Let’s assume the digit at unit’s place is a and at ten’s place is b. Thus from the question, the number we need to find is $10b+a$.
From the question since the number is $4$ times the sum of the two digits. We can write,
$10b+a=4(a+b)$
⇒ $10b+a=4a+ 4b$
⇒ $4a+4b–10b-a=0$
⇒ $3a–6b=0$
⇒ $3(a–2b)=0$
⇒ $a–2b=0$ ……………… (i)
After reversing the digits, the new number formed is $10a+b$.
It is also given that if $18$ is added to the original number, the digits are reversed. Thus, we have
$(10b+a)+18=10a+b$
⇒ $10a+b-10b–a=18$
⇒ $9a–9b=18$
⇒ $9(a-b)=18$
⇒ $a–b=18/9$
⇒ $a-b=2$ …………. (ii)
Now, on solving equation (i) and (ii) we can find the value of a and b and thus the number.
On subtracting the equation (i) from equation (ii), we get;
$(a-b)–(a–2b)=2-0$
⇒ $a–b–a+2b=2$
⇒ $b=2$
Putting the value of b in the equation (i) to find a, we get
$a-2\times 2=0$
⇒ $a–4=0$
⇒ $a=4$
Thus, the required number is $10\times 2+4=24$.