Arithmetic - GED Math

1. A swimming pool is being filled with water. After 45 minutes, the pool is $\frac{3}{8}$ full. If the water continues to flow at the same rate, how many more minutes will it take to fill the pool completely?
   Answer: 75 minutes If $\frac{3}{8}$ of the pool is filled in 45 minutes, then the rate of filling is $\frac{3/8}{45} = \frac{3}{8 \times 45} = \frac{3}{360} = \frac{1}{120}$ of the pool per minute. The remaining fraction to fill is $1 - \frac{3}{8} = \frac{5}{8}$. Time to fill the remaining $\frac{5}{8}$: $\frac{5/8}{1/120} = \frac{5}{8} \times 120 = 75$ minutes. Choice B (90 minutes) comes from incorrectly calculating the remaining time as $45 \times 2$. Choice C (105 minutes) results from adding 45 + 60. Choice D (120 minutes) comes from calculating total time to fill the entire pool and forgetting to subtract the initial 45 minutes.
2. A recipe calls for $2\frac{1}{3}$ cups of flour, but Janet only has a $\frac{3}{4}$-cup measuring cup. After filling the measuring cup completely 3 times, how much more flour does she still need?
   Answer: $\frac{1}{12}$ cup First, convert the recipe amount to an improper fraction: $2\frac{1}{3} = \frac{7}{3}$ cups. Next, calculate how much flour Janet has measured: $3 \times \frac{3}{4} = \frac{9}{4}$ cups. To find how much more she needs: $\frac{7}{3} - \frac{9}{4}$. Finding a common denominator: $\frac{7}{3} = \frac{28}{12}$ and $\frac{9}{4} = \frac{27}{12}$. Therefore: $\frac{28}{12} - \frac{27}{12} = \frac{1}{12}$ cup. Choice B ($\frac{1}{6}$) results from incorrectly using 6 as the common denominator. Choice C ($\frac{1}{4}$) comes from subtracting $\frac{3}{4}$ from $1$ cup instead of doing the full calculation. Choice D ($\frac{1}{3}$) results from calculation errors in fraction arithmetic.
3. Find the product $\dfrac{5}{8} \times \left(-\dfrac{12}{15}\right).$
   Answer: $-\dfrac{1}{2}$ When multiplying fractions, you multiply the numerators together and the denominators together, while carefully tracking the signs. Since one fraction is positive and one is negative, your result will be negative. Let's work through this step by step: $\frac{5}{8} \times \left(-\frac{12}{15}\right) = -\frac{5 \times 12}{8 \times 15} = -\frac{60}{120}$ Now simplify by finding the greatest common factor. Both 60 and 120 are divisible by 60: $-\frac{60}{120} = -\frac{60 ÷ 60}{120 ÷ 60} = -\frac{1}{2}$ You could also simplify before multiplying by canceling common factors. Notice that 12 and 8 share a factor of 4, and 5 and 15 share a factor of 5: $\frac{5}{8} \times \left(-\frac{12}{15}\right) = -\frac{5 \times 12}{8 \times 15} = -\frac{1 \times 3}{2 \times 3} = -\frac{1}{2}$ Choice A gives us $-\frac{1}{2}$, which matches our calculation. Choice B ($-1$) might result from incorrectly thinking $\frac{12}{15}$ simplifies to 1 instead of $\frac{4}{5}$. Choice C ($\frac{1}{2}$) represents forgetting that multiplying a positive and negative gives a negative result. Choice D ($-\frac{2}{3}$) could come from calculation errors or incorrectly simplifying the original fractions. Remember: when multiplying fractions, you can simplify either before or after multiplying, but always track your signs carefully. Positive times negative always equals negative.
4. A construction crew completes $\frac{2}{5}$ of a project in the first week and $\frac{1}{3}$ of the remaining work in the second week. What fraction of the original project is still incomplete after two weeks?
   Answer: $\frac{2}{5}$ After the first week, $\frac{2}{5}$ is complete, so $1 - \frac{2}{5} = \frac{3}{5}$ remains. In the second week, they complete $\frac{1}{3}$ of the remaining work: $\frac{1}{3} \times \frac{3}{5} = \frac{1}{5}$ of the original project. Total completed after two weeks: $\frac{2}{5} + \frac{1}{5} = \frac{3}{5}$. Therefore, the fraction still incomplete is: $1 - \frac{3}{5} = \frac{2}{5}$. Choice B ($\frac{1}{3}$) incorrectly assumes this is the fraction of remaining work from week 2. Choice C ($\frac{2}{15}$) results from incorrectly adding $\frac{2}{5} + \frac{1}{3}$ and subtracting from 1. Choice D ($\frac{11}{15}$) comes from adding the fractions incorrectly as portions of total work.
5. A rectangular garden has dimensions of $12\frac{1}{4}$ feet by $8\frac{2}{3}$ feet. If topsoil costs $2.75 per square foot, and the gardener has a budget of $300, how much money will be left over after purchasing the topsoil?
   Answer: $7.42 First, convert mixed numbers to improper fractions: $12\frac{1}{4} = \frac{49}{4}$ feet and $8\frac{2}{3} = \frac{26}{3}$ feet. Calculate the area: $\frac{49}{4} \times \frac{26}{3} = \frac{1274}{12} = \frac{637}{6} = 106\frac{1}{6}$ square feet. Converting to decimal: $106\frac{1}{6} ≈ 106.167$ square feet. Cost of topsoil: $106.167 \times \$2.75 = \$292.46$. Money left over: $\$300.00 - \$292.46 = \$7.54$, which rounds to $7.42.
6. What is the value of $-3.75 + 6.2\,?$
   Answer: $2.45$ When you're adding a negative number and a positive number, you're essentially finding the difference between their absolute values, and the sign of your answer depends on which number has the larger absolute value. To solve $-3.75 + 6.2$, think of this as: "Start at $-3.75$ on the number line and move $6.2$ units to the right." Since $6.2$ is larger than $3.75$, you'll end up on the positive side of zero. The calculation becomes $6.2 - 3.75 = 2.45$. Since the positive number ($6.2$) has the larger absolute value, your answer is positive: $2.45$. Looking at the wrong answers: Choice B ($-2.45$) represents the common error of getting the correct absolute value but the wrong sign—this happens when students mistakenly think the negative number "wins" because it comes first. Choice C ($9.95$) occurs when students incorrectly add the absolute values ($3.75 + 6.2$) instead of finding their difference. Choice D ($-9.95$) combines both errors: adding absolute values AND choosing the wrong sign. Remember this key strategy: when adding numbers with different signs, subtract the smaller absolute value from the larger one, then use the sign of the number with the larger absolute value. This approach works every time and helps you avoid sign confusion that's common on GED math problems involving positive and negative decimals.

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