Solve mixture and concentration problems.
Free, timed SHSAT practice sets on mixture problems— every question includes a full worked explanation, and there's no account required.
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Mixture problems are weighted averages: the ingredient you use more of pulls the mix price toward its own price.
Price of a mixture: price per pound = total cost \(\div\) total weight. Example: 2 lb at 3 dollars plus 3 lb at 8 dollars costs \(6 + 24 = 30\) dollars for 5 lb, so 6 dollars per lb, not 5.50 dollars (the plain average of the two prices).
How much of each to use: let \(x\) be the pounds of the cheaper item, so the rest is total \(- x\). Then cost of the parts = cost of the mix. To make 9 lb at 5 dollars from 4-dollar and 7-dollar items: \(4x + 7(9 - x) = 45\), so \(63 - 3x = 45\) and \(x = 6\) lb.
Traps:
Check: the mix price must fall between the two prices, closer to the one you use more of.
Total cost \(= 9 \times \$6.75 + 11 \times \$4.75 = \$113.00\). Total weight \(= 9 + 11 = 20\) lbs. Price per lb \(= \$113.00 \div 20 = \$5.65\). (Averaging the two prices without weighting by amount gives $5.75. Multiplying each price by the other item's amount (\((11 \times 6.75 + 9 \times 4.75) \div 20\)) gives $5.85. Leaving out the cost of the jasmine rice gives about $3.04.)
By alligation, pounds of cheddar popcorn \(= \dfrac{30}{120} \times 12 = 3\) lbs. (Reporting the other ingredient's amount (the caramel popcorn) gives 9 lbs. Assuming an even split gives 6 lbs. Reporting the total weight of the mixture gives 12 lbs.)