2024 Fryer Contest
Solutions
(Grade 9)
Thursday, April 4, 2024
(in North America and South America)
Friday, April 5, 2024
(outside of North American and South America)
Thursday, April 4, 2024
(in North America and South America)
Friday, April 5, 2024
(outside of North American and South America)
©2024 University of Waterloo
The 5th term is obtained by adding \(6\) to the 4th term. Thus, the 5th term is \(21+6=27\).
Solution 1:
The 6th term is obtained by adding \(6\) to the 5th term. Thus, the 6th term is \(27+6=33\), and so the average of the 4th, 5th and 6th terms is \(\frac{21+27+33}{3}=\frac{81}{3}=27\).
Solution 2:
The 4th term is \(6\) less than the 5th term, and the 6th term is \(6\) more than the 5th term, and so the average of the 4th, 5th and 6th terms is the 5th term, which is \(27\).
The \(n\)th term (\(n\geq2\)) is obtained by adding \(n-1\) 6s to the first term, \(3\). For example, the 2nd term is \(3+1\times6\), the 3rd term is \(3+2\times6\), the 4th term is \(3+3\times6\), and so on.
In general, the \(n\)th term is given
by \(3+(n-1)\times6\).
Therefore, the 20th term is \(3+19\times6=3+114=117\).
Solution 1:
Since each new term is obtained by adding \(6\) to the previous term and \(\frac{1000}{6}\approx166.7\), then it makes
sense to begin by determining the 166th term. The 166th term is \(3+165\times6=993\), the next term is \(993+6=999\) (still less than \(1000\)), and so the smallest term that is
greater than 1000 is \(999+6=1005\).
(We note that \(1005\) is the 168th
term and is equal to \(3+167\times6\).)
Solution 2:
From part (c), the \(n\)th term is
given by the expression \(3+(n-1)\times6\).
We want the smallest term that is greater than \(1000\). To begin, we find the smallest
possible value of \(n\) for which \(3+(n-1)\times6>1000\).
Solving this inequality, we get \[\begin{align*}
3+(n-1)\times6&>1000\\
3+6n-6&>1000\\
6n-3&>1000\\
6n&>1003\\
n&>\tfrac{1003}{6}\approx167.2\end{align*}\] Since \(n\) must be an integer, the first term
number to exceed \(1000\) is the 168th
term, and its value is \(3+167\times6=1005\).
At Store 2, \(50\%\) of the
shirts dropped off were red, and so the other \(50\%\) were blue.
That is, Ella dropped off an equal number of red and blue shirts at
Store 2.
All 200 blue shirts were dropped off at Store 2, and thus \(200\) red shirts were also dropped off at
Store 2. Ella began the day with \(800\) red shirts, dropped off \(200\) at Store 2, and so she dropped off
\(800-200=600\) red shirts at Store
1.
At Store 1, Ella dropped off \(40\%\) of \(5x\) red shirts or \(0.40\times5x=2x\) red shirts.
Ella dropped off the remaining \(5x-2x=3x\) red shirts at Store 2, in
addition to the \(5x\) blue shirts. At
Store 2, \(\dfrac{5x}{3x+5x}=\dfrac{5}{8}\) of the
shirts dropped off were blue, which is \(\frac58\times100\%=62.5\%\) of the
shirts.
Ella dropped off no blue shirts at Store 1, and so she dropped
off all \(y\) blue shirts at Store 2.
Since there were equal numbers of red, blue and green shirts dropped off
at Store 2, then \(y\) green shirts
(and \(y\) red shirts) were dropped off
at Store 2.
On Wednesday, Ella dropped off \(3y\)
red shirts, \(y\) blue shirts, and
\(y\) green shirts, for a total of
\(5y\) shirts.
Of all the shirts dropped off on Wednesday, \(\dfrac{y}{5y}=\dfrac{1}{5}\) were green,
which is \(\frac15\times100\%=20\%\) of
the shirts.
In Figure 2, each of the smaller pieces has the same length of
crust.
Thus, we must determine the length of \(MN\) so that each of the smaller pieces has
the same area.
The area of square \(ABCD\) is \(30\times30=900\), and so the area of each
of the \(3\) smaller pieces is \(\frac{900}{3}=300\).
Since \(ABCD\) is a square, then \(AD=BC=30\) and so \(AM=15\).
The smaller piece \(AMNB\) is a
trapezoid (\(AB\) and \(MN\) are perpendicular to \(AM\) and thus parallel to one another), and
so \(AMNB\) has area \(\frac{AM}{2}(MN+AB)\) or \(\frac{15}{2}(MN+30)\).
Solving, we get \(\frac{15}{2}(MN+30)=300\) or \(MN+30=\frac{300\times2}{15}\) or \(MN+30=40\), and so \(MN=10\).
Alternately, \(\triangle BNC\) also has area 300 with base \(BC=30\), and so has height \(h=\frac{300}{15}=20\). Thus, \(MN=AB-h\) or \(MN=30-20=10\).
In Figure 3, each of the \(5\)
smaller pieces must have the same length of crust.
Since the slice of bread has crusts on \(3\) of its edges, the total length of crust
is \(3\times30=90\).
Thus the length of crust for each of the \(5\) smaller pieces is \(\frac{90}{5}=18\).
Since \(\triangle TPQ\) is one of these
smaller pieces, and its length of crust is \(PQ\), then \(PQ=18\).
The area of square \(ABCD\) is
\(30\times30=900\), and so the area of
each of the \(5\) smaller pieces is
\(\frac{900}{5}=180\).
Consider a point \(W\) on \(BC\) so that \(MW\) is perpendicular to \(BC\).
Since \(MW\) must pass through \(S\) and \(T\), then \(TW\) is the altitude of \(\triangle TPQ\).
The area of \(\triangle TPQ\) is 180,
and so \(\frac12\times PQ\times
TW=180\) or \(\frac12\times18\times
TW=180\), and so \(TW=\frac{180}{9}=20\).
Since \(MS\) and \(AU\) are both perpendicular to \(AM\), then \(MS\) and \(AU\) are parallel to one another, and so
\(AMSU\) is a trapezoid.
Each of the \(5\) smaller pieces has
the same length of crust, and so \(AU=PQ=18\).
The area of \(AMSU\) is \(180\). Solving \(\frac{AM}{2}(MS+AU)=180\), we get \(\frac{15}{2}(MS+18)=180\) or \(MS+18=24\), and so \(MS=6\).
Since \(MW=AB=30\), then \(ST=30-MS-TW\) or \(ST=30-6-20\), and so \(ST=4\).
Since each player has \(3\)
distinct integers that they may spin, then the total number of possible
outcomes is \(3\times3=9\), and each
outcome is equally probable.
If Alice spins a \(5\), she wins if
Binh spins a \(1\), and loses if Binh
spins \(8\) or \(10\).
If Alice spins a \(9\), she wins if
Binh spins \(1\) or \(8\), and loses if Binh spins a \(10\).
If Alice spins an \(11\), she wins if
Binh spins \(1\), \(8\) or \(10\).
We summarize these results in the table below, using an A to indicate
that Alice wins and a B where Binh wins.
| Alice's Spin | ||||
|---|---|---|---|---|
| \(5\) | \(9\) | \(11\) | ||
| Binh's Spin | \(1\) | A | A | A |
| \(8\) | B | A | A | |
| \(10\) | B | B | A | |
We see that Alice wins \(6\) of the \(9\) possible outcomes, and thus the probability that Alice wins is \(\frac69=\frac23\).
Carole’s spinner is \(\{1,5,10\}\). Suppose that Darsh’s spinner
is \(\{a,b,c\}\) with \(a<b<c\) and where \(a\), \(b\), \(c\)
are chosen from \(2\), \(3\), \(4\), \(6\), \(7\), \(8\), and \(9\).
Since each player has \(3\) distinct
integers that they may spin, then the total number of possible outcomes
is \(3\times3=9\), and each outcome is
equally probable.
Thus, Darsh’s probability of winning is greater than Carole’s
probability of winning if the integer he spins is greater than Carole’s
for at least \(5\) of the \(9\) possible outcomes.
If Carole spins a \(1\), then Darsh
wins since each of his integers must be greater than \(1\).
When Carole spins a \(1\), there are
\(3\) winning outcomes for Darsh since
he can spin \(a\), \(b\) or \(c\) (and each is greater than \(1\)).
If Carole spins a \(10\), then Darsh
loses since each of his integers must be less than \(10\).
There are similarly \(3\) outcomes when
Carole spins a \(10\), except each of
these is a losing outcome for Darsh.
To this point, Darsh wins \(3\) of the
possible outcomes if Carole spins a \(1\) or a \(10\).
This tells us that the probability of Darsh winning is determined by
comparing \(a\), \(b\), \(c\)
to Carole’s spin of a \(5\).
Specifically, if at least two of Darsh’s integers are greater than \(5\), then Darsh’s probability of winning is
greater than Carole’s probability of winning (and the opposite is true
if at least two of Darsh’s integers are less than \(5\)).
Why is this true? When Carole spins a \(1\), Darsh wins all \(3\) possible outcomes. When Carole spins a
\(10\), Darsh wins \(0\) possible outcomes. When Carole spins a
\(5\), Darsh wins at least \(2\) possible outcomes exactly when at least
two of his integers are greater than \(5\), and thus Darsh wins at least \(3+0+2=5\) of the \(9\) possible outcomes (and Carole wins
\(4\) or fewer).
To determine the number of different spinners for which at least two of
Darsh’s integers are greater than \(5\), we consider the following two
cases.
Case 1: All three of Darsh’s integers are greater than \(5\).
In this case, Darsh wins \(6\) of the \(9\) possible outcomes, as shown.
| Darsh's Spin | ||||
|---|---|---|---|---|
| \(a\) | \(b\) | \(c\) | ||
| Carole's Spin | \(1\) | D | D | D |
| \(5\) | D | D | D | |
| \(10\) | C | C | C | |
Case 2: Two of Darsh’s integers are greater than \(5\) and one of his integers is less than \(5\).
In this case, Darsh wins \(5\) of the \(9\) possible outcomes, as shown.
| Darsh's Spin | ||||
|---|---|---|---|---|
| \(a\) | \(b\) | \(c\) | ||
| Carole's Spin | \(1\) | D | D | D |
| \(5\) | C | D | D | |
| \(10\) | C | C | C | |
Therefore, Darsh can make \(4+18=22\) different spinners so that his probability of winning is greater than Carole’s probability of winning.
We begin by determining the value of \(p\), the probability that Fynn beats
Erin.
We summarize these results in the table below, using an F to indicate
that Fynn wins and an E where Erin wins.
| Fynn's Spin | ||||
|---|---|---|---|---|
| \(2\) | \(10\) | \(18\) | ||
| Erin's Spin | \(5\) | E | F | F |
| \(8\) | E | F | F | |
| \(15\) | E | E | F | |
We see that Fynn wins \(5\) of the
\(9\) possible outcomes, and thus \(p=\frac59\).
In the question, we are given that \(p=q=r\) and thus \(q=r=\frac59\), or the probability that Erin
beats Gina is \(\frac59\) and the
probability that Gina beats Fynn is also \(\frac59\).
This means that Erin wins exactly \(5\)
of the \(9\) possible outcomes when
playing against Gina, and similarly, Gina wins exactly \(5\) of the \(9\) possible outcomes when playing against
Fynn.
We begin by considering the game in which Erin plays Gina.
Erin’s spinner is \(\{5,8,15\}\).
Gina’s spinner is {\(x,y,z\)} with
\(x<y<z\) and where \(x\), \(y\), \(z\)
are chosen from \[1,3,4,6,7,9,11,12,13,14,16,17,19,20\] Each
of Erin’s \(3\) integers \((5, 8, 15)\) wins \(0\), \(1\), \(2\), or \(3\) possible outcomes.
For example, if Erin spins \(15\) and
\(15>z\), then Erin wins all \(3\) possible outcomes (\(15\) beats each of \(x,y\) and \(z\)).
Conversely, if Erin spins \(15\) and
\(15<x\), then Erin wins \(0\) possible outcomes.
We note that if a player’s spinner is {\(p,q,r\)} with \(p<q<r\), then the number of winning
outcomes when spinning \(r\) must be
greater than or equal to the number of winning outcomes when spinning
\(q\), which must be greater than or
equal to the number of winning outcomes when spinning \(p\). Can you explain why this is?
There are three different ways that Erin can win exactly \(5\) of the \(9\) outcomes against Gina.
These are:
Erin’s largest integer wins exactly \(3\) outcomes, her second largest integer wins exactly \(2\) outcomes, and her smallest integer wins \(0\) outcomes, or
Erin’s largest integer wins exactly \(3\) outcomes, and her remaining two integers each win exactly \(1\) outcome, or
Erin’s largest two integers each win exactly \(2\) outcomes, and her smallest integer wins exactly \(1\) outcome.
We will refer to the first of these bullets as the \(3/2/0\) result, the second bullet as the
\(3/1/1\) result, and the final bullet
as the \(2/2/1\) result.
We note that in each case, the sum of the number of winning outcomes is
\(5\), and further, these are the only
possible ways for exactly \(5\) winning
outcomes to occur.
If the result in the game between Erin and Gina is \(3/2/0\), then Erin wins all \(3\) outcomes when spinning \(15\), wins \(2\) outcomes when spinning \(8\), and wins \(0\) outcomes when spinning \(5\).
Recalling that Gina’s spinner is {\(x,y,z\)} with \(x<y<z\), this means that if the
result is \(3/2/0\), then the two
spinners’ six integers are ordered as follows, \[5<x<y<8<z<15\] That is,
since \(15\) is greater than each of
\(x\), \(y\), and \(z\), then Erin wins \(3\) outcomes when spinning \(15\). Since \(8\) is greater than \(x\) and \(y\), then Erin wins \(2\) outcomes when spinning \(8\), and since \(5\) is less than \(x\), \(y\)
and \(z\), then Erin wins \(0\) outcomes when spinning \(5\).
Thus, if Gina makes her spinner with \(x=6\), \(y=7\) and \(z=9\), \(11\), \(12\), \(13\), or \(14\), then the probability that Erin beats
Gina is \(\frac59\) since Erin will win
exactly \(5\) of the \(9\) possible outcomes.
If the result is \(3/1/1\), then the
two spinners’ six integers are ordered as follows, \[x<5<8<y<z<15\] In this
case, a spin of \(15\) wins \(3\) times and spins of \(5\) and \(8\) each win exactly once.
Thus, if Gina makes her spinner with \(x=1,3\) or 4, and \(y\) and \(z\) (with \(y<z\)) are chosen from \(9\), \(11\), \(12\), \(13\), \(14\), then the probability that Erin beats
Gina is also \(\frac59\).
Finally, if Erin beats Gina with a \(2/2/1\) result, then the two spinners’ six
integers are ordered as follows, \[x<5<y<8<15<z\] and so \(x=1,3 \text{ or } 4\), \(y=6 \text{ or } 7\), and \(z=16, 17, 19, \text{ or } 20\).
We summarize these results in the table below.
| Case | Result | Integer Ordering | Possible \(\boldsymbol{x}\) | Possible \(\boldsymbol{y}\) | Possible \(\boldsymbol{z}\) |
|---|---|---|---|---|---|
| 1 | \(3/2/0\) | \(5<x<y<8<z<15\) | \(6\) | \(7\) | \(9\), \(11\), \(12\), \(13\), \(14\) |
| 2 | \(3/1/1\) | \(x<5<8<y<z<15\) | \(1\), \(3\), \(4\) | \(9\), \(11\), \(12\), \(13\), \(14\) | \(9\), \(11\), \(12\), \(13\), \(14\) |
| 3 | \(2/2/1\) | \(x<5<y<8<15<z\) | \(1\), \(3\), \(4\) | \(6, 7\) | \(16\), \(17\), \(19\), \(20\) |
(Note that in Case 2, the values of \(y\) are only possible provided that \(y<z\).)
We also require Gina’s spinner to be made so that the probability that
Gina beats Fynn is \(\frac59\).
In the game between Gina and Fynn, we use the same process and notation
as was used in the game between Erin and Gina.
In addition, we see from the table above that \(x,y,z\) must be chosen so that \[x=1,3,4, \text{or } 6, \text{ and
}y=6,7,9,11,12,13, \text{or }14, \text{ and }
z=9,11,12,13,14,16,17,19,\text{or }20\] again noting that we
require \(x<y<z\).
In the table below, we exclude all other possible values of \(x,y,z\) not listed above.
| Case | Result | Integer Ordering | Possible \(\boldsymbol{x}\) | Possible \(\boldsymbol{y}\) | Possible \(\boldsymbol{z}\) |
|---|---|---|---|---|---|
| 4 | \(3/2/0\) | \(x<2<10<y<18<z\) | \(1\) | \(11\), \(12\), \(13\), \(14\) | \(19,20\) |
| 5 | \(3/1/1\) | \(2<x<y<10<18<z\) | \(3\), \(4\), \(6\) | \(6\), \(7\), \(9\) | \(19,20\) |
| 6 | \(2/2/1\) | \(2<x<10<y<z<18\) | \(3\), \(4\), \(6\) | \(11\), \(12\), \(13\), \(14\) | \(11\), \(12\), \(13\), \(14\), \(16\), \(17\) |
(Note that in Cases 5 and 6, the values of \(x\) and \(y\) are only possible if \(x<y<z\).)
To make a spinner for which Erin beats Gina and Gina beats Fynn, each
with probability \(\frac59\), then we
must determine values for \(x,y,z\)
that satisfy at least one of the Cases 1, 2 or 3 while simultaneously
satisfying at least one of the Cases 4, 5 or 6.
We begin by determining if there are values for \(x,y,z\) that satisfy Case 1 while
simultaneously satisfying at least one of the Cases 4, 5 or 6.
To satisfy Case 1, \(y=7\) and \(z=9,11,12,13, \text{ or }14\), however,
each of the Cases 4, 5 and 6 do not overlap the restrictions on \(y\) and \(z\), and so there are no spinners for which
Erin beats Gina with the \(3/2/0\)
result.
Next, we determine if there are values for \(x,y,z\) that satisfy Case 2 while
simultaneously satisfying at least one of the Cases 4, 5 or 6.
To satisfy Case 2, \(z=9,11,12,13, \text{ or }
14\), however, each of the Cases 4 and 5 do not overlap the
restrictions on \(z\), and so we are
left to consider Case 2 and Case 6.
If \(x=3 \text{ or }4\), and \(y=11,12,13, \text{ or } 14\), and \(z=11,12,13, \text{ or } 14\), then the
restrictions on both Case 2 and 6 are satisfied.
Recalling that \(y<z\), there are 6
pairs \((y,z)\) satisfying these
restrictions:
\((y,z)=(11,12), (11,13), (11, 14), (12,13),
(12,14), (13,14)\).
For each of these \(6\) pairs, there
are \(2\) possible choices for \(x\) (\(x\)
can equal \(3\) or \(4\)), and thus there are \(6\times2=12\) spinners that Gina can
make.
Finally, we determine if there are values for \(x,y,z\) that satisfy Case 3 while
simultaneously satisfying at least one of the Cases 4, 5 or 6.
To satisfy Case 3, \(y=6 \text{ or }
7\), however, each of the Cases 4 and 6 do not overlap the
restrictions on \(y\), and so we are
left to consider Case 3 and Case 5.
If \(x=3 \text{ or }4\), and \(y=6\text{ or }7\), and \(z=19\text{ or } 20\), then the restrictions
on both Case 3 and Case 5 are satisfied.
Since there are \(2\) choices for each
of \(x\), \(y\) and \(z\), then there are \(2\times2\times2=8\) spinners that Gina can
make (we note that each of the \(8\)
spinners has \(x<y<z\)).
Recognizing that the \(12+8=20\) spinners are different from one another, Gina can make \(20\) different spinners so that \(p=q=r=\frac59\).