2006 SeniorPrelimBoth2006 Mar08
BRITISH COLUMBIA SECONDARY SCHOOL
MATHEMATICS CONTEST, 2006
Senior Preliminary Round Problems & Solutions
1.
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)
9999
(B)
3333
(C)
1111
(D)
111
(E)
11
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
=
5724
1
5724 10000
,
=
10000 1 − 10−4
10000 9999
it follows that
5724
636
=
.
9999
1111
Hence the minimum number of people that could have been asked is 1111.
57.245724% =
Answer is (C).
2.
An object moved in a straight line for 8 seconds at a constant speed. It then reversed direction and
travelled back along the same straight line for 3 seconds at twice its initial speed. It then reversed
direction again and travelled for one second at 40% of its initial speed. If the object is 9.6 m from where
it started at the end of the 12 seconds, its initial speed was:
(A)
3 m/s
(B)
4 m/s
(C)
4.5 m/s
(D)
5 m/s
(E)
6 m/s
Solution:
Let v be the initial speed. Recalling that an object moving with speed v for time t goes a distance
d = vt, and using the given information, we have
v(8) − (2v)(3) + (0.4v)(1) = 8v − 6v + 0.4v = 2.4v = 9.6 ⇒ v = 4
So the initial speed is 4 m/s.
Answer is (B).
3.
The value of (123456789) (123456789) − (123456794) (123456784) is:
(A)
-1000
(B)
-25
(C)
25
(D)
1000
(E)
125
Solution:
Let x = 123456789, then the expression above is
x2 − (x + 5)(x − 5) = x2 − x2 − 25 = x2 − x2 + 25 = 25
Answer is (A).
BCSSMC 2006 Senior Preliminary Round
Problems & Solutions
4.
Page 2
A cylindrical round of cheese is cut into several pieces using seven cuts.
Five cuts are vertical, perpendicular to the top of the round, and two at
a diagonal, as shown in the diagram. Each cut is made with a straight
blade and the diagonal cuts are made through the lines formed by the
vertical cuts. The total number of pieces of cheese that result is:
(A)
21
(B)
24
(D) 54
(E)
60
(C)
36
Solution:
The five vertical cuts divide the block of cheese into twelve pieces. The first of the diagonal cuts passes
through a single row of pieces and hence adds three pieces to the total. The second diagonal cut passes
through two rows of pieces and adds six pieces to the total. Therefore there are 12 + 3 + 6 = 21 pieces
of cheese.
Answer is (A).
5.
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:
√
√
√
(A) 4 2
(B) 4 5
(C) 16 2
√
√
9 3
(E) 8 2
(D)
4
Solution:
B
C
h
A
E O
D
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
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
which gives h =
d. It follows that the area of the trapezoid is given by
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
h+ h+
h
2 3
3
2 6
d
2
d
= h + h = dh
3
3
3
A=
as above.
Answer is (E).
BCSSMC 2006 Senior Preliminary Round
Problems & Solutions
6.
Page 3
The difference between the sum of the first 100 positive multiples of 3 and the sum of the first 100
positive even integers is:
(A)
5000
(B)
5050
(C)
10100
(D)
2525
(E)
None of these
Solution:
Let D denote this difference, then
D=
100
X
k=1
Now, since
Pn
k=1
3k −
100
X
k=1
2k =
100
X
k.
k=1
k = n(n + 1)/2,
D=
100 × 101
= 5050 .
2
Answer is (B).
7.
The maximum number of pieces of a circular pie that can be cut by six straight cuts of a knife is:
(A)
6
(B)
7
(C)
12
(D)
16
(E)
22
Solution:
The first cut always produces two pieces of pie. If the remaining five cuts pass through the centre of the
pie, then each cut passes through exactly two pieces. This means that each cut adds two more pieces
of pie for a total of twelve pieces of pie. However, if we have k − 1 lines in the plane, we can always add
a k th line l such that:
1. l intersects each of the k − 1 lines, and
2. l intersects no more than two lines intersect at a single point.
Moreover, if the k − 1 lines divide the plane into a total of m regions, then line l adds k + 1 regions
to the total. If we ensure that all intersection points are in the interior of the pie, then the maximum
number of pieces M that can be cut from the pie is
M = 2 + 2 + 3 + 4 + 5 + 6 = 22 .
Answer is (E).
8.
A certain set S has 24 more subsets than a second set T . The number of elements in set S is:
(A)
2
(B)
3
(C)
4
(D)
5
(E)
6
Solution:
Suppose that S has n elements, then S has a total of 2n subsets. Since χ = 2n − 24 is the total number
of subsets of the set T , χ must be a power of 2. Since 25 − 24 = 8 = 23 , the set S must have 5 elements.
Answer is (D).
BCSSMC 2006 Senior Preliminary Round
Problems & Solutions
9.
Page 4
Without using any number more than once, a selection of numbers from 25, 27, 3, 12, 6, 15, 9, 30, 21,
and 19 gives a sum of 50. The number of numbers needed is:
(A)
3
(B)
4
(C)
5
(D)
6
(E)
7
Solution:
Every number in the given set of numbers is a multiple of 3 except for 25 and 19. Further, subtracting
either 25 or 19 from 50 gives a result that is not a multiple of three. So both 25 and 19 must be included
in the sum. Since 25 + 19 = 44, the other number is 6. This is the only selection of numbers from the
given set of numbers that adds to 50. So there are three numbers in the selection.
Answer is (A).
10. If
x=
r
(C)
4
2+
then x is equal to:
(A)
2
(B)
2+
√
2
q
2+
√
2 + ···
(D)
6
(E)
Answer is
infinite
Solution:
√
We have x = 2 + x, which implies that x must satisfy the quadratic equation x2 − x − 2 = 0. Since
x2 − x − 2 = (x − 2)(x + 1), the solutions of this quadratic equation are x1,2 = 2, −1. However x > 0
so x = 2.
Answer is (A).
11. There are 21 roads connecting a certain number of towns. Each road connects exactly two towns. Six
of the towns can be reached by exactly three roads. The rest can be reached by exactly four roads. The
number of towns in total is:
(A)
6
(B)
8
(C)
10
(D)
12
(E)
16
Solution:
If we think of the towns as nodes in a graph, then the roads are the edges. The degree of a node is the
number of edges connected to it and since each edge is connected to two nodes, the sum of the degrees
of all of the nodes in the graph is twice the number of edges. Let n be the number of towns that can
be reached by four roads (i.e. the number of nodes of degree 4). Then we find that 3 × 6 + 4n = 2 × 21
or 4n = 24. There are 6 + 24/4 = 12 towns in total.
Answer is (D).
BCSSMC 2006 Senior Preliminary Round
Problems & Solutions
Page 5
12. Consider the fraction
1630
4542
If you swap two digits in the numerator with two digits in the denominator a fraction equalling
The sum of the digits in the final numerator is:
(A)
10
(B)
11
(C)
12
(D)
13
(E)
1
3
results.
15
Solution:
Since the new denominator must be 3 times the new numerator, the new numerator must be a multiple
of 3. Since no multiple of 3 ends in 0, the 0 must be swapped into the denominator and it cannot be
put into the left most position. Further, it cannot be swapped with the 5, since then there would have
to be a 5 in the right most position in the new numerator and then there would also be a 5 in the right
most position of the new denominator. Thus, we must swap the 0 in the numerator with the second 4 in
the denominator. (The new denominator cannot start with 0). We cannot swap the 1 in the numerator
with any of the remaining digits in the denominator, otherwise 3 times the new numerator will be too
large. Also, swapping the 1 into the denominator will mean that the new denominator will not be a
multiple of 3. This means that there are only four possibilities to try: swap the 6 and the 4, swap the
6 and the 5, swap the 3 and the 4, or swap the 3 and the 5. Only the first and last of these leaves the
denominator a multiple of 3. Now we just check the two remaining possibilities:
3 × 1654 = 4962 6= 4302
and 3 × 1534 = 4602
So the new numerator is 1534 and the sum of the digits is 1 + 5 + 3 + 4 = 13.
Answer is (D).
MATHEMATICS CONTEST, 2006
Senior Preliminary Round Problems & Solutions
1.
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)
9999
(B)
3333
(C)
1111
(D)
111
(E)
11
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
=
5724
1
5724 10000
,
=
10000 1 − 10−4
10000 9999
it follows that
5724
636
=
.
9999
1111
Hence the minimum number of people that could have been asked is 1111.
57.245724% =
Answer is (C).
2.
An object moved in a straight line for 8 seconds at a constant speed. It then reversed direction and
travelled back along the same straight line for 3 seconds at twice its initial speed. It then reversed
direction again and travelled for one second at 40% of its initial speed. If the object is 9.6 m from where
it started at the end of the 12 seconds, its initial speed was:
(A)
3 m/s
(B)
4 m/s
(C)
4.5 m/s
(D)
5 m/s
(E)
6 m/s
Solution:
Let v be the initial speed. Recalling that an object moving with speed v for time t goes a distance
d = vt, and using the given information, we have
v(8) − (2v)(3) + (0.4v)(1) = 8v − 6v + 0.4v = 2.4v = 9.6 ⇒ v = 4
So the initial speed is 4 m/s.
Answer is (B).
3.
The value of (123456789) (123456789) − (123456794) (123456784) is:
(A)
-1000
(B)
-25
(C)
25
(D)
1000
(E)
125
Solution:
Let x = 123456789, then the expression above is
x2 − (x + 5)(x − 5) = x2 − x2 − 25 = x2 − x2 + 25 = 25
Answer is (A).
BCSSMC 2006 Senior Preliminary Round
Problems & Solutions
4.
Page 2
A cylindrical round of cheese is cut into several pieces using seven cuts.
Five cuts are vertical, perpendicular to the top of the round, and two at
a diagonal, as shown in the diagram. Each cut is made with a straight
blade and the diagonal cuts are made through the lines formed by the
vertical cuts. The total number of pieces of cheese that result is:
(A)
21
(B)
24
(D) 54
(E)
60
(C)
36
Solution:
The five vertical cuts divide the block of cheese into twelve pieces. The first of the diagonal cuts passes
through a single row of pieces and hence adds three pieces to the total. The second diagonal cut passes
through two rows of pieces and adds six pieces to the total. Therefore there are 12 + 3 + 6 = 21 pieces
of cheese.
Answer is (A).
5.
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:
√
√
√
(A) 4 2
(B) 4 5
(C) 16 2
√
√
9 3
(E) 8 2
(D)
4
Solution:
B
C
h
A
E O
D
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
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
which gives h =
d. It follows that the area of the trapezoid is given by
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
h+ h+
h
2 3
3
2 6
d
2
d
= h + h = dh
3
3
3
A=
as above.
Answer is (E).
BCSSMC 2006 Senior Preliminary Round
Problems & Solutions
6.
Page 3
The difference between the sum of the first 100 positive multiples of 3 and the sum of the first 100
positive even integers is:
(A)
5000
(B)
5050
(C)
10100
(D)
2525
(E)
None of these
Solution:
Let D denote this difference, then
D=
100
X
k=1
Now, since
Pn
k=1
3k −
100
X
k=1
2k =
100
X
k.
k=1
k = n(n + 1)/2,
D=
100 × 101
= 5050 .
2
Answer is (B).
7.
The maximum number of pieces of a circular pie that can be cut by six straight cuts of a knife is:
(A)
6
(B)
7
(C)
12
(D)
16
(E)
22
Solution:
The first cut always produces two pieces of pie. If the remaining five cuts pass through the centre of the
pie, then each cut passes through exactly two pieces. This means that each cut adds two more pieces
of pie for a total of twelve pieces of pie. However, if we have k − 1 lines in the plane, we can always add
a k th line l such that:
1. l intersects each of the k − 1 lines, and
2. l intersects no more than two lines intersect at a single point.
Moreover, if the k − 1 lines divide the plane into a total of m regions, then line l adds k + 1 regions
to the total. If we ensure that all intersection points are in the interior of the pie, then the maximum
number of pieces M that can be cut from the pie is
M = 2 + 2 + 3 + 4 + 5 + 6 = 22 .
Answer is (E).
8.
A certain set S has 24 more subsets than a second set T . The number of elements in set S is:
(A)
2
(B)
3
(C)
4
(D)
5
(E)
6
Solution:
Suppose that S has n elements, then S has a total of 2n subsets. Since χ = 2n − 24 is the total number
of subsets of the set T , χ must be a power of 2. Since 25 − 24 = 8 = 23 , the set S must have 5 elements.
Answer is (D).
BCSSMC 2006 Senior Preliminary Round
Problems & Solutions
9.
Page 4
Without using any number more than once, a selection of numbers from 25, 27, 3, 12, 6, 15, 9, 30, 21,
and 19 gives a sum of 50. The number of numbers needed is:
(A)
3
(B)
4
(C)
5
(D)
6
(E)
7
Solution:
Every number in the given set of numbers is a multiple of 3 except for 25 and 19. Further, subtracting
either 25 or 19 from 50 gives a result that is not a multiple of three. So both 25 and 19 must be included
in the sum. Since 25 + 19 = 44, the other number is 6. This is the only selection of numbers from the
given set of numbers that adds to 50. So there are three numbers in the selection.
Answer is (A).
10. If
x=
r
(C)
4
2+
then x is equal to:
(A)
2
(B)
2+
√
2
q
2+
√
2 + ···
(D)
6
(E)
Answer is
infinite
Solution:
√
We have x = 2 + x, which implies that x must satisfy the quadratic equation x2 − x − 2 = 0. Since
x2 − x − 2 = (x − 2)(x + 1), the solutions of this quadratic equation are x1,2 = 2, −1. However x > 0
so x = 2.
Answer is (A).
11. There are 21 roads connecting a certain number of towns. Each road connects exactly two towns. Six
of the towns can be reached by exactly three roads. The rest can be reached by exactly four roads. The
number of towns in total is:
(A)
6
(B)
8
(C)
10
(D)
12
(E)
16
Solution:
If we think of the towns as nodes in a graph, then the roads are the edges. The degree of a node is the
number of edges connected to it and since each edge is connected to two nodes, the sum of the degrees
of all of the nodes in the graph is twice the number of edges. Let n be the number of towns that can
be reached by four roads (i.e. the number of nodes of degree 4). Then we find that 3 × 6 + 4n = 2 × 21
or 4n = 24. There are 6 + 24/4 = 12 towns in total.
Answer is (D).
BCSSMC 2006 Senior Preliminary Round
Problems & Solutions
Page 5
12. Consider the fraction
1630
4542
If you swap two digits in the numerator with two digits in the denominator a fraction equalling
The sum of the digits in the final numerator is:
(A)
10
(B)
11
(C)
12
(D)
13
(E)
1
3
results.
15
Solution:
Since the new denominator must be 3 times the new numerator, the new numerator must be a multiple
of 3. Since no multiple of 3 ends in 0, the 0 must be swapped into the denominator and it cannot be
put into the left most position. Further, it cannot be swapped with the 5, since then there would have
to be a 5 in the right most position in the new numerator and then there would also be a 5 in the right
most position of the new denominator. Thus, we must swap the 0 in the numerator with the second 4 in
the denominator. (The new denominator cannot start with 0). We cannot swap the 1 in the numerator
with any of the remaining digits in the denominator, otherwise 3 times the new numerator will be too
large. Also, swapping the 1 into the denominator will mean that the new denominator will not be a
multiple of 3. This means that there are only four possibilities to try: swap the 6 and the 4, swap the
6 and the 5, swap the 3 and the 4, or swap the 3 and the 5. Only the first and last of these leaves the
denominator a multiple of 3. Now we just check the two remaining possibilities:
3 × 1654 = 4962 6= 4302
and 3 × 1534 = 4602
So the new numerator is 1534 and the sum of the digits is 1 + 5 + 3 + 4 = 13.
Answer is (D).