Algebra Puzzle – Challenge 57

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Algebra Puzzle – Challenge 57

Challenge:

The sum of three positive integers is 4000. The ratio of the first number to the second number is \(\frac{2}{3}\) and the ratio of the first number to the third number is \(\frac{6}{5}\). What is the third number?

A- 800

B- 1000

C- 1200

D- 1600

E- 2000

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The correct answer is B.

Let X, Y and Z represent the three numbers. So,
X + Y + Z = 4000 and
\(X = \frac{2}{3} Y → Y = \frac{3}{2}X\)
\(X = \frac{6}{5} Z → Z = \frac{5}{6}X\)
Replace the values of Y and Z in the first equations with their values in the second and third equations:
\(X + \frac{3}{2}X + \frac{5}{6}X = 4000 →X = 1200\)
\(Y = \frac{3}{2}X → Y = \frac{3}{2} (1200) = 1800\)
\(Z = \frac{5}{6}X → Z = \frac{5}{6}(1200) = 1000\)
The third number, Z, is 1000.

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