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# 1.2: Practice Questions

Difficulty Level: At Grade Created by: CK-12

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.

1. (Numbers and Operations) — A racecar driver has completed 1212\begin{align*}12\frac{1}{2}\end{align*} laps of a 50lap\begin{align*}50-\text{lap}\end{align*} race. What fractional part of the race remains?
1. 1/4\begin{align*}1/4\end{align*}
2. 1/5\begin{align*}1/5\end{align*}
3. 3/4\begin{align*}3/4\end{align*}
4. 4/5\begin{align*}4/5\end{align*}
5. 75/2\begin{align*}75/2\end{align*}

ANSWER EXPLANATIONS — The driver has 501212\begin{align*}50 - 12\frac{1}{2}\end{align*} or 3712\begin{align*}37 \frac{1}{2}\end{align*} laps to go out of the 50 laps\begin{align*}50 \ \text{laps}\end{align*}. Therefore, 371250\begin{align*}\frac{37{\frac{1}{2}}}{50}\end{align*} is the fractional part of the race remaining. This is equivalent to 34\begin{align*}\frac{3}{4}\end{align*}, which is answer C.

1. This is the fraction of the race that the driver has completed.
2. This comes from mistakenly dividing 1212\begin{align*}12\frac{1}{2}\end{align*} by 6212\begin{align*}62\frac{1}{2}\end{align*}, the latter number coming from adding both lap amounts. The race is only 50 laps\begin{align*}50 \ \text{laps}\end{align*}.
3. Correct
4. This comes from taking the fraction in response B and subtracting it from 1.
5. This comes from 3712\begin{align*}37 \frac{1}{2}\end{align*}, but that’s how many laps are left - the driver is only in one race, not 50\begin{align*}50\end{align*}!
2. (Numbers and Operations) — If M\begin{align*}\text{M}\end{align*} is the set of positive multiples of 2\begin{align*}2\end{align*} less than 150\begin{align*}150\end{align*} and N\begin{align*}\text{N}\end{align*} is the set of positive multiples of 9\begin{align*}9\end{align*} less than 150\begin{align*}150\end{align*}, how many members are there in M \ n \ N\begin{align*}\text{M \ n \ N}\end{align*}?
1. 0\begin{align*}0\end{align*}
2. 8\begin{align*}8\end{align*}
3. 9\begin{align*}9\end{align*}
4. 18\begin{align*}18\end{align*}
5. 74\begin{align*}74\end{align*}

ANSWER EXPLANATIONS — The only elements in both sets are multiples of 2×9\begin{align*}2 \times 9\end{align*} or 18\begin{align*}18\end{align*}. The multiples of 18\begin{align*}18\end{align*} less than 150\begin{align*}150\end{align*} are 18,36,54,72,90,108,126\begin{align*}18, 36, 54, 72, 90, 108, 126\end{align*} and 144\begin{align*}144\end{align*}. These are the 8\begin{align*}8\end{align*} elements that are in the intersection of both sets. Thus, B is the correct response.

1. There are elements that are in both sets.
2. Correct
3. There are 8\begin{align*}8\end{align*} elements in both sets, not 9\begin{align*}9\end{align*}.
4. This is the number that the elements are multiples of.
5. This is the number of multiples of 2\begin{align*}2\end{align*} less than 150\begin{align*}150\end{align*}
6. .
3. (Algebra and Functions) — If x2y\begin{align*}x \neq 2 y\end{align*}, then x2y2yx+2yxx2y=\begin{align*}\frac{x - 2y}{2y - x} + \frac{2y - x}{x - 2y} = \end{align*}
1. 2(x2y)\begin{align*}2(x - 2y)\end{align*}
2. 2yx\begin{align*}2y - x\end{align*}
3. 1\begin{align*}1\end{align*}
4. 0\begin{align*}0\end{align*}
5. 2\begin{align*}-2\end{align*}

ANSWER EXPLANATIONS — Both fractions each r1\begin{align*}\text{r} - 1\end{align*}, so adding 1\begin{align*}-1\end{align*} twice gives an answer of 2\begin{align*}-2\end{align*}. This is choice E.

1. Both fractions are not equal to (x2y)\begin{align*}(x - 2y)\end{align*}, so we do not have 2\begin{align*}2\end{align*} like terms.
2. The other parts of the fractions are not eliminated, so this is not possible.
3. Each fraction equals 1\begin{align*}-1\end{align*}, so this result is not possible.
4. Each fraction equals 1\begin{align*}-1\end{align*}, so adding 1\begin{align*}-1\end{align*} twice gives 2\begin{align*}-2\end{align*}.
5. Correct
4. (Algebra and Functions) — If 2x=10\begin{align*}2x = 10\end{align*}, what is the value of 625x\begin{align*}\frac{625}{x}\end{align*}? ANSWER EXPLANATION The x\begin{align*}x\end{align*} value is 5\begin{align*}5\end{align*}, so 6255=125\begin{align*}\frac{625}{5} = 125\end{align*}.
5. (Algebra and Functions) — If Dave drove one-third of the distance of his trip on the first day, and 60 miles\begin{align*}60 \ \text{miles}\end{align*} on the second day, he figured out that he still had 12\begin{align*}\frac{1}{2}\end{align*} of the trip to drive. What was the total length, in miles, of his trip?
1. 360\begin{align*}360\end{align*}
2. 180\begin{align*}180\end{align*}
3. 120\begin{align*}120\end{align*}
4. 60\begin{align*}60\end{align*}
5. 90\begin{align*}90\end{align*}

ANSWER EXPLANATIONS — We can model this situation by the equation ?x+60=12x\begin{align*}?x + 60 = \frac{1}{2} x\end{align*}. Solving for x\begin{align*}x\end{align*}, we get x=360\begin{align*}x = 360\end{align*}, which is answer A.

1. Correct
2. This is the number of miles he drove before he figured out how much of his trip was left.
3. This amount is how much he drove the first day.
4. This is how much he drove the second day.
5. This amount is half of the distance he covered by the end of the second day.
6. (Geometry) — What is f(2)\begin{align*}f(2)\end{align*} for the graph of f(x)\begin{align*}f(x)\end{align*} below?
1. 1\begin{align*}1\end{align*}
2. 12\begin{align*}\frac{1}{2}\end{align*}
3. 0\begin{align*}0\end{align*}
4. 2\begin{align*}2\end{align*}
5. 1\begin{align*}-1\end{align*}

ANSWER EXPLANATIONS — Asking for the value of f(2)\begin{align*}f(2)\end{align*} is just a way of saying “what does y\begin{align*}y\end{align*} equal when x=2\begin{align*}x = 2\end{align*}?” So you look for when x=2\begin{align*}x=2\end{align*}. On the graph, we have the point (2,12)\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 2\begin{align*}2\end{align*} tick marks to the right, and then the line crosses through a point on the y\begin{align*}y\end{align*} axis between 0\begin{align*}0\end{align*} and 1\begin{align*}1\end{align*}. Therefore, f(2)\begin{align*}f(2)\end{align*} is equal to 12\begin{align*}\frac{1}{2}\end{align*}. Choice B is the only possible choice.

7. (Geometry) — Find the perimeter of lsosceles triangle ABC\begin{align*}\text{ABC}\end{align*} if mAD=3\begin{align*}\text{mAD} = 3\end{align*} and m<BAC=55\begin{align*}\text{m}<\text{BAC} = 55^\circ\end{align*}. Round to the nearest hundredth.
1. 5.21\begin{align*}5.21\end{align*}
2. 10.42\begin{align*}10.42\end{align*}
3. 13.48\begin{align*}13.48\end{align*}
4. 16.46\begin{align*}16.46\end{align*}
5. 13.39\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 AB\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 cos55=3(AB)\begin{align*}\cos 55^\circ= \frac{3}{(\text{AB})}\end{align*} Multiplying AB\begin{align*}\text{AB}\end{align*} to the left gives (AB)cos55=3\begin{align*}(\text{AB}) \cos 55 = 3\end{align*} Dividing by cos55\begin{align*}\cos 55\end{align*} gives AB=3cos55\begin{align*}\text{AB} = \frac{3}{\cos 55^\circ}\end{align*} Using a calculator gives us AB=5.23\begin{align*}\text{AB} = 5.23\end{align*} Since AB = BC\begin{align*}\text{AB = BC}\end{align*}, we know that AB + BC=10.46\begin{align*}\text{AB + BC} = 10.46\end{align*}. Also, since AD = DC\begin{align*}\text{AD = DC}\end{align*}, we know that AD + DC=6\begin{align*}\text{AD + DC} = 6\end{align*} Therefore, the sum of all three sides is 16.46\begin{align*}16.46\end{align*}.

8. (Geometry) — Which segment is congruent to BE¯¯¯¯¯\begin{align*}\overline{BE}\end{align*}?
1. DE\begin{align*}\text{DE}\end{align*}
2. AC\begin{align*}\text{AC}\end{align*}
3. BD\begin{align*}\text{BD}\end{align*}
4. CE\begin{align*}\text{CE}\end{align*}
5. AE\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

9. (Probability and Statistics) According to the graph, the greatest change in the profit of the Sports Shack occurred between which two consecutive months?
1. January and February
2. February and March
3. March and April
4. April and May
5. May and June

ANSWER EXPLANATION — The profit in January was 20,000\begin{align*}\20,000\end{align*}. The profit in February was20,000\begin{align*}-\20,000\end{align*}. This represents a change in profit of \$40,000\begin{align*}-\40,000\end{align*}, which is the greatest that the graph shows.

10. (Probability and Statistics) — An automobile company made 36,000\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?
1. 39\begin{align*}39\end{align*}
2. 9,000\begin{align*}9,000\end{align*}
3. 5,040\begin{align*}5,040\end{align*}
4. 14,040\begin{align*}14,040\end{align*}
5. 18,000\begin{align*}18,000\end{align*}

ANSWER EXPLANATION — SUVs and vans, together, made up 25%+14%=39%\begin{align*}25 \% + 14 \% = 39 \%\end{align*} of the total production.

39% of 36,000 autos is found by 0.39×36000=14040.

## Date Created:

Feb 23, 2012

Jul 07, 2015
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