Percentages

Choice A is correct. It's given that \(10 percent sign\) of the \(320\) marbles in a container are red. Therefore, the number of red marbles can be calculated by multiplying the number of marbles in the container by \(10/100\), which gives \(320(10/100)\), or \(32\). Choice B is incorrect. This is the number of marbles in the container that aren't red. Choice C is incorrect. This is the total number of marbles in the container. Choice D is incorrect and may result from conceptual or calculation errors.

Choice C is correct. It's given that the price of the collectible card was \($ 15.50\) in March and \($ 17.36\) in April. It's also given that the price of the collectible card in April was \(p percent sign\) of the price in March. It follows that \($ 17.36\) is \(p percent sign\) of \($ 15.50\). Therefore, the value of \(p\) can be calculated by solving the equation \(17.36 =(p/100)(15.50)\), or \(17.36 = 15.50 p/100\). Multiplying each side of this equation by \(100\) yields \(1,736 = 15.50 p\). Dividing each side of this equation by \(15.50\) yields \(112 = p\). Therefore, the value of \(p\) is \(112\). Choice A is incorrect. \(12 percent sign\) is the percent increase in the price of the collectible card from March to April. Choice B is incorrect and may result from conceptual or calculation errors. Choice D is incorrect and may result from conceptual or calculation errors.

Choice A is correct. It’s given that the value of the comic book increased by \(167 percent sign\) from the end of \(2011\) to the end of \(2012\). Therefore, if the value of the comic book at the end of \(2011\) was \(x\) dollars, then the value, in dollars, of the comic book at the end of \(2012\) was \(x +(167/100)x\), which can be rewritten as \(1 x + 1.67 x\), or \(2.67 x\). It’s also given that the value of the comic book decreased by \(16 percent sign\) from the end of \(2012\) to the end of \(2013\). Therefore, the value, in dollars, of the comic book at the end of \(2013\) was \(2.67 x -2.67 x(16/100)\), which can be rewritten as \(2.67 x -(2.67 x)(0.16)\), or \(2.2428 x\). Thus, if the value of the comic book at the end of \(2011\) was \(x\) dollars, and the value of the comic book at the end of \(2013\) was \(2.2428 x\) dollars, then from the end of \(2011\) to the end of \(2013\), the value of the comic book increased by \(2.2428 x -1 x\), or \(1.2428 x\), dollars. Therefore, the increase in the value of the comic book from the end of \(2011\) to the end of \(2013\) is equal to \(1.2428\) times the value of the comic book at the end of \(2011\). It follows that from the end of \(2011\) to the end of \(2013\), the net percentage increase in the value of the comic book was \((1.2428)(100)percent sign\), or \(124.28 percent sign\). Choice B is incorrect and may result from conceptual or calculation errors. Choice C is incorrect. This is the difference between the net percentage increase in the value of the comic book from the end of \(2011\) to the end of \(2012\) and the net percentage decrease in the value of the comic book from the end of \(2012\) to the end of \(2013\), not the net percentage increase in the value of the comic book from the end of \(2011\) to the end of \(2013\). Choice D is incorrect. This is the net percentage increase in the value of the comic book from the end of \(2011\) to the end of \(2013\), if the value of the comic book increased by \(167 percent sign\) from the end of \(2011\) to the end of \(2012\) and then increased, not decreased, by \(16 percent sign\) from the end of \(2012\) to the end of \(2013\).

The correct answer is \(.88\). It’s given that the number \(a\) is \(60 percent sign\) greater than the positive number \(b\). Therefore, \(a =(1 + 60/100)b\), which is equivalent to \(a =(1 + 0.60)b\), or \(a = 1.60 b\). It’s also given that the number \(c\) is \(45 percent sign\) less than \(a\). Therefore, \(c =(1 -45/100)a\), which is equivalent to \(c =(1 -0.45)a\), or \(c = 0.55 a\). Since \(a = 1.60 b\), substituting \(1.60 b\) for \(a\) in the equation \(c = 0.55 a\) yields \(c = 0.55(1.60 b)\), or \(c = 0.88 b\). Thus, the number \(c\) is \(0.88\) times the number \(b\). Note that .88 and 22/25 are examples of ways to enter a correct answer.

Choice C is correct. It's given that \(37 percent sign\) of the items in a box are green. Let \(x\) represent the total number of items in the box. It follows that \(37/100 x\), or \(0.37 x\), items in the box are green. It's also given that of those, \(37 percent sign\) are also rectangular. Therefore, \(37/100(0.37 x)\), or \(0.1369 x\), items in the box are green rectangular items. It's also given that of the green rectangular items, \(42 percent sign\) are also metal. Therefore, \(42/100(0.1369 x)\), or \(0.057498 x\), items in the box are rectangular green metal items. The number of the items in the box that are not rectangular green metal items is the total number of items in the box minus the number of rectangular green metal items in the box. Therefore, the number of items in the box that are not rectangular green metal items is \(x -0.057498 x\), or \(0.942502 x\). The percentage of items in the box that are not rectangular green metal items is the percentage that \(0.942502 x\) is of \(x\). If \(p percent sign\) represents this percentage, the value of \(p\) is \(100(.942502 x/x)\), or \(94.2502\). Of the given choices, \(94.25 percent sign\) is closest to the percentage of items in the box that are not rectangular green metal items. Choice A is incorrect and may result from conceptual or calculation errors. Choice B is incorrect and may result from conceptual or calculation errors. Choice D is incorrect and may result from conceptual or calculation errors.

The correct answer is \(.0014\). It's given that a certain type of substance can contain a maximum of \(0.001 percent sign\) phosphorus by mass to meet a set of standards. If a sample of the substance has a mass of \(140\) grams, it follows that the maximum mass, in grams, of phosphorus the sample can contain to meet the standards is \(0.001 percent sign\) of \(140\), or \(.001/100(140)\), which is equivalent to \((0.00001)(140)\), or \(0.0014\). Note that .0014 and 0.001 are examples of ways to enter a correct answer.

The correct answer is \(240\). It’s given that \(80 percent sign\) of the \(300\) seeds sprouted. Therefore, the number of seeds that sprouted can be calculated by multiplying the number of seeds that were planted by \(80/100\), which gives \(300(80/100)\), or \(240\).

Choice B is correct. The decimal equivalent of 20% is 0.2. The number that is 20% greater than 60 is also 120% of 60. The decimal equivalent of 120% is 1.2, and .Alternate approach: 10% of 60 is 6, and 20% of 60 is double that amount, or 12. It follows that the number that is 20% greater than 60 is 12 more than 60, or .Choice A is incorrect and may result from dividing, instead of multiplying, 60 by 1.2. Choice C is incorrect because it’s 25% greater than 60, rather than 20% greater than 60. Choice D is incorrect and may result from multiplying 60 by 2.2 instead of 1.2.

The correct answer is \(1.8\). It’s given that the regular price of a shirt at a store is \($ 11.70\), and the sale price of the shirt is \(80 percent sign\) less than the regular price. It follows that the sale price of the shirt is \($ 11.70(1 -80/100)\), or \($ 11.70(1 -0.8)\), which is equivalent to \($ 2.34\). It’s also given that the sale price of the shirt is \(30 percent sign\) greater than the store’s cost for the shirt. Let \(x\) represent the store’s cost for the shirt. It follows that \(2.34 =(1 + 30/100)x\), or \(2.34 = 1.3 x\). Dividing both sides of this equation by \(1.3\) yields \(x = 1.80\). Therefore, the store’s cost, in dollars, for the shirt is \(1.80\). Note that 1.8 and 9/5 are examples of ways to enter a correct answer.

Choice D is correct. It's given that the number of coins in the collection increased from \(9\) to \(90\). It follows that the number of coins in the collection increased by \(90 -9\), or \(81\). Let \(x percent sign\) represent the percentage that \(81\) is of \(9\). The value of \(x\) can be found using the proportion \(81/9 = x/100\), or \(9 = x/100\). Multiplying both sides of this equation by \(100\) yields \(900 = x\). Thus, when the number of coins in the collection increased from \(9\) to \(90\), the percent increase was \(900 percent sign\). Choice A is incorrect and may result from conceptual or calculation errors. Choice B is incorrect and may result from conceptual or calculation errors. Choice C is incorrect and may result from conceptual or calculation errors.

Choice D is correct. The total bill was $25.00. The percentage 20% is equivalent to the decimal 0.2. The tip is the product of the percentage and the total bill; therefore,, so the tip was $5.00.Choices A, B, and C are incorrect and may be the result of incorrectly converting the given percentage or a calculation error.

Choice B is correct. If the stove’s value depreciates by 20% of the previous year’s estimated value, then each year it retains 100% – 20% = 80%, or 0.80, of the previous year’s estimated value. Since the stove’s value was $800 when Thomas installed it, the estimated value after two years would be (0.80)(0.80)($800) = $512.Choice A is incorrect. This is the value of the stove if each year it had depreciated by 20% of the original value rather than by 20% of the previous year’s estimated value. Choice C is incorrect and may be the result of a computational error. Choice D is incorrect. This is the estimated value of the stove 1 year after Thomas installed it, not 2 years.

Choice B is correct. \(10 percent sign\) of a quantity means \(10/100\) times the quantity. Therefore, \(10 percent sign\) of \(370\) can be represented as \(10/100(370)\), which is equivalent to \(0.10(370)\), or \(37\). Therefore, \(10 percent sign\) of \(370\) is \(37\). Choice A is incorrect. This is \(10 percent sign\) of \(270\), not \(10 percent sign\) of \(370\). Choice C is incorrect. This is \(90 percent sign\) of \(370\), not \(10 percent sign\) of \(370\). Choice D is incorrect. This is \(370 -10\), not \(10 percent sign\) of \(370\).

Choice A is correct. It’s given that a scientist studying the life cycle of dragonflies counted the number of dragonflies in a certain habitat each day for \(46\) days. It’s also given that on February \(15\), there were \(99\) dragonflies in the habitat and that the percent increase in the number of dragonflies in the habitat from January \(1\) to February \(15\) was \(12.50 percent sign\). This can be represented by the equation\(99 =(1 + 12.50/100)x\), where \(x\) represents the number of dragonflies in the habitat on January \(1\). This equation can be rewritten as \(99 = 1.125 x\). Dividing both sides of this equation by \(1.125\) yields \(88 = x\). Therefore, there were \(88\) dragonflies in the habitat on January \(1\). Choice B is incorrect and may result from conceptual or calculation errors. Choice C is incorrect and may result from conceptual or calculation errors. Choice D is incorrect and may result from conceptual or calculation errors.

The correct answer is \(4.06\). It's given that the retail price is \(290 percent sign\) of the wholesale price of \($ 7.00\). Thus, the retail price is \($ 7.00(290/100)\), which is equivalent to \($ 7.00(2.9)\), or \($ 20.30\). It's also given that the discounted price is \(80 percent sign\) off the retail price. Thus, the discounted price is \($ 20.30(1 -80/100)\), which is equivalent to \($ 20.30(0.20)\), or \($ 4.06\).

The correct answer is \(27\). It’s given that \(6 percent sign\) of the \(450\) tiles in a box are black. Therefore, the number of black tiles in the box can be calculated by multiplying the number of tiles in the box by \(6/100\), which is equivalent to \(450(6/100)\), or \(27\).

Choice D is correct. It’s given that the original price of the jacket has been reduced by 20%. Multiplying the original price, $50, by 20% gives the amount, in dollars, that the price of the jacket is reduced by: . Subtracting this value from the original price results in the sale price of the jacket:, or $40.Choices A, B, and C are incorrect and may result from a conceptual or calculation error.

The correct answer is \(52\). It's given that \(13\) is \(p percent sign\) of \(25\). It follows that \(13/25 = p/100\). Multiplying both sides of this equation by \(100\) gives \(52 = p\). Therefore, the value of \(p\) is \(52\).

The correct answer is \(161\). For a number to be \(40 percent sign\) greater than \(115\), it follows that the number is \((100 percent sign of 115)+(40 percent sign of 115)\), which can be written as \(100/100(115)+ 40/100(115)\). This expression is equivalent to \(1(115)+ 0.4(115)\), or \(1.4(115)\), which is equal to \(161\). Therefore, \(161\) is \(40 percent sign\) greater than \(115\).

The correct answer is \(79\). Let \(x\) represent the number of zebras in the population in \(2014\) and let \(y\) represent the number of zebras in the population in \(2018\). It’s given that the number of zebras in this population in \(2018\) was \(1.27\) times the number of zebras in this population in \(2014\). It follows that the equation \(y = 1.27 x\) represents this situation. Dividing both sides of this equation by \(1.27\) yields \(y/1.27 = x\), or \((1/1.27)y = x\). Therefore, the number of zebras in this population in \(2014\) is \(1/1.27\) times the number of zebras in this population in \(2018\). If the number of zebras in this population in \(2014\) is \(p percent sign\) of the number of zebras in this population in \(2018\), then \(x = p/100 y\). It follows that \(1/1.27 = p/100\), or \(100/1.27 = p\), which means \(p\) is approximately equal to \(78.74\). Therefore, the value of \(p\), to the nearest whole number, is \(79\).

The correct answer is \(450\). Let \(x\) represent the number that \(432\) is \(96 percent sign\) of. This can be written as \((96/100)x = 432\), or \(0.96 x = 432\). Dividing both sides of this equation by \(0.96\) yields \(x = 450\). Therefore, \(432\) is \(96 percent sign\) of \(450\).

The correct answer is \(47.25\). It’s given that the number \(w\) is \(110 percent sign\) greater than the number \(z\). It follows that \(w =(1 + 110/100)z\), or \(w = 2.1 z\). It’s also given that the number \(z\) is \(55 percent sign\) less than \(50\). It follows that \(z =(1 -55/100)(50)\), or \(z = 0.45(50)\), which yields \(z = 22.5\). Substituting \(22.5\) for \(z\) in the equation \(w = 2.1 z\) yields \(w = 2.1(22.5)\), which is equivalent to \(w = 47.25\). Therefore, the value of \(w\) is \(47.25\). Note that 47.25 and 189/4 are examples of ways to enter a correct answer.

Choice C is correct. If the bill is $78.20, 15% of the bill can be found by multiplying 78.20 by the decimal conversion of 15%, 78.20 × 0.15 = $11.73. The exact amount $11.73 is closest in value to $12.00.Choices A, B, and D are incorrect and may be the result of errors when calculating 15% of the total $78.20.

Choice D is correct. The decimal equivalent of 29% is 0.29. It’s given that the 113,000 occupational therapy jobs in the United States in 2012 are projected to increase by 29% by 2022. Increasing 113,000 by 29% can be expressed as, or . Evaluating this expression yields 145,770. The closest number is 145,800 in choice D.Choice A is incorrect and may result from increasing 113,000 by 2,900 instead of by 29%. Choice B is incorrect and may result from increasing 113,000 by 2.9% instead of by 29%. Choice C is incorrect and may result from increasing 113,000 by 29,000 instead of by 29%.

Choice B is correct. It's given that \(140\) is \(p percent sign\) greater than \(10\). It follows that \(140 = 10 +(p/100)10\), which is equivalent to \(140 = 10 + 10/100 p\), or \(140 = 10 + 0.1 p\). Subtracting \(10\) from each side of this equation yields \(130 = 0.1 p\). Dividing each side of this equation by \(0.1\) yields \(1,300 = p\), or \(p = 1,300\). Choice A is incorrect. This would be the value of \(p\) if \(140\) were \(p percent sign\) of \(10\), not \(p percent sign\) greater than \(10\). Choice C is incorrect and may result from conceptual or calculation errors. Choice D is incorrect and may result from conceptual or calculation errors.

Choice B is correct. The percent of Employee A’s calls that received a rating of 1 star is the number of Employee A’s 1-star calls divided by the total number of Employee A’s calls. This quotient,, is approximately equal to, or . To the nearest whole percent, this is 11%. Choice A is incorrect. This is the percent of all calls taken by Employee A that received a rating of 1 star. Choice C is incorrect and may result from a conceptual error. For example, 16 is the number, not the percent, of calls taken by Employee A that received a rating of 1 star. Choice D is incorrect. This is the percent of all calls that received a rating of 1 star that were taken by Employee A.

The correct answer is \(17\). It's given that there are \(170\) blocks in a container, and of these blocks, \(10 percent sign\) are green. Since \(10 percent sign\) can be rewritten as \(10/100\), or \(0.1\), the number of green blocks in the container is \(0.1(170)\), or \(17\).

The correct answer is \(9.87\). It’s given that the number \(a\) is \(110 percent sign\) greater than the number \(b\). It follows that \(a =(1 + 110/100)b\), or \(a = 2.1 b\). It’s also given that the number \(b\) is \(90 percent sign\) less than \(47\). It follows that \(b =(1 -90/100)(47)\), or \(b = 0.1(47)\), which yields \(b = 4.7\). Substituting \(4.7\) for \(b\) in the equation \(a = 2.1 b\) yields \(a = 2.1(4.7)\), which is equivalent to \(a = 9.87\). Therefore, the value of \(a\) is \(9.87\).

Choice B is correct. It's given that \(20 percent sign\) of the original number of marbles were removed from the group. Let \(x\) represent the original number of marbles in the group. It follows that \(20/100 x\), or \(0.20 x\), marbles were removed from the group. Therefore, \(x -0.20 x\) marbles remained in the group. It's also given that \(360\) marbles remained in the group. Thus, \(x -0.20 x = 360\), or \(0.80 x = 360\). Dividing both sides of this equation by \(0.80\) yields \(x = 450\). Substituting \(450\) for \(x\) in the expression \(0.20 x\) yields \(0.20(450)\), or \(90\). Therefore, \(90\) marbles were removed from the group. Choice A is incorrect. This is \(20 percent sign\) of the remaining number of marbles. Choice C is incorrect. This is the original number of marbles, not the number of marbles that were removed. Choice D is incorrect and may result from conceptual or calculation errors.

Choice D is correct. It’s given that the result of increasing the quantity \(x\) by \(1,800 percent sign\) is \(684\). It follows that \(x +(1,800/100)x = 684\), which is equivalent to \(x + 18 x = 684\), or \(19 x = 684\). Dividing each side of this equation by \(19\) yields \(x = 36\). Therefore, the value of \(x\) is \(36\). Choice A is incorrect. The result of increasing the quantity \(12,996\) by \(1,800 percent sign\) is \(246,924\), not \(684\). Choice B is incorrect. The result of increasing the quantity \(12,312\) by \(1,800 percent sign\) is \(233,928\), not \(684\). Choice C is incorrect. The result of increasing the quantity \(38\) by \(1,800 percent sign\) is \(722\), not \(684\).

The correct answer is \(.48\). It's given that the number \(a\) is \(70 percent sign\) less than the positive number \(b\). Therefore, \(a =(1 -70/100)b\), which is equivalent to \(a =(1 -0.70)b\), or \(a = 0.30 b\). It's also given that the number \(c\) is \(60 percent sign\) greater than \(a\). Therefore, \(c =(1 + 60/100)a\), which is equivalent to \(c =(1 + 0.60)a\), or \(c = 1.60 a\). Since \(a = 0.30 b\), substituting \(0.30 b\) for \(a\) in the equation \(c = 1.60 a\) yields \(c = 1.60(0.30 b)\), or \(c = 0.48 b\). Thus, \(c\) is \(0.48\) times \(b\). Note that .48 and 12/25 are examples of ways to enter a correct answer.

Choice D is correct. There are 60 minutes in one hour, so an 8-hour workday has minutes. To calculate 15% of 480, multiply 0.15 by 480: . Therefore, Lani spent 72 minutes of her workday in meetings.Choice A is incorrect because 1.2 is 15% of 8, which gives the time Lani spent of her workday in meetings in hours, not minutes. Choices B and C are incorrect and may be the result of computation errors.

Choice A is correct. The percentage of adults aged 18–50 who responded “warm” is, or about 58.9%. The percentage of adults aged 51 years or greater who responded “warm” is, or about 55.0%. The difference between 58.9% and 55.0% is 3.9%. Of the answer choices, 4% is closest to this number.Choices B, C, and D are incorrect and may result from calculation errors.

Choice B is correct. The number of harvested potatoes Isabel saved to plant next year can be calculated by multiplying the total number of potatoes Isabel harvested, \(760\), by the proportion of potatoes she saved. Since she saved \(10 percent sign\) of the potatoes she harvested, the proportion of potatoes she saved is \(10/100\), or \(0.1\). Multiplying \(760\) by this proportion gives \(760(0.1)\), or \(76\), potatoes that she saved to plant next year. Choice A is incorrect and may result from conceptual or calculation errors. Choice C is incorrect and may result from conceptual or calculation errors. Choice D is incorrect and may result from conceptual or calculation errors.

The correct answer is \(10\). It’s given that the amount of Hanna's food order was \($ 50\) and that Hanna gave a tip of \(20 percent sign\) of the amount of the bill. \(20 percent sign\) of \(50\) can be calculated as \((20/100)(50)\), which yields \(1000/100\), or \(10\). Therefore, the amount, in dollars, of the tip Hanna gave is \(10\).

Choice C is correct. It's given that during an assembly, students occuπed \(50 percent sign\) of the \(840\) seats in the school auditorium. Therefore, the number of seats that the students occuπed during this assembly can be calculated by multiplying the number of seats in the school auditorium by \(50/100\). Thus, the students occuπed \(840(50/100)\), or \(420\), seats during this assembly. Choice A is incorrect. This is approximately \(3 percent sign\), not \(50 percent sign\), of \(840\). Choice B is incorrect. This is approximately \(6 percent sign\), not \(50 percent sign\), of \(840\). Choice D is incorrect. This is approximately \(94 percent sign\), not \(50 percent sign\), of \(840\).

Choice D is correct. To find the percent increase of the customer’s water bill, the absolute increase of the bill, in dollars, is divided by the original amount of the bill, and the result is multiplied by 100%, as follows: ; .Choice A is incorrect. This choice is the difference rounded to the nearest tenth, which is the (absolute) increase of the bill’s amount, not its percent increase. Choice B is incorrect and may be the result of some calculation errors. Choice C is incorrect and is the result of dividing the difference between the two bill amounts by the new bill amount instead of the original bill amount.

Choice A is correct. It's given that the result of increasing the quantity \(x\) by \(400 percent sign\) is \(60\). This can be written as \(x +(400/100)x = 60\), which is equivalent to \(x + 4 x = 60\), or \(5 x = 60\). Dividing each side of this equation by \(5\) yields \(x = 12\). Therefore, the value of \(x\) is \(12\). Choice B is incorrect. The result of increasing the quantity \(15\) by \(400 percent sign\) is \(75\), not \(60\). Choice C is incorrect. The result of increasing the quantity \(240\) by \(400 percent sign\) is \(1,200\), not \(60\). Choice D is incorrect. The result of increasing the quantity \(340\) by \(400 percent sign\) is \(1,700\), not \(60\).

Choice C is correct. The decimal equivalent of 6% is 0.06. Since increasing the number n by 6% yields the number 318, this situation can be represented by the equation, or . Dividing both sides of this equation by 1.06 yields .Choice A is incorrect. This is the result when n is increased by 60%, not by 6%. Choice B is incorrect. This is the approximate result of decreasing 318 by 6%. Choice D is incorrect. This is the approximate result of increasing 318 by 6%.

Choice B is correct. It’s given that in a group, \(40 percent sign\) of the items are red. It follows that the number of red items in the group can be represented by \(0 period 4 x\), where \(x\) represents the total number of items in the group. It’s also given that of all the red items in the group, \(30 percent sign\) also have stripes. It follows that the number of items in the group that are red and have stripes can be represented by \(0 period 3(0 period 4 x)\), or \(0 period 12 x\). The expression \(0 period 12 x\) represents \(12 percent sign\) of \(x\). Since \(x\) represents the total number of items in the group, it follows that \(12 percent sign\) of the items in the group are red and have stripes. Choice A is incorrect and may result from subtracting \(30 percent sign\) from \(40 percent sign\) rather than calculating \(30 percent sign\) of \(40 percent sign\). Choice C is incorrect and may result from adding \(30 percent sign\) and \(40 percent sign\) rather than calculating \(30 percent sign\) of \(40 percent sign\). Choice D is incorrect and may result from calculating the percentage that \(30 percent sign\) is of \(40 percent sign\) rather than calculating \(30 percent sign\) of \(40 percent sign\).

Choice C is correct. Let \(x\) be the 2015 population of Greenville. It's given that the population increased by \(7 percent sign\) from 2015 to 2016. The increase in population can be written as \((0.07)x\). The 2016 population of Greenville is given as the sum of the 2015 population of Greenville and the increase in population from 2015 to 2016. This can be rewritten as \(x +(0.07)x\), or \(1.07 x\). Therefore, the value of \(k\) is \(1.07\). Choice A is incorrect. This is the percent, represented as a decimal, that the population increased from 2015 to 2016, not the value of \(k\). Choice B is incorrect and may result from conceptual or calculation errors. Choice D is incorrect. This is the value of \(k\) if the population increased by \(70 percent sign\), not \(7 percent sign\), from 2015 to 2016.

Choice B is correct. It's given that there are \(250\) trees in a park and of these trees, \(6 percent sign\) are birch trees. The number of birch trees in the park can be calculated by multiplying the number of trees in the park by \(6/100\). Therefore, the number of birch trees in the park is \(250(6/100)\), or \(15\). Choice A is incorrect. This is the percentage of trees in the park that are birch trees, not the number of birch trees in the park. Choice C is incorrect. This is \(30 percent sign\), not \(6 percent sign\), of \(250\). Choice D is incorrect and may result from conceptual or calculation errors.

Choice D is correct. The proportion of the beads that are silver can be written as \(828,000/900,000\), or \(0.92\). Therefore, the percentage of the beads that are silver is \(0.92(100)\), or \(92 percent sign\). Choice A is incorrect. This is the percentage of the beads that are not silver. Choice B is incorrect and may result from conceptual or calculation errors. Choice C is incorrect and may result from conceptual or calculation errors.

Choice C is correct. According to the table, 63% of survey respondents get most of their medical information from a doctor and 13% get most of their medical information from the Internet. Therefore, 76% of the 1,200 survey respondents get their information from either a doctor or the Internet, and 76% of 1,200 is 912.Choices A, B, and D are incorrect. According to the table, 76% of survey respondents get their information from either a doctor or the Internet. Choice A is incorrect because 865 is about 72% (the percent of survey respondents who get most of their medical information from a doctor or from magazines/brochures), not 76%, of 1,200. Choice B is incorrect because 887 is about 74%, not 76%, of 1,200. Choice D is incorrect because 926 is about 77%, not 76%, of 1,200.

Choice D is correct. It's given that the length of the base of the parallelogram is \(89 percent sign\) of the height of the parallelogram. Since \(h\) is the height of the parallelogram, it follows that the length of the base of the parallelogram can be represented by the expression \(89/100 h\), or \(0.89 h\). Choice A is incorrect. This expression represents \(8,900 percent sign\), not \(89 percent sign\), of the height of the parallelogram. Choice B is incorrect. This expression represents \(8.9 percent sign\), not \(89 percent sign\), of the height of the parallelogram. Choice C is incorrect. This expression represents \(890 percent sign\), not \(89 percent sign\), of the height of the parallelogram.

Choice B is correct. It’s given that the value of \(z\) is \(1.13\) times \(100\). This can be written as \(z =(1.13)(100)\), which is equivalent to \(z =(1 + 0.13)(100)\), or \(z =(1 + 13/100)(100)\). It follows that the value of \(z\) is \(100 percent sign\) of \(100\) plus \(13 percent sign\) of \(100\). Therefore, the value of \(z\) is \(13 percent sign\) greater than \(100\). Choice A is incorrect. This gives a value of \(z\) that is \(1.113\), not \(1.13\), times \(100\). Choice C is incorrect. This gives a value of \(z\) that is \(2.30\), not \(1.13\), times \(100\). Choice D is incorrect. This gives a value of \(z\) that is \(3.13\), not \(1.13\), times \(100\).

The correct answer is \(600\). It’s given that \(210\) is \(p percent sign\) greater than \(30\). It follows that \(210 =(1 + p/100)(30)\). Dividing both sides of this equation by \(30\) yields \(7 = 1 + p/100\). Subtracting \(1\) from both sides of this equation yields \(6 = p/100\). Multiplying both sides of this equation by \(100\) yields \(p = 600\). Therefore, the value of \(p\) is \(600\).

The correct answer is \(.54\). It's given that the number \(a\) is \(70 percent sign\) less than the positive number \(b\). Therefore, \(a =(1 -70/100)b\), which is equivalent to \(a =(1 -0.70)b\), or \(a = 0.30 b\). It's also given that the number \(c\) is \(80 percent sign\) greater than \(a\). Therefore, \(c =(1 + 80/100)a\), which is equivalent to \(c =(1 + 0.80)a\), or \(c = 1.80 a\). Since \(a = 0.30 b\), substituting \(0.30 b\) for \(a\) in the equation \(c = 1.80 a\) yields \(c = 1.80(0.30 b)\), or \(c = 0.54 b\). Thus, \(c\) is \(0.54\) times \(b\). Note that .54 and 27/50 are examples of ways to enter a correct answer.

Choice D is correct. It’s given that in \(2008\) Zinah earned \(14 percent sign\) more than in \(2007\). Let \(h\) represent the amount Zinah earned in \(2007\) and let \(j\) represent the amount that Zinah earned in \(2008\). This situation can be represented by the equation \(j =(1 + 14/100)h\), or \(j = 1.14 h\). It’s also given that in \(2009\) Zinah earned \(4 percent sign\) more than in \(2008\). Let \(k\) represent the amount Zinah earned in \(2009\). This situation can be represented by the equation \(k =(1 + 4/100)j\), or \(k = 1.04 j\). Substituting \(1.14 h\) for \(j\) in the equation \(k = 1.04 j\) yields \(k =(1.04)(1.14 h)\), or \(k = 1.1856 h\). If Zinah earned \(y\) times as much in \(2009\) as in \(2007\), then the value of \(y\) is \(1.1856\). Choice A is incorrect and may result from conceptual or calculation errors. Choice B is incorrect and may result from conceptual or calculation errors. Choice C is incorrect and may result from conceptual or calculation errors.

Choice C is correct. It’s given that the number k is 36% greater than 50. Therefore, the value of k is the number 50 plus 36% of 50. This can be rewritten as . Multiplying the terms yields 18, so, or . It’s also given that k is the product of 50 and r, which can be rewritten as . Substituting 68 for k yields . Dividing both sides of this equation by 50 yields .Choice A is incorrect. This is the percentage that k is greater than 50. Choice B is incorrect and may result from a calculation error. Choice D is incorrect. This would be the value of r if k were 36% of 50, instead of 36% greater than 50.

Choice B is correct. \(20 percent sign\) of \(440\) can be calculated as \((20/100)(440)\), which is equivalent to \(8,800/100\), or \(88\). Choice A is incorrect. This is \(10 percent sign\), not \(20 percent sign\), of \(440\). Choice C is incorrect. This is \(200 percent sign\), not \(20 percent sign\), of \(440\). Choice D is incorrect. This is \(400 percent sign\), not \(20 percent sign\), of \(440\).

Choice A is correct. Increasing the positive quantity x by 9% is the result of adding 9% of x to x. 9% of x can be represented algebraically as, or . Adding this expression to x yields, or .Choice B is incorrect. This represents 9% of x. Choice C is incorrect. This represents increasing x by 9, not by 9%. Choice D is incorrect. This represents increasing x by 0.09, not by 9%.

Choice B is correct. \(10 percent sign\) of a quantity means \(10/100\) times the quantity. Therefore, \(10 percent sign\) of \(470\) can be represented as \(10/100(470)\), which is equivalent to \(0.10(470)\), or \(47\). Therefore, \(10 percent sign\) of \(470\) is \(47\). Choice A is incorrect. This is \(10 percent sign\) of \(370\), not \(10 percent sign\) of \(470\). Choice C is incorrect. This is \(90 percent sign\) of \(470\), not \(10 percent sign\) of \(470\). Choice D is incorrect. This is \(470 -10\), not \(10 percent sign\) of \(470\).

Choice C is correct. It's given that the population of City A increased by \(7 percent sign\) from \(20 15\) to \(20 16\). Therefore, the population of City A in \(20 16\) includes \(100 percent sign\) of the population of City A in \(20 15\) plus an additional \(7 percent sign\) of the population of City A in \(20 15\). This means that the population of City A in \(20 16\) is \(107 percent sign\) of the population in \(20 15\). Thus, the population of City A in \(20 16\) is \(107/100\), or \(1.07\), times the \(20 15\) population. Therefore, the value of \(k\) is \(1.07\). Choice A is incorrect. This would be the value of \(k\) if the population in \(20 16\) was \(7 percent sign\) of the population in \(20 15\). Choice B is incorrect. This would be the value of \(k\) if the population in \(20 16\) was \(70 percent sign\) of the population in \(20 15\). Choice D is incorrect. This would be the value of \(k\) if the population increased by \(70 percent sign\), not \(7 percent sign\), from \(20 15\) to \(20 16\).

Choice A is correct. The result of increasing a positive quantity \(w\) by \(x percent sign\) can be represented by the expression \((1 + x/100)w\). Therefore, the result of increasing a positive quantity \(w\) by \(43 percent sign\) can be found by substituting \(43\) for \(x\) in the expression \((1 + x/100)w\), which gives \((1 + 43/100)w\), or \(1.43 w\). Thus, the expression \(1.43 w\) represents the result of increasing a positive quantity \(w\) by \(43 percent sign\). Choice B is incorrect. This is the result of decreasing a positive quantity \(w\) by \(43 percent sign\). Choice C is incorrect. This is the result of increasing a positive quantity \(w\) by \(4,200 percent sign\). Choice D is incorrect. This is the result of decreasing a positive quantity \(w\) by \(57 percent sign\).

Choice B is correct. It's given that the number \(b\) is \(80 percent sign\) less than \(24\). It follows that \(b\) is equal to \(24\) minus \(80 percent sign\) of \(24\), which can be written as \(b = 24 -(80/100)24\). This is equivalent to \(b = 24 -0.8(24)\), or \(b = 4.8\). It's also given that the number \(a\) is \(190 percent sign\) greater than the number \(b\). It follows that \(a\) is equal to \(b\) plus \(190 percent sign\) of \(b\). Since \(b = 4.8\), this can be written as \(a = 4.8 +(190/100)4.8\). This is equivalent to \(a = 4.8 + 1.9(4.8)\), or \(a = 13.92\). Choice A is incorrect. This would be the value of \(a\) if \(a\) were \(190 percent sign\) of \(b\), not \(190 percent sign\) greater than \(b\). Choice C is incorrect. This is \((190 -80)percent sign\) of \(24\). Choice D is incorrect. This would be the value of \(a\) if \(b\) were \(80 percent sign\) of \(24\), not \(80 percent sign\) less than \(24\), and \(a\) were \(190 percent sign\) of \(b\), not \(190 percent sign\) greater than \(b\).

Choice A is correct. \(23 percent sign\) of \(100\) can be calculated by multiplying \(23/100\) by \(100\), which yields \((23/100)100\), or \(23\). Choice B is incorrect. This is \(46 percent sign\), not \(23 percent sign\), of \(100\). Choice C is incorrect. This is \(23 percent sign\) less than \(100\), not \(23 percent sign\) of \(100\). Choice D is incorrect. This is \(23 percent sign\) greater than \(100\), not \(23 percent sign\) of \(100\).

The correct answer is \(56\). It’s given that Cedric had \(35\) plants in his garden last year and that the number of plants in Cedric's garden this year is \(60 percent sign\) greater than the number of plants in his garden last year. It follows that the number of plants in Cedric’s garden this year is \(35\) plus \(60 percent sign\) of \(35\), which is equal to \(35 + 35(60/100)\), or \(35 + 35(0.6)\). This expression is equivalent to \(35 + 21\), or \(56\). Therefore, Cedric has \(56\) plants in his garden this year.

Choice D is correct. Let \(x\) represent the number that \(21\) is \(21 percent sign\) of. It follows that \(21/x = 21/100\). Multiplying each side of this equation by \(x\) yields \(21 = 21 x/100\). Multiplying each side of this equation by \(100\) yields \(2,100 = 21 x\). Dividing each side of this equation by \(21\) yields \(100 = x\). Therefore, \(21\) is \(21 percent sign\) of \(100\). Choice A is incorrect. \(21 percent sign\) of \(0\) is \(0\), not \(21\). Choice B is incorrect. \(21 percent sign\) of \(1\) is \(0.21\), not \(21\). Choice C is incorrect. \(21 percent sign\) of \(42\) is \(8.82\), not \(21\).

Choice B is correct. It's given that the number of students enrolled in an interior design program this year is \(147 percent sign\) of last year's number, which is \(200\). \(147 percent sign\) of \(200\) can be expressed as \((147/100)(200)\), or \((1.47)(200)\), which is equivalent to \(294\). Therefore, \(294\) students are enrolled in the interior design program this year. Choice A is incorrect and may result from conceptual or calculation errors. Choice C is incorrect and may result from conceptual or calculation errors. Choice D is incorrect and may result from conceptual or calculation errors.

Choice C is correct. Of the 1,300,000 units sold during the first month, 15% were returned, so units were returned during the first month. If the units were sold and returned at the same rate for the next 5 months, then a total of smartphone units were returned during the 6-month period.Choice A is incorrect. This is the number of units that were returned in 1 month. Choice B is incorrect. This is the number of units that were returned in 5 months. Choice D is incorrect. This is the number of units sold and not returned during the first 6 months.

Choice D is correct. It is given that Contestant 2 earned 70% of the votes cast using social media and 40% of the votes cast using a text message. Based on this information, viewers voting by social media were more likely to prefer Contestant 2 than were viewers voting by text message.Choices A, B, and C are incorrect. There is not enough information about the viewers to reach these conclusions.

Choice C is correct. The total cost to purchase the shirt is the $20 cost of the shirt plus the 5% sales tax. The value of the 5% sales tax on the $20 shirt is equivalent to, or $1. Therefore, the total cost to purchase the shirt is, or $21.Choice A is incorrect and may result from neglecting to multiply by $20 when finding the value of the sales tax. Choice B is incorrect and may result from dividing by 10, instead of by 100, and then neglecting to multiply by $20 when finding the sales tax. Choice D is incorrect and may result from interpreting the sales tax of 5% as $5.

Choice A is correct. Let \(x\) represent the percentage of \(300\) that is \(75\). This can be written as \(x/100(300)= 75\), or \(3 x = 75\). Dividing both sides of this equation by \(3\) yields \(x = 25\). Therefore, \(25 percent sign\) of \(300\) is \(75\). Choice B is incorrect. \(50 percent sign\) of \(300\) is \(150\), not \(75\). Choice C is incorrect. \(75 percent sign\) of \(300\) is \(225\), not \(75\). Choice D is incorrect. \(225 percent sign\) of \(300\) is \(675\), not \(75\).

Choice D is correct. It's given that \(a\) is \(230 percent sign\) of \(b\). It follows that \(a = 230/100 b\). It's also given that \(a\) is \(60 percent sign\) of \(c\). It follows that \(a = 60/100 c\). Since \(a = 230/100 b\) and \(a = 60/100 c\), it follows that \(230/100 b = 60/100 c\). Multiplying each side of this equation by \(100/60\) yields \(23/6 b = c\). If \(c\) is \(p percent sign\) of \(b\), then \(c = p/100 b\). It follows that \(23/6 = p/100\). Multiplying each side of this equation by \(100\) yields \(2,300/6 = p\). It follows that the value of \(p\) is approximately \(383.33\). Therefore, of the given choices, \(383\) is closest to the value of \(p\). Choice A is incorrect. This is closest to the value of \(p\) if \(b\) is \(230 percent sign\) of \(a\), rather than if \(a\) is \(230 percent sign\) of \(b\), and if \(b\) is \(p percent sign\) of \(c\), rather than if \(c\) is \(p percent sign\) of \(b\). Choice B is incorrect. This is closest to the value of \(p\) if \(a\) is \(230 percent sign\) greater than \(b\), rather than \(230 percent sign\) of \(b\). Choice C is incorrect. This is closest to the value of \(p\) if \(c\) is \(p percent sign\) greater than \(b\), rather than \(p percent sign\) of \(b\).

The correct answer is 75. The percent decrease between two values is found by dividing the difference between the two values by the original value and multiplying by 100. The line graph shows that the annual snowfall in 2003 was 40 inches, and the annual snowfall in 2007 was 10 inches. Therefore, the percent decrease in the annual snowfall from 2003 to 2007 is, or 75. It’s given that this is equivalent to, so the value of p is 75.

Choice B is correct. The percent increase from 2012 to 2013 was, or 5%. Since the percent increase from 2012 to 2013 was estimated to be double the percent increase from 2013 to 2014, the percent increase from 2013 to 2014 was expected to be 2.5%. Therefore, the number of subscriptions sold in 2014 is expected to be the number of subscriptions sold in 2013 multiplied by, or .Choice A is incorrect and is the result of adding half of the value of the increase from 2012 to 2013 to the 2013 result. Choice C is incorrect and is the result adding twice the value of the increase from 2012 to 2013 to the 2013 result. Choice D is incorrect and is the result of interpreting the percent increase from 2013 to 2014 as double the percent increase from 2012 to 2013.

Choice B is correct. According to the table, the minimum wage in 1960 was $1.00 per hour, and in 1970 it was $1.60 per hour. The percentage change is therefore .Choice A is incorrect and may result from averaging the two wages before calculating the percentage change. Choice C is incorrect. This is the 1960 wage expressed as a percentage of the 1970 wage, not the percentage change between the two. Choice D is incorrect and may result from a calculation error.

Choice D is correct. It’s given that \(a\) is \(2,241 percent sign\) of the sum of \(b\) and \(c\). This can be represented by the equation \(a =(2,241/100)(b + c)\), or \(a = 22.41(b + c)\). It’s also given that that \(b\) is \(83 percent sign\) of \(c\). This can be represented by the equation \(b =(83/100)c\), or \(b = 0.83 c\). Dividing both sides of this equation by \(0.83\) yields \(b/0.83 = c\). Substituting \(b/0.83\) for \(c\) in the equation \(a = 22.41(b + c)\) yields \(a = 22.41(b + b/0.83)\), or \(a = 22.41(1.83 b/0.83)\), which is equivalent to \(a = 41.0103 b/0.83\), or \(a = 49.41 b\). This equation is equivalent to \(a =(4,941/100)b\); therefore, \(a\) is \(4,941 percent sign\) of \(b\). Choice A is incorrect and may result from conceptual or calculation errors. Choice B is incorrect and may result from conceptual or calculation errors. Choice C is incorrect and may result from conceptual or calculation errors.

Choice D is correct. The proportion of the paper clips that are size large can be written as \(234,000/300,000\), or \(0.78\). Therefore, the percentage of the paper clips that are size large is \(0.78(100)\), or \(78 percent sign\). Choice A is incorrect. This is the percentage of the paper clips that are not size large. Choice B is incorrect and may result from conceptual or calculation errors. Choice C is incorrect and may result from conceptual or calculation errors.

The correct answer is \(110\). Let \(x\) represent the total number of items in the sample. It’s given that \(80 percent sign\) of the items are faulty and that there are \(88\) faulty items in the sample. Therefore, \(80 percent sign\) of \(x\) is \(88\). Since \(80 percent sign\) can be rewritten as \(80/100\), it follows that \(80/100 x = 88\). Multiplying both sides of this equation by \(100\) yields \(80 x = 8,800\). Dividing both sides of this equation by \(80\) yields \(x = 110\). Therefore, there are \(110\) total items in the sample.

The correct answer is 8. The 6 students represent of those invited to join the committee. If x people were invited to join the committee, then . Thus, there were people invited to join the committee. It follows that there were teachers and school and district administrators invited to join the committee. Therefore, there were 8 more teachers than school and district administrators invited to join the committee.

Choice D is correct. Let \(n percent sign\) represent the percent by which the positive quantity \(x\) is decreased to result in \(0.35 x\). The value of \(n\) can be found by solving the equation \(x -(n/100)x = 0.35 x\). Since \(x\) is a common factor of each of the terms on the left-hand side of this equation, the equation can be rewritten as \(x(1 -n/100)= 0.35 x\). Dividing each side of this equation by \(x\) yields \(1 -n/100 = 0.35\). Multiplying each side of this equation by \(100\) yields \(100 -n = 35\). Subtracting \(100\) from each side of this equation yields \(-n = -65\). Dividing each side of this equation by \(-1\) yields \(n = 65\). Therefore, the expression \(0.35 x\) represents the result of decreasing the positive quantity \(x\) by \(65 percent sign\). Choice A is incorrect. Decreasing the quantity \(x\) by \(3.5 percent sign\) yields \(x -0.035 x\), or \(0.965 x\), not \(0.35 x\). Choice B is incorrect. Decreasing the quantity \(x\) by \(35 percent sign\) yields \(x -0.35 x\), or \(0.65 x\), not \(0.35 x\). Choice C is incorrect. Decreasing the quantity \(x\) by \(6.5 percent sign\) yields \(x -0.065 x\), or \(0.935 x\), not \(0.35 x\).

Choice B is correct. It’s given that each serving of Crunchy Grain cereal provides 5% of an adult’s daily allowance of potassium, so x servings would provide x times 5%. The percentage of an adult’s daily allowance of potassium, p, is 5 times the number of servings, x. Therefore, the percentage of an adult’s daily allowance of potassium can be expressed as .Choices A, C, and D are incorrect and may result from incorrectly converting 5% to its decimal equivalent, which isn’t necessary since p is expressed as a percentage. Additionally, choices C and D are incorrect because the context should be represented by a linear relationship, not by an exponential relationship.