GMAT Math : Problem-Solving Questions

Study concepts, example questions & explanations for GMAT Math

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Example Questions

Example Question #1061 : Problem Solving Questions

The graph of a function \(\displaystyle f\) has \(\displaystyle y\,\)-intercept \(\displaystyle (0,8)\). Which of the following could be the definition of \(\displaystyle f\) ?

Possible Answers:

\(\displaystyle f(x)= \log _{2}x+6\)

\(\displaystyle f(x)= \log _{2}(x+8) +6\)

\(\displaystyle f(x)= \log _{2}(x+2) +6\)

\(\displaystyle f(x)= \log _{2}(x+4) +6\)

\(\displaystyle f(x)= \log _{2}(x+1) +6\)

Correct answer:

\(\displaystyle f(x)= \log _{2}(x+4) +6\)

Explanation:

All of the functions take the form 

\(\displaystyle f(x)= \log _{2}(x+A) +6\)

for some integer \(\displaystyle A\). To find the choice that has \(\displaystyle y\,\)-intercept \(\displaystyle (0,8)\), set \(\displaystyle x = 0\) and \(\displaystyle f(x)= 8\), and solve for \(\displaystyle A\):

\(\displaystyle 8= \log _{2}(0+A) +6\)

\(\displaystyle 8= \log _{2}A +6\)

\(\displaystyle 2= \log _{2}A\)

In exponential form:

\(\displaystyle A =2^{2}=4\)

The correct choice is \(\displaystyle f(x)= \log _{2}(x+4) +6\).

Example Question #824 : Geometry

Define a function \(\displaystyle g\) as follows:

\(\displaystyle g(x) =-3 \log_{8}(x+2)+ 2\)

Give the \(\displaystyle x\)-intercept of the graph of \(\displaystyle g\).

Possible Answers:

\(\displaystyle (-2, 0)\)

\(\displaystyle (6,0)\)

\(\displaystyle \left ( -2 + 16 \sqrt{2}, 0 \right )\)

\(\displaystyle (2,0)\)

\(\displaystyle \left ( -2 - 16 \sqrt{2}, 0 \right )\)

Correct answer:

\(\displaystyle (2,0)\)

Explanation:

Set \(\displaystyle g(x) =0\) and evaluate \(\displaystyle x\) to find the \(\displaystyle x\)-coordinate of the \(\displaystyle x\)-intercept.

\(\displaystyle g(x) =-3 \log_{8}(x+2)+ 2 = 0\)

\(\displaystyle -3 \log_{8}(x+2)= - 2\)

\(\displaystyle \log_{8}(x+2)= \frac{-2}{-3}= \frac{2}{3}\)

Rewrite in exponential form:

\(\displaystyle x+ 2 = 8^{\frac{2}{3}}\)

\(\displaystyle x+ 2 =\left ( \sqrt[3]{8} \right )^{2} = 2^{2} = 4\)

\(\displaystyle x=2\).

The \(\displaystyle x\)-intercept is \(\displaystyle (2,0)\).

Example Question #825 : Geometry

Define functions \(\displaystyle f\) and \(\displaystyle g\) as follows:

\(\displaystyle f(x) = 2 \log (x-4)\)

\(\displaystyle g(x) = \log (x+3)+ \log (x-2)\)

Give the \(\displaystyle y\)-coordinate of a point at which the graphs of the functions intersect.

Possible Answers:

\(\displaystyle 2 \frac{4}{9}\)

\(\displaystyle \log 2\frac{34}{81}\)

The graphs of \(\displaystyle f\) and \(\displaystyle g\) do not intersect.

\(\displaystyle \log 1\frac{5}{9}\)

\(\displaystyle 5\)

Correct answer:

The graphs of \(\displaystyle f\) and \(\displaystyle g\) do not intersect.

Explanation:

Since \(\displaystyle a \log N = \log( N^{a})\), the definition of \(\displaystyle f\) can be rewritten as follows:

\(\displaystyle f(x) = 2 \log (x-4)\)

\(\displaystyle f(x) = \log (x-4) ^{2}\)

Since \(\displaystyle \log M + \log N = \log MN\), the definition of \(\displaystyle g\) can be rewritten as follows:

\(\displaystyle g(x) = \log (x+3)+ \log (x-2)\)

\(\displaystyle g(x) = \log (x+3) (x-2)\)

First, we need to find the \(\displaystyle x\)-coordinate of the point at which the graphs of \(\displaystyle f\) and \(\displaystyle g\) meet by setting 

\(\displaystyle f(x)= g(x)\)

\(\displaystyle \log (x-4) ^{2} = \log (x+3) (x-2)\)

Since the common logarithms of the two polynomials are equal, we can set the polynomials themselves equal, then solve:

\(\displaystyle (x-4) ^{2} = (x+3) (x-2)\)

\(\displaystyle x ^{2}- 8x+16 = x^{2} +x-6\)

\(\displaystyle x ^{2}- 8x+16 - x^{2} -x - 16 = x^{2} +x-6 - x^{2} -x - 16\)

\(\displaystyle -9x = - 22\)

\(\displaystyle x = \frac{22}{9} = 2\frac{4}{9}\)

However, if we evaluate \(\displaystyle f \left ( 2\frac{4}{9} \right )\), the expression becomes

\(\displaystyle f(x) = 2 \log (x-4)\)

\(\displaystyle f\left ( 2\frac{4}{9} \right ) = 2\log \left ( 2\frac{4}{9} -4\right ) = 2 \log \left ( -1\frac{5}{9} \right )\),

which is undefined, since a negative number cannot have a logarithm.

Consequently, the two graphs do not intersect.

 

Example Question #826 : Geometry

The graph of a function \(\displaystyle f\) has \(\displaystyle x\)-intercept \(\displaystyle \left ( \frac{1}{32}, 0 \right )\). Which of the following could be the definition of \(\displaystyle f\) ?

Possible Answers:

\(\displaystyle f(x)= \log_{5}x-2\)

None of the other responses gives a correct answer.

\(\displaystyle f(x)= \log_{2}x+5\)

\(\displaystyle f(x)= \log_{2}x-5\)

\(\displaystyle f(x)= \log_{5}x+2\)

Correct answer:

\(\displaystyle f(x)= \log_{2}x+5\)

Explanation:

All of the functions are of the form \(\displaystyle f(x) = \log_{b}x - C\). To find the \(\displaystyle x\)-intercept of a function \(\displaystyle f(x) = \log_{b}x - C\),  we can set \(\displaystyle f(x) = 0\) and solve for \(\displaystyle x\):

\(\displaystyle \log_{b}x - C = 0\)

\(\displaystyle \log_{b}x =C\)

\(\displaystyle b^{C}= x\).

Since we are looking for a function whose graph has \(\displaystyle x\)-intercept \(\displaystyle \left ( \frac{1}{32}, 0 \right )\), the equation here becomes \(\displaystyle b^{C}= \frac{1}{32}\), and we can examine each of the functions by finding the value of \(\displaystyle b^{C}\) and seeing which case yields this result.

 

\(\displaystyle f(x)= \log_{5}x-2\):

\(\displaystyle b = 5, C=2\)

\(\displaystyle b^{C} = 5^{2} = 25\)

 

\(\displaystyle f(x)= \log_{2}x-5\):

\(\displaystyle b = 2, C=5\)

\(\displaystyle b^{C} = 2^{5} = 32\)

 

\(\displaystyle f(x)= \log_{5}x+2\):

\(\displaystyle b = 5, C=-2\)

\(\displaystyle b^{C} = 5^{-2} = \frac{1}{5^{2}}= \frac{1}{25}\)

 

\(\displaystyle f(x)= \log_{2}x+5\):

\(\displaystyle b = 2, C=-5\)

\(\displaystyle b^{C} = 2^{-5} = \frac{1}{2^{5}}= \frac{1}{32}\)

The graph of \(\displaystyle f(x)= \log_{2}x+5\) has \(\displaystyle x\)-intercept \(\displaystyle \left ( \frac{1}{32}, 0 \right )\) and is the correct choice.

Example Question #827 : Geometry

Define a function \(\displaystyle g\) as follows:

\(\displaystyle g(x) =2 \log_{9}(x+3)- 3\)

A line passes through the \(\displaystyle x\)- and \(\displaystyle y\,\)-intercepts of the graph of \(\displaystyle g\). Give the equation of the line.

Possible Answers:

\(\displaystyle y = 12x+24\)

\(\displaystyle y= -\frac{1}{12} x +2\)

\(\displaystyle y= \frac{1}{12} x -2\)

\(\displaystyle y =- 12x+30\)

\(\displaystyle y= -\frac{1}{12} x -4\)

Correct answer:

\(\displaystyle y= \frac{1}{12} x -2\)

Explanation:

The \(\displaystyle x\)-intercept of the graph of \(\displaystyle g\) can befound by setting \(\displaystyle g(x) = 0\) and solving for \(\displaystyle x\):

\(\displaystyle 2 \log_{9}(x+3)- 3 = 0\)

\(\displaystyle 2 \log_{9}(x+3) = 3\)

\(\displaystyle \log_{9}(x+3) = \frac{3}{2}\)

Rewritten in exponential form:

\(\displaystyle x+3 = 9 ^{\frac{3}{2}}\)

\(\displaystyle x+3 = (\sqrt{9} )^{3}\)

\(\displaystyle x+3 = 3^{3}\)

\(\displaystyle x+3 = 27\)

\(\displaystyle x = 24\)

The \(\displaystyle x\)-intercept of the graph of \(\displaystyle g\) is \(\displaystyle (24,0)\).

 

The \(\displaystyle y \,\)-intercept of the graph of \(\displaystyle g\) can be found by evaluating \(\displaystyle g(0):\)

\(\displaystyle g(x) =2 \log_{9}(x+3)- 3\)

\(\displaystyle g(0) =2 \log_{9}(0+3)- 3\)

\(\displaystyle =2 \log_{9}3- 3\)

\(\displaystyle = \log_{9}3^{2}- 3\)

\(\displaystyle = \log_{9}9- 3\)

\(\displaystyle = 1-3\)

\(\displaystyle = -2\)

The \(\displaystyle y \,\)-intercept of the graph of \(\displaystyle g\) is \(\displaystyle (0, -2)\).

 

If \(\displaystyle (a,0)\) and \(\displaystyle (0,b)\) are the \(\displaystyle x\)- and \(\displaystyle y\,\)-intercepts, respectively, of a line, the slope of the line is \(\displaystyle m=-\frac{b}{a}\). Substituting \(\displaystyle a = 24\) and \(\displaystyle b = -2\), this is

\(\displaystyle m=-\frac{-2}{24} = \frac{1}{12}\).

Setting \(\displaystyle m= \frac{1}{12}\) and \(\displaystyle b = -2\) in the slope-intercept form of the equation of a line:

\(\displaystyle y = mx+b\)

\(\displaystyle y= \frac{1}{12} x -2\)

Example Question #1061 : Problem Solving Questions

Define functions \(\displaystyle f\) and \(\displaystyle g\) as follows:

\(\displaystyle f(x) = 2 \log (x+5)\)

\(\displaystyle g(x) = \log (2x+13)\)

Give the \(\displaystyle y\)-coordinate of a point at which the graphs of the functions intersect.

Possible Answers:

\(\displaystyle \log 6\)

The graphs of \(\displaystyle f\) and \(\displaystyle g\) do not intersect.

\(\displaystyle 0\)

\(\displaystyle -2\)

\(\displaystyle \log 9\)

Correct answer:

\(\displaystyle \log 9\)

Explanation:

Since \(\displaystyle a \log N= \log( N^{a})\), the definition of \(\displaystyle f\) can be rewritten as follows:

 \(\displaystyle f(x) = \log \left [(x+5)^{2} \right ]\)

Find the \(\displaystyle x\)-coordinate of the point at which the graphs of \(\displaystyle f\) and \(\displaystyle g\) meet by setting 

\(\displaystyle f(x)= g(x)\)

\(\displaystyle \log \left [(x+5)^{2} \right ]= \log (2x+13)\)

Since the common logarithms of the two polynomials are equal, we can set the polynomials themselves equal, then solve:

\(\displaystyle (x+5)^{2} = 2x+13\)

\(\displaystyle x^{2} +10x+25= 2x+13\)

\(\displaystyle x^{2} +10x+25- 2x - 13 = 2x+13 - 2x - 13\)

\(\displaystyle x^{2} +8x+12= 0\)

The quadradic trinomial can be "reverse-FOILed" by noting that 2 and 6 have  product 12 and sum 8:

\(\displaystyle (x+2)(x+6 ) = 0\)

Either \(\displaystyle x+2 = 0\), in which case \(\displaystyle x=-2\)

or

\(\displaystyle x+6 = 0\), in which case \(\displaystyle x = -6\)

Note, however, that we can eliminate \(\displaystyle x = -6\) as a possible \(\displaystyle x\)-value, since

\(\displaystyle f(-6) = 2 \log (-6+5) = 2 \log (-1)\),

an undefined quantity since negative numbers do not have logarithms. 

Since 

\(\displaystyle f(-2) = 2 \log (-2+5) = 2 \log 3 = \log (3^{2}) = \log 9\)

and 

\(\displaystyle g(-2) = \log [2(-2)+13] = \log (-4+13) = \log 9\),

\(\displaystyle -2\) is the correct \(\displaystyle x\)-value, and \(\displaystyle \log 9\) is the correct \(\displaystyle y\)-value.

Example Question #2 : How To Graph Complex Numbers

Give the \(\displaystyle x\)-intercept(s) of the parabola with equation \(\displaystyle f(x) = x^{2}+4x+7\). Round to the nearest tenth, if applicable.

Possible Answers:

\(\displaystyle (-4.6,0), (0.6,0)\)

The parabola has no  \(\displaystyle x\)-intercept.

\(\displaystyle (7,0)\)

\(\displaystyle (-2,0)\)

\(\displaystyle (-0.6,0), (4.6,0)\)

Correct answer:

The parabola has no  \(\displaystyle x\)-intercept.

Explanation:

The \(\displaystyle x\)-coordinate(s) of the \(\displaystyle x\)-intercept(s) are the real solution(s) to the equation \(\displaystyle f(x) = x^{2}+4x+7 = 0\). We can use the quadratic formula to find any solutions, setting \(\displaystyle a=1, b=4, c=7\) - the coefficients of the expression.

An examination of the discriminant \(\displaystyle b^{2} - 4ac\), however, proves this unnecessary.

\(\displaystyle b^{2} - 4ac = 4^{2} - 4\cdot 1\cdot 7 = -12\)

The discriminant being negative, there are no real solutions, so the parabola has no  \(\displaystyle x\)-intercepts.

Example Question #1 : Graphing Complex Numbers

In which quadrant does the complex number  \(\displaystyle -5+3i\)  lie?

Possible Answers:

\(\displaystyle I\)

\(\displaystyle II\)

\(\displaystyle y\)-axis

\(\displaystyle III\)

\(\displaystyle IV\)

Correct answer:

\(\displaystyle II\)

Explanation:

When plotting a complex number, we use a set of real-imaginary axes in which the x-axis is represented by the real component of the complex number, and the y-axis is represented by the imaginary component of the complex number. The real component is  \(\displaystyle -5\)  and the imaginary component is  \(\displaystyle 3\),  so this is the equivalent of plotting the point  \(\displaystyle (-5,3)\)  on a set of Cartesian axes.  Plotting the complex number on a set of real-imaginary axes, we move  \(\displaystyle 5\)  to the left in the x-direction and  \(\displaystyle 3\)  up in the y-direction, which puts us in the second quadrant, or in terms of Roman numerals:

\(\displaystyle Answer: II\)

Example Question #1 : How To Graph Complex Numbers

In which quadrant does the complex number  \(\displaystyle 9-4i\)  lie?

Possible Answers:

\(\displaystyle IV\)

\(\displaystyle II\)

\(\displaystyle III\)

\(\displaystyle I\)

Correct answer:

\(\displaystyle IV\)

Explanation:

If we graphed the given complex number on a set of real-imaginary axes, we would plot the real value of the complex number as the x coordinate, and the imaginary value of the complex number as the y coordinate. Because the given complex number is as follows:

\(\displaystyle 9-4i\)

We are essentially doing the same as plotting the point  \(\displaystyle (9,-4)\)  on a set of Cartesian axes.  We move  \(\displaystyle 9\)  units right in the x direction, and  \(\displaystyle 4\)  units down in the y direction, which puts us in the fourth quadrant, or in terms of Roman numerals:

\(\displaystyle Answer:IV\)

Example Question #2 : Graphing Complex Numbers

In which quadrant does the complex number  \(\displaystyle -13-5i\)  lie?

Possible Answers:

\(\displaystyle I\)

\(\displaystyle II\)

\(\displaystyle III\)

\(\displaystyle IV\)

Correct answer:

\(\displaystyle III\)

Explanation:

If we graphed the given complex number on a set of real-imaginary axes, we would plot the real value of the complex number as the x coordinate, and the imaginary value of the complex number as the y coordinate. Because the given complex number is as follows:

\(\displaystyle -13-5i\)

We are essentially doing the same as plotting the point  \(\displaystyle (-13,-5)\)  on a set of Cartesian axes.  We move  \(\displaystyle 13\)  units left of the origin in the x direction, and  \(\displaystyle 5\)  units down from the origin in the y direction, which puts us in the third quadrant, or in terms of Roman numerals:

\(\displaystyle Answer:III\)

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