www.InyaTrust.com BELTHANGADY TALUK MATHS TEACHERS WORKSHOP MINIMUM STUDY LEVEL QUESTION PAPER Time: 2hour TOTAL: 50 1x7=7 SUBJECT: MATHEMATICS I . Choose the best alternative from following options 1) The Arithmetic Mean between 12 and 30. (A) 12 (B) 30 (C) (D) 42 n 2) If P2 = 90,then the value of n ‘ is ---(A) 10 (B) 9 (C) 90 (D) 45 3) Three coins are tossed at a time, the probability of getting all heads. (A) (B) (C) (D) 4) If The Standard deviation of 10 data is 8, then the variance is ---(A) 16 (B) 4 (C) 80 (D) 5) If 2sin = 1, then he value of θ is ---(A) 300 (B) 450 (C) 600 (D) 900 6) If in = , then, ec = ----- (B) (A) 3 (C) (D) 2 7) If f(x) = 3x + 3x – 11x - 6 ,then f(1) is ---(A) -11 (B) 22 (C) 0 (D) -6 II. Answer the following 8) Express 120 as the product of prime factors... 1x5=5 2 120 2 60 2 30 3 15 5 ∴ 120 = 2x2x2x3x5 9) If A and B are disjoint sets then what is n (A∩B)? n(AUB) = n(A) + n (B) 10) The corresponding heights of two similar triangles are 3cm and 5cm. Find The ratio of their areas. Ratio of their areas are 32 : 52 ⇒ 9 : 25 11) What are concentric circles? Circles having the same centre but different radii are called concentric circles. Typed By: Yakub S., GHS Nada, Belthangady for MSTF Mangalore (Belthangady wing) www.InyaTrust.com Page 1 www.InyaTrust.com 12) Write a formula to find the total surface area of a hemisphere. A = III. Answer the following 2x 10 = 20 13) Cons ruct a pair of tangents to a circle of radius 5cm, and the angle between them Is 500 14) Rationalize the denominator and simplify = = = = = √ √ √ √ √ √ x √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ − √ Typed By: Yakub S., GHS Nada, Belthangady for MSTF Mangalore (Belthangady wing) www.InyaTrust.com Page 2 www.InyaTrust.com 15) Find the product. √2 x √3 √2 = 2 = 2 = √2 √3 = 3 = 3 √2 x3 √16x27 = √3 √432 16) In a group of passengers, 100 know Kannada, 50 know English and 25 know both. If passengers know either Kanada or English. Drawing Venn diagram find the total number of passengers are in a group. A – { Know Kannada}, B – { Know English} n(AUB) = ?, n(A) = 100, n(B) = 50 , n(A∩B) = 25 n(AUB) = n(A) + n(B) – n(A∩B) = 100 + 50 - 25 = 125 17) How many triangles to be drawn through 12 points on a circle The number of triangles to be drawn = 12C3 ! n Cr = ( )! 12 C3 = ( 12 C2 = 12 ! ! )! ! C2 = 18) Solve by using the formula. x2 - 2x + 4 =0 a = 1, b = -2 , c = 4 −(−2) ± (−2) − 4x1x4 = 2x1 2 ± √4 − 16 = 2 2 ± √−12 = 2 2 ± 2√−3 = 2 = 1 ± √−3 Typed By: Yakub S., GHS Nada, Belthangady for MSTF Mangalore (Belthangady wing) www.InyaTrust.com Page 3 www.InyaTrust.com 19) If tan = ,then find sin In the Figure, AC = √ + AC = √7 + 24 AC = √49 + 576 AC = √625 AC = 25 ∴ and cos = = 20) Find the slope of a line forming by joining the points (-4, 1) and (-5, 2). m= − − m= ( ) m= m= m= − 21) Verify Euler’s formula for the given graph N = 4, R = 6, A = 8 N+R = 4+ 6 = 10 A+2 = 8 + 2 = 10 ∴ N+R = A+2 22) Sketch out the field to the following notes from the field book (1 cm = 25m.) To D 100 To E 200 125 75 25 75 To C 50 To B From A Typed By: Yakub S., GHS Nada, Belthangady for MSTF Mangalore (Belthangady wing) www.InyaTrust.com Page 4 www.InyaTrust.com Scale 1cm = 25m ⇒75m = 3cm, 125m=5cm, 50m=2cm,100m=4cm,200m=8cm IV. Answer the following 23) Calculate the standard deviation of the following data X 5-15 15-25 25-35 35-45 f 8 12 CI f 5-15 15-25 25-35 35-45 45-55 55-65 8 12 20 10 7 3 20 10 Assumed Mean Methdo d=x-30 fd X 10 20 30 40 50 60 -20 -10 0 10 20 30 -160 -120 0 100 140 90 60 50 = ∑ 45-55 55-65 7 3 d2 fd2 400 100 0 100 400 900 3200 1200 0 1000 2800 2700 10900 ∑ − = 3 x 2= 6 − = . = √ – . . = . Typed By: Yakub S., GHS Nada, Belthangady for MSTF Mangalore (Belthangady wing) www.InyaTrust.com Page 5 www.InyaTrust.com 24) prove that “If two circles touch each other internally, the centers and the point of Contact Collinear” If two circles touch each other internally the centres and the point of contact are collinear. Given:A and B are centres of touching circles. P is point of contact. To prove : A,P,and B are collinear Construction: Draw the common tangent XPY . Join AP and BP. Proof:In the figure ∠APX = 900……………..(1) ∵Radius drawn at the point of contact ∠BPX = 900 ………… ..(2) is perpendicular to the tangent. ∠APX = ∠BPX = 900 [ From (1) and (2) AP and BP lie on the same line ∴ APB is a straight line ∴ A, P and B are collinear. V) Answer the following 4 x 3= 12 25) Draw transverse common tangents to two circles of radii 5cm and 2cm having Their centers 9cm 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) Solve graphically: x2 +3 x + 2 =0 Typed By: Yakub S., GHS Nada, Belthangady for MSTF Mangalore (Belthangady wing) www.InyaTrust.com Page 7 www.InyaTrust.com 27) Prove that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides. Theorem:Phythagoras Therem 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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