Free math worksheets

- Both decimals name tenths, so compare the tenths digits. - Think about which value contains more tenths.

1. Both decimals are written in tenths. 2. Four tenths is less than seven tenths. 3. Therefore, \(0.4<0.7\).

\(0.4<0.7\)

- Compare the whole-dollar amounts first. - If those are equal, compare the tenths digits. - If the tenths digits are also equal, compare the hundredths digits.

1. All four prices have \(1\) in the ones place. 2. Compare the tenths digits. Prices with \(4\) tenths are less than prices with \(9\) tenths. 3. Within each pair, compare the hundredths digits. 4. The order is \(\$1.44 < \$1.49 < \$1.94 < \$1.99\).

\(\$1.44 < \$1.49 < \$1.94 < \$1.99\)

- Compare the whole-number parts before looking at decimal places. - If whole-number parts match, compare tenths next. - Use hundredths only when the earlier places are equal.

1. Compare whole-dollar parts first. The two prices beginning with \(3\) dollars are greater than all prices below \(1\) dollar. 2. Among the \(3\)-dollar prices, compare tenths: \(3.30>3.03\). 3. Among the prices below \(1\), compare tenths and then hundredths: \(0.33>0.30>0.03\). 4. Combine the two groups.

\(\$3.30>\$3.03>\$0.33>\$0.30>\$0.03\)

- The whole-dollar digit is the same in every value. - Compare tenths before hundredths. - When the tenths match, use the hundredths digit to decide the order.

1. The whole-number part is \(5\) for every price. 2. Compare tenths: the prices with \(0\) tenths come first, then those with \(1\) tenth, then \(5\) tenths. 3. Within equal tenths, compare hundredths: \(5.01<5.05\) and \(5.10<5.15\). 4. Therefore, \(5.01<5.05<5.10<5.15<5.50\).

\(\$5.01<\$5.05<\$5.10<\$5.15<\$5.50\)

- Compare the whole-dollar amounts first. - If the dollars are equal, compare the cents. - Pay attention to place value.

1. For a), compare the whole-dollar amounts. Since \(14 < 18\), \(\$14.80 < \$18.40\). 2. For b), the whole-dollar amounts are equal. Compare the cents: \(7 < 70\), so \(\$6.07 < \$6.70\). 3. For c), the amounts are identical, so they are equal.

a) \(<\) b) \(<\) c) \(=\)

- Write each price to hundredths so the place values line up. - First look for two decimal forms that name the same value. - For the remaining order, compare tenths before hundredths.

1. Rewrite \(1.5\) as \(1.50\). This shows that \(\$1.5\) and \(\$1.50\) are equal. 2. The three distinct decimal values are \(1.05\), \(1.15\), and \(1.50\). 3. Compare tenths first: \(1.05\) has \(0\) tenths, \(1.15\) has \(1\) tenth, and \(1.50\) has \(5\) tenths. 4. Therefore, \(1.05<1.15<1.50\). 5. A zero added to the right of the last decimal digit adds zero hundredths, so it does not change the value.

a) \(\$1.5\) and \(\$1.50\) b) \(\$1.05<\$1.15<\$1.50\) c) The added zero represents zero hundredths, so \(1.5=1.50\).

- Write both decimals with the same number of decimal places. - Compare the whole-number parts first, then tenths. - Do not treat all digits after the decimal point as one whole number.

1. Rewrite \(\$4.7\) as \(\$4.70\). 2. The whole-dollar parts are both \(4\). 3. Compare tenths: \(7\) tenths is greater than \(6\) tenths, so \(4.70>4.65\). 4. Noah compared the digit strings after the decimal as whole numbers instead of comparing place values.

No. \(\$4.70>\$4.65\). The tenths place decides the comparison because \(7>6\).

- Try writing all the numbers with the same number of digits after the decimal point by adding zeros. - Would it help to think about these numbers as amounts of money? - Compare the digits in each place value, starting from left to right. - Remember that adding a zero to the end of a decimal doesn't change its value. - Look for any numbers that might actually represent the same value.

1. Align the given numbers to the same number of decimal places to facilitate comparison: \(0.6 = 0.60\), \(0.58 = 0.58\), \(0.60 = 0.60\), and \(0.62 = 0.62\). 2. Compare the resulting values: \(0.58\) is less than \(0.60\), and \(0.60\) is less than \(0.62\). 3. Identify the equality between the original values: \(0.6 = 0.60\). 4. Arrange the original values from least to greatest using the appropriate symbols: \(0.58 < 0.6 = 0.60 < 0.62\).

\(0.58 < 0.6 = 0.60 < 0.62\)

- Read both runner positions from the number line. - Both measurements lie on the same side of the goal. - For two values below the same target, consider which value lies farther to the right.

1. The marked distances are \(2.86\) miles and \(2.91\) miles. 2. Both values are below \(3\), and \(2.91>2.86\), so \(2.91\) is farther along toward the common endpoint. 3. Therefore, the runner at \(2.91\) miles is closer to \(3\) miles.

The runner at \(2.91\) miles is closer to \(3\) miles.

- Read the three marked values from left to right. - Identify which marked value is exactly one whole. - Compare the other two points by their distance from one whole.

1. The marked values are \(0.99\), \(1.00\), and \(1.01\). 2. \(0.99\) is one hundredth less than \(1\), \(1.00\) equals \(1\), and \(1.01\) is one hundredth greater than \(1\). 3. Therefore, \(0.99<1.00<1.01\).

\(0.99<1.00<1.01\)

- Compare the tenths digits before the hundredths digits. - The same digit can represent different amounts in different places. - Identify the first place where the decimals differ.

1. In \(0.37\), the digit \(3\) represents three tenths. 2. In \(0.73\), the digit \(7\) represents seven tenths. 3. Since seven tenths is greater than three tenths, \(0.73>0.37\). The digits have different values when they occupy different places.

\(0.73>0.37\) because seven tenths is greater than three tenths; swapping the digits changes their place values.

- Align the decimal places in the two times. - Compare the hundredths after the whole-number and tenths places match. - For race times, the smaller value represents the faster swimmer.

1. Rewrite \(12.5\) as \(12.50\). 2. Compare the times: \(12.48<12.50\). 3. A smaller race time is faster, so the swimmer with \(12.48\) seconds is faster.

The swimmer with a time of \(12.48\) seconds is faster.

- Compare corresponding digits from left to right. - Stop at the first place where the digits are different. - Use that place to determine the inequality.

1. Both decimals have \(5\) in the ones place. 2. The tenths digits are \(2\) and \(3\), so the tenths place is the first place where they differ. 3. Since \(2<3\), \(5.27<5.31\).

They first differ in the tenths place, and \(5.27<5.31\).

- Compare the two lengths one place at a time. - The first difference appears in the hundredths place. - Consecutive hundredths are separated by \(0.01\).

1. The ones and tenths digits match in both measurements. 2. In the hundredths place, \(7>6\), so \(1.27>1.26\). 3. The difference is \(1.27-1.26=0.01\) meter, or one hundredth of a meter.

a) The \(1.27\,\text{m}\) ribbon b) \(0.01\,\text{m}\)

- Write every decimal to the hundredths place. - First identify the greatest two values. - The smaller of those two is the second greatest overall.

1. Rewrite \(3.8\) as \(3.80\). 2. The two greatest values are \(3.81\) and \(3.80\). 3. Since \(3.81>3.80\), the second-greatest decimal is \(3.8\).

\(3.8\)

- Since the whole-number and tenths places already match, focus on the hundredths place. - Use the lower bound to determine which side of its hundredths digit the box must lie on. - Use the upper bound separately, remembering that “strictly between” excludes both endpoints.

1. The whole-number and tenths digits already match both bounds. 2. To be above \(3.42\), the hundredths digit must be greater than \(2\). 3. To be below \(3.47\), the hundredths digit must be less than \(7\). 4. The possible digits are \(3,4,5,6\).

\(3,4,5,6\)

- Convert the words into decimal notation first. - Compare corresponding places from left to right. - The hundredths place decides the comparison after the earlier places match.

1. The place-value description represents \(4.36\). 2. The decimals have the same ones and tenths digits. 3. Since \(6>5\) in the hundredths place, \(4.36>4.35\).

The described number, \(4.36\), is greater than \(4.35\).

- Express the benchmark to the hundredths place. - “Strictly greater” rules out the benchmark itself. - Move to the very next hundredth.

1. Rewrite \(0.8\) as \(0.80\). 2. A decimal strictly greater than \(0.80\) cannot equal \(0.80\). 3. The next hundredth is \(0.81\), so it is the least possible decimal.

\(0.81\)

- Read the three values from the labeled number line. - Count hundredth-sized intervals from the middle point to each endpoint. - A midpoint must be the same distance from both endpoints.

1. The marked values are A \(=0.58\), B \(=0.60\), and C \(=0.62\). 2. The distance from A to B is \(0.60-0.58=0.02\). 3. The distance from B to C is \(0.62-0.60=0.02\). 4. Since the two distances are equal, B is exactly halfway between A and C.

Yes. A \(=0.58\), B \(=0.60\), C \(=0.62\), and both distances are \(0.02\).

- Apply the comparison condition before the digit condition. - Examine only the tenths and hundredths digits. - Both requirements must hold for the same decimal.

1. The decimals above \(0.50\) are \(0.51, 0.52, 0.53\). 2. Their decimal-digit sums are \(6, 7, 8\), respectively. 3. Only \(0.53\) satisfies both conditions.

\(0.53\)

- Think of the values as consecutive hundredths. - Count the hundredths steps from the lower endpoint to the upper endpoint. - The endpoints are excluded by the word “strictly.”

1. At the hundredths scale, \(1.34\), \(1.35\), and \(1.36\) are consecutive values. 2. The only hundredths value greater than \(1.34\) and less than \(1.36\) is \(1.35\). 3. No other two-place decimal can fit because adjacent hundredths differ by \(0.01\).

\(1.35\). It is the only possibility because \(1.34\), \(1.35\), and \(1.36\) are consecutive hundredths and the endpoints are excluded.

- Read the two rod values and the benchmark from the number line. - Order the three points from left to right. - Count hundredths from each rod point to the benchmark.

1. The marked values are \(0.86\,\text{m}\), \(0.90\,\text{m}\), and \(0.91\,\text{m}\). 2. Their order is \(0.86<0.90<0.91\). 3. Rod A is \(0.90-0.86=0.04\,\text{m}\) below the benchmark. 4. Rod B is \(0.91-0.90=0.01\,\text{m}\) above the benchmark.

a) \(0.86<0.90<0.91\) b) Rod A: \(0.04\,\text{m}\) below; Rod B: \(0.01\,\text{m}\) above

- Interpret \(0.10\) as ten hundredths. - Move backward from the known decimal by that exact amount. - Match the resulting decimal with the box pattern.

1. Ten hundredths less than \(1.52\) is \(1.42\). 2. Matching \(1.\square2\) with \(1.42\) shows that the missing tenths digit is \(4\). 3. Check: \(1.52-1.42=0.10\).

\(4\); \(1.52-1.42=0.10\)

- Read the starting point, benchmark, and tick size from the number line. - Pay attention to the word “greater”; landing exactly on the benchmark is not enough. - Count one additional tick after first reaching the benchmark.

1. The blue starting point is \(0.76\), the gray benchmark is \(0.80\), and each tick is \(0.01\). 2. Four steps reach \(0.80\), but the condition requires a value greater than the benchmark. 3. A fifth step reaches \(0.81\), which is the first value greater than \(0.80\).

a) \(5\) steps b) \(0.81\)

- Write each decimal to the hundredths place. - Decide which labels are equal before comparing the remaining value. - Distance is the positive difference between the two distinct values.

1. Rename \(6.4\) as \(6.40\), so \(6.4\) and \(6.40\) name the same value. 2. The remaining value is \(6.04\), which is less than \(6.40\). 3. The distance is \(6.40-6.04=0.36\).

a) \(6.4\) and \(6.40\) b) \(0.36\)

- Use the labeled tenths and the hundredth-sized tick spacing to read each point. - Compare the two decimals after writing both to the hundredths place. - Count the tick intervals between the points for the distance.

1. Point A is at \(0.36\), and point B is at \(0.44\). 2. Since \(0.36<0.44\), point B has the greater value. 3. The distance is \(0.44-0.36=0.08\), or eight hundredths.

a) A: \(0.36\); B: \(0.44\) b) B c) \(0.08\), or eight hundredths

- Rewrite each decimal in hundredths. - Once the denominators match, compare the numerators. - Pay close attention to place value and zeros.

1. For a), \(0.3 = \frac{30}{100}\) and \(0.03 = \frac{3}{100}\). Since \(30 > 3\), \(0.3 > 0.03\). 2. For b), \(0.05 = \frac{5}{100}\), so the values are equal. 3. For c), \(1.2 = \frac{12}{10} = \frac{120}{100}\), so the values are equal. 4. For d), \(0.08 = \frac{8}{100}\) and \(0.01 = \frac{1}{100}\). Since \(8 > 1\), \(0.08 > 0.01\).

a) \(0.3 > 0.03\) b) \(0.05 = \frac{5}{100}\) c) \(1.2 = \frac{120}{100}\) d) \(0.08 > 0.01\)

- Read the two decimal values from the marked points. - Write both decimals to the same number of decimal places before comparing. - Use the left-to-right positions on the number line as a check.

1. The blue point is \(0.73\), and the orange point is \(0.9\). 2. Rename \(0.9\) as \(0.90\) so both decimals are compared to the hundredths place. 3. \(0.73=\frac{73}{100}\) and \(0.90=\frac{90}{100}\), so \(0.73<0.9\). 4. Ben compared the digit strings as whole numbers instead of matching their place values.

Ben did not align the place values. \(0.73<0.9\).

- Separate the fraction of each board from the actual area of each board. - Compare the decimals, then consider whether the wholes are equal. - A smaller fraction of a larger whole can still represent a larger amount.

1. The decimal comparison is \(0.7=0.70>0.65\), so Board A has the greater painted fraction. 2. However, the boards are different sizes, and Board B is larger. 3. Without the actual board areas, the painted areas cannot be compared.

No. Board A has the greater painted fraction, but the actual painted area cannot be determined because the boards are different sizes.

- Compare the tenths place before looking at the hundredths place. - Test the case where the tenths digit matches the benchmark’s tenths digit. - After finding the first working digit, count it and every larger digit.

1. Both values have the same whole-number part, \(2\). 2. If the tenths digit is less than \(6\), the decimal is less than \(2.63\). 3. If the tenths digit is \(6\), the decimal is \(2.67\), which is greater than \(2.63\). Therefore, the smallest working digit is \(6\). 4. The working digits are \(6,7,8,9\), so \(4\) digits work altogether.

a) \(6\) b) \(4\) digits

- Align the decimal places in each pair. - Compare corresponding place values from left to right. - Remember that a trailing zero does not change a decimal’s value.

1. For a), rewrite \(0.8\) as \(0.80\). Since \(0.80>0.79\), the statement is false. 2. For b), rewrite \(1.4\) as \(1.40\). Since \(1.04<1.40\), the statement is false. 3. For c), a trailing zero does not change the value, so \(2.30=2.3\) is true.

a) False; \(0.8>0.79\). b) False; \(1.04<1.4\). c) True.

- Translate each decimal into tenths or hundredths before checking the chain. - Use the number line to find two labels at the same point. - A correct comparison chain must be true at every symbol.

1. The decimal \(0.07\) is seven hundredths, so it is less than \(0.7\), not greater. 2. The decimals \(0.7\) and \(0.70\) are equal, so a less-than sign cannot go between them. 3. The correct chain is \(0.07<0.7=0.70<0.77\).

Devon placed \(0.07\) incorrectly and treated \(0.7\) and \(0.70\) as unequal. The correct chain is \(0.07<0.7=0.70<0.77\).

- Choose the greatest possible tenths digit without crossing the bound. - After fixing the tenths place, maximize the hundredths place. - Check that the completed decimal still satisfies the strict inequality.

1. To stay below \(0.70\), the greatest possible tenths digit is \(6\). 2. With \(6\) in the tenths place, choose the greatest remaining digit, \(8\), for the hundredths place. 3. The decimal is \(0.68\), and it is less than \(0.70\).

\(0.68\)

- First compare the two decimal portions without using the poster sizes. - Then count the shaded equal-size squares in each model. - Identify how the sizes of the two wholes affect the area comparison.

1. As decimals, \(0.6>0.5\), so Poster A has the greater shaded fraction of its own whole. 2. Poster A has \(6\) shaded small squares. Poster B has \(10\) shaded small squares. 3. Because all small squares are the same size, Poster B has the greater shaded area. 4. The decimals describe portions of different-size wholes, so the greater decimal does not necessarily represent the greater actual area.

a) \(0.6>0.5\) b) Poster B c) The decimals compare portions of different-size wholes; Poster B has \(10\) shaded equal-size squares, while Poster A has \(6\).

- Notice that the same digit has a different place value in the two decimals. - Use the first comparison to restrict the possible digits. - Use the second comparison independently, then keep only digits that satisfy both.

1. In the first comparison, the unknown is in the hundredths place. It must make \(0.5\square\) greater than \(0.54\), so the digit must be greater than \(4\). 2. In the second comparison, the same digit is in the tenths place. It must make \(0.\square5\) less than \(0.75\), so the digit must be less than \(7\). 3. The digits that satisfy both conditions are \(5\) and \(6\).

\(5\) or \(6\)

- Compare the tenths places before looking at the hundredths places. - Test what happens when the shared digit is below, equal to, or above \(7\). - Remember that the comparison must be strictly less than, not equal.

1. If the shared digit is less than \(7\), the right-hand decimal has fewer tenths than the left-hand decimal, so the comparison cannot be true. 2. If the shared digit is \(7\), both decimals are \(0.77\), so the strict inequality is still false. 3. Digits \(8\) and \(9\) make the right-hand decimal have more tenths than \(0.7\square\), so both work.

\(8\) or \(9\)

- Write every decimal to the hundredths place. - Check each neighboring comparison rather than sorting the entire list from scratch. - For the incorrect pair, identify the first place where the values differ.

1. Rename \(2.6\) as \(2.60\). 2. Since \(2.39<2.60\), the adjacent pair \(2.6\) and \(2.39\) is reversed. 3. Swapping that pair gives \(2.36<2.39<2.6<2.63\). 4. The error treats \(2.6\) as though it were less than \(2.39\), but \(2.6=2.60\).

a) Swap \(2.6\) and \(2.39\). b) \(2.36<2.39<2.6<2.63\) c) \(2.6=2.60\), so it is greater than \(2.39\).

- Compare the original lengths by aligning decimal places. - Consider what subtracting the same amount does to the order and the gap. - Use the calculated new lengths only to verify the prediction.

1. Initially, \(8.40>8.37\), with a difference of \(0.03\,\text{cm}\). 2. Subtracting the same amount from both lengths preserves their order, so Ribbon A will remain longer. 3. The new lengths are \(8.40-0.20=8.20\,\text{cm}\) and \(8.37-0.20=8.17\,\text{cm}\). 4. Their difference is \(8.20-8.17=0.03\,\text{cm}\), unchanged from the original difference.

a) Ribbon A b) Ribbon A: \(8.20\,\text{cm}\); Ribbon B: \(8.17\,\text{cm}\) c) \(0.03\,\text{cm}\)

- Establish the order before computing any gaps. - Compare tenths first, then hundredths within each tenths group. - Subtract only neighboring values in the final order.

1. Comparing tenths and then hundredths gives \(0.22<0.29<0.92<0.99\). 2. The neighboring gaps are \(0.29-0.22=0.07\), \(0.92-0.29=0.63\), and \(0.99-0.92=0.07\). 3. The largest gap is \(0.63\), between \(0.29\) and \(0.92\).

a) \(0.22<0.29<0.92<0.99\) b) \(0.63\), between \(0.29\) and \(0.92\)

- Focus on the boundary value shared by the two conditions. - Decide which side of that boundary each card must lie on. - Exact card values are unnecessary because the two allowed intervals do not overlap.

1. The first card's value is less than \(0.74\). 2. The second card's value is greater than \(0.74\). 3. Therefore, the first card is less than \(0.74\), which is less than the second card, so the first card has the smaller value.

The first card's value is less than the second card's value.

- Read the two endpoints from the number line and locate the value halfway between them. - Decide which hundredths digits put the unknown decimal to the right of that halfway point. - Treat the exact halfway value separately.

1. The endpoints are \(4.60\) and \(4.70\), which are ten hundredths apart. 2. The midpoint is \(4.65\). 3. Digits \(6,7,8,9\) make \(4.6\square\) greater than \(4.65\), so those values are closer to \(4.70\). 4. Digit \(5\) makes \(4.65\), which is equally distant from both endpoints.

a) \(6,7,8,9\) b) \(5\)

- Compare the two decimals from left to right by place value. - Write both decimals as counts of hundredths to check the actual difference. - Decide whether a difference in one digit alone determines the difference between the complete decimals.

1. The hundredths digits \(7\) and \(8\) differ by \(1\), but the tenths digits also differ. 2. The actual difference is \(0.68-0.57=0.11\). 3. Therefore, Sofia's claim is false. The tenths place must also be considered when comparing the two decimals.

The claim is false. \(0.68-0.57=0.11\), not \(0.01\). The tenths place must also be checked.

- Begin with the greatest place value where the numbers might differ. - A later place cannot overturn a difference already found in an earlier place. - Use the hundredths digits only if the tenths digits match.

1. The ones digits are equal, so compare the tenths next. 2. In the tenths place, \(4<5\). This decides the order before the hundredths place is reached. 3. Therefore, \(3.49<3.52\).

\(3.49<3.52\). Decimal places are compared from greatest to least place value, and the tenths digits \(4<5\) decide the order before the hundredths are considered.

- Convert the two boundary fractions to decimals. - List the tenths near those decimals. - Remember that the fraction must be strictly greater than one boundary and strictly less than the other.

1. Write the boundary fractions as decimals: \(\frac{1}{4}=0.25\) and \(\frac{2}{5}=0.4\). 2. Fractions with denominator \(10\) represent tenths. The tenths near this interval are \(\frac{2}{10}=0.2\), \(\frac{3}{10}=0.3\), and \(\frac{4}{10}=0.4\). 3. Only \(0.3\) is strictly between \(0.25\) and \(0.4\). Therefore, the only solution is \(\frac{3}{10}\).

\(\frac{3}{10}\). It is the only tenth strictly between \(0.25\) and \(0.4\).