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If two pipes function simultaneously , a reservoir will be filled in 12 hrs.one pipe fills the reservoir 10 hrs faster than the other.how many hrs will the second pipe take to fill the reservoir?

let the reservoir be filled by first pipe in x hours.

  Then ,second pipe fill it in (x+10)hrs.

  Therefore (1/x)+(1/x+10)=(1/12)  <=>(x+10+x)/(x(x+10))=(1/12).

  x^2 –14x-120=0  <=> (x-20)(x+6)=0

  <=> x=20  [neglecting the negative value of x]

  so, the second pipe will take (20+10)hrs. (i.e) 30 hours to fill the reservoir

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 let the work done by a pipe be x 

so


1/x + 1/ x + 10  =  1/12 

12(x)(x+10)/x + 12(x)(x+10)/x+10 = 12(x)(x+10) /12

12(x+10) + 12x = (x)(x+10)

12x + 120 + 12x = x^2 +10x

x^2 - 14x - 120 = 0

if U solve this u will get x 6 or -20 ..

so another pipe = 6 +10 = 16 !!!
 

I THINK THiS IS CORRCT !!!

SRY IF IT IS NOT !

AND PLZ PROVIDE THE CORRCT ANS TO(O !

 

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The  answer iz wrong so don't copy . The answer iz 30hrs

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