Free math worksheets

- Rename the tenths using hundredths before adding. - Once both addends use hundredths, combine the numerators. - Read the resulting hundredths as a decimal.

1. Rename \(\frac{4}{10}\) as \(\frac{40}{100}\). 2. Add the hundredths: \(\frac{40}{100}+\frac{23}{100}=\frac{63}{100}\). 3. \(\frac{63}{100}=0.63\).

\(\frac{63}{100}=0.63\)

- Express both the tenths addend and the decimal target in hundredths. - Treat the equation as a missing-addend problem using hundredth-sized parts. - Find how many more hundredths are needed to reach the target.

1. Rename \(\frac{7}{10}\) as \(\frac{70}{100}\), and write \(0.86\) as \(\frac{86}{100}\). 2. The missing hundredths must make the numerator grow from \(70\) to \(86\). 3. \(86-70=16\), so the missing fraction is \(\frac{16}{100}\).

\(16\)

- Read the shaded amount in each hundred grid. - Express both addends using the same hundredth-sized unit. - Read the final number of hundredths as a decimal.

1. Grid a) shows \(60\) hundredths, which is \(\frac{60}{100}=\frac{6}{10}\). 2. Grid b) shows \(\frac{7}{100}\). 3. Add the hundredths: \(\frac{60}{100}+\frac{7}{100}=\frac{67}{100}\). 4. The decimal form is \(0.67\).

\(\frac{67}{100}=0.67\)

- Express the whole and the tenths fraction in hundredths. - Find the gap between the known amount and one whole. - Use that gap as the missing numerator.

1. Rewrite \(\frac{6}{10}\) as \(\frac{60}{100}\) and \(1\) as \(\frac{100}{100}\). 2. The missing number of hundredths is \(100-60=40\). 3. Therefore, the missing fraction is \(\frac{40}{100}\).

\(40\)

- Rename the tenths addend as hundredths before adding. - Add the hundredths, then write the result as a decimal. - Compare the decimal with the neighboring tenths.

1. Rename \(\frac{6}{10}\) as \(\frac{60}{100}\). 2. Add: \(\frac{60}{100}+\frac{17}{100}=\frac{77}{100}\). 3. Write the decimal: \(\frac{77}{100}=0.77\). 4. Since \(0.7<0.77<0.8\), the sum lies between \(0.7\) and \(0.8\).

\(\frac{77}{100}=0.77\); it lies between \(0.7\) and \(0.8\).

- Rewrite the tenths addend as hundredths in both expressions. - Compare the same two quantities in the two different orders. - Addition can be checked by evaluating each expression.

1. The first sum is \(\frac{20}{100}+\frac{59}{100}=\frac{79}{100}\). 2. The second sum is \(\frac{59}{100}+\frac{20}{100}=\frac{79}{100}\). 3. Reversing the addends does not change the sum.

No. Both sums equal \(\frac{79}{100}=0.79\).

- Look for two addends whose hundredths numerators total \(100\). - Reorder the addends to make a whole number first. - Add the remaining fraction after forming the whole.

1. Rename \(\frac{6}{10}\) as \(\frac{60}{100}\). 2. Combine \(\frac{60}{100}+\frac{40}{100}=1\). 3. Then \(1+\frac{17}{100}=1\frac{17}{100}=1.17\).

Combine \(\frac{6}{10}=\frac{60}{100}\) with \(\frac{40}{100}\) first to make \(1\). The final sum is \(1.17\).

- Use the grid itself to determine how many complete tenths are currently shaded. - Express the target decimal as hundredths. - Compare the current number of shaded hundredths with the target number.

1. The grid has \(9\) complete shaded rows out of \(10\), so the current amount is \(\frac{9}{10}=\frac{90}{100}\). 2. The target is \(0.96=\frac{96}{100}\). 3. The additional amount is \(\frac{96}{100}-\frac{90}{100}=\frac{6}{100}\), so \(6\) more small squares must be shaded. 4. The completed equation is \(\frac{9}{10}+\frac{6}{100}=\frac{96}{100}=0.96\).

a) \(\frac{9}{10}=\frac{90}{100}\) b) \(6\) small squares c) \(\frac{9}{10}+\frac{6}{100}=\frac{96}{100}=0.96\)

- Compare the sizes of tenths and hundredths before combining them. - Rename the tenths fraction using hundredths. - Denominators name the size of the parts; adding quantities does not mean adding those part sizes.

1. The addends use different-sized parts, so first rename \(\frac{5}{10}=\frac{50}{100}\). 2. Add hundredths: \(\frac{50}{100}+\frac{18}{100}=\frac{68}{100}\). 3. Tessa incorrectly added the denominators instead of expressing both fractions with the same unit. 4. \(\frac{68}{100}=0.68\).

Tessa incorrectly added unlike denominators. The correct sum is \(\frac{68}{100}=0.68\).

- Convert the tenths to hundredths before adding. - Notice whether the total number of hundredths passes one hundred. - Separate one whole from the remaining hundredths.

1. Rename \(\frac{8}{10}\) as \(\frac{80}{100}\). 2. Add: \(\frac{80}{100}+\frac{47}{100}=\frac{127}{100}\). 3. The result is \(1\frac{27}{100}=1.27\).

\(\frac{127}{100}=1\frac{27}{100}=1.27\)

- Think about how many hundredths are in one tenth. - Renaming a fraction must preserve its value. - Check whether the rewritten fraction is the same size as the original.

1. One tenth equals ten hundredths, so \(\frac{3}{10}=\frac{30}{100}\), not \(\frac{3}{100}\). 2. Add the correct hundredths: \(\frac{30}{100}+\frac{19}{100}=\frac{49}{100}\). 3. The correct decimal sum is \(0.49\).

Change \(\frac{3}{100}\) to \(\frac{30}{100}\). The sum is \(\frac{49}{100}=0.49\).

- Express the sum and the benchmark using hundredths. - Compare the total number of hundredths with \(50\). - For part b), subtract the smaller value from the larger value.

1. Rename \(\frac{2}{10}\) as \(\frac{20}{100}\). 2. The sum is \(\frac{20}{100}+\frac{34}{100}=\frac{54}{100}\). 3. Since \(\frac{1}{2}=\frac{50}{100}\), the sum is greater than \(\frac{1}{2}\). 4. The difference is \(\frac{54}{100}-\frac{50}{100}=\frac{4}{100}\).

a) Greater than \(\frac{1}{2}\) b) \(\frac{4}{100}\)

- Express the benchmark \(\frac{1}{2}\) in hundredths. - Test nearby tenths digits around the point where the sum changes from below the benchmark to above it. - Use the fact that increasing the box digit always increases the sum.

1. Rename \(\frac{1}{2}\) as \(\frac{50}{100}\). 2. With \(4\) in the box, the sum is \(\frac{4}{10}+\frac{7}{100}=\frac{47}{100}\), which is still less than \(\frac{50}{100}\). 3. With \(5\) in the box, the sum is \(\frac{57}{100}\), which is greater than \(\frac{50}{100}\). 4. Every digit below \(4\) gives a smaller sum than \(\frac{47}{100}\), and every digit above \(5\) gives a larger sum than \(\frac{57}{100}\). Therefore, the working digits are \(0,1,2,3,4\).

\(0,1,2,3,4\)

- Combine addends with the same denominator first. - Rename the tenths total as hundredths. - Add the final hundredths amount and write its decimal form.

1. Combine the tenths: \(\frac{2}{10}+\frac{3}{10}=\frac{5}{10}=\frac{50}{100}\). 2. Add the remaining hundredths: \(\frac{50}{100}+\frac{14}{100}=\frac{64}{100}\). 3. The decimal form is \(0.64\).

\(\frac{64}{100}=0.64\)

- Evaluate both expressions in hundredths. - Compare the two totals before answering part a). - For part b), subtract the smaller total from the larger total.

1. The first sum is \(\frac{30}{100}+\frac{42}{100}=\frac{72}{100}\). 2. The second sum is \(\frac{50}{100}+\frac{19}{100}=\frac{69}{100}\). 3. Since \(\frac{72}{100}>\frac{69}{100}\), the first sum is greater. 4. The difference is \(\frac{72}{100}-\frac{69}{100}=\frac{3}{100}\).

a) \(\frac{3}{10}+\frac{42}{100}\) b) \(\frac{3}{100}\)

- A single counterexample is enough to disprove an “always” statement. - Add the given fractions after expressing them in the same unit. - Compare the resulting number of hundredths with one hundred.

1. Rename \(\frac{7}{10}\) as \(\frac{70}{100}\). 2. The sum is \(\frac{70}{100}+\frac{44}{100}=\frac{114}{100}=1.14\). 3. Both addends are less than \(1\), but their sum is greater than \(1\), so Omar’s statement is false.

The statement is false because \(\frac{7}{10}+\frac{44}{100}=1.14\).

- Convert each tenths card to hundredths. - Compare how much larger one tenths card is than the other. - Balance the larger tenths card with the smaller hundredths card.

1. Convert the tenths cards to hundredths: \(\frac{2}{10}=\frac{20}{100}\) and \(\frac{4}{10}=\frac{40}{100}\). 2. Pair \(\frac{20}{100}\) with \(\frac{33}{100}\) to get \(\frac{53}{100}\). 3. Pair \(\frac{40}{100}\) with \(\frac{13}{100}\) to get \(\frac{53}{100}\). 4. Therefore, the two sums are equal.

\(\frac{2}{10}+\frac{33}{100}=\frac{4}{10}+\frac{13}{100}=\frac{53}{100}\)

- Addition combines values, not written digit strings. - Express both fractions as hundredths. - Use the total number of hundredths to write the decimal.

1. Rename \(\frac{2}{10}\) as \(\frac{20}{100}\). 2. Add the hundredths: \(\frac{20}{100}+\frac{47}{100}=\frac{67}{100}\). 3. The correct decimal is \(0.67\); joining digits does not combine place values correctly.

The correct sum is \(\frac{67}{100}=0.67\).

- Convert every tenths addend to hundredths. - Evaluate the three expressions before ordering them. - Compare the resulting numerators because the denominators match.

1. The first sum is \(\frac{30}{100}+\frac{28}{100}=\frac{58}{100}\). 2. The second sum is \(\frac{40}{100}+\frac{23}{100}=\frac{63}{100}\). 3. The third sum is \(\frac{50}{100}+\frac{11}{100}=\frac{61}{100}\). 4. Since \(58<61<63\), the order is first sum, third sum, second sum.

\(\frac{3}{10}+\frac{28}{100}<\frac{5}{10}+\frac{11}{100}<\frac{4}{10}+\frac{23}{100}\)

- For part a), combine the strip lengths using hundredths of a meter. - Express one meter in the same fractional unit. - For part b), subtract the combined length from one whole meter.

1. Rename \(\frac{7}{10}\) as \(\frac{70}{100}\). 2. The combined length is \(\frac{70}{100}+\frac{26}{100}=\frac{96}{100}=0.96\) meter. 3. One meter is \(\frac{100}{100}\) meter, so the gap is \(\frac{100}{100}-\frac{96}{100}=\frac{4}{100}\) meter.

a) \(0.96\,\text{m}\) b) \(\frac{4}{100}\,\text{m}\)

- An equivalent fraction must keep the same value. - Compare the proposed rewrite with one whole. - Use the same scale factor for the numerator and denominator.

1. Multiplying the denominator by \(10\) requires multiplying the numerator by \(10\), so \(\frac{5}{10}=\frac{50}{100}\). 2. The fraction \(\frac{500}{100}=5\), which is not equal to \(\frac{5}{10}\). 3. The correct sum is \(\frac{50}{100}+\frac{8}{100}=\frac{58}{100}=0.58\).

The correct rewrite is \(\frac{50}{100}\), and the sum is \(0.58\).

- Distinguish the amount added from the amount removed. - Express every change in the same fractional unit. - Follow the events in chronological order.

1. Rename \(\frac{4}{10}\) as \(\frac{40}{100}\). 2. The amount is \(\frac{15}{100}+\frac{40}{100}-\frac{8}{100}=\frac{47}{100}\). 3. As a decimal, the amount is \(0.47\) inch.

\(\frac{15}{100}+\frac{40}{100}-\frac{8}{100}=\frac{47}{100}=0.47\) inch

- Remove the known hundredths addend from the target first. - Express what remains in tenths. - Look for two digits with the required total and difference.

1. Remove the known \(\frac{7}{100}\) from \(\frac{87}{100}\). The two tenths together must equal \(\frac{80}{100}=\frac{8}{10}\). 2. Therefore, the two box digits must add to \(8\). 3. Among digit pairs that add to \(8\), the pair with the first digit exactly \(2\) greater than the second is \(5\) and \(3\). 4. Check: \(\frac{5}{10}+\frac{3}{10}+\frac{7}{100}=\frac{80}{100}+\frac{7}{100}=\frac{87}{100}\).

First box: \(5\); second box: \(3\)

- Combine the forward and backward movements first. - Express both movements in hundredths before finding the net change. - Convert the starting position to hundredths before adding the net movement.

1. The net movement is \(\frac{38}{100}-\frac{10}{100}=\frac{28}{100}\) meter forward. 2. The starting position is \(\frac{2}{10}=\frac{20}{100}\) meter. 3. The final position is \(\frac{20}{100}+\frac{28}{100}=\frac{48}{100}=0.48\) meter.

a) \(\frac{28}{100}\,\text{m}\) forward b) \(\frac{48}{100}\,\text{m}=0.48\,\text{m}\)

- Rename each tenths addend in hundredths before checking the totals. - Compare how many tenths the two decompositions use. - Identify the equal-value exchange that connects the expressions.

1. Ava's decomposition is \(\frac{7}{10}+\frac{4}{100}=\frac{70}{100}+\frac{4}{100}=\frac{74}{100}\). 2. Ben's decomposition is \(\frac{6}{10}+\frac{14}{100}=\frac{60}{100}+\frac{14}{100}=\frac{74}{100}\). 3. Ben uses one fewer tenth than Ava but ten more hundredths. Since \(\frac{1}{10}=\frac{10}{100}\), this exchange does not change the total.

a) Both are correct. b) Replacing one tenth with ten hundredths preserves the total amount.

- Combine the masses before changing the unit. - Express the tenths amount in hundredths of a kilogram. - Relate one hundredth of a kilogram to grams.

1. Rename \(\frac{8}{10}\) as \(\frac{80}{100}\). 2. The combined mass is \(\frac{80}{100}+\frac{16}{100}=\frac{96}{100}=0.96\,\text{kg}\). 3. Since \(1\,\text{kg}=1000\,\text{g}\), \(0.96\,\text{kg}=960\,\text{g}\).

\(0.96\,\text{kg}=960\,\text{g}\)

- Express the fixed tenths addend and both bounds in hundredths. - Compare how many hundredths are still needed to reach the lower bound. - Separately determine the greatest additional number of hundredths allowed by the upper bound.

1. Rename \(\frac{9}{10}\) as \(\frac{90}{100}\). 2. To reach the lower bound \(1.05=\frac{105}{100}\), the box must contribute at least \(\frac{15}{100}\), so the least possible numerator is \(15\). 3. To stay at or below \(1.10=\frac{110}{100}\), the box can contribute at most \(\frac{20}{100}\), so the greatest possible numerator is \(20\).

Least: \(15\) Greatest: \(20\)

- Find the combined amount contributed by both boxes. - The two missing addends are equal, so split their combined amount equally. - Substitute the result into both positions to check it.

1. Rename \(\frac{4}{10}\) as \(\frac{40}{100}\). 2. The two equal missing addends together must be \(\frac{34}{100}\). 3. Each missing addend is \(\frac{17}{100}\), and \(\frac{40}{100}+\frac{17}{100}+\frac{17}{100}=\frac{74}{100}\).

The numerator is \(17\). Verification: \(\frac{40}{100}+\frac{17}{100}+\frac{17}{100}=\frac{74}{100}\).

- Convert each tenths movement into a count of hundredths. - Combine the moves within each route separately. - Compare the total number of equal-sized parts.

1. Route A totals \(\frac{30}{100}+\frac{27}{100}+\frac{10}{100}=\frac{67}{100}\) meter. 2. Route B totals \(\frac{50}{100}+\frac{17}{100}=\frac{67}{100}\) meter. 3. The robot ends the same distance from its starting point because each route contains \(67\) hundredths of a meter.

Yes. Route A totals \(\frac{67}{100}\) meter, and Route B also totals \(\frac{67}{100}\) meter.

- Compare the original hundredths addend with its nearby replacement. - Decide whether the replacement increased or decreased the sum. - Adjust the estimate by exactly that difference.

1. The estimate is \(\frac{1}{10}+\frac{60}{100}=\frac{70}{100}=0.70\). 2. The substituted addend is \(\frac{2}{100}\) too large, so the estimate is high by \(\frac{2}{100}\). 3. Correcting gives \(\frac{70}{100}-\frac{2}{100}=\frac{68}{100}=0.68\).

Estimate: \(0.70\), high by \(0.02\); exact sum: \(0.68\).

- Name the unit counted by each numerator before subtracting. - Convert the tenths addend to hundredths. - Check whether the corrected addend rebuilds the target total.

1. The numerator \(5\) counts tenths, while \(82\) counts hundredths, so they cannot be subtracted as if they name the same-sized parts. 2. Rename \(\frac{5}{10}\) as \(\frac{50}{100}\). 3. The missing addend is \(\frac{82}{100}-\frac{50}{100}=\frac{32}{100}\).

Eli mixed tenths with hundredths. Since \(\frac{5}{10}=\frac{50}{100}\), the missing addend is \(\frac{32}{100}\).

- First find the fraction in either named group. - Use the fact that the two groups do not share any photos. - Compare the combined fraction with one whole collection.

1. Rename \(\frac{3}{10}\) as \(\frac{30}{100}\). 2. Landscapes and portraits together make \(\frac{30}{100}+\frac{42}{100}=\frac{72}{100}\). 3. The remaining fraction is \(1-\frac{72}{100}=\frac{28}{100}=0.28\).

\(\frac{28}{100}=0.28\)

- Use the same digit in all three boxes. - Try values near the point where the total changes from below one whole to above one whole. - Once you know what happens for neighboring boundary digits, use the fact that increasing the shared digit increases every addend.

1. Test the largest likely value below one. With \(8\) in every box, the sum is \(\frac{8}{10}+\frac{8}{100}+\frac{8}{100}=\frac{96}{100}\), which is less than \(1\). 2. Every smaller digit gives a smaller sum, so digits \(0\) through \(8\) all make the sum less than \(1\). 3. With \(9\) in every box, the sum is \(\frac{90}{100}+\frac{9}{100}+\frac{9}{100}=\frac{108}{100}\), which is greater than \(1\). 4. The sum jumps from \(\frac{96}{100}\) at digit \(8\) to \(\frac{108}{100}\) at digit \(9\), so no digit makes exactly \(1\).

a) \(0,1,2,3,4,5,6,7,8\) b) \(9\) c) No.

- Remove the known tenths amount from the mixed-number target. - Express the remainder as hundredths. - Look for two consecutive whole-number numerators whose sum matches that remainder.

1. Rename \(\frac{8}{10}\) as \(\frac{80}{100}\) and the target as \(\frac{113}{100}\). 2. The two missing addends must total \(\frac{33}{100}\), so their numerators are consecutive whole numbers with sum \(33\). 3. Those consecutive numerators are \(16\) and \(17\). 4. Check: \(\frac{80}{100}+\frac{16}{100}+\frac{17}{100}=\frac{113}{100}=1\frac{13}{100}\).

\(16\) and \(17\)

- Interpret the denominator as the size of each part. - Ask how many tenths make one whole. - Convert the whole to hundredths before adding the remaining fraction.

1. \(\frac{10}{10}\) means ten parts of size one tenth, which make \(1\), not \(10\). 2. Rename the addends as \(\frac{100}{100}+\frac{23}{100}=\frac{123}{100}\). 3. The correct sum is \(1\frac{23}{100}=1.23\).

The numerator \(10\) counts ten tenths, which make \(1\). The correct sum is \(1.23\).

- Read point S by using the hundredth-sized tick marks between labeled tenths. - Rename the \(\frac{2}{10}\) move in hundredths before combining the moves. - Your addition equation should include the starting value and both rightward moves.

1. Point S is eight hundredth-sized ticks to the right of \(0.60\), so \(S=\frac{68}{100}=0.68\). 2. Rename the tenths move: \(\frac{2}{10}=\frac{20}{100}\). 3. Write and evaluate the addition: \(\frac{68}{100}+\frac{20}{100}+\frac{5}{100}=\frac{93}{100}\). 4. Therefore, the marker finishes at \(\frac{93}{100}=0.93\).

a) \(\frac{68}{100}=0.68\) b) \(\frac{68}{100}+\frac{20}{100}+\frac{5}{100}=\frac{93}{100}\) c) \(\frac{93}{100}=0.93\)

- Express the tenths addend and both benchmarks in hundredths. - Find the exact total before calculating either distance. - The closer benchmark is the one with the smaller difference from the total.

1. The total is \(\frac{40}{100}+\frac{39}{100}=\frac{79}{100}\) meter. 2. Rename \(\frac{3}{4}=\frac{75}{100}\). The distance from that benchmark is \(\frac{79}{100}-\frac{75}{100}=\frac{4}{100}\) meter. 3. Rename \(\frac{8}{10}=\frac{80}{100}\). The distance from that benchmark is \(\frac{80}{100}-\frac{79}{100}=\frac{1}{100}\) meter. 4. Since \(\frac{1}{100}<\frac{4}{100}\), the total is closer to \(\frac{8}{10}\) meter.

a) \(\frac{79}{100}\,\text{m}\) b) \(\frac{4}{100}\,\text{m}\) from \(\frac{3}{4}\,\text{m}\); \(\frac{1}{100}\,\text{m}\) from \(\frac{8}{10}\,\text{m}\) c) Closer to \(\frac{8}{10}\,\text{m}\)

- Convert each tenths addend to hundredths. - Subtract each converted addend from the fixed row total. - Compare consecutive rows to identify the compensation pattern.

1. Rename the tenths addends as \(\frac{20}{100}\), \(\frac{40}{100}\), and \(\frac{60}{100}\). 2. Subtract each from \(\frac{73}{100}\). The missing addends are \(\frac{53}{100}\), \(\frac{33}{100}\), and \(\frac{13}{100}\). 3. Each increase of \(\frac{2}{10}=\frac{20}{100}\) in the first addend requires a decrease of \(\frac{20}{100}\) in the second addend, keeping the total fixed.

a) \(\frac{53}{100}\), \(\frac{33}{100}\), \(\frac{13}{100}\) b) The hundredths addend decreases by \(\frac{20}{100}\) each row.

- Rename one tenth as hundredths before describing the change. - For the first example, choose numerators whose total in hundredths is below \(100\). - For the second example, choose numerators whose total in hundredths is above \(100\).

1. Increasing the tenths numerator by \(1\) adds one tenth. Since \(\frac{1}{10}=\frac{10}{100}\), the total increases by \(\frac{10}{100}\). 2. For example, \(\frac{2}{10}+\frac{20}{100}=\frac{40}{100}<1\). 3. For example, \(\frac{9}{10}+\frac{20}{100}=\frac{110}{100}>1\).

a) One additional tenth equals \(\frac{10}{100}\), so the sum increases by \(\frac{10}{100}\). b) Example below \(1\): \(\frac{2}{10}+\frac{20}{100}=\frac{40}{100}\). Example above \(1\): \(\frac{9}{10}+\frac{20}{100}=\frac{110}{100}\).

- Convert the tenths step to hundredths before generating the sequence. - Keep the two step sizes in the stated alternating order. - Compare the starting and fourth values to check the total increase.

1. Rename the first step as \(\frac{1}{10}=\frac{10}{100}\). 2. The next four values are \(\frac{53}{100}\), \(\frac{55}{100}\), \(\frac{65}{100}\), and \(\frac{67}{100}\). 3. One complete cycle increases the value by \(\frac{10}{100}+\frac{2}{100}=\frac{12}{100}\). 4. Two complete cycles increase it by \(2\times\frac{12}{100}=\frac{24}{100}\).

a) \(\frac{53}{100},\frac{55}{100},\frac{65}{100},\frac{67}{100}\) b) \(\frac{24}{100}\)

- Rename the tenths addend in hundredths before adding. - Keep track of which digit represents tenths and which represents hundredths. - Check whether the proposed result is large enough for the sum of two positive addends.

1. Rename \(\frac{3}{10}\) as \(\frac{30}{100}\). 2. The sum is \(\frac{30}{100}+\frac{46}{100}=\frac{76}{100}\), not \(\frac{67}{100}\). 3. In \(0.76\), the digit \(7\) represents tenths and the digit \(6\) represents hundredths, so reversing them changes the value.

\(\frac{76}{100}=0.76\). The digit order matters because \(7\) is in the tenths place and \(6\) is in the hundredths place.

- First determine how many different arrangements are possible. - Group arrangements by the digit in the hundredths place of the two-digit numerator. - Convert the tenths addend to hundredths and evaluate each arrangement. - Check that your list accounts for all six arrangements before choosing the greatest valid sum.

1. There are \(3\times2=6\) possible arrangements because any of the three digits can be the tenths numerator and the remaining two digits can be ordered in two ways. 2. The arrangements with \(7\) in the hundredths place are \(\frac{2}{10}+\frac{47}{100}=\frac{67}{100}\) and \(\frac{4}{10}+\frac{27}{100}=\frac{67}{100}\). 3. The arrangements with \(4\) in the hundredths place are \(\frac{2}{10}+\frac{74}{100}=\frac{94}{100}\) and \(\frac{7}{10}+\frac{24}{100}=\frac{94}{100}\). 4. The arrangements with \(2\) in the hundredths place are \(\frac{4}{10}+\frac{72}{100}=\frac{112}{100}\) and \(\frac{7}{10}+\frac{42}{100}=\frac{112}{100}\), so they are not less than \(1\). 5. All six arrangements have been checked. The four valid arrangements are listed above, and the greatest resulting sum is \(\frac{94}{100}=0.94\).

\(\frac{2}{10}+\frac{47}{100}=0.67\) \(\frac{4}{10}+\frac{27}{100}=0.67\) \(\frac{2}{10}+\frac{74}{100}=0.94\) \(\frac{7}{10}+\frac{24}{100}=0.94\) Greatest sum: \(0.94\)

- Express each possible blue tenths amount in hundredths. - For each possible blue numerator, determine the remaining hundredths needed to reach the total. - Apply the positive-multiple-of-nine condition to those possible remainders.

1. Try the possible blue tenths from \(0\) through \(4\), because \(5\) tenths would already exceed \(\frac{47}{100}\). 2. The corresponding yellow hundredths needed to reach \(\frac{47}{100}\) are \(47,37,27,17,7\). 3. Only \(27\) is a positive multiple of \(9\), so the blue numerator is \(2\) and the yellow numerator is \(27\). 4. Check: \(\frac{2}{10}+\frac{27}{100}=\frac{20}{100}+\frac{27}{100}=\frac{47}{100}\).

Blue numerator: \(2\) Yellow numerator: \(27\)

- Express the target entirely in hundredths. - Determine which digit must occupy the ones place of the two-digit hundredths numerator. - Check both orders of the remaining two digits.

1. Rewrite the target as \(\frac{101}{100}\). 2. The two-digit hundredths numerator contributes its ones digit directly to the ones place of the total numerator. To end in \(1\), that ones digit must be the card \(1\). 3. The remaining digits \(3\) and \(7\) each contribute tens of hundredths, whether one is used as tenths or as the tens digit of the hundredths numerator. 4. Therefore, both arrangements work: \(\frac{3}{10}+\frac{71}{100}=\frac{101}{100}\) and \(\frac{7}{10}+\frac{31}{100}=\frac{101}{100}\).

\(\frac{3}{10}+\frac{71}{100}=1.01\) \(\frac{7}{10}+\frac{31}{100}=1.01\)