Free math worksheets

- Count the shaded tiles in the diagram. - Count the unshaded tiles separately. - Keep the counts in the same order as the words “shaded to unshaded.”

1. Count \(6\) shaded tiles. 2. Count \(4\) unshaded tiles. 3. The phrase “shaded tiles to unshaded tiles” fixes the order, so the ratio is \(6:4\).

\(6:4\)

- Keep the quantities in the order named in the question. - The three forms should describe the same comparison. - In the “for every” form, make clear which count goes with each marker color.

1. The requested order is blue markers first and black markers second. 2. The counts are 4 blue markers and 7 black markers, so the colon form is \(4:7\). 3. The same ordered comparison can be written as “4 to 7.” 4. In words, it is “for every 4 blue markers, there are 7 black markers.”

a) \(4:7\) b) 4 to 7 c) For every 4 blue markers, there are 7 black markers.

- 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.

- Read the words before deciding which count comes first. - Write each unsimplified ratio before reducing it. - Compare the order in Ava’s ratio with the order requested in the sentence.

1. Ava used the counts in the order blue to red, not red to blue. 2. Red to blue is \(8:12\), which simplifies to \(2:3\). 3. Blue to red is \(12:8\), which simplifies to \(3:2\). 4. Reversing the order of the quantities reverses the ratio.

Ava reversed the requested order. Red to blue: \(2:3\) Blue to red: \(3:2\)

- Count the filled and unfilled circles separately. - For part-to-part, compare the two club groups only. - For part-to-whole, compare one club group with every circle in the diagram.

1. Count 5 filled circles and 7 unfilled circles, for 12 circles in all. 2. Music-club badges to robotics-club badges compares one part with another part, so the ratio is \(5:7\). 3. Music-club badges to all badges compares a part with the whole, so the ratio is \(5:12\). 4. The first ratio is part-to-part, while the second is part-to-whole.

a) \(5:7\) b) \(5:12\) c) The first ratio is part-to-part; the second ratio is part-to-whole.

- Count the filled counters, the unfilled counters, and the total separately. - Read each requested ratio from left to right before writing the numbers. - Do not simplify; the question asks you to preserve the displayed counts.

1. The diagram has 6 filled counters and 9 unfilled counters, for 15 counters total. 2. Filled to unfilled is \(6:9\). 3. Reversing the order gives unfilled to filled as \(9:6\). 4. Unfilled to all compares 9 unfilled counters with all 15 counters, so it is \(9:15\).

a) \(6:9\) b) \(9:6\) c) \(9:15\)

- Identify which quantity must appear first in the requested ratio. - Match each number with its unit before writing the ratio. - The question asks you to keep the original quantities rather than reduce the ratio.

1. The requested order is hours first and miles second. 2. The situation gives 2 hours and 8 miles. 3. Mateo placed the mile value first, so he reversed the requested order. 4. The correct ratio of hours to miles is \(2:8\).

Mateo reversed the requested order. The correct ratio of hours to miles is \(2:8\).

- 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 \(b=10\).

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

- Convert the two squirrel measurements to the same unit. - Divide the jump distance by the body length. - Multiply the child's height by the same factor.

1. Convert \(5\,\text{m}\) to centimeters: \(5\,\text{m}=500\,\text{cm}\). 2. Find the ratio: \(500\,\text{cm}\div 25\,\text{cm}=20\). The character jumps \(20\) times its body length. 3. Apply the same factor to the child's height: \(1.45\,\text{m}\times 20=29\,\text{m}\).

a) \(20\) times its body length b) \(29\,\text{m}\)

- 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 \(1200\,\text{lb}\).

- 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.

- 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.

- 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 \(11\).

- 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.

- 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.

- For each bar, compare the number of filled parts with the number of unfilled parts. - Equivalent ratios multiply both quantities by the same factor. - Check whether each visual can be grouped into copies of the same filled-to-unfilled pattern.

1. In bar a), 3 of the 5 parts are filled, leaving 2 unfilled, so its ratio is \(3:2\). 2. In bar b), 6 of the 10 parts are filled, leaving 4 unfilled. The ratio \(6:4\) is equivalent to \(3:2\). 3. In bar c), 8 of the 15 parts are filled, leaving 7 unfilled. The ratio \(8:7\) is not equivalent to \(3:2\). 4. Therefore bars a) and b) show the required multiplicative relationship.

Bars a) and b). Their filled-to-unfilled ratios are \(3:2\) and \(6:4\), which are equivalent. Bar c) has ratio \(8:7\).

- Track both the quantity and its unit before deciding which number comes first. - “Border length to planting sections” and “planting sections to border length” reverse the order. - Explain what each ordered comparison says, not just how its numbers look.

1. Border length to planting sections keeps the border quantity first, so the ratio is \(6\,\text{yd}:4\,\text{sections}\). 2. In words, the same comparison is “6 yards of border for every 4 planting sections.” 3. Reversing the order gives \(4\,\text{sections}:6\,\text{yd}\). 4. The reversed ratio is not the same requested comparison because it describes sections per amount of border rather than border length per number of sections.

a) \(6\,\text{yd}:4\,\text{sections}\) b) 6 yards of border for every 4 planting sections c) \(4\,\text{sections}:6\,\text{yd}\); it compares the same quantities in the opposite order, so it answers a different question.

- 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.