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Equivalent ratios in tables

Learning goals

  • Interprets a ratio as a multiplicative comparison of two quantities.
  • Writes equivalent ratios in appropriate notation.
  • States which quantity comes first and second in a ratio.
  • Represents ratios with tables, diagrams, or equations when useful.

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Equivalent ratios in tables 1
A smoothie recipe keeps the same ratio of fruit to yogurt. <table> <tr><th>Fruit (cups)</th><th>Yogurt (cups)</th></tr> <tr><td>2</td><td>3</td></tr> <tr><td>4</td><td>?</td></tr> </table> What number belongs in the missing table entry?

Hints

- Compare the two fruit entries first. - Equivalent-ratio rows scale both quantities by the same factor. - Apply that same change to the yogurt entry.

Solution

1. The fruit amount changes from \(2\) cups to \(4\) cups, which is a factor of \(2\). 2. Multiply the yogurt amount by the same factor: \(3 \times 2 = 6\).

Answer

\(6\) cups
Equivalent ratios in tables 2
Each lunch bag holds \(2\) sandwiches. Fill in the missing sandwich count. <table><tr><th>Lunch bags</th><td>\(1\)</td><td>\(3\)</td><td>\(7\)</td></tr><tr><th>Sandwiches</th><td>\(2\)</td><td>\(6\)</td><td>\(\square\)</td></tr></table>

Hints

- What stays the same when the number of lunch bags changes? - How can a complete table column help you find a missing one?

Solution

1. Every lunch bag has two sandwiches, so each row uses the same number of sandwiches per bag. 2. Seven bags have \(7\times2=14\) sandwiches.

Answer

\(14\) sandwiches
Equivalent ratios in tables 3
Three robots move at constant rates. Complete the table. <table> <tr> <td>Robot</td> <td>Distance in \(1\,\text{min}\)</td> <td>Distance in \(2\,\text{min}\)</td> <td>Distance in \(5\,\text{min}\)</td> </tr> <tr> <td>Racer</td> <td>\(50\,\text{ft}\)</td> <td></td> <td></td> </tr> <tr> <td>Walker</td> <td>\(10\,\text{ft}\)</td> <td></td> <td></td> </tr> <tr> <td>Crawler</td> <td>\(1\,\text{ft}\)</td> <td></td> <td></td> </tr> </table>

Hints

- Work across one row at a time. - Double the one-minute distance to find the two-minute distance. - Multiply the one-minute distance by \(5\) to find the five-minute distance.

Solution

1. Double each one-minute distance to find the two-minute distances. 2. Multiply each one-minute distance by \(5\) to find the five-minute distances. 3. The completed table is: <table> <tr><td>Robot</td><td>Distance in \(1\,\text{min}\)</td><td>Distance in \(2\,\text{min}\)</td><td>Distance in \(5\,\text{min}\)</td></tr> <tr><td>Racer</td><td>\(50\,\text{ft}\)</td><td>\(100\,\text{ft}\)</td><td>\(250\,\text{ft}\)</td></tr> <tr><td>Walker</td><td>\(10\,\text{ft}\)</td><td>\(20\,\text{ft}\)</td><td>\(50\,\text{ft}\)</td></tr> <tr><td>Crawler</td><td>\(1\,\text{ft}\)</td><td>\(2\,\text{ft}\)</td><td>\(5\,\text{ft}\)</td></tr> </table>

Answer

<table> <tr><td>Robot</td><td>Distance in \(1\,\text{min}\)</td><td>Distance in \(2\,\text{min}\)</td><td>Distance in \(5\,\text{min}\)</td></tr> <tr><td>Racer</td><td>\(50\,\text{ft}\)</td><td>\(100\,\text{ft}\)</td><td>\(250\,\text{ft}\)</td></tr> <tr><td>Walker</td><td>\(10\,\text{ft}\)</td><td>\(20\,\text{ft}\)</td><td>\(50\,\text{ft}\)</td></tr> <tr><td>Crawler</td><td>\(1\,\text{ft}\)</td><td>\(2\,\text{ft}\)</td><td>\(5\,\text{ft}\)</td></tr> </table>
Equivalent ratios in tables 4
Complete the table for movie tickets. <table> <tr> <td>\(1\) movie ticket</td> <td>\(\$9.00\)</td> </tr> <tr> <td>\(3\) movie tickets</td> <td>?</td> </tr> <tr> <td>\(7\) movie tickets</td> <td>?</td> </tr> </table>

Hints

- What is the price of one ticket? - Multiply the one-ticket price by the number of tickets. - Which operation represents equal prices repeated several times?

Solution

1. Find the cost of \(3\) tickets: \(3 \times \$9.00 = \$27.00\). 2. Find the cost of \(7\) tickets: \(7 \times \$9.00 = \$63.00\). 3. Complete the table. <table> <tr><td>\(1\) movie ticket</td><td>\(\$9.00\)</td></tr> <tr><td>\(3\) movie tickets</td><td>\(\$27.00\)</td></tr> <tr><td>\(7\) movie tickets</td><td>\(\$63.00\)</td></tr> </table>

Answer

<table> <tr><td>\(1\) movie ticket</td><td>\(\$9.00\)</td></tr> <tr><td>\(3\) movie tickets</td><td>\(\$27.00\)</td></tr> <tr><td>\(7\) movie tickets</td><td>\(\$63.00\)</td></tr> </table>
Equivalent ratios in tables 5
At a garden store, \(5\) seed packets cost \(\$4\). Find the cost of \(10\), \(15\), and \(25\) seed packets. Show the equivalent ratios in a table.

Hints

- Determine how many groups of \(5\) are in each requested quantity. - Multiply the cost of one group by the same factor. - Keep the packet and cost values aligned in the table.

Solution

1. Ten packets are \(2\) groups of \(5\), so they cost \(2 \times \$4 = \$8\). 2. Fifteen packets are \(3\) groups of \(5\), so they cost \(3 \times \$4 = \$12\). 3. Twenty-five packets are \(5\) groups of \(5\), so they cost \(5 \times \$4 = \$20\). <table> <tr><th>Seed packets</th><th>Cost</th></tr> <tr><td>\(5\)</td><td>\(\$4\)</td></tr> <tr><td>\(10\)</td><td>\(\$8\)</td></tr> <tr><td>\(15\)</td><td>\(\$12\)</td></tr> <tr><td>\(25\)</td><td>\(\$20\)</td></tr> </table>

Answer

<table> <tr><th>Seed packets</th><th>Cost</th></tr> <tr><td>\(5\)</td><td>\(\$4\)</td></tr> <tr><td>\(10\)</td><td>\(\$8\)</td></tr> <tr><td>\(15\)</td><td>\(\$12\)</td></tr> <tr><td>\(25\)</td><td>\(\$20\)</td></tr> </table>
Equivalent ratios in tables 6
Copy and complete the ratio table for purchasing notebooks for a school. <table> <tr><th>Number of notebooks</th><th>Price</th></tr> <tr><td>\(8\)</td><td>\(\$11.20\)</td></tr> <tr><td>\(1\)</td><td></td></tr> <tr><td>\(15\)</td><td></td></tr> </table>

Hints

- Start with the cost of \(8\) notebooks. - Divide to find the cost of one notebook. - Multiply the unit price by \(15\).

Solution

1. Find the price of one notebook: \(\$11.20 \div 8 = \$1.40\). 2. Find the price of \(15\) notebooks: \(15 \times \$1.40 = \$21.00\). 3. Complete the table. <table><tr><th>Number of notebooks</th><th>Price</th></tr><tr><td>\(8\)</td><td>\(\$11.20\)</td></tr><tr><td>\(1\)</td><td>\(\$1.40\)</td></tr><tr><td>\(15\)</td><td>\(\$21.00\)</td></tr></table>

Answer

<table><tr><th>Number of notebooks</th><th>Price</th></tr><tr><td>\(8\)</td><td>\(\$11.20\)</td></tr><tr><td>\(1\)</td><td>\(\$1.40\)</td></tr><tr><td>\(15\)</td><td>\(\$21.00\)</td></tr></table>
Equivalent ratios in tables 7
A snack mix uses \(3\) cups of cereal for every \(2\) cups of pretzels. One row in the table is not equivalent to this ratio. <table> <tr><th>Cereal (cups)</th><th>Pretzels (cups)</th></tr> <tr><td>\(3\)</td><td>\(2\)</td></tr> <tr><td>\(6\)</td><td>\(4\)</td></tr> <tr><td>\(9\)</td><td>\(5\)</td></tr> <tr><td>\(12\)</td><td>\(8\)</td></tr> </table> Identify the incorrect row and correct it.

Hints

- Compare each row with the original \(3:2\) relationship. - In an equivalent ratio, both quantities change by the same scale factor. - Test every row by checking whether both quantities use the same scale factor.

Solution

1. Equivalent rows must multiply both parts of \(3:2\) by the same factor. 2. The row \(6:4\) uses a factor of \(2\), and \(12:8\) uses a factor of \(4\). 3. For \(9\) cups of cereal, the factor is \(3\), so the pretzel amount must be \(2 \times 3 = 6\) cups. 4. The row \(9:5\) is incorrect; it should be \(9:6\).

Answer

The row \(9:5\) is incorrect. The corrected row is \(9:6\).
Equivalent ratios in tables 8
A banner design uses 3 gold stars for every 4 blue stars. Construct an equivalent-ratio table with four rows for \(3\), \(6\), \(9\), and \(12\) gold stars. Include the matching number of blue stars in each row.

Hints

- Use the given “3 for every 4” relationship as the first row. - Each new row must multiply both entries by the same scale factor. - Check that every row describes the same gold-to-blue comparison.

Solution

1. Start with the given ratio \(3:4\), so the first row is \((3,4)\). 2. Scale both quantities by \(2\) to get \((6,8)\). 3. Scale both quantities by \(3\) to get \((9,12)\). 4. Scale both quantities by \(4\) to get \((12,16)\).

Answer

<table> <tr><th>Gold stars</th><th>Blue stars</th></tr> <tr><td>3</td><td>4</td></tr> <tr><td>6</td><td>8</td></tr> <tr><td>9</td><td>12</td></tr> <tr><td>12</td><td>16</td></tr> </table>
Equivalent ratios in tables 9
A rice recipe uses \(2\) cups of rice for every \(3\) cups of water. Mira and Noah made tables for larger batches. Do both tables keep this ratio? Explain. Mira's table: <table><tr><th>Rice (cups)</th><th>Water (cups)</th></tr><tr><td>\(2\)</td><td>\(3\)</td></tr><tr><td>\(6\)</td><td>\(9\)</td></tr></table> Noah's table: <table><tr><th>Water (cups)</th><th>Rice (cups)</th></tr><tr><td>\(6\)</td><td>\(4\)</td></tr><tr><td>\(12\)</td><td>\(8\)</td></tr></table>

Hints

- Check the two headings in each table before reading any pair. - Can you express a row from each table as rice to water? - Would both amounts in a larger batch be the same multiple of the recipe amounts?

Solution

1. Mira's table places rice first. Her pairs \((2,3)\) and \((6,9)\) keep the rice-to-water ratio \(2:3\). 2. Noah's table places water first. His pairs \((6,4)\) and \((12,8)\) have rice-to-water ratios \(4:6=2:3\) and \(8:12=2:3\). 3. Both tables show the same recipe, despite the different column order.

Answer

Yes. Both tables keep rice to water at \(2:3\), even though Noah lists water first.
Equivalent ratios in tables 10
A class buys pencils and erasers in the ratio \(3:2\). Which row cannot belong in its equivalent-ratio table? Explain how you checked the row. <table><tr><th>Pencils</th><th>Erasers</th></tr><tr><td>\(3\)</td><td>\(2\)</td></tr><tr><td>\(6\)</td><td>\(4\)</td></tr><tr><td>\(12\)</td><td>\(6\)</td></tr><tr><td>\(15\)</td><td>\(10\)</td></tr><tr><td>\(21\)</td><td>\(14\)</td></tr></table>

Hints

- Compare what the pencil and eraser counts mean in each row. - How could you test one row against the original \(3:2\) comparison?

Solution

1. Each correct row multiplies both parts of \(3:2\) by the same number. For example, \((15,10)=(5\times3,5\times2)\). 2. In \((12,6)\), the pencils are \(4\times3\), but the erasers are \(3\times2\). 3. The two factors do not match, so that row is incorrect.

Answer

The row with \(12\) pencils and \(6\) erasers; the scale factors differ.
Equivalent ratios in tables 11
A paint mix has \(4\) parts blue to \(3\) parts yellow. Complete the table for the same shade. <table><tr><th>Blue (cups)</th><th>Yellow (cups)</th></tr><tr><td>\(4\)</td><td>\(3\)</td></tr><tr><td>\(12\)</td><td>\(\square\)</td></tr><tr><td>\(\square\)</td><td>\(15\)</td></tr></table>

Hints

- Treat each row as a pair of amounts in the same mixture. - What factor changes the known blue amount in the second row? - In the third row, start from the yellow amount instead.

Solution

1. Blue increases from \(4\) to \(12\) by a factor of \(3\), so yellow increases from \(3\) to \(9\). 2. Yellow increases from \(3\) to \(15\) by a factor of \(5\), so blue increases from \(4\) to \(20\).

Answer

Missing yellow: \(9\) cups; missing blue: \(20\) cups
Equivalent ratios in tables 12
Each plotted point pairs the number of gift bags with the number of candies packed. Complete a table for \(1\), \(2\), \(3\), and \(4\) bags using the same rule. How many candies are in \(3\) bags?
Figure for problem 559293

Hints

- Read the axes before reading the plotted pairs. - Compare how the candy count changes with the bag count. - Use the shared rate to fill the missing table entry.

Solution

1. The graph has points at \((1,3)\), \((2,6)\), and \((4,12)\). 2. Each bag adds \(3\) candies, so the table pairs are \((1,3)\), \((2,6)\), \((3,9)\), and \((4,12)\). 3. Three bags contain \(9\) candies.

Answer

<table><tr><th>Bags</th><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td></tr><tr><th>Candies</th><td>\(3\)</td><td>\(6\)</td><td>\(9\)</td><td>\(12\)</td></tr></table> \(9\) candies in three bags
Equivalent ratios in tables 13
A painter mixes blue and yellow paint to make one shade of green. The table shows mixtures that should all have the same color. <table> <tr><td>Blue paint, in fluid ounces</td><td>\(6\)</td><td>\(9\)</td><td>\(18\)</td></tr> <tr><td>Yellow paint, in fluid ounces</td><td>\(10\)</td><td>\(15\)</td><td>\(?\)</td></tr> </table> a) Show with calculations that blue paint \(\to\) yellow paint is a proportional relationship. b) Find the amount of yellow paint needed for \(18\) fluid ounces of blue paint. c) The painter wants exactly \(40\) fluid ounces of the green mixture. How many fluid ounces of blue paint should be used?

Hints

- Compare the ratio of yellow paint to blue paint in the complete columns. - Use the constant ratio for the missing table value. - Add the parts in the blue-to-yellow mixing ratio.

Solution

1. The ratios are \(10\div 6=\frac{5}{3}\) and \(15\div 9=\frac{5}{3}\), so the relationship is proportional. 2. For \(18\) fluid ounces of blue paint, the yellow amount is \(18\times \frac{5}{3}=30\) fluid ounces. 3. The blue-to-yellow ratio is \(3\) to \(5\), for \(8\) total parts. 4. Each part of a \(40\)-fluid-ounce mixture is \(40\div 8=5\) fluid ounces, so the blue amount is \(3\times 5=15\) fluid ounces.

Answer

a) The ratio of yellow to blue is constantly \(\frac{5}{3}\), so the relationship is proportional. b) \(30\) fluid ounces. c) \(15\) fluid ounces of blue paint.
Equivalent ratios in tables 14
Admission to a climbing park costs \(\$18\) per person. a) Make a price table for groups of \(1, 2, 3, 4, 5, 6, 7, 8, 9,\) and \(10\) people. b) A group pays \(\$126\). How many people are in the group?

Hints

- How does the price change when one more person is added? - Use your table to locate \(\$126\). - Finding the cost for \(5\) people first may help you continue the pattern.

Solution

1. Multiply each group size by \(\$18\) to complete the table. <table> <tr><td>People</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td><td>\(6\)</td><td>\(7\)</td><td>\(8\)</td><td>\(9\)</td><td>\(10\)</td></tr> <tr><td>Price</td><td>\(\$18\)</td><td>\(\$36\)</td><td>\(\$54\)</td><td>\(\$72\)</td><td>\(\$90\)</td><td>\(\$108\)</td><td>\(\$126\)</td><td>\(\$144\)</td><td>\(\$162\)</td><td>\(\$180\)</td></tr> </table> 2. Locate \(\$126\) in the table, or calculate \(\$126 \div \$18 = 7\).

Answer

a) <table> <tr><td>People</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td><td>\(6\)</td><td>\(7\)</td><td>\(8\)</td><td>\(9\)</td><td>\(10\)</td></tr> <tr><td>Price</td><td>\(\$18\)</td><td>\(\$36\)</td><td>\(\$54\)</td><td>\(\$72\)</td><td>\(\$90\)</td><td>\(\$108\)</td><td>\(\$126\)</td><td>\(\$144\)</td><td>\(\$162\)</td><td>\(\$180\)</td></tr> </table> b) The group has \(7\) people.
Equivalent ratios in tables 15
At the Mountain View Inn, breakfast costs \(\$14\) for an adult and \(\$9\) for a child. a) What is the total breakfast cost for a family with \(2\) adults and \(4\) children? b) Make a price table for \(1, 2, 3, 4,\) and \(5\) child breakfasts.

Hints

- Find the total cost for the adults. - Find the total cost for the children. - Add the two amounts. - Use multiples of \(9\) for the table.

Solution

1. Find the cost for the adults: \(2 \times \$14 = \$28\). 2. Find the cost for the children: \(4 \times \$9 = \$36\). 3. Add the costs: \(\$28 + \$36 = \$64\). 4. Complete the child-breakfast table using multiples of \(\$9\). <table> <tr><td>Child breakfasts</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td></tr> <tr><td>Price</td><td>\(\$9\)</td><td>\(\$18\)</td><td>\(\$27\)</td><td>\(\$36\)</td><td>\(\$45\)</td></tr> </table>

Answer

a) The family’s breakfast costs \(\$64\). b) <table> <tr><td>Child breakfasts</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td></tr> <tr><td>Price</td><td>\(\$9\)</td><td>\(\$18\)</td><td>\(\$27\)</td><td>\(\$36\)</td><td>\(\$45\)</td></tr> </table>
Equivalent ratios in tables 16
A hardware store sells rope for \(\$2.40\) per foot. Make a table showing the cost of \(5\,\text{ft}\), \(10\,\text{ft}\), and \(20\,\text{ft}\) of rope. A customer wants to buy \(25\,\text{ft}\) of rope and has \(\$50\). Is that enough? Justify your answer with a calculation.

Hints

- Use the price per foot to find each value in the table. - How can the cost for \(10\,\text{ft}\) help you find the cost for \(20\,\text{ft}\)? - Find the total cost of \(25\,\text{ft}\), and compare it with \(\$50\).

Solution

1. Multiply each length by \(\$2.40\) per foot to complete the table. <table> <tr><td>Rope length</td><td>\(5\,\text{ft}\)</td><td>\(10\,\text{ft}\)</td><td>\(20\,\text{ft}\)</td></tr> <tr><td>Cost</td><td>\(\$12.00\)</td><td>\(\$24.00\)</td><td>\(\$48.00\)</td></tr> </table> 2. Find the cost of \(25\,\text{ft}\): \(25 \times \$2.40 = \$60.00\). 3. Since \(\$60.00 > \$50.00\), the customer does not have enough money.

Answer

<table> <tr><td>Rope length</td><td>\(5\,\text{ft}\)</td><td>\(10\,\text{ft}\)</td><td>\(20\,\text{ft}\)</td></tr> <tr><td>Cost</td><td>\(\$12.00\)</td><td>\(\$24.00\)</td><td>\(\$48.00\)</td></tr> </table> No. The \(25\,\text{ft}\) of rope costs \(\$60.00\), which is more than \(\$50.00\).
Equivalent ratios in tables 17
Two tables describe batches of the same kind of drink mix, but the column order is different. <table> <tr><th colspan="2">Table A</th></tr> <tr><th>Concentrate (cups)</th><th>Water (cups)</th></tr> <tr><td>\(2\)</td><td>\(6\)</td></tr> <tr><td>\(5\)</td><td>\(15\)</td></tr> <tr><td>\(7\)</td><td>\(21\)</td></tr> </table> <table> <tr><th colspan="2">Table B</th></tr> <tr><th>Water (cups)</th><th>Concentrate (cups)</th></tr> <tr><td>\(9\)</td><td>\(3\)</td></tr> <tr><td>\(18\)</td><td>\(6\)</td></tr> <tr><td>\(24\)</td><td>\(8\)</td></tr> </table> Do the two tables represent the same concentrate-to-water relationship? Explain.

Hints

- Check the headings before comparing the ordered pairs. - Express both tables in the same concentrate-to-water order. - Simplify at least one row from each table, then check that the pattern continues.

Solution

1. In Table A, concentrate to water is \(2:6 = 1:3\). The other rows also simplify to \(1:3\). 2. Table B lists water first. Reorder a row as concentrate to water: \(3:9 = 1:3\). The other rows also give \(6:18 = 1:3\) and \(8:24 = 1:3\). 3. Both tables therefore represent the same concentrate-to-water relationship.

Answer

Yes. After accounting for the reversed column order, every row represents concentrate to water as \(1:3\).
Equivalent ratios in tables 18
The rows in the table must all be equivalent ratios. <table> <tr><th>Tickets</th><th>Cost</th></tr> <tr><td>6</td><td>\(\$15\)</td></tr> <tr><td>14</td><td>?</td></tr> <tr><td>?</td><td>\(\$5\)</td></tr> </table> Find both missing entries using multiplicative scale factors between rows.

Hints

- A scale factor can be greater than \(1\) or less than \(1\). - Compare a known entry with the target entry in the same column. - Whatever multiplicative factor changes one entry in a row must change the other entry by the same factor.

Solution

1. From \(6\) tickets to \(14\) tickets, the scale factor is \(\frac{14}{6}=\frac{7}{3}\). 2. Apply the same factor to the cost: \(\$15 \times \frac{7}{3}=\$35\). 3. From \(\$15\) to \(\$5\), the scale factor is \(\frac{1}{3}\). 4. Apply the same factor to the ticket count: \(6 \times \frac{1}{3}=2\).

Answer

The missing cost is \(\$35\), and the missing ticket count is \(2\).
Equivalent ratios in tables 19
Jordan says every row in this paintbrush price table is equivalent because the number of brushes increases by 4 each time. <table> <tr><th>Paintbrushes</th><th>Cost</th></tr> <tr><td>4</td><td>\(\$14\)</td></tr> <tr><td>8</td><td>\(\$28\)</td></tr> <tr><td>12</td><td>\(\$32\)</td></tr> </table> Is Jordan correct? Identify any incorrect row, correct it, and explain why adding the same amount in one column is not enough to make ratios equivalent.

Hints

- Check each row against the first complete row rather than only looking at differences down one column. - Ask what factor changes the number of brushes from one known row to another. - An equivalent-ratio table preserves multiplication by a common factor in both columns.

Solution

1. The first row scales by a factor of \(2\) to the second row: \(4 \times 2=8\) and \(\$14 \times 2=\$28\). 2. From \(4\) brushes to \(12\) brushes, the scale factor is \(3\). 3. The matching cost must therefore be \(\$14 \times 3=\$42\), not \(\$32\). 4. Equivalent ratios require both quantities to be multiplied by the same factor. A repeated additive change in one column does not by itself preserve the ratio.

Answer

Jordan is not correct. The row \((12,\$32)\) is incorrect and should be \((12,\$42)\). Equivalent-ratio rows preserve a multiplicative relationship between both columns.
Equivalent ratios in tables 20
Two drink mixes use the equivalent-ratio tables below. Mix P: <table> <tr><th>Water (cups)</th><th>Concentrate (cups)</th></tr> <tr><td>6</td><td>2</td></tr> <tr><td>12</td><td>4</td></tr> <tr><td>18</td><td>6</td></tr> </table> Mix Q: <table> <tr><th>Water (cups)</th><th>Concentrate (cups)</th></tr> <tr><td>9</td><td>2</td></tr> <tr><td>18</td><td>4</td></tr> <tr><td>27</td><td>6</td></tr> </table> Which mix has the greater concentrate-to-water ratio? Use the tables to justify your answer.

Hints

- Look for rows in the two tables that have the same concentrate amount. - Compare how much water is paired with that same amount of concentrate. - Keep the requested order as concentrate to water when you write each ratio.

Solution

1. Compare rows with the same amount of concentrate. With \(2\) cups of concentrate, Mix P uses \(6\) cups of water and Mix Q uses \(9\) cups of water. 2. The concentrate-to-water ratios are \(2:6=1:3\) for Mix P and \(2:9\) for Mix Q. 3. For the same amount of concentrate, Mix P has less water, so concentrate makes up a greater share of the mix.

Answer

Mix P. Its concentrate-to-water ratio is \(1:3\), which is greater than Mix Q’s ratio of \(2:9\).
Equivalent ratios in tables 21
A drink recipe keeps the same ratio of syrup to water. The table shows two equivalent mixtures. The graph shows three labeled points that could represent another mixture. <table> <tr><th>Syrup (cups)</th><th>Water (cups)</th></tr> <tr><td>\(1\)</td><td>\(2\)</td></tr> <tr><td>\(3\)</td><td>\(6\)</td></tr> </table> a) What water-to-syrup ratio does the table show? b) Read the coordinates of points A, B, and C from the graph, using syrup as the x-coordinate and water as the y-coordinate. c) Which labeled point can be added as a new row of the table while keeping the same ratio? Explain.
Figure for problem 554197

Hints

- Determine the ordered relationship shown by the two table rows. - Use the graph axes to read each labeled point as \((\text{syrup},\text{water})\). - Test each graph point against the same ordered ratio as the table.

Solution

1. The table shows \(2\) cups of water for each \(1\) cup of syrup, so the water-to-syrup ratio is \(2:1\). 2. From the graph, \(A=(4,8)\), \(B=(4,6)\), and \(C=(5,8)\). 3. Point A has water-to-syrup ratio \(8:4 = 2:1\), which matches the table. 4. Points B and C do not preserve that ratio, so only A can be added as an equivalent row.

Answer

a) \(2:1\) b) \(A=(4,8)\), \(B=(4,6)\), \(C=(5,8)\) c) Point A, because \(8:4 = 2:1\).
Equivalent ratios in tables 22
Every row in this partial table must represent the same red-ribbon-to-gold-ribbon ratio. <table> <tr><th>Red ribbon (ft)</th><th>Gold ribbon (ft)</th></tr> <tr><td>12</td><td>18</td></tr> <tr><td>?</td><td>7.5</td></tr> <tr><td>22</td><td>?</td></tr> </table> Find both missing entries, state the simplest red-to-gold ratio, and justify why all three rows are equivalent.

Hints

- Start with the only row in which both entries are known. - Preserve the same multiplicative relationship even when the scale factor is not a whole number. - Check each completed row against the same simplest ratio.

Solution

1. The complete row gives \(12:18\), which simplifies to \(2:3\). 2. To pair with \(7.5\) feet of gold ribbon, use the red-to-gold relationship \(2:3\): \(7.5 \times \frac{2}{3}=5\) feet of red ribbon. 3. To pair with \(22\) feet of red ribbon, multiply by the gold-to-red factor \(\frac{3}{2}\): \(22 \times \frac{3}{2}=33\) feet of gold ribbon. 4. The completed rows \(12:18\), \(5:7.5\), and \(22:33\) each reduce to the same ratio \(2:3\).

Answer

The missing entries are \(5\) ft of red ribbon and \(33\) ft of gold ribbon. The simplest red-to-gold ratio is \(2:3\), and every completed row is equivalent to \(2:3\).

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