Question

The sum of a two digit number and the number obtained by reversing the order of its digits is 121, and the two digits differ by 3. Find the number.

Answer 1

Let one no. be x
Another no=x+3
Let x be the tens digit and x+3 be the ones digit
actual no=11x+3
When interchanged no.=11x+30
ATQ
=>11x+3+11x+30=121
=>22x+33=121
=>22x=88
=>x=4
Actual no.=11x4+3=44+3=47

Answer 2

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

Assumption

$\textbf{\underline{Ones\;digit\;=\;p}}$

$\textbf{\underline{Tens\;digit\;=\;c}}$

${\boxed{\sf\:{Before\;Reversing}}}$

= 10c + p

${\boxed{\sf\:{After\;Reversing}}}$

= 10p + c

$\textbf{\underline{First\;Situation}}$

⇒ (10c + p) + (10p + c) = 121

⇒ 11p + 11c = 121

⇒ p + c = 11 …….(1)

$\textbf{\underline{Second\;Situation}}$

${\boxed{\sf\:{Digit\;differ\;by\;three}}}$

p - c = 3 ……….(2)

${\boxed{\sf\:{Adding\;(1)\;and\;(2)}}}$

⇒ 2p = 14

⇒ p = 7

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

⇒ p + c = 11

⇒ 7 + c = 11

⇒ c = 4

$\textbf{\underline{Hence,\;we\;get:-}}$

p = 4

c = 7

$\textbf{\underline{Therefore}}$

$\Large{\boxed{\sf\:{Two\;digit\;No.=74\;or\;47}}}$