Question

One pipe can fill a cistern in 3 hours less than the other.The two pipes together can fill the cistern in 6 hours 40 minutes.Find the time each pipe will take to fill the cistern
@sepeario

Answer 1

assume the two pipes are x and y. Hence \[x=y+3\] and \[x+y=400\] (I have converted 6hrs 40 minutes into 400 minutes)

Answer 2

Can you use simultaneous equations to solve this?

Answer 3

????????

Answer 4

From the second equation, we can also find that \[x=400-y\]

Answer 5

by rearranging the values, yes?

Answer 6

You can then substitute that value for x into the first equation, leaving only one variable in the equation. This is known as solving through simultaneous equations.

Answer 7

No 6hrs 40 mins will be

6 + 40/60

As per this book :/

Answer 8

Yeah but I like to simplify it into a mixed number.. It's just easier working with that.

Answer 9

Anyway let me complete the equation for you. \[400-y=y+\] and then \[2y=400, y=200\] If y = 200 minutes, then converted back to hours, it is 3 hours and 20 minutes.

Answer 10

It's suppose to be 13 hrs and 15 hrs :/
@perl @Sepeario

Answer 11

12hrs and 15 hrs**

Answer 12

1/x + 1/x-3 = 20/3??
@perl

Answer 13

Never mind, I stuffed up :/

Answer 14

Sorry lol

Answer 15

u mean r^1 and r^2 right??

Answer 16

they are the unknown rates of the pipes

Answer 17

what r subscripts ??????

Answer 18

its a way to label our unknown rates

Answer 19

you can use x and y if you like instead

Answer 20

1 = (2x + 3)(20/3)????

Answer 21

what next
explain please

Answer 22

i think we should do this over again

Answer 23

Alright @perl

Answer 24

lets do this carefully

Answer 25

Can't we do
1/x + 1/x-3 = 20/3?

Answer 26

yes that works

Answer 27

i used t, and you used x thats the only difference

Answer 28

Let x and y be the rates of the first and second pipe respectively.

One pipe can fill a cistern in 3 hours less than the other.
I am going to use : distance = rate * time, but distance is number of cisterns completed (like number of miles completed by a runner)

Lets assume the time it takes for the first pipe with rate x to fill one cistern is t. Then the time it takes for the second pipe with rate y is t - 3.

1 cistern = x * t
1 cistern = y * (t - 3)

1 = x*t
1 = y ( t - 3)

x = 1/t
y = 1/ ( t - 3)

Now it says if we have the two pipes working together, they can complete one cistern in
6 and 40/60 hour

1 = ( x + y ) ( 60 + 40/60)

substitute

1 = ( 1/t + 1/(t-3) ) ( 60 + 40/60)

1 = ( 1/t + 1/(t-3) ) ( 20/3)

3/20 = 1/t + 1 / (t-3)

Answer 29

1/x + 1/x-3 = 20/3
x-3+x/x^2-3x = 20/3?

Answer 30

20x^2 - 66x + 9 = 0?

Answer 31

That's what I wrote in a comment above.
so now?

Answer 32

theres an arithmetic mistake

Answer 33

\(\color{blue}{\text{Originally Posted by}}\) @AaronAndyson
1/x + 1/x-3 = 20/3
x-3+x/x^2-3x = 20/3?
\(\color{blue}{\text{End of Quote}}\)
this is wrong
it should be
1/x + 1/(x-3) = 3/20

Answer 34

?

Answer 35

your equation is incorrect

Answer 36

correct equation: 1/x + 1/(x-3) = 3/20

Answer 37

1/x + 1/(x-3) = 3/20

multiply both sides by x*(x-3)

[x*(x-3)] (1/x + 1/(x-3)] =x*(x-3) 3/20

distribute

x*(x-3)*1/x + x(x-3)*1/(x-3) =x*(x-3) 3/20

cancel

(x-3) + x = x(x-3) * 3/20

2x -3 = (x^2 -3x) * 3/20

Multiply both sides by 20

20(2x-3) = (x^2 -3x)*3

40x - 60 = 3x^2 - 9x

3x^2 - 9x - 40x + 60 = 0

3x^2 - 49x + 60 = 0

(3x -4) ( x-15) = 0

x = 15, 4/3

Answer 38

"(3x -4) ( x-15) = 0 "
How???????????

Answer 39

I factored
3x^2 -49x + 60 = 0

Answer 40

omg, im so good with these, but @perl is better. >.<

Answer 41

Explain the factoring process @perl

Answer 42

I will make that look better for you. I hope Perl will not mind it. :)

\[\cfrac{1}{x} + \cfrac{1}{x-3} = \cfrac{3}{20} \\
\cfrac{1}{x} \times x(x-3) + \cfrac{1}{x-3} \times x(x-3) = \cfrac{3(x)(x-3)}{20} \\
x-3 + x = \cfrac{3(x)(x-3)}{20} \\
2x - 3 = \cfrac{3x(x-3)}{20} \\
(2x -3)(20) = 3x(x-3) \\
40 x - 60 = 3x^2 -9x \\
3x^2 -49x + 60 = 0 \\
3x^2 - 45x - 4x + 60 = 0 \\
3x(x-15) - 4(x-15) = 0 \\
(3x-4) (x-15) = 0 \\
3x = 4 \ \text{OR} \ x = 15 \\
x = 4/3 \ \text{OR} \ x = 15 \]

Just thought to include LaTeX to make the equations look attractive ;)

Answer 43

3x^2 -49x + 60 = 0

We can break down the middle number -49x , then factor by grouping

3x^2 - 4x -45x + 60 = 0

x( 3x - 4) - 15( 3x - 4) = 0

(3x-4) ( x - 15) = 0

Answer 44

you can also use quadratic formula if you're stumped how to factor the quadratic, or it does not factor at all (that can happen)

Answer 45

And he actually found those numbers by :

product of roots should be 3 * 60 = 180
sum of roots should be -49

and so, by guessing the roots, he found the two numbers that satisfy these two conditions are : -45 and -4

:)

Answer 46

Thanks. now I need help with 2 more questions like this one

Answer 47

whats strange about this problem, there are two solutions?

Answer 48

Yes.

Answer 49

oh wait, one of the solutions do not work

Answer 50

we found that
x = 15, and x = 4/3

but if you plug in x = 4/3 you get a problem, (a negative rate for the second pipe).
so the only solution that works is x = 15

Answer 51

@mathslover
\(\color{blue}{\text{Originally Posted by}}\) @mathslover
And he actually found those numbers by :

product of roots should be 3 * 60 = 180
sum of roots should be -49

and so, by guessing the roots, he found the two numbers that satisfy these two conditions are : -45 and -4

:)
\(\color{blue}{\text{End of Quote}}\)


i think theres an issue here.

$$ \Large {
ax^2 + bx + c = 0
\\~\\ \text{sum of roots } r_1, r_2
\\ r_1\cdot r_2 = \frac{c}{a}
\\ r_1 + r_2 = \frac{-b}{a}




}
$$

Answer 52

what im saying is, the product of the roots is not a*c

Answer 53

Pardon me for that, perl! Didn't get it right. Thanks for pointing it out.

Answer 54

actually this isn't bad, we can go back and derive an interesting result

what if we were to do something

Answer 55

\[
\Large {
ax^2 + bx + c = 0
\\~\\ \text{sum of roots } r_1, r_2
\\ r_1\cdot r_2 = \frac{c}{a}
\\ r_1 + r_2 = \frac{-b}{a}

}
\] multiply both sides of first equation by a^2, second equation by a

\[
\Large {
\\ a^2 \cdot r_1\cdot r_2 = a^2 \cdot \frac{c}{a}
\\ a \cdot( r_1 + r_2) = a \cdot \frac{-b}{a}
}\] now rearrange and simplify
\[
\Large {
\\ (a \cdot r_1)(a \cdot r_2) = ac
\\ a \cdot r_1 + a \cdot r_2 = -b
}\] Now we have the form
\[
\Large {
\\ u\cdot v = ac
\\ u + v = -b
}\] So we see that if we can find two numbers that multiply to the product of the first and last number, a*c and add up to the middle number -b, we can factor the original quadratic.

Answer 56

that negative sign in front of the b is bugging me ...

Answer 57

That looks neat. Why does that negative sign bug you? Does it create any issues with the mathematical point of view? :O