Question

45 orange and 108 bananas are to be distributed among some boys equally what would be the maximum number of the boys?​

Answer 1

Answer

9 boys

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Step-by-step explanation:

Orange = 45

banana = 108

We will take HCF

45: 3×3×5

108: 2×2×3×3×3

HCF: 3×3 =9

hcf of 45 and 108 = 9

Well orange and banana can be divide equally in 9 boys

45÷9 = 5 oranges each

108÷9 = 12 banana each

All the best

Answer 2

Answer:

HCF of 45 and 108

Step-by-step explanation:

Factor of 45 = 3×3×5

Factor of 108 = 2×2×3×3×3

HCF of 45 and 108 is 9

Therefore there are maximum of 9 boys

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