Question 1125550
Here is a very different way to solve mixture problems like this which, if you understand it, will get you to the answer much faster and with less work than the usual algebraic method shown by the other tutor.<br>
Think of starting with the lower priced candy and adding the more expensive candy.  The price of the mixture moves from $1.64 per pound towards $3.36 per pound.  You want to stop adding the more expensive candy when the cost of the mixture is $2.33 per pound.<br>
The fraction comparing how far you go from the starting point of $1.64 to $2.33, compared to the total difference between $1.64 and $3.36, tells you how much of the more expensive candy you need to add.<br>
The explanation is probably a bit confusing.  So let's look at the actual calculations to see how easy this method is.<br>
2.33-1.64 = 0.69; 3.36-1.64 = 1.72; 0.69/1.72 = 69/172<br>
The cost per pound of the mixture is $0.69 more than the cost of the less expensive candy; the cost per pound of the more expensive candy is $1.72 more than the price of the less expensive candy.  So the cost of the mixture is 69/172 of the way from the cost of the lower priced candy to the cost of the higher priced candy.<br>
That calculation tells you 69/172 of the mixture must be the more expensive candy.  So<br>
{{{(69/172)*30 = 12.035}}}<br>
The mixture should have 12.035 pounds of the more expensive candy and 30-12.035 = 17.965 pounds of the less expensive candy.