Question

the sum of a two digit number and the number by interchanging its digits is 110.if 10 is subtracted from the first number ,the new number is 4 more than 5 times the sum of the digits of the first number.find the number

Answer 1

Let the number is 10x+y. Then by the given condition,
10x+y+10y+x=110
or, 11x+11y=110
or, x+y=10 --------------(1) and
10x+y-10=4+5(x+y)
or, 10x-5x+y-5y=4+10
or, 5x-4y=14 ------------(2)
Multiplying (1) with 4 and adding with (2) we get,
9x=54
or, x=6
Putting in (1),
y=10-x
or, y=10-6
or, y=4
Therefore the required number is -
(10×6)+4=60+4=64 Ans.

Answer 2

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

Assumption

Number is 10p + n

Hence

$\textbf{\underline{Situation}}$

10p + n + 10n + p = 110

$\textbf{\underline{We\;may\;also\;write}}$

11p + 11n = 110

11(p + n) = 110

$\tt{\rightarrow p+n=\dfrac{110}{11}}$

p + n = 10 ........ (1)

10p + n - 10 = 4 + 5(p + n)

10p - 5 p + n - 5n = 4 + 10

5p - 4n = 14 ....... (2)

$\textbf{\underline{Multiply\;(1)\;by\;(4)\;and\;add\;with\;(2)}}$

9p = 54

p = 6

$\textbf{\underline{Substitute\;the\;value\;in\;(1)}}$

n = 10 - p

n = 10 - 6

n = 4

$\textbf{\underline{Therefore}}$

(10 × 6) + 4

= 60 + 4

= 64

${\boxed{\sf\:{Required\;no\;=64}}}$