Section A
Multiple Choice Questions — 15 Questions
(+4 / -1)
1
arithmetic
easy
What is the value of 80 × 3 − 70 × 4 + 60 × 5 − 50 × 6 + 40 × 7 − 30 × 8 + 20 × 9?
A
150
B
180
C
220
D
250
E
300
Show answer
Answer: B
Notice that 80 × 3 = 30 × 8, 70 × 4 = 40 × 7, and 60 × 5 = 50 × 6. Thus the first six products have a sum of 0. Then 20 × 9 = 180 and 0 + 180 = 180.
2
arithmetic
easy
Find the following sum
150 + 195 + 255 + 210 + 120
A
600
B
900
C
930
D
330
E
1200
Show answer
Answer: C
150 + 195 + 255 + 210 + 120 = 150 + (195 + 255) + 210 + 120
= 150 + 450 + 210 + 210 = (150 + 450) + (210 + 120) = 600 + 330
= 930.
3
ratio-word-problem
medium
Faye and Ingrid are fruit vendors. Faye's total fruits is 21 more than 1½ of Ingrid's total fruits. Their total number of fruits is 216. How many fruits does Faye have?
A
195
B
39
C
138
D
117
E
78
Show answer
Answer: C
Let the number of Ingrid's fruits be 2 units. Then Faye has 3 units and 21.
Ingrid: 1 unit + 1 unit
Faye: 1 unit + 1 unit + 1 unit + 21
Total: 216
5 units + 21 = 216
5 units = 216 − 21 = 195
1 unit = 195 ÷ 5 = 39
Hence Faye has 3 × 39 + 21 = 138 fruits.
4
repeating-pattern
medium
In the repeating pattern below, how many letter O's will be written before the 1808th letter?
ODAKOOODAKOOODAK...
Source PDF defect: the Solutions page for this question prints only the bare heading '4' with no worked reasoning at all (page 13 ends immediately after that heading, and page 14 jumps straight to heading '5'). The answer below (904, option E) is taken from the printed Answers-page list and was independently re-derived from the pattern itself (see solution).
A
900
B
1200
C
1204
D
903
E
904
Show answer
Answer: E
The repeating unit of the pattern is "ODAKOO", which has period 6 and contains 3 O's per period (at positions 1, 5 and 6 of each period). Since 1807 = 6 × 301 + 1, the first 1807 letters (i.e. all letters before the 1808th letter) consist of 301 complete periods, contributing 301 × 3 = 903 O's, plus 1 extra letter which is the first letter of the next period, namely another O. So the total number of O's written before the 1808th letter is 903 + 1 = 904.
5
number-pattern
medium
Consider the figures below. How many dots will the 22nd figure contain?
A
761
B
841
C
925
D
1013
E
1105
Show answer
Answer: C
From the table below, we can notice that the differences are consecutive multiples of 4.
Figure 1: 1 dot
Figure 2: 5 dots (+4)
Figure 3: 13 dots (+8)
Figure 4: 25 dots (+12)
Figure 5: 41 dots (+16)
Figure 6: 61 dots (+20)
By listing the number of dots in the figures in the same way, we will find that there are 925 dots in Figure 22.
6
digit-sum
hard
How many 4-digit numbers have a digit sum of 5?
A
31
B
32
C
34
D
35
E
36
Show answer
Answer: D
Using the digit 1 and 2: 1112, 1121, 1211, 2111 - 4 numbers.
Using the digits 3, 0 and 1: 1130, 1103, 1310, 1013, 1031, 1301, 3110, 3101, 3011 - 9 numbers
Using the digits 2, 0 and 1: 1220, 1202, 1022, 2210, 2201, 2120, 2102, 2021, 2012 - 9 numbers.
Using the digits 3 and 2: 2003, 2030, 2300, 3002, 3020, 3200 - 6 numbers.
Using the digits 1 and 4: 1004, 1040, 1400, 4001, 4010, 4100 - 6 numbers.
Using the digit 5: 5000, 1 number.
4 + 9 + 9 + 6 + 6 + 1 = 35.
7
area
medium
Jessica has a rectangle made of construction paper. She folds the rectangle in half to form another rectangle. She folds the resulting rectangle in half to form a 6-cm-by-6-cm square. What is the area (in cm²) of Jessica's original rectangle?
A
36
B
72
C
144
D
288
E
324
Show answer
Answer: C
Draw pictures. The last square is 6 × 6, which means the original rectangle was either a square or a rectangle. The dimensions of the original rectangle were either 12 × 12 or 6 × 24. In either case, the area is 144 cm².
8
number-theory
hard
Jericho has a collection of stickers. When the stickers are arranged in piles of 6, there are 5 stickers left over. When the stickers are arranged in piles of 22, there are 21 stickers left over. When the stickers are arranged in piles of 14, there are 13 stickers left over. What is the least number of stickers Jericho can have in his collection?
A
461
B
462
C
923
D
924
E
131
Show answer
Answer: A
If we add 1 to the total number of chips, then the new number N is divisible by 6, 14 and 22. The smallest possible value of N is the least common multiple of 6, 14 and 22, which is 462. Hence the least number of chips that Jericho can have is 462 − 1 = 461.
9
digit-counting
medium
The first 73 odd whole numbers are written. How many times does '3' appear as a digit?
A
23
B
24
C
25
D
26
E
27
Show answer
Answer: C
The first 73 odd counting numbers begin with 1 and end with 145.
The number '3' will appear in the tens place: 10 times (31, 33, 35, 37, 39, 131, 133, 135, 137 and 139)
and it will appear in the ones place: 15 times (3, 13, 23, ..., 123, 133, 143).
Therefore the number '3' will appear 10 + 15 = 25 times in the first 73 odd whole numbers.
10
algebra-system-of-equations
medium
Given □ + Δ + ◇ = 60, Δ + ◇ + ★ = 44, ◇ + ★ + □ = 52, and
Δ + ◇ + ★ + □ = 74, find the value of ★ + □ + Δ.
A
30
B
22
C
44
D
66
E
11
Show answer
Answer: D
Since Δ + ◇ + ★ + □ = 74 and
• Δ + ◇ + ★ = 44, 44 + □ = 74 so □ = 30
• □ + Δ + ◇ = 60, 60 + ★ = 74 so ★ = 14
• ◇ + ★ + □ = 52, 52 + Δ = 74 so Δ = 22
Thus, ★ + □ + Δ = 14 + 30 + 22 = 66.
11
perimeter
easy
Remi had a square garden that had an area of 16 square meters. She extends the length of the garden by 3 meters and the width by 2 meters. What is the new perimeter, in meters, of the garden?
Answer cross-checked against the Solutions page: the Answers page for Section A skips directly from Question 10 to Question 12, omitting an explicit printed answer for this question (the same generator defect the grade4/grade5 sibling papers show at other question numbers).
A
26
B
24
C
22
D
20
E
18
Show answer
Answer: A
The side length of the square garden is 4 meters since the area is 16 m². The new length is 4 + 3 = 7 meters and the new width is 4 + 2 = 6 meters. Hence the new perimeter is 2 × (7 + 6) = 26 meters.
12
combinatorics
hard
The place cards shown are folded along the dotted line so that only a number or letter is visible. Chrissy enters the room and sees all six place cards, some with numbers showing and some with letters showing. The numbers that she sees add up to 11. How many different sets of numbers are possible?
Source PDF prints only 4 answer choices for this question (5, 4, 8, 7) instead of the usual 5 — verified against the page layout itself, not a rendering artifact of this transcription.
A
5
B
4
C
8
D
7
Show answer
Answer: A
Since 6 is the greatest number and can be used only once, find other numbers that add up to 5 so that together they will equal 11. The sets (6, 5), (6, 4, 1) and (6, 3, 2) add up to 11. Then we use 5 as the greatest number and get (5, 4, 2) as one possible set and (5, 3, 2, 1) as another set. There are 5 sets of different numbers possible.
13
percentage
medium
The length of each edge of a cube is increased by 10%. By what percent is the volume increased?
A
33.1%
B
3.31%
C
331%
D
0.331%
E
1.331%
Show answer
Answer: A
Let the edge be 10 so the volume is 10³ = 1000. Increase the edge by 10% to get 11. The volume is now 11³ = 1331. The percent increase is (1331 − 1000)/1000 = 0.331 = 33.1%.
14
consecutive-integers
medium
Phineas forms an ordered list consisting of seven consecutive whole numbers. The sum of the first, third, and sixth of these numbers is 175. Find the sum of the remaining four whole numbers.
A
238
B
176
C
181
D
178
E
299
Show answer
Answer: A
Let the seven consecutive whole numbers be: x−3, x−2, x−1, x, x+1, x+2, and x+3. The statement of the problem suggests the following equation: (x−3) + (x−1) + (x+2) = 175 → 3x−2 = 175 → 3x = 177 and finally x = 177/3 = 59. The seven consecutive numbers are 56, 57, 58, 59, 60, 61, and 62. The sum of the four requested numbers is 57 + 59 + 60 + 62 = 238.
15
multiples
medium
How many 4-digit multiples of 37 are there in total?
A
270
B
243
C
210
D
209
E
195
Show answer
Answer: B
Use what you know about remainders to find the first and the last number.
The largest 4-digit number is 9999. Divide 9999 by 37 and the quotient is 270 r 9. Subtract 9 from 9999: 9999 − 9. So the largest 4-digit multiple of 37 is 9,990 = 37 × 270.
The smallest 4-digit number is 1000. Divide 1000 by 37 and the quotient is 27 r 1. Subtract 1 from 1000: 1000 − 1 = 999. So the smallest 4-digit multiple of 37 is 999 + 37 = 1036 = 37 × 28.
So, there are 270 − 27 = 243 4-digit multiples of 37.
Section B
Open-Ended (Integer) Questions — 10 Questions
(+5)
16
age-problems
hard
Lucy is Sherry's mother. Lucy's age and Sherry's age have the same digits but in reverse order. In 13 years, Lucy will be twice as old as Sherry. How old will Lucy be when Sherry reaches her age?
Show answer
Answer: 68
Make a table for all possible ages. Since Sherry's age and Lucy's age have the same digits but in reverse order, they could be 12 and 21; 13 and 31; 14 and 41; etc. Looking at the ages in 13 years, there is only one pair where Lucy's age would be twice Sherry's age, which is 27 and 54. Therefore, Sherry is currently 14 years old and Lucy is 41. When Sherry is 41, Lucy will be 41 + (41 − 14) = 68 years old.
17
rate-and-distance
hard
The figure shown is created by overlapping three 2 cm-by-6 cm rectangles. Suppose an ant moves around the perimeter travelling at 1 cm/second on each horizontal segment and 2 cm/second on each vertical segment. How long, in seconds, did the ant take to complete its entire journey?
Show answer
Answer: 22
Each 'smaller' line segment has a length of 2 cm. There are 6 horizontal segments with a total length of 12 cm and 10 vertical segments with a total length of 20 cm. The total time to complete the entire journey is 12 × 1 + 10 × 1 = 22 seconds.
18
tiling
hard
Ananya wants to tile a floor that is 24 m by 40 m. There are two types of tiles: a square that is 2 m on a side and an L-shape as shown. The L-shaped tile can be turned over or rotated if needed. What is the least number of tiles Ananya needs to tile the floor completely?
Show answer
Answer: 240
Let us show that the least number of tiles Ananya needs to tile a 12 m by 20 m floor completely is 60. Then the least number needed to tile a 24 m by 40 m floor is 4 × 60 = 240.
First, note that both the 2×2 square and the L-shape fill 4 units of area. Because the floor is 20 × 12 = 240 m², the best we can hope for is 240 ÷ 4 = 60 tiles. Now, let's put a few tiles together at a time. Two L-shapes can fill a 2×4 rectangle if one L is flipped and rotated. Two 2×2 squares can also join to become a 2×4 rectangle. Either way, the 12 rows of the floor's grid can be filled by all squares, all L-shapes, or a combination of both. That would yield 6 row blocks by 5 column blocks, each with 2 matching tiles. That is, 6 × 5 × 2 = 60 tiles, the optimal answer.
19
cryptarithm
hard
In the following cryptarithm, each different letter represents a different digit. Assuming a number cannot start with 0, find the last three digits of the smallest possible sum. (For example, if your answer is 1234, then write 234.)
MINED
+ DENIM
Show answer
Answer: 833
Since M and D have the highest place values, let them represent 1 and 2. Since I and E have the second highest place values, let them represent 0 and 3. Since N has the next highest place value, let it represent 4. Note that M and D share the same place value, therefore each letter could represent either digit (1 or 2) and would yield the same sum. The same applies to I and E. Therefore, the last three digits of the largest possible sum [sample representation: 10432 + 23401] is 833.
20
work-rate
medium
A painting job is shared by four sisters who all paint at the same rate. After Alice finishes 1/4 of the job, Beth comes in and finishes 2/3 of what is left. Then Cathy comes in and finishes 3/4 of what remains. The last part of the job is finished by Dee, who completes her part in 20 minutes. What was the total length of time, in minutes, for the whole job to be completed?
Show answer
Answer: 320
Let the painting job be 16 units. After Alice finishes 1/4 of the job, Beth comes in and finishes 2/3 of what is left. Then Cathy comes in and finishes 3/4 of what remains. The last part (or 1 unit) of the job is finished by Dee, who completes her part in 20 minutes. Hence the total length of time for the whole job to be completed is 20 × 16 = 320 minutes.
21
logic-puzzle
hard
The numbers 1 through 12 are placed in the diagram, one in each circle, so that the sum of the numbers along each line is the same. What is the largest possible value of this sum per line?
Show answer
Answer: 37
Let the sum of the numbers along each line be S and the numbers on the vertices (corner circles) of the triangle be a, b and c. Next, let us add all the numbers along the 3 lines. We can notice that the numbers 1 through 12 except a, b and c were added only once whereas a, b and c were added twice. Thus, 3S = (1 + 2 + 3 + ⋯ + 12) + a + b + c = 78 + a + b + c, so S = (78 + a + b + c) / 3. The largest possible value of S is obtained when a + b + c is as large as possible. Hence S must be (78 + 10 + 11 + 12) / 3 = 111 / 3 = 37.
22
combinatorics
hard
Each of the whole numbers 1 through 9 is written on 9 cards, one number per card. Three cards are given to each of the three players. How many ways can the cards be given such that the sum of Player One's cards is 8 and the sum of Player Two's cards is 15?
Show answer
Answer: 3
There are only 2 ways to get 8 as the sum of 3 numbers: 1 + 2 + 5 = 8, and 1 + 3 + 4 = 8.
When Player One gets 1, 2 and 5, there is only 1 way to get 15 as the sum of 3 numbers: 3 + 4 + 8 = 15.
When Player One gets 1, 3 and 4, there are only 2 ways to get 15 as the sum of 3 numbers: 2 + 5 + 8 = 15, and 2 + 6 + 7 = 15.
In total, there are 3 ways for Players One and Two to get their cards.
23
tiling
hard
A MOEMS-tile is shaped like an M as shown. It is a 5 by 5 square with two 4 by 1 rectangles removed. Jimmy is playing a game where the object is to place as many MOEMS-tiles as possible on a 6 by 42 game board without any overlap. What is the maximum number of tiles Jimmy can place on this board?
Show answer
Answer: 14
Flip one MOEMS-tile upside down and then fit the tile together with a second MOEMS-tile to form a 6 by 6 square tile with two of the corners missing. Since the game board is 6 by 42, we can arrange 7 of these square tiles on the board. Therefore, there will be 14 of the MOEMS-tiles on the board.
24
cryptarithm
hard
In the cryptarithm shown, different letters represent different digits. If two letters are the same, they represent the same digit. What is the least value that GOO could be?
DUCK
+DUCK
————
GOOSE
Show answer
Answer: 166
Notice that the only possible value for G is 1. The least possible value that GOO could be is 100. Let us check whether it is possible.
If O=0, then D=5 and U=0, which is impossible since O≠U.
If O=1, then D=5 and U=5, which is impossible since D≠U.
Similarly, O=2, 3, 4 and 5 are not possible.
Thus, O=6, DUCK=8327 or 8345, GOOSE=16654 or 16690. The least value of GOO is 166.
25
cryptarithm
hard
In the cryptarithm shown, different letters represent different digits. If two letters are the same they represent the same digit. When A = 5 and O = 4, what is the greatest value that WIN could be?
TIC
TAC
+TOE
————
WIN
Show answer
Answer: 796
Make a list of the possible numbers to be used and cross them off once used: 0, 1, 2, 3, 6, 7, 8, and 9.
Since the goal is to make WIN as great as possible we want W to be 9. This can only happen if T = 3.
Consider the tens column. Since 'I' is both an addend and the sum, the tens column must add to a 2-digit number so the regrouping will add 1 to the hundreds column, forcing a 4-digit final sum. Therefore, we reject W = 9.
If W = 8, then T = 2 and we need I to make the sum I + 5 + 4 be at least 20. This is not possible so reject W = 8.
If W = 7, then T = 2 and I + 5 + 4 must be greater than 9 so that regrouping adds a 1 to the hundreds column. Since we want WIN to be as great as possible let I = 9. We now have 29C + 25C + 24E = 79N. We need 9 < C + C + E < 20 and we want N to be as great as possible. The remaining choices for C, E and N are 0, 1, 3, 6, and 8.
If N = 8, we need C + C + E = 18 but there are no numbers remaining that satisfy that condition.
If N = 6, we would need C + C + E = 16. This can occur when C = 8 and E = 0.
Therefore, the greatest value for WIN is 796. This occurs when we add the numbers 298, 258, and 240.