Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Successfully reported this slideshow.

Like this presentation? Why not share!

Weekly Dose 19 - Maths Olympiad Practice

No Downloads

Total views

1,054

On SlideShare

0

From Embeds

0

Number of Embeds

837

Shares

0

Downloads

21

Comments

3

Likes

1

No notes for slide

- 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

No public clipboards found for this slide

Login to see the comments