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Candy bars DS (GPREP)

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by aj5105 » Wed Jul 01, 2009 7:08 am
R bought two kinds of candy bars, C and T, that came in packages of 2 bars each. He handed out 2/3 of C bars and 3/5 of the T bars. How many packages of C bars did R buy?

1). R bought 1 fewer package of C bars than T bars

2). R handed out the same number of each kind of candy bars
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Source: — Data Sufficiency |
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Re: Candy bars DS

by ssmiles08 » Wed Jul 01, 2009 7:24 am
aj5105 wrote:R bought two kinds of candy bars, C and T, that came in packages of 2 bars each. He handed out 2/3 of C bars and 3/5 of the T bars. How many packages of C bars did R buy?

1). R bought 1 fewer package of C bars than T bars

2). R handed out the same number of each kind of candy bars
IMO C.

since 1 package contains 2 bars, we can write it as:

1) C = T - 2

INSUFF, 3 unknowns. C, T and total number of bars.

2) (2/3)*C = (3/5)*T
INSUFF 2 unknowns.

together: sufficient.

(2/3)*C = (3/5)*T

C = T - 2

T = C + 2

substitue C for T and we get C = 18.

since each package has 2 bars each, R bought 9 packages of C.
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by cramya » Wed Jul 01, 2009 8:42 pm
Let CP be the no of C packages
Let TP be the no of T packages

Stmt I

CP = TP -1 (I)
Many values posible
INSUFF


Stmt II
2/3 (CP * 2) = 3/5 (TP*2) (II)
(No of packages fof each * 2 gives the total no of candies)

Can get a ratio of CP:TP but not an actual number
INSUFF


Together:
Substitute I in II to find CP

SUFF

C
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