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Q1. Find the number of ways to choose an ordered pair ((a,b)) of numbers from the set (1...10) such that (|a-b| <= 5). S1. Answer: 80 Method 1: Direct counting 1: 1,2,3,4,5,6 => count:6 Similarly the count is 6 for 2,3,4,5. So total: 30*2 = 60 pairs. Then for 6,7,8,9,10 there are 5,4,3,2,1 options. So 15*2 = 30 pairs. Total: 90 But we have counted same number pairs twice. So 90-10=80 = answer Method 2: faster and better: complementary counting Total: 10x10 = 100 Subtract invalid: 1: 7,8,9,10 2: 8,9,10 3: 9,10 4: 10 Total: 10*2 = 20 100-20 = 80 = Answer Q2. Suppose that in a poll made of 150 people, the following information was obtained: 70 of them read The Hindu , 80 read The Indian Express and 50 read Deccan Herald . 30 read both The Hindu and The Indian Express ; 20 read both The Hindu and Deccan Herald and 25 read both The Indian Express and Deccan Herald . Find at most how many of them read all the three. S2. Incorrect solution first: Initially I applied PIE without thi...
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practice problems pending starting Q3

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Q1. Ten 1's and ten 2's are written on a blackboard. In one turn, a player may erase any two figures. If the two figures erased are identical, they are replaced with a 2. If they are different, they are replaced with a 1. The first player wins if a 1 is left at the end, and the second player wins if a 2 is left. S1. Only P2 can win. Why? 1. There are total 19 moves since in every move the count of total numbers goes down by 1. At the end we will have a single number. 2. Let's say P1 plays the first move, then he will also play the 19th move. Let C1, C2 denote count of 1s and 2s at any given time. 3. For P1 to win, after the 18th move we need C1 = 1, C2 = 1. For P2 to win C1 = 0, C2 = 2 or C1 = 2, C2 = 0. 4. Let's focus on P1's win. That means that in the initial 18 moves both C1 and C2 should decrease by 9 each. 5. What happens in every move? Case 1: Two 1s replaced by 2 => C1 -= 2, C2 += 1 Case 2: Two 2s replaced by 2 => C1 += 0, C2 -= 1 Case 3: One 1 and One...
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practice problems

Q1. On a board, the numbers a, b, c, d, e, f are written in a circle (say counter-clockwise). You may increase two neighbouring numbers by 1. For which case is it impossible to equalize all of them by a sequence of such steps? Select any one option a = 4, b = 10, c = 5, d = 6, e = 9, f = 2 a = 3, b = 9, c = 7, d = 6, e = 9, f = 4 a = 4, b = 11, c = 7, d = 7, e = 10, f = 2 a = 7, b = 10, c = 5, d = 8, e = 9, f = 3 S1. First a partially correct approach which works in this case but will not work always. Final sum of all numbers is even. All the increments are also even.(multiple of 2) So the original sum of all the numbers should also be even. Only option 3 fails. Hence that's the answer. But that approach fails for: a=2,b=4,c=2,d=4,e=2,f=4 So the better way to solve it is this: Let there be 2 groups:  Group 1: a,c,e Group 2: b,d,f Each time we increment, we increment by 1 in each group. So the original sum in both the groups should be same. Else it won't work. Q2. On a 10x10 boa...
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Q1. While applying Menelaus on a triangle ABC, can the traversal pass through one of its vertices? S1. No. In one of the fractions numerator would become zero. In another denominator. Q2. The points (X, Y) are taken on (CA, AB) respectively of triangle ABC. If (BX) and (CY) meet at (P) and AX/CX = BY/AY = 1/2 then find BP/PX. S2. Triangle XAB, Traversal YPC. Apply Menelaus. Answer: 3/4 Q3. In △ABC the points E, F, G are on AB, BC & CA respectively, such that AE/EB = BF/FC = CG/GA = 1/3. The (K, L, M) are the intersections of (AF) & (CE); (BG) & (AF); (CE) & (BG) respectively. If ([ABC] = 1), find ([KLM] = ?) S3. Here we need to use mass points on 2 Cevians at a time. We can't apply on all 3 together since they are not concurrent. Our aim is to find [KLM] like this:  [KLM] = [ABC] - [AKC] - [ALB] - [BMC] For [AKC], use mass points on cevians AF,CE to get the ratio AK:KF. Then [CKA]:[CKF] = AK:KF But [CKA] + [CKF] = [AFC] And [AFC]/[ABC] = FC/BC So [AFC] = 3/4 So...
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Q0. In a ||gram ABCD, does the diagonal AC bisect the angle DAB? Q1. M,N,D are midpoints of AB,AC,BC. X,Y are midpoints of NC,ND. Find [MNXY]/[ABC]. S1. Let [ABC] = f => [AMN] = f/4 => [BCNM] = 3f/4 [CND] = f/4 => [NYX] = f/16 [MNXY] = [NYX] + [MNY] => we need to find [MNY] now. [BDNM] = [BCNM] - [NDC] = f/2 [BDM] = [NDM] = f/4 [MNY] = [DMY] = f/8 => [MNXY] = [NYX] + [MNY] = f/16 + f/8 = 3f/16 Q2. ABCD is a parallelogram. X divides AB in the ratio 3:2. Y divides CD in the ratio 4:1. XY intersects AC at Z. Find (AZ : AC). S2.
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practice problems pending

 1. Show that length of direct common tangent of 2 non intersecting circles(touching is fine) is sqrt(d^2 - (r1-r2)^2) and that of the transverse tangent is sqrt(d^2 - (r1+r2)^2) where 'd' is the distance between the centers of those 2 circles. 2. Now show that if the 2 circles touch each other than DCT length = 2.sqrt(r1r2). 3.  If two circles with radii (a) and (b) touch each other externally. Let (c) be the radius of a circle that touches these two circles as well as a common tangent of these two circles. Prove 1/sqrt(c) = 1/sqrt(a) + 1/sqrt(b)
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homework practice problems

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 Q1. In △ABC, AD, BE and CF are concurrent lines. P, Q, R are points on EF, FD, DE such that DP, EQ and FR are concurrent. Prove that AP, BQ and CR are also concurrent. S1. Prerequisite: Standard and trigonometric form of Ceva's theorem. Standard form gives us: AF/FB * BD/DC * CE/EA = 1 and EP/PF * FQ/QD * DR/RE = 1 Using the trigonometric form, we need to prove that: Sin(CAP)/Sin(PAB) * Sin(ABQ)/Sin(QBC) * Sin(BCR)/Sin(RCA) = 1 Look at triangles EAP and PAF and compute their area ratio: [EAP]/[PAF] = AE * AP * Sin(CAP)/AF*AP*Sin(PAB) = EP/PF Similarly do for others. Multiply both sides of all 3. You will get your desired relation. H.P. Q2. If (X) and (Y) are variable points on the sides (CA, AB) of (\triangle ABC) such that CX/XA + AB/AY = 1 prove that (XY) passes through a fixed point. S2. CX/XA > 0 since everything is positive. => AB/AY < 1 => AY > AB => Y lies on AB extended towards B. => AB = AY - BY => AB/AY = 1 - BY/AY From here: we have 2 cases for X...
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