Question

Anyone answer this question

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Answer 1

Answer:

13,14,15

Step-by-step explanation:

Let the smallest of the three number be n.

Then the other numbers are n+1 and n+2 since the numbers are consecutive.

Now according to the question

(n/3)+((n+1)/4)+((n+2)/5) should be at most 20

that means,

n/3 + (n+1)/4+ (n+2)/5 <= 20

Taking LCM we get

((60*n)+15*(n+1)+12*(n+2))/60 <= 20

(60*n + 15*n +15+ 12*n +24) <= 20*60

(87*n+ 39) <= 1200

87*n <= 1161

n<= 13.34

n=13 since we need an integer.

Therefore the required numbers are:

13,14,15