Rasheed bought two kinds of candy bars

[guest]

Can someone explain the solution to this data sufficiency problem?

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 bar

[RonPurewal]

since both types of bars come 2 to a box, we can rephrase the question to just this: how many chocolate bars did he buy? (this works because the answer is just going to be double the number of packages; since it's data sufficiency, there's no good reason to waste the time doing the extra step).

so, in other words, we can ignore 'packages' and think in terms of bars instead.

--

(1)
rasheed bought 2 fewer chocolate bars than toffee bars.
so chocolate bars = x (this is what we want to find)
and toffee bars = x + 2
all we know is:
he hands out (2x/3) chocolate bars and (3x/5) toffee bars
INSUFFICIENT

(2) if c = # of chocolate bars and t = # of toffee bars, we know 2c/3 = 3t/5, which cross-multiplies to 10c = 9t.
this means that the ratio c : t is 9 : 10, but we have no idea what the actual numbers in the ratio are.
INSUFFICIENT

together:
taking both statements together, we have 10(x) = 9(x + 2), or (x) : (x + 2) = 9 : 10.
this is an equation with a unique solution.
SUFFICIENT (note that there is no need to solve; simply noting that there is 1 solution is good enough)

answer = c

[zarak_khan]

Hi Ron,

This is how I solved this 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 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 bar

Question stem --> Hands out 2/6 C and 3/10 T. What is C?

1) C + 1 = T (Insufficient)
2) 2/6 C = 3/10 T --> 20/18 = T/C --> 10/9 = T/C (Insufficient)
C) 10/9 = C+1/C --> 10C = 9C + 9 --> C = 9 (Sufficient)

Thanks!

[RonPurewal]

Hi Ron,

This is how I solved this 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 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 bar

Question stem --> Hands out 2/6 C and 3/10 T. What is C?

1) C + 1 = T (Insufficient)
2) 2/6 C = 3/10 T --> 20/18 = T/C --> 10/9 = T/C (Insufficient)
C) 10/9 = C+1/C --> 10C = 9C + 9 --> C = 9 (Sufficient)

Thanks!

that works as well.

the difference between your solution and mine is that you used variables to stand for the number of packages, while i used variables to stand for the number of bars.

both are good.

[natarajan.suman]

Ron,

I realize this post is rather old, but just in case I thought I would try.

I am confused about why everyone is so quick to eliminate statement one alone. Yes, we only have 1 equation and 2 variables. HOWEVER, we know that the numbers of packages must be whole numbers, and we know that the number of chocolate packages is a multiple of 3 and the number of toffee packages is a multiple of 5. Therefore, the only set of 2 numbers which also satisfy the condition that C+1 = T, must be C=9, T=10.

I encountered another case like this in another problem dealing with whole numbers. Given two prices for pencils and the total cost of both pencils. See link: marta-bought-several-pencils-t5144.html

Isn't this the same type of reasoning? Please let me know if that is not the case.

Thanks!

[lj6871849]

Ron,

I realize this post is rather old, but just in case I thought I would try.

I am confused about why everyone is so quick to eliminate statement one alone. Yes, we only have 1 equation and 2 variables. HOWEVER, we know that the numbers of packages must be whole numbers, and we know that the number of chocolate packages is a multiple of 3 and the number of toffee packages is a multiple of 5. Therefore, the only set of 2 numbers which also satisfy the condition that C+1 = T, must be C=9, T=10.

I encountered another case like this in another problem dealing with whole numbers. Given two prices for pencils and the total cost of both pencils. See link: marta-bought-several-pencils-t5144.html

Isn't this the same type of reasoning? Please let me know if that is not the case.

Thanks!

Hey - I guess by your logic 24 = 25-1 ( C = T-1) will also work.

so we have (9,10) or (24,25) so Statement B forces (9,10) for (c, t)

[jlucero]

Ron,

I realize this post is rather old, but just in case I thought I would try.

I am confused about why everyone is so quick to eliminate statement one alone. Yes, we only have 1 equation and 2 variables. HOWEVER, we know that the numbers of packages must be whole numbers, and we know that the number of chocolate packages is a multiple of 3 and the number of toffee packages is a multiple of 5. Therefore, the only set of 2 numbers which also satisfy the condition that C+1 = T, must be C=9, T=10.

I encountered another case like this in another problem dealing with whole numbers. Given two prices for pencils and the total cost of both pencils. See link: marta-bought-several-pencils-t5144.html

Isn't this the same type of reasoning? Please let me know if that is not the case.

Thanks!

Hey - I guess by your logic 24 = 25-1 ( C = T-1) will also work.

so we have (9,10) or (24,25) so Statement B forces (9,10) for (c, t)

Good response. Notice that there will be an infinite number of values that could work as we could add a multiple of 3 and 5 (15) to your original values forever- 9,10; 24,25; 39,40; etc.

And that's the trick to identifying why this is different than the pencil problem you linked to. That problem gives us a cap on the total amount spent. Limits (either high or low) are often a hint that you'll need to methodically solve and see if only one possible solution exists.