SAT Practice Test 1 College Board Answer Explanations

Section 3: Math Test - No Calculator

Solution of Question 1 (No Calculator)

QUESTION 1 Choice D is correct. Since $k=3$, one can substitute 3 for $k$ in the equation $\frac{x-1}{3}=k$, which gives $\frac{x-1}{3}=3$. Multiplying both sides of $\frac{x-1}{3}=3$ by 3 gives $x-1=9$ and then adding 1 to both sides of $x-1=9$ gives $x=10$. Choices A, B, and C are incorrect because the result of subtracting 1 from the value and dividing by 3 is not the given value of $k$, which is 3 .

Solution of Question 2 (No Calculator)

QUESTION 2 Choice A is correct. To calculate $(7+3 i+(-8+9 i)$, add the real parts of each complex number, $7+(-8)=-1$, and then add the imaginary parts, $3 i+9 i=12 i$. The result is $-1+12 i$. Choices B, C, and D are incorrect and likely result from common errors that arise when adding complex numbers. For example, choice B is the result of adding $3 i$ and $-9 i$, and choice $\mathrm{C}$ is the result of adding 7 and $8 .$

Solution of Question 3 (No Calculator)

QUESTION 3 Choice $\mathbf{C}$ is correct. The total number of text messages sent by Armand can be found by multiplying his rate of texting, in number of text messages sent per hour, by the total number of hours he spent sending them; that is $m$ texts/hour $\times 5$ hours $=5 m$ texts. Similarly, the total number of text messages sent by Tyrone is his hourly rate of texting multiplied by the 4 hours he spent texting: $p$ texts/hour $\times 4$ hours $=4 p$ texts. The total number of text messages sent by Armand and Tyrone is the sum of the total number of messages sent by Armand and the total number of messages sent by Tyrone: $5 m+4 p$. Choice A is incorrect and arises from adding the coefficients and multiplying the variables of $5 \mathrm{~m}$ and $4 p$. Choice B is incorrect and is the result of multiplying $5 \mathrm{~m}$ and $4 p$. The total number of messages sent by Armand and Tyrone should be the sum of $5 \mathrm{~m}$ and $4 p$, not the product of these terms. Choice D is incorrect because it multiplies Armand's number of hours spent texting by Tyrone's hourly rate of texting, and vice versa. This mix-up results in an expression that does not equal the total number of messages sent by Armand and Tyrone.

Solution of Question 4 (No Calculator)

QUESTION 4 Choice $\mathrm{B}$ is correct. The value 108 in the equation is the value of $P$ in $P=108-23 d$ when $d=0$. When $d=0$, Kathy has worked 0 days that week. In other words, 108 is the number of phones left before Kathy has started work for the week. Therefore, the meaning of the value 108 in the equation is that Kathy starts each week with 108 phones to fix. Choice A is incorrect because Kathy will complete the repairs when $P=0$. Since $P=108-23 d$, this will occur when $0=108-23 d$ or when $d=\frac{108}{23}$, not when $d=108$. Therefore, the value 108 in the equation does not represent the number of days it will take Kathy to complete the repairs. Choices $\mathrm{C}$ and $\mathrm{D}$ are incorrect because the number 23 in $P=108-23 d$ indicates that the number of phones left will decrease by 23 for each increase in the value of $d$ by 1 ; in other words, Kathy is repairing phones at a rate of 23 per day, not 108 per hour (choice C) or 108 per day (choice D).

Solution of Question 5 (No Calculator)

QUESTION 5 Choice $C$ is correct. Only like terms, with the same variables and exponents, can be combined to determine the answer as shown here: $$ \begin{aligned} &\left(x^{2} y-3 y^{2}+5 x y^{2}-\left(-x^{2} y+3 x y^{2}-3 y^{2}\right)\right.\\ =&\left(x^{2} y-\left(-x^{2} y+\left(-3 y^{2}-\left(-3 y^{2}\right)\right)+\left(5 x y^{2}-3 x y^{2}\right)\right.\right.\\ =& 2 x^{2} y+0+2 x y^{2} \\ =& 2 x^{2} y+2 x y^{2} \end{aligned} $$ Choices A, B, and D are incorrect and are the result of common calculation errors or of incorrectly combining like and unlike terms.

Solution of Question 6 (No Calculator)

QUESTION 6 Choice $\mathrm{A}$ is correct. In the equation $h=3 a+28.6$, if $a$, the age of the boy, increases by 1 , then $h$ becomes $h=3(a+1)+28.6=3 a+3+28.6=$ $3 a+28.6)+3$. Therefore, the model estimates that the boy's height increases by 3 inches each year. Alternatively: The height, $h$, is a linear function of the age, $a$, of the boy. The coefficient 3 can be interpreted as the rate of change of the function; in this case, the rate of change can be described as a change of 3 inches in height for every additional year in age. Choices B, C, and D are incorrect and are likely the result of dividing $28.6$ by 5,3 , and 2 , respectively. The number $28.6$ is the estimated height, in inches, of a newborn boy. However, dividing $28.6$ by 5,3 , or 2 has no meaning in the context of this question.

Solution of Question 7 (No Calculator)

QUESTION 7 Choice $B$ is correct. Since the right-hand side of the equation is $P$ times the expression $\frac{\left(\frac{r}{1,200}\right)\left(1+\frac{r}{1,200}\right)^{N}}{\left(1+\frac{r}{1,200}\right)^{N}-1}$, multiplying both sides of the equation by the reciprocal of this expression results in $\frac{\left(1+\frac{r}{1,200}\right)^{N}-1}{\left(\frac{r}{1,200}\right)\left(1+\frac{r}{1,200}\right)^{N}} m=P$. Choice A is incorrect and is the result of multiplying both sides of the equation by the rational expression $\frac{\left(\frac{r}{1,200}\right)\left(1+\frac{r}{1,200}\right)^{N}}{\left(1+\frac{r}{1,200}\right)^{N}-1}$ rather than by the reciprocal of this expression $\frac{\left(1+\frac{r}{1,200}\right)^{N}-1}{\left(\frac{r}{1,200}\right)\left(1+\frac{r}{1,200}\right)^{N}}$. Choices C and $\mathrm{D}$ are incorrect and are likely the result of errors while trying to solve for $P$.

Solution of Question 8 (No Calculator)

QUESTION 8 Choice $\mathbf{C}$ is correct. Since $\frac{a}{b}=2$, it follows that $\frac{b}{a}=\frac{1}{2}$. Multiplying both sides of the equation by 4 gives $4\left(\frac{b}{a}\right)=4\left(\frac{1}{2}\right)$, or $\frac{4 b}{a}=2$. Choice A is incorrect because if $\frac{4 b}{a}=0$, then $\frac{a}{b}$ would be undefined. Choice B is incorrect because if $\frac{4 b}{a}=1$, then $\frac{a}{b}=4$. Choice D is incorrect because if $\frac{4 b}{a}=4$, then $\frac{a}{b}=1$.

Solution of Question 9 (No Calculator)

QUESTION 9 Choice $B$ is correct. Adding $x$ and 19 to both sides of $2 y-x=-19$ gives $x=2 y+19$. Then, substituting $2 y+19$ for $x$ in $3 x+4 y=-23$ gives $3(2 y+19)+4 y=-23$. This last equation is equivalent to $10 y+57=-23$. Solving $10 y+57=-23$ gives $y=-8$. Finally, substituting $-8$ for $y$ in $2 y-x=-19$ gives $2(-8)-x=-19$, or $x=3$. Therefore, the solution $(x, y)$ to the given system of equations is $(3,-8)$. Choices A, C, and D are incorrect because when the given values of $x$ and $y$ are substituted in $2 y-x=-19$, the value of the left side of the equation does not equal $-19$.

Solution of Question 10 (No Calculator)

QUESTION 10 Choice $\mathrm{A}$ is correct. Since $g$ is an even function, $g-4)=g(4)=8$. Alternatively: First find the value of $a$, and then find $g(-4)$. Since $g(4)=8$, substituting 4 for $x$ and 8 for $g(x)$ gives $8=a(4)^{2}+24=16 a+24$. Solving this last equation gives $a=-1$. Thus $g\left(x=-x^{2}+24\right.$, from which it follows that $\left.g-4)=-(-4)^{2}+24 ; g-4\right)=-16+24 ;$ and $\left.g-4\right)=8$. Choices B, C, and D are incorrect because $g$ is a function and there can only be one value of $g-4$ ).

Solution of Question 11 (No Calculator)

QUESTION 11 Choice $\mathbf{D}$ is correct. To determine the price per pound of beef when it was equal to the price per pound of chicken, determine the value of $x$ (the number of weeks after July 1 ) when the two prices were equal. The prices were equal when $b=c$; that is, when $2.35+0.25 x=1.75+0.40 x$. This last equation is equivalent to $0.60=0.15 x$, and so $x=\frac{0.60}{0.15}=4$. Then to determine $b$, the price per pound of beef, substitute 4 for $x$ in $b=2.35+0.25 x$, which gives $b=2.35+0.25(4)=3.35$ dollars per pound. Choice $A$ is incorrect. It results from substituting the value 1 , not 4 , for $x$ in $b=2.35+0.25 x$. Choice B is incorrect. It results from substituting the value 2 , not 4 , for $x$ in $b=2.35+0.25 x$. Choice $C$ is incorrect. It results from substituting the value 3 , not 4 , for $x$ in $c=1.75+0.40 x$.

Solution of Question 12 (No Calculator)

QUESTION 12 Choice $\mathbf{D}$ is correct. In the $x y$-plane, all lines that pass through the origin are of the form $y=m x$, where $m$ is the slope of the line. Therefore, the equation of this line is $y=\frac{1}{7} x$, or $x=7 y$. A point with coordinates $(a, b)$ will lie on the line if and only if $a=7 b$. Of the given choices, only choice $\mathrm{D},(14,2)$, satisfies this condition: $14=7(2)$. Choice $A$ is incorrect because the line determined by the origin $(0,0)$ and $(0,7)$ is the vertical line with equation $x=0$; that is, the $y$-axis. The slope of the $y$-axis is undefined, not $\frac{1}{7}$. Therefore, the point $(0,7)$ does not lie on the line that passes the origin and has slope $\frac{1}{7}$. Choices B and C are incorrect because neither of the ordered pairs has a $y$-coordinate that is $\frac{1}{7}$ the value of the corresponding $x$-coordinate.

Solution of Question 13 (No Calculator)

QUESTION 13 Choice $\mathbf{B}$ is correct. To rewrite $\frac{1}{\frac{1}{x+2}+\frac{1}{x+3}}$, multiply by $\frac{(x+2)(x+3)}{(x+2)(x+3)}$. This results in the expression $\frac{(x+2)(x+3)}{(x+2)+(x+3)}$, which is equivalent to the expression in choice $\mathrm{B}$. Choices A, C, and D are incorrect and could be the result of common algebraic errors that arise while manipulating a complex fraction.

Solution of Question 14 (No Calculator)

QUESTION 14 Choice A is correct. One approach is to express $\frac{8^{x}}{2^{y}}$ so that the numerator and denominator are expressed with the same base. Since 2 and 8 are both powers of 2 , substituting $2^{3}$ for 8 in the numerator of $\frac{8^{x}}{2^{y}}$ gives $\frac{\left(2^{3}\right)^{x}}{2^{y}}$, which can be rewritten as $\frac{2^{3 x}}{2^{y}}$. Since the numerator and denominator of $\frac{2^{3 x}}{2^{y}}$ have a common base, this expression can be rewritten as $2^{3 x-y}$. It is given that $3 x-y=12$, so one can substitute 12 for the exponent, $3 x-y$, given that the expression $\frac{8^{x}}{2^{y}}$ is equal to $2^{12}$. Choice B is incorrect. The expression $\frac{8^{x}}{2^{y}}$ can be rewritten as $\frac{2^{3 x}}{2^{y}}$, or $2^{3 x-y}$. If the value of $2^{3 x-y}$ is $4^{4}$, which can be rewritten as 28 , then $2^{3 x-y}=2^{8}$, which results in $3 x-y=8$, not 12 . Choice $\mathrm{C}$ is incorrect. If the value of $\frac{8^{x}}{2^{y}}$ is $8^{2}$, then $2^{3 x-y}=8^{2}$, which results in $3 x-y=6$, not 12 . Choice D is incorrect because the value of $\frac{8^{x}}{2^{y}}$ can be determined.

Solution of Question 15 (No Calculator)

QUESTION 15 Choice $\mathbf{D}$ is correct. One can find the possible values of $a$ and $b$ in $(a x+2)(b x+7)$ by using the given equation $a+b=8$ and finding another equation that relates the variables $a$ and $b$. Since $(a x+2)(b x+7)=15 x^{2}+c x+14$, one can expand the left side of the equation to obtain $a b x^{2}+7 a x+2 b x+14=15 x^{2}+c x+14$. Since $a b$ is the coefficient of $x^{2}$ on the left side of the equation and 15 is the coefficient of $x^{2}$ on the right side of the equation, it must be true that $a b=15$. Since $a+b=8$, it follows that $b=8-a$. Thus, $a b=15$ can be rewritten as $a 8-a)=15$, which in turn can be rewritten as $a^{2}-8 a+15=0 .$ Factoring gives $(a-3)(a-5)=0$. Thus, either $a=3$ and $b=5$, or $a=5$ and $b=3$. If $a=3$ and $b=5$, then $(a x+2)(b x+7)=(3 x+2)(5 x+7)=15 x^{2}+31 x+14$. Thus, one of the possible values of $c$ is 31 . If $a=5$ and $b=3$, then $(a x+2)(b x+7)=(5 x+2)(3 x+7)=15 x^{2}+41 x+14$. Thus, another possible value for $c$ is 41 . Therefore, the two possible values for $c$ are 31 and 41 . Choice A is incorrect; the numbers 3 and 5 are possible values for $a$ and $b$, but not possible values for $c$. Choice B is incorrect; if $a=5$ and $b=3$, then 6 and 35 are the coefficients of $x$ when the expression $(5 x+2)(3 x+7)$ is expanded as $15 x^{2}+35 x+6 x+14$. However, when the coefficients of $x$ are 6 and 35, the value of $c$ is 41 and not 6 and 35. Choice $\mathrm{C}$ is incorrect; if $a=3$ and $b=5$, then 10 and 21 are the coefficients of $x$ when the expression $(3 x+2)(5 x+7)$ is expanded as $15 x^{2}+21 x+10 x+14$. However, when the coefficients of $x$ are 10 and 21 , the value of $c$ is 31 and not 10 and 21 .

Solution of Question 16 (No Calculator)

QUESTION 16 The correct answer is 2 . To solve for $t$, factor the left side of $t^{2}-4=0$, giving $(t-2)(t+2)=0$. Therefore, either $t-2=0$ or $t+2=0$. If $t-2=0$, then $t=2$, and if $t+2=0$, then $t=-2$. Since it is given that $t>0$, the value of $t$ must be 2 . Another way to solve for $t$ is to add 4 to both sides of $t^{2}-4=0$, giving $t^{2}=4$. Then, taking the square root of the left and the right side of the equation gives $t=\pm \sqrt{4}=\pm 2$. Since it is given that $t>0$, the value of $t$ must be 2 .

Solution of Question 17 (No Calculator)

QUESTION 17 The correct answer is 1600 . It is given that $\angle A E B$ and $\angle C D B$ have the same measure. Since $\angle A B E$ and $\angle C B D$ are vertical angles, they have the same measure. Therefore, triangle $E A B$ is similar to triangle $D C B$ because the triangles have two pairs of congruent corresponding angles (angle-angle criterion for similarity of triangles). Since the triangles are similar, the corresponding sides are in the same proportion; thus $\frac{C D}{x}=\frac{B D}{E B} .$ Substituting the given values of 800 for $C D$, 700 for $B D$, and 1400 for $E B$ in $\frac{C D}{x}=\frac{B D}{E B}$ gives $\frac{800}{x}=\frac{700}{1400}$. Therefore, $x=\frac{800)(1400)}{700}=1600$.

Solution of Question 18 (No Calculator)

QUESTION 18 The correct answer is 7. Subtracting the left and right sides of $x+y=-9$ from the corresponding sides of $x+2 y=-25$ gives $(x+2 y-(x+y=-25-(-9)$, which is equivalent to $y=-16$ Substituting $-16$ for $y$ in $x+y=-9$ gives $x+(-16)=-9$, which is equivalent to $x=-9-(-16)=7$.

Solution of Question 19 (No Calculator)

QUESTION 19 The correct answer is $\frac{4}{5}$ or $0.8$. By the complementary angle relationship for sine and cosine, $\sin \left(x^{\circ}\right)=\cos \left(90^{\circ}-x^{\circ}\right)$. Therefore, $\cos \left(90^{\circ}-x^{\circ}\right)=\frac{4}{5}$. Either the fraction $4 / 5$ or its decimal equivalent, $0.8$, may be gridded as the correct answer. Alternatively, one can construct a right triangle that has an angle of measure $x^{\circ}$ such that $\sin \left(x^{\circ}\right)=\frac{4}{5}$, as shown in the figure below, where $\sin \left(x^{\circ}\right)$ is equal to the ratio of the length of the side opposite the angle measuring $x^{\circ}$ to the length of the hypotenuse, or $\frac{4}{5}$. Since two of the angles of the triangle are of measure $x^{\circ}$ and $90^{\circ}$, the third angle must have the measure $180^{\circ}-90^{\circ}-x^{\circ}=90^{\circ}-x^{\circ}$. From the figure, $\cos \left(90^{\circ}-x^{\circ}\right)$, which is equal to the ratio of the length of the side adjacent to the angle measuring $90^{\circ}-x^{\circ}$ to the hypotenuse, is also $\frac{4}{5}$.

Solution of Question 20 (No Calculator)

QUESTION 20 The correct answer is 100 . Since $a=5 \sqrt{2}$, one can substitute $5 \sqrt{2}$ for $a$ in $2 a=\sqrt{2 x}$, giving $10 \sqrt{2}=\sqrt{2 x}$. Squaring each side of $10 \sqrt{2}=\sqrt{2 x}$ gives $10 \sqrt{2})^{2}=(\sqrt{2 x})^{2}$, which simplifies to $(10)^{2}(\sqrt{2})^{2}=(\sqrt{2 x})^{2}$, or $200=2 x$. This gives $x=100$. To verify, substitute 100 for $x$ and $5 \sqrt{2}$ for $a$ in the equation $2 a=\sqrt{2 x}$, which yields $2(5 \sqrt{2})=\sqrt{2)(100)}$; this is true since $2(5 \sqrt{2})=10 \sqrt{2}$ and $\sqrt{2)(100)}=\sqrt{2} \sqrt{100}=10 \sqrt{2}$.

Section 4: Math Test - Calculator

Solution of Question 1 (Calculator)

QUESTION 1 Choice $B$ is correct. On the graph, a line segment with a positive slope represents an interval over which the target heart rate is strictly increasing as time passes. A horizontal line segment represents an interval over which there is no change in the target heart rate as time passes, and a line segment with a negative slope represents an interval over which the target heart rate is strictly decreasing as time passes. Over the interval between 40 and 60 minutes, the graph consists of a line segment with a positive slope followed by a line segment with a negative slope, with no horizontal line segment in between, indicating that the target heart rate is strictly increasing then strictly decreasing. Choice A is incorrect because the graph over the interval between 0 and 30 minutes contains a horizontal line segment, indicating a period in which there was no change in the target heart rate. Choice C is incorrect because the graph over the interval between 50 and 65 minutes consists of a line segment with a negative slope followed by a line segment with a positive slope, indicating that the target heart rate is strictly decreasing then strictly increasing. Choice D is incorrect because the graph over the interval between 70 and 90 minutes contains horizontal line segments and no segment with a negative slope.

Solution of Question 2 (Calculator)

QUESTION 2 Choice $\mathrm{C}$ is correct. Substituting 6 for $x$ and 24 for $y$ in $y=k x$ gives $24=(k)(6)$, which gives $k=4$. Hence, $y=4 x$. Therefore, when $x=5$, the value of $y$ is $(4)(5)=20$. None of the other choices for $y$ is correct because $y$ is a function of $x$, and so there is only one $y$-value for a given $x$-value. Choices A, B, and D are incorrect. Choice A is the result of substituting 6 for $y$ and substituting 5 for $x$ in the equation $y=k x$, when solving for $k$. Choice B results from substituting 3 for $k$ and 5 for $x$ in the equation $y=k x$, when solving for $y$. Choice D results from using $y=k+x$ instead of $y=k x$.

Solution of Question 3 (Calculator)

QUESTION 3 Choice D is correct. Consider the measures of $\angle 3$ and $\angle 4$ in the figure below. The measure of $\angle 3$ is equal to the measure of $\angle 1$ because they are corresponding angles for the parallel lines $\ell$ and $m$ intersected by the transversal line $t$. Similarly, the measure of $\angle 3$ is equal to the measure of $\angle 4$ because they are corresponding angles for the parallel lines $s$ and $t$ intersected by the transversal line $m$. Since the measure of $\angle 1$ is $35^{\circ}$, the measures of $\angle 3$ and $\angle 4$ are also $35^{\circ}$. Since $\angle 4$ and $\angle 2$ are supplementary angles, the sum of the measures of these two angles is $180^{\circ}$. Therefore, the measure of $\angle 2$ is $180^{\circ}-35^{\circ}=145^{\circ}$. Choice $\mathrm{A}$ is incorrect because $35^{\circ}$ is the measure of $\angle \mathrm{l}$, and $\angle \mathrm{l}$ is not congruent to $\angle 2$. Choice $\mathrm{B}$ is incorrect because it is the measure of the complementary angle of $\angle 1$, and $\angle 1$ and $\angle 2$ are not complementary angles. Choice $\mathrm{C}$ is incorrect because it is double the measure of $\angle 1$, which cannot be inferred from the information given.

Solution of Question 4 (Calculator)

QUESTION 4 Choice $C$ is correct. The description " $16+4 x$ is 10 more than 14 " can be written as the equation $16+4 x=10+14$, which is equivalent to $16+4 x=24$. Subtracting 16 from each side of $16+4 x=24$ gives $4 x=8$. Since $8 x$ is 2 times $4 x$, multiplying both sides of $4 x=8$ by 2 gives $8 x=16$. Therefore, the value of $8 x$ is 16 . Choice $\mathrm{A}$ is incorrect because it is the value of $x$, not $8 \mathrm{x}$. Choices $\mathrm{B}$ and D are incorrect and may be the result of errors made when solving the equation $16+4 x=10+14$ for $x$. For example, choice D could be the result of subtracting 16 from the left side of the equation and adding 16 to the right side of the equation $16+4 x=10+14$, giving $4 x=40$ and $8 x=80$.

Solution of Question 5 (Calculator)

QUESTION 5 Choice $\mathbf{D}$ is correct. A graph with a strong negative association between $d$ and $t$ would have the points on the graph closely aligned with a line that has a negative slope. The more closely the points on a graph are aligned with a line, the stronger the association between $d$ and $t$, and a negative slope indicates a negative association. Of the four graphs, the points on graph D are most closely aligned with a line with a negative slope. Therefore, the graph in choice D has the strongest negative association between $d$ and $t$. Choice $\mathrm{A}$ is incorrect because the points are more scattered than the points in choice D, indicating a weaker negative association between $d$ and $t$. Choice B is incorrect because the points are aligned to either a curve or possibly a line with a small positive slope. Choice $C$ is incorrect because the points are aligned to a line with a positive slope, indicating a positive association between $d$ and $t$.

Solution of Question 6 (Calculator)

QUESTION 6 Choice D is correct. Since there are 10 grams in 1 decagram, there are $2 \times 10=20$ grams in 2 decagrams. Since there are 1,000 milligrams in 1 gram, there are $20 \times 1,000=20,000$ milligrams in 20 grams. Therefore, 20,0001 -milligram doses of the medicine can be stored in a 2 -decagram container. Choice A is incorrect; $0.002$ is the number of grams in 2 milligrams. Choice B is incorrect; it could result from multiplying by 1,000 and dividing by 10 instead of multiplying by both 1,000 and 10 when converting from decagrams to milligrams. Choice $\mathrm{C}$ is incorrect; 2,000 is the number of milligrams in 2 grams, not the number of milligrams in 2 decagrams.

Solution of Question 7 (Calculator)

QUESTION 7 Choice $\mathbf{C}$ is correct. Let $x$ represent the number of installations that each unit on the $y$-axis represents. Then $9 x, 5 x, 6 x, 4 x$, and $3.5 x$ are the number of rooftops with solar panel installations in cities A, B, C, D, and E, respectively. Since the total number of rooftops is 27,500 , it follows that $9 x+5 x+6 x+4 x+3.5 x=27,500$, which simplifies to $27.5 x=27,500$. Thus, $x=1,000$. Therefore, an appropriate label for the $y$-axis is "Number of installations (in thousands)." Choices A, B, and D are incorrect and may result from errors when setting up and calculating the units for the $y$-axis.

Solution of Question 8 (Calculator)

QUESTION 8 Choice $\mathbf{D}$ is correct. If the value of $|n-1|+1$ is equal to 0 , then $|n-1|+1=0$. Subtracting 1 from both sides of this equation gives $|n-1|=-1$. The expression $|n-1|$ on the left side of the equation is the absolute value of $n-1$, and the absolute value of a quantity can never be negative. Thus $|n-1|=-1$ has no solution. Therefore, there are no values for $n$ for which the value of $|n-1|+1$ is equal to 0 . Choice $\mathrm{A}$ is incorrect because $|0-1|+1=1+1=2$, not 0 . Choice $\mathrm{B}$ is incorrect because $|1-1|+1=0+1=1$, not 0 . Choice $C$ is incorrect because $|2-1|+1=1+1=2$, not 0 .

Solution of Question 9 (Calculator)

QUESTION 9 Choice A is correct. Subtracting 1,052 from both sides of the equation $a=1,052+1.08 t$ gives $a-1,052=1.08 t$. Then dividing both sides of $a-1,052=1.08 t$ by $1.08$ gives $t=\frac{a-1,052}{1.08}$. Choices B, C, and D are incorrect and could arise from errors in rewriting $a=1,052+1.08 t$. For example, choice B could result if 1,052 is added to the left side of $a=1,052+1.08 t$ and subtracted from the right side, and then both sides are divided by $1.08$.

Solution of Question 10 (Calculator)

QUESTION 10 Choice $B$ is correct. The air temperature at which the speed of a sound wave is closest to 1,000 feet per second can be found by substituting 1,000 for $a$ and then solving for $t$ in the given formula. Substituting 1,000 for $a$ in the equation $a=1,052+1.08 t$ gives $1,000=1,052+1.08 t$. Subtracting 1,052 from both sides of the equation $1,000=1,052+1.08 t$ and then dividing both sides of the equation by $1.08$ yields $t=\frac{-52}{1.08} \approx-48.15$. Of the choices given, $-48^{\circ} \mathrm{F}$ is closest to $-48.15^{\circ} \mathrm{F}$. Choices A, C, and D are incorrect and might arise from errors made when substituting 1,000 for $a$ or solving for $t$ in the equation $a=1,052+1.08 t$ or in rounding the result to the nearest integer. For example, choice C could be the result of rounding $-48.15$ to $-49$ instead of $-48$.

Solution of Question 11 (Calculator)

QUESTION 11 Choice A is correct. Subtracting $3 x$ and adding 3 to both sides of $3 x-5 \geq 4 x-3$ gives $-2 \geq x$. Therefore, $x$ is a solution to $3 x-5 \geq 4 x-3$ if and only if $x$ is less than or equal to $-2$ and $x$ is NOT a solution to $3 x-5 \geq 4 x-3$ if and only if $x$ is greater than $-2$. Of the choices given, only $-1$ is greater than $-2$ and, therefore, cannot be a value of $x$. Choices B, C, and D are incorrect because each is a value of $x$ that is less than or equal to $-2$ and, therefore, could be a solution to the inequality.

Solution of Question 12 (Calculator)

QUESTION 12 Choice $\boldsymbol{C}$ is correct. The average number of seeds per apple is the total number of seeds in the 12 apples divided by the number of apples, which is 12. On the graph, the horizontal axis is the number of seeds per apple and the height of each bar is the number of apples with the corresponding number of seeds. The first bar on the left indicates that 2 apples have 3 seeds each, the second bar indicates that 4 apples have 5 seeds each, the third bar indicates that 1 apple has 6 seeds, the fourth bar indicates that 2 apples have 7 seeds each, and the fifth bar indicates that 3 apples have 9 seeds each. Thus, the total number of seeds for the 12 apples is $2 \times 3)+(4 \times 5)+(1 \times 6)+(2 \times 7)+(3 \times 9)=73$, and the average number of seeds per apple is $\frac{73}{12}=6.08$. Of the choices given, 6 is closest to $6.08$. Choice A is incorrect; it is the number of apples represented by the tallest bar but is not the average number of seeds for the 12 apples. Choice B is incorrect; it is the number of seeds per apple corresponding to the tallest bar, but is not the average number of seeds for the 12 apples. Choice D is incorrect; a student might choose this value by correctly calculating the average number of seeds, $6.08$, but incorrectly rounding up to 7 .

Solution of Question 13 (Calculator)

QUESTION 13 Choice $\mathrm{C}$ is correct. From the table, there was a total of 310 survey respondents, and $19 \%$ of all survey respondents is equivalent to $\frac{19}{100} \times 310=58.9$ respondents. Of the choices given, 59 , the number of males taking Geometry, is closest to $58.9$ respondents. Choices A, B, and D are incorrect because the number of males taking Geometry is closer to $58.9$ (which is $19 \%$ of 310 ) than the number of respondents in each of these categories.

Solution of Question 14 (Calculator)

QUESTION 14 Choice $\mathbf{C}$ is correct. The range of the lengths of the 21 fish represented in the table is $24-8=16$ inches, and the range of the remaining 20 lengths after the 24 -inch measurement is removed is $16-8=8$ inches. Therefore, after the 24 -inch measurement is removed, the change in range, 8 inches, is much greater than the change in the mean or median. Choice A is incorrect. Let $m$ be the mean of the lengths, in inches, of the 21 fish. Then the sum of the lengths, in inches, of the 21 fish is $21 \mathrm{~m}$. After the 24 -inch measurement is removed, the sum of the lengths, in inches, of the remaining 20 fish is $21 m-24$, and the mean length, in inches, of these 20 fish is $\frac{21 m-24}{20}$, which is a change of $\frac{24-m}{20}$ inches. Since $m$ must be between the smallest and largest measurements of the 21 fish, it follows that $8 < m < 24$, from which it can be seen that the change in the mean, in inches, is between $\frac{24-24}{20}=0$ and $\frac{24-8}{20}=\frac{4}{5}$, and so must be less than the change in the range, 8 inches. Choice $B$ is incorrect because the median length of the 21 fish represented in the table is 12 , and after the 24 -inch measurement is removed, the median of the remaining 20 lengths is also 12 . Therefore, the change in the median $(0)$ is less than the change in the range (8). Choice D is incorrect because the changes in the mean, median, and range of the measurements are different.

Solution of Question 15 (Calculator)

QUESTION 15 Choice $\mathrm{A}$ is correct. The total cost $C$ of renting a boat is the sum of the initial cost to rent the boat plus the product of the cost per hour and the number of hours, $h$, that the boat is rented. The $C$-intercept is the point on the $C$-axis where $h$, the number of hours the boat is rented, is 0. Therefore, the $C$-intercept is the initial cost of renting the boat. Choice B is incorrect because the graph represents the cost of renting only one boat. Choice $\mathrm{C}$ is incorrect because the total number of hours of rental is represented by $h$-values, each of which corresponds to the first coordinate of a point on the graph not the $C$-intercept of the graph. Choice D is incorrect because the increase in cost for each additional hour is given by the slope of the line, not by the $C$-intercept.

Solution of Question 16 (Calculator)

QUESTION 16 Choice $\mathbf{C}$ is correct. If $m$ is the slope and $b$ is the $C$-intercept of the line, the relationship between $h$ and $C$ can be represented by $C=m h+b$. The $C$-intercept of the line is 5. Since the points $(0,5)$ and $(1,8)$ lie on the line, the slope of the line is $\frac{8-5}{1-0}=\frac{3}{1}=3$. Therefore, the relationship between $h$ and $C$ can be represented by $C=3 h+5$, the slope-intercept equation of the line. Choices A and D are incorrect because each of these equations represents a line that passes through the origin $(0,0)$. However, $C$ is not equal to zero when $h=0$. Choice B is incorrect and may result from errors made when reading the scale on each axis as related to calculating the slope.

Solution of Question 17 (Calculator)

QUESTION 17 Choice $B$ is correct. The minimum value of the function corresponds to the $y$-coordinate of the point on the graph that has the smallest $y$-coordinate on the graph. Since the smallest $y$-coordinate belongs to the point with coordinates $(-3,-2)$, the minimum value of the graph is $f(-3)=-2$. Therefore, the minimum value of $f(x)$ is at $x=-3$. Choice $\mathrm{A}$ is incorrect; $-5$ is the least value for an $x$-coordinate, not the $y$-coordinate, of a point on the graph of $y=f(x)$. Choice $\mathrm{C}$ is incorrect; it is the minimum value of $f$, not the value of $x$ that corresponds to the minimum of $f$. Choice $\mathrm{D}$ is incorrect; it is the value of $x$ for which the value of $f(x)$ has its maximum, not minimum.

Solution of Question 18 (Calculator)

QUESTION 18 Choice $A$ is correct. Since $(0,0)$ is a solution to the system of inequalities, substituting 0 for $x$ and 0 for $y$ in the given system must result in two true inequalities. After this substitution, $y < -x+a$ becomes $0 < a$, and $y>x+b$ becomes $0>b$. Hence, $a$ is positive and $b$ is negative. Therefore, $a>b$. Choice B is incorrect because $b>a$ cannot be true if $b$ is negative and $a$ is positive. Choice $\mathrm{C}$ is incorrect because it is possible to find an example where $(0,0)$ is a solution to the system, but $|a| < |b|$; for example, if $a=6$ and $b=-7$. Choice D is incorrect because the equation $a=-b$ doesn't have to be true; for example, $(0,0)$ is a solution to the system of inequalities if $a=1$ and $b=-2$.

Solution of Question 19 (Calculator)

QUESTION 19 Choice $B$ is correct. To determine the number of salads sold, write and solve a system of two equations. Let $x$ equal the number of salads sold and let $y$ equal the number of drinks sold. Since a total of 209 salads and drinks were sold, the equation $x+y=209$ must hold. Since salads cost $\$ 6.50$ each, drinks cost $\$ 2.00$ each, and the total revenue from selling $x$ salads and $y$ drinks was $\$ 836.50$, the equation $6.50 x+2.00 y=836.50$ must also hold. The equation $x+y=209$ is equivalent to $2 x+2 y=418$, and subtracting $(2 x+2 y)$ from the left-hand side and subtracting 418 from the right-hand side of $6.50 x+2.00 y=836.50$ gives $4.5 x=418.50$. Therefore, the number of salads sold, $x$, was $x=\frac{418.50}{4.50}=93$. Choices A, C, and D are incorrect and could result from errors in writing the equations and solving the system of equations. For example, choice $\mathrm{C}$ could have been obtained by dividing the total revenue, $\$ 836.50$, by the total price of a salad and a drink, \$8.50, and then rounding up.

Solution of Question 20 (Calculator)

QUESTION 20 Choice $\mathbf{D}$ is correct. Let $x$ be the original price of the computer, in dollars. The discounted price is 20 percent off the original price, so $x-0.2 x=0.8 x$ is the discounted price, in dollars. The sales tax is 8 percent of the discounted price, so $0.08(0.8 x)$ represents the sales tax Alma paid. The price $p$, in dollars, that Alma paid the cashiers is the sum of the discounted price and the tax: $p=0.8 x+(0.08)(0.8 x)$ which can be rewritten as $p=1.08(0.8 x)$. Therefore, the original price, $x$, of the computer, in dollars, can be written as $\frac{p}{0.8)(1.08)}$ in terms of $p$. Choices A, B, and C are incorrect. The expression in choice A represents $88 \%$ of the amount Alma paid to the cashier, and can be obtained by subtracting the discount of $20 \%$ from the original price and adding the sales $\operatorname{tax}$ of $8 \%$. However, this is incorrect because $8 \%$ of the tax is over the discounted price, not the original one. The expression in choice $B$ is the result of adding the factors associated with the discount and sales tax, $0.8$ and .08, rather than multiplying them. The expression in choice $\mathrm{C}$ results from assigning $p$ to represent the original price of the laptop, rather than to the amount Alma paid to the cashier.

Solution of Question 21 (Calculator)

QUESTION 21 Choice $\mathbf{C}$ is correct. The probability that a person from Group $\mathrm{Y}$ who recalled at least 1 dream was chosen at random from the group of all people who recalled at least 1 dream is equal to the number of people in Group $\mathrm{Y}$ who recalled at least 1 dream divided by the total number of people in the two groups who recalled at least 1 dream. The number of people in Group $Y$ who recalled at least 1 dream is the sum of the 11 people in Group $\mathrm{Y}$ who recalled 1 to 4 dreams and the 68 people in Group $\mathrm{Y}$ who recalled 5 or more dreams: $11+68=79$. The total number of people who recalled at least 1 dream is the sum of the 79 people in Group $Y$ who recalled at least 1 dream, the 28 people in Group $X$ who recalled 1 to 4 dreams, and the 57 people in Group $X$ who recalled 5 or more dreams: $79+28+57=164$. Therefore, the probability is $\frac{79}{164}$. Choice A is incorrect; it is the probability of choosing at random a person from Group $Y$ who recalled 5 or more dreams. Choice B is incorrect; it is the probability of choosing at random a person from Group $\mathrm{Y}$ who recalled at least 1 dream. Choice $\mathrm{D}$ is incorrect; it is the probability of choosing at random a person from the two groups combined who recalled at least 1 dream.

Solution of Question 22 (Calculator)

QUESTION 22 Choice B is correct. The amounts given in the table are in thousands of dollars. Therefore, the amount in the annual budget for agriculture/natural resources is actually $\$ 488,106,000$ in 2010 and $\$ 358,708,000$ in 2008 . Therefore, the change in the budgeted amount is $\$ 488,106,000-\$ 358,708,000=\$ 129,398,000$. Hence, the average change in the annual budget for agriculture/natural resources from 2008 to 2010 is $\frac{\$ 129,398,000}{2}=\$ 64,699,000$ per year. Of the options given, this average rate of change is closest to $\$ 65,000,000$ per year. Choices $\mathrm{A}$ and $\mathrm{C}$ are incorrect and may result from errors in setting up or calculating the average rate of change. Choice D is incorrect; $\$ 130,000,000$ is the approximate total change in the annual budget for agriculture/natural resources from 2008 to 2010 , not the average rate of change from 2008 to 2010 .

Solution of Question 23 (Calculator)

QUESTION 23 Choice $B$ is correct. The human resources budget in 2007 was $4,051,050$ thousand dollars, and the human resources budget in 2010 was $5,921,379$ thousand dollars. Therefore, the ratio of the 2007 budget to the 2010 budget is slightly greater than $\frac{4}{6}=\frac{2}{3}$. Similar estimates for agriculture/natural resources give a ratio of the 2007 budget to the 2010 budget of slightly greater than $\frac{3}{4}$; for education, a ratio of slightly greater than $\frac{2}{3}$; for highways and transportation, a ratio of slightly less than $\frac{5}{6}$; and for public safety, a ratio of slightly greater than $\frac{5}{9}$. Therefore, of the given choices, education's ratio of the 2007 budget to the 2010 budget is closest to that of human resources. Choices A, C, and D are incorrect because the ratio of the 2007 budget to 2010 budget for each of the programs given in these choices is further from the corresponding ratio for human resources than the corresponding ratio for education.

Solution of Question 24 (Calculator)

QUESTION 24 Choice $A$ is correct. The equation of a circle can be written as $(x-h)^{2}+(y-k)^{2}=r^{2}$ where $(h, k)$ are the coordinates of the center of the circle and $r$ is the radius of the circle. Since the coordinates of the center of the circle are $(0,4)$, the equation of the circle is $x^{2}+(y-4)^{2}=r^{2}$. The radius of the circle is the distance from the center, $0,4)$, to the given endpoint of a radius, $\left(\frac{4}{3}, 5\right)$. By the distance formula, $r^{2}=\left(\frac{4}{3}-0\right)^{2}+(5-4)^{2}=\frac{25}{9} .$ Therefore, an equation of the given circle is $x^{2}+(y-4)^{2}=\frac{25}{9}$. Choices B and D are incorrect. The equations given in these choices represent a circle with center $(0,-4)$, not $(0,4)$. Choice $C$ is incorrect; it results from using $r$ instead of $r^{2}$ in the equation for the circle.

Solution of Question 25 (Calculator)

QUESTION 25 Choice D is correct. When the ball hits the ground, its height is 0 meters. Substituting 0 for $h$ in $h=-4.9 t^{2}+25 t$ gives $0=-4.9 t^{2}+25 t$, which can be rewritten as $0=t-4.9 t+25$ ). Thus, the possible values of $t$ are $t=0$ and $t=\frac{25}{4.9} \approx 5.1$. The time $t=0$ seconds corresponds to the time the ball is launched from the ground, and the time $t \approx 5.1$ seconds corresponds to the time after launch that the ball hits the ground. Of the given choices, $5.0$ seconds is closest to $5.1$ seconds, so the ball returns to the ground approximately $5.0$ seconds after it is launched. Choice $\mathrm{A}, \mathrm{B}$, and $\mathrm{C}$ are incorrect and could arise from conceptual or computation errors while solving $0=-4.9 t^{2}+25 t$ for $t$.

Solution of Question 26 (Calculator)

QUESTION 26 Choice B is correct. Let $x$ represent the number of pears produced by the Type B trees. Type A trees produce 20 percent more pears than Type B trees, or $x$, which can be represented as $x+0.20 x=1.20 x$ pears. Since Type A trees produce 144 pears, it follows that $1.20 x=144$. Thus $x=\frac{144}{1.20}=120$. Therefore, the Type B trees produced 120 pears. Choice A is incorrect because while 144 is reduced by approximately 20 percent, increasing 115 by 20 percent gives 138 , not 144. Choice C is incorrect; it results from subtracting 20 from the number of pears produced by the Type A trees. Choice D is incorrect; it results from adding 20 percent of the number of pears produced by Type A trees to the number of pears produced by Type A trees.

Solution of Question 27 (Calculator)

QUESTION 27 Choice $C$ is correct. The area of the field is 100 square meters. Each 1-meter-by-1-meter square has an area of 1 square meter. Thus, on average, the earthworm counts to a depth of 5 centimeters for each of the regions investigated by the students should be about $\frac{1}{100}$ of the total number of earthworms to a depth of 5 centimeters in the entire field. Since the counts for the smaller regions are from 107 to 176 , the estimate for the entire field should be between 10,700 and 17,600 . Therefore, of the given choices, 15,000 is a reasonable estimate for the number of earthworms to a depth of 5 centimeters in the entire field. Choice A is incorrect; 150 is the approximate number of earthworms in 1 square meter. Choice $\mathrm{B}$ is incorrect; it results from using 10 square meters as the area of the field. Choice D is incorrect; it results from using 1,000 square meters as the area of the field.

Solution of Question 28 (Calculator)

QUESTION 28 Choice $\mathrm{C}$ is correct. To determine which quadrant does not contain any solutions to the system of inequalities, graph the inequalities. Graph the inequality $y \geq 2 x+1$ by drawing a line through the $y$-intercept $(0,1)$ and the point $(1,3)$, as shown. The solutions to this inequality are all points contained on and above this line. Graph the inequality $y>\frac{1}{2} x-1$ by drawing a dashed line through the $y$-intercept $(0,-1)$ and the point $(2,0)$, as shown. The solutions to this inequality are all points above this dashed line. The solution to the system of inequalities is the intersection of the regions above the graphs of both lines. It can be seen that the solutions only include points in quadrants I, II, and III and do not include any points in quadrant IV. Choices A and B are incorrect because quadrants II and III contain solutions to the system of inequalities, as shown in the figure above. Choice D is incorrect because there are no solutions in quadrant IV.

Solution of Question 29 (Calculator)

QUESTION 29 Choice $\mathbf{D}$ is correct. If the polynomial $p(x)$ is divided by $x-3$, the result can be written as $\frac{p(x)}{x-3}=q(x)+\frac{r}{x-3}$, where $q(x)$ is a polynomial and $r$ is the remainder. Since $x-3$ is a degree 1 polynomial, the remainder is a real number. Hence, $p(x)$ can be written as $p(x)=(x-3) q(x)+r$, where $r$ is a real number. It is given that $p(3)=-2$ so it must be true that $-2=p 3)=(3-3) q 3)+r=(0) q 3)+r=r$. Therefore, the remainder when $p(x)$ is divided by $x-3$ is $-2$. Choice $\mathrm{A}$ is incorrect because $p 3)=-2$ does not imply that $p(5)=0$. Choices $B$ and $C$ are incorrect because the remainder $-2$ or its opposite, 2 , need not be a root of $p(x)$.

Solution of Question 30 (Calculator)

QUESTION 30 Choice D is correct. Any quadratic function $q$ can be written in the form $q(x)=a(x-h)^{2}+k$, where $a, h$, and $k$ are constants and $(h, k)$ is the vertex of the parabola when $q$ is graphed in the coordinate plane. This form can be reached by completing the square in the expression that defines $q$. The equation of the graph is $y=x^{2}-2 x-15$. Since the coefficient of $x$ is $-2$, this equation can be written in terms of $(x-1)^{2}=x^{2}-2 x+1$ as follows: $y=x^{2}-2 x-15=\left(x^{2}-2 x+1\right)-16=$ $(x-1)^{2}-16$. From this form of the equation, the coefficients of the vertex can be read as $(1,-16)$. Choices $\mathrm{A}$ and $\mathrm{C}$ are incorrect because the coordinates of the vertex $A$ do not appear as constants in these equations. Choice B is incorrect because it is not equivalent to the given equation.

Solution of Question 31 (Calculator)

QUESTION 31 The correct answer is any number between 4 and 6 , inclusive. Since Wyatt can husk at least 12 dozen ears of corn per hour, it will take him no more than $\frac{72}{12}=6$ hours to husk 72 dozen ears of corn. On the other hand, since Wyatt can husk at most 18 dozen ears of corn per hour, it will take him at least $\frac{72}{18}=4$ hours to husk 72 dozen ears of corn. Therefore, the possible times it could take Wyatt to husk 72 dozen ears of corn are 4 hours to 6 hours, inclusive. Any number between 4 and 6 , inclusive, can be gridded as the correct answer.

Solution of Question 32 (Calculator)

QUESTION 32 The correct answer is 107. Since the weight of the empty truck and its driver is 4500 pounds and each box weighs 14 pounds, the weight, in pounds, of the delivery truck, its driver, and $x$ boxes is $4500+14 x$. This weight is below the bridge's posted weight limit of 6000 pounds if $4500+14 x < 6000$. Subtracting 4500 from both sides of this inequality and then dividing both sides by 14 yields $x < \frac{1500}{14}$ or $x < 107 \frac{1}{7}$. Since the number of packages must be an integer, the maximum possible value for $x$ that will keep the combined weight of the truck, its driver, and the $x$ identical boxes below the bridge's posted weight limit is 107 .

Solution of Question 33 (Calculator)

QUESTION 33 The correct answer is $\frac{5}{8}$ or .625. Based on the line graph, the number of portable media players sold in 2008 was 100 million, and the number of portable media players sold in 2011 was 160 million. Therefore, the number of portable media players sold in 2008 is $\frac{100 \text { million }}{160 \text { million }}$ of the portable media players sold in 2011 . This fraction reduces to $\frac{5}{8}$. Either $5 / 8$ or its decimal equivalent, $.625$, may be gridded as the correct answer.

Solution of Question 34 (Calculator)

QUESTION 34 The correct answer is 96 . Since each day has a total of 24 hours of time slots available for the station to sell, there is a total of 48 hours of time slots available to sell on Tuesday and Wednesday. Each time slot is a 30-minute interval, which is equal to a $\frac{1}{2}$-hour interval. Therefore, there are $\frac{48 \text { hours }}{\frac{1}{2} \text { hours/time slot }}=96$ time slots of 30 minutes for the station to sell on Tuesday and Wednesday.

Solution of Question 35 (Calculator)

QUESTION 35 The correct answer is 6 . The volume of a cylinder is $\pi r^{2} h$, where $r$ is the radius of the base of the cylinder and $h$ is the height of the cylinder. Since the storage silo is a cylinder with volume $72 \pi$ cubic yards and height 8 yards, it follows that $72 \pi=\pi r^{2}(8)$, where $r$ is the radius of the base of the cylinder, in yards. Dividing both sides of the equation $72 \pi=\pi r^{2}(8)$ by $8 \pi$ gives $r^{2}=9$, and so the radius of the base of the cylinder is 3 yards. Therefore, the diameter of the base of the cylinder is 6 yards.

Solution of Question 36 (Calculator)

QUESTION 36 The correct answer is 3 . The function $h(x)$ is undefined when the denominator of $\frac{1}{(x-5)^{2}+4(x-5)+4}$ is equal to zero. The expression $(x-5)^{2}+4(x-5)+4$ is a perfect square: $(x-5)^{2}+4(x-5)+4=$ $((x-5)+2)^{2}$, which can be rewritten as $(x-3)^{2}$. The expression $(x-3)^{2}$ is equal to zero if and only if $x=3$. Therefore, the value of $x$ for which $h(x$ is undefined is 3 .

Solution of Question 37 (Calculator)

QUESTION 37 The correct answer is $1.02$. The initial deposit earns 2 percent interest compounded annually. Thus at the end of 1 year, the new value of the account is the initial deposit of $\$ 100$ plus 2 percent of the initial deposit: $\$ 100+\frac{2}{100}(\$ 100)=\$ 100(1.02)$. Since the interest is compounded annually, the value at the end of each succeeding year is the sum of the previous year's value plus 2 percent of the previous year's value. This is again equivalent to multiplying the previous year's value by $1.02$. Thus, after 2 years, the value will be $\$ 100(1.02)(1.02)=\$ 100(1.02)^{2}$; after 3 years, the value will be $\$ 100(1.02)^{3}$; and after $t$ years, the value will be $\$ 100(1.02)^{t}$. Therefore, in the formula for the value for Jessica's account after $t$ years, $\$ 100(x)^{t}$, the value of $x$ must be $1.02$.

Solution of Question 38 (Calculator)

QUESTION 38 The correct answer is $6.11$. Jessica made an initial deposit of $\$ 100$ into her account. The interest on her account is 2 percent compounded annually, so after 10 years, the value of her initial deposit has been multiplied 10 times by the factor $1+0.02=1.02$. Hence, after 10 years, Jessica's deposit is worth $\$ 100(1.02)^{10}=\$ 121.899$ to the nearest tenth of a cent. Tyshaun made an initial deposit of $\$ 100$ into his account. The interest on his account is $2.5$ percent compounded annually, so after 10 years, the value of his initial deposit has been multiplied 10 times by the factor $1+0.025=1.025$. Hence, after 10 years, Tyshaun's deposit is worth $\$ 100(1.025)^{10}=\$ 128.008$ to the nearest tenth of a cent. Hence, Jessica's initial deposit earned $\$ 21.899$ and Tyshaun's initial deposit earned $\$ 28.008$. Therefore, to the nearest cent, Tyshaun's initial deposit earned $\$ 6.11$ more than Jessica's initial deposit.