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23 Jun 2022, 14:00
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Difficulty:
95%
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Question Stats:
18% (02:48) correct
82%
(02:42)
wrong
based on 38
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Chocolates are packed in boxes that measure six inches long, three inches wide, and two inches high. If four individual boxes are packaged together in bulk by either arranging them in a single row or stacking them, what could be the total surface area of the bulk unit?
I. 168
II. 180
III. 192
A. I only
B. II only
C. III only
D. I and III only
E. I, II and III
I. 168
II. 180
III. 192
A. I only
B. II only
C. III only
D. I and III only
E. I, II and III
24 Jun 2022, 09:56
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Bunuel
Chocolates are packed in boxes that measure six inches long, three inches wide, and two inches high. If four individual boxes are packaged together in bulk by either arranging them in a single row or stacking them, what could be the total surface area of the bulk unit?
I. 168
II. 180
III. 192
A. I only
B. II only
C. III only
D. I and III only
E. I, II and III
I. 168
II. 180
III. 192
A. I only
B. II only
C. III only
D. I and III only
E. I, II and III
Breaking Down the Info:
There are 3 formations by stacking them or arranging them in a single row. Connect the lengths to get 24 * 3 * 2, connect the widths to get 6 * 12 * 2, or stack to get 6 * 3 * 8.
Formation 1 gives SA of: 2*(24*(3 + 2) + 6) = 2*126 = 252
Formation 2: 2*(12*(6 + 2) + 12) = 2*12*9 = 216
Formation 3: 2*(8*(6+3) + 6*3) = 2*81 = 162
I got none of the values but maybe the question meant "3 boxes" stacked together as I found values I and III working with 3 boxes.
Answer: ?
24 Jun 2022, 13:39
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Bunuel
Chocolates are packed in boxes that measure six inches long, three inches wide, and two inches high. If four individual boxes are packaged together in bulk by either arranging them in a single row or stacking them, what could be the total surface area of the bulk unit?
I. 168
II. 180
III. 192
A. I only
B. II only
C. III only
D. I and III only
E. I, II and III
I. 168
II. 180
III. 192
A. I only
B. II only
C. III only
D. I and III only
E. I, II and III
So, I went with the stacking approach first and I think the key here is that its not mentioned explicitly in the question how boxes have been stacked. As far as rectangular boxes go, they could be stacked all 4 on top of each other or in pairs of 2 (like 2 rows of 2 boxes each)
Case 1: All 4 stacked on top of each other
So, in this case, the Length and Width remain the same but height increases to 4 times 2 so 8
Therefore, A=2(wl+hl+hw) = 2(3x6 + 8x6 + 8x3) = 180 So, options A,C,D are out
Case 2: 2 rows of 2 boxes stacked on top of each other
In this case, Length becomes twice of original and height becomes twice of original and width remains same
Therefore, A=2(wl+hl+hw) = 2(3x12 + 4x12 + 4x3) = 192 So. option B is also out
We are left with Option E, and we did not need to test for the last value so we can save some time as well
