## Average Commuter Speed

A commuter has a $20$-mile trip each way from home to work. It takes her $30$ minutes to get to work and $60$ minutes to get home. What is her average (mean) speed?
Source: NCTM Mathematics Teacher, March 2008

Solution
By definition, average speed = distance traveled divided by travel time.
Distance traveled in miles = $20+20=40$
Travel time in hours = $.5+1=1.5$
Average speed = $40/1.5=26.7$ mph

Answer: $26.7$ mph

## Diagonals of Rhombus

Quadrilateral $ABCD$ is a rhombus. Its diagonals are $12$ and $20$ units in length. Find the diameter of the circle inscribed in rhombus $ABCD$. Round your answer to the nearest hundredth.

Source: NCTM Mathematics Teacher, February 2008

Solution

In a rhombus the diagonals are perpendicular bisectors of each other. In right triangle $EAD$, the length of the hypotenuse $\overline{AD}$ is given by
$AD^2=EA^2+ED^2$
$=6^2+10^2$
$=136$
$AD=\sqrt{136}=2\sqrt{34}$

Since $\overline{AD}$ is tangent to the circle at point $F$, it is perpendicular to radius $\overline{EF}$ the length of which we now calculate. Altitude $\overline{EF}$ creates two smaller triangles $FAE$ and $FED$ that are similar to the original triangle $EAD$ and to each other.
$EA^2=AF\times AD$  — side lengths of similar triangles are in proportion
$6^2=AF\times 2\sqrt{34}$
$AF=36/(2\sqrt{34})$
$=18/\sqrt{34}$
In right triangle $FAE$
$EF^2+AF^2=EA^2$
Substitute the value of $AF$
$EF^2+(18/\sqrt{34})^2=6^2$
$EF^2=36-324/34$
$=900/34$
$EF=\sqrt{900/34}$
Diameter of circle
$2EF=2\sqrt{900/34}$
$=10.29$

Answer: $10.29$ units

Alternative solution
The diagonals divide the rhombus into four congruent triangles. Using the diagonal lengths of $12$ and $20$, the area of rhombus is given by
$1/2(12)(20)=120$
The area of triangle EAD is given by
$1/2(EF)(AD)=30$  — one fourth the area of the rhombus
Substitute the value of $AD$
$EF=60/(2\sqrt{34})$
Diameter of circle
$2EF=60/\sqrt{34}=10.29$

## Even Sum

Two students, Katelyn and Carly, are playing a game that uses ten different numbered cards lying face down on a table. The faces of the cards are labeled with one of the numbers from $1$ to $10$ (each card has a different number). If Katelyn and Carly each turn over a different card, what is the probability that the sum of the two cards is even?
Source: NCTM Mathematics Teacher, February 2008

Solution
Suppose the first card is $3$. There are four possible ways to get an even sum from a second card
$3\:\:1$
$3\:\:5$
$3\:\:7$
$3\:\:9$
This is true for any of the five odd numbers. Hence, there are $5\times 4=20$ possible ways to make the sum even.
Suppose the first card is $4$. There are four possible ways to get an even sum from a second card
$4\:\:2$
$4\:\:6$
$4\:\:8$
$4\:\:10$
This is true for any of the five even numbers. Hence, there are $5\times 4=20$ possible ways to make the sum even.
Count of desirable outcomes = $20+20=40$
Count of possible outcomes = $10\times 9=90$
Probability = $40/90=4/9$

Answer: $4/9$

Alternative solution
For ease in discussion let’s call a card with an odd number an odd card and a card with an even number an even card. To get an even sum we need either two odd cards or two even cards. The probability of getting a first card odd equals $5/10$. Given that the first card is odd, the probability of getting a second card odd equals $4/9$. The probability of getting two odd cards equals $(5/10)(4/9)=20/90$.
By the same argument, the probability of getting two even cards equals $20/90$.
The probability of getting an even sum = $20/90+20/90=40/90=4/9$.

## Complement and Supplement

Determine the measure (in degrees) of an angle whose complement is four-ninths its supplement.
Source: NCTM Mathematics Teacher, February 2008

Solution
Let $x$ represent the measure of the angle in degrees. By definition of complement and supplement, $(90-x)$ is the measure of the complement of $x$ and $(180-x)$ is the measure of the supplement of $x$. We have the following equation
$(90-x)=4/9(180-x)$
Solving for $x$ yields $x=18^\circ$.

Answer: $18^\circ$

Alternative solution
If we graph the two linear functions $y_1=90-x$ and $y_2=4/9(180-x)$, the two lines intersect at $(x,y)=(18,72)$.

## Birthday Dinner

A group of friends went to dinner and received a bill totaling $\288$. The group decided to treat two people whose birthdays were that month and split the charges equally among the rest of the group. This resulted in each person having to pay $\4.80$ more than each would have paid had the bill split equally among the entire group. How many people were in the dinner group?
Source: NCTM Mathematics Teacher, February 2008

Solution
Suppose there are $x$ people in the group. Then, $288/x$ represents the amount each person in the entire group pays and $288/(x-2)$ the greater amount each person pays to treat two birthday people. We have the following equation
$288/(x-2)-288/x=4.8$
Multiply both sides by $x(x-2)$
$288x-288(x-2)=4.8x(x-2)$
Simplify
$576=4.8x^2-9.6x$
$4.8x^2-9.6x-576=0$
Solve for $x$ using the quadratic formula
$x=\dfrac{-(-9.6)\pm\sqrt{(-9.6)^2-4(4.8)(-576)}}{2(4.8)}$
$x=12$
There are $12$ people in the group.

Answer: $12$

Alternative solution
If we plot the functions $y_1=288/(x-2)$ and $y_2=288/x$, the table of values show that when $x=12, y_1-y_2=28.80-24.00=4.80$.
$x\quad\quad\quad y_1\quad\quad\:\: y_2$
——————————-
$3.00\quad 288.00\quad 96.00$
$3.75\quad 164.57\quad 76.80$
$4.50\quad 115.20\quad 64.00$
$5.25\quad\:\: 88.62\quad 54.86$
$6.00\quad\:\: 72.00\quad 48.00$
$6.75\quad\:\: 60.63\quad 42.67$
$7.50\quad\:\: 52.36\quad 38.40$
$8.25\quad\:\: 46.08\quad 34.91$
$9.00\quad\:\: 41.14\quad 32.00$
$9.75\quad\:\: 37.16\quad 29.54$
$10.50\quad 33.88\quad 27.43$
$11.25\quad 31.14\quad 25.60$
$12.00\quad 28.80\quad 24.00$

## Floor Function

If $\lfloor x\rfloor$ (or floor function) represents the greatest integer less than or equal to $x$, solve for $x$: $x\times\lfloor x\rfloor=40$.
Source: NCTM Mathematics Teacher, February 2008

Solution
Suppose $x=n$ for some integer $n$, then $x\times\lfloor x\rfloor=n\times n=n^2=40$; not possible because $40$ is not a perfect square. Thus, $x$ cannot be an integer.

Suppose $x$ is an irrational number such that $n for some integer $n$. Then, $x\times \lfloor x\rfloor=x\times n=40$; not possible because the left hand side of the equation is an irrational number and the right hand side is an integer. Thus, $x$ cannot be an irrational number.

So the solution to $x\times \lfloor x\rfloor=40$ must be a rational number. Since $x\times\lfloor x\rfloor$ is close to $x^2$, and $6^2<40<7^2$, $x$ is larger than $6$ and smaller than $7$. We are looking for a rational number $40/a$ such that $6<40/a<7$.
$40/1=40$
$40/2=20$
$40/3=13.33$
$40/4=10$
$40/5=8$
$40/6=6.67$
$40/7=5.71$
$40/8=5$
$x=40/6$ is the rational number between $6$ and $7$ and is the solution to $x\times \lfloor x\rfloor=40$.

Answer: $40/6$

Arrange the digits $1,1,2,2,3$, and $3$ to form a six-digit number such that the $1\mathrm{s}$ have one digit between them, the $2\mathrm{s}$ have two digits between them, and the $3\mathrm{s}$ have three digits between them. What is the sum of all possible six-digit numbers formed under these conditions?
Source: NCTM Mathematics Teacher February 2008

Solution
Two possible ways to satisfy the $3\mathrm{s}$ digits condition: $3\!-\!-\!-\!3-$ and $-\!3\!-\!-\!-\!3$.
$3\!-\!-1\:3\:1$      if we sandwich the $3$ digit at the end with two $1\mathrm{s}$
$3\: 2\: 2\: 1\: 3\: 1$      the $2\mathrm{s}$ have no place to go but here; not a solution
$3\: 1\: 2\: 1\: 3\: 2$      putting the two $1\mathrm{s}$ here provides a solution

$1\: 3\: 1\!-\!-3$      if we sandwich the $3$ digit at the beginning with two $1\mathrm{s}$
$1\: 3\: 1\: 2\: 2\: 3$      the $2\mathrm{s}$ have no place to go but here; not a solution
$2\: 3\: 1\: 2\: 1\: 3$      putting the two $1\mathrm{s}$ here provides a solution

Sum of all possible six-digit numbers = $312132+231213=543345$

Answer: $543345$

Extension
The original problem specifies the logical AND condition, i.e. the $1\mathrm{s}$ have one digit between them AND the $2\mathrm{s}$ have two digits between them AND the $3\mathrm{s}$ have three digits between them. In this extension, we use the logical exclusive OR condition, i.e. the $1\mathrm{s}$ have one digit between them OR the $2\mathrm{s}$ have two digits between them OR the $3\mathrm{s}$ have three digits between them. What is the sum of all possible six-digit numbers formed under these conditions?

Solution
The $1\mathrm{s}$ have one digit between them in four possible cases
$1\!-\!1\!-\!--$
$-\!1\!-\!1\!-\!-$
$-\!-\!1\!-\!1\!-$
$-\!-\!-\!1\!-\!1$
In each case we have to place the remaining four digits $2,2,3,3$ in the four empty slots. If the digits $2,2,3,3$ were distinguishable, we could have formed $4\times 3\times 2\times 1=24$ possible numbers. Since they are not, we overcount by $2!2!$ and so the true count is $24/(2!2!)=6$. The $6+6+6+6=24$ numbers of the four cases are listed below
$121233\quad 212133\quad 221313\quad 223131$
$121323\quad 213123\quad 231213\quad 232131$
$121332\quad 213132\quad 231312\quad 233121$
$131223\quad 312123\quad 321213\quad 322131$
$131232\quad 312132\quad 321312\quad 323121$
$131322\quad 313122\quad 331212\quad 332121$
$---\:\:\:\:---\:\:\:\:---\:\:\:\:\:---$
$757665\quad 1575765\:\: 1657575\:\;1665756$
Sub-total = $757665+1575765+1657575+1665756=5656761$

The $2\mathrm{s}$ have two digits between them in three possible cases
$2\!-\!-\!2\!-\!-$
$-2\!-\!-2-$
$-\!-\!2\!-\!-\!2$
As before we need to place the remaining four digits $1,1,3,3$ in the four empty slots. The $6+6+6=18$ numbers are listed below
$211233\quad 121323\quad 112332$
$213213\quad 123123\quad 132132$
$213231\quad 123321\quad 132312$
$231213\quad 321123\quad 312132$
$231231\quad 321321\quad 312312$
$233211\quad 323121\quad 332112$
$---\:\:\:\:---\:\:\:\:---$
$1333332\:\: 1333332\:\: 1333332$
Sub-total = $1333332+1333332+1333332=3999996$

The 3s have three digits between them in two possible cases
$3\!-\!-\!-\!3-$
$-\!3\!-\!-\!-\!3$
We need to place the remaining four digits $1,1,2,2$ in the four empty slots. The $6+6=12$ numbers are listed below
$311232\quad 131223$
$312132\quad 132123$
$312231\quad 132213$
$321132\quad 231123$
$321231\quad 231213$
$322131\quad 232113$
$---\quad ---$
$1900089\:\: 1090008$
Sub-total = $1900089+1090008=2990097$
Grand total = $5656761+3999996+2990097=12646854$

Answer: $12646854$

## Difference of Sequences

For the sequence $1,-2,3,-4,5,-6,7,\cdots$ , what is the difference between the mean of the sequence’s first $400$ terms and the mean of its first $200$ terms?
Source: NCTM Mathematics Teacher February 2008

Solution
If we add the first $400$ terms by pairs $1+(-2),3+(-4),5+(-6),\cdots$ we end up with $200(-1)=-200$. The mean of the first $400$ terms is $-200/400=-1/2$.
Likewise, if we add the first $200$ terms by pairs, we get $100(-1)=-100$. The mean of the first $200$ terms is $-100/200=-1/2$.
The difference between the two means is $-1/2-(-1/2)=0$.

Answer: $0$

## Sums of Reciprocals

The sum of the positive divisors of $480$ is $1512$. Find the sum of the reciprocals of the positive divisors of $480$.
Source: NCTM Mathematics Teacher, February 2008

Solution
Given that $480=2^5\cdot 3^1\cdot 5^1$, the number of divisors is $(5+1)(1+1)(1+1)=24$. The first twelve divisors are small and easy to guess: $1,2,3,4,5,6,8,10,12,15,16$, and $20$ and as a bonus we get the last twelve by dividing $480$ by the first twelve divisors. For example, $480/1=480,480/2=240,480/3=160$, etc. : $480,240,160,120,96,80,60,48,40,32,30,24$.
When we add the reciprocals of the divisors, the divisors appear as denominators in a set of 24 fractions
$\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{240}+\dfrac{1}{480}$
When we reduce the fractions to $480$ (the least common denominator), the $24$ divisors appear as numerators
$\dfrac{480}{480}+\dfrac{240}{480}+\cdots+\dfrac{2}{480}+\dfrac{1}{480}=\dfrac{480+240+\cdots+2+1}{480}=\dfrac{1512}{480}$

Answer: $1512/480$

## Divisible by Units Digit

The number $64$ is divisible by its units digit $(4)$. How many whole numbers less than $64$ are divisible by their respective units digit?
Source: NCTM Mathematics Teacher, February 2008

Solution
Nine numbers: $1,2,3,4,5,6,7,8,9$ divisible by themselves
Six numbers: $11,21,31,41,51,61$ divisible by $1$
Six numbers: $12,22,32,42,52,62$ divisible by $2$
Five numbers: $15,25,35,45,55$ divisible by $5$
Six numbers: $33,63,24,44,36,48$ divisible by their respective units digit
Total $9+6+6+5+6=32$

Answer: $32$