Free math worksheets

- Use the grids to see how many hundredths each addend represents. - Rename the tenths as hundredths before adding. - Read the final number of hundredths as a decimal.

1. Rename \(\frac{6}{10}\) as \(\frac{60}{100}\). 2. Add: \(\frac{60}{100}+\frac{7}{100}=\frac{67}{100}\). 3. The decimal form of \(\frac{67}{100}\) 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 \(\frac{40}{100}\).

\(40\)

- 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 \(1.17\).

- Relate each complete row to one tenth of the grid. - Express the target decimal as hundredths. - Find the difference between the current shaded hundredths and the target.

1. Nine complete rows represent \(\frac{9}{10}=\frac{90}{100}\). 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) \(\frac{9}{10}+\frac{6}{100}=\frac{96}{100}=0.96\)

- Use the grids to compare the sizes of tenths and hundredths. - Rewrite the tenths fraction using hundredths before combining. - Denominators name the part size; they are not added in this situation.

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.

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 \(\frac{1}{2}\) in hundredths. - Each possible tenths digit contributes a multiple of ten hundredths. - Find the first digit that makes the total reach or pass \(50\) hundredths. - Explain why every larger digit must also fail.

1. Rewrite the inequality in hundredths: \(\frac{10\square+7}{100}<\frac{50}{100}\). 2. The digits \(0\), \(1\), \(2\), \(3\), and \(4\) give totals of \(7\), \(17\), \(27\), \(37\), and \(47\) hundredths, all less than \(50\) hundredths. 3. The digit \(5\) gives \(57\) hundredths, which is too large. 4. Any digit greater than \(5\) gives an even larger total, so the complete list is \(0,1,2,3,4\).

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

- 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 \(0.64\).

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

- Find the exact sum in hundredths first. - Locate that hundredths value on the number line. - Identify the labeled tenths immediately below and above it.

1. Rename \(\frac{6}{10}\) as \(\frac{60}{100}\). 2. The sum is \(\frac{60}{100}+\frac{17}{100}=\frac{77}{100}=0.77\). 3. Since \(0.7<0.77<0.8\), the sum lies between seven tenths and eight tenths.

Between \(0.7\) and \(0.8\)

- 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 part from the target first. - Translate the remaining tenths into a condition on the two numerators. - Find two digits with the required sum and difference.

1. The two tenths addends must total \(\frac{80}{100}=\frac{8}{10}\), so \(a+b=8\). 2. The second condition gives \(a-b=2\). 3. The digits that satisfy both conditions are \(a=5\) and \(b=3\).

\(a=5\), \(b=3\)

- 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 the number of tenths in the two decompositions. - Identify the equal-value exchange that connects the two 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: \(\frac{70}{100}+\frac{4}{100}=\frac{74}{100}\) and \(\frac{60}{100}+\frac{14}{100}=\frac{74}{100}\). b) Replacing one tenth with ten hundredths preserves the shaded 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 addend and both bounds in hundredths. - Use the lower bound to determine the least possible contribution from \(b\). - Use the upper bound to determine the greatest possible contribution from \(b\).

1. Rename \(\frac{9}{10}\) as \(\frac{90}{100}\). 2. To reach \(1.05=\frac{105}{100}\), the least possible numerator is \(b=105-90=15\). 3. To stay at or below \(1.10=\frac{110}{100}\), the greatest possible numerator is \(b=110-90=20\).

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

- 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 \(1-\frac{72}{100}=\frac{28}{100}=0.28\).

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

- Express all three addends in hundredths using the same digit. - Compare the combined numerator with \(100\). - Check the boundary between the last sum below one and the first sum above one.

1. The sum is \(\frac{10d+d+d}{100}=\frac{12d}{100}\). 2. For \(d=0\) through \(8\), \(12d<100\), so the sum is less than \(1\). 3. For \(d=9\), the sum is \(\frac{108}{100}>1\). 4. No digit makes \(12d=100\), so no sum is exactly \(1\).

a) \(d=0,1,2,3,4,5,6,7,8\) b) \(d=9\) c) No digit makes the sum exactly \(1\).

- Remove the known tenths amount from the mixed-number target. - Translate the remainder into a sum of two consecutive numerators. - Check both missing addends in the original order.

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

\(x=16\). Verification: \(\frac{80}{100}+\frac{16}{100}+\frac{17}{100}=\frac{113}{100}=1\frac{13}{100}\).

- 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 \(1.23\).

- Write each destination in hundredths. - Count the hundredths from the starting value to each destination separately. - Compare the two destination values, not just the step counts.

1. The next tenth is \(\frac{9}{10}=\frac{90}{100}\). The distance from \(\frac{88}{100}\) is \(\frac{2}{100}\), so it takes \(2\) steps. 2. The next whole number is \(1=\frac{100}{100}\). The distance from \(\frac{88}{100}\) is \(\frac{12}{100}\), so it takes \(12\) steps. 3. The next tenth and the next whole are different landmarks. They are \(10\) hundredths apart.

a) \(2\) steps b) \(12\) steps c) The next tenth is \(\frac{90}{100}\), while the next whole is \(\frac{100}{100}\); those landmarks are \(10\) hundredths apart.

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

1. The total is \(\frac{40}{100}+\frac{39}{100}=\frac{79}{100}\) meter. 2. The distance from \(\frac{3}{4}=\frac{75}{100}\) is \(\frac{79}{100}-\frac{75}{100}=\frac{4}{100}\) meter. 3. The distance from \(\frac{8}{10}=\frac{80}{100}\) 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) \(\frac{24}{100}\)

- Convert the tenths addend before combining numerators. - Keep track of which digit represents tenths and which represents hundredths. - Check the size of the result against the 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. The digit \(7\) counts tenths and the digit \(6\) counts hundredths, so the correct decimal is \(0.76\).

\(\frac{76}{100}=0.76\). 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 the tenths section in hundredths. - List the possible remainder for the yellow section as the tenths digit changes. - Apply the positive-multiple-of-nine condition, then verify the total.

1. Express the blue section in hundredths. The numerator equation is \(10a+b=47\). 2. As \(a\) takes the possible values \(0,1,2,3,4\), the possible values of \(b\) are \(47,37,27,17,7\). 3. Only \(27\) is a positive multiple of \(9\), so \(b=27\) and \(a=2\). 4. Check: \(\frac{2}{10}+\frac{27}{100}=\frac{20}{100}+\frac{27}{100}=\frac{47}{100}\).

\(a=2\), \(b=27\)

- Convert the complete equation to hundredths. - Determine which card must occupy the ones place of the combined numerator. - Test both orders of the remaining digits.

1. If the tenths digit is \(a\) and the two-digit numerator is \(10b+c\), the total numerator in hundredths is \(10a+10b+c\). 2. The ones digit of \(101\) requires \(c=1\). 3. The remaining digits \(3\) and \(7\) can occupy the two tens positions in either order, giving both valid arrangements.

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