Solving problem with a middle number between 2 numbers

[java9999]

Hi guys,
I have a question relating to finding a middle number between 2 numbers as followed:
You're given 2 numbers A and B and 2 other numbers X and Y
The objective is to find number C so that A+/- x=C And
B +/- y = C (condition is B<C<A or C is always between A and B)

The rule for x and y is: x could be X or x could be a number < X
y could be Y or y could be a number < Y
x or y can be multiplied by 2, 4 however only one of them multiplied at the time

Example : A = 9, B = 13.5, X = 0.375, Y=3.1875, the result C is 10.6875

However I play with the guessing like this: with Y = 3.1875, y could be 3.1875 (1.5 + 1.5 +0.1875) so I assign y = 1.5 + 0.1875 =1.6875 then 1.6875 + 9 = 10.6875
13.5 - 3.1875 + 0.375 = 10.6875

The problem is that it took a lot of guessing and playing around with those numbers AND there could other missing result as well, 10.6875 is NOT the only result. I'm wondering if there is a FORMULA or equation or function to solve this problem, I really appreciate anyone help me with this!

 

[jvicens]

Hi there,

Thank you for your question. There is indeed a formula that can help you find the middle number between two numbers A and B, given two other numbers X and Y. This formula is known as the "weighted average" formula.

The formula is as follows:

C = (A*X + B*Y) / (X+Y)

In your example, A = 9, B = 13.5, X = 0.375, Y = 3.1875. Plugging these values into the formula, we get:

C = (9*0.375 + 13.5*3.1875) / (0.375+3.1875) = 10.6875

As you can see, this formula gives us the same result as the one you found by guessing and playing around with the numbers. This formula works because it takes into account the "weight" of each number. In this case, the weight is determined by the value of X or Y. So if X is a smaller number, it will have less weight and the final result will be closer to A. Similarly, if Y is a smaller number, the final result will be closer to B.

I hope this helps you solve your problem more efficiently. Let me know if you have any other questions. Good luck!