How Old Are They?

Mary’s and Bob’s ages combined are twice Jane’s age. Mary is $8$ years older than Bob. Jane’s age plus Bob’s age is $20$ years. How old are they?
Source: NCTM Mathematics Teacher, September 2006

Solution
Let $b,j$, and $m$ represent the ages of Bob, Jane, and Mary respectively.
$b+m=2j\qquad (1)$
$m-b=8\qquad\; (2)$
$b+j=20\qquad\; (3)$
Add Eq. $(1)$ and Eq. $(2)$
$b+m=2j$
$m-b=8$
——————
$2m=2j+8$
$m=j+4\qquad (4)$
Add Eq. $(2)$ and Eq. $(3)$
$m-b=8$
$b+j=20$
——————
$m+j=28\quad\;\; (5)$
Substitute the value of $m$ in Eq. $(4)$  into Eq. $(5)$
$(j+4)+j=28$
$2j=24$
$j=12$
Jane’s age equals $12$.
Substitute the value of Jane’s age into Eq. $(4)$
$m=12+4=16$
Mary’s age equals $16$.
Substitute the value of Jane’s age into Eq. $(3)$
$b+12=20$
$b=8$
Bob’s age equals $8$.

Answer: Bob is $8$, Jane $12$, Mary $16$

Arc of a Circle

$\overline{AB}$ is a diameter of a circle of radius $1$ unit. $\overline{CD}$ is a chord perpendicular to $\overline{AB}$ that cuts $\overline{AB}$ at $E$. If the arc $CAD$ is $2/3$ of the circumference of the circle, what is the length of the segment $\overline{AE}$?
Source: NCTM Mathematics Teacher, September 2006

Solution

Circumference = $2\pi(r)=2\pi(1)=2\pi$. Since measure of major arc $CAD=2/3(\mathrm{circumference})$, measure of minor arc $CBD=1/3(\mathrm{circumference})=2\pi/3$. Central angle $COD$ which tends arc $CBD$ therefore measures $2\pi/3$ radians or $120^\circ$. Triangle $COD$ is isosceles, hence altitude $OE$ is also the angle bisector of central angle $COD$. Triangle $OEC$ is a $30^\circ\!\textrm{-}60^\circ\!\textrm{-}90^\circ$ triangle with hypotenuse $OC$ equal $1$. Thus, side $\overline{OE}$ has length equal $1/2$.
$AE=AO+OE=1+1/2=3/2$ units

Answer: $3/2$ units

Alternative solution
Since chord $\overline{AB}$ is a diameter and chord $\overline{CD}$ is perpendicular to $\overline{AB}$, arc $AD$ is a mirror image of arc $AC$ in the $\overleftrightarrow{AB}$ line of reflection. Points $A,C$, and $D$ divide the circumference into three equal arcs each measuring $120^\circ$. Hence, $ACD$ is an equilateral triangle where the circumcenter $O$ is also the centroid of the triangle.
By the Concurrency of Medians of a Triangle theorem
$AO=(2/3)AE$
$AE=(3/2)AO=(3/2)(1)=3/2$

Multiplicative Inverse

If $\alpha=2+i$, then in the form of $a+bi$, what does $\alpha^{-1}$ equal?
Source: NCTM Mathematics Teacher, September 2006

Solution
By definition of multiplication inverse
$\alpha\alpha^{-1}=1$
$\alpha^{-1}=\dfrac{1}{2+i}$
$=\dfrac{1}{2+i}\times\dfrac{2-i}{2-i}$
$=\dfrac{2-i}{2^2-i^2}$
$=\dfrac{2-i}{4+1}$
$=\dfrac{2-i}{5}$
$=\dfrac{2}{5}-\dfrac{1}{5}i$

Answer: $\dfrac{2}{5}-\dfrac{1}{5}i$

Second Number Larger Than The First

Suppose a six-sided die has sides numbered $1$ through $6$. If a person throws the die two times, what is the probability that the second number will be larger than the first?
Source: NCTM Mathematics Teacher, September 2006

Solution
The desirable outcomes are
$1:\left \{2,3,4,5,6\right \}$ where $1$ is the first number and $2,3,4,5,6$ are the second number
$2:\left \{3,4,5,6\right \}$
$3:\left \{4,5,6\right \}$
$4:\left \{5,6\right \}$
$5:\left \{6\right \}$
$6:\left \{\textrm{not possible}\right\}$
Total number of desirable outcomes
$5+4+3+2+1=15$
Number of possible outcomes
$6\times 6=36$
Probability that the second number is larger than the first = $15/36$

Answer: $15/36$

Alternative solution
First, the die is thrown and the first number $\left \{1,2,3,4,5,6\right \}$ comes up each with a probability equal $1/6$. Then, the die is thrown again and we want the second number larger than the first.
$1:\left \{2,3,4,5,6\right \}$ probability = $5/6$
$2:\left \{3,4,5,6\right \}$ probability = $4/6$
$3:\left \{4,5,6\right \}$ probability = $3/6$
$4:\left \{5,6\right \}$ probability =$2/6$
$5:\left \{6\right \}$ probability = $1/6$
$6:\left \{\textrm{not possible}\right \}$ probability = $0/6$
Probability of rolling the die two times and the second number is larger than the first
$\dfrac{1}{6}\times \dfrac{5}{6}+\dfrac{1}{6}\times \dfrac{4}{6}+\dfrac{1}{6}\times \dfrac{3}{6}+\dfrac{1}{6}\times \dfrac{2}{6}+\dfrac{1}{6}\times \dfrac{1}{6}+\dfrac{1}{6}\times \dfrac{0}{6}=\dfrac{1}{6}\left (\dfrac{5+4+3+2+1+0}{6}\right )$
$=\dfrac{1}{6}\times \dfrac{15}{6}$
$=15/36$

Values of Integers

If $m$ and $n$ are integers such that $2m-n=3$, what are the possible values of $m-2n$?
(a) $-3$ only
(b) $0$ only
(c) only multiples of $3$
(d) any integer
(e) none of these
Source: NCTM Mathematics Teacher, September 2006

Solution
Let $x$ be an integer such that $x=m-2n$. We have the following two equations
$2m-n=3\qquad\qquad (1)$
$m-2n=x\qquad\qquad (2)$
Multiply Eq. $(1)$ by $-2$ and add to Eq. $(2)$
$-4m+2n=-6$
$m-2n=x$
————————
$-3m=x-6$
Divide both sides by $-3$
$m=-x/3+2$
For $m$ to be an integer, $x$ must be a multiple of $3$.

Answer: (c) only multiples of $3$

Alternative solution
$m-2n=m-2n+(3)-3$
$=m-2n+(2m-n)-3$
$=3m-3n-3$
$=3(m-n-1)$

Length of Side of Square

If the numerical value of the area of a square plus two times the numerical value of its perimeter is equal to $20$, what is the length of one side of the square?
Source: NCTM Mathematics Teacher, September 2006

Solution
Let $a$ represent the length of one side of a square. The following equation is true
$a^2+2(4a)=20$
$a^2+8a-20=0$
$(a-2)(a+10)=0$
$a=2$ or $a=-10$ (extraneous solution)
The length of one side of the square equals $2$ units.

Answer: $2$ units

John has a lemonade stand. He sells a small lemonade for $50$ cents and a large lemonade for $\1$. A small serving contains $1$ cup of lemonade; a large contains $1.5$ cups. At the end of the day, John has made $\9$ and sold $15.5$ cups of lemonade. How many small and large lemonades has he sold?
Source: NCTM Mathematics Teacher, September 2006

Solution
Let $x$ represent the number of small lemonades and $y$ the number of large lemonades sold. We have the following system of equations
$x(.50)+y(1)=9\qquad\qquad\; (1)$
$x(1)+y(1.5)=15.5\qquad\quad\, (2)$
Multiply Eqs. $(1)$ and $(2)$ by $2$
$x+2y=18\qquad\quad\:\: (3)$
$2x+3y=31\qquad\quad (4)$
Multiply Eq. $(3)$ by $-2$ and add to Eq. $(4)$
$-2x-4y=-36$
$2x+3y=31$
————————-
$-y=-5$
$y=5$
Substitute the value of $y=5$ into Eq. $(3)$
$x+2(5)=18$
$x=8$
John has sold $8$ small and $5$ large lemonades.

Answer: $8$ small and $5$ large

Red and Yellow Plums

One store sold red plums at $4$ for $\1$ and yellow plums at $3$ for $\1$. A second store sold red plums at $4$ for $\1$ and yellow plums at $6$ for $\1$. You bought $m$ red plums and $n$ yellow plums from each store, spending a total of $\10$. How many plums in all did you buy?
Source: NCTM Mathematics Teacher, September 2006

Solution
In dollars the following equation depicts how you spent $\10$ at the two stores
$m(1/4)+n(1/3)+m(1/4)+n(1/6)=10$
Collect like terms and simplify
$m(1/2)+n(1/2)=10$
$m+n=20$
At each store you purchase $20$ plums; in all you bought $40$ plums.

Answer: $40$

Value of Sum

What is the value of the following sum?
$2006[(-1)^1+(-1)^2+(-1)^3+(-1)^4+\cdots+(-1)^{2006}]$
Source: NCTM Mathematics Teacher, September 2006

Solution
The square bracket in the sum expression contains 1003 pairs of consecutive powers $(-1)^n+(-1)^{n+1}$ which pairwise sum to zero
$(-1)^1+(-1)^2=-1+1=0$
$(-1)^3+(1-)^4=-1+1=0$
$(-1)^5+(-1)^6=-1+1=0$
$\cdots$
$(-1)^{2005}+(-1)^{2006}=-1+1=0$
$2006[(-1)^1+(-1)^2+(-1)^3+\cdots+(-1)^{2006}]=2006[0+0+0+\cdots+0]=0$

Answer: $0$

Biggest Pizza Slice

An 8-inch pizza is cut into $3$ equal slices. A 10-inch pizza is cut into $4$ equal slices. A 12-inch pizza is cut into $6$ equal slices. A 14-inch pizza is cut into $8$ equal slices. From which pizza would you take a slice if you want as much pizza as possible?
Source: NCTM Mathematics Teacher, September 2006

Solution
Areas of pizza slices
8-inch: $\dfrac{\pi\times 4^2}{3}=\dfrac{\pi\times 16}{3}=\dfrac{\pi\times 128}{24}$
10-inch: $\dfrac{\pi\times 5^2}{4}=\dfrac{\pi\times 25}{4}=\dfrac{\pi\times 150}{24}$
12-inch: $\dfrac{\pi\times 6^2}{6}=\dfrac{\pi\times 6^2}{6}=\dfrac{\pi\times 144}{24}$
14-inch: $\dfrac{\pi\times 7^2}{8}=\dfrac{\pi\times 49}{8}=\dfrac{\pi\times 147}{24}$
Since $\dfrac{150}{24}$ is the largest fraction, you want to take a slice from the 10-inch pizza.