two water taps can fill a tank in 100/9 min if one pipe takes 5min 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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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 = 100/9 minutes
Portion of tank filled by faster pipe in 1 minute = 1/x
Portion of tank filled by faster pipe in 1 minute = 1/(x+5)
Portion of tank filled by both pipe in one 1 minute = 1/(100/9) = 9/100
A/q,
1/x + 1/(x+5) = 9/100
⇒ (x+5 + x)/{x(x+5)} = 9/100
⇒ (2x+5)/(x2+5x) = 9/100
⇒ 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 = -25/9 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.
Time taken by both pipe to fill the tank = 100/9 minutes
Portion of tank filled by faster pipe in 1 minute = 1/x
Portion of tank filled by faster pipe in 1 minute = 1/(x+5)
Portion of tank filled by both pipe in one 1 minute = 1/(100/9) = 9/100
A/q,
1/x + 1/(x+5) = 9/100
⇒ (x+5 + x)/{x(x+5)} = 9/100
⇒ (2x+5)/(x2+5x) = 9/100
⇒ 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 = -25/9 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.
Sushantss:
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