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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.

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Presentation on theme: "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."β€” Presentation transcript:

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 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 hour, therefore B’s speed =20Γ· 1 2 = 40km/h When they started at the same time and met after 4 hours, B has traveled 4Γ—40=160π‘˜π‘š. A has travelled 370βˆ’160=210π‘˜π‘š. Therefore, A’s speed =210Γ·4=____π‘˜π‘š/h. Answer: π‘˜π‘š/β„Ž

2 There are 5 circles with 3 different diameters
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 π‘Ž and the area (let’s name it 𝑆) as π‘Ž2πœ‹. The radius for big circle is 3π‘Ž and the area (let’s name it 𝐡) is 9π‘Ž2πœ‹. The radius for medium circle is 2π‘Ž and the area (let’s name it 𝑀) is 4 π‘Ž2πœ‹. The area of unshaded parts + the area of shaded parts = 𝐡 βˆ’βˆ’β†’ β‘  Given the area of unshaded parts = 𝐡 – 𝑀 + 2𝑆 – 𝑆 = 20π‘π‘š2 𝐡 – 𝑀 + 𝑆 = 20π‘π‘š2 9π‘Ž2πœ‹β€“ 4π‘Ž2πœ‹+ π‘Ž2πœ‹= 20π‘π‘š2 6π‘Ž2πœ‹= 20π‘π‘š2 βˆ’βˆ’β†’ β‘‘ From β‘ , the area of shaded parts = 𝐡 – area of unshaded parts = 9π‘Ž2πœ‹β€“ 6π‘Ž2πœ‹ = 3π‘Ž2πœ‹ From β‘‘, = ____π‘π‘š2 Answer: 10 π‘π‘š2

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 = ____ βˆ’βˆ’β†’ β‘  The average of remaining ten numbers is 6.1, their sum must be = 6.1Γ—10=61 Hence the number which is removed is β‘  βˆ’ 61 = ____ Answer: 5

4 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? Solution: The difference between the last number (n) and 64 = the difference between 37 and 1 π‘›βˆ’64=37 βˆ’1 𝑛=____ Answer: 100


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