If two pipes are put in operation simultaneously, the tank is filled i

1.	If two pipes are put in operation simultaneously, the tank is filled in 24 min. One pipe fills the tank in 20 min faster than the other. How many hours does the faster pipe take to fill the tank? (a) 60 min (b) 45 min (c) 40 min (d) 30 min
Answer» (c) 40 min Let one pipe take x minutes to fill the tank, then the other will take (x + 20) minutes. Given, \(\frac{1}{X}+\frac{1}{X+20}\) = \(\frac{1}{24}\) \(\Rightarrow\) \(\frac{X+20+X}{X^2+20X}\)= \(\frac{1}{24}\) \(\Rightarrow\) (2x + 20)24 = x² + 20x \(\Rightarrow\) x² – 28x – 480 = 0 \(\Rightarrow\) x 2 – 40x + 12x – 480 = 0 \(\Rightarrow\) x(x – 40) + 12(x – 40) = 0 \(\Rightarrow\) (x – 40) (x +12) = 0 \(\Rightarrow\) x = 40 or –12 Neglecting negative values x = 40 min

If two pipes are put in operation simultaneously, the tank is filled in 24 min. One pipe fills the tank in 20 min faster than the other. How many hours does the faster pipe take to fill the tank? (a) 60 min (b) 45 min (c) 40 min (d) 30 min

Answer»

(c) 40 min

Let one pipe take x minutes to fill the tank, then the other will take (x + 20) minutes.

Given, \(\frac{1}{X}+\frac{1}{X+20}\) = \(\frac{1}{24}\)

\(\Rightarrow\) \(\frac{X+20+X}{X^2+20X}\)= \(\frac{1}{24}\) \(\Rightarrow\) (2x + 20)24 = x² + 20x

\(\Rightarrow\) x² – 28x – 480 = 0 \(\Rightarrow\) x 2 – 40x + 12x – 480 = 0

\(\Rightarrow\) x(x – 40) + 12(x – 40) = 0

\(\Rightarrow\) (x – 40) (x +12) = 0

\(\Rightarrow\) x = 40 or –12

Neglecting negative values x = 40 min