Question

In a two digit number, the digit in ten's twice the digit in units number formed by reversing the digits 36 less than the number. Find the number?​

Answer 1

let the no=10x +y

x=2y

x-2y=0.....1

reversed=10y+x

A.T.Q

10x+y+36=10y+x

9x-9y=-36

x-y=4....2

hope you like it bro

Answer 2

Answer

As per the question :-

• The digit in ten's place is twice the digit in ones place

• Number formed by reversing is 36 less than The real number

♦ Let number at ten's place be 10x and number at ones place be y

♦ By using aboves information

x = 2y ( Consider only Face value)

and

(10x + y) - (10y + x) = 36

By taking value of x = 2y from above

$\implies [10(2y) + y ] - [10y + (2y) ] = 36$

$\implies [20y + y ] - [10y + 2y] = 36$

$\implies [21y] - [12y] = 36$

$\implies 21y - 12y = 36$

$\implies 9y = 36$

$\implies y = \dfrac{36}{9}$

$\implies y = 4$

♦ Then the value of x = 2y

= 2(4)

= 8

So x = 8

♦ So number = 10x + y

= 80 + 4

= 84

♦ Verification

(10x + y) - (10y + x) = 36

$\implies (80 + 4) - (40+8) = 36$

$\implies 84 - 48 = 36$

$\implies 36 = 36$

♦ Hence as RHS = LHS

So the values are correct .