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Question

Two pipes running together can fill a tank in 1119 minutes. If one tap takes 5 mintues more than the other to fill the tank separately, find the time in which each pipe would fill the tank separately.

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Solution

open parentheses n o t e rightwards arrow 11 1 over 9 equals 100 over 9 close parentheses
Let one of the pipe take x minutes to then the slower pipe will take (x+5) minutes.

Time taken by both pipe to fill the tank = 100 over 9 minutes

Portion of tank filled by faster pipe in 1 minute = 1 over x

Portion of tank filled by faster pipe in 1 minute = fraction numerator 1 over denominator x plus 5 end fraction

Portion of tank filled by both pipe in one 1 minute = fraction numerator 1 over denominator begin display style 100 over 9 end style end fraction = 9 over 100

A/q,

1 over x plus fraction numerator 1 over denominator x plus 5 end fraction equals 9 over 100 fraction numerator x plus 5 plus x over denominator x open parentheses x plus 5 close parentheses end fraction equals 9 over 100 fraction numerator 2 x plus 5 over denominator x squared plus 5 x end fraction equals 9 over 100

⇒ 200x + 500 = 9x2 + 45x

⇒ 9x2 + 45x - 200x - 500 = 0

⇒ 9x2 - 155x - 500 = 0

⇒ 9x2 - 180x + 25x - 500 = 0

⇒ 9x (x - 20) + 25 (x - 20) = 0

⇒ (9x + 25) (x - 20) = 0

x = negative 25 over 9 and x = 20

But x can't be negative as it represents time.

Therefor, X = 20 minutes.

Time taken by faster pipe to fill the tank = 20 minutes.

Time taken by slowerr pipe to fill the tank = 20+5 = 25 minutes.


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