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MPE Practice Problems for Math 142

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Math Placement Exam

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MPE Practice MPE Practice Solutions
Directions. The following are review problems for the MPE. We recommend you work the problems yourself, and then click "Answer" to check your answer. If you do not understand a problem, you can click "More Examples" to see similar examples. While there are no videos for the MPE problems, the similar examples do have videos explaining the solution.
 
  1. Rationalize the denominator. \[\displaystyle \frac{14}{3+\sqrt{2}}\]

    \[
    \begin{align}
    \left(\dfrac{14}{3+\sqrt{2}}\right) \left(\dfrac{3-\sqrt{2}}{3-\sqrt{2}}\right)&=\dfrac{42-14\sqrt{2}}{9-2}\\
    &=\dfrac{7\left(6-2\sqrt{2}\right)}{7}\\
    &=6-2\sqrt{2}
    \end{align}
    \]
    Follow the link below to see more examples. 
    Section A.2 – Exponents and Radicals

  2. Find the sum or difference as indicated, and write your answer in simplified form.\[\frac{x+2a-3}{x+a}-\frac{x+6}{2x}\]

    \[\begin{align*}
    \frac{x+2a-3}{x+a}-\frac{x+6}{2x} &=\left(\frac{x+2a-3}{x+a}\right) \left(\frac{2x}{2x}\right) - \left(\frac{x+6}{2x}\right)\left(\frac{x+a}{x+a}\right)\\
    &=\frac{(x+2a-3)(2x)-(x+6)(x+a)}{(x+a)(2x)}\\
    &=\frac{2x^2+4xa-6x-x^2-6x-xa-6a}{(x+a)(2x)}\\
    &=\frac{x^2+3xa-12x-6a}{(x+a)(2x)}
    \end{align*}\]

    Follow the link below to see more examples. 
    Section A.4 – Rational Expressions

  3. Factor and reduce to simplest form.\[\frac{6x^2+11xy-10y^2}{3x^2+10xy-8y^2}\]

    \[ \dfrac{6x^2+11xy-10y^2}{3x^2+10xy-8y^2} = \dfrac{(3x-2y)(2x+5y)}{(3x-2y)(x+4y)}=\dfrac{2x+5y}{x+4y}\]
    Follow the link below to see more examples. 
    Section A.4 – Rational Expressions

  4. Solve the following equation: \(5(x-7)-13(x-7)-6=0\).

    \[\begin{align*}
    5(x-7)-13(x-7)-6&=0\\
    (x-7)(5-13)-6&=0\\
    (x-7)(-8)-6&=0\\
    -8x+56-6&=0\\
    -8x+50&=0\\
    -8x&=-50\\
    x&=\dfrac{50}{8}
    \end{align*}\]

  5. Find the point \((x,y)\) which satisfies both equations. What is the value of \(x+y\)?\[\begin{align*} -2x+4y&=12\\ 3x-5y&=-3\end{align*}\]

    Multiply \(-2x+4y=12\) by \(3\) and \(3x-5y=-3\) by 2. This gives
    \[\begin{align*}
    -6x+12y&=36\\
    6x-10y&=-6\\
    \end{align*}\]
    Add these two equations and we get \(2y=30\) \(\enspace \Longrightarrow \enspace\) \(y=15\). Substituting \(y=15\) into \(3x-5y=-3\) gives \(x=24\). So the point that satisfies both equations is \((24,15)\), and the value of \(x+y\) is \(24+15 =39\). 
    Follow the link below to see more examples. 
    Section 7.1&2 – Systems of Equations

  6. Two investments are made, totaling \($10,000\). In one year, these investments yield \($650\) in simple interest. Part of the \($10,000\) is invested at \(5\frac{1}{2}\%\), and the rest at \(6\frac{3}{4}\%\). How much more money is invested at \(6\frac{3}{4}\%\)?

    Let \(x\) be the amount invested at \(5\frac{1}{2}\%\) and \(y\) be the amount invested at \(6\frac{3}{4}\%\). The resulting system of equations is 
    \[\begin{align*}
    x+y&=10,000\\
    0.055x+0.0675y&=650
    \end{align*}\]
    Solve the system of equations to find \(x\) and \(y\). 
    \(x=10,000-y\) \(\enspace\Longrightarrow\enspace\) \(0.055(10,000-y)+0.0675y=650\) \(\enspace\Longrightarrow\enspace\) \(550-0.055y+0.0675y=650\) \(\enspace\Longrightarrow\enspace\) \(0.0125y = 100\)  \(\enspace\Longrightarrow\enspace\) \(y=8000\). Since \(x+y=10,000\), we know that \(x=2000\), and thus \($6,000\) more is invested at \(6\frac{3}{4}\%\).
    Follow the link below to see more examples. 
    Section 7.1&2 – Systems of Equations

  7. Given the linear equation \(2ax + 3by = 7c\), where \(a\), \(b\) and \(c > 0\), if \(x\) decreases by \(10\) units, what is the corresponding change in \(y\)?

    Since \(2ax+3by=7c\) \(\enspace\Longrightarrow\enspace\) \(3by=7c-2ax\) \(\enspace\Longrightarrow\enspace\) \(y=\dfrac{7c-2ax}{3b}=\dfrac{-2a}{3b}x +\dfrac{7c}{3b}\). If \(x\) decreases by \(10\), then we know \(y=\dfrac{-2a}{3b}(x-10)+\dfrac{7c}{3b}=\dfrac{-2ax}{3b}+\dfrac{20a}{3b}+\dfrac{7c}{3b}\), so \(y\) will increase by \(\dfrac{20a}{3b}\).

  8.  Perform the indicated operations and simplify. \[\frac{8}{x+1}-\left(\frac{y}{z+2} \div \frac{y-4}{w}\right)\]

    \[\begin{align*}
    \frac{8}{x+1}-\left( \frac{y}{z+2} \div \frac{y-4}{w}\right) &= \frac{8}{x+1}-\left( \frac{y}{z+2} \cdot \frac{w}{y-4}\right)\\
    &=\frac{8}{x+1} - \left( \frac{yw}{(z+2)(y-4)}\right) \\
    &=\frac{ 8(z+2)(y-4)-yw(x+1)}{(x+1)(z+2)(y-4)}\\
    &=\frac{8zy-32z+16y-64-ywx-yw}{(x+1)(z+2)(y-4)}
    \end{align*}\]
    Follow the link below to see more examples. 
    Section A.4 – Rational Expressions

  9. Find the equation of the line passing through the point \((5,1)\) with a slope of 7. Use the equation you find to determine the value of \(y\) when \(x=-4.\)

    Using the point-slope equation, we get \(y-1=7(x-5)\) \(\enspace \Longrightarrow\enspace \) \(y=7x-35+1\) \(\enspace \Longrightarrow\enspace \) \(y=7x-34\). Using this to find \(y\) when \(x=-4\) gives \(y=7(-4)-34\) \(\enspace \Longrightarrow\enspace \) \(y=-62\). 
    Follow the link below to see more examples. 
    Section 2.1 – Review of Lines

  10. Line \(A\) passes through the points \((2k+3, 4k−6)\) and \((−2, 16)\). Find the value of \(k\) if line \(A\) has a slope of \(0\).

    \(\dfrac{4k-6-16}{2k+3+2}=0\) \(\enspace\Longrightarrow\enspace\) \(\dfrac{4k-22}{2k+5}=0\) \(\enspace\Longrightarrow\enspace\) \(4k-22=0\) \(\enspace\Longrightarrow\enspace\) \(k=\dfrac{22}{4}=5.5\)
    Follow the link below to see more examples. 
    Section 2.1 – Review of Lines

  11. Beginning with the function \(f(x)=\sqrt{x}\), find the function \(g(x)\) that shows \(f(x)\) shifted left 2 units, reflected about the \(x\)-axis, and then shifted up 7 units.

    \(f(x)=-\sqrt{x+2}+7\)
    Follow the link below to see more examples. 
    Section 1.6&7 – Parent Functions and Transformations

  12. Solve for \(x\) in the inequality \[x+2-(5x-10)\geq 3\]

    \[\begin{align*}
    x+2-(5x-10)&\geq 3\\
    x+2-5x+10 &\geq 3\\
    -4x+12 &\geq 3\\
    -4x&\geq -9\\
    x&\leq \frac{9}{4}
    \end{align*}
    \]
    Follow the link below to see more examples. 
    Section A.6 – Linear Inequalities

  13. Find the domain of \[f(x)=\dfrac{x^2-3x-2}{6x^2-54}\]

    \(f(x)=\dfrac{x^2-3x-2}{6x^2-54}=\dfrac{x^2-3x-2}{6(x+3)(x-3)},\) so the domain is all real numbers such that \(x+3\neq 0\) and \(x-3\neq 0\). This gives the domain \((-\infty,-3)\cup (-3,3)\cup(3,\infty)\). 
    Follow the link below to see more examples. 
    Section 1.4 – Functions

  14. Find the domain of the function below. \[f(x)=\left\{\begin{array}{cc}\dfrac{2x^2+13}{x^2-1}, & x<0\\ \dfrac{5x-26}{x+2}, & x\geq 0\end{array} \right.\]

    For the function \(\dfrac{2x^2+13}{(x+1)(x-1)}\) on \( x<0\) the domain is \((-\infty,-1)\cup(-1,0)\). For the function \(\dfrac{5x-26}{x+2}\) on \(x\geq 0\) the domain is \([0,\infty)\). So for the function \(f(x)\), the domain is \( (-\infty,-1)\cup (-1,\infty)\).
    Follow the link below to see more examples. 
    Section 1.4 – Functions

  15. Find the \(x\)- and \(y\)-intercept(s) of the function \(2x+3y=10\), if any exist.

    When \(x = 0\) we get \(2(0)+3y = 10\) \(\ \Longrightarrow\ \) \(y = \dfrac{10}{3}\). When \(y = 0\) we get \(2x+3(0) = 10\) \(\ \Longrightarrow\ \) \(x = 5\). So the \(x\)-intercept is \((5, 0)\) and the \(y\)-intercept is \(\left(0, \dfrac{10}{3}\right)\).
    Follow the link below to see more examples. 
    Section 2.1 – Review of Lines

  16. Simplify the expression \( \dfrac{2}{\sqrt{x^5}} \left(\sqrt[3]{4x}\right)\).

    \[\begin{align*}
    \left( \dfrac{2}{\sqrt{x^5}}\right) \left(\sqrt[3]{4x}\right) &= \left(2x^{-5/2} \right) (4x)^{1/3}\\
    &=\left(2x^{-5/2}\right) \left(4^{1/3}\cdot x^{1/3}\right) \\
    &=\left(2x^{-5/2}\right)\left( \left(2^2\right)^{1/3}x^{1/3}\right)\\
    &=\left(2x^{-5/2}\right)\left(2^{2/3}x^{1/3}\right)\\
    &=\dfrac{2^{5/3}}{x^{13/6}}
    \end{align*}\]
    Follow the link below to see more examples. 
    Section A.2 – Exponents and Radicals

  17. If we begin with the graph of \(f(x) = x^2\) and shift \(f (x)\) 4 units to the right, shrink f (x) vertically by a factor of \(\frac{1}{2}\), and then shift \(f(x)\) upward 10 units, write the equation for the transformed graph.

    The transformed graph is \(g(x)=\dfrac{1}{2}(x-4)^2+10\).
    Follow the link below to see more examples. 
    Section 1.6&7 – Parent Functions and Transformations

  18. Find \(f\circ g\) (also denoted \(f(g(x))\)) if \(f(x)=\dfrac{x}{x+1}\) and \(g(x)=\dfrac{2}{x}\). Simplify.

    \[\begin{align*}
    f\circ g &= \frac{ \frac{2}{x}}{\frac{2}{x}+1}\\
    &= \frac{ \frac{2}{x}}{\frac{2}{x}+\frac{x}{x}}\\
    &=\frac{ \frac{2}{x}}{\frac{2+x}{x}}\\
    &=\frac{2}{x}\cdot \frac{x}{2+x} \\
    &=\frac{2}{2+x}
    \end{align*}\]
    Follow the links below to see more examples. 
    Section 1.8 – Combinations of Functions

    Section A.4 – Rational Expressions

  19. Write the following inequalities in interval notation. 
    1. \(x\geq 2\)
    2. \(-4\leq x < 7\)
    3. \(x<-5\)
    4. All \(x\) such that \(x\) is a real number.

      1. \([2,\infty)\)
      2. \([-4,7)\)
      3. \((-\infty,-5)\)
      4. \(-\infty,\infty)\)

  20. Write the following intervals using inequality notation.
    1. \([0,2)\)
    2. \((-\infty,4)\)
    3. \([7,\infty)\)

      1. \(0\leq x<2\)
      2. \(x<4\)
      3. \(x\geq 7\)

      ​
  21. If \(f(x)=\sqrt{x+4}\), find and simplify \(\dfrac{f(2+h)-f(2)}{h}\).

    \[\begin{align*}
    \dfrac{f(2+h)-f(2)}{h} &=\dfrac{\sqrt{2+h+4}-\sqrt{2+4}}{h}\\
    &=\left(\dfrac{\sqrt{6+h}-\sqrt{6}}{h}\right) \cdot \left( \dfrac{\sqrt{6+h}+\sqrt{6}}{\sqrt{6+h}+\sqrt{6}}\right)\\
    &=\dfrac{6+h-6}{h\left(\sqrt{6+h}+\sqrt{6}\right)}\\
    &=\dfrac{1}{\sqrt{6+h}+\sqrt{6}}
    \end{align*}\]
    Follow the link below to see more examples. 
    Section 1.4 – Difference Quotient

  22. Perform the operations indicated and simplify.\[\frac{x^2}{x^2-x-2}-\frac{4}{x^2+x-6}+\frac{x}{x^2+4x+3}\]

    \[\begin{align*}
    \frac{x^2}{x^2-x-2}&-\frac{4}{x^2+x-6}+\frac{x}{x^2+4x+3}\\
    &=\frac{x^2}{(x-2)(x+1)}-\frac{4}{(x+3)(x-2)}+\frac{x}{(x+1)(x+3)}\\
    &=\left(\frac{x+3}{x+3}\right)\left(\frac{x^2}{(x-2)(x+1)}\right) -\left(\frac{x+1}{x+1}\right) \left(\frac{4}{(x+3)(x-2)}\right)+\left(\frac{x-2}{x-2}\right)\left(\frac{x}{(x+1)(x+3)}\right)\\
    &= \frac{x^3+3x^2-4x-4+x^2-2x}{(x+3)(x-2)(x+1)}\\
    &=\frac{x^3+4x^2-6x-4}{(x+3)(x-2)(x+1)}
    \end{align*}\]
    Follow the link below to see more examples. 
    Section A.4 – Rational Expressions

  23. Evaluate \( f(2)-f(-3)\) given \( f(x)=\left\{ \begin{array}{ll} x^3+1, & x>1\\2x^2-3,&x\leq 1\end{array}\right.\)

    \(f(2)=(2)^3+1=9\) and \(f(-3)=2(-3)^2-3=18-3=15\), so \(f(2)-f(-3)=9-15=-6\). 
    Follow the link below to see more examples.
    Section 1.4 – Functions

  24. Given the points \(A(4,5)\), \(B(-3,2)\), and \(C(1,-4)\), do the following: 
    1. Plot points \(A\), \(B\), and \(C.\)
    2. Find the equations of the horizontal line going through the point \(B.\)
    3. Find the equation of the vertical line going through point \(C.\)
    4. Find the equation of the line containing points \(A\) and \(C\) and write the equation in point-slope form.
    5. Find an equation of the line that contains point \(B\), but is perpendicular to the line containing points \(A\) and \(C\).

      2. \(y=2\)
      3. \(x=1\)
      4. To find the slope we calculate \(m=\dfrac{5-(-4)}{4-1}=\dfrac{9}{3}=3\). So in point-slope form the equation is \(y-5=3(x-4).\)
      5. The slope of the line perpendicular to the line containing points \(A\) and \(C\) is \(-\dfrac{1}{3}\). Using this slope with point \(B\) we get \(y-2=-\dfrac{1}{3}(x+3)\) \(\ \Longrightarrow\ \) \(y=-\dfrac{1}{3}x+1\)
      Follow the link below to see more examples. 
      Section 2.1 – Review of Lines

  25. Graph the inequality \(4x+2y \leq 10\). Does the point \( \left( -3, \dfrac{8}{3}\right)\) lie in the solution set?


    Yes, the point \(\left(-3,\dfrac{8}{3}\right)\) lies in the solution set.
    Follow the link below to see more examples.
    Section 3.2 – Graphing Systems of Linear Inequalities in Two Variables

  26. Simplify the following expressions
    1. \(\dfrac{2x}{xy+xz+5x}\)
    2. \(\dfrac{(24)(.4)}{(2.5)(.4)+(.6)(.4)+(1.9)(.4)}\)

      1.  \[\begin{align*} \dfrac{2x}{xy+xz+5x} &= \dfrac{2x}{x(y+z+5)} \\ &= \dfrac{2}{(y+z+5)}\end{align*}\]
      2.  \[\begin{align*}\dfrac{(2.4)(.4)}{(2.5)(.4)+(.6)(.4)+(1.9)(.4)} &=\dfrac{(2.4)(.4)}{(.4)((2.5)+(.6)+(1.9))}\\ &= \dfrac{2.4}{(2.5)+(.6)+(1.9)}\\ &=0.48\end{align*} \]
      Follow the link below to see more examples. 
      Section A.4 – Rational Expressions

  27. Russ set up a lemonade stand where he sold 16 ounce cups of freshly squeezed lemonade for $2 per cup. It cost Russ $0.25 in supplies (lemons, sugar, water, paper cups, and ice) to make each cup of lemonade, and he spent a total of $52.50 on other necessary supplies (e.g., table, lemonade dispenser, signs). If \(x\) is the number of cups of lemonade that Russ makes and sells on the day of the lemonade stand, find the following: 
    1. The cost function, \(C(x)\), to make \(x\) cups of lemonade.
    2. The revenue function, \(R(x)\), generated from selling \(x\) cups of lemonade. 
    3. The profit function, \(P(x)\), made from selling \(x\) cups of lemonade. 
    4. How many cups of lemonade must Russ sell to break even on his lemonade stand?

      1. \(C(x)+0.25x+52.50\)
      2. \(R(x)=2x\)
      3.  \[\begin{align*} P(x)&=R(x)-C(x)\\ &=2x-(0.25x+52.50)\\ &=2x-0.25x-52.50\\ &=1.75x-52.50\end{align*}\]
      4. \(P(x)=0\) when \(1.75x-52.50=0\) \(\ \Longrightarrow\ \) \(x=\dfrac{52.50}{1.75}=30.\) So Russ must sell 30 cups of lemonade to break even on his lemonade stand.
      Follow the links below to see more examples. 
      Section 2.2 – Modeling with Linear Functions
      MATH 140 WIR3 #4

  28. Find the roots of the function \(f(x)=3x^2+7x-2.\)

    Apply the quadratic formula with \(a=3\), \(b=7\), and \(c=-2.\) This gives 
    \[\begin{align*}
    x&=\dfrac{-7 \pm \sqrt{7^2-4(3)(-2)}}{2(3)}\\
    &= \dfrac{-7 \pm \sqrt{49+24}}{6}\\
    &=\dfrac{-7 \pm \sqrt{73}}{6}
    \end{align*}\]
    So one root is at \(x=\dfrac{-7+\sqrt{73}}{6}\) and the other root is at \(x=\dfrac{-7-\sqrt{73}}{6}.\)
    Follow the link below to see more examples. 
    Section A.5 – Solving Equations

  29. There are 5 white balls, 8 red balls, 7 yellow balls, and 4 green balls in a container. A ball is chosen at random. What is the probability a red ball is chosen?

    \(\dfrac{8}{24}=\dfrac{1}{3}\)
    Follow the link below to see more examples. 
    Exam 2 Review

  30. Sue has 7 different books to put on a shelf. How many different ways can she arrange the books on the shelf?

    \(7!=7\cdot 6 \cdot 5\cdot 4\cdot 3 \cdot 2\cdot 1 =5,040\)

  31. Jay wants to make a box, with no lid (or top), out of a \(10'' \times 6''\) rectangular piece of cardboard. If Jay cuts squares with dimensions \(x\) by \(x\) out of each corner of the cardboard, and then folds up the corners to make an open box, find a function that represents:
    1. The volume of the box.
    2. The surface area of the box.

      1. \(V=(10-2x)(6-2x)x=4x^3-32x^2+60x\)
      2.  \[\begin{align*} SA&=2x(6-2x)+2x(10-2x)+(6-2x)(10-2x)\\&=12x-4x^2+20x-4x^2+60-32x+4x^2\\&=-4x^2+60\end{align*}\]

  32. A classroom of 100 students has 65 females, 10 seniors, and 6 females who are seniors. How many students in this classroom are not female and also not a senior?

    31
    Follow the link below to see more examples. 
    Exam 2 Review

  33. When making his 6-period class schedule, Mark has 2 options of classes to take for first period, 4 options for second period, 1 option for third period, 1 option for fourth period, 3 options for fifth period, and 5 options for sixth period. How many different schedules can Mark possibly have if he takes one class per period?

  34. Suppose an object is moving at 66 feet per second. How fast, in miles per hour, would a car have to travel to keep pace with this object?

    \(\dfrac{66 \text{ feet}}{1 \text{ sec}}\cdot \dfrac{60 \text{ sec}}{1 \text{ min}} \cdot \dfrac{60 \text{ min}}{1 \text{ hour}} \cdot \dfrac{1 \text{ mile}}{5280 \text{ feet}} = 45 \dfrac{\text{ miles}}{\text{ hour}}\)