Biker B traveled at 40km/h. When they started at the same time and met after 4 hours, Biker A must have traveled at 52.5km/h. The total area of the unshaded parts of the circles is 20cm^2. Letting the radius of the smallest circle be a, the medium circle be 2a, and the largest be 3a, the area of the shaded parts is calculated to be 10cm^2. To get an average of 6.1 from 11 numbers, the number 5 must be removed. If house number 37 is opposite 64, and numbers increase by 1 on each side of the street, the total number of houses

1. Two bikers A and B were 370km apart traveling towards each other at a
constant speed. They started at the same time, meeting after 4 hours.
If biker B started
1
2
hour later than biker A, they would be 20km apart 4
hours after A started. At what speed was biker A traveling?
Solution:
B can cover 20km in
1
2
hour, therefore B’s speed = 20 ÷
1
2
= 40km/h
When they started at the same time and met after 4 hours, B has traveled
4 × 40 = 160𝑘𝑚. A has travelled 370 − 160 = 210𝑘𝑚.
Therefore, A’s speed = 210 ÷ 4 = ____𝑘𝑚/h.
Answer: 52
1
2
𝑘𝑚/ℎ

2. There are 5 circles with 3 different diameters. Some of the circles touch
each other as shown in the figure below. If the total area of the unshaded
parts is 20cm2, find the total area of the shaded parts, in cm2.
Solution:
Let’s name the radius of the smallest circle as 𝑎 and
the area (let’s name it 𝑆) as 𝑎2 𝜋.
The radius for big circle is 3𝑎 and the area (let’s name it 𝐵) is 9𝑎2 𝜋.
The radius for medium circle is 2𝑎 and the area (let’s name it 𝑀) is 4 𝑎2 𝜋.
The area of unshaded parts + the area of shaded parts = 𝐵 −−→ ①
Given the area of unshaded parts = 𝐵 – 𝑀 + 2𝑆 – 𝑆 = 20𝑐𝑚2
𝐵 – 𝑀 + 𝑆 = 20𝑐𝑚2
9𝑎2 𝜋– 4𝑎2 𝜋 + 𝑎2 𝜋 = 20𝑐𝑚2
6𝑎2 𝜋 = 20𝑐𝑚2 −−→ ②
From ①, the area of shaded parts = 𝐵 – area of unshaded parts
= 9𝑎2 𝜋– 6𝑎2 𝜋
= 3𝑎2 𝜋
From ②, = ____𝑐𝑚2
Answer: 10 𝑐𝑚2

3. Which number should be removed from: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 and 11 so
that the average of the remaining numbers is 6.1?
Solution:
The sum of eleven numbers is = ____ −−→ ①
The average of remaining ten numbers is 6.1, their sum must be = 6.1 × 10 = 61
Hence the number which is removed is ① − 61 = ____
Answer: 5

4. Solution:
The houses in a street are located in such a way that each house is directly
opposite another house. The houses are numbered 1, 2, 3, … up one side,
continuing down the other side of the street. If number 37 is opposite
number 64, how many houses are there in the street altogether?
The difference between the last number (n) and 64
= the difference between 37 and 1
𝑛 − 64 = 37 − 1
𝑛 = ____
Answer: 100