Converting Forms Strategically - Pre Algebra

1. A bakery uses a mixture that is $60\%$ flour and $40\%$ other ingredients by weight. If they want to create a new mixture that is $75\%$ flour by adding pure flour, how much pure flour must they add to $20$ pounds of the original mixture?
   Answer: $12$ pounds Original mixture has $20 \times 0.6 = 12$ pounds of flour. Let $x$ = pounds of pure flour added. New mixture will have $(12 + x)$ pounds of flour out of $(20 + x)$ total pounds. Setting up equation: $\frac{12 + x}{20 + x} = 0.75$. Solving: $12 + x = 0.75(20 + x) = 15 + 0.75x$, so $0.25x = 3$, thus $x = 12$. Choice B calculates the final amount of flour needed, not the amount to add. Choice C uses incorrect percentage calculations. Choice D results from setting up the proportion incorrectly.
2. A store offers a discount such that customers pay $\frac{4}{5}$ of the marked price. During a special sale, an additional $15\%$ is taken off the already discounted price. If a customer pays $68 for an item during the special sale, what was the original marked price?
   Answer: $100 When you encounter multi-step discount problems, work backwards from the final price through each discount in reverse order. This systematic approach prevents calculation errors and helps you track the original value. Let's trace backwards from the $68 final price. First, the customer received an additional $15\%$ off the already discounted price, meaning they paid $85\%$ (or $\frac{85}{100} = \frac{17}{20}$) of the discounted price. So the discounted price before the special sale was: $\$68 \div \frac{17}{20} = \$68 \times \frac{20}{17} = \$80$. Next, this $80 represents the price after the original discount where customers pay $\frac{4}{5}$ of the marked price. Therefore: $\$80 \div \frac{4}{5} = \$80 \times \frac{5}{4} = \$100$. The original marked price was $100, confirming answer D. Answer A ($95) likely comes from incorrectly applying the $15\%$ additional discount to the original price instead of the already discounted price. Answer B ($120) results from miscalculating one of the fractions, possibly treating $\frac{4}{5}$ as $\frac{3}{5}$. Answer C ($85) appears to be the result of working forward incorrectly or confusing which discount applies first. Remember: when solving backwards through multiple discounts, divide by each discount rate (expressed as a decimal or fraction). When working forward, you multiply. Always identify which direction you're working before starting your calculations.
3. A car's value depreciates by $\frac{1}{8}$ each year. After two years, the car is worth $19,250. What was the car's original value?
   Answer: $24,640 If the car depreciates by $\frac{1}{8}$ each year, it retains $\frac{7}{8}$ of its value each year. After two years, it retains $(\frac{7}{8})^2 = \frac{49}{64}$ of its original value. If $\frac{49}{64} \times \text{original value} = 19,250$, then original value = $19,250 \times \frac{64}{49} = \$24,640$. Choice A results from incorrectly treating depreciation as simple rather than compound. Choice B uses wrong fraction calculations. Choice D results from using $\frac{1}{8}$ appreciation instead of depreciation.
4. The ratio of boys to girls in a class is $7:5$. If $25\%$ of the boys and $40\%$ of the girls wear glasses, what fraction of the entire class wears glasses?
   Answer: $\frac{5}{16}$ In the ratio $7:5$, there are $7$ boys and $5$ girls out of $12$ total students. Boys with glasses: $7 \times 0.25 = 1.75$. Girls with glasses: $5 \times 0.40 = 2$. Total with glasses: $1.75 + 2 = 3.75$ out of $12$, so $\frac{3.75}{12} = \frac{15}{48} = \frac{5}{16}$. Choice B averages the two percentages incorrectly. Choice C results from calculation errors. Choice D uses the wrong total.
5. A rectangular garden has dimensions in the ratio $3:4$. If the area is increased by $44\%$ while keeping the same ratio, and the new perimeter is $42$ meters, what were the original dimensions?
   Answer: $7.5$ m by $10$ m When you encounter problems involving ratios and area changes, you need to work systematically through the scaling relationships. Since the garden maintains a $3:4$ ratio, you can express the dimensions as $3x$ and $4x$ for some scale factor $x$. The original area is $3x \times 4x = 12x^2$. After a $44\%$ increase, the new area becomes $12x^2 \times 1.44 = 17.28x^2$. Since the ratio stays $3:4$, the new dimensions are $3y$ and $4y$ where $12y^2 = 17.28x^2$. Solving this: $y^2 = 1.44x^2$, so $y = 1.2x$. The new dimensions are $3.6x$ and $4.8x$. With a perimeter of $42$ meters: $2(3.6x + 4.8x) = 42$, which gives $16.8x = 42$, so $x = 2.5$. Therefore, the original dimensions were $7.5$ m by $10$ m. Choice A ($9$ m by $12$ m) would give $x = 3$, resulting in a new perimeter of $50.4$ m, not $42$ m. Choice B ($10.5$ m by $14$ m) corresponds to $x = 3.5$, yielding a new perimeter of $58.8$ m. Choice C ($6$ m by $8$ m) has $x = 2$, producing a new perimeter of $33.6$ m. Strategy tip: In ratio problems with area scaling, always establish your variable relationships first, then use the constraint (here, the perimeter) to solve for the scale factor. Double-check by verifying both the ratio and the given measurements work out correctly.
6. A solution is $15\%$ salt by volume. If $2.4$ liters of water is added to $3.6$ liters of this solution, what percent of the new mixture is salt?
   Answer: $9\%$ When you encounter mixture problems involving percentages, focus on tracking the actual amount of the substance (salt) and the total volume separately, then find the new percentage. Start with what you know: $3.6$ liters of $15\%$ salt solution contains $3.6 \times 0.15 = 0.54$ liters of pure salt. When you add $2.4$ liters of water, you're adding zero salt but increasing the total volume. The new mixture has $0.54$ liters of salt in a total volume of $3.6 + 2.4 = 6.0$ liters. The new percentage is $\frac{0.54}{6.0} = 0.09 = 9\%$, making C correct. Looking at the wrong answers: A) $10\%$ likely comes from incorrectly assuming the percentage drops proportionally to the volume increase ($15\% \times \frac{3.6}{6.0} = 9\%$ would be correct, but students might miscalculate). B) $7.5\%$ results from the common error of thinking the percentage is simply halved because you're roughly doubling the volume ($15\% \div 2 = 7.5\%$). D) $12\%$ might come from subtracting the percentage of water added relative to original solution ($15\% - 3\% = 12\%$), which is mathematically meaningless. For mixture problems, always calculate the actual amount of the key substance first, then find the new total, and finally compute the new percentage. Don't try to work directly with percentages—track the concrete quantities.
7. A store marks up items by $37.5\%$ above cost. If an item costs the store $24, what fraction of the selling price represents the store's profit?
   Answer: $\frac{3}{11}$ First convert the markup percentage: $37.5\% = \frac{3}{8} = 0.375$. The selling price is $\$24 + \$24 \times 0.375 = \$24 + \$9 = \$33$. The profit is $9, so the fraction of selling price that is profit is $\frac{9}{33} = \frac{3}{11}$. Choice A represents the markup rate as a fraction of cost, not selling price. Choice C represents cost as a fraction of selling price. Choice D results from incorrectly calculating $37.5\%$ as $\frac{5}{13}$.
8. Maria's test scores decreased by $12\%$ from her first test to her second test, then increased by $25\%$ from her second test to her third test. If her third test score was $77$, what was her first test score?
   Answer: $70$ Working backwards: if the third test score is $77$ and this represents a $25\%$ increase from the second test, then the second test score was $77 ÷ 1.25 = 61.6$. If the second test score of $61.6$ represents a $12\%$ decrease from the first test, then the first test score was $61.6 ÷ 0.88 = 70$. Choice A results from incorrectly applying the percentages in forward direction. Choice C comes from calculation errors in the decimal conversions. Choice D assumes simple arithmetic relationships rather than percentage calculations.
9. A recipe calls for ingredients in the ratio $2:3:5$ by weight. If the total weight of these three ingredients is $4.8$ kg, and one ingredient makes up $30\%$ of the mixture, what is the weight of the heaviest ingredient?
   Answer: $2.4$ kg The ratio $2:3:5$ means the parts are $\frac{2}{10} = 20\%$, $\frac{3}{10} = 30\%$, and $\frac{5}{10} = 50\%$ of the total. The ingredient that makes up $30\%$ corresponds to the middle ratio part (3). The heaviest ingredient corresponds to ratio part 5, which is $50\%$ of $4.8$ kg = $2.4$ kg. Choice B assumes the $30\%$ ingredient is the heaviest. Choice C miscalculates $30\%$ of $4.8$. Choice D calculates $30\%$ of $4.8$, identifying the wrong ingredient as heaviest.

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