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```2016 State Competition Solutions
Are you wondering how we could have possibly thought that a Mathlete® would be able
to answer a particular Sprint Round problem without a calculator?
Are you wondering how we could have possibly thought that a Mathlete would be able
to answer a particular Target Round problem in less 3 minutes?
Are you wondering how we could have possibly thought that a particular Team Round
problem would be solved by a team of only four Mathletes?
The following pages provide solutions to the Sprint, Target and Team Rounds of the
2016 MATHCOUNTS® State Competition. These solutions provide creative and concise
ways of solving the problems from the competition.
There are certainly numerous other solutions that also lead to the correct answer,
some even more creative and more concise!
We encourage you to find a variety of approaches to solving these fun and challenging
MATHCOUNTS problems.
Special thanks to solutions author
for graciously and voluntarily sharing her solutions
with the MATHCOUNTS community
for many, many years!
2016 State Competition
Sprint Round
1. Given: ππ@ππ =
Find: [email protected]
5
ππ
2ππ+ππ
[email protected] = (2×5)+3 =
5
13
Ans.
There are 2 rectangles that are two rows
high by two columns wide.
1 here
2. Find: the number of rectangles displayed in
the diagram.
and 1 here
There are 2 rectangles that are three rows
high by one column wide.
There are 6 rectangles that are one row
high by one column wide.
Finally, there is 1 rectangle three rows high
by 2 columns wide.
There are 3 rectangles that are one row
high by two columns wide.
There are 4 rectangles that are two rows
high by one column wide.
2 here
and 2 here
The total number of rectangles is
6 + 3 + 4 + 2 + 2 + 1 = 18 Ans.
3. Given: 7π₯π₯ + 13 = 328
Find: the value of 14π₯π₯ + 13
Multiply the equation by 2.
14π₯π₯ + 26 = 656
14π₯π₯ + 26 β 13 = 656 β 13 = 643 Ans.
4. Find: the median of the positive perfect
squares less than 250.
The largest square less than 250 is 225 or
152 . That would be 15 integers so the
median is the eighth one. 82 = 64 Ans.
5. Given:
π₯π₯+5
π₯π₯β2
=
2
3
Find: π₯π₯
π₯π₯ + 5 2
=
π₯π₯ β 2 3
3π₯π₯ + 15 = 2π₯π₯ β 4
π₯π₯ = β19 Ans.
6. Given: WX=4, XY=2, YV=1 and UV=6.
Find: the absolute difference between the
areas of triangles TXZ and UYZ.
The area of triangle TXZ is equal to the area
of triangle TXU minus TZU. The area of
triangle UYZ is equal to the area of triangle
TYU minus TZU. This means the difference
in the area of triangles TXZ and UYZ is equal
to the difference between TXU and TYU.
The base of both triangles is TU. The height
of both triangles is the same β a
perpendicular from WV to TU. Therefore,
the area of each triangle is the same. The
difference between the two areas is 0. Ans.
7. Given: A bag has 4 blue, 5 green and 3 red
marbles.
Find: How many green marbles must be
added to the bag so that 75% of the
marbles are green?
There are a total of 4 + 5 + 3 = 12
marbles.
Let π₯π₯ = the number of green marbles added.
3
5 + π₯π₯
=
12 + π₯π₯ 4
20 + 4π₯π₯ = 36 + 3π₯π₯
π₯π₯ = 16 Ans.
8. Given: MD swaps out the tires on his trike
so that all four will have been used for the
same distance as he drives 25,000 miles.
Find: How many miles does each tire drive.
3
4
Each tire will be on for of the trip.
3
4
× 25000 = 18750 Ans.
9. Given: Lucy and her dad have the same
birthday. When Lucy was 15 her father was
3 times her age. On this birthday, Lucyβs
father turned twice as old and Lucy did.
Find: How old is Lucy?
When Lucy was 15 her father was 45.
Currently, her Dad is twice her age. Let π₯π₯
represent the number of years since Lucy
was 15. Then
2 × (15 + π₯π₯) = 45 + π₯π₯
30 + 2π₯π₯ = 45 + π₯π₯
π₯π₯ = 15
Her Dad is now 45 + 15 = 60, and Lucy is
15 + 15 = 30 Ans.
10. Given: The sum of three distinct 2-digit
primes is 53. Two of the primes have a
units digit of 3 and the other prime has a
units digit of 7.
Find: the greatest of the three primes.
Letβs list the 2-digit primes less than 53 that
end in 3: 13, 23, 43.
Now list the 2-digit primes less than 53 that
end in 7: 17, 37, 47.
Remove 43 and 47 because they are too
large to be one of the three primes.
13 + 23 = 36
53 β 36 = 17
So the primes are 13, 17 and 23.
The greatest of the primes is 23. Ans.
11. Given: Ross and Max weigh 184 pounds.
Ross and Seth weigh 197 pounds. Max and
Seth weigh 189 pounds.
Find: Rossβ weight.
Let ππ = Rossβ weight.
Let ππ = Maxβs weight.
Let π π  = Sethβs weight.
ππ + ππ = 184
ππ + π π  = 197
ππ + π π  = 189
2ππ + 2ππ + 2π π  = 184 + 197 + 189 = 570
Doubling the third equations, we get
2ππ + 2π π  = 189 × 2 = 378
Substituting this into the equation,
2r + 2m + 2s = 570, we get 378 + 2ππ = 570
2ππ = 570 β 378 = 192
ππ = 96 Ans.
12. Find: the least possible denominator of a
positive rational number 0. οΏ½οΏ½οΏ½οΏ½
π΄π΄π΄π΄ where π΄π΄ and
π΅π΅ are distinct digits.
οΏ½οΏ½οΏ½οΏ½
Let π₯π₯ = 0. π΄π΄π΄π΄
Then 100π₯π₯ = π΄π΄π΄π΄. οΏ½οΏ½οΏ½οΏ½
π΄π΄π΄π΄
99π₯π₯ = π΄π΄π΄π΄
π΄π΄π΄π΄
π₯π₯ =
99
99 = 11 × 9
Suppose the denominator is 9. An example
11
1
οΏ½οΏ½οΏ½
is = β 0. οΏ½11
99
9
No, this doesnβt work since π΄π΄ = π΅π΅ and that
violates the requirements. What about 11?
An example is:
9
1
οΏ½οΏ½οΏ½
= β 0. οΏ½09
99
11
Yes, in this case π΄π΄ β  π΅π΅.
11 Ans.
13. Given: It costs \$3.25 for the first mile and
\$0.45 for each
1
4
mile.
Find: How many miles can be travelled for
\$13.60
13.60 β 3.25 = 10.35
So, after the first mile there is \$10.35 left.
10.35
= 23
0.45
So thatβs an additional 23 quarter miles.
1+
23
4
3
4
=1+5 =6
3
4
Ans.
14. Given: 2 m3 of soil containing 35% sand is
mixed with 6 m3 of soil containing 15%
sand.
Find: the percent of the new mixture that is
sand
Of the 2 m3 mixture, 0.35 × 2 = 0.7 m3 is
sand, and of the 6 m3 mixture, 0.15 × 6 =
0.9 m3 is sand.
Putting the two mixtures together gives us
2 + 6 = 8 m3 of mixture, of which 0.7 + 0.9 =
1.6 m3 is sand.
1.6
8
1
5
= = 20% Ans.
15. Given: Alex runs a lap in 1 minute, 28
seconds. Becky runs a complete lap in 1
minute, 16 seconds.
Find: How many complete laps will Alex
have run when Becky passes him for the
first time?
Alex runs at a rate of 88 seconds per lap
and Becky at a rate of 76 seconds per lap.
Let ππ = the length of the lap.
After one second Alex has run
1
88
of a lap.
1
Becky has run of a lap. The difference
76
1
1
88β76
12
3
3
β =
=
=
=
76
88
76×88
76×88
19×88
1672
3
laps per second it
At a separation of
1672
1672
will take
seconds to reach one lap
3
difference.
1672
3
557
88
1
3
1
3
= 557 seconds
1
3
= 6 laps.
Therefore, Alex will have run 6 complete
laps when Becky passes him. 6 Ans.
is
16. Given: Four teams, the Beavers, Ducks,
Platypuses and Narwhals, are in a singleelimination tournament.
Find: The probability that the Ducks and the
Beavers will play each other in one of the
two rounds.
Letβs start out with the Ducks and the
Beavers playing each other in the first
round.
There are 2 possibilities for the outcome of
the Ducks versus Beavers game and 2
possibilities for the outcome of the
Platypuses versus Narwhals game. There
are 2 × 2 = 4 match-up possibilities for the
final round. In 4 out of 4 of these
possibilities the Ducks and the Beavers will
play since they meet in the first round.
If the Ducks and the Beavers do not play in
the first round, there are two different first
round scenarios. First, letβs look at the
Beavers playing the Platypuses and the
Ducks playing the Narwhals in the first
round. Again, there are 2 × 2 = 4 possible
outcomes of these matchups. Of these
possibilities, only 1 will the Beavers and the
Ducks play each other, shown here.
In the other first round scenario the
Beavers play the Narwhals and the Ducks
play the Platypuses. Again, there are 4
possible outcomes and only 1, shown here,
where the Beavers play the Ducks.
Of the three possible first round lineups,
there are 4 + 4 + 4 = 12 possible second
round matchups. Out of these matchups
the Ducks and the Beavers will end up
playing each other in 4 + 1 + 1 = 6 of the
scenarios. So the probability is
6
12
=
1
2
Ans.
17. Given: ππ(π₯π₯) is defined for positive integers
and ππ(ππ) + ππ(ππ) = ππ(ππππ) and ππ(3) = 5.
Find: ππ(27)
ππ(3) + ππ(3) = ππ(3 × 3)
ππ(9) = 5 + 5 = 10
ππ(3) + ππ(9) = ππ(3 × 9)
ππ(27) = 5 + 10 = 15 Ans.
18. Given: In rectangle ABCD, AB = 6 and
AD = 5. AC is extended to point E so that
οΏ½οΏ½οΏ½οΏ½
οΏ½οΏ½οΏ½οΏ½ .
π΄π΄π΄π΄ is congruent to πΆπΆπΆπΆ
Find: the length of οΏ½οΏ½οΏ½οΏ½
π·π·π·π·
Since we extend C to E and π΄π΄π΄π΄ = πΆπΆπΆπΆ, then
CE is just another diagonal of a 5 × 6
rectangle.
it. This tells us that 2 must be added, which
will be carried over from the thousands
column. So B + 7B + N = 21, where N
represents a single digit integer number
carried over from the hundreds column.
Rearranging we get
8B + N = 21
If B is 1, N would be 13, if B is 2, N would be
5, and if B is 3, N would be β3. The only way
to get N to be a single digit positive integer
is if B = 2. This means our multiplication
would be 127 × 721 which equals 91567.
721 Ans.
20. Given: Diagonal XZ of rectangle WXYZ is
divided into 3 segments each of length 2.
MW and NY are parallel and perpendicular
to XZ.
οΏ½οΏ½οΏ½οΏ½ is 12 and the length of οΏ½οΏ½οΏ½οΏ½
The length of π·π·π·π·
πΉπΉπΉπΉ
is 5. Triangle DEF is a 5-12-13 right triangle.
οΏ½οΏ½οΏ½οΏ½ is
Therefore, the length of π·π·π·π·
13. Ans.
19. Given: The digits of a 3-digit number are
reversed. The product of the new and
original integers is 91567.
Find: the new integer.
Since the last digit of our product is a 7, letβs
look at multiply a 7 and a 1. The product
will look like this:
Looking at the far left column, the tenthousands place, the product of our two
numbers has a 9 here but we see a 7 above
Find: the area of WXYZ.
First letβs draw in a line parallel to XZ from
point R to point P, as shown, to create
rectangle MRPN and triangle RPY.
We know mβ MXR β mβ PRY, RP = MN = 2
and MR = NP. Triangles XMR and RPY are
congruent (by ASA), and, therefore, XR = RY
and MR = PY.
We also can determine, by AAA similarity,
that triangle RPY is similar to triangle YNZ.
Using this information, we can set up two
ratios to solve for unknown distances.
2/NY = PY/2 and NY = 2PY
Solving we get:
4 = NY × PY
4 = 2PY × PY
2 = PY2
PY = β2
And therefore
NY = 2β2
Now we can use the Pythagorean theorem
to find XY and YZ.
PY2 + 22 = RY2
(β2)2 + 4 = RY2
6 = RY2
RY = β6
Therefore, XY = 2RY = 2β6.
NY2 + 22 = YZ2
(2β2)2 + 4 = YZ2
12 = YZ2
YZ = 2β3.
The area of rectangle WXYZ is
XY × YZ = 2β6 × 2β3 = 4β18 = 12β2 Ans.
21. Given: A spinner is divided into 5 sectors.
The angles of sections 1 through 3 measure
60°. The angles of sections 4 and 5
measure 90°.
Find: the probability that at least one spin
lands on an even number if the spinner is
spun twice.
There are two even numbers, 2 and 4. The
section labeled 2 is 60°, and the section
labeled 4 is 90°. The probability of one spin
landing on an even number is, therefore,
60+90
360
=
150
360
5
.
12
=
The probability of one spin landing on an
odd number is 1 β
5
12
=
7
.
12
If you are spinning twice, there are three
ways to get at least one even. You can spin
two evens, an even and then and odd or an
odd and then an even.
The probability of spinning two evens is
5
12
×
5
12
=
25
.
144
The probability of spinning an even then an
odd is
5
12
×
7
12
=
35
.
144
The probability of spinning an odd then an
even is
7
12
×
5
12
=
35
.
144
So the probability of spinning at least one
even is
25
35
35
+
+
144
144
144
=
95
144
Ans.
22. Given: The student council has 8 members
consisting of 4 officers and 4 other
members. The 4 officers must sit together
at the circular table.
Find: How many distinguishable circular
seating orders are possible?
Let o1 through o4 be the officers and m1
through m4 be the members.
There are 4! = 24 ways that o1 through o4
can be ordered. Similarly, there are 24
ways that m1 through m4 can be ordered.
24 × 24 = 576 Ans.
23. Given: a chip is placed in the upper-left
corner square of a 15 × 10 grid. The chip
can move in an L-shaped pattern.
Find: the minimum number of L-shaped
moves to move the chip to the square
marked βXβ.
You need to move 9 down and 14 to the
right.
Moving 14 to the right will take at least
7 L-shaped patterns that move 2 to the
right and 1 down.
But, if we do 7 of those we end up at (15, 8)
given that we started at (1, 1) and must end
up at (15, 10).
So, can we do it with 8 L-shaped moves?
We can do 6 moves of (2 right, 1 down) to
get us to (13, 7).
One move of (1 right, 2 down) will get us to
(14, 9) leaving us with having to go (1, 1)
more but thereβs no L-shaped move to get
us there.
Again, after 6 moves, we can move
(1 left, 2 down) to (12, 9). From there we
are still going to need 2 more moves
because
15 β 12 = 3.
It looks like it will take us 9 moves to get
this to work.
We start at (1, 1).
Now letβs do 6 (2 right, 1 down) to get to
(13, 7)
This has positioned us to move 2 more Lshaped moves of 2 right and 1 down.
9 Ans.
24. Given: On line segment π΄π΄π΄π΄, π΅π΅ is the
midpoint of segment π΄π΄π΄π΄ and π·π· is the
midpoint of segment πΆπΆπΆπΆ. π΄π΄π΄π΄ = 17 and
π΅π΅π΅π΅ = 21.
Find: the length of segment π΄π΄π΄π΄.
π΄π΄π΄π΄ = π΅π΅π΅π΅
πΆπΆπΆπΆ = π·π·π·π·
Let π₯π₯ = π΄π΄π΄π΄
Let π¦π¦ = πΆπΆπΆπΆ
Then the length of segment π΄π΄π΄π΄ is 2π₯π₯ + π¦π¦
and the length of π΅π΅π΅π΅ is π₯π₯ + 2π¦π¦.
Since AD = 17 and BE = 21, we have
2π₯π₯ + π¦π¦ = 17
π₯π₯ + 2π¦π¦ = 21
3π₯π₯ + 3π¦π¦ = 38
3(π₯π₯ + π¦π¦) = 38
38
π₯π₯ + π¦π¦ =
3
The length of π΄π΄π΄π΄ is 2π₯π₯ + 2π¦π¦.
2π₯π₯ + 2π¦π¦ = 2(π₯π₯ + π¦π¦) = 2 ×
Now the 7th L-shaped move is 2 left and 1
down getting us to (11, 8).
38
3
=
76
Ans.
3
25. Given: There are 12 different mixed
numbers that can be created by substituting
three of the numbers 1, 2, 3 and 5 for a, b,
ππ
ππ
and c in the expression ππ where ππ < ππ.
Find: the mean of the 12 numbers.
These are the 12 numbers:
2 2 3
1 ,1 ,1
3 5 5
1 1 3
2 ,2 ,2
3 5 5
1 1 2
3 ,3 ,3
2 5 5
1 1 2
5 ,5 ,5
2 3 3
Add up all the mixed numbers whose
denominator is 2.
1
1
3 +5 =9
2
2
Add up all the mixed numbers whose
denominator is 3.
2
1
1
2
6
1 + 2 + 5 + 5 = 13 = 15
3
3
3
3
3
Add up all the mixed numbers whose
denominator is 5.
2
3
1
3
1
2
1 +1 +2 +2 +3 +3 =
5
5
5
5
5
5
2
12
= 14
12
5
5
2
2 192
9 + 15 + 14 = 38 =
5
5
5
Now determine the mean.
192
5
12
=
192
5×12
=
192
60
=3
12
60
=3
1
5
Ans.
26. Given: 738 consecutive integers are added.
The 178th number in the sequence is
4,256,815.
Find: the remainder when the sum is
divided by 6.
First, notice 738 is evenly divisible by 6
because it is even and its digits sum to 18, a
multiple of 3.
The formula for adding the first n
consecutive integers is (n)(n + 1)/2.
Let x represent the first integer in this
sequence.
The sum of these 738 integers will be
(738)(738 + 1)/2 + 738x
Because we know 738 is divisible by 6, we
know 738x will also be divisible by 6 and
therefore have remainder 0. We donβt need
to even find the value of x.
Now lets look at what still remains in the
sum
(738)(738 + 1)/2 = (369)(738 + 1) =
369 × 738 + 369
We know 6 evenly divides 369 × 738
So the remainder for this integer sum is
equal to the remainder when 369 is divided
by 6.
369 ÷ 6 = 61 remainder 3. Ans.
27. Given: A(-5,0) and B(5,0)
Find: How many points X are there such
that XA and XB are both positive integer
distances, each less than or equal to 10?
Letβs assume that the point X is not on the
x-axis. Then itβs all about creating a triangle
where one side is 10 and the other two are
10 or smaller.
being 10. The max length for the third side
would be 19 but we are limited to lengths
β€ 10. So the third side can be 1 through 10
for a total of 10 triangles.
Now let one of the other two sides be 9.
The max length for the third side is 18.
Again, with the limitations, and taking into
account that the first two lines differ by
one, the third side can be 2 through 10 or 9
possibilities.
Similarly for the second side having a length
of 8, the third side can be 3 through 10 or 8
possibilities.
Thereβs the pattern:
Second side is 7; third side is 4 through 10.
Second side is 6; third side is 5 through 10.
Second side is 5; third side is 6 through 10.
Second side is 4; third side is 7 through 10.
Second side is 3; third side is 8 through 10.
Second side is 2; third side is 9 through 10.
Second side is 1, third side is 10.
1 + 2 + β― + 9 + 10 = 55
All of these triangles have been above the
x-axis. But we can have the same triangles
reflected over the x-axis.
That gives us, so far, 55 + 55 = 110
possibilities.
All of these triangles assumed that X was
not on the x-axis. But if it is, you could have
X at (β4,0), (β3,0), β¦ , (3,0), (4,0)
Thatβs another 9 points.
110 + 9 = 119 Ans.
28. Given: ππ(ππ) = ππ β ππ! + ππ, where ππ and ππ are
positive integers. The range of ππ contains
two numbers that differ by 20.
Find: the least possible value of ππ(1).
Two values, f(y) and f(x) must differ by 20.
f(y) β f(x) = a β y! + b β (a β x! + b)
f(y) β f(x) = a β (y! β x!)
20 = a β (y! β x!)
Since all the variable are positive integers,
letβs look at pairs integers that multiply to
be 20: 1 × 20, 2 × 10, 4 × 5.
We want to minimize the value of a in order
to minimize the value of f(1).
Letβs look at the value of the first 5
factorials:
1! = 1, 2! = 2, 3! = 6, 4! = 24, 5! = 120
If a is 1, then we need two factorials that
differ by 20. There arenβt two.
If a is 2, then we need two factorials that
differ by 10. There arenβt two.
If a is 4, then the factorials 6! β 1! = 5 will
work.
So we know a is 4, b can be any positive
integer but we want to minimize f(1) so lets
pick b to be 1.
f(1) = 4 β 1! + 1 = 5 Ans.
29. Given: A list of numbers 1, 2, β¦ 9999 where
the digits 0 through 9 are replaced with the
letters A through J. The resulting list of
9999 strings is sorted alphabetically.
Find: How many strings appear before
βCHAIβ?
Alphabetic sorting is just what you find in
the dictionary.
- Single digit numbers
1 and 2 (i.e., B and C) will be before CHAI.
Thatβs two numbers.
- Double digit numbers.
with B. 10-19 (BA-BJ) will appear before
CHAI. For double digit numbers starting
with C you have 20-27 (CA-CH) that will
appear before CHAI. Thatβs a total of 18
numbers.
- Three digit numbers.
before CHAI. 200-269 (CAA-CGJ) will appear
before CHAI. Moving to 270, thatβs the only
one we can have (CHA). Thatβs a total of
100 + 70 + 1 = 171
- Four digit numbers.
As usual we start with B β we have 10001999 (BAAA-BJJJ) for 1000 numbers.
Then we have 2000-2699 (CAAA-CGJJ) or
700 numbers. Finally we have 2700-2707)
(CHAA-CHAH) or 8 numbers.
Thatβs a total of 1000 + 700 + 8 = 1708
Now letβs add them all up:
2 + 18 + 171 + 1708 = 1899 Ans.
30. Given: A 12-sided die has the shape of a
hexagonal bipyramid. Each has a hexagonal
base with side 1 and height 1. The die is
rolled and lands on one of its triangular
faces.
Find: How high off the ground is the
opposite face?
This is what a hexagonal bipyramid looks
like.
β=1
π₯π₯ is just half of the height of a hexagon, or
Now the problem says tht the bipyramid
has landed on one of its sides.
So letβs rotate it onto its side.
Doesnβt it look, in 2-dimensional space, like
a rhombus?
So what we have is a rhombus and what we
are looking for is the height of the rhombus.
If you think of one of the triangles of one of
β3
2
So now we have to find π¦π¦, which is the
height of the triangles of the pyramid and
the hypotenuse of the triangle formed with
1 and π₯π₯.
12 + π₯π₯ 2 = π¦π¦ 2
2
β3
1 + οΏ½ οΏ½ = π¦π¦ 2
2
3 7
π¦π¦ 2 = 1 + =
4 4
7 β7
π¦π¦ = οΏ½ =
2
4
The sides of the rhombus are, therefore,
the pyramids lying flat on the surface the
height of the triangle is the length of the
top and bottom of the rhombus.
The triangles that connect the bottom
triangle (lying flat on the surface) to its
opposite side also use the height of the
triangle as the length of the other two
sides. Thatβs why this is a rhombus.
So we are given that the height of the
hexagonal pyramid is 1 and that the
hexagonal βconnectorβ between the two
pyramids has sides of length 1.
β7
2
Great! Now we just need the height of the
rhombus. But thatβs not so easy. Or is it.
We do know of two equations to get the
area of the rhombus.
π΄π΄ = ππβ where ππ is the side length of the
rhombus and β is the height.
1
π΄π΄ = ππ1 ππ2
2
where ππ1 and ππ2 are the diagonals of the
rhombus. Hmm, maybe we do know those.
One of those diagonals, if you look back at
the picture of the hexagonal bipyramid, is
just the heights of each pyramid or
2×1=2
The other one is just the height of the
hexagon or β3.
Now we can finally find the height!
1
ππβ = ππ1 ππ2
2
1
β7
β = × 2 × β3 = β3
2
2
2 × β3 2 × β3 × β7
=
β=
β7
β7 × β7
β=
2β21
7
Ans.
Target Round
1. Given: The bug crawls 1 unit up, 2 units
right, 3 units down and 4 units left. The
bug repeats this set of moves 2015 more
times.
Find: If the coordinates of the bugβs final
location are (ππ, ππ), find ππ + ππ.
Starting at (0, 0) moving up 1 unit puts the
bug at (0, 1). Moving to the right 2 units
puts the bug at (2,1). Moving down three
units puts the bug at (2, β2). Moving 4
units to the left puts the bug at (β2, β2).
So the bug started at (0, 0) and ended at
(β2, β2).
Since the bug will continue to move in the
same pattern, we can see that for each
move β2 is added to both the x and y
coordinates.
At the end of 2016 moves the bug can be
found at the coordinates (β4032, β4032).
ππ + ππ = β4032 + β4032 = β8064 Ans.
2. Given: A rectangular piece of cardboard
measures 6 by 8. It is trimmed identically
on all four corners. Each corner is a
quarter-circle.
Find: the perimeter of the resulting figure.
The four quarter-circles form a complete
circle with radius 3. Its perimeter is
2ππππ = 6ππ
Since the length of the longer side is 8 and
we have 2 quarter-circles with radius 3,
what remains of the original length is
8 β (2 × 3) = 2
Therefore, the perimeter of the resulting
figure is 2 + 2 + 6ππ = 6ππ + 4 Ans.
3. Given: A coin is flipped four times.
Find: the probability of getting 2 heads and
two tails.
This one is simple enough to just list the
variations.
HHTT, HTHT, HTTH, TTHH, THTH, THHT
1 4
2
6×οΏ½ οΏ½ =
6
24
=
6
16
3
8
= = 37.5% Ans.
4. Given: The areas of three faces of a
rectangular prism are 54, 24 and 36.
Find: the length of the space diagonal.
Let ππ = the length of the prism.
Let π€π€ = the width of the prism.
Let β = the height of the prism. Then
ππππ = 54
π€π€β = 24
ππβ = 36
ππ 2 π€π€ 2 β2 = 54 × 24 × 36 =
9×6×4×6×9×4=
92 × 62 × 42
So, we have a 9 × 6 × 4 rectangular prism.
There are less ways to get sums of 10 or
more than less than 10, so letβs look at the
probability of drawing those.
We can draw 4 + 6, 5 + 5, 5 + 6 or 6 + 6
The probability of drawing two of the same
specific card is
First letβs find the hypotenuse of the
rectangle with sides 4 and 9.
Let π₯π₯ = the length of the red line.
42 + 92 = π₯π₯ 2
π₯π₯ 2 = 16 + 81 = 97
Now we can find the length of the space
diagonal.
Let π¦π¦ = the length of the space diagonal.
π¦π¦ 2 = π₯π₯ 2 + 62
π¦π¦ 2 = 97 + 36 = 133
π¦π¦ = β133 Ans.
5. Given: Two 8 by 10 inch sheets are placed
on top of a 2 by 3 foot table. The area of
the table not covered by the paper is
708 in2.
Find: the area of the overlap of the two
sheets of paper.
A 2 foot by 3 foot rectangular table is a
24 inch by 36 inch rectangular table. The
area of the table is 24 × 36 = 864 in2. The
area not covered by the paper is 708 in2
leaving 864 β 708 = 156 in2 that is
covered by the paper. If each of the sheets
of paper did not overlap each other, the
amount of area taken by the paper would
be 8 × 10 × 2 = 160 in2. That is 4 more
than 156. 4 Ans.
6. Given: Each card in a deck of cards contains
a number between 2 and 6. The deck has 4
cards of each value. Two of the cards are
chosen at random without replacement.
Find: the probability that the sum of their
values is less than 10.
4
20
×
3
19
=
3
95
The probability of drawing a specific pair of
8
20
×
4
19
8
95
=
22
95
two different cards is
=
8
95
So for the four possible sums of 10 or more,
we have two that are the same number and
two that are different numbers. The
probability of picking one of these four
sums is 2 ×
3
95
+2×
So the probability of drawing two cards
whose sum is less than 10 is
1β
22
95
=
73
Ans.
95
7. Given: (π₯π₯ 2 + 3π₯π₯ + 6)(π₯π₯ 2 + ππππ + ππ) =
π₯π₯ 4 + πππ₯π₯ 2 + ππ for integers ππ, ππ, ππ, ππ.
Find: the product of ππ and ππ.
(π₯π₯ 2 + 3π₯π₯ + 6)(π₯π₯ 2 + ππππ + ππ) =
π₯π₯ 4 + πππ₯π₯ 3 + πππ₯π₯ 2 + 3π₯π₯ 3 + 3πππ₯π₯ 2 + 3π₯π₯π₯π₯ +
6π₯π₯ 2 + 6ππππ + 6ππ =
π₯π₯ 4 + (ππ + 3)π₯π₯ 3 + (ππ + 3ππ + 6)π₯π₯ 2 +
(3ππ + 6ππ)π₯π₯ + 6ππ =
π₯π₯ 4 + πππ₯π₯ 2 + ππ
The coefficient of π₯π₯ 3 must be 0. Therefore,
ππ + 3 = 0
ππ = β3
The coefficient of π₯π₯ 2 is ππ. Therefore,
ππ = ππ + 3ππ + 6 = ππ + (3 × β3) + 6 =
ππ β 9 + 6 = ππ β 3
The coefficient of π₯π₯ must be 0. Therefore,
3ππ + 6ππ = 0
3ππ + (6 × β3) = 0
3ππ β 18 = 0
3ππ = 18
ππ = 6
ππ = ππ β 3 = 6 β 3 = 3
Finally, ππ = 6ππ = 36
ππ × ππ = 3 × 36 = 108 Ans.
8. Given: The polygon is constructed from two
squares and six equilateral triangles, each of
side length 6. The polygon can be folded
into a polyhedron by creasing along the
dotted lines and joining adjacent edges as
indicated by the arrows.
Find: the volume of the resulting
polyhedron.
2
β2 + οΏ½3β2οΏ½ = 62
β2 + 18 = 36
β2 = 18
β = 3β2
The created polyhedron is twice the volume
of the right square pyramid.
β
3β2
= 2(6)2
=
3
3
72β2 Ans.
2ππ = 2ππ 2
Team Round
When you fold the polygon what you get is
two square pyramids connected at an
angle.
The volume of the polyhedron is the sum of
the volume of two identical square
pyramids. The volume of a square pyramid
is ππ = ππ 2
β
3
where ππ is a base edge and β is the height.
The base edge, ππ, is just 6.
To find β, we need to find half the diagonal
of the square.
Let π₯π₯ = half the diagonal of the square.
Then 62 + 62 = (2π₯π₯)2
36 + 36 = 4π₯π₯ 2
4π₯π₯ 2 = 72
π₯π₯ 2 = 18
π₯π₯ = 3β2
Now, we can determine the height.
β2 + π₯π₯ 2 = 62
1. Given: The end of each blade of a fan is two
feet from the center of the fan. Each blade
makes three full rotations per second.
Find: the speed, in miles per hour, of the
end of one of the blades.
One rotation is 2ππππ = 4ππ feet.
The blade travels 12ππ feet in three
rotations (one second) or 720ππ feet in one
minute. It travels 43,200ππ feet in one hour.
In miles per hour this is
43,200ππ
β 25.7
5280
25.7 Ans.
2. Given: π΄π΄, π΅π΅, πΆπΆ, π·π·, πΈπΈ in 0. π΄π΄π΄π΄π΄π΄ and 0. π·π·π·π·
represent the digits 1, 2, 3, 4 and 5, in some
order.
Find: the least possible absolute difference
between the numbers 0. π΄π΄π΄π΄π΄π΄ and 0. π·π·π·π·.
To minimize the difference, we need to
make 0.ABC as close to 0.DE as close in
value as possible.
The smallest possible difference between A
and D is 1.
If D < A, then we want to maximize the
value of E and minimize the value of BC. The
largest possible value for E is 5 and the
smallest value for BC is 12. This makes the
absolute difference 0.412 β 0.35 = 0.062.
Similarly, if A < D, then we want to
maximize BC and minimize E. The largest
possible value for BC is 54 and the smallest
possible value for E is 1. This makes the
absolute difference 0.31 β 0.254 = 0.056.
The least possible absolute difference is
0.056 Ans.
3. Given: 70.8% of Earthβs surface is covered
with water. Mars is entirely covered by land
and the amount is approximately the same
as the amount of land on Earth.
Find: What percent of Earthβs radius is
The surface area of a sphere is 4ππππ 2
Let πππΈπΈ = the radius of Earth.
Let ππππ = the radius of Mars.
Then
0.292 × 4ππ(πππΈπΈ )2 = 4ππ(ππππ )2
0.292(πππΈπΈ )2 = (ππππ )2
ππππ β 0.54037πππΈπΈ
As a percentage, the radius of Mars is 54%
of the radius of earth. 54% Ans.
4. Given: Billy wrote a sequence of 5 numbers,
each an integer between 0 and 4. Penny
wrote a sequence of five numbers that
included the number of 0βs, 1βs, 2βs, 3βs and
4βs in Billyβs sequence. Pennyβs sequence is
the same as Billyβs.
Find: the sequence.
The sum of all 5 numbers must be 5
because the sequence is a count of how
many times each number appears and we
know there are only 5 numbers total
present in the sequence. You canβt have 5
of any number because 5 is not one of the
numbers. You also canβt have 4 of any
number since only one other number could
be used. And weβre going to have to have
some values that we will have none of. For
example, 11112 would translate to 04100.
Similarly, you canβt have 3 of one number
and 2 of another number or 3 of one
number, one of another number and one of
a third number. For example, 01112 would
translate to 13100.
Letβs look at 2 of one number, 2 of another
number and 1 of a third number.
2 1 2 0 0 would translate to 2 1 2 0 0.
2 1 2 0 0 Ans.
5. Given: 3π₯π₯ + 3π₯π₯+2 = 9π₯π₯ + 9π₯π₯+2
Find: 3π₯π₯
3x + (3x32) = (3 × 3)x + (3 × 3)x(3 × 3)2
3x(1 + 32) = 3x3x + 3x3x3232
3x(10) = 3x3x(1 + 34)
10 = 3x(82)
3x = 10/82 = 5/41 Ans.
6. Given: A dime has a mass of 2,268 grams. A
nickel has a mass of 5,000 grams. Diane has
a bag of dimes and Nick has a bag of nickels.
Both bags of coins have the same mass.
Find: how many times the total value of
Nickβs coins is Dianeβs coins.
Let π₯π₯ = the number of dimes.
Let π¦π¦ = the number of nickels.
2268π₯π₯ = 5000π¦π¦
π₯π₯ 5000
=
β 2.2045
π¦π¦ 2268
There are 2.2045 times as many dimes as
there are nickels. The value of dimes to
nickels is
10π₯π₯
5π¦π¦
π₯π₯
π¦π¦
= 2 β 4.409 β 4.4 Ans.
7. Given: For each positive integer ππ,
ππππ = 9ππ + 2 and
ππππ = 7ππ + 3
The values common to both sequences are
written as a sequence. The nth term can be
expressed as pn+q.
Find: the value of ππ β ππ
Since an is multiples of 9 shifted by 2 and bn
is multiples of 7 shifted by 3. We just need
to find the first common term, then we
know all common terms following that will
just be 9 × 7 = 63 more than the previous.
Letβs compare the first few values of ππππ :
11, 20, 29, 38, 47 to the first few values of
ππππ : 10, 17, 24, 31, 38, 45.
The first common value is 38. So the
sequence of common values will be
represented by the formula 38 + 63(n β 1) =
38 + 63n β 63 = 63n β 25.
Therefore, the value of p is 63 and the value
of q is β25, and p β q = 63 β (β25) = 88 Ans.
8. Given: Quadrilateral APBQ has vertices
π΄π΄(0,0) and π΅π΅(8,0). Vertices P and Q lie on
the line 4π₯π₯ + 3π¦π¦ = 19. ππππ = 3
Find: the area of quadrilateral π΄π΄π΄π΄π΄π΄π΄π΄.
The slope of the line 4x + 3y = 19 is β4/3.
If we draw in a right triangle with
hypotenuse PQ, then we know the ratio of
its height to its base is equal to the slope,
rise over run, of the line (note: we can
ignore the negative because we are talking
h/b = 4/3 β h = 4/3 × b.
We also know, since it is a right triangle,
that b2 + h2 = 32.
We have two equations and two variables,
so we can solve for the values of h and b.
b2 + (4/3 × b)2 = 9
b2 + 16/9 × b2 = 9
25/9 × b2 = 9
b2 = 81/25
b = 9/5
Therefore, h = 4/3 × 9/5 = 12/5.
The area of the quadrilateral is the area of
the triangle APB minus the area of triangle
AQB. The difference in the height of these
two triangles heights is h. So the area of the
1/2 × 8 × h = 4 × h = 4 × 12/5 = 48/5 Ans.
9. Given: 8 blue and 5 orange tiles are
arranged in an ordered line such that the
tile on the left must be blue and every tile
must be adjacent to at least one tile of the
same color.
Find: How many different arrangements
are possible if all 13 tiles must be used?
The requirements are that B (for blue) must
me the first tile (starting from the left) and
that Bβs and Oβs (for orange) must always
come in a minimum of two tiles of the same
color. We have 5 orange tiles so weβll be
able to have them in one group of 2 and
one group of 3 (since we canβt have a group
of 1) and, of course, a group of 5. So letβs
- BBBBBBBBOOOOO (#1)
- BBBBBBOOOOOBB (#2)
- BBBBBOOOOOBBB (#3)
- BBBBOOOOOBBBB (#4)
- BBBOOOOOBBBBB (#5)
- BBOOOOOBBBBBB (#6)
So much for the group of 5 tiles.
Now letβs work with 2 orange tiles always
come before the group of 3 orange tiles.
- BBOOBBOOOBBBB (#7)
- BBOOBBBOOOBBB (#8)
- BBOOBBBBOOOBB (#9)
- BBOOBBBBBBOOO (#10)
Move the 2 orange tiles to the right by 1.
- BBBOOBBOOOBBB (#11)
- BBBOOBBBOOOBB (#12)
- BBBOOBBBBBOOO (#13)
And one more move to the right.
- BBBBOOBBOOOBB (#14)
- BBBBOOBBBBOOO (#15)
And one more to the right
- BBBBBOOBBBOOO (#16)
And one more
- BBBBBBOOBBOOO (#17)
Now letβs work with 3 orange tiles always
before the 2 orange tiles. Inste
- BBOOOBBOOBBBB (#18)
- BBOOOBBBOOBBB (#19)
- BBOOOBBBBOOBB (#20)
- BBOOOBBBBBBOO (#21)
Now move the group of 3 Oβs to the right by
one tile.
- BBBOOOBBOOBBB (#22)
- BBBOOOBBBOOBB (#23)
- BBBOOOBBBBBOO (#24)
Now move the three tiles one more
position to the right.
- BBBBOOOBBOOBB (#25)
- BBBBOOOBBBBOO (#26)
And one more to the right.
- BBBBBOOOBBBOO (#27)
And one more
- BBBBBBOOOBBOO (#28)
28 Ans.
10. Given: A fountainβs diameter is 10 feet. The
grass extends out from the edge of the
fountain 20 feet. Grass seed comes in bags
that cover 300 sq. ft. of grass.
Find: How many whole bags of grass seed
are needed to seed around the fountain?
The radius of the fountain itself is
10
2
= 5.
The radius of the entire area including the
grass around the fountain is
20+10+20
2
=
50
2
= 25
The area of the fountain is
ππππ 2 = ππ52 = 25ππ
The area of the fountain and the grass is
ππ252 = 625ππ
This makes the area of the grass
625ππ β 25ππ = 600ππ β 1963.49375
square feet. Each bag of grass covers 300
square feet.
1963.49375
300
β 6.554979 bags.
This means we must have 7 bags to cover
the area completely.
7 Ans.
```
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