SAT Explanations_6-10

pay for n tickets is given by 15 n + 12. The value 12 represents only a portion of this total amount. Therefore, the value 12 does not represent the total amount, in dollars, for any number of tickets. QUESTION 2 Choice B is correct . Since Fertilizer A contains 60% filler materials by weight, it follows that x pounds of Fertilizer A consists of 0.6 x pounds of filler materials. Similarly, y pounds of Fertilizer B consists of 0.4 y pounds of filler materials. When x pounds of Fertilizer A and y pounds of Fertilizer B are combined, the result is 240 pounds of filler materials. Therefore, the total amount, in pounds, of filler materials in a mixture of x pounds of Fertilizer A and y pounds of Fertilizer B can be expressed as 0.6 x + 0.4 y = 240. Choice A is incorrect. This choice transposes the percentages of filler materials for Fertilizer A and Fertilizer B. Fertilizer A consists of 0.6 x pounds of filler materials and Fertilizer B consists of 0.4 y pounds of filler materials. Therefore, 0.6 x + 0.4 y is equal to 240, not 0.4 x + 0.6 y . Choice C is incorrect. This choice incorrectly represents how to take the percentage of a value mathematically. Fertilizer A consists of 0.6 x pounds of filler materials, not 60 x pounds of filler materials, and Fertilizer B consists of 0.4 y pounds of filler materials, not 40 y pounds of filler materials. Choice D is incorrect. This choice transposes the percentages of filler materials for Fertilizer A and Fertilizer B and incorrectly represents how to take the percentage of a value mathematically. QUESTION 3 Choice C is correct . For a complex number written in the form a + bi , a is called the real part of the complex number and b is called the imaginary part. The sum of two complex numbers, a + bi and c + di , is found by adding real parts and imaginary parts, respectively; that is, ( a + bi ) + ( c + di ) = ( a + c ) + ( b + d ) i . Therefore, the sum of 2 + 3 i and 4 + 8 i is (2 + 4) + (3 + 8) i = 6 + 11 i . Choice A is incorrect and is the result of disregarding i and adding all parts of the two complex numbers together, 2 + 3 + 4 + 8 = 17. Choice B is incorrect and is the result of adding all parts of the two complex numbers together and multiplying the sum by i . Choice D is incorrect and is the result of multiplying the real parts and imaginary parts of the two complex numbers, (2)(4) = 8 and (3)(8) = 24, instead of adding those parts together. QUESTION 4 Choice A is correct . The right side of the equation can be multiplied using the distributive property: ( px + t )( px о t ) = p 2 x 2 о ptx + ptx о t 2 . Combining like terms gives p 2 x 2 о t 2 . Substituting this expression for the right side of the equation gives 4 x 2 о ϵ с p 2 x 2 о t 2 , where p and t are

Jamie will bicycle during this time is the sum of the distances he has completed and has yet to complete. Thus T = 240 + 310 + 320 + x . Substituting this expression into the inequality T ш 4(280) gives 240 + 310 + 320 + x ш ϰ;ϮϴϬͿ͘ Therefore͕ choice D iƐ ƚhe correcƚ anƐǁer͘ Choices A, B, and C are incorrect because they do not correctly capture the relationships between the total number of miles Jaime will ride his bicycle (240 + 310 + 320 + x ) and the minimum number of miles he is attempting to bicycle for the four weeks (280 + 280 + 280 + 280). QUESTION 11 Choice B is correct . Since the shown parabola opens upward, the coefficient of x 2 in the equation y = ax 2 + c must be positive. Given that a is positive, ʹ a is negative, and therefore the graph of the equation y = о a ( x о b ) 2 + c will be a parabola that opens downward. The vertex of this parabola is ( b , c ), because the maximum value of y , c , is reached when x = b . Therefore, the answer must be choice B. Choices A and C are incorrect. The coefficient of x 2 in the equation y с о a ( x о b ) 2 + c is negative. Therefore, the parabola with this equation opens downward, not upward. Choice D is incorrect because the vertex of this parabola is ( b , c Ϳ͕ noƚ ;о b , c ), because the maximum value of y , c , is reached when x = b . QUESTION 12 Choice D is correct. Dividing 4 x 2 + 6 x by 4 x + 2 gives: 2 1 4 2 4 6 (4 2 ) 4 (4 2) 2 x x x x x x x x ³ ³ ³ ´ ³ ´ ³ ´ Therefore, the expression 2 4 6 4 2 x x x ³ ³ is equivalent to 2 1 4 2 x x ³ ´ ³ . Alternate approach: The numerator of the given expression, 4 x 2 + 6 x , can be rewritten in terms of the denominator, 4 x + 2, as follows: 4 x 2 + 2 x + 4 x + 2 о 2, or x (4 x + 2) + (4 x н ϮͿ о Ϯ͘ So ƚhe given expression can be rewritten as (4 2) (4 2) 2 2 1 4 2 4 2 x x x x x x ³ ³ ³ ´ ³ ´ ³ ³ .

଺ ଵ଴ ൌ ଵ଼ ஼ா . Multiplying each side by CE, and then multiplying by ଵ଴ ଺ yields CE = 30. Therefore, the length of ࠵?࠵? തതതത is 30. QUESTION 19 The correct answer is 1.5 or ૜ ૛ . The total amount, in liters, of a saline solution can be expressed as the liters of each type of saline solution multiplied by the percent of the saline solution. This gives 3(0.10), x (0.25), and ( x + 3)(0.15), where x is the amount, in liters, of a 25% saline solution and 10%, 15%, and 25% are represented as 0.10, 0.15, and 0.25, respectively. Thus, the equation 3(0.10) + 0.25 x = 0.15( x + 3) must be true. Multiplying 3 by 0.10 and distributing 0.15 to ( x + 3) yields 0.30 + 0.25 x = 0.15 x + 0.45. Subtracting 0.15 x and 0.30 from each side of the equation gives 0.10 x = 0.15. Dividing each side of the equation by 0.10 yields x = 1.5, or x = ଷ ଶ . QUESTION 20 The correct answer is ࠵? ૟ , .166, or .167. The circumference, C , of a circle is C = 2 ʋƌ , where r is the radius of the circle. For the given circle with a radius of 1, the circumference is C = 2( ʋ )(1), or C = 2 ʋ . To find what fraction of the circumference the length of arc AB is, divide the length of the arc by the circumference, which gives గ ଷ ൊ 2࠵? . This division can be represented by గ ଷ ∙ ଵ ଶగ ൌ ଵ ଺ . The fraction ଵ ଺ can also be rewritten as .166 or .167. Section 4: Math Test - Calculator QUESTION 1 Choice A is correct . The given expression (2 x 2 о 4) о ( о 3 x 2 + 2 x о 7) can be rewritten as 2 x 2 о 4 + 3 x 2 о 2 x + 7. Combining like terms yields 5 x 2 о 2 x + 3. Choices B, C, and D are incorrect because they are the result of errors when applying the distributive property. QUESTION 2 Choice C is correct . The lines shown on the graph give the positions of Paul and Mark during the race. At the start of the race, 0 seconds have elapsed, so the y -intercept of the line that repreƐenƚƐ Mark͛Ɛ poƐiƚion dƵring ƚhe race repreƐenƚƐ ƚhe nƵmber of yards Mark was from PaƵl͛Ɛ poƐiƚion ;aƚ Ϭ LJardƐͿ aƚ ƚhe Ɛƚarƚ of ƚhe race͘ BecaƵƐe ƚhe y -intercept of the line that

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, 92.1 о 11.3 = 80.8 square miles. The population density is the population divided by the land area, or 621,000 7,685 80.8 , 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. QUESTION 10 Choice B is correct . Let x represent the number of days the second voyage lasted. The number of days the first voyage lasted is then x + 43. Since the two voyages combined lasted a total of 1,003 days, the equation x + ( x + 43) = 1,003 must hold. Combining like terms yields 2 x + 43 = 1,003, and solving for x gives x = 480. Choice A is incorrect because 460 + (460 + 43) = 963, not 1,003 days. Choice C is incorrect because 520 + (520 + 43) = 1,083, not 1,003 days. Choice D is incorrect because 540 + (540 + 43) = 1,123, not 1,003 days. QUESTION 11 Choice B is correct . Adding the equations side-by-side eliminates y , as shown below. 7 3 8 6 3 5 13 0 13 x y x y x ³ ´ ³ Solving the obtained equation for x gives x = 1. Substituting 1 for x in the first equation gives 7(1) + 3 y = 8. Subtracting 7 from both sides of the equation yields 3 y = 1, so y = 1 3 . Therefore, the value of x о y is 1 о 1 3 , or 2 3 . Choice C is incorrect because 1 4 1 3 3 ³ is the value of x + y , not x о y . Choices A and D are incorrect and may be the result of some computational errors. QUESTION 12

opposite to the angle with measure 32° and AB is the hypotenuse in right triangle ABC , the ratio BC AB is equal to DF DE . Alternate approach: The trigonometric ratios can be used to answer this question. In right triangle ABC , the ratio sin(32 ) BC AB q . The angle E in triangle DEF has measure 32° because m( ) m( ) 90 D E ³ q . In triangle DEF , the ratio sin(32 ) DF DE q . Therefore, DF BC DE AB . Choice A is incorrect because DE DF is the inverse of the ratio BC AB . Choice C is incorrect because DF BC EF AC , not BC AB . Choice D is incorrect because EF AC DE AB , not BC AB . QUESTION 17 Choice B is correct . Isolating the term that contains the riser height, h , in the formula 2 h + d = 25 gives 2 h = 25 о d . Dividing both sides of this equation by 2 yields 25 2 d h ´ , or 1 (25 ) 2 h d ´ . Choices A, C, and D are incorrect and may result from incorrect transformations of the riser- tread formula 2 h + d = 25 when expressing h in terms of d . QUESTION 18 Choice C is correct . Since the tread depth, d , must be at least 9 inches, and the riser height, h , must be at least 5 inches, it follows that d ш ϵ and h ш ϱ͕ reƐpecƚiǀelLJ͘ Solǀing for d in the riser- tread formula 2 h + d = 25 gives d = 25 о 2 h . Thus the first inequality, d ш ϵ͕ is equivalent to 25 о 2 h ш ϵ͘ ThiƐ ineqƵaliƚLJ can be Ɛolǀed for h as follows: о 2 h ш 9 о 25 2 h ч 25 о 9 2 h ч 16 h ч 8

Choice B is correct . A linear function has a constant rate of change, and any two rows of the shown table can be used to calculate this rate. From the first row to the second, the value of x is increased by 2 and the value of f ( x ) is increased by 6 = 4 о ( о 2). So the values of f ( x ) increase by 3 for every increase by 1 in the value of x . Since f (2) = 4, it follows that f (2 + 1) = 4 + 3 = 7. Therefore, f (3) = 7. Choice A is incorrect. This is the third x -value in the table, not f (3). Choices C and D are incorrect and may reƐƵlƚ from errorƐ ǁhen calcƵlaƚing ƚhe fƵncƚion͛Ɛ raƚe of change . QUESTION 26 Choice C is correct . Since Gear A has 20 teeth and Gear B has 60 teeth, the gear ratio for Gears A and B is 20:60. Thus the ratio of the number of revolutions per minute (rpm) for the two gears is 60:20, or 3:1. That is, when Gear A turns at 3 rpm, Gear B turns at 1 rpm. Similarly, since Gear B has 60 teeth and Gear C has 10 teeth, the gear ratio for Gears B and C is 60:10, and the ratio of the rpms for the two gears is 10:60. That is, when Gear B turns at 1 rpm, Gear C turns at 6 rpm. Therefore, if Gear A turns at 100 rpm, then Gear B turns at 100 3 rpm, and Gear C turns at 100 6 200 3 u rpm. Alternate approach: Gear A and Gear C can be considered as directly connected since their ͞conƚacƚ͟ ƐpeedƐ are ƚhe Ɛame͘ Gear A has twice as many teeth as Gear C, and since the ratios of the number of teeth are equal to the reverse of the ratios of rotation speeds, in rpm, Gear C would be rotated at a rate that is twice the rate of Gear A. Therefore, Gear C will be rotated at a rate of 200 rpm since Gear A is rotated at 100 rpm. Choice A is incorrect and may result from using the gear ratio instead of the ratio of the rpm when calculating the rotational speed of Gear C. Choice B is incorrect and may result from comparing the rpm of the gears using addition instead of multiplication. Choice D is incorrect and may be the result of multiplying the 100 rpm for Gear A by the number of teeth in Gear C. QUESTION 27 Choice A is correct . One way to find the radius of the circle is to put the given equation in standard form, ( x о h ) 2 + ( y о k ) 2 = r 2 , where ( h, k ) is the center of the circle and the radius of the circle is r . To do this, divide the original equation, 2 x 2 о 6 x + 2 y 2 + 2 y = 45, by 2 to make the leading coefficients of x 2 and y 2 each equal to 1: x 2 о 3 x + y 2 + y = 22.5. Then complete the square to put the equation in standard form. To do so, first rewrite x 2 о 3 x + y 2 + y = 22.5 as ( x 2 о 3 x + 2.25) о 2.25 + ( y 2 + y + 0.25) о 0.25 = 22.5. Second, add 2.25 and 0.25 to both sides of the equation: ( x 2 о 3 x + 2.25) + ( y 2 + y + 0.25) = 25. Since x 2 о 3 x + 2.25 = ( x о 1.5) 2 , y 2 о x + 0.25 = ( y

2(5 x о 20) о 15 о 8 x = 7 2(5 x ) о 2(20) о 15 о 8 x = 7 (Apply the distributive property.) 10 x о 40 о 15 о 8 x = 7 (Multiply.) 2 x о 55 = 7 (Combine like terms.) 2 x = 62 (Add 55 to both sides of the equation.) x = 31 (Divide both sides of the equation by 2.) QUESTION 33 The possible correct answers are 97, 98, 99, 100, and 101. The volume of a cylinder can be found by using the formula V = ʋ r 2 h , where r is the radius of the circular base and h is the height of the cylinder. The smallest possible volume, in cubic inches, of a graduated cylinder produced by the laboratory supply company can be found by substituting 2 for r and 7.75 for h , giving V = ʋ (2 2 )(7.75). This gives a volume of approximately 97.39 cubic inches, which rounds to 97 cubic inches. The largest possible volume, in cubic inches, can be found by substituting 2 for r and 8 for h , giving V = ʋ (2 2 )(8). This gives a volume of approximately 100.53 cubic inches, which rounds to 101 cubic inches. Therefore, the possible volumes are all the integers greater than or equal to 97 and less than or equal to 101, which are 97, 98, 99, 100, and 101. Any of these numbers may be gridded as the correct answer. QUESTION 34 The correct answer is 5. The intersection points of the graphs of y = 3 x 2 о 14 x and y = x can be found by solving the system consisting of these two equations. To solve the system, substitute x for y in the first equation. This gives x = 3 x 2 о 14 x . Subtracting x from both sides of the equation gives 0 = 3 x 2 о 15 x. Factoring 3 x out of each term on the left-hand side of the equation gives 0 = 3 x ( x о 5). Therefore, the possible values for x are 0 and 5. Since y = x , the two intersection points are (0, 0) and (5, 5). Therefore, a = 5. QUESTION 35 The correct answer is 1.25 or 5 4 . The y -coordinate of the x -intercept is 0, so 0 can be substituted for y , giving 4 5 x + 1 3 (0) = 1. This simplifies to 4 5 x = 1. Multiplying both sides of 4 5 x

= 1 by 5 gives 4 x = 5. Dividing both sides of 4 x = 5 by 4 gives x = 5 4 , which is equivalent to 1.25. Either 5/4 or 1.25 may be gridded as the correct answer. QUESTION 36 The correct answer is 2.6 or 13 5 . Since the mean of a set of numbers can be found by adding the numbers together and dividing by how many numbers there are in the set, the mean mass, in kilograms, of the rocks Andrew collected is 2.4 2.5 3.6 3.1 2.5 2.7 6 ³ ³ ³ ³ ³ = 16.8 6 = 2.8. Since the mean mass of the rocks Maria collected is 0.1 kilogram greater than the mean mass of rocks Andrew collected, the mean mass of the rocks Maria collected is 2.8 + 0.1 = 2.9 kilograms. The value of x can be found by using the algorithm for finding the mean: 3.1 2.7 2.9 3.3 2.8 2.9 6 x ³ ³ ³ ³ ³ . Solving this equation gives x = 2.6, which is equivalent to 13 5 . Either 2.6 or 13/5 may be gridded as the correct answer. QUESTION 37 The correct answer is 30. The situation can be represented by the equation x (2 4 ) = 480, 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 x (2 4 ) = 480 gives 16 x = 480. Therefore, x = 30. QUESTION 38 The correct answer is 8. The 6 students represent (100 о 15 о 45 о 25)% = 15% of those invited to join the committee. If x people were invited to join the committee, then 0.15 x = 6. Thus, there were 6 0.15 = 40 people invited to join the committee. It follows that there were 0.45(40) = 18 teachers and 0.25(40) = 10 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.

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