Question

the sum of a two digit number and the number obtained by reversing the digit is 66 if the digit of the number differ by 2 find the number how many such numbers are there

Answer 1

Answer:Number is 42.

Solution:

Let the two digits number is xy,this can be represented as 10x+y

on reversing the number the number become yx, now this can be represented as 10y+x

according to the question the sum of two digit number and the number reversing the digits is equal to 66

$10x + y + 10y + x = 66 \\ \\ 11x + 11y = 66 \\ \\ x + y = 6....eq1$

the digit of the number differ by 2,hence it can be represented algebraically

$x - y = 2.....eq2$
Now solve these two equations,here I am using Elimination method,add both equations

$x + y = 6 \\ x - y = 2 \\ - - - - - - - \\ 2x = 8\\ \\ x = 4 \\ \\4+ y = 6 \\\\y=2$

Hence number is 42.
it's reverse number is 24.

Verification:
42+24=66

Answer 2

Answer:

Numbers are 42  & 24

& reverse of numbers are 24 & 42

Step-by-step explanation:

The sum of a two digit number and the number obtained by reversing the digit is 66 if the digit of the number differ by 2 find the number how many such numbers are there

Let say NUmber is XY

Value = 10X + Y

Reverse of number = YX

Value = 10Y + X

10X + Y + 10Y + X = 66

=> 11X + 11Y =  66

=> X + Y = 6   - eq1

|X - Y| = 2

Case 1

X - Y = 2    - eq2

Adding eq 1 & eq 2

=> 2X = 8 => X  = 4  & Y = 2

Case 2

X - Y = -2    eq 3

Adding Eq 1 & eq 3

=> 2X = 4 => X = 2  & Y = 4

Numbers are 42  & 24

& reverse of numbers are 24 & 42