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A case contains c cartons. Each carton contains b boxes, and each box contains 100 paper clips. How many paper clips are contained in 2 cases?
(A) 100bc
(B) 100b/c
(C) 200bc
(D) 200b/c
(E) 200/bc
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(A) 100bc
(B) 100b/c
(C) 200bc
(D) 200b/c
(E) 200/bc
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SOLUTION
A case contains c cartons. Each carton contains b boxes, and each box contains 100 paper clips. How many paper clips are contained in 2 cases?
(A) 100bc
(B) 100b/c
(C) 200bc
(D) 200b/c
(E) 200/bc
Each case contains \(100bc\) paper clips, hence 2 cases contain \(200bc\) paper clips.
Answer: C.
_________________
A case contains c cartons. Each carton contains b boxes, and each box contains 100 paper clips. How many paper clips are contained in 2 cases?
(A) 100bc
(B) 100b/c
(C) 200bc
(D) 200b/c
(E) 200/bc
Each case contains \(100bc\) paper clips, hence 2 cases contain \(200bc\) paper clips.
Answer: C.
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General Discussion
picking numbers
c = 5
b = 2
200 paperclips per carton
200 * 5 = 1000 per case
A. 1000
B. 200/5 = 40
C. 2000
D. 400/5 = 80
E. 20
c = 5
b = 2
200 paperclips per carton
200 * 5 = 1000 per case
A. 1000
B. 200/5 = 40
C. 2000
D. 400/5 = 80
E. 20
A.
To complete your answer, there are 2 cases so answer (C) is correct.
arthuro69
Ah yes. Disappointing on my part
PareshGmat
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Case >> Carton >> Box >> Clips
Just multiplication required
Its 2 * (100 * b * c) = 200bc
Answer = C
Just multiplication required
Its 2 * (100 * b * c) = 200bc
Answer = C
Posts: 6,727
Bunuel
Bunuel has already demonstrated an algebraic approach, so let's try solving it by picking some numbers (aka using the INPUT-OUTPUT approach).
Let b = 3 and c = 5
So, a case contains 5 cartons. Each carton contains 3 boxes, and each box contains 100 paper clips.
So, the number of paper clips in 1 case = (100)(5)(3) = 1500
How many paper clips are contained in 2 cases?
The number of paper clips in 2 cases = 3000
So, we're looking for the answer choice so that when we INPUT b = 3 and c = 5, we get an OUTPUT of 3000
Now check the answer choices:
(A) 100(3)(5) = 1500 ELIMINATE
(B) 100(3)/(5) = 60 ELIMINATE
(C) 200(3)(5) = 3000 KEEP!!
(D) 200(3)/(5) = 120 ELIMINATE
(E) 200/(3)(5) = 200/15 ELIMINATE
Answer: C
RELATED RESOURCES
As you can see, Bunuel and I have demonstrated the two methods (Algebraic and Input-Output) for solving a question type I call Variables in the Answer Choices.
If you'd like more information on these approaches, we have some free videos:
- - Variables in the Answer Choices - https://www.gmatprepnow.com/module/gmat- ... /video/933
- Tips for the Algebraic Approach - https://www.gmatprepnow.com/module/gmat- ... /video/934
- Tips for the Input-Output Approach - https://www.gmatprepnow.com/module/gmat- ... /video/935
Cheers,
Brent
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Expert reply
Attached is a visual that should help.
Attachments
Screen Shot 2016-04-11 at 5.06.44 PM.png [ 71.7 KiB | Viewed 45470 times ]
ScottTargetTestPrep
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Expert reply
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Bunuel
Solution:
We are given that a case contains c cartons. Each carton contains b boxes, and each box contains 100 paper clips. Thus we can say the following:
c x b x 100 = 100bc paper clips per case. Thus, 2 cases would contain 2 x 100bc = 200bc paper clips.
The answer is C.
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I think you can draw this out (really helped me)
1 case --> 2 cartons --> 1 carton = 3 boxes --> 1 box = 100 pc
>> From here, there are 300 pc in 3 boxes (AKA 1 carton). Since you have 2 cartons here, in 1 case, you have 600 pc.
>> You want 2 cases, so you have 1,200 pc.
Plug in A/C w/ variables you defined above:
c=2 (cartons)
b=3 (boxes)
When you plug in, should give you: 1,200.
C) 200bc --> 200(3)(2) --> 200(6) --> 1,200
1 case --> 2 cartons --> 1 carton = 3 boxes --> 1 box = 100 pc
>> From here, there are 300 pc in 3 boxes (AKA 1 carton). Since you have 2 cartons here, in 1 case, you have 600 pc.
>> You want 2 cases, so you have 1,200 pc.
Plug in A/C w/ variables you defined above:
c=2 (cartons)
b=3 (boxes)
When you plug in, should give you: 1,200.
C) 200bc --> 200(3)(2) --> 200(6) --> 1,200
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