am
And for a similarly themed official question, see here: https://www.beatthegmat.com/og-12-ps-65-t35672.html
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by Jay@ManhattanReview » Thu Apr 06, 2017 5:57 am
ziyuenlau wrote:Mary bought a number of pencils. If each pencil was either a 24-cent pencil or a 30-cent pencil, how many 24-cent pencils did Mary buy?

1) The total price of the pencils Mary bought was 102 cents
2) Mary bought a total of 4 pencils

Source : Math Revolution
Official Answer : A
Hi ziyuenlau,

Rich presented a great solution. Here's my take on this one by algebraic route.

Say # of 24-cent pencil = x and # of 30-cent pencil = y

Thus, 24x + 30y = 102

=> 12x + 15y = 51

=> y = (51 - 12x)/15

Since x and y each is a positive integer, we can plug-in values for x = 1, 2, 3.. and find where we get a unique integer value for y.

@x = 1 and 2, y is not an integer, but @ x = 3, y = 1 an integer (A solution!)

We must see whether there are more solutions.

@x = 4, y is not an integer.

We must not try larger than 4 values for x as this would result in negative values for y (Unqualified)

Thus, we have x = 3 and y = 1. Suffcient.

Needless to discuss statement 2 as it is clearly insufficient.

The correct answer: A

Hope this helps!

Relevant book: Manhattan Review GMAT Math Essentials Guide

-Jay
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