Question

The Sum of digits of a two-digit number is 11. The given number is less than the number obtained by reversing the digits by 9. Find the number

Answer 1

Answer: 65

Step-by-step explanation:

Given:

Sum of two digit number = 11

Let unit’s digit be ‘x’

and tens digit be ‘y’,

then x+y=11…(i)

and number = x+10y

By reversing the digits,

Unit digit be ‘y’

and tens digit be ‘x’

and number =y+10x+9

Now by equating both numbers,

y+10x+9=x+10y

10x+y–10y–x=−9

9x–9y=−9

x–y=−1…(ii)

Adding (i) and (ii), we get

2x=10

x=10/2

=5

∴y=1+5=6

By substituting the vales of x and y, we get

Number = x+10y

=5+10×6

=5+60

=65

∴ The number is 65.

Nope that this helps

Pls mark me as the brainliest

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

The sum of the digits of two digit number is 11.

$\begin{gathered}\begin{gathered}\bf\: Let-\begin{cases} &\sf{digit \: at \: tens \: place \: be \: x} \\ \\ &\sf{digits \: at \: ones \: place \: be \: 11 - x} \end{cases}\end{gathered}\end{gathered}$

So,

Number formed = 10 × x + 11 - x = 10x + 11 - x = 9x + 11

Reverse number = 10 × (11 - x) + x = 110 - 10x + x = 110 - 9x

$\begin{gathered}\begin{gathered}\bf\: So-\begin{cases} &\sf{number \: formed = 9x + 11} \\ \\ &\sf{reverse \: number = 110 - 9x} \end{cases}\end{gathered}\end{gathered}$

According to statement

The given number is less than the number obtained by reversing the digits by 9.

$\rm :\longmapsto\:110 - 9x - (11 + 9x) = 9$

$\rm :\longmapsto\:110 - 9x - 11 - 9x = 9$

$\rm :\longmapsto\:99 - 18x = 9$

$\rm :\longmapsto\: - 18x = 9 - 99$

$\rm :\longmapsto\: - 18x = - 90$

$\bf\implies \:x = 5$

$\begin{gathered}\begin{gathered}\bf\: So-\begin{cases} &\sf{number \: formed = 9x + 11 = 45 + 11 = 56} \\ \\ &\sf{reverse \: number = 110 - 9x = 110 - 45 = 65} \end{cases}\end{gathered}\end{gathered}$

Hence,

• Two digit number is 56.