Question

The sum of the didgit of 2 digit no. Is 11 the number obtained by interchanging the digits exceeds the original no. By 27 find the number

Answer 1

Answer:

let the digit in unit place be x

let the digit in tens place be y

the number is written as 10x + y

interchanging the digits it becomes 10y + x

which is 27 more than the original number

sum of the two digits x + y = 11

\begin{gathered}10x + y + 27 = 10y + x \\ 10x + y - (10y + x) = - 27 \\ 10x + y - 10y - x = - 27 \\ 9x - 9y = - 27 \\ 9(x - y) = - 27 \\ x - y = \frac{ - 27}{9} \\ x - y = - 3\end{gathered}

10x+y+27=10y+x

10x+y−(10y+x)=−27

10x+y−10y−x=−27

9x−9y=−27

9(x−y)=−27

x−y=

9

−27

x−y=−3

x + y = 11

x - y = -3

adding the above equation we get

2x = 8

x = 8/2

x = 4

substitute x in x + y = 11

x + y = 11

4 + y = 11

y = 11 - 4

y = 7

the required number is 10x + y

= 10(4) + 7

= 40 + 7

= 47

hope you get your answer