{"id":89266,"date":"2021-07-06T09:07:38","date_gmt":"2021-07-06T03:37:38","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=89266"},"modified":"2021-07-06T09:07:38","modified_gmt":"2021-07-06T03:37:38","slug":"1st-puc-basic-maths-model-question-paper-3","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-basic-maths-model-question-paper-3\/","title":{"rendered":"1st PUC Basic Maths Model Question Paper 3 with Answers"},"content":{"rendered":"

Students can Download 1st PUC Basic Maths Model Question Paper 3 with Answers, Karnataka 1st PUC Basic Maths Model Question Papers with Answers<\/a> helps you to revise the complete syllabus.<\/p>\n

Karnataka 1st PUC Basic Maths Model Question Paper 3 with Answers<\/h2>\n

Time: 3.15 Hours
\nMax. Marks: 100<\/p>\n

Instructions:<\/p>\n

    \n
  1. The questions paper consists of five parts A, B, C, D, and E.<\/li>\n
  2. Part – A carries 10 marks, Part – B carries 20 marks, Part – C curries 30 marks, Part – D carries 30, and Part – E carries 10 marks.<\/li>\n
  3. Write the question numbers properly as indicated in the questions paper<\/li>\n<\/ol>\n

    PART – A<\/span><\/p>\n

    I. Answer any ten questions (10 \u00d7 1 = 10)<\/span><\/p>\n

    Question 1.
    \nGive the canonical representation of 156.
    \nAnswer:
    \n\"1st
    \nCanonical representation is 156 = 22 \u00d7 31 \u00d7 13′<\/p>\n

    Question 2.
    \nIf A = {1, 2, 3, 4, 5} B = {1, 2, 3, 4, 5, 6^ 7} find a relation from A to B defined by R = {(x, y)\/x > y}
    \nAnswer:
    \nR = {(2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3), (5, 4)}<\/p>\n

    Question 3.
    \nIf f(x) = x + 1 and g(x) = x2<\/sup> + 1 find fog (1)
    \nAnswer:
    \n(fog) (1) = f[g(l)] = f[l2 + 1] = f(2) = 2+1 = 3<\/p>\n

    Question 4.
    \nSimplify \\(\\left(\\frac{5 x^{3}}{2 y}\\right)^{2}\\)
    \nAnswer:
    \n\\(\\frac{5^{2} x^{6}}{2^{2} y^{2}}=\\frac{25 x^{6}}{4 . y^{2}}\\)<\/p>\n

    \"KSEEB<\/p>\n

    Question 5.
    \nFind x if logx<\/sub> 625 = 4
    \nAnswer:
    \n625 =x4<\/sup>
    \n54<\/sup> = x4<\/sup> \u21d2 x = 5<\/p>\n

    Question 6.
    \nFind the 6th term of the GP. 3, 6,12 …….
    \nAnswer:
    \na = 3, r = 6\/3 = 2
    \nT6<\/sub> = ar6-1<\/sup> = ar5<\/sup> = 3.(2)5<\/sup> = 3.32 = 96<\/p>\n

    Question 7.
    \nSolve for x if (x + 2) (x + 3) = (x – 2) (x – 4) + 20.
    \nAnswer:
    \n\"1st
    \n5x + 6 = -6x + 28
    \n5x + 6x =28-6
    \n11x = 22
    \nx = 2<\/p>\n

    Question 8.
    \nWhat percent is 64m of t 2km?
    \nAnswer:
    \n12 km = 12 \u00d7 1000m = 12000m
    \nLet the required number be x
    \nx% of 12000 = 64
    \n\\(\\frac{\\mathrm{x}}{100}\\) \u00d7 12000 = 64
    \n120x = 64
    \nx = \\(\\frac{64}{120}\\) = 0.5333%<\/p>\n

    Question 9.
    \nDefine an annuity.
    \nAnswer:
    \nAn annuity is a fixed sum paid at regular intervals of time under certain conditions.
    \nExample: LIC premia<\/p>\n

    Question 10.
    \nExpress 3TC\/4 in degrees
    \nAnswer:
    \n\\(\\frac{3 \\pi^{c}}{4}=\\frac{3 \\pi}{4}\\) \u00d7 = 3 \u00d7 45 = 135\u00b0<\/p>\n

    \"KSEEB<\/p>\n

    Question 11.
    \nThe average age of 10 students is 6 years. The sum of the ages of 9 of them is 52 years. Find the age of the 10th student.
    \nAnswer:
    \nAverage age of 10 students = 6 years
    \n\u2234 Total age of 10 students = 10(6) = 60 years
    \nTotal age of 9 students = 52 years
    \n\u2234 age of the 10th student = 60 – 52 = 8 years<\/p>\n

    Question 12.
    \nFind the slope of the line joining the points (1, 2) and (-1, -2)
    \nAnswer:
    \nm = \\(\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\\frac{-2-2}{-1-1}=\\frac{-4}{-2}\\) = 2<\/p>\n

    PART – B<\/span><\/p>\n

    II. Answer any Ten questions. (10 \u00d7 2 = 20)<\/span><\/p>\n

    Question 13.
    \nFind the greatest number which when divides 989 and i327 leaves the remainder of 5 and 7 respectively.
    \nAnswer:
    \n989 – 5 = 984
    \n1327 – 7= 1320
    \nHCF of 984 and 1320
    \n984 = 23 \u00d7 31 \u00d7 41
    \n\"1st
    \n\"1st
    \n1320 = 23<\/sup> \u00d7 31<\/sup> \u00d7 51<\/sup> \u00d7 111<\/sup>
    \nHCF = 23<\/sup> \u00d7 31<\/sup> = 8 \u00d7 3 = 24<\/p>\n

    Question 14.
    \nIfA={x: x \u2208 N and x < 3} and B={x: x2<\/sup> – 16 = 0 and x < 0} find B \u00d7 A
    \nAnswer:
    \nA= {1, 2), B={-4}
    \nB \u00d7 A= {(-4, 1),(-4,2)}<\/p>\n

    Question 15.
    \nFind the number of positive divisors and the sum of all positive divisors of 825.
    \nAnswer:
    \n825 = 31<\/sup> \u00d7 52<\/sup> \u00d7 111
    \nP1<\/sub> = 3, \u03b11<\/sub> = 1, P2<\/sub> = 5, \u03b12<\/sub> = 2, P3<\/sub> =11, \u03b13<\/sub> = 11
    \nT (n) = (1 + \u03b11<\/sub>) (1 + \u03b12<\/sub>) (1 + \u03b13<\/sub>)
    \n=(1 + 1)(1 +2)(1 + 1)
    \n= (2) (3) (2)
    \n= 12
    \n\"1st
    \nS(n) = \\(\\)
    \n= \\(\\)
    \n= 4 x 31 x 12 = 1488<\/p>\n

    Question 16.
    \nSimplify \\(\\left(\\frac{x^{b}}{x^{c}}\\right)^{\\frac{1}{b c}}\\left(\\frac{x^{e}}{x^{2}}\\right)^{\\frac{1}{c u}}\\left(\\frac{x^{a}}{x^{b}}\\right)^{\\frac{1}{2 b}}\\)
    \nAnswer:
    \n\"1st<\/p>\n

    Question 17.
    \nFind the value of k so that 2\/3, k (5\/3)k are three consecutive terms of A.P.
    \nAnswer:
    \n\"1st<\/p>\n

    Question 18.
    \nDetermine the nature of the roots of the equation 2x2<\/sup> – 9x + 7 = 0
    \nAnswer:
    \na = 2, b = -9, c = l
    \n\u0394 = b2<\/sup> – 4ac = 81 – 4(2) (7) = 81 – 56 = 25 > 0 and is perfect square.
    \n\u2234 The roots are rational and distinct.<\/p>\n

    Question 19.
    \nFind the amount on \u20b9 500 for 10 years at the rate of 15% compound interest?
    \nAnswer:
    \nP = r, 500, n= 10, r = 15%
    \nA = P(1 + r)n<\/sup> = 500(1 + 0.15)10<\/sup> = 500(1.15)10<\/sup> = \u20b9 2022.77<\/p>\n

    \"KSEEB<\/p>\n

    Question 20.
    \nSolve for x if 4x – 5 < 27 and represent on a number line.
    \nAnswer:
    \n4x – 5 \u2264 27
    \n4x \u2264 27 + 5
    \n4x \u2264 32
    \nx \u2264 8
    \n\"1st<\/p>\n

    Question 21.
    \nThe average score of 65 boys is 60 and the average score of 15 girls is 65. Find the combined average score.
    \nAnswer:
    \nX1<\/sub> = 60, N1<\/sub> = 65
    \nX2<\/sub> = 65, N2<\/sub> = 15
    \nX\u0304 = \\(\\frac{\\mathrm{X}_{1} \\mathrm{~N}_{1}+\\mathrm{X}_{2} \\mathrm{~N}_{2}}{\\mathrm{~N}_{1}+\\mathrm{N}_{2}}=\\frac{(60)(65)+(65)(15)}{65+15}=\\frac{4875}{80}\\) = 60.93<\/p>\n

    Question 22.
    \nA shopkeeper buys 50 pencils for \u20b9 80 and sells them at 40 pencils for \u20b9 90. Find his gain or loss percent?
    \nAnswer:
    \nCP of 1 pencil = \\(\\frac{80}{50}\\) = \u20b9 1.6
    \nSPof 1 pencil = \\(\\frac{90}{40}\\) = \u20b9 2.25
    \nProfit on 1 pencil = 2.25 – 1.6 = \u20b9 0.65
    \nProfit % = \\(\\frac{\\text { Profit }}{\\mathrm{CP}}\\) \u00d7 100 = \\(\\frac{0.65}{1.6}\\) \u00d7 100 = 40,625%<\/p>\n

    Question 23.
    \nProve that \\(\\frac{1}{1+\\cos A}+\\frac{1}{1-\\cos A}\\) = 2cosec2A
    \nAnswer:
    \n\\(\\frac{1}{1+\\cos \\mathrm{A}}+\\frac{1}{1-\\cos \\mathrm{A}}=\\frac{1-\\cos \\mathrm{A}+1+\\cos \\mathrm{A}}{(1+\\cos \\mathrm{A})(1-\\cos \\mathrm{A})}=\\frac{2}{1-\\cos ^{2} \\mathrm{~A}}=\\frac{2}{\\sin ^{2} \\mathrm{~A}}\\) =2cosec2A<\/p>\n

    Question 24.
    \nIf \u03b8 = 45 show that \\(\\frac{\\tan \\theta}{1+\\tan \\theta}-\\frac{1+\\tan \\theta}{\\tan \\theta}=\\frac{-3}{2}\\)
    \nAnswer:
    \nLHS = \\(\\frac{\\tan 45^{\\circ}}{1+\\tan 45^{\\circ}}-\\frac{1+\\tan 45}{\\tan 45^{\\circ}}=\\frac{1}{1+1}-\\frac{1+1}{1}=\\frac{1}{2}-\\frac{2}{1}=\\frac{1-4}{2}=\\frac{-3}{2}\\) = RHS<\/p>\n

    \"KSEEB<\/p>\n

    Question 25.
    \nDerive the equation of the straight line in the slope point form.
    \nAnswer:
    \n\"1st
    \nSlope of AP = m = tan \u03b8 = \\(\\frac{P N}{A N}=\\frac{P M-M N}{L M}=\\frac{P M-M N}{O M-O L}\\)
    \nm = \\(\\frac{y-y_{1}}{x-x_{1}}\\)
    \n\u2234 y – y1<\/sub> = m(x – x1<\/sub>)<\/p>\n

    PART – C<\/span><\/p>\n

    III. Answer any ten questions. (10 \u00d7 3 = 30)<\/span><\/p>\n

    Question 26.
    \nIn a group of 65 people, 40 were found to like hockey, 10 like both tennis and hockey. How many like only tennis but not hockey? How many like tennis?
    \nAnswer:
    \nn(C\u222aH) = n(C) + n (H) -n(C \u2229H)
    \n65 = 40 + n(H) – 10
    \n65 = 30 + n(H)
    \nn(H) = 35
    \nSince n (C \u2229 H) = 10 number of people like hockey 35.<\/p>\n

    Question 27.
    \nGiven A = {2, 4, 6, 8} and R = {(2, 4), (4, 2), (4, 6), (6, 4)} Show that R is not reflexive, symmetric and not transitive.
    \nAnswer:
    \n(2. 2) \u2209 R \u2234 R is not reflexive
    \n(2, 4) \u2208 R = (4, 2) \u2208 R (4, 6) \u2208 R = (6, 4) \u2208 R
    \n(a, b) \u2208 R = (b, a) \u2208 R \u2234 R is symmetric
    \n(2, 4) \u2208 R, (4, 2) \u2208 R But (2, 2) \u2209 R \u2234 (a, b) \u2208 R, (b, c) \u2208 R But (a, c) \u2209 R
    \n\u2234 R is not transitive<\/p>\n

    Question 28.
    \nThe sum of the two numbers is 528 and their HCF is 33. Find the number of pairs satisfying the given condition?
    \nAnswer:
    \nLet the two numbers be 33a, 33b
    \n33a + 33b = 529
    \n-33 a + b= 16
    \nThe possible value of a and b are (1, 15), (3, 13), (5, 11), (7, 9)
    \nNote that a and b must be relatively prime i.e., they do not have any common factor other than 1.
    \n\u2234 The possible pairs of numbers are (33, 495), (99, 429), (165, 363), (231, 297)<\/p>\n

    \"KSEEB<\/p>\n

    Question 29.
    \nProve that zyz = 1 if \\(\\frac{\\log x}{a-b}=\\frac{\\log y}{b-c}=\\frac{\\log z}{c-a}\\)
    \nAnswer:
    \n\\(\\frac{\\log x}{a-b}=\\frac{\\log y}{b-c}=\\frac{\\log z}{c-a}\\) = k
    \n\u2234 log x = k(a – b)
    \nlogy = k(b – c)
    \nlogz = k(c – a)
    \n\u2234 logx + logy + logz = k(a – b) + k(b – c) + k(c – a)
    \n= k(a – b + b – c + c – a) = k( 0) = 0
    \n\u2234 log (xyz) = 0 \u2234 xyz = 1<\/p>\n

    Question 30.
    \nIf the first term of GP is 729 and the 7th term is 64 find the sum of first seven terms of the GP
    \nAnswer:
    \na = 729, T7<\/sub> = 64, S7<\/sub> = ?
    \nT7<\/sub> = ar6<\/sup>
    \n64 =729. r6<\/sup> \u21d2 r6<\/sup> = \\(\\frac{64}{729}=\\frac{2^{6}}{3^{6}}=\\left(\\frac{2}{3}\\right)^{6}\\) \u21d2 r = \\(\\frac{2}{3}\\)
    \nSn<\/sub> = \\(\\frac{a\\left(1-r^{n}\\right)}{1-r}\\)
    \n\u2234 Sn<\/sub> = \\(=\\frac{729\\left(1-(2 \/ 3)^{7}\\right)}{1-2 \/ 3}=\\frac{729\\left(1-\\frac{128}{2187}\\right)}{\\frac{3-2}{3}}=\\frac{729\\left(\\frac{2187-128}{2187}\\right)}{\\frac{1}{3}}=\\frac{729\\left(\\frac{2059}{2187}\\right)}{\\frac{1}{3}}\\)
    \n= \\(\\frac{2059}{3}{\\frac{1}{3}}\\) = 2059<\/p>\n

    Question 31.
    \nFind the present value of the annuity immediate for \u20b9 3000 for 5 years at 10% p.a
    \nAnswer:
    \nP = \\(\\frac{A\\left[(1-r)^{n}-1\\right]}{r(1+r)^{n}}\\)
    \nA = 3,000, n = 5, r = 0.10
    \n\u2234 P = \\(\\frac{3000\\left[(1+0.1)^{5}-1\\right]}{0.1(1+0.1)^{5}}=\\frac{3000\\left[(1.1)^{5}-1\\right]}{0.1(1.1)^{5}}=\\frac{3000[1.61051-14]}{0.1[1.61051]}\\)
    \n= \\(\\frac{3000(0.61051)}{0.1(1.61051)}=\\frac{1831.53}{0.161051}\\)
    \nP = 11,372.30<\/p>\n

    Question 32.
    \nSolve the linear inequalities graphically
    \n3x + 2y \u2264 6,
    \n4x – y \u2264 6
    \nAnswer:
    \n3x + 2y = 6
    \nPut y = 0 3x = 6 \u2234 x = 2 \u2234 A = (2, 0)
    \nx = 0 2y = 6 \u2234 y = 3 B = (0, 3)<\/p>\n

    4x – y = 6
    \nx = 0 -y = 6 \u21d2 y = -6 C(0, -6)
    \ny = 0 4x = 6 \u21d2 x = 3\/2 D(3\/2, 0)
    \n\"1st
    \nThe shaded region is the feasible region which is the intersection of the solution set of each inequality 3x + 2y \u2264 6 and 4x – y \u2264 6.<\/p>\n

    Question 33.
    \nThe average temperature for the first four days of the week was 39\u00b0C. The average for the whole week was 40\u00b0C. What was the average temperature during the last three days of the week?
    \nAnswer:
    \nThe average temperature of first four days of the week = 39\u00b0C
    \n\u2234 The total temperature of the first four days of the week = 39 \u00d7 4= 156
    \nThe average temperature of the whole week = 40\u00b0
    \n\u2234 Total temperature of the whole week = 40 \u00d7 7 = 280\u00b0
    \n\u2234 The total temperature of the last three days of the week = 280\u00b0 – 156\u00b0 = 124\u00b0
    \n\u2234 Average temperature of the last three days of the week = \\(\\frac{124}{3}\\) = 41\u00b0C.<\/p>\n

    \"KSEEB<\/p>\n

    Question 34.
    \nFind the circumcentre of the triangle with vertices A(-3, 4), B (3, 4), and C(-4, 3). Also, find the circumradius and the area of the circle.
    \nAnswer:
    \nLet S(x, y) be the circumcentre of \u0394ABC
    \nSA = SB => SA2<\/sup> = SB2<\/sup>
    \n(x – 3)2<\/sup> + (y – 4)2<\/sup> = (x + 4)2<\/sup> + (y + 3)2<\/sup>
    \n\"1st
    \n-6x – 8x – 8y – 6y =0
    \n-14x- 14y = 0 +-14
    \nx + y =0
    \n0 + y = 0
    \ny = 0
    \n\u2234 circumcentre = (0, 0)
    \nCircumradius = \\(\\sqrt{(0+3)^{2}+(0-4)^{2}}=\\sqrt{9+16}=\\sqrt{25}\\) = 5 r = 5
    \n\u2234 Area of the circle = rcr2<\/sup> = TC(25) = 25K sq. units.<\/p>\n

    Question 35.
    \nThe rate of a movie ticket was Rs. 150. This was reduced by 20% due to the discount in price the revenue increased by 20%, what was the percentage increase in the number of viewers?
    \nAnswer:
    \nLet the no. of initial viewers be 100
    \nCost of each ticket = Rs. 150
    \nRevenue generated initially = 150 \u00d7 100 =Rs. 15000
    \nAfter the reduction in price cost of each ticket
    \n= 150 – 20% of 150 = 150 – \\(\\frac{20}{100}\\)\u00d7 150 = 150 – 30 = Rs 120<\/p>\n

    New revenue generated = 15000 + 20% of 15000
    \n= 15000+ \\(\\frac{20}{100}\\) \u00d7 15000 = 15000 + 3000 = Rs. 18000
    \nNo. of viewers = \\(\\frac{\\text { Total revenue }}{\\text { Cost of each ticket }}\\) = 150
    \nPercentage increase in the no. of viewers = 150 – 100 = 50%<\/p>\n

    Question 36.
    \nFind the value of x if x sin 45\u00b0 tan 60\u00b0 = \\(\\frac{\\sin 30^{\\circ} \\cot 30^{\\circ}}{\\cos 60^{\\circ}{cosec} 45}\\)
    \nAnswer:
    \n\"1st<\/p>\n

    Question 37.
    \nFind the third vertex of the triangle if two of its vertices are A(-2,4) and B(7, -3) and the centroid is (3, 2)
    \nAnswer:
    \nA = (-2, 4), B = (7, -3), O = (a, b), G = (3, 2)
    \nG = \\(\\left(\\frac{x_{1}+x_{2}+x_{3}}{3}, \\frac{y_{1}+y_{2}+y_{3}}{3}\\right)\\)
    \n(3, 2) = \\(\\left(\\frac{-2+7+a}{3}, \\frac{4-3+b}{3}\\right)\\)
    \n(3, 2) = \\(\\left(\\frac{5+a}{3}, \\frac{1+b}{3}\\right)\\)
    \n\\(\\frac{5+a}{3}\\) = 3 \u21d2 5 + a = 9 \u21d2 a = 4
    \n\\(\\frac{1+b}{3}\\) = 2 \u21d2 1 + b = 6 \u21d2 b = 5
    \n\u2234 Third vertex = C = (4, 5)<\/p>\n

    Question 38.
    \nA number which when decreased by 20 is equal to 69 times the reciprocal of the number. Find the number.
    \nAnswer:
    \nLet the number be x
    \nx – 20 = 69\\(\\frac{1}{x}\\)
    \nx2<\/sup> – 20x – 69 =0
    \nx2<\/sup> – 23x + 3x -69 = 0
    \nx(x – 23) + 3(x – 23) = 0
    \nx = 23 or x = – 3<\/p>\n

    PART – D<\/span><\/p>\n

    IV. Answer any six questions. (6 \u00d7 5 = 30)<\/span><\/p>\n

    Question 39.
    \nIn a survey it was found that 31 people liked a product A, 36 liked a product B and 39 liked the product C. If 24<people liked products A and B, 22 people liked product C and A, 24 people liked products B and C, 18 liked all the three products, then find how many people liked product C only?
    \nAnswer:
    \nn(A) = 31
    \nn(B) = 36
    \nn(C) = 39
    \nn(A\u2229B) = 24 n(B\u2229C) = 24 n(C\u2229A) = 22 n(A\u2229B\u2229C) = 18
    \nn(A\u222aB\u222aC) = n(A) + n(B) + n(C) – n(A\u2229B) – n(A\u2229C) – n(C\u2229A) + n(A\u2229B\u2229C)
    \n= 31 + 36 + 39 – 24 – 24 – 22 + 18 = 54
    \n\"1st
    \nNumber of people who like the product C only = 11<\/p>\n

    \"KSEEB<\/p>\n

    Question 40.
    \nThe sum of three numbers which are in GP is 43 and their product is 216. Find the numbers.
    \nAnswer:
    \n\\(\\frac{a}{r}\\), a, ar
    \n\\(\\frac{a}{r}\\) \u00d7 a \u00d7 ar = 216
    \na3<\/sup> = 216 \u21d2 a =6
    \n\\(\\frac{a}{r}\\)+ a + ar = 43
    \n\\(\\frac{6}{r}\\) + 6 + 6r = 43
    \n\\(\\frac{6}{r}\\) + 6r = 43 – 6
    \n\\(\\frac{6}{r}\\) + 6r = 37
    \n6 + 6r2<\/sup> = 37r
    \n6r – 37r + 6 =0
    \n6r2<\/sup> – 36r – r + 6 = 0
    \n6r(r – 6) – (r – 6) =0
    \n(r- 6) (6r- 1) = 0
    \n\u2234 r = 6 or r = \\(\\frac{1}{6}\\)
    \n\u2234 The numbers are \\(\\frac{6}{6}\\), 6, 6×6
    \n1, 6, 36<\/p>\n

    Question 41.
    \nIf \u03b1 and \u03b2 are the roots of the equation 3x2<\/sup> – 4x + 15 = 0 then find the value of \\(\\frac{\\alpha^{2}}{\\beta}+\\frac{\\beta^{2}}{\\alpha}\\)
    \nAnswer:
    \n\u03b1 + \u03b2 = –\\(\\left(\\frac{-4}{3}\\right)=\\frac{4}{3}\\)
    \n\u03b1\u03b2 = \\(\\frac{15}{3}\\) = 5
    \n\\(\\frac{\\alpha^{2}}{\\beta}+\\frac{\\beta^{2}}{\\alpha}=\\frac{\\alpha^{3}+\\beta^{3}}{\\alpha \\beta}=\\frac{(\\alpha+\\beta)^{3}-3 \\alpha \\beta(\\alpha+\\beta)}{\\alpha \\beta}\\)<\/p>\n

    = \\(\\frac{\\left(\\frac{4}{3}\\right)^{3}-3(5)\\left(\\frac{4}{3}\\right)}{5}=\\frac{\\frac{64}{27}-20}{5}=\\frac{64-540}{27(5)}=\\frac{-476}{135}\\)<\/p>\n

    Question 42.
    \nIf the present price of a bike is \u20b9 32,290. It is value decreases every year by 10% then find its value before 3 years?
    \nAnswer:
    \nP = 32,290, r = 0.10, n = 3
    \nA = P( 1 – r)n<\/sup>
    \nA =32,290(1 -0.1)3<\/sup>
    \n= 32290 (0.9)3<\/sup>
    \n= 32290 (0.729)
    \n= 23,539.41
    \n\u2234 The value of the bike after 3 years = \u20b9 23,539.41<\/p>\n

    \"KSEEB<\/p>\n

    Question 43.
    \nIn what time will a sum of \u20b9 2000 becomes \u20b9 3900 at 5% p.a compound interest payable half yearly?
    \nAnswer:
    \nP = \u20b9 2,000, A= 3900, n = ?, R= 5%, q = 2, r = ?
    \nr = \\(\\left(1+\\frac{\\mathrm{R}}{q}\\right)^{q}\\) – 1 = \\(\\left(1+\\frac{0.05}{2}\\right)^{2}\\) – 1
    \nr = 0.050625
    \nA = P(1 + r)n<\/sup>
    \n3900 = 2000 (1 + 0.050625)n<\/sup>
    \n3900 =2000(1.050625)
    \n\\(\\frac{39}{20}\\) = (1.050625)
    \nlog 39 – log 20 = n log 1.050625 \u21d2 1.5910 – 1.3010 = n(0.02 144)
    \n\u21d2 0.29 = n(0.02144)
    \nn = \\(\\frac{0.29}{0.02144}\\) =13.52years.<\/p>\n

    Question 44.
    \nFind the lengths of the altitudes of the triangle whose vertices are (5,2) (3, -3) and (-4, 3).
    \nAnswer:
    \nA = (5,2) B = (3, -3) C = (-4, 3)
    \nArea of AABC = \\(\\frac{1}{2}\\) [x1(y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)]
    \n= \\(\\frac{1}{2}\\)[5 (-3-3) + 3(3-2)-4(2 + 3)]= \\(\\frac{1}{2}\\) [5 (-6) + 3 (1)-4 (5)]
    \n= \\(\\frac{1}{2}\\)[-30 + 3 – 20] = \\(\\frac{1}{2}\\)[-47] = \\(\\left|\\frac{-47}{2}\\right|=\\frac{47}{2}\\) Square units<\/p>\n

    Area of the \u0394ABC = \\(\\frac{1}{2}\\) \u00d7 BC \u00d7 AD
    \nAD = \\(\\frac{2 \\Delta \\mathrm{ABC}}{\\mathrm{BC}}\\)
    \nBC = \\(\\sqrt{(-4-3)^{2}+(3+3)^{2}}=\\sqrt{49+36}=\\sqrt{85}\\)
    \n\"1st
    \nAC = \\(\\sqrt{(-4-5)^{2}+(3-2)^{2}}=\\sqrt{81+1}=\\sqrt{82}\\)
    \n\u0394ABC = \\(\\frac{1}{2}\\) \u00d7 AC \u00d7 BE
    \nBE = \\(\\frac{2 \\times \\Delta \\mathrm{ABC}}{\\mathrm{AC}}=\\frac{2 \\times 47}{2 \\times \\sqrt{82}}\\)
    \nBE = \\(\\)
    \nAB = \\(\\sqrt{(3-5)^{2}+(-3-2)^{2}}=\\sqrt{4+25}=\\sqrt{29}\\)
    \n\u0394ABC = \\(\\frac{1}{2}\\) \u00d7 AB \u00d7 CF
    \n\"1st<\/p>\n

    Question 45.
    \n(a) Form a quadratic equation whose roots are -2 and 5
    \nAnswer:
    \n(a) (x + 2) (x – 5) = 0
    \nx2<\/sup> – 5x + 2x – 10 = 0
    \nx2<\/sup> – 3x – 10 = 0<\/p>\n

    (b) Evaluate using log. tables \\(\\frac{(25.36)^{2} \\times(0.4569)^{2}}{(847.5)}\\)
    \nAnswer:
    \nLet x = \\(\\frac{(25.36)^{2} \\times(0.4569)^{2}}{(847.5)}\\)
    \nlog x = 2 log 25.36 + 2 log 0.4569 – log 847.5
    \n= 2(1.4041) + 2(1.6598) – (2.9281)
    \n= 2.8082 + 2(-1 + 0.6598)-(2.9281)
    \n=-0.8003 = -1 + 1 -0.8003 = -1 + 0.1997
    \nlog x = 1.1997
    \nx = AL[1.1997] = 0.1584<\/p>\n

    \"KSEEB<\/p>\n

    Question 46.
    \nFind the equation of the straight line passing through the point of intersection of 2x + 4y = 3 and x + 5y = 1 and making equal positive intercepts on the coordinate axes.
    \nAnswer:
    \n2x + 4y = 3 \u00d7 1
    \nx + 5y = 1 \u00d7 2
    \n\"1st
    \ny = –\\(\\frac{1}{6}\\)
    \nx + 5y = 1
    \nx + 5(-1\/6) = 1
    \nx = 1 + \\(\\frac{5}{6}=\\frac{6+5}{6}=\\frac{11}{6}\\)<\/p>\n

    Point of intersection \\(\\left(\\frac{11}{6}, \\frac{-1}{6}\\right)\\)
    \n\\(\\frac{x}{a}+\\frac{y}{b}\\) = 1
    \n\\(\\frac{x}{a}+\\frac{y}{a}\\) = 1 \u21d2 x + y = a (1)
    \nIt passes through\\(\\left(\\frac{11}{6}, \\frac{-1}{6}\\right)\\) put x = \\(\\frac{11}{6}\\) , y = -1\/6 in(1)
    \n\\(\\frac{11}{6}-\\frac{1}{6}\\) = a \u21d2 \\(\\frac{10}{6}\\) = a
    \na = \\(\\frac{5}{3}\\)
    \nb = \\(\\frac{5}{3}\\)
    \n\\(\\frac{x}{a}+\\frac{y}{b}\\) = 1
    \n\\(\\frac{x}{\\frac{5}{3}}+\\frac{y}{\\frac{5}{3}}\\) = 1
    \n\\(\\frac{3 x}{5}+\\frac{3 y}{5}\\) = 1
    \n3x + 3y = 5
    \n3x + 3y – 5 = 0<\/p>\n

    Question 47.
    \nFind the locus of a point equidistant from (1, 0) and (-1, 0).
    \nAnswer:
    \nLet A = (I, O), B = (-1, 0)
    \nLet P(x, y) be any point on the locus
    \n\u2234 PA = PB
    \n\\(\\)
    \n(x – 1)2<\/sup> + y2<\/sup> = (x + 1)2<\/sup> + y2<\/sup>
    \n\"1st
    \n-4x = 0
    \n\u2234 x = 0<\/p>\n

    Question 48.
    \nFind the sum of the following series: 1 + (1 + 2) + (1 + 2 + 3)………..n terms.
    \nAnswer:
    \nt =1 + 2 + 3 + ………….. + n = \\(\\frac{n(n+1)}{2}\\)
    \nSn<\/sub> = \u03a3tn<\/sub> = \\(\\frac{\\Sigma n(n+1)}{2}=\\frac{1}{2}\\)[\u03a3n2<\/sup> + \u03a3n] = \\(\\frac{1}{2}\\left[\\frac{n(n+1)(2 n+1)}{6}+\\frac{n(n+1)}{2}\\right]\\)
    \n= \\(\\frac{1}{2}\\left[n(n+1)\\left\\{\\frac{2 n+1}{6}+\\frac{1}{2}\\right\\}\\right]=\\frac{n(n+1)}{2}\\left\\{\\frac{2 n+1+3}{6}\\right\\}=\\frac{n(n+1)\\{2 n+4\\}}{2.6}\\)
    \n\"1st<\/p>\n

    PART-E<\/span><\/p>\n

    V. Answer any one equestion. (1 \u00d7 10 = 10)<\/span><\/p>\n

    Question 49.
    \n(a) Prove that the lines x + y + 4 = 0, 2x = 3y + 7 and 3x + y + 6 = 0 are concurrent. Also find the point of concurrency.
    \nAnswer:
    \n(a) x + y = -4 \u00d7 2
    \n2x – 3y = 7 \u00d7 1
    \n\"1st
    \ny = -3
    \nx + y = -4
    \nx – 3 = -4
    \nx = -4 + 3
    \nx = -1
    \n\u2234 point of intersection = (-1, -3)
    \nPut x = -1 y = -3 in 3x + y + 6 = 0
    \nwe get 3(-1) -3 + 6 = -3 – 3 + 6 = 0
    \n\u2234 The three lines are concurrent
    \npoint of concurrency = (-1, -3)<\/p>\n

    (b) If 12 cot2<\/sup> A – 31 cosecA + 32 = 0, find the valu&of sinA
    \nAnswer:
    \n12cot2<\/sup> A – 31 cosec A + 32 = 0
    \n12(cose2c A – 1) – 31 cosec A + 32 = 0
    \n12 cosec2A- 12 – 31 cosec A + 32 = 0
    \n12 cosec2 A – 31 cosec A + 19 = 0
    \n12 cosec2A- 12 cosecA- 19 cosec A + 19 = 0
    \n12 cosecA (cosecA – 1) – 19 (cosecA – 1) = 0
    \n(cosecA – 1) (12 cosecA – 19) = 0
    \ncosec A – 1 =0
    \ncosecA = 1
    \nsinA = 1<\/p>\n

    12 cosecA – 19 = 0
    \ncosec A = \\(\\frac{19}{12}\\)
    \nsin A = \\(\\frac{12}{19}\\)<\/p>\n

    (c) Find the number of digits in 2M
    \nAnswer:
    \nLet x = 2M
    \nlog x = log2<\/sub> 64 = 64 log2 = 64 (0.3010) = 19.264
    \nAs the characteristic of log x is 19 the number of digits = 19 + 1 = 20. 50<\/p>\n

    \"KSEEB<\/p>\n

    Question 50.
    \n(a) Find the sum to n terms of the series:
    \n.3 + .33 + .333 + .3333 +…….. n terms
    \nAnswer:
    \n(a) Let Sn<\/sub> = 0.3 + 0.33 + 0.333 +………..n terms
    \n= \\(\\frac{3}{10}+\\frac{33}{100}+\\frac{333}{1000}\\) + ……………
    \n\\(\\frac{S_{n}}{3}=\\frac{1}{10}+\\frac{11}{100}+\\frac{111}{1000}\\) + ……………
    \n\\(\\frac{9 \\mathrm{~S}_{n}}{3}=\\frac{9}{10}+\\frac{99}{100}+\\frac{999}{1000}\\) + …………
    \n\"1st
    \n\\(\\frac{9 \\mathrm{~S}_{\\mathrm{n}}}{3}\\) = n – \\(\\frac{1}{9}\\left(1-\\frac{1}{10^{8}}\\right)\\)
    \nSn<\/sub> = \\(\\frac{3}{9}\\left[n-\\frac{1}{9}\\left(1-\\frac{1}{10^{n}}\\right)\\right]=\\frac{1}{3}\\left[n-\\frac{1}{9}\\left(1-\\frac{1}{10^{n}}\\right)\\right]\\) n \u2208 N<\/p>\n

    (b) A publishing house finds that the production of each book and the cost of the book are directly attributed, if the cost of each book is 30 and the fixed costs are 15000, selling price of each book is 45 then determine
    \n(i) Revenue function
    \n(ii) Cost function
    \n(iii) Break even point function
    \nAnswer:
    \nFC = \u20b9 15,000, VC = 30xm
    \nTotal cost function = C(x) = VC + FC = 3Ox + 15,000
    \nRevenue function = R(x) = 45x
    \nFor Break even point, C(x) = R(x)
    \n30x+ 15000 = 45x
    \n15000 = 45x – 30x 15000 = 15x
    \nx = 1000 units.<\/p>\n

    (c) If n(\u222a) = 700, n(A) = 200, n(B) = 300 and n (A \u2229 B) = 100 Find n (A’ \u2229 B’)
    \nAnswer:
    \nn(\u222a) = 700, n(A) – 200, n(B) = 300, n(A\u2229B) = 100
    \nn(A’\u2229B’) = n(A\u222aB)’ – n(\u222a) – n(A\u222aB) = 700 – [n(A) + n(B) – n(A\u2229B)]
    \n= 700 – [200 + 300 – 100] = 700 – [400] = 300<\/p>\n","protected":false},"excerpt":{"rendered":"

    Students can Download 1st PUC Basic Maths Model Question Paper 3 with Answers, Karnataka 1st PUC Basic Maths Model Question Papers with Answers helps you to revise the complete syllabus. Karnataka 1st PUC Basic Maths Model Question Paper 3 with Answers Time: 3.15 Hours Max. Marks: 100 Instructions: The questions paper consists of five parts …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[81],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/89266"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=89266"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/89266\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=89266"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=89266"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=89266"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}