{"id":89644,"date":"2021-07-16T11:21:29","date_gmt":"2021-07-16T05:51:29","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=89644"},"modified":"2021-07-16T11:21:29","modified_gmt":"2021-07-16T05:51:29","slug":"1st-puc-basic-maths-model-question-paper-7","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-basic-maths-model-question-paper-7\/","title":{"rendered":"1st PUC Basic Maths Model Question Paper 7 with Answers"},"content":{"rendered":"

Students can Download 1st PUC Basic Maths Model Question Paper 7 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 7 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.
    \nWrite the canonical representation of 96.
    \nAnswer:
    \n\"1st<\/p>\n

    Question 2.
    \nDefine an equivalence relation.
    \nAnswer:
    \nA relation R on a set A is called an equivalence relation if it is reflexive. symmetric and transitive.<\/p>\n

    Question 3.
    \nSimplify: 5(5\u00b0) + (52)\u00b0
    \nAnswer:
    \n51<\/sup> + (25)\u00b0 = 5 + 1 = 6.<\/p>\n

    Question 4.
    \nFind \u2018x\u2019 if log 625 = 4.
    \nAnswer:
    \nx4<\/sup> = 625 \u21d2 x4<\/sup> = 54<\/sup> \u21d2 x = 5.<\/p>\n

    Question 5.
    \nFind the 30th term of the AP -2, -5, 8,………………..
    \nAnswer:
    \na = -2 d = 5 + 2 – 3
    \nT30<\/sub> = a + (30 – 1)d
    \n= -2 + (29)(-3) = -2 – 87 = -89.<\/p>\n

    \"KSEEB<\/p>\n

    Question 6.
    \nForm the quadratic equation whose roots are 1, 2.
    \nAnswer:
    \n(x – 1)(x – 2) = 0 \u21d2 x2<\/sup> – 3x + 2 = 0.<\/p>\n

    Question 7.
    \nCalculate the S.I. on Rs. 18,000 for 4 months at 12\\(\\frac{1}{2}\\)% per annum.
    \nAnswer:
    \nP= 18,000 T = \\(\\frac{4}{12}=\\frac{1}{3}\\)R = \\(\\frac{25}{2}\\)%
    \n\"1st<\/p>\n

    Question 8.
    \nWrite the formula to find the future value of Annuity Due.
    \nAnswer:
    \nF = \\(\\frac{\\mathrm{A}\\left[(1+r)^{n}-1\\right]}{r}\\)(1 + r)<\/p>\n

    Question 9.
    \nA person bought a cycle for Rs. 3000. For what price should he sell it to gain 10%?
    \nAnswer:
    \nSP = \\(\\frac{100+\\text { gain } \\%}{100}\\) \u00d7 CP = \\(\\frac{100+10}{100}\\) \u00d7 3000 = \\(\\frac{110}{100}\\) \u00d7 3000 = Rs. 3300<\/p>\n

    Question 10.
    \nConvert 15\u00b0 into radian measure.
    \nAnswer:
    \n15\u00b0 = \\(\\frac{15 \\pi}{180}=\\frac{\\pi^{c}}{12}\\)<\/p>\n

    Question 11.
    \nIf A = 45\u00b0, then prove that sin 2A = 2 A. Cos A.
    \nAnswer:
    \nSin 2A = Sin 90\u00b0 = 1
    \n2 sin A cos A = 25 in 45\u00b0 cos 450 = 2\\(\\frac{1}{\\sqrt{2}} \\frac{1}{\\sqrt{2}}\\) = 1
    \n\u2234 LHS = RHS.<\/p>\n

    \"KSEEB<\/p>\n

    Question 12.
    \nFind the equation to the locus of a point moving in a plane such that the sum of the squares of its distance from the coordinate axes is 2.
    \nAnswer:
    \nx2<\/sup> + y2<\/sup> = 2.<\/p>\n

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

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

    Question 13.
    \nFind the imaginary part of (1 + i) (4 – 3i).
    \nAnswer:
    \n(1 + i) (4 – 3i) = 4 – 3i + 4i + 3 = 7 + i
    \nimaginary’ part = i.<\/p>\n

    Question 14.
    \nFind the HCF \\(\\frac{8}{9}, \\frac{32}{81}, \\frac{16}{27}\\)
    \nAnswer:
    \nHCF of Numerators 8, 32, 16 = 8
    \nLCM of Denominators 9, 81, 27 = 81
    \n\u2234 HCF of fractions = \\(\\frac{\\text { HCF of Numerators }}{\\text { LCM of Denominators }}=\\frac{8}{81}\\)<\/p>\n

    Question 15.
    \nIf A = {3, 5, 7},B = {5, 7, 9} and U = {1, 2, 3, 4, 5, 6, 7, 8, 9} then find(A – B) \u00d7 B’.
    \nAnswer:
    \nA – B = {3}
    \nB’ = {1, 2, 3, 4, 6, 7, 8)
    \n(A – B) \u00d7 B = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 6), (3, 7), (3. 8)}.<\/p>\n

    Question 16.
    \nIf the 5th term of a HP is \\(\\frac{1}{5}\\) and its 7th term is \\(\\frac{6}{5}\\), find the 10th term.
    \nAnswer:
    \n5th term of HP = \\(\\frac{1}{5}\\)
    \n5th term of AP = 5
    \na + 4d = 5 ………………..(1)
    \nSeventh term of HP = \\(\\frac{6}{5}\\)
    \nSeventh term of AP = \\(\\frac{5}{6}\\)
    \na + 6d = \\(\\frac{5}{6}\\) ………….(2)
    \n\"1st
    \n= –\\(\\frac{25}{6}\\)
    \nd = –\\(\\frac{25}{12}\\)<\/p>\n

    a + 4d = 5 \u21d2 a + 4\\(\\left(-\\frac{25}{12}\\right)\\) = 5 \u21d2 a – \\(\\frac{25}{3}\\) = 5
    \na = 5 + \\(\\frac{25}{3}=\\frac{15+25}{3}=\\frac{40}{3}\\)
    \nTenth term of AP = a + 9d
    \n= \\(\\frac{40}{3}+9\\left(\\frac{-25}{12}\\right)=\\frac{40}{3}+\\frac{3(-25)}{4}=\\frac{160-225}{12}=\\frac{-65}{12}\\)
    \nTenth term of the corresponding HP = –\\(\\frac{-12}{65}\\) .<\/p>\n

    \"KSEEB<\/p>\n

    Question 17.
    \nNine tables and eight chairs cost Rs. 456. Eight tables and nine chairs cost Rs. 462. Find the cost of each chair and each table.
    \nAnswer:
    \n\"1st
    \n9x + 8 (30) = 456
    \n9x + 240 = 456
    \n9x = 456 – 240
    \n9x = 216
    \nx = 24.
    \n\u2234 Cost of 1 table = Rs. 24
    \nCost of 1 chair = Rs. 30.<\/p>\n

    Question 18.
    \nSolve: \\(\\frac{x-5}{10}+\\frac{x+5}{5}\\) = 5
    \nAnswer:
    \n\\(\\frac{x-5}{10}+\\frac{x+5}{5}\\) = 5
    \nx – 5 + 2x + 10 = 50
    \n3x + 5 = 50
    \n3x + 45
    \nx= 15<\/p>\n

    Question 19.
    \nSolve the following system of linear inequalities. 2 (x – 1) < x + 5 and 3 (x + 2) > 2 – x.
    \nAnswer:
    \n2x – 2 < x + 5 and
    \n2x – x < 5 + 2
    \nx < 7……………(1)<\/p>\n

    3 (x + 2) > 2 – x
    \n4x > 2 – 6
    \n4x > – 4
    \nx > \\(\\frac{-4}{45}\\)
    \n\u2234 x > -1 ……………(2)
    \nFrom (1) and (2)
    \n-1 < x < 7.<\/p>\n

    Question 20.
    \nThe average age of Ashok and Abdul is 45 years, the average age of Abdul and Anthony is 50years and the average age of Anthony and Ashok is 35 years. Find the ages of Ashok and Anthony.
    \nAnswer:
    \nAge of Ashok = x
    \nAge of Abdul =y
    \nAge of Anthony = z
    \nGiven,
    \n\\(\\frac{x+y}{2}\\) = 45 \u21d2 x + y = 90 ………………….(1)
    \n\\(\\frac{y+z}{2}\\) = 50 \u21d2 y + z = 100 …………………(2)
    \n\\(\\frac{z+x}{2}\\) = 35 \u21d2 z + x = 70 2 ………………….(3)<\/p>\n

    Add (1), (2), (3)
    \n2x + 2y + 2z = 260
    \nx + y – z = 130
    \nx+ 100 = 130
    \nx = 30<\/p>\n

    x + y = 90
    \n30 + y = 90
    \ny = 60
    \nz = 100<\/p>\n

    60 + z = 100
    \nz = 40<\/p>\n

    Age of Ashok = 30 years.
    \nAge of Abdul = 60 years.
    \nAge of Anthony = 40 years.<\/p>\n

    Question 21.
    \nProve that cos 189\u00b0 + cos 9\u00b0 = 0.
    \nAnswer:
    \nCos (180 + 9) + cos 9 = – cos 9 + cos 9 = 0.<\/p>\n

    \"KSEEB<\/p>\n

    Question 22.
    \nThree consecutive vertices of a parallelogram are (3, 0), (5, 2) and (-2, 6). Find the fourth vertex.
    \nAnswer:
    \nA = (3, 0) B = (5, 2) C = (-2, 6) D = (x, y)
    \nMid point of AC = mid point of BD
    \n\\(\\)
    \n3 – 2 = 5 + x
    \n1 = 5 + x
    \nx = -4<\/p>\n

    0 + 6 = 2 + y
    \ny = 4
    \nD = (- 4, 4).<\/p>\n

    Question 23.
    \nFind the equation to the locus of a point moving in a plane that is equidistant from (- 1, – 1) and (4, 2).
    \nAnswer:
    \nLet P(x, y) be a point which is equidistant from A (-1, -1) and B (4, 2)
    \nThen |PA| = |PB|
    \nSquaring PA2 = PB2
    \n(x + 1)2<\/sup> + (y + 1)2<\/sup> = (x – 4)2<\/sup> + (y – 2)2<\/sup>
    \ni,e. x2<\/sup> + 2x + 1 + y2<\/sup> + 2y + 1 = x2<\/sup> – 8x + 16 + y2<\/sup> – 4y + 4
    \ni.e. 2x + 2y + 2 = – 8x- 4y + 20
    \ni.e. 5x + 3y = 9
    \nHence the locus of P is the straight-line 5x + 3y = 9.<\/p>\n

    Question 24.
    \nFind the equation of a line perpendicular to 3x – 2y + 1 = 0 and passing through the point ( 1, -2).
    \nAnswer:
    \nSlope of the given line 3x – 2y + 1 = 0 is \\(\\frac{3}{2}\\) = m1<\/sub>
    \nSince the two lines are perpendicular m1<\/sub> m2<\/sub> = – 1
    \nm2<\/sub> = \\(\\frac{-1}{m_{1}}=\\frac{-1}{3 \/ 2}=\\frac{-2}{3}\\)
    \nThus the equation of the line with slope m = \\(\\frac{-2}{3}\\) and passing through
    \n(x1<\/sub>, y1<\/sub>) = (1, -2) is given by
    \ny – y1<\/sub> = m(x – x1<\/sub>)
    \ny + 2 = \\(\\frac{-2}{3}\\)(x – 1)
    \n\u21d2 3y + 6 = -2x + 2
    \n\u21d2 2x + 3y + 4 = 0.<\/p>\n

    Question 25.
    \nIf the lines x – 2y + 1 = 0, 2x – 5y + 3 = 0 and 5x – 9y + k = 0 are concurrent, find k.
    \nAnswer:
    \n\"1st
    \ny = 1
    \n2x – 4 + 2 = 0 \u21d2 2x – 2 = 0 \u21d2 x = 1 ,
    \n5 (1) – 9 (1) + k= 0
    \n– 4 + k = 0
    \nk = 4
    \nPoint of concurrency = (1, 1).<\/p>\n

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

    III. Answer any TEN questions: (10 \u00d7 3 = 20)<\/span><\/p>\n

    Question 26.
    \nThree bells toll at intervals of 30 seconds, 40 seconds, and 50 seconds respectively. They start together. After how many minutes will the next bell toll together?
    \nAnswer:
    \nFind the LCM of 30, 40 and 50
    \n\"1st
    \nLCM = 30 \u00d7 20 = 600
    \nHence the bells will fall together after 600 sec.
    \nOr \\(\\frac{600}{60}\\) = 10 minutes.<\/p>\n

    Question 27.
    \nIf f(x) = x and g (x) = x3<\/sup> + 1, find fog (1), gof (1) and fof (2) .
    \nAnswer:
    \n(fog)(1) = f[g(1)] = f[2] = 2
    \n(gof)(1) = g[f(1)] = g[1] = 1 + 1 = 2
    \n(fof)(2) = f[f(2)] = f(2) = 2.<\/p>\n

    \"KSEEB<\/p>\n

    Question 28.
    \nIf 21\/a<\/sup> = 31\/b<\/sup> = 541\/c<\/sup> show that a + 3b = c.
    \nAnswer:
    \n\u2234 21\/a<\/sup> = k \u21d2 2 = ka<\/sup>
    \n31\/b<\/sup> = k \u21d2 3 = kb<\/sup>
    \n541\/c<\/sup> = k \u21d2 54 = kc<\/sup>
    \nNow 2 \u00d7 33<\/sup> = 54
    \nka<\/sup>.(kb<\/sup>)3<\/sup> = kc<\/sup> \u21d2 ka<\/sup>k3b<\/sup> = kc<\/sup>
    \n\u2234 ka+3b<\/sup>\u00a0= kc<\/sup> \u21d2 a + 3b = c.<\/p>\n

    Question 29.
    \nProve : logy<\/sub> x2<\/sup>. logz<\/sub> y3<\/sup>. logx<\/sub> z5<\/sup> = 30.
    \nAnswer:
    \nLHS = 2 logy<\/sub> x. 3logz<\/sub> y. 5logx<\/sub> z
    \n\"1st<\/p>\n

    Question 30.
    \n30. Find three numbers in GP. whose sum is 39 and their product is 729.
    \nLet the three numbers be \\(\\frac{a}{r}\\), a, ar
    \nProduct = 729
    \n\u21d2 a3<\/sup> = 729 \u21d2 a3<\/sup> = 91 \u21d2 a = 9
    \nSum = 39
    \n\\(\\frac{a}{r}\\) + a + ar = 39
    \n\\(\\frac{a}{r}\\) + 9 + 9r = 39
    \n\\(\\frac{a}{r}\\) + 9r = 39 – 9
    \n\\(\\frac{a}{r}\\) + 9r = 30 \u21d2 9 + 9r2<\/sup> = 30r \u21d2 9r2<\/sup> – 30r + 9 = 0
    \n\u21d2 9r2<\/sup> – 27r – 3r + 9 = 0
    \n9r(r – 3) – 3(r – 3) = 0
    \n(r – 3)(9r – 3) = 0
    \nr = 3 or 9r = 3
    \nr = 3 or r = \\(\\frac{3}{9}=\\frac{1}{3}\\)
    \nr = 3 or r = \\(\\frac{1}{3}\\)<\/p>\n

    The numbers are \\(\\frac{a}{r}\\),a,ar
    \n\\(\\frac{9}{3}\\), 9, 9(3)
    \n3, 9, 27.<\/p>\n

    Question 31.
    \nIf \u03b1 and \u03b2 are the roots of the equation 2x – 5x + 7 = 0, then evaluate \\(\\frac{\\alpha^{2}}{\\beta}+\\frac{\\beta^{2}}{\\alpha}\\)
    \nAnswer:
    \n\\(\\frac{\\alpha^{2}}{\\beta}+\\frac{\\beta^{2}}{a}=\\frac{\\alpha^{3}+\\beta^{3}}{a \\beta}\\)
    \n\"1st<\/p>\n

    Question 32.
    \nRs. 16,000 invested at 10% p.a. interest compounded semiannually amounts to Rs. 18,522. Find the period of investment.
    \nAnswer:
    \nP = Rs. 16,000
    \nA = Rs. 18,522
    \nR = 10%
    \nq = 2
    \nr = \\(\\left(1+\\frac{\\mathrm{R}}{q}\\right)^{q}\\) – 1 = \\(\\left(1+\\frac{0.10}{2}\\right)^{2}\\) – 1 = (1 + 0.05)2<\/sup> – 1 = (1.05)2<\/sup> – 1 = 1.1025 – 1
    \n= 0.1025 = 10.25%
    \nA = P (1 + r)n<\/sup>
    \n18522 = 16000(1 + 0.1025)n<\/sup>
    \n\\(\\frac{18522}{16000}\\) = (1.10025)n<\/sup>
    \n1.157625 = (1.1025)n<\/sup>
    \nlog (1.157625) = n log 1.1025
    \nn = \\(\\frac{\\log (1.157625)}{\\log 1.1025}=\\frac{0.0635}{0.0423}\\) = 1.501 \u2248 1.5
    \n\u2234 n = 1.5 years.<\/p>\n

    Question 33.
    \nThe rate of a movie ticket was Rs. 150. This was reduced by 20%, due to which the revenue increased by 20%.
    \nWhat was the percentage increase in the number of viewers?
    \nAnswer:
    \nLet the no. of initial viewers be 100 Cost 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
    \nNew revenue generated = 15000 + 20% of 15000 20
    \n= 15000 + \\(\\frac{20}{100}\\) \u00d7 15000 = 15000 + 3000 = Rs. 18000 100
    \nNo. of viewers = \\(\\frac{\\text { Total revenue }}{\\text { Cost of each ticket }}=\\frac{18000}{120}\\) = 150
    \nPercentage increase in the no. of viewers = 150 – 100 = 50%.<\/p>\n

    \"KSEEB<\/p>\n

    Question 34.
    \nFor the first year of production, the fixed cost for setting up a new company is Rs. 3,00,000 and the variable cost is 70. The company expects revenue from sales to be Rs. 270 per unit of the product. Find the revenue function, cost function, and the break-even output.
    \n(i) Construct the revenue function.
    \n(ii) Construct the cost function.
    \n(iii) Find the Break-Even output
    \n(iv) Find the number of Calculator produced for which the company will suffere loss.
    \nAnswer:
    \nC (x) = 70x + 3,00,000,
    \nR(x)=270x
    \nP(x) = R(x) – C(x)
    \n270 x = (70x + 3,00,000) = 200x – 3,00,000<\/p>\n

    For BEP P(x)= 0
    \nR(x) = C(x)
    \n270x = 70x + 3,00,000, 20x = 3,00,000
    \nx = 1500 BEP
    \nless than x = 1500 units the company will suffer loss.<\/p>\n

    Question 35.
    \nThe angles of a triangle are in the ratio 2: 3: 5. Find them in degrees and radians.
    \nAnswer:
    \n2 : 3 : 5 or 2x, 3x, 5x.
    \n2x + 3x + 5x – 180
    \n10x = 180 \u21d2 x – 18\u00b0
    \n2x = 2 (18) = 36\u00b0 36 \u00d7 \\(\\frac{\\pi}{180}=\\frac{\\pi^{c}}{5}\\)
    \n3x = 3 (18) = 54\u00b0
    \n= \\(\\frac{54 \\pi}{180}=\\frac{3 \\pi}{10}\\)
    \n5x = 5(18) = 90\u00b0 = \\(\\frac{\\pi}{2}\\).<\/p>\n

    Question 36.
    \nFind \u2018x\u2019 given : \\(\\frac{x \\cdot \\sin ^{2} 300^{\\circ} \\sec ^{2} 240^{\\circ}}{\\cos ^{2} 225^{\\circ} \\cdot \\cos e c^{2} 240^{\\circ}}\\) = cot 315.tan2 300\u00b0.
    \nAnswer:
    \nsin300 = sin(360 – 60) = -sin60 = \\(\\frac{-\\sqrt{3}}{2}\\)
    \nsec 240 = sec (180 + 60) = – sec 60\u00b0 = – 2
    \ncos225 = cos(180 + 45\u00b0) = -cos45\u00b0 = –\\(\\frac{1}{\\sqrt{2}}\\)
    \ncosec 240 = cosec (180 + 600) = cosec 60 = –\\(\\frac{2}{\\sqrt{3}}\\)
    \ncot 315\u00b0 = cot(360 – 45\u00b0) – cot45\u00b0 = -1
    \ntan 300 = tan (360 – 600) = – tan 600 = –\\(\\sqrt{3}\\)
    \n\u2234 The given equation becomes:
    \n\\(\\frac{x \\cdot \\frac{3}{4} \\cdot 4}{\\frac{1}{2} \\cdot \\frac{4}{3}}\\) = 1.3
    \n\u21d2 x = \\(\\frac{6}{9}=\\frac{2}{3}\\)<\/p>\n

    Question 37.
    \nShow that the points A(2, 4), B(2, 6) and C(2 + \\(\\sqrt{3}\\), 5) form the vertices of an equilateral triangle.
    \nAnswer:
    \nAB = \\(\\sqrt{(2-2)^{2}+(6-4)^{2}}=\\sqrt{0+4}=\\sqrt{4}\\) = 2
    \nBC = \\(\\sqrt{(2+\\sqrt{3}-2)^{2}+(5-6)^{2}}=\\sqrt{3+1}=\\sqrt{4}\\) = 2
    \nCA = \\(\\sqrt{(2+\\sqrt{3}-2)^{2}+(5-4)^{2}}=\\sqrt{3+1}=\\sqrt{4}\\) = 2
    \nAB = BC = CA
    \n\u2234 A, B, C form an equilateral triangle.<\/p>\n

    \"KSEEB<\/p>\n

    Question 38.
    \nDerive the equation of a straight line in slope-intercept form.
    \nAnswer:
    \n\"1st
    \nSlope of AP = m = tan \u03b8 = \\(\\frac{P N}{\\mathrm{AN}}=\\frac{P M-M N}{O M}\\)
    \n\\(\\frac{\\mathrm{PM}-\\mathrm{OA}}{\\mathrm{OM}}=\\frac{\\mathrm{Y}-\\mathrm{C}}{x}\\)
    \n\u2234 y – c = mx \u21d2 y = mx + c.<\/p>\n

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

    Question 39.
    \nIn a certain college with 500 students, 300 take the milk and 251 take tea. Assuming that every student takes at least one of the two, find how many take.
    \n(a) both milk and tea
    \n(b) milk only
    \n(c) tea only
    \nAnswer:
    \nn (M \u222a T) = 500
    \nn (M) = 300
    \nn (T) = 250
    \nn (M \u222a T) = n (M) + n (T) – n (M \u2229 T)
    \nn (M \u2229 T) = n (M) + n (T) – n (M \u222a T) = 300 + 250 – 500
    \nn (M \u2229 T) = 50
    \nn (M – T) = n (M) – n (M \u2229 T) = 300 – 50 = 250
    \nn(T – M) = n(T) – n(M \u2229 T) = 250 – 50 = 200.<\/p>\n

    Question 40.
    \nEvaluate using log tables: \\(\\frac{25.36 \\times 0.4569}{847.5}\\)
    \nAnswer:
    \nLet x = \\(\\frac{25.36 \\times 0.4569}{847.5}\\)
    \nlog x = log 25.36 + log 0.4569 – log 847.5
    \n= 1.4041 + 1.6599 – 2.9282
    \n= 1.4041 – 1 + 0.6599 – 2.9282
    \n= – 1.8642
    \n= – 1 – 0.8642 = – 1 + 1 -0. 8642
    \n= -2 + 0.1358
    \n= 2.1358
    \n= 01368<\/p>\n

    Question 41.
    \nFind the sum to \u2018n\u2019 terms of the series 4 + 44 + 444 +……………..
    \nAnswer:
    \nSn<\/sub> = 4 + 44 + 444 + …………………….. to n terms
    \nSn<\/sub> = 4(1 + 11 + 111 + …………………….. to n terms)
    \n\\(\\frac{9 \\mathrm{~S}_{n}}{4}\\) = 9 + 99 + 999 + ……………. to n terms
    \n\\(\\frac{9 \\mathrm{~S}_{n}}{4}\\) = (10 – 1) + (102<\/sup> – 1) + (103<\/sup> – 1) +……………………
    \n= (10 + 102<\/sup> + 103<\/sup> + ………………. ) – (1 + 1 + 1 +………………… )
    \n\\(\\frac{9 S_{n}}{4}=\\frac{10\\left(10^{n}-1\\right)}{10-1}\\) – n
    \nSn<\/sub> = \\(\\frac{4}{9}\\left[\\frac{10\\left(10^{n}-1\\right)}{9}-n\\right]\\)<\/p>\n

    \"KSEEB<\/p>\n

    Question 42.
    \nSolve the cubic equation x3<\/sup> + 6x2<\/sup> + 9x + 4 = 0 using synthetic division, given that there is an integral root between – 3 and 3.
    \n(1) x3<\/sup> + 6x2<\/sup> + 9x + 4 = 0
    \nAnswer:
    \nPut x = – 1
    \n(-1)3<\/sup> + 6(-1)2<\/sup> + 9 (-1) + 4
    \n= -1 + 6 – 9 + 4
    \n= 10 – 10 = 0
    \n\u2234 x = – 1 is a root of the equation
    \n\u2234 x + 1 is the factor (factor theorem)
    \n\"1st
    \nQuotient = x2<\/sup> + 5x + 4
    \nremainder = 0
    \nx3<\/sup> + 6x2<\/sup> + 9x + 4 = 0.
    \n(x + 1)(x2<\/sup> + 5x + 4) = 0
    \nx + 1 = 0, x2<\/sup> + 5x + 4 = 0
    \nx2<\/sup> + 4x + x + 4 = 0
    \nx(x + 4) + 1 (x + 4) = 0
    \n(x + 4)(x + 1) = 0
    \nx = -4, x = -1
    \nThe roots are -1, -1, -4.<\/p>\n

    Question 43.
    \nIf the difference between simple interest and compound interest for 3 years at 2.5% p.a. is Rs. 625, find the sum invested.
    \nAnswer:
    \nLet the sum invested = P
    \nr = 2.5% p.a. = 0.025 p.a.
    \nn = 3 years
    \nA = P(1 +r)n<\/sup> = P(1 +0.025)3<\/sup> = P (1.025)3<\/sup>
    \nCI = A – P = P (1.025)3<\/sup> – P = P [(1.025)3 – 1] = P (0.07689)
    \nSI
    \nr = 2.5
    \nT = 3
    \nSI = \\(\\frac{P \\times T \\times R}{100}=\\frac{P \\times 3 \\times 2.5}{100}\\) = 0.075P
    \n\u2234 CI – SI = 625
    \n0.07689P – 0.075 P = 625
    \n\u2234 0.00189 P = 625
    \nP = 3,30687.83
    \n\u2234 P = Rs. 3,30,687.83.<\/p>\n

    Question 44.
    \nHow much should you invest if you want to receive Rs.5000 at the beginning of each year for the next 5 years and the Cl is 16% p.a. compounded quarterly.
    \nAnswer:
    \nA = Rs. 5,000 n = 5 years R = 0.16 q = 4
    \nr = \\(\\left(1+\\frac{\\mathrm{R}}{q}\\right)^{q}\\) – 1 = \\(\\left(1+\\frac{0.16}{4}\\right)^{4}\\) – 1 = 0.1698 = 16.98
    \nEffective rate of interest = 16.98%
    \nP = \\(\\frac{A\\left[(1+r)^{n}-1\\right]}{r(1+r)^{n}}\\)(1 + r)
    \n= \\(\\frac{5000\\left[(1+0.1698)^{5}-1\\right]}{0: 1698(1.1698)^{5}}\\)(1.1698) = \\(\\frac{6963}{0.3719}\\) = 18,722.77<\/p>\n

    \"KSEEB<\/p>\n

    Question 45.
    \nRam Singh purchased 2 camels for Rs. 18,000 and Rs. 15,000 respectively. He sold them at a loss of 15% and a gain of 19% respectively. Find the selling price of each camel. Also, find the overall loss or gain percent in this transaction.
    \nAnswer:
    \nC.P of camel = 18000
    \nS.P while sold at loss of 15% = \\(\\frac{85}{100}\\) \u00d7 18000 = 15300
    \nC.P. of second camel = 15000
    \nS.P. while sold at gain of 19% = \\(\\frac{119}{100}\\) \u00d7 15000 = 17,850
    \nTotal CP = 18000 + 15000 = 33000
    \nTotal S.P. = 15,300 + 17,850 = 33150
    \n\u2234 Profit = 33,150 – 33000 = 150
    \nProfit % = \\(\\frac{150}{33000}\\) \u00d7 100 = 0.45%.<\/p>\n

    Question 46.
    \nProve: \\(\\frac{\\tan \\theta}{\\sec \\theta-1}+\\frac{\\tan \\theta}{\\sec \\theta+1}\\) = 2 cosec \u03b8
    \nAnswer:
    \n\"1st<\/p>\n

    Question 47.
    \nFind the area of the quadrilateral whose vertices are A (1, 2), B (6, 2), C (5, 3) and D (3,4).
    \nAnswer:
    \n(1,2) (6,2) (5, 3) (3,4)
    \n\u2206 ABC = 1\/2 [1 (2 – 3) + 6(3 – 2) + 5 (2 – 2)]
    \n= 1\/2 [1 (-1) + 6 (1) + 5 (0)] = 1\/2 [-1+6 + 0] = \\(\\frac{5}{2}\\)
    \nACD = 1\/2 [1 (3 – 4) + 5 (4 – 2) + 3 (2 – 3)]
    \n= 1\/2 [1 (-1) + 5 (2) + 3 (-1 )] = 1\/2 [-1 + 10 – 3] = 1\/2 [10-4 ] = 1\/2 (6) = 3
    \nArea of the quadrilateral ABCD
    \n= |Area of \u2206ABC| + | Area of \u2206ACD| = \\(\\left|\\frac{5}{2}\\right|\\) + |3| = \\(\\frac{5}{2}\\) + 3 = \\(\\frac{5+6}{2}\\) = \\(\\frac{11}{2}\\)sq. unit.<\/p>\n

    Question 48.
    \nFind the equations of lines passing through the point (3, 4) and having intercepts on the axes such that their sum is 14.
    \nAnswer:
    \nLet the x and y intercepts of the line be a and b respectively so that a + b = 14 or b = 14 – a
    \nThus the equation of the line in the intercept form is
    \n\\(\\frac{x}{a}+\\frac{y}{14-a}\\) = 1
    \nBut it passes through the point(3,4).
    \ntherefore \\(\\frac{3}{a}+\\frac{4}{14-a}\\) – 1
    \n3 (14 – a) + 4a = a (14 – a)
    \ni.e., a2<\/sup> – 13a + 42 = 0
    \ni.e., (a – 6) (a – 7) = 0
    \nor a = 6 or a = 7
    \nIf a = 6 then b = 8; If a = 7 then 6 = 7<\/p>\n

    Therefore equations of the lines are \\(\\frac{x}{6}+\\frac{y}{8}\\) = 1 or \\(\\frac{x}{7}+\\frac{y}{7}\\) = 1
    \ni.e., 4x + 3y – 24 = 0 or x + y – 7 = 0.<\/p>\n

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

    III. Answer any ONE question. (1 \u00d7 1 = 10)<\/span><\/p>\n

    Question 49.
    \n(a) If A {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {4, 5, 6, 7, 8}, then verify:
    \n(i) A \u222a (B \u2229 C) = ( A \u222a B) \u2229 ( A \u222a C)
    \n(ii) A\u2229(B\u222aC) = (A\u2229B)\u222a(A\u2229C)
    \nAnswer:
    \n(i)Let A = {1, 2, 3, 4}
    \nB = {3, 4, 5, 6}
    \nC = {4, 5, 6, 7, 8}
    \nB\u2229C = {4,5,6}<\/p>\n

    A\u222a(B\u2229C) = {1, 2, 3, 4} \u222a {4, 5, 6}
    \n= {1,2, 3, 4, 5, 6}<\/p>\n

    A\u222aB = {1, 2, 3, 4, 5, 6}
    \nA\u222aC = {1, 2, 3, 4, 5, 6, 7, 8}
    \n(A\u222aB) \u2229 (A\u222aC)
    \n= {1, 2, 3, 4, 5,6} \u2229 {1, 2, 3,4, 5, 6, 7, 8}
    \n= {1, 2, 3, 4, 5, 6}
    \nA\u222a(B\u2229C) = (A\u222aB) \u2229 (A\u222aC)<\/p>\n

    (ii) B\u222aC = {3, 4, 5, 6, 7, 8}
    \nA\u2229(B\u222aC)= {1, 2, 3, 4} \u2229 {3, 4, 5, 6, 7, 8}
    \n= {3, 4}
    \nA\u2229B = {3, 4}
    \nA\u2229C = {4}
    \n(A\u2229B)\u222a(A\u2229C) = {3, 4}
    \nThus we have
    \nA\u2229(B\u222aC) = (A\u2229B) \u222a (A\u2229C).<\/p>\n

    (b) Find 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+r)^{n}\\right]}{r}=\\frac{5000\\left[(1+0.12)^{6}-1\\right]}{0.12}=\\frac{4869.11}{0.12}\\) =40,575.9<\/p>\n

    (c) The average age of 12 boys is 8 years. Another boy of 21 years joins the group. Find the average age of the new group.
    \nAnswer:
    \nAverage age of 12 boys = 8 years
    \n\u2234 Total age of boys = 8 \u00d7 12 = 96 years
    \nwhen a boy of 21 yrs joins the group,
    \nthe total age of 13 boys = 96 + 21 = 117 years
    \n\u2234 Average age of the new group = \\(\\frac{117}{13}\\) = 9 years.<\/p>\n

    \"KSEEB<\/p>\n

    50. (a) Solve the linear inequalities graphically :
    \n3x + 3y \u2264 6,
    \nx + 4y \u2264 4,
    \nx \u2265 0,y \u2265 0.
    \nAnswer:
    \n\"1st<\/p>\n

    (b) A watch manufacturer finds that his total cost function is linear. The total cost for 100 watches is Rs. 20,000 and for 200 watches, the total cost increased to Rs. 30,000. What are the fixed cost and variable cost per unit?
    \nAnswer:
    \nTotal cost = T.V.C. + T.F.C.
    \nC (x) = ax + b
    \nWhen x = 100 watches
    \nT.C. = a (100) + b
    \ni.e.20,000 = 100a+ 6 …………………. (1)<\/p>\n

    Similarly when x = 200 watches
    \n30.0 = 200 a + b ……………….(2)
    \n(2) – (1)30,000 = 200 a + b
    \n-20,000 = 100 a + b
    \n10,000 = 100 a
    \na = \\(\\frac{10,000}{100}\\) = 100 substitute in (1)
    \n20,000= 100 (100) + b
    \n20,000 = 10,000 + b
    \nb =10,000<\/p>\n

    \u2234 When x = 150, TC = (100) (150)+ 10,000
    \n= 15,000+10,000,
    \nTC = \u20b9 25,000.<\/p>\n

    (c) The HCF of 2 numbers is 16 and their LCM is 160. If one of the numbers is 32, then find the other num ber.
    \nAnswer:
    \nHCF =16 LCM =160 a = 32 b = ?
    \na \u00d7 b = H \u00d7 L
    \nb = \\(\\frac{\\mathrm{H} \\times \\mathrm{L}}{a}=\\frac{16 \\times 160}{32}=\\frac{2560}{32}\\) = 80
    \n\u2234 The other number is 80.<\/p>\n","protected":false},"excerpt":{"rendered":"

    Students can Download 1st PUC Basic Maths Model Question Paper 7 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 7 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\/89644"}],"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=89644"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/89644\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=89644"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=89644"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=89644"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}