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22 Oct 2025, 00:40
Kudos
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
55% (03:07) correct
45%
(02:29)
wrong
based on 73
sessions
History
Date
Time
Result
Not Attempted Yet
Six different sweet varieties of count 32, 216, 136, 88, 184, 120 were ordered for a particular occasion. They need to be packed in such a way that each box has the same variety of sweet and the same number of sweets in each box. What is the minimum number of boxes required to pack?
A. 48
B. 64
C. 97
D. 120
E. 129
A. 48
B. 64
C. 97
D. 120
E. 129
22 Oct 2025, 01:07
Kudos
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The core of this question is figuring out the biggest possible number of sweets you can put in each box, while following the rules. Since every box must have the same number of sweets, that number has to be a factor of all the different sweet counts (32, 216, 136, etc.). To use the minimum number of boxes, we need to put the maximum number of sweets in each one.
This means we're looking for the Highest Common Factor (HCF) of all the numbers, which is 8.
32 \ 8 = 4
216 \ 8 = 27
136 \ 8 = 17
88 \ 8 = 11
184 \ 8 = 23
120 \ 8 = 15
It works for all of them! So, this means each box will contain exactly 8 sweets.
Now that we know each box has 8 sweets, we just need to find out how many boxes are needed for each variety and add them all up
Total Boxes = 4 + 27 + 17 + 11 + 23 + 15 = 97
So, the minimum number of boxes required is 97. This corresponds to option C.
This means we're looking for the Highest Common Factor (HCF) of all the numbers, which is 8.
32 \ 8 = 4
216 \ 8 = 27
136 \ 8 = 17
88 \ 8 = 11
184 \ 8 = 23
120 \ 8 = 15
Now that we know each box has 8 sweets, we just need to find out how many boxes are needed for each variety and add them all up
Total Boxes = 4 + 27 + 17 + 11 + 23 + 15 = 97
So, the minimum number of boxes required is 97. This corresponds to option C.
Bunuel
22 Oct 2025, 12:04
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Bunuel
The six different sweets count 32, 216, 136, 88, 184, 120.
The only criteria is : same number of sweets in each box.
we need to find the Highest Common Factor (HCF).
32 = 8 * 4
216 = 8 * 27
136 = 8 * 17
88 = 8 * 11
184 = 8 * 23
120 = 8 * 15
The HCF is 8.
The number of boxes are the bolded ones. Hence, the total minimum number of Boxes = 4+27+17+11+23+15 = 97
Option C
