{"id":89583,"date":"2021-07-14T17:24:22","date_gmt":"2021-07-14T11:54:22","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=89583"},"modified":"2021-07-15T16:14:39","modified_gmt":"2021-07-15T10:44:39","slug":"1st-puc-basic-maths-previous-year-question-paper-march-2016-north","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-basic-maths-previous-year-question-paper-march-2016-north\/","title":{"rendered":"1st PUC Basic Maths Previous Year Question Paper March 2016 (North)"},"content":{"rendered":"

Students can Download 1st PUC Basic Maths Previous Year Question Paper March 2016 (North), 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 Previous Year Question Paper March 2016 (North)<\/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.
    \nWrite the imaginary part of the complex number \\(\\frac{-3+4 i}{5}\\)
    \nAnswer:
    \nImaginary part of \\(\\frac{-3}{5}\\) + i\\(\\frac{4}{5}\\) is \\(\\frac{4}{5}\\).<\/p>\n

    Question 2.
    \nIf set A has 16 subsets, find the number of elements in A.
    \nAnswer:
    \nA has 4 elements.<\/p>\n

    Question 3.
    \nIf a function f(x) is defined by f(x) = 3x – 1, find f(\\(\\frac{1}{3}\\)) .
    \nAnswer:
    \n\"1st<\/p>\n

    Question 4.
    \nFind the 8th term of the sequence \\(\\frac{1}{2}\\), \\(\\frac{1}{4}\\), \\(\\frac{1}{6}\\)………..
    \nAnswer:
    \n8th term = T8<\/sub> = \\(\\frac{1}{16}\\)<\/p>\n

    Question 5.
    \nFind the value of log01 100
    \nAnswer:
    \nLet log00.1<\/sub> 100 = x
    \n100 = (0.1)x
    \n102 = 10 – x \u21d2 x = -2
    \nlog0.1<\/sub> 100 = -2<\/p>\n

    \"KSEEB<\/p>\n

    Question 6.
    \nSimplify \\(\\left(\\frac{64}{81}\\right)^{-1 \/ 4}\\)
    \nAnswer:
    \n\\(\\left(\\frac{64}{81}\\right)^{-1 \/ 4}=\\left(\\frac{81}{64}\\right)^{1 \/ 4}=\\frac{\\left(3^{4}\\right)^{1 \/ 4}}{\\left(2^{6}\\right)^{1 \/ 4}}=\\frac{3}{2^{3 \/ 2}}=\\frac{3}{2 \\sqrt{2}}\\)<\/p>\n

    Question 7.
    \nSolve for x : x + a(x + b) = ax + b.
    \nAnswer:
    \n\"1st
    \nx = b – ab
    \nx = b(1 – a)<\/p>\n

    Question 8.
    \nWhat is the present value of an income of Rs. 10,000 a year to be received forever at 10% p.a.?
    \nAnswer:
    \nP\u221e<\/sub> = \\(\\frac{a}{r}=\\frac{10,000}{0.1}\\) = 1,00000<\/p>\n

    Question 9.
    \nConvert 315\u00b0 into radians.
    \nAnswer:
    \n\"1st<\/p>\n

    Question 10.
    \nFind the slope of a line perpendicular to the line 3x – 4y + 5 = 0.
    \nAnswer:
    \nSlope of the given line = \\(\\frac{3}{4}\\)
    \nSlope of the perpendicular line = –\\(\\frac{4}{3}\\)<\/p>\n

    Question 11.
    \nFind the coordinates of the centroid of the triangle formed by the points (1, 1), (-2, 1), and (-5, -5).
    \nAnswer:
    \nCentroid = G(x,y) = \\(\\left[\\frac{x_{1}+x_{2}+x_{3}}{3}, \\frac{y_{1}+y_{2}+y_{3}}{3}\\right]=\\left[\\frac{1-2-5}{3}, \\frac{1+1-5}{3}\\right]=\\left[\\frac{-6}{3}, \\frac{-3}{3}\\right]\\) = [-2, -1]<\/p>\n

    \"KSEEB<\/p>\n

    Question 12.
    \nMr. Prem bought a cycle at Rs. 5000 and sold the same at a loss of 20% to Mr. Bheem. Find the cost price of Bheem.
    \nAnswer:
    \nCost price = 5000 – 20% of 5000 = 5000 – 1000 = 4000 Rs.<\/p>\n

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

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

    Question 13.
    \nFind the sum of positive divisors of 198.
    \nAnswer:
    \n198 = 21<\/sup> \u00d7 32<\/sup> \u00d7 111<\/sup>
    \nT(198) = (1 + 1)(2 + 1)(1 + 1) = 2 \u00d7 3 \u00d7 2 = 12.<\/p>\n

    Question 14.
    \nFind greatest number which can divide 60,72 and 84 without leaving any remainder.
    \nAnswer:
    \n60 = 22<\/sup> \u00d7 31<\/sup> \u00d7 51<\/sup>
    \n72 = 23<\/sup> \u00d7 32<\/sup>
    \n84 = 22<\/sup> \u00d7 31<\/sup> \u00d7 71<\/sup>
    \n\u2234 greatest number is nothing but HCF = 22 \u00d7 31 = 12.<\/p>\n

    Question 15.
    \nIf A = {x \/ x \u2208 N, x < 4}, B = {x – 25 = 0, x \u2208 N} find B x A.
    \nAnswer:
    \nA = (1, 2, 3} B = {5}
    \nB \u00d7 A = {(5, 1) (5, 2) (5, 3)<\/p>\n

    Question 16.
    \nProve that \\(\\frac{1}{1+\\log _{b} a}+\\frac{1}{1+\\log _{a}}\\) = 2
    \nAnswer:
    \nL.H.S = \\(\\frac{1}{\\log _{b}^{b}+\\log _{b}^{a}}+\\frac{1}{\\log _{a}^{a}+\\log _{a}^{b}}=\\frac{1}{\\log _{b}^{b a}}+\\frac{1}{\\log _{a}^{a b}}\\) = logab b + laga ab = logab ab = 1 = R.H.S<\/p>\n

    Question 17.
    \nInsert 3 geometric means between \\(\\frac{1}{4}\\) and \\(\\frac{1}{64}\\)
    \nAnswer:
    \nInsert 3 Gms G1<\/sub>, G2<\/sub>, G3<\/sub>. between \\(\\frac{1}{4}\\) & \\(\\frac{1}{64}\\)
    \n\\(\\frac{1}{4}\\), G1<\/sub>, G2<\/sub>, G3<\/sub>, \\(\\frac{1}{64}\\) are in G.P
    \na = \\(\\frac{1}{4}\\), ar4<\/sup> = \\(\\frac{1}{64}\\) \u21d2 \\(\\frac{1}{4}\\)r4<\/sup> = \\(\\frac{1}{64}\\) \u21d2 r4<\/sup> = \\(\\frac{4}{64}\\) \u21d2 r4<\/sup> = \\(\\frac{1}{16}\\) = \\(\\left(\\frac{1}{2}\\right)^{4}\\) \u21d2 r = \\(\\frac{1}{2}\\)<\/p>\n

    Question 18.
    \nIf sum of 2 numbers is 107 and their difference is 17, find the numbers.
    \nAnswer:
    \nLet the two number be x and y
    \nGiven x + y = 107 …. (1)
    \nx – y = 17 …. (2) solving we get
    \n2x = 124 \u21d2 x = 62, y = 107 – 62,y = 45
    \n\u2234 the 2 numbers are x = 62 and y = 45.<\/p>\n

    \"KSEEB<\/p>\n

    Question 19.
    \nSolve : 3x – 2 \u2264 2x + 1, x \u2208 R and represent on the number line.
    \nAnswer:
    \n3x – 2 \u2264 2x + 1, x \u2208 R .
    \n\"1st
    \n3x – 2x \u2264 2 +1
    \nx \u2264 3<\/p>\n

    Question 20.
    \nIn how many years will a sum of money double itself at 10% simple interest p.a.?
    \nAnswer:
    \nLet principle = 100 Rs. \u2234 Amount = 200 Rs.
    \nP = 100, F = 200, r = 10% = 0.1 T = ?
    \nSI = F – P= 100 Rs.
    \nw.k.t S.I = \\(\\frac{\\text { PTR }}{100}\\)
    \n\"1st
    \n\u2234 In 10 years sum of money double itself.<\/p>\n

    Question 21.
    \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 = 6yrs
    \n\u2234 Total age of 10 students = 10 \u00d7 6 = 60yrs
    \nTotal age of 9 students = 52 yrs
    \n\u2234 Age of the 10th students = 60 – 52 = 8yrs.<\/p>\n

    Question 22.
    \nThe difference between the cost price and the selling price is Rs. 225. If the profit percentage is 15% find the selling price.
    \nAnswer:
    \nLet the cost price be x
    \nAs there is profit SP > CP
    \nSPx + 225
    \nSP = \\(\\frac{100+\\text { Profit } \\%}{100}\\) \u00d7 CP
    \nx + 225 = \\(\\left(\\frac{100+15}{100}\\right)\\)x
    \n100x + 22500 = 100x + 15x
    \n15x = 22500
    \nx = = 1500
    \nSP = x + 225 = 1500+225 = 1725<\/p>\n

    Question 23.
    \nProve that cot\u03b8 + tan\u03b8 = sec\u03b8 cosec\u03b8.
    \nAnswer:
    \nLHS = \\(\\frac{\\cos \\theta}{\\sin \\theta}+\\frac{\\sin \\theta}{\\cos \\theta}=\\frac{\\cos ^{2} \\theta+\\sin ^{2} \\theta}{\\sin \\theta \\cos \\theta}=\\frac{1}{\\sin \\theta \\cdot \\cos \\theta}\\) = cosec\u03b8 sec \u03b8 = RHS<\/p>\n

    Question 24.
    \nIfA, B, C are angles of a triangle, then prove that cos\\(\\left(\\frac{\\mathbf{A}+\\mathbf{B}}{\\mathbf{2}}\\right)\\) = sinC\/2.
    \nAnswer:
    \nW.K.T A + B + C = 180\u00b0
    \n\\(\\frac{A+B+C}{2}\\) = 90 \u21d2 \\(\\frac{A+B}{2}\\) = 90 – \\(\\frac{\\mathrm{C}}{2}\\)
    \nTaking sin on both sides
    \nWe get sin\\(\\left(\\frac{A+B}{2}\\right)\\) = sin \\(\\left(90-\\frac{\\mathrm{C}}{2}\\right)\\)
    \nsin\\(\\left(\\frac{A+B}{2}\\right)\\) = cos\\(\\frac{\\mathrm{C}}{2}\\)<\/p>\n

    \"KSEEB<\/p>\n

    Question 25.
    \nDerive two point form of equation of straight line in the form y – y1<\/sub> = \\(\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\)(x – x1<\/sub>)
    \nAnswer:
    \nLet A(x1<\/sub>, y1<\/sub>), B(x2<\/sub>, y2<\/sub>) be the two given points, let p(x, y) be any point on the line. Here A, B, and P tie on the same line
    \n\u2234 The slope of AP = Slope of AB
    \n\"1st
    \ni.e., \\(\\frac{y-y_{1}}{x-x_{1}}=\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\)
    \n(y – y1<\/sub>) = \\(\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\)(x – x1<\/sub>)
    \nThis is called two point formula.<\/p>\n

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

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

    Question 26.
    \nProve that \\(\\sqrt{5}\\) is irrational.
    \nAnswer:
    \nIf possible let \\(\\sqrt{5}\\) is a rational number.
    \n\u2234 .\\(\\sqrt{5}\\) = p, q\u2208z, q \u2260 0
    \nwe shall also assume that p and q do not have any common factor
    \n\u2234 5 = \\(\\frac{\\mathrm{p}^{2}}{\\mathrm{q}^{2}}\\)
    \n\u2234 p2<\/sup> is a multiple of 5
    \n\u2234 5 divides p2<\/sup>
    \n\u2234 5 divides p
    \n\u2234 p = 5k where k\u2208z, k\u2260O
    \np2<\/sup> = 25k2<\/sup>
    \n5q2<\/sup> = 25k2<\/sup> \u2235 p2<\/sup> = 5q2<\/sup>
    \nq2<\/sup> = 5k2<\/sup>
    \n\u2234 q2 <\/sup>is a multiple of 5
    \n\u2234 5 divides q2<\/sup>
    \n\u2234 5 divides q
    \n\u2234 5 divides both p and q
    \nThis is a contradiction
    \n\u2234 Our assumption is wrong
    \n\u2234 \\(\\sqrt{5}\\) is irrational.<\/p>\n

    Question 27.
    \nLet f {(1, 1), (2, 3), (0, -1)) be a function from Z to Z defined by f(x) = ax + b a, b \u2208 Z determine a and b.
    \nAnswer:
    \nGiven f(x) = ax + b
    \nWhen x = 1, f(x) = 1 \u21d2 a + b = 1
    \nWhen x = 0, f(x) = -1 \u21d2 0 + b = -1 \u21d2 b = -1
    \nif b = 1 then a – 1 = 1 \u21d2 a = 2
    \n\u2234 a = 2, b = -1.<\/p>\n

    Question 28.
    \nIf A = {0, 1,2,3), B = {2, 3, 4), C = {4,5,6) then prove that A\u2229(B\u222aC)=(A\u2229B)\u222a(A\u2229C)
    \nAnswer:
    \nA\u2229(B\u2229C) = {0, 1, 2, 3} n {2, 3, 4, 5, 6} {2, 3} ………………(1)
    \n(A \u2229 B) \u222a (A \u2229 C) = {2, 3} \u222a {Q} = {2, 3) …………………..(2)
    \nFrom 1 and 2 we have
    \nA\u2229(B\u222aC) = (A\u2229B)\u222a(A\u2229C)<\/p>\n

    Question 29.
    \nIf x2<\/sup> + y2<\/sup> = 12xy. Show that 2 log (x – y) = log 10 + log (xy)
    \nAnswer:
    \nGiven x2<\/sup> + y2<\/sup> = 12xy
    \nx2<\/sup> + y2<\/sup> – 2xy = 10xy
    \n(x – y)2<\/sup> = 10xy
    \nTaking logm both sides
    \nlog(x – y)2<\/sup> = log(10xy)
    \n2 log (x – y) = log 10 + log xy<\/p>\n

    \"KSEEB<\/p>\n

    Question 30.
    \nFind three numbers in A.P, whose sum is 24 and product is 440.
    \nAnswer:
    \nLet the 3 number in AP is {a – d) a, a + d
    \nGiven Sum = 24
    \n\"1st
    \n3a = 24 \u21d2 a = 8
    \nAlso given product = 440
    \n(a – d) (a) (a + d) = 440
    \na2<\/sup> – d2<\/sup> =55 \u21d2 d2<\/sup> =64 – 65 = 9 \u21d2d = \u00b13
    \n\u2234 the 3 number are 5, 8,11 of a = 8, d = 3 the 3 numbers are 11,8, 5 if a = 8, d = -3<\/p>\n

    Question 31.
    \nIf \u03b1 and \u03b2 are the roots of the equation 3x2<\/sup> – 6x + 4 = 0 find the value of
    \n(i) \u03b13<\/sup> + \u03b23<\/sup>
    \n(ii) \u03b1\/\u03b22<\/sup>\u00a0+ \u03b2\/\u03b12<\/sup>
    \nAnswer:
    \nGiven 3x2<\/sup> – 6x + 4 = 0
    \nHere \u03b1 + \u03b2= \\(\\frac{-b}{a}=\\frac{6}{3}\\) = 2
    \n\u03b1\u03b2 = \\(\\frac{c}{a}=\\frac{4}{3}\\)
    \n\"1st<\/p>\n

    Question 32.
    \nSolve the system of inequalities 2x + y \u2265 8, and x + y \u2265 10 graphically.
    \nAnswer:
    \nConsider 2x + y = 8
    \nPut x = 0, y = 8 (0, 8)
    \nPut y = 0, x = 4 (4, 0)<\/p>\n

    x + y = 10
    \nput x = 0, y = 10 (0, 10)
    \nput y =0, x = 10 (10, 0)<\/p>\n

    Plot the above point on the graph.
    \n\"1st<\/p>\n

    Question 33.
    \nMr X bought a Walkman for Rs. 1800. If it depreciates at the rate of 15% per year, how much is it worth after 3 years?
    \nAnswer:
    \nC = Rs 1800, r = 15% = 0.15, n = 3yrs, B = ?
    \nB = C(1 – r)n
    \nB = 1800 (1 – 0.15)3 = (1800) \u00d7 (0.85)3 = 1105.425
    \nworth after 3 years = 1105.425<\/p>\n

    Question 34.
    \nA batsman finds that, by scoring a century in the 11th innings of his test matches, he has bettered his average of preview test innings by 5 runs. What is the average after the 11 innings?
    \nAnswer:
    \nLet the average of 10 inninngs = x
    \n\u2234 Total runs after 10 inninngs = lOx
    \nAverage after 11 inninngs x +5
    \n\u2234 Total runs after 11 inninngs = 11 (x + 5)
    \n11 (x + 5) – 10x = 100
    \n11x + 55 – 10x = 100
    \nx + 55 = 100
    \nx = 100 – 55 = 45
    \nAverage after the 11th inninngs = x + 5 = 45 + 5 = 50 runs<\/p>\n

    Question 35.
    \nBy how many percent should the use of coffee be increased if the price of coffee is decreased by 10% so that the expenditure remains unchanged.
    \nAnswer:
    \nLet the price of coffee =100
    \nLet the quantity used be 100
    \n\u2234 Total expenditure = 100 x 100 = 10,000
    \n\u2234 New price = 100 – 10 = 90
    \nLet the new quantity used = x
    \nthe total expenditure = 90x
    \nFor unchanged expenditure 90x = 10000
    \n\"1st
    \n\u2234 the increase in consanption of coffee =111.11 – 100 = 11.11<\/p>\n

    Question 36.
    \nFind x, if xsin\\(\\frac{\\pi}{4}\\) xtan \\(\\frac{\\pi}{3}\\) = \\(\\frac{\\sin \\frac{\\pi}{6} \\times \\cot \\frac{\\pi}{6}}{3 \\cos \\frac{\\pi}{6} \\times \\cos s e \\frac{\\pi}{4}}\\)
    \nAnswer:
    \n\"1st<\/p>\n

    Question 37.
    \nFind the equation of the locus of the point which moves such that its distance from the line 3x – 4y + 1 = 0 is equal to its distance from (1, -1).
    \nAnswer:
    \nA = (1,-1)
    \nLet p(x, y) be any point on the locus
    \nGiven: PA = distance of P from the line 3x- 4y + 1 = 0
    \n\\(\\sqrt{(x-1)^{2}+(y+1)^{2}}=\\left|\\frac{3 \\dot{x}-4 y+1}{\\sqrt{3^{2}+(-4)^{2}}}\\right|\\)
    \n\\(\\sqrt{(x-1)^{2}+(y+1)^{2}}=\\frac{3 x-4 y+1}{\\sqrt{25}}\\) S.B.S
    \n(x – 1)2<\/sup> + (y – 1)2<\/sup> = \\(\\frac{(3 x-4 y+1)^{2}}{25}\\)
    \n25[(x – 1)2<\/sup> + (y + 1)2<\/sup>] = (3x – 4y + 1)2<\/sup>
    \n25[x2<\/sup> + y2<\/sup> – 2x + 2y + 2] = 9x2<\/sup> + 16y2<\/sup> +1 – 34xy – 8y + 6x
    \n\u21d2 25x2<\/sup> + 25y2<\/sup> – 50x + 50y + 50 = 9x2<\/sup> + 16y2<\/sup> + 1 – 24xy – 8y + 6x
    \n16x2<\/sup> + 9y2<\/sup> + 24xy – 42y – 44x + 49 – 0 is the required equation.<\/p>\n

    \"KSEEB<\/p>\n

    Question 38.
    \nFind the future value of an annuity of Rs. 5000 at 12% p.a. for 6 years.
    \nAnswer:
    \nA = Rs. 5,000 r = 0.12 n = 6
    \nF = \\(\\frac{\\mathrm{A}\\left[(1+\\mathrm{r})^{\\mathrm{n}}-1\\right]}{\\mathrm{r}}=\\frac{5000\\left[(1+0.12)^{6}-1\\right]}{0.12}=\\frac{4869.11}{0.12}\\) = 40,575.9<\/p>\n

    PART-D<\/span>
    \nIV. Answer any six questions : (6 \u00d7 5 = 30)<\/span><\/p>\n

    Question 39.
    \nIn a class of 50 students, 15 do not participate in any game, 25 play cricket, and 20 play football. Find the number of students who play.
    \n(i) both cricket and football.
    \n(ii) Only football.
    \nRepresent the results through the Venn diagram.
    \nAnswer:
    \nGiven n(u) = 50
    \n\"1st
    \nn(C \u222a F) = 50 – 15 = 35, r(C) = 25, n(F) = 20
    \nn(C \u2229 F) = n(C) + r(F) – r(C \u222a F) = 25 + 20 – 35 = 10
    \nFrom venn diagram we get only
    \nNumber of students who play only foot ball = 10.<\/p>\n

    Question 40.
    \nEvaluate \\(\\frac{123.4 \\times 54.8}{0.562 \\times 7346}\\) using logarithmic tables.
    \nAnswer:
    \nLet x = \\(\\frac{123.4 \\times 54.8}{0.562 \\times 7346}\\)
    \nThen log x = log\\(\\left[\\frac{123.4 \\times 54.8}{0.562 \\times 7540}\\right]\\)
    \n= log123.4 + log54.8- log0.562 – log7346
    \nx = Anti log (value)<\/p>\n

    Question 41.
    \nThe first term of G.P exceeds the second term by \\(\\frac{1}{2}\\) and the sum to infinity is 2. Find the GP.
    \nAnswer:
    \na, ar, ar2<\/sup> ……………. GP
    \na – ar = 1\/2
    \nS\u221e<\/sub> = 2 …….(2)
    \n\"1st
    \na2<\/sup> = 1 \u21d2 a = 1
    \n\\(\\frac{\\mathrm{a}}{1-\\mathrm{r}}\\) = 2 \\(\\frac{1}{1-r}\\) = 2
    \n1 – r = 1\/2
    \n1 – 1\/2 = r
    \nr = 1\/2<\/p>\n

    \"KSEEB<\/p>\n

    Question 42.
    \nTwo years ago, a father’s age was three times the square of his son’s age. After 3 years, his age will be four times his son’s age. Find the present age of father and son.
    \nAnswer:
    \nLet the age of the father be x yrs and age of the son be y yrs.
    \nGiven: x-2 = 3(y-2)2<\/sup> \u21d2 x-2 = 3(y2<\/sup> + 4 – 4y)
    \nAlso given x + 3 = 4(y + 3)
    \nx + 3 = 4y + 12
    \n\"1st
    \n3y2<\/sup> – 16y + 5 = 0
    \n3y2<\/sup> – 15y – y + 5 = 0
    \n3y(y – 5) – 1(y – 5) = 0
    \n(3y – 1)(y – 5) = 0
    \ny = 51<\/p>\n

    Sons age is 5 yrs and fathers age is
    \nx = 4y + 9 = 20 + 9
    \nx = 29yrs<\/p>\n

    Question 43.
    \nA certain amount invested at 4% p.a. compounded semi-annually amounts to Rs. 78,030 at the end of one year. Find the sum.
    \nAnswer:
    \nF = 78030, i = 0.04, n= 1
    \nF = P\\(\\left(1+\\frac{i}{2}\\right)^{n \\times 2}\\)
    \n78030 = P\\(\\left(1+\\frac{0.04}{2}\\right)^{1 \\times 2}\\)
    \nP = \\(\\frac{78030}{(1.02)^{2}}\\) = 75,000
    \nsum = 75,000.<\/p>\n

    Question 44.
    \nIf a person wants to have Rs. 80,000 after 5 years, how much should he deposit every year if the bank offers 12% p.a. interest compounded quarterly.
    \nAnswer:
    \nF = 80,000, n = 5yrs, r = 12% = 0.12, i = \\(\\left(1+\\frac{0.12}{4}\\right)^{4}\\) – 1 = 0.125
    \nF = \\(\\frac{a\\left[(1+i)^{n}-1\\right]}{i}\\)
    \na = \\(\\frac{\\mathrm{F} \\times i}{(1+i)^{n}-1}=\\frac{80,000 \\times 10.125}{(1.125)^{5}-1}=\\frac{80,000}{6.416}\\) = \u20b912,468.8<\/p>\n

    Question 45.
    \nIn a dance competition, 70% of the participants were girls, 35% of the boys, and 65% of the girls got qualified for the next round. If 49 girls were eliminated find the number of boys who were eliminated.
    \nAnswer:
    \nLet the tota! number of participants = x
    \nTotal number of girls = \\(\\frac{70}{100}\\)x = 0.7x \u21d2 Total number of boys = \\(\\frac{30}{100}\\) \u00d7 x = 0.3x
    \nIf 65% of girls got qualified then 35% of girls got eliminated
    \n\u2234 The total girls eliminated = \\(\\frac{35}{100}\\) \u00d7 0.7x = 0.245x
    \nHence 0.245. x = 49
    \nx = \\(\\frac{49}{0.245}\\)
    \nHence total number of boys eleminated = \\(\\frac{35}{1000}\\) \u00d7 0.3 \u00d7 200 = 2<\/p>\n

    Question 46.
    \nProve that \\(\\frac{\\operatorname{cosec}\\left(180^{\\circ}+0\\right) \\sin \\left(360^{\\circ}-0\\right) \\tan \\left(36^{\\circ}+\\theta\\right)}{\\sin \\left(90^{\\circ}+\\theta\\right) \\cos \\left(180^{\\circ}-\\theta\\right) \\tan (-\\theta)}\\) = sec2 \u03b8
    \nAnswer:
    \n\"1st
    \n\\(\\frac{1}{\\cos ^{2} \\theta}\\) = sec2\u03b8 = RHS<\/p>\n

    \"KSEEB<\/p>\n

    Question 47.
    \nShow that the points (4, 4), (3, 5), and (-1, -1) are the vertices of a right-angled triangle. Also, find its area.
    \nAnswer:
    \nLet A= (4, 4) B = (3, 5) and C = (-1, -1)
    \nAB = \\(\\sqrt{(4-3)^{2}+(4-5)^{2}}=\\sqrt{1^{2}+(-1)^{2}}\\) \u21d2 AB2<\/sup> = 2
    \nBC = \\(\\sqrt{(4-3)^{2}+(4-5)^{2}}=\\sqrt{1^{2}+(-1)^{2}}\\) \u21d2 AB2<\/sup> = 2
    \nCA = \\(\\sqrt{(3+1)^{2}+(5+1)^{2}}=\\sqrt{16+36}=\\sqrt{50}\\) \u21d2 CA2<\/sup> = 50<\/p>\n

    From the above AB2 + CA2 = BC2
    \n\"1st<\/p>\n

    Question 48.
    \nFind the equation of line passing through (3, 4) and having intercepts on the axes whose sum is 14.
    \nAnswer:
    \nReq. Eqn is \\(\\frac{x}{a}+\\frac{y}{b}\\) = 1
    \nGive a + 6=14 \u21d2 b = 14 – a
    \neqn becomes \\(\\frac{x}{a}+\\frac{y}{14-a}\\)
    \n(14 – a)x + ay = a(14 – a)
    \nBy data this must pass the (3, 4)
    \n(14 – a) 3 + 4a = a(14 – a)
    \n42 – 3a + 4a = 14a – a2<\/sup>
    \na2<\/sup> + a – 14a + 42 = 0
    \na2<\/sup> -13a + 42 = 0
    \n(a – 7)(a – 6) = 0 \u21d2 a = 7
    \nIf a = 7 then b = 7 If a = 6 then b = 8.
    \n\u2234 the required equations are
    \n\\(\\frac{x}{7}+\\frac{y}{7}\\) = 1 & \\(\\frac{x}{6}+\\frac{y}{8}\\) = 1<\/p>\n

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

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

    Question 49.
    \n(a) Find the equation of a line passing through the point of intersection of the lines x – 2y + 4 = 0 and 4x – 3y + 1 = 0 inclined at angle of 135\u00b0 with positive x-axis. Also write the x intercept of the line. (4M)
    \nAnswer:
    \nPoint of inter section of x – 2y + 4 = 0 and 4x – 3y + 1 = 0
    \nMultiply equation one by 4 we get
    \n\"1st
    \ny = \\(\\frac{15}{-5}\\) = 3
    \nx – 2y + 4 = 0
    \nx = 2y – 4
    \nx = 6 – 4 = 2
    \n\u2234 point of intersection is (2,3)
    \nSlope = m = tan (135\u00b0) = – 1
    \nThe required equation of the line with slope = -1 and the point (2,3) is given by
    \ny – y1<\/sub> = m(x – x1<\/sub>).
    \ny- 3 = -1(x – 2)
    \n\u2234 x + y – 5 = 0<\/p>\n

    (b) If sin \u03b8 = -8\/17, \u03c0< \u03b8 < \\(\\frac{3 \\pi}{2}\\) find the value of \\(\\frac{\\tan \\theta-\\cos \\theta}{\\sec \\theta+{cosec} \\theta}\\)
    \nAnswer:
    \nIf sin\u03b8 = \\(\\frac{-8}{17}=\\frac{O p p}{1+y p}\\) Here adj = \\(\\sqrt{17^{2}-8^{2}}=\\sqrt{225}\\) = 15
    \n\u03b8 lies in Quadrant
    \n\"1st<\/p>\n

    (c) Two numbers are in the ratio 7:5 and their difference is 17. Find the numbers. 2
    \nAnswer:
    \nLet the two numbers be 7x & 5x
    \nGiven 7x – 5x = 17
    \n2x = 17
    \nx = 8.5
    \n\u2234 the two number are 7 \u00d7 8.5 & 5 \u00d7 8.5 = 59.5 and 92.5.<\/p>\n

    \"KSEEB<\/p>\n

    Question 50.
    \n(a) A confectioner makes and sells biscuits. 11e sells one pack of biscuits at 80. His cost of manufacturing is 40 per pack as a variable cost and 3000 as a fixed cost.
    \nFind (i) his profit function.
    \n(ii) break-even point. also
    \n(iii) if he limits his Production o 100 packets, can be.make profit? 4
    \nAnswer:
    \nT.R (x) = Price x Quantity = 80x
    \nT.C C(x) = ax + b = 40x + 3000
    \nProfit function = P(x) = R(x) – C(x) = 80x – 40x – 3000 = 40x – 3000
    \nIf x = 100 then P(100) = 40 \u00d7 100 – 3000 = \u20b91 ,000<\/p>\n

    (b) Find the sum of n terms the series 5 + 35 + 555 + ………….
    \nAnswer:
    \nLet S = 5 + 55 + 555 +…………… to n terms
    \n\\(\\frac{\\mathrm{S}}{5}\\) = 1+ 11 + 111+………..to n terms
    \n\\(\\frac{9 \\mathrm{~S}}{5}\\) = 9 + 99 + 999 +…………… to n terms
    \n=(10 – 1) + (100 – 1) + (1000 – 1) +……….. to n terms
    \n= (10 + 100 + 1000 +………… to n terms) – (1 + 1 + 1………. to n ternis)
    \n= \\(\\frac{10\\left(10^{n}-1\\right)}{9}\\) – 1 \u00d7 n
    \nS = \\(\\frac{5}{9}\\left[\\frac{10\\left(10^{n}-1\\right)}{9}-n\\right]\\)<\/p>\n

    (c) If 3x<\/sup> = 5y<\/sup> = 15z<\/sup>, Show that z(x + y) = xy.
    \nAnswer:
    \nLet 3x<\/sup> = 5y<\/sup> = 15z<\/sup> = k(say)
    \n3x<\/sup> = k \u21d2 3 = k1\/x<\/sup>, 5y<\/sup> = k \u21d2 5 = k1\/y<\/sup>, 15z<\/sup> = k\u21d2 15 = k1\/z<\/sup>
    \nNow 3 x 5 = 15
    \nk1\/x<\/sup>. k1\/y<\/sup> = k1\/z<\/sup> \u21d2 K1\/x+1\/y<\/sup> = K1\/z<\/sup>
    \n\u21d2 \\(\\frac{1}{x}+\\frac{1}{y}=\\frac{1}{z}\\)
    \n\u21d2 \\(\\frac{\\mathrm{y}+\\mathrm{x}}{\\mathrm{xy}}=\\frac{1}{\\mathrm{z}}\\)
    \n\u21d2 z (y + x) = xy \u21d2 z (x + y) = xy<\/p>\n","protected":false},"excerpt":{"rendered":"

    Students can Download 1st PUC Basic Maths Previous Year Question Paper March 2016 (North), Karnataka 1st PUC Basic Maths Model Question Papers with Answers helps you to revise the complete syllabus. Karnataka 1st PUC Basic Maths Previous Year Question Paper March 2016 (North) Time: 3.15 Hours Max. Marks: 100 Instructions: The questions paper consists of …<\/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\/89583"}],"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=89583"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/89583\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=89583"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=89583"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=89583"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}