Ratios and Proportions — Free Pre-Algebra Review Games.

Divide both by the GCF of 3: 6/3 = 2, 9/3 = 3, so 2:3.

3/5 = 15/x, so x = 15 * 5/3 = 25.

Unit rate = 150/3 = 50 miles per hour.

Set up proportion: 2/12 = x/30, so x = 60/12 = 5 cups.

8/12 simplified by dividing by 4 gives 2/3.

Cross multiply: 4 * 14 = 8 * x, so 56 = 8x, x = 7.

$8.75 / 5 = $1.75 per notebook.

3.5 * 50 = 175 miles.

$5.40/3 = $1.80/lb vs $8.50/5 = $1.70/lb. The 5 pound option is cheaper per pound.

4/100 = x/350, x = 1400/100 = 14 gallons.

Cross multiply: 30 = 6(x+1), so 30 = 6x+6, 24 = 6x, x = 4.

5k + 8k = 78, 13k = 78, k = 6. Larger = 8*6 = 48.

Real length = 6 * 72 = 432 inches = 432/12 = 36 feet.

y = 3x/4. Then x + 3x/4 = 35, 7x/4 = 35, x = 20, y = 15.

A: 200/5 = 40/hr. B: 360/8 = 45/hr. B is faster by 5 parts/hr.

Dividing both terms of \(15:20\) by their greatest common factor, \(5\), gives \(3:4\), which is the simplest form of the ratio. The choice \(4:5\) is wrong because it reverses the division result and does not come from dividing \(15\) and \(20\) by the same factor. When simplifying any ratio, always divide both terms by their greatest common divisor to reach lowest terms.

The word has vowels A, E, A (3 vowels) and consonants L, G, B, R (4 consonants), giving the ratio \(3:4\). The choice \(3:7\) incorrectly compares vowels to the total number of letters rather than to consonants alone. When writing a ratio, always identify exactly which two quantities are being compared before forming the ratio.

Dividing \(240\) miles by \(4\) hours gives \(60\) miles per hour, the unit rate. The choice \(50\) mph is incorrect because it does not result from dividing \(240\) by \(4\). A unit rate always expresses a quantity per single unit of another quantity, found by division.

Multiplying both terms of \(2:3\) by \(2\) gives \(4:6\), which represents the same relationship between the quantities. The choice \(3:2\) is wrong because it reverses the order of the original ratio, changing its meaning. Equivalent ratios are formed by multiplying or dividing both terms by the same nonzero number.

Dividing both \(10\) and \(25\) by their greatest common factor, \(5\), gives \(\frac{2}{5}\). The choice \(\frac{5}{2}\) is incorrect because it inverts the fraction, reversing which quantity is being compared to which. A ratio written as a fraction keeps the first quantity in the numerator and the second in the denominator.

Dividing the total cost, $3, by the number of apples, \(6\), gives \(0.50\) per apple. The choice \(2.00\) is wrong because it does not come from dividing the cost by the quantity purchased. Unit price is always found by dividing total cost by the number of items.

\(9:4\) reverses the order of the quantities, so it represents the ratio of \(9\) to \(4\), not \(4\) to \(9\). The choice \(\frac{4}{9}\) is a correct representation because it preserves the same order as the original ratio. The order of terms in a ratio matters and must match the order stated in the problem.

Dividing both terms by their greatest common factor, \(6\), gives \(2:3\), the fully reduced ratio. The choice \(4:6\) is not in simplest form because it can still be divided further by \(2\). A ratio is only fully simplified when the two terms share no common factor other than \(1\).

Dividing \(300\) words by \(5\) minutes gives \(60\) words per minute. The choice \(50\) is incorrect because it does not equal \(300\) divided by \(5\). To find a unit rate in words per minute, always divide the total words by the total minutes.

Dividing both terms by their greatest common factor, \(3\), gives \(3:1\). The choice \(1:3\) is wrong because it reverses the order of the original ratio. Simplifying a ratio never changes the order of the compared quantities, only their relative size.

Multiplying both terms of \(5:6\) by \(2\) gives \(10:12\), preserving the same relationship. The choice \(6:5\) is incorrect because it swaps the order of the terms, creating a different ratio. Equivalent ratios must be scaled by the same factor on both terms.

Dividing $6.00 by \(4\) pens gives $1.50 per pen. The choice $0.67 is wrong because it results from dividing \(4\) by \(6\) instead of \(6\) by \(4\). Unit price is always total cost divided by quantity, not the reverse.

Dividing both \(25\) and \(100\) by their greatest common factor, \(25\), gives \(\frac{1}{4}\). The choice \(\frac{4}{1}\) is incorrect because it inverts the fraction, reversing the comparison. A simplified ratio fraction must keep the numerator and denominator in the original order while reducing to lowest terms.

Cross-multiplying gives \(20x = 60\), so \(x = 3\), since cross multiplication turns a proportion into a solvable linear equation. The choice \(4\) is wrong because substituting it back gives \(\frac{4}{5} \neq \frac{12}{20}\). Whenever solving a proportion, cross multiply first, then isolate the variable by division.

Cross-multiplying gives \(15x = 150\), so \(x = 10\), which satisfies the proportion since \(\frac{6}{10} = \frac{15}{25}\) after simplification. The choice \(9\) is incorrect because \(\frac{6}{9}\) does not equal \(\frac{15}{25}\) when checked. Always verify a proportion answer by substituting it back into the original ratio.

Setting up the proportion \(\frac{3}{2} = \frac{x}{10}\) and cross-multiplying gives \(2x = 30\), so \(x = 15\) eggs. The choice \(12\) is wrong because it does not maintain the same \(3:2\) ratio scaled to \(10\) cups. Proportional reasoning requires keeping the same rate constant as quantities scale up.

Dividing \(240\) miles by \(8\) gallons gives \(30\) miles per gallon, the unit rate of fuel efficiency. The choice \(32\) is incorrect because it does not result from dividing \(240\) by \(8\). Fuel efficiency, like any unit rate, is found by dividing distance by the amount of fuel used.

The unit price for \(16\) oz is \(4.16 \div 16 = \\)0.26$ per ounce, while \(12\) oz costs \(3.60 \div 12 = \\)0.30$ per ounce, making the \(16\) oz option cheaper per unit. The choice 'They cost the same per ounce' is wrong because \(0.26 \neq 0.30\). To find the better deal, always compute and compare unit prices, not total prices.

Setting up the proportion \(\frac{2}{5} = \frac{8}{x}\) and cross-multiplying gives \(2x = 40\), so \(x = 20\) m. The choice \(16\) m is incorrect because it results from multiplying \(8\) by \(2\) instead of correctly applying the scale ratio. Scale drawing problems require setting up a proportion between the drawing measurement and the actual measurement.

Cross-multiplying gives \(9x = 315\), so \(x = 35\), since \(45\) is \(5\) times \(9\) and \(x\) must be \(5\) times \(7\). The choice \(40\) is wrong because it does not preserve the same scale factor of \(5\) from the original ratio. Recognizing the scale factor between denominators can be a quick shortcut for solving proportions.

The ratio \(4:9\) has \(13\) total parts, and \(39 \div 13 = 3\) marbles per part, so blue marbles equal \(9 \times 3 = 27\). The choice \(12\) is wrong because it corresponds to the red marbles, not blue. When a total is given with a part-to-part ratio, first find the value of one part before multiplying by each ratio term.

Setting up the proportion \(\frac{2}{6} = \frac{x}{15}\) and cross-multiplying gives \(6x = 30\), so \(x = 5\) cups. The choice \(4\) cups is incorrect because it does not maintain the original \(2:6\) ratio of sugar to servings. Scaling a recipe requires keeping every ingredient's ratio to the number of servings constant.

The ratio \(3:5\) has \(8\) total parts, and \(40 \div 8 = 5\) liters per part, so yellow paint equals \(5 \times 5 = 25\) liters. The choice \(15\) liters is wrong because it corresponds to the blue paint amount, not yellow. Solving part-to-part ratio problems with a known total always starts by finding the total number of parts.

Dividing \(45\) pages by \(3\) minutes gives \(15\) pages per minute, the unit rate of the printer. The choice \(12\) is incorrect because it does not result from dividing \(45\) by \(3\). A unit rate always compares a quantity to exactly one unit of time or another measure.

The unit price is \(2.40 \div 8 = \\)0.30$ per pencil, so \(20\) pencils cost \(0.30 \times 20 = \\)6.00$. The choice $4.80 is wrong because it does not use the correct unit price found by dividing total cost by quantity. Finding the unit rate first makes scaling to any new quantity straightforward.

With \(5\) shaded and \(15\) unshaded squares, the ratio \(5:15\) simplifies by dividing both terms by \(5\) to get \(1:3\). The choice \(5:15\) is correct in value but is not in simplest form, so it is not the best answer. Ratios should always be reduced to lowest terms unless the problem specifically asks for the unreduced form.

Dividing \(180\) km by \(2.5\) hours gives \(72\) km per hour, the cyclist's unit rate. The choice \(75\) is incorrect because it does not equal \(180\) divided by \(2.5\). Dividing distance by time always produces the speed unit rate, regardless of whether the time value is a whole number.

Since \(56 \div 8 = 7\), the numerator must also scale by \(7\), giving \(x = 5 \times 7 = 35\). The choice \(40\) is wrong because it does not preserve the same scale factor between the two proportion terms. Recognizing scale factors between denominators offers a fast alternative to cross multiplication.

The ratio \(2:7\) has \(9\) total parts, and \(18 \div 9 = 2\) students per part, so girls equal \(7 \times 2 = 14\). The choice \(4\) is wrong because it corresponds to the number of boys, not girls. Part-to-part ratio problems with a known total always require dividing the total by the sum of the ratio parts first.

The unit price for \(5\) shirts is \(70 \div 5 = \\)14$ per shirt, while \(3\) shirts cost \(45 \div 3 = \\)15$ per shirt, making the \(5\)-shirt deal cheaper per item. The choice 'They cost the same' is wrong because \(14 \neq 15\). Comparing unit prices, not total prices, is the correct way to determine the better deal.

Converting \(300\) m to \(30{,}000\) cm and dividing by the scale factor \(1500\) gives \(30{,}000 \div 1500 = 20\) cm for the model. The choice \(30\) cm is wrong because it results from an incorrect division that does not use the full converted length. Scale problems often require converting units before applying the proportion.

Dividing \(15\) feet by \(6\) minutes gives \(2.5\) feet per minute, the turtle's unit rate. The choice \(3\) is incorrect because it does not equal \(15\) divided by \(6\). Unit rates for movement are always found by dividing distance by time.

Letting the numbers be \(3x\) and \(7x\), the difference \(7x - 3x = 4x = 24\) gives \(x = 6\), so the larger number is \(7 \times 6 = 42\). The choice \(18\) is wrong because it equals the smaller number, \(3 \times 6\), not the larger one. When ratio problems give a sum or difference, always express both quantities in terms of a single variable before solving.

Cross-multiplying gives \(5x = 2(x+3) = 2x + 6\), so \(3x = 6\) and \(x = 2\). The choice \(6\) is wrong because substituting it back gives \(\frac{6}{9} = \frac{2}{3} \neq \frac{2}{5}\). When the variable appears in both the numerator and part of the denominator, cross multiplication followed by careful distribution is required to isolate it.

Converting \(12.5\) km to \(1{,}250{,}000\) cm and dividing by the scale factor \(25{,}000\) gives \(1{,}250{,}000 \div 25{,}000 = 50\) cm on the map. The choice \(500\) cm is wrong because it results from a division error that misplaces a decimal point. Scale problems involving different units always require careful unit conversion before dividing by the scale factor.

With \(10\) total parts, each part is \(30\) mL, giving \(30\) mL of salt and \(270\) mL of water; adding \(50\) mL of water makes it \(320\) mL, so the new ratio \(30:320\) simplifies to \(3:32\). The choice \(1:9\) is wrong because it ignores the added water that changes the original ratio. Adding a quantity to only one part of a ratio always requires recalculating the ratio from the new totals, not reusing the original one.

Car A's speed is \(150 \div 2.5 = 60\) mph, while Car B's speed is \(240 \div 3 = 80\) km/h, which converts to \(80 \div 1.6 = 50\) mph, so Car A is faster. The choice \(80\) mph for Car B is wrong because it fails to convert km/h into mph before comparing. When comparing rates given in different units, always convert to the same unit before making a comparison.

Letting \(a = 3x\) and \(b = 5x\), the sum \(3x + 5x = 8x = 64\) gives \(x = 8\), so \(a = 24\) and \(b = 40\), making \(a - b = -16\). The choice \(16\) is wrong because it reverses the sign, mistakenly treating \(a\) as the larger value. When a ratio and a sum are given together, express both variables in terms of one unknown before computing any other relationship.

Since the smaller triangle's perimeter corresponds to the ratio value \(4\), one unit equals \(36 \div 4 = 9\), so the larger perimeter equals \(7 \times 9 = 63\). The choice \(49\) is wrong because it does not come from scaling the perimeter by the correct factor derived from the side ratio. In similar figures, all corresponding linear measurements, including perimeter, scale by the same ratio as the sides.

Cross-multiplying gives \(8(x-1) = 2(x+5)\), which simplifies to \(8x - 8 = 2x + 10\), so \(6x = 18\) and \(x = 3\). The choice \(4\) is wrong because substituting it back gives \(\frac{3}{2} \neq \frac{9}{8}\). When solving proportions with binomials on both sides, distribute carefully after cross multiplying before combining like terms.

Setting up the proportion \(\frac{3/4}{2} = \frac{1}{t}\) and solving gives \(t = 2 \div \frac{3}{4} = 2 \times \frac{4}{3} = \frac{8}{3}\) hours. The choice \(3\) hours is wrong because it does not correctly divide the full tank's rate by the constant fill rate. When a partial amount and its time are known, dividing time by the fractional amount gives the time for the whole quantity.

Letting Sam be \(4x\) and Tom be \(5x\), the equation \(\frac{4x+6}{5x+6} = \frac{5}{6}\) leads to \(24x + 36 = 25x + 30\), giving \(x = 6\), so Sam's current age is \(4 \times 6 = 24\). The choice \(20\) is wrong because it does not satisfy the future ratio condition when \(6\) is added to both ages. Age ratio problems require setting up an equation that accounts for the same amount of time added to both quantities.

The ratio \(5:3:2\) has \(10\) total parts, and \(200 \div 10 = 20\) g per part, so sugar equals \(3 \times 20 = 60\) g. The choice \(40\) g is wrong because it corresponds to twice the butter amount, not the sugar amount. Three-part ratio problems are solved the same way as two-part ones: find the value of one part, then multiply by each term.

Since the trains move apart, their combined speed times \(2\) hours equals \(210\) miles, so combined speed is \(105\) mph; dividing by the \(7\) total ratio parts gives \(15\) mph per part, and the faster train's speed is \(4 \times 15 = 60\) mph. The choice \(70\) mph is wrong because it does not result from correctly dividing the combined speed by the sum of the ratio parts. Opposite-direction distance problems require adding the two speeds before applying the distance-time relationship.

Letting the common ratio equal \(k\), then \(x=4k\), \(y=5k\), \(z=6k\), and their sum \(15k = 45\) gives \(k = 3\), so \(z = 6 \times 3 = 18\). The choice \(15\) is wrong because it equals \(y\), not \(z\), from mismatching the ratio term to the variable. When three quantities share a common ratio, express each as a multiple of one variable before using the given sum to solve.