Question

On reversing the digits of a two digit number , the number obtained is 9 less than three times the original number. If difference of these two numbers is 45. Then find the original number.​

Answer 1

Answer:

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Step-by-step explanation:

hope this help full for you

Answer 2

Answer:

Let the ones place digit be x

and the tens place digit be y

Thus, the number be 10y+x

Reversed number be 10x+y

ATQ,

$10x + y = 3 \times (10y + x) - 9 \\ \\ 10x + y= 30y + 3x - 9 \\ \\ 7x - 29y + 9 = 0 - - - - - (i)$

$10x + y - (10y + x) = 45 \\ \\ 10x + y - 10y - x = 45 \\ \\ 9x - 9y = 45 \\ \\ x = 5 + y - - - - - (ii)$

$7x - 29y + 9 = 0 \\ \\ 7(5 + y) - 29y + 9 =0 \\ \\ 35 + 7y - 29y + 9 = 0 \\ \\ 22y = 44 \\ \\ { \large{ \boxed{y = 2}}} \\ \\{ \large{ \boxed{ x = 7}}}$

The original number is 27