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Two pipes A and B together can fill a cistern in 4 hours. Had they been opened separately, then B would have taken 6 hours more than A to fill the cistern. How much time will be taken by A to fill the cistern separately?
2 hours
6 hours
8 hours
1 hour
6 hours
Use the quadratic root property for work problems: x2−(T−D)x−T×D=0, where T=4 (total time) and D=6 (difference). Thus x2−(−2)x−24=0⟹x2+2x−24=0, giving x=6.
Pipes A and B fill a cistern in 4 hours. Pipe B takes 6 hours more than pipe A to fill it.
x1+x+61=41
Use the quadratic root property for work problems: x2−(T−D)x−T×D=0, where T=4 (total time) and D=6 (difference). Thus x2−(−2)x−24=0⟹x2+2x−24=0, giving x=6.
Students often misinterpret '6 hours more than A' as B=A+6 for rates, but it applies to the time taken, so TimeB=TimeA+6.
Define Variables
Let the time taken by pipe A be x hours. Therefore, the time taken by pipe B is x+6 hours.
TimeA=x,TimeB=x+6
Set up the Rate Equation
Since they fill the cistern together in 4 hours, the sum of their work rates in 1 hour is 1/4.
x1+x+61=41
Solve the Quadratic Equation
Combine the fractions and solve: x(x+6)x+6+x=41⟹x2+6x2x+6=41⟹x2+6x=8x+24.
x2−2x−24=0
Factorization and Final Result
Factor the quadratic x2−2x−24=0 as (x−6)(x+4)=0. Since time cannot be negative, x=6.
x=6
B is correct because solving the quadratic equation derived from the work rates gives the time taken by pipe A as 6 hours.
This problem uses the same logic as 'Distance = Speed × Time' problems where the variable represents a duration influenced by a difference constraint.
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