Free math worksheets

- A unit rate tells how much ribbon is cut in exactly \(1\) minute. - Compare the given half-minute interval with a full minute. - You can also write the rate as length divided by time.

1. Divide the ribbon length by the time: \(\frac{3}{4} \div \frac{1}{2}\). 2. Multiply by the reciprocal: \(\frac{3}{4} \cdot \frac{2}{1} = \frac{3}{2}\). 3. Therefore, the unit rate is \(\frac{3}{2}\,\text{yd/min} = 1.5\,\text{yd/min}\).

\(1.5\,\text{yd/min}\)

- The unit price asks for the cost of exactly \(1\) pound. - Rewrite the mixed number as an improper fraction before dividing. - Divide total cost by total weight.

1. Write \(1\frac{1}{2}\) as \(\frac{3}{2}\). 2. Divide the cost by the number of pounds: \(\$9.60 \div \frac{3}{2}\). 3. Multiply by the reciprocal: \(\$9.60 \cdot \frac{2}{3} = \$6.40\).

\(\$6.40\) per pound

- Compare each requested time with the given quarter-minute interval. - Keep the same scale factor for time and water amount. - The one-minute entry is the unit rate.

1. Doubling the time from \(\frac{1}{4}\) minute to \(\frac{1}{2}\) minute doubles the water amount to \(\frac{6}{5}\,\text{gal}\). 2. Four quarter-minute intervals make \(1\) minute, so the one-minute amount is \(4 \cdot \frac{3}{5} = \frac{12}{5}\,\text{gal}\). 3. The completed table is: <table> <tr><th>Time (min)</th><td>\(\frac{1}{4}\)</td><td>\(\frac{1}{2}\)</td><td>\(1\)</td></tr> <tr><th>Water (gal)</th><td>\(\frac{3}{5}\)</td><td>\(\frac{6}{5}\)</td><td>\(\frac{12}{5}\)</td></tr> </table> 4. The unit rate is \(\frac{12}{5}\,\text{gal/min} = 2.4\,\text{gal/min}\).

<table> <tr><th>Time (min)</th><td>\(\frac{1}{4}\)</td><td>\(\frac{1}{2}\)</td><td>\(1\)</td></tr> <tr><th>Water (gal)</th><td>\(\frac{3}{5}\)</td><td>\(\frac{6}{5}\)</td><td>\(\frac{12}{5}\)</td></tr> </table> Unit rate: \(2.4\,\text{gal/min}\)

- First determine how much of a spool is used in \(1\) hour. - Then compare the target spool amount with that hourly rate. - Keep track of whether your rate is spools per hour or hours per spool.

1. Find the unit rate: \(\frac{3}{8} \div \frac{5}{6} = \frac{3}{8} \cdot \frac{6}{5} = \frac{9}{20}\) spool per hour. 2. Divide the target amount by the unit rate: \(\frac{9}{10} \div \frac{9}{20}\). 3. \(\frac{9}{10} \cdot \frac{20}{9} = 2\), so the printer would take \(2\) hours.

\(2\) hours

- Find each robot’s distance for exactly \(1\) second. - Use the same unit, feet per second, for both rates before comparing. - After deciding which rate is larger, subtract the smaller rate from the larger one.

1. Robot A's unit rate is \(\frac{5}{8} \div \frac{1}{4} = \frac{5}{2}\,\text{ft/s}\). 2. Robot B's unit rate is \(\frac{7}{9} \div \frac{1}{3} = \frac{7}{3}\,\text{ft/s}\). 3. Since \(\frac{5}{2} > \frac{7}{3}\), Robot A is faster. 4. The difference is \(\frac{5}{2} - \frac{7}{3} = \frac{15}{6} - \frac{14}{6} = \frac{1}{6}\,\text{ft/s}\).

Robot A is faster by \(\frac{1}{6}\,\text{ft/s}\).

- For part a), compare the amount of seed with exactly \(1\) acre. - Keep the unit “bags per acre” attached to the rate. - Use the unit rate from part a) for the \(2\)-acre application.

1. Find the unit rate: \(\frac{3}{5} \div \frac{3}{4} = \frac{3}{5} \cdot \frac{4}{3} = \frac{4}{5}\) bag per acre. 2. For \(2\) acres, multiply the unit rate by \(2\): \(\frac{4}{5} \cdot 2 = \frac{8}{5} = 1.6\) bags.

a) \(\frac{4}{5}\) bag per acre b) \(1.6\) bags

- Use the requested unit, miles per hour, to decide which quantity belongs in the numerator. - Check what units Jordan’s quotient would have. - After choosing the quotient, divide the two fractions carefully.

1. Miles per hour means distance divided by time, so the required quotient is \(\frac{2}{3} \div \frac{4}{5}\), not time divided by distance. 2. \(\frac{2}{3} \div \frac{4}{5} = \frac{2}{3} \cdot \frac{5}{4} = \frac{5}{6}\). 3. The correct walking rate is \(\frac{5}{6}\,\text{mph}\). Jordan’s calculation instead has units of hours per mile.

Jordan reversed the quotient. The correct rate is \(\frac{5}{6}\,\text{mph}\).

- Find each pump’s rate in tanks per hour before comparing completion times. - Use the same target amount, \(\frac{3}{2}\) tanks, with each unit rate. - Compare the two times, then convert their difference from hours to minutes.

1. Pump A's unit rate is \(\frac{5}{6} \div \frac{3}{4} = \frac{10}{9}\) tank per hour. 2. Pump B's unit rate is \(\frac{7}{8} \div \frac{5}{6} = \frac{21}{20}\) tank per hour. 3. Pump A needs \(\frac{3}{2} \div \frac{10}{9} = \frac{27}{20}\) hour. 4. Pump B needs \(\frac{3}{2} \div \frac{21}{20} = \frac{10}{7}\) hour. 5. The time difference is \(\frac{10}{7} - \frac{27}{20} = \frac{11}{140}\) hour. 6. Convert to minutes: \(\frac{11}{140} \cdot 60 = \frac{33}{7}\) minutes, which is about \(4.7\) minutes. 7. Pump A finishes first.

Pump A finishes first by \(\frac{33}{7}\) minutes, or about \(4.7\) minutes.