Using sin 32 = cos (90-32), cos68= sin(90-68) = + - as cosec52= sec(90-52), tan 18= cot(90-18), tan 35= cot(90-35) = 1 + 1 - = { as sec = 1/ cos & cot = 1/ tan ) 1 1 13 For correct figure steps 2 1 14 Mean= 5275/25= 211 Thus, the mean daily expenditure is Rs. 211 2 1
X= -3/2 Thus, required point = (-3/2, 0) 1 12 = + - Using sin 32 = cos (90-32), cos68= sin(90-68) = + - as cosec52= sec(90-52), tan 18= cot(90-18), tan 35= cot(90-35) = 1 + 1 - = { as sec = 1/ cos & cot = 1/ tan ) 1 1 1 13 For correct figure steps 2 1 14 2
8 CIRCLES --- 3(1) --- 3(1) 9 CONSTRUCTIONS --- ---- ---- 4(1) 4(1) 10 INTRODUCTION TO TRIGONOMETRY 1(1) 3(1)* 4(1) 8(3) 11 SOME APPLICATIONS OF TRIGONOMETRY ---- ------ ---- 4(1) 4(1) 12 AREAS RELATED TO CIRCLES --- 3(1) 3(1) 13 SURFACE AREAS AND VOLUMES --- ---- 3(1)* 4(1) 7(2) 14 STATISTICS --- ---- 3(1) 4(1)* 7(2) 15 PROBABILITY 4(2) ---- 4(2) 16 TOTAL 6(6) 12(6) 30(10) 32(8) 80(30) Prepared By : A. N. Borle, Rakesh Kumar, Pramodkumar F. Shende KENDRIYA VIDYALAYA SANGATHAN MUMBAI REGION KV AFS OJHAR 2 - DAYS WORKSHOP CBSE MODEL QUESTION PAPER SECTION (A) [ 1 Mark each] Q.1 Given that HCF (306, 657) = 9, find LCM (306, 657). Q.2 Find the nature of the roots of the given quadratic equations 2x2 – 3x + 5 = 0 Q.3 Find 30th term of the AP: 10, 7, 4, . . . ,. Q.4 Find the distance between the given pairs of points (– 5, 7), (– 1, 3) Q.5 If 15 cot A = 8 then find sin A. Q.6 IF DE ll BC, find EC
Marks Below 10 Below 20 Below 30 Below 40 Below 50 Below 60 Below 70 Below 80 Below 90 Below 100 No. of Students 5 9 17 29 45 60 70 78 83 85 KENDRIYA VIDYALAYA SANGATHAN – MUMBAI REGION KENDRIYA VIDYALAYA, A.F.S. OJHAR Two Days Workshop HOTS Question Paper #61 ~ Marking Scheme Class: X Subject: Mathematics Time allowed: 2 hours Maximum Marks: 50 Q. No. Description Marks 1 NO 1m 2 10th term = 3 1m 3 3cm 1m 4 P(x,y) PA = PA 3 x + y = 5 ½m ½m 5 , where a and b are coprime, b≠0 Proof of Contradiction 1m 1m 6 Figure Given Proof ½m ½m 1m 7 Centroid divides in 2:1 Section Formula D(5.5, 8.5) ½m ½m 1m
8a 8b P(Face)=12/52=4/13; P(not Face) = 1 – 4/13 =9/13 P(Red Ace) = 2/52 = 1/26 ½m ½m 1m 9 (x2-2) is a factor. Long Division Method Other two zeros are ½ and 1 1m 1m 1m 10 x + y = 4 ; x – y = 4 x=4, y=0 2m 1m 11 n=0 1m 2m 12 Statement Proof 1m 2m 13 Correct Proof 3m 14 Construction of correct triangle Construction of Circle Construction of Tangents 1m 1m 1m 15 y= and simplification of Quadratic Eqs y = 8/7 and 4/3 x= 7, 1/7 and 3, 1/3 1m 1m 2m 16 Correct Figure Height = 180m 1m 3m 17 Area of Triangle = 98cm2 Area of Sector = 49=154cm2 Area of Segment= 154-98=56cm2 Area of semi circle= 154cm2 1m 1m 1m
6 Figure Pythagoras theorem Length of Chord = 48cm ½m ½m 1m 7 Section Formula 2:7 1m 1m 8a 8b P(E)=25/36 P(F)=35/36 1m 1m 9 (x2-4x+1) is a factor. Long Division Method Other two zeros are -5 and 7 1m 1m 1m 10 Correct Graph Finding Coordinates 1m 2m 11 3m 12 Statement Proof 1m 2m 13 Correct Proof 3m 14 Construction of correct triangle Construction of Corresponding Triangle 1m 2m 15 y= and simplification of Quadratic Eqs y = 14 and -6 x= 7, -2 and 2, 3 1m 1m 2m 16 Correct Figure Height = 6250m 1m 3m 17 Area of Cake = 13013cm2 Area of Each piece of cake = 2168.83cm2 1½m 1½m
Value: Sharing and Caring 1m 18 Volume of Frustum = h (r12+r22+r1.r2) Find (h)=15cm Find Slant height (l)=16.55 CSA = 2546.84 TSA = CSA + Base = 2546.84 +1386 =3932.84 1m 1m 1m 1m 19 Find x and y Mode Mean 2m 1m 1m KENDRIYA VIDYALAYA SANGATHAN MUMBAI REGION K.V.AFS OJHAR TWO DAYS WORKSHOP 2017-2018 SUB- MATHEMATICS(X) HOT QP TIME: 2hrs. BLUE PRINT (Set-I) -71 Max.Mark-50 Unit Chapter VSA (1 mark) SA -I (2 mark) SA-II (3 mark) LA (4 mark) Total 50(18) Unit Total Number System Real Numbers 3(1) 3(1) 3 Algebra Polynomials 1(1) 1(1) 10 Pair of Linear Equations in Two Variables 4(1) 4(1) Arithmetic Progression 3(1) 3(1) Quadratic Equation 2(1) 2(1)
10 AP = PB AP2 = PB2 ( + 4)2 + ( - 3)2 = 2 +( + 2)2 2 + 8 + 16 + 2 - 6 + 9 = 2 + 2 + 4 + 4 8 - 6 - 4 + 25 – 4 = 0 8 - 10 + 21 = 0 1 1 1 11 From P draw PQ AB AQ = QB = 3cm Join PB. Since arc APC is described by a circle with center B, so BA = BP=BC =6cm. In PQB Cos = = = 60o Area of sector BPA = (62) = 18.84cm Area of BPQ = (QB) (PQ) = (3)( 6 Sin 60) = 7.794Sq.cm Area of portion APQ = Area of sector BPA – Area of BPQ = 18.84 – 7.794 = 11.046 Sq.cm Area of shaded portion = 2 x Area of Quadrant ABC – 2 Area APQ = [2 x (6)2 - 2 x 11.046] =34.428 Sq.cm 1 1
17 Neat and clean correct construction. Steps of Construction 3 1 18 Let r2 & r2 be the two ends of the frustum = tan 30 volume of frustum = h ( r2 1 + r 22 + r1 r2 ) = 10 cm Since the frustum is drawn into a wire of length x Volume of frustum = volume of cylinder 10 = ()2 x x = 7168000 cm 9 x = 7964.4m ½ ½ ½ ½ ½ 1 ½ KENDRIYA VIDYALAYA SANGATHAN MUMBAI REGION K.V.AFS OJHAR TWO DAYS WORKSHOP 2017-2018 SUB- MATHEMATICS(X) HOT QP
TIME: 2hrs. BLUE PRINT 72 Max.Mark-50 Unit Chapter VSA (1 mark) SA -I (2 mark) SA-II (3 mark) LA (4 mark) Total 50(18) Unit Total Number System Real Numbers 1(1) 1(1) 1 Algebra Polynomials 3(1) 3(1) 10 Pair of Linear Equations in Two Variables 4(1) 4(1) Arithmetic Progression 1(1) 1(1) Quadratic Equation 2(1) 2(1) Coordinate Geometry Coordinate Geometry 3(1) 3(1) 3 Trigonometry Introduction to Trigonometry 2(1) 2(1) 6 Application of Trigonometry 1(1) 3(1) 4(1) Geometry Triangles 2(1) 4(1) 6(2) 12 Circles 4(1) 4(1) Construction 2(1) 2(1) Mensuration Areas Related to Circles 3(1) 4(1) 7(2) 15 Surface Areas and Volumes 1(1) 3(1) 4(1) 8(3) Statistics and Probability Statistics 3(1) 3(1) 3 Probability TOTAL 4(4) 8(4) 18(6) 20(5) 50(19) 50 Prepared By 1.Mr.C.J.N.RAO TGT-MATHS K.V.NO2 MIRC,AHMEDNAGAR 2.MR.T.S.SONAWANE TGT-MATHS K.V. ISP NASIK 3.MR.AJAYKUMAR PATOLE TGT-MATHS K.V.NO.1,AHMEDNAGAR
2 The probability of an event that cannot happen 0 3 The probability of an event that is certain to happen 1 4 An event that cannot happen is called impossible event 5 An event that is certain to happen is called sure event 6 The sum of the probabilities of all elementary events of an experiment 1 7 A die thrown once. Probability of getting a prime number prime no.-(2,3,5); P(E)=3/6=1/2 8 In well shuffled deck of 52 cards, probability of getting- a) a king of red colour 2/52 b) the jack of hearts 1/52 c) a spade 13/52 d) the queen of diamonds 1/52 9 A box contains 90 discs which are numbered from 1 to 90, one disc is drawn . Find the probability of getting a) a two digit number 81/90 b)a perfect square number 9/90 c) a number divisible by 5 18/90 10 The probability expressed as a percentage of a particular occurrence can never be Greater than 1 SURFACE AREAS AND VOLUMS S.NO FILL IN THE BLANKS Answer 1 A surahi is the combination of _________________ a sphere and a cylinder 2 The shape of a glass is usually in the form of ________________ frustum of a cone 3 Surface ara of solid cylinder of radius r and height h is _______ 2πr(r+h) 4 A solid cylinder of radius r and height h is placed over same cylinder. Total surface area of so formed shape is__________ 4πrh+2πr.r 5 The curved surface area of a frustum of a con is_____________ πl(R+r) 6 Volume of cylinder is __________ πr.r.h 7 Volume of vertical cylinder is base area ×_______ height 8 Volume of hemispher is____________ 2/3(π.r.r.r)
In triangle ABC, Sin --------- Cos = -------- tan --------- Cosec -------- Sec = -------- Cot = -------- Fill in the blanks: ratio 0 30 45 60 90 sin cos tan cosec Sec Cot PREPARED BY:- 1 . Mr. K. Selvaraju PGT (Maths) K. V. Ganeshkhind, Pune 2. Mr. Deepak G. Meshram PGT (Maths) K.V. Pulgaon 3. Mr. S. A. Azad TGT (Maths) K.V. Ojhar 4. Mrs. Rajani Dabhade TGT (Maths) K.V. Ojhar