Estimating Decimals & Percents - Pre Algebra

1. In a recent survey, $0.237$ of respondents preferred brand A, $0.184$ preferred brand B, and the rest were undecided. If there were $1,847$ total respondents, approximately how many more people preferred brand A than brand B?
   Answer: $98$ people Estimate by rounding: $1,847 \approx 1,800$ and $0.237 - 0.184 = 0.053 \approx 0.05$. So the difference is about $0.05 \times 1,800 = 90$ people, closest to $98$. Choice B uses exact decimals with rounded population. Choice C uses original population with rounded difference. Choice D uses both exact values.
2. A recipe calls for $2.75$ cups of flour, but you want to make $130\%$ of the recipe. You only have a $\frac{1}{3}$ cup measuring tool. Approximately how many scoops will you need?
   Answer: $11$ scoops This problem combines three key pre-algebra skills: converting percentages to decimals, multiplying decimals, and dividing fractions. When you see a recipe scaling problem, break it down into clear steps. First, calculate how much flour you need for 130% of the recipe. Convert the percentage: $130\% = 1.30$. Then multiply: $2.75 \times 1.30 = 3.575$ cups of flour. Next, determine how many $\frac{1}{3}$-cup scoops you need by dividing: $3.575 \div \frac{1}{3}$. To divide by a fraction, multiply by its reciprocal: $3.575 \times 3 = 10.725$ scoops. Since you can't make partial scoops when measuring, you need to round up to 11 complete scoops. Looking at the wrong answers: Choice A (10 scoops) gives you $10 \times \frac{1}{3} = 3.33$ cups, which is slightly less than the 3.575 cups needed. Choice B (13 scoops) likely comes from incorrectly rounding 10.725 to the nearest whole number above 12, or from calculation errors. Choice C (12 scoops) results from rounding 10.725 to the nearest whole number instead of considering that you need at least enough flour, so you must round up to ensure you have sufficient ingredients. Remember: in cooking problems involving measurements, always round up when the exact amount falls between whole measuring units. You want to have enough ingredients, and having slightly more is better than running short mid-recipe.
3. A investment account loses $15.3\%$ of its value in the first year, then gains $18.7\%$ in the second year. If the account started with $8,500, approximately what is its value after two years?
   Answer: $8,800 Round to $15\%$ loss and $19\%$ gain on $8,500. After year 1: $\$8,500 \times 0.85 \approx \$7,225$. After year 2: $\$7,225 \times 1.19 \approx \$8,598$, closest to $8,800. Choice A stops after the first year loss. Choice B assumes the changes cancel out. Choice D adds the percentages linearly to the original amount.
4. A factory produces widgets at a rate where $12.7\%$ are defective. If the factory aims to ship $2,450$ non-defective widgets, approximately how many total widgets must they produce?
   Answer: $2,800$ widgets If $12.7\%$ are defective, then $87.3\%$ are good. Round to $87\%$ good and $2,450$ to $2,500$. We need $0.87x = 2,500$, so $x \approx 2,500 \div 0.87 \approx 2,500 \div 0.9 \approx 2,778$, closest to $2,800$. Choice A adds the defective amount incorrectly. Choice C uses $85\%$ good rate. Choice D significantly overestimates the defective rate impact.
5. The price of gasoline increased from $3.47 per gallon to $4.23 per gallon over six months. What is the best estimate for the percent increase?
   Answer: $22\%$ When you encounter percent increase problems, you're finding how much a quantity has grown relative to its original value. The formula is: percent increase = $\frac{\text{new value} - \text{original value}}{\text{original value}} \times 100\%$. Here, gasoline increased from $3.47 to $4.23. The change is $\$4.23 - \$3.47 = \$0.76$. So the percent increase is $\frac{0.76}{3.47} \times 100\%$. To estimate this quickly, round $3.47 to $3.50 and $0.76 to $0.75. Now you have $\frac{0.75}{3.50} = \frac{3}{14} \approx 0.21 = 21\%$. This is closest to $22\%$, making D the correct answer. Choice A ($18\%$) is too low—it underestimates the substantial price jump. Choice B ($30\%$) represents a common error where students might incorrectly calculate $\frac{0.76}{4.23}$ using the new price as the denominator instead of the original price. Choice C ($26\%$) could result from rounding errors or miscalculation, perhaps from estimating $\frac{0.76}{3.00}$ instead of using the actual denominator. For percent increase problems, always remember: divide the change by the original amount, not the new amount. Practice estimating by rounding to friendly numbers—turning $\frac{0.76}{3.47}$ into $\frac{3}{14}$ makes the mental math much easier and gets you close enough to identify the right answer choice.
6. A store marks up items by $40\%$ above wholesale cost. During a sale, they offer $25\%$ off the retail price. If an item costs $60 wholesale, approximately what profit does the store make during the sale?
   Answer: $9 Wholesale cost: $60. For estimation, round the $40\%$ markup to $50\%$: Retail price ≈ $\$60 \times 1.5 = \$90$. Sale price: $\$90 \times 0.75 = \$67.50$. Profit: $\$67.50 - \$60 = \$7.50 ≈ \$9$. Choice A uses exact calculations without rounding. Choice B underestimates the markup. Choice D overestimates the final profit.
7. A baseball player's batting average dropped from $0.347$ to $0.298$ during a slump. By approximately what percent did their batting average decrease?
   Answer: $14\%$ Round $0.347$ to $0.35$ and $0.298$ to $0.30$. The decrease is $0.05$. Percent decrease = $\frac{0.05}{0.35} \times 100\% \approx 14.3\%$, closest to $14\%$. Choice B uses $\frac{0.05}{0.30}$ (wrong denominator). Choice C uses the difference $0.049$ without proper context. Choice D uses $\frac{0.05}{0.25}$ from misreading.
8. In a chemistry lab, a solution is $0.0847$ molar concentration. If you need to dilute it to $65\%$ of its original concentration and you start with $250$ mL, approximately how much water should you add?
   Answer: $130$ mL For dilution, if final concentration is $65\%$ of original, then $C_1V_1 = C_2V_2$ gives us $250 \text{ mL} = 0.65 \times V_2$, so $V_2 = \frac{250}{0.65} \approx \frac{250}{0.65} \approx 385$ mL total volume. Water to add: $385 - 250 = 135$ mL, closest to $130$ mL. Choice B uses $\frac{250}{0.6}$. Choice C uses $250 \times 0.6$. Choice D uses $250 \times 0.65$.
9. A restaurant bill comes to $47.83 before tax. The tax rate is $8.25\%$ and you want to leave a $18\%$ tip on the pre-tax amount. Which of the following is the best estimate for the total amount you will pay?
   Answer: $62 To estimate: Round $47.83 to $48. Tax is about $8\% \times \$48 = \$3.84 \approx \$4$. Tip is about $18\% \times \$48 = \$8.64 \approx \$9$. Total: $\$48 + \$4 + \$9 = \$61 \approx \$62$. Choice A forgets either tax or tip. Choice C uses tip on post-tax amount. Choice D overestimates both percentages significantly.

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