One tank can be filled up by two taps in 6 hours. The Smaller tap alone takes 5 hours more than the bigger tap alone. Find the time required (in hours) by the smaller tap to fill the tank separately.

Solution

Let time taken to fill tank by small tap be x hours

Time taken to fill tank by big tap be y hours

Then, x−y=5 or x=5+y

Smaller tap can fill in one hour = 1/x

Bigger tap can fill in one hour = 1/y

Then, 1/x+1/y = 1/6

x+y/xy = 1/6

xy=6(x+y)

y(5+y)=6(5+y+y)

5y+y² =30+12y

y² −7y−30=0

\(y=\frac{7\pm\sqrt{49+120}}{2}\)

\(y=\frac{7\pm13}{2}\)

y=10,−3

x=5+10=5+10=15 hours

Hence, Small tap can fill the tank in 15 hours and bigger tap can fill in 10 hours

\(Let time taken to fill tank by small tap be x hoursTime taken to fill tank by big tap be y hoursThen, x−y=5 or x=5+ySmaller tap can fill in one hour =x1​Bigger tap can fill in one hour =y1​Then,x1​+y1​=61​xyx+y​=61​xy=6(x+y)y(5+y)=6(5+y+y)5y+y2=30+12yy2−7y−30=0y=27±49+120​​y=27±13​y=10,−3x=5+10=5+10=15 hoursHence, Small tap can fill the tank in 15 hours and bigger tap can fill in 10 hours\)