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

C- \(\frac{18}{15}\)

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

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

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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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