Free Math Worksheets

- Identify the digit in the place you are rounding to. - Look only at the digit immediately to its right. - Digits \(0\) through \(4\) round down; digits \(5\) through \(9\) round up.

1. For a), the tenths digit is \(3\), so \(12.3549\) rounds to \(12\). 2. For b), the hundredths digit is \(5\), so the tenths digit rounds up: \(12.4\). 3. For c), the thousandths digit is \(4\), so the hundredths digit stays the same: \(12.35\). 4. For d), the ten-thousandths digit is \(9\), so the thousandths digit rounds up: \(12.355\).

a) \(12\) b) \(12.4\) c) \(12.35\) d) \(12.355\)

- Look at the digit immediately to the right of the rounding place. - When a \(9\) rounds up, regroup into the place to its left. - Preserve trailing zeros that show the requested precision.

1. For a), the thousandths digit is \(6\), so the hundredths digit rounds up. Regrouping gives \(0.40\). 2. For b), the hundredths digit is \(5\), so the tenths digit rounds up. Regrouping gives \(2.0\). 3. For c), the thousandths digit is \(4\), so the hundredths digit stays the same. The result is \(7.00\). 4. For d), the hundredths digit is \(9\), so the tenths digit rounds up. Regrouping gives \(1.0\).

a) \(0.40\) b) \(2.0\) c) \(7.00\) d) \(1.0\)

- Look at the digit immediately to the right of the requested place. - Use the same place-value rule for each decimal. - Regroup when a \(9\) must round up.

1. For a), the hundredths digit is \(6\), so the tenths digit rounds up. The result is \(13.0\). 2. For b), the thousandths digit is \(5\), so the hundredths digit rounds up. The result is \(0.05\). 3. For c), the ten-thousandths digit is \(6\), so the thousandths digit rounds up. Regrouping across the \(9\)s gives \(6.000\).

a) \(13.0\) b) \(0.05\) c) \(6.000\)

- Which digit in the original number controls rounding to the nearest tenth? - Do not round to an intermediate place first. - Compare the direct rounding result with the two-stage result.

1. To round directly to the nearest tenth, look at the hundredths digit in the original number. 2. In \(0.445\), the hundredths digit is \(4\). 3. Since \(4<5\), the tenths digit stays \(4\), so \(0.445\) rounds to \(0.4\). 4. Jordan rounded in stages. The first rounding changed the hundredths digit and caused the second rounding to produce a different result. Rounding should be done directly from the original number to the requested place.

Jordan’s method is incorrect because rounding in stages can change the final result. Rounded directly to the nearest tenth, \(0.445\) is \(0.4\).

- To round to the nearest hundredth, inspect the thousandths digit. - Test each number independently. - Digits beyond the thousandths place do not change which way the hundredths digit rounds.

1. For A, the thousandths digit is \(5\), so \(3.415\) rounds to \(3.42\). 2. For B, the thousandths digit is \(4\), so \(3.4249\) rounds to \(3.42\). 3. For C, the thousandths digit is \(5\), so \(3.425\) rounds to \(3.43\). 4. For D, the thousandths digit is \(4\), so \(3.4149\) rounds to \(3.41\). 5. For E, the thousandths digit is \(0\), so \(3.4201\) rounds to \(3.42\).

A, B, and E

- In Situation A, ask whether buying too little would meet the need. - In Situation B, ask whether you may spend more than the budget. - Decide whether the context requires at least a certain amount or allows at most a certain amount.

1. In Situation A, \(2.1\) packages means that \(2\) whole packages are not enough. You must round up and buy \(3\) packages. 2. In Situation B, the budget is a maximum. Three tickets would cost \(3 \times \$6.80 = \$20.40\), which is more than \(\$20.00\). You must round down and buy \(2\) tickets.

Situation A: Round up and buy \(3\) packages. Situation B: Round down and buy \(2\) tickets.

- The hundredths digit controls rounding to the nearest tenth. - Find the lower endpoint of the interval that rounds to \(4.7\). - Apply the rounding rule directly to \(4.75\).

1. In \(4.649\), the hundredths digit is \(4\), so it rounds to \(4.6\). 2. In \(4.65\), the hundredths digit is \(5\), so it rounds to \(4.7\). 3. The least two-decimal-place number that rounds to \(4.7\) is \(4.65\). 4. The number \(4.75\) has a hundredths digit of \(5\), so its tenths digit rounds up from \(7\) to \(8\). Therefore, it rounds to \(4.8\), not \(4.7\).

\(4.649\) rounds to \(4.6\), and \(4.65\) rounds to \(4.7\). a) \(4.65\) b) \(4.75\) rounds to \(4.8\) because its hundredths digit is \(5\).

- Identify the digit immediately to the right of the tenths place. - Recall which digits cause rounding up. - Do not round in stages from right to left.

1. When rounding to the nearest tenth, the hundredths digit controls the rounding decision. 2. To round \(4.3\square9\) up to \(4.4\), the digit in the box must be \(5\), \(6\), \(7\), \(8\), or \(9\). 3. The final digit \(9\) is in the thousandths place. It does not affect direct rounding to the nearest tenth because only the hundredths digit is inspected.

\(5, 6, 7, 8, 9\). The thousandths digit does not affect direct rounding to the nearest tenth.

- Which digit determines whether the rounding digit stays the same or increases? - What happens when increasing a \(9\) requires regrouping? - Think about what the number of decimal places communicates.

1. For \(0.4951\), the thousandths digit is \(5\), so \(0.4951\) rounds to \(0.50\) to the nearest hundredth. To the nearest tenth, it rounds to \(0.5\). 2. For \(12.997\), the thousandths digit is \(7\), so rounding causes regrouping through both decimal places. It rounds to \(13.00\) to the nearest hundredth and \(13.0\) to the nearest tenth. 3. The zeros in \(13.00\) show that the value was rounded to the hundredths place. Without them, the stated precision would not be clear.

Rounded values: \(0.4951 \rightarrow 0.50\) to the nearest hundredth and \(0.5\) to the nearest tenth. \(12.997 \rightarrow 13.00\) to the nearest hundredth and \(13.0\) to the nearest tenth. The trailing zeros in \(13.00\) indicate that the number was rounded to the hundredths place.

- Find the smallest hundredths digit that makes \(5.3\) round up to \(5.4\). - Find the largest hundredths digit that keeps \(5.4\) from rounding up to \(5.5\). - Remember that the number must have exactly two decimal places.

1. A number rounds to \(5.4\) to the nearest tenth when it is at least \(5.35\) but less than \(5.45\). 2. With exactly two decimal places, the least possible number is \(5.35\). 3. The greatest possible number is \(5.44\).

Least: \(5.35\) Greatest: \(5.44\)

- Identify the lower endpoint of each set of numbers that rounds to the given result. - A hundredth-sized rounding interval is narrower than a tenth-sized interval. - Subtract the smaller lower endpoint from the larger one.

1. Numbers that round to \(5.0\) to the nearest tenth begin at \(4.95\). Thus, the least possible number in Case A is \(4.95\). 2. Numbers that round to \(5.00\) to the nearest hundredth begin at \(4.995\). Thus, the least possible number in Case B is \(4.995\). 3. Their difference is \(4.995-4.95=0.045\).

1. Case A: \(4.95\); Case B: \(4.995\) 2. \(0.045\)

- Find the lower endpoint of the rounding interval for each number. - A tenth has a half-unit of \(0.05\), and a hundredth has a half-unit of \(0.005\). - Add the two least possible values.

1. The least number that rounds to \(7.4\) to the nearest tenth is \(7.35\). 2. The least number that rounds to \(2.15\) to the nearest hundredth is \(2.145\). 3. The least possible sum is \(7.35+2.145=9.495\).

The lower bounds are \(x=7.35\) and \(y=2.145\), so the least possible sum is \(9.495\).

- Identify the lower and upper boundaries for numbers that round to \(2.46\). - The lower boundary is included, but the upper boundary is not. - Count by thousandths from the least value to the greatest value.

1. The rounding interval begins at \(2.455\) and ends just before \(2.465\). 2. With exactly three decimal places, the least possible number is \(2.455\), and the greatest is \(2.464\). 3. The possible values are \(2.455, 2.456, 2.457, 2.458, 2.459, 2.460, 2.461, 2.462, 2.463, 2.464\). 4. There are \(10\) values.

a) Least: \(2.455\); greatest: \(2.464\) b) \(10\) decimals

- First determine the full interval of numbers that round to \(5.0\). - Consider separately numbers beginning with \(4.9\) and \(5.0\). - Use the digit-sum condition together with the rounding condition. - Each digit must be from \(0\) through \(9\).

1. A number that rounds to \(5.0\) to the nearest tenth must satisfy \(4.950 \le x < 5.050\). 2. A number beginning with \(4.9\) would need a hundredths digit of at least \(5\) to round up. But then the three decimal digits could not have sum \(10\), because the tenths digit is already \(9\). 3. Therefore, the number begins with \(5.0\). Let the hundredths and thousandths digits be \(a\) and \(b\). Then \(a + b = 10\), and \(a < 5\) so the number rounds down to \(5.0\). 4. The possible digit pairs are \((1, 9)\), \((2, 8)\), \((3, 7)\), and \((4, 6)\). 5. The possible numbers are \(5.019\), \(5.028\), \(5.037\), and \(5.046\).

\(5.019\), \(5.028\), \(5.037\), or \(5.046\)

- Find the lower and upper boundaries for values that round to \(2.4\). - Convert tons to pounds after finding the interval. - The upper rounding boundary is not included.

1. Values that round to \(2.4\) tons lie from \(2.35\) tons, inclusive, to \(2.45\) tons, exclusive. 2. Convert the lower boundary: \(2.35\times2000=4700\,\text{lb}\). 3. Convert the upper boundary: \(2.45\times2000=4900\,\text{lb}\), but this boundary is not included. 4. Therefore, the possible whole-pound weights are from \(4700\,\text{lb}\) through \(4899\,\text{lb}\).

From \(4700\,\text{lb}\) through \(4899\,\text{lb}\)

- Find the rounding interval for each reported distance. - Use the overlap of the two intervals. - Convert the interval endpoints from miles to feet, then account for the excluded upper boundary.

1. Values that round to \(4.5\) miles to the nearest tenth lie in \([4.45, 4.55)\). 2. Values that round to \(4.53\) miles to the nearest hundredth lie in \([4.525, 4.535)\). This entire interval also satisfies the first condition. 3. Convert the lower boundary: \(4.525\times5280=23{,}892\,\text{ft}\). 4. Convert the upper boundary: \(4.535\times5280=23{,}944.8\,\text{ft}\), which is not included. 5. Therefore, the least possible whole-number length is \(23{,}892\,\text{ft}\), and the greatest is \(23{,}944\,\text{ft}\).

Least: \(23{,}892\,\text{ft}\) Greatest: \(23{,}944\,\text{ft}\)

- Find the lower and upper boundaries for values that round to \(7\) acres. - Convert each boundary using \(43{,}560\,\text{ft}^2\) per acre. - The upper boundary is excluded.

1. Areas from \(6.5\) acres, inclusive, to \(7.5\) acres, exclusive, round to \(7\) acres. 2. Convert the lower boundary: \(6.5\times43{,}560=283{,}140\,\text{ft}^2\). 3. Convert the upper boundary: \(7.5\times43{,}560=326{,}700\,\text{ft}^2\). 4. Therefore, the least possible area is \(283{,}140\,\text{ft}^2\), and the area must be less than \(326{,}700\,\text{ft}^2\).

a) \(283{,}140\,\text{ft}^2\) b) The area must be less than \(326{,}700\,\text{ft}^2\).

- Compare the restrictions “exactly three decimal places” and “any number of decimal places.” - Try adding digits after \(8.249\) while staying below \(8.25\). - Ask whether a number below \(8.25\) can always be replaced by a slightly larger one that is still below \(8.25\).

1. With exactly three decimal places, \(8.249\) is the greatest number below \(8.250\), so Julia is correct for that restriction. 2. Examples greater than \(8.249\) that still round to \(8.2\) are \(8.2491\), \(8.2499\), and \(8.24999\). Each is less than \(8.25\). 3. All numbers \(x\) satisfying \(8.15\le x<8.25\) round to \(8.2\). For any such number below \(8.25\), another number can be chosen between it and \(8.25\). Therefore, the interval has no greatest number.

a) Julia is correct when exactly three decimal places are required. b) Possible examples are \(8.2491\), \(8.2499\), and \(8.24999\). c) There is no greatest number because values can get arbitrarily close to \(8.25\) without reaching it, and \(8.25\) rounds to \(8.3\).