1. Solution:
Try to find the common factor in the statement. E.g.
- all are divided by 23,
- all have factors which are multiple of 11
11 × 5 ÷ 23 + 2 × 11 × 6 ÷ 23 + 3 × 11 × 7 ÷ 23 + 4 × 11 × 8 ÷ 23 + 5 × 11 × 9 ÷ 23
=
11
23
(5 + 2 × 6 + 3 × 7 + 4 × 8 + 5 × 9)
=
11
23
(5 + 12 + 21 + 32 + 45)
=
11
23
(115)
= ____
Answer: 55
What is the simplified value of 11 × 5 ÷ 23 + 22 × 6 ÷ 23 + 33 × 7 ÷ 23 +
44 × 8 ÷ 23 + 55 × 9 ÷ 23?
2. Joaquin and Leo walk to school from their homes every morning at the
same time. They walk towards each other. Joaquin’s walking speed is 70
m per minute while that of Leo is 50 m per minute. They meet each other
on time every day. One morning, Leo leaves home earlier to school.
Therefore, he meets Joaquin 8 minutes earlier than usual. How many
minutes earlier did Leo leave home?
Solution:
Let the distance between them = 𝑑 meter.
The total distance remains the same, which is the first y minutes, Leo
walks alone, and the distance covered is 50𝑦. And for the remaining (𝑥 −
8) minutes, Jaoquin and Leo walk toward each other, and the distance
covered is (𝑥 − 8) × (70 + 50)
𝑑 = 70 + 50 𝑥
𝑑 = 50𝑦 + (𝑥 − 8) × (70 + 50)
50𝑦 + (𝑥 − 8) × (70 + 50) = 70 + 50 𝑥
50𝑦 + 120𝑥 − 960 = 120𝑥
50𝑦 = 960
Answer:19
1
5
3. The given figure shows a triangle ABC where AB is 5 times the length of AD, AC
is 3 times the length of AE. How many times greater is the shaded part of the
triangle ABC than the area of triangle ADE?
Solution:
Given AB = 5AD, AC = 3AE
Shaded area =△ 𝐴𝐵𝐶 −△ 𝐴𝐷𝐸 = 15𝑥 − 𝑥 = _____
Answer: shaded area is 14 times greater than triangle ADE
Assume area for △ 𝐴𝐷𝐸 = 𝑥,
drawing a line from B to E, we can derive the
area for △ 𝐴𝐵𝐸 = 5𝑥 (same base [line AB]same
height as △ 𝐴𝐷𝐸)
Then, we can continue to derive area for
△ 𝐴𝐵𝐶 = 3 △ 𝐴𝐵𝐸(same base [line AC]same height as △ 𝐴𝐵𝐸)
= 15𝑥,
4. Solution:
Write down the natural numbers continuously from 1:
12345678910111213141516171819202122232425….
Counting from left to right, the three 2s appear successively for the first
time starting with the 34th digit. Starting with which digit will the five 2s
appear successively for the first time?
The five 2s appear when counting 222, then 223
12345678910111213141516171819202122232425…221222223….
↑
One digit: from 1 till 9, 1 x 9 = 9 digits
Two digits: from 10 till 99, (99 – 10 + 1) x 2 = 180 digits
Three digits: from 100 till 221, (221 – 100 + 1) x 3 = 366
The number 222 starts at 9 + 180 + 366 + 1 = ___
Note: remember to add 1
Answer: 556
