- Choose a number of chairs and subtract their legs from \(38\). - Check whether the remaining number is a multiple of \(3\). - Try different numbers of chairs.

1. With \(2\) chairs, the chairs have \(2 \times 4 = 8\) legs. The remaining \(38 - 8 = 30\) legs make \(30 \div 3 = 10\) stools. 2. With \(5\) chairs, the chairs have \(5 \times 4 = 20\) legs. The remaining \(38 - 20 = 18\) legs make \(18 \div 3 = 6\) stools.

Two possible combinations are: - \(10\) stools and \(2\) chairs - \(6\) stools and \(5\) chairs

- Try the possible numbers of tricycles systematically. - Subtract the tricycle wheels from \(18\). - Check whether the remaining wheels can be divided into groups of \(2\).

1. Try positive numbers of tricycles and check whether the remaining wheels can form bicycles. 2. With \(2\) tricycles, \(2 \times 3 = 6\) wheels are used. The remaining \(12\) wheels make \(12 \div 2 = 6\) bicycles. 3. With \(4\) tricycles, \(4 \times 3 = 12\) wheels are used. The remaining \(6\) wheels make \(6 \div 2 = 3\) bicycles. 4. Six tricycles use all \(18\) wheels and leave no bicycles, so that case is excluded. These are all the possible combinations.

The two possible combinations are: - \(6\) bicycles and \(2\) tricycles - \(3\) bicycles and \(4\) tricycles

- First find the new total after Julia receives \(8\) cards. - Divide the new total equally between the two children. - Reverse Julia’s increase to find her starting amount.

1. After Julia receives the cards, they will have \(72+8=80\) cards altogether. 2. At that point, each child will have \(80\div 2=40\) cards. 3. Mark’s number did not change, so he began with \(40\) cards. 4. Julia began with \(40-8=32\) cards.

Julia began with \(32\) cards, and Mark began with \(40\) cards.

- “Exactly halfway” means the distance from \(150\) to each unknown marker is the same. - Relate that equal distance to the full \(64\)-mile separation. - Check that your two marker numbers are equally far from \(150\).

1. Half of the \(64\)-mile separation is \(64 \div 2=32\) miles. 2. The lower marker is \(150-32=118\), and the higher marker is \(150+32=182\).

The mile markers are \(118\) and \(182\).

- First find the equal number of marbles in each jar at the end. - For each jar, decide whether marbles were added or removed. - Reverse all the moves made from the first jar. - A before-and-after table may help organize the changes.

1. After the moves, each jar contains \(90\div 3=30\) marbles. 2. The second jar received \(5\) marbles, so it began with \(30-5=25\). 3. The third jar received \(7\) marbles, so it began with \(30-7=23\). 4. The first jar gave away \(5+7=12\) marbles, so it began with \(30+12=42\).

The first jar had \(42\) marbles, the second had \(25\), and the third had \(23\).

- Find how many apples remain after the removals. - Divide the remaining total equally among the three crates. - Add the removed apples back to the appropriate starting amounts.

1. A total of \(6+3=9\) apples are removed. 2. The crates then contain \(75-9=66\) apples altogether. 3. Each crate has \(66\div 3=22\) apples after the removals. 4. Reverse the changes: the first crate began with \(22+6=28\), the second with \(22+3=25\), and the third with \(22\).

The first crate had \(28\) apples, the second had \(25\), and the third had \(22\).

- First find the new total after Ethan’s and Maya’s changes. - Divide the new total equally among four children. - Reverse each change to recover Ethan’s and Maya’s starting amounts.

1. After the changes, the total is \(100-8+4=96\). 2. Each child then has \(96\div 4=24\) acorns. 3. Reverse Ethan’s change: he began with \(24+8=32\) acorns. 4. Reverse Maya’s change: she began with \(24-4=20\) acorns. 5. The other two children each began with \(24\) acorns because their amounts did not change.

Ethan had \(32\) acorns, Maya had \(20\), and each of the other two children had \(24\).

- First determine the total after Tom loses the marbles. - Equal amounts for three children call for division by \(3\). - Work backward from Tom's final amount to find his starting amount.

1. Find the total number of marbles remaining: \(80 - 8 = 72\). 2. Find the number each child has now: \(72 \div 3 = 24\). 3. Find Tom's original number of marbles: \(24 + 8 = 32\). 4. The other two children did not lose any marbles, so each had \(24\) at first. Since \(24 < 32\), each had fewer marbles than Tom.

a) Each child has \(24\) marbles now. b) Tom had \(32\) marbles at first. c) Each of the other children had fewer marbles than Tom.

- Find how many students remain after the two boys leave. - When two groups are equal and their total is known, divide the total by \(2\). - The number of girls does not change. - Check that your two starting groups add to \(22\).

1. After two boys leave, \(22 - 2 = 20\) students remain. 2. The remaining students are split equally between girls and boys: \(20 \div 2 = 10\). 3. The number of girls did not change, so there were \(10\) girls at first. 4. Add back the two boys who left: \(10 + 2 = 12\).

There were \(10\) girls and \(12\) boys at first.

- Find how many fruits remain after one pear is eaten. - If the number of apples is twice the number of pears, how many equal parts are there altogether? - Divide the remaining fruits into those equal parts. - Add the eaten pear back at the end.

1. After one pear is eaten, \(16 - 1 = 15\) fruits remain. 2. The apples make up two equal parts and the pears make up one equal part, for \(3\) parts total. 3. Each part is \(15 \div 3 = 5\). Therefore, \(10\) apples and \(5\) pears remain. 4. Add back the pear that was eaten: there were originally \(5 + 1 = 6\) pears and \(10\) apples.

There were \(10\) apples and \(6\) pears at first.

- Find how many more sets Class 4B orders. - Use the bill difference to find the price of one set. - Multiply the unit price by each order, then add the two totals.

1. Class 4B orders \(15 - 12 = 3\) more sets. 2. One set costs \(\$36 \div 3 = \$12\). 3. Class 4A pays \(12 \times \$12 = \$144\). 4. Class 4B pays \(15 \times \$12 = \$180\). 5. Together, the classes pay \(\$144 + \$180 = \$324\).

a) One watercolor set costs \(\$12\). b) Class 4A pays \(\$144\), and Class 4B pays \(\$180\). c) Together, the classes pay \(\$324\).

- The total refers only to the flour, not the bags. - How much flour is in each bag after the amounts become equal? - Reverse the transfer to recover the original amounts.

1. After the transfer, each bag holds \(100\div2=50\,\text{oz}\) of flour. 2. The first bag originally held \(50+10=60\,\text{oz}\). 3. The second bag originally held \(50-10=40\,\text{oz}\). 4. The amounts check because \(60+40=100\) and transferring \(10\,\text{oz}\) leaves \(50\,\text{oz}\) in each bag.

The first bag originally held \(60\,\text{oz}\); the second held \(40\,\text{oz}\).

- First find how many stickers each person has when the amounts are equal. - Reverse the transfer of \(4\) stickers. - Decide who had more and who had fewer before the transfer. - Use a sketch or table to check that the starting amounts total \(64\).

1. After the transfer, each person has \(64 \div 2 = 32\) stickers. 2. Before giving away \(4\), Tim had \(32 + 4 = 36\) stickers. 3. Before receiving \(4\), Lisa had \(32 - 4 = 28\) stickers.

Tim had \(36\) stickers, and Lisa had \(28\) stickers.

- Imagine first that both shelves had the same number of books. - Temporarily remove the difference of \(18\) from the total. - Divide the remaining total equally between the two shelves. - Add the \(18\) extra books back to the second shelf.

1. Remove the extra \(18\) books from the total: \(92 - 18 = 74\). 2. Divide the remaining books equally: \(74 \div 2 = 37\). This is the number on the first shelf. 3. The second shelf has \(37 + 18 = 55\) books.

The first shelf has \(37\) books, and the second shelf has \(55\) books.

- Find Class C's amount first. - Then focus on the combined amount for Classes A and B. - Temporarily remove their difference before dividing equally. - Check that all three amounts total \(450\,\text{kg}\).

1. Class C collects \(450 - 310 = 140\), so it collects \(140\,\text{kg}\). 2. Remove the \(20\,\text{kg}\) difference from the combined amount for Classes A and B: \(310 - 20 = 290\). 3. Divide equally to find Class A's amount: \(290 \div 2 = 145\). 4. Class B collects \(145 + 20 = 165\), so it collects \(165\,\text{kg}\).

Class A collects \(145\,\text{kg}\), Class B collects \(165\,\text{kg}\), and Class C collects \(140\,\text{kg}\).

- Compare each two-person subtotal with the overall total. Which person is missing from each subtotal? - Find Lucas’s and Tom’s individual amounts before comparing them. - Check that the three individual amounts would be consistent with both given pair totals.

1. Tom’s amount is the part not included in the Lucas-and-Mia total: \(750-480=270\). 2. Lucas’s amount is the part not included in the Mia-and-Tom total: \(750-510=240\). 3. Since \(270-240=30\), Tom collected \(30\) more acorns than Lucas.

Tom collected \(30\) more acorns than Lucas.

- The two ticket counts have both a known total and a known difference. - Think about how you could temporarily make the two groups equal without changing which one is larger. - Check any proposed pair against both the total and the difference.

1. Set aside the \(180\)-ticket difference: \(1250-180=1070\). 2. The remaining tickets form two equal groups, so the child-ticket count is \(1070\div2=535\). 3. The adult-ticket count is \(535+180=715\). 4. Therefore the principal is correct.

Yes. The fair sold \(535\) child tickets and \(715\) adult tickets, so the principal’s \(715\)-ticket claim is correct.

- Find the total weight remaining after the apples are sold. - Divide that remaining weight equally between apples and pears. - The pear amount did not change. - Add the sold apples back to the remaining apples.

1. After the sale, \(400 - 240 = 160\), so \(160\,\text{kg}\) of fruit remain. 2. Equal amounts of apples and pears remain, so each amount is \(160 \div 2 = 80\), or \(80\,\text{kg}\). 3. No pears were sold, so the original pear amount was \(80\,\text{kg}\). 4. The original apple amount was \(80 + 240 = 320\), or \(320\,\text{kg}\).

There were \(320\,\text{kg}\) of apples and \(80\,\text{kg}\) of pears at first.

- Find how many campers remain after the canoe group leaves. - Divide the remaining campers into two equal groups. - All the girls stayed at camp. - Add the boys at camp and the boys canoeing.

1. The number of campers remaining is \(750 - 350 = 400\). 2. Equal numbers of boys and girls remain, so each group has \(400 \div 2 = 200\) campers. 3. All the girls stayed, so there are \(200\) girls altogether. 4. There are \(200 + 350 = 550\) boys altogether.

There are \(550\) boys and \(200\) girls enrolled.

- Find the total length left after the piece is cut. - Divide the remaining total into two equal lengths. - Decide which roll was longer at first.

1. After the cut, the total length is \(870 - 150 = 720\), or \(720\,\text{m}\). 2. The two remaining rolls are equal, so each is \(720 \div 2 = 360\), or \(360\,\text{m}\). 3. The second roll was not cut, so it was originally \(360\,\text{m}\). The first roll was originally \(360 + 150 = 510\), or \(510\,\text{m}\).

1. The first roll was \(510\,\text{m}\), and the second roll was \(360\,\text{m}\). 2. After the cut, each roll is \(360\,\text{m}\).

- Two of the three weights are equal, while the third has one known extra part. - Represent the total as three equal base amounts plus the extra weight. - Check your results against both the total and Tim’s \(16\,\text{lb}\) difference.

1. Set aside Tim’s extra \(16\,\text{lb}\): \(286-16=270\,\text{lb}\). 2. The three equal base weights are \(270\div3=90\,\text{lb}\). 3. Max and Leo each weigh \(90\,\text{lb}\), and Tim weighs \(90+16=106\,\text{lb}\). 4. Check: \(90+90+106=286\).

Tim weighs \(106\,\text{lb}\), Max weighs \(90\,\text{lb}\), and Leo weighs \(90\,\text{lb}\).

- Focus on the point in the story when the two crate masses become equal. - Relate the final combined mass to those two equal amounts. - After finding the final state, work backward only for the crate that changed.

1. After the removal, the two crates contain \(42-6=36\,\text{kg}\) altogether. 2. Equal final masses are \(36\div2=18\,\text{kg}\) per crate. 3. The second crate originally had \(18\,\text{kg}\). The first originally had \(18+6=24\,\text{kg}\).

The first crate originally had \(24\,\text{kg}\), and the second crate originally had \(18\,\text{kg}\).

- How many rolls are not whole-wheat rolls? - The white and rye counts add to that remainder and differ by the stated amount. - Can you imagine removing the extra white rolls to create two equal groups?

1. The white and rye rolls total \(1250-450=800\). 2. Remove the difference: \(800-120=680\). 3. Half the remaining amount is the number of rye rolls: \(680\div2=340\). 4. There are \(340+120=460\) white rolls.

The bakery made \(460\) white rolls and \(340\) rye rolls.

- Organize your attempts in a table. - Choose a number of large towers and find the blocks remaining. - Check whether the remainder can be divided into groups of \(4\).

1. Try even numbers of large towers so the remaining blocks can be grouped by \(4\). 2. With \(2\) large towers, \(2 \times 6 = 12\) blocks are used. The remaining \(48 - 12 = 36\) blocks make \(36 \div 4 = 9\) small towers. 3. With \(4\) large towers, \(4 \times 6 = 24\) blocks are used. The remaining \(24\) blocks make \(24 \div 4 = 6\) small towers. 4. With \(6\) large towers, \(6 \times 6 = 36\) blocks are used. The remaining \(12\) blocks make \(12 \div 4 = 3\) small towers.

Three combinations are: - \(9\) small towers and \(2\) large towers - \(6\) small towers and \(4\) large towers - \(3\) small towers and \(6\) large towers

- The number of \(\$10\) bills must be twice the number of \(\$5\) bills. - Keep track of both the total number of bills and their total value. - Try the possible small counts systematically and reject any choice that breaks one of the three conditions.

1. Test possible numbers of \(\$5\) bills while keeping the number of \(\$10\) bills twice as large. 2. With \(2\) five-dollar bills, there must be \(4\) ten-dollar bills. That uses \(6\) of the \(8\) bills, so the remaining \(2\) are twenty-dollar bills. 3. Their value is \(2 \times \$5+4 \times \$10+2 \times \$20=\$90\), so all conditions are met. 4. Smaller possible numbers of five-dollar bills make the total value too large, so this solution is unique.

There are \(2\) five-dollar bills, \(4\) ten-dollar bills, and \(2\) twenty-dollar bills. The \(4\) ten-dollar bills satisfy the “twice as many” condition, all three counts total \(8\) bills, and their value is \(\$90\). Checking the other possible small counts of five-dollar bills shows that none satisfies both the bill-count and value conditions, so the solution is unique.

- The two animal types contribute different numbers of legs but each counts as one animal. - Try organizing what would happen in one simple extreme case, then compare it with the actual leg total. - Check any proposed pair of counts against both the total number of animals and the total number of legs.

1. If all \(10\) animals were beetles, they would have \(10 \times 6=60\) legs. 2. The actual total has \(68-60=8\) more legs. 3. Replacing one beetle with one spider adds \(8-6=2\) legs, so \(8 \div 2=4\) of the animals are spiders. 4. The remaining \(10-4=6\) animals are beetles. 5. Check: \(4 \times 8+6 \times 6=32+36=68\).

There are \(6\) beetles and \(4\) spiders.

- Test the daughter’s proposed counts against both conditions before trying to replace them. - The two animal types have different leg counts, so changing one type into the other changes the total in a predictable way. - Check your final counts against both \(25\) animals and \(74\) legs.

1. The claim would give \(15 \times 4+10 \times 2=60+20=80\) legs, so it is incorrect. 2. If all \(25\) animals were chickens, they would have \(25 \times 2=50\) legs. 3. The actual total has \(74-50=24\) more legs. 4. Replacing one chicken with one pig adds \(4-2=2\) legs, so there are \(24 \div 2=12\) pigs. 5. There are \(25-12=13\) chickens. Check: \(12 \times 4+13 \times 2=48+26=74\).

The daughter’s claim is incorrect. There are \(12\) pigs and \(13\) chickens.

- Think about how the difference changes when one box gains \(1\) and the other loses \(1\). - Use the total and the original difference together once you have determined that difference. - Check your pair using the hypothetical change described in the problem.

1. Adding \(1\) to the smaller box and removing \(1\) from the larger box reduces the original difference by \(2\). Since the adjusted difference is \(5\), the original difference is \(7\). 2. Subtract the difference from the total: \(27-7=20\). 3. Splitting \(20\) equally gives \(20 \div 2=10\), so the smaller box has \(10\) markers and the larger has \(10+7=17\). 4. Check: after the changes the boxes would have \(11\) and \(16\) markers, a difference of \(5\).

The boxes contain \(10\) markers and \(17\) markers.

- What does “four times as much” mean? - In part b), distinguish between “more than” and “exactly.” - For part c), test possible amounts used and compare each amount with what remains.

1. Subtract to find the remaining flour: \(20 - 4 = 16\,\text{kg}\). 2. Four times the amount used is \(4 \times 4 = 16\,\text{kg}\). The remaining amount is exactly four times, not more than four times, the amount used. 3. For part c), the used and remaining amounts total \(20\,\text{kg}\), and the remaining amount must include an extra \(10\,\text{kg}\). Remove that difference: \(20\,\text{kg} - 10\,\text{kg} = 10\,\text{kg}\). Split the \(10\,\text{kg}\) equally between the used amount and the matching part of the remaining amount: \(10\,\text{kg} \div 2 = 5\,\text{kg}\). Then \(15\,\text{kg} - 5\,\text{kg} = 10\,\text{kg}\).

a) \(16\,\text{kg}\) b) No. The remaining amount is exactly four times the amount used because \(4 \times 4 = 16\). c) The baker would need to use \(5\,\text{kg}\) altogether.

- The transfer does not change the total number of apples in the two baskets. - First reason about the two amounts after the transfer, when both their total and their difference are known. - Reverse the transfer to recover the starting amounts.

1. Moving apples between the baskets keeps the total at \(60\). 2. After the move, the amounts differ by \(10\), so they are \(25\) and \(35\). 3. Reversing the move gives \(25+5=30\) apples in the first basket and \(35-5=30\) in the second. 4. The baskets began with equal amounts. The transfer decreased one basket by \(5\) and increased the other by \(5\), creating the final difference of \(10\).

a) Each basket began with \(30\) apples. b) The baskets began with equal numbers of apples.

- The total amount stays the same when money moves between jars. - First determine the two amounts after the move, when their difference is \(\$4\). - Then reverse the transfer of \(\$8\).

1. Moving money between the jars does not change the total of \(\$60\). 2. After the move, subtract the \(\$4\) difference from the total: \(\$60 - \$4 = \$56\). 3. Split \(\$56\) equally: \(\$56 \div 2 = \$28\). After the move, the first jar has \(\$28\), and the second has \(\$28 + \$4 = \$32\). 4. Reverse the move. The first jar originally had \(\$28 + \$8 = \$36\), and the second originally had \(\$32 - \$8 = \$24\).

Originally, the first jar contained \(\$36\), and the second jar contained \(\$24\).

- Use the equal amounts after Mia gives away stickers to work backward to the starting amounts. - For part c), a direct transfer changes both people’s amounts at the same time. - Check the proposed transfer by calculating both final sticker counts.

1. After Mia gives away \(12\) stickers, the total is \(100-12=88\). 2. Equal shares of \(88\) are \(88 \div 2=44\), so Ben began with \(44\) stickers. 3. Mia began with \(44+12=56\) stickers. 4. Their original difference was \(56-44=12\). A direct transfer of \(6\) stickers gives each person \(50\) stickers.

a) Mia had \(56\) stickers. b) Ben had \(44\) stickers. c) Mia would need to give Ben \(6\) stickers.

- Compare the two combinations and notice what stays the same and what changes. - The difference between the two total prices can isolate the value of the items that changed. - Once one item price is known, use either combination to determine the other item price.

1. Combination A has two more juice boxes than Combination B while the pretzel count is unchanged. 2. The \(\$8-\$6=\$2\) price difference is therefore the cost of two juice boxes, so one juice box costs \(\$2 \div 2=\$1\). 3. In Combination B, two juice boxes cost \(2 \times \$1=\$2\), leaving \(\$6-\$2=\$4\) for two pretzels. 4. One pretzel costs \(\$4 \div 2=\$2\).

One juice box costs \(\$1\), and one pretzel costs \(\$2\).

- Represent the larger price as the smaller price plus the stated difference. - Use both the total and the difference together rather than treating either fact alone. - Check your two prices against both conditions in the problem.

1. Removing the \(\$80\) difference from the total leaves \(\$110-\$80=\$30\), which represents two equal pencil-case amounts. 2. The pencil case costs \(\$30 \div 2=\$15\). 3. The backpack costs \(\$15+\$80=\$95\). 4. Check: \(\$15+\$95=\$110\).

The pencil case costs \(\$15\), and the backpack costs \(\$95\).

- Use the combined mass and the stated difference together to determine the original package masses. - Check that your original masses add to \(750\,\text{g}\) and differ by \(150\,\text{g}\). - For part b), track only the material moved into Package B.

1. Removing the \(150\,\text{g}\) difference from the combined mass leaves \(750\,\text{g}-150\,\text{g}=600\,\text{g}\), representing two equal lighter-package amounts. 2. Package B has mass \(600\,\text{g} \div 2=300\,\text{g}\), and Package A has mass \(300\,\text{g}+150\,\text{g}=450\,\text{g}\). 3. After the transfer, Package B has mass \(300\,\text{g}+50\,\text{g}=350\,\text{g}\).

a) Package A has mass \(450\,\text{g}\), and Package B has mass \(300\,\text{g}\). b) Package B would have mass \(350\,\text{g}\).

- Use both the fixed total and the stated difference to determine the two current amounts. - For part b), think about what must happen to the smaller share if the same total is split less evenly. - You may test one larger difference as a check after reasoning generally.

1. Mia’s amount is Jordan’s amount plus \(40\). Removing that extra \(40\) from the total leaves \(240-40=200\), representing two equal Jordan amounts. 2. Jordan collected \(200 \div 2=100\) acorns, and Mia collected \(100+40=140\). 3. If the total stays fixed while the difference grows, the larger amount must increase and the smaller amount must decrease. Jordan would have fewer than \(100\) acorns.

a) Jordan collected \(100\) acorns, and Mia collected \(140\). b) Jordan would have fewer acorns.

- The combined savings and the difference between the two savings amounts must both be satisfied. - Solve for the two individual amounts before checking the trampoline shortfall. - Verify that the individual savings add to \(\$370\) and differ by \(\$50\).

1. Removing the \(\$50\) difference from the combined savings leaves \(\$370-\$50=\$320\), representing two equal Lucas amounts. 2. Lucas saved \(\$320 \div 2=\$160\), and Sophie saved \(\$160+\$50=\$210\). 3. They still need \(\$450-\$370=\$80\).

Lucas saved \(\$160\), Sophie saved \(\$210\), and together they still need \(\$80\).

- The total amount does not change when water is moved between barrels. - First find the amounts after the transfer using the \(20\,\text{L}\) difference. - Then reverse the \(30\,\text{L}\) transfer.

1. After the transfer, the total is still \(240\,\text{L}\), and the second barrel has \(20\,\text{L}\) more than the first. 2. Remove the extra \(20\,\text{L}\): \(240\,\text{L} - 20\,\text{L} = 220\,\text{L}\). 3. Split the remaining amount equally: \(220\,\text{L} \div 2 = 110\,\text{L}\). After the transfer, the first barrel has \(110\,\text{L}\), and the second has \(130\,\text{L}\). 4. Reverse the transfer: the first barrel originally had \(110\,\text{L} + 30\,\text{L} = 140\,\text{L}\), and the second originally had \(130\,\text{L} - 30\,\text{L} = 100\,\text{L}\).

The first barrel originally contained \(140\,\text{L}\), and the second barrel originally contained \(100\,\text{L}\).

- Compare the total weight with the combined weight of the dog and cat. - Once one animal's weight is known, use another pair total. - Check that all three weights add to \(45\) pounds.

1. Subtract the combined weight of the dog and cat from the total to find the rabbit's weight: \(45 - 38 = 7\) pounds. 2. Subtract the rabbit's weight from the combined weight of the cat and rabbit: \(15 - 7 = 8\) pounds for the cat. 3. Subtract the cat's weight from the combined weight of the dog and cat: \(38 - 8 = 30\) pounds for the dog.

The dog weighs \(30\) pounds, the cat weighs \(8\) pounds, and the rabbit weighs \(7\) pounds.

- Identify which fruit is missing from each two-fruit total. - Subtract the apples-and-pears weight from the total to find the plums. - Once two individual weights are known, use one pair total to find the third. - A diagram or table may help organize the information.

1. The plums are the part not included in the apples-and-pears total: \(950\,\text{g} - 700\,\text{g} = 250\,\text{g}\). 2. The apples are the part not included in the pears-and-plums total: \(950\,\text{g} - 550\,\text{g} = 400\,\text{g}\). 3. Find the pears: \(700\,\text{g} - 400\,\text{g} = 300\,\text{g}\). 4. Check: \(400 + 300 + 250 = 950\).

Apples: \(400\,\text{g}\) Pears: \(300\,\text{g}\) Plums: \(250\,\text{g}\)

- Convert \(1\,\text{kg}\) to grams. - Subtract each pair total from the overall total to find the missing bag. - Use the three individual amounts to check both pair totals.

1. Convert the total mass to grams: \(1\,\text{kg} = 1000\,\text{g}\). 2. Find the amount in the third bag: \(1000 - 650 = 350\,\text{g}\). 3. Find the amount in the first bag: \(1000 - 750 = 250\,\text{g}\). 4. Find the amount in the second bag: \(1000 - 350 - 250 = 400\,\text{g}\).

The first bag contains \(250\,\text{g}\), the second contains \(400\,\text{g}\), and the third contains \(350\,\text{g}\).

- Subtract the first-and-second total from the full length to find the third section. - Subtract the first-and-third total from the full length to find the second section. - Find the remaining section, then compare all three lengths.

1. The third section is the part not included in the first-and-second total: \(600\,\text{m} - 380\,\text{m} = 220\,\text{m}\). 2. The second section is the part not included in the first-and-third total: \(600\,\text{m} - 420\,\text{m} = 180\,\text{m}\). 3. Find the first section: \(600 - 220 - 180 = 200\), so it is \(200\,\text{m}\). 4. Compare: \(220\,\text{m} > 200\,\text{m} > 180\,\text{m}\).

The third section is longest at \(220\,\text{m}\).

- First determine how many crates were sold altogether from the revenue information. - The red and green counts then have a known total and a known difference. - Check your two counts against both the revenue total and the \(12\)-crate difference.

1. The orchard sells \(\$336\div\$8=42\) crates altogether. 2. Remove the \(12\)-crate difference: \(42-12=30\). 3. Splitting the remaining \(30\) crates equally gives \(30\div2=15\) green-apple crates. 4. The red-apple count is \(15+12=27\).

The orchard sells \(15\) crates of green apples and \(27\) crates of red apples.

- Represent the full route as two parts whose sum is known. - One of those two parts is \(36\,\text{miles}\) longer than the other. - Look for a way to compare or equalize the two parts before finding either distance.

1. The Pine Ridge distance is the longer part. Adding the \(36\)-mile difference to the total gives \(782+36=818\). 2. Splitting \(818\) equally gives \(818\div2=409\,\text{miles}\) from Pine Ridge. 3. The distance to Cedar Bay is \(409-36=373\,\text{miles}\), and \(409+373=782\,\text{miles}\), so the distances check.

Lakeview is \(409\,\text{miles}\) from Pine Ridge.

- First use the total and the \(\$150\) difference to determine the two class amounts. - Think about how the two class amounts could be made equal temporarily. - After the class amounts are known, interpret each amount as groups of \(\$50\).

1. Set aside the \(\$150\) difference: \(\$1350-\$150=\$1200\). 2. Split the remaining amount equally: \(\$1200\div2=\$600\). Class 4B earns \(\$600\). 3. Class 4A earns \(\$600+\$150=\$750\). 4. Since \(12 \times \$50=\$600\), Class 4B fills \(12\) envelopes. 5. Since \(15 \times \$50=\$750\), Class 4A fills \(15\) envelopes.

Class 4A fills \(15\) envelopes, and Class 4B fills \(12\) envelopes.

- Two crates are equal, while the third differs from that shared amount by a known quantity. - Think about how you could represent the total as three equal base amounts plus one extra part. - Check any proposed three masses against both the total and the \(15\,\text{kg}\) difference.

1. If the extra \(15\,\text{kg}\) is set aside, the remaining mass is \(75-15=60\,\text{kg}\). 2. The three equal base amounts are \(60\div3=20\,\text{kg}\) each. 3. The first and second crates each have \(20\,\text{kg}\), and the third has \(20+15=35\,\text{kg}\). 4. Check: \(20+20+35=75\,\text{kg}\).

The first crate has \(20\,\text{kg}\), the second has \(20\,\text{kg}\), and the third has \(35\,\text{kg}\).

- Subtract the combined load of two trucks from the total to find the third truck's load. - Use each given pair to find another truck. - Compare the three loads, then subtract the least from the greatest.

1. Start with the total of \(15{,}000\,\text{kg}\) and subtract the first-and-second-truck total of \(10{,}400\,\text{kg}\). The third truck carries \(4600\,\text{kg}\). 2. Start with the total of \(15{,}000\,\text{kg}\) and subtract the second-and-third-truck total of \(9100\,\text{kg}\). The first truck carries \(5900\,\text{kg}\). 3. Start with the first-and-second-truck total of \(10{,}400\,\text{kg}\) and subtract the first truck's \(5900\,\text{kg}\). The second truck carries \(4500\,\text{kg}\). 4. The greatest load is \(5900\,\text{kg}\), and the least load is \(4500\,\text{kg}\). 5. Find the difference: \(5900\,\text{kg} - 4500\,\text{kg} = 1400\,\text{kg}\).

a) The first truck carries \(5900\,\text{kg}\), the second carries \(4500\,\text{kg}\), and the third carries \(4600\,\text{kg}\). b) The difference is \(1400\,\text{kg}\).

- The two unknown amounts have both a known total and a known difference. - Think about how the larger amount can be viewed as an equal base amount plus one extra part. - Check any proposed pair against both conditions.

1. Set aside the \(100\)-notebook difference: \(700-100=600\). 2. The remaining \(600\) notebooks form two equal base amounts, so \(600\div2=300\). 3. The smaller room has \(300\) notebooks and the larger room has \(300+100=400\). 4. Check: \(300+400=700\) and \(400-300=100\).

The rooms hold \(300\) notebooks and \(400\) notebooks.

- Two ribbons are equal, and the third differs from that shared length by a known amount. - Think about a representation with equal base lengths and one missing piece. - Check that the three lengths have both the required total and the required difference.

1. If Ribbon B were \(30\,\text{cm}\) longer, all three ribbons would be equal and the adjusted total would be \(150+30=180\,\text{cm}\). 2. Each equal length would be \(180\div3=60\,\text{cm}\). 3. Ribbons A and C are each \(60\,\text{cm}\), while Ribbon B is \(60-30=30\,\text{cm}\).

Ribbon A: \(60\,\text{cm}\) Ribbon B: \(30\,\text{cm}\) Ribbon C: \(60\,\text{cm}\)

- The four bags can be described using equal base amounts plus extra pieces on two bags. - Think about how the total would change if those two extra pieces were temporarily removed. - Check all four final weights against both equality conditions and the total.

1. Bags 3 and 4 together contain \(20+20=40\,\text{lb}\) of extra weight beyond the common base amount. 2. Removing that extra part gives \(220-40=180\,\text{lb}\) for four equal base amounts. 3. Each base amount is \(180\div4=45\,\text{lb}\). 4. Bags 1 and 2 each weigh \(45\,\text{lb}\), while Bags 3 and 4 each weigh \(45+20=65\,\text{lb}\). 5. Check: \(45+45+65+65=220\).

Bag 1: \(45\,\text{lb}\) Bag 2: \(45\,\text{lb}\) Bag 3: \(65\,\text{lb}\) Bag 4: \(65\,\text{lb}\)

- The two center amounts have both a known total and a known difference. - Look for a way to combine those two pieces of information so that one unknown amount appears twice. - Verify both the total and the difference when you have two candidate amounts.

1. Adding the total and the difference gives \(752{,}300+48{,}100=800{,}400\), which represents two copies of the larger amount. 2. The larger amount is \(800{,}400\div2=400{,}200\,\text{lb}\). 3. The smaller amount is \(400{,}200-48{,}100=352{,}100\,\text{lb}\). 4. Check: \(400{,}200+352{,}100=752{,}300\) and \(400{,}200-352{,}100=48{,}100\).

The centers process \(400{,}200\,\text{lb}\) and \(352{,}100\,\text{lb}\).

- First determine the combined distance for Schools B and C. - Their two distances must add to that amount and differ by \(300\,\text{miles}\). - After finding the two distances, compare all three schools with the stated goal.

1. Find the distance covered by Schools B and C together: \(2450\,\text{miles}-750\,\text{miles}=1700\,\text{miles}\). 2. Remove the \(300\)-mile difference: \(1700\,\text{miles}-300\,\text{miles}=1400\,\text{miles}\). 3. Divide equally to find School C's distance: \(1400\,\text{miles}\div2=700\,\text{miles}\). 4. Find School B's distance: \(700\,\text{miles}+300\,\text{miles}=1000\,\text{miles}\). 5. Compare the three distances with \(900\,\text{miles}\). Only School B reaches the goal.

a) School B covers \(1000\,\text{miles}\), and School C covers \(700\,\text{miles}\). b) Only School B reaches the \(900\)-mile goal.

- Start with the child whose row has the most information. - Test number pairs that satisfy all conditions, then use Anna's blue count to begin Bo's row.

1. Anna's \(10\) stamps split into \(6\) red and \(4\) blue because the red count is \(2\) greater. 2. Bo therefore has \(4\) red stamps. 3. Bo has \(9 - 4 = 5\) blue stamps.

Child	Red stamps	Blue stamps
Anna	\(6\)	\(4\)
Bo	\(4\)	\(5\)

- Start with the relationship between Room 12 and the combined amount for the other two rooms. - After finding how many seedlings Rooms 14 and 16 plant together, use their multiplicative relationship to separate that combined amount. - Check that all three class amounts add to \(120\) and satisfy both comparison statements.

1. Let the combined number planted by Rooms 14 and 16 be one amount. Room 12 plants that amount plus \(40\). 2. Remove the extra \(40\) from the total: \(120-40=80\). The remaining \(80\) represents two equal combined amounts, so Rooms 14 and 16 together plant \(80 \div 2=40\) seedlings. 3. Room 12 plants \(40+40=80\) seedlings. 4. Room 14 plants three times as many as Room 16. Their combined \(40\) seedlings therefore consist of four equal shares, so one share is \(40 \div 4=10\). 5. Room 16 plants \(10\) seedlings, and Room 14 plants \(3 \times 10=30\) seedlings.

Room 12 plants \(80\) seedlings, Room 14 plants \(30\), and Room 16 plants \(10\).