Section A
Multiple Choice Questions — 15 Questions
(+4)
1
rate-problems
easy
Four kids can paint four balls in four minutes. If everyone works at the same speed, how many kids can paint 12 balls in 12 minutes?
Answer: A
Since 4 kids can paint 4 balls in 4 minutes, 4 kids can paint 4 × 3 = 12 balls in 4 × 3 = 12 minutes.
2
arithmetic
easy
Find the value of 25 × 48 + 32 × 25 + 25 × 20.
Answer: B
Using the regrouping strategy:
25 × 48 + 32 × 25 + 25 × 20 = 25 × (48 + 32 + 20) = 25 × 100 = 2500.
3
number-theory
medium
How many whole numbers from 1 to 100 are not divisible by 5 or 7?
Answer: A
100 ÷ 5 = 20, so 20 numbers from 1 to 100 are divisible by 5.
100 ÷ 7 = 14 R 2, so 14 numbers are divisible by 7.
Among these, 35 and 70 are common multiples of 5 and 7, so they are counted twice.
Numbers divisible by 5 or 7 = 20 + 14 − 2 = 32.
Thus, there are 100 − 32 = 68 numbers which are not divisible by 5 or 7.
4
magic-square
hard
The following 3 × 3 grid is to be filled with numbers from 1 to 9. The numbers in each row and column add up to 15. Find the value of *.
Answer: D
The bottom row must be filled with 8 and 4 (alongside the given 3), leaving 1, 5, 6 and 7 for the remaining empty cells.
* cannot be 1, as it is too small for any row or column containing it to reach 15.
* cannot be 5, since the middle row would then need another 8 to reach 15, but 8 is already used.
* cannot be 6, since the right column would then need another 6 to reach 15.
Thus, * must be 7. (The completed grid is: Row 1: 1, 9, 5; Row 2: 6, 2, 7; Row 3: 8, 4, 3.)
5
cryptarithm
hard
Find the sum of A + B + C if different letters stand for different digits, given the multiplication A8 × 8 = BCC, where A8 is a two-digit number and BCC is the three-digit product.
Answer: A
Since 8 × 8 = 64, the ones digit C = 4.
Checking each possible value of A from 1 to 6 in A8 × 8, only A = 1 or A = 6 gives a tens digit of 4 (needed so the product's tens and ones digits both equal C).
If A = 1, then B would also be 1, which is not allowed since all letters must stand for different digits.
So A = 6, and 68 × 8 = 544, giving B = 5 and C = 4.
A + B + C = 6 + 5 + 4 = 15.
6
number-theory
easy
What is the largest 3-digit number which is divisible by 24?
Answer: C
The largest 3-digit number is 999.
However, 999 ÷ 24 = 41 R 15, so 999 is not divisible by 24.
Hence, the largest 3-digit multiple of 24 is 41 × 24 = 984.
7
counting-figures
medium
The following picture shows a 4 × 4 square. How many squares of any size are there comprised of the one marked with X?
Answer: D
Counting all squares of every size that contain the cell marked X:
1×1 squares: 1
2×2 squares: 4
3×3 squares: 4
4×4 squares: 1
Total = 1 + 4 + 4 + 1 = 10.
8
cryptarithm
hard
Find the value of A5B + C9, given the subtraction A5B − C9 = 299, where A5B is a three-digit number and C9 is a two-digit number, and different letters stand for different digits.
Answer: B
In the ones column, B − 9 gives 9, so B must borrow: (10 + B) − 9 = 9, giving B = 8.
In the tens column, after lending 1, (5 − 1) − C = 9 requires borrowing again: (10 + 4) − C = 9, so C = 5.
Since the tens column borrowed 1 from the hundreds column, A − 1 = 2, so A = 3.
Thus A5B = 358 and C9 = 59.
A5B + C9 = 358 + 59 = 417.
9
fractions
medium
Sarah and Rose have a few pencils. Two-fifths of Sarah's pencils and eight-ninths of Rose's pencils are both 40 each. Who has more pencils and by how many?
Answer: E
Two-fifths of Sarah's pencils = 40, so Sarah has 40 ÷ 2 × 5 = 100 pencils.
Eight-ninths of Rose's pencils = 40, so Rose has 40 ÷ 8 × 9 = 45 pencils.
Thus, Sarah has more pencils, and she has 100 − 45 = 55 more pencils than Rose.
10
pigeonhole-principle
medium
There are 5 green, 9 orange, 6 red and 4 blue candies in a bag. At least how many candies must Sam take out without looking to be sure that he gets the first red candy?
Answer: A
There are 5 + 9 + 4 = 18 candies in total that are not red.
In the worst-case scenario, Sam could take out all 18 non-red candies before drawing a red one.
Therefore, he must take out 18 + 1 = 19 candies to be sure of getting his first red candy.
11
3d-geometry
medium
The following 4 × 4 × 4 cube is painted red on all its faces. Find the number of smaller cubes with at most two faces painted.
Answer: E
There are 4 × 4 × 4 = 64 small cubes in total.
Only the corner cubes have 3 faces painted, and a cube has 8 corners, so 8 small cubes have 3 faces painted.
The number of smaller cubes with at most 2 faces painted = 64 − 8 = 56.
12
algebra
medium
In five years from now, the combined age of Derrick and Stacy will be 37. Derrick was twice as old as Stacy 3 years ago. How old is Derrick now?
Answer: D
Combined age now = 37 − 5 − 5 = 27.
Combined age 3 years ago = 27 − 3 − 3 = 21.
Let Stacy's age 3 years ago be 1 unit, so Derrick's age 3 years ago is 2 units (since Derrick was twice as old).
1 unit + 2 units = 3 units = 21, so 1 unit = 21 ÷ 3 = 7 (Stacy's age 3 years ago).
Derrick's current age = (7 × 2) + 3 = 17 years old.
13
logic-puzzle
medium
Two regular 6-sided dice were rolled. All the numbers visible on the dice were added. The sum of numbers on all the visible faces was 31. What is the difference between the 2 numbers facing the ground?
Answer: B
The sum of all numbers on one die is 1 + 2 + 3 + 4 + 5 + 6 = 21, so on two dice it is 21 + 21 = 42.
Since the sum of the visible faces was only 31, the two faces facing the ground have a sum of 42 − 31 = 11.
The only pair of faces on a die (values 1 to 6) that adds up to 11 is 5 and 6.
Therefore, the difference between the 2 numbers facing the ground is 6 − 5 = 1.
14
number-theory
hard
Ron and Vanda each thought of a two-digit number. The product of their digits was 24 and 35 respectively. The difference between their numbers was 37. What is the sum of their numbers?
Answer: C
Ron's number could be 38, 46, 64 or 83, since the product of the digits in each of these is 24.
Vanda's number could be 57 or 75, since the product of the digits in each of these is 35.
Comparing every combination of Ron's and Vanda's possible numbers, only 75 and 38 have a difference of 37 (75 − 38 = 37).
So Ron's number is 38 and Vanda's number is 75, giving a sum of 38 + 75 = 113.
15
counting-digits
medium
Ann read a book with 1000 pages. How many digits were used to print the page numbers on the book?
Answer: E
Pages 1 to 9: 9 one-digit page numbers use 9 digits.
Pages 10 to 99: 90 two-digit page numbers use 90 × 2 = 180 digits.
Pages 100 to 999: 900 three-digit page numbers use 900 × 3 = 2700 digits.
Page 1000: 1 four-digit page number uses 4 digits.
Total digits used = 9 + 180 + 2700 + 4 = 2893.
Section B
Open-Ended (Integer) Questions — 10 Questions
(+5)
16
algebra
medium
Sam filled up some boxes with sweets. If he put 5 sweets in each box, he would have been left with 4 extra sweets. If he put 6 sweets in each box, the last box would have only 2 sweets. Find the number of sweets Sam had.
Answer: 44
Let n be the number of boxes.
Putting 5 sweets in each box with 4 left over gives a total of 5n + 4 sweets.
Putting 6 sweets in each box, with the last box short by 4 (holding only 2 instead of 6), gives a total of 6(n − 1) + 2 = 6n − 4 sweets.
Setting these equal: 5n + 4 = 6n − 4, so n = 8 boxes.
Total sweets = 5 × 8 + 4 = 44.
17
algebra
medium
When the product of 9 and a number is divided by 4 and then multiplied by 25, the result is 19800. Find the number.
Answer: 352
(Number × 9) ÷ 4 × 25 = 19800.
Number = (19800 ÷ 25) × 4 ÷ 9 = 792 × 4 ÷ 9 = 3168 ÷ 9 = 352.
18
divisibility
medium
What is the largest possible value of m in the 5-digit number 653m2 such that it is divisible by 3?
Answer: 8
A number is divisible by 3 when the sum of its digits is a multiple of 3.
6 + 5 + 3 + m + 2 = 16 + m must be a multiple of 3.
16 + m = 18, 21 or 24 (the multiples of 3 reachable with a single digit m), so m = 2, 5 or 8.
The largest possible value of m is 8.
19
number-theory
hard
A five-digit number is formed such that it satisfies the following conditions:
• It is a multiple of 3 and 5.
• The third digit is half of the first digit and one less than the second digit.
• The sum of the first three digits is 13 and the sum of the last three digits is 8.
• The fourth digit is the second-largest digit of that number.
Find the sum of digits of that number.
Answer: 18
Since the third digit is half of the first digit and one less than the second digit, the possible (1st, 2nd, 3rd) digit triples are (4,3,2), (6,4,3) or (8,5,4).
Only (6,4,3) has a sum of 13, so the first three digits must be 6, 4, 3.
Since the sum of the last three digits is 8 and the third digit is 3, the last two digits must sum to 8 − 3 = 5.
6 must be the largest digit of the number, so the fourth digit (the second-largest digit) is 5, which makes the last digit 5 − 5 = 0.
The number is 64350, and the sum of its digits is 6 + 4 + 3 + 5 + 0 = 18.
20
patterns
medium
Find the sum of the numbers in the 12th group of the following sequence: (1,4,6), (2,6,10), (3,8,14), …
Answer: 88
The sum of the numbers in each group forms a pattern: Group 1 = 1+4+6 = 11, Group 2 = 2+6+10 = 18, Group 3 = 3+8+14 = 25.
Each group's sum increases by 7 from the previous one, so the sum of Group n = 11 + (n − 1) × 7.
For Group 12: sum = 11 + 11 × 7 = 11 + 77 = 88.
21
counting
easy
16 trees are planted on one side of the road. Two cars are parked between every two trees. How many cars are there?
Answer: 30
Between 16 trees, there are 16 − 1 = 15 gaps.
Since 2 cars are parked in each gap, the total number of cars = 15 × 2 = 30.
22
measurement
medium
Ten small ropes of the same length were knotted together to form a big rope of 81 cm. If the knotted area was 1 cm for every rope, how long was each rope?
Answer: 9
When 10 ropes are knotted together, there are 10 − 1 = 9 knots joining them, and each knot uses up 1 cm of overlap.
So the total length of rope before the overlaps were counted was 81 + 9 = 90 cm.
The length of each rope = 90 ÷ 10 = 9 cm.
23
number-theory
hard
Find the value of the digit at the ones place of the following expression: 100 × 102 × 104 × 106 × 108 − 99 × 101 × 103 × 105 × 107.
Answer: 5
The product 99 × 101 × 103 × 105 × 107 has the same ones digit as 9 × 1 × 3 × 5 × 7 (9×1×3 = 27, 27×5 = 135, 135×7 = 945), which is 5.
The product 100 × 102 × 104 × 106 × 108 has '0' at the ones place, since 100 itself ends in 0.
Therefore, the ones digit of the expression is 10 − 5 = 5.
24
logic-puzzle
hard
Among Jack, Jade and James, one of them is a painter, the other is a lawyer and the third is a firefighter. Jack is older than the firefighter while the lawyer is younger than James. Also, Jade and the lawyer are not the same age. Find who the firefighter is. (Put Jack=1, Jade=2 and James=3 in your answer)
Answer: 2
The lawyer is younger than James, so James is not the lawyer. Also, Jade is not the same age as the lawyer, so Jade is not the lawyer either. Thus, Jack must be the lawyer.
Jack (the lawyer) is older than the firefighter, so Jack is not the firefighter, which is already known. Since Jack is younger than James, and Jack is older than the firefighter, James cannot be the firefighter (James is older than Jack, who is already older than the firefighter).
Therefore, Jade must be the firefighter, so the answer is 2.
25
averages
easy
Patrick scored 82 and 85 marks respectively in English and Science. If the total marks of these papers are 100, how many marks should he score in Mathematics so that his average marks in these three subjects becomes 87?
Answer: 94
If the average of the three subjects is 87, the total marks of the three subjects = 87 × 3 = 261.
Total marks in English and Science = 82 + 85 = 167.
Marks to be scored in Mathematics = 261 − 167 = 94.