Two pipe running together can fill a cistern in 3.5mins and one pipe takes 3 mins more than the other find the time in which each pipe would fill cistern

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- anonymous

- katieb

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- anonymous

- perl

Let x be the rate of the first pipe and y is the rate of the second pipe.
And t be the time it takes for first pipe to fill cistern.
Then we have the following equations, use d = r* t
1 cistern = (x+y) * 3.5 min
1 cistern = x * t
1 cistern = y * (t + 3)

- anonymous

I can't understand the above.

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## More answers

- perl

Have you used the formula
distance = rate * time

- anonymous

Yes..

- perl

I am using a similar equation.
1 cistern = rate of pipe * time

- anonymous

What next?

- perl

Let x be the rate of the first pipe and y is the rate of the second pipe.
And t be the time it takes for first pipe to fill cistern.
1 cistern = (x+y) * 3.5 min
1 cistern = x * t
1 cistern = y * (t + 3)

- anonymous

3 cistern?

- perl

they are three different equations

- perl

"Two pipe running together can fill a cistern in 3.5mins"
That gives us
1 cistern = (x+y) * 3.5 min , where x and y are the rates of the pipes

- anonymous

Help PLEASE?

- anonymous

- Michele_Laino

I think that @perl gave you the right explanation

- Michele_Laino

nevertheless I can give you another way to solve your problem.
The working rates of the two pipes are:
\[\Large\frac{W}{x},\quad \frac{W}{{x + 3}}\]

- Michele_Laino

where W is the work to be done

- Michele_Laino

now, following the text of your problem, we can write:
\[\Large \frac{W}{x} + \frac{W}{{x + 3}} = \frac{W}{{3.5}}\]
and, simplifying that expression for W, we get:
\[\Large \frac{1}{x} + \frac{1}{{x + 3}} = \frac{1}{{3.5}}\]
Please solve that equation for x

- anonymous

x+3+x/x^2+3x = 1/3.5?

- Michele_Laino

first step:
we have:
\[\Large \frac{1}{x} + \frac{1}{{x + 3}} = \frac{2}{7}\]

- Michele_Laino

yes! since 3.5= 7/2

- Michele_Laino

now, the least common multiple, of
x, x+3 and 7, is:
x*(x+3)*7
am I right?

- anonymous

Yes

- Michele_Laino

is that equation is equivalent to this one?
\[\Large 1 \times 7 \times \left( {x + 3} \right) + 7x = 2x\left( {x + 3} \right)\]

- Michele_Laino

oops..is that equation equivalent to this one?

- anonymous

I guess..

- Michele_Laino

more steps:
\[\Large \frac{{1 \times 7\left( {x + 3} \right)}}{{7x\left( {x + 3} \right)}} + \frac{{1 \times 7x}}{{7x\left( {x + 3} \right)}} = \frac{{2x\left( {x + 3} \right)}}{{7x\left( {x + 3} \right)}}\]

- Michele_Laino

now?

- anonymous

I'm lost...

- Michele_Laino

as denominator, of each fraction, we find the least common multiple

- anonymous

Getting it..
Now?

- Michele_Laino

now, if we consider the first fraction, for example, we have:
|dw:1436624598950:dw|

- Michele_Laino

similarly for the other 2 fractions

- Michele_Laino

after that, we have to simplify the last equation, so we get:
\[\Large 7x + 21 + 7x = 2{x^2} + 6x\]
I have applied the distributive property of multiplication over addition
Please rewrite that equation in this form:
\[\Large bA{x^2} + Bx + C = 0\]

- Michele_Laino

oops..
in this form:
\[\Large A{x^2} + Bx + C = 0\]

- Michele_Laino

what do you get?

- anonymous

It should be 7x^2, right?

- Michele_Laino

hint:
here are more steps:
\[\Large \begin{gathered}
7x + 21 + 7x = 2{x^2} + 6x \hfill \\
14x + 21 = 2{x^2} + 6x \hfill \\
\end{gathered} \]

- Michele_Laino

now I subtract 14 x from both sides so I can write:
\[\Large 14x + 21 - 14x = 2{x^2} + 6x - 14x\]
please simplify

- Michele_Laino

hint:
\[\Large \begin{gathered}
14x - 14x = 0 \hfill \\
6x - 14x = - 8x \hfill \\
\end{gathered} \]

- anonymous

2x^2-8x-21=0

- Michele_Laino

perfect!

- Michele_Laino

now you have to solve that equation, using the standard formula:
\[\Large x = \frac{{ - b \pm \sqrt {{b^2} - 4 \times a \times c} }}{{2 \times a}}\]
where
\[\Large \begin{gathered}
a = 2, \hfill \\
b = - 8, \hfill \\
c = - 21. \hfill \\
\end{gathered} \]

- anonymous

Um?
I can't use this formula as it's npt stated in the question
*not

- Michele_Laino

have you studied that formula in your course of mathematics?

- anonymous

Yep..
But my teacher would add
"correct to 2 or 3 decimals"
ONLY THEN can I use this formula.

- Michele_Laino

then we can use that formula, here is the next step:
\[\Large \begin{gathered}
x = \frac{{8 \pm \sqrt {{{\left( { - 8} \right)}^2} - 4 \times 2 \times \left( { - 21} \right)} }}{{2 \times 2}} = \hfill \\
\hfill \\
= \frac{{8 \pm \sqrt {64 + 168} }}{4} = ... \hfill \\
\end{gathered} \]
please continue

- anonymous

I would get this marked wrong if I use the formula...
Please understand....

- Michele_Laino

ok! I understand

- anonymous

So
I'll HAVE TO factor it out..

- Michele_Laino

ok! I will write your next steps

- Michele_Laino

If I divide both sides of your equation by 2, I get:
\[\Large {x^2} - 4x - \frac{{21}}{2} = 0\]

- Michele_Laino

Then I add and subtract 4 at left side, so I can write:
\[\Large {x^2} - 4x + 4 - 4 - \frac{{21}}{2} = 0\]

- Michele_Laino

is it ok, for you?

- anonymous

Yes

- Michele_Laino

now we note that:
\[\Large {x^2} - 4x + 4 = {\left( {x - 2} \right)^2}\]

- Michele_Laino

so we can rewrite my last expression as follows:
\[\Large {\left( {x - 2} \right)^2} - 4 - \frac{{21}}{2} = 0\]

- Michele_Laino

and, being:
\[\Large - 4 - \frac{{21}}{2} = \frac{{ - 8 - 21}}{2} = - \frac{{29}}{2}\]

- Michele_Laino

we have:
\[\Large {\left( {x - 2} \right)^2} - \frac{{29}}{2} = 0\]

- Michele_Laino

now I add -29/2 to both sides of that equation, so I get:
\[\Large \begin{gathered}
{\left( {x - 2} \right)^2} - \frac{{29}}{2} + \frac{{29}}{2} = 0 + \frac{{29}}{2} \hfill \\
\hfill \\
{\left( {x - 2} \right)^2} = \frac{{29}}{2} \hfill \\
\end{gathered} \]

- Michele_Laino

am I right?

- anonymous

Yes

- Michele_Laino

ok! now I ask you, what is:
\[\Large \sqrt {\frac{{29}}{2}} = ...?\]
nearest to tenth?

- anonymous

3.8

- anonymous

wow! @Michele_Laino perfect explanation :)

- Michele_Laino

that's right!
Now if I take the square rooth of both sides of my last equation, I get:
\[\Large x - 2 = \pm 3.8\]
please remember we have 2 square roots

- Michele_Laino

thanks!! @behappy

- Michele_Laino

so we have 2 equations:
\[\Large \begin{gathered}
x - 2 = 3.8 \hfill \\
\hfill \\
x - 2 = - 3.8 \hfill \\
\end{gathered} \]

- anonymous

5.8 and -1.8

- Michele_Laino

yes! Nevertheless -1.8 can not be acceptable, since x represents a time. So only x=5.8 is the acceptable solution

- Michele_Laino

finally the requested time are:
\[\Large \begin{gathered}
{t_1} = 5.8\min \hfill \\
{t_2} = 5.8 + 3 = ...\min \hfill \\
\end{gathered} \]

- Michele_Laino

times*

- anonymous

8.8

- Michele_Laino

that's right!

- anonymous

THANKS!!!!!!
CAN YOU PLEASE HELP ME MORE?!

- Michele_Laino

ok!

- Michele_Laino

thanks again for your appreciation to my work @behappy :)

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