Question

Four boxes of chocolate contain 10,12,15 and 20 chocolates respectively. For his birthday Raman wants to buy equal number of chocolates each kind.He needs to buy complete boxes.What is the minimum number of chocolates of each kind he should buy?How many boxes of each kind should he buy?

Answer 1

Answers:

6 boxes of type A

5 boxes of type B

4 boxes of type C

3 boxes of type D

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Step-by-step explanation:

We need to find the LCM (lowest common multiple)

Find the prime factorization of 10,12,15 and 20:

• 10 = 2*5
• 12 = 2*2*3
• 15 = 3*5
• 20 = 2*2*5

The unique prime factors are: 2, 3, 5

The prime 2 shows up at most twice when we had 2*2 as the piece. So 2*2 = 4 is one of the factors of the LCM. The other pieces 3 and 5 are also factors of the LCM

LCM = 2*2*3*5 = 4*3*5 = 12*5 = 60

The LCM of {10,12,15,20} is 60. This can be confirmed by listing out the multiples of 10,12,15, and 20 to note that 60 is the smallest common multiple of the four values.

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From here, divide the LCM 60 over each of 10,12,15 and 20.

• 60/10 = 6
• 60/12 = 5
• 60/15 = 4
• 60/20 = 3

Those results tell us how many boxes of each type are needed so you buy 60 chocolates of each type.

Example: Type A has 10 chocolates per box. Buying 6 boxes gives 6*10 = 60 chocolates of type A.