We need to compare each answer choice to the given range of \(\frac{1}{5}\) to \(\frac{3}{5}\). Using decimals can be helpful here: \(\frac{1}{5} = 0.2\) and \(\frac{3}{5} = 0.6\). The question asks for the fraction that is NOT between 0.2 and 0.6.\n\n* (A) \(\frac{1}{4} = 0.25\). This is between 0.2 and 0.6.\n* (B) \(\frac{1}{3} \approx 0.333\). This is between 0.2 and 0.6.\n* (C) \(\frac{1}{2} = 0.5\). This is between 0.2 and 0.6.\n* (D) \(\frac{2}{3} \approx 0.666\). This is greater than 0.6.\n\nTherefore, \(\frac{2}{3}\) is the fraction that is not between \(\frac{1}{5}\) and \(\frac{3}{5}\).

We need to find a fraction between \(\frac{2}{5}\) and \(\frac{2}{3}\). Let's find a common denominator for the given fractions and the answer choices. A common denominator for 5, 3, 4, 2 is 60.\n\n* The lower bound is \(\frac{2}{5} = \frac{2 \times 12}{5 \times 12} = \frac{24}{60}\).\n* The upper bound is \(\frac{2}{3} = \frac{2 \times 20}{3 \times 20} = \frac{40}{60}\).\n\nWe need to find a fraction between \(\frac{24}{60}\) and \(\frac{40}{60}\).\n\n* (A) \(\frac{1}{3} = \frac{20}{60}\). This is less than \(\frac{24}{60}\).\n* (B) \(\frac{3}{4} = \frac{45}{60}\). This is greater than \(\frac{40}{60}\).\n* (C) \(\frac{1}{2} = \frac{30}{60}\). This is between \(\frac{24}{60}\) and \(\frac{40}{60}\).\n* (D) \(\frac{4}{5} = \frac{48}{60}\). This is greater than \(\frac{40}{60}\).\n\nTherefore, the fraction Liam could be thinking of is \(\frac{1}{2}\).

First, we need to determine the number of squares that represent \(\frac{1}{4}\) and \(\frac{1}{2}\) of the 12-square bar.\n\n* The lower limit is \(\frac{1}{4}\) of 12, which is \(\frac{1}{4} \times 12 = 3\) squares.\n* The upper limit is \(\frac{1}{2}\) of 12, which is \(\frac{1}{2} \times 12 = 6\) squares.\n\nThe problem states that Maya ate more than 3 squares and less than 6 squares. This means the number of squares eaten must be a whole number greater than 3 and less than 6. The possible whole numbers are 4 and 5.\n\nLooking at the answer choices:\n* (A) 3 is not more than 3.\n* (B) 5 is a possible number of squares.\n* (C) 6 is not less than 6.\n* (D) 7 is more than 6.\n\nTherefore, Maya could have eaten 5 squares.

We are looking for a fraction between \(\frac{2}{7}\) and \(\frac{2}{5}\). When comparing fractions with the same numerator (in this case, 2), the fraction with the smaller denominator is larger. So, \(\frac{2}{5}\) is larger than \(\frac{2}{7}\). We need a fraction, let's call it \(\frac{2}{x}\), where the denominator x is between 5 and 7. The whole number between 5 and 7 is 6. So, the fraction \(\frac{2}{6}\) is between \(\frac{2}{7}\) and \(\frac{2}{5}\). The fraction \(\frac{2}{6}\) simplifies to \(\frac{1}{3}\).\nLet's check the choices:\n(A) \(\frac{2}{9}\): Since 9 is greater than 7, \(\frac{2}{9}\) is less than \(\frac{2}{7}\).\n(B) \(\frac{1}{4} = \frac{2}{8}\): Since 8 is greater than 7, \(\frac{2}{8}\) is less than \(\frac{2}{7}\).\n(C) \(\frac{1}{2} = \frac{2}{4}\): Since 4 is less than 5, \(\frac{2}{4}\) is greater than \(\frac{2}{5}\).\n(D) \(\frac{1}{3} = \frac{2}{6}\): Since 6 is between 5 and 7, \(\frac{2}{6}\) is between \(\frac{2}{7}\) and \(\frac{2}{5}\).

We need to check each pair of fractions to see if \(\frac{4}{9}\) lies between them. Using decimals is an efficient method. \(\frac{4}{9} = 4 \div 9 \approx 0.444\).\n\n(A) \(\frac{1}{3} \approx 0.333\) and \(\frac{2}{5} = 0.4\). The number 0.444 is not between 0.333 and 0.4.\n(B) \(\frac{2}{5} = 0.4\) and \(\frac{1}{2} = 0.5\). The number 0.444 is between 0.4 and 0.5. This is the correct answer.\n(C) \(\frac{1}{2} = 0.5\) and \(\frac{3}{5} = 0.6\). The number 0.444 is not between 0.5 and 0.6.\n(D) \(\frac{3}{8} = 0.375\) and \(\frac{7}{16} = 0.4375\). The number 0.444 is not between 0.375 and 0.4375.\n\nTherefore, \(\frac{4}{9}\) is between \(\frac{2}{5}\) and \(\frac{1}{2}\).

We are looking for a fraction between \(\frac{3}{5}\) (Box B) and \(\frac{2}{3}\) (Box A). Let's convert these to decimals to make comparison easier.\n\n* \(\frac{3}{5} = 0.6\)\n* \(\frac{2}{3} \approx 0.667\)\n\nWe need to find an answer choice with a value between 0.6 and 0.667.\n\n* (A) \(\frac{4}{7} \approx 0.571\). This is less than 0.6.\n* (B) \(\frac{5}{8} = 0.625\). This is between 0.6 and 0.667.\n* (C) \(\frac{3}{4} = 0.75\). This is greater than 0.667.\n* (D) \(\frac{7}{10} = 0.7\). This is greater than 0.667.\n\nThe fraction \(\frac{5}{8}\) is the only one that falls between \(\frac{3}{5}\) and \(\frac{2}{3}\).

The whole number part of the mixed numbers is 2 for both boundaries and all answer choices, so we only need to compare the fractional parts. We need to find a fraction that is between \(\frac{1}{3}\) and \(\frac{1}{2}\).\nLet's convert \(\frac{1}{3}\) and \(\frac{1}{2}\) to decimals to compare: \(\frac{1}{3} \approx 0.333\) and \(\frac{1}{2} = 0.5\). We are looking for a fraction between 0.333 and 0.5.\n\n* (A) The fractional part is \(\frac{1}{4} = 0.25\), which is less than 0.333.\n* (B) The fractional part is \(\frac{2}{5} = 0.4\), which is between 0.333 and 0.5.\n* (C) The fractional part is \(\frac{3}{5} = 0.6\), which is greater than 0.5.\n* (D) The fractional part is \(\frac{1}{6} \approx 0.167\), which is less than 0.333.\n\nTherefore, the class could have collected \(2\frac{2}{5}\) boxes.

When you see a problem asking which value falls within a given range, you need to compare fractions by finding a common way to evaluate them all.

The water amount must be between $$\frac{3}{4}$$ and $$\frac{4}{5}$$ of a liter. To compare these fractions with the answer choices, convert everything to decimals or find a common denominator. Using decimals: $$\frac{3}{4} = 0.75$$ and $$\frac{4}{5} = 0.80$$, so you need a value between 0.75 and 0.80.

Let's check each option:

Choice A: $$\frac{31}{40} = 0.775$$. This falls perfectly between 0.75 and 0.80, so this works.

Choice B: $$\frac{2}{3} = 0.667$$. This is less than 0.75, so it's too small to be in our range.

Choice C: $$\frac{15}{20} = \frac{3}{4} = 0.75$$. The problem states the amount must be more than $$\frac{3}{4}$$, so this exact value doesn't qualify.

Choice D: $$\frac{7}{10} = 0.70$$. This is also less than 0.75, making it too small.

Only choice A gives us a value that's actually between the two boundaries, not equal to them or outside the range.

Strategy tip: When comparing fractions, converting to decimals often makes the relationships clearer than finding common denominators. Also, pay close attention to whether the problem uses "more than/less than" versus "at least/at most" — this determines whether boundary values are included.

We need to find a fraction between \(\frac{5}{8}\) and \(\frac{3}{4}\). Let's convert the fractions to have a common denominator, such as 16.\n\n* Jenny's distance: \(\frac{5}{8} = \frac{5 \times 2}{8 \times 2} = \frac{10}{16}\) mile.\n* Leo's distance: \(\frac{3}{4} = \frac{3 \times 4}{4 \times 4} = \frac{12}{16}\) mile.\n\nThe park's distance must be between \(\frac{10}{16}\) and \(\frac{12}{16}\) of a mile. Let's check the answer choices.\n\n* (A) \(\frac{1}{2} = \frac{8}{16}\), which is less than \(\frac{10}{16}\).\n* (B) \(\frac{9}{16}\) is less than \(\frac{10}{16}\).\n* (C) \(\frac{11}{16}\) is between \(\frac{10}{16}\) and \(\frac{12}{16}\).\n* (D) \(\frac{7}{8} = \frac{14}{16}\), which is greater than \(\frac{12}{16}\).\n\nThe correct answer is \(\frac{11}{16}\) mile.

We need a fraction with a decimal value between 0.65 and 0.8. Let's convert each answer choice to a decimal.\n\n* (A) \(\frac{3}{5} = 0.6\), which is less than 0.65.\n* (B) \(\frac{4}{5} = 0.8\), which is equal to the upper boundary, not between the boundaries.\n* (C) \(\frac{2}{3} = 2 \div 3 \approx 0.667\), which is between 0.65 and 0.8.\n* (D) \(\frac{5}{8} = 5 \div 8 = 0.625\), which is less than 0.65.\n\nTherefore, \(\frac{2}{3}\) is the only fraction that falls between 0.65 and 0.8.