www.InyaTrust.com BELTHANGADY TALUK MATHS TEACHERS WORKSHOP MINIMUM STUDY LEVEL QUESTION PAPER Time: 2hour TOTAL: SUBJECT: MATHEMATICS I. Choose the best alternative from following options 1) If Tn = 2n + 1 ,The first sum of first three terms --(A) 3 (B) 7 (C) 10 (D) 2) If n! =24 then the value of n’ is ---(A) 24 (B) 6 (C) 4 (D) 1 3) Probability of sure event is--(A) 0 (B) (C) (D) 1 4) The formula to find coefficient of variation--(A) x100 (B x 100 (C) . 50 1x7=7 (D) . 5) The slope of the equation 3x + 2y +1 is ----(A) (B) (C) (D) 6) If the standard deviation of a certain data is 4 , its variance is-------(A) 2 (B) 4 (C) 8 (D) 7) The zero of the polynomial f(x) = 3x – 6 is ---(A) 2 (B) -2 (C) (D) II. Answer the following 8) Express 210 in prime factors 1x5=5 2 x 3x5x7 9) If A = { 1, 3, 4, 5 }, B = { 1, 4, 9 } ,find A – B A-B = { 3, 5 } 10) State converse of Thales Theorem. If a straight line divides two sides of a triangle proportionally,then the straight line is parallel to the third side. 11) The two circles are intersecting, what is the maximum number of common tangents to be drawn? 2 12) Find the curved surface area of a cone, if its radius is 7cm and slant height is 10cm. A= A= 7x10 A= . . Typed by: Yakub S,GHS Nada,Belthangady.For MSTF Mangalore(Belthangady wing) www.InyaTrust.com Page 1 www.InyaTrust.com III. Answer the following 2x 10 = 20 13) In a circle of radius 3cm draw two radii such that the angle between them is60 0 Construct tangents at the non-centre ends of the radii. 14) Rationalize the denominator and simplify √ = = = √ (√ x √ √ √ √ √ ) √ (√ √ √ ) (√ √ ) =√ +√ 15) Simplify: 4√63 + 5√7 - 8√28 4√63 + 5√7 - 8√28 = 4√9x7 + 5√7 - 8√4x7 = 4x3√7 + 5√7 – 8x2√7 = 12√7 + 5√7 – 16√7 = √7 (12+ 5 – 16) =√ Typed by: Yakub S,GHS Nada,Belthangady.For MSTF Mangalore(Belthangady wing) www.InyaTrust.com Page 2 www.InyaTrust.com 16) There are 50 students in a school. Out o them 30 students are participating in Dance,17 are in drama, 5 members are participating in both in a school day . Find the number of students who are not participating in any one of these. A-{Students Participating in Dance} B-{Students participating in drama} n(AUB) = n(A) + n(B) – n(A∩B) n(AUB) = 30 + 17 – 5 n(AUB) = 42 ∴ The number of students who are not participating in any = 50 – 42 = 8 17) Classify the following as permutation and combination. a). Five different subject books to be arranged in a shelf. Permutation b).There is 8 chairs and 8 people to occupy them. Permutation c). A collection of 10 toys are to be divided equally between two children. Combination d).Five keys are to be arranged in a circular key ring. permuation OR How many committees of five with a given chairperson can be selected from 12 persons? ! Formula: nCr = ( )! − 12 C1x 11C4 = 12 x = 12 x 330 ! = 3960 18) Solve by using the formula x2 + x = 12 a = 1, b = 1 , c = -12 −1 ± √1 − 4x1x − 12 = 2x1 −1 ± √1 + 48 = 2 −1 ± √49 = 2 −1 ± 7 2 = Or = = 19) = Or = = Or =− Typed by: Yakub S,GHS Nada,Belthangady.For MSTF Mangalore(Belthangady wing) www.InyaTrust.com Page 3 www.InyaTrust.com 20) Prove that(1 + = sec .cos = ).cos = 1. .cos = 21) Find the distance between the points (2, 3) and (6, 6). d = (6 − 2) + (6 − 3) d = √4 + 3 d = √16 + 9 d = √25 d = units 22) Verify Euler’s formula for the given graph N = 4, R = 4, A = 6 N+R = 4+ 4 = 8 A+2 = 6 + 2 = 8 ∴ N+R = A+2 22) Sketch out the field to the following notes from the field book (1 cm = 25m.) To D 75To E 225 150 100 50 75 To C 50 To B From A 1cm = 25m, 50m=2cm,75m=3cm,100m=4cm,150m=6cm,225m=9cm Typed by: Yakub S,GHS Nada,Belthangady.For MSTF Mangalore(Belthangady wing) www.InyaTrust.com Page 4 www.InyaTrust.com IV. Answer the following 23) Calculate the standard deviation of the following data X 0-10 10-20 20-30 f 1 2 3 3 x 2= 6 30-40 4 CI X f d=x-25 fd d2 fd2 0-10 10-20 20-30 30-40 5 15 25 35 1 2 3 4 -2 -1 0 1 -2 -2 0 4 4 1 0 1 4 2 0 4 10 = = ∑ − 0 10 ∑ − = √ = Typed by: Yakub S,GHS Nada,Belthangady.For MSTF Mangalore(Belthangady wing) www.InyaTrust.com Page 5 www.InyaTrust.com 23) Prove that the tangents drawn from an external point to a circle are equal. Given:A is the centre.B is an external point.. BP and BQ are the tangents.AP, AQ and AB are joined To prove : (a). BP = BQ Proof: In ∆APB and ∆AQB, [ ∵ Radius of the same circle AP = AQ ∠APB = ∠AQB =900 [ ∵ Radius drawn at the point of contact is perpendicular to the tangent AB = ∴ ∴ ∆APB ≡ ∆AQB (a). AB [ ∵ RHS postulates BP = BQ V) Answer the following 4 x 3= 12 25) Draw direct common tangents to two circles of radii 5cm and 2.5cm having their Centers 10cm apart and measure their lengths. Typed by: Yakub S,GHS Nada,Belthangady.For MSTF Mangalore(Belthangady wing) www.InyaTrust.com Page 6 www.InyaTrust.com 26) Draw the graph of y = x2 - 6x - 7 x 3 1 5 y -16 -12 -12 0 -7 6 -7 -2 9 8 9 Roots are x= -1 and x=7 Typed by: Yakub S,GHS Nada,Belthangady.For MSTF Mangalore(Belthangady wing) www.InyaTrust.com Page 7 www.InyaTrust.com 27) State and prove converse of Pythagoras theorem. In a right angled triangle,the square of the hypotenuse is equal to the sum of the square of the other two sides. Given: ∆ABC In which ∠ABC = 900 To Prove : AB2 + BC2 = CA2 Construction: Draw BD ⟘ AC . Proof: In ∆ABC and ∆ADB , ∠ABC = ∠ADB = 900 [ ∵ Given and Construction ∠BAD =∠BAD [∵ Common angle ∴ ∆ABC ~ ∆ADB [∵ AA criteria ⇒ AB AD = AC AB ⇒ AB2 = AC.AD……..(1) In ∆ABC and ∆BDC , ∠ABC = ∠BDC = 900 [ ∵ Given and construction ∠ACB = ∠ACB [∵ Common angle ∴ ∆ABC ~ ∆BDC [∵ AA criteria ⇒ BC DC = AC BC ⇒ BC2 = AC.DC……..(2) (1) + (2) AB2+ BC2 = (AC.AD) + (AC.DC) AB2+ BC2 = AC.(AD + DC) AB2+ BC2 = AC.AC AB2+ BC2 = AC2 [ ∵AD + DC = AC] Typed by: Yakub S,GHS Nada,Belthangady.For MSTF Mangalore(Belthangady wing) www.InyaTrust.com Page 8

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