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30 Dec 2020, 03:00
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
19% (01:57) correct
81%
(02:41)
wrong
based on 96
sessions
History
Date
Time
Result
Not Attempted Yet
Tom is going to move, he has three kinds of boxes: large, medium, and small. Now he has already 11 large boxes, each of which is either empty or filled with 6 medium boxes. Medium boxes are filled in the large boxes and each of them is either empty or filled with 6 small boxes. Small boxes are empty and filled in the medium boxes. How many boxes does Tom have?
(1) He has 56 empty boxes
(2) He has 42 small boxes
Are You Up For the Challenge: 700 Level Questions
(1) He has 56 empty boxes
(2) He has 42 small boxes
Are You Up For the Challenge: 700 Level Questions
30 Dec 2020, 09:56
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Bunuel
Large boxes = 11.
Let n of them be filled with 6 medium boxes. So medium boxes = 6n
Let m of the medium boxes be filled with 6 small boxes. So small boxes = 6m
Total boxes = \(11+6n+6m = 11+6(n+m)\)
So we are looking for value of n+m
(1) He has 56 empty boxes
Empty boxes are : 11-n large boxes, 6n-m medium boxes and all 6m small boxes.
So \(11-n+6n-m+6m=54.........11+5n+5m=11+5(n+m)=56.....n+m=9\)
Sufficient
(2) He has 42 small boxes
so 6m=42....m=7, bt nothing about n.
Insufficient
A
30 Dec 2020, 17:06
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Bunuel
We may use a table (see attachment) to clarify the quantities, let us assume x large boxes are filled and y medium boxes are filled, then we can extrapolate the data by filling in the table below. We can see in order to find the total amount of boxes, we would like to know \(x + y\).
Statement 1:
We have \(56 = 11 + 5x + 5y\) so we can solve for \(x + y\), sufficient.
Statement 2:
We can find y but not x. Insufficient.
Ans: A
Attachments
Table.png [ 10.99 KiB | Viewed 3400 times ]
