Question

The sum of 3 consecutive multiples of 9 is 999.Find these multiples...

Answer 1

Answer :

Let, Ist consecutive multiple of 9 = x
IInd consecutive multiple of 9 = x + 9
IIIrd consecutive multiple be 9 = x + 18

A. T. Q

x + x + 9 + x + 18 = 999

3x + 27 = 999

3x = 999 - 27

3x = 972

x = 972/3

x = 324

Hence, Ist consecutive multiple of 9 = x = 324
IInd consecutive multiple of 9 = x + 9 = 324 + 9 = 333
IIIrd consecutive multiple of 9 = x + 18 = 324 + 18 = 342

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Answer 2

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$Given \: that : \\ Sum \: of \: three \: consecutive \: multiples \: of \: 9 \: is \: 999 \\ \\ \: Let \: the \: first \: number \: = x \\ \\ Let \: the \: 2nd \: number \: = x + 1 \\ \\ Let \: the \: 3rd \: number \: = x + 2 \\ \\ Now ,\: Acc.to \: statement : \\ \\ = > 9(x) + 9(x + 1) + 9(x + 2) = 999 \\ \\ = > 9x + 9x + 9 + 9x + 18 = 999 \\ \\ = > 27x + 27 = 999 \\ \\ = > 27x = 999 - 27 \\ \\ = > 27x = 972 \\ \\ = > x = 36 \\ \\ So, \\ The \: first \: number \: of \: multiple \: of \: 9 \: is \: 9(x) = 9(36) = 324 \\ \\ The \: 2nd \: number \: of \: multiple \: of \: 9 \: is \: 9(x + 1) = 333 \\ \\ The \: 3rd \: number \: of \: multiple \: of \: 9 \: is \: 9(x + 2) = 342$
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