Three company of soldiers containing 120, 192, and 144 soldiers are to

Question

Three company of soldiers containing 120, 192, and 144 soldiers are to be broken down into smaller groups such that each group contains soldiers from one company only and all the groups have equal number of soldiers. What is the least number of total groups formed?

A. 8
B. 12
C. 19
D. 24
E. 38

A. 8
B. 12
C. 19
D. 24
E. 38

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nick1816 · Answer

HCF (120,192,144)= 24

Least number of such groups= (120/24) + (192/24)+ (144/24)= 5+8+6= 19

Archit3110 · Answer

for 120,192 & 144 l HCF ; 2^3*3 ; 24
120/24+192/24+144/24 = 5+8+6 ; 19
IMO C

Kinshook · Answer

HCF (120,192,144) = 24

Number of groups (min) = 120/24 + 192/24 + 144/24 = 5 + 8 + 6 = 19

IMO C

ScottTargetTestPrep · Answer

We need to find the greatest common factor (GCF) of 120, 192, and 144. First we can prime factorize them:

120 = 12 x 10 = 2^3 x 3^1 x 5^1

192 = 2 x 96 = 2 x 2 x 48 = 2^6 x 3^1

144 = 12 x 12 = 2^4 x 3^2

So, the GCF of 120, 192, and 144 is 2^3 x 3^1 = 24, so there will be 24 soldiers in each group.

Thus, the smallest number of groups is 120/24 + 192/24 + 144/24 = 5 + 8 + 6 = 19.

Answer: C

jkolachi · Answer

We need to find the biggest number that divides all three. Find prime factors and find numbers that are common i.e....

120 --> (2^3) * 3 * 5. (5 not common so ignore) 
144 --> (2^4) * 3^2
192 --> (2^5) * 3

What's common amongst all three? 2^3 * 3 = 24.

Divide 120, 144, & 192 by 24. 
5 + 6 + 8 = 19.

bumpbot · Answer

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