If the sum of the digits of a two digit number is 9 and the difference...
From given question
x + y = 9 ...(i )
(10x + y) – (10y + x) = 45
9 (x – y) = 45
x – y = 5 ...(ii )
Solving equations (i ) and (ii ), we get
∴ x = 7, y = 2
So the number
= 10x + y
= 10 x 7 + 2
= 72
∴ 72 is the number
If the sum of the digits of a two digit number is 9 and the difference...
To solve this problem, we need to find a two-digit number where the sum of its digits is 9 and the difference between the number and its reverse is 45.
Let's assume the two-digit number is represented as '10x + y', where x and y are the digits of the number.
Finding the sum of the digits:
The sum of the digits can be represented as x + y = 9.
Finding the difference between the number and its reverse:
The number can be represented as '10x + y', and its reverse can be represented as '10y + x'. The difference between the number and its reverse is (10x + y) - (10y + x) = 9x - 9y = 45.
From the equation x + y = 9, we can rearrange it to get y = 9 - x.
Substituting y = 9 - x in the equation 9x - 9y = 45, we get:
9x - 9(9 - x) = 45
Simplifying the equation,
9x - 81 + 9x = 45
18x - 81 = 45
18x = 45 + 81
18x = 126
x = 126/18
x = 7
Substituting the value of x in the equation y = 9 - x, we get:
y = 9 - 7
y = 2
Therefore, the two-digit number is 10x + y = 10(7) + 2 = 72.
Hence, the correct answer is option B, 72.