Rasheed bought two kinds of candy bars, chocolate and toffee, that cam

Question

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.

The term that thew me off in this question is word " Packages ".

Bunuel · Accepted Answer

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.

EMPOWERgmatRichC · Answer

Hi All,

This question can certainly be solved as a "system" question (as Bunuel showed). If you don't see that "pattern" though, there IS another approach (although it will take more note-taking and some extra work to get to the correct answer) and there are some useful Number Properties in this prompt that would save you some time.

We're told that candy bars come in packs of 2; this is important because it means that the TOTAL number of each type of candy bar will be EVEN.

Rasheed handed out 2/3 of the chocolate bars; since we can't have an ODD number of chocolate bars, we know that the total number of chocolate bars will be a MULTIPLE of 6 (and the total that he HANDED OUT will be a MULTIPLE OF 4). The number of PACKAGES of chocolate bars MUST be a MULTIPLE of 3.

Rasheed handed out 3/5 of the toffee bars; since we can't have an ODD number of toffee bars, we know that the total number of toffee bars will be a MULTIPLE of 10 (and the total that he HANDED OUT will be a MULTIPLE OF 6). The number of PACKAGES of toffee bars MUST be a MULTIPLE of 5.

To make the above easier to reference, here's the essential info:

Chocolate:
Total packages = multiple of 3
Total bars = Multiple of 6
Total handed out = multiple of 4

Toffee:
Total packages = multiple of 5
Total bars = multiple of 10
Total handed out = multiple of 6

We're asked how many PACKAGES of chocolate bars Rasheed bought?

Fact 1: Rasheed bought 1 fewer package of toffee bars than chocolate bars.

Using the deductions from earlier, the number of packages of chocolate MUST be a multiple of 3 and the number of packages of toffee MUST be a multiple of 5...

9 packages of chocolate and 10 packages of toffee
OR
24 packages of chocolate and 25 packages of toffee
Fact 1 is INSUFFICIENT

Fact 2: Rasheed handed out the same NUMBER of each type of candy bar.

Again, using the deductions from earlier, we need a MULTIPLE of 4 that EQUALS a MULTIPLE of 6....so we need a MULTIPLE of 12 bars given of each type....

We could have (among many options)....
12 bars given --> 9 packages of chocolate
24 bars given --> 18 packages of chocolate
Fact 2 is INSUFFICIENT

Combined, we can use the limitations in each of the two Facts against one another:

Using the "multiple of 12s" from Fact 2, we can do a quick comparison...
12 bars given --> 9 packs chocolate, 10 packs toffee --> difference is 1 PACKAGE
24 bars given --> 18 packs chocolate, 20 packs toffee --> difference is 2 PACKAGES
36 bars given --> 27 packs chocolate, 30 packs toffee --> difference is 3 PACKAGES
Etc.
The pattern here proves that as Rasheed gives out MORE chocolate bars, the DIFFERENCE in the number of packages INCREASES.

Thus, there's only 1 situation in which both of the Facts combine....When Rasheed buys 9 packs of chocolate bars.
Combined, SUFFICIENT

Final Answer:  C

GMAT assassins aren't born, they're made,
Rich

keats · Answer

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!

MathRevolution · Answer

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 [ 3.66 KiB | Viewed 63857 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.

Ndkms · Answer

Best answer hands down.... Thanks Bunuel

devctg · Answer

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

nikitamaheshwari · Answer

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

HaileyCusimano · Answer

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!

Basshead · Answer

(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.

ueh55406 · Answer

(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.

GulfTube · Answer

Let choc bars = x & toffee bars = y
given out: (2/3)x & (3/5)y
 Q: What is x/2 = ?

i) x/2 = (y/2)-1
y is unknown so not sufficient

ii) (2/3)x = (3/5)y
y is unknown so insufficient

together: two simultaneous equations with 2 unknowns is sufficient.  C

Rasheed bought two kinds of candy bars, chocolate and toffee, that cam

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