Chris paid $\81$ for a dress that had been discounted $25\%$ and then marked down an additional $10\%$. Taking both discounts into consideration, determine the original price of the dress.
Source: NCTM Mathematics Teacher, March 2008

Solution
Let $x$ represent the original price. We apply the first discount then the additional markdown off the first discount.
$81=(x-.25x)-(x-.25x)(.10)$
$=.75x-.075x$
$=.675x$
$x=81/.675=120$

Answer: $\120$

Alternative solution 1
This solution does not make any subtraction.
$81=.9(x \times .75)$
$90=.75x$
$x=90/.75=120$

Andy is taking a multiple-choice test consisting of $10$ questions. All the items have four answer choices but only one correct answer. Unfortunately, Andy did not study for the test and plans on guessing the answer for each item. What is the probability that Andy will guess the answer for every item correctly?
Source: NCTM Mathematics Teacher, March 2008

Solution
Suppose the test has $1$ question
Number of choices = $4$
$A\;B\;C\;D$
Probability of guessing every item correctly = $1/4$

$2$ questions
Number of choices = $4^2$
$AA\;AB\;AC\;AD$
$BA\;BB\;BC\;BD$
$CA\;CB\;CC\;CD$
$DA\;DB\;DC\;DD$
Probability = $1/4^2$

$3$ questions
Number of choices = $4^3$
$AAA\;AAB\;AAC\;AAD$
$ABA\;ABB\;ABC\;ABD$
etc.
Probability = $1/4^3$

$10$ questions
Number of choices = $4^{10}$
$AAAAAAAAAA\;AAAAAAAAAB\;AAAAAAAAAC\;AAAAAAAAAD$
$AAAAAAAAB\!A\;AAAAAAAAB\!B\;AAAAAAAABC\;AAAAAAAABD$

etc.
Probability = $1/4^{10}=1/1048576\approx 0.0001\%$ or about $1$ in a million

Answer: $\approx 0.0001\%$

Special Operation

If $m\,\phi\,n=m^2-6mn+3n^3$, what is $-2\,\phi\,3$?
Source: NCTM Mathematics Teacher, March 2008

Solution
$-2\,\phi\,3=(-2)^2-6(-2)(3)+3(3)^3=4+36+81=121$

Answer: $121$

Identify Two Integers

The sum of two integers is $-5$ and their product is $-24$. Identify both integers.
Source: NCTM Mathematics Teacher, March 2008

Solution
$24$ is small enough that we can exhaustively enumerate its factors
$1\times 24$
$2\times 12$
$3\times 8$
$4\times 6$
The possible negative sums of the factors are
$1+(-24)=-23$
$2+(-12)=-11$
$3+(-8)=-5$
$4+(-6)=-2$
The two integers are $3$ and $-8$.

Answer: $3$ and $-8$

Alternative solution 1
Let $a$ and $b$ represent the two integers. Given $a+b=-5$ and $ab=-24$, we have
$(a+b)^2=a^2+2ab+b^2$
$(-5)^2=a^2+2(-24)+b^2$
$25=a^2-48+b^2$
$a^2+b^2=73$
Consider the consecutive squares less than $73: 1,4,9,16,25,36,49,64$. The possible $a^2+b^2$ odd values are
$1+64=65$
$9+64=73$
$25+36=61$
$49+64=113$
$49+36=85$
$49+16=65$
Only $a^2+b^2=9+64=73$. Hence $a=\pm 3$ and $b=\pm 8$. The two integers are $3$ and $-8$.

Alternative solution 2
If two integers $a$ and $b$ are the solutions of a quadratic equation, we can write $x^2-(a+b)x+ab=0$.
Substitute the values of $a+b$ and $ab$
$x^2+5x-24=0$
Solve the quadratic equation by factoring
$(x-3)(x+8)=0$
$x=3$ and $x=-8$

Alternative solution 3
Let $x$ and $y$ represent the two integers. We have
$x+y=-5\qquad (1)$
$xy=-24\qquad\:\:\,\, (2)$
Solving for $y$ in Eq. $(1)$
$y=-x-5$
Substitute the value of $y$ into Eq. $(2)$
$x(-x-5)=-24$
$x^2+5x-24=0$

Distance Graph

For the graph shown below, create a graph showing distance traveled versus time

Source: NCTM Mathematics Teacher, March 2008

Solution
Distance = $\mathrm{speed}\times\mathrm{time}$. When speed is constant, distance increases linearly; the graph representing distance is a line. When speed is not constant and increases as a linear function of time, distance increases quadratically; the distance graph has the curvature of a parabola. A possible graph is shown below

Story of Graph

Write a story to accompany the graph below

Source: NCTM Mathematics Teacher, March 2008

Solution
Minute $0$ to $3$: speed increases rapidly
$3$ to $15$: speed remains constant
$15$ to $18$: speed increases again though at a lesser rate than before
$18$ to $24$: speed remains constant

Alternative solution
From a complete stop, you begin driving your car at a steadily increasing rate. After approximately $2$ minutes, you reach your desired speed and set the cruise control. For the next $15$ minutes, you travel at a steady speed. At this point you notice that the speed limit has changed, so you steadily increase the speed of your car for the next $2$ minutes. Then you set the cruise control at the desired speed.

Right Cylinder

What is the volume and surface area of the right cylinder capped by two hemispheres picture below

Source: NCTM Mathematics Teacher, March 2008

Solution
Total volume = volume of cylinder + volume of two hemispheres
$=\pi r^2 h+(4/3)\pi r^3$
$=\pi 6^2(12)+(4/3)\pi 6^3$
$=\pi (432+288)$
$=720\pi$
$=2262\:\mathrm{ft}^3$
Total surface area = surface area of cylinder + surface area of two hemispheres
$=2\pi r h+4\pi r^2$
$=2\pi(6)(12)+4\pi 6^2$
$=\pi(144+144)$
$=288\pi$
$=905\:\mathrm{ft}^2$

Answer: volume = $2262\:\mathrm{ft}^3$; surface area = $905\:\mathrm{ft}^2$

Apples and Oranges

Juanita is reviewing her grocery bills. Three oranges and four apples cost $\4.33$. Five oranges and three apples cost $\5.42$. What was the cost of each orange and each apple?
Source: NCTM Mathematics Teacher, March 2008

Solution
Let $r$ represent the cost of an orange and $a$ represent the cost of an apple. We have the following equations
$3r+4a=4.33\qquad (1)$
$5r+3a=5.42\qquad (2)$
Multiply Eq. $(1)$ by $-3$ and Eq. $(2)$ by $4$ and add the equations
$-9r-12a=-12.99$
$20r+12a=21.68$
——————————-
$11r=8.69$
$r=.79$
Substitute the value of $r$ into Eq. $(1)$
$3(.79)+4a=4.33$
$2.37+4a=4.33$
$4a=1.96$
$a=.49$

Answer: oranges cost $79$ cents each; apples $49$ cents each

Photograph Five Students

A Math Olympiad team is posing for its annual media photograph. How many ways are there to arrange the five members of the team in one line?
Source: NCTM Mathematics Teacher, March 2008

Solution
Choose one for the first position – $5$ ways
Choose one for the second – $4$ ways
Choose one for the third – $3$ ways
Choose one for the fourth – $2$ ways
Choose one for the fifth – $1$ way
$5\times 4\times 3\times 2\times 1=120$ ways to arrange the five members of the team in one line.

Answer: $120$

Mean, Median, and Mode

Using the frequency table below, determine the mean, median, and modal income.
$\mathrm{\underline{Income}}\quad\quad\:\:\mathrm{\underline{Frequency}}$
$\1,\!400,\!000\quad\quad\: 1$
$\520,\!000\quad\quad\quad 3$
$\125,\!000\quad\quad\quad 6$
$\85,\!000\quad\quad\quad\:\: 8$
$\45,\!000\quad\quad\quad\: 12$
Source: NCTM Mathematics Teacher, March 2008

Solution
mean = $(1400000+(3)520000+(6)125000+(8)85000+(12)45000)/(1+3+6+8+12)$
$=164333.33$
We order the $30$ values from smallest to greatest. Since the count of numbers is even, the median equals the average of the fifteenth and sixteenth values.
median = $(85000+85000)/2=85000$
mode = $45000$ because this value occurs the most — twelve times.

Answer: mean=$\164,\!333.33;$ median = $\85,\!000$; mode = $\45,\!000$