5-QUESTION
CHALLENGE 3
Name
Calculators may be used.
patterns A bracelet is made by stringing together four beads. Each bead is
1.��������
either red or green. How many different color patterns are possible for
the bracelet, where patterns are considered the same if turning one will
produce the other, as shown here? R
G
R
G
the
same
pattern
R
G
R
G
2.�������� Place a digit 0 through 9 in each blank so the equation below is true. What is the sum of the six digits you placed in the blanks? ( _6_,3_1 ) – ( 76,_35 ) = ( _9,28_ ) 3.�������� In the sequence 16, 80, 48, 64, A, B, each term beyond the second term
is the arithmetic mean (average) of the previous two terms. What is the
value of B? seats Each row of passenger seats on an airplane is set up as shown, where
4.��������
“(“ and “)” indicate a window, “ ” represents a seat, and an aisle is
between the second and third seat in every row. There are 96 passenger
seats that are not along the aisle. How many passenger seats are along
the aisle? window
aisle
seat
sq units Triangle ABC has sides with lengths 10, 23 and 27 units. The perimeter
5.��������
of a particular square is 60% of triangle ABC’s perimeter. What is the
area of the square? Copyright MATHCOUNTS Inc. 2013. All rights reserved. The National Math Club: 5 Question Challenges
s
n
io
5-QUESTION
CHALLENGE 3
t
u
l
o
S
Name
Calculators may be used.
6 patterns A bracelet is made by stringing together four beads. Each bead is
1.��������
either red or green. How many different color patterns are possible for
the bracelet, where patterns are considered the same if turning one will
produce the other, as shown here? R
G
R
G
the
same
pattern
R
G
R
G
Our bracelet can have 4 red beads (1 pattern), 4 green beads (1 pattern), 3 red/1 green (1 pattern), 3 green/1 red (1 pattern) or 2 red/2 green (2 patterns:
alternating colors or as shown in the example). This is a total of 6 different
patterns for the beads.
22
2.��������
Place a digit 0 through 9 in each blank so the equation below is true. What is the sum of the six digits you placed in the blanks? ( _6_,3_1 ) – ( 76,_35 ) = ( _9,28_ )
This problem may be easier if we rewrite the subtraction problem as an addition
problem. Our new equation is then ( _9,28_ ) + ( 76,_35 ) = ( _6_,3_1). From here,
we can fill in the missing ones digit knowing that 6 + 5 will get us a 1 in the sum’s
ones digit: ( _9,286 ) + ( 76,_35 ) = ( _6_,3_1). We’ll carry the other 1, and then
have 1 + 8 + 3 in our tens column to give us a 2 in the sum’s tens column while
carrying another 1: ( _9,286 ) + ( 76,_35 ) = ( _6_,321). In the hundreds column, we
have 1 + 2 + _ = 3 or 13. Only a 0 in the blank will work: ( _9,286 ) + ( 76,035 ) = ( _6_,321). We did not carry a 1, so the thousands column is simply 9 + 6, which
gives us a 5 and a carried 1: ( _9,286 ) + ( 76,035 ) = ( _65,321). Now we have 1 +
_ + 7 = 6 or 16, so the missing digit is an 8, which gives us 16, and the hundredthousands digit of the sum is a 1: ( 89,286 ) + ( 76,035 ) = ( 165,321). The sum of
the digits we filled in is 8 + 6 + 0 + 1 + 5 + 2 = 22.
Copyright MATHCOUNTS Inc. 2013. All rights reserved. The National Math Club: 5 Question Challenges
60
3.��������
In the sequence 16, 80, 48, 64, A, B, each term beyond the second term
is the arithmetic mean (average) of the previous two terms. What is the
value of B? The value of A is the average of 48 and 64, which is (48 + 64) ÷ 2 = 56. Now B is
the average of 64 and 56, which is (64 + 56) ÷ 2 = 60.
64 seats Each row of passenger seats on an airplane is set up as shown, where
4.��������
“(“ and “)” indicate a window, “ ” represents a seat, and an aisle is
between the second and third seat in every row. There are 96 passenger
seats that are not along the aisle. How many passenger seats are along
the aisle? window
aisle seat In each row the ratio of “aisle seats” to “non-aisle seats” is 2:3. If we let y be the
total number of aisle seats, the ratio of “aisle seats” to “non-aisle seats” for all of
the passenger seats is y:96. Because every row has the same configuration, the
ratio for each row is equal to the ratio for all of the passenger seats, so 2:3 = y:96. Our first ratio becomes our second ratio by multiplying each value by
32. Therefore, y = 2 × 32 = 64. Similarly, we can see that each row has 3 non-aisle
seats. If there are a total of 96 non-aisle seats, there must be 96 ÷ 3 = 32 rows.
Each row has 2 aisle seats, so there are 32 × 2 = 64 aisle seats.
81 sq units Triangle ABC has sides with lengths 10, 23 and 27 units. The perimeter
5.��������
of a particular square is 60% of triangle ABC’s perimeter. What is the
area of the square? The perimeter of triangle ABC is 10 + 23 + 27 = 60 units. If the perimeter of a
particular square is 60% of 60 units, then its perimeter is 0.60 × 60 = 36 units.
This means each side of the square is 36 ÷ 4 = 9 units, and its area is 9 × 9 = 81 square units.
Copyright MATHCOUNTS Inc. 2013. All rights reserved. The National Math Club: 5 Question Challenges