-
The expression $6a - 5a + 4a - 3a + 2a - a$ is equal to
- $3a$
- $3a^6$
- $3$
- $-21a$
- $-21a^6$
Answer: A
Solution:
Simplifying the terms in pairs, $6a-5a+4a-3a+2a-a=a+a+a=3a$.-
When $x=9$, which of the following has the largest value?
- $\sqrt{x}$
- $\displaystyle\frac{x}{2}$
- $x-5$
- $\displaystyle\frac{40}{x}$
- $\displaystyle\frac{x^2}{20}$
Answer: B
Solution:
Evaluating each of the given choices with $x=9$, \[ \sqrt{9}=3~~~~~~~\frac{9}{2}=4\textstyle\frac{1}{2}~~~~~~~9-5=4~~~~~~~\frac{40}{9} = 4 \textstyle\frac{4}{9}~~~~~~~\frac{9^2}{20}=\frac{81}{20}=4 \textstyle\frac{1}{20}\] Since $\frac{1}{2}$ is larger than either $\frac{4}{9}$ or $\frac{1}{20}$, then the largest of the possibilities when $x=9$ is $\frac{x}{2}$.-
In the diagram, what is the area of $\triangle ABC$?
- $36$
- $54$
- $108$
- $72$
- $48$
Answer: B
Solution:
We think of $BC$ as the base of $\triangle ABC$. Its length is 12.Since the $y$-coordinate of $A$ is 9, then the height of $\triangle ABC$ from base $BC$ is 9.
Therefore, the area of $\triangle ABC$ is $\frac{1}{2}(12)(9)=54$.
-
In the diagram, $JLMR$ and $JKQR$ are rectangles. Also, $JR=2$, $RQ=3$ and $JL=8$. What is the area of rectangle $KLMQ$?
- $6$
- $16$
- $10$
- $15$
- $24$
Answer: C
Solution:
Solution 1Since $JLMR$ is a rectangle and $JR=2$, then $LM=2$.
Similarly, since $JL=8$, then $RM=8$.
Since $RM=8$ and $RQ=3$, then $QM=8-3=5$.
Since $KLMQ$ is a rectangle with $QM=5$ and $LM=2$, its area is $5(2)=10$.
Solution 2
Since $JL=8$ and $JR=2$, then the area of rectangle $JLMR$ is $2(8)=16$.
Since $RQ=3$ and $JR=2$, then the area of rectangle $JKQR$ is $2(3)=6$.
The area of rectangle $KLMQ$ is the difference between these areas, or $16-6=10$.
-
The surface area of a large cube is 5400 cm$^2$. This cube is cut into a number of identical smaller cubes. Each smaller cube has a volume of 216 cm$^3$. How many smaller cubes are there?
- $25$
- $125$
- $164$
- $180$
- $216$
Answer: B
Solution:
Solution 1The large cube has a total surface area of $5400\mbox{ cm}^2$ and its surface is made up of 6 identical square faces. Thus, the area of each face, in square centimetres, is $5400 \div 6 = 900$.
Because each face is square, the side length of each face is $\sqrt{900}=30$ cm.
Therefore, each edge of the cube has length 30 cm and so the large cube has a volume of $30^3 = 27\,000\mbox{ cm}^3$.
Because the large cube is cut into small cubes each having volume $216\mbox{ cm}^3$, then the number of small cubes equals $27\,000\div 216=125$.
Solution 2
Since the large cube has 6 square faces of equal area and the total surface area of the cube is $5400\mbox{ cm}^2$, then the surface area of each face is $5400 \div 6 = 900\mbox{ cm}^2$.
Since each face is square, then the side length of each square face of the cube is $\sqrt{900}=30\mbox{ cm}$, and so the edge length of the cube is 30 cm.
Since each smaller cube has a volume of $216\mbox{ cm}^3$, then the side length of each smaller cube is $\sqrt[3]{216}=6\mbox{ cm}$.
Since the side length of the large cube is 30 cm and the side length of each smaller cube is 6~cm, then $30 \div 6 = 5$ smaller cubes fit along each edge of the large cube.
Thus, the large cube is made up of $5^3 = 125$ smaller cubes.
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If $10\%$ of $s$ is $t$, then $s$ equals
- $0.1t$
- $0.9t$
- $9t$
- $10t$
- $90t$
Answer: D
Solution:
The percentage $10\%$ is equivalent to the fraction $\frac{1}{10}$.Therefore, $t = \frac{1}{10}s$, or $s=10t$.
-
John lists the integers from 1 to 20 in increasing order. He then erases the first half of the integers in the list and rewrites them in order at the end of the second half of the list. Which integer in the new list has exactly 12 integers to its left?
- $1$
- $2$
- $3$
- $12$
- $13$
Answer: C
Solution:
John first writes the integers from 1 to 20 in increasing order.When he erases the first half of the numbers, he erases the numbers from 1 to 10 and rewrites these at the end of the original list.
Therefore, the number 1 has 10 numbers to its left. (These numbers are $11, 12, \ldots, 20$.)
Thus, the number 2 has 11 numbers to its left, and so the number 3 has 12 numbers to its left. (We could write out the new list to verify this.)
-
Two circles are centred at the origin, as shown. The point $P(8,6)$ is on the larger circle and the point $S(0,k)$ is on the smaller circle. If $QR=3$, what is the value of $k$?
- $3.5$
- $4$
- $6$
- $6.5$
- $7$
Answer: E
Solution:
We can determine the distance from $O$ to $P$ by dropping a perpendicular from $P$ to $T$ on the $x$-axis. We have $OT=8$ and $PT=6$, so by the Pythagorean Theorem, \[ OP^2 = OT^2 + PT^2 = 8^2+6^2=64+36=100 \] Since $OP>0$, then $OP = \sqrt{100}=10$.Therefore, the radius of the larger circle is $10$.
Thus, $OR=10$.
Since $QR=3$, then $OQ = OR - QR = 10 - 3 = 7$.
Therefore, the radius of the smaller circle is $7$.
Since $S$ is on the positive $y$-axis and is 7 units from the origin, then the coordinates of $S$ are $(0,7)$, which means that $k=7$.
-
In the diagram, the horizontal distance between adjacent dots in the same row is 1. Also, the vertical distance between adjacent dots in the same column is 1. What is the perimeter of quadrilateral $PQRS$?
- $12$
- $13$
- $14$
- $15$
- $16$
Answer: C
Solution:
The perimeter of quadrilateral $PQRS$ equals $PQ + QR + RS + SP$.Since the dots are spaced 1 unit apart horizontally and vertically, then $PQ = 4$, $QR=4$, and $PS=1$.
Thus, the perimeter equals $4 + 4 + RS + 1$ which equals $RS + 9$.
We need to determine the length of $RS$.
If we draw a horizontal line from $S$ to point $T$ on $QR$, we create a right-angled triangle $STR$ with $ST = 4$ and $TR=3$.
By the Pythagorean Theorem, $RS^2 = ST^2 + TR^2 = 4^2 + 3^2 = 25$.
Since $RS>0$, then $RS = \sqrt{25}=5$.
Thus, the perimeter of quadrilateral $PQRS$ is $5+9=14$.
-
If $x=11$, $y=-8$, and $2x-3z=5y$, what is the value of $z$?
- $-6$
- $13$
- $54$
- $\frac{62}{3}$
- $-\frac{71}{3}$
Answer: D
Solution:
Since $x=11$, $y=-8$ and $2x-3z=5y$, then $2\times 11 - 3z = 5\times(-8)$ or $22 - 3z = -40$.Therefore, $3z=22+40=62$ and so $z=\frac{62}{3}$.
-
Sam rolls a fair four-sided die containing the numbers 1, 2, 3, and 4. Tyler rolls a fair six-sided die containing the numbers 1, 2, 3, 4, 5, and 6. What is the probability that Sam rolls a larger number than Tyler?
- $\frac{1}{8}$
- $\frac{5}{12}$
- $\frac{3}{5}$
- $\frac{3}{4}$
- $\frac{1}{4}$
Answer: E
Solution:
We make a chart that shows the possible combinations of the number that Sam rolls and the number that Tyler rolls. Since Sam rolls a fair four-sided die and Tyler rolls a fair six-sided die, then there are 4 possible numbers that Sam can roll and 6 possible numbers that Tyler can roll and so there are $4\times 6 = 24$ equally likely combinations in total. In the chart, we put a Y when Sam's roll is larger than Tyler's and an N otherwise. Since there are 24 equally likely possibilities and Sam's roll is larger in 6 of these, then the probability that Sam's roll is larger than Tyler's is $\frac{6}{24}=\frac{1}{4}$.-
A rectangular flag is divided into four triangles, labelled Left, Right, Top, and Bottom, as shown. Each triangle is to be coloured one of red, white, blue, green, and purple so that no two triangles that share an edge are the same colour. How many different flags can be made?
- $180$
- $200$
- $220$
- $240$
- $260$
Answer: E
Solution:
Since there are four triangular sections in each flag, then at most four colours can be used in a single flag.Since no two adjacent triangles are the same colour, then at least two colours must be used. (For example, the sections Top and Left must be different colours.)
Therefore, the number of colours used is 2, 3 or 4.
We count the number of possible flags in each case.
Case 1: 2 colours
We call the colours A and B.
Assign the colour A to Top.
Since Left and Right cannot be coloured A and there is only one other colour, then Left and Right are both coloured B.
Bottom cannot be coloured B (since it shares an edge with Left and Right) so must be coloured~A.
This configuration does not violate the given rule.
There are 5 possible colours for A (red, white, blue, green, purple).
For each of these 5 choices, there are 4 possible colours for B (any of the remaining 4 colours).
Therefore, there are $5(4) = 20$ possible flags in this case.
Case 2: 4 colours
We call the colours A, B, C, and D.
Since there are 4 sections and 4 colours used, then each section is a different colour.
This configuration does not violate the given rule.
There are 5 possible colours for A. For each of these 5 choices, there are 4 possible colours for B. For each of these combinations, there are 3 possible colours for C and 2 possible colours for~D.
Therefore, there are $5(4)(3)(2)= 120$ possible flags in this case.
Case 3: 3 colours
We call the colours A, B and C.
Assign the colour A to Top.
Since Left cannot be coloured A, we assign it the colour B.
Section Right cannot be coloured A, so could be B or C.
If Right is coloured B, then in order to use all three colours, Bottom must be coloured C.
If Right is coloured C, then Bottom (which shares an edge with each of Left and Right) must be coloured A.
This gives two possible configurations: Neither configuration violates the given rule.
In each configuration, there are 5 possible colours for A. For each of these 5 choices, there are 4 possible colours for B. For each of these combinations, there are 3 possible colours for C.
Since there are two such configurations, then there are $2(5)(4)(3) = 120$ possible flags in this case.
In total, there are $20+120+120=260$ possible flags.
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The expression $4 + \frac{3}{10} + \frac{9}{1000}$ is equal to
- $4.12$
- $4.309$
- $4.039$
- $4.012$
- $4.39$
Answer: B
Solution:
Converting from fractions to decimals, \(4 + \tfrac{3}{10} + \tfrac{9}{1000} = 4+0.3+0.009 = 4.309\).-
The average age of Andras, Frances and Gerta is 22 years. What is Gerta's age?
- $19$
- $20$
- $21$
- $22$
- $23$
Answer: A
Solution:
Since the average of the three ages is 22, then the sum of the three ages is \(3\cdot 22 = 66\).Since Andras’ age is 23 and Frances’ age is 24, then Gerta’s age is \(66-23-24=19\).
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The first four rows of a table with columns $V$, $W$, $X$, $Y$, and $Z$ are shown. For each row, whenever integer $n$ appears in column $V$, column $W$ contains the integer $2n + 1$, column $X$ contains $3n + 1$, column $Y$ contains $5n+1$, and column $Z$ contains $7n+1$. For every row after the first, the number in column $V$ is the smallest positive integer that does not yet appear in any previous row. The integer 2731 appears in column $W$. The complete list of columns in which 2731 appears is
- $W$
- $W$, $X$, $Y$, and $Z$
- $W$, $X$ and $Z$
- $W$, $Y$ and $Z$
- $W$ and $Z$
Answer: D
Solution:
We begin by noting that an integer either appears in column \(V\) or it appears in one or more of the columns \(W\), \(X\), \(Y\), \(Z\):If an integer \(v\) appears in column \(V\), then \(v\) cannot have appeared in an earlier row in the table. In particular, \(v\) cannot have appeared in any of columns \(W\), \(X\), \(Y\), \(Z\) earlier in the table.
Furthermore, \(v\) cannot appear again later in the table, since each entry in its row is larger than it and each entry in later rows is again larger (since the entries in \(V\) in those later rows will be larger).
If an integer \(a\) appears in one or more of the columns \(W\), \(X\), \(Y\), \(Z\), then it will not appear in column \(V\) later in the table, since every entry in \(V\) is one that has not yet appeared in the table.
Furthermore, \(a\) cannot have appeared in \(V\) earlier in the table since every entry in \(V\) before \(a\) appears in \(W\), \(X\), \(Y\), or \(Z\) will be smaller than \(a\).
Therefore, an integer that appears in the table either appears in column \(V\) or it appears in one or more of the columns \(W\), \(X\), \(Y\), \(Z\).
If an integer \(v\) has not appeared in the table, then it will eventually be the smallest positive integer that has not appeared in the table so far, and so \(v\) will appear in column \(V\) in the next row.
We check to see if 2731 appears in column \(Z\).
Since \(2731 = 7(390)+1\), then 2731 appears in \(Z\) if 390 appears in \(V\).
Since 389 is not a multiple of 2, 3, 5 or 7 (we can check by dividing by each of these), then 390 cannot appear in \(W,X,Y,Z\) (because \(390\) cannot be written in the form \(2n+1\), \(3n+1\), \(5n+1\), or \(7n+1\) for some positive integer \(n\)).
This means that 390 appears in \(V\) and so \(2731=7(390)+1\) appears in \(Z\) in this row.
This eliminates answer (A).
We note also that, since 2731 appears in \(Z\), it cannot appear in \(V\).
Next, we show that 2731 appears in column \(Y\).
Since \(2731 = 5(546)+1\), then 2731 appears in \(Y\) if 546 appears in \(V\).
We now show that 546 does appear in \(V\).
Since 545 is a multiple of 5, but not of 2, 3 or 7, then either 546 appears in column \(V\) or in column \(Y\), as explained above.
We will show that 546 does not appear in \(Y\).
Since \(546 = 5(109)+1\), then \(546\) appears in \(Y\) if 109 appears in \(V\).
We will show that 109 does not appear in \(V\).
Since 108 is a multiple of 2 and 3, but not of 5 or 7, then 109 could appear in \(V\), \(W\) or \(X\).
We will show that 109 appears in \(X\).
Since \(109 = 3(36)+1\), then \(109\) appears in \(X\) if 36 appears in \(V\).
We will show that 36 appears in \(V\).
Since 35 is a multiple of 5 and 7, but not of 2 or 3, then 36 could appear in \(V\), \(Y\) or \(Z\).
If 36 appears in \(Z\), then the \(V\) entry in this row would be 5.
If 36 appears in \(Y\), then the \(V\) entry in this row would be 7.
Neither 5 nor 7 is in column \(V\). (Each appears in the second row.)
Therefore, 36 appears in \(V\), which means that 109 appears in \(X\).
This means that 109 does not appear in \(V\) and so \(546\) does not appear in \(Y\).
Since 546 is in \(V\) or \(Y\), then 546 is in \(V\), which means that 2731 is in \(Y\).
This eliminates answers (C) and (E).
Lastly, we show that 2731 does not appear in column \(X\).
Since 29 is not a multiple of 2, 3, 5 or 7, then 30 must appear in \(V\).
Since 30 appears in \(V\), then \(151=5(30)+1\) appears in \(Y\).
Since 151 appears in \(Y\), it does not appear in \(V\).
Since 151 does not appear in \(V\), then \(303=2(151)+1\) does not appear in \(W\).
Since 303 does not appear in \(W\), then 303 appears in \(V\). (This is because \(302\) is not a multiple of 3, 5 or 7 and so \(303\) cannot appear in \(X\), \(Y\) or \(Z\).)
Since 303 appears in \(V\), then \(910=3(303)+1\) appears in \(X\).
Since 910 appears in \(X\), then 910 does not appear in \(V\).
Since 910 does not appear in \(V\), then \(2731=3(910)+1\) does not appear in \(X\).
Therefore, 2731 appears in \(W\), \(Y\) and \(Z\).
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A rectangle with height $x$ and width $2x$ has the same perimeter as an equilateral triangle with side length 10. What is the area of the rectangle?
- 18
- 50
- 25
- 200
- 100
Answer: B
Solution:
Since the given equilateral triangle has side length 10, its perimeter is \(3 \times 10 = 30\).
In terms of \(x\), the perimeter of the given rectangle is \(x + 2x + x + 2x = 6x\).
Since the two perimeters are equal, then \(6x = 30\) which means that \(x=5\).
Since the rectangle is \(x\) by \(2x\), its area is \(x(2x)=2x^2\).
Since \(x=5\), its area is \(2(5^2)=50\).
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In the diagram, $\triangle PQR$ has $\angle PQR = 120^\circ$. Also, $\angle QPS = \angle RPS$ and
$\angle QRS = \angle PRS$. (In other words, $SP$ and $SR$ bisect $\angle QPR$ and $\angle QRP$, respectively.) What is the measure of $\angle PSR$?
- $130^\circ$
- $120^\circ$
- $140^\circ$
- $160^\circ$
- $150^\circ$
Answer: E
Solution:
Consider \(\triangle PQR\).
Since the sum of the angles in a triangle is \(180^\circ\), then \[\angle QPR + \angle QRP = 180^\circ - \angle PQR = 180^\circ - 120^\circ = 60^\circ\] Since \(\angle QPS = \angle RPS\), then \(\angle RPS = \frac{1}{2}\angle QPR\).
Since \(\angle QRS = \angle PRS\), then \(\angle PRS = \frac{1}{2}\angle QRP\).
Therefore, \[\begin{aligned}
\angle RPS + \angle PRS
& = \tfrac{1}{2}\angle QPR + \tfrac{1}{2}\angle QRP\\
& = \tfrac{1}{2}(\angle QPR + \angle QRP)\\
& = \tfrac{1}{2}\times 60^\circ\\
& = 30^\circ\end{aligned}\] Finally, \(\angle PSR = 180^\circ - (\angle RPS + \angle PRS) = 180^\circ - 30^\circ = 150^\circ\).
-
If $2x + 6 = 16$, the value of $x+4$ is
- 7
- 8
- 9
- 15
- 13
Answer: C
Solution:
Solution 1
Since \(2x+6=16\), then \(\dfrac{2x+6}{2} = \dfrac{16}{2}\) and so \(x+3 = 8\).
Since \(x+3 = 8\), then \(x+4 = (x+3)+1 = 8+1 = 9\).
Solution 2
Since \(2x+6=16\), then \(2x = 16 - 6 = 10\).
Since \(2x = 10\), then \(\dfrac{2x}{2} = \dfrac{10}{2}\) and so \(x=5\).
Since \(x=5\), then \(x+4=5+4=9\).
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In the diagram, three lines intersect at a point. What is the value of $x$?
- 30
- 45
- 60
- 90
- 120
Answer: C
Solution:
Since opposite angles are equal, then the three unmarked angles around the central point each has measure \(x^\circ\).
Since the angles around a point add to \(360^\circ\), then \(6 \times x^\circ = 360^\circ\).
From this, \(6x = 360\) and so \(x = 60\).
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Ali, Bea, Che, and Deb compete in a checkers tournament. Each player plays each other player exactly once. At the end of each game, either the two players tie or one player wins and the other player loses. A player earns 5 points for a win, 0 points for a loss, and 2 points for a tie. Exactly how many of the following final point distributions are possible?
\begin{array}{c|c} Player & Points \\ \hline Ali & 15 \\ Bea & 7 \\ Che & 4 \\ Deb & 2 \end{array} \begin{array}{c|c} Player & Points \\ \hline Ali & 10 \\ Bea & 10 \\ Che & 4 \\ Deb & 4 \end{array} \begin{array}{c|c} Player & Points \\ \hline Ali & 15 \\ Bea & 5 \\ Che & 5 \\ Deb & 2 \end{array} \begin{array}{c|c} Player & Points \\ \hline Ali & 12 \\ Bea & 10 \\ Che & 5 \\ Deb & 0 \end{array} - 0
- 1
- 2
- 3
- 4
Answer: B
Solution:
Since there are 4 players in the tournament and each player plays each other player once, then each player plays 3 games.
Since each win earns 5 points and each tie earns 2 points, the possible results for an individual player are:
3 wins, 0 losses, 0 ties: 15 points
2 wins, 0 losses, 1 tie: 12 points
2 wins, 1 loss, 0 ties: 10 points
1 win, 0 losses, 2 ties: 9 points
1 win, 1 loss, 1 tie: 7 points
1 win, 2 losses, 0 ties: 5 points
0 wins, 0 losses, 3 ties: 6 points
0 wins, 1 loss, 2 ties: 4 points
0 wins, 2 losses, 1 tie: 2 points
0 wins, 3 losses, 0 ties: 0 points
In the third table given, Deb has 2 points which means that Deb had 1 tie. If one player has a tie, then another player must also have a tie. But neither 15 points nor 5 points is a possible total to obtain with a tie. Therefore, the third table is not possible.
Similarly, in the fourth table, Ali with 12 points must have had a tie, but none of the other players’ scores allow for have a tie, so the fourth table is not possible.
In the second table, each of Che and Deb must have 2 ties and neither Ali nor Bea can have a tie because of their totals of 10 points each. Since Che and Deb only played each other once, then each of them must have a tie against another player, which is not possible. Therefore, the second table is not possible.
The first table is possible:
| Result | Ali | Bea | Che | Deb |
|---|---|---|---|---|
| Ali wins against Bea | 5 points | 0 points | ||
| Ali wins against Che | 5 points | 0 points | ||
| Ali wins against Deb | 5 points | 0 points | ||
| Bea ties Che | 2 points | 2 points | ||
| Bea wins against Deb | 5 points | 0 points | ||
| Che ties Deb | 2 points | 2 points | ||
| TOTAL | 15 points | 7 points | 4 points | 2 points |
Therefore, exactly one of the four given final point distributions is possible.
3 wins, 0 losses, 0 ties: 15 points
2 wins, 0 losses, 1 tie: 12 points
2 wins, 1 loss, 0 ties: 10 points
1 win, 0 losses, 2 ties: 9 points
1 win, 1 loss, 1 tie: 7 points
1 win, 2 losses, 0 ties: 5 points
0 wins, 0 losses, 3 ties: 6 points
0 wins, 1 loss, 2 ties: 4 points
0 wins, 2 losses, 1 tie: 2 points
0 wins, 3 losses, 0 ties: 0 points
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The expression \(\dfrac{20+22}{2}\) is equal to
- \(1\)
- \(4\)
- \(20\)
- \(21\)
- \(22\)
Answer: D
Solution:
Evaluating, \(\dfrac{20+22}{2} = \dfrac{42}{2} = 21\).
-
Alvin, Bingyi and Cheska play a two-player game that never ends in a tie. In a recent tournament between the three players, a total of 60 games were played and each pair of players played the same number of games. How many games did Bingyi win?
- \(12\)
- \(24\)
- \(28\)
- \(30\)
- \(36\)
Answer: C
Solution:
Since 60 games are played and each of the 3 pairs plays the same
number of games, each pair plays \(60 \div 3 =
20\) games.
Alvin wins 20% of the 20 games that Alvin and Bingyi play, so Alvin wins
\(\frac{20}{100} \times 20 = \frac{1}{5}
\times 20 = 4\) of these 20 games and Bingyi wins \(20 - 4 = 16\) of these 20 games.
Bingyi wins 60% of the 20 games that Bingyi and Cheska play, so Bingyi
wins a total of \(\frac{60}{100} \times 20 =
\frac{3}{5} \times 20 = 12\) of these 20 games.
The games played by Cheska and Alvin do not affect Bingyi’s total
number of wins.
In total, Bingyi wins \(16 + 12 = 28\)
games.
-
Jurgen is travelling to Waterloo by bus. He packs for 25 minutes. He then walks to the bus station, which takes 35 minutes. He arrives 60 minutes before his bus leaves. His bus leaves at 6:45 p.m. At what time did he start packing?
- 4:45 p.m.
- 4:40 p.m.
- 4:35 p.m.
- 4:55 p.m.
- 4:50 p.m.
Answer: A
Solution:
Jurgen takes \(25 + 35 = 60\) minutes to pack and then walk to the bus station. Since Jurgen arrives 60 minutes before the bus leaves, he began packing \(60 + 60 = 120\) minutes, or 2 hours, before the bus leaves. Since the bus leaves at 6:45 p.m., Jurgen began packing at 4:45 p.m.-
A Pretti number is a seven-digit positive integer with the following properties:
-
The integer formed by its leftmost three digits is a perfect square.
-
The integer formed by its rightmost four digits is a perfect cube.
-
Its ten thousands digit and ones (units) digit are equal.
-
Its thousands digit is not zero.
How many Pretti numbers are there?
-
Answer: 30
Solution:
Since a Pretti number has 7 digits, it is of the form \(a\,bcd\,efg\).
From the given information, the integer with digits \(abc\) is a perfect square.
Since a Pretti number is a seven-digit positive integer, then \(a > 0\), which means that \(abc\) is between 100 and 999,
inclusive.
Since \(9^2 = 81\) (which has two
digits) \(10^2 = 100\) (which has three
digits) and \(31^2 = 961\) (which has
three digits) and \(32^2 = 1024\)
(which has four digits), then \(abc\)
(which has three digits) must be one of \(10^2, 11^2, \ldots, 30^2, 31^2\), since
\(32^2\) has 4 digits..
From the given information, the integer with digits \(defg\) is a perfect cube.
Since the thousands digit of a Pretti number is not 0, then \(d>0\).
Since \(9^3 = 729\) and \(10^3 = 1000\) and \(21^3 = 9261\) and \(22^3 = 10\,648\), then \(defg\) (which has four digits) must be one
of \(10^3, 11^3, \ldots, 20^3, 21^3\),
since \(22^3\) has 5 digits.
Since the ten thousands digit and units digit of the original number are
equal, then \(c = g\).
In other words, the units digits of \(abc\) and \(defg\) are equal.
The units digit of a perfect square depends only on the units digit of
the integer being squared, since in the process of multiplication no
digit to the left of this digit affects the resulting units digit.
The squares \(0^2\) through \(9^2\) are \(0, 1,
4, 9, 16, 25, 36, 49, 64, 81\).
This gives the following table:
| Units digit of \(n^2\) | Possible units digits of \(n\) |
|---|---|
| 0 | 0 |
| 1 | 1, 9 |
| 4 | 2, 8 |
| 5 | 5 |
| 6 | 4, 6 |
| 9 | 3, 7 |
Similarly, the units digit of a perfect cube depends only on the
units digit of the integer being cubed.
The cubes \(0^3\) through \(9^3\) are \(0, 1,
8, 27, 64, 125, 216, 343, 512, 729\).
This gives the following table:
| Units digit of \(m^3\) | Possible units digits of \(m\) |
|---|---|
| 0 | 0 |
| 1 | 1 |
| 2 | 8 |
| 3 | 7 |
| 4 | 4 |
| 5 | 5 |
| 6 | 6 |
| 7 | 3 |
| 8 | 2 |
| 9 | 9 |
We combine this information to list the possible values of \(c=g\) (from the first table, these must be 0, 1, 4, 5, 6, 9), the squares between \(10^2\) and \(31^2\), inclusive, with this units digit, and the cubes between \(10^3\) and \(21^3\) with this units digit:
| Digit \(c=g\) | Possible squares | Possible cubes | Pretti numbers |
|---|---|---|---|
| 0 | \(10^2\), \(20^2\), \(30^2\) | \(10^3\), \(20^3\) | \(3 \times 2 = 6\) |
| 1 | \(11^2\), \(19^2\), \(21^2\), \(29^2\), \(31^2\) | \(11^3\), \(21^3\) | \(5 \times 2 = 10\) |
| 4 | \(12^2\), \(18^2\), \(22^2\), \(28^2\) | \(14^3\) | \(4 \times 1 = 4\) |
| 5 | \(15^2\), \(25^2\) | \(15^3\) | \(2 \times 1 = 2\) |
| 6 | \(14^2\), \(16^2\), \(24^2\), \(26^2\) | \(16^3\) | \(4 \times 1 = 4\) |
| 9 | \(13^2\), \(17^2\), \(23^2\), \(27^2\) | \(19^3\) | \(4 \times 1 = 4\) |
For each square in the second column, each cube in the third column
of the same row is possible. (For example, \(19^2\) and \(11^3\) give the Pretti number \(3\,611\,331\) while \(19^2\) and \(21^3\) give the Pretti number \(3\,619\,261\).) In each case, the number of
Pretti numbers is thus the product of the number of possible squares and
the number of possible cubes.
Therefore, the number of Pretti numbers is \(6
+ 10 + 4 + 2 + 4 + 4 = 30\).
-
A \(3 \times 3\) table starts with every entry equal to \(0\) and is modified using the following steps:
(i) adding \(1\) to all three numbers in any row;
(ii) adding \(2\) to all three numbers in any column.After step (i) has been used a total of \(a\) times and step (ii) has been used a total of \(b\) times, the table appears as shown.
\(7\) \(1\) \(5\) \(9\) \(3\) \(7\) \(8\) \(2\) \(6\) What is the value of \(a+b\)?
Answer: 11
Solution:
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, \(a + b = 11\).