Solving for a Variable in Formulas - Pre Algebra

1. The formula $C = \frac{5}{9}(F - 32)$ converts temperature from Fahrenheit to Celsius. On a day when the temperature increased from $41°F$ to $68°F$, what was the increase in Celsius degrees?
   Answer: $15°C$ Convert both temperatures to Celsius. For 41°F: $C_1 = \frac{5}{9}(41 - 32) = \frac{5}{9}(9) = 5°C$. For 68°F: $C_2 = \frac{5}{9}(68 - 32) = \frac{5}{9}(36) = 20°C$. The increase is $20 - 5 = 15°C$. Choice A results from calculation errors in the conversions. Choice C gives the final Celsius temperature instead of the increase. Choice D gives the Fahrenheit increase instead of the Celsius increase.
2. The formula $F = \frac{9}{5}C + 32$ converts Celsius to Fahrenheit. During a science experiment, the temperature needs to be maintained between $68°F$ and $86°F$. What is the Celsius temperature range?
   Answer: $20°C$ to $30°C$ When you encounter temperature conversion problems with ranges, you need to convert both endpoints of the Fahrenheit range to find the corresponding Celsius range. This involves solving the conversion formula for C, then applying it to both boundary values. To find the Celsius equivalents, rearrange the formula $F = \frac{9}{5}C + 32$ to solve for C. Subtract 32 from both sides: $F - 32 = \frac{9}{5}C$. Then multiply both sides by $\frac{5}{9}$: $C = \frac{5}{9}(F - 32)$. For the lower bound (68°F): $C = \frac{5}{9}(68 - 32) = \frac{5}{9}(36) = 20°C$ For the upper bound (86°F): $C = \frac{5}{9}(86 - 32) = \frac{5}{9}(54) = 30°C$ The Celsius range is 20°C to 30°C, making D correct. Looking at the wrong answers: A (15°C to 25°C) represents temperatures that are too cold - this range converts to roughly 59°F to 77°F. B (30°C to 40°C) gives temperatures that are too hot, converting to 86°F to 104°F. C (25°C to 35°C) is also too warm, converting to 77°F to 95°F. Study tip: When working with temperature conversion ranges, always solve the conversion formula for the unknown variable first, then substitute both endpoint values. Double-check by converting your answer back to the original units - your Celsius range should give you back the original Fahrenheit range.
3. The formula $d = rt$ relates distance, rate, and time. Two cars start from the same point and travel in opposite directions. Car A travels at $55$ mph and Car B travels at $65$ mph. After how many hours will they be $300$ miles apart?
   Answer: $2.5$ hours Since the cars travel in opposite directions, their combined rate of separation is $55 + 65 = 120$ mph. Using $d = rt$: $300 = 120t$, so $t = \frac{300}{120} = 2.5$ hours. Choice A results from using only one car's speed in the calculation. Choice C comes from arithmetic errors. Choice D results from incorrectly subtracting the speeds instead of adding them: $\frac{300}{65} \approx 4.6$.
4. The formula $P = 2l + 2w$ gives the perimeter of a rectangle. A rectangular garden has a perimeter of $84$ feet and a length that is $6$ feet more than twice the width. What is the width of the garden?
   Answer: $12$ feet Let $w$ be the width. Then the length is $l = 2w + 6$. Substitute into the perimeter formula: $84 = 2(2w + 6) + 2w = 4w + 12 + 2w = 6w + 12$. Solving: $84 - 12 = 6w$, so $72 = 6w$, thus $w = 12$ feet. Choice B results from solving $84 = 2l + 2w$ with $l = 2w + 6$ but making arithmetic errors. Choice C gives the length instead of the width. Choice D results from setting up the equation incorrectly.
5. The formula for the volume of a triangular prism is $V = \frac{1}{2}bhl$, where $b$ is the base of the triangle, $h$ is the height of the triangle, and $l$ is the length of the prism. A triangular prism has a volume of $180$ cubic centimeters, a triangular base with height $6$ cm, and the prism length is $5$ cm. What is the length of the triangle's base?
   Answer: $12$ cm Substitute the known values: $180 = \frac{1}{2}b(6)(5)$. This simplifies to $180 = \frac{1}{2}b(30) = 15b$. Solving for b: $b = \frac{180}{15} = 12$ cm. Choice A results from using the wrong formula $V = bhl$ without the $\frac{1}{2}$. Choice B comes from confusing which variable represents which measurement. Choice D results from calculation errors when solving $15b = 180$.
6. The formula $A = P(1 + rt)$ gives the amount $A$ after $t$ years when principal $P$ earns simple interest at rate $r$. If Maria invests money for $6$ years at $4\%$ annual interest and the final amount is $1860, what was her initial investment?
   Answer: $1500 Substitute known values: $1860 = P(1 + 0.04 \times 6)$. This gives $1860 = P(1 + 0.24) = P(1.24)$. Solving for P: $P = \frac{1860}{1.24} = 1500$. Choice A results from incorrectly using the compound interest formula or calculation errors. Choice C comes from solving $1860 = P + P(0.04)(6)$ but making arithmetic mistakes. Choice D results from confusing the problem and thinking the interest earned was $60.
7. The formula $I = Prt$ calculates simple interest. If $2400 in interest is earned on a principal of $8000 over $5$ years, what is the annual interest rate as a percentage?
   Answer: $6\%$ When you encounter a simple interest problem, you're working with the formula $I = Prt$, where $I$ is interest earned, $P$ is principal (starting amount), $r$ is the annual interest rate (as a decimal), and $t$ is time in years. Your goal is to solve for the unknown variable by substituting the given values. Here, you know $I = \$2400$, $P = \$8000$, and $t = 5$ years. Substitute these into the formula: $2400 = 8000 \times r \times 5$. This simplifies to $2400 = 40000r$. Solving for $r$: $r = \frac{2400}{40000} = 0.06$. Since interest rates are expressed as percentages, multiply by 100: $0.06 \times 100 = 6\%$. Looking at the wrong answers: Choice A ($4\%$) would only generate $1600 in interest over 5 years ($$8000 \times 0.04 \times 5$$), which is too low. Choice B ($$10\%$$) would generate $4000 in interest, nearly double what was actually earned. Choice C ($$8\%$$) would produce $3200 in interest, also more than the given $2400. The key strategy for simple interest problems is to carefully substitute known values and solve algebraically for the unknown. Always remember to convert your final decimal answer to a percentage when the question asks for a rate. Double-check by plugging your answer back into the original formula to verify it produces the given interest amount.

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