Algebra Puzzle – Challenge 40

Challenge:

What is the average of three numbers if the sum of the first and second number is (frac{11}{5}), the sum of the second and third number is (frac{31}{15}), and the sum of the first and third umber is (frac{7}{3})?

A- (frac{13}{15})

B- (1frac{1}{10})

C- (frac{18}{15})

D- (2frac{1}{5})

E- (6frac{3}{5})

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Let X, Y and Z represent the three numbers. Therefore:
(X + Y = frac{11}{5}, Y + Z = frac{31}{15}, X + Z = frac{31}{15})
Add the three questions:
((X + Y) + (Y + Z) + (X + Z) = frac{11}{15} + frac{31}{15} + frac{7}{3})
(2X + 2Y + 2Z = frac{99}{15} → 2 (X + Y + Z) = frac{99}{15} → X + Y + Z = frac{99}{30 }= frac{33}{1})
The average of the three numbers is:
(frac{X + Y + Z}{3 }= frac{33}{10} ÷ 3 = frac{33}{30} = frac{11}{10})

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