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If two pipes function simultaneously ,a reservior will be filled inb 12 hrs .one pipe fills the reservior ao hrs faster than the other. how many hrs will

the second pipe take to fill the reservior?


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^2 –14x-120=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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