A number consists of 3 digits hose sum is 10 . The middle digit is e3qual to the3 sum of the other two and the number will be increased by 99 if its digits are recersed. The number is-

Let the three digits of the number be A, B, and C, where A is the hundreds digit, B is the tens digit, and C is the units digit.

We are given three pieces of information:

The sum of the digits is 10, so we have: A + B + C = 10.

The middle digit is equal to the sum of the other two, which gives us: B = A + C.

Reversing the digits increases the number by 99. This means that if we reverse the digits, the new number minus the original number equals 99. In mathematical terms, this can be written as: 100C + 10B + A - (100A + 10B + C) = 99.

Now, let's use these equations to find the values of A, B, and C:

From equation 1 (A + B + C = 10), we can substitute B with A + C: A + (A + C) + C = 10 2A + 2C = 10 A + C = 5

Now, we know that A + C = 5. Let's consider the equation for reversing the digits: 100C + 10B + A - (100A + 10B + C) = 99

Substitute B with A + C: 100C + 10(A + C) + A - (100A + 10(A + C) + C) = 99

Simplify and solve for C: 100C + 10A + 10C + A - (100A + 10A + 10C + C) = 99 100C + 10A + 10C + A - 100A - 10A - 10C - C = 99 99C - 99A = 99 C - A = 1

Now we have a system of two equations:

1. A + C = 5
2. C - A = 1

Solve this system of equations: Adding the two equations together: (A + C) + (C - A) = 5 + 1 2C = 6 C = 3

Now, substitute C back into equation 1 to find A: A + 3 = 5 A = 5 - 3 A = 2

So, A = 2, B = A + C = 2 + 3 = 5, and C = 3.

Therefore, the three-digit number is ABC = 253.