## Real Solutions (x,y)

How  many real solutions $(x,y)$ are there that satisfy the two equations $x^2+y^2=30$ and $4y^2-x^2=100$?
Source: NCTM Mathematics Teacher, September 2006

Solution

The graph of $x^2+y^2=30$ is a circle centered at the origin and radius = $\sqrt{30}$. The graph of $4y^2-x^2=100$ is a hyperbola with vertices at $(0,5)$ and $(0,-5)$. The two graphs intersect at four points. There are four real solutions $(x,y)$ that satisfy the two equations.

Answer: $4$

Alternate solution
Add the two equations
$x^2+y^2=30$
$4y^2-x^2=100$
————————
$5y^2=130$
$y^2=26$
$y=\pm\sqrt{26}$
Substitute the value of $y^2$ into the first equation
$x^2+26=30$
$x^2=4$
$x=\pm 2$

## Twin Liars

Kerry and Kelly are twins. One of them lies on Mondays, Tuesdays, and Wednesdays and tells the truth on the other days of the week. The other lies on Thursdays, Fridays, and Saturdays and tells the truth on the other days of the week. On which day of the week did they have this conversation?
Kerry: I lie on Saturdays.
Kelly: I will lie tomorrow.
Kerry: I lie on Sundays.
Source: NCTM Mathematics Teacher, September 2006

Solution
There are two types of liars, the Mon-Tue-Wed liar and the Thu-Fri-Sat liar. We need to determine who is who? Clearly, Kerry is lying when he said that “I lie on Sundays” since both of them tell the truth on Sundays. Therefore, his statement “I lie on Saturdays” is false; he tells the truth on Saturdays and so Kerry is the Mon-Tue-Wed liar and Kelly is the Thu-Fri-Sat liar.

Since we know Kerry is lying, the conversation can only take place on Monday, Tuesday, or Wednesday.

Suppose it was Monday. Since Kelly is telling the truth on Mondays, the statement “I will lie tomorrow” (Tuesday) is true, which characterizes her as a Tue. liar. But that is not possible, because she is a Thu-Fri-Sat liar.

Suppose it was Tuesday. Since Kelly is telling the truth on Tuesdays, the statement “I will lie tomorrow” (Wednesday) is true, which characterizes her as a Wed. liar. But that is not possible, because she is a Thu-Fri-Sat liar.

Suppose it was Wednesday. Since Kelly is telling the truth on Wednesdays, the statement “I will lie tomorrow” (Thursday) is true, which characterizes her as a Thu. liar. There is no contradiction here, she is indeed a Thu-Fri-Sat liar.

## 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: $40$