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1.2: Practice Questions

Difficulty Level: At Grade Created by: CK-12
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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) — At Bruno’s Video World, the regular price for a DVD is \begin{align*}d\end{align*} dollars. How many DVDs can be purchased for \begin{align*}x\end{align*} dollars when the DVDs are on sale at \begin{align*}20 \%\end{align*}off the regular price?
    1. \begin{align*}\frac{4}{5x}\end{align*}
    2. \begin{align*}\frac{5}{4x}\end{align*}
    3. \begin{align*}\frac{4}{5d}\end{align*}
    4. \begin{align*}\frac{4x}{5d}\end{align*}
    5. \begin{align*}\frac{5x}{4d}\end{align*}

    EXPLANATION — Since you have \begin{align*}x\end{align*} dollars to spend, you must divide \begin{align*}x\end{align*} by the sale price of a CD, which is \begin{align*}(100\% - 20\%)d\end{align*} or \begin{align*}80\%d\end{align*}. \begin{align*}80\%d\end{align*} can be written as a fraction in lowest terms as \begin{align*}\frac{4}{5d}\end{align*}. Therefore, you are dividing. \begin{align*}\frac{x}{\left( \frac{4}{5d} \right)} = \frac{5x}{4d}\end{align*} This is answer E.

    1. This is the percent of the original price times \begin{align*}x\end{align*}.
    2. This is the reciprocal of the sale fraction times \begin{align*}x\end{align*}.
    3. This is the fraction of the original price, which represents the sale price.
    4. This comes from incorrectly handling the fraction in the denominator.
    5. Correct
  2. (Numbers and Operations) — If a geometric sequence starts with a first term of \begin{align*}2\end{align*} and grows exponentially by a factor of \begin{align*}3\end{align*}, what is the sum of the 4th and 5th terms?
    1. \begin{align*}216\end{align*}
    2. \begin{align*}162\end{align*}
    3. \begin{align*}108\end{align*}
    4. \begin{align*}54\end{align*}
    5. \begin{align*}27\end{align*}

    EXPLANATION — This sequence can be shown as \begin{align*}2 \ X \ 3n-1\end{align*}. The 4th term is \begin{align*}2 \ X \ 33\end{align*}, which is \begin{align*}54\end{align*}. The 5th term is \begin{align*}2 \ X \ 34\end{align*}, which is \begin{align*}162\end{align*}. The sum of \begin{align*}54\end{align*} and \begin{align*}162\end{align*} is \begin{align*}216\end{align*}. This is answer A.

    1. Correct
    2. This is the 5th term.
    3. This is the difference of the two terms.
    4. This is the 4th term.
    5. This is part of the 4th term.
  3. (Algebra and Functions) — If \begin{align*}x^2-y^2 = 48\end{align*}, then \begin{align*}\frac{2}{3}(x+y)(x-y)=\end{align*}
    1. \begin{align*}16\end{align*}
    2. \begin{align*}72\end{align*}
    3. \begin{align*}96\end{align*}
    4. \begin{align*}32\end{align*}
    5. \begin{align*}64\end{align*}

    EXPLANATION — The difference of two squares \begin{align*}(x^{2} - y^{2})\end{align*} equals \begin{align*}48\end{align*}, and the factors are \begin{align*}(x + y)(x - y)\end{align*}, so we can just replace both factors in the second expression by \begin{align*}48\end{align*} and multiply by \begin{align*}\frac{2}{3}\end{align*} to get \begin{align*}32\end{align*}. This is choice D.

    1. This involves canceling the \begin{align*}3\end{align*} into the \begin{align*}48\end{align*}, but forgetting to multiply by \begin{align*}2\end{align*}.
    2. This comes from multiplying \begin{align*}\frac{3}{2}\end{align*} by \begin{align*}48\end{align*}.
    3. This is \begin{align*}48\end{align*} \begin{align*}X\end{align*} \begin{align*}2\end{align*}, but it was not divided by \begin{align*}3\end{align*}.
    4. Correct
    5. This is \begin{align*}\frac{4}{3}\end{align*} times \begin{align*}48\end{align*}.
  4. (Algebra and Functions) — Eddie is \begin{align*}7\end{align*} years older than Brian. If Brian is \begin{align*}x\end{align*} years old, then how old was Eddie \begin{align*}11\end{align*}years ago?
    1. \begin{align*}x - 18\end{align*}
    2. \begin{align*}x - 4\end{align*}
    3. \begin{align*}x - 7\end{align*}
    4. \begin{align*}7x - 11\end{align*}
    5. \begin{align*}x + 18\end{align*}

    EXPLANATION — We can model this by getting Eddie’s age now and then figuring out how old he was \begin{align*}11\end{align*} years ago. Right now, he is \begin{align*}x + 7\end{align*}. Eleven years ago, Eddie was \begin{align*}x + 7 - 11 = x - 4\end{align*} years old. This is answer B.

    1. The correct calculation should be \begin{align*}x + 7\end{align*} (to get Eddie’s age now) \begin{align*}- 11\end{align*}, which is \begin{align*}x - 4\end{align*}.
    2. Correct
    3. This is an incorrect calculation of Eddie’s present age.
    4. This response does not relate their ages properly.
    5. This also does not express their age relationship properly.

Directions: The following question (5) is an example of a grid-in 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 grid-in questions, please visit sat.collegeboard.com/practice/sat-practice-questions-math/student-produced-response.

  1. (Algebra and Functions) — If \begin{align*}1.5y = 30\end{align*}, what is the value of \begin{align*}\frac{y}{(y+30)}\end{align*}?
    EXPLANATION — From the equation given \begin{align*}y=20\end{align*}. This is because basic algebra teaches us that with the given equation \begin{align*}1.5y=30\end{align*}, we must divide each side by \begin{align*}1.5\end{align*} to isolate \begin{align*}y\end{align*}. By doing \begin{align*}y=\frac{(30)}{1.5}\end{align*}, we get \begin{align*}y=20\end{align*}. Substituting \begin{align*}y=20\end{align*} into the given expression \begin{align*}\frac{y}{(y+30)}\end{align*} provides \begin{align*}\frac{(20)}{(20+30)}\end{align*} or \begin{align*}\frac{20}{50}\end{align*}, which must be reduced to \begin{align*}\frac{2}{5}\end{align*} as a fraction or \begin{align*}0.4\end{align*} as a decimal. \begin{align*}0.4\end{align*} is the correct answer.
  2. (Geometry) — Which line has a slope of \begin{align*}-3\end{align*}?
    1. \begin{align*}3x + 2y = 4\end{align*}
    2. \begin{align*}-3x + y = 5\end{align*}
    3. \begin{align*}3x + y = 2\end{align*}
    4. \begin{align*}6x + 3y = 9\end{align*}
    5. \begin{align*}-3x + 2y = 10\end{align*}

    EXPLANATION — To find the slope of a line, the equation must be solved for \begin{align*}y\end{align*}. The coefficient of \begin{align*}x\end{align*} is the slope when the equation is solved for \begin{align*}y\end{align*}. The only equation above that gives a coefficient for \begin{align*}x\end{align*} of \begin{align*}-3\end{align*} is C.

  3. (Geometry) — Which of the following would shift \begin{align*}f(x)\end{align*} right \begin{align*}3 \ units\end{align*}?
    1. \begin{align*}\frac{1}{3}f(x)\end{align*}
    2. \begin{align*}f(x+3)\end{align*}
    3. \begin{align*}f(x-3)\end{align*}
    4. \begin{align*}f(x) + 3\end{align*}
    5. \begin{align*}f(x) - 3\end{align*}

    EXPLANATION — Remember that horizontal (left or right) changes are changes to the \begin{align*}x-\end{align*}values and changes to the \begin{align*}x-\end{align*}values happen inside the parentheses. Also, remember that they do the opposite of what they look like they should. That means that the \begin{align*}x - 3\end{align*} in the parentheses is going to cause the graph to shift to the right. Therefore, C is the solution.

  4. (Geometry)\begin{align*}\overline{AD} || \overline{BC}. \ mAC=13. \ mBC =5\end{align*}. If \begin{align*}mBD = 15\end{align*}, what is \begin{align*}mAD\end{align*}?  
    1. \begin{align*}8\end{align*}
    2. \begin{align*}9\end{align*}
    3. \begin{align*}10\end{align*}
    4. \begin{align*}11\end{align*}
    5. \begin{align*}12\end{align*}

    EXPLANATION — Correct answer: B To find \begin{align*}AD\end{align*}, we first need to know \begin{align*}AB\end{align*}. Since \begin{align*}AB\end{align*} belongs to both triangles, we’ll use what we know about triangle \begin{align*}ABC\end{align*} to find \begin{align*}AB\end{align*}, using the Pythagorean Theorem. \begin{align*}AB\end{align*} is a short side, so we’ll call it a. The Pythagorean Theorem is \begin{align*}a(^{2}) + b(^{2}) = c(^{2}) \ldots\end{align*} (\begin{align*}a\end{align*} and \begin{align*}b\end{align*} are the shorter sides, \begin{align*}c\end{align*} is the hypotenuse) Plugging in gives \begin{align*}a(^{2}) + 52 = 132\end{align*} Squaring gives \begin{align*}a(^{2}) + 25 = 169\end{align*} Subtracting \begin{align*}25\end{align*} on each side gives \begin{align*}a(^{2}) = 144\end{align*}. Taking the square roots gives \begin{align*}a = 12\end{align*}. This means \begin{align*}AB = 12\end{align*}. To find \begin{align*}AD\end{align*}, we’ll use the Pythagorean Theorem again, with a representing \begin{align*}AD\end{align*} this time. Plugging in gives \begin{align*}a(^{2}) + 122 = 152\end{align*} Squaring gives \begin{align*}a(^{2}) + 144 = 225\end{align*} Subtracting \begin{align*}144\end{align*} from each side gives \begin{align*}a(^{2}) = 81\end{align*}. Taking the square root gives \begin{align*}a = 9\end{align*}, which is length of \begin{align*}AD\end{align*}.

  5. (Probability and Statistics)The following chart gives the graduation ages of 10 students. What is the mean age of the graduating students?
    1. \begin{align*}17\end{align*}
    2. \begin{align*}17.1\end{align*}
    3. \begin{align*}17.2\end{align*}
    4. \begin{align*}17.3\end{align*}
    5. \begin{align*}17.4\end{align*}

    EXPLANATION — According to the graph, there are: Two \begin{align*}16\end{align*} year-olds Five \begin{align*}17\end{align*} year-olds Three \begin{align*}18\end{align*} year-olds That means their ages, added up, will be \begin{align*}(16 + 16) + (17 + 17 + 17 + 17 + 17) + (18 + 18 + 18) = 171\end{align*}. Note that because there are \begin{align*}10\end{align*} elements (the number of students), we must now divide the total by \begin{align*}10\end{align*}. \begin{align*}171\end{align*} divided by \begin{align*}10 = 17.1\end{align*}.

  6. (Probability and Statistics) — A high school of \begin{align*}350\end{align*} students offers biology and chemistry courses. \begin{align*}213\end{align*} students are enrolled in biology while \begin{align*}155\end{align*} students are studying chemistry. \begin{align*}78\end{align*}students are studying both subjects. How many students are not studying either subject?
    1. \begin{align*}18\end{align*}
    2. \begin{align*}60\end{align*}
    3. \begin{align*}70\end{align*}
    4. \begin{align*}50\end{align*}
    5. \begin{align*}78\end{align*}

    EXPLANATION — The correct answer is B. \begin{align*}60\end{align*} students. To solve this problem, first find the total number of students in the combined biology and chemistry courses \begin{align*}[213+155=368]\end{align*} Subtract from this total the number of students taking both courses, as they should only be counted once \begin{align*}[368-78=290]\end{align*}. This leaves you with the number of students enrolled in biology and/or chemistry. The difference between this total of students and the number of students in the high school will be the number of students not studying either subject \begin{align*}[350-290=60]\end{align*}. \begin{align*}60\end{align*} students are not enrolled in Biology and/or Chemistry. B is the correct answer.

    1. \begin{align*}18\end{align*} is the difference between the actual number of students in the high school class and the total number of enrolled students in the combined science courses. This is a trap answer that faulty logic may lead one to choose.
    2. Correct.
    3. \begin{align*}70\end{align*} would be an answer resulting from faulty arithmetic or other human errors.
    4. \begin{align*}50\end{align*} would be an answer resulting from faulty arithmetic or other human errors.
    5. \begin{align*}78\end{align*} is the number of students studying both subjects.

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Date Created:
Feb 23, 2012
Last Modified:
Jun 08, 2015
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