Free Compass Math Test Practice Questions
Solutions with Explanations - Sample 3

Solutions with detailed explanations to compass math test practice questions in sample 3.

1. If number \(x\) is increased by one 1/4 of itself, which of the following expressions represents the new number?

   A) \(x + \frac{1}{4}\)
   B) \(x + 0.25\)
   C) \(x + 0.25x\)
   D) \(4x - 0.25\)
   E) \(\frac{x}{4}\)

   Solution

   \(x\) is the number and one quarter of itself is \(\frac{1}{4}x\).
   \(x\) increased by a quarter of itself is written as: \[ x + \frac{1}{4}x = x + 0.25x \] since \(\frac{1}{4} = 0.25\).
   Answer C.

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2. \((5 + \frac{1}{5})x = 6\), find \(x\).

   Solution

   Multiply all terms by \(5\) to eliminate the denominator: \[ 5 \left[(5 + \frac{1}{5})x\right] = 5 \cdot 6 \] \[ 25x + x = 30 \implies 26x = 30 \] \[ x = \frac{30}{26} = \frac{15}{13} = 1\frac{2}{13} \]

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3. If \(f(x) = 3x + 2\), then \(f(2a + b) =\)

   Solution

   Substitute \(x\) in \(f(x)\) by \(2a + b\): \[ f(2a + b) = 3(2a + b) + 2 = 6a + 3b + 2 \]

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4. If \(f(x) = 3x + 2\) and \(g(x) = x^2 - 2\), then \(f(2) - g(3) =\)

   Solution

   \[ f(2) = 3(2) + 2 = 8,\quad g(3) = 3^2 - 2 = 7 \] \[ f(2) - g(3) = 8 - 7 = 1 \]

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5. For all \(x\), \((x + 3)(-x + 3) =\)

   Solution

   Expand the expression: \[ (x + 3)(-x + 3) = -x^2 + 3x - 3x + 9 = 9 - x^2 \]

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6. What is the tenth term of the geometric series if the first, second and third terms are \(0.5,\; 1.0,\; 2.0,\; \ldots\)?

   Solution

   The \(n\)th term of a geometric series is \(a_n = a_1 r^{n-1}\), where \(a_1\) is the first term and \(r\) the common ratio. \[ r = \frac{1.0}{0.5} = 2 \] \[ a_{10} = 0.5 \cdot 2^{9} = 0.5 \cdot 512 = 256 \]

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7. If \((2^{x})(2^{-4x}) = \frac{1}{8}\), then \(x = ?\)

   Solution

   \[ (2^{x})(2^{-4x}) = 2^{x - 4x} = 2^{-3x} \] \[ \frac{1}{8} = 2^{-3} \] Equating exponents: \(-3x = -3 \implies x = 1\).

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8. The solution of the equation \(2x - 3 = 5x - 2\) falls between two consecutive integers

   A) \(0\) and \(1\)
   B) \(-1\) and \(0\)
   C) \(-2\) and \(-1\)
   D) \(1\) and \(2\)
   E) \(-3\) and \(-2\)

   Solution

   \[ 2x - 3 = 5x - 2 \implies -3x = 1 \implies x = -\frac{1}{3} \] \(-\frac{1}{3}\) lies between \(-1\) and \(0\). Answer B.

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9. What is the tenth term of the geometric sequence \(1,\; -\frac{1}{2},\; \frac{1}{4},\; -\frac{1}{8},\; \ldots\)

   Solution

   First term \(a_1 = 1\), common ratio \(r = \frac{-1/2}{1} = -\frac{1}{2}\). \[ a_{10} = 1 \cdot \left(-\frac{1}{2}\right)^{9} = -\frac{1}{512} \]

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10. If \(f(x) = 3x^3 + 2x^2 + 3\) and \(g(x) = x^2 + 2\), then \(\frac{f(x)}{g(x)} = ?\)

    Solution

    Use polynomial long division to divide \(f(x)\) by \(g(x)\):

    	\(3x + 2\)
    \(x^2 + 2\)	\(3x^3 + 2x^2 + 0x + 3\)
    	\(3x^3 + 0x^2 + 6x\)
    	\(2x^2 - 6x + 3\)
    	\(2x^2 + 0x + 4\)
    	\(-6x - 1\)

    Using the relationship: dividend / divisor = quotient + remainder / divisor, we write: \[ \frac{f(x)}{g(x)} = 3x + 2 + \frac{-6x - 1}{x^2 + 2} \]

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11. If \(i\) is the imaginary unit such that \(\sqrt{-1} = i\), then \((7i)^2 = ?\)

    Solution

    \[ (7i)^2 = 49 i^2 = 49(-1) = -49 \]

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12. Which of these is a complete factorization of \(f(x) = 3x^3 + x^2 + (3x + 1)(2x - 3)\)?

    Solution

    Factor \(3x^3 + x^2 = x^2(3x + 1)\). \[ f(x) = x^2(3x + 1) + (3x + 1)(2x - 3) = (3x + 1)(x^2 + 2x - 3) \] Factor the quadratic: \(x^2 + 2x - 3 = (x - 1)(x + 3)\). \[ f(x) = (3x + 1)(x - 1)(x + 3) \]

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13. Find \(k\) if \(8! = 6! \, k\).

    Solution

    \(8! = 8 \cdot 7 \cdot 6!\), so \(8 \cdot 7 \cdot 6! = 6! \, k \implies k = 56\).

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14. Find \(f(-2)\) if \(f(x) = 2x^2 + kx + 2\) and \(f(1) = 3\).

    Solution

    \(f(1) = 2(1)^2 + k(1) + 2 = 4 + k = 3 \implies k = -1\). \[ f(x) = 2x^2 - x + 2 \implies f(-2) = 2(4) - (-2) + 2 = 8 + 2 + 2 = 12 \]

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15. In which interval does \(f(x) = -2x^2 - 3x - 4\) have a maximum value?

    A) \((0, 1)\)
    B) \((-3, -2)\)
    C) \((-0.5, 0)\)
    D) \((-2, -1)\)
    E) \((-1, 0)\)

    Solution

    The quadratic has a maximum at \(x = -\frac{b}{2a} = -\frac{-3}{2(-2)} = -\frac{3}{4} = -0.75\), which lies in \((-1, 0)\). Answer E.

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16. Simplify \(x^{3/4} \cdot x^{1/3} \cdot x^{-2/3}\).

    Solution

    \[ x^{3/4 + 1/3 - 2/3} = x^{3/4 - 1/3} = x^{\frac{9}{12} - \frac{4}{12}} = x^{5/12} \]

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17. Find \(x\) if \(\left(\frac{2}{7}\right)^{2x} = \left(\frac{7}{2}\right)^{3x + 5}\).

    Solution

    \(\left(\frac{7}{2}\right)^{3x+5} = \left(\frac{2}{7}\right)^{-(3x+5)}\). Thus, \[ \left(\frac{2}{7}\right)^{2x} = \left(\frac{2}{7}\right)^{-3x-5} \] Equating exponents: \(2x = -3x - 5 \implies 5x = -5 \implies x = -1\).

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18. \(2x^2 + 3x - 5\) is the product of \((2x + 5)\) and another factor. What is the other factor?

    Solution

    Factor the quadratic: \(2x^2 + 3x - 5 = (2x + 5)(x - 1)\). The other factor is \(x - 1\).

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19. If the operator \(**\) is defined by \(x ** y = 2xy + x + y\). What is \(2 ** 3\)?

    Solution

    \[ 2 ** 3 = 2(2)(3) + 2 + 3 = 12 + 5 = 17 \]

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20. If \(i = \sqrt{-1}\), then \(i^2 + i^3 + i^4 + i^5 =\)

    Solution

    \[ i^2 = -1,\; i^3 = -i,\; i^4 = 1,\; i^5 = i \] \[ (-1) + (-i) + 1 + i = 0 \]

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21. For all \(x\), \((2x - 4)^2 =\)

    Solution

    \[ (2x - 4)^2 = 4x^2 - 16x + 16 \]

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