GMAT Math : Problem-Solving Questions

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Example Questions

Example Question #46 : Understanding Functions

$\left \lfloor N \right \rfloor$ is defined to be the greatest integer less than or equal to .

Define $f (x) = \left \lfloor x ^{2} \right \rfloor - 4 x + 7$

Evaluate $f \left ( \frac{7}{5} \right )$

Possible Answers:

$f \left ( \frac{7}{5} \right ) = 2$

$f \left ( \frac{7}{5} \right ) = 3$

$f \left ( \frac{7}{5} \right ) = 2 \frac{2}{5}$

$f \left ( \frac{7}{5} \right ) = 1 \frac{2}{5}$

$f \left ( \frac{7}{5} \right ) = 3 \frac{2}{5}$

Correct answer:

$f \left ( \frac{7}{5} \right ) = 2 \frac{2}{5}$

Explanation:

$f (x) = \left \lfloor x ^{2} \right \rfloor - 4 x + 7$

$f \left ( \frac{7}{5} \right ) = \left \lfloor\left ( \frac{7}{5} \right )^{2} \right \rfloor - 4\cdot \frac{7}{5}+ 7$

$= \left \lfloor \frac{49}{25} \right \rfloor - \frac{28}{5}+ 7$

$= \left \lfloor 1 \frac{24}{25} \right \rfloor -5 \frac{3}{5}+ 7$

$=1 -5 \frac{3}{5}+ 7 = 2 \frac{2}{5}$

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Example Question #1271 : Problem Solving Questions

$f (x) = \left\{\begin{matrix} 2x + 1 & x < 0 \\ 3x - 1 & x \geq 0 \end{matrix}\right.$

$\small g (x) = \left\{\begin{matrix} 3x - 1 & x < 0 \\ 4x +7 & x \geq 0 \end{matrix}\right.$

Evaluate $\small \small (f \circ g) (4) - (g \circ f) (-4)$ .

Possible Answers:

162

113

Correct answer:

Explanation:

$\small \small \small \small (f \circ g) (4) = f (g (4)) = f (4 \cdot 4 + 7) = f (23) = 3 \cdot 23 - 1 = 68$

$\small \small \small \small (g \circ f) (-4) = g (f (-4)) = g (2 \cdot (-4) + 1)$ $\small \small = g (-7) = 3 (-7) -1 = -22$

$\small \small \small \small (f \circ g) (4) - (g \circ f) (-4) = 68 - (-22) = 90$

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Example Question #1272 : Problem Solving Questions

For any real a,b , define $a\odot b =(1-a)(1-b)$ .

For what value or values of would $x\odot x = 4$ ?

Possible Answers:

No such value of exists.

-1, 3

-2,2

$-\frac{1}{2}, \frac{1}{2}$

$1- \sqrt{2}, 1+ \sqrt{2}$

Correct answer:

-1, 3

Explanation:

For such an to exist, it must hold that $x\odot x = (1 -x )(1-x) =(1-x)^{2}=4$ .

Take the square root of both sides:

1-x = -2 or 1-x = 2

Case 1:

1-x = -2

-x = -3

x=3

Case 2:

1-x = 2

-x = 1

x=-1

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Example Question #1273 : Problem Solving Questions

Define an operation $\ddagger$ on the set of real numbers as follows:

$a \ddagger b = 2ab - a - b$

Evaluate $4 \ddagger (3\ddagger 2 )$ .

Possible Answers:

Correct answer:

Explanation:

First, evaluate $3\ddagger 2$ by substituting a = 3,b= 2 :

$a \ddagger b = 2ab - a - b$

$3 \ddagger 2 = 2\cdot 3 \cdot 2 - 3 - 2 = 12 - 3 - 2 = 7$

Second, evaluate $4\ddagger 7$ in the same way.

$a \ddagger b = 2ab - a - b$

$4\ddagger 7 = 2 \cdot 4 \cdot 7 - 4 - 7 = 56 - 4 - 7 = 45$

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Example Question #1274 : Problem Solving Questions

Define an operation $\bigstar$ as follows:

For any real a,b , $a \; \bigstar \; b = a^{2} +b^{2} -8$ .

For what value or values of is it true that $N \; \bigstar \; N = 0$ ?

Possible Answers:

N = 4

No such value of exists.

N = 2

$N = -4 \textrm{ or } N =4$

$N = -2 \textrm{ or } N = 2$

Correct answer:

$N = -2 \textrm{ or } N = 2$

Explanation:

Substitute a = N, b = N into the definition, and then set the expression equal to 0 to solve for :

$a \; \bigstar \; b = a^{2} +b^{2} -8$

$N \; \bigstar \; N = N^{2} +N^{2} -8 = 2N^{2} -8$

$2N^{2} -8 = 0$

$2N^{2} =8$

$N^{2} =4$

$N = -2 \textrm{ or } N = 2$

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Example Question #1275 : Problem Solving Questions

Consider the function $f(x) = x^{3} - 2$ .

State whether this function is even, odd, or neither, and give the reason for your answer.

Possible Answers:

is not odd, because there exists at least one value of for which $f(-x) \neq -f(x)$ ; is not even, because there exists at least one value of for which $f(-x) \neq f(x)$ .

is odd because f(-x) = -f(x) for each value of in the domain.

is odd because it is a polynomial of degree 3.

is even because f(-x) = f(x) for each value of in the domain.

is even because it is a polynomial of degree 3.

Correct answer:

is not odd, because there exists at least one value of for which $f(-x) \neq -f(x)$ ; is not even, because there exists at least one value of for which $f(-x) \neq f(x)$ .

Explanation:

A function is odd if and only if f(-x) = -f(x) for each value of in the domain; it is even if and only if f(-x) = f(x) for each value of in the domain. To disprove a function is odd or even, we need only find one value of for which the appropriate statement fails to hold.

Consider x = 1 :

$f(x) = x^{3} - 2$

$f(1) = 1^{3} - 2 = 1-2 = -1$

$f(x) = x^{3} - 2$

$f(-1) = (-1)^{3} - 2 = -1 -2 = -3$

$f(-1 ) \neq - f(1)$ , so is not an odd function; $f(-1 ) \neq f(1)$ , so is not an even function.

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Example Question #1276 : Problem Solving Questions

$f (x) = \left\{\begin{matrix} 1\textrm{ if } x< 0\\ x \textrm{ if } x \geq 0 \end{matrix}\right.$

$g (x) = \left\{\begin{matrix} 2x-5 \textrm{ if } x< 0\\ 2\; \; \; \; \; \; \; \; \textrm{ if } x \geq 0 \end{matrix}\right.$ .

Evaluate $(f \circ g) (-4)$ .

Possible Answers:

Correct answer:

Explanation:

$(f \circ g) (-4) = f (g(-4))$

First we evaluate g (-4) . Since the parameter is negative, we use the first half of the definition of :

g (x) = 2x-5

$g (-4) = 2 \left ( -4\right )-5 = -8 - 5 = -13$

f (g(-4)) = f(-13) ; since the parameter here is again negative, we use the first half of the definition of :

f(-13) = 1

Therefore, $(f \circ g) (-4) = 1$ .

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Example Question #1277 : Problem Solving Questions

$\left \lfloor N \right \rfloor$ is defined to be the greatest integer less than or equal to .

Define $f (x) = \left \lfloor x ^{2} \right \rfloor - \left \lfloor 4 x \right \rfloor + 8$ .

Evaluate $f \left ( \frac{3}{5} \right )$ .

Possible Answers:

$f \left ( \frac{3}{5} \right ) = 8$

$f \left ( \frac{3}{5} \right ) = 6$

$f \left ( \frac{3}{5} \right ) = 5$

$f \left ( \frac{3}{5} \right ) = 9$

$f \left ( \frac{3}{5} \right ) = 7$

Correct answer:

$f \left ( \frac{3}{5} \right ) = 6$

Explanation:

$f (x) = \left \lfloor x ^{2} \right \rfloor - \left \lfloor 4 x \right \rfloor + 8$

$f \left ( \frac{3}{5} \right ) = \left \lfloor \left (\frac{3}{5} \right ) ^{2} \right \rfloor - \left \lfloor 4 \cdot \frac{3}{5}\right \rfloor + 8$

$= \left \lfloor \frac{9}{25} \right \rfloor - \left \lfloor \frac{12}{5}\right \rfloor + 8$

$= \left \lfloor \frac{9}{25} \right \rfloor - \left \lfloor 2 \frac{2}{5}\right \rfloor + 8$

=0 - 2 + 8 = 6

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Example Question #54 : Understanding Functions

If $f(x)=x^{2}-\sqrt{x}$ and g(x)=3x-2 , what is f(g(2)) ?

Possible Answers:

$9-\sqrt{3}$

$10-3\sqrt{2}$

Correct answer:

Explanation:

We start by finding g(2):

$g(2)=3\cdot2-2 = 6-2 = 4$

Then we find f(g(2)) which is f(4):

$f(4)=4^{2}-\sqrt{4}=16-2=14$

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Example Question #55 : Understanding Functions

Define two real-valued functions as follows:

$p (x)= x^{2}$

$q(x )= \left\{\begin{matrix} x+1 & x< 0\\ x-1 & x \ge 0 \end{matrix}\right.$

Determine $\left ( q \circ p\right )(x)$ .

Possible Answers:

$\left ( q \circ p\right )(x) = \left\{\begin{matrix} x^{2}+2x + 1 & x < 0\\ x^{2}-2x + 1&x \ge 0 \end{matrix}\right.$

$\left ( q \circ p\right )(x)= x^{2}-2x + 1$

$\left ( q \circ p\right )(x) = \left\{\begin{matrix} x^{2}+ 1 & x < 0\\ x^{2}- 1&x \ge 0 \end{matrix}\right.$

$\left ( q \circ p\right )(x)= x^{2}+1$

$\left ( q \circ p\right )(x)= x^{2}- 1$

Correct answer:

$\left ( q \circ p\right )(x)= x^{2}- 1$

Explanation:

$\left ( q \circ p\right )(x) = q(p(x))= q(x^{2})$ by definition. is piecewise defined, with one defintion for negative values of the domain and one for nonnegative values. However, $p(x)= x^{2}$ is nonnegative for all real numbers, so the defintion for nonnegative numbers, q(x )= x-1 , is the one that will always be used. Therefore,

$\left ( q \circ p\right )(x) = q(p(x))= q(x^{2}) = x^{2}-1$ for all values of .

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GMAT Math : Problem-Solving Questions

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #46 : Understanding Functions

Example Question #1271 : Problem Solving Questions

Example Question #1272 : Problem Solving Questions

Example Question #1273 : Problem Solving Questions

Example Question #1274 : Problem Solving Questions

Example Question #1275 : Problem Solving Questions

Example Question #1276 : Problem Solving Questions

Example Question #1277 : Problem Solving Questions

Example Question #54 : Understanding Functions

Example Question #55 : Understanding Functions

All GMAT Math Resources

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