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Re: Rasheed bought two kinds of candy bars, chocolate and toffee, that cam [#permalink]
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Bunuel
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 chocolates bar did Rasheed buy?

Let C be # of packages of chocolate bars and T be # of packages of toffee bars. Rasheed handed out 2/3*2C of the chocolate bars and 3/5*2T of the toffee bars. Question: C=?

(1) Rasheed bought 1 fewer package of chocolate bars than toffee bars --> C=T-1, two variables one equation. Not sufficient.

(2) Rasheed handed out the same number of each kind of candy bar --> 2/3*2C=3/5*2T, two variables one equation. Not sufficient.

(1)+(2) We have two distinct equations and two variables, hence we can solve for them. Sufficient.

Answer: C.

Bunuel -

You are amazing, your explanation makes it so easy. I really had a problem understanding 'packages' and 'bars'.

The question is pretty simple, it is only the language based questions that bog me down some times !

Thank You again!
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Re: Rasheed bought two kinds of candy bars, chocolate and toffee, that cam [#permalink]
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

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 chocolates bar 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 bar.

Looking at the original condition, we can easily figure out that this is a “2 by 2” question, a common type of question in GMAT math. We can represent the information using a table as below:
Attachment:
GCDS enigma123 Rasheed bought(20151007).png
GCDS enigma123 Rasheed bought(20151007).png [ 3.66 KiB | Viewed 58938 times ]

From above, you can see that there are 2 variables (C,T) and 2 equations from the 2 equations; the number of variables match that of the equations, so there is high chance that (C) is going to be our answer.
Combining the 2 equations,
C=T-1, and 2C/3=3T/5 are sufficient to solve for the variables, so the answer becomes (C).

For cases where we need 2 more equation, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Re: Rasheed bought two kinds of candy bars, chocolate and toffee, that cam [#permalink]
Bunuel
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 chocolates bar did Rasheed buy?

Let C be # of packages of chocolate bars and T be # of packages of toffee bars. Rasheed handed out 2/3*2C of the chocolate bars and 3/5*2T of the toffee bars. Question: C=?

(1) Rasheed bought 1 fewer package of chocolate bars than toffee bars --> C=T-1, two variables one equation. Not sufficient.

(2) Rasheed handed out the same number of each kind of candy bar --> 2/3*2C=3/5*2T, two variables one equation. Not sufficient.

(1)+(2) We have two distinct equations and two variables, hence we can solve for them. Sufficient.

Answer: C.

Best answer hands down.... Thanks Bunuel
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Re: Rasheed bought two kinds of candy bars, chocolate and toffee, that cam [#permalink]
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Let us assume
chocolate bar = c ; toffee bar = t
Package of chocolate bar = Pc ; Package of toffee bar = Pt

As per question:
Pc = 2c ; Pt = 2t

1) Pc = Pt - 1 (1 package = 2 toffee bar)
=> 2c = 2t - 2
=> c = t -1 - - - eq 1
Not Sufficient since we can not find the packages of chocolate bar (Pc)

2) => (2/3)c = (3/5)t
=> 10c = 9t - - - eq 2
Not Sufficient since we can not find the packages of chocolate bar (Pc)

1 + 2
=> two variable and two equation , we will get the value of c , and Pc = 2c , so sufficient ; Ans => C
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Re: Rasheed bought two kinds of candy bars, chocolate and toffee, that cam [#permalink]
if we consider x as the no of chocolate bars and y as the no of toffee bars
then the first equation can be
x-2 (1 less package means 2 chocolates less) =y
and the other equation is
2/3x = 3/5y

But, the value of y is coming as negative

can you please correct my understanding
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Re: Rasheed bought two kinds of candy bars, chocolate and toffee, that cam [#permalink]
Expert Reply
Hi nikitamaheshwari! Happy to clarify for you!

It looks like you misinterpreted or erroneously swapped the algebraic translation of "Rasheed bought 1 fewer package of chocolate bars than toffee bars."

Here, if the number of chocolate bars is two less than the number of toffee bars, according to the variables you established, you would have "x = y-2," or, "the number of chocolate bars" "is" "two less than the number of toffee bars."

Keep in mind, when you build an equation, you're really just saying the same thing two different ways. So, we're expressing the number of chocolate bars in terms of x, and the number of chocolate bars in terms of y (y-2).

One quick aside - I'd strongly suggest having your variables indicate what they correspond to. So, as you've seen in the explanations above, by calling the number of chocolate bars "c" and the number of toffee bars "t," it becomes easier to ensure you're correctly translating the algebra and answering the question being asked. :)

I hope this helps! Let me know if you have any follow-up questions!
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Re: Rasheed bought two kinds of candy bars, chocolate and toffee, that cam [#permalink]
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(1) C = T - 1

Insufficient

(2) 2C/3 = 3T/5
9T = 10C

Insufficient

(1&2)

9(C-1) = 10C
9C - 9 = 10C
C = 9

SUFFICIENT.
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Re: Rasheed bought two kinds of candy bars, chocolate and toffee, that cam [#permalink]
enigma123
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 chocolates bar 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 bar.

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

Let the number of toffee packages =\( x\) ; then number of choc. packages = \(x-1\)

#Total number of Choc= \(2(x-1)\) ( as there are 2 bars in every package)

Insufficient.


(2) Rasheed handed out the same number of each kind of candy bar
# Of choc package x, # of toffee package y

\(2/3(x)= 3/5(y) \) -------------- > clearly, we can't find anything from this.


1+2

now we have a relation we can use.

#total of choc. = \(2(x-1)\)
#total of toffee = \(2(x)\)

\(2/3(2(x-1)= 3/5(2x)\) --- we can find x from here, once we have x, we can plug that value to 2(x-1)= #total number of choc.
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Re: Rasheed bought two kinds of candy bars, chocolate and toffee, that cam [#permalink]
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Re: Rasheed bought two kinds of candy bars, chocolate and toffee, that cam [#permalink]
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