Question

The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.[4 MARKS]

Answer 1

Concept : 1 Mark
Application : 1 Mark
Calculation : 2 Marks

Let the two-digit number be “ab”

Then the number is of the form 10a + b

According to given condition,

$a+b=9\dots \dots \left(i\right)$

Also,

$9(10a+b)=2(10b+a)$

$90a+9b=20b+2a$

$\Rightarrow 8a–b=0\dots \dots \left(ii\right)$

Solving (i) and (ii),

$a=9-b$ [From (i)]

Substituting value of a in (ii)

$8a–b=0$

$8(9-b)-b=0$

$-9b=-72\Rightarrow b=8$

Now, $a=9-b$

$\Rightarrow a=9-8=1$

$\therefore a=1,b=8$

$\therefore$ required number is 18