SAT - Free Full Practice Tests and Questions by Category <- SAT Math Full Practice Test <- SAT Ratios, Rates, Proportional relationships, and Units type Practice Questions SAT Ratios, Rates, Proportional relationships, and Units type Practice Questions Share Quiz Get Embed Code Copy the code below to embed this quiz on your website: <iframe src="https://tutorone.ca/practice-test/?embed=true" width="100%" height="800" style="border: none; max-width: 100%;" data-source="tutorone" allowfullscreen></iframe> Copy Code 1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859606162636465666768697071727374757677787980818283 SAT Ratios, Rates, Proportional relationships, and Units type Practice Questions SAT Ratios, Rates, Proportional relationships, and Units type Practice Questions 1 / 83 How many meters are equivalent to \(2,300\) centimeters? \((100 centimeters = 1 meter)\) A) \(0.043\) B) \(23\) C) \(2,400\) D) \(230,000\) Choice B is correct. It's given that \(100\) centimeters is equal to \(1\) meter. Therefore, \(2,300\) centimeters is equivalent to \((2,300 centimeters)(1 meter/100 centimeters)\), or \(23\) meters. Choice A is incorrect. \(0.043\) meters is equivalent to \(4.3\), not \(2,300\), centimeters. Choice C is incorrect. \(2,400\) meters is equivalent to \(240,000\), not \(2,300\), centimeters. Choice D is incorrect. \(230,000\) meters is equivalent to \(23,000,000\), not \(2,300\), centimeters. 2 / 83 The ratio \(x\) to \(y\) is equivalent to the ratio \(12\) to \(t\). When \(x = 156\), what is the value of \(y\) in terms of \(t\)? A) \(13 t\) B) \(12 t\) C) \(144 t\) D) \(168 t\) Choice A is correct. It's given that the ratio \(x\) to \(y\) is equivalent to the ratio \(12\) to \(t\). This can be represented by \(x/y = 12/t\). Substituting \(156\) for \(x\) in this equation yields \(156/y = 12/t\). This can be rewritten as \(12 y = 156 t\). Dividing both sides of this equation by \(12\) yields \(y = 13 t\). Therefore, when \(x = 156\), the value of \(y\) in terms of \(t\) is \(13 t\). 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. 3 / 83 For a certain rectangular region, the ratio of its length to its width is \(35\) to \(10\). If the width of the rectangular region increases by \(7\) units, how must the length change to maintain this ratio? A) It must decrease by \(24.5\) units. B) It must increase by \(24.5\) units. C) It must decrease by \(7\) units. D) It must increase by \(7\) units. Choice B is correct. It’s given that the ratio of the rectangular region’s length to its width is \(35\) to \(10\). This can be written as a proportion: \(length/width = 35/10\), or \(\ell/w = 35/10\). This proportion can be rewritten as \(10 ell = 35 w\), or \(ell = 3.5 w\). If the width of the rectangular region increases by \(7\), then the length will increase by some number \(x\) in order to maintain this ratio. The value of \(x\) can be found by replacing \(ell\) with \(ell + x\) and \(w\) with \(w + 7\) in the equation, which gives \(ell + x = 3.5(w + 7)\). This equation can be rewritten using the distributive property as \(ell + x = 3.5 w + 24.5\). Since \(ell = 3.5 w\), the right-hand side of this equation can be rewritten by substituting \(ell\) for \(3.5 w\), which gives \(ell + x = ell + 24.5\), or \(x = 24.5\). Therefore, if the width of the rectangular region increases by \(7\) units, the length must increase by \(24.5\) units in order to maintain the given ratio. Choice A is incorrect. If the width of the rectangular region increases, the length must also increase, not decrease. Choice C is incorrect. If the width of the rectangular region increases, the length must also increase, not decrease. Choice D is incorrect. Since the ratio of the length to the width of the rectangular region is \(35\) to \(10\), if the width of the rectangular region increases by \(7\) units, the length would have to increase by a proportional amount, which would have to be greater than \(7\) units. 4 / 83 How many teaspoons are equivalent to \(44\) tablespoons? \((3 teaspoons = 1 tablespoon)\) A) \(47\) B) \(88\) C) \(132\) D) \(176\) Choice C is correct. It's given that \(3\) teaspoons is equivalent to \(1\) tablespoon. Therefore, \(44\) tablespoons is equivalent to \((44 tablespoons)(3 teaspoons/1 tablespoon)\), or \(132\) teaspoons. Choice A is incorrect. This is equivalent to approximately \(15.66\) tablespoons, not \(44\) tablespoons. Choice B is incorrect. This is equivalent to approximately \(29.33\) tablespoons, not \(44\) tablespoons. Choice D is incorrect. This is equivalent to approximately \(58.66\) tablespoons, not \(44\) tablespoons. 5 / 83 A certain park has an area of \(11,863,808\) square yards. What is the area, in square miles, of this park? \((1 mile = 1,760 yards)\) A) \(1.96\) B) \(3.83\) C) \(3,444.39\) D) \(6,740.8\) Choice B is correct. Since \(1\) mile is equal to \(1,760\) yards, \(1\) square mile is equal to \(1,760 ^2\), or \(3,097,600\), square yards. It’s given that the park has an area of \(11,863,808\) square yards. Therefore, the park has an area of \((11,863,808 square yards)(1 square mile/3,097,600 square yards)\), or \(11,863,808/3,097,600\) square miles. Thus, the area, in square miles, of the park is \(3.83\). Choice A is incorrect and may result from conceptual or calculation errors. Choice C is incorrect. This is the square root of the area of the park in square yards, not the area of the park in square miles. Choice D is incorrect and may result from converting \(11,863,808\) yards to miles, rather than converting \(11,863,808\) square yards to square miles. 6 / 83 How many yards are equivalent to \(1,116\) inches? \((1 yard = 36 inches)\) A) \(31\) B) 0.5 C) 1 D) 2 The correct answer is \(31\). It's given that \(1\) yard is equal to \(36\) inches. Therefore, \(1,116\) inches is equivalent to \((1,116 inches)(1 yard/36 inches)\), or \(31\) yards. 7 / 83 Anita created a batch of green paint by mixing 2 ounces of blue paint with 3 ounces of yellow paint. She must mix a second batch using the same ratio of blue and yellow paint as the first batch. If she uses 5 ounces of blue paint for the second batch, how much yellow paint should Anita use? A) Exactly 5 ounces B) 3 ounces more than the amount of yellow paint used in the first batch C) 1.5 times the amount of yellow paint used in the first batch D) 1.5 times the amount of blue paint used in the second batch Choice D is correct. It’s given that Anita used a ratio of 2 ounces of blue paint to 3 ounces of yellow paint for the first batch. For any batch of paint that uses the same ratio, the amount of yellow paint used will be, or 1.5, times the amount of blue paint used in the batch. Therefore, the amount of yellow paint Anita will use in the second batch will be 1.5 times the amount of blue paint used in the second batch.Alternate approach: It’s given that Anita used a ratio of 2 ounces of blue paint to 3 ounces of yellow paint for the first batch and that she will use 5 ounces of blue paint for the second batch. A proportion can be set up to solve for x, the amount of yellow paint she will use for the second batch: . Multiplying both sides of this equation by 3 yields, and multiplying both sides of this equation by x yields . Dividing both sides of this equation by 2 yields . Since Anita will use 7.5 ounces of yellow paint for the second batch, this is times the amount of blue paint (5 ounces) used in the second batch.Choices A, B, and C are incorrect and may result from incorrectly interpreting the ratio of blue paint to yellow paint used. 8 / 83 If, which of the following is equivalent to ? A) B) C) D) Choice A is correct. It’s given that . Multiplying both sides of this equation by 2 yields, or .Choice B is incorrect and may result from dividing, instead of multiplying, the right-hand side of the equation by 2. Choices C and D are incorrect and may result from calculation errors. 9 / 83 An insect moves at a speed of \(3 twentieths\) feet per second. What is this speed, in yards per second? \((3 feet = 1 yard)\) A) \(1 twentieth\) B) \(9 twentieths\) C) \(6\) D) \(20\) Choice A is correct. It’s given that \(3 feet = 1 yard\). It follows that a speed of \(3 twentieths\) feet per second is equivalent to \((3 twentieths feet/1 second)(1 yard/3 feet)\), which is equivalent to \((3 twentieths)(1 third)\), or \(1 twentieth\), yards per second. Choice B is incorrect. This is the speed, in feet per second, that's equivalent to \(3 twentieths\) yards per second. Choice C is incorrect. This is the speed, in yards per second, that's equivalent to \(18\), not \(3 twentieths\), feet per second. Choice D is incorrect. This is the speed, in yards per second, that's equivalent to \(60\), not \(3 twentieths\), feet per second. 10 / 83 The area of a rectangular region is increasing at a rate of \(250\) square feet per hour. Which of the following is closest to this rate in square meters per minute? \((Use 1 meter = 3.28 feet period)\) A) \(0.39\) B) \(1.27\) C) \(13.67\) D) \(23.24\) Choice A is correct. It’s given that \(1 meter = 3.28 feet\). It follows that \(1 ^2 square meter = 3.28 ^2 square feet\), or \(1 square meter = 10.7584 square feet\). Since \(1 hour = 60 minutes\), it follows that \(250\) square feet per hour is equivalent to \((250 square feet/1 hour)(1 square meter/10.7584 square feet)(1 hour/60 minutes)\), or \(250 square meters/645.504 minutes\), which is approximately \(0.3873\) square meters per minute. Of the given choices, \(0.39\) is closest to \(0.3873\). 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. 11 / 83 A competition consisted of four different events. One participant completed the first event with an average speed of \(20.300\) miles per hour. What was this average speed, in yards per hour? \((1 mile = 1,760 yards)\) A) \(\)35 B) 0.5 C) 1 D) 2 The correct answer is \(35,728\). It's given that \(1 mile = 1,760 yards\). It follows that an average speed of \(20.300\) miles per hour is equivalent to \((20.300 miles/1 hour)(1,760 yards/1 mile)\), or \(35,728\) yards per hour. 12 / 83 A product costs \(11.00\) dollars per pound. What is the cost, in dollars, for \(6\) pounds of the product? A) \(66\) B) 3 C) 5.5 D) 6 The correct answer is \(66\). It’s given that a product costs \(11.00\) dollars per pound. Therefore, the cost for \(6\) pounds of the product is \((11.00 dollars/1 pound)(6 pounds)\), which is equivalent to \(66.00\), or \(66\), dollars. 13 / 83 Makayla is planning an event in a 5,400-square-foot room. If there should be at least 8 square feet per person, what is the maximum number of people that could attend this event? A) 588 B) 675 C) 15,274 D) 43,200 Choice B is correct. It’s given that the event will be in a 5,400-square-foot room and that there should be at least 8 square feet per person. The maximum number of people that could attend the event can be found by dividing the total square feet in the room by the minimum number of square feet needed per person, which gives .Choices A and C are incorrect and may result from conceptual or computational errors. Choice D is incorrect and may result from multiplying, rather than dividing, 5,400 by 8. 14 / 83 An object travels at a constant speed of \(6\) centimeters per second. At this speed, what is the time, in seconds, that it would take for the object to travel \(24\) centimeters? A) 1 B) 2 C) \(4\) D) 2.8 The correct answer is \(4\). It’s given that the object travels at a constant speed of \(6\) centimeters per second. The speed of the object can be written as \(6 centimeters/1 second\). Let \(x\) represent the time, in seconds, it would take for the object to travel \(24\) centimeters. The value of \(x\) can be calculated by solving the equation \(6 centimeters/1 second = 24 centimeters/x seconds\), which can be written as \(6 oneths = 24/x\), or \(6 = 24/x\). Multiplying each side of this equation by \(x\) yields \(6 x = 24\). Dividing each side of this equation by \(6\) yields \(x = 4\). Therefore, it would take the object \(4\) seconds to travel \(24\) centimeters. 15 / 83 A landscaper uses a hose that puts \(88 x\) ounces of water in a bucket in \(5 y\) minutes. Which expression represents the number of ounces of water the hose puts in the bucket in \(9 y\) minutes at this rate? A) \(9 x/440\) B) \(440 x/9\) C) \(5 x/792\) D) \(792 x/5\) Choice D is correct. It’s given that a hose puts \(88 x\) ounces of water in a bucket in \(5 y\) minutes. Therefore, the rate at which the hose puts water in the bucket, in ounces per minute, can be represented by the expression \(88 x/5 y\). Let \(w\) represent the number of ounces of water the hose puts in the bucket in \(9 y\) minutes at this rate. It follows that the rate at which the hose puts water in the bucket, in ounces per minute, can be represented by the expression \(w/9 y\). The expressions \(88 x/5 y\) and \(w/9 y\) represent the same rate, so it follows that \(88 x/5 y = w/9 y\). Multiplying both sides of this equation by \(9 y\) yields \(792 x y/5 y = w\), or \(792 x/5 = w\). Therefore, the number of ounces of water the hose puts in the bucket in \(9 y\) minutes can be represented by the expression \(792 x/5\). 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. 16 / 83 For an electric field passing through a flat surface perpendicular to it, the electric flux of the electric field through the surface is the product of the electric field’s strength and the area of the surface. A certain flat surface consists of two adjacent squares, where the side length, in meters, of the larger square is \(3\) times the side length, in meters, of the smaller square. An electric field with strength \(29.00\) volts per meter passes uniformly through this surface, which is perpendicular to the electric field. If the total electric flux of the electric field through this surface is \(4,640 volts dot meters\), what is the electric flux, in \(volts dot meters\), of the electric field through the larger square? A) 1 B) 1.5 C) \(\)4 D) 2 The correct answer is \(4,176\). It’s given that the side length of the larger square is \(3\) times the side length of the smaller square. This means that the area of the larger square is \(3 ^2\), or \(9\), times the area of the smaller square. If the area of the smaller square is represented by \(x\), then the area of the larger square can be represented by \(9 x\). Therefore, the flat surface of the two adjacent squares has a total area of \(x + 9 x\), or \(10 x\). It’s given that an electric field with strength \(29.00\) volts per meter passes uniformly through this surface and the total electric flux of the electric field through this surface is \(4,640 volts dot meters\). Since it's given that the electric flux is the product of the electric field’s strength and the area of the surface, the equation \(29.00(10 x)= 4,640\), or \(290 x = 4,640\), can be used to represent this situation. Dividing each side of this equation by \(290\) yields \(x = 16\). Substituting \(16\) for \(x\) in the expression for the area of the larger square, \(9 x\), yields \(9(16)\), or \(144\), square meters. Since the area of the larger square is \(144\) square meters, the electric flux, in \(volts dot meters\), of the electric field through the larger square can be determined by multiplying the area of the larger square by the strength of the electric field. Thus, the electric flux is \((144 square meters)(29.00 volts/meter)\), or \(4,176 volts dot meters\). 17 / 83 A participant in a bicycle race completes the race with an average speed of \(24,816\) yards per hour. What is this average speed, in miles per hour? \((1 mile = 1,760 yards)\) A) 0.5 B) 1 C) 2 D) \(\)14 The correct answer is \(14.1\). It’s given that a participant completes the bicycle race with an average speed of \(24,816\) yards per hour and \(1 mile = 1,760 yards\). It follows that this average speed is equivalent to \((24,816 yards/1 hour)(1 mile/1,760 yards)\), which yields \(14.1 miles/1 hour\), or \(14.1\) miles per hour. 18 / 83 One side of a flat board has an area of \(874\) square inches. If a pressure of \(19\) pounds per square inch of area is exerted on this side of the board, what is the total force, in pounds, exerted on this side of the board? A) 4 B) 8 C) \(\)16 D) 9.5 The correct answer is \(16,606\). It's given that one side of a flat board has an area of \(874\) square inches. If a pressure of \(19\) pounds per square inch of area is exerted on this side of the board, the total force exerted on this side of the board is \((874 square inches)(19 pounds/1 square inch)\), or \(16,606\) pounds. 19 / 83 A distance of \(354\) furlongs is equivalent to how many feet? \((1 furlong = 220 yards and 1 yard = 3 feet)\) A) \(306\) B) \(402\) C) \(25,960\) D) \(233,640\) Choice D is correct. It's given that \(1 furlong = 220 yards\) and \(1 yard = 3 feet\). It follows that a distance of \(354\) furlongs is equivalent to \((354 furlongs)(220 yards/1 furlong)(3 feet/1 yard)\), or \(233,640\) feet. 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. 20 / 83 The equation above can be used to calculate the distance d, in miles, traveled by a car moving at a speed of 55 miles per hour over a period of t hours. For any positive constant k, the distance the car would have traveled after hours is how many times the distance the car would have traveled after hours? A) 3 B) 6 C) D) Choice A is correct. Since the distance is equal to the amount of time multiplied by a constant, the given equation represents a proportional relationship between distance and time in this situation. Since, the time when hours is 3 times the time when hours. Therefore, the distance traveled after hours is 3 times the distance after hours.Choices B and D are incorrect and may result from interpreting the proportional relationship between time and distance as additive rather than multiplicative. Choice C is incorrect and may result from an arithmetic error. 21 / 83 The total area of a coastal city is 92.1 square miles, of which 11.3 square miles is water. If the city had a population of 621,000 people in the year 2010, which of the following is closest to the population density, in people per square mile of land area, of the city at that time? A) 6,740 B) 7,690 C) 55,000 D) 76,000 Choice B is correct. The land area of the coastal city can be found by subtracting the area of the water from the total area of the coastal city; that is, square miles. The population density is the population divided by the land area, or, which is closest to 7,690 people per square mile.Choice A is incorrect and may be the result of dividing the population by the total area, instead of the land area. Choice C is incorrect and may be the result of dividing the population by the area of water. Choice D is incorrect and may be the result of making a computational error with the decimal place. 22 / 83 Tilly earns \(p\) dollars for every \(w\) hours of work. Which expression represents the amount of money, in dollars, Tilly earns for \(39 w\) hours of work? A) \(39 p\) B) \(p/39\) C) \(p + 39\) D) \(p -39\) Choice A is correct. It’s given that Tilly earns \(p\) dollars for every \(w\) hours of work. This can be represented by the proportion \(p/w\). The amount of money, \(x\), Tilly earns for \(39 w\) hours of work can be found by setting up the proportion \(p/w = x/39 w\). This can be rewritten as \(39 p w = x w\). Dividing both sides by \(w\) results in \(x = 39 p\). Choice B is incorrect. This is the amount of money Tilly earns in dollars per hour, not the amount of money Tilly earns for \(39 w\) hours of work. Choice C is incorrect. This is the amount of money Tilly earns for \(w\) hours of work plus \(39\), not the amount of money Tilly earns for \(39 w\) hours of work. Choice D is incorrect. This is the amount of money Tilly earns for \(w\) hours of work minus \(39\), not the amount of money Tilly earns for \(39 w\) hours of work. 23 / 83 An object's speed is \(64 yards\) per second. What is the object's speed, in feet per second? \((1 yard = 3 feet)\) A) \(61\) B) \(67\) C) \(94\) D) \(192\) Choice D is correct. Since \(1\) yard is equal to \(3\) feet, \(64\) yards is equal to \(64 yards(3 feet/1 yard)\), or \(192\) feet. It follows that \(64\) yards per second is equivalent to \(192\) feet per second. Therefore, the object's speed is \(192\) feet per second. Choice A is incorrect. A speed of \(61\) feet per second is equivalent to \(61/3\), not \(64\), yards per second. Choice B is incorrect. A speed of \(67\) feet per second is equivalent to \(67/3\), not \(64\), yards per second. Choice C is incorrect. A speed of \(94\) feet per second is equivalent to \(94/3\), not \(64\), yards per second. 24 / 83 Objects R and S each travel at a constant speed. The speed of object R is half the speed of object S. Object R travels a distance of \(4 x\) inches in \(y\) seconds. Which expression represents the time, in seconds, it takes object S to travel a distance of \(24 x\) inches? A) \(12 y\) B) \(3 y\) C) \(16 y\) D) \(6 y\) Choice B is correct. It's given that object R travels a distance of \(4 x\) inches in \(y\) seconds. This speed can be written as \(4 x inches/y seconds\). It's given that the speed of object R is half the speed of object S. It follows that the speed of object S is twice the speed of object R, which is \(2(4 x inches/y seconds)\), or \(8 x inches/y seconds\). Let \(n\) represent the time, in seconds, it takes object S to travel a distance of \(24 x\) inches. The value of \(n\) can be found by solving the equation \(8 x inches/y seconds = 24 x inches/n seconds\), which can be written as \(8 x/y = 24 x/n\). Multiplying each side of this equation by \(n y\) yields \(8 x n = 24 x y\). Dividing each side of this equation by \(8 x\) yields \(n = 3 y\). Therefore, the expression \(3 y\) represents the time, in seconds, it takes object S to travel a distance of \(24 x\) inches. Choice A is incorrect. This expression represents the time, in seconds, it would take object S to travel a distance of \(24 x\) inches if the speed of object R were twice, not half, the speed of object S. Choice C is incorrect. This expression represents the time, in seconds, it takes object S to travel a distance of \(128 x\) inches, not \(24 x\) inches. Choice D is incorrect. This expression represents the time, in seconds, it takes object R, not object S, to travel a distance of \(24 x\) inches. 25 / 83 A triathlon is a multisport race consisting of three different legs. A triathlon participant completed the cycling leg with an average speed of \(19.700\) miles per hour. What was the average speed, in yards per hour, of the participant during the cycling leg? \((1 mile = 1,760 yards)\) A) \(\)34 B) 0.5 C) 1 D) 2 The correct answer is \(34,672\). It's given that \(1 mile = 1,760 yards\). It follows that an average speed of \(19.700\) miles per hour is equivalent to \((19.700 miles/1 hour)(1,760 yards/1 mile)\), or \(34,672\) yards per hour. 26 / 83 A special camera is used for underwater ocean research. The camera is at a depth of \(39\) fathoms. What is the camera's depth in feet? \((1 fathom = 6 feet)\) A) \(234\) B) \(117\) C) \(45\) D) \(7\) Choice A is correct. It’s given that a special camera is used for underwater ocean research, and this camera is at a depth of \(39\) fathoms. It's also given that \(1\) fathom is equal to \(6\) feet. Thus, \(39\) fathoms is equivalent to \((39 fathoms)(6 feet/1 fathom)\), or \(234\) feet. Therefore, the camera's depth, in feet, is \(234\). Choice B is incorrect. This is the camera's depth, in feet, if the camera is at a depth of \(19.5\) fathoms. Choice C is incorrect. This is the camera's depth, in feet, if the camera is at a depth of \(7.5\) fathoms. Choice D is incorrect and may result from conceptual or calculation errors. 27 / 83 A certain bird species can fly at an average speed of \(16\) meters per second when in continuous flight. At this rate, how many meters would this bird species fly in \(4\) seconds? A) \(64\) B) \(20\) C) \(16\) D) \(12\) Choice A is correct. It's given that a certain bird species can fly at an average speed of \(16\) meters per second when in continuous flight. At this rate, in \(4\) seconds this bird species would fly \((16 meters/second)(4 seconds)\), or \(64\) meters. Choice B is incorrect. This is the value of \(16 + 4\), not \(16(4)\). Choice C is incorrect. This is the distance the bird would fly in \(1\) second, not \(4\) seconds. Choice D is incorrect. This is the value of \(16 -4\), not \(16(4)\). 28 / 83 How many tablespoons are equivalent to \(14\) teaspoons? \((3 teaspoons = 1 tablespoon)\) A) \(21/5\) B) \(14/3\) C) \(11/3\) D) \(77/15\) The correct answer is \(14/3\). It's given that \(3\) teaspoons is equivalent to \(1\) tablespoon. Therefore, \(14\) teaspoons is equivalent to \((14 teaspoons)(1 tablespoon/3 teaspoons)\), or \(14/3\) tablespoons. Note that 14/3, 4.666, and 4.667 are examples of ways to enter a correct answer. 29 / 83 To study fluctuations in composition, samples of pumice were taken from \(29\) locations and cut in the shape of a cube. The length of the edge of one of these cubes is \(3.000\) centimeters. This cube has a density of \(0.230\) grams per cubic centimeter. What is the mass of this cube, in grams? A) \(\)6 B) 0.1 C) 0.2 D) 0.5 The correct answer is \(6.21\). It’s given that the samples of pumice were cut in the shape of a cube. It's also given that the length of the edge of one of these cubes is \(3.000\) centimeters. Therefore, the volume of this cube is \((3.000 centimeters)cubed\), or \(27\) cubic centimeters. Since the density of this cube is \(0.230\) grams per cubic centimeter, it follows that the mass of this cube is \(( .230 grams/1 cubic centimeter)(27 cubic centimeters)\), or \(6.21\) grams. 30 / 83 Marta has 7,500 pesos she will convert to US dollars using a currency exchange service. At this time, the currency exchange rate is 1 peso = 0.075 US dollars. The exchange service will charge Marta a 2% fee on the converted US dollar amount. How many US dollars will Marta receive from the currency exchange after the 2% fee is applied? A) $551.25 B) $562.50 C) $5,625.00 D) $98,000.00 Choice A is correct. At the exchange rate of 1 peso = 0.075 US dollars, 7,500 pesos would be converted to 7,500 × 0.075 = $562.50. However, since Maria pays a 2% fee on the converted US dollar amount, she receives only (100 – 2)%, or 98%, of the converted US dollars, and 562.50 × 0.98 = $551.25.Choice B is incorrect. This is the number of US dollars Maria would receive if the exchange service did not charge a 2% fee. Choice C is incorrect and may result from a decimal point error made when calculating the conversion to US dollars and from not assessing the 2% fee. Choice D is incorrect and may result from reversing the units of the exchange rate. 31 / 83 For the values \(j\) and \(k\), the ratio of \(j\) to \(k\) is \(11\) to \(12\). If \(j\) is multiplied by \(17\), what is \(k\) multiplied by in order to maintain the same ratio? A) \(17\) B) 4.1 C) 5.5 D) 6 The correct answer is \(17\). If one value is multiplied by a number, then the other value must be multiplied by the same number in order to maintain the same ratio. It’s given that \(j\) is multiplied by \(17\). Therefore, in order to maintain the same ratio, \(k\) must also be multiplied by \(17\). 32 / 83 A distance of \(112\) furlongs is equivalent to how many feet? \((1 furlong = 220 yards and 1 yard = 3 feet)\) A) 0.5 B) 1 C) \(\)73 D) 2 The correct answer is \(73,920\). It's given that \(1 furlong = 220 yards\) and \(1 yard = 3 feet\). It follows that a distance of \(112\) furlongs is equivalent to \((112 furlongs)(220 yards/1 furlong)(3 feet/1 yard)\), or \(73,920\) feet. 33 / 83 A certain town has an area of \(4.36\) square miles. What is the area, in square yards, of this town? \((1 mile = 1,760 yards)\) A) \(404\) B) \(7,674\) C) \(710,459\) D) \(13,505,536\) Choice D is correct. Since the number of yards in \(1\) mile is \(1,760\), the number of square yards in \(1\) square mile is \((1,760)(1,760)= 3,097,600\). Therefore, if the area of the town is \(4.36\) square miles, it is \(4.36(3,097,600)= 13,505,536\), in square yards. Choice A is incorrect and may result from dividing the number of yards in a mile by the square mileage of the town. Choice B is incorrect and may result from multiplying the number of yards in a mile by the square mileage of the town. Choice C is incorrect and may result from dividing the number of square yards in a square mile by the square mileage of the town. 34 / 83 What length, in centimeters, is equivalent to a length of \(51\) meters? \((1 meter = 100 centimeters)\) A) \(0.051\) B) \(0.51\) C) \(5,100\) D) \(51,000\) Choice C is correct. Since \(1\) meter is equal to \(100\) centimeters, \(51\) meters is equal to \(51 meters(100 centimeters/1 meter)\), or \(5,100\) centimeters. Choice A is incorrect and may result from conceptual or calculation errors. Choice B is incorrect and may result from dividing, rather than multiplying, \(51\) by \(100\). Choice D is incorrect. This is the length, in millimeters rather than centimeters, that is equivalent to a length of \(51\) meters. 35 / 83 The population density of Cedar County is \(230\) people per square mile. The county has a population of \(85,100\) people. What is the area, in square miles, of Cedar County? A) 19.2 B) 42.5 C) \(370\) D) 50 The correct answer is \(370\). It’s given that the population density of Cedar County is \(230\) people per square mile and the county has a population of \(85,100\) people. Based on the population density, it follows that the area of Cedar County is \((85,100 people)(1 square mile/230 people)\), or \(370\) square miles. 36 / 83 A fish swam a distance of \(5,104\) yards. How far did the fish swim, in miles? \((1 mile = 1,760 yards)\) A) \(0.3\) B) \(2.9\) C) \(3,344\) D) \(6,864\) Choice B is correct. It’s given that the fish swam \(5,104\) yards and that \(1\) mile is equal to \(1,760\) yards. Therefore, the fish swam \(5,104 yards(1 mile/1, 760 yards)\), which is equivalent to \(5, 104/1, 760 miles\), or \(2.9\) miles. 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. 37 / 83 Food Protein Cost 1 large egg 6 grams $0.36 1 cup of milk 8 grams $0.24 The table above shows the amount of protein in two foods and the cost of each food. Based on the table, what is the ratio of the cost per gram of protein in a large egg to the cost per gram of protein in a cup of milk? A) 1 : 2 B) 2 : 3 C) 3 : 4 D) 2 : 1 Choice D is correct. The cost per gram of protein in 1 large egg is $0.36 ÷ 6 = $0.06. The cost per gram of protein in 1 cup of milk is $0.24 ÷ 8 = $0.03. It follows that the ratio of the cost per gram of protein in a large egg to the cost per gram of protein in a cup of milk is 0.06:0.03, which can be rewritten as 2:1.Choice A is incorrect and may result from finding the ratio of the cost per gram of protein in a cup of milk to the cost per gram of protein in a large egg (the reciprocal of the ratio specified in the question). Choices B and C are incorrect and may result from incorrectly calculating the unit rates or from errors made when simplifying the ratio. 38 / 83 A distance of \(61\) furlongs is equivalent to how many feet? \((1 furlong = 220 yards and 1 yard = 3 feet)\) A) \(\)40 B) 0.5 C) 1 D) 2 The correct answer is \(40,260\). It's given that \(1 furlong = 220 yards\) and \(1 yard = 3 feet\). It follows that a distance of \(61\) furlongs is equivalent to \((61 furlongs)(220 yards/1 furlong)(3 feet/1 yard)\), or \(40,260\) feet. 39 / 83 Which of the following speeds is equivalent to 90 kilometers per hour? (1 kilometer = 1,000 meters) A) 25 meters per second B) 32 meters per second C) 250 meters per second D) 324 meters per second Choice A is correct. Since 1 kilometer is equal to 1,000 meters, it follows that 90 kilometers is equal to meters. Since 1 hour is equal to 60 minutes and 1 minute is equal to 60 seconds, it follows that 1 hour is equal to seconds. Now is equal to, which reduces to or 25 meters per second.Choices B, C, and D are incorrect and may result from conceptual or calculation errors. 40 / 83 To make a bakery’s signature chocolate muffins, a baker needs 2.5 ounces of chocolate for each muffin. How many pounds of chocolate are needed to make 48 signature chocolate muffins? (1 pound = 16 ounces) A) 7.5 B) 10 C) 50.5 D) 120 Choice A is correct. If 2.5 ounces of chocolate are needed for each muffin, then the number of ounces of chocolate needed to make 48 muffins is ounces. Since 1 pound = 16 ounces, the number of pounds that is equivalent to 120 ounces is pounds. Therefore, 7.5 pounds of chocolate are needed to make the 48 muffins.Choice B is incorrect. If 10 pounds of chocolate were needed to make 48 muffins, then the total number of ounces of chocolate needed would be ounces. The number of ounces of chocolate per muffin would then be ounces per muffin, not 2.5 ounces per muffin. Choices C and D are also incorrect. Following the same procedures as used to test choice B gives 16.8 ounces per muffin for choice C and 40 ounces per muffin for choice D, not 2.5 ounces per muffin. Therefore, 50.5 and 120 pounds cannot be the number of pounds needed to make 48 signature chocolate muffins. 41 / 83 How many feet are equivalent to \(34\) yards? \((1 yard = 3 feet)\) A) 0.5 B) \(102\) C) 1 D) 1.5 The correct answer is \(102\). It’s given that \(1\) yard is equivalent to \(3\) feet. Therefore, \(34\) yards is equivalent to \((34 yards)(3 feet/1 yard)\), or \(102\) feet. 42 / 83 The density of a certain type of wood is \(353\) kilograms per cubic meter. A sample of this type of wood is in the shape of a cube and has a mass of \(345\) kilograms. To the nearest hundredth of a meter, what is the length of one edge of this sample? A) \(0.98\) B) \(0.99\) C) \(1.01\) D) \(1.02\) Choice B is correct. It’s given that the density of a certain type of wood is \(353\) kilograms per cubic meter \((kg slash m cubed)\), and a sample of this type of wood has a mass of \(345 kg\). Let \(x\) represent the volume, in \(m Superscript 3\), of the sample. It follows that the relationship between the density, mass, and volume of this sample can be writtenas \(353 kg/1 m cubed = 345 kg/x m cubed\), or \(353 = 345/x\). Multiplying both sides of this equation by \(x\) yields \(353 x = 345\). Dividing both sides of this equation by \(353\) yields \(x = 345/353\). Therefore, the volume of this sample is \(345/353 m cubed\). Since it’s given that the sample of this type of wood is a cube, it follows that the length of one edge of this sample can be found using the volume formula for a cube, \( V = s cubed\), where \( V\) represents the volume, in \(m Superscript 3\), and \(s\) represents the length, in m, of one edge of the cube. Substituting \(345/353\)for \( V\) in this formula yields \(345/353 = s cubed\). Taking the cube root of both sides of this equation yields \(RootIndex 3 √345/353 = s\), or \(s almost = 0 period 99\). Therefore, the length of one edge of this sample to the nearest hundredth of a meter is \(0.99\). Choices A, C, and D are incorrect and may result from conceptual or calculation errors. 43 / 83 One of a planet's moons orbits the planet every \(252\) days. A second moon orbits the planet every \(287\) days. How many more days does it take the second moon to orbit the planet \(29\) times than it takes the first moon to orbit the planet \(29\) times? A) 0.5 B) \(\)1 C) 0.7 D) 1.5 The correct answer is \(1,015\). It’s given that the first moon orbits the planet every \(252\) days. Therefore, it takes the first moon \(252(29)\), or \(7,308\), days to orbit the planet \(29\) times. It’s also given that the second moon orbits the planet every \(287\) days. Therefore, it takes the second moon \(287(29)\), or \(8,323\), days to orbit the planet \(29\) times. Since it takes the first moon \(7,308\) days and the second moon \(8,323\) days, it takes the second moon \(8,323 -7,308\), or \(1,015\), more days than it takes the first moon to orbit the planet \(29\) times. 44 / 83 The weight of an object on Venus is approximately of its weight on Earth. The weight of an object on Juπter is approximately of its weight on Earth. If an object weighs 100 pounds on Earth, approximately how many more pounds does it weigh on Juπter than it weighs on Venus? A) 90 B) 111 C) 140 D) 230 Choice C is correct. The weight of an object on Venus is approximately of its weight on Earth. If an object weighs 100 pounds on Earth, then the object’s weight on Venus is approximately pounds. The same object’s weight on Juπter is approximately of its weight on Earth; therefore, the object weighs approximately pounds on Juπter. The difference between the object’s weight on Juπter and the object’s weight on Venus is approximately pounds. Therefore, an object that weighs 100 pounds on Earth weighs 140 more pounds on Juπter than it weighs on Venus.Choice A is incorrect because it is the weight, in pounds, of the object on Venus. Choice B is incorrect because it is the weight, in pounds, of an object on Earth if it weighs 100 pounds on Venus. Choice D is incorrect because it is the weight, in pounds, of the object on Juπter. 45 / 83 If \(4 a/b = 6.7\) and \(a/b n = 26.8\), what is the value of \(n\)? A) \(\) B) 2 C) 2.5 D) 3.2 The correct answer is \(.0625\). It's given that \(4 a/b = 6.7\) and \(a/b n = 26.8\). The equation \(4 a/b = 6.7\) can be rewritten as \((4)(a/b)= 6.7\). Dividing both sides of this equation by \(4\) yields \(a/b = 1.675\). The equation \(a/b n = 26.8\) can be rewritten as \((a/b)(1/n)= 26.8\). Substituting \(1.675\) for \(a/b\) in this equation yields \((1.675)(1/n)= 26.8\), or \(1.675/n = 26.8\). Multiplying both sides of this equation by \(n\) yields \(1.675 = 26.8 n\). Dividing both sides of this equation by \(26.8\) yields \(n = 0.0625\). Therefore, the value of \(n\) is \(0.0625\). Note that .0625, 0.062, 0.063, and 1/16 are examples of ways to enter a correct answer. 46 / 83 The table shows the distribution, by location and power capacity (maximum rate of power generation) of the twenty largest wind projects in the United States in 2013. The total power capacity of the nine wind projects located in Texas was 4,952 megawatts (MW), and the total power capacity of the twenty wind projects was 11,037 MW in 2013. The amount of energy produced in one hour at a rate of one megawatt is one megawatt-hour. If each of the nine Texas wind projects in 2013 had operated continuously for 24 hours at the maximum rate of power generation, approximately how many megawatt-hours of energy would the nine projects have produced? A) 200 B) 5,000 C) 11,000 D) 120,000 Choice D is correct. It’s given that the total power capacity of the nine wind projects in Texas was 4,952 megawatts. Therefore, if all nine Texas projects operated continuously for 1 hour, the amount of energy produced would be 4,952 megawatt-hours. It follows that, if all nine Texas projects operated continuously for 24 hours, the amount of energy produced, in megawatt-hours, would be, which is closest to 120,000.Choice A is incorrect. This is approximately the amount of energy produced for the nine projects divided by 24 hours. Choice B is incorrect. This is approximately the amount of energy produced for the nine projects. Choice C is incorrect. This is approximately the given amount of energy produced for all twenty projects in the table. 47 / 83 An object travels at a constant speed of \(12\) centimeters per second. At this speed, what is the time, in seconds, that it would take for the object to travel \(108\) centimeters? A) \(9\) B) \(96\) C) \(120\) D) \(972\) Choice A is correct. If the object travels \(108\) centimeters at a speed of \(12\) centimeters per second, the time of travel can be determined by dividing the total distance by the speed. This results in \(108 centimeters/12 centimeters slash second\), which is \(9\) seconds. 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. 48 / 83 The glassπctured above can hold a maximum volume of 473 cubic centimeters, which is approximately 16 fluid ounces. Jenny has aπtcher that contains 1 gallon of water. How many times could Jenny completely fill the glass with 1 gallon of water? A) 16 B) 8 C) 4 D) 3 Choice B is correct. It is given that the volume of the glass is approximately 16 fluid ounces. If Jenny has 1 gallon of water, which is 128 fluid ounces, she could fill the glass times.Choice A is incorrect because Jenny would need fluid ounces = 256 fluid ounces, or 2 gallons, of water to fill the glass 16 times. Choice C is incorrect because Jenny would need only fluid ounces = 64 fluid ounces of water to fill the glass 4 times. Choice D is incorrect because Jenny would need only fluid ounces = 48 fluid ounces to fill the glass 3 times. 49 / 83 A kangaroo has a mass of \(28\) kilograms. What is the kangaroo's mass, in grams? \((1 kilogram = 1,000 grams)\) A) \(28,000\) B) \(1,028\) C) \(972\) D) \(784\) Choice A is correct. It's given that a kangaroo has a mass of \(28\) kilograms and that \(1\) kilogram is equal to \(1,000\) grams. Therefore, the kangaroo's mass, in grams, is \(28 kilograms(1,000 grams/1 kilogram)\), which is equivalent to \(28,000\) grams. 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. 50 / 83 A cherryπtting machineπts \(12\) pounds of cherries in \(3\) minutes. At this rate, how many minutes does it take the machine toπt \(96\) pounds of cherries? A) \(8\) B) \(15\) C) \(24\) D) \(36\) Choice C is correct. It's given that the cherryπtting machineπts \(12\) pounds of cherries in \(3\) minutes. This rate can be written as \(12 pounds of cherries/3 minutes\). If the number of minutes it takes the machine toπt \(96\) pounds of cherries is represented by \(x\), the value of \(x\) can be calculated by solving the equation \(12 pounds of cherries/3 minutes = 96 pounds of cherries/x minutes\), which can be rewritten as \(12/3 = 96/x\), or \(4 = 96/x\). Multiplying each side of this equation by \(x\) yields \(4 x = 96\). Dividing each side of this equation by \(4\) yields \(x = 24\). Therefore, it takes the machine \(24\) minutes toπt \(96\) pounds of cherries. Choice A is incorrect. This is the number of minutes it takes the machine toπt \(32\), not \(96\), pounds of cherries. Choice B is incorrect. This is the number of minutes it takes the machine toπt \(60\), not \(96\), pounds of cherries. Choice D is incorrect. This is the number of minutes it takes the machine toπt \(144\), not \(96\), pounds of cherries. 51 / 83 Tanya earns $13.50 per hour at her part-time job. When she works z hours, she earns dollars. Which of the following expressions gives the amount, in dollars, Tanya will earn if she works hours? A) B) C) D) Choice A is correct. It’s given that when Tanya works z hours, she earns dollars. Since her hourly rate is constant, if she works 3 times as many hours, or hours, she will earn 3 times as many dollars, or .Choice B is incorrect. This expression represents adding 3 dollars to the dollars Tanya will earn. Choice C is incorrect. This expression can be rewritten as, which implies that Tanya earns $16.50 per hour, not $13.50. Choice D is incorrect. This expression adds 3 to the number of hours Tanya works, rather than multiplying the hours she works by 3. 52 / 83 Jeremy deposited x dollars in his investment account on January 1, 2001. The amount of money in the account doubled each year until Jeremy had 480 dollars in his investment account on January 1, 2005. What is the value of x ? A) 33 B) 30 C) 29 D) 31 The correct answer is 30. The situation can be represented by the equation, where the 2 represents the fact that the amount of money in the account doubled each year and the 4 represents the fact that there are 4 years between January 1, 2001, and January 1, 2005. Simplifying gives . Therefore, . 53 / 83 How many yards are equivalent to \(77\) rods? \((5.5 yards = 1 rod)\) A) 0.5 B) 1 C) \(\)423 D) 2 The correct answer is \(423.5\). It's given that \(5.5 yards = 1 rod\). Therefore, \(77\) rods is equivalent to \((77 rods)(5.5 yards/1 rod)\), or \(423.5\) yards. Note that 423.5 and 847/2 are examples of ways to enter a correct answer. 54 / 83 A giant armadillo has a mass of \(39\) kilograms. What is the giant armadillo's mass in grams? \((1 kilogram = 1,000 grams)\) A) \(\)39 B) 0.5 C) 1 D) 2 The correct answer is \(39,000\). It’s given that the giant armadillo has a mass of \(39\) kilograms. Since \(1\) kilogram is equal to \(1,000\) grams, \(39\) kilograms is equal to \(39 kilograms(1,000 grams/1 kilogram)\), or \(39,000\) grams. Therefore, the giant armadillo’s mass, in grams, is \(39,000\). 55 / 83 An object has a mass of \(168\) grams and a volume of \(24\) cubic centimeters. What is the density, in grams per cubic centimeter, of the object? A) \(7\) B) \(144\) C) \(192\) D) \(4,032\) Choice A is correct. It's given that the object has a mass of \(168\) grams and a volume of \(24\) cubic centimeters. Dividing the mass, in grams, of the object by the volume, in cubic centimeters, of the object gives the density, in grams per cubic centimeter, of the object. It follows that the density of the object is \(168 grams/24 cubic centimeters\), which is equivalent to \(168/24\) grams per cubic centimeter, or \(7\) grams per cubic centimeter. 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. 56 / 83 Shaquan has 7 red cards and 28 blue cards. What is the ratio of red cards to blue cards that Shaquan has? A) 1 to 4 B) 4 to 1 C) 1 to 7 D) 7 to 1 Choice A is correct. It’s given that Shaquan has 7 red cards and 28 blue cards. Therefore, the ratio of red cards to blue cards that Shaquan has is 7 to 28. This ratio can be reduced by dividing both parts of the ratio by 7, which yields the ratio 1 to 4.Choice B is incorrect. This is the ratio of blue cards to red cards that Shaquan has. Choice C is incorrect and may result from a calculation error when reducing the ratio. Choice D is incorrect. This may result from finding the ratio of blue cards to red cards, or 28 to 7, and then making a calculation error when reducing the ratio. 57 / 83 Rectangle A has length 15 and width w. Rectangle B has length 20 and the same length-to-width ratio as rectangle A. What is the width of rectangle B in terms of w ? A) B) C) D) Choice A is correct. It’s given that rectangle A has length 15 and width w. Therefore, the length-to-width ratio of rectangle A is 15 to w. It’s also given that rectangle B has length 20 and the same length-to-width ratio as rectangle A. Let x represent the width of rectangle B. The proportion can be used to solve for x in terms of w. Multiplying both sides of this equation by x yields, and then multiplying both sides of this equation by w yields . Dividing both sides of this equation by 15 yields . Simplifying this fraction yields .Choices B and D are incorrect and may result from interpreting the difference in the lengths of rectangle A and rectangle B as equivalent to the difference in the widths of rectangle A and rectangle B. Choice C is incorrect and may result from using a length-to-width ratio of w to 15, instead of 15 to w. 58 / 83 The speed of a vehicle is increasing at a rate of \(7.3\) meters per second ^2. What is this rate, in miles per minute ^2, rounded to the nearest tenth? \((Use 1 mile = 1,609 meters period)\) A) \(0.3\) B) \(16.3\) C) \(195.8\) D) \(220.4\) Choice B is correct. It's given that the speed of a vehicle is increasing at a rate of \(7.3\) meters per second ^2. It's given to use \(1 mile = 1,609 meters\). There are \(60\) seconds in \(1\) minute; therefore, \(60 ^2\) or \(3,600\) seconds ^2 is equal to \(1\) minute ^2. It follows that the rate of \(7.3\) meters per second ^2 is equivalent to \((7.3 meters/1 second ^2)(1 mile/1,609 meters)(3,600 seconds ^2/1 minute ^2)\), or approximately \(16.33 miles per minute ^2\). The rate, \(in miles per minute ^2\), rounded to the nearest tenth is \(16.3\). 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. 59 / 83 How many yards are equivalent to \(612\) inches? \((1 yard = 36 inches)\) A) \(0.059\) B) \(17\) C) \(576\) D) \(22,032\) Choice B is correct. It’s given that \(1 yard = 36 inches\). Therefore, \(612\) inches is equivalent to \(612 inches(1 yard/36 inches)\), which can be rewritten as \(612 yards/36\), or \(17\) yards. Choice A is incorrect. This is the number of yards that are equivalent to \(2.124\) inches. Choice C is incorrect. This is the number of yards that are equivalent to \(20,736\) inches. Choice D is incorrect. This is the number of yards that are equivalent to \(793,152\) inches. 60 / 83 On April 18, 1775, Paul Revere set off on his midnight ride from Charlestown to Lexington. If he had ridden straight to Lexington without stopπng, he would have traveled 11 miles in 26 minutes. In such a ride, what would the average speed of his horse have been, to the nearest tenth of a mile per hour? A) 25 B) \(45/2\) C) 24 D) 26 The correct answer is 25.4. The average speed is the total distance divided by the total time. The total distance is 11 miles and the total time is 26 minutes. Thus, the average speed is miles per minute. The question asks for the average speed in miles per hour, and there are 60 minutes in an hour; converting miles per minute to miles per hour gives the following: Therefore, to the nearest tenth of a mile per hour, the average speed of Paul Revere’s ride would have been 25.4 miles per hour. Note that 25.4 and 127/5 are examples of ways to enter a correct answer. 61 / 83 At a particular track meet, the ratio of coaches to athletes is \(1\) to \(26\). If there are \(x\) coaches at the track meet, which of the following expressions represents the number of athletes at the track meet? A) \(x/26\) B) \(26 x\) C) \(x + 26\) D) \(26/x\) Choice B is correct. It’s given that at a particular track meet, the ratio of coaches to athletes is \(1\) to \(26\). If one number in a ratio is multiplied by a value, the other number must be multiplied by the same value in order to maintain the same ratio. If there are \(x\) coaches at the track meet, multiplying both numbers in the ratio by \(x\) yields \(1(x)\) to \(26(x)\), or \(x\) to \(26 x\). Therefore, the expression \(26 x\) represents the number of athletes at the track meet. 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. 62 / 83 A sample of oak has a density of \(807\) kilograms per cubic meter. The sample is in the shape of a cube, where each edge has a length of \(0.90\) meters. To the nearest whole number, what is the mass, in kilograms, of this sample? A) \(588\) B) \(726\) C) \(897\) D) \(1,107\) Choice A is correct. It’s given that the sample is in the shape of a cube with edge lengths of \(0.9\) meters. Therefore, the volume of the sample is \(0.90 cubed\), or \(0.729\), cubic meters. It’s also given that the sample has a density of \(807\) kilograms per \(1\) cubic meter. Therefore, the mass of this sample is \(0.729 cubic meters(807 kilograms/1 cubic meter)\), or \(588.303\) kilograms. Rounding this mass to the nearest whole number gives \(588\) kilograms. Therefore, to the nearest whole number, the mass, in kilograms, of this sample is \(588\). 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. 63 / 83 A special camera is used for underwater ocean research. When the camera is at a depth of \(58\) fathoms, what is the camera's depth in feet? \((1 fathom = 6 feet)\) A) 0.5 B) 1 C) 2 D) \(348\) The correct answer is \(348\). It's given that \(1\) fathom is equivalent to \(6\) feet. Therefore, \(58\) fathoms is equivalent to \((58 fathoms)(6 feet/1 fathom)\), or \(348\) feet. Thus, when the camera is at a depth of \(58\) fathoms, the camera's depth, in feet, is \(348\). 64 / 83 The ratio of t to u is 1 to 2, and . What is the value of u ? A) 2 B) 5 C) 10 D) 20 Choice D is correct. It’s given that the ratio of t to u is 1 to 2. Since, it follows that the ratio of 10 to u is also 1 to 2. The relationship between these ratios can be represented by the proportion . Multiplying both sides of this equation by 2 and then by u yields .Choice A is incorrect. This is the value of u when . Choice B is incorrect. This would be the value of u if the ratio of t to u were 2 to 1. Choice C is incorrect. This is the value of t, not u. 65 / 83 One method of calculating the approximate age, in years, of a tree of a particular species is to multiply the diameter of the tree, in inches, by a constant called the growth factor for that species. The table above gives the growth factors for eight species of trees. If a white birch tree and aπn oak tree each now have a diameter of 1 foot, which of the following will be closest to the difference, in inches, of their diameters 10 years from now? (1 foot = 12 inches) A) 1.0 B) 1.2 C) 1.3 D) 1.4 Choice C is correct. According to the given information, multiplying a tree species’ growth factor by the tree’s diameter is a method to approximate the age of the tree. A white birch with a diameter of 12 inches (or 1 foot) has a given growth factor of 5 and is approximately 60 years old. Aπn oak with a diameter of 12 inches (or 1 foot) has a given growth factor of 3 and is approximately 36 years old. The diameters of the two trees 10 years from now can be found by dividing each tree’s age in 10 years, 70 years, and 46 years, by its respective growth factor. This yields 14 inches and inches. The difference between and 14 is, or approximately 1.3 inches.Alternate approach: Since a white birch has a growth factor of 5, the age increases at a rate of 5 years per inch or, equivalently, the diameter increases at a rate of of an inch per year. Likewise, theπn oak has a growth factor of 3, so its diameter increases at a rate of of an inch per year. Thus, theπn oak grows of an inch per year more than the white birch. In 10 years it will grow of an inch more, which is approximately 1.3 inches.Choices A, B, and D are incorrect and a result of incorrectly calculating the diameters of the two trees in 10 years. 66 / 83 The ratio \(x\) to \(y\) is equivalent to the ratio \(9\) to \(5\). If the value of \(x\) is \(162\), what is the value of \(y\)? A) \(90\) B) 2.5 C) 4.5 D) 5 The correct answer is \(90\). It’s given that the ratio of \(x\) to \(y\) is equivalent to the ratio \(9\) to \(5\). It follows that \(x/y = 9 fifths\). Multiplying each side of this equation by \(5 y\) yields \((5 y)x/y = 9(5 y)/5\), or \(5 x = 9 y\). Dividing each side of this equation by \(9\) yields \(5 x/9 = y\). Substituting \(162\) for \(x\) in this equation yields \(5(162)/9 = y\), which is equivalent to \(810/9 = y\), or \(90 = y\). Therefore, if the value of \(x\) is \(162\), the value of \(y\) is \(90\). 67 / 83 The ratio of the length of line segment \( X Y\) to the length of line segment \( Z V\) is \(6\) to \(1\). If the length of line segment \( X Y\) is \(102\) inches, what is the length, in inches, of line segment \( Z V\)? A) \(17\) B) \(96\) C) \(102\) D) \(612\) Choice A is correct. It’s given that the ratio of the length of line segment \( X Y\) to the length of line segment \( Z V\) is \(6\) to \(1\), which means \( X Y/ Z V = 6 oneths\). It’s given that the length of line segment \( X Y\) is \(102\) inches. If the length, in inches, of line segment \( Z V\) is represented by \(ell\), the value of \(ell\) can be calculated by solving the equation \(102/ell = 6 oneths\), or \(102/ell = 6\). Multiplying each side of this equation by \(ell\) yields \(102 = 6 ell\). Dividing each side of this equation by \(6\) yields \(17 = ell\). Therefore, the length of line segment \( Z V\) is \(17\) inches. Choice B is incorrect. This is the length, in inches, of line segment \( Z V\) if the length of line segment \( X Y\) is \(576\), not \(102\), inches. Choice C is incorrect. This is the length, in inches, of line segment \( X Y\), not line segment \( Z V\). Choice D is incorrect. This is the length, in inches, of line segment \( Z V\) if the ratio of the length of line segment \( X Y\) to the length of line segment \( Z V\) is \(1\) to \(6\), not \(6\) to \(1\). 68 / 83 In the table above, the ratio of y to x for each ordered pair is constant. What is the value of k ? A) 28 B) 36 C) 80 D) 160 Choice D is correct. Since the ratio of y to x is constant for each ordered pair in the table, the first row can be used to determine that the ratio of y to x is 4 to 1. The proportion can be used to solve for k. Multiplying each side of the equation by 40 yields .Choice A is incorrect. This is the value of y when the value of x is 7, not 40. Choice B is incorrect and may result from subtracting 4 from 40 instead of multiplying 40 by 4. Choice C is incorrect and may result from incorrectly setting up the proportion. 69 / 83 The number of raccoons in a \(131\)-square-mile area is estimated to be \(2,358\). What is the estimated population density, in raccoons per square mile, of this area? A) \(18\) B) \(131\) C) \(149\) D) \(2,376\) Choice A is correct. It’s given that there are \(2,358\) raccoons in a \(131\)-square-mile area. The estimated population density, in raccoons per square mile, is the estimated number of raccoons divided by the number of square miles. Therefore, the estimated population density of this area is \(2,358 raccoons/131 square miles\), or \(18\) raccoons per square mile. Choice B is incorrect. This is the number of square miles in the area, not the estimated number of raccoons per square mile in this area. Choice C is incorrect and may result from conceptual or calculation errors. Choice D is incorrect and may result from conceptual or calculation errors. 70 / 83 A wind turbine completes \(900\) revolutions in \(50\) minutes. At this rate, how many revolutions per minute does this turbine complete? A) \(18\) B) \(850\) C) \(950\) D) \(1,400\) Choice A is correct. Dividing the number of revolutions by the number of minutes gives the number of revolutions the turbine completes per minute. It’s given that the wind turbine completes \(900\) revolutions in \(50\) minutes. Therefore, at this rate, this turbine completes \(900/50\), or \(18\), revolutions per minute. 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. 71 / 83 The population density of Worthington is \(290\) people per square mile. Worthington has a population of \(92,800\) people. What is the area, in square miles, of Worthington? A) \(102,400\) B) \(93,090\) C) \(320\) D) \(32\) Choice C is correct. It’s given that the population density of Worthington is \(290\) people per square mile and Worthington has a population of \(92,800\) people. Therefore, the area of Worthington is \(92,800 people(1 square mile/290 people)\), which is equivalent to \(92,800 square miles/290\), or \(320\) square miles. 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. 72 / 83 Pure beeswax has a density of 0.555 ounce per cubic inch. An online company sells pure beeswax at a price of $8.00 per ounce. What is the selling price, in dollars per cubic inch, for pure beeswax purchased from this company? A) 4 B) \(22/5\) C) \(18/5\) D) 3 The correct answer is 4.44. The selling price, in dollars per cubic inch, is found by multiplying the density, in ounces per cubic inch, by the unit price, in dollars per ounce: yields . Thus, the selling price, in dollars per cubic inch, is 4.44. 73 / 83 In a box of pens, the ratio of black pens to red pens is \(8\) to \(1\). There are \(40\) black pens in the box. How many red pens are in the box? A) \(5\) B) \(8\) C) \(40\) D) \(320\) Choice A is correct. It’s given that the ratio of black pens to red pens is \(8\) to \(1\). Therefore, there are \(1 eighth\) as many red pens as black pens in the box. It’s also given that there are \(40\) black pens in the box. Therefore, the number of red pens is \(1 eighth\) of the \(40\) black pens. Thus, the number of red pens is \(40(1 eighth)\), or \(5\). Choice B is incorrect. This is the number of black pens in the box for every red pen. Choice C is incorrect. This is the number of black pens in the box. Choice D is incorrect. This is the number of red pens in the box if the ratio of black pens to red pens is \(1\) to \(8\). 74 / 83 If \(x/y = 4\) and \(24 x/n y = 4\), what is the value of \(n\)? A) 2 B) 4 C) \(24\) D) 4.9 The correct answer is \(24\). The equation \(24 x/n y = 4\) can be rewritten as \((24/n)(x/y)= 4\). It's given that \(x/y = 4\). Substituting \(4\) for \(x/y\) in the equation \((24/n)(x/y)= 4\) yields \((24/n)(4)= 4\). Multiplying both sides of this equation by \(n\) yields \((24)(4)= 4 n\). Dividing both sides of this equation by \(4\) yields \(24 = n\). Therefore, the value of \(n\) is \(24\). 75 / 83 A group of monarch butterflies migrated from Chicago, Illinois, to Michoacán, Mexico, flying a total of 2,100 miles. It took a single butterfly in the group 120 days to travel this route one way. On average, how many miles did the butterfly travel per day? A) 0.057 B) 0.729 C) 17.5 D) 24 Choice C is correct. If the butterfly traveled 2,100 miles in 120 days, then it traveled, on average, miles per day.Choice A is incorrect. This is approximately the average amount of time, in days, it took the butterfly to fly one mile: days per mile. Choice B is incorrect and may result from an arithmetic error. Choice D is incorrect. This is the number of hours in a day rather than the number of miles flown per day. 76 / 83 How many fluid ounces are equivalent to \(76\) quarts? \((8 fluid ounces = 1 cup and 4 cups = 1 quart)\) A) 0.5 B) 1 C) \(\)2 D) 1.4 The correct answer is \(2,432\). It's given that \(4 cups = 1 quart\). It follows that \(76\) quarts is equivalent to \((76 quarts)(4 cups/1 quart)\), or \(304\) cups. It's also given that \(8 fluid ounces = 1 cup\). It follows that \(304\) cups is equivalent to \((304 cups)(8 fluid ounces/1 cup)\), or \(2,432\) fluid ounces. 77 / 83 A mechanical device in a workshop produces items at a constant rate of \(60\) items per hour. At this rate, how many items will the mechanical device produce in \(3\) hours? A) 1.5 B) 3 C) \(180\) D) 6 The correct answer is \(180\). It’s given that a mechanical device produces items at a constant rate of \(60\) items per hour. This rate can be written as \(60 items/1 hour\). Let \(x\) represent the number of items the mechanical device will produce in \(3\) hours at the given rate. It follows that \(60 items/1 hour = x items/3 hours\), which can be written as \(60/1 = x/3\), or \(60 = x/3\). Multiplying each side of this equation by \(3\) yields \(180 = x\). Therefore, at the given rate, the mechanical device will produce \(180\) items in \(3\) hours. Alternate approach: It's given that a mechanical device produces items at a constant rate of \(60\) items per hour. At this rate, the mechanical device will produce \((60 items/1 hour)(3 hours)\), or \(180\) items in \(3\) hours. 78 / 83 A customer spent \($ 27\) to purchase oranges at \($ 3\) per pound. How many pounds of oranges did the customer purchase? A) \(9\) B) 1.5 C) 3 D) 4.5 The correct answer is \(9\). It’s given that the customer spent \($ 27\) to purchase oranges at \($ 3\) per pound. Therefore, the number of pounds of oranges the customer purchased is \($ 27(1 pound/$ 3)\), or \(9\) pounds. 79 / 83 The ratio \(140\) to \(m\) is equivalent to the ratio \(4\) to \(28\). What is the value of \(m\)? A) 2 B) 4 C) \(980\) D) 8 The correct answer is \(980\). It's given that the ratio \(140\) to \(m\) is equivalent to the ratio \(4\) to \(28\). Therefore, the value of \(m\) can be found by solving the equation \(140/m = 4 20 eighths\). Multiplying each side of this equation by \(m\) yields \(140 = 4 m/28\). Multiplying each side of this equation by \(28\) yields \(3,920 = 4 m\). Dividing each side of this equation by \(4\) yields \(980 = m\). Therefore, the value of \(m\) is \(980\). 80 / 83 The population density of Iceland, in people per square kilometer of land area, increased from 2.5 in 1990 to 3.3 in 2014. During this time period, the land area of Iceland was 100,250 square kilometers. By how many people did Iceland’s population increase from 1990 to 2014? A) 330,825 B) 132,330 C) 125,312 D) 80,200 Choice D is correct. The increase in Iceland’s population can be found by multiplying the increase in population density, in people per square kilometer, by the area, in square kilometers. It’s given that the population density of Iceland was 2.5 people per square kilometer in 1990 and 3.3 people per square kilometer in 2014. The increase in population density can be found by subtracting 2.5 from 3.3, which yields 0.8. It’s given that the land area of Iceland was 100,250 square kilometers. Thus, the increase in population is, or 80,200.Alternate approach: It’s given that the population density of Iceland, in people per square kilometer of land area, in 1990 was 2.5. Since the land area of Iceland was 100,250 square kilometers, it follows that the population of Iceland in 1990 was, or 250,625. Similarly, the population of Iceland in 2014 was, or 330,825. The population increase is the difference in the population from 1990 to 2014, or, which yields 80,200. Therefore, Iceland’s population increased by 80,200 from 1990 to 2014.Choice A is incorrect. This is the population of Iceland in 2014. Choice B is incorrect and may result from dividing 3.3 by 2.5, instead of subtracting 2.5 from 3.3. Choice C is incorrect and may result from dividing the population of Iceland in 1990 by 2. 81 / 83 For a person m miles from a flash of lightning, the length of the time interval from the moment the person sees the lightning to the moment the person hears the thunder is k seconds. The ratio of m to k can be estimated to be 1 to 5. According to this estimate, the person is how many miles from a flash of lightning if the time interval is 25 seconds? A) 10 B) 9 C) 6 D) 5 Choice D is correct. It’s given that the ratio of m to k is estimated to be 1 to 5. Therefore, when, the relationship between these ratios can be expressed by the proportion . Multiplying both sides of this equation by 25 yields .Choices A, B, and C are incorrect and may result from calculation errors. 82 / 83 The International Space Station orbits Earth at an average speed of 4.76 miles per second. What is the space station’s average speed in miles per hour? A) 285.6 B) 571.2 C) 856.8 D) 17,136.0 Choice D is correct. Since 1 minute = 60 seconds and 1 hour = 60 minutes, it follows that 1 hour = (60)(60), or 3,600 seconds. Using this conversion factor, the space station’s average speed of 4.76 miles per second is equal to an average speed of, or 17,136 miles per hour.Choice A is incorrect. This is the space station’s average speed in miles per minute. Choice B is incorrect. This is double the space station’s average speed in miles per minute, or the number of miles the space station travels on average in 2 minutes. Choice C is incorrect. This is triple the space station’s average speed in miles per minute, or the number of miles the space station travels on average in 3 minutes. 83 / 83 A printer produces posters at a constant rate of \(42\) posters per minute. At what rate, in posters per hour, does the printer produce the posters? A) 1 B) \(\)2 C) 1.4 D) 3 The correct answer is \(2,520\). There are \(60\) minutes in one hour. At a rate of \(42\) posters per minute, the number of posters produced in one hour can be determined by \((42 posters/1 minute)(60 minutes/1 hour)\), which is \(2,520\) posters per hour. Your score is Follow us on socials! LinkedIn Facebook Twitter 0% Restart quiz Send feedback About This Quiz SAT Ratios, Rates, Proportional relationships, and Units type Practice Questions More Quizzes Linear Equations in one Variable SAT Full Practice Test Take Quiz SAT Randomized Questions - 1 Full Math Practice Test - Answers and Detailed Explanations at the END Take Quiz Evaluating statistical claims: Observational studies and experiments Take Quiz Lines, Angles and Triangles Take Quiz Probability and Conditional Probability Take Quiz Two variable Data - Models and Scatter Plots Take Quiz Circles on the SAT Take Quiz Linear Inequalities Take Quiz
