Triangle practice problems

 


Answer: 92



Answer: 92

Answer: 10






















Answer: 75,150,90,150,75 = A,B,C,D,E


Answer: 108





Answer: 540 degrees




















Solution link. Answer: 72 degrees.





























Answer: 71

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IOQM 2024 Paper solutions (Done 1-21, 29)

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Q1 (number-theory). The smallest positive integer that does not divide 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 is: Solution : This product has all the integers from 1 to 9. So smallest integer not dividing it would be the next prime after 9. Which is 11. Answer: 11 Q2 (combinatorics): The number of four-digit odd numbers having digits 1, 2, 3, 4, each occurring exactly once, is: Solution: Last digit has to be 1 or 3: 2C1 Choosing remaining 3 digits: 3C3 Arranging those 3 digits: 3! Answer: 2C1*3C3*3! = 12 Q3 (number-theory): The number obtained by taking the last two digits of \(5^{2024}\) in the same order is: Solution: To get last 2 digits from number, divide by 100 and take the remainder. So we have to compute mod 100. 25 mod 100 = 25 25.5 mod 100 = 125 mod 100 = 25 5^3.5 mod 100 = 25.5 mod 100 = 25 5^4.5 mod 100 = 25.5 mod 100 = 25 ..... So: 5^2024 mod 100 = 25. Q4 (geometry): Let ABCD be a quadrilateral with \(\angle ADC = 70^\circ\), \(\angle ACD = 70^\circ\), \(\angle ACB = 10^\circ\...
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Combinatorics DPP - RACE 6 - Q16 pending discussion

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Q1 . A question paper consists of two sections. Section A has 7 questions and section B has 8 questions. A student has to answer 10 questions out of these 15 questions. The number of ways in which he can answer if he must answer at least 3 of section A and at least 4 of the section B is: Solution: One wrong way to solve this: For the mandatory questions: 7C3*8C4 Now 3 more questions need to be answered from 4 of A and 4 of B. Cases: 3A,0B | 2A,1B | 1A,2B | 0A,3B 4C3*4C0 + 4C2*4C1 + 4C1*4C2 + 4C0*4C1 = 4 + 24 + 24 + 4 = 56 Finally: 7C3*8C4*56 = 35*70*56 = 137200 Why is it wrong? Let's say we chose A1,A2,A3 as mandatory from A and then chose A4,A5,A6 when we did 4C3*4C0 for (3A,0B). But we could have done it the other way round also, i.e. choose A4,A5,A6 as mandatory and then choose A1,A2,A3. So there is overcounting. Correct Solution : Simply create the cases for A,B: (3,7),(4,6),(5,5),(6,4) 7C3*8C7 + 7C4*8C6 + 7C5*8C5 + 7C6*8C4 = 2926. Q2 . A kindergarten teacher has 25 kids in her...
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Algebra DPP 1.3 Quadratics

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