Question

If a, b, c, d are four different positive integers selected from 1 to 25, then the highest possible value of
(a + b)+(c+d)/
(a+b)+(c-d)
would be
Given that (a+b)+(c-d) is not equal to zero)​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

51

Step-by-step explanation:

Given a, b, c, d are distinct positive integers.

The least possible value of denominator ((a + b) + (c - d)) can be only 1.

We have to maximize the value of the numerator ((a + b) + (c + d)) simultaneously.

We can observe that in numerator 'd' is added and in denominator 'd' is subtracted.

So, if d is given the highest value we can simultaneously maximize the numerator and minimize the denominator.

The obvious choice for is d = 25

If d = 25 and from the denominator we know ((a + b) + (c - d)) = 1

(a + b) + c = 26

Substituting (a + b) + c = 26 and d = 25 in the numerator

we get ((a + b) + (c + d)) = 26 + 25 = 51

Maximum value of the numerator is 51 and the minimum value of the denominator is 1.

(a+b)+(c+d)(a+b)+(c−d) = 511 = 51