Question

the sum of the digit of a number is 11 and the difference between the number and that formed by reversing the digit is 45 find the number​

Answer 1

Answer:let the no. be 10x+y

X+y=11. (I)

10x+y-(10y+X)=45

9x-9y=45

x-y=5. (II)

From (I) and (II) we get

2x=16

x=8

8+y=11

y=3

No. = 8*10+3=83

Answer 2

Answer :-

The required number is 83.

Solution :-

Let the digits of the number be x and y

Sum of digits of a number = 11

⇒ x + y = 11 ---(1)

Number = 10x + y

Number formed when digits are reversed = 10y + x

Given

Difference between a number and number formed when digits are reversed = 45

⇒ 10x + y - (10y + x) = 45

⇒ 10x + y - 10y - x = 45

⇒ 9x - 9y = 45

⇒ 9(x - y) = 45

⇒ x - y = 45/9

⇒ x - y = 5 --(2)

Adding (1) & (2)

⇒ x + y + (x - y) = 11 + 5

⇒ x + y + x - y = 16

⇒ 2x = 16

⇒ x = 16/2

⇒ x = 8

Substitute x = 8 in (1)

⇒ x + y = 11

⇒ 8 + y = 11

⇒ y = 11 - 8

⇒ y = 3

Substitute x = 8, y = 3 in 10x + y to know the number

Number = 10x + y

= 10(8) + 3

= 80 + 3

= 83

Therefore the required number is 83.