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by GMATGuruNY » Sat Apr 25, 2015 2:34 am
Rasheed bought two kinds of candy bars, chocolate and toffee, that came in packages of 2 bars each. He handed out 2/3 of the chocolate bars and 3/5 of the toffee bars. How many packages of chocolate bars did Rasheed buy?

1) Rasheed bought 1 fewer package of chocolate bars than toffee bars.

2) Rasheed handed out the same number of each kind of candy bars.
If we know the number of chocolate BARS that were purchased, we can determine the number of chocolate bar PACKAGES that were purchased, since each package contains 2 bars.
Let c = the number of chocolate bars purchased and t = the number of toffee bars purchased.
Question rephrased: What is the value of c?

Statement 1: Rasheed bought 1 fewer package of chocolate bars than toffee bars.
Since each package contains 2 bars, the number of chocolate bars purchased is 2 less than the number of toffee bars purchased.
c = t - 2.
No way to solve for c.
INSUFFICIENT.

Statement 2: Rasheed handed out the same number of each kind of candy bar.
Thus, the 2/3 of the chocolate bars handed out are equal to the 3/5 of the toffee bars handed out:
(2/3)c = (3/5)t.
10c = 9t.
No way to solve for c.
INSUFFICIENT.

Statements combined:
Since we have 2 variables (c and t) and 2 distinct linear equations (c = t-2 and 10c = 9t), we can solve for each variable and thus determine the value of c.
SUFFICIENT.

The correct answer is C.

One way to solve:
c = t-2
9c = 9t - 18
9c + 18 = 9t.

Since 9c + 18 = 9t and 9t = 10c, we get:
9c + 18 = 10c
c = 18.

Since 18 chocolate bars are handed out, and each package contains 2 bars, the number of packages of chocolate bars = 9.
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by [email protected] » Sat Apr 25, 2015 10:54 am
Hi binaras,

This is a "thick" question that will likely take more time and effort than you'll spend on a typical DS question. You can certainly TEST Values on this question and get the solution, but there are some hidden Number Properties that would make the work considerably easier:

We're told that candy bars are purchased in packs of 2, so we're going to end up with an even number of chocolate bars and an even number of toffee bars. We're also told that he handed out 2/3 of the chocolate and 3/5 of the toffee bars. Since those denominators are "odd", we need to convert the fractions to make the math easier.

2/3 = 4/6 of the chocolate handed out

3/5 = 6/10 of the toffee handed out

This also tells us that he bought the items in "multiples":
Chocolate was bought in multiples of 3 "packs"
Toffee was bought in multiples of 5 "packs"

These limitations will severely limit the possibilities....

The question asks how many packs of toffee he bought?

Fact 1: 1 fewer pack of chocolate than toffee.

In other words:
Number of chocolate packs + 1 = Number of toffee packs.

Keep in mind that chocolate is bought in sets of 3-packs and toffee is bought in sets of 5-packs
So a (multiple of 3) + 1 = (multiple of 5)

TESTING Values gives us some options...
Chocolate = 9 packs, Toffee = 10 packs
Chocolate = 24 packs, Toffee = 25 packs
There are other options, but you don't need to find them.
Fact 1 is INSUFFICIENT

Fact 2: Same NUMBER of each kind of bar were handed out.

So, the number of chocolate bars given = the number of toffee bars given

Let's choose some variables:
X = TOTAL number of chocolate bars
Y = TOTAL number of toffee bars

X(4/6) = Y(6/10)

4X/6 = 6Y/10

40X = 36Y

10X = 9Y

This equation has multiple solutions...
X = 9, Y = 10
X = 18, Y = 20
X = 27, Y = 30,
Etc.
Fact 2 is INSUFFICIENT

Combined, there's an overlap worth noting:
In Fact 1, the numbers must differ by 1
In Fact 2, the numbers differ by an ever increasing value (first by 1, then by 2, then by 3, etc.)
So, there's only one set of values that fits BOTH: X = 9, Y = 10
Combined, SUFFICIENT

Final Answer: C

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