Multiple choice

A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank $5$ hours faster than the first pipe and $4$ hours slower than the third pipe. The time required by the first pipe is.

  1. $6$ hours
  2. $10$ hours
  3. $15$ hours
  4. $30$ hours
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the rates of the pipes be 1/x, 1/(x-5), and 1/(x-9). The problem states the rate of the third pipe equals the sum of the first two: 1/(x-9) = 1/x + 1/(x-5). Solving this quadratic equation x^2 - 19x + 45 = 0 leads to x = 15 as the valid solution.

AI explanation

Let the first pipe take x hours, meaning the second pipe takes x minus 5 hours and the third takes x minus 9 hours. The first two pipes combined equal the third pipe, so 1 divided by x plus 1 divided by x minus 5 equals 1 divided by x minus 9. Cross-multiplying gives 2x minus 5 multiplied by x minus 9 equals x multiplied by x minus 5, yielding the quadratic equation x squared minus 18x plus 45 equals 0. Solving this by factorization gives x equals 15 hours as the valid root.

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