Successfully reported this slideshow.
We use your LinkedIn profile and activity data to personalize ads and to show you more relevant ads. You can change your ad preferences anytime.

Weekly Dose 11 - Maths Olympiad Practice

Weekly Dose 11 - Maths Olympiad Practice

  • Login to see the comments

  • Be the first to like this

Weekly Dose 11 - Maths Olympiad Practice

  1. 1. Each of the numbers from 1 to 9 is placed, one per circle, into the pattern shown. The sums along each of the four sides are equal. How many different numbers can be placed in the middle circle to satisfy these conditions? Solution: Because the sums along each side must be equal, therefore sums of the 8 numbers must be divisible by 4. Note that 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 = 4 Γ— 11 + 1. Therefore the numbers which have a remainder of 1 when divided by 4 are the candidates to be placed in the middle. Answer:
  2. 2. When Anura was 8 years old his father was 31 years old. Now his father is twice as old as Anura is. How old is Anura now? Solution: Now, after π‘₯ year, - Anura’s father’s age is 31 + π‘₯ and Anura’s age is 8 + π‘₯ - Anura’s father’s age is 2 Γ— Anura’s age 31 + π‘₯ = 2 Γ— (8 + π‘₯) = 16 + 2π‘₯ π‘₯ = 15 Anura’s age = 8 + 15 = ____ Answer: 23
  3. 3. In rectangle 𝐴𝐡𝐢𝐷, 𝐴𝐡 = 12π‘π‘š and 𝐴𝐷 = 5π‘π‘š. Point P, Q, R and S are all on diagonal AC, so that 𝐴𝑃 = 𝑃𝑄 = 𝑄𝑅 = 𝑅𝑆 = 𝑆𝐢. What is the total area of the shaded region, in cm2? Solution: The area for 𝐴𝐡𝐢𝐷 = 5 Γ— 12 = 60 π‘π‘š2 The area for β–³ 𝐴𝐡𝐢 and β–³ 𝐴𝐷𝐢 = 60 Γ· 2 = 30 π‘π‘š2 Since 𝐴𝑃 = 𝐢𝑆 = 1 5 𝐴𝐢, therefore the area for β–³ 𝐴𝑃𝐷 =β–³ 𝐴𝑃𝐡 =β–³ 𝐢𝑆𝐷 =β–³ 𝐢𝑆𝐡 = 1 5 β–³ 𝐴𝐡𝐢 = 1 5 Γ— 30 = 6 π‘π‘š2 Total area of shaded region = 6 + 6 + 6 + 6 = ____ Answer: 24 π‘π‘š2
  4. 4. Solution: 𝐴𝑃 = 𝐴𝑄 β‡’ βˆ π΄π‘ƒπ‘„ = βˆ π΄π‘„π‘ƒ = ∝ Β° 𝐡𝑄 = 𝐡𝑅 β‡’ βˆ π΅π‘…π‘„ = βˆ π΅π‘„π‘… = 𝛽° βˆ π‘ƒπ΄π‘„ = 180Β° βˆ’ 2 ∝ Β° βˆ π‘…π΅π‘„ = 180Β° βˆ’ 2𝛽° From β–³ 𝐴𝐡𝐢, 70Β° + (180Β° βˆ’ 2 ∝ Β°) + (180Β° βˆ’ 2𝛽°) = 180Β° 2 ∝ Β° + 2𝛽° = 250Β° ∝ Β° + 𝛽° = 125Β° βˆ π‘ƒπ‘„π‘… = 180Β° βˆ’ ∝ Β° βˆ’ 𝛽° = 180Β° βˆ’ (∝ Β° + 𝛽°) = 180Β° βˆ’ 125Β° = 55 Β° Answer: 55Β° In triangle 𝐴𝐡𝐢, 𝐴𝑃 = 𝐴𝑄 and 𝐡𝑄 = 𝐡𝑅. Determine angle 𝑃𝑄𝑅 in degrees.

Γ—