<PRE><B>Question 824374</B>
let large tap fill in x minutes
small tap will fill in (x-40) minutes


They together fill in 48 minutes
1/48 of the job is done by both iin 1 minute

large tap does 1/x of the job in 1 minute
small tap does 1/(x-40) of the job in 1 minute

1/x + 1/(x-40) = 1/48

LCM

(x-40+x)/x(x-40) = 1/48

(2x-40)=(x^2-40x)/48

48(2x-40)= x^2-40x

96x-1920=x^2-40x

x^2-136x+1920=0

x^2-120x-16x+1920=0

x(x-120)-16(x-120)=0

(x-16)(x-120)=0

x=16 OR 120 minutes

large tap cannot take 16 minutes as difference is 40

large tap 120 minutes
small tap = 80 minutes


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