Two pipes running together can fill the tank in 11(1÷9) minutes. If one pipe tales 5 minutes 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

Suppose the faster pipe take x minute to fill the tank. Therefore slower pipe will take ( x + 5) minutes to fill the tank. Since the faster pipe take x minutes to fill tank.

Therefore portion of tank filled by faster pipe in one minute =
Similarly portion of tank filled by slower pipe in one minute = $fraction numerator 1 over denominator x plus 5 end fraction$

Now both the pipe fill together the tank in

Therefore portion of tank filled by both the pipe in one minute = $fraction numerator 1 over denominator begin display style 100 over 9 end style end fraction equals 9 over 100$

According to the given condition.

$1 over x plus fraction numerator 1 over denominator x plus 5 end fraction equals 9 over 100$
$T a k i n g space L C M space a n d space c r o s s space m u l t i p l i n g comma w e space g e t space fraction numerator left parenthesis x plus 5 right parenthesis plus left parenthesis x right parenthesis over denominator x left parenthesis x plus 5 right parenthesis end fraction equals 9 over 100 100 left parenthesis 2 x plus 5 right parenthesis equals 9 left parenthesis x squared plus 5 x right parenthesis 200 x plus 500 equals 9 x squared plus 45 x t a k i n g space L H S space t o space r i g h t comma 9 x squared plus 45 x minus 200 x minus 500 equals 0 i. e space space 9 x squared minus 155 x minus 500 equals 0 i. e space space 9 x squared minus 180 x plus 25 x minus 500 equals 0 i. e space space 9 x left parenthesis x minus 20 right parenthesis plus 25 left parenthesis x minus 20 right parenthesis equals 0 i. e space space left parenthesis 9 x plus 25 right parenthesis left parenthesis x minus 20 right parenthesis equals 0 i. e space space x space equals space fraction numerator negative 25 over denominator 9 end fraction space a n d space x space equals 20 space B u t space x space c a n apostrophe t space b e space n e g a t i v e space a s space i t space r e p r e s e n t s space t i m e. space space T h e r e f o r comma space x space equals space 20 space m i n u t e s.$

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

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

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