{"id":89392,"date":"2021-07-08T12:24:32","date_gmt":"2021-07-08T06:54:32","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=89392"},"modified":"2021-07-13T10:16:42","modified_gmt":"2021-07-13T04:46:42","slug":"1st-puc-basic-maths-previous-year-question-paper-march-2020-north","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-basic-maths-previous-year-question-paper-march-2020-north\/","title":{"rendered":"1st PUC Basic Maths Previous Year Question Paper March 2020 (North)"},"content":{"rendered":"

Students can Download 1st PUC Basic Maths Previous Year Question Paper March 2020 (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 2020 (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 3-i.
    \nAnswer:
    \nImaginary part = – 1.<\/p>\n

    Question 2.
    \nIf A has 4 elements, how many elements will P(A) have?
    \nAnswer:
    \nP(A) = 24 = 16 element<\/p>\n

    Question 3.
    \nSimplify: ax+y<\/sup> a2x-y<\/sup>
    \nAnswer:
    \n\"1st<\/p>\n

    Question 4.
    \nFind x if logx 25 = 2
    \nAnswer:
    \nx2<\/sup> = 25
    \n\u2234 x = \u00b15<\/p>\n

    Question 5.
    \nFind the sum to infinity of GP 3,\\(\\frac{1}{1}, \\frac{1}{3}\\)…………..
    \nAnswer:
    \na = 3, r = \\(\\frac{1}{3}\\)
    \n\u2234 S\u221e<\/sub> = \\(\\frac{a}{1-r}=\\frac{3}{1-\\frac{1}{3}}=\\frac{9}{2}\\)<\/p>\n

    \"KSEEB<\/p>\n

    Question 6.
    \nSolve for x : 7(x – 2) + 8(x – 3) = x + 10.
    \nAnswer:
    \n7x – 14 + 8x – 24 = n + 10
    \n\u2234 14x = 48<\/p>\n

    Question 7.
    \nCalculate SI for 2 years for \u20b9 6000 at 8% pa.
    \nAnswer:
    \n\"1st<\/p>\n

    Question 8.
    \nFind the present value of a perpetuity of 3,000 to be received forever at 14% pa.
    \nAnswer:
    \nPerpetuity P\u221e<\/sub> = \\(\\frac{A}{r}=\\frac{3000}{0.14}\\) = \u20b921,428<\/p>\n

    Question 9.
    \nConvert 0.32 into a percentage.
    \nAnswer:
    \n0.32 = 0.32 \u00d7 100 = 32%<\/p>\n

    Question 10.
    \nConvert 45\u00b0 into radians.
    \nAnswer:
    \n\"1st<\/p>\n

    Question 11.
    \nIf A = 45\u00b0 then show that sin 2 A = 2 sin A cos A.
    \nAnswer:
    \nsin(2 \u00d7 45) = 2sin45\u00b0cos45\u00b0
    \nsin90\u00b0= 2 \u00d7 \\(\\frac{1}{\\sqrt{2}} \\times \\frac{1}{\\sqrt{2}}\\)
    \n1 = 1<\/p>\n

    \"KSEEB<\/p>\n

    Question 12.
    \nFind the slope of the line 2x + 5y – 11 = 0.
    \nAnswer:
    \nSlope = m = \\(\\frac{-2}{5}\\)<\/p>\n

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

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

    Question 13.
    \nFind the number of positive divisors of 34<\/sup> \u00d7 53<\/sup> \u00d7 72<\/sup>
    \nAnswer:
    \n34<\/sup> \u00d7 53<\/sup> \u00d7 72<\/sup> = p1<\/sub>\u03b11<\/sub><\/sup>\u00d7 p2<\/sub>\u03b12<\/sub><\/sup> \u00d7 p3<\/sub>\u03b13<\/sub>
    \n<\/sup>p1<\/sub> = 3, p2<\/sub> = 5, p3<\/sub> = 7, \u03b11<\/sub> = 4, \u03b12<\/sub> = 3, \u03b13<\/sub> = 2
    \nT(n) = (1 + \u03b11<\/sub>)(1 + \u03b12<\/sub>)(1 + \u03b13<\/sub>) = (1 +4)(1 + 3)(1 + 2) = (5)(4)(3) = 60<\/p>\n

    Question 14.
    \nThree bells call at intervals 30 sec, 40 sec, 50 sec respectively. They start together. After how many minutes will the next bell call together.
    \nAnswer:
    \n\"1st
    \nHence the bills will fall together after 600 second or \\(\\frac{600}{60}\\) = 10 minutes<\/p>\n

    Question 15.
    \nIf (x + y, x – y) = (5, 1) find x and y.
    \nAnswer:
    \n\"1st<\/p>\n

    Question 16.
    \nProve that (xb-c)a.(xc-a)b.(xa-b)c = 1
    \nAnswer:
    \n\"1st<\/p>\n

    \"KSEEB<\/p>\n

    Question 17.
    \nWhich term ofan AP \\(\\frac{1}{2}\\), 1, \\(\\frac{1}{2}\\) is 5?
    \nAnswer:
    \na = \\(\\frac{1}{2}\\), d = 1 – \\(\\frac{1}{2}\\)= \\(\\frac{1}{2}\\), an = 5, n =?
    \nan<\/sub> = a + (n – 1)d
    \n5 = \\(\\frac{1}{2}\\) + (n – 1)\\(\\frac{1}{2}\\)
    \n\"1st<\/p>\n

    Question 18.
    \nSolve x + 2y = 1
    \n3x – 2y = 5
    \nAnswer:
    \n\"1st
    \nx = 3\/2
    \nx + 2y = 1
    \n\\(\\frac{1}{2}\\) + 2y = 1
    \n2y = 1 – \\(\\frac{3}{2}\\)
    \n2y = –\\(\\frac{1}{2}\\)
    \n\u2234 y = – \\(\\frac{1}{4}\\)<\/p>\n

    Question 19.
    \nIf a and b are the roots of the equation 3x2<\/sup> – 6x +4 = 0. Find \u03b12<\/sup> + \u03b22<\/sup>.
    \nAnswer:
    \n\u03b1 + \u03b2 = \\(-\\frac{b}{a}=\\frac{-3}{-6}=\\frac{1}{2}\\)
    \n\u2234 \u03b12<\/sup> + \u03b22<\/sup> = (\u03b1 + \u03b2)2<\/sup> – \u03b1\u03b2
    \n\u03b1\u03b2 = \\(\\frac{c}{a}=\\frac{4}{5}\\)
    \n= \\(\\left(\\frac{1}{2}\\right)^{2}-\\frac{4}{3}=\\frac{1}{4}-\\frac{4}{3}-\\frac{3-16}{12}=\\frac{-13}{12}\\)<\/p>\n

    Question 20.
    \nSolve linear inequalities and represent on the number line
    \n7x + 3 < 5x + 9, X\u2208 R
    \nAnswer:
    \n\"1st
    \n7x + 3 < 5x + 9
    \n7x – 5x < 9 – 3
    \n2x < 6
    \n\u2234 x < 3<\/p>\n

    \"KSEEB<\/p>\n

    Question 21.
    \nFind the future value of an annuity of 600 at 4% pa payable for 2 years,
    \nAnswer:
    \nF=? u= 600, r = 0.04, n = 2ys
    \nF = \\(\\frac{a\\left[(1+r)^{n}-1\\right]}{r}=\\frac{600\\left[(1+0.04)^{2}-1\\right]}{0.04}=\\frac{600\\left[(1.04)^{2}-1\\right]}{0.04}\\) = \u20b9 1224<\/p>\n

    Question 22.
    \nProve that (i + tan2<\/sup> \u03b8)(1 – sin2<\/sup> \u03b8 ) = 1
    \nAnswer:
    \nLHS= (1 + tan2<\/sup>\u03b8)(1 – sin2<\/sup>\u03b8)
    \n\"1st<\/p>\n

    Question 23.
    \nFind the value of cos 60\u00b0 – sin 30\u00b0 – cots 450
    \nAnswer:
    \n\"1st<\/p>\n

    Question 24.
    \nOne end of a diameter of a circle is (1, 3) and its center is (4, -2). Find the coordinates of the other end of this diameter.
    \nAnswer:
    \nBy midsection formula:
    \n(4, -2) = \\(\\left[\\frac{x_{1}+1}{2}, \\frac{y_{1}+3}{2}\\right]\\)
    \n\u2234 \\(\\frac{x+1}{2}\\) = 4
    \n\u2234 x = 7
    \n\\(\\frac{y_{1}+3}{2}\\) = -2
    \n\u2234 y = -7
    \n\u2234 the other end of diameter = (x, y) = (7, -7)<\/p>\n

    Question 25.
    \nFind the equation of the locus of all points equidistant from the point (2,4) and y-axis.
    \nAnswer:
    \nLocus = P(x,y).A(2,4) and y-sinB (0, y)
    \nGiven PA = PB
    \n\\(\\sqrt{(x-2)^{2}+(y-4)^{2}}=\\sqrt{(x-0)^{2}+(y-y)^{2}}\\)
    \n\"1st
    \ny2 – 4x – 8y + 20 = 0 equation of the locus<\/p>\n

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

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

    Question 26.
    \nProve that 3 + \\(\\sqrt{5}\\) is an irrational number.
    \nAnswer:
    \nIf possible let 3 + \\(\\sqrt{5}\\)be a rational number
    \n\u2234 3 + \\(\\sqrt{5}\\) = \\(\\frac{p}{q}\\) where p, q \u2208 Z, q \u2260 0
    \n\u2234 \\(\\sqrt{5}\\) = \\(\\frac{p}{q}\\) – 3 = rational – rational = rational number
    \n\u2234 \\(\\sqrt{5}\\) is a rational number which is a contradiction.
    \n\u2234 our assumption is wrong
    \n\u2234 3 + \\(\\sqrt{5}\\) is irrational.<\/p>\n

    Question 27.
    \nIf f(x) = 2x + 1 g(x) = x2<\/sup> + 2x + 1.
    \nFind (i) fog (2)
    \n(ii) gof (3)
    \n(iii) fog (1)
    \nAnswer:
    \n(i) fog(1) = f[g(1)] = f[13+1] = f(2) = 2
    \n(ii) fog(2) = f[g(2)] = f[23+1] = f(9) = 9
    \n(iii) gof(1) = g[f(1)] = g[1] = 13 + 1 = 2
    \n(iv) gof(-1) = g[f(-1)] = g[-1] = (-1)3 + 1 = -1 + 10<\/p>\n

    Question 28.
    \nIf ax<\/sup> = by<\/sup> = cx<\/sup> and b2<\/sup> = ac show that \\(\\frac{1}{x} \\times \\frac{1}{z}=\\frac{2}{y}\\).
    \nAnswer:
    \nLet a = b = c = k (say)
    \n\u2234 ax<\/sup> = k \u21d2 a = k1\/x<\/sup>, by<\/sup> = k \u21d2 b = k1\/y<\/sup>, cz<\/sup> = k\u21d2 cz<\/sup> = k1\/z<\/sup>
    \nNow, b2<\/sup> = ac
    \n\u2234 (k1\/y<\/sup>)2<\/sup> = k1\/x<\/sup>.k1\/z<\/sup>
    \n\u2234 k2\/x<\/sup> = k1\/x+1\/z<\/sup><\/p>\n

    Bases are same \u2234 Equating powers on both the sides, we get.
    \n\\(\\frac{2}{y}=\\frac{1}{x}+\\frac{1}{z}\\)<\/p>\n

    \"KSEEB<\/p>\n

    Question 29.
    \nIf x2<\/sup> + y2<\/sup> = 12xy show that 2log(x-y) = log2 + log5 + logx + logy.
    \nAnswer:
    \nLHS = 2 log (x -y)
    \n= log(x – y)2<\/sup> = log(x2<\/sup> + y2<\/sup> – 2xy) Put x2<\/sup> + y2<\/sup> = 12xy
    \n= log(12xy – 2xy) = log(10xy) = log(2 \u00d7 5 \u00d7 x \u00d7 y)
    \n= log2 + log5 + logx + logy = RHS<\/p>\n

    Question 30.
    \nThe 3 HM’s between \\(\\frac{1}{4}\\) and \\(\\frac{1}{12}\\).
    \nAnswer:
    \nInsert 3 H.M.\u2019s between and \\(\\frac{1}{4}\\) and \\(\\frac{1}{12}\\)
    \n\\(\\frac{1}{4}\\), h1, h2, \\(\\frac{1}{12}\\)………. HP
    \n4, \\(\\frac{1}{\\mathrm{~h}_{1}}, \\frac{1}{\\mathrm{~h}_{2}}, \\frac{1}{\\mathrm{~h}_{3}}\\), 12……… AP
    \na = 4 d =? T = 12 n = 5
    \nTn = a + (n – 1)d
    \n12 = 4 + (5 – 1)d
    \n12 – 4 = 4d
    \n4d = 8d = 2
    \n4,6,8,10,12,……… AP
    \n\\(\\frac{1}{4}, \\frac{1}{6}, \\frac{1}{8}, \\frac{1}{10}, \\frac{1}{12}\\) …………..HP .
    \n\u2234 Harmonic means are \\(\\frac{1}{6}, \\frac{1}{8}, \\frac{1}{10}\\)<\/p>\n

    Question 31.
    \nFive years ago, the father’s age was 5 times as old as his son and after 3 years he will be 3 times as old as his son. Find their present ages.
    \nAnswer:
    \nLet the present age of the son = x ys
    \n5 yrs ago son’s age = (x – 5) yr
    \nFather’s age = y yrs
    \n5 year ago father age = (y – 5)
    \nGiven (y – 5) = 5(x – 5)
    \ny – 5 = 5x – 25
    \n5x – y – 20 = 0 …(1)
    \nAfter 3 year Son’s age = (x + 3)
    \nFather’s age = (y + 3)
    \n\u2234 (y + 3) = 3(x + 3)
    \ny + 3 = 3x + 9
    \n\u2234 3x- y + 6 = 0 …(2)
    \n\"1st
    \n\u2234 x = 13yrs
    \ny = 45 yrs<\/p>\n

    Question 32.
    \nSolve the following graphically x + y < 5.
    \nAnswer:
    \nx + y < 5
    \n\"1st
    \n\"1st<\/p>\n

    \"KSEEB<\/p>\n

    Question 33.
    \nIf \u20b9 500 amounts to \u20b9 725 at 9% simple interest in sometimes what will \u20b9 600 amount to at 11% in the same time.
    \nAnswer:
    \nGiven P = \u20b9 500, A = \u20b9 725
    \nSI = 225, R = 9%, T = ?
    \nSI = \\(\\frac{\\mathrm{PTR}}{100}\\) = 225 = \\(\\frac{500 \\times 7 \\times 9}{100}\\)
    \n\u2234 T = \u20b9 5yrs
    \nAgain is P = 600, R = 11% T = 5yrs I = ?
    \nI = \\(\\frac{600 \\times 5 \\times 4}{100}\\) = \u20b9330
    \n\u2234 A = P + I = (600 + 300) = 930<\/p>\n

    Question 34.
    \n5 kg of Sugar at the rate of \u20b9 15\/kg
    \n8 kg of Wheat at a rate of \u20b9 22\/kg
    \n7 kg of Rice at a rate of \u20b9 351kg and
    \n4 kg of oil at a rate of \u20b9 85\/kg. What is the average price\/kg of all the commodities?
    \nAnswer:
    \nAverage price \/kg (x\u0304w) = \\(\\frac{n_{1} w_{1}+n_{2} w_{2}+n_{3} w_{3}+n_{4} w_{4}}{n_{1}+n_{2}+n_{3}+n_{4}}\\)
    \nn = Quality, w = weight
    \n= \\(\\frac{(5 \\times 15)+(8 \\times 22)+(7 \\times 35)+(4 \\times 85)}{(5+8+7+4)}\\)
    \nx\u0304w = \\(\\frac{836}{24}\\) = \u20b9 34.83<\/p>\n

    Question 35.
    \nBy how many percent should the use of tea be increased if the price of tea is decreased by 10% so that expenditure remains unchanged.
    \nAnswer:
    \nLet the price of tea = \u20b9 100
    \nQuantity = 100 unit
    \n\u2234 Total expenditure = (100 \u00d7 100) = 10,000
    \nNew price = 100 – 10 = 90 Net quantity = y
    \n\u2234 Total expenditure = 90 y
    \n\u2234 90y= 10,000
    \n\u2234 y = \\(\\frac{10000}{90}\\) = 111.11
    \n\u2234 % Increase = \\(\\frac{\\text { Increase }}{\\mathrm{CP}}\\) \u00d7 100 = \\(\\frac{11.11}{100}\\) \u00d7 100 = 11.11%<\/p>\n

    \"KSEEB<\/p>\n

    Question 36.
    \nFind x : x sin45\u00b0 cos260\u00b0 = \\(\\frac{\\tan ^{2} 60^{\\circ}{cosec} 30^{\\circ}}{\\sec 45^{\\circ} \\cdot \\cot ^{2} 30^{\\circ}}\\)
    \nAnswer:
    \n\"1st<\/p>\n

    Question 37.
    \nThree comers of a parallelogram ABCD taken in order are A(-1, -6) B(2, 5) C(7, 2). Find the fourth vertex.
    \nAnswer:
    \nMid point of the diagonal AC = midpt. of diagonal BD
    \nFormal D (x, y) \\(\\left[\\frac{-1+7}{2}, \\frac{-6+2}{2}\\right]=\\left[\\frac{x+2}{2}, \\frac{y-5}{2}\\right]\\)
    \n\"1st
    \n4th vertex D = (x, y) = (4,1)<\/p>\n

    Question 38.
    \nFind the equation of a line that passes through the point (-4 5) and whose intercepts are equal in magnitude but opposite in sign.
    \nAnswer:
    \nLet a = a, b = – a \u2234 \\(\\frac{x}{a}+\\frac{y}{b}\\) = 1 (Intercept form)
    \nat(-4, 5), \\(\\frac{-4}{a}+\\frac{5}{a}\\) = 1
    \n\u2234 a = 1, b = -1<\/p>\n

    Equation of the line is \\(\\frac{x}{1}+\\frac{y}{1}\\) = 1
    \n\u2234 x – y – 1 = 0<\/p>\n

    Part-D<\/span><\/p>\n

    IV. 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 250 take tea. Find how many take (a) Milk only (b) tea only (c) both milk and tea. Also, represent this by Venn diagram.
    \nAnswer:
    \nn(M\u222aT) = 500
    \nn(M) = 300 n(T) = 250
    \nn(M\u222aT)= n(M) + n(T) – n(M\u2229T)
    \nn(M\u222aT) = n(M)+n(T) – n(M\u222aT) = 300 + 250 – 500
    \nn (M \u2229 T) = 50
    \nn(M – T)=n(M) – n(M\u2229T) = 300 – 50= 250
    \nn(T – M)= n(T) – n(M \u2229 T) = 250 – 50 = 200<\/p>\n

    Question 40.
    \nSolve using log table : \\(\\frac{5.6348 \\times 25.645}{12.75 \\times 19.78}\\)
    \nAnswer:
    \nLet x = \\(\\frac{5.6348 \\times 25.645}{12.75 \\times 19.78}\\)
    \napply log on both sides
    \nlog x = log\\(\\left[\\frac{5.6348 \\times 25.845}{120+2 \\times 1 \\times 19.7}\\right]\\)
    \nlogx = log 5.6348 + log 25. 645 – log (12.72)- log (19.78)
    \n= (0.7508) + (1.4090) -(1.1044) – (1.2962)
    \nlog x = 0.2408
    \nx = 1.4237<\/p>\n

    Question 41.
    \nThe first and last term of the GP is 3 and 96 respectively sum to n terms is 189. Find the common ratio and the number of terms.
    \nAnswer:
    \nGiven a = 3, an<\/sub> = 96, Sn<\/sub> = 189, r = ?
    \nan <\/sub>= a.rn-1<\/sup>
    \n96 = 3.(r)n-1<\/sup>
    \n\u2234 (r)n-1<\/sup> = 32 ………(1)
    \nIf Sn<\/sub> = \\(\\frac{a\\left(l-r^{n}\\right)}{(l-r)}\\)
    \n189 = \\(\\frac{3\\left(1-r^{n}\\right)}{1-r}=\\frac{3\\left(1-r \\cdot r^{n-1}\\right)}{1-r}\\)
    \n189 = \\(\\frac{3(1-r(32))}{1-r}\\)
    \n\"1st
    \n1 – 32r = 63 – 63r
    \n63r – 32r = 63 – 1
    \n31r = 62
    \n\u2234 r = 62\/31 = 2
    \n\u2234 rn-1<\/sup> = 32 = 2n-1<\/sup> = 32
    \n2n-1<\/sup> = 25
    \n\u2234 n – 1 = 5
    \n\u2234 n = 6<\/p>\n

    \"KSEEB<\/p>\n

    Question 42.
    \nFind the quotient and remainder obtained x3<\/sup> + 4x2<\/sup> – 7x – 10 by (x + 1) dividing and solve.
    \nAnswer:
    \nGiven x + 1 = 0
    \n\"1st
    \nRemeinder = 0, Quotient \u2234 x2<\/sup> – 3x -10<\/p>\n

    Question 43.
    \nFind the difference between simple interest and compound interest on \u20b9 18,000 invested for 4 years at 8% pa. where Cl is compounded annually.
    \nAnswer:
    \nGiven P = \u20b9 18,000, T = 4yr, R = 8%
    \nSI – \\(\\frac{\\mathrm{PTR}}{100}=\\frac{18,000 \\times 4 \\times 8}{100}\\) = \u20b95,760
    \nCI = A = P[(1 + i)]n<\/sup>
    \n= 18,000[1 + 0.08]4<\/sup>
    \nA = 24,488.8
    \n\u2234 CI = 24.488.8 – 18,000 = 6488.8
    \n\u2234 Difference between Cl and SI = 6488.8 – 5760 = \u20b9728.8<\/p>\n

    Question 44.
    \nMrs. Geetha deposits 4,00,000 on retirement in a bank that pays 10% p.a. interest. How much can be drawn annually for a period of 10 years?
    \nAnswer:
    \nP = 4,00,000, i = 0.10, n = 10 ys, a = ?
    \nP = \\(\\frac{a\\left[(1+i)^{n}-1\\right]}{i(1+i)^{n}}\\)
    \n4,00,000 = \\(\\frac{a\\left[(1+0.10)^{10}-1\\right]}{0.1(1+0.1) 10}\\)
    \n4,00,000 = \\(\\frac{a[2.5937-1]}{0.1 \\times 2.5937}\\)
    \n\u2234 a = \\(\\frac{4,00,000}{6.1445}\\)
    \na = \u20b965,098.8<\/p>\n

    Question 45.
    \nA shopkeeper sold a watch at a 5% loss. Had he purchased it at 10% p.a. sold it for 140 more his gain would have been one-fourth of the original cost price. Find the cost price of the watch.
    \nAnswer:
    \nLet x be the original C.P of the watch
    \nthen 5% of x = \\(\\frac{95 x}{100}\\) = SP
    \ngiven caritian
    \n(x – 10 %g.x) + \\(\\frac{x}{4}=\\frac{95 x}{100}\\) + 140
    \n\\(\\frac{90 x}{100}=\\frac{x}{4}=\\frac{95 x+14000}{100}\\)
    \n\"1st
    \n90x + 25x = 95x + 14000
    \n20x = 14000
    \n\u2234 x = \\(\\frac{14000}{20}\\)
    \nx = \u20b9700<\/p>\n

    Question 46.
    \nIf then prove that x = arsin AcosB, y = brsin AsinBr and z = crossA then prove that \\(\\) = r2
    \nAnswer:
    \n\\(\\frac{\\mathrm{x}}{\\mathrm{a}}\\) = rsinAcosB \\(\\frac{\\mathrm{y}}{\\mathrm{b}}\\) = rsinAstnB
    \n\\(\\frac{\\mathrm{z}}{\\mathrm{c}}\\) = rcos A
    \n\\(\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}+\\frac{z^{2}}{c^{2}}\\) = r2<\/sup>2sin2Acos2 B + r2<\/sup> sin A sin2 B + r2<\/sup> cos2A
    \n= r2<\/sup> sin2 A (cos2<\/sup> B + sin2<\/sup> B) + r2<\/sup> cos2<\/sup> A = r2<\/sup> sin2<\/sup> A + r2<\/sup> cos2<\/sup> A = r2 (sin2<\/sup> A + cos2<\/sup> A) = r2<\/sup> (1) = r2<\/p>\n

    Question 47.
    \nFind the circumcenter of the triangle whose vertices are (-1,2) (3,0) (-2,4).
    \nAnswer:
    \nAssume the circum centre in p(x, y)
    \nA (-1,2) B(3, 0), C(-2, 4)
    \nPA = PB = PC
    \nIf PA = PB
    \n\u2234 \\(\\sqrt{(x+1)^{2}+(y-2)^{2}}=\\sqrt{(x-3)^{2}+(y-0)^{2}}\\)
    \n\"1st
    \n\u2234 2x – 4y + 5 = -6x + 3
    \n\u2234 8x – 4y – 4 = 0 + 4
    \n\u2234 2x – y – 1 = 0 ………….(1)
    \nLet PB = PC
    \n\\(\\sqrt{(x-3)^{2}+(y-0)^{2}}=\\sqrt{(x+2)^{2}+(y-4)^{2}}\\)
    \n\u2234 (x + 1)2<\/sup> + (y-2)2<\/sup> = (x-3)2 <\/sup>+ y2<\/sup>
    \n-6x – 4x + 8 v + 9 – 20 = 0
    \n-10x + 8y- 11 = 0 …(2)
    \nSolve (1) and (2)
    \n2x – y – 1=0
    \nxy by 8
    \n\"1st
    \nx = 2
    \ny = 3
    \nthe circum centre = (2,3)<\/p>\n

    Question 48.
    \nFind the foot of the perpendicular drawn from the point (-2, -1) on the line 3x + 2y – 5 = 0.
    \nAnswer:<\/p>\n

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

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

    Question 49.
    \n(a) If A = {x : x2<\/sup> – 5x + 6 = 0; x \u2208 N}
    \nB = {x : x2<\/sup> – 7x + 12 = 0 x \u2208 N}
    \nFind (A – B) \u00d7 B. 4
    \nAnswer:
    \nA= {2, 3},B = {3,4}
    \n\u2234 A – B ={2}
    \n\u2234 (A – B) \u00d7 B = {2} \u00d7 {3 – 4} = {(2, 3), (2, 4)}<\/p>\n

    (b) Find the sum to V terms of a GP. 0.6 + 0.66 + 0.66 +
    \nAnswer:
    \nLet Sn<\/sub> = 0.6 + 0.66 + 0.666 +………….n term
    \nSn<\/sub> = 6{0.1 + 0.11 + 0.111 +……………n term}
    \n\u2234 \\(\\frac{S n}{6}\\) = 0.1 +0.11 +0.11 +…………..+ n term
    \nxly<\/sup> both side by 9
    \n\u2234 \\(\\frac{9 S n}{6}\\) = 0.9 + 0.99 + 0.999 +……………. + n term
    \n\u2234 \\(\\frac{9 S n}{6}\\) = (1-0.1) + (1 – 0.01) + (1 – 0.001) +………..+ n term
    \n\\(\\frac{9 S n}{6}=\\left(1-\\frac{1}{10}\\right)+\\left(1-\\frac{1}{100}\\right)+\\left(1-\\frac{1}{1000}\\right)\\) + ……+ n term
    \n\"1st
    \n\"1st<\/p>\n

    (c) The average marks of 15 students of a class is 45. A student also has secured 50 marks leaves the class room. Find the average marks of the remaining 14 students.
    \nAnswer:
    \nA \u00d7 B = LCM \u00d7 HCF
    \nHCF 16, hem = 160, A = 64
    \n\"1st
    \n\u2234 another no. = 4<\/p>\n

    \"KSEEB<\/p>\n

    Question 50.
    \n(a) If the cost function C(x) of producing \u2018x\u2019 unit of a product is given by C(x)= 500x2<\/sup> + 2500 x + 5000 and if each unit of the product is sold at 6000 then find BEP.
    \nAnswer:
    \nGiven C(x) = 500x2<\/sup> + 2500 x \u00b1 5000 (x = output)
    \n\u2234 Total Revenue = R(x) = price x quantity = 6000x
    \nFor B \u2208 P: C(x) = R(x)
    \n\u2234 500x2<\/sup> + 2500x + 5000 = 6000x
    \n\u2234 500x2<\/sup> – 3500x + 50000
    \n\u00f7500 \u2234 x2<\/sup> – 7x + 10 = 0
    \nx = 5 or 2 units<\/p>\n

    (b) Find \u2019a\u2019 so that the lines x – 6y + a = 0, 2x + 3y + 4 = 0 and x + 4y + 1 = 0 are concurrent.
    \nAnswer:
    \nSolve 2x + 3y + 4 = 0 ….(1) and x + 4y + 1 = 0 xly by 2
    \n2x + 8y + 2 = 0 …(2)
    \n(1) and (2)
    \n\"1st
    \n2x + 3(2\/5) + 4 = 0
    \n2x = \\(\\frac{-6}{5}\\) – 4 = \\(\\frac{-26}{5}\\)
    \n\u2234 x = -13\/5
    \nSub in x – 6y + a = 0
    \na = 6y – x
    \n\u2234 a = 6(2\/5) – (-13\/5)
    \na = \\(\\frac{12+13}{5}\\) = 5<\/p>\n

    (c) The average marks of 15 students of a class is 45. A student also has secured 50 marks leaves the class room. Find the average marks of the remaining 14 students.
    \nAnswer:
    \nTotal marks and A. 15 students = 45 \u00d7 15 = 675
    \n\"1st
    \n\u2234 Average marks of 14 student = \\(\\frac{625}{14}\\) = 44.64<\/p>\n","protected":false},"excerpt":{"rendered":"

    Students can Download 1st PUC Basic Maths Previous Year Question Paper March 2020 (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 2020 (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\/89392"}],"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=89392"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/89392\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=89392"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=89392"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=89392"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}