Free Math Worksheets

- Convert each part and whole to matching units. - Write each comparison as \(\frac{\text{part}}{\text{whole}}\). - Use equivalent fractions with a common denominator to compare.

1. For a), convert \(1.5\,\text{kg}\) to \(1500\,\text{g}\). The fraction is \(\frac{450}{1500}=\frac{3}{10}\). 2. For b), convert \(0.5\,\text{hr}\) to \(30\,\text{min}\). The fraction is \(\frac{12}{30}=\frac{2}{5}=\frac{4}{10}\). 3. Since \(\frac{4}{10}>\frac{3}{10}\), the fraction in b) is greater.

The fraction in b), \(12\,\text{min}\) out of \(0.5\,\text{hr}\), is greater.

- Write the ratio of water to concentrate. - How many equal ratio parts make up the entire mixture? - The concentrate corresponds to one of those equal parts.

1. The water-to-concentrate ratio is \(4:1\). 2. The mixture therefore contains \(4 + 1 = 5\) equal ratio parts. 3. The concentrate is one ratio part: \(7.5 \div 5 = 1.5\,\text{gal}\).

The mixture requires \(1.5\,\text{gal}\) of concentrate.

- Write the ratio of the shorter length to the longer length. - How many equal ratio parts make up the entire rope? - Find the length of one ratio part before finding both pieces.

1. The shorter-to-longer length ratio is \(1:4\). 2. The whole rope contains \(1 + 4 = 5\) equal ratio parts. 3. Find the length of one part: \(60 \div 5 = 12\,\text{cm}\). 4. Find the longer piece: \(4 \times 12 = 48\,\text{cm}\).

The two pieces are \(12\,\text{cm}\) and \(48\,\text{cm}\) long.

- Express the three amounts as a ratio. - Determine how many equal ratio parts make up the total. - Find the value of one ratio part, then scale it for each class. - Verify the three amounts by adding them.

1. Class B collects twice as much as Class A, and Class C collects six times as much as Class A. The ratio \(A:B:C\) is \(1:2:6\). 2. The ratio has \(1 + 2 + 6 = 9\) equal parts. 3. Find one ratio part: \(1800 \div 9 = 200\,\text{lb}\). 4. Class A collects \(200\,\text{lb}\), Class B collects \(2 \times 200 = 400\,\text{lb}\), and Class C collects \(6 \times 200 = 1200\,\text{lb}\). 5. Check: \(200 + 400 + 1200 = 1800\).

Class A collected \(200\,\text{lb}\), Class B collected \(400\,\text{lb}\), and Class C collected \(1200\,\text{lb}\).

- Represent the two dimensions as equal-sized ratio parts. - Use the perimeter formula for a rectangle. - Count the total number of ratio parts in the perimeter. - Find the value of one ratio part first.

1. Represent the length and width as \(5x\) feet and \(3x\) feet. 2. Use the perimeter formula: \(2(5x + 3x) = 64\). 3. Simplify: \(16x = 64\). 4. Divide by \(16\): \(x = 4\). 5. The length is \(5 \times 4\,\text{ft} = 20\,\text{ft}\), and the width is \(3 \times 4\,\text{ft} = 12\,\text{ft}\).

The garden bed is \(20\,\text{ft}\) long and \(12\,\text{ft}\) wide.

- Think of the ratio as equal-sized parts. - How many more parts of Arabica are there than Robusta? - Use the weight difference to find one part, then add all the ratio parts.

1. Let \(x\) be the weight of one ratio part in ounces. 2. The difference between the two amounts is \(7x - 2x = 15\). 3. Combine like terms: \(5x = 15\), so \(x = 3\). 4. The Arabica beans weigh \(7 \times 3 = 21\,\text{oz}\), and the Robusta beans weigh \(2 \times 3 = 6\,\text{oz}\). 5. The total weight is \(21 + 6 = 27\,\text{oz}\).

The bag contains \(27\,\text{oz}\) of coffee.

- Express all volumes in the same unit. - Find the scale factor from the pitcher to the dispenser. - Determine the fraction of the pitcher that is apple juice. - Apply the same ratio to the larger container.

1. Convert the total volumes to milliliters: \(1.5\,\text{L}=1500\,\text{mL}\) and \(6\,\text{L}=6000\,\text{mL}\). 2. The apple-juice fraction in the pitcher is \(\frac{600}{1500}=\frac{2}{5}\). 3. The dispenser has the same ratio, so its apple-juice amount is \(6000\times\frac{2}{5}=2400\,\text{mL}\). Equivalently, the dispenser is \(4\) times as large, so \(600\times4=2400\).

The dispenser contains \(2400\,\text{mL}\), or \(2.4\,\text{L}\), of apple juice.

- Rewrite the fraction equality as a ratio between \(a\) and \(b\). - Simplify the ratio \(12:30\). - Count how many equal ratio parts make the total of \(14\).

1. Rewrite the equality as a ratio: \(\frac{a}{b}=\frac{12}{30}=\frac{2}{5}\). Thus \(a:b=2:5\). 2. Let \(a=2k\) and \(b=5k\). 3. Use the sum: \(2k+5k=14\), so \(7k=14\) and \(k=2\). 4. Therefore, \(a=4\) and \(b=10\).

\((a, b)=(4, 10)\)

- How many more ratio parts of gravel are there than sand? - Use the \(60\,\text{lb}\) difference to find the weight of one part. - Add all the ratio parts to find the total number of parts.

1. Let \(x\) be the weight of one ratio part in pounds. 2. Gravel weighs \(8x\), and sand weighs \(4x\). Their difference gives \(8x - 4x = 60\). 3. Simplify: \(4x = 60\), so \(x = 15\). 4. Cement is one part, so its weight is \(15\,\text{lb}\). 5. The batch contains \(1 + 4 + 8 = 13\) parts. 6. The total weight is \(13 \times 15 = 195\,\text{lb}\).

The batch uses \(15\,\text{lb}\) of cement and weighs \(195\,\text{lb}\) in all.

- Add the ratio parts. - Convert the total amount from quarts to cups. - Divide the total cups equally among the ratio parts, then multiply for each ingredient.

1. The ratio has \(5+3+2=10\) total parts. 2. Convert the total volume: \(2.5\,\text{qt}\times 4=10\) cups. 3. Each ratio part is \(10\div 10=1\) cup. 4. The amounts are \(5\) cups of apple juice, \(3\) cups of sparkling water, and \(2\) cups of cranberry juice.

Apple juice: \(5\) cups Sparkling water: \(3\) cups Cranberry juice: \(2\) cups

- Add the ratio parts. - For a), divide the total weight by the total number of parts. - For b), identify how many parts the sand represents.

1. The ratio has \(4+3+1=8\) total parts. 2. In a), one part weighs \(120\div 8=15\,\text{lb}\). Therefore, the mixture contains \(4\times 15=60\,\text{lb}\) of compost, \(3\times 15=45\,\text{lb}\) of topsoil, and \(1\times 15=15\,\text{lb}\) of sand. 3. In b), the sand represents one part, so one part weighs \(18\,\text{lb}\). The entire eight-part mixture weighs \(8\times 18=144\,\text{lb}\).

a) Compost: \(60\,\text{lb}\) Topsoil: \(45\,\text{lb}\) Sand: \(15\,\text{lb}\) b) \(144\,\text{lb}\) of mixture

- Add the three ratio parts. - Find the weight represented by one part. - Determine the cement weight before finding the number of bags.

1. The ratio has \(1+3+5=9\) total parts. 2. One part weighs \(540\div 9=60\,\text{lb}\). 3. The sand weighs \(3\times 60=180\,\text{lb}\), and the gravel weighs \(5\times 60=300\,\text{lb}\). 4. The cement weighs \(1\times 60=60\,\text{lb}\). Since \(60\div 20=3\), three bags are needed.

a) Sand: \(180\,\text{lb}\) Gravel: \(300\,\text{lb}\) b) \(3\) bags of cement

- Simplify the given ratio first. - Express one number as a multiple of the other. - Add the two expressions and set their sum equal to \(44\).

1. Simplify the ratio: \(3.6 \div 1.2 = 3\), so the ratio is \(3:1\). 2. Let the smaller number be \(y\). Then the larger number is \(3y\). 3. Write the sum equation \(3y + y = 44\). 4. Combine like terms: \(4y = 44\). 5. Divide by \(4\): \(y = 11\). 6. The larger number is \(3 \times 11 = 33\).

The two numbers are \(33\) and \(11\).

- Find the value of one ratio part from the difference between the two days. - Add the original distances to find the total. - A \(1:1\) split gives each day half of the total. - Compare the first-day distance with the equal-share distance.

1. Let \(x\) be the number of miles in one ratio part. The difference gives \(3x - 2x = 15\). 2. Thus \(x = 15\). 3. The first day is \(3 \times 15 = 45\,\text{miles}\), and the second day is \(2 \times 15 = 30\,\text{miles}\). 4. The total distance is \(45 + 30 = 75\,\text{miles}\). 5. An equal split gives \(75 \div 2 = 37.5\,\text{miles}\) per day. 6. The first day must be shortened by \(45 - 37.5 = 7.5\,\text{miles}\), which are added to the second day.

a) The first day is \(45\,\text{miles}\), and the second day is \(30\,\text{miles}\). b) \(7.5\,\text{miles}\) must be shifted from the first day to the second day.

- Write one fourth as a percent. - Subtract the known percentages from \(100\%\). - A ratio of \(2:1\) divides the remaining amount into three equal parts. - A full circle is \(360^\circ\).

1. Apple juice represents \(\frac{1}{4}=25\%\). 2. Soda and tea together represent \(100\%-25\%-45\%=30\%\). 3. A ratio of \(2:1\) has \(3\) equal parts. Each part represents \(30\%\div3=10\%\). 4. Tea represents \(10\%\), and soda represents \(2\times10\%=20\%\). 5. The tea angle is \(360^\circ\times0.10=36^\circ\). The soda angle is \(360^\circ\times0.20=72^\circ\).

Soda: \(20\%\) and \(72^\circ\) Tea: \(10\%\) and \(36^\circ\)

- Temporarily remove the extra \(100\,\text{mL}\) of water from the total. - Represent the remaining amounts as equal parts based on the lemon-juice amount. - Find the size of one part, then add the extra water back.

1. Remove the extra water amount from the total: \(1000\,\text{mL} - 100\,\text{mL} = 900\,\text{mL}\). 2. Let the lemon juice be one equal part. The orange juice is \(4\) parts, and the water without the extra \(100\,\text{mL}\) is \(1\) part. There are \(1 + 4 + 1 = 6\) equal parts. 3. Find one part: \(900\,\text{mL} \div 6 = 150\,\text{mL}\). 4. Find each amount: lemon juice is \(150\,\text{mL}\), orange juice is \(4 \times 150\,\text{mL} = 600\,\text{mL}\), and water is \(150\,\text{mL} + 100\,\text{mL} = 250\,\text{mL}\). 5. Check: \(600\,\text{mL} + 150\,\text{mL} + 250\,\text{mL} = 1000\,\text{mL}\).

The punch uses \(600\,\text{mL}\) of orange juice, \(150\,\text{mL}\) of lemon juice, and \(250\,\text{mL}\) of water.