Question

Two pipes together can fill a water tank is 6hours and 40 minutes. Find the time each will take to fill the tank if one of the two pipes can fill it in 3 hours less than the other

Answer 1

therefore time taken by one pipe=15 hours or 4/3 hours
time taken by other pipe= 15-3=12 hours or
4/3-3= 4-9/3=-5/3
as time cannot be negative therefore time taken by other pipe is 12 hours.

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Answer 2

Answer:

Time taken by the pipe-A to fill the tank itself is = 15 hr

Time taken by the pipe-B to fill the tank itself is = 18 hr

Step-by-step explanation:

Total time taken by the two pipes to fill the tank is = 6 hr 40 min=$6+\frac{40}{60}=6+\frac{2}{3}=\frac{(18+2)}{3}=\frac{20}{3}$ hr

Consider the total volume of the tank V

Time taken by the pipe-A to fill the tank itself is = x hr

Time taken by the pipe-B to fill the tank itself is = (x+3) hr

We know that $Discharge (Q) = \frac{Volume}{Time}$

Discharge of Pipe-A = $\frac{V}{x}$

Discharge of Pipe-B = $\frac{V}{x-3}$

Total volume filled by pipe A in 6 hr 40 min =$\frac{V}{x}\times\frac{20}{3}$

Total volume filled by pipe B in 6 hr 40 min =$\frac{V}{x-3}\times\frac{20}{3}$

As per problem;

$\frac{V}{x}\times\frac{20}{3}+\frac{V}{x-3}\times\frac{20}{3}=V\\\\\Rightarrow\frac{20V}{3x}+\frac{20V}{(x-3)3}=V\\\\\Rightarrow\frac{20}{3x}+\frac{20}{(x-3)3}=1\\\\\Rightarrow\frac{20}{3}[\frac{1}{x}+\frac{1}{(x-3)}]=1\\\\\Rightarrow[\frac{1}{x}+\frac{1}{(x-3)}]=\frac{3}{20}\\\\\Rightarrow[\frac{(x-3)+x}{x(x-3)}]=\frac{3}{20}\\\\\Rightarrow\frac{2x-3}{x^{2}-3x}=\frac{3}{20}\\\\\Rightarrow3x^{2}-9x=40x-60\\\\\Rightarrow3x^{2}-9x-40x+60=0\\\\\Rightarrow3x^{2}-49x+60=0$

For Quadratic Equation $ax^{2}+bx+c=0 \text{ [where x is the variable and a, b and c are known values]}$

The value of $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

$b^{2}-4ac \text{ is called Discriminant (D)}$

when Discriminant (D) is positive, we get two Real solutions  for x

when Discriminant (D) is zero we get just ONE real solution (both answers are the same)

when Discriminant (D) is negative we get a pair of Complex solutions

In equation 1 the Discriminant (D) is = $b^{2}-4ac= (-49)^{2}-4\times3\times(60)=2401-720=1681$

As the Discriminant (D) is positive, we get two Real solutions  for x

therefore the value of x is either $x=\frac{-b+\sqrt{D}}{2a}\text{or}\text{ } x=\frac{-b-\sqrt{D}}{2a}$

$x=\frac{-b+\sqrt{D}}{2a}=\frac{-(-49)+\sqrt{1681}}{2\times3}=\frac{49+41}{6}=\frac{90}{6}=15$

$x=\frac{-b-\sqrt{D}}{2a}=\frac{-(-49)-\sqrt{1681}}{2\times3}=\frac{49-41}{6}=\frac{8}{6}=1.33$

As we get two positive number for the value of x then the value of x either 15 or 1.33

We consider the value of x=15 hr because time taken to fill the tank by individual pipe is always greater than the 6 hrs 40 min.

therefore,

Time taken by the pipe-A to fill the tank itself is = 15 hr

Time taken by the pipe-B to fill the tank itself is = (15+3)=18 hr