2006 JuniorPrelimBoth2006 Mar08
BRITISH COLUMBIA SECONDARY SCHOOL
MATHEMATICS CONTEST, 2006
Junior Preliminary Round Problems & Solutions
1.
In the diagram, rectangle P QRS is divided into three identical
squares. If P QRS has a perimeter of 120 cm, then the area of
one of the squares, in cm2 , is:
(A)
675
(B)
400
(D) 141
(E)
45
(C)
225
P
Q
S
R
Solution:
Let s be the side length of the one of the squares. The perimeter of the rectangle P QRS is 8s = 120 ⇒
s = 15. Thus, the area of one of the squares is s2 = 225 cm.
Answer is (C).
2.
Alice has four friends who visit her regularly. Betty visits every three days, Charles visits every six
days, and Dorothy visits every seven days. If the four friends all show up at once only every 84 days
and Efran visits less often than any of the others, then Efran could visit every
(A)
12 days
(B)
14 days
(C)
18 days
(D)
21 days
(E)
42 days
Solution:
The frequency of Efran’s visits must be a divisor of 84 that is larger than seven. The divisors of 84 are:
1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, and 84. If Efran visits every 14, 21, or 42 days, then all four of the
friends would show up every 42 days. Since 18 is not a divisor of 84, of the choices given the only choice
is that Efran visits every 12 days. Note that 28 or 84 days are also possible.
Answer is (A).
3.
A
In triangle ABC segments AD, BD, and CD are equal.
The value of the difference (y − x), in degrees, is:
(A)
0
(D) 22
(B)
8
(E)
34
(C)
14
x
112◦
y
B
C
D
Triangles ABD and ADC are both isosceles. It follows that the angles x and y must satisfy the equations
2x + 112 = 180 and 2y + (180 − 112) = 180. It follows that x = 34◦ and y = 56◦ . The difference in
these angles is y − x = 22◦ .
Solution:
Answer is (D).
BCSSMC 2006 Junior Preliminary Round
Problems & Solutions
4.
Page 2
The number 10100 is a googol and number 10000n is also a googol. The value of n is:
(A)
10
(B)
25
(C)
30
(D)
75
(E)
100
Solution:
If 10100 = 10000n , then using the laws of exponents
n
10000n = 104 = 104n = 10100
so that 4n = 100 ⇒ n = 25.
5.
Answer is (B).
Three blocks and one top balance 15 marbles. One top balances one block and seven marbles. The
number of marbles that balance one top is:
(A)
3
(B)
5
(C)
9
(D)
11
(E)
12
Solution:
Let b be the mass of one block, m the mass of one marble, and t the mass of one top. Then we have
the following two equations for b, m, and t:
3b + t = 15m and t = b + 7m
Substituting t from the second equation into the first gives 3b + b + 7m = 15m ⇒ 4b = 8m ⇒ b = 2m.
Substituting this into the second equation gives t = 2m + 7m = 9m. Thus, it takes 9 marbles to balance
one top.
Answer is (C).
6.
The smallest whole number x that has exactly 12 distinct divisors, including 1 and x, can be found in
the interval:
(A)
45 ≤ x < 55 (B)
55 ≤ x < 65 (C)
65 ≤ x < 75 (D)
75 ≤ x < 85 (E)
85 ≤ x ≤ 90
Solution:
If the prime factorization of the whole number x is x = pα q β rγ , then the number of divisors x has is
(α + 1)(β + 1)(γ + 1). If x has exactly 12 divisors, this product must equal 12. So we must factor 12
into a product of whole number factors. The possible factorings are
12 = 12 · 1 = 6 · 2 = 4 · 3 = 3 · 2 · 2
Consider each possible factoring:
i) For the first factoring there is only one prime factor with α = 11 for p = 2, giving x = 2 11 = 2048.
ii) For the second there are two prime factors with α = 5 and β = 1 for p = 2 and q = 3, giving
x = 25 · 31 = 96.
iii) For the third there are two prime factors with α = 3 and β = 2 for p = 2 and q = 3, giving
x = 23 · 32 = 72.
iv) For the fourth there are three prime factors with α = 2, β = 1, and γ = 1 for p = 2, q = 3, and
r = 5 giving x = 22 · 31 · 51 = 60.
The 60 is the smallest whole number that has exactly 12 distinct divisors, which is in the interval
55 ≤ x ≤ 65.
Answer is (B).
BCSSMC 2006 Junior Preliminary Round
Problems & Solutions
7.
Page 3
The semicircle centred at O has a diameter of 6 units. The chord
BC is parallel to the diameter AD and is one third the length. The
area of the trapezoid ABCD, in square units, is:
√
√
√
(B) 4 5
(C) 16 2
(A) 4 2
√
√
9 3
(E) 8 2
(D)
4
B
C
h
A
E O
D
Solution:
Let a = AD, b = BC and h be the perpendicular distance between the chord BC and the diameter
AD. The area A of the trapezoid ABCD is given by
A=
1
(a + b)h.
2
Since a = d and b = d/3, where d is the diameter of the circle, we have
A=
1 4d
2
h = dh .
2 3
3
We need only determine h. Consider the right triangle BOE formed by dropping a perpendicular from B
to the point E on the diameter AD. The length of this perpendicular is h, and BO = d/2 since BO is a
radius of the circle. Further, by symmetry 2AE + BC = d ⇒ AE = d/3, so that EO = d/2 − d/3 = d/6.
Then using Pythagoras’ Theorem gives
d2
d2
+ h2 = ,
36
4
√
2
d. It follows that the area of the trapezoid is given by
which gives h =
3
√
√
√
2 2 2
2 2
A=
d =
36 = 8 2 .
9
9
If you don’t remember the formula for the area of a trapezoid, we can break the area into the sum of
the areas of two triangles and a rectangle. Then we get
A=
1 d
d
1 d
d
d
2
h+ h+
h = h + h = dh
2 3
3
2 6
3
3
3
as above.
Answer is (E).
8.
The value of
s
√
√
3
3
1+
1 −
2
2
s
in simplified form is:
3
2
Solution:
(A)
Recall that
(B)
1
4
(C)
1
2
(D)
√
3
4
(E)
1
√ √
√
x y = xy, so that
s
s
v
u
√
√
3
3 u
= t1 −
1+
1−
2
2
r
√ !2 r
1
3
3
1
= 1− =
=
2
4
4
2
Answer is (C).
BCSSMC 2006 Junior Preliminary Round
Problems & Solutions
9.
Page 4
Exactly 57.245724% of the people replied ‘yes’ when asked if they used BLEU-OUT face cream. The
fewest number of people who could have been asked is:
(A)
11
(B)
3333
(C)
9999
(D)
111
(E)
1111
Solution:
We want to write 57.245724% as a rational number in lowest terms. Since
57.245724% = 0.5724 + 0.00005724 + 0.000000005724 + . . .
!
4
1
1
+ ...
= 0.5724 1 + 4 +
10
104
= 0.5724
1
10000
= 0.5724
,
−4
1 − 10
9999
it follows that
57.245, 724% =
5, 724
636
=
.
9, 999
1, 111
Hence the minimum number of people that could have been asked is 1111.
Answer is (E).
10. My front lawn is in the shape of an equilateral triangle, of area A square metres. I plan to tether Sadie
the goat to a post at one corner of the triangle. I want Sadie to be able to eat exactly half the grass.
The length of the tether, in metres, must be:
r
r
r
3A
6A
3A
6A
π
(B)
(C)
(D)
(E)
(A)
π
π
π
π
3A
Solution:
The area that can be grazed is the sector of a circle. Since the
π
of the sector is θ = radians. The area of this sector must be
3
triangle is equilateral, the central angle
1
A so
2
θ
A
= r2 ,
2
2
where r is the radius of the circle. If follows that
r
r
A
3A
r=
=
m.
θ
π
If you are not familiar with radian measure the angle determining the sector is 60 ◦ , since all of the
angles in an equilateral triangle are 60◦ . Since a sector with a central angle of 60◦ is one sixth of the
circle, if the length of the tether is r, the radius of the sector, then
Giving the same result as above.
1
1
3A
πr2 = A ⇒ r2 =
6
2
π
Answer is (A).
BCSSMC 2006 Junior Preliminary Round
Problems & Solutions
Page 5
11. Starting with the 2 in the grid shown, the number 2006 can be formed
by moving horizontally, vertically, or diagonally from square to square in
the grid, without backtracking. The number of distinct paths that can
be followed to form 2006 in this way is:
(A)
24
(B)
48
(D) 88
(E)
96
(C)
6
6
6
6
6
6
0
0
0
6
6
0
2
0
6
6
0
0
0
6
6
6
6
6
6
64
Solution:
We only need to consider two initial moves: up to the centre 0 in the second row, or diagonally up
and to the left to the left most 0 in the second row. Then by symmetry we just need to multiply by
4. After moving up we can move right or left to one of the other 0’s on the second row, or diagonally
down to the centre 0 in the second or fourth column. From either of the 0’s in the second row there
are 5 possible moves. For example, from the left most 0 they are: down left, left, up left, up, and up
right. From either of the centre 0’s there are 3 possible moves. For example, from the centre 0 in the
second column they are: down left, left, and up left. This gives a total of 16 distinct paths with initial
move up to the centre 0 in the second row. After moving diagonally up and to the left, there are two
possible moves: down to the centre 0 in the second column or right to the centre 0 in the second row.
In either case, there are three possible moves, as described above. This gives a 6 more possible paths
with initial move up to the centre 0 in the second row, for a total of 22 distinct paths. So the total
number of distinct paths is 88.
Answer is (D).
12. A six-team league has a schedule that requires each team to play every other team four times. The
total number of games in the league schedule is:
(A)
36
(B)
60
(C)
72
(D)
120
(E)
144
Solution:
There are
6×5
6
=
= 15
2
2
distinct pairings of teams. Since each team must play every other four times, the total number of games
is 4 × 15 = 60.
Answer is (B).
MATHEMATICS CONTEST, 2006
Junior Preliminary Round Problems & Solutions
1.
In the diagram, rectangle P QRS is divided into three identical
squares. If P QRS has a perimeter of 120 cm, then the area of
one of the squares, in cm2 , is:
(A)
675
(B)
400
(D) 141
(E)
45
(C)
225
P
Q
S
R
Solution:
Let s be the side length of the one of the squares. The perimeter of the rectangle P QRS is 8s = 120 ⇒
s = 15. Thus, the area of one of the squares is s2 = 225 cm.
Answer is (C).
2.
Alice has four friends who visit her regularly. Betty visits every three days, Charles visits every six
days, and Dorothy visits every seven days. If the four friends all show up at once only every 84 days
and Efran visits less often than any of the others, then Efran could visit every
(A)
12 days
(B)
14 days
(C)
18 days
(D)
21 days
(E)
42 days
Solution:
The frequency of Efran’s visits must be a divisor of 84 that is larger than seven. The divisors of 84 are:
1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, and 84. If Efran visits every 14, 21, or 42 days, then all four of the
friends would show up every 42 days. Since 18 is not a divisor of 84, of the choices given the only choice
is that Efran visits every 12 days. Note that 28 or 84 days are also possible.
Answer is (A).
3.
A
In triangle ABC segments AD, BD, and CD are equal.
The value of the difference (y − x), in degrees, is:
(A)
0
(D) 22
(B)
8
(E)
34
(C)
14
x
112◦
y
B
C
D
Triangles ABD and ADC are both isosceles. It follows that the angles x and y must satisfy the equations
2x + 112 = 180 and 2y + (180 − 112) = 180. It follows that x = 34◦ and y = 56◦ . The difference in
these angles is y − x = 22◦ .
Solution:
Answer is (D).
BCSSMC 2006 Junior Preliminary Round
Problems & Solutions
4.
Page 2
The number 10100 is a googol and number 10000n is also a googol. The value of n is:
(A)
10
(B)
25
(C)
30
(D)
75
(E)
100
Solution:
If 10100 = 10000n , then using the laws of exponents
n
10000n = 104 = 104n = 10100
so that 4n = 100 ⇒ n = 25.
5.
Answer is (B).
Three blocks and one top balance 15 marbles. One top balances one block and seven marbles. The
number of marbles that balance one top is:
(A)
3
(B)
5
(C)
9
(D)
11
(E)
12
Solution:
Let b be the mass of one block, m the mass of one marble, and t the mass of one top. Then we have
the following two equations for b, m, and t:
3b + t = 15m and t = b + 7m
Substituting t from the second equation into the first gives 3b + b + 7m = 15m ⇒ 4b = 8m ⇒ b = 2m.
Substituting this into the second equation gives t = 2m + 7m = 9m. Thus, it takes 9 marbles to balance
one top.
Answer is (C).
6.
The smallest whole number x that has exactly 12 distinct divisors, including 1 and x, can be found in
the interval:
(A)
45 ≤ x < 55 (B)
55 ≤ x < 65 (C)
65 ≤ x < 75 (D)
75 ≤ x < 85 (E)
85 ≤ x ≤ 90
Solution:
If the prime factorization of the whole number x is x = pα q β rγ , then the number of divisors x has is
(α + 1)(β + 1)(γ + 1). If x has exactly 12 divisors, this product must equal 12. So we must factor 12
into a product of whole number factors. The possible factorings are
12 = 12 · 1 = 6 · 2 = 4 · 3 = 3 · 2 · 2
Consider each possible factoring:
i) For the first factoring there is only one prime factor with α = 11 for p = 2, giving x = 2 11 = 2048.
ii) For the second there are two prime factors with α = 5 and β = 1 for p = 2 and q = 3, giving
x = 25 · 31 = 96.
iii) For the third there are two prime factors with α = 3 and β = 2 for p = 2 and q = 3, giving
x = 23 · 32 = 72.
iv) For the fourth there are three prime factors with α = 2, β = 1, and γ = 1 for p = 2, q = 3, and
r = 5 giving x = 22 · 31 · 51 = 60.
The 60 is the smallest whole number that has exactly 12 distinct divisors, which is in the interval
55 ≤ x ≤ 65.
Answer is (B).
BCSSMC 2006 Junior Preliminary Round
Problems & Solutions
7.
Page 3
The semicircle centred at O has a diameter of 6 units. The chord
BC is parallel to the diameter AD and is one third the length. The
area of the trapezoid ABCD, in square units, is:
√
√
√
(B) 4 5
(C) 16 2
(A) 4 2
√
√
9 3
(E) 8 2
(D)
4
B
C
h
A
E O
D
Solution:
Let a = AD, b = BC and h be the perpendicular distance between the chord BC and the diameter
AD. The area A of the trapezoid ABCD is given by
A=
1
(a + b)h.
2
Since a = d and b = d/3, where d is the diameter of the circle, we have
A=
1 4d
2
h = dh .
2 3
3
We need only determine h. Consider the right triangle BOE formed by dropping a perpendicular from B
to the point E on the diameter AD. The length of this perpendicular is h, and BO = d/2 since BO is a
radius of the circle. Further, by symmetry 2AE + BC = d ⇒ AE = d/3, so that EO = d/2 − d/3 = d/6.
Then using Pythagoras’ Theorem gives
d2
d2
+ h2 = ,
36
4
√
2
d. It follows that the area of the trapezoid is given by
which gives h =
3
√
√
√
2 2 2
2 2
A=
d =
36 = 8 2 .
9
9
If you don’t remember the formula for the area of a trapezoid, we can break the area into the sum of
the areas of two triangles and a rectangle. Then we get
A=
1 d
d
1 d
d
d
2
h+ h+
h = h + h = dh
2 3
3
2 6
3
3
3
as above.
Answer is (E).
8.
The value of
s
√
√
3
3
1+
1 −
2
2
s
in simplified form is:
3
2
Solution:
(A)
Recall that
(B)
1
4
(C)
1
2
(D)
√
3
4
(E)
1
√ √
√
x y = xy, so that
s
s
v
u
√
√
3
3 u
= t1 −
1+
1−
2
2
r
√ !2 r
1
3
3
1
= 1− =
=
2
4
4
2
Answer is (C).
BCSSMC 2006 Junior Preliminary Round
Problems & Solutions
9.
Page 4
Exactly 57.245724% of the people replied ‘yes’ when asked if they used BLEU-OUT face cream. The
fewest number of people who could have been asked is:
(A)
11
(B)
3333
(C)
9999
(D)
111
(E)
1111
Solution:
We want to write 57.245724% as a rational number in lowest terms. Since
57.245724% = 0.5724 + 0.00005724 + 0.000000005724 + . . .
!
4
1
1
+ ...
= 0.5724 1 + 4 +
10
104
= 0.5724
1
10000
= 0.5724
,
−4
1 − 10
9999
it follows that
57.245, 724% =
5, 724
636
=
.
9, 999
1, 111
Hence the minimum number of people that could have been asked is 1111.
Answer is (E).
10. My front lawn is in the shape of an equilateral triangle, of area A square metres. I plan to tether Sadie
the goat to a post at one corner of the triangle. I want Sadie to be able to eat exactly half the grass.
The length of the tether, in metres, must be:
r
r
r
3A
6A
3A
6A
π
(B)
(C)
(D)
(E)
(A)
π
π
π
π
3A
Solution:
The area that can be grazed is the sector of a circle. Since the
π
of the sector is θ = radians. The area of this sector must be
3
triangle is equilateral, the central angle
1
A so
2
θ
A
= r2 ,
2
2
where r is the radius of the circle. If follows that
r
r
A
3A
r=
=
m.
θ
π
If you are not familiar with radian measure the angle determining the sector is 60 ◦ , since all of the
angles in an equilateral triangle are 60◦ . Since a sector with a central angle of 60◦ is one sixth of the
circle, if the length of the tether is r, the radius of the sector, then
Giving the same result as above.
1
1
3A
πr2 = A ⇒ r2 =
6
2
π
Answer is (A).
BCSSMC 2006 Junior Preliminary Round
Problems & Solutions
Page 5
11. Starting with the 2 in the grid shown, the number 2006 can be formed
by moving horizontally, vertically, or diagonally from square to square in
the grid, without backtracking. The number of distinct paths that can
be followed to form 2006 in this way is:
(A)
24
(B)
48
(D) 88
(E)
96
(C)
6
6
6
6
6
6
0
0
0
6
6
0
2
0
6
6
0
0
0
6
6
6
6
6
6
64
Solution:
We only need to consider two initial moves: up to the centre 0 in the second row, or diagonally up
and to the left to the left most 0 in the second row. Then by symmetry we just need to multiply by
4. After moving up we can move right or left to one of the other 0’s on the second row, or diagonally
down to the centre 0 in the second or fourth column. From either of the 0’s in the second row there
are 5 possible moves. For example, from the left most 0 they are: down left, left, up left, up, and up
right. From either of the centre 0’s there are 3 possible moves. For example, from the centre 0 in the
second column they are: down left, left, and up left. This gives a total of 16 distinct paths with initial
move up to the centre 0 in the second row. After moving diagonally up and to the left, there are two
possible moves: down to the centre 0 in the second column or right to the centre 0 in the second row.
In either case, there are three possible moves, as described above. This gives a 6 more possible paths
with initial move up to the centre 0 in the second row, for a total of 22 distinct paths. So the total
number of distinct paths is 88.
Answer is (D).
12. A six-team league has a schedule that requires each team to play every other team four times. The
total number of games in the league schedule is:
(A)
36
(B)
60
(C)
72
(D)
120
(E)
144
Solution:
There are
6×5
6
=
= 15
2
2
distinct pairings of teams. Since each team must play every other four times, the total number of games
is 4 × 15 = 60.
Answer is (B).