2024 Pascal Contest
Solutions
(Grade 9)
Wednesday, February 28, 2024
(in North America and South America)
Thursday, February 29, 2024
(outside of North American and South America)
Wednesday, February 28, 2024
(in North America and South America)
Thursday, February 29, 2024
(outside of North American and South America)
©2023 University of Waterloo
Calculating, \(2 - 0 + 2 - 4 = 2 + 2 - 4 = 0\).
Answer: (B)
The distance between two numbers on the number line is equal to
their positive difference.
Here, this distance is \(6 - (-5) =
11\).
Answer: (D)
Since a turn of \(180\degree\)
is a half-turn, the resulting figure is
.
(Note that we would obtain the same result rotating by \(180\degree\) clockwise or \(180\degree\) counterclockwise.)
Answer: (C)
Since July 1 is a Wednesday, then July 8 and July 15 are both
Wednesdays.
Since July 15 is a Wednesday, then July 17 is a Friday.
Answer: (D)
The first rhombus and the last rhombus each have three edges that
form part of the exterior of the figure, and so they each contribute
\(3\) to the perimeter.
The inner four rhombi each have two edges that form part of the exterior
of the figure, and so they each contribute \(2\) to the perimeter.
Thus, the perimeter is \(2 \times 3 + 4 \times
2 = 14\).
Answer: (B)
On Monday, Narsa ate \(4\)
cookies.
On Tuesday, Narsa ate \(12\)
cookies.
On Wednesday, Narsa ate \(8\)
cookies.
On Thursday, Narsa ate \(0\)
cookies.
On Friday, Narsa ate \(6\)
cookies.
This means that Narsa ate \(4 + 12 + 8 + 0 + 6
= 30\) cookies.
Since the package started with \(45\)
cookies, there are \(45 - 30 = 15\)
cookies left in the package after Friday.
Answer: (D)
For there to be equal numbers of each colour of candy, there must
be at most \(3\) red candies and at
most \(3\) yellow candies, since there
are \(3\) blue candies to start.
Thus, Shuxin ate at least \(7\) red
candies and at least \(4\) yellow
candies.
This means that Shuxin ate at least \(7 + 4 =
11\) candies.
We note that if Shuxin eats \(7\) red
candies, \(4\) yellow candies, and
\(0\) blue candies, there will indeed
be equal numbers of each colour.
Answer: (C)
Since \(10\) students have black hair and \(3\) students have black hair and wear glasses, then a total of \(10 - 3 = 7\) students have black hair but do not wear glasses.
Answer: (A)
Since \(25\%\) is equivalent to
\(\frac{1}{4}\), then the fraction of
the trail covered by the section along the river and the section through
the forest is \(\frac{1}{4} + \frac{5}{8} =
\frac{2}{8} + \frac{5}{8} = \frac{7}{8}\).
This means that the final section up a hill represents \(1 - \frac{7}{8} = \frac{1}{8}\) of the
trail.
Since \(\frac{1}{8}\) of the trail is
\(3 \text{ km}\) long, then the entire
trail is \(8 \times 3\text{ km} = 24\text{
km}\) long.
Answer: (A)
Using the definition, \((5 \nabla 2) \nabla 2 = (4 \times 5 + 2) \nabla 2 = 22 \nabla 2 = 4 \times 22 + 2 = 90\).
Answer: (E)
Solution 1:
If all of Lauren’s \(10\) baskets
are worth \(2\) points, she would have
\(10 \times 2 = 20\) points in
total.
Since she has \(26\) points in total,
then she scores \(26 - 20 = 6\) more
points than if all of her baskets are worth \(2\) points.
This means that, if \(6\) of her
baskets are worth \(3\) points, she
would gain \(1\) point for each of
these \(6\) baskets and so have \(20 + 6 = 26\) points.
Thus, she makes \(6\) baskets worth
\(3\) points.
(We note that \(6 \times 3 + 4 \times 2 =
26\).)
Solution 2:
Suppose that Lauren makes \(x\)
baskets worth \(3\) points each.
Since she makes \(10\) baskets, then
\(10-x\) baskets that she made are
worth \(2\) points each.
Since Lauren scores \(26\) points, then
\(3x + 2(10-x) = 26\) and so \(3x + 20 - x = 26\) which gives \(x = 6\).
Therefore, Lauren makes \(6\) baskets
worth \(3\) points.
Answer: (B)
From the given list, the numbers \(11\) and \(13\) are the only prime numbers, and so
must be Karla’s and Levi’s numbers in some order.
From the given list, \(16\) is the only
perfect square; thus, Glen’s number was \(16\).
The remaining numbers are \(12\), \(14\), \(15\).
Since Hao’s and Julia’s numbers were even, then their numbers must be
\(12\) and \(14\) in some order.
Thus, Ioana’s number is \(15\).
Answer: (B)
Each of the \(4\) lines can
intersect each of the other \(3\) lines
at most once.
This might appear to create \(4 \times 3 =
12\) points of intersection, but each point of intersection is
counted twice – one for each of the \(2\) lines.
Thus, the maximum number of intersection points is \(\dfrac{4 \times 3}{2} = 6\).
The diagram below demonstrates that 6 intersection points are indeed
possible:
Answer: (D)
When \(10\) numbers have an
average of \(17\), their sum is \(10 \times 17 = 170\).
When \(9\) numbers have an average of
\(16\), their sum is \(9 \times 16 = 144\).
Therefore, the number that was removed was \(170 - 144 = 26\).
Answer: (A)
Since \(CD = DE = EC\), then
\(\triangle CDE\) is equilateral, which
means that \(\angle DEC =
60\degree\).
Since \(\angle DEB\) is a straight
angle, then \(\angle CEB = 180\degree - \angle
DEC = 180\degree - 60\degree = 120\degree\).
Since \(CE = EB\), then \(\triangle CEB\) is isosceles with \(\angle ECB = \angle EBC\).
Since \(\angle ECB + \angle CEB + \angle EBC =
180\degree\), then \(2 \times \angle
EBC + 120\degree = 180\degree\), which means that \(2 \times \angle EBC = 60\degree\) or \(\angle EBC = 30\degree\).
Therefore, \(\angle ABC = \angle EBC =
30\degree\).
Answer: (A)
Since \(x^2 < x\) and \(x^2 \geq 0\), then \(x > 0\) and so it cannot be the case
that \(x\) is negative.
Thus, neither (D) nor (E) is the answer.
Since \(x^2 < x\), then we cannot
have \(x > 1\). This is because when
\(x > 1\), we have \(x^2 > x\).
Thus, (A) is not the answer and so the answer is (B) or (C).
If \(x = \dfrac{1}{3}\), then \(x^2 = \dfrac{1}{3} \times \dfrac{1}{3} =
\dfrac{1}{9}\) and \(\dfrac{x}{2} =
\dfrac{1/3}{2} = \dfrac{1}{6}\).
Since \(\dfrac{1}{6} >
\dfrac{1}{9}\), then (B) cannot be the answer.
Therefore, the answer must be (C).
Checking, when \(x = \dfrac{3}{4}\), we
have \(x^2 = \dfrac{9}{16}\) and \(\dfrac{x}{2} = \dfrac{3}{8}\).
Since \(\dfrac{x}{2} = \dfrac{3}{8} =
\dfrac{6}{16} < \dfrac{9}{16} = x^2\), then \(\dfrac{x}{2} < x^2\).
Also, \(x^2 = \dfrac{9}{16} <
\dfrac{12}{16} = \dfrac{3}{4} = x\).
This confirms that \(x = \dfrac{3}{4}\)
does satisfy the required conditions.
Answer: (C)
In \(2\) hours travelling at
\(100 \text{ km/h}\), Melanie travels
\(2\text{ h} \times 100\text{ km/h} =
200\text{ km}\).
When Melanie travels \(200 \text{ km}\)
at \(80 \text{ km/h}\), it takes \(\dfrac{200\text{ km}}{80\text{ km/h}} = 2.5\text{
h}\).
Melanie travels a total of \(200\text{ km} +
200\text{ km} = 400\text{ km}\).
Melanie travels for a total of \(2\text{ h} +
2.5\text{ h} = 4.5\text{ h}\).
Therefore, Melanie’s average speed is \(\dfrac{400\text{ km}}{4.5\text{ h}} \approx
88.89\text{ km/h}\).
Of the given choices, this is closest to \(89
\text{ km/h}\).
Answer: (B)
From the given information, we know that \[\begin{aligned}
\text{S}+\text{E}+\text{T}&=2\\
\text{H}+\text{A}+\text{T}&=7\\
\text{T}+\text{A}+\text{S}+\text{T}+\text{E}&=3\\
\text{M}+\text{A}+\text{T}&=4 \end{aligned}\] Since \(\text{T}+\text{A}+\text{S}+\text{T}+\text{E}=3\)
and \(\text{S}+\text{E}+\text{T}=2\),
then \(\text{T}+\text{A}=3-2=1\).
Since \(\text{H}+\text{A}+\text{T}=7\)
and \(\text{T}+\text{A}=1\), then \(\text{H} = 7 - 1 = 6\).
Since \(\text{M}+\text{A}+\text{T}=4\)
and \(\text{H}=7\), then \(\text{M} + (\text{A} + \text{T}) + \text{H} = 4 +
6 = 10\).
Therefore, the value of the word MATH is \(10\).
We note that it is also possible to find specific values for S, E, T, A
that give the correct values to the words. One such set of values is
\(\text{A} = 1\), \(\text{T}=0\), \(\text{S} = 4\), and \(\text{E} = -2\). These values are not
unique, even though the values assigned to \(\text{M}\) and \(\text{H}\) (namely, \(3\) and \(6\)) are unique.
Answer: (E)
The perimeter of \(\triangle
ABC\) is equal to \((3x+4) + (3x+4) +
2x = 8x+8\).
The perimeter of rectangle \(DEFG\) is
equal to \[2 \times (2x-2) + 2 \times (3x-1)
= 4x - 4 + 6x - 2 = 10x - 6\] Since these perimeters are equal,
we have \(10x - 6 = 8x + 8\) which
gives \(2x = 14\) and so \(x = 7\).
Thus, \(\triangle ABC\) has \(AC = 2 \times 7 = 14\) and \(AB=BC=3 \times 7 + 4 25\).
We drop a perpendicular from \(B\) to
\(T\) on \(AC\).
Since \(\triangle ABC\) is
isosceles, then \(T\) is the midpoint
of \(AC\), which gives \(AT = TC = 7\).
By the Pythagorean Theorem, \(BT = \sqrt{BC^2
- TC^2} = \sqrt{25^2 - 7^2} = \sqrt{625 - 49} = \sqrt{576} =
24\).
Therefore, the area of \(\triangle
ABC\) is equal to \(\frac{1}{2} \cdot
AC \cdot BT = \frac{1}{2} \times 14 \times 24 = 168\).
Answer: (C)
Since \(N\) is between \(1\,000\,000\) and \(10\,000\,000\), inclusive, then \(25 \times N\) is between \(25\,000\,000\) and \(250\,000\,000\), inclusive, and so \(25 \times N\) has \(8\) digits or it has \(9\) digits.
We consider the value of \(25 \times
N\) as having \(9\) digits, with
the possibility that the first digit could be \(0\).
Since \(25 \times N\) is a multiple of
\(25\), its final two digits must be
\(00\), \(25\), \(50\), or \(75\).
For a fixed set of leftmost three digits, \(xyz\), the multiple of \(25\) that has the largest sum of digits
must be \(xyz\,999\,975\) since the
next four digits are as large as possible (all \(9\)s) and the rightmost two digits have the
largest possible sum among the possible endings for multiples of \(25\).
So to answer the question, we need to find the integer of the form \(xyz\,999\,975\) which is between \(25\,000\,000\) and \(250\,000\,000\) and has the maximum
possible sum \(x+y+z\).
We know that the maximum possible value of \(x\) is \(2\), the maximum possible value of \(y\) is 9, and the maximum possible value of
\(z\) is 9.
This means that \(x + y + z \leq 2 + 9 + 9 =
20\).
We cannot have \(299\,999\,975\) since
it is not in the given range.
However, we could have \(x + y + z =
19\) if \(x = 1\) and \(y = 9\) and \(z =
9\).
Therefore, the integer \(199\,999\,975\) is the multiple of \(25\) in the given range whose sum of digits
is as large as possible. This sum is \(1 + 6
\times 9 + 7 + 5 = 67\).
We note that \(199\,999\,975 = 25 \times
7\,999\,999\) so it is a multiple of \(25\). Note that \(N = 7\,999\,999\) is between \(1\,000\,000\) and \(10\,000\,000\).
Answer: (C)
Since the second column includes the number \(1\), then step (ii) was never used on the
second column, otherwise each entry would be at least \(2\).
To generate the \(1\), \(3\) and \(2\) in the second column, we thus need to
have used step (i) \(1\) time on row
\(1\), \(3\) times on row \(2\), and \(2\) times on row \(3\).
This gives:
| \(1\) | \(1\) | \(1\) |
| \(3\) | \(3\) | \(3\) |
| \(2\) | \(2\) | \(2\) |
We cannot use step (i) any more times, otherwise the entries in
column \(2\) will increase. Thus, \(a = 1 + 3 + 2 = 6\).
To obtain the final grid from this current grid using only step (ii), we
must increase each entry in column \(1\) by \(6\) (which means using step (ii) \(3\) times) and increase each entry in
column \(3\) by \(4\) (which means using step (ii) \(2\) times). Thus, \(b = 3 + 2 = 5\).
Therefore, \(a + b = 11\).
Answer: \(11\)
The \(27\) small cubes that make up the larger \(3 \times 3 \times 3\) can be broken into \(4\) categories: \(1\) small cube in the very centre of the larger cube (not seen in the diagram), \(8\) small cubes at the vertices of the larger cube (an example is marked with \(V\)), \(12\) small cubes on the edges not at vertices (an example is marked with \(E\)), and \(6\) small cubes at the centre of each face (an example is marked with \(F\)).
The centre cube contributes \(0\) to
the surface area of the cube.
Each of the \(8\) vertex cubes
contributes \(3\) to the surface area
of the larger cube, as \(3\) of the
\(6\) faces of each such cube are on
the exterior of the larger cube.
Each of the \(12\) edge cubes
contributes \(2\) to the surface area
of the larger cube.
Each of the \(6\) face cubes
contributes \(1\) to the surface area
of the larger cube.
There are \(10\) small red cubes that
need to be placed as part of the larger cube.
To minimize the surface area that is red, we place the red cubes in
positions where they will contribute the least to the overall surface
area. To do this, we place 1 red cube at the centre (contributing \(0\) to the surface area), \(6\) red cubes at the centres of the faces
(each contributing \(1\) to the surface
area), and the remaining \(3\) red
cubes on the edges (each contributing \(2\) to the surface area).
In total, the surface area that is red is \(1
\times 0 + 6 \times 1 + 3 \times 2 = 12\).
Answer: \(12\)
We want to count the number of four-digit codes \(abcd\) that satisfy the given rules.
From the first rule, at least one of the digits must be \(4\), but \(b \neq
4\) and \(d \neq 4\).
Therefore, either \(a=4\) or \(c=4\). The fourth rule tells us that we
could have both \(a=4\) and \(c=4\).
Suppose that \(a=4\) and \(c=4\).
The code thus has the form \(4b4d\).
The second and third rules tell us that the remaining digits are \(2\) and \(7\), and that there are no further
restrictions on where the \(2\) and
\(7\) are placed.
Therefore, in this case, the code is either \(4247\) or \(4742\), and so there are \(2\) possible codes.
Suppose that \(a=4\) and \(c \neq 4\). (Recall that \(b \neq 4\) and \(d \neq 4\).)
The code thus has the form \(4bcd\).
The remaining digits include a \(2\)
(which can be placed in any of the remaining positions), a \(7\), and either a \(1\) or a \(6\).
There are \(3\) positions in which the
\(2\) can be placed, after which there
are \(2\) positions in which the \(7\) can be placed, after which there are
\(2\) digits that can be placed in the
remaining position.
Therefore, in this case, there are \(3 \times
2 \times 2 = 12\) possible codes.
Suppose that \(c = 4\) and \(a \neq 4\).
The code thus has the form \(ab4d\).
The remaining digits include a \(2\)
(with the restriction that \(a \neq
2\)), a \(7\), and either a
\(1\) or a \(6\).
There are \(2\) positions in which the
\(2\) can be placed, after which the
\(7\) can be placed in either of the
\(2\) remaining positions, after which
there are \(2\) digits that can be
placed in the remaining position.
Therefore, in this case, there are \(2 \times
2 \times 2 = 8\) possible codes.
In total, there are \(2 + 12 + 8 = 22\) possible codes.
Answer: \(22\)
We label the two other regions \(w\) and \(z\) as shown:
If we start with the area of the larger quarter circle (which is
equal to \(y + w + z\)) and then
subtract the area of the smaller quarter circle (which is equal to \(w\)), we are left \(y + z\).
If we then subtract the area of the rectangle (which is equal to \(x + z\)), we are left with \(y - x\).
In other words, \(y - x\) is equal to
the area of the larger quarter circle minus the area of the smaller
quarter circle minus the area of the rectangle.
The larger quarter circle has radius \(30\) and so its area is \(\frac{1}{4} \pi \times 30^2 =
225\pi\).
The radius of the smaller quarter circle is half of that of the larger
quarter circle, because \(F\) is the
midpoint of \(CE\).
Thus, the smaller quarter circle has radius 15 and so its area is \(\frac{1}{4}\pi \times 15^2 =
\frac{225}{4}\pi\).
The width of the rectangle is equal to \(FC\), which is half of \(CE\) or \(15\).
The height of the rectangle is \(30\),
and so its area is \(15 \times 30 =
450\).
Therefore, \(y - x = 225\pi - \frac{225}{4}\pi
- 450 = \frac{900}{4}\pi -
\frac{225}{4}\pi - 450 = \frac{675}{4}\pi - 450 \approx
80.1\).
This tells us that \(y - x\) is
positive (the diagram certainly makes it look positive), which means
that \(d = y - x\) and the closest
integer to \(d\) is \(80\).
Answer: \(80\)
We write \(a = 3^r\), \(b = 3^s\) and \(c
= 3^t\) where each of \(r\),
\(s\), \(t\) is between 1 and 8, inclusive.
Since \(a \leq b \leq c\), then \(r \leq s \leq t\).
Next, we note that \[\begin{aligned}
\dfrac{ab}{c} &= \dfrac{3^r 3^s}{3^t} = 3^{r+s-t}\\
\dfrac{ac}{b} &= \dfrac{3^r 3^t}{3^s} = 3^{r+t-s} \\
\dfrac{bc}{a} &= \dfrac{3^s 3^t}{3^r} = 3^{s+t-r}
\end{aligned}\]
Since \(t \geq s\), then \(r + t - s = r + (t-s) \geq r > 0\) and
so \(\dfrac{ac}{b}\) is always an
integer.
Since \(t \geq r\), then \(s + t - r = s + (t-r)\geq s > 0\) and so
\(\dfrac{bc}{a}\) is always an
integer.
Since \(\dfrac{ab}{c} = 3^{r+s-t}\),
then \(\dfrac{ab}{c}\) is an integer
exactly when \(r+s-t \geq 0\) or \(t \leq r+s\).
This means that we need to count the number of triples \((r,s,t)\) where \(r \leq s \leq t\), each of \(r\), \(s\), \(t\)
is an integer between \(1\) and \(8\), inclusive, and \(t \leq r + s\).
Suppose that \(r = 1\). Then \(1 \leq s \leq t \leq 8\) and \(t \leq s+1\).
If \(s = 1\), \(t\) can equal \(1\) or \(2\). If \(s =
2\), \(t\) can equal \(2\) or \(3\). This pattern continues so that when
\(s = 7\), \(t\) can equal \(7\) or \(8\). When \(s=8\), though, \(t\) must equal \(8\) since \(t
\leq 8\).
In this case, there are \(2 \times 7 + 1 =
15\) pairs of values for \(s\)
and \(t\) that work, and so \(15\) triples \((r,s,t)\).
Suppose that \(r = 2\). Then \(2 \leq s \leq t \leq 8\) and \(t \leq s + 2\).
This means that, when \(2 \leq s \leq
6\), \(t\) can equal \(s\), \(s+1\) or \(s+2\).
When \(s = 7\), \(t\) can equal \(7\) or \(8\), and when \(s
= 8\), \(t\) must equal \(8\).
In this case, there are \(5 \times 3 + 2 + 1 =
18\) triples.
Suppose that \(r = 3\). Then \(3 \leq s \leq t \leq 8\) and \(t \leq s + 3\).
This means that, when \(3 \leq s \leq
5\), \(t\) can equal \(s\), \(s+1\), \(s+2\), or \(s+3\).
When \(s = 6, 7, 8\), there are \(3\), \(2\), \(1\)
values of \(t\), respectively.
In this case, there are \(3 \times 4 + 3 + 2 +
1 = 18\) triples.
Suppose that \(r = 4\). Then \(4 \leq s \leq t \leq 8\) and \(t \leq s + 4\).
This means that when \(s = 4\), there
are \(5\) choices for \(t\).
As in previous cases, when \(s= 5, 6, 7,
8\), there are \(4\), \(3\), \(2\), \(1\)
choices for \(t\), respectively.
In this case, there are \(5 + 4 + 3 + 2 + 1 =
15\) triples.
Continuing in this way, when \(r=5\), there are \(4+3+2+1=10\) triples, when \(r= 6\), there are \(3 + 2 + 1 = 6\) triples, when \(r = 7\), there are \(2+1 = 3\) triples, and when \(r = 8\), there is \(1\) triple.
The total number of triples \((r,s,t)\)
is \(15 + 18 + 18 + 15 + 10 + 6 + 3 + + 1 =
86\).
Since the triples \((r,s,t)\)
correspond with the triples \((a,b,c)\), then the number of triples \((a,b,c)\) is \(N
= 86\).
Answer: \(86\)