Question

The sum of a 2-digit number is 7. If the digits are reversed, the number formed is 9 less than the original number. Find the number.

Answer 1

$\underline{\textbf{Given:}}$

$\textsf{Sum of digits of 2 digit number is 7}$

$\textsf{When the digits are reversed, the number formed}$

$\textsf{is 9 less than the original number}$

$\underline{\textbf{To find:}}$

$\textsf{The number}$

$\underline{\textbf{Solution:}}$

$\textsf{Let the two digit number be 10x+y}$

$\textsf{As per given dat,}$

$\textsf{Sum of digits is 7}$

$\implies\mathsf{x+y=7}$

$\implies\mathsf{y=7-x}$

$\textsf{When the digits are reversed,}$

$\mathsf{10y+x=10x+y-9}$

$\mathsf{-9x+9y=-9}$

$\mathsf{-x+(7-x)=-1}\;\;\;\;\textsf{(Using y=7-x)}$

$\mathsf{-2x=-8}}$

$\implies\mathsf{x=4}$

$\mathsf{when\;x=4,\;y=7-x}$

$\mathsf{y=7-4}$

$\implies\mathsf{y=3}$

$\textbf{Now,}$

$\mathsf{10x+y=10(4)+3=40+3=43}$

$\therefore\textbf{The required number is 43}$

Answer 2

As we know, two digits numbers can be written as,

$xy=10\times x+y$

Given: The sum of a 2-digit number is 7.

$x+y=7$.....(1)

If the digits are reversed, the number formed is 9 less than the original number.

Therefore,

$yx=xy-9$

$10\times y+x=10\times x+y-9$

$9y=9x-9\\y=x-1$

$x-y=1$.....(2)

Adding eq(1) and eq(2),

$x+y=7\\x-y=1$

             

$2x=8$

 $x=4$

$y=7-x\\y=7-4\\y=3$

Hence the required number is 43.

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