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Add and subtract decimals fluently

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5542686
Use the standard addition algorithm to find \(27.46+8.9\). Show how place value determines where the digits are aligned.

Hints

- Line up digits by place value, not by the final digit in each number. - A trailing zero can be added to a decimal without changing its value. - Add from the smallest place value toward the largest.

Solution

1. Align the decimal points. Rewrite \(8.9\) as \(8.90\) so hundredths line up with hundredths. 2. Add from right to left, regrouping when needed. 3. The sum is \(36.36\).

Answer

\(36.36\)
5106806
A hiker carries four packages with masses of \(0.85\,\text{kg}\), \(2.14\,\text{kg}\), \(3.15\,\text{kg}\), and \(1.86\,\text{kg}\). Explain how to find the total mass mentally by making convenient pairs. Then give the total.

Hints

- Look closely at the hundredths digits. - Which pairs add to a whole number? - You do not need to add the four numbers in the order given.

Solution

1. Pair numbers that add to whole numbers: \(0.85 + 3.15\) and \(2.14 + 1.86\). 2. Evaluate the first pair: \(0.85 + 3.15 = 4.00\). 3. Evaluate the second pair: \(2.14 + 1.86 = 4.00\). 4. Add the pair sums: \(4.00\,\text{kg} + 4.00\,\text{kg} = 8.00\,\text{kg}\).

Answer

The total mass is \(8.00\,\text{kg}\). One efficient grouping is \((0.85 + 3.15) + (2.14 + 1.86) = 4.00 + 4.00 = 8.00\).
5113356
Evaluate each expression efficiently by rearranging or grouping the decimal numbers. a) \(12.35 + 7.8 + 7.65 + 2.2\) b) \(24.7 - 3.9 - 6.1 - 4.7\)

Hints

- Look for decimal parts that combine to make whole numbers. - Several subtractions can be rewritten as subtracting the sum of the amounts. - In b), find two numbers with the same tenths digit and two numbers whose decimal parts make a whole.

Solution

1. For a), group numbers that add to whole numbers: \((12.35 + 7.65) + (7.8 + 2.2) = 20 + 10 = 30\). 2. For b), rewrite the subtractions as subtracting a total and group conveniently: \((24.7 - 4.7) - (3.9 + 6.1) = 20 - 10 = 10\).

Answer

a) \(30\) b) \(10\)
5122356
Evaluate the expression mentally by grouping the decimal numbers efficiently. \(12.4 - 5.8 - 2.4 - 1.2\)

Hints

- Look for decimal parts that combine or cancel to make whole numbers. - Rewrite repeated subtraction as subtracting a sum before regrouping. - Keep the signs attached to the terms when you rearrange.

Solution

1. Group the terms to create easy differences and sums: \((12.4 - 2.4) - (5.8 + 1.2)\). 2. Evaluate the groups: \(12.4 - 2.4 = 10\) and \(5.8 + 1.2 = 7\). 3. Subtract: \(10 - 7 = 3\).

Answer

\(3\)
5351996
A data logger recorded \(8.03\) in one trial and \(5.47\) in another. Use the standard subtraction algorithm to find the difference between the readings.

Hints

- Write the two decimals with their decimal points aligned. - Check whether regrouping is needed before subtracting the hundredths. - Subtract by place value from right to left.

Solution

1. Align the decimal points: subtract \(5.47\) from \(8.03\). 2. Regroup across the zero so the hundredths place can be subtracted. 3. The standard subtraction gives \(8.03-5.47=2.56\).

Answer

\(2.56\)
5412086
Find the missing decimal \(x\) in \(x+17.806=50.2\). Verify your result by addition.

Hints

- Use the inverse relationship between addition and subtraction. - Align the decimal places before subtracting. - Substitute the result into the original equation.

Solution

1. Rewrite \(50.2\) as \(50.200\). 2. Subtract to find the missing addend: \(x=50.200-17.806=32.394\). 3. Verification: \(32.394+17.806=50.200=50.2\).

Answer

\(x=32.394\)
5542696
Use the standard subtraction algorithm to find \(50.2-7.86\). Explain where regrouping is needed.

Hints

- Rewrite the minuend with a trailing zero so both numbers show hundredths. - Start with the hundredths place and decide whether regrouping is necessary. - Keep the decimal point aligned in the difference.

Solution

1. Align the decimal points and write \(50.2\) as \(50.20\). 2. Regroup across place values so the hundredths and tenths can be subtracted. 3. Subtract by place value from right to left. 4. The difference is \(42.34\).

Answer

\(42.34\)
5106466
A hiker starts with \(1.5\) liters of water in a bottle. During the hike, the hiker drinks \(\frac{3}{8}\) liter, adds \(0.5\) liter at a spring, and then drinks another \(0.4\) liter. a) How much water is in the bottle at the end? b) What is the minimum bottle capacity needed so that it does not overflow when refilled?

Hints

- Decide whether fractions or decimals will be easier to use. - Subtract when water is consumed and add when water is refilled. - For the capacity, identify the greatest amount in the bottle at any time.

Solution

1. Convert \(\frac{3}{8}\) to a decimal: \(\frac{3}{8}=0.375\). 2. The final amount is \(1.5-0.375+0.5-0.4=1.225\) liters. 3. After the first drink, the bottle contains \(1.5-0.375=1.125\) liters. After refilling, it contains \(1.125+0.5=1.625\) liters. This is the greatest amount in the bottle. 4. Therefore, the minimum capacity is \(1.625\) liters.

Answer

a) \(1.225\) liters b) At least \(1.625\) liters
5412096
Estimate \(103.005-27.89\) by rounding each number to the nearest whole number. Then find the exact difference and explain how the estimate supports it.

Hints

- Round both values to the same place for a quick check. - Add a trailing zero to the number with fewer decimal places. - Compare the exact answer's size with the estimate.

Solution

1. Rounding gives \(103-28=75\) as the estimate. 2. Align decimal places: \(103.005-27.890=75.115\). 3. The exact difference \(75.115\) is close to the estimate \(75\), so the result is reasonable.

Answer

Estimate: \(75\); exact difference: \(75.115\)
5412106
Four audio-guide sections have these lengths. <table> <tr><th>Section</th><th>Length in minutes</th></tr> <tr><td>Origins</td><td>\(2.475\)</td></tr> <tr><td>Materials</td><td>\(3.08\)</td></tr> <tr><td>Restoration</td><td>\(1.995\)</td></tr> <tr><td>Display</td><td>\(2.61\)</td></tr> </table> a) Find the total length. b) Find the difference between the longest and shortest sections.

Hints

- Use trailing zeros only to align all four values while calculating. - Identify the greatest and least entries separately from finding the total. - Remove padding zeros that do not communicate stated measurement precision.

Solution

1. Align decimal places and add: \(2.475+3.080+1.995+2.610=10.160\), so the total is \(10.16\) minutes. 2. The longest section is \(3.08\) minutes, and the shortest is \(1.995\) minutes. 3. Their difference is \(3.080-1.995=1.085\) minutes.

Answer

a) \(10.16\) minutes b) \(1.085\) minutes
5412116
A student claims \(40-6.785=34.785\). Use place value to explain why the claim is incorrect and find the correct difference.

Hints

- Express the whole number to the thousandths place. - Regroup across the zeros carefully. - Use a nearby whole-number estimate to test the claimed result.

Solution

1. Write the whole number as \(40.000\). 2. Subtract with regrouping: \(40.000-6.785=33.215\). 3. The claim is unreasonable because subtracting nearly \(7\) from \(40\) should give a result near \(33\), not near \(35\).

Answer

The claim is incorrect because \(40\) must be written as \(40.000\) and regrouped by place value. Also, subtracting nearly \(7\) from \(40\) should give a result near \(33\). The correct difference is \(33.215\).
5412126
A school garden water tank starts with \(100\,\text{L}\). Students use \(48.907\,\text{L}\), and then \(3.28\,\text{L}\) is added. How much water is in the tank at the end?

Hints

- Follow the events in the order they occur. - Treat water used and water added as different operations. - Add zeros only as needed to align decimal places.

Solution

1. Align the decimals as \(100.000\), \(48.907\), and \(3.280\). 2. After the water is used, \(100.000-48.907=51.093\,\text{L}\). 3. After more water is added, \(51.093+3.280=54.373\,\text{L}\).

Answer

\(54.373\,\text{L}\)
5412136
Route A has three sections measuring \(4.78\,\text{km}\), \(6.095\,\text{km}\), and \(2.4\,\text{km}\). Route B is \(13.06\,\text{km}\) long. a) Find the length of Route A. b) Which route is longer? c) By how much?

Hints

- Find the full length of the multi-section route first. - Rewrite both totals with equal decimal places before comparing. - Subtract the shorter route from the longer one.

Solution

1. Add the Route A sections: \(4.780+6.095+2.400=13.275\,\text{km}\). 2. Compare \(13.275\) and \(13.060\); Route A is longer. 3. The difference is \(13.275-13.060=0.215\,\text{km}\).

Answer

a) \(13.275\,\text{km}\) b) Route A c) \(0.215\,\text{km}\)
5412146
Solve \(63.5-z=27.846\). Then explain whether \(z\) should be greater or less than \(27.846\).

Hints

- Use the relationship among minuend, subtrahend, and difference. - Align decimal places before finding the missing value. - Substitute the result to check the comparison statement.

Solution

1. Rewrite \(63.5\) as \(63.500\). 2. The missing subtrahend is \(z=63.500-27.846=35.654\). 3. The value \(35.654\) is greater than \(27.846\); subtracting it from \(63.5\) leaves the smaller amount \(27.846\).

Answer

\(z=35.654\), and \(z\) is greater than \(27.846\).
5412166
Three sections of a bike route are \(6.375\,\text{km}\), \(5.8\,\text{km}\), and \(7.025\,\text{km}\). The completed route will be \(25\,\text{km}\). a) Find the combined length of the three finished sections. b) How much additional route is needed?

Hints

- Find the current total before comparing it with the planned length. - Add zeros only as needed to align place values. - Report the final distances without unnecessary padding zeros.

Solution

1. Align the decimals and add: \(6.375+5.800+7.025=19.200\), so the finished length is \(19.2\,\text{km}\). 2. Subtract from the planned total: \(25.000-19.200=5.800\), so \(5.8\,\text{km}\) remains.

Answer

a) \(19.2\,\text{km}\) b) \(5.8\,\text{km}\)
5412176
Evaluate \(45.8-(12.675+9.04)\).

Hints

- Complete the operation inside the parentheses first. - Align decimal places in both calculations. - Estimate the final subtraction to check the result's size.

Solution

1. Evaluate the parentheses: \(12.675+9.040=21.715\). 2. Rewrite \(45.8\) as \(45.800\). 3. Subtract: \(45.800-21.715=24.085\).

Answer

\(24.085\)
5412196
Decide whether each equation is correct. Replace any incorrect result with the correct one. a) \(7.05+2.995=10.045\) b) \(12.4-5.678=6.822\) c) \(0.906+0.94=1.846\)

Hints

- Check each equation independently. - Add trailing zeros to align place values. - Use an estimate to notice any result with an unreasonable size.

Solution

1. Part a) is correct: \(7.050+2.995=10.045\). 2. Part b) is incorrect: \(12.400-5.678=6.722\). 3. Part c) is correct: \(0.906+0.940=1.846\).

Answer

a) Correct. b) Incorrect; \(12.4-5.678=6.722\). c) Correct.
5412216
Evaluate \(18.675+2.325+4.09\). Choose an order that makes the addition efficient, and explain your choice.

Hints

- Look for two addends whose decimal parts combine conveniently. - Addition allows the addends to be grouped in a helpful order. - Use zeros for alignment when needed, but do not pad the final result.

Solution

1. Add \(18.675+2.325=21.000\), which is \(21\). 2. Then add \(21+4.09=25.09\). 3. Grouping the first two addends reduces regrouping and gives the same total.

Answer

\(25.09\)
5542706
Jordan wants to add \(14.6+3.75\). Jordan rewrites \(14.6\) as \(14.06\) before using the standard algorithm. Explain the place-value error and find the correct sum.

Hints

- Compare the place value of the digit \(6\) in \(14.6\) and \(14.06\). - A zero may be appended to the right of a decimal without changing its value. - After correcting the equivalent decimal, align decimal points before adding.

Solution

1. Adding a zero immediately after the decimal changes \(14.6\) to \(14.06\), so Jordan changed the value. 2. To show hundredths without changing the number, write \(14.6\) as \(14.60\). 3. Align the decimal points and add: \(14.60+3.75=18.35\).

Answer

Jordan placed the zero in the wrong place. The correct rewrite is \(14.60\), and the correct sum is \(18.35\).
5412156
Find the missing digit \(d\) in \(7.d4+2.68=10.12\). Explain how the tenths column determines the digit.

Hints

- Start in the hundredths column and note any regrouping. - Use the tenths digit in the sum together with the regrouped amount. - Check the completed addition.

Solution

1. In the hundredths column, \(4+8=12\), so write \(2\) and regroup \(1\) tenth. 2. In the tenths column, \(d+6+1\) must end in \(1\) and regroup \(1\) whole. 3. Thus, \(d+7=11\), so \(d=4\). 4. Check: \(7.44+2.68=10.12\).

Answer

\(d=4\)
5412186
Find the missing digit \(d\) in \(15.2-7.d65=7.835\).

Hints

- Treat the number containing the missing digit as an unknown subtrahend. - Use the minuend and difference to recover that number. - Match the recovered decimal to the given digit pattern.

Solution

1. Find the missing subtrahend by subtraction: \(15.200-7.835=7.365\). 2. Compare \(7.365\) with \(7.d65\). 3. The missing tenths digit is \(d=3\). 4. Check: \(15.200-7.365=7.835\).

Answer

\(d=3\)
5412206
Two swimmers compare their times with a target of \(25\,\text{s}\). Swimmer A finishes in \(24.683\,\text{s}\), and Swimmer B finishes in \(25.412\,\text{s}\). a) How far below the target time is Swimmer A? b) How far above the target time is Swimmer B? c) Which swimmer's time is closer to the target, and by how much?

Hints

- Find each positive difference from the target time. - Compare the two distances from the target. - Subtract those distances to find how much closer one time is.

Solution

1. Swimmer A differs from the target by \(25.000-24.683=0.317\,\text{s}\). 2. Swimmer B differs from the target by \(25.412-25.000=0.412\,\text{s}\). 3. Swimmer A is closer because \(0.317<0.412\). 4. The difference in their distances from the target is \(0.412-0.317=0.095\,\text{s}\).

Answer

a) \(0.317\,\text{s}\) b) \(0.412\,\text{s}\) c) Swimmer A, by \(0.095\,\text{s}\)

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