SOLUTION: 2 taps (one large and one small) fill a tank together in total of 48 minutes. difference between time taken to fill the tank by small tap (only) and large tap(only) but (separatel

Question 824374: 2 taps (one large and one small) fill a tank together in total of 48 minutes.
difference between time taken to fill the tank by small tap (only) and large tap(only) but (separately) is 40 minutes.
find the time taken by each tap separately

Found 2 solutions by mananth, MathTherapy:
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
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

Answer by MathTherapy(10552) (Show Source): You can put this solution on YOUR website!
2 taps (one large and one small) fill a tank together in total of 48 minutes.
difference between time taken to fill the tank by small tap (only) and large tap(only) but (separately) is 40 minutes.
find the time taken by each tap separately

Time larger tap will take to fill tank, separately: minutes
Time smaller tap will take to fill tank, separately: 80 + 40, or minutes
You can do the check!!
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