Free Math Worksheets

- First find the total number of notebooks in the new shipment. - Then add the notebooks that were already on the shelves. - Line up the place values when you add.

1. Add the notebooks in the new shipment: \(145 + 128 + 64 = 337\). 2. Add the notebooks already in the store: \(337 + 35 = 372\).

The store has \(372\) notebooks in all.

- How much does the first addend need to reach the next multiple of \(10\) or \(100\)? - Can you break apart the second addend to reach that number first? - Then add the remaining amount.

1. For \(146 + 7\), add \(4\) to reach \(150\), then add the remaining \(3\): \(150 + 3 = 153\). 2. For \(378 + 5\), add \(2\) to reach \(380\), then add the remaining \(3\): \(380 + 3 = 383\). 3. For \(594 + 8\), add \(6\) to reach \(600\), then add the remaining \(2\): \(600 + 2 = 602\). 4. For \(895 + 9\), add \(5\) to reach \(900\), then add the remaining \(4\): \(900 + 4 = 904\).

a) \(153\) b) \(383\) c) \(602\) d) \(904\)

- A related fact such as \(26 + 6\) can help with \(260 + 60\). - How many tens are needed to reach the next hundred? - Break apart the second addend so you can make the next hundred first.

1. For \(260 + 60\), add \(40\) to reach \(300\), then add the remaining \(20\): \(300 + 20 = 320\). 2. For \(480 + 40\), add \(20\) to reach \(500\), then add the remaining \(20\): \(500 + 20 = 520\). 3. For \(750 + 70\), add \(50\) to reach \(800\), then add the remaining \(20\): \(800 + 20 = 820\). 4. For \(890 + 30\), add \(10\) to reach \(900\), then add the remaining \(20\): \(900 + 20 = 920\).

a) \(320\) b) \(520\) c) \(820\) d) \(920\)

- How much does the first addend need to reach the next multiple of \(100\)? - Can you break apart the second addend to reach that hundred first? - Solve each problem in two smaller steps.

1. Break \(6\) into \(3 + 3\): \(497 + 3 = 500\), then \(500 + 3 = 503\). 2. Break \(8\) into \(5 + 3\): \(295 + 5 = 300\), then \(300 + 3 = 303\). 3. Break \(5\) into \(2 + 3\): \(698 + 2 = 700\), then \(700 + 3 = 703\). 4. Break \(7\) into \(1 + 6\): \(199 + 1 = 200\), then \(200 + 6 = 206\).

a) \(503\) b) \(303\) c) \(703\) d) \(206\)

- Focus first on the tens and ones of the three-digit number. - How do you use the hundreds digit after solving the related fact? - You may first add enough to reach the next ten, then add the rest.

1. \(48 + 7 = 55\). Keeping the \(2\) hundreds gives \(248 + 7 = 255\). 2. \(15 + 6 = 21\). Keeping the \(6\) hundreds gives \(615 + 6 = 621\). 3. \(79 + 4 = 83\). Keeping the \(4\) hundreds gives \(479 + 4 = 483\).

a) \(248 + 7 = 255\); related fact: \(48 + 7 = 55\) b) \(615 + 6 = 621\); related fact: \(15 + 6 = 21\) c) \(479 + 4 = 483\); related fact: \(79 + 4 = 83\)

- Is one addend close to a number that is easier to use? - Can you make the next hundred first? - Look for ones or tens that combine to make a friendly number.

1. For a), use compensation: \(400 + 256 = 656\), then subtract \(2\). The sum is \(654\). 2. For b), make a hundred first: \(420 + 80 = 500\), then \(500 + 200 = 700\). 3. For c), add by place value: \(500 + 200 = 700\), \(40 + 10 = 50\), and \(3 + 5 = 8\). Then \(700 + 50 + 8 = 758\). 4. For d), combine parts that make friendly numbers: \((200 + 100) + (75 + 25) = 300 + 100 = 400\).

a) \(654\) b) \(700\) c) \(758\) d) \(400\)

- Can you break an addend into smaller parts? - Which place values change in each sum? - Break the numbers into hundreds, tens, and ones. - Add the hundreds first and then the tens when helpful.

1. \(560 + 30 = 590\). 2. \(240 + 500 = 740\). 3. \(310 + 280 = 590\). 4. \(470 + 6 = 476\). 5. \(620 + 150 = 770\).

a) \(590\) b) \(740\) c) \(590\) d) \(476\) e) \(770\)

- Evaluate both sides of each comparison. - You may also compare how the addends change from one side to the other. - Use \(<\) or \(>\) when one sum is smaller or greater. - Use \(=\) when the two sums have the same value.

1. For a), \(340 + 50 = 390\) and \(320 + 70 = 390\), so the values are equal. 2. For b), \(600 + 230 = 830\) and \(500 + 340 = 840\), so \(830 < 840\). 3. For c), \(180 + 400 = 580\) and \(200 + 380 = 580\), so the values are equal. 4. For d), \(450 + 120 = 570\) and \(460 + 100 = 560\), so \(570 > 560\).

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

- Identify all three kinds of drinks. - Add the two juice amounts first. - Then add the bottles of water.

1. Add the two kinds of juice: \(88 + 54 = 142\). 2. Add the bottles of water: \(145 + 142 = 287\).

There are \(287\) bottles of drinks altogether.

- How far is the changed addend from the next hundred? - If you use a larger addend, subtract the extra amount afterward. - Which addend is close to a multiple of \(100\)?

1. Use \(400\) instead of \(398\), which adds \(2\) too much: \(400 + 154 - 2 = 554 - 2 = 552\). 2. Use \(500\) instead of \(499\), which adds \(1\) too much: \(270 + 500 - 1 = 770 - 1 = 769\). 3. Use \(600\) instead of \(595\), which adds \(5\) too much: \(600 + 130 - 5 = 730 - 5 = 725\).

a) \(398 + 154 = 400 + 154 - 2 = 552\) b) \(270 + 499 = 270 + 500 - 1 = 769\) c) \(595 + 130 = 600 + 130 - 5 = 725\)

- The numbers being added are called addends. - The result of addition is called the sum. - Use a place-value addition strategy.

1. Part a: \(470 + 360 = 830\). 2. Part b: \(280 + 540 = 820\).

a) The sum is \(830\). b) The sum is \(820\).

- Subtract the known addend from the sum to find the missing addend. - Calculate the original sum before changing the addends. - Account for the increase in both addends.

1. Part a: Subtract the known addend from the sum: \(820 - 550 = 270\). 2. Part b: The original sum is \(190 + 430 = 620\). 3. Increasing both addends by \(10\) gives \(200 + 440 = 640\). 4. The sum increases by \(20\) because each of the two addends increases by \(10\).

a) The other addend is \(270\). b) The original sum is \(620\). The new sum is \(640\), so the sum increases by \(20\).

- Break each target sum into hundreds and tens. - Think of numbers that appear when counting by tens. - Choose one simple multiple of \(10\), then subtract it from the target. - Many answers are possible, but the two addends must be different.

1. a) One possible pair is \(200\) and \(250\), because \(200 + 250 = 450\). 2. b) One possible pair is \(400\) and \(420\), because \(400 + 420 = 820\). 3. c) One possible pair is \(250\) and \(350\), because \(250 + 350 = 600\). 4. Other pairs of different multiples of \(10\) are also possible.

a) One possible answer is \(200\) and \(250\). b) One possible answer is \(400\) and \(420\). c) One possible answer is \(250\) and \(350\).

- Find both sums separately. - Compare the two results. - You can estimate first to predict which sum will be greater. - Pay attention when an addition crosses a hundred.

1. The first sum is \(470 + 80 = 550\). 2. The second sum is \(390 + 150 = 540\). 3. Since \(550 > 540\), the first sum is greater.

\(470 + 80\) is greater. It equals \(550\), while \(390 + 150 = 540\).

- Break apart the second addend to reach the next hundred first. - How much does the first addend need to reach the next multiple of \(100\)? - After reaching that hundred, add the remaining amount.

1. Break \(50\) into \(30 + 20\): \(270 + 30 = 300\), then \(300 + 20 = 320\). 2. Break \(60\) into \(20 + 40\): \(480 + 20 = 500\), then \(500 + 40 = 540\). 3. Break \(70\) into \(40 + 30\): \(760 + 40 = 800\), then \(800 + 30 = 830\). 4. Break \(40\) into \(10 + 30\): \(590 + 10 = 600\), then \(600 + 30 = 630\).

a) \(320\) b) \(540\) c) \(830\) d) \(630\)

- Add by place value, and record your work if the sums are not easy to do mentally. - Check any regrouping in the ones and tens places.

1. The left brick in the second row is \(245 + 138 = 383\). 2. The right brick in the second row is \(138 + 304 = 442\). 3. The top brick is \(383 + 442 = 825\).

Second row: \(383\), \(442\) Top: \(825\)

- Break the second addend into hundreds, tens, and ones. - Add the largest place-value part first, then the smaller parts. - Watch for regrouping when adding the ones.

1. For \(435 + 258\): \(435 + 200 = 635\), \(635 + 50 = 685\), and \(685 + 8 = 693\). 2. For \(167 + 544\): \(167 + 500 = 667\), \(667 + 40 = 707\), and \(707 + 4 = 711\). 3. For \(329 + 482\): \(329 + 400 = 729\), \(729 + 80 = 809\), and \(809 + 2 = 811\).

a) \(693\) b) \(711\) c) \(811\)

- Each result becomes the starting number in the next line. - Notice whether each step adds hundreds, tens, or ones.

1. For part a, \(374 + 200 = 574\), \(574 + 40 = 614\), and \(614 + 7 = 621\). 2. For part b, \(586 + 100 = 686\), \(686 + 30 = 716\), and \(716 + 5 = 721\).

a) \(574\), \(614\), \(621\) b) \(686\), \(716\), \(721\)

- Look for an addend that is close to a multiple of \(100\). - If you add too much at first, how can you adjust the sum? - Can you break an addend into hundreds and a small number?

1. For part a, replace \(399\) with \(400\): \(256 + 400 = 656\). Since \(400\) is \(1\) too much, subtract \(1\): \(656 - 1 = 655\). 2. For part b, replace \(198\) with \(200\): \(437 + 200 = 637\). Since \(200\) is \(2\) too much, subtract \(2\): \(637 - 2 = 635\). 3. For part c, break \(202\) into \(200 + 2\): \(524 + 200 = 724\), and \(724 + 2 = 726\).

a) \(655\) b) \(635\) c) \(726\)

- Break \(47\) into tens and ones. - What do you get after adding only the tens? - Then add the ones to the intermediate result.

1. Add the tens: \(384 + 40 = 424\). 2. Add the remaining ones: \(424 + 7 = 431\).

\(431\)

- How many hundreds, tens, and ones are in \(265\)? - Add one place-value part at a time. - What do you get after adding the hundreds first?

1. Add the hundreds: \(456 + 200 = 656\). 2. Add the tens: \(656 + 60 = 716\). 3. Add the ones: \(716 + 5 = 721\).

\(721\)

- The second addend is close to a multiple of \(100\). - Could you add a little too much first and then adjust? - How far is \(198\) from \(200\)?

1. Replace \(198\) with the nearby multiple of \(100\), \(200\). 2. Add: \(647 + 200 = 847\). 3. Since \(200\) is \(2\) more than \(198\), subtract \(2\): \(847 - 2 = 845\).

\(845\)

- Add the tens of the second addend first, then the ones. - Check whether adding the tens changes the hundreds digit. - Use the first intermediate result as the starting number for the second step.

1. \(367 + 20 = 387\), then \(387 + 5 = 392\). 2. \(584 + 30 = 614\), then \(614 + 8 = 622\). 3. \(749 + 60 = 809\), then \(809 + 3 = 812\).

a) \(392\) b) \(622\) c) \(812\)

- Work from left to right, one step at a time. - Pay attention when the chain reaches \(400\). - Can you break \(7\) into two parts to reach a multiple of \(10\) or \(100\)?

1. \(386 + 7 = 393\). 2. \(393 + 7 = 400\). 3. \(400 + 7 = 407\). 4. \(407 + 7 = 414\).

\(393\), \(400\), \(407\), \(414\)

- Use the number at the start of each part as the first addend. - What pattern do you notice between adding \(8\) and adding \(9\)? - Break apart the second addend to reach the next hundred first.

1. For a), \(594 + 8 = 602\) and \(594 + 9 = 603\). 2. For b), \(296 + 8 = 304\) and \(296 + 9 = 305\). 3. For c), \(795 + 8 = 803\) and \(795 + 9 = 804\). Each sum can be found by breaking apart the ones to reach the next hundred first.

a) \(602\), \(603\) b) \(304\), \(305\) c) \(803\), \(804\)

- Identify the first and second addends. - How many hundreds, tens, and ones are in \(287\)? - Add those parts one at a time. - Watch for regrouping across tens and hundreds.

1. Add the hundreds: \(345 + 200 = 545\). 2. Add the tens: \(545 + 80 = 625\). 3. Add the ones: \(625 + 7 = 632\).

\(632\)

- Break \(354\) into hundreds, tens, and ones. - Use the first result as the starting number in the second line. - Use the second result as the starting number in the third line.

1. Add the hundreds: \(468 + 300 = 768\). 2. Add the tens: \(768 + 50 = 818\). 3. Add the ones: \(818 + 4 = 822\).

\(768\), \(818\), \(822\)

- Break the second addend into hundreds, tens, and ones. - Add the hundreds first, then the tens, and finally the ones. - Watch for regrouping across tens and hundreds.

1. For part a, \(357 + 200 = 557\), \(557 + 60 = 617\), and \(617 + 4 = 621\). 2. For part b, \(482 + 100 = 582\), \(582 + 50 = 632\), and \(632 + 9 = 641\). 3. For part c, \(576 + 300 = 876\), \(876 + 30 = 906\), and \(906 + 8 = 914\).

a) \(621\) b) \(641\) c) \(914\)

- Break \(457\) into hundreds, tens, and ones. - Record each intermediate result. - Check whether adding the tens reaches a new hundred. - Remember to add the ones in the final step.

1. Add the hundreds: \(386 + 400 = 786\). 2. Add the tens: \(786 + 50 = 836\). 3. Add the ones: \(836 + 7 = 843\).

\(386 + 457 = 843\)

- First add only the hundreds from the second class’s amount. - Break \(385\) into hundreds, tens, and ones. - What total does the problem ask you to find?

1. Break \(385\) into \(300 + 80 + 5\). 2. Add the hundreds: \(429 + 300 = 729\). 3. Add the tens: \(729 + 80 = 809\). 4. Add the ones: \(809 + 5 = 814\).

The two classes have collected \(814\) bottles altogether.

- Notice which part of \(348\) is added in each line. - Use each result as the starting number in the next line. - Watch for the new hundred when adding the tens.

1. Add the hundreds: \(574 + 300 = 874\). 2. Use \(874\) as the next starting number and add the tens: \(874 + 40 = 914\). 3. Use \(914\) as the next starting number and add the ones: \(914 + 8 = 922\).

\(574 + 348 = 922\) \(574 + 300 = 874\) \(874 + 40 = 914\) \(914 + 8 = 922\)

- Which nearby multiple of \(100\) would be easier to add? - If you add too much at first, how can you adjust the sum? - Find how far the second addend is from the next hundred.

1. Replace \(199\) with \(200\): \(236 + 200 = 436\). Since \(200\) is \(1\) too much, subtract \(1\): \(436 - 1 = 435\). 2. Replace \(298\) with \(300\): \(458 + 300 = 758\). Since \(300\) is \(2\) too much, subtract \(2\): \(758 - 2 = 756\). 3. Replace \(399\) with \(400\): \(375 + 400 = 775\). Since \(400\) is \(1\) too much, subtract \(1\): \(775 - 1 = 774\).

a) \(435\) b) \(756\) c) \(774\)

- Break the second addend into hundreds, tens, and ones. - Add each part to the first addend in order. - Record the intermediate results.

1. For \(423 + 354\): \(423 + 300 = 723\), \(723 + 50 = 773\), and \(773 + 4 = 777\). 2. For \(267 + 125\): \(267 + 100 = 367\), \(367 + 20 = 387\), and \(387 + 5 = 392\). 3. For \(538 + 246\): \(538 + 200 = 738\), \(738 + 40 = 778\), and \(778 + 6 = 784\).

a) \(777\) b) \(392\) c) \(784\)

- Can you make one addend a multiple of \(100\)? - What happens to the sum if you move an amount from one addend to the other? - Is there a nearby addition problem that is easier to solve?

1. For \(295 + 165\), move \(5\) from \(165\) to \(295\): \(300 + 160 = 460\). 2. For \(398 + 244\), use \(400 + 244 = 644\). Since \(400\) is \(2\) more than \(398\), subtract \(2\): \(644 - 2 = 642\). 3. For \(499 + 301\), move \(1\) from \(301\) to \(499\): \(500 + 300 = 800\).

a) \(460\) b) \(642\) c) \(800\)

- Which addend should be broken apart? - Break \(286\) into hundreds, tens, and ones. - How does the running total change at each step?

1. Add the hundreds: \(457 + 200 = 657\). 2. Add the tens: \(657 + 80 = 737\). 3. Add the ones: \(737 + 6 = 743\).

\(457 + 200 = 657\), \(657 + 80 = 737\), and \(737 + 6 = 743\). The sum is \(743\).

- Break the second addend into hundreds, tens, and ones. - Add the place-value parts from largest to smallest. - Watch for regrouping across tens and hundreds.

1. For part a, \(524 + 300 = 824\), \(824 + 80 = 904\), and \(904 + 7 = 911\). 2. For part b, \(675 + 100 = 775\), \(775 + 40 = 815\), and \(815 + 8 = 823\).

a) \(911\) b) \(823\)

- Did Lucas add too much or too little at first? - What adjustment will undo that extra amount? - Which nearby numbers are easier to add than \(99\) and \(19\)?

1. Lucas used \(300\) instead of \(295\), so he added \(5\) too much. Subtract \(5\) from \(348\): \(348 - 5 = 343\). 2. For \(57 + 99\), use \(57 + 100 = 157\), then subtract \(1\): \(157 - 1 = 156\). 3. For \(146 + 19\), use \(146 + 20 = 166\), then subtract \(1\): \(166 - 1 = 165\).

a) Lucas must subtract \(5\). The sum is \(343\). b) \(57 + 99 = 156\) \(146 + 19 = 165\)

- What do you notice when you add \(34\) and \(66\)? - Is \(91 + 66\) or \(100 + 57\) easier to find mentally? - In the new problems, look for addends that make a multiple of \(100\).

1. Method B is easier because \(34 + 66 = 100\), and adding to \(100\) is simple. 2. For part a, first add \(123 + 77 = 200\). Then \(200 + 49 = 249\). 3. For part b, first add \(250 + 150 = 400\). Then \(400 + 368 = 768\).

Method B is easier because \(34 + 66 = 100\). a) \(249\) b) \(768\)

- Evaluate the left sum first. - Evaluate the right sum next. - Compare the two results. - Before calculating exactly, look for compensating changes in the addends.

1. For a), \(430 + 280 = 710\) and \(520 + 190 = 710\), so the sums are equal. 2. For b), \(360 + 470 = 830\) and \(250 + 590 = 840\), so \(830 < 840\). 3. For c), \(180 + 740 = 920\) and \(630 + 290 = 920\), so the sums are equal.

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

- Look for two addends that make a multiple of \(100\). - Can you form two convenient pairs from the four addends? - Find each pair’s sum, then add the two results.

1. For part a, pair \(57 + 43 = 100\) and \(34 + 66 = 100\). Then \(100 + 100 = 200\). 2. For part b, pair \(215 + 85 = 300\) and \(88 + 12 = 100\). Then \(300 + 100 = 400\). 3. For part c, pair \(46 + 54 = 100\) and \(127 + 73 = 200\). Then \(100 + 200 = 300\).

a) \(200\) b) \(400\) c) \(300\)

- Break the numbers into hundreds and tens. - What happens if you add the tens first? - Could you first make the next hundred? - Break the calculation into two addition steps.

1. Add the first two numbers: \(350 + 70 = 420\). 2. Add the third number: \(420 + 120 = 540\).

The sum is \(540\).

- Recall the names of the numbers and result in an addition equation. - Calculate both the original and changed equations. - Compare the equal-sized changes to the two addends.

1. Part a: The numbers being added are called addends. 2. Part b: \(340 + 180 = 520\). The result of addition is called the sum. 3. Part c: The new addends are \(340 + 20 = 360\) and \(180 - 20 = 160\). 4. The new sum is \(360 + 160 = 520\). 5. The sum stays the same because increasing one addend by \(20\) and decreasing the other by \(20\) are equal and opposite changes.

a) The numbers are addends. b) The result is \(520\), and it is called the sum. c) The sum remains \(520\) because the increase of \(20\) in one addend is offset by the decrease of \(20\) in the other addend.

- Start in the ones place and remember that you may need to regroup. - Then work one place at a time from right to left.

1. Ones place: \(* + 8 = 17\), so the missing ones digit is \(9\). Regroup \(1\) ten. 2. Tens place: \(* + 8 + 1 = 12\), so the missing tens digit is \(3\). Regroup \(1\) hundred. 3. The regrouped \(1\) becomes the hundreds digit of the sum. The completed equation is \(39 + 88 = 127\).

The missing digits are \(3\) and \(9\). The completed equation is \(39 + 88 = 127\).

- Read all four bar values. - Add the values in convenient pairs. - Check that the total is reasonable compared with the bars.

1. The four values are \(120\), \(200\), \(160\), and \(160\). 2. Add them: \(120 + 200 + 160 + 160 = 640\).

The graph shows \(640\) rolls altogether.

- Read all three bar values. - Add the values. - Compare the sum with \(200\).

1. The bars show \(70\), \(50\), and \(90\) tickets. 2. Their sum is \(70 + 50 + 90 = 210\). 3. Since \(210 \ne 200\), Leah is not correct.

No. The three booths sold \(210\) raffle tickets altogether.

- A digit in the hundreds place affects the sum more than a digit in the tens or ones place. - A three-digit number cannot have \(0\) in the hundreds place. - For the least sum, place the least possible digits in the highest-value places; reverse that idea for the greatest sum.

1. To minimize the sum, place the least nonzero digits, \(1\) and \(2\), in the hundreds places. Place the next least digits, \(0\) and \(3\), in the tens places, and place \(4\) and \(5\) in the ones places. 2. One possible arrangement is \(104+235=339\). Any arrangement with the same pairs of place-value digits has the same least sum. 3. To maximize the sum, place \(5\) and \(4\) in the hundreds places, \(3\) and \(2\) in the tens places, and \(1\) and \(0\) in the ones places. 4. One possible arrangement is \(531+420=951\).

a) \(339\) b) \(951\)

- Consider whether a carry is possible in any place-value column. - Find pairs of available digits that have a sum of \(9\). - You need three such pairs to make the two numbers.

1. Since the greatest possible sum of two different digits from the set is less than \(19\), no column can produce a carry while the total is \(999\). 2. Therefore, the two digits in each place must have a sum of \(9\). 3. The available pairs are \((1,8)\), \((2,7)\), \((3,6)\), and \((4,5)\). 4. Choose any three of these pairs for the hundreds, tens, and ones places. The unused pair is left over. 5. For example, \(123+876=999\), leaving \(4\) and \(5\).

The possible leftover pairs are \(1\) and \(8\), \(2\) and \(7\), \(3\) and \(6\), or \(4\) and \(5\). Examples: \(234+765=999\), leaving \(1\) and \(8\) \(134+865=999\), leaving \(2\) and \(7\) \(124+875=999\), leaving \(3\) and \(6\) \(123+876=999\), leaving \(4\) and \(5\)

- Choose ones digits whose sum ends in \(0\), and account for any regrouped ten. - Use the regrouped ten when choosing the tens digits. - In the hundreds place, include the regrouped hundred so that the total is \(5\).

1. To make \(0\) in the ones place of the sum, choose two ones digits with a sum of \(10\), such as \(6\) and \(4\). Regroup \(1\) ten. 2. The tens digits and the regrouped ten must total \(10\), so the two tens digits must have a sum of \(9\), such as \(2\) and \(7\). Regroup \(1\) hundred. 3. The hundreds digits and the regrouped hundred must total \(5\), so the two hundreds digits must have a sum of \(4\), such as \(1\) and \(3\). 4. This gives \(126+374=500\), using six different cards.

One possible equation is \(126+374=500\). Other answers are possible.