Aimathic – Equivalent ratios in tables

Equivalent ratios in tables

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Difficulty: 1 – 5

5162886

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: Racer $100\,\text{ft}$, Walker $20\,\text{ft}$, and Crawler $2\,\text{ft}$. 2. Multiply each one-minute distance by $5$ to find the five-minute distances: Racer $250\,\text{ft}$, Walker $50\,\text{ft}$, and Crawler $5\,\text{ft}$.

Answer

Racer: $100\,\text{ft}$ in $2$ minutes and $250\,\text{ft}$ in $5$ minutes Walker: $20\,\text{ft}$ in $2$ minutes and $50\,\text{ft}$ in $5$ minutes Crawler: $2\,\text{ft}$ in $2$ minutes and $5\,\text{ft}$ in $5$ minutes

5168116

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: $7 \times \$9.00 = \$63.00$.

Answer

$3$ movie tickets cost $\$27.00$, and $7$ movie tickets cost $\$63.00$.

5204256

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

$10$ packets cost $\$8$, $15$ packets cost $\$12$, and $25$ packets cost $\$20$.

5128116

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.

5167256

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 $\$18$: $1: \$18$, $2: \$36$, $3: \$54$, $4: \$72$, $5: \$90$, $6: \$108$, $7: \$126$, $8: \$144$, $9: \$162$, $10: \$180$. 2. Locate $\$126$ in the table, or calculate $\$126 \div \$18 = 7$.

Answer

a) The prices are $\$18, \$36, \$54, \$72, \$90, \$108, \$126, \$144, \$162,$ and $\$180$. b) The group has $7$ people.

5167266

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. For the child-breakfast table, multiply each number of breakfasts by $\$9$: $\$9, \$18, \$27, \$36,$ and $\$45$.

Answer

a) The family’s breakfast costs $\$64$. b) The child-breakfast prices are $\$9, \$18, \$27, \$36,$ and $\$45$.

5190946

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. Find the table values: - $5\,\text{ft}$: $5 \times \$2.40 = \$12.00$ - $10\,\text{ft}$: $10 \times \$2.40 = \$24.00$ - $20\,\text{ft}$: $20 \times \$2.40 = \$48.00$ 2. Find the cost of $25\,\text{ft}$: $\$48.00 + \$12.00 = \$60.00$. Equivalently, $25 \times \$2.40 = \$60.00$. 3. Since $\$60.00 > \$50.00$, the customer does not have enough money.

Answer

The table values are: $5\,\text{ft}$: $\$12.00$ $10\,\text{ft}$: $\$24.00$ $20\,\text{ft}$: $\$48.00$ No. The $25\,\text{ft}$ of rope costs $\$60.00$, which is more than $\$50.00$.

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Equivalent ratios in tables

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