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GRE Subject Test: Math : Algebra

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Example Questions

Example Question #171 : Algebra

The rate of decrease of the number of concert attendees to former teen heartthrob Justice Beaver is proportional to the population. The population decreased by 34 percent between 2013 and 2015. What is the constant of proportionality?

Possible Answers:

0.15

-0.08

-0.40

-0.21

-1.11

Correct answer:

-0.21

Explanation:

We're told that the rate of growth of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

$p(t)=p_0e^{kt}$

Where p_0 is an initial population value, and is the constant of proportionality.

Since the population decreased by 34 percent between 2013 and 2015, we can solve for this constant of proportionality:

$(1-0.34)y_0=y_0e^{k(2015-2013)}$

$0.66=e^{2k}$

2k=ln(0.66)

$k=\frac{ln(0.66)}{2}=-0.21$

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Example Question #32 : Exponential Functions

The rate of growth of the Land of Battlecraft players is proportional to the population. The population increased by 72 percent between February and October of 2015. What is the constant of proportionality?

Possible Answers:

0.059

0.012

0.118

0.093

0.068

Correct answer:

0.068

Explanation:

We're told that the rate of growth of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

$p(t)=p_0e^{kt}$

Where p_0 is an initial population value, and is the constant of proportionality.

Since the population increased by 72 percent between February and October, we can solve for this constant of proportionality. It'll help to represent the months by their number in the year:

$(1+0.72)y_0=y_0e^{k(10-2)}$

$1.72=e^{8k}$

8k=ln(1.72)

$k=\frac{ln(1.72)}{8}=0.068$

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Example Question #33 : Exponential Functions

The rate of decrease of the gluten-eating demographic of the US is proportional to the population. The population decreased by 8 percent between 2014 and 2015. What is the constant of proportionality?

Possible Answers:

-0.177

-0.194

-0.109

-0.097

-0.083

Correct answer:

-0.083

Explanation:

We're told that the rate of growth of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

$p(t)=p_0e^{kt}$

Where p_0 is an initial population value, and is the constant of proportionality.

Since the population decreased by 8 percent between 2014 and 2015, we can solve for this constant of proportionality:

$(1-0.08)y_0=y_0e^{k(2015-2014)}$

$0.92=e^{k}$

k=ln(0.92)

k=-0.083

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Example Question #31 : Exponential Functions

Bob invests $\$10,000$ in a bank that compounds interest continuously at a rate of $12\%$ . How much money will Bob have in his account after years? (Round answer to decimal places.)

Possible Answers:

$\$14,333$

$\$15,000$

$\$14,000$

$\$14,333.29$

Correct answer:

$\$14,333.29$

Explanation:

Step 1: Recall the formula for continuously compounded interest

The formula is: $A=Pe^{rt}$ , where:

is the Final balance after years.

is the original investment balance.

is the exponential function

is the interest rate, usually written as a decimal

is the time, usually in years

Step 2: Plug in all the information that we have into the formula

$r=12\%=0.12$

$A=(\$10,000)e^{0.12(3)}$

Simplify:

$A=(\$10,000)e^{.36}$

Step 3: Evaluate.

$e^{.36}\approx1.433329$

$(\$10,000)(1.433329)=\$14,333.29$

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Example Question #172 : Algebra

Perform the following operation.

$\begin{bmatrix} 2& 4& 1\\ 5& 3& -2\\ -3& -1&0 \end{bmatrix}$ $\cdot \begin{bmatrix} 2\\ 4\\ 1 \end{bmatrix}$ $+\begin{bmatrix} -7\\ 4\\ 11 \end{bmatrix}$

Possible Answers:

$\begin{bmatrix} 7\\ 20\\ 7 \end{bmatrix}$

$\begin{bmatrix} 14\\ 24 \\ 1 \end{bmatrix}$

$\begin{bmatrix} 34\\ -25\\ 7 \end{bmatrix}$

$\begin{bmatrix} 14& 24& 1 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 14\\ 24 \\ 1 \end{bmatrix}$

Explanation:

The first step to solving this operation is to do the multiplication:

$\begin{bmatrix} 2& 4& 1\\ 5& 3& -2\\ -3& -1&0 \end{bmatrix}$ $\cdot \begin{bmatrix} 2\\ 4\\ 1 \end{bmatrix}$ $=\begin{bmatrix} 2(2)& 4(4)& 1(1)\\ 5(2)& 3(4)& -2(1)\\ -3(2)& -1(4)&0(1) \end{bmatrix}$ $=\begin{bmatrix} 4+16+1\\ 10+12-2\\ -6-4+0 \end{bmatrix}$ $=\begin{bmatrix} 21\\ 20\\ -10 \end{bmatrix}$

Once we have multiplied the matrices, we can perform the addition portion:

$\begin{bmatrix} 21\\ 20\\ -10 \end{bmatrix} + \begin{bmatrix} -7\\ 4\\ 11 \end{bmatrix} = \begin{bmatrix} 14\\ 24\\ 1 \end{bmatrix}$

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Example Question #173 : Algebra

Perform the following operation.

$3\left(\begin{bmatrix} 7\\ -3\\ 2 \end{bmatrix}+\begin{bmatrix} -1\\ 4\\ 6 \end{bmatrix}\right)$

Possible Answers:

$\begin{bmatrix} 18\\ 3\\ 24 \end{bmatrix}$

$\begin{bmatrix} 4\\ 9\\ 20 \end{bmatrix}$

$\begin{bmatrix} 18& 3& 24 \end{bmatrix}$

$\begin{bmatrix} 20\\ -5\\ 12 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 18\\ 3\\ 24 \end{bmatrix}$

Explanation:

The first step is to solve whatever is in the parentheses, in this case it is addition:

$\begin{bmatrix} 7\\ -3\\ 2 \end{bmatrix}+\begin{bmatrix} -1\\ 4\\ 6 \end{bmatrix}=\begin{bmatrix} 6\\ 1\\ 8 \end{bmatrix}$

We then substitute our solution into the parentheses:

$3\left(\begin{bmatrix} 6\\ 1\\ 8 \end{bmatrix}\right)$

Our next, and final step in this problem, is to carry out the multiplication:

$3\left(\begin{bmatrix} 6\\ 1\\ 8 \end{bmatrix}\right)=\begin{bmatrix} 18\\ 3\\ 24 \end{bmatrix}$

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Example Question #1 : Inverses

Find the inverse of the following matrix, if possible.

$A=\begin{bmatrix} 4&-4 \\ -4&-4 \end{bmatrix}$

Possible Answers:

The inverse does not exist.

$\begin{bmatrix} \frac{1}{8}&-\frac{1}{8} \\ -\frac{1}{8}& -\frac{1}{8} \end{bmatrix}$

$-\frac{1}{32}$

$\begin{bmatrix} \frac{1}{4}&-\frac{1}{4} \\ -\frac{1}{4}& -\frac{1}{4} \end{bmatrix}$

Correct answer:

$\begin{bmatrix} \frac{1}{8}&-\frac{1}{8} \\ -\frac{1}{8}& -\frac{1}{8} \end{bmatrix}$

Explanation:

Write the formula to find the inverse of a matrix.

$\begin{bmatrix} a&b\\ c&d \end{bmatrix}^-^1=\frac{1}{ad-bc}\begin{bmatrix} d& -b\\ -c& a \end{bmatrix}$

Substituting in the given matrix we are able to find the inverse matrix.

$\begin{bmatrix} 4&-4 \\ -4&-4 \end{bmatrix}^-^1=\frac{1}{(4)(-4)-(-4)(-4)}\begin{bmatrix} -4&4 \\ 4&4 \end{bmatrix}$

$\frac{1}{-32}\begin{bmatrix} -4&4 \\ 4&4 \end{bmatrix}=\begin{bmatrix} \frac{1}{8}&-\frac{1}{8} \\ -\frac{1}{8}& -\frac{1}{8} \end{bmatrix}$

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Example Question #2 : Inverses

Find the inverse of the following matrix, if possible.

$A=\begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix}$

Possible Answers:

The inverse does not exist.

$\begin{bmatrix} 0& 1\\ 1&0 \end{bmatrix}$

$\begin{bmatrix} 0&-1 \\-1&0 \end{bmatrix}$

$\begin{bmatrix} 1& 0\\ 0&1 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 0& 1\\ 1&0 \end{bmatrix}$

Explanation:

Write the formula to find the inverse of a matrix.

$\begin{bmatrix} a&b\\ c&d \end{bmatrix}^-^1=\frac{1}{ad-bc}\begin{bmatrix} d& -b\\ -c& a \end{bmatrix}$

Using the given information we are able to find the inverse matrix.

$\begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix}^-^1=\frac{1}{(0)(0)-(1)(1)}\begin{bmatrix} 0&-1 \\-1&0 \end{bmatrix}$

$-1\begin{bmatrix} 0&-1 \\ -1&0 \end{bmatrix}=\begin{bmatrix} 0& 1\\ 1&0 \end{bmatrix}$

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Example Question #3 : Inverses

Find the inverse of the function.

$g(x)= \frac{x+3}{2x-4}$

Possible Answers:

$g^{-1}(x)= \frac{3x+4}{x-2}$

$g^{-1}(x)= \frac{x+4}{2-x}$

$g^{-1}(x)= \frac{x+3}{-4x+1}$

$g^{-1}(x)= \frac{4x+3}{2x-1}$

Correct answer:

$g^{-1}(x)= \frac{4x+3}{2x-1}$

Explanation:

To find the inverse function, first replace g(x) with :

$y= \frac{x+3}{2x-4}$

Now replace each with an and each with a :

$x= \frac{y+3}{2y-4}$

Solve the above equation for :

x(2y-4)=y+3

2xy-4x=y+3

2xy-y=4x+3

(2x-1)y=4x+3

$y= \frac{4x+3}{2x-1}$

Replace with $g^{-1}(x)$ . This is the inverse function:

$g^{-1}(x)= \frac{4x+3}{2x-1}$

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Example Question #3 : Find The Inverse Of A Relation

Find the inverse of the function y=2x^2+1 .

Possible Answers:

$f^-^1(x)=\pm \sqrt{\frac{x-1}{2}}$

$f^-^1(x)=\pm \sqrt{x-1}$

f^-^1(x)=x+1

$f^-^1(x)=\pm \sqrt{\frac{x+1}{2}}$

f^-^1(x)=-x+1

Correct answer:

$f^-^1(x)=\pm \sqrt{\frac{x-1}{2}}$

Explanation:

To find the inverse of y=2x^2+1 , interchange the and terms and solve for .

y=2x^2+1

x=2y^2+1

x-1=2y^2

$\frac{x-1}{2}=y^2$

$f^-^1(x)=\pm \sqrt{\frac{x-1}{2}}$

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GRE Subject Test: Math : Algebra

Study concepts, example questions & explanations for GRE Subject Test: Math

All GRE Subject Test: Math Resources

Example Questions

Example Question #171 : Algebra

Example Question #32 : Exponential Functions

Example Question #33 : Exponential Functions

Example Question #31 : Exponential Functions

Example Question #172 : Algebra

Example Question #173 : Algebra

Example Question #1 : Inverses

Example Question #2 : Inverses

Example Question #3 : Inverses

Example Question #3 : Find The Inverse Of A Relation

All GRE Subject Test: Math Resources

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