5-QUESTION
CHALLENGE 32
Calculators may not be used.
Name
feet
1.����������Suzanne
has determined that swimming the length of a rectangular pool
60 times is the same distance as swimming its perimeter 18 times. If the
length of the swimming pool is 120 feet, what is the width of the pool? 2. ��������� What is the positive difference in the sums of the letters in the words
MATH and COUNTS, if each letter is assigned a value as follows: A = 1,
B = −2, C = 3, D = −4, …, Y = 25, Z = −26? assort3.����������
ments
Maraya is going to buy six cookies at the bakery. She will
choose from sugar, ginger, chocolate and peanut butter cookies.
How many different assortments of cookies can she buy if she
will buy at least one of each of the four kinds of cookies? D
inches
4.����������The
radius of circle O is 12 inches, and segments AB and
CD are tangent to the circle at B and D, respectively.
AB = 16 inches, and CD = 5 inches. What is the sum of
the number of inches in OC + OA? A
C
O
B
ordered
5.����������How
many ordered triples (a, b, c) of positive integers satisfy the equation
triples
a + b + c = 6?
Copyright MATHCOUNTS Inc. 2013. All rights reserved. The National Math Club: 5 Question Challenges
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n
o
uti
5-QUESTION
CHALLENGE 32
ol
S
Name
Calculators may not be used.
80 feet
1.����������Suzanne
has determined that swimming the length of
a rectangular pool 60 times is the same distance as
swimming its perimeter 18 times. If the length of the
swimming pool is 120 feet, what is the width of the pool? The perimeter of the pool can be expressed as 2L + 2W → 2(120) + 2W → 240 + 2W,
where L is the length and W is the width of the pool. From the given information, we can
now write 60(120) = 18(240 + 2W) → 7200 = 4320 + 36W → 36W = 2880 → W = 80 feet.
38
2. ���������
What is the positive difference in the sums of the letters in the words
MATH and COUNTS, if each letter is assigned a value as follows: A = 1,
B = −2, C = 3, D = −4, …, Y = 25, Z = −26? First, we can determine the value for the letters through U:
A
B
C
D
E
F
G
H
I
J
K
L
M
N
O
P
Q
R
S
T
U
1
−2
3
−4
5
−6
7
−8
9
−10
11
−12
13
−14
15
−16
17
−18
19
−20
21
The value of MATH is 13 + 1 + (−20) + (−8) = −14. The value of COUNTS is 3 + 15 + 21 + (−14) + (−20) + 19 = 24. The positive difference is |(−14) − 24| = |−38| = 38.
10 assort3.����������
ments
Maraya is going to buy six cookies at the bakery. She will
choose from sugar, ginger, chocolate and peanut butter cookies.
How many different assortments of cookies can she buy if she
will buy at least one of each of the four kinds of cookies? We know four of her six cookies: S, G, C, PB, __, __. The question is really, “how many
different assortments of two cookies can she buy if there are four different kinds from
which to choose?” If the two cookies are the same kind as each other, there are four
possibilities (S,S; G,G; C,C; PB,PB). If the two cookies are different from each other,
there are “four choose two” = 4C2 = (4!)/[(2!)(4−2)!] = 6 combinations (S,G; S,C; S,PB; G,C;
G,PB; C,PB). This is a total of 4 + 6 = 10 assortments.
Copyright MATHCOUNTS Inc. 2013. All rights reserved. The National Math Club: 5 Question Challenges
33 inches
4.����������The
radius of circle O is 12 inches, and segments
AB and CD are tangent to the circle at B and D,
respectively. AB = 16 inches, and CD = 5 inches. What
is the sum of the number of inches in OC + OA? D
C
O
A
The figure shown here includes the measures we know and the right angles
formed by the segments tangent to the circle. Looking at right triangle CDO,
we can determine its side lengths are the 5-12-13 Pythagorean Triple, so
OC = 13 inches. Similarly, looking at right triangle ABO, we see each of
the known sides are multiples of 4. Dividing them by 4, they would be
3 and 4, which are the lengths of the legs of a 3-4-5 right triangle.
16
Thus, OA = 4 × 5 = 20 inches. Finally, OC + OA = 13 + 20 = 33 inches.
B
5
12
12
10 ordered
5.����������How
many ordered triples (a, b, c) of positive integers satisfy the
triples
equation a + b + c = 6?
The sum of 6 is relatively low if a, b and c must be positive integers. So making an
ordered list may be a good strategy. We’ll start with a = 1, and use values of b that
increase from 1: (1, 1, 4), (1, 2, 3), (1, 3, 2), (1, 4, 1). When b = 5, c would be zero, which is
not a positive integer. Now let’s move to a = 2 and follow the same process: (2, 1, 3),
(2, 2, 2), (2, 3, 1). When a = 3, we have: (3, 1, 2), (3, 2, 1). When a = 4: (4, 1, 1). These are
the 4 + 3 + 2 + 1 = 10 possible ordered triples.
Copyright MATHCOUNTS Inc. 2013. All rights reserved. The National Math Club: 5 Question Challenges