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# A number consists of two digits, whose sum is five. When the digits are reversed, the original number become greater by nine. Find the number.

A two-digit number has 2 digits, one at the unit's place and one at the tens place.

## Answer: A two-digit number whose sum of the digits is five and when the digits are reversed, the original number becomes greater by 9 is 32.

The general form of a two-digit number with 'a' as the digit at the units place and 'b' as the digit at the tens place is:

10 × b + a = 10b + a

**Explanation:**

Let the digit at the unit's place be 'a' and the unit's at the ten's place be 'b', so the original number = 10b+a.

According to the conditions given in the question:

Sum of the two digits is five, that is:

a + b = 5 ---------------------- (1)

If the original number is 10b + a, numbers formed when the digits are reversed = 10a + b.

So the original number is greater than the reversed number by nine, that is:

(10b + a) - (10a + b) = 9

or, 10b + a - 10a - b = 9

or, 9b - 9a = 9

or, b - a = 1 ------------------ (2)

Now let us use the two equations (1) and (2) that we have got to find the values of a and b:

Form the first equation:

a + b = 5

or, a = 5 - b

Let's substitute this value of a in the second equation, as:

b - (5 - b) = 1

or, b - 5 + b = 1

or, 2b - 5 = 1

or, 2b = 1 + 5

or, 2b = 6

or, b = 3

Let's substitute this value of b in a = 5 - b to find the value of a:

a = 5 - 3 = 2

Substitute the values of a and b in the original number = 10b + a = 10 × 3 + 2 = 30 + 2 = 32.

### Therefore, the original number is 32.

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