1.2: Practice Questions
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) — A racecar driver has completed \begin{align*}12\frac{1}{2}\end{align*}
1212 laps of a \begin{align*}50\text{lap}\end{align*}50−lap race. What fractional part of the race remains?
\begin{align*}1/4\end{align*}
1/4 
\begin{align*}1/5\end{align*}
1/5 
\begin{align*}3/4\end{align*}
3/4 
\begin{align*}4/5\end{align*}
4/5 
\begin{align*}75/2\end{align*}
75/2
ANSWER EXPLANATIONS — The driver has \begin{align*}50  12\frac{1}{2}\end{align*}
50−1212 or \begin{align*}37 \frac{1}{2}\end{align*} laps to go out of the \begin{align*}50 \ \text{laps}\end{align*}. Therefore, \begin{align*}\frac{37{\frac{1}{2}}}{50}\end{align*} is the fractional part of the race remaining. This is equivalent to \begin{align*}\frac{3}{4}\end{align*}, which is answer C. This is the fraction of the race that the driver has completed.
 This comes from mistakenly dividing \begin{align*}12\frac{1}{2}\end{align*} by \begin{align*}62\frac{1}{2}\end{align*}, the latter number coming from adding both lap amounts. The race is only \begin{align*}50 \ \text{laps}\end{align*}.
 Correct
 This comes from taking the fraction in response B and subtracting it from 1.
 This comes from \begin{align*}37 \frac{1}{2}\end{align*}, but that’s how many laps are left  the driver is only in one race, not \begin{align*}50\end{align*}!

\begin{align*}1/4\end{align*}

(Numbers and Operations) — If \begin{align*}\text{M}\end{align*} is the set of positive multiples of \begin{align*}2\end{align*} less than \begin{align*}150\end{align*} and \begin{align*}\text{N}\end{align*} is the set of positive multiples of \begin{align*}9\end{align*} less than \begin{align*}150\end{align*}, how many members are there in \begin{align*}\text{M \ n \ N}\end{align*}?
 \begin{align*}0\end{align*}
 \begin{align*}8\end{align*}
 \begin{align*}9\end{align*}
 \begin{align*}18\end{align*}
 \begin{align*}74\end{align*}
ANSWER EXPLANATIONS — The only elements in both sets are multiples of \begin{align*}2 \times 9\end{align*} or \begin{align*}18\end{align*}. The multiples of \begin{align*}18\end{align*} less than \begin{align*}150\end{align*} are \begin{align*}18, 36, 54, 72, 90, 108, 126\end{align*} and \begin{align*}144\end{align*}. These are the \begin{align*}8\end{align*} elements that are in the intersection of both sets. Thus, B is the correct response.
 There are elements that are in both sets.
 Correct
 There are \begin{align*}8\end{align*} elements in both sets, not \begin{align*}9\end{align*}.
 This is the number that the elements are multiples of.
 This is the number of multiples of \begin{align*}2\end{align*} less than \begin{align*}150\end{align*}
 .

(Algebra and Functions) — If \begin{align*}x \neq 2 y\end{align*}, then \begin{align*}\frac{x  2y}{2y  x} + \frac{2y  x}{x  2y} = \end{align*}
 \begin{align*}2(x  2y)\end{align*}
 \begin{align*}2y  x\end{align*}
 \begin{align*}1\end{align*}
 \begin{align*}0\end{align*}
 \begin{align*}2\end{align*}
ANSWER EXPLANATIONS — Both fractions each \begin{align*}\text{r}  1\end{align*}, so adding \begin{align*}1\end{align*} twice gives an answer of \begin{align*}2\end{align*}. This is choice E.
 Both fractions are not equal to \begin{align*}(x  2y)\end{align*}, so we do not have \begin{align*}2\end{align*} like terms.
 The other parts of the fractions are not eliminated, so this is not possible.
 Each fraction equals \begin{align*}1\end{align*}, so this result is not possible.
 Each fraction equals \begin{align*}1\end{align*}, so adding \begin{align*}1\end{align*} twice gives \begin{align*}2\end{align*}.
 Correct
 (Algebra and Functions) — If \begin{align*}2x = 10\end{align*}, what is the value of \begin{align*}\frac{625}{x}\end{align*}? ANSWER EXPLANATION The \begin{align*}x\end{align*} value is \begin{align*}5\end{align*}, so \begin{align*}\frac{625}{5} = 125\end{align*}.

(Algebra and Functions) — If Dave drove onethird of the distance of his trip on the first day, and \begin{align*}60 \ \text{miles}\end{align*} on the second day, he figured out that he still had \begin{align*}\frac{1}{2}\end{align*} of the trip to drive. What was the total length, in miles, of his trip?
 \begin{align*}360\end{align*}
 \begin{align*}180\end{align*}
 \begin{align*}120\end{align*}
 \begin{align*}60\end{align*}
 \begin{align*}90\end{align*}
ANSWER EXPLANATIONS — We can model this situation by the equation \begin{align*}?x + 60 = \frac{1}{2} x\end{align*}. Solving for \begin{align*}x\end{align*}, we get \begin{align*}x = 360\end{align*}, which is answer A.
 Correct
 This is the number of miles he drove before he figured out how much of his trip was left.
 This amount is how much he drove the first day.
 This is how much he drove the second day.
 This amount is half of the distance he covered by the end of the second day.

(Geometry) — What is \begin{align*}f(2)\end{align*} for the graph of \begin{align*}f(x)\end{align*} below?
 \begin{align*}1\end{align*}
 \begin{align*}\frac{1}{2}\end{align*}
 \begin{align*}0\end{align*}
 \begin{align*}2\end{align*}
 \begin{align*}1\end{align*}
ANSWER EXPLANATIONS — Asking for the value of \begin{align*}f(2)\end{align*} is just a way of saying “what does \begin{align*}y\end{align*} equal when \begin{align*}x = 2\end{align*}?” So you look for when \begin{align*}x=2\end{align*}. On the graph, we have the point \begin{align*}\left (2, \frac{1}{2} \right )\end{align*}. Note that exact numbers are not provided in the graph, but standard practice indicates that each tick mark is one unit. So we moved \begin{align*}2\end{align*} tick marks to the right, and then the line crosses through a point on the \begin{align*}y\end{align*} axis between \begin{align*}0\end{align*} and \begin{align*}1\end{align*}. Therefore, \begin{align*}f(2)\end{align*} is equal to \begin{align*}\frac{1}{2}\end{align*}. Choice B is the only possible choice.

(Geometry) — Find the perimeter of lsosceles triangle \begin{align*}\text{ABC}\end{align*} if \begin{align*}\text{mAD} = 3\end{align*} and \begin{align*}\text{m}<\text{BAC} = 55^\circ\end{align*}. Round to the nearest hundredth.
 \begin{align*}5.21\end{align*}
 \begin{align*}10.42\end{align*}
 \begin{align*}13.48\end{align*}
 \begin{align*}16.46\end{align*}
 \begin{align*}13.39\end{align*}
ANSWER EXPLANATIONS — Correct answer: D — The first step is to find one of the other sides of the triangle. The easiest is \begin{align*}\text{AB}\end{align*}. Since we have the side adjacent (next to) the angle we know, we can use cosine. Cosine of an angle \begin{align*}=\end{align*} (adjacent) \begin{align*}\div\end{align*} (hypotenuse) The setup is \begin{align*}\cos 55^\circ= \frac{3}{(\text{AB})}\end{align*} Multiplying \begin{align*}\text{AB}\end{align*} to the left gives \begin{align*}(\text{AB}) \cos 55 = 3\end{align*} Dividing by \begin{align*}\cos 55\end{align*} gives \begin{align*}\text{AB} = \frac{3}{\cos 55^\circ}\end{align*} Using a calculator gives us \begin{align*}\text{AB} = 5.23\end{align*} Since \begin{align*}\text{AB = BC}\end{align*}, we know that \begin{align*}\text{AB + BC} = 10.46\end{align*}. Also, since \begin{align*}\text{AD = DC}\end{align*}, we know that \begin{align*}\text{AD + DC} = 6\end{align*} Therefore, the sum of all three sides is \begin{align*}16.46\end{align*}.

(Geometry) — Which segment is congruent to \begin{align*}\overline{BE}\end{align*}?
 \begin{align*}\text{DE}\end{align*}
 \begin{align*}\text{AC}\end{align*}
 \begin{align*}\text{BD}\end{align*}
 \begin{align*}\text{CE}\end{align*}
 \begin{align*}\text{AE}\end{align*}
ANSWER EXPLANATION Correct answer: D Congruent segments (or shapes of any kind) are exactly equal in length. The only segment in the diagram that we know is the same as BE is CE. Incorrect answers: A, E: Incorrectly assuming diagonals of a parallelogram are all equal B, C: Incorrectly assuming that the sides of the parallelogram are the same as BE just because they look like they are  never assume two segments are congruent just because they look like it. D: Correct answer

(Probability and Statistics) According to the graph, the greatest change in the profit of the Sports Shack occurred between which two consecutive months?
 January and February
 February and March
 March and April
 April and May
 May and June
ANSWER EXPLANATION — The profit in January was \begin{align*}\$20,000\end{align*}. The profit in February was \begin{align*}\$20,000\end{align*}. This represents a change in profit of \begin{align*}\$40,000\end{align*}, which is the greatest that the graph shows.

(Probability and Statistics) — An automobile company made \begin{align*}36,000\end{align*} automobiles in the proportions shown in the table below. How many of the automobiles produced were either SUVs or vans?
 \begin{align*}39\end{align*}
 \begin{align*}9,000\end{align*}
 \begin{align*}5,040\end{align*}
 \begin{align*}14,040\end{align*}
 \begin{align*}18,000\end{align*}
ANSWER EXPLANATION — SUVs and vans, together, made up \begin{align*}25 \% + 14 \% = 39 \%\end{align*} of the total production. \begin{align*}39 \% \ \text{of} \ 36,000 \ \text{autos is found by}\ 0.39 \times 36000 = 14040.\end{align*}
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