Adding & Subtracting Fractions - Pre Algebra

1. A tank is $\frac{5}{8}$ full of water. After using $\frac{1}{6}$ of the tank's total capacity for irrigation, rain adds water equal to $\frac{1}{12}$ of the tank's total capacity. What fraction of the tank is now filled with water?
   Answer: $\frac{13}{24}$ of the tank Starting with $\frac{5}{8}$ of the tank full, then subtracting $\frac{1}{6}$ (used for irrigation) and adding $\frac{1}{12}$ (from rain): $\frac{5}{8} - \frac{1}{6} + \frac{1}{12}$. Converting to common denominator 24: $\frac{15}{24} - \frac{4}{24} + \frac{2}{24} = \frac{15-4+2}{24} = \frac{13}{24}$. Choice A ($\frac{7}{12} = \frac{14}{24}$) might result from computational errors. Choice C ($\frac{11}{24}$) could result from adding instead of subtracting the irrigation amount. Choice D ($\frac{5}{12} = \frac{10}{24}$) might result from multiple calculation errors.
2. A garden sprinkler system has three zones. Zone A uses $\frac{3}{8}$ of the total daily water allocation, Zone B uses $\frac{1}{6}$ of the total allocation, and Zone C uses the remaining water. If the system malfunctions and Zone A uses an additional $\frac{1}{12}$ of the total daily allocation, what fraction of the total daily allocation does Zone C now receive?
   Answer: $\frac{3}{8}$ of the allocation Originally, Zone C gets: $1 - \frac{3}{8} - \frac{1}{6}$. Converting to common denominator 24: $1 - \frac{9}{24} - \frac{4}{24} = \frac{24-9-4}{24} = \frac{11}{24}$. After malfunction, Zone A uses $\frac{3}{8} + \frac{1}{12} = \frac{9}{24} + \frac{2}{24} = \frac{11}{24}$. Zone B still uses $\frac{1}{6} = \frac{4}{24}$. Zone C now gets: $1 - \frac{11}{24} - \frac{4}{24} = \frac{9}{24} = \frac{3}{8}$. Choice B ($\frac{1}{3} = \frac{8}{24}$) might result from minor calculation errors. Choice C ($\frac{5}{12} = \frac{10}{24}$) could result from not accounting for the malfunction properly. Choice D ($\frac{1}{4} = \frac{6}{24}$) might result from various computational mistakes.
3. A recipe for trail mix calls for $\frac{3}{5}$ cup of nuts and $\frac{1}{4}$ cup of dried fruit. Lisa wants to make $\frac{2}{3}$ of this recipe, but she accidentally adds $\frac{1}{3}$ cup of nuts instead of the correct amount. How much more nuts does she need to add to have the right amount for $\frac{2}{3}$ of the recipe?
   Answer: $\frac{1}{15}$ cup more For $\frac{2}{3}$ of the recipe, Lisa needs $\frac{2}{3} \times \frac{3}{5} = \frac{6}{15} = \frac{2}{5}$ cup of nuts. She added $\frac{1}{3} = \frac{5}{15}$ cup. The correct amount needed is $\frac{6}{15}$ cup. Additional nuts needed: $\frac{6}{15} - \frac{5}{15} = \frac{1}{15}$ cup. Choice B ($\frac{2}{15}$) might result from calculation errors. Choice C ($\frac{1}{5} = \frac{3}{15}$) could result from finding the difference between the full recipe amount and what she added. Choice D ($\frac{4}{15}$) might result from various computational mistakes.
4. A carpenter cuts a board into three pieces. The first piece is $\frac{2}{5}$ of the original length, and the second piece is $\frac{1}{3}$ of the original length. If $\frac{1}{15}$ of the original board length was lost to sawdust during cutting, what fraction of the original length is the third piece?
   Answer: $\frac{1}{5}$ of the original length The total must equal 1 (the whole board). We have: first piece + second piece + third piece + sawdust = 1. So: $\frac{2}{5} + \frac{1}{3} + \text{third piece} + \frac{1}{15} = 1$. Converting to common denominator 15: $\frac{6}{15} + \frac{5}{15} + \text{third piece} + \frac{1}{15} = 1$. This gives us: $\frac{12}{15} + \text{third piece} = 1$. Therefore: third piece = $1 - \frac{12}{15} = \frac{15}{15} - \frac{12}{15} = \frac{3}{15} = \frac{1}{5}$. Choice A is correct. Choice B ($\frac{4}{15}$) would result if students forgot to account for sawdust. Choice C ($\frac{1}{15}$) might result from calculation errors. Choice D ($\frac{2}{15}$) could result from misreading the sawdust amount.

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