1. Each duck weights the same, and each duckling weighs the same. If the
total weight of 3 ducks and 2 ducklings is 32 kilograms, the total weight of
4 ducks and 3 ducklings is 44 kilograms, what is the total weight, in
kilograms, of 2 ducks and 1 duckling?
Solution:
Let the weight of a duck as 𝑥 𝑘𝑔 and weight of a duckling as 𝑦 𝑘𝑔
3𝑥 + 2𝑦 = 32 −− − ①
4𝑥 + 3𝑦 = 44 −− − ②
① × 3 9𝑥 + 6𝑦 = 96 −− −③
② × 2 8𝑥 + 6𝑦 = 88 −− −④
③ − ④ 𝑥 = ____ 𝑘𝑔
From ①, 𝑦 = ____ 𝑘𝑔
∴ 2𝑥 + 𝑦 = ____ 𝑘𝑔
Answer: 20 𝑘𝑔
2. A wooden rectangular block, 4 cm x 5 cm x 6 cm, is painted red and then
cut into several 1 cm x 1 cm x 1 cm cubes. What is the ratio of the
number of cubes with two red faces to the number of cubes with three red
faces?
Solution:
Only the cubes at the 8 corners of the block have three red faces.
The cubes with two red faces lie along the 12 edges of the block
(excluding the corners):
- 4 of them have 4 cubes = 16 cubes;
- 4 of them have 3 cubes = 12 cubes;
- 4 of them have 2 cubes = 8 cubes
⇒ total of cubes with two faces = 36
The ratio is = 36 : 8 = __ : __
Answer: 9 : 2
3. What is the smallest amount of numbers in the product
1 × 2 × 3 × 4 × ⋯ × 26 × 27
that should be removed so that the product of the remaining numbers is a
perfect square?
Solution:
Rewrite the equation after prime factorization:
1 × 2 × 3 × 4 × ⋯ × 26 × 27 = 223 × 313 × 56 × 73 × 112 × 132 × 17 × 19 × 23
To get a square number of an integer, the power of the factors must be even.
Therefore, we need to
remove 2 × 3 × 7 × 17 × 19 × 23, 𝑂𝑅
remove 6 × 7 × 17 × 19 × 23, 𝑂𝑅
remove 3 × 14 × 17 × 19 × 23, 𝑂𝑅
remove 2 × 21 × 17 × 19 × 23, 𝑂𝑅
The smallest amount of numbers to be removed is ___
Answer: 5
4. We need to find
𝑥(𝑥+1)
2
< 2008 <
(𝑥+1)(𝑥+2)
2
From calculation, we know that
62×63
2
= 1953 < 2008 <
63×64
2
= 2016
Solution:
In the figure below, the positive numbers are arranged in the grid follow by
the arrows’ direction. For example, “8” is placed in Row 2, Column 3 and
“9” is placed in Row 3, Column 2.
Which Row and which Column that “2008” is placed?
In Row 1, the 3rd number is 6 = 1+2+3,
the 5th number is 15 = 1+2+3+4+5
Therefore, where the number of term is odd, the
number in Row n is: 1+2+3+…+n =
𝑛(𝑛+1)
2
In Column 1, the 2nd number is 3 = 1+2, the 4th number is 10 = 1+2+3+4
Therefore, where the number of term is even, the number in Column n is:
1+2+3+…+n =
𝑛(𝑛+1)
2
Answer: Row 9, Column 55
2008 is located at Column (2008 – 1954) + 1 = 55,
and Row (63 – 55) + 1 = 9
