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iii mock paper maths 2016-2017

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Udayan Sathe
Mumbai University (MU), Mumbai
Msc and M.Ed physics

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SECTION A (40 marks) (Solve all questions) Question 1:a) The price of an article inclusive of sales tax of 12% is Rs 2016. Find its market price. If the sales tax is reduced to 7%., how much less does the customer pay for the article? [4] b) Find the remainder when (2x3 3x2 + 7x 8) is divided by (x -1) [3] c) Construct a ABC in which BC = 6.4 cm ABC = 700 and CA = 5.8 cm. Draw the inscribed circle of ABC. Measure the radius of circle. [3] Question 2:a) b) A = 3 1 -1 2 B = -1 6 5 -3 C= 6 2 -4 -3 Verify that (AB)C = A(BC) In the circle O is centre of circle. Find value of a, b, c, A D O [3] [3] C B c) If P = 4xy ; find the value of x+y P + 2x P + 2y P 2x P 2y [4] Question 3:a) b) A die is thrown once. What is the probability of getting a number : [3] i) greater than 2 ? ii) less than or equal to 2 ? Calculate compound interest for the second year and third year on Rs 16000 invested for 5 years at 10% per annum. c) ABCD is a square of side 5 cm inscribed in a circle. Find. i) the radius of circle ii) The area of shaded region. D C D C A A B B [3] [4] Question 4:a) b) Solve the equation. (x + 3)2 4 (x + 3) 5 = 0. The arithmetic mean is 118 marks. Find the missing number x. Marks Frequency c) Prove that: sin A sin (90 A) 40 16 80 20 x 40 160 30 [3] [4] 200 14 [3] cos A , sec A. cosec A cos (90 A) SECTION B (40 Marks) (Solve any Four questions.) Question 5:a) b) c) Given x {whole number}. Find the solution set of : -1 3 + 4x < 23. [3] Find the lengths of the medians of a ABC whose vertices are A(-1, 3), B(1, -1), C(5, 1) [3] Using rules and compass construct [4] i) a triangle ABC in which AB = 5.5 cm, BC = 3.4 cm and CA = 4.9 cm. ii) the locus of points equidistant from A and C. iii) a circle touching AB at A and passing through C. Question 6:a) The length of the shadow of a vertically tower on level ground increases by 10 m, when the altitude of the sun changes from 450 to 300 calculate the height of the tower correct to two decimal places. ( 3 = 1.732) [4] b) Rajesh deposits Rs 300 per month in a recurring deposits account for 2 years. If the rate of interest is 10% per year, calculate the amount that Rajesh will receive at the end of 2 years i.e.at the time of maturity. c) Find the mode of following data using a histogram. Class Frequency 0- 10 5 10 -20 12 20 - 30 20 [3] [3] 30- 40 9 40 50 4 Question 7:a) i)PA = 4cm. PB = 6 cm, PC = 5 cm, Find PD. D A [2] B C ii)PA = 20 cm, PB = 16 cm. Find BC C [2] B P A Find the positive value of k for which the equations x2 + kx + 64 = 0 and x2 8x + k = 0 will b) have real roots. [3] c) A rectangular tank has l = 4 m, width= 3 m and capacity = 30 m3. A small model of the tank is made with capacity 240 cm3. Find the dimension of the model. [3] Question 8:a) From the following frequency distribution table, find. i) Lower quartile ii) Upper quartile iii) Inter quartile range C.I. Frequency b) 5 - 10 3 10 - 15 4 15 - 20 6 [4] 20 - 25 9 25 - 30 7 30 - 35 1 Solve using formula. 6x2 35x + 50 = 0 If A (-4, 3) and B = ( 8, -6) i) Find length of AB ii) In what ratio is the line joining AB, divided by the x-axis? c) [3] [3] Question 9:a) i)The point P (2, -4) is reflected about the line x = 0 the get the image Q. Find the co-ordinates of Q. ii)The point Q is reflected about the line y = 0 to get the image R. Find the co-ordinates of R. iii) Name the figure PQR. iv) Find the area of figure PQR. [4] A two-digit number is such that the product of its digits is 12. When 36 is added to this number b) the digits interchange their places. Find the numbers. [3] c) ABC ~ PQR, area ( ABC) = 36 cm2 and area ( PQR) = 28 cm2. If QR = 6 cm, find BC. [3] Question 10:a) Mr. Ashok has an account in the central Bank of India. The following are the entries from his pass book. Date 1.01. 05 7. 01.05 17.01.05 [5] Particulars B/F By cash To cheque Withdrawals 400 Deposits .. 500 Balance 1200 10.02.05 25.09.05 20.09.05 21.11.05 5.12.05 By cash To cheque By cash To cheque By cash 500 .. 600 . 800 .. 700 .. 300 If Mr. Ashok gets Rs. 83.75 as interest at the end of the year where the interest is compounded annually, calculate the rate of interest paid by the bank in his Saving Bank account on 31 st December 05. b) Use step deviation method and calculate the mean of the following frequency distribution. [5] Class- interval Frequency 10 - 15 5 15- 20 6 20 - 25 8 25 - 30 12 30 - 35 6 35 40 3 Question 11:a) By purchasing Rs 50 shares of a company for Rs 75 each, Sonal gets 8% per annum on her investment. i) What rate per cent is the company paying? ii) What is her annual dividend, if she buys 350 shares? b) Find the equation of the line passing through (-2, -4) and perpendicular to the line. 3x y + 5 = 0 c) Calculate the mean, the median and the mode of the following number. 17, 19, 11, 23, 19 SOLUTION Answer 1. a) Let marked price be Rs. x x + 12% of x = 2016 x + 12 x = 2016 100 112x = 2016 100 x = 2016 x 100 112 x = 1800 Marked price of the article Rs.1800 [4] [3] [3] b) New sales tax = 7% Now the customer will pay = 1800 + 7% of Rs. 1800 = 1800 + 7 x 1800 100 = 1800 + 126 = 1926 Customer will pay for the article = 2016 1926 Rs.= 90 By remainder Thm if x = 1 f(x) = 2x3 3x2 + 7x 8 f(x) = 2(1)3 3(1)2 + 7(1) 8 =2 3+7 8 = -11 + 9 = -2 Remainder is -2 c) Answer 2:a) H.S. = (AB) C = 3 1 -1 2 -1 5 6 -3 6 -4 2 -3 = -3 + 6 15 3 +1+12 -5 -6 6 4 2 -3 = 3 12 13 -11 6 -4 2 -3 = 18 + 24 - 12 36 78 22 - 52 + 33 = 42 - 48 56 - 19 R.H.S. = A (BC) = 3 1 -1 5 6 -4 -1 2 6 -3 2 -3 = 3 1 -1 2 -6 + 10 4 - 15 36 6 -24 + 9 = 3 1 -1 2 4 30 = -11 -15 12 + 30 -33 15 -4 + 60 +11 30 = 42 -48 56 -19 L.H.S. = R.H.S. b) In ABC A + B + C = 180 .. sum of angles of a triangle a + 70 + 35 = 180. a = 180 105 a = 75 DBC = 900 . angle subtended in semicircle. DBA + ABC = 90 c + 70 = 90 c = 90 70 c = 20 BDC = BAC = 75 b = 75 c) P = 4xy x+ y p = 2y 2x x + y By componendo devidendo P + 2x = 2y + x + y P 2x 2y x y = x + 3y (i) P = 2x 2y x + y P + 2y = 2x + x + y P 2y 2x x y = 3x + y ..(ii) x-y P + 2x + P + 2y . From i & ii P 2x P 2y = x + 3y + 3x + y y x x y = x + 3y - 3x + y y x y-x = x + 3y 3x y y -x = -2x + 2y = 2 (y x) = 2 y x (y x) Answer 3. a) Number of total possible outcomes = 6 a) total favourable outcomes = 4 P (greater than 2) = 4 = 2 6 3 b) total favourable outcomes = 2 P (less than or equal to 2) = 2 - 1 6 3 b) P = 16000 Rs. Rs. 10% per annum T = 1 year S.I. 1 = P x N x R 100 = 16000 x 1 x 10 100 = 1600 Rs. P for second year = 16000 + 1600 = 17600 Rs. S.I. 2 = 17600 x 1 x 10 100 = 1760 Rs. P for third year = 17600 + 1760 = 19360 Rs. S.I. = 19360 x 1 x 10 100 = 1936 Rs. C.I. for 2nd year = 1760 Rs. 3rd year = 1936 Rs. c) In ABD A = 90 AB = 5 cm AD = 5 cm by Pythagoras thm. BD2 = AD2 + AB2 = 5 2 + 52 = 2 x 25 = 5 2 cm Diameter = 5 2 cm radius = 5 2/2 cm Area of circle = r2 = 22 x 5 2 x 5 2 7 2 2 = 22 x 25 x 2 7 2x2 = 275 cm2 7 Area of square = (AB)2 52 = 25 cm2 A of shaded region = A of circle = A of square = 39 29 25 = 14.29 cm2 Answer 4 a) Let (x + 3) = P p2 4p 5 = 0 p2 5p + p 5 = 0 p(p 5) + 1(p 5) = 0 (p 5) (p + 1) = 0 P 5=0 OR p + 1 = 0 P=5 p = -1 Reconstitute (x + 3) = 5 OR (x + 3) = -1 x=5 3 x = -1 -3 2 OR -4 Sol . {2, -4} b) mean = fixi fi 118 = (40 x 16) + (80 x 20) + (x x 40) + (160 x 30) + (200 x 14) 16 + 20 + 40 + 30 + 14 118 = 640 + 1600 + 40 x+ 4800 + 2800 120 118 x 120 = 40x + 9840 14160 = 40x = 9840 40x = 4820 x = 4320 40 = 108 c) L.H.S. = sin A + cos A cos A sin A = sin2A + cos2A cos A. sin A = 1 cos A. sin A = sec A. cosec A = R.H.S. Hence proved Answer 5 a) Whole numbers = {0, 1, 2, 3 } -1 3 + 4x 3 + 4x < 23 -1 -3 4x 4x < 23 3 -4 4x 4x < 20 -4 x x < 20 4 4 -1 x -1 x < 5 Sol set = { 0, 1, 2, 3, 4} b) A (-1, 3) F (0, 1) E(2, 2) B(1, -1) D(3, 0) C (5,1) D, E, F are mid-points of BC, AC and AB respectively Co-ordinates of midpoints = x1 + x2 , y1 + y2 2 2 Co-ordinates of D = 1 + 5 , -1 + 1 2 2 = (3, 0) Co-ordinates of E = 5 + (-1) , 1 + 3 2 2 = 4 , 4 2 2 Co-ordinates of F = 1 + (-1) , (-1) + 3 2 2 = 0 ,2 2 2 Distance between points = (x2 x1)2 + (y2 y1)2 l(AD) = (-1 3)2 + (3 0)2 = 16 + 9 = 25 = 5 cm l(BE) = (1 2)2 + (-1 -2)2 = (-1)2 + (-3)2 = 1 + 9 = 10 cm l(CF) = (5- 0)2 + (1 1)2 = 25 + 0 = 5 cm c) Q C ii) Locus of points equidistant from A and C is a perpendicular bisector of AC i.e. line PQ. iii) The circle touches at A and passing through C mean the radius of circle is perpendicular 5 line AB. Draw perpendicular at A it interests at O 5 PQ. Which is the centre. Answer 6 a) P Q x R 10 m S Let (QR) = x m (QS) = (x + 10) m tan 30 = PQ QS 1 = h 3 x + 10 tan 45 = h x 1 = h x x=h put x = h in I 1 = x 3 x + 10 x + 10 = 3x 3 x x = 10 ..(i) ..(ii) b) ( 3 1) x = 10 x = 10 3 - 1 x = 10 1.732 1 = 10 0.732 = 10000 732 x ~ 13.66 m h =13.66 m N = 2 years = 24 months 300 Rs per month I = P x n (n + 1) x r 2 x 12 100 P = 300 x 24 = 7200 Rs 2 1 I = 300 x 24 (24 + 1) x 10 2 x 12 100 1 = 3 x 10 x 25 = 750 A= P+A = 7200 + 750 = 7950 Rs. c) Answer 7 a) i) PA x PB = PC x PD 4 x 6 = 5 x PD PD = 4 x 6 5 = 24/5 = 4.8 cm ii) PA2 = BP x PC (20)2 = 16 x PC PC= 25 BC = PC PB = 25 16 = 9 cm b) x2 + kx + 64 = 0 a = 1 b = k c = 64 b2 4ac = (k)2 4 x 1 x 64 = k2 256 The roots are real b2 4 ac 0 R = 10% p.a. c) k2 256 0 k2 256 k + 16 i 2 x 8x + k = 0 a=1 b = -8 c=l b2 4 ac = (-8)2 4 x 1 x k = 64 4k The roots are real b2 4 ac 0 64 4k 0 64 4k k 64 4 k 16 ..ii k + 16 k 16 k -16 From i & ii k = 16 = 4cm width = 3m V + 30 m3 v=lxbxL h = 30 10 5 4x3 2 1 = 5/2 = 2.5 m Scale factor = Length of model Length of object K3 = Volume of model Volume of object = 240 30 x 100 x 100 x 100 K3 = 8 (10)6 K = 2 taking cube root 102 = 2 100 K = lm lo 2 = 4 100 lo l0 = 4 x 100 2 = 200 m width = 3 lw lw = 3 x 100 50 2 1 = 150 m h = 2.5 m

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