Question 999050
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Four numbers, taken two at a time, give the sums 84, 88, 100, 100, 112 and 116. What are the four numbers?
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GIVEN:
A+B=84
A+C=88
A+D=100
B+C=100
B+D=112
C+D=116

Let us consider those sums that relate to 3 numbers A, B and C:

A+B=84     (1)
A+C=88     (2)
B+C=100    (3)

Sum up the left and the right sides. You will get

2*(A + B + C) = 84 + 88 + 100 = 272. It gives 

A + B + C = {{{272/2}}} = 136   (4).

Now, distract (1) from (4). You will get

C = 136 - 84 = 52.

Distract (2) from (4). You will get

B = 136 - 88 = 48.

Distract (3) from (4). You will get

A = 136 - 100 = 36.

Now you can easily determine D from A+D=100, for instance:

D = 100 - 36 = 64.

<U>Answer</U>. A=36, B=48, C=52, D=64.

<U>Notice</U>. We have 4 unknowns, therefore 4 equations is enough to find the unknowns. The rest of equations are excessive. 
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