1.2: Practice Questions
Directions: The following question (1) is an example of a gridin math problem. On the SAT, you will solve the problem and indicate your answer by darkening the ovals on the special grid provided. Since you do not have this type of answer sheet to practice on, simply write your response. For more information about gridin questions, please visit sat.collegeboard.com/practice/satpracticequestionsmath/studentproducedresponse.

(Numbers and Operations) — The average of five positive odd numbers is \begin{align*}73\end{align*}
73 . If \begin{align*}x\end{align*}x is the greatest of these integers, what is the greatest possible value of \begin{align*}x\end{align*}x ? EXPLANATION — If we use the lowest four odd numbers (\begin{align*}1, \ 3, \ 5\end{align*}1, 3, 5 and \begin{align*}7\end{align*}7 ) and then solve for \begin{align*}x\end{align*}x in the equation \begin{align*}1 + 3 + 5 + 7 + x = (5)*(73)\end{align*}1+3+5+7+x=(5)∗(73) , we can get this solution. The lowest four odd numbers add up to \begin{align*}16\end{align*}16 . This leaves \begin{align*}x = (365  16)\end{align*}x=(365−16) , which means that \begin{align*}349\end{align*}349 is the greatest possible value of \begin{align*}x\end{align*}x .
Directions: For this section, solve each problem and decide which is the best of the choices given. You may use any available space for scratchwork.

(Numbers and Operations) — How many unique real roots does the equation \begin{align*}y = x^2  10x + 25\end{align*}
y=x2−10x+25 have? No solutions.

\begin{align*}1\end{align*}
1 
\begin{align*}2\end{align*}
2 
\begin{align*}3\end{align*}
3
EXPLANATION — Factoring this quadratic equations gives the factors \begin{align*}(x  5)(x  5) = 0\end{align*}
(x−5)(x−5)=0 . The solutions are both \begin{align*}5\end{align*}5 , so there is just one unique root. This is answer C. There are two equal solutions at \begin{align*}x = 5\end{align*}
x=5 .  See A.
 Correct
 There are \begin{align*}2\end{align*}
2 solutions, but they are not unique.  A quadratic equation can have at most \begin{align*}2\end{align*}
2 roots.

(Algebra and Functions) — If \begin{align*}g(x) = 4x  4x\end{align*}
g(x)=4x−4x , what is the value of \begin{align*}g \left(\frac{3}{2}\right)\end{align*}? \begin{align*}2\end{align*}
 \begin{align*}6\end{align*}
 \begin{align*}8\end{align*}
 \begin{align*}2\end{align*}
EXPLANATION — Substituting \begin{align*}\frac{3}{2}\end{align*} into the function, we get \begin{align*}(4)\left(\frac{3}{2}\right)  4^{\left(\frac{3}{2}\right)} = 6  (\sqrt{ 4})^{3} = 6  2^{3} = 6  8 = 2\end{align*}. This is response A.
 Correct
 This is the value of \begin{align*}4\left( \frac{3}{2}\right)\end{align*}.
 This is the value of \begin{align*}4\frac{3}{2}\end{align*}.
 This is the result of an incorrect subtraction.
 The two terms do not cancel out!

(Algebra and Functions) — If \begin{align*}f(x) = x + 9\end{align*}, which of the following is a solution of \begin{align*}f(5a) + 3 = f(3a) + 11\end{align*}? In other words, what might be the value of ‘\begin{align*}a\end{align*}’?
 There are no solutions.
 \begin{align*}6\end{align*}
 \begin{align*}8\end{align*}
 \begin{align*}\frac{5}{3}\end{align*}
 \begin{align*}4\end{align*}
EXPLANATION — Substituting both \begin{align*}5a\end{align*} and \begin{align*}3a\end{align*} into \begin{align*}f(x)\end{align*} and then using the second equation, we get \begin{align*}5a + 9 + 3 = 3a + 9 + 11\end{align*}. This simplifies to \begin{align*}5a + 12 = 3a + 20\end{align*}, and then we get \begin{align*}2a = 8\end{align*}. Finally, we get \begin{align*}a = 4\end{align*}. This is response E.
 Not true! \begin{align*}a = 4\end{align*} is a solution.
 This does not solve the equation.
 This is the difference of \begin{align*}11\end{align*} and \begin{align*}3\end{align*}, but it does not solve the equation.
 This is the quotient of \begin{align*}5a\end{align*} and \begin{align*}3a\end{align*}.
 Correct

(Algebra and Functions) — If the mean of \begin{align*}x\end{align*} and \begin{align*}4x\end{align*} is \begin{align*}10\end{align*}, then \begin{align*}x =\end{align*}
 \begin{align*}20\end{align*}
 \begin{align*}16\end{align*}
 \begin{align*}4\end{align*}
 \begin{align*}10\end{align*}
 \begin{align*}5\end{align*}
EXPLANATION — First off, what is the question asking? The mean is a form of central tendency that is looking for the average. So if we are finding the mean, we will add up the elements (\begin{align*}x\end{align*} and \begin{align*}4x\end{align*}) and then since there are only \begin{align*}2\end{align*}, we will divide by \begin{align*}2\end{align*}. If I asked you what the average, or mean, of \begin{align*}10, \ 20\end{align*} and \begin{align*}30\end{align*}, you would sum up the numbers to get \begin{align*}60\end{align*} and then divide by \begin{align*}3\end{align*} to get \begin{align*}20\end{align*}the mean. We can write this as \begin{align*}\frac{(x+4x)}{2} = 10\end{align*}. So, \begin{align*}5x = 20\end{align*}, and \begin{align*}x = 4\end{align*}. This is choice C.
 This is the sum of \begin{align*}x\end{align*} and \begin{align*}4x\end{align*}.
 This is equal to \begin{align*}4x\end{align*}.
 Correct
 This is the average of \begin{align*}x\end{align*} and \begin{align*}4x\end{align*}.
 This is the coefficient of the average equation.

(Geometry) — If a line is perpendicular to the line \begin{align*}y = 3x  4\end{align*} and passes through the point \begin{align*}(0, 6)\end{align*}, what is the equation of the perpendicular line?
 \begin{align*}y = 3x + 6\end{align*}
 \begin{align*}y = 3x + 6\end{align*}
 \begin{align*}y = \frac{1}{3x + 6}\end{align*}
 \begin{align*}y = \frac{1}{3x  6}\end{align*}
 \begin{align*}y = \frac{1}{3x + 6}\end{align*}
EXPLANATION — The slopes of two perpendicular lines are negative reciprocals of each other. This also means that the product of the two slopes equals \begin{align*}1\end{align*}. A perpendicular line has the negative reciprocal slope of another line. Since the slope of the given line is \begin{align*}3\end{align*}, the slope of a perpendicular line is \begin{align*}\frac{1}{3}\end{align*}. The basic equation is then \begin{align*}y = \frac{1}{3x + b}\end{align*}. We need to use the point \begin{align*}(0, 6)\end{align*} to find \begin{align*}b\end{align*}. Let’s do that: \begin{align*}6 =\frac{1}{3(0)+ b}\end{align*}. Therefore, \begin{align*}b = 6\end{align*}, and the complete perpendicular line equation is \begin{align*}y = \frac{1}{3x+ 6}\end{align*}.
 This is a parallel line.
 A perpendicular line has a negative reciprocal slope, not just a negative slope.
 Correct
 This is an incorrect calculation of the \begin{align*}y\end{align*}intercept.
 A perpendicular line has the negative reciprocal slope, not just a reciprocal slope.

(Geometry) — What is the height of a building if the angle to the top is \begin{align*}27^\circ\end{align*} when you are standing \begin{align*}150 \ feet\end{align*} away from the building’s base? Round to the nearest whole number.
 \begin{align*}68\end{align*}
 \begin{align*}134\end{align*}
 \begin{align*}76\end{align*}
 \begin{align*}330\end{align*}
 \begin{align*}168\end{align*}
EXPLANATION — If you draw a triangle where the base is \begin{align*}150 \ ft\end{align*} and the height is the height of the building, this can be easy to visualize. The angle where you are “standing” in the problem is \begin{align*}27 \ degrees\end{align*}. From that angle, we have the adjacent (next to the angle) side and the opposite side. That information makes this a tangent problem. Tangent of an angle = (opposite) \begin{align*}\div\end{align*} (adjacent) The setup is \begin{align*}\tan 27^\circ= \left(\frac{height}{150}\right)\end{align*} Multiplying the \begin{align*}150\end{align*} to the left and using a calculator gives \begin{align*}76 = height\end{align*} for the answer. Incorrect answers could result from using the wrong trigonometric function for the situation or dividing by \begin{align*}150\end{align*} or \begin{align*}\tan 27\end{align*} instead of multiplying.

(Geometry) — Which of the following equations defines \begin{align*}g(x)\end{align*} in terms of \begin{align*}f(x)\end{align*}?
 \begin{align*}g(x) = f(x  2) + 3\end{align*}
 \begin{align*}g(x) = f(x + 2) + 3\end{align*}
 \begin{align*}g(x) = f(x  2) 3\end{align*}
 \begin{align*}g(x) = f(x + 2)  3\end{align*}
 \begin{align*}g(x) = f(x  2)\end{align*}
EXPLANATION — A \begin{align*}g(x)\end{align*} is a shift \begin{align*}2 \ units\end{align*} to the right and \begin{align*}3 \ units\end{align*} up (to find this easily, just follow the vertex – the lowest point on the parabola; it is the easiest point to find and imagine moving). The function that does this is the one with \begin{align*}x  2\end{align*} in the parentheses, because if you remember, changes to the \begin{align*}x\end{align*}values happen inside the parentheses and do the opposite of what they look like they do. Changes to the \begin{align*}y\end{align*}value (up or down changes) happen outside the parentheses but they do exactly what they look like they do. So we need \begin{align*}a + 3\end{align*} to move the graph \begin{align*}3 \ units\end{align*} up. Therefore, with all that in mind, the only choice that has everything we need is A.

(Probability and Statistics) — The following chart gives the graduation ages of 10 students? What is the median age of the graduating students?
 \begin{align*}16\end{align*}
 \begin{align*}16.5\end{align*}
 \begin{align*}17\end{align*}
 \begin{align*}17.5\end{align*}
 \begin{align*}18\end{align*}
EXPLANATION — C The median is the middle of a number distribution when the numbers are ordered in an either ascending or descending order. Unlike the other \begin{align*}2\end{align*} measures of central tendency (mean is the average, mode is the value that occurs the most), the median is basically the center number. According to the graph, we have: Two \begin{align*}16\end{align*} yearolds, five \begin{align*}17\end{align*} yearolds and three \begin{align*}18\end{align*} yearolds. If you wrote these ages in a list it would look like this: \begin{align*}16, \ 16, \ 17, \ 17, \ 17, \ 17, \ 17, \ 18, \ 18, \ 18\end{align*}. In a list of \begin{align*}9\end{align*} terms, the 5th term (that leaves \begin{align*}4\end{align*} terms below it and \begin{align*}4\end{align*} terms above it) is the median. However, with \begin{align*}10\end{align*} terms like in this problem you must average the 5th and 6th terms, which both separate the distribution. Since the 5th and 6th term are both \begin{align*}17\end{align*}, the average is \begin{align*}17\end{align*} and this is your median. Therefore, \begin{align*}17\end{align*} is the median age.

(Probability and Statistics) — Assume that the edge of the smaller shaded square is \begin{align*}1 \ inch\end{align*} and the edge of the larger square is \begin{align*}2 \ inches\end{align*}. What percent of the diagram is unshaded?
 \begin{align*}35\%\end{align*}
 \begin{align*}30\%\end{align*}
 \begin{align*}15\%\end{align*}
 \begin{align*}10\%\end{align*}
 \begin{align*}25\%\end{align*}
EXPLANATION — E The area of the unshaded region can be found as follows: First, find the area of the shaded region: The smaller square has an edge of \begin{align*}1\end{align*}. It’s \begin{align*}area = e(^{2}) = 12 = 1\end{align*} The four triangles have \begin{align*}area = \frac{1}{2} \ bh = \frac{1}{2} \ (1)(1) = \frac{1}{2}\end{align*} All four together have an area of \begin{align*}4\left ( \frac{1}{2} \right ) = 2\end{align*} The shaded areas together have an area of \begin{align*}1 + 2 = 3\end{align*}. The total area of the figure is \begin{align*}area = e(^{2}) = 22 = 4\end{align*}. That also means that the unshaded region has an area of \begin{align*}1\end{align*} (because \begin{align*}4  3 = 1\end{align*}) To find the probability, divide the unshaded area by the whole area. That means we have \begin{align*}\left(\frac{unshaded}{whole} = \frac{1}{4}\right)\end{align*} which is the same as \begin{align*}25\%\end{align*}. Note: ‘\begin{align*}e\end{align*}’ means edge.
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