Question

A number consist of two digits whose product is 8. If the digits are interchanged the resulting number will exceed the original one by 18. Find the number

Answer 1

Answer: 24

Step-by-step explanation:

number = 10x + y

10y +x = 10x+y +18

⇒ 9y -9x = 18  ⇒ y - x = 2 ⇒ y = x+2

xy= 8 ⇒ x(x+2) = 8 ⇒ $x^2 + 2x - 8 = 0$ ⇒ (x+4)(x-2) =0

x-2 = 0 or x +4 =0

x = 2 or x = -4, x can not be negative so

x = 2 then y = x+2 = 4

number = 24

Answer 2

Solution :

$\bf{\green{\underline{\bf{Given\::}}}}$

A number consist of two digit whose product is 8. If the digits are interchanged the resulting number will exceed the original one by 18.

$\bf{\green{\underline{\bf{To\:find\::}}}}$

The number.

$\bf{\green{\underline{\bf{Explanation\::}}}}$

Let the number be;

$\boxed{\bf{10r+m}}}}$

Let the ten's place be r

Let the one's place be m

$\dag\underline{\bf{The\:reversed\:number=10m+r}}}}$

A/q

$\implies\sf{rm=8.....................(1)}$

&

$\implies\sf{10r+m+18=10m+r}\\\\\\\implies\sf{10r-r+m-10m=-18}\\\\\\\implies\sf{9r-9m=-18}\\\\\\\implies\sf{9(r-m)=-18}\\\\\\\implies\sf{r-m=\cancel{\dfrac{-18}{9} }}\\\\\\\implies\sf{r-m=-2}\\\\\\\implies\sf{r=-2+m....................(2)}$

Putting the value of r in equation (1),we get;

$\implies\sf{-2+m(m)=8}\\\\\implies\sf{-2m+m^{2} =8}\\\\\implies\sf{m^{2} -2m-8=0}\\\\\implies\sf{m^{2} -4m+2m-9=0\:\:\:\:[factorise]}\\\\\implies\sf{m(m-4)+2(m-4)=0}\\\\\implies\sf{(m-4)(m+2)=0}\\\\\implies\sf{m-4=0\:\:\:Or\:\:\:m+2=0}\\\\\implies\sf{\pink{m=4\:\:\:Or\:\:\:m\neq -2}}$

Putting the value of m in equation (2),we get;

$\implies\sf{r=-2+4}\\\\\implies\sf{\pink{r=2}}$

Thus;

$\underbrace{\sf{The\:original\:number=(10r+m)=10(2)+4=20+4=\boxed{24}}}}$