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PART 4 QUESTION 7 Choice D is correct. Multiplying both sides of the given equation by [ yields 160[ = 8. Dividing both sides of the equation 160[ = 8 by 160 results in [ = 8
\160 . Reducing 8
\160 to its simplest form gives [ = 1
\20 , or its decimal equivalent 0.05. Choice A is incorrect and may result from multiplying, instead of dividing, the left-hand side of the given equation by 160. Choice B is incorrect and may result from a computational error. Choice C is incorrect. This is the value of 1 _[ . QUESTION 8 Choice C is correct. Applying the distributive property of multiplication to the right-hand side of the given equation gives (3[+ 15) + (5[
? RU [ + 10. An equation in the form F[ + G = U[ + V will have no solutions if F = U and G 
?V . Therefore, it follows that the equation 2D[
? [ + 10 will have no solutions if 2D= 8 , or D = 4 . Choice A is incorrect. If D = 1, then the given equation could be written as 2[
? [6LQFH 
?WKLVHTXDWLRQKDVH[DFWO\RQHVROXWLRQ  Choice B is incorrect. If D = 2, then the given equation could be written as 4[
? [ 6LQFH 
?WKLVHTXDWLRQKDVH[DFWO\RQHVROXWLRQ  Choice D is incorrect. If D = 8, then the given equation could be written as 16[ 
? [ 6LQFH 
?WKLVHTXDWLRQKDVH[DFWO\RQHVROXWLRQ QUESTION 9 Choice B is correct. A solution to the system of three equations is any ordered pair ([, \) that is a solution to each of the three equations. Such an ordered pair ([ , \) must lie on the graph of each equation in the [\ SODQHLQRWKHUZRUGVLWPXVWEHDSRLQWZKHUHDOOWKUHHJUDSKV  intersect. The graphs of all three equations intersect at exactly one SRLQW 
? 7KHUHIRUHWKHV\VWHPRIHTXDWLRQVKDVRQHVROXWLRQ Choice A is incorrect. A system of equations has no solutions when there is no point at which all the graphs intersect. Because the graphs RIDOOWKUHHHTXDWLRQVLQWHUVHFWDWWKHSRLQW 
? WKHUHLVDVROXWLRQ  Choice C is incorrect. The graphs of all three equations intersect DWRQO\RQHSRLQW 
? 6LQFHWKHUHLVQRRWKHUVXFKSRLQWWKHUH  cannot be two solutions. Choice D is incorrect and may result from counting the number of points of inter section of the graphs of any two equations, including the point of intersection of all three equations. QUESTION 10 Choice C is correct. If the equation is true for all [ , then the expressions on both sides of the equation will be equivalent. Multiplying the polynomials on the left-hand side of the equation gives 5D[ 3
?DE[ 2 + 4D[+ 15[ 2
?E[2QWKHULJKWKDQGVLGHRI the equation, the only [ 2WHUPLV 
?[ 2. Since the expressions on both



PART 4 _(LJKW2
?FLDO3UDFWLFH7HVWVZLWK$QVZHU([SODQDWLRQV 538 are (3, 1), so h = 3 and k 7KHUHIRUHDQHTXDWLRQWKDWGH
?QHVf can be written as f([ ) = D ([ 
? 27R
?QGD , substitute a value for [ and its corresponding value for \ , or f([ ). For example, (4, 5) is a point on the graph of f . So D must satisfy the equation 5 = D 
? 2 + 1, which can be rewritten as 4 = D (1) 2, or D $QHTXDWLRQWKDWGH
?QHVf is therefore f([ ) = 4([ 
? 2 + 1. Choice B is incorrect and may result from a sign error when writing the equation of the parabola in vertex form. Choice C is incorrect and may result from omitting the constant D from the vertex form of the equation of the parabola. Choice D is incorrect and may result from a sign error when writing the equation of the parabola in vertex form as well as by miscalculating the value of D . QUESTION 14 Choice B is correct.7KHVROXWLRQVRIWKH
?UVWLQHTXDOLW\ \
?[ + 2, lie on or above the line \ = [ + 2, which is the line that passes through
? DQG   7KHVHFRQGLQHTXDOLW\FDQEHUHZULWWHQLQVORSH intercept form by dividing the second inequality, 2[ + 3\ 
?E\on both sides, which yields 2
\ 3 [ + \ 
? DQGWKHQVXEWUDFWLQJ 2
\ 3 [ from both sides, which yields \ 
?
? 2
\ 3 [ + 2. The solutions to this inequality lie on or below the line \ 
? 2
\ 3 [ + 2, which is the line that passes through (0, 2) and (3, 0). The only graph in which the shaded region meets these criteria is choice B. Choice A is incorrect and may result from reversing the inequality sign LQWKH
?UVWLQHTXDOLW\&KRLFH&LVLQFRUUHFWDQGPD\UHVXOWIURPUHYHUVLQJ the inequality sign in the second inequality. Choice D is incorrect and may result from reversing the inequality signs in both inequalities. QUESTION 15 Choice B is correct. Squaring both sides of the given equation yields [ + 2 = [ 2. Subtracting [ and 2 from both sides of [ + 2 = [ 2 yields [ 2
?[
? )DFWRULQJWKHOHIWKDQGVLGHRIWKLVHTXDWLRQ\LHOGV  ([
? [+ 1) = 0. Applying the zero product property, the solutions to ([
? [+ 1) = 0 are [
? RU [= 2 and [ + 1 = 0, or [ 
?  Substituting [ = 2 in the given equation gives
?  
?__ ZKLFKLV  false because
?  __ E\WKHGH
?QLWLRQRIDSULQFLSDOVTXDUHURRW  So, [ = 2 isn?t a solution. Substituting [  
? LQWRWKHJLYHQHTXDWLRQ  gives
?  
?
? __ ZKLFKLVWUXHEHFDXVH 
?
?  6R [ 
? LVWKH  only solution. Choices A and C are incorrect. The square root symbol represents the principal, or nonnegative, square root. Theref ore, in the equation
? [  
?[_ WKHYDOXHRI
?[ must be zero or positive. If [ = 2, then
?[ 
?ZKLFKLVQHJDWLYHVRFDQ
^WEHLQWKHVHWRIVROXWLRQV  Choice D is incorrect and may r esult from incorrectly reasoning that
? [ always has a negative value and therefore can?t be equal to a value of a principal square root, which cannot be negative.

ANS WER EX PLANATIONS _6$73UDFWLFH7HVW 539 QUESTION 16 The correct answer is 360. The volume of a right rectangular prism is calculated by multiplying its dimensions: length, width, and height. Multiplying the values given for these dimensions yields a volume of      FXELFFHQWLPHWHUV QUESTION 17 The correct answer is 2. The left-hand side of the given equation contains a common factor of 2 and can be rewritten as 2(2[+ 1). Dividing both sides of this equation by 2 yields 2[ + 1 = 2. Therefore, the value of 2[+ 1 is 2. Alternate approach: Subtracting 2 from both sides of the given equation yields 4[ = 2. Dividing both sides of this equation by 4 yields [ = 1
\2 . Substituting 1
\2 for [ in the expression 2[ + 1 yields 2 ( 1
\ 2) + 1 = 2 . QUESTION 18 The correct answer is 8. The graph shows that the maximum value of f([ ) is 2. Since g([ ) = f ([ ) + 6, the graph of g is the graph of f shifted up by 6 units. Therefore, the maximum value of g ([ ) is 2 + 6 = 8. QUESTION 19 The correct answer is 3?4 , or .75.%\GH
?QLWLRQRIWKHVLQHUDWLRVLQFH  sin R = 4
\ 5, 34 _ 35 = 4
\5 . Ther efore, if 34= 4 Q , then 35= 5 Q , where Q is a positive constant. Then 45 = NQ, where k is another positive constant. Applying the Pythagorean theorem, the following relationship holds: (NQ) 2 + (4Q ) 2 = (5Q) 2, or k 2Q2 + 16Q 2 = 25Q 2. Subtracting 16Q 2 from both sides of this equation yields k 2Q2 Q 2. Taking the square root of both sides of k 2Q2 Q 2 yields NQ = 3Q . It follows that k = 3. Therefore, if 34 = 4Q and 35 = 5Q , then 45 = 3Q DQGE\GH
?QLWLRQRIWKHWDQJHQWUDWLR  tan 3 = 3 Q
\4Q , or 3
\4 (LW KHU  RU PD\EHHQWHUHGDVWKHFRUUHFWDQVZHU QUESTION 20 The correct answer is 2.5. The graph of the linear function f passes through the points (0, 3) and (1, 1). The slope of the graph of the function f is therefore 
? _ 
?  
? ,W
^VJLYHQWKDWWKHJUDSKRIWKHOLQHDU  function g is perpendicular to the graph of the function f . Therefore, the slope of the graph of the function gLVWKHQHJDWLYHUHFLSURFDORI 
?ZKLFK  LV
? 1
\
? = 1
\2  DQGDQHTXD  WLRQWKDWGH
?QHVWKHIXQFWLRQg is g([ ) = 1
\ 2 [ + b , where b is a constant. Since it?s given that the graph of the function g passes through the point (1, 3), the value of b can be found using the equation 3 = 1
\ 2 (1) + b . Solving this equation for b yields b = 5
\2 , so an HTXDWLRQWKDWGH
?QHVWKHIXQFWLRQg is g([ ) = 1
\ 2 [ + 5
\2 . F inding the value of g (0) by substituting 0 for [ into this equation yields g(0) = 1
\ 2 (0) + 5
\2 , or 5
\2 . (LWKHU  RU PD\EHHQWHUHGDVWKHFRUUHFWDQVZHU

PART 4 _(LJKW2
?FLDO3UDFWLFH7HVWVZLWK$QVZHU([SODQDWLRQV 540 Section 4: Math Test ? Calculator QUESTION 1 Choice B is correct. Subtracting 3 from both sides of the equation yields 3[ = 24. Dividing both sides of this equation by 3 yields [ = 8. &KRLFH$LVLQFRUUHFWDQGPD\UHVXOWIURP
?QGLQJDFRPPRQIDFWRU DPRQJWKHWKUHHJLYHQWHUPVLQVWHDGRI
?QGLQJ[ . Choice C is incorrect and may result from incorrectly adding 3 to, instead of subtracting 3 from, the right-hand side of the equation. Choice D is incorrect. This is the value of 3[ + 3, not the value of [ . QUESTION 2 Choice D is correct. Since 1 cubit is equivalent to 7 palms, 140 cubits DUHHTXLYDOHQWWR   SDOPVRUSDOPV Choice A is incorrect and may result from dividing 7 by 140. Choice B is incorrect and may result from dividing 140 by 7. Choice C is incorrect. This is the length of the Great Sphinx statue in cubits, not palms. QUESTION 3 Choice B is correct. Multiplying both sides of the given equation by 5 yields 2Q = 50. Substituting 50 for 2Q in the expression 2Q
? \LHOGV  
?  Alternate approach: Dividing both sides of 2Q = 50 by 2 yields Q = 25. (YDOXDWLQJWKHH[SUHVVLRQ Q 
? IRU Q  \LHOGV   
?  &KRLFH$LVLQFRUUHFWDQGPD\UHVXOWIURP
?QGLQJWKHYDOXHRI Q 
?  instead of 2Q 
?&KRLFH&LVLQFRUUHFWDQGPD\UHVXOWIURP
?QGLQJWKH  value of 2Q instead of 2Q
?&KRLFH'LVLQFRUUHFWDQGPD\UHVXOWIURP 
?QGLQJWKHYDOXHRI Q 
? LQVWHDGRI Q
? QUESTION 4 Choice A is correct. The square root symbol represents the principal, or nonnegative, square root. Therefore, the equa tion
? _ [2= [ is only true for values of [ JUHDWHUWKDQRUHTXDOWR7K XV 
? LVQ
^WDVROXWLRQWR the giv en equation. Choices B, C, and D are incorrect because these values of [ are solutions to the equa tion
? _ [2= [ . Choosing one of t hese as a value of [ that isn?t a solution may result from incorrectly using the rules of exponents or incorrectly evaluating these values in the given equation.

ANS WER EX PLANATIONS _6$73UDFWLFH7HVW 541 QUESTION 5 Choice D is correct. The [-axis of the graph represents the time, in PLQXWHVDIWHUWKHFR
?HHZDVUHPRYHGIURPWKHKHDWVRXUFHDQGWKH \ -axis of the graph represents the temperature, in degrees Fahrenheit, RIWKHFR
?HH7KHFR
?HHZDV
?UVWUHPRYHGIURPWKHKHDWVRXUFH when [ = 0. The graph shows that when [ = 0, the \ -value was DOLWWOHOHVVWKDQ r)2IWKHDQVZHUFKRLFHVJLYHQLVWKH  best approximation. &KRLFH$LVLQFRUUHFWDQGPD\UHVXOWIURP
?QGLQJWKHWHPSHUDWXUHDIWHU PLQXWHV&KRLFH%LVLQFRUUHFWDQGPD\UHVXOWIURP
?QGLQJWKH temperature after 50 minutes. Choice C is incorrect and may result from
?QGLQJWKHWHPSHUDWXUHDIWHUPLQXWHV QUESTION 6 Choice A is correct. The average rate of change in temperature of WKHFR
?HHLQGHJUHHV)DKUHQKHLWSHUPLQXWHLVFDOFXODWHGE\GLYLGLQJ WKHGL
?HUHQFHEHWZHHQWZRUHFRUGHGWHPSHUDWXUHVE\WKHQXPEHURI minutes in the corresponding interval of time. Since the time intervals given are all 10 minutes, the average rate of change is greatest for the SRLQWVZLWKWKHJUHDWHVWGL
?HUHQFHLQWHPSHUDWXUH2IWKHFKRLFHVWKH JUHDWHVWGL
?HUHQFHLQWHPSHUDWXUHRFFXUVEHWZHHQDQGPLQXWHV Choices B, C, and D are incorrect and may result from misinterpreting the average rate of change from the graph. QUESTION 7 Choice C is correct. It?s given that [ WKHUHIRUHVXEVWLWXWLQJ  for [ in triangle $%& gives two known angle measures for this triangle. The sum of the measures of the interior angles of any triangle equals 180?. Subtracting the two known angle measures of triangle $%& from r JLYHVWKHWKLUGDQJOHPHDVXUH r
?r
?r r7KLVLV  the measure of angle %&$ . Since vertical angles are congruent, the measure of angle '&( is also 60?. Subtracting the two known angle measures of triangle &'( from 180? gives the third angle measure: r
?r
?r r7KHUHIRUHWKHYDOXHRI\ is 80. Choice A is incorrect and may result from a calculation error. Choice B is incorrect and may result from classifying angle &'( as a right angle. &KRLFH'LVLQFRUUHFWDQGPD\UHVXOWIURP
?QGLQJWKHPHDVXUHRIDQJOH  %&$ or '&( instead of the measure of angle &'(. QUESTION 8 Choice A is correct. The cost of each additional mile traveled is represented by the slope of the given line. The slope of the line can be calculated by identifying two points on the line and then calculating the ratio of the change in \ to the change in [ between the two points. Using the points (1, 5) and (2, 7), the slope is equal to 
?_ 
? , or 2. Ther efore, the cost f or each additional mile traveled of the cab ride is $2.00 .

PART 4 _(LJKW2
?FLDO3UDFWLFH7HVWVZLWK$QVZHU([SODQDWLRQV 542 Choice B is incorrect and may result from calculating the slope of the line that passes through the points (5, 13) and (0, 0). However, (0, 0) does not lie on the line shown. Choice C is incorrect. This is the \-coordinate of the \ LQWHUFHSWRIWKHJUDSKDQGUHSUHVHQWVWKH
?DWIHHIRU a cab ride before the charge for any miles traveled is added. Choice D is incorrect. This value represents the total cost of a 1-mile cab ride. QUESTION 9 Choice D is correct. The total number of gas station customers on Tuesday was 135. The table shows that the number of customers who did not purchase gasoline was 50. Finding the ratio of the number of customers who did not purchase gasoline to the total number of customers gives the probability that a customer selected at random on that day did not purchase gasoline, which is 50
\135 . &KRLFH$LVLQFRUUHFWDQGPD\UHVXOWIURP
?QGLQJWKHSUREDELOLW\WKDWD customer did not purchase a beverage, given that the customer did not SXUFKDVHJDVROLQH&KRLFH%LVLQFRUUHFWDQGPD\UHVXOWIURP
?QGLQJ the probability that a customer did not purchase gasoline, given that the customer did not purchase a beverage. Choice C is incorrect and PD\UHVXOWIURP
?QGLQJWKHSUREDELOLW\WKDWDFXVWRPHUGLGSXUFKDVHD beverage, given that the customer did not purchase gasoline. QUESTION 10 Choice D is correct. It is given that the number of students surveyed was 336. Finding 1
\4 of 336 yields (1
\ 4) (336) = 84 , the number of IUHVKPHQDQG
?QGLQJ 1
\3 of 336 yields (1
\ 3) (336) = 112 , the number of sophomores. Subtracting these numbers from the total number of VHOHFWHGVWXGHQWVUHVXOWVLQ 
?
? WKHQXPEHURIMXQLRUV  and seniors combined. Finding half of this total yields (1
\ 2) (140) = 70 , WKHQXPEHURIMXQLRUV6XEWUDFWLQJWKLVQXPEHUIURPWKHQ XPEHURIMXQLRUV DQGVHQLRUVFRPELQHG\LHOGV 
? WKHQXPEHURIVHQLRUV Choices A and C are incorrect and may result from calculation errors. &KRLFH%LVLQFRUUHFW7KLVLVWKHWRWDOQXPEHURIMXQLRUVDQGVHQLRUV QUESTION 11 Choice A is correct. It?s given that the ratio of the heights of Plant A to Plant B is 20 to 12 and that the height of Plant C is 54 centimeters. Let [ be the height of Plant D. The proportion 20
\ 12 = 54
\[ can be used to solve for the value of [ . Multiplying bot h sides of this equation by [ yields 20[
\ 12 = 54 and then multiplying both sides of this equation by 12 yields 20[ = 648 . Dividing both sides of this equation by 20 yields [ = 32.4 centimeters.

ANS WER EX PLANATIONS _6$73UDFWLFH7HVW 543 Choice B is incorrect and may result from a calculation error. Choice C LVLQFRUUHFWDQGPD\UHVXOWIURP
?QGLQJWKHGL
?HUHQFHLQKHLJKWV between Plant A and Plant B and then adding that to the height of Plant C. Choice D is incorrect and may result from using the ratio 12 to 20 rather than 20 to 12. QUESTION 12 Choice D is correct. It?s given that 1 kilometer is approximately equivalent to 0.6214 miles. Let [ be the number of kilometers equivalent to 3.1 miles. The proportion 1 kilometer
\ 0.6214 miles = [ kilometers
\3.1 miles can be used to solve for the value of [ . Multiplying both sides of this equation by 3.1 yields 3.1
\ 0.6214 = [, or [ 
? 7KLVLVDSSUR[LPDWHO\NLORPHWHUV Choice A is incorrect and may result from misidentifying the ratio of kilometers to miles as miles to kilometers. Choice B is incorrect and may result from calculation errors. Choice C is incorrect and may result from calculation and rounding errors. QUESTION 13 Choice C is correct. Let D equal the number of 120-pound packages, and let b equal the number of 100-pound packages. It?s given that the total weight of the packages can be at most 1,100 pounds: the inequality 120D + 100b 
? UHSUHVHQWVWKLVVLWXDWLRQ,W
^VDOVR  given that the helicopter must carry at least 10 packages: the inequality D + b 
? UHSUHVHQWVWKLVVLWXDWLRQ9DOXHVRID and b that satisfy these two inequalities represent the allowable numbers of 120-pound packages and 100-pound packages the helicopter can transport. To maximize the number of 120-pound packages, D , in the helicopter, the n umber of 100-pound packages, b , in the helicopter needs to be PLQLPL]HG([SUHVVLQJb in terms of D in the second inequality yields b 
?
?D , so the minimum value of b LVHTXDOWR 
?D. Substituting 
?D for bLQWKH
?UVWLQHTXDOLW\UHVXOWVLQ D  
?D 
?  Using the distributive property to rewrite this inequality yields 120D
?D 
?RU D
?6XEWUDFWLQJ  from both sides of this inequality yields 20D 
?'LYLGLQJERWKVLGHV  of this inequality by 20 results in D
?7KLVPHDQVWKDWWKHPD[LPXP  number of 120-pound packages that the helicopter can carry per trip is 5. Choices A, B, and D are incorrect and may result from incorrectly cr eating or solving the system of inequalities. QUESTION 14 Choice B is correct.7KHGL
?HUHQFHEHWZHHQWKHPDFKLQH
^VVWDUWLQJ  value and its value after 10 years can be found by subtracting $30,000 IURP  
? ,W
^VJLYHQWKDWWKHYDOXHRI  the machine depreciates by t he same amount each year for 10 years. 'LYLGLQJ  E\JL YHV ZKLFKLVWKHDPRXQWE\ZKLFK  the value depreciates each year. Therefore, over a period of t years,

PART 4 _(LJKW2
?FLDO3UDFWLFH7HVWVZLWK$QVZHU([SODQDWLRQV 544 WKHYDOXHRIWKHPDFKLQHGHSUHFLDWHVE\DWRWDORI t dollars. The value v of the machine, in dollars, t years after it was purchased is the starting value minus the amount of depreciation after t years, or v 
?t . Choice A is incorrect and may result from using the value of the machine after 10 years as the machine?s starting value. Choice C is incorrect. This equation shows the amount the machine?s value changes each year being added to, rather than subtracted from, the starting value. Choice D is incorrect and may result from multiplying the machine?s value after 10 years by t instead of multiplying the amount the machine depreciates each year by t . QUESTION 15 Choice D is correct. The slope-intercept form of a linear equation is \ = D[ + b , where D is the slope of the graph of the equation and b is the \-coordinate of the \ -intercept of the graph. Two ordered pairs ([1, \ 1) and ([2, \ 2) can be used to compute the slope of the line with the formula D = \ 2
?\ 1
\[2
?[ 1 . Substituting t he two ordered pairs (2, 4) and (0, 1) into this formula gives D = 
? _ 
?  ZKLFKVLPSOL
?HVWR   3
\2 . Substituting this value for D in the slope-intercept form of the equation yields \ = 3
\ 2 [ + b . Substituting values from the ordered pair (0, 1) into this equation yields 1 = 3
\ 2 (0) + b , so b = 1. Substituting this value for b in the slope-intercept equation yields \ = 3
\ 2 [ + 1 . Choice A is incorrect. This may result from misinterpreting the change in [-values as the slope and misinterpreting the change in \ -values as the \-coordinate of the \ -intercept of the graph. Choice B is incorrect and may result from using the [ - and \-values of one of the given points as t he slope and \ -coordinate of the \ -intercept, respectively. Choice C is incorrect. This equation has the correct slope but the incorrect \-coordinate of the \ -intercept. QUESTION 16 Choice B is correct. Multiplying the binomials in the given expression results in 4D[ 2 + 4D[
?[ 
?
?[ 2 + 4. Combining like terms yields 4 D[ 2 + 4D[
?
?[ 2. Grouping by powers of [ and factoring out their greatest common factors yields (4D
? [ 2 + (4D 
? [. It?s given that this expression is equivalent to E[, so (4D
? [ 2 + (4D 
? [ = E[. Since the right-hand side of the equation has no [ 2WHUPWKHFRH
?FLHQWRI  the [ 2 term on the left-hand side must be 0. This gives 4D 
?  DQG  4D 
? b . Since 4D
?  D = 1. Substituting the value of 4D into the VHFRQGHTXDWLRQJLYHV 
? b , so b 
? Choices A, C, and D are incorrect and may result from a calculation error.

ANS WER EX PLANATIONS _6$73UDFWLFH7HVW 545 QUESTION 17 Choice C is correct. Multiplying both sides of 2Z + 4t = 14 by 2 yields 4Z + 8t = 28. Subtracting the second given equation from 4Z + 8t = 28 yields (4Z
?Z ) + (8t 
?t   
? RUt = 3. Dividing both sides of this equation by 3 yields t = 1. Substituting 1 for t in the equation 2Z + 4t = 14 yields 2Z + 4(1) = 14, or 2Z + 4 = 14. Subtracting 4 from both sides of this equation yields 2Z = 10, and dividing both sides of this equation by 2 yields Z = 5. Substituting 5 for Z and 1 for t in the expression 2Z + 3t yields 2(5) + 3(1) = 13. Choices A, B, and D are incorrect and may result from incorrectly calculating the values of Z and t, or from correctly calculating the values of Z and tEXW
?QGLQJWKHYDOXHRIDQH[SUHVVLRQRWKHUWKDQ 2 Z + 3t . For instance, choice A is the value of Z + t , choice B is the value of 2Z, and choice D is the value of 2t + 3Z . QUESTION 18 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 [ servings would provide [ times 5%. The percentage of an adult?s daily allowance of potassium, S , is 5 times the number of servings, [ . Therefore, the percentage of an adult?s daily allowance of potassium can be expressed as S = 5[ . Choices A, C, and D are incorrect and may result from incorrectly converting 5% to its decimal equivalent, which isn?t necessary since S is expressed as a percentage. Additionally, choices C and D are incorrect because the context should be r epresented by a linear relationship, not by an exponential relationship. QUESTION 19 Choice B is correct. It?s given that a 3
\4 -cup serving of Crunch y Grain cereal provides 210 calories. The total number of calories per cup can be found by dividing 210 by 3
\4 , which gi ves 210 ? 3
\ 4 = 280 calories per cup. Let F be the number of cups of Crunch y Grain cereal and V be the number of cups of Super Grain cereal. The expression 280F represents the number of calories in F cups of Crunch y Grain cereal, and 240V represents the number of cal ories in V cups of Super Grain cereal. The equation 280F + 240V = 270 gives t he total number of calories in one cup of the mixture. Since F + V = 1 cup, F 
?V 6XEVWLWXWLQJ 
?V f or F in the equation 280F + 240V  \LHOGV 
?V ) + 240V = 270, RU
?V + 240V = 270. Simplifying this equation yields 
?V  6XEWUDFWLQJIURPERWKVLGHVUHVXOWVLQ
?V  
? 'LYLGLQJERWKVLGHVRIWKHHTXDWLRQE\
?UHVXOWVLQV = 1
\4 , so t here is 1
\4 cup of Super Grain cereal in one cup of the mixture. Choices A, C, and D are incorrect and may result from incorrectly creating or solving the system of equations.

PART 4 _(LJKW2
?FLDO3UDFWLFH7HVWVZLWK$QVZHU([SODQDWLRQV 546 QUESTION 20 Choice A is correct. There are 0 calories in 0 servings of Crunchy Grain cereal so the line must begin at the point (0, 0). Point (0, 0) is t he origin, labeled O . Additionally, each serving increases the calories by 250. Therefore, the number of calories increases as the number RIVHUYLQJVLQFUHDVHVVRWKHOLQHPXVWKDYHDSRVLWLYHVORSH2IWKH choices, only choice A shows a graph with a line that begins at the origin and has a positive slope. Choices B , C, and D are incorrect. These graphs don?t show a line that passes through the origin. Additionally, choices C and D may result from misidentifying the slope of the graph. QUESTION 21 Choice D is correct. Since the function h is exponential, it can be writt en as h([ ) = DE [, where D is the \-coordinate of the \ -intercept and b is t he growth rate. Since it?s given that the \ -coordinate of the \-int ercept is G , the exponential function can be written as h([ ) = GE [. These conditions ar e only met by the equation in choice D. Choice A is incorrect. For this function, the value of h ([ ) when [ = 0 LV 
?QRWG . Choice B is incorrect. This function is a linear function, not an exponential function. Choice C is incorr ect. This function is a polynomial function, not an exponential function. QUESTION 22 Choice B is correct. The median weight is found by ordering the horses? weights from least to greatest and then determining the middle value from this list of weights. Decreasing the value for the horse ZLWKWKHORZHVWZHLJKWGRHVQ
^WD
?HFWWKHPHGLDQVLQFHLW
^VVWLOOWKH lowest value. &KRLFH$LVLQFRUU HFW7KHPHDQLVFDOFXODWHGE\
?QGLQJWKHVXPRIDOO the weights of the horses and then dividing by the number of horses. Decreasing one of the weights would decrease the sum and therefore GHFUHDVHWKHPHDQ&KRLFH&LVLQFRUUHFW5DQJHLVWKHGL
?HUHQFH between the highest and lowest weights, so decreasing the lowest weight would increase the range. Choice D is incorrect. Standard deviation is calculated based on the mean weight of the horses. Decreasing one of the weights decreases the mean and therefore would D
?HFWWKHVWDQGDUGGHYLDWLRQ QUESTION 23 Choice B is correct. In order for the poll results from a sample of a population to represent the entire population, the sample must be representative of the population. A sample that is randomly selected from a population is more likely than a sample of the type described to represent the population. In this case, the people who responded were people with access to cable television and websites,

ANS WER EX PLANATIONS _6$73UDFWLFH7HVW 547 which aren?t accessible to the entire population. Moreover, the people who responded also chose to watch the show and respond to the poll. The people who made these choices aren?t representative of the entire population of the United States because they were not a random sample of the population of the United States. Choices A, C, and D are incorrect because they present reasons unrelated to whether the sample is representative of the population of the United States. QUESTION 24 Choice C is correct. Substituting [ + D for [ in f( [ ) = 5[ 2
? \LHOGV  f( [ + D ) = 5([ + D) 2
?([SDQGLQJWKHH[SUHVVLRQ  [ + D ) 2 by multiplication yields 5[ 2 + 10D[ +5D 2, and thus f ( [ + D ) = 5[ 2 + 10D[ + 5D 2
?6HWWLQJ  the expression on the right-hand side of t his equation equal to the given expression f or f ( [ + D ) yields 5[ 2 + 30[ + 42 = 5[ 2 + 10D[ + 5D 2
?  Because this equality must be true for all values of [ WKHFRH
?FLHQWVRI each power of [DUHHTXDO6HWWLQJWKHFRH
?FLHQWVRI[ equal to each other gives 10D = 30. Di viding each side of this equation by 10 yields D = 3. Choices A, B, and D ar e incorrect and may result from a calculation error. QUESTION 25 Choice C is correct. The sine of an angle is equal to the cosine of the angle?s complement. This relationship can be expressed by the equation sin [ r FRV r
? [?). Therefore, if sin [ ? = D, then FRV r
?[ ?) must also be equal to D . Choices A and B ar e incorrect and may result from misunderstanding the r elationship between the sine and cosine of complementary angles. Choice D is incorrect and ma y result from misinterpreting sin ( [ 2)? as sin 2 ( [ )?. QUESTION 26 Choice D is correct. The positive [ -intercept of the graph of \ = h ([ ) is a point ([, \) for which \ = 0. Since \ = h ([ ) models the height above W KHJURXQGLQIHHWRIWKHSURMHFWLOHD\ -value of 0 must correspond WRWKHKHLJKWRIWKHSURMHFWLOHZKHQLWLVIHHWDERYHJURXQGRULQ RWKHUZRUGVZKHQWKHSURMHFWLOHLVRQWKHJURXQG6LQFH[ represents WKHWLPHVLQFHWKHSURMHFWLOHZDVODXQFKHGLWIROORZVWKDWWKHSRVLWLYH  [ -int ercept, ([ UHSUHVHQWVWKHWLPHDWZKLFKWKHSURMHFWLOHKLWV  t he ground. Choice A is incorrect and may result from misidentifying the \-intercept as a positive [ -intercept. Choice B is incorrect and may result from misidentifying the \ -value of the vertex of the graph of the function as an [-intercept. Choice C is incorrect and may result from misidentifying the [-value of the vertex of the graph of the function as an [ -intercept.

PART 4 _(LJKW2
?FLDO3UDFWLFH7HVWVZLWK$QVZHU([SODQDWLRQV 548 QUESTION 27 Choice A is correct. Since where the graph of f crosses the [-axis, it must be true that and f ( b ) = 0 and that f( [ ) the given choices, choice A is the only function for which this is true. If f( [ ) = ([ 
?D)([
?b ) , then as f ( D ) = 0(D 
?b) , or f( D ) = 0 rewritten as f( b ) = (b 
?D)(0) is equal to 0, then it follows that either each of these equations by adding D WRERWKVLGHVRIWKH
?UVWHTXDWLRQ and adding b to both sides of the second equation yields Therefore, the graph of two points, (D , 0) and (b , 0) Choice B is incorrect because because it?s given that D and b are positive. Choice C is incorrect because f ( b ) = (b 
?D)(2b cross the [ -axis at only one point, since same point. Choice D is incorrect because its graph crosses the [ -axis at (0, 0) as well as at (D , 0) (D , 0) is not equal to 0 for any other value of [ 2I f( D ) = (D 
?D)(D
?b ) . Also, , or f ( [ ) = ([ 
?D)([
?b ) . LWVJUDSKFRXOGRQO\EHLI and and (b , 0) are the only two points f ( D ) = 0  , which can be rewritten f ( b ) = (b 
?D)(b
?b ), which can be f ( b ) = 0. Furthermore, if f( [ ) = ([ 
?D)([
?b ) [ 
?D = 0 or [
?b = 0. Solving  [ = D or [ = b . crosses the [ -axis at exactly f ( D ) = (2D )(D + b ), which can?t be 0  b = D , but it would (D , 0) and (b, 0) would be the (b , 0). QUESTION 28 Choice C is correct. Substituting 0 for [ in the given equation yields 3(0) 2 + 6(0) + 2 = 2. Therefore, the graph of the given equation passes through the point (0, 2), which is the \-intercept of the graph. The right-hand side of the given equation, \ = 3[ 2 + 6[ + 2, displays the constant 2, which directly corresponds to the \ -coordinate of the \-intercept of the graph of this equation in the [\ -plane. Choice A is incorrect. The \ -coordinate of the vertex of the graph LV
?QRWRU&KRLFH%LVLQFRUUHFW7KH[ -coordinates of the [LQWHUFHSWVRIWKHJUDSKDUHDWDSSUR[LPDWHO\ 
? DQG 
?QRW  6, or 2. Choice D is incorrect. The [ -coordinate of the [-intercept of the OLQHRIV\PPHWU\LVDW 
?QRWRU QUESTION 29 Choice A is correct. The given equation is in slope-intercept form, or \ = P[ + b , where m LVWKHYDOXHRIWKHVORSHRIWKHOLQHRIEHVW
?W  7KHUHIRUHWKHVORSHRIWKHOLQHRIEHVW
?WLV )URPWKHGH
?QLWLRQ  of slope, it follows that an increase of 1 in t he [ -value corresponds WRDQLQFUHDVHRI  LQWKH\ Y DOXH7KHUHIRUHWKHOLQHRIEHVW
?W  SUHGLFWVWKDWI RUHDFK\HDUEHWZHHQDQGWKHPLQLPXPZDJH  ZLOOLQFUHDVHE\   GROODUSHUKRXU Choice B is incorrect and may result from using the \ -coordinate of the \-intercept as the average increase, instead of the slope. Choice C is incorrect and may result from using the 10-year increments given on the [ D[LVWRLQFRUUHFWO\LQWHUSUHWWKHVORSHRIWKHOLQHRIEHVW
?W Choice D is incorrect and may result from using the \ -coordinate

ANS WER EX PLANATIONS _6$73UDFWLFH7HVW 549 of the \-intercept as the average increase, instead of the slope, and from using the 10-year increments given on the [ -axis to incorrectly LQWHUSUHWWKHVORSHRIWKHOLQHRIEHVW
?W QUESTION 30 Choice D is correct.2QWKHOLQHRIEHVW
?WG increases from approximately 480 to 880 between t = 12 and t = 24. The slope of the OLQHRIEHVW
?WLVWKHGL
?HUHQFHLQG YDOXHVGLYLGHGE\WKHGL
?HUHQFHLQ  t-values, which gives 
? _ 
? = 400
\12 , or approximately 33. Writing the HTXDWLRQRIWKHOLQHRIEHVW
?WLQVORSHLQWHUFHSWIRUPJLYHV  G = 33t + b, where b is the \-coordinate of the \ -intercept. This equation LVVDWLV
?HGE\DOOSRLQWVRQWKHOLQHVR G = 480 when t = 12. Thus, 480 = 33(12) + b ZKLFKLVHTXLYDOHQWWR  b6XEWUDFWLQJ  from both sides of this equation gives b = 84. Therefore, an equation IRUWKHOLQHRIEHVW
?WFRXOGEH G = 33t + 84. Choice A is incorrect and may result from an error in calculating the slope and misidentifying the \ -coordinate of the \-intercept of the graph as the value of G at t = 10 rather than the value of G at t = 0. Choice B is incorrect and may result from using the smallest value of t on the graph as the slope and misidentifying the \ -coordinate of the \-intercept of the graph as the value of G at t = 10 rather than the value of G at t = 0. Choice C is incorrect and may result from misidentifying the \ -coordinate of the \ -intercept as the smallest value of G on the graph. QUESTION 31 The correct answer is 6. Circles are symmetric with respect to any given diameter through the center (h, k 2QHGLDPHWHURIWKHFLUFOH  is perpendicular to the [ -axis. Therefore, the value of h is the mean of the [-coordinates of the circle?s two [ -intercepts: h = 20 + 4
\ 2 = 12. The radius of the circle is given as 10, so the point (h, k) must be a distance of 10 units from any point on the circle. The equation of any circle can be written as ([ 
?h ) 2 + (\ 
?k) 2 = U 2, where (h, k) is the center of the circle and U is the length of the radius of the circle. Substituting 12 for h and 10 for U into this equation gives ([
? 2 + (\ 
?k) 2 = 10 2. Substituting the [ -coordinate and \-coordinate of a point on the circle,  JLYHV  
? 2 
?k) 2 = 10 2, or 64 + k 2 = 100. Subtracting 64 from both sides of this equation yields k 2 = 36. Therefore, k = ?
? 36_ . Since the graph shows the point (h , k LQWKH
?UVWTXDGUDQWk must be the positive square root of 36, so k = 6. QUESTION 32 The correct answer is 2.,W
^VJLYHQWKDWOLQH 
}LVSHUSHQGLFXODUWRWKH  line with equation \ 
? 2
\ 3 [ . Since the equation \ 
? 2
\ 3 [ is written in slope-intercept form, the slope of the line is
? 2
\3  7KHVORSHRIOLQH 
} must be the negativ e reciprocal of
? 2
\3 , which is  3
\2 . It? s also given that

PART 4 _(LJKW2
?FLDO3UDFWLFH7HVWVZLWK$QVZHU([SODQDWLRQV 550 the \-coordinate of the \ LQWHUFHSWRIOLQH 
}LV
?VRWKHHTXDWLRQ  RIOLQH 
}LQVORSHLQWHUFHSWIRUPLV  3 \ =
\ [ 
? 2 . If \ = b when [ = 10, b = 3
\ 2  
?, which is equivalent to b = 15
? 13, or b = 2. QUESTION 33 The correct answer is 8. In this group, 1
\  th of the people who are rhesus negative have blood type B. The total number of people who are rhesus negative in the group is 7 + 2 + 1 + [ , and there are 2 people who are rhesus negative with blood type B. Therefore, 2
\ (7 + 2 + 1 + [ )= 1
\ . Combining like terms on the left-hand side of the equation yields 2
\ (10 + [ )= 1
\ . Multipl ying both sides of this HTXDWLRQE\\LHOGV  18 _ (10 + [ ) = 1 , and multiplying both sides of this equation by (10 + [) yields 18 = 10 + [ . Subtracting 10 from both sides of this equation yields 8 = [. QUESTION 34 The correct answer is 9. The median number of goals scored is found by ordering the number of goals scored from least to greatest and then determining the middle value in the list. If the number of goals scored LQHDFKRIWKHJDPHVZHUHOLVWHGLQRUGHUIURPOHDVWWRJUHDWHVWWKH PHGLDQZRXOGEHWKH
?IWHHQWKQXPEHURIJRDOV7KHJUDSKVKRZVWKHUH ZHUHJDPHVZLWKJRDOVFRUHGDQGJDPHVZLWKJRDOVVFRUHG 7KHUHIRUHWKH
?IWHHQWKQXPEHURUWKHPHGLDQQXPEHURIJRDOVVFRUHG PXVWEH$FFRUGLQJWRWKHJUDSKWKHVRFFHUWHDPVFRUHGJRDOVLQ of the games played. QUESTION 35 The correct answer is 15. It?s given that the deductions reduce the original amount of taxes owed by $2,325.00. Since the deductions reduce the original amount of taxes owed by G%, the equation 2,325
\ 15,500 = G
\100 FDQEHXVHGWR
?QGWKLVSHUFHQWGHFUHDVHG . Multiplying both sides of this equation by 100 yields 232,500
\ 15,500 = G , or 15 = G . Thus, the tax deductions reduce the original amount of taxes owed by 15%. QUESTION 36 The correct answer is 1.5. It?s given that the system of linear equations has no solutions. Therefore, the lines represented by t he two HTXDWLRQVDUHSDUDOOHO(DFKRIWKHHTXDWLRQVFDQEHZULWWHQLQVORSH intercept form, or \ = P[ + b , where m is the slope of t he line and b is the \ -coordinate of the line ?s \ -intercept. Subtracting 3
\ 4 [ from both sides of 3
\ 4 [ 
? 1
\ 2 \ = 12 yiel ds
?1
\ 2 \  
? 3
\ 4 [ + 12  . Dividing both sides of

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