Question

35. Two water taps together can fill a tank in 1*7/8 hours. The tap with longer diameter takes 2 hours less
than the tap with smaller one to fill the tank separately. Find the time in which tap can fill the tank
separately.
1​

Answer 1

The time in which tap with longer diameter can fill the tank in 3 hrs and the tap with a smaller diameter can fill the tank in 5 hrs separately.

Step-by-step explanation:

The two taps can together fill the tank in = 1$\frac{7}{8} hrs = \frac{15}{8} hrs$

So, in 1 hr, the two taps together will fill = $\frac{8}{15}$ of the tank

Let the tap with smaller diameter alone fill the tank in "x" hrs, then the tap with longer diameter will fill the tank in "(x - 2)" hrs.

So, the part of the tank filled by each of the taps separate in 1 hr will be,

Smaller tap: $\frac{1}{x}$

Longer tap: $\frac{1}{x - 2}$

Therefore, we have the final eq. as,

$\frac{1}{x} + \frac{1}{x -2} = \frac{8}{15}$

⇒ $\frac{x - 2 + x}{x(x -2)} = \frac{8}{15}$

⇒ $\frac{2x - 2}{x^2 - 2x} = \frac{8}{15}$

⇒ $\frac{x - 1}{x^2 - 2x} = \frac{4}{15}$

⇒ 15x - 15 = 4x² - 8x

⇒ 4x² - 23x + 15 = 0

⇒ 4x² - 20x - 3x + 15 = 0

⇒ 4x(x - 5) - 3(x-5) = 0

⇒ (x - 5)(4x -3) = 0

⇒ x = 5 or $\frac{3}{4}$

neglecting the x = $\frac{3}{4}$ because the the time taken by longer tap will be negative

⇒ x = 5 hrs ← time taken by the tap with smaller diameter to fill the tank separately

∴ x - 2 = 5 - 2 = 3 hrs ← time taken by the tap with longer diameter to fill the tank separately

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Answer 2

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