The tank is filled by three pipes with different uniform flow rates. While the first two pipes are operating simultaneously, they fill the tank in the same duration that the third pipe takes to fill it alone. The second pipe can fill the tank 5 hours quicker than the first pipe, yet 4 hours slower than the third pipe. What is the time (in hours) needed for the first pipe to fill the tank?

Correct — D. Let the first pipe take x hours.

The second is 5 hours quicker than the first: x − 5
The second is 4 hours slower than the third, so the third takes x − 9

The first two together match the third alone:
1⁄x + 1⁄(x − 5) = 1⁄(x − 9)
(2x − 5)⁄[x(x − 5)] = 1⁄(x − 9)

(2x − 5)(x − 9) = x(x − 5)
2x² − 23x + 45 = x² − 5x
x² − 18x + 45 = 0
(x − 15)(x − 3) = 0

x = 3 would make the second pipe take −2 hours, so x = 15 → option (d).

Check: 1⁄15 + 1⁄10 = 1⁄6, and the third pipe takes 15 − 9 = 6 hours.

Why the others are wrong

• (a)
  18 — 18 makes the pair too fast. The times would be 18, 13 and 9, and 1⁄18 + 1⁄13 is about 0.132 against the third pipe's 1⁄9 ≈ 0.111, so the two sides are not equal.
• (b)
  9 — 9 breaks the chain before any equation. The second pipe would take 4 hours and the third 9 − 9 = 0 hours, which is a pipe that fills the tank instantly.
• (c)
  12 — 12 makes the pair too slow. The times would be 12, 7 and 3, and 1⁄12 + 1⁄7 ≈ 0.226 against the third pipe's 1⁄3 ≈ 0.333, so the equality in the stem fails.

Work problems add rates, never times. A pipe that fills a tank in t hours contributes 1⁄t of the tank per hour, and pipes running together have their rates summed.

The stem gives no absolute time at all — only two gaps, 5 hours and 4 hours. Those chain the three unknowns onto one variable: x, x − 5 and x − 9.

A rate equation in x with denominators clears into a quadratic. Both roots satisfy the algebra, so the physical check matters: any root that makes a filling time zero or negative is discarded.

Key facts

• A pipe filling in t hours works at 1⁄t of the tank per hour, and simultaneous rates add.
• The three times chain as x, x − 5 and x − 9 from the two stated gaps.
• x² − 18x + 45 = 0 gives x = 15 or x = 3, and x = 3 is rejected as it makes a time negative.
• At x = 15: 1⁄15 + 1⁄10 = 1⁄6, matching the third pipe's 6 hours.

Common traps

• Adding times instead of rates and writing 15 + 10 = 6 in some rearranged form.
• Reading the third pipe as 4 hours slower than the second, which gives x − 1 and a quadratic with no real root.
• Keeping x = 3 because it satisfies the quadratic, without checking that x − 5 must be positive.

No directly related past PYQ was found.