Confusing

[scoot212]

At a footbal game, 200 souvenirs were sold, some at $1.50 each and the rest $2.50 each. if the proceeds from the sales totaled $380, how many of each kind were sold?

[Mr. Hui]

Try guess and check or make a table. There is a method using linear programming also.

Solution: (highlight to read)
120 * $1.50 + 80 *$2.50 = $380

[hunter34]

Mr. Hui, you didn't follow one the problems parts.

Let x be the number of $1.50 souvenirs sold, and y be the number of $2.50. So x+y=200, this means that y=200-x. Now solving the equation:

\begin{eqnarray*}
1.50x+2.50(200-x)&=&380\\
1.50x-2.50x+500&=&380\\
-x&=&-120\\
x&=&120
\end{eqnarray*}

Then, you can easily find y which is 80. So the Solution to set is that there are 120 $1.50 souvenirs and 80 $2.50 souvenirs.


Edit:You have to fix the Hide feature so it can cover latex.

Edit by Temperal: You can use [more] instead.

[Mr. Hui]

Thanks hunter34. I completely ignored the item count constraint.

I was using Excel's Solver. :cool: