Question

A certain number between 10 and 100 is 8 times the sum of its digits and if 45 be subtracted from it the digits will be reversed .find the number

Answer 1

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Answer 2

Answer:

$\red { Required \: number }\green {= 72 }$

Step-by-step explanation:

$Required \: number \\= number \: between \:10\:and \: 100 \\= two \:digited \: number$

$Let \: ten's \: place \: digit = x$

$Units \: place \: digit = y$

$Original \: number = 10x + y$

$Reversing \: the \: digits \\ the \: new \\number \: so \: formed = 10y+x$

/* According to the problem given

$10x + y = 8( x + y )$

$\implies 10x + y = 8x + 8y$

$\implies 10x - 8x = 8y - y$

$\implies 2x = 7y \: ---(1)$

$10x + y - 45 = 10y + x$

$\implies 10x - x + y - 10y = 45$

$\implies 9x - 9y = 45$

/* Divide each term by 9 , we get

$\implies x - y = 5$

$\implies x - 5 = y \: ---(2)$

/* Substitute value of y in equation (1) , we get

$2x = 7(x - 5)$

$\implies 2x = 7x - 35$

$\implies 35 = 7x - 2x$

$\implies 35 = 5x$

$\implies \frac{35}{5} = x$

$\implies x = 7$

/* Put x = 7 in eqution (2), we get

$y = 7 - 5 = 2$

$\red { Required \: number } \\= 10x + y\\= 10\times 7 + 2 \\= 70 + 2 = 72$

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