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practice problems pending - easy

Q1). A number N in base 10, is 503 in base b and 305 in base b+2. What is the product of the digits of N? S1. N = 248 Answer = 64 Q2).  Let (ABCD) be a rectangle, in which (AB + BC + CD = 20) and (AE = 9), where (E) is the midpoint of the side (BC). Find the area of the rectangle. S2. 2x + y = 20______[1] 81 = x^2 + y^2/4______[2] We just need to find xy. Square the first equation and you will 4x^2 + y^2, replace it using 2 and you will get x.y = 19. Q3). What is the least positive integer by which 2^5. 3^6. 4^3. 5^3.6^7 should be multiplied so that the product is a perfect square? S3. Write it as 2^(5+6+7).3^(6+7).5^3 = 2^18.3^13.5^3 So the answer is 15. Q4). Find the number of integer solutions to ||x| - 2020| < 5 S4. -5 < |x| - 2020 < 5 2015 < |x| < 2025 => x has 18 possible integer solutions when |x| = 2016 to 2024 Q5). The product (55 * 60 * 65) is written as the product of five distinct positive integers. What is the least possible value of the largest of th...
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 Q1. ABC is a triangle and D and E are interior points of the sides AB and BC respectively such that: AD/DB = 1/3 CE/EB = 3 If AE and CD intersect at F, find CF/FD. S1. Approach 1: Using mass points : AD/DB = 1/3 => A has higher mass since it's closer to D. Let mA = 3 mB = 1 => mC = 1/3 => mE = 4/3 Now mF = 3 + 4/3 = 13/3 and it matches 4 + 1/3 = 13/3 So we have assigned masses correctly. => CF/FD = 4/(1/3) = 12 Approach 2: Menelaus theorem Typically you would notice that if we have a solution using mass points, we can also solve it using Ceva's theorem or Menelaus' theorem. Here, in triangle BDC, AFE is the traversal intersecting all sides. BE/EC * CF/FD * DA/AB = 1 => 1/3 * CF/FD * 1/4 = 1 => CF/FD = 12 = Answer Q2.  L and M are the mid-points of the diagonals BD and AC respectively of the quadrilateral ABCD. Through D, draw DE equal and parallel to AB. Show that EC || LM EC = 2LM S2. It's straightforward with co-ordinate geometry. A (0,0) B(x1,0) C(...
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 Q1. In quadrilateral ABCD, the diagonals AC and BD meet at O. Suppose the four triangles AOB,BOC,COD and DOA are equal in area, prove that ABCD is a parallelogram. S1. [AOB] = [COD] => 1/2.OA.OB.Sin(AOB) = 1/2.OC.OD.Sin(COD) Angles AOB and COD are same. => OA.OB = OC.OD Similarly, OA.OD = OC.OB Multiply both => OA = OC and OB = OD => O bisects both diagonals. If the diagonals bisect each other, the quadrilateral is a ||gram. Why? Let the quadrilateral be A (0,0), B(x,0), C(a,b) D(p,q) O = Midpoint of AC = (a/2,b/2) = Midpoint of BD = ((p+x)/2,q/2) b/2 = q/2 => b = q => CD || AB and a = p+x Slope of AD = q/p Slope of BC = b/(a-x) = q/p => AD || BC Opposite sides are || => ABCD is a ||gram. H.P. Q2. In a parallelogram  A B C D A BC D , a point  P P on the segment  A B A B is taken such that  A P A B = A B A P ​ =   61 2022 2022 61 ​  and a point  Q Q on the segment  A D A D is taken such that  A Q A D = 61 2065...
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