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HuiLi · Feb 12, 2010, 06:52 PM

A trader bought some paraffin for $500.He paid $x for each litre of paraffin.
Qn:
a)Due to a leak,he lost 3 litres of paraffin.He sold the remainder of the paraffin for $1 per litre more than he paid for it.Write down an expression,in terms of x,for the sum of money he received.
b)He made a profit of $20.Write an equation in x to represent this information.

galactus · Feb 13, 2010, 11:37 AM

Since he paid $500 for all the parafin and paid $x for each liter, then he bought 500/x liters.

But 3 leaked out and he sold it for $1 more than what he paid.

$(x+1)(\frac{500}{x}-3)$

Now, you try to do part b. If he made $20 profit, then how much did he gross? Set the expression equal to that and solve for x.

HuiLi · Feb 14, 2010, 12:27 AM

I think I get something wrong here...
The equation I get:
523xsquare+23x-500=0
The answer:
3xsquare+23x-500=0

I can't find the problem with it

galactus · Feb 14, 2010, 04:57 AM

You made a mistake then. What I showed you reduces to $3x^{2}+23x-500=0$

Just expand

$(x+1)(\frac{500}{x}-3)-520=0$

and multiply by -x.

Post how you came up with what you have and maybe we can help. As is, we have no way of knowing how you got that.

HuiLi · Feb 14, 2010, 08:52 PM

Here is how I get it:

(x+1)(500/x-3)=520

(500-3x(x+1)/xsquare+x)-520=0

500x+500-3xsquare-3x/xsquare+x

(497x+500-3xsqiare/xsquare+x)-520=0

(497x+500-3xsqiare/xsquare+x)-(520xsquare+520x)/xsquare+x

both sides times -x

23x-500+523xsquare/-xsquare-x =0

How do I make it an equation from here on?

HuiLi · Feb 17, 2010, 04:19 AM

(520/x-3)(x-1)

520=(500x+500/xsquare+x)-3x+3

520square+520x=500x+500-3x+3

520xsquare+520x-503x+497=0

520xsquare+17x+497=0

this is the new method I can think of,but I still cannto get the answer?is anyone here?

Unknown008 · Feb 17, 2010, 07:28 AM

I think you have got a little problem in expansion:

$(x+1)(\frac{500}{x}-3) = 520$

I'll take only the left hand side for now.

$\begin{array}{rcl} (x+1)(\frac{500}{x}-3) &=& (x+1)(\frac{500-3x}{x}) \\ &=& \frac{(x+1)(500-3x)}{x} \\ &=& \frac{(500x+500-3x^2-3x)}{x} \\ &=& \frac{(497x+500-3x^2)}{x}\end{array}$

Now, I put back the 520 from before:

$\begin{array}{rcl} \frac{(497x+500-3x^2)}{x}&=& 520 \\ 497x+500-3x^2 &=& 520x\\ 0 &=& 3x^2 + 23x - 500\end{array}$

HuiLi · Feb 19, 2010, 05:09 AM

Thanks!that is very clear & helpful!
I finally understand maths & what I did wrong!yay for me :D

Unknown008 · Feb 19, 2010, 07:34 AM

You're welcome :)