(1)
Team Level 4 | 09/11/2010
In each stage of the pattern shown, the number of black tiles is one more
than the number of white tiles. Further, the side length at each stage increases by 2. How
many black tiles will the 16th stage contain?
Stage 1
(2)
Stage 2
Stage 3
A dart is thrown at the dart board shown and is equally likely to land
anywhere inside the large circle of radius 10. What is the probability that it will land inside
the square but not inside the inner circle? Express your answer rounded to the nearest
whole percent.
10
(3)
What is the greatest number of non-overlapping T-shaped tetrominoes, as
shown below, that can be placed on the 4 5 grid shown: that is, how many can be placed
so there is no overlap and no overhang? The small squares are all congruent.
(tetromino)
(4)
Unit cubes are glued together to make a cube several units on each side.
Some of the faces of this large cube are painted. When the cube is taken apart, there are
exactly 45 unit cubes without any paint. How many unit cubes were used to create the
larger cube?
(5)
Right 4ABC with legs AB = 3 cm and CB = 4 cm is rotated about one of
its legs. What is the greatest possible number of cubic centimeters in the volume of the
resulting solid? Express your answer to the nearest whole number.
(6)
Marianna has only nickels and quarters in her piggy bank. Their combined
value is $9.15. Their combined weight is one pound. Ninety nickels weigh one pound.
Eighty quarters weigh one pound. How many nickels does Marianna have in her piggy
bank?
(7)
The XY Z company has to pay $5,000 for rent each month. In addition,
their monthly electricity bill is $1.45 per kilowatt of power used. If the total cost for both
rent and electricity in January was $16,520.25, how many kilowatts of electricity did they
use?
(8)
One night two cylindrical wax candles of dierent heights and dierent
diameters were lit. One of the candles was 20 cm taller than the other. They were both lit
at the same time and each burned at a steady rate. Five hours after they were lit they were
both the same height. The taller one burned all of its wax six hours after it was lit, and the
shorter one burned all of its wax 10 hours after it was lit. What was the ratio of the
original height of the shorter candle to the original height of the taller candle? Express your
answer as a common fraction.
(9)
If n = 23 32 5, how many even positive factors does n have?
(10)
Circle T has a circumference of 12 inches, and segment XY is a diameter.
If the measure of angle T XZ is 60 , what is the length, in inches, of segment XZ ?
Y
T
X
Z
(11)
Sue is 12 years old. Sue says, I'm thinking of a positive three-digit multiple
of 65 whose digits add to my younger sister's age. Carlos correctly complains, I can't gure
out what your number is. Sue then says, My number is divisible by the sum of its digits.
Carlos says, Thanks, I've got it now. What is Sue's number?
(12)
The areas of the two adjacent squares are 256 square inches and 16 inches,
respectively, and their bases lie on the same line. What is the number of inches in the
length of the segment that joins the centers of the two inscribed circles? Express your
answer as a decimal to the nearest tenth.
(13)
On the game board below Kendra will start at the center of the board. For
each turn she will spin this spinner with four congruent sectors once, and then she will
move one space in the direction indicated on the spinner. The \Start" square does not
have a numerical value, but Kendra may land on it during her turns. What is the probability
that the sum of the numbers in the spaces on which she will land will be exactly 30 after
her third complete turn? Express your answer as a common fraction.
Move Move
Right Left
Move Move
Down Up
(14)
10
5
20
5
5
10
5
20
20
15
10
10
10
15
20
5
5
10
Start
10
5
5
20
15
10
10
10
15
20
20
5
5
5
20
10
5
10
What is the number of square centimeters in the area of the quadrilateral?
Express your answer rounded to the nearest whole number.
5 cm
10 cm
4 cm
(15)
A group of 25 friends were discussing a large positive integer. \It can be
divided by 1," said the rst friend. \It can be divided by 2," said the second friend. \And
by 3," said the third friend. \And by 4," added the fourth friend. This continued until
everyone had made such a comment. If exactly two friends were incorrect, and those two
friends said consecutive numbers, what was the least possible integer they were discussing?
(16)
A cereal company sets its prices by charging a xed rate per ounce of cereal
and a xed price for the empty box, regardless of its size. A box containing 12 ounces of
cereal costs $3:35 and a box with 18 ounces of cereal costs $4:67. What is the number of
cents in the xed price of the empty box?
(17)
A spiral staircase turns 270 as it rises 10 feet. The radius of the staircase is
3 feet. What is the number of feet in the length of the handrail? Express your answer as a
decimal to the nearest tenth.
(18)
If a ladybug walks on the segments of the diagram from point A to point B
moving only to the right or downward, how many distinct paths are possible?
A
B
(19)
If x and y are two dierent digits and z is any natural number, for how
many combinations of x and y will (10z )x + y be divisible by 9?
(20)
The mean of a set of ve positive integers is 1.5 times the median. If three
of the integers in the set are 24, 52 and 86, and one of them is the median, what is the
sum of all distinct possible sums of the other two integers?
Copyright MATHCOUNTS Inc. All rights reserved
Answer Sheet
Number
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Answer
481 black tiles
14
4 tetrominoes
125
50
63 nickels
7945 kilowatts
1/3
18
6 inches
910
11.7 inches
5/16
40 square centimeters
787386600
71 cents
17.3 feet
55 paths
10
729
Problem ID
CB44
CA041
B0131
1A231
BBB01
A043
42522
43AC
104C
2B5C
AA5
B2011
2DC
D5A11
35041
45D21
3C44
31312
340D1
51312
Copyright MATHCOUNTS Inc. All rights reserved
Solutions
(1)
481 black tiles
(2)
14
(3)
4 tetrominoes
(4)
125
(5)
50
(6)
63 nickels
ID: [CB44]
No solution is available at this time.
ID: [CA041]
No solution is available at this time.
ID: [B0131]
No solution is available at this time.
ID: [1A231]
No solution is available at this time.
ID: [BBB01]
No solution is available at this time.
ID: [A043]
Let the number of nickels Marianna has be n and the number of quarters she has be q . If
ninety nickels weigh one pound, then each nickel weighs 1=90 pounds. If eighty quarters
weigh one pound, then each quarter weighs 1=80 pounds. We have the two equations
:05n + :25q = 9:15
1
1
n+ q =1
90
80
First, we make the two above equations look nicer by multiplying both sides of the rst
equation by 20 and multiplying both sides of the second equation by 720. The equations
become
n + 5q = 183
8n + 9q = 720
To avoid large numbers, we will solve for q rst and then solve for n as desired. From the
rst equation, we have n = 183 5q . Substituting this into the second equation to
eliminate n, we have 8(183 5q ) + 9q = 720 ) q = 24. Plugging this value of q in to the
rst equation to solve for n, we get n = 183 5(24) = 63. Thus, Marianna has 63 nickels.
(7)
7945 kilowatts
(8)
1/3
(9)
ID: [42522]
If x is the number of kilowatts of electricity that the company used in January, then the
company paid 5000 + 1:45x dollars for that month. Setting 5000 + 1:45x = 16520:25; we
nd x = (16520:25 5000)=1:45 = 7945 .
ID: [43AC]
If we let x equal the height of the shorter candle, then x + 20 will be the height of the
larger candle. We know the taller candle burned all of its wax six hours after it was lit, so it
burns at a rate of x +20
6 cm/hour. The shorter candle burned all of its wax 10 hours after it
was lit, so it burns at a rate of x=10 cm/hour. Five hours after the shorter candle was lit,
it was exactly half burned, so x=2 cm remained. Exactly x +20
6 of the taller candle remained.
We know the candles are the same height at this point, so
x x + 20
=
:
2
6
Cross-multiplying gives 6x = 2x + 40, so x = 10. Therefore, the shorter candle was 10 cm
to start, and the taller candle was 10+20=30 cm to start. The ratio is 10=30 = 1=3 .
18
ID: [104C]
A positive integer is a factor of n if and only if its prime factorization is of the form
2a 3b 5c where 0 a 3, 0 b 2, and 0 c 1. An integer is even if and only if the
exponent of 2 in its prime factorization is at least 1. Therefore, we have 3 choices for a, 3
choices for b and 2 choices for c , for a total of (3)(3)(2) = 18 ways to form an even
positive factor of n.
(10)
6 inches
ID: [2B5C]
We can begin by using the circumference to solve for the radius of the circle. If the
circumference is 12, then 2r = 12 which implies r = 6. Now, we can draw in the radius
T Z as shown:
Y
T
X
Z
We know that T X = T Z , since both are radii of length 6. We are given \T XZ = 60 ,
so \T ZX = 60 , and triangle T XZ is equilateral. Thus, T X = T Z = XZ = 6 .
(11)
(12)
910
ID: [AA5]
Let ABC be Sue's number. From the given information, we have
ABC = 65m = 5 13m = (A + B + C )n, where m 2 and n 10 are positive integers, and
A + B + C 11 (Sue's younger sister is at most 11 years old). If A + B + C = 11, then
since (11; 65) = 1, ABC = 65 11 = 715. But 7 + 1 + 5 = 13 6= 11, so A + B + C 6= 11.
Next, suppose A + B + C = 10. Then, ABC = 65 2k = 130k for some positive integer k .
k = 7 gives ABC = 910, the sum of whose digits is in fact 9 + 1 + 0 = 10. Sue's number is
determined by the information given in the problem, so her number is 910 .
11.7 inches
ID: [B2011]
No solution is available at this time.
(13)
5/16
ID: [2DC]
On her rst turn, Kendra must get 10 points. If she wants a total of 30 after three turns,
she must then either get two tens in a row or a 5 and then a 15. To get three tens in a
row, she can move any direction on her rst move, go in two possible directions for her
second move, and go in two possible directions for her third spin, meaning her probability of
success is 14 . On the other hand, if she wants to get a 10, a 5, and a 15, she can only move
left or right on her rst move, further out on her second move, and then up or down on her
third move, leading to a probability of success of 21 41 21 = 161 . Adding, her total probability
5
is 41 + 161 =
.
16
(14)
40 square centimeters
(15)
787386600
(16)
71 cents
ID: [45D21]
(17)
17.3 feet
ID: [3C44]
ID: [D5A11]
No solution is available at this time.
ID: [35041]
The two incorrect numbers are consecutive numbers. So, one must be even, and one must
be odd. To get the least possible integers, we must maximize the incorrect numbers. As
such, we should start with the highest possible incorrect number and work down.
24 = 6 4, so 24 cannot be an incorrect number if 6 and 4 are correct. The largest even
number that can be incorrect is 16, which is 24 . As such, 16 and 17 are the largest
incorrect numbers. 23 32 52 7 11 13 19 23 = 787386600 .
No solution is available at this time.
No solution is available at this time.
(18)
55 paths
ID: [31312]
Beginning with A, we label each vertex V with the number of distinct paths from A to V .
The number we write at a vertex V is the sum of the numbers written at the vertices above
and to the left of V , since every path from A to V must either go through the vertex above
V or through the vertex to the left of V , and all these paths are distinct. When this
process carried out, the number assigned to B is 55, so there are 55 paths from A to B .
(19)
10
(20)
729
A 1
1
1
1
1
2
3
4
1
3
6
10
3
9
19
3
12
31
12
43
12
55
B
ID: [340D1]
Since z can be any natural number, we can choose to work with the simplest case, when
z = 1. Now we have to nd the number of combinations of x and y such that 10x + y is
divisible by 9 and x and y are dierent. This is the same as asking for the total number of
one- and two-digit multiples of 9 where the two digits are dierent (in the case when the
multiple of 9 is a one-digit number, we consider the tens digit to be 0). Since
100 9 = 11 R1, we know that there are 11 one- and two-digit multiples of 9, with the
smallest being 9 and the largest being 99. Of these multiples, only 99 has tens and ones
digits that are the same, so we exclude it. Therefore, there are 11 1 = 10 combinations
of x and y that satisfy the given conditions.
ID: [51312]
The sum of the ve integers is 5 times their mean. Since the mean is either 1:5(24) = 36,
1:5(52) = 78 or 1:5(86) = 129, the possible values for the sum of all ve integers are
5 36 = 180, 78 5 = 390 and 5 129 = 645. Since the three integers 24, 52, and 86 sum
to 162, the remaining two integers sum to 180 162 = 18, 390 162 = 228, or
645 162 = 483. We have to make sure that there exist pairs of missing integers which
give the correct mean and median in each of these cases. In the rst case, the two missing
integers could be 9 and 9. In the second case they could be 1 and 227. In the third case,
they could be 240 and 243. Thus all three possible sums are attainable. Adding these sums
together, we get 18 + 228 + 483 = 729 .
Copyright MATHCOUNTS Inc. All rights reserved