BBB Heat Round
Primary 4 Session 1
Section A:
1. In the figure below, at most how many triangles are there?
2. A 3-digit number is divided by 7 and 11 to get a remainder of 6. Determine the
largest possible value of this 3-digit number.
3. How many 2-digit positive integers are there that are divisible by 2 or 3?
4. While doing division, Mike accidentally wrote the dividend 496 to 694. He got a
quotient of 86 and a remainder of 6. What should be the correct quotient?
5. Find the positive difference between the highest common factor and the least
common multiple of 36 and 64.
6. Find the average of the following numbers below.
13, 27, 15, 25, 17, 23, 19, 21
7. Box A weighs 2 kg while Box B weighs 3 kg. Now we have a total of 25 boxes of A and
B weighing 64 kg. How many Box As are there?
8. Determine the value of 22 × 6 + 13 × 12 + 15 × 6.
9. The day after the day after tomorrow is a Monday. Which day of the week will 146
days after today be?
10. The sum of A and B is 522. A is 28 times of B. Determine the value of A.
Section B:
11. Determine the value of 17 ÷ 9 + 33 ÷ 18 + 46 ÷ 36.
12. Numbers are drawn from 50 integers 1 to 50. At least how many draws should be
done to ensure you have 2 numbers whose sum is 36?
13. Refer to the figure below, how many rectangles are there with the black dot?
14. Determine the last 3 digits of 4 × 8 × 12 × 16 × … × 72 × 76 × 80.
15. Observe the operation being done below. Determine 3 5.
2 4 = 2 + 6 + 10 + 14
6 3 = 6 + 9 + 12
4 5 = 4 + 9 + 14 + 19 + 24
16. If the number ̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅
20232𝐴240𝐵 is divisible by 36, what is the value of A + B?
17. Find the value of 8 + 14 + 20 + 26 + ⋯ + 374 + 380.
18. In a mathematics test, 40 students got an average score of 65. If we are not to
include the 20 students who got 20 marks each, what will be the new average?
19. Given the dimensions given in the figure below, determine the surface area in square
centimeters of the figure.
̅̅̅̅̅̅̅̅?
20. Refer to the puzzle below. What is the value of 𝐴𝐵𝐶𝐷
Section C:
21. Between 1:30 PM and 3:45 AM, how many times will the hour hand and minute
hand of a clock overlap?
22. Refer to the puzzle below. 𝐴, 𝐵, ̅̅̅̅
𝐴𝐵, 𝑎𝑛𝑑 ̅̅̅̅
𝐵𝐴 are all prime numbers. Determine the
value of C + A + D+ E.
23. There are 65 of blue, 58 of red, and 74 of yellow socks in a bag mixed randomly. Each
time, you draw one sock from the bag randomly. At least how many draws should
you make to ensure you have 11 pairs of every color?
24. If Mitch can only go upward or rightwards, how many ways can she go from point A
to point B according to the pathway below?
25. The length of the red line, which is the diagonal of the square, is 15√2 cm.
Determine the circumference in cm of the circle which is inscribed the square.
Use π = 3.14.
BBB Heat Round
Primary 4 Session 1
Section A:
1. In the figure below, at most how many triangles are there?
Answer: 15
Solution:
2. A 3-digit number is divided by 7 and 11 to get a remainder of 6. Determine the
largest possible value of this 3-digit number.
Answer: 930
Solution:
𝐿𝐶𝑀(7,11) = 77
999 ÷ 77 = 12 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 75.
𝑆𝑜, 𝑡ℎ𝑒 𝑙𝑎𝑟𝑔𝑒𝑠𝑡 3 − 𝑑𝑖𝑔𝑖𝑡 𝑛𝑢𝑚𝑏𝑒𝑟 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑏𝑦 77 𝑖𝑠
77 × 12 = 924.
𝑇𝑜 ℎ𝑎𝑣𝑒 𝑎 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑓 6, 𝑤𝑒 𝑛𝑒𝑒𝑑 𝑡𝑜 𝑎𝑑𝑑 924 + 6 = 930.
3. How many 2-digit positive integers are there that are divisible by 2 or 3?
Answer: 60
Solution:
90 90 90
[ ]+[ ]−[ ]
2 3 6
= 45 + 30 − 15
= 60
4. While doing division, Mike accidentally wrote the dividend 496 to 694. He got a
quotient of 86 and a remainder of 6. What should be the correct quotient?
Answer: 62
Solution:
694 − 6 = 688
688 ÷ 86 = 8 (𝑑𝑖𝑣𝑖𝑠𝑜𝑟)
496 ÷ 8 = 62
5. Find the positive difference between the highest common factor and the least
common multiple of 36 and 64.
Answer: 572
Solution:
𝑃𝑟𝑖𝑚𝑒 𝑓𝑎𝑐𝑡𝑜𝑟𝑖𝑧𝑒 𝑏𝑜𝑡ℎ 𝑛𝑢𝑚𝑏𝑒𝑟𝑠:
36 = 2 × 2 × 3 × 3
64 = 2 × 2 × 2 × 2 × 2 × 2
𝑇ℎ𝑒 𝐿𝐶𝑀 𝑖𝑠 (2 × 2 × 2 × 2 × 2 × 2 × 3 × 3) = 576.
𝑇ℎ𝑒 𝐻𝐶𝐹 𝑖𝑠 (2 × 2) = 4.
𝑇ℎ𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑖𝑠 576 − 4 = 572.
6. Find the average of the following numbers below.
13, 27, 15, 25, 17, 23, 19, 21
Answer: 20
Solution:
𝑅𝑒𝑎𝑟𝑟𝑎𝑛𝑔𝑖𝑛𝑔 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟𝑠, 𝑤𝑒 ′𝑙𝑙 𝑔𝑒𝑡
13, 15, 17, 19, 21, 23, 25, 27.
𝑇ℎ𝑖𝑠 𝑖𝑠 𝑎𝑛 𝑎𝑟𝑖𝑡ℎ𝑚𝑒𝑡𝑖𝑐 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒.
𝑇𝑜 𝑔𝑒𝑡 𝑡ℎ𝑒 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑜𝑓 𝑎𝑟𝑖𝑡ℎ𝑚𝑒𝑡𝑖𝑐 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒, 𝑤𝑒 𝑗𝑢𝑠𝑡 𝑛𝑒𝑒𝑑 𝑡𝑜
𝑓𝑖𝑟𝑠𝑡 𝑡𝑒𝑟𝑚 + 𝑙𝑎𝑠𝑡 𝑡𝑒𝑟𝑚
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 =
2
13 + 27
=
2
= 20
7. Box A weighs 2 kg while Box B weighs 3 kg. Now we have a total of 25 boxes of A and
B weighing 64 kg. How many Box As are there?
Answer: 11
Solution:
𝐴𝑠𝑠𝑢𝑚𝑖𝑛𝑔 𝑎𝑙𝑙 𝑏𝑜𝑥𝑒𝑠 𝑎𝑟𝑒 𝐵𝑜𝑥 𝐵, 𝑡ℎ𝑒 𝑡𝑜𝑡𝑎𝑙 𝑤𝑒𝑖𝑔ℎ𝑡 𝑤𝑜𝑢𝑙𝑑 𝑏𝑒
25 × 3 = 75.
𝑆𝑢𝑏𝑡𝑟𝑎𝑐𝑡𝑖𝑛𝑔 𝑡ℎ𝑖𝑠 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑎𝑐𝑡𝑢𝑎𝑙 𝑤𝑒𝑖𝑔ℎ𝑡 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 64, 𝑤𝑒 ℎ𝑎𝑣𝑒
75 − 64 = 11.
𝐷𝑖𝑣𝑖𝑑𝑖𝑛𝑔 𝑡ℎ𝑖𝑠 𝑏𝑦 𝑡ℎ𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑤𝑒𝑖𝑔ℎ𝑡𝑠 (3 − 2 = 1), 𝑤𝑒 ℎ𝑎𝑣𝑒 11 ÷ 1 = 11.
8. Determine the value of 22 × 6 + 13 × 12 + 15 × 6.
Answer: 378
Solution:
𝐹𝑎𝑐𝑡𝑜𝑟𝑖𝑛𝑔 𝑡𝑜 𝑔𝑒𝑡 𝑎 𝑐𝑜𝑚𝑚𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟, 𝑤𝑒 ℎ𝑎𝑣𝑒
22 × 6 + 13 × 2 × 6 + 15 × 6.
6 × (22 + 13 × 2 + 15)
= 6 × (22 + 26 + 15)
= 6 × 63
= 378
9. The day after the day after tomorrow is a Monday. Which day of the week will 146
days after today be?
Answer: Thursday
Solution:
𝐵𝑦 𝑎𝑛𝑎𝑙𝑦𝑧𝑖𝑛𝑔 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑠𝑡𝑎𝑡𝑒𝑚𝑒𝑛𝑡, 𝑡𝑜𝑑𝑎𝑦 𝑖𝑠 𝑎 𝐹𝑟𝑖𝑑𝑎𝑦.
146
= 20 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 6
7
𝑇ℎ𝑖𝑠 𝑚𝑒𝑎𝑛𝑠 20 𝑤𝑒𝑒𝑘𝑠 𝑎𝑛𝑑 6 𝑑𝑎𝑦𝑠 𝑎𝑓𝑡𝑒𝑟 𝐹𝑟𝑖𝑑𝑎𝑦 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝑎 𝑇ℎ𝑢𝑟𝑠𝑑𝑎𝑦.
10. The sum of A and B is 522. A is 28 times of B. Determine the value of A.
Answer: 504
Solution:
𝐴 + 𝐵 = 522
𝐴 = 28𝐵
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝐴 𝑡𝑜 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛, 𝑤𝑒 ℎ𝑎𝑣𝑒
28𝐵 + 𝐵 = 522
29𝐵 = 522
522
𝐵=
29
𝐵 = 18
𝐴 = 28 × 18
𝐴 = 504
Section B:
11. Determine the value of 17 ÷ 9 + 33 ÷ 18 + 46 ÷ 36.
Answer: 5
Solution:
17 33 46 68 + 66 + 46
+ + =
9 18 36 36
180
=
36
=5
12. Numbers are drawn from 50 integers 1 to 50. At least how many draws should be
done to ensure you have 2 numbers whose sum is 36?
Answer: 34
Solution:
𝑑𝑟𝑎𝑤𝑠 = 𝑝𝑎𝑖𝑟𝑒𝑑 + 𝑢𝑛𝑝𝑎𝑖𝑟𝑒𝑑 + 1
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑢𝑛𝑝𝑎𝑖𝑟𝑒𝑑 𝑖𝑛𝑡𝑒𝑔𝑒𝑟𝑠:
𝑎. 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑓𝑟𝑜𝑚 36 𝑡𝑜 50 = 15 𝑑𝑟𝑎𝑤𝑠
𝑏. 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 18 = 1 𝑑𝑟𝑎𝑤
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑖𝑎𝑟𝑒𝑑 𝑖𝑛𝑡𝑒𝑔𝑒𝑟𝑠:
𝑎. 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 1 𝑡𝑜 17 𝑜𝑟 19 𝑡𝑜 35 = 17 𝑑𝑟𝑎𝑤𝑠
𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑑𝑟𝑎𝑤𝑠 = 15 + 17 + 1 + 1 = 34 𝑑𝑟𝑎𝑤𝑠.
13. Refer to the figure below, how many rectangles are there with the black dot?
Answer: 48
Solution:
2 × 4 × 3 × 2 = 48
14. Determine the last 3 digits of 4 × 8 × 12 × 16 × … × 72 × 76 × 80.
Answer: 000
Solution:
𝐼𝑓 𝑤𝑒 𝑒𝑥𝑝𝑎𝑛𝑑 𝑡ℎ𝑒 𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛, 𝑤𝑒 ′ 𝑙𝑙 ℎ𝑎𝑣𝑒 20, 40, 𝑎𝑛𝑑 60 𝑎𝑠 𝑓𝑎𝑐𝑡𝑜𝑟𝑠.
𝑇ℎ𝑖𝑠 𝑤𝑜𝑢𝑙𝑑 𝑚𝑒𝑎𝑛 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑒𝑛𝑑𝑠 𝑤𝑖𝑡ℎ 𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 4 𝑧𝑒𝑟𝑜𝑒𝑠.
15. Observe the operation being done below. Determine 3 5.
2 4 = 2 + 6 + 10 + 14
6 3 = 6 + 9 + 12
4 5 = 4 + 9 + 14 + 19 + 24
Answer: 65
Solution:
3 😊 5 =3 + 8 + 13 + 18 + 23
= 65
̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅ is divisible by 36, what is the maximum value of A + B?
16. If the number 20232𝐴240𝐵
Answer: 12
Solution:
𝑇𝑜 𝑏𝑒 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑏𝑦 36, 𝑖𝑡 𝑠ℎ𝑜𝑢𝑙𝑑 𝑏𝑒 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑏𝑦 4 𝑎𝑛𝑑 9.
𝐵𝑦 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑦 𝑟𝑢𝑙𝑒 𝑜𝑓 4, 𝐵 𝑐𝑎𝑛 𝑜𝑛𝑙𝑦 𝑏𝑒 4 𝑜𝑟 8.
𝐵𝑦 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑦 𝑟𝑢𝑙𝑒 𝑜𝑓 9, 𝑤𝑒 ℎ𝑎𝑣𝑒
2 + 0 + 2 + 3 + 2 + 2 + 4 + 0 + 𝐴 + 𝐵 = 15 + 𝐴 + 𝐵.
𝐼𝑓 𝐵 = 4, 𝑡ℎ𝑒𝑛 𝐴 𝑠ℎ𝑜𝑢𝑙𝑑 𝑏𝑒 8.
𝐼𝑓 𝐵 = 8, 𝑡ℎ𝑒𝑛 𝐴 𝑠ℎ𝑜𝑢𝑙𝑑 𝑏𝑒 4.
𝐸𝑖𝑡ℎ𝑒𝑟 𝑤𝑎𝑦, 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝐴 + 𝐵 = 12.
17. Find the value of 8 + 14 + 20 + 26 + ⋯ + 374 + 380.
Answer: 12222
Solution:
𝑇ℎ𝑖𝑠 𝑖𝑠 𝑎𝑛 𝑎𝑟𝑖𝑡ℎ𝑚𝑒𝑡𝑖𝑐 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒. 𝑇𝑜 𝑔𝑒𝑡 𝑖𝑡𝑠 𝑠𝑢𝑚, 𝑤𝑒 ℎ𝑎𝑣𝑒
𝑠𝑢𝑚 = 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 × 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑒𝑟𝑚𝑠
𝑓𝑖𝑟𝑠𝑡 + 𝑙𝑎𝑠𝑡 380 + 8
𝑎𝑣𝑒𝑟𝑎𝑔𝑒 = = = 194
2 2
𝑙𝑎𝑠𝑡 − 𝑓𝑖𝑟𝑠𝑡 380 − 8
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑒𝑟𝑚𝑠 = +1= + 1 = 63
𝑐𝑜𝑚𝑚𝑜𝑛 𝑑𝑖𝑓𝑓𝑒𝑟𝑛𝑐𝑒 6
𝑠𝑢𝑚 = 194 × 63
= 12222
18. In a mathematics test, 40 students got an average score of 65. If we are not to
include the 20 students who got 20 marks each, what will be the new average?
Answer: 110
Solution:
𝑇ℎ𝑒 𝑡𝑜𝑡𝑎𝑙 𝑠𝑐𝑜𝑟𝑒 𝑜𝑓 𝑡ℎ𝑒 40 𝑠𝑡𝑢𝑑𝑒𝑛𝑡𝑠 𝑖𝑠 40 × 65 = 2600.
𝑇ℎ𝑒 𝑠𝑐𝑜𝑟𝑒 𝑜𝑓 𝑡ℎ𝑒 20 𝑠𝑡𝑢𝑑𝑒𝑛𝑡𝑠 𝑤ℎ𝑜 𝑔𝑜𝑡 20 𝑚𝑎𝑟𝑘𝑠 𝑖𝑠 20 × 20 = 400.
2600 − 400 2200
𝑇ℎ𝑒 𝑛𝑒𝑤 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑤𝑖𝑙𝑙 𝑏𝑒 = = 110.
20 20
19. Given the dimensions given in the figure below, determine the surface area in square
centimeters of the figure.
Answer: 1152
Solution:
𝑇ℎ𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑟𝑒𝑎𝑠 𝑜𝑓 𝑒𝑎𝑐ℎ 𝑓𝑎𝑐𝑒.
𝐹𝑜𝑟 𝑡ℎ𝑒 𝑡𝑟𝑖𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑓𝑎𝑐𝑒𝑠, 𝑤𝑒 ℎ𝑎𝑣𝑒 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒.
12 × 16
𝐴𝑟𝑒𝑎 = = 96 𝑐𝑚2 .
2
𝑆𝑖𝑛𝑐𝑒 𝑤𝑒 ℎ𝑎𝑣𝑒 𝑡𝑤𝑜 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒𝑠, 𝐴𝑟𝑒𝑎 = 96 × 2 = 192 𝑐𝑚2 .
𝐹𝑜𝑟 𝑡ℎ𝑒 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒, 𝑤𝑒 ℎ𝑎𝑣𝑒
𝐴𝑟𝑒𝑎 = 12 × 20 = 240 𝑐𝑚2 .
𝐹𝑜𝑟 𝑡ℎ𝑒 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒, 𝑤𝑒 ℎ𝑎𝑣𝑒
𝐴𝑟𝑒𝑎 = 16 × 20 = 320 𝑐𝑚2 .
𝐹𝑜𝑟 𝑡ℎ𝑒 𝑠𝑙𝑎𝑛𝑡𝑒𝑑 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒, 𝑡ℎ𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑖𝑠 20. 𝑊𝑒 𝑠𝑡𝑖𝑙𝑙 𝑛𝑒𝑒𝑑 𝑡𝑜 𝑠𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑤𝑖𝑑𝑡ℎ.
𝐵𝑦 𝑡ℎ𝑒 𝑝𝑦𝑡ℎ𝑎𝑔𝑜𝑟𝑒𝑎𝑛 𝑡ℎ𝑒𝑜𝑟𝑒𝑚, 𝑡ℎ𝑒 3𝑟𝑑 𝑠𝑖𝑑𝑒 𝑖𝑠
𝑐 = √122 + 162 .
𝑐 = √400 = 20 𝑐𝑚 = 𝑤𝑖𝑑𝑡ℎ
𝑇ℎ𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑙𝑎𝑛𝑡𝑒𝑑 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒 𝑖𝑠
𝐴𝑟𝑒𝑎 = 20 × 20 = 400 𝑐𝑚2 .
𝑇𝑜𝑡𝑎𝑙 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐴𝑟𝑒𝑎 = 192 + 240 + 320 + 400 = 1152 𝑐𝑚2 .
̅̅̅̅̅̅̅̅?
20. Refer to the puzzle below. What is the value of 𝐴𝐵𝐶𝐷
Answer: 2437
Solution:
𝐼𝑛 𝑡ℎ𝑒 𝑡ℎ𝑜𝑢𝑠𝑎𝑛𝑑𝑠 𝑝𝑙𝑎𝑐𝑒, 𝐴 + 𝐴 + 𝐴 + 𝐴 = 9. 𝑇ℎ𝑖𝑠 𝑚𝑒𝑎𝑛𝑠 𝐴 = 2.
𝐼𝑛 𝑡ℎ𝑒 ℎ𝑢𝑛𝑑𝑟𝑒𝑑𝑠 𝑝𝑙𝑎𝑐𝑒, 𝐴 + 𝐴 + 𝐴 + 𝐷 = 14 𝑜𝑟 𝐴 + 𝐴 + 𝐴 + 𝐷 = 13.
𝑇ℎ𝑖𝑠 𝑚𝑒𝑎𝑛𝑠 𝐷 = 8 𝑜𝑟 𝐷 = 7.
𝐹𝑟𝑜𝑚 𝑡ℎ𝑖𝑠 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛, 𝑖𝑡 𝑠𝑖𝑚𝑝𝑙𝑦 𝑚𝑒𝑎𝑛𝑠 𝑡ℎ𝑎𝑡 𝐴 + 𝐵 + 𝐶 + 𝐷 = 16
𝐼𝑛 𝑡ℎ𝑒 𝑡𝑒𝑛𝑡ℎ𝑠 𝑝𝑙𝑎𝑐𝑒, 𝐴 + 𝐴 + 𝐶 + 𝐷 = 14. 𝑇ℎ𝑖𝑠 𝑚𝑒𝑎𝑛𝑠 𝐶 = 2 𝑜𝑟 3.
𝑆𝑖𝑛𝑐𝑒 𝐴 = 2, 𝑡ℎ𝑒𝑛 𝐶 ≠ 2. 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, 𝐶 = 3.
𝑇ℎ𝑖𝑠 𝑙𝑒𝑎𝑣𝑒𝑠 𝑢𝑠 𝐵 = 4.
̅̅̅̅̅̅̅̅ = 2437.
𝑆𝑜, 𝐴𝐵𝐶𝐷
Section C:
21. Between 1:30 PM and 3:45 AM, how many times will the hour hand and minute
hand of a clock overlap?
Answer: 13
Solution:
𝑅𝑒𝑚𝑒𝑚𝑏𝑒𝑟 𝑡ℎ𝑎𝑡 𝑖𝑛 𝑎 𝑠𝑝𝑎𝑛 𝑜𝑓 12 ℎ𝑜𝑢𝑟𝑠, 𝑡ℎ𝑒𝑦 𝑤𝑖𝑙𝑙 𝑜𝑣𝑒𝑟𝑎𝑙𝑝 11 𝑡𝑖𝑚𝑒𝑠.
𝑆𝑜, 𝑓𝑟𝑜𝑚 1: 30 𝑃𝑀 𝑡𝑜 1: 30 𝐴𝑀, 𝑡ℎ𝑒𝑦 𝑜𝑣𝑒𝑟𝑙𝑎𝑝𝑝𝑒𝑑 11 𝑡𝑖𝑚𝑒𝑠.
𝐵𝑒𝑡𝑤𝑒𝑒𝑛 1: 30 𝐴𝑀 𝑎𝑛𝑑 2: 30 𝐴𝑀, 𝑡ℎ𝑒𝑦 𝑤𝑖𝑙𝑙 𝑜𝑣𝑒𝑟𝑙𝑎𝑝 𝑜𝑛𝑐𝑒.
𝐵𝑒𝑡𝑤𝑒𝑒𝑛 2: 30 𝐴𝑀 𝑎𝑛𝑑 3: 45 𝐴𝑀, 𝑡ℎ𝑒𝑦 𝑤𝑖𝑙𝑙 𝑜𝑣𝑒𝑟𝑙𝑎𝑝 𝑜𝑛𝑐𝑒.
𝐴 𝑡𝑜𝑡𝑎𝑙 𝑜𝑓 13 𝑜𝑣𝑒𝑟𝑙𝑎𝑝𝑠.
22. Refer to the puzzle below. 𝐴, 𝐵, ̅̅̅̅
𝐴𝐵, 𝑎𝑛𝑑 ̅̅̅̅
𝐵𝐴 are all prime numbers. Determine the
value of C + A + D+ E.
Answer: 10
Solution:
𝑆𝑖𝑛𝑐𝑒 𝑎𝑙𝑙 𝑑𝑖𝑔𝑖𝑡𝑠 𝑎𝑟𝑒 𝑝𝑟𝑖𝑚𝑒, 𝑡ℎ𝑒 𝑜𝑛𝑙𝑦 𝑜𝑝𝑡𝑖𝑜𝑛𝑠 𝑎𝑟𝑒 3, 5, 𝑎𝑛𝑑 7.
𝑇𝑎𝑘𝑒 𝑛𝑜𝑡𝑒 𝑡ℎ𝑎𝑡 𝑤𝑒 𝑐𝑎𝑛𝑛𝑜𝑡 𝑖𝑛𝑐𝑙𝑢𝑑𝑒 2 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 𝑖𝑓 𝐵 = 2, 𝐴𝐵 𝑤𝑖𝑙𝑙 𝑛𝑜𝑡 𝑏𝑒 𝑝𝑟𝑖𝑚𝑒.
𝐺𝑒𝑡𝑡𝑖𝑛𝑔 𝑎𝑙𝑙 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛𝑠, 𝑜𝑛𝑙𝑦 37 𝑎𝑛𝑑 73 𝑤𝑖𝑙𝑙 𝑓𝑖𝑡 𝑡ℎ𝑒 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑠.
𝑇ℎ𝑖𝑠 𝑤𝑖𝑙𝑙 𝑔𝑖𝑣𝑒 𝑢𝑠 𝐴 = 7 𝑎𝑛𝑑 𝐵 = 3.
73 × 37 = 2701
𝑆𝑜, 2 + 7 + 0 + 1 = 10.
23. There are 65 of blue, 58 of red, and 74 of yellow socks in a bag mixed randomly. Each
time, you draw one sock from the bag randomly. At least how many draws should
you make to ensure you have 11 pairs of every color?
Answer: 161
Solution:
𝐴𝑠𝑠𝑢𝑚𝑖𝑛𝑔 𝑡ℎ𝑒 𝑤𝑜𝑟𝑠𝑡 − 𝑐𝑎𝑠𝑒 𝑠𝑐𝑒𝑛𝑎𝑟𝑖𝑜:
𝑇ℎ𝑎𝑡 𝑤𝑜𝑢𝑙𝑑 𝑏𝑒 𝑖𝑓 𝑤𝑒 𝑑𝑟𝑎𝑤 𝑓𝑖𝑟𝑠𝑡 𝑎𝑙𝑙 𝑡ℎ𝑒 𝑦𝑒𝑙𝑙𝑜𝑤 𝑠𝑜𝑐𝑘𝑠, 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑏𝑙𝑢𝑒 𝑠𝑜𝑐𝑘𝑠.
74 + 65 + 22(11 𝑝𝑎𝑖𝑟𝑠 𝑜𝑓 𝑟𝑒𝑑 ) = 161 𝑑𝑟𝑎𝑤𝑠.
24. If Mitch can only go upward or rightwards, how many ways can she go from point A
to point B according to the pathway below?
Answer: 30
Solution:
25. The length of the red line, which is the diagonal of the square, is 15√2 cm.
Determine the circumference in cm of the circle which is inscribed the square.
Use π = 3.14.
Answer: 47.1
Solution:
𝑇ℎ𝑒 𝑠𝑖𝑑𝑒 𝑙𝑒𝑛𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑞𝑢𝑎𝑟𝑒 𝑐𝑎𝑛 𝑏𝑒 𝑠𝑜𝑙𝑣𝑒𝑑 𝑢𝑠𝑖𝑛𝑔 𝑖𝑡𝑠 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙.
𝑈𝑠𝑖𝑛𝑔 𝑡ℎ𝑒 𝑝𝑦𝑡ℎ𝑎𝑔𝑜𝑟𝑒𝑎𝑛 𝑡ℎ𝑒𝑜𝑟𝑒𝑚,
2
(15√2) = 𝑎2 + 𝑏2 .
𝐼𝑛 𝑡ℎ𝑖𝑠 𝑐𝑎𝑠𝑒, 𝑎 = 𝑏. 𝑆𝑜, 𝑤𝑒 ℎ𝑎𝑣𝑒
2
(15√2) = 𝑎2 + 𝑎2 .
450 = 2𝑎2
𝑎2 = 225
𝑎 = 15 = 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑖𝑟𝑐𝑙𝑒
𝐶𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑜𝑓 𝑎 𝑐𝑖𝑟𝑐𝑙𝑒 = 2𝜋𝑟 = 𝜋𝑑
= 15 × 3.14
= 47.1 cm
