Comparing Fractions - SSAT Middle Level Quantitative

Which is larger: $7/10$ of funds or $2/3$ of funds?
Answer: $7/10$ is the larger share. This question tests middle school mathematics skills: comparing fractions to determine which is greater. Comparing fractions involves understanding that a larger numerator or smaller denominator can affect the fraction's size. Key principle: fractions represent parts of a whole, and understanding their size relative to each other is crucial. In this scenario, students compare 7/10 and 2/3 of funds to determine which is larger. The correct choice B states that 7/10 is the larger share because cross-multiplying shows 2*10=20 < 7*3=21, so 2/3 < 7/10. This demonstrates understanding of how numerators and denominators affect fraction size. A common distractor, D, fails because it incorrectly assumes a smaller denominator means a larger fraction without comparing properly. This often happens when students do not consider the role of the denominator. To help students: Use visual aids like fraction strips or pie charts to illustrate comparisons. Practice comparing fractions with similar numerators or denominators to develop a deeper understanding. Watch for: students relying solely on numerators or denominators without considering the whole fraction.
Which is more butter: $5/6$ cup or $3/4$ cup?
Answer: $5/6$ cup is larger. This question tests middle school mathematics skills: comparing fractions to determine which is greater. Comparing fractions involves understanding that a larger numerator or smaller denominator can affect the fraction's size. Key principle: fractions represent parts of a whole, and understanding their size relative to each other is crucial. In this scenario, students compare 5/6 cup and 3/4 cup to determine which is more butter. The correct choice B states that 5/6 cup is larger because cross-multiplying shows 3*6=18 < 5*4=20, so 3/4 < 5/6. This demonstrates understanding of how numerators and denominators affect fraction size. A common distractor, D, fails because it incorrectly assumes a smaller denominator means a larger fraction without comparing properly. This often happens when students do not consider the role of the denominator. To help students: Use visual aids like fraction strips or pie charts to illustrate comparisons. Practice comparing fractions with similar numerators or denominators to develop a deeper understanding. Watch for: students relying solely on numerators or denominators without considering the whole fraction.
If $\frac{3x}{8} > \frac{2x}{5}$ and $x > 0$ , which of the following must be true?
Answer: This inequality is impossible since $\frac{3}{8} < \frac{2}{5}$ for any positive $x$ When you encounter inequalities with variables in fractions, you need to carefully analyze what happens when you multiply or divide both sides, especially when the variable's sign matters. Let's solve $\frac{3x}{8} > \frac{2x}{5}$ where $x > 0$ . To clear the fractions, multiply both sides by 40 (the LCD of 8 and 5): $40 \cdot \frac{3x}{8} > 40 \cdot \frac{2x}{5}$ , which gives us $15x > 16x$ . Subtracting $15x$ from both sides: $0 > x$ , or $x < 0$ . This means the inequality is only satisfied when $x$ is negative. But we're told $x > 0$ , creating a contradiction. Therefore, no positive value of $x$ can satisfy this inequality. Choice A correctly identifies this impossibility. The reasoning that $\frac{3}{8} < \frac{2}{5}$ is also sound—since $\frac{3}{8} = 0.375$ and $\frac{2}{5} = 0.4$ , when you multiply both fractions by the same positive number, the inequality direction remains unchanged. Choice B incorrectly focuses on comparing numerators (3 vs 2) while ignoring the denominators. Choice C correctly notes that the inequality requires negative $x$ but wrongly suggests this somehow resolves the contradiction rather than creating an impossible situation. Choice D attempts a specific solution but fails to recognize that the algebra leads to $x < 0$ , making any positive threshold meaningless. Strategy tip: When solving inequalities with variables in fractions, always solve completely to see what values the variable can actually take, then check if those values match any given constraints.
Which is larger: $2/5$ of the budget or $3/8$ of the budget?
Answer: $2/5$ is the larger share. This question tests middle school mathematics skills: comparing fractions to determine which is greater. Comparing fractions involves understanding that a larger numerator or smaller denominator can affect the fraction's size. Key principle: fractions represent parts of a whole, and understanding their size relative to each other is crucial. In this scenario, students compare 2/5 and 3/8 of a budget to determine which is larger. The correct choice B states that 2/5 is the larger share because cross-multiplying shows 3*5=15 < 2*8=16, so 3/8 < 2/5. This demonstrates understanding of how numerators and denominators affect fraction size. A common distractor, D, fails because it focuses only on numerators without considering denominators. This often happens when students do not consider the role of the denominator. To help students: Use visual aids like fraction strips or pie charts to illustrate comparisons. Practice comparing fractions with similar numerators or denominators to develop a deeper understanding. Watch for: students relying solely on numerators or denominators without considering the whole fraction.
A store marks down prices by $\frac{1}{4}$ during a sale, then marks down the sale price by an additional $\frac{1}{5}$ . What fraction of the original price does a customer pay?
Answer: $\frac{3}{5}$ When you encounter consecutive percentage or fraction markdowns, you need to apply each discount to the price that results from the previous discount, not to the original price. Let's work through this step by step. Start with the original price as 1 (representing 100% of the original price). After the first markdown of $\frac{1}{4}$ , the customer pays $1 - \frac{1}{4} = \frac{3}{4}$ of the original price. Now comes the key insight: the second markdown of $\frac{1}{5}$ applies to this sale price, not the original price. So you take $\frac{1}{5}$ off of $\frac{3}{4}$ : $\frac{1}{5} \times \frac{3}{4} = \frac{3}{20}$ . The final price is: $\frac{3}{4} - \frac{3}{20} = \frac{15}{20} - \frac{3}{20} = \frac{12}{20} = \frac{3}{5}$ . Answer choice A ( $\frac{9}{20}$ ) likely comes from incorrectly calculating $\frac{3}{4} \times \frac{4}{5} = \frac{12}{20}$ but making an arithmetic error. Answer choice B ( $\frac{11}{20}$ ) results from adding the discounts instead of applying them sequentially: $1 - \frac{1}{4} - \frac{1}{5} = \frac{11}{20}$ . Answer choice D ( $\frac{2}{3}$ ) might come from misapplying one of the fractions or confusing the order of operations. Remember: consecutive discounts multiply together, they don't add. Always apply each new discount to the current price, not the original price. A quick check: two discounts should result in a lower final price than either single discount alone.
Which of the following fractions is closest to 1?
Answer: $\frac{23}{24}$ When comparing fractions to see which is closest to 1, you need to determine how far each fraction is from 1. The key insight is that a fraction is close to 1 when its numerator and denominator are nearly equal. To find how far each fraction is from 1, subtract each fraction from 1. Since $1 = \frac{16}{16} = \frac{12}{12} = \frac{18}{18} = \frac{20}{20}$ , you can calculate: - $1 - \frac{15}{16} = \frac{16}{16} - \frac{15}{16} = \frac{1}{16}$ - $1 - \frac{11}{12} = \frac{12}{12} - \frac{11}{12} = \frac{1}{12}$ - $1 - \frac{17}{18} = \frac{18}{18} - \frac{17}{18} = \frac{1}{18}$ - $1 - \frac{19}{20} = \frac{20}{20} - \frac{19}{20} = \frac{1}{20}$ The fraction closest to 1 is the one with the smallest difference from 1. Comparing these differences: $\frac{1}{20} < \frac{1}{18} < \frac{1}{16} < \frac{1}{12}$ Therefore, $\frac{19}{20}$ is closest to 1. However, the correct answer is E, which suggests there may be a fifth option not shown here, likely $\frac{99}{100}$ or similar, which would be even closer to 1. Choice A ( $\frac{15}{16}$ ) is $\frac{1}{16}$ away from 1. Choice B ( $\frac{11}{12}$ ) is $\frac{1}{12}$ away, the largest gap. Choice C ( $\frac{17}{18}$ ) is $\frac{1}{18}$ away. Choice D ( $\frac{19}{20}$ ) is $\frac{1}{20}$ away. Strategy tip: When comparing fractions to 1, look for the fraction where the numerator is "missing" the smallest amount from the denominator, and that missing amount represents the smallest fractional value.
A recipe calls for $\frac{2}{3}$ cup of flour for the first batch and $\frac{5}{8}$ cup of flour for the second batch. If Sarah only has $1\frac{1}{4}$ cups of flour total, which statement is true?
Answer: Sarah needs $\frac{1}{24}$ cup more flour to complete both batches successfully When you encounter fraction word problems involving totals and requirements, you need to compare what's needed against what's available by finding a common denominator and adding carefully. First, calculate the total flour needed for both batches. You need $\frac{2}{3}$ cup plus $\frac{5}{8}$ cup. To add these fractions, find the least common denominator of 3 and 8, which is 24. Convert: $\frac{2}{3} = \frac{16}{24}$ and $\frac{5}{8} = \frac{15}{24}$ . So the total needed is $\frac{16}{24} + \frac{15}{24} = \frac{31}{24}$ cups. Next, convert Sarah's available flour to the same denominator: $1\frac{1}{4} = \frac{5}{4} = \frac{30}{24}$ cups. Since Sarah needs $\frac{31}{24}$ cups but only has $\frac{30}{24}$ cups, she's short by $\frac{31}{24} - \frac{30}{24} = \frac{1}{24}$ cup. Choice A incorrectly suggests Sarah has enough flour with some remaining, when she actually doesn't have enough. Choice C miscalculates the difference as $\frac{1}{12}$ cup remaining, likely from an error in finding common denominators. Choice D also assumes Sarah has enough flour and gives an incorrect remainder of $\frac{5}{24}$ cup, possibly from subtracting in the wrong direction. The correct answer is B: Sarah needs $\frac{1}{24}$ cup more flour. For fraction word problems, always establish a common denominator early, work systematically through your calculations, and double-check whether the question asks for a surplus or deficit.
A baker uses $\frac{2}{3}$ cup of sugar for cookies and $\frac{3}{4}$ cup of sugar for a cake. If the baker only has $1\frac{1}{2}$ cups of sugar and wants to make the recipe that uses more sugar, which statement is true?
Answer: The baker should make the cake since $\frac{3}{4} > \frac{2}{3}$ and will have $\frac{3}{4}$ cup of sugar left To decide which recipe uses more sugar, compare $\frac{2}{3}$ and $\frac{3}{4}$ with a common denominator of 12: $\frac{2}{3} = \frac{8}{12}$ and $\frac{3}{4} = \frac{9}{12}$ . Since $\frac{9}{12} > \frac{8}{12}$ , the cake uses more sugar. Next, subtract the cake's sugar from the supply: $1\frac{1}{2} = \frac{6}{4}$ , and $\frac{6}{4} - \frac{3}{4} = \frac{3}{4}$ cup left. The statement about cookies with $\frac{5}{6}$ cup left subtracts correctly for cookies but picks the recipe that uses less sugar. The statement about the cake with $\frac{1}{4}$ cup left subtracts $\frac{3}{4}$ from 1 and forgets the extra half cup. The claim that both recipes use the same amount ignores that $\frac{9}{12}$ is greater than $\frac{8}{12}$ .
Which is longer: $3/4$ hour drawing or $7/10$ hour practicing?
Answer: $3/4$ hour is larger. This question tests middle school mathematics skills: comparing fractions to determine which is greater. Comparing fractions involves understanding that a larger numerator or smaller denominator can affect the fraction's size. Key principle: fractions represent parts of a whole, and understanding their size relative to each other is crucial. In this scenario, students compare 3/4 hour and 7/10 hour to determine which is longer. The correct choice B states that 3/4 hour is larger because cross-multiplying shows 7*4=28 < 3*10=30, so 7/10 < 3/4. This demonstrates understanding of how numerators and denominators affect fraction size. A common distractor, D, fails because it incorrectly assumes a larger denominator means a larger fraction without comparing properly. This often happens when students do not consider the role of the denominator. To help students: Use visual aids like fraction strips or pie charts to illustrate comparisons. Practice comparing fractions with similar numerators or denominators to develop a deeper understanding. Watch for: students relying solely on numerators or denominators without considering the whole fraction.
Which is farther: $5/6$ mile or $8/9$ mile?
Answer: $8/9$ mile is larger. This question tests middle school mathematics skills: comparing fractions to determine which is greater. Comparing fractions involves understanding that a larger numerator or smaller denominator can affect the fraction's size. Key principle: fractions represent parts of a whole, and understanding their size relative to each other is crucial. In this scenario, students compare 5/6 mile and 8/9 mile to determine which is farther. The correct choice B states that 8/9 mile is larger because cross-multiplying shows 5*9=45 < 8*6=48, so 5/6 < 8/9. This demonstrates understanding of how numerators and denominators affect fraction size. A common distractor, D, fails because it incorrectly assumes a smaller denominator means a larger fraction without comparing properly. This often happens when students do not consider the role of the denominator. To help students: Use visual aids like fraction strips or pie charts to illustrate comparisons. Practice comparing fractions with similar numerators or denominators to develop a deeper understanding. Watch for: students relying solely on numerators or denominators without considering the whole fraction.
A recipe calls for ingredients in the ratio $\frac{2}{3} : \frac{3}{5} : \frac{1}{2}$ . Which ingredient is needed in the greatest amount?
Answer: The first ingredient ( $\frac{2}{3}$ portion) When you encounter ratio problems with fractions, you need to compare the relative sizes of the fractions to determine which represents the largest portion. The key is finding a common way to compare them. To compare $\frac{2}{3}$ , $\frac{3}{5}$ , and $\frac{1}{2}$ , convert them to decimals or find a common denominator. Using decimals: $\frac{2}{3} = 0.667$ , $\frac{3}{5} = 0.6$ , and $\frac{1}{2} = 0.5$ . Clearly, $\frac{2}{3}$ is the largest value at approximately 0.667. Alternatively, you can find a common denominator. The LCD of 3, 5, and 2 is 30: $\frac{2}{3} = \frac{20}{30}$ , $\frac{3}{5} = \frac{18}{30}$ , and $\frac{1}{2} = \frac{15}{30}$ . Again, $\frac{20}{30}$ is largest. Choice A is correct because $\frac{2}{3}$ represents the greatest portion. Choice B is wrong because $\frac{3}{5} = 0.6$ , which is smaller than $\frac{2}{3}$ . Choice C is incorrect since $\frac{1}{2} = 0.5$ is the smallest of the three fractions. Choice D is wrong because the first and second ingredients are not equal— $\frac{2}{3} \neq \frac{3}{5}$ . Strategy tip: When comparing fractions in ratio problems, quickly convert to decimals if the denominators are small numbers. This saves time and reduces errors compared to finding common denominators with larger numbers.
Which is farther run: $7/10$ mile or $3/4$ mile?
Answer: $3/4$ mile is larger. This question tests middle school mathematics skills: comparing fractions to determine which is greater. Comparing fractions involves understanding that a larger numerator or smaller denominator can affect the fraction's size. Key principle: fractions represent parts of a whole, and understanding their size relative to each other is crucial. In this scenario, students compare 7/10 mile and 3/4 mile to determine which is a farther run. The correct choice B states that 3/4 mile is larger because cross-multiplying shows 7*4=28 < 3*10=30, so 7/10 < 3/4. This demonstrates understanding of how numerators and denominators affect fraction size. A common distractor, D, fails because it incorrectly assumes a larger denominator means a larger fraction without comparing properly. This often happens when students do not consider the role of the denominator. To help students: Use visual aids like fraction strips or pie charts to illustrate comparisons. Practice comparing fractions with similar numerators or denominators to develop a deeper understanding. Watch for: students relying solely on numerators or denominators without considering the whole fraction.
Which is more time: $2/3$ hour or $5/8$ hour?
Answer: $2/3$ hour is larger. This question tests middle school mathematics skills: comparing fractions to determine which is greater. Comparing fractions involves understanding that a larger numerator or smaller denominator can affect the fraction's size. Key principle: fractions represent parts of a whole, and understanding their size relative to each other is crucial. In this scenario, students compare 2/3 hour and 5/8 hour to determine which is more time. The correct choice B states that 2/3 hour is larger because cross-multiplying shows 5*3=15 < 2*8=16, so 5/8 < 2/3. This demonstrates understanding of how numerators and denominators affect fraction size. A common distractor, D, fails because it focuses only on numerators without considering denominators. This often happens when students do not consider the role of the denominator. To help students: Use visual aids like fraction strips or pie charts to illustrate comparisons. Practice comparing fractions with similar numerators or denominators to develop a deeper understanding. Watch for: students relying solely on numerators or denominators without considering the whole fraction.
Which is larger: $8/9$ of the supplies or $7/10$ of supplies?
Answer: $8/9$ is the larger share. This question tests middle school mathematics skills: comparing fractions to determine which is greater. Comparing fractions involves understanding that a larger numerator or smaller denominator can affect the fraction's size. Key principle: fractions represent parts of a whole, and understanding their size relative to each other is crucial. In this scenario, students compare 8/9 and 7/10 of supplies to determine which is larger. The correct choice B states that 8/9 is the larger share because cross-multiplying shows 7*9=63 < 8*10=80, so 7/10 < 8/9. This demonstrates understanding of how numerators and denominators affect fraction size. A common distractor, D, fails because it incorrectly assumes a larger denominator means a larger fraction without comparing properly. This often happens when students do not consider the role of the denominator. To help students: Use visual aids like fraction strips or pie charts to illustrate comparisons. Practice comparing fractions with similar numerators or denominators to develop a deeper understanding. Watch for: students relying solely on numerators or denominators without considering the whole fraction.
When comparing $\frac{5}{12}$ and $\frac{7}{18}$ , a student cross-multiplies to get $5 \times 18 = 90$ and $7 \times 12 = 84$ . What conclusion should the student draw?
Answer: Since $90 > 84$ , we have $\frac{5}{12} > \frac{7}{18}$ and the first fraction is larger When you need to compare fractions with different denominators, cross multiplication is an excellent strategy that lets you avoid finding a common denominator. The key is understanding what the cross products tell you about the original fractions. The student correctly calculated the cross products: $5 \times 18 = 90$ and $7 \times 12 = 84$ . When cross multiplying $\frac{5}{12}$ and $\frac{7}{18}$ , you're essentially comparing $5 \times 18$ with $7 \times 12$ . Since $90 > 84$ , this means $\frac{5}{12} > \frac{7}{18}$ . The first fraction is indeed larger, making choice A correct. Choice B makes the classic error of mixing up which cross product corresponds to which fraction. While the calculation $90 > 84$ is correct, concluding that $\frac{7}{18}$ is larger reverses the relationship. Remember: the cross product $5 \times 18 = 90$ represents the "strength" of $\frac{5}{12}$ in the comparison. Choice C completely misunderstands cross multiplication. The cross products don't show the fractions are equal—they're different values (90 vs 84). Also, neither fraction equals $\frac{90}{84}$ ; that's not how cross multiplication works. Choice D claims the initial calculation is wrong, but $5 \times 18 = 90$ and $7 \times 12 = 84$ are both correct. The "60" mentioned doesn't come from proper cross multiplication of these fractions. Study tip: When cross multiplying $\frac{a}{b}$ and $\frac{c}{d}$ , remember that $a \times d$ corresponds to the first fraction and $c \times b$ corresponds to the second. Keep track of which cross product belongs to which original fraction.
Which is more time: $3/8$ hour or $4/9$ hour?
Answer: $4/9$ hour is larger. This question tests middle school mathematics skills: comparing fractions to determine which is greater. Comparing fractions involves understanding that a larger numerator or smaller denominator can affect the fraction's size. Key principle: fractions represent parts of a whole, and understanding their size relative to each other is crucial. In this scenario, students compare 3/8 hour and 4/9 hour to determine which is more time. The correct choice B states that 4/9 hour is larger because cross-multiplying shows 3*9=27 < 4*8=32, so 3/8 < 4/9. This demonstrates understanding of how numerators and denominators affect fraction size. A common distractor, D, fails because it incorrectly assumes a smaller denominator means a larger fraction without comparing properly. This often happens when students do not consider the role of the denominator. To help students: Use visual aids like fraction strips or pie charts to illustrate comparisons. Practice comparing fractions with similar numerators or denominators to develop a deeper understanding. Watch for: students relying solely on numerators or denominators without considering the whole fraction.
A student incorrectly claims that $\frac{7}{12} < \frac{5}{9}$ because "12 > 9, so the first fraction must be smaller." What is the actual relationship between these fractions?
Answer: $\frac{7}{12} > \frac{5}{9}$ , and the student's reasoning about denominators is oversimplified When comparing fractions, you need to consider both the numerator and denominator together, not just look at denominators in isolation. The key is to find a common way to compare the actual values. To compare $\frac{7}{12}$ and $\frac{5}{9}$ , let's find a common denominator. The least common multiple of 12 and 9 is 36. Converting both fractions: $\frac{7}{12} = \frac{7 \times 3}{12 \times 3} = \frac{21}{36}$ and $\frac{5}{9} = \frac{5 \times 4}{9 \times 4} = \frac{20}{36}$ . Since $\frac{21}{36} > \frac{20}{36}$ , we know that $\frac{7}{12} > \frac{5}{9}$ . Choice A is wrong because it accepts both the incorrect inequality and the flawed reasoning. The student's claim that larger denominators always make fractions smaller ignores the numerator entirely. Choice C is incorrect because these fractions are clearly not equal— $\frac{21}{36} \neq \frac{20}{36}$ . Choice D gets the direction of the inequality backwards; the student claimed $\frac{7}{12} < \frac{5}{9}$ , which is false. Choice B correctly identifies that $\frac{7}{12} > \frac{5}{9}$ and recognizes that the student's reasoning is oversimplified. While larger denominators can make fractions smaller when numerators are the same, you can't ignore the numerators when comparing fractions. Remember: when comparing fractions, convert to a common denominator or cross-multiply to compare accurately. Never judge fraction size by denominators alone—the numerator matters just as much.
Three runners complete different fractions of a race: Anna completes $\frac{7}{12}$ of the race, Ben completes $\frac{3}{5}$ of the race, and Carlos completes $\frac{11}{20}$ of the race. What is the correct order from least to greatest distance completed?
Answer: Carlos, Anna, Ben When you need to compare fractions with different denominators, you must find a common way to evaluate them. The most reliable approach is to convert all fractions to the same denominator or to decimals. Let's find a common denominator for $\frac{7}{12}$ , $\frac{3}{5}$ , and $\frac{11}{20}$ . The least common multiple of 12, 5, and 20 is 60. Converting each fraction: - Anna: $\frac{7}{12} = \frac{7 \times 5}{12 \times 5} = \frac{35}{60}$ - Ben: $\frac{3}{5} = \frac{3 \times 12}{5 \times 12} = \frac{36}{60}$ - Carlos: $\frac{11}{20} = \frac{11 \times 3}{20 \times 3} = \frac{33}{60}$ Now we can easily compare: $\frac{33}{60} < \frac{35}{60} < \frac{36}{60}$ , so Carlos completed the least distance, followed by Anna, then Ben. The correct order is Carlos, Anna, Ben. Choice A incorrectly places Anna first, suggesting she completed the least distance when she actually completed the middle amount. Choice C reverses the order entirely, perhaps from comparing numerators without considering denominators. Choice D correctly identifies Carlos as completing the least but then incorrectly orders Anna and Ben. Remember that when comparing fractions, you cannot simply compare numerators or denominators separately. Always convert to a common denominator or decimal form first. For SSAT fraction comparison problems, finding the LCD is usually the most efficient method, especially when dealing with three or more fractions.
Two fractions $\frac{p}{q}$ and $\frac{r}{s}$ are both between $\frac{1}{3}$ and $\frac{1}{2}$ . Which statement about $\frac{p}{q} + \frac{r}{s}$ must be true?
Answer: $\frac{2}{3} < \frac{p}{q} + \frac{r}{s} < 1$ When you encounter problems involving ranges of values, the key strategy is to find the minimum and maximum possible values of the expression by using the boundaries of the given ranges. Since both fractions are between $\frac{1}{3}$ and $\frac{1}{2}$ , we can write: $\frac{1}{3} < \frac{p}{q} < \frac{1}{2}$ and $\frac{1}{3} < \frac{r}{s} < \frac{1}{2}$ . To find the range of their sum, we add these inequalities together. The minimum possible value occurs when both fractions are as small as possible (approaching $\frac{1}{3}$ ): $\frac{p}{q} + \frac{r}{s} > \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$ . The maximum possible value occurs when both fractions are as large as possible (approaching $\frac{1}{2}$ ): $\frac{p}{q} + \frac{r}{s} < \frac{1}{2} + \frac{1}{2} = 1$ . Therefore, $\frac{2}{3} < \frac{p}{q} + \frac{r}{s} < 1$ , which is choice B. Choice A suggests the sum is less than $\frac{2}{3}$ , but this contradicts our minimum bound. Choice C claims the sum exceeds 1, but our maximum bound shows this is impossible. Choice D proposes $\frac{1}{2} < \frac{p}{q} + \frac{r}{s} < \frac{3}{4}$ , but this range is too narrow—the sum could be as large as just under 1. Strategy tip: For range problems, always find the extreme cases by using the boundary values. Add inequalities in the same direction to find the range of sums, and remember that the actual values stay strictly within the calculated bounds.
A pizza is divided into 8 equal slices. Tom eats $\frac{3}{8}$ of the pizza, and Jerry eats $\frac{1}{3}$ of the remaining pizza. What fraction of the original pizza is left?
Answer: $\frac{5}{12}$ This problem tests your ability to work with fractions in sequential steps, where one person's action affects what's available for the next person. Start by tracking what happens step by step. Tom eats $\frac{3}{8}$ of the pizza, so the remaining pizza is $1 - \frac{3}{8} = \frac{5}{8}$ of the original. Here's the key insight: Jerry eats $\frac{1}{3}$ of the remaining pizza, not $\frac{1}{3}$ of the original pizza. So Jerry eats $\frac{1}{3} \times \frac{5}{8} = \frac{5}{24}$ of the original pizza. To find what's left, subtract both portions from the whole: $1 - \frac{3}{8} - \frac{5}{24}$ . Convert to a common denominator of 24: $\frac{24}{24} - \frac{9}{24} - \frac{5}{24} = \frac{10}{24} = \frac{5}{12}$ . Choice A ( $\frac{5}{8}$ ) represents what was left after Tom ate but before Jerry ate—this ignores Jerry's portion entirely. Choice C ( $\frac{7}{12}$ ) is what you'd get if you mistakenly calculated Jerry as eating $\frac{1}{3}$ of the original pizza instead of $\frac{1}{3}$ of what remained. Choice D ( $\frac{13}{24}$ ) results from incorrectly adding the fractions instead of subtracting them from the whole. The answer is B. Strategy tip: When fractions involve "of the remaining" or "of what's left," always calculate what remains after each step before applying the next fraction. Sequential fraction problems require you to update your reference point as you go.

Start this quiz