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04 Feb 2020, 00:54
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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
Updated on: 04 Feb 2020, 02:28
Originally posted by GMATinsight on 04 Feb 2020, 01:07.
Last edited by GMATinsight on 04 Feb 2020, 02:28, edited 1 time in total.
Last edited by GMATinsight on 04 Feb 2020, 02:28, edited 1 time in total.
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DisciplinedPrep
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
Six different sweet varieties of count 32, 216, 136, 88, 184, 120
Since every pack should have count of each varieties and the boxes should have equal number of sweets in each box
So the number of box must be a divisor of each count 32, 216, 136, 88, 184, 120
In order to minimize the number of boxes, we must find Greatest Common divisor
i.e. HCF (count 32, 216, 136, 88, 184, 120 )
\(32 = 2^5\)
\(216 = 2^3*3^3\)
\(136 = 4*34 = 2^3*17\)
\(88 = 8*1 = 2^3*11\)
\(184 = 4*46 = 2^3*23\)
\(120 = 8*15 = 2^3*3*5\)
i.e. HCF (count 32, 216, 136, 88, 184, 120 )\(= 2^3 = 8\)
Total (32/8) + (216/8) + (136/8) + (88/8) + (184/8) + (120/8) = 4 + 27 + 17 + 11 + 23 + 15 = 97
ANswer: Option C
04 Feb 2020, 01:35
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GMATinsight
DisciplinedPrep
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
Six different sweet varieties of count 32, 216, 136, 88, 184, 120
Since every pack should have count of each varieties and the boxes should have equal number of sweets in each box
So the number of box must be a divisor of each count 32, 216, 136, 88, 184, 120
In order to minimize the number of boxes, we must find Greatest Common divisor
i.e. HCF (count 32, 216, 136, 88, 184, 120 )
\(32 = 2^5\)
\(216 = 2^3*3^2\)
\(136 = 4*34 = 2^3*17\)
\(88 = 8*1 = 2^3*11\)
\(184 = 4*46 = 2^3*23\)
\(120 = 8*15 = 2^3*3*5\)
i.e. HCF (count 32, 216, 136, 88, 184, 120 )\(= 2^3 = 8\)
Total (32/8) + (216/8) + (136/8) + (88/8) + (184/8) + (120/8) = 4 +
ANswer: Option C
DisciplinedPrep There seems a typo error in option C
GMATinsight ,
You have made calculation mistake. I have marked it. Pls correct.
IMO-C
Regards!!
