2023 Gauss Contests
Solutions
(Grade 7 and 8)
Wednesday, May 17, 2023
(in North America and South America)
Thursday, May 18, 2023
(outside of North American and South America)
Wednesday, May 17, 2023
(in North America and South America)
Thursday, May 18, 2023
(outside of North American and South America)
©2023 University of Waterloo
Half of 24 is \(24\div2=12\). Kiyana gives her friend 12 grapes.
Answer: (D)
Reading from the graph, Friday had the highest temperature.
Answer: (C)
At a cost of $16.50 a basket, the cost to buy 4 baskets of strawberries is \(4\times\$16.50=\$66.00\).
Answer: (B)
The difference between \(3\) and \(-5\) is \(3-(-5)=3+5=8\). Therefore, it is now \(8\)℃ warmer.
Answer: (A)
Since \(5\times5=25\) and each
of the given answers is greater than 25, then the integer that Sarah
multiplied by itself must have been greater than 5.
Further, \(6\times6=36\) and each of
the given answers is less than or equal to 36.
Thus, of the given answers, only 36 could be the result of multiplying
an integer by itself.
Alternatively, we may have noted that the result of multiplying an
integer by itself is a perfect square, and of the answers given, 36 is
the only perfect square.
Answer: (E)
Since the perimeter of \(PQRS\)
is 40 cm and \(SR=16\) cm, then the
combined length of the remaining three sides is \(40 \text{ cm}-16 \text{ cm}=24\) cm.
Each of the remaining three sides is equal in length, and so \(PQ=\dfrac{24 \text{ cm}}{3}=8\) cm.
Answer: (C)
Dividing 52 by each of the given denominators, we get that \(\dfrac{52}{4}=13\) is the only whole number.
Answer: (C)
The line segment with greatest length that joins two points on a circle is a diameter of the circle. Since the circle has a radius of 4 cm, then its diameter has length \(2\times4 \text{ cm}=8\) cm, and so the greatest possible length of the line segment is 8 cm.
Answer: (B)
The number of integers in the list is 10. Of these integers, 5 are even (they are 10, 12, 14, 16, and 18). Thus, the probability that the chosen integer is even is \(\dfrac{5}{10}\).
Answer: (C)
Before adding tax, the combined cost of the three items is \(\$4.20+\$7.60+\$3.20=\$15.00\).
The 5% tax on $15.00 is \(0.05\times
\$15.00=\$0.75\), and so the total cost of the three items, after
tax is added, is \(\$15.00+\$0.75=\$15.75\).
Note that we could have calculated the 5% tax on each individual item,
however doing so is less efficient than calculating tax on the $15.00
total.
Answer: (D)
Since \(BCD\) is a straight line
segment, then \(\angle BCD
=180\degree\).
Therefore, \(\angle ACB =\angle BCD-\angle
ACD=180\degree-75\degree=105\degree\).
Since the sum of the three angles in \(\triangle ABC\) is \(180\degree\), then \(\angle
ABC=180\degree-105\degree-35\degree=40\degree\).
Answer: (B)
Of the 100 small identical squares, 28 are presently unshaded,
and so \(100-28=72\) are shaded.
So that 75% of the area of \(WXYZ\) is
shaded, 75 of the 100 small squares must be shaded.
Therefore, \(75-72=3\) more of the
small squares must be shaded.
Answer: (A)
Suppose we call the unknown vertex \(V\).
The side of the rectangle joining the points \((2,1)\) and \((2,5)\) is vertical, and so the opposite
side of the rectangle (the side joining \((4,1)\) to \(V\)) must also be vertical.
This means that \(V\) has the same
\(x\)-coordinate as \((4,1)\), which is 4.
Similarly, the side of the rectangle joining the points \((2,1)\) and \((4,1)\) is horizontal, and so the opposite
side of the rectangle (the side joining \((2,5)\) to \(V\)) must also be horizontal.
This means that \(V\) has the same
\(y\)-coordinate as \((2,5)\), which is 5.
Therefore, the coordinates of the fourth vertex of the rectangle are
\((4,5)\).
Answer: (D)
The prime numbers that are less than 10 are \(2,3,5\), and 7.
Thus, the only two different prime numbers whose sum is 10 are 3 and
7.
The product of these two numbers is \(3\times7=21\).
Answer: (D)
The given list, \(2,9,4,n,2n\)
contains 5 numbers.
The average of these 5 numbers is 6, and so the sum of the 5 numbers is
\(5\times 6=30\).
That is, \(2+9+4+n+2n=30\) or \(15+3n=30\), and so \(3n=15\) or \(n=5\).
Answer: (D)
The sum of \(P\) and \(Q\) is equal to 5, and so \(P\) and \(Q\) are (in some order) either equal to 1
and 4, or they are equal to 2 and 3.
The difference between \(R\) and \(S\) is equal to 5, and so \(R\) and \(S\) must be (in some order) equal to 1 and
6.
Since \(R\) and \(S\) are equal to 1 and 6, then neither
\(P\) nor \(Q\) can be 1, which means that \(P\) and \(Q\) cannot be equal to 1 and 4, and so they
must be equal to 2 and 3.
The only numbers from 1 to 6 not accounted for are 4 and 5.
Since \(T\) is greater than \(U\), then the number that replaces \(T\) is 5.
Answer: (E)
Solution 1
The area of \(\triangle AED\) is
equal to one-half its base times its height.
Suppose the base of \(\triangle AED\)
is \(AE\), then its height is \(BD\) (\(AE\) is perpendicular to \(BD\)).
Since \(AB=BC=24\) cm and \(E\) and \(D\) are the midpoints of their respective
sides, then \(AE=12\) cm and \(BD=12\) cm.
Thus, the area of \(\triangle AED\) is
\(\frac12\times12\text{ cm}\times12\text{
cm}=72 \text{ cm}^2\).
Solution 2
The area of \(\triangle AED\) is
equal to the area of \(\triangle ABD\)
minus the area of \(\triangle
EBD\).
Suppose the base of \(\triangle EBD\)
is \(BD\), then its height is \(EB\).
Since \(AB=BC=24\) cm and \(E\) and \(D\) are the midpoints of their respective
sides, then \(EB=12\) cm and \(BD=12\) cm.
Thus, the area of \(\triangle EBD\) is
\(\frac12\times12\text{ cm}\times12\text{
cm}=72 \text{ cm}^2\).
The area of \(\triangle ABD\) is equal to \(\frac12\times BD\times AB=\frac12\times 12\text{ cm} \times 24\text{ cm}=144\text{ cm}^2\).
Thus, the area of \(\triangle AED\) is \(144\text{ cm}^2-72\text{ cm}^2=72\text{ cm}^2\).
Answer: (C)
The water is in the shape of a rectangular prism with a 2 cm by 5
cm base and depth 6 cm.
Therefore, the volume of water is \(2 \text{
cm} \times 5 \text{ cm} \times 6\text{ cm} = 60\text{
cm}^3\).
A face of the prism having the greatest area has dimensions 5 cm by 8
cm.
When the prism is tipped so that it stands on a 5 cm by 8 cm face, the
water is once again in the shape of a rectangular prism with a 5 cm by 8
cm base and unknown depth.
Suppose that after the prism is tipped, the water’s depth is \(d\) cm.
Since the volume of water is still \(60\text{
cm}^3\) when the prism is tipped, then \(5 \text{ cm} \times 8 \text{ cm} \times d\text{
cm} = 60\text{ cm}^3\) or \(40d\text{
cm}^3 = 60\text{ cm}^3\), and so \(d=\dfrac{60}{40}=\dfrac32\).
When the prism is tipped so that it stands on a face with the greatest
area, the depth of the water is \(\dfrac32
\text{ cm}=1.5\) cm.
Answer: (D)
Solution 1
We begin by completing a table in which the ones digit of each
possible product is listed.
For example, when the number on the first die is 3 and the number on the
second die is 6, the entry in the table is 8 since \(3\times6=18\) and the ones digit of 18 is
8.
| \(\times\) | Number on the second die | ||||||
|---|---|---|---|---|---|---|---|
| 1 | 2 | 3 | 4 | 5 | 6 | ||
| Number on the First Die | 1 | 1 | 2 | 3 | 4 | 5 | 6 |
| 2 | 2 | 4 | 6 | 8 | 0 | 2 | |
| 3 | 3 | 6 | 9 | 2 | 5 | 8 | |
| 4 | 4 | 8 | 2 | 6 | 0 | 4 | |
| 5 | 5 | 0 | 5 | 0 | 5 | 0 | |
| 6 | 6 | 2 | 8 | 4 | 0 | 6 | |
Of the 36 possible outcomes in the table above, 6 outcomes have a
ones digit that is equal to 0.
Thus, the probability that the ones digit of the product is 0 is \(\dfrac{6}{36}=\dfrac{1}{6}\).
Solution 2
Since the ones digit of the product is 0, then the product is
divisible by 5 and is even.
Since the possible numbers in the product are 1, 2, 3, 4, 5, 6, then one
of the numbers rolled must be 5.
Since the product is even (and 5 is not), then the other number rolled
must be one of the three even numbers, namely 2, 4, 6.
Thus, the possible pairs of numbers that can be rolled to give a product
whose ones digit is 0, are \((5,2)\),
\((5,4)\), \((5,6)\) or \((2,5)\), \((4,5)\), \((6,5)\). (We note that the first number in
the ordered pair represents the first number rolled, while the second
number in the pair is the second number rolled.)
Since there are 6 possible rolls for each of the two dice, then there
are \(6\times6=36\) possible ordered
pairs representing all possible outcomes.
Since 6 of these ordered pairs represent a product whose ones digit is
0, then the required probability is \(\dfrac{6}{36}=\dfrac{1}{6}\).
Answer: (D)
Since \(a\) and \(b\) are positive integers, then each of
\(\dfrac{a}{7}\) and \(\dfrac{2}{b}\) is greater than 0.
The sum of \(\dfrac{a}{7}\) and \(\dfrac{2}{b}\) is equal to 1, and so each
is less than 1.
Since \(\dfrac{a}{7}\) is greater than
0 and less than 1, then the possible values of \(a\) are \(1,2,3,4,5,6\).
By substituting each of these values for \(a\) into the equation one at a time, we can
determine if there is a positive integer value of \(b\) for which the equation is true.
Substituting \(a=1\), we get \(\dfrac{1}{7}+\dfrac{2}{b}=1\) or \(\dfrac{2}{b}=1-\dfrac{1}{7}\), and so \(\dfrac{2}{b}=\dfrac{6}{7}\).
Since \(\dfrac{2}{b}=\dfrac{6}{7}\), we can multiply the numerator and denominator of the first fraction by 3 (which is \(6\div2\)) to get \(\dfrac{6}{3b}=\dfrac{6}{7}\).
This gives \(3b=7\) which does not
have an integer solution (\(b=\dfrac73\)).
Thus when \(a=1\), there is no positive
integer value of \(b\) that satisfies
the equation.
Substituting \(a=2\), we get \(\dfrac{2}{7}+\dfrac{2}{b}=1\) or \(\dfrac{2}{b}=1-\dfrac{2}{7}\), and so \(\dfrac{2}{b}=\dfrac{5}{7}\).
Since \(\dfrac{2}{b}=\dfrac{5}{7}\), we
can multiply the numerator and denominator of the first fraction by 5,
and the numerator and denominator of the second fraction by 2 to get
\(\dfrac{10}{5b}=\dfrac{10}{14}\).
This gives \(5b=14\) which does not
have an integer solution.
Thus when \(a=2\), there is no positive
integer value of \(b\) that satisfies
the equation.
Substituting \(a=3\), we get \(\dfrac{3}{7}+\dfrac{2}{b}=1\) or \(\dfrac{2}{b}=1-\dfrac{3}{7}\), and so \(\dfrac{2}{b}=\dfrac{4}{7}\).
Since \(\dfrac{2}{b}=\dfrac{4}{7}\), we
can multiply the numerator and denominator of the first fraction by 2 to
get \(\dfrac{4}{2b}=\dfrac{4}{7}\).
This gives \(2b=7\) which does not have
an integer solution. Thus when \(a=3\),
there is no positive integer value of \(b\) that satisfies the equation.
Substituting \(a=4\) and
simplifying, we get \(\dfrac{2}{b}=\dfrac{3}{7}\).
Since \(\dfrac{2}{b}=\dfrac{3}{7}\), we
can multiply the numerator and denominator of the first fraction by 3,
and the numerator and denominator of the second fraction by 2 to get
\(\dfrac{6}{3b}=\dfrac{6}{14}\).
This gives \(3b=14\) which does not have an integer solution.
Thus when \(a=4\), there is no positive integer value of \(b\) that satisfies the equation.
Substituting \(a=5\) and
simplifying, we get \(\dfrac{2}{b}=\dfrac{2}{7}\).
Since the numerators are equal, then the denominators must be equal, and
so \(b=7\) satisfies the equation.
Finally, substituting \(a=6\) and
simplifying, we get \(\dfrac{2}{b}=\dfrac{1}{7}\).
Since \(\dfrac{2}{b}=\dfrac{1}{7}\), we
can multiply the numerator and denominator of the second fraction by 2
to get \(\dfrac{2}{b}=\dfrac{2}{14}\),
and so \(b=14\).
Thus, there are two pairs of positive integers \(a\) and \(b\) that satisfy the given equation: \(a=5, b=7\) and \(a=6, b=14\).
Answer: (E)
Since \(ABCD\) is a square and
its side lengths are integers, then its area is equal to a perfect
square.
Since the product of the areas of \(ABCD\) and \(EFGH\) (the rectangle) is equal to 98, then
the area of \(ABCD\) is a divisor of
98.
The positive divisors of 98 are 1, 2, 7, 14, 49, and 98.
There are exactly two divisors of 98 that are perfect squares, namely 1
and 49.
Since the area of \(ABCD\) is greater
than the area of \(EFGH\), then the
area of \(ABCD\) is 49, and so the area
of \(EFGH\) is 2 (since \(49\times2=98\)).
Square \(ABCD\) has area 49, and so
\(AB=BC=CD=DA=7\).
The perimeter of \(ABCDEFGH\) is
equal to \[\begin{align*}
&AB+BC+CD+DE+EF+FG+GH+HA & \\
= &7+7+7+DE+EF+EH+GH+HA & \text{(since $EH=FG$)}\\
= &21+DE+EH+HA+EF+GH & \text{(reorganizing)}\\
= &21+DA+EF+GH & \text{(since $DE+EH+HA=DA$)}\\
= &21+7+EF+GH & \text{(since $DA=7$)}\\
= &28+2\times GH & \text{(since $EF=GH$)}\end{align*}\]
Since the side lengths are integers and the area of \(EFGH\) is 2, then either \(GH=1\) (and \(FG=2\)), or \(GH=2\) (and \(FG=1\)).
If \(GH=1\), then the perimeter of
\(ABCDEFGH\) is \(28+2\times1=30\).
Since 30 is not given as a possible answer, then \(GH=2\) and the perimeter is \(28+2\times2=32\).
Answer: (B)
If a Gareth sequence begins 10, 8, then the 3rd number in the
sequence is \(10-8=2\), the 4th is
\(8-2=6\), the 5th is \(6-2=4\), the 6th is \(6-4=2\), the 7th is \(4-2=2\), the 8th is \(2-2=0\), the 9th is \(2-0=2\), the 10th is \(2-0=2\), and the 11th is \(2-2=0\).
Thus, the resulting sequence is \(10,8,2,6,4,2,2,0,2,2,0,\dots\).
The first 5 numbers in the sequence are \(10,8,2,6,4\), the next 3 numbers are \(2,2,0\), and this block of 3 numbers
appears to repeat.
Since each new number added to the end of this sequence is determined by
the two previous numbers in the sequence, then this block of 3 numbers
will indeed continue to repeat. (That is, since the block repeats once,
then it will continue repeating.)
The first 30 numbers of the sequence begins with the first 5 numbers,
followed by 8 blocks of \(2,2,0\),
followed by one additional 2 (since \(5+8\times 3+1=30\)).
The sum of the first 5 numbers is \(10+8+2+6+4=30\).
The sum of each repeating block is \(2+2+0=4\), and so the sum of 8 such blocks
is \(8\times 4=32\).
Thus, the sum of the first 30 numbers in the sequence is \(30+32+2=64\).
Answer: (E)
Suppose that the length, or the width, or the height of the
rectangular prism is equal to 5.
The product of 5 with any of the remaining digits has a units (ones)
digit that is equal to 5 or it is equal to 0.
This means that if the length, or the width, or the height of the
rectangular prism is equal to 5, then at least one of the two-digit
integers (the area of a face) has a units digit that is equal to 5 or
0.
However, 0 is not a digit that can be used, and each digit from 1 to 9
is used exactly once (that is, 5 cannot be used twice), and so it is not
possible for one of the dimensions of the rectangular prism to equal
5.
Thus, the digit 5 occurs in one of the two-digit integers (the area of a
face).
The digit 5 cannot be the units digit of the area of a face, since this
would require that one of the dimensions be 5.
Therefore, one of the areas of a face has a tens digit that is equal to
5.
The two-digit integers with tens digit 5 that are equal to the product
of two different one-digit integers (not equal to 5) are \(54=6\times 9\) and \(56=7\times 8\).
Suppose that two of the dimensions of the prism are 7 and 8, and so one
of the areas is 56.
In this case, the digits \(5,6,7\), and
8 have been used, and so the digits \(1,2,3,4\), and 9 remain.
Which of these digits is equal to the remaining dimension of the
prism?
It cannot be 1 since the product of 1 and 7 does not give a two-digit
area, nor does the product of 1 and 8.
It cannot be 2 since the product of 2 and 8 is 16 and the digit 6 has
already been used.
It cannot be 3 since \(3\times 7=21\)
and \(3\times 8=24\), and so the areas
of two faces share the digit 2.
It cannot be 4 since \(4\times 7=28\)
and the digit 8 has already been used.
Finally, it cannot be 9 since \(9\times7=63\) and the digit 6 has already
been used.
Therefore, it is not possible for 7 and 8 to be the dimensions of the
prism, and thus 6 and 9 must be two of the three dimensions.
Using a similar systematic check of the remaining digits, we determine
that 3 is the third dimension of the prism.
That is, when the dimensions of the prism are \(3,6\) and 9, the areas of the faces are
\(3\times 6=18\), \(3\times 9=27\), and \(6\times9=54\), and we may confirm that each
of the digits from 1 to 9 has been used exactly once.
Since the areas of the faces are 18, 27 and 54, the surface area of the
rectangular prism is \(2\times(18+27+54)\) or \(2\times99=198\).
Answer: (D)
Solution 1
Begin by colouring the section at the top blue.
Since two circles have the same colouring if one can be rotated to match
the other, it does not matter which section is coloured blue, so we
arbitrarily choose the top section.
There are now 5 sections which can be coloured green.
After choosing the section to be coloured green, there are 4 sections
remaining which can be coloured yellow.
Each of the remaining 3 sections must then be coloured red.
Thus, the total number of different colourings of the circle is \(5 \times 4 = 20\).
Solution 2
We begin by considering the locations of the three sections coloured
red, relative to one another.
The three red sections could be adjacent to one another, or exactly two
red sections could be adjacent to one another, or no red section could
be adjacent to another red section.
We consider each of these 3 cases separately.
Case 1: All three red sections are adjacent to one another.
Begin by colouring any three adjacent sections red.
Since two circles have the same colouring if one can be rotated to
match the other, it does not matter which three adjacent sections are
coloured red.
Consider the first section that follows the three red sections as we
move clockwise around the circle.
There are 3 choices for the colour of this section: blue, green or
yellow.
Continuing to move clockwise to the next section, there are now 2
choices for the colour of this section.
Finally, there is 1 choice for the colour of the final section, and thus
there are \(3\times2\times1=6\)
different colourings of the circle in which all three red sections are
adjacent to one another.
These 6 colourings are shown below.
Case 2: Exactly two red sections are adjacent to one another.
There are two different possible arrangements in which exactly two
red sections are adjacent to one another.
In the first of these, the next two sections that follow the two
adjacent red sections as we move clockwise around the circle, are both
not red. We call this Case 2a.
In the second of these, the section that follows the two adjacent red
sections as we move clockwise around the circle is not red, but the next
section is. We call this Case 2b.
The arrangements for Cases 2a and 2b are shown below.
Notice that the first of these two circles cannot be rotated to match
the second.
The number of colourings in Case 2a and in Case 2b are each equal to the
number of colourings as in Case 1.
That is, there are 3 choices for the first uncoloured section that
follows the two red sections as we move clockwise around the
circle.
Continuing to move clockwise to the next uncoloured section, there are
now 2 choices for the colour of this section.
Finally, there is 1 choice for the colour of the final section, and thus
there are \(3\times2\times1=6\)
different colourings of the circle in Case 2a as well as in Case
2b.
These 12 colourings are shown below.
Case 3: No red section is adjacent to another red section.
Begin by colouring any three non-adjacent sections red.
Since two circles have the same colouring if one can be rotated to
match the other, it does not matter which three non-adjacent sections
are coloured red.
In this case, there are 2 possible colourings as shown below.
A circle with any other arrangement of the green, yellow and blue sections can be rotated to match one of the two circles above.
The total number of different colourings of the circle is \(6+12+2=20\).
Answer: (E)
We can represent the given information in a Venn diagram by first
introducing some variables.
Let \(x\) be the number of students
that participated in hiking and canoeing, but not swimming.
Let \(y\) be the number of students
that participated in hiking and swimming, but not canoeing.
Let \(z\) be the number of students
that participated in canoeing and swimming, but not hiking.
Since 10 students participated in all three activities and no students
participated in fewer than two activities, we complete the Venn diagram
as shown.
Suppose that the total number of students participating in the school
trip was \(n\).
Since 50% of all students participated in at least hiking and canoeing,
then \(\dfrac{50}{100}n\) or \(\dfrac{n}{2}\) participated in at least
hiking and canoeing.
Since this number of students, \(\dfrac{n}{2}\), is an integer, then \(n\) must be divisible by 2.
Similarly, \(\dfrac{60}{100}n\) or
\(\dfrac{3n}{5}\) students participated
in at least hiking and swimming.
Since this number of students, \(\dfrac{3n}{5}\), is an integer, then \(n\) must be divisible by 5 (since 3 and 5
have no factors in common).
This means that \(n\) is divisible by
both 2 and 5, and thus \(n\) is
divisible by 10.
From the Venn diagram, we see that \(x+10=\dfrac{n}{2}\), and \(y+10=\dfrac{3n}{5}\).
Since the total number of participants is \(n\), we also get that \(x+y+z+10=n\) or \(z=n-10-x-y\).
We may now use these equations, \[x=\dfrac{n}{2}-10,\ y=\dfrac{3n}{5}-10, \text{
and } z=n-10-x-y\] and the fact that \(n\) is divisible by 10, to determine all
possible values of \(z\).
We can then use the values of \(z\) to
determine all possible values of the positive integer \(k\), where \(k\%\) participated in at least canoeing and
swimming.
Since \(n\) is a positive integer
that is divisible by 10, its smallest possible value is 10.
However, substituting \(n=10\) into
\(x=\dfrac{n}{2}-10\), we get \(x=5-10\) and so \(x=-5\) which is not possible. (Recall that
\(x\) is the number of students that
participated in hiking and canoeing, but not swimming, and so \(x\geq0\).)
Next, we try \(n=20\).
When \(n=20\), \(x=10-10\) and so \(x=0\).
When \(n=20\), \(y=\dfrac{3\times20}{5}-10\) or \(y=12-10\), and so \(y=2\).
Finally, when \(n=20\), \(x=0\), and \(y=2\), we get \(z=20-10-0-2=8\).
When \(z=8\), the number of students
who participated in at least canoeing and swimming is \(8+10=18\) (since 10 students participated
in all three), and so the percentage of students who participated in at
least canoeing and swimming is \(\dfrac{18}{20}\times 100\%=90\%\), and so
\(k=90\).
In the table below, we continue in this way by using successively
greater multiples of 10 for the value of \(n\).
| \(n\) | \(x=\dfrac{n}{2}-10\) | \(y=\dfrac{3n}{5}-10\) | \(z=n-10-x-y\) | \(k=\dfrac{z+10}{n}\times100\) |
|---|---|---|---|---|
| 20 | 0 | 2 | 8 | \(k=\dfrac{8+10}{20}\times100=90\) |
| 30 | 5 | 8 | 7 | \(k=\dfrac{7+10}{30}\times100\approx56.7\) |
| 40 | 10 | 14 | 6 | \(k=\dfrac{6+10}{40}\times100=40\) |
| 50 | 15 | 20 | 5 | \(k=\dfrac{5+10}{50}\times100=30\) |
| 60 | 20 | 26 | 4 | \(k=\dfrac{4+10}{60}\times100\approx23.3\) |
| 70 | 25 | 32 | 3 | \(k=\dfrac{3+10}{70}\times100\approx18.6\) |
| 80 | 30 | 38 | 2 | \(k=\dfrac{2+10}{80}\times100=15\) |
| 90 | 35 | 44 | 1 | \(k=\dfrac{1+10}{90}\times100\approx12.2\) |
| 100 | 40 | 50 | 0 | \(k=\dfrac{0+10}{100}\times100=10\) |
For values of \(n\) that are greater
than 100, we get that \(z<0\), which
is not possible.
Therefore, the sum of all such positive integers \(k\) is \(90+40+30+15+10=185\).
Answer: (B)
The fraction \(\frac14\) is equivalent to \(1\div4=0.25\).
Answer: (B)
Reading from the graph, the forecast wind speed is less than 20
km/h on Monday, Tuesday, Wednesday, and Sunday.
Thus, Jack will be able to sail alone on 4 days during this 7-day
period.
Answer: (A)
We note that \(15\times
10=150\), \(15\times 2=30\),
\(15\times 3=45\), and \(15\times 4=60\).
Since there is no integer \(n\) for
which \(15\times n=25\), then 25 is not
a multiple of 15.
Answer: (B)
Ordering the given list of integers from least to greatest, we
get \(-9,-7,0,9,10\).
The third integer in the ordered list is 0.
Answer: (D)
Solution 1
If \(2n=14\), then \(n=\dfrac{14}{2}=7\).
When \(n=7\), the value of \(10n\) is \(10\times 7=70\).
Solution 2
Multiplying both sides of the given equation by 5, we get \(5\times 2n=5\times14\), and so \(10n=70\).
Answer: (C)
There are 6 possible outcomes when Tallulah rolls a single
standard die once.
She loses if she rolls 2 of these 6 outcomes, and so the probability
that she loses is \(\dfrac26=\dfrac13\).
Answer: (A)
Solution 1
We may convert the given addition problem to a subtraction
problem.
That is, since \(1013+PQPQ=2023\), then
\(PQPQ=2023-1013\).
The difference between 2023 and 1013 is \(2023-1013=1010\), and so \(P=1\), \(Q=0\), and \(P+Q=1+0=1\).
Solution 2
The ones (units) digit of the sum 2023 is 3.
Thus, the ones digit of the sum \(3+Q\)
must equal 3.
Since \(Q\) is a digit, the only
possible value of \(Q\) is 0.
The tens digit of the sum 2023 is 2.
Since there is no “carry" from the ones column to the tens column, the
ones digit of the sum \(1+P\) must
equal 2.
Since \(P\) is a digit, the only
possible value of \(P\) is 1.
We may confirm that when \(P=1\) and
\(Q=0\), we get \(1013+1010=2023\) as required.
The value of \(P+Q\) is \(1+0=1\).
Answer: (B)
Suppose the salad dressing initially contains 300 mL of
oil.
Since the ratio of oil to vinegar is \(3:1\), then the salad dressing initially
contains one-third as much vinegar as oil, or \(\dfrac13\times 300 \text{ mL}=100\) mL of
vinegar (note that \(300:100=3:1\)).
If the volume of vinegar is doubled, the new volume of vinegar is
\(2\times 100\) mL \(=200\) mL, and so the new ratio of oil to
vinegar is \(300:200=3:2\).
Note: We chose to begin with 300 mL of oil, but performing the above
calculations with any starting volume of oil will give the same \(3:2\) ratio.
Answer: (A)
Before including tax, the combined cost of the three items is
\(\$4.20+\$7.60+\$3.20=\$15.00\).
The 5% tax on $15.00 is \(0.05\times
\$15.00=\$0.75\), and so the total cost of the three items,
including tax, is \(\$15.00+\$0.75=\$15.75\).
Note that we could have calculated the 5% tax on each individual item,
however doing so is less efficient than calculating tax on the $15.00
total.
Answer: (D)
When \((1,3)\) is reflected in
the \(y\)-axis, the reflected point is
\((-1,3)\).
In general, when a point is reflected in the \(y\)-axis, its \(x\)-coordinate changes sign, and its \(y\)-coordinate does not change.
Thus, the vertices of the reflected rectangle are \((-1,3)\), \((-1,1)\), \((-4,1)\), and \((-4,3)\).
Of the given possibilities, \((-3,4)\)
is not a vertex of the reflected rectangle.
Answer: (C)
In the leftmost rectangle, the length of the path along the
rectangle’s diagonal, \(d\), and the
sides with lengths 3 and 4, form a right-angled triangle.
Using the Pythagorean Theorem, we get \(d^2=3^2+4^2\), and so \(d=\sqrt{3^2+4^2}=\sqrt{25}=5\) (these are
“3-4-5" right-angled triangles).
The path from \(A\) to \(B\) includes one such diagonal, two
vertical sides each of length 4, and three horizontal sides each of
length 3, and thus has length \(5+(2\times4)+(3\times3)=22\).
Answer: (A)
Since \(\angle PQR\) is a
straight angle, its measure is \(180\degree\), and so \(\angle
SQR=180\degree-125\degree=55\degree\).
Since \(SQ=SR\), then \(\angle SRQ=\angle SQR=55\degree\).
The sum of the angles inside \(\triangle
SQR\) is \(180\degree\), and so
\(x=180-55-55=70\).
Answer: (B)
The number of peaches in the original pile is two more than a
multiple of three.
Of the choices given, 29 is the only number which is two more than a
multiple of three (\(29=3\times9+2\)).
Answer: (D)
The sum of each block of 5 repeating integers is \(4-3+2-1+0=2\).
In the first 23 integers, there are 4 such blocks of 5 integers, plus 3
additional integers (since \(23=4\times5+3\)).
The sum of the 4 blocks of 5 repeating integers is \(4\times2=8\), and the next three integers
in the list following these 4 blocks are \(4,-3,2\).
Thus, the sum of the first 23 integers is \(8+4-3+2=11\).
Answer: (D)
The circumference of each of Bindu’s bike tires is \(2\times\pi\times30\text{ cm}=60\pi\)
cm.
If the bike tires rotate exactly five times, then the distance travelled
by Bindu’s bike is \(5\times60\pi\text{
cm}=300\pi\text{ cm}\).
Answer: (D)
Solution 1
The sum of all 8 numbers in the list is \(41+35+19+9+26+45+13+28=216\).
When the 8 numbers are arranged in pairs, there are 4 pairs.
The sum of the numbers in each pair is the same, and so this sum is
\(\dfrac{216}{4}=54\).
Therefore, the number paired with 13 is \(54-13=41\).
Note: We may confirm that \(45+9=54\),
\(41+13=54\), \(35+19=54\), and \(28+26=54\).
Solution 2
Since the sum of the numbers in each pair is the same, then the
largest number in the list must be paired with the smallest number, the
second largest with the second smallest, and so on.
(Can you reason why this must be true?)
That is, the largest and smallest numbers in the list, 45 and 9, must be
paired.
The second largest and second smallest numbers, 41 and 13, must be
paired, and so the number paired with 13 is 41.
Note: We may confirm that \(45+9=54\),
\(41+13=54\), \(35+19=54\), and \(28+26=54\).
Answer: (E)
The mean (average) is determined by adding the 30 recorded
temperatures, and dividing the sum by 30.
The sum of the temperatures for the first 25 days was \((25\times
21)\text{℃}=525\text{℃}\).
The sum of the temperatures for the last 5 days was \((5\times
15)\text{℃}=75\text{℃}\).
Thus, the mean of the recorded temperatures was \(\dfrac{525\text{℃}+75\text{℃}}{30}=\dfrac{600\text{℃}}{30}=20\text{℃}\).
Answer: (C)
We begin by listing, in order, the smallest 2-digit positive
divisors of 630.
We note that by first writing 630 as a product of its prime factors
(\(630=2\times3^2\times5\times7\)), it
may be easier to determine these divisors.
The smallest five 2-digit positive divisors of 630 are \(10,14,15,18\), and 21.
The next largest 2-digit positive divisor of 630 is 30, and we notice
that \(21\times30=630\).
That is, 21 and 30 are a pair of 2-digit positive integers whose product
is 630, and they are consecutive in the ordered list of positive
divisors.
Thus, each of \(10,14,15\), and 18 must
be paired with a divisor of 630 that is greater than 30.
We may check that the divisors that pair with each of \(10,14,15\), and 18 is a 2-digit positive
integer by dividing 630 by the smaller divisor.
That is, \(\dfrac{630}{18}=35\), \(\dfrac{630}{15}=42\), \(\dfrac{630}{14}=45\), and \(\dfrac{630}{10}=63\).
Thus, the pairs of 2-digit positive integers whose product is 630 are 21 and 30, 18 and 35, 15 and 42, 14 and 45, and 10 and 63, and so there are 5 such pairs.
Answer: (D)
Between 9 a.m. and 10 a.m., Ryan cut \(\dfrac78-\dfrac12=\dfrac78-\dfrac48=\dfrac38\)
of his lawn.
Ryan cut \(\dfrac38\) of his lawn in 1
hour (60 minutes), and so he cut \(\dfrac18\) of his lawn in \(\dfrac{60}{3}\) minutes or 20
minutes.
At 10 a.m., Ryan had cut \(\dfrac78\)
of his lawn, and thus had \(1-\dfrac78=\dfrac18\) of his lawn left to
cut.
Since Ryan cuts \(\dfrac18\) of his
lawn in 20 minutes, then he finished at 10:20 a.m.
Answer: (C)
Begin by placing four tiles in the squares of the first
row.
The only restriction is that the row must contain one tile of each
colour.
Thus in the first row, there are 4 choices of tile colour for the first
column, 3 choices for the second, 2 for the third, and 1 choice for the
fourth.
That is, there are \(4\times3\times2\times1=24\) different ways
to cover the squares in the first row using one tile of each of the four
colours.
For example, using \(R\) for red, \(B\) for black, \(G\) for green, and \(Y\) for yellow, the first row could contain
tiles coloured \(G~Y~R~B\), in that
order.
Suppose that the first row does contain tiles coloured \(G~Y~R~B\), in that order.
We will show that there is only one way to arrange the remaining 12
tiles in the grid.
Consider the first square in the second row, that is, the square
directly below the tile coloured \(G\).
The tile placed in this square cannot be coloured \(G\) since it shares an edge with the tile
coloured \(G\) in row 1.
Also, the tile placed in this square cannot be coloured \(Y\) since it touches the corner of the
square containing the tile coloured \(Y\) in row 1. Assume that the tile in the
first square of row 2 is coloured \(B\), as shown.
Next, consider the colour of the tiles that could be placed in the
second square of row 2.
The tile in this square cannot be coloured \(Y\) since it shares an edge with the tile
coloured \(Y\) in row 1.
Also, the tile in this square cannot be coloured \(G\) since it touches the corner of the
square containing the tile coloured \(G\) in row 1. Further, the tile in this
square cannot be coloured \(R\) since
it touches the corner of the square containing the tile coloured \(R\) in row 1.
Since the first square in this row contains a tile coloured \(B\), then we have no possible tile that can
be placed in the second square of row 2.
This means that the tile in the first square of row 2 cannot be coloured
\(B\), and thus it must be coloured
\(R\), as shown.