# Too thick to solve this problem!!!!

• Jan 19th 2006, 12:24 PM
bonbhoy
Too thick to solve this problem!!!!
A housewife buys a certain amount of beef at £4 per kilo and the same amount of sausages at £3 per kilo. If she had split the money she spent equally between the beef and sausages she would have got 2 kilos more. How much did she spend
• Jan 19th 2006, 01:06 PM
ThePerfectHacker
Quote:

Originally Posted by bonbhoy
A housewife buys a certain amount of beef at £4 per kilo and the same amount of sausages at £3 per kilo. If she had split the money she spent equally between the beef and sausages she would have got 2 kilos more. How much did she spend

It is worded in a confusing sense but I think I understand it. Let $x$ represent the money she spend on sausages. Also $x$ is the money she spend on beef because she spilt the money equally. Since she can buy more sausages then beef (they are less expensive) the difference between them is 2 (because she gets 2 more kilos). Now place that into an equation. If she spends $x$ yen on sausages then she gets $\frac{x}{3}$kilos because they are worth (3 yen per kilo). Similarily she gets $\frac{x}{4}$ kilos of beef. Thus, their difference is 2:
$\frac{x}{3}-\frac{x}{4}=2$
To solve this you need to find the common denominator. Which is 12, and multiply both sides by 12:
$12\frac{x}{3}-12\frac{x}{4}=12\times 2$
thus,
$4x-3x=24$
thus,
$x=24kg$

Thus, she brought 24 kilos for both of them. Thus, a total of 48 kilograms of meat where brough.
• Jan 19th 2006, 02:14 PM
CaptainBlack
Quote:

Originally Posted by bonbhoy
A housewife buys a certain amount of beef at £4 per kilo and the same amount of sausages at £3 per kilo. If she had split the money she spent equally between the beef and sausages she would have got 2 kilos more. How much did she spend

Let $x$ denote the mass of beef bought and $s$ be the amount spent.

Then:

$4x+3x=s$

Now if instead $s/2$ is spent on beef and also on sausage these
buy $(s/2)/4\ kg$ of beef and $(s/2)/3\ kg$ of sausage. As the sum
of these is $2\ kg$ more than previously we have:

$s/8+s/6 = 2x+2$,

but $s=7x$ so:

$\frac{7x}{8}+\frac{7x}{6}=2x+2$,

which has solution $x=48\ kg$, or $s= 336$pounds

These numbers seem implausibly large given the scenario.

RonL