{"id":89174,"date":"2021-07-05T16:24:42","date_gmt":"2021-07-05T10:54:42","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=89174"},"modified":"2021-07-05T16:24:42","modified_gmt":"2021-07-05T10:54:42","slug":"1st-puc-basic-maths-model-question-paper-1","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-basic-maths-model-question-paper-1\/","title":{"rendered":"1st PUC Basic Maths Model Question Paper 1 with Answers"},"content":{"rendered":"

Students can Download 1st PUC Basic Maths Model Question Paper 1 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 1 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.
    \nDefine an imaginary number.
    \nAnswer:
    \nA number of the form a + ib where a, b \u2208 R, the set of real numbers and i = \\(\\sqrt{-1}\\) is called a complex number or an imaginary number.
    \nExample : 2 + 3i<\/p>\n

    Question 2.
    \nIf A = {1, 2} and B = {a, b} then find B \u00d7 A.
    \nAnswer:
    \nB \u00d7 A={(a, 1),(a, 2),(b, 1 ),(b, 2)}<\/p>\n

    Question 3.
    \nIf f: R \u2192 R is defined by f(x) = 2x + 3 then find f(\\(\\frac{1}{2}\\))
    \nAnswer:
    \nf(\\(\\frac{1}{2}\\)) = 2(\\(\\frac{1}{2}\\)) + 3 = 1 + 3 = 4<\/p>\n

    Question 4.
    \nSimplify \\(\\left(\\frac{81}{256}\\right)^{\\frac{1}{4}}\\)
    \nAnswer:
    \n\\(\\left(\\frac{8}{256}\\right)^{1 \/ 4}=\\left(\\frac{3^{4}}{4^{4}}\\right)^{1 \/ 4}=\\left\\{\\left(\\frac{3}{4}\\right)^{4}\\right\\}^{1 \/ 4}=\\frac{3}{4}\\)<\/p>\n

    \"KSEEB<\/p>\n

    Question 5.
    \nFind the value of log0.1<\/sub> 10
    \nAnswer:
    \nLet log0.1<\/sub> 10 = x \u21d2(0.1)x<\/sup> \u21d2 \\(\\left(\\frac{1}{10}\\right)^{x}\\) = 10
    \n\u21d2 10-x<\/sup> = 101<\/sup> \u21d2 -x = 1 \u21d2 x = -1
    \n\u2234 log0.1<\/sub> = 10 = -1<\/p>\n

    Question 6.
    \nFind the nature of the roots without solving the equation x2<\/sup> – x + 1 = 0
    \nAnswer:
    \na = 1, b = -1, c = 1
    \n\u0394 = b2<\/sup> – 4a = (-1)2<\/sup> – 4(1)(1) = 1 – 4 = -3 < 0
    \n\u0394 < 0
    \n\u2234 The roots are complex.<\/p>\n

    Question 7.
    \nWhat is the present value of an income of 1000 a year to be received forever? Assume the discount rate to be 12%.
    \n12%
    \nAnswer:
    \nA = \u20b9 1,000, r = 12% = \\(\\frac{12}{100}\\) = 0.12,
    \nP\u221e<\/sub> = A\/r = \\(\\frac{1000}{0.12}\\) = \u20b9 8333.33<\/p>\n

    Question 8.
    \nConvert \\(\\frac{1}{4}\\) into percentage
    \nAnswer:
    \n\\(\\frac{1}{4}=\\frac{1}{4}\\) \u00d7 100% = 25%<\/p>\n

    Question 9.
    \nDefine a radian
    \nAnswer:
    \nA radian is an angle subtended at the center of a circle by an arc of the circle such that the length of the arc is equal to the radius of the die circle.<\/p>\n

    \"KSEEB<\/p>\n

    Question 10.
    \nConvert 315\u00b0 into radians
    \nAnswer:
    \n315\u00b0 = 315 \u00d7 \\(\\frac{\\pi}{180}=\\frac{7 \\pi^{c}}{4}\\)<\/p>\n

    Question 11.
    \nIf the slope of the line AB is \\(\\) and AB \u22a5 CD then find the slope of CD.
    \nAnswer:
    \nm1<\/sub> = \\(\\frac{3}{4}\\), m2<\/sub> = ?
    \nAB \u22a5 CD
    \n\u2234 m1<\/sub> \u00d7 m2<\/sub> = -1 \u21d2 \\(\\frac{3}{4}\\) \u00d7 m2<\/sub> = 1 \u21d2 m2<\/sub> = – \\(\\frac{4}{3}\\) = slope of CD<\/p>\n

    Question 12.
    \nFind the centroid of the triangle formed by the points (2, 4), (5, 3), and (8, 3)
    \nAnswer:
    \nG = \\(\\left(\\frac{2+5+8}{3}, \\frac{4+3+3}{3}\\right)=\\left(\\frac{15}{3}, \\frac{10}{3}\\right)=\\left(5, \\frac{10}{3}\\right)\\)<\/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 960
    \nAnswer:
    \n960
    \n960 = 26 \u00d7 31 \u00d7 51
    \np1<\/sub> = 2, \u03b11<\/sub> = 6, p2<\/sub> = 3, \u03b12<\/sub> = 1, p3<\/sub> = 5, \u03b13<\/sub> = 1
    \nT(n) = (1 + \u03b11<\/sub>)(1 + \u03b12<\/sub>)(1 + \u03b13<\/sub>)
    \nT(960) = (1 + 6)(1 + 1)(1 + 1) = (7)(2)(2) = 28
    \n\"1st<\/p>\n

    Question 14.
    \nIf A =(1, 3,5, 7,9) and B = (2,4,6,8, 10, 12) define a function f: A \u2192 B by f(x) = x + 1 \u2200 x \u2208 A, verfiy that the function is one – one and onto
    \nAnswer:
    \nf: A \u2192 B is one-one but not onto as 12 has no pre-image.<\/p>\n

    \"KSEEB<\/p>\n

    Question 15.
    \nSimplify \\(\\left(\\frac{x^{2}}{x^{b}}\\right)^{n+b}\\left(\\frac{x^{b}}{x^{c}}\\right)^{b+c}\\left(\\frac{x^{c}}{x^{2}}\\right)^{c+x}\\)
    \nAnswer:
    \n(xa-b<\/sup>)a+b<\/sup>(xb-c<\/sup>)b+c<\/sup>(xc-a<\/sup>)c+a<\/sup> = xa2<\/sup>-b2<\/sup><\/sup><\/sup>.xb2<\/sup>-c2<\/sup><\/sup><\/sup>.xc2<\/sup>-a2<\/sup><\/sup><\/sup>
    \n= \\(x^{a^{2}-b^{2}+b^{2}-c^{2}+c^{2}-a^{2}}\\) = x\u00b0 = 1<\/p>\n

    Question 16.
    \nProve that log2<\/sub> [log2<\/sub> [log2<\/sub> 16]] = 1
    \nAnswer:
    \nLog2<\/sub> [log2<\/sub>[log2<\/sub>16]] = log2<\/sub> [log2<\/sub>[log2<\/sub> 24]] = log2<\/sub>[log2<\/sub>[4log2<\/sub> 2]] = log2<\/sub> [log2<\/sub> [4(1)]]
    \n= log2<\/sub> [log2 4] = log2<\/sub> [log2<\/sub> 22] = log2<\/sub> [2 log2<\/sub> 2] = log2<\/sub> [2(1)] = log2<\/sub> 2 = 1 [\u2235 loga<\/sub> o = 1]<\/p>\n

    Question 17.
    \nIs -300 a term of the A.P. 10, 7, 4….. ?
    \nAnswer:
    \na =10, d = 7 – 10 = -3
    \nTn<\/sub> = a + (n- 1) \u21d2 -300= 10 + (n – 1) (-3)
    \n\u21d2 -300 = 10 – 3n + 3 \u21d2 3n = 10 + 3 + 300
    \n\u2234 3n = 313
    \n\u2234 n = \\(\\frac{313}{3}\\) \u2209 N
    \n\u2234 n being number of terms cannot be a fraction.
    \n\u2234 -300 is not a term of the A.P<\/p>\n

    Question 18.
    \nIf k + 9, k – 6, 4 are in GP then find the value of K.
    \nAnswer:
    \n(k – 6)2<\/sup> = (k + 9)4
    \nk2<\/sup> – 12k + 36 = 4k + 36 \u21d2 k2<\/sup> – 12k – 4k + 36 – 36 = 0
    \nk2<\/sup> – 16k =0 \u21d2 k(k – 16) = 0
    \nk =0, k= 16<\/p>\n

    \"KSEEB<\/p>\n

    Question 19.
    \nIf a and p are the roots of the equation x2<\/sup> – x + 2 = 0 then show that \u03b12<\/sup>\u03b2 + \u03b22<\/sup>\u03b1 = 2
    \nAnswer:
    \n\u03b1 + \u03b2 = \\(\\frac{-b}{a}=\\frac{-(-1)}{1}\\) = 1 and \u03b1\u03b2 = \\(\\frac{c}{a}=\\frac{2}{1}\\) = 2
    \n\u03b12<\/sup>\u03b2 + \u03b1\u03b22<\/sup> = \u03b1\u03b2(\u03b1 + \u03b2) = 1(2) = 2<\/p>\n

    Question 20.
    \nSove 5x – 3 < 3x + 1 when x is an integer and x is a real number
    \nAnswer:
    \n5x – 3 < 3x + 1
    \n2x <4<\/p>\n

    5x – 3x < 3 + 1
    \n\u21d2 x < 2<\/p>\n

      \n
    1. x = {…………..-2, -1,0, 1} x \u2208 Z<\/li>\n
    2. x \u2208 (-h, 2) x \u2208 R<\/li>\n<\/ol>\n

      Question 21.
      \nSowmya invested \u20b9 1500 for 8 years and Anisha invested \u20b9 7500 for 3 years at the same rate of interest. If altogether they received \u20b9 1725 as interest find the rate of simple interest charged.
      \nAnswer:
      \nSowmya,
      \nP1<\/sub> = 1500, T1<\/sub> = 8, R1<\/sub> = R
      \nI1<\/sub> = \\(\\frac{1500 \\times 8 \\times \\mathrm{R}}{100}\\) = 120R<\/p>\n

      Anisha,
      \nP2<\/sub> = 7500 T2<\/sub> = 3 R, = R
      \nI2<\/sub> = \\(\\frac{7500 \\times 3 \\times R}{100}\\) = 225R
      \nI1<\/sub> + I2<\/sub> = 1725
      \n120 R + 225 R = 1725
      \n345 R = 1725 \u21d2 R = \\(\\frac{1725}{345}\\) = 5%<\/p>\n

      Question 22.
      \nIf the cost price of a machine is \u20b9 150 and selling price is \u20b9 100 find the loss percentage?
      \nAnswer:
      \nCP = \u20b9 150 . SP = \u20b9 100
      \nCP > SP
      \n\u2234 loss
      \nloss = CP – SP = 150- 100 = \u20b9 50
      \nloss% = \\(\\frac{\\text { loss }}{\\mathrm{CP}}\\) \u00d7 100 = \\(\\frac{50}{150}\\) \u00d7 100 = 33.3%<\/p>\n

      Question 23.
      \nThe average score of 20 boys is 60% and the average score of 30 girls is 70%. Find the combined average of boys and girls?
      \nAnswer:
      \nX1<\/sub> = 60%, N1<\/sub> = 20; X2<\/sub> = 70%, N2<\/sub> = 30
      \nX\u030412<\/sub> = \\(\\frac{\\mathrm{X}_{1} \\mathrm{~N}_{1}+\\mathrm{X}_{2} \\mathrm{~N}_{2}}{\\mathrm{~N}_{1}+\\mathrm{N}_{2}}=\\frac{(60)(20)+(70)(30)}{20+30}=\\frac{3300}{50}\\) = 66%<\/p>\n

      \"KSEEB<\/p>\n

      Question 24.
      \nDerive the equation of the line in one point form. i.e. y – y1<\/sub> = m (x – x1<\/sub>) geometrically, where m is the slope and p(x1<\/sub> y1<\/sub>) is the given point
      \nAnswer:
      \n\"1st
      \nSlope of AP = m = tan \u03b8 = \\(\\frac{\\mathrm{PL}}{\\mathrm{AL}}=\\frac{\\mathrm{PN}-\\mathrm{NL}}{\\mathrm{MN}}=\\frac{\\mathrm{PN}-\\mathrm{AM}}{\\mathrm{ON}-\\mathrm{OM}}=\\frac{y-y_{1}}{x-x_{1}}\\)<\/p>\n

      \u2234 y – y1<\/sub> = m(x – x1<\/sub>)<\/p>\n

      Question 25.
      \nFind the value of x if the distance between (x, 3) and (4, 5) is “5 units”.
      \nAnswer:
      \nA = (x, 3), B = (4, 5)
      \nAB =5
      \n\\(\\sqrt{(x-4)^{2}+(3-5)^{2}}\\) =5
      \n(x- 4)2<\/sup> + 4= 25
      \nx2<\/sup> – 8x + 16 + 4 – 25 =0
      \nx2<\/sup> – 8x – 5 = 0
      \nx = \\(\\frac{-(-8) \\pm \\sqrt{64-4(1)(-5)}}{2(1)}=\\frac{8 \\pm \\sqrt{64+20}}{2}=\\frac{8 \\pm \\sqrt{84}}{2}=\\frac{8 \\pm 2 \\sqrt{21}}{2}=\\frac{2(4 \\pm \\sqrt{21})}{2}\\)
      \nx = 4 \u00b1 \\(\\sqrt{21}\\)<\/p>\n

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

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

      Question 26.
      \nProve the \\(\\sqrt{2}\\) is an irrational number. Sol. We shall prove it by contradiction.
      \nAnswer:
      \nWe shall prove it by contradiction
      \nIf possible, Let \\(\\sqrt{2}\\) be a rational number
      \nLet \\(\\sqrt{2}\\) = \\(\\frac{\\mathrm{p}}{\\mathrm{q}}\\) where p and q are integers and q \u2260 0.
      \nFurther let p and q are coprime i.e. H.C.F. of p and q = 1.
      \n\\(\\sqrt{2}\\) = \\(\\frac{\\mathrm{p}}{\\mathrm{q}}\\) \u21d2 \\(\\sqrt{2}\\)q = p ,
      \n\u21d2 2q2<\/sup> = p2<\/sup>
      \n\u21d2 2 divides p2<\/sup> \u21d2 2 divides p \u21d2 p is even
      \nLet p = 2k where k is an integer p2<\/sup> = 4k2<\/sup>
      \n2q2<\/sup> = 4k2<\/sup>
      \nq2<\/sup> = 2k2<\/sup> q2<\/sup> is even
      \n\u21d2 q is even
      \nNow p is even and q is even which implies p and q have a common factor 2. which is a contradiction of the fact that p and q are co-prime.
      \n\u2234 our assumption that \\(\\sqrt{2}\\) is rational is wrong and hence \\(\\sqrt{2}\\) is irrational.<\/p>\n

      Question 27.
      \nDefine an equivalence relation with an example. Also, give an example of a relationship that is only symmetric.
      \nAnswer:
      \nA relation R on a non-empty set A is called an equivalence relation if it is reflexive, symmetric, and transitive.
      \nExample: On a set of straight lines in a plane the relation “parallel to” is an equivalence relation. Let A= {1, 2, 3} R = {(1, 1), (3, 3), (2, 1), (1, 2)}
      \nR is symmetric but not reflexive and not transitive.<\/p>\n

      \"KSEEB<\/p>\n

      Question 28.
      \nIf ax<\/sup> = by<\/sup> = cz<\/sup> and b2<\/sup> = ac. Show that \\(\\frac{1}{x}+\\frac{1}{x}=\\frac{2}{y}\\)
      \nAnswer:
      \nLet ax<\/sup> = by<\/sup> = cz<\/sup> = k (say)
      \n\u2234 ax<\/sup> = k \u21d2 a = k 1\/x,
      \nby<\/sup> = k \u21d2 b = k 1\/y,
      \ncz<\/sup> = k \u21d2 c = k’*
      \nNow, b2<\/sup> = ac
      \n\u2234 (k 1\/y<\/sup>)2<\/sup> = ,k1\/x<\/sup>. k1\/z<\/sup>
      \n\u2234 k2\/y<\/sup> = k1\/x + 1\/z<\/sup>.
      \nBases are same
      \n\u2234 Equating powers on both the sides, we get.
      \n\\(\\frac{2}{y}=\\frac{1}{x}+\\frac{1}{z}\\)<\/p>\n

      Question 29.
      \nFind the number of zeroes in (0.2)100 after the decimal point and the first significant figure.
      \nAnswer:
      \nlog (0.2)100<\/sup> = 100 log 0.2= 100 (T. 0.3010)= 100(-1 + 0.3010) = – 100 + 30.10
      \n= -100 + 30 + 0.10 = -70 + 0.10 = 70.10
      \ncharacteristic = 70
      \n\u2234 number of zeroes = 70 – 1 = 69.<\/p>\n

      Question 30.
      \nFind the sum of the following series : 7 + 77 + 777 + 7777 +……… n terms.
      \nAnswer:
      \nLet Sn<\/sub> = 7 + 77 + 777…………..n terms
      \n= 7 [1 + 11+ 111 +………..to n terms]
      \n= \\(\\frac{7}{9}\\) [9 + 99 + 999 + …………..to n terms]
      \n= \\(\\frac{7}{9}\\) [(10- 1) + (100 – 1) + (1000 – 1)…………..to n terms]
      \n= \\(\\frac{7}{9}\\left[\\frac{10\\left(10^{n}-1\\right)}{10-1}-n\\right]\\) [\u2235 10 + 102<\/sup> + 103<\/sup> ……… to n terms is in G.P
      \nwhere a = 10, r = 10]<\/p>\n

      Sn<\/sub> = \\(\\frac{7}{9}\\left[\\frac{10\\left(10^{n}-1\\right)}{9}-\\mathrm{n}\\right]\\)<\/p>\n

      Question 31.
      \nA father is 28 years older than a son. After 5 years the father’s age will be 7 years more than twice that of the son. Find their present ages.
      \nAnswer:
      \nLet father’s age be x years and son’s age be y years.
      \nGiven: x – y = 28……………….(1)
      \nAfter five years father’s age is x + 5 and son’s age is y + 5.
      \nAlso given: x + 5 = 2(y + 5) + 7
      \nx + 5 = 2y + 10 + 7
      \nSolving (1) and (2), we get: x = 44 and y = 16
      \nThus father’s age is 44 years and the son’s age is 16.<\/p>\n

      \"KSEEB<\/p>\n

      Question 32.
      \nSolve the linear inequalities graphically: x +3y \u2265 3, 2x + y \u2265 2, x \u2265 10, y \u2265 0
      \nAnswer:
      \n\"1st<\/p>\n

      Question 33.
      \nIn what time a sum of \u20b9 500 will earn \u20b9 975 at the rate pf 6% p.a if the compound interest is payable half yearly?
      \nAnswer:
      \nR = 6% = \\(\\frac{6}{100}\\) = 0.06
      \nq = 4
      \nr = \\(\\left(1+\\frac{R}{q}\\right)^{q}\\) – 1 = 0.0607
      \nA = P(1 + r)n<\/sup>
      \nlog A= log P + n log (1 + r)
      \nn = \\(\\frac{\\log A-\\log P}{\\log (1+r)}=\\frac{\\log 975-\\log 500}{\\log (1+0.0607)}=\\frac{2.9890-2.6990}{0.0256}\\) = 11.33 years<\/p>\n

      Question 34.
      \nInsert 4 arithmetic means between 5 and 10.
      \nAnswer:
      \nLet A1<\/sub>, A2<\/sub>, A3<\/sub>, A4<\/sub> be the 4 A.M’s between 5 and 10
      \n\u2234 5, A1<\/sub>, A2<\/sub>, A3<\/sub>, A4<\/sub>, 10 are in A.P,
      \nLast element is 6th element a + 5d = 10
      \n5 + 5d =10
      \n5d = 10-5
      \n5d = 5
      \n\u2234 A. M\u2019s are 6,7, 8, 9<\/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 the expenditure remains unchanged.
      \nAnswer:
      \nLet Price be 100 Let Quantity be 100
      \nTotal expenditure = 100 \u00d7 100 = 10,000
      \nThen New Price = 100 – 10 = 90
      \nLet New Qty = y
      \nTotal expenditure = 90 y
      \n90 y = 10,000
      \nIncrease = 111.11 – 100 = 11.11
      \n\u2234 % increase = \\(\\frac{\\text { increase }}{\\text { Original value }}\\) \u00d7 100 = \\(\\frac{11.11}{100}\\) \u00d7 100 = 11.11%<\/p>\n

      \"KSEEB<\/p>\n

      Question 36.
      \nIf tan A = \\(\\frac{12}{13}\\) and 180 < A < 270. Find the value of \\(\\frac{3 \\sin A-2 \\cos A}{9 \\cos A+4 \\sin A}\\)
      \nAnswer:
      \nOP2<\/sup> = OM2<\/sup> + MP2<\/sup> = 169 + 144 = 313
      \nOP = \\(\\sqrt{313}\\)
      \nSinA = \\(\\frac{-12}{\\sqrt{313}}\\)
      \nCosA = \\(\\frac{-13}{\\sqrt{313}}\\)
      \n\"1st
      \n\"1st<\/p>\n

      Question 37.
      \nShow that the straight lines 2x – 3y = 7,3x – 4y =13, 8x – 11y = 33 are concurrent. Also find the point of concurrency. .
      \nAnswer:
      \n2x – 3y = 7 \u00d7 3
      \n3x – 4y = 7 \u00d7 2
      \n\"1st
      \ny = 5
      \n2x-3y = 7 \u21d2 2x – 3(5) = 7 \u21d2 2x – 15 = 7 \u21d2 2x = 22 \u21d2 x = 11
      \nPut x =11 y = 5 in 8x- 11y = 33
      \n8 (11) – 11 (5) = 88 – 55 =33 .
      \n\u2234 The three lines are concurrent. Point of concurrency = (11, 5)<\/p>\n

      Question 38.
      \nFind the distance between two parallel lines 3x + 4y + 5 = 0 and 6x + 8y + 20 = 0
      \nAnswer:
      \n3x + 4y + 5 = 0 \u00d7 2
      \n6x + 8y + 10 = 0 ……(1)
      \n6x + 8y + 20 = 0 ………(2)
      \n\"1st<\/p>\n

      PART – D<\/span><\/p>\n

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

      Question 39.
      \nIn a group of 150 people, 70 like Cricket, 30 like hockey and cricket both. How many like Cricket only and not hockey? How many like hockey? Show the result using the Venn diagram.
      \nAnswer:
      \nn(C\u222aA)=150 n(C) = 70 n(H) = ? n(C\u2229H) = 30
      \nn(C\u222aH) = n(C) + n(H) – n(C\u2229H)
      \n150 = 70 + n(H) – 30
      \n150 = 40 + n(H) [\u2235 n(H) = 110]
      \nn(C – H) = n(C) – n(C\u2229H) = 70 – 30 = 40 .
      \n\"1st
      \n\"1st<\/p>\n

      Question 40.
      \nEvaluate using log, tables \\(\\frac{42.15 \\times(0.2713)^{2}}{0.8932}\\)
      \nAnswer:
      \nLet x = \\(\\frac{42.15(0.2713)^{2}}{0.8932}\\)
      \nlog x = log 42.15+ 21og 0.2713-log 0.8932
      \n= 1.6247+ 2(1.433)-(). 9509)= 1.6247+ 2(-l + 0.433)-(1.9509)
      \n= 1.6247-2 +0.866-(-1 + 0.9509)= 1.6247-2+ 0.866-(-1 + 0.9509)
      \n= 1.6247-2 + 0:866+ 1 -0,9509
      \nlog x = 0.5398 =AL [0.5398] = 3.4733
      \n\u2234 x = 3.4733<\/p>\n

      Question 41.
      \nFind the sum of the series : 4 + 44 + 444 + ….. n terms,
      \nAnswer:
      \nLet Sn<\/sub> = 4 + 44 + 444 + n terms
      \nSn<\/sub> = 4 (1 + 11 + 111 + ……….)
      \n\\(\\frac{9 \\mathrm{~S}_{\\mathrm{n}}}{4}\\) = 1 + 11 + 111 + ……
      \n\\(\\frac{9 \\mathrm{~S}_{\\mathrm{n}}}{4}\\) = 9 + 99 + 999
      \n\\(\\frac{9 \\mathrm{~S}_{\\mathrm{n}}}{4}\\) = (10 – 1) + (102 – 1) + (103 – 1) + ………
      \n\\(\\frac{9 \\mathrm{~S}_{\\mathrm{n}}}{4}\\) = (10 + 102 + 103 + ………) – (1 + 1 + 1 + …………….)
      \n\\(\\frac{9 S_{n}}{4}=\\frac{10\\left(10^{n}-1\\right)}{10-1}\\) – n
      \n\\(\\frac{9 S_{n}}{4}=\\frac{10\\left(10^{n}-1\\right)}{9}\\) – n
      \nSn<\/sub> = \\(\\frac{4}{9}\\left[\\frac{10}{9}\\left(10^{n}-1\\right)-n\\right]\\), n\u2208N<\/p>\n

      \"KSEEB<\/p>\n

      Question 42.
      \nThe cost of 2 kgs of sugar and 5 bags of dhal is \u20b9 90. The cost of 5 kgs of sugar and 2 kgs of dhal is \u20b9 120. Find the cost of sugar and dhal per kg.
      \nAnswer:
      \nLet the cost of 1kg of sugar = x and cost of 1 kg of dhal = y
      \n2x + 5y = 90 \u00d7 2
      \n5x + 2y – 120 \u00d7 5
      \n\"1st
      \n2x + 5 = 90
      \n2(20) + 5y = 90
      \n5y = 90 – 40 = 50
      \ny = 10<\/p>\n

      \u2234 cost of 1 kg of sugar = \u20b9 20
      \ncost of 1 kg of dhal = \u20b9 10<\/p>\n

      Question 43.
      \nGovindappa bought 51 bags in the wholesale market at an average price of \u20b9 318 each. The price of 33 leather bags was \u20b9 426 each. The price of the remaining cotton bags forms an increasing Arithmetic progression. If the price of the costliest cotton bag was \u20b9 150\/ – find the price of the cheapest cotton bag.
      \nAnswer:
      \nAverage price of 11 bags = Rs.450
      \nTotal price of 11 bages = 11 \u00d7 450 = 4950
      \nAverage price of 7 leather bags = Rs. 575
      \nTotal price of 7 leather bags = 575 \u00d7 7 = 4025
      \nThe price of the remaining 4 cotton bags from an increasing A.P, Whose first term be a and common difference d,
      \n1 = 300, n = 4, Sn = 925.
      \nSum of remaining 4 bags = 4950 – 4025 = 925
      \nSn =925
      \n\\(\\frac{\\mathrm{n}}{2}\\) (a + 1) = 925
      \n\\(\\frac{4}{2}\\) (a + 300) = 925
      \n2 (a + 300) = 925
      \n2a + 600 = 925
      \n2a = 325
      \na = \\(\\frac{325}{2}\\) =162.50
      \nPrice of the cheapest cotton bag = Rs. 162.50<\/p>\n

      Question 44.
      \nThe daily cost of production ‘C is 1\u20b9 and ‘x’ unit of an assembly is C(x) = 3.5x + 1200. If each unit is sold for 6 then find the minimum number of units that should be produced and sold to ensure no loss. If the selling price is increased by half a rupee a unit then what would be the Break-Even Point.
      \nAnswer:
      \nC(x) = 3.5x + 1200
      \nR(x) = C(x) for a break even point
      \nR(x) = 6x
      \n6x = 3.5x + 1200
      \n6x – 3.5x = 1200
      \n2.5x = 1200
      \nx = \\(\\frac{1200}{2.5}\\) = 480 units<\/p>\n

      If the S.P is increased by \u20b9 0.5 a unit then
      \nR(x) = 6.5x
      \nR(x) = C(x)
      \n6.5x = 3.5x + 1200
      \n6.5x – 3.5x = 1200
      \n3x = 1200
      \nx = 400 units<\/p>\n

      \"KSEEB<\/p>\n

      Question 45.
      \nProve that sec2<\/sup>\\(\\frac{5 \\pi}{4}\\) cosec2<\/sup>\\(\\frac{5 \\pi}{4}\\) – sin2<\/sup>\\(\\frac{3 \\pi}{4}\\)cos2<\/sup>\\(\\frac{4 \\pi}{3}\\) = 31\/8
      \nAnswer:
      \nsec\\(\\frac{5 \\pi}{4}\\) = sec 225 = sec(180+45\u00b0) = Sec45\u00b0 = – \\(\\sqrt{2}\\)
      \ncosec\\(\\frac{5 \\pi}{4}\\) = coscc 225\u00b0 = cosec (180 + 45) = cosec 45\u00b0 = – \\(\\sqrt{2}\\)
      \nsin \\(\\frac{3 \\pi}{4}\\) = sin 135\u00b0= sin(180 – 45\u00b0)sin45\u00b0 = \\(\\frac{1}{\\sqrt{2}}\\)
      \ncos \\(\\frac{4 \\pi}{3}\\) = cos 240 = cos (180 + 60) = – cos 60 = –\\(\\frac{1}{2}\\)
      \n\u2234 LHS = sec2<\/sup>\\(\\frac{5 \\pi}{4}\\) cosec2<\/sup>\\(\\frac{5 \\pi}{4}\\) – sin2<\/sup>\\(\\frac{3 \\pi}{4}\\) cos2<\/sup>\\(\\frac{4 \\pi}{3}\\)<\/p>\n

      = \\((\\sqrt{2})^{2} \\cdot(\\sqrt{2})^{2}-\\left(\\frac{1}{\\sqrt{2}}\\right)^{2} \\cdot\\left(\\frac{-1}{2}\\right)^{2}\\)
      \n= 2.2 – \\(\\frac{1}{2}\\).\\(\\frac{1}{4}\\) = 4 – \\(\\frac{1}{8}\\) = \\(\\frac{32-1}{8}=\\frac{31}{8}\\) = R.H.S<\/p>\n

      Question 46.
      \nDerive the section formula for the internal division if A (x1<\/sub>, y1<\/sub>) and B (x2<\/sub>, y2<\/sub>) are the two given points and P(x, y) be the point that divides the join of AB in the ratio m:n Sol. Let X\u2019 OX and YOY\u2019 be the axes.
      \nAnswer:
      \nLet P (x1<\/sub>, y1<\/sub>), Q (x2<\/sub>, y2<\/sub>) be the given points?
      \nLet R (x, y) be the point that divides PQ internally in the ratio m:n.
      \n\"1st
      \nThe \u0394les PRT and RQS are similar
      \n\u2234 \\(\\frac{\\mathrm{PT}}{\\mathrm{RS}}=\\frac{\\mathrm{RT}}{\\mathrm{QS}}=\\frac{\\mathrm{PR}}{\\mathrm{RQ}}\\)
      \n\\(\\frac{\\mathrm{PT}}{\\mathrm{RS}}=\\frac{\\mathrm{PR}}{\\mathrm{RQ}}\\)
      \n\u21d2 \\(\\frac{\\mathrm{PT}}{\\mathrm{RS}}=\\frac{\\mathrm{m}}{\\mathrm{n}}\\) (from (1))
      \n\u21d2 \\(\\frac{\\mathrm{ML}}{\\mathrm{LN}}=\\frac{\\mathrm{m}}{\\mathrm{n}}\\)
      \n\u21d2 \\(\\frac{\\mathrm{OL}-\\mathrm{OM}}{\\mathrm{ON}-\\mathrm{OL}}=\\frac{\\mathrm{m}}{\\mathrm{n}}\\)
      \n\\(\\frac{x-x_{1}}{x_{2}-x}=\\frac{m}{n}\\)<\/p>\n

      n (x – x1<\/sub>) = m (x2<\/sub> – x)
      \n\u21d2 nx – nx1<\/sub> = mx2<\/sub> – mx
      \n\u21d2 mx – nx = mx2<\/sub> – nx1<\/sub>
      \n\u21d2 (m + n) x = mx2<\/sub> – nx1<\/sub>
      \n\u21d2 x = \\(\\frac{m x_{2}+n x_{1}}{m+n}\\)
      \n\\(\\frac{\\mathrm{RT}}{\\mathrm{QS}}=\\frac{\\mathrm{PR}}{\\mathrm{RQ}}\\)
      \n\u21d2 \\(\\frac{\\mathrm{RT}}{\\mathrm{QS}}=\\frac{\\mathrm{m}}{\\mathrm{n}}\\)
      \n\u21d2 \\(\\frac{\\mathrm{RL}-\\mathrm{TL}}{\\mathrm{QN}-\\mathrm{SN}}=\\frac{\\mathrm{m}}{\\mathrm{n}}\\)
      \n\u21d2 \\(\\frac{\\mathrm{RL}-\\mathrm{PM}}{\\mathrm{QN}-\\mathrm{RL}}=\\frac{\\mathrm{m}}{\\mathrm{n}}\\)
      \n\u21d2 \\(\\frac{y-y_{1}}{y_{2}-y}=\\frac{m}{n}\\)
      \n\u21d2 ny- ny1<\/sub> = my2<\/sub> – my
      \n\u21d2 my – ny = my2<\/sub> – ny1<\/sub>
      \n\u21d2 (m – n)y = my2<\/sub> – ny1<\/sub>
      \n\u21d2 y = \\(\\frac{\\mathrm{my}_{2}+\\mathrm{ny}_{1}}{\\mathrm{~m}+\\mathrm{n}}\\)
      \nThe coordinates of R are
      \n(x, y) = \\(\\left(\\frac{m y_{2}+n x_{1}}{m+n}, \\frac{m y_{2}+n y_{1}}{m+n}\\right)\\)
      \nThis is called the section formula for the internal point of division.<\/p>\n

      Question 47.
      \nFind the equation of the locus of the point which moves such that the ratio of its distances (2, -3) and (4, -2) is 2 : 3.
      \nAnswer:
      \nA = (2,-3), B = (4,-2)
      \nLet P(x, y) be any point on the locus
      \n\\(\\frac{\\mathrm{PA}}{\\mathrm{PB}}=\\frac{2}{3}\\) \u21d2 3PA = 2PB<\/p>\n

      3\\(\\sqrt{(x-2)^{2}+(y+3)^{2}}\\) = 2\\(\\sqrt{(x-4)^{2}+(y+2)^{2}}\\)
      \n9[(x – 2)2<\/sup> + (y + 3)2<\/sup>] = 4[(x – 4)2<\/sup> + (y + 2)2<\/sup>]
      \n9[x2<\/sup> – 4x + 4 + y2<\/sup> + 6y + 9] = 4[x2<\/sup> – 8x + 16 + y2<\/sup> + 4y + 4]
      \n9[x2<\/sup> – y2<\/sup> – 4x + 6y + 13] = 4[x2<\/sup> + y2<\/sup> – 8x + 4y + 20]
      \n9x2<\/sup> + 9y2<\/sup> – 36x + 54y + 117 = 4x2<\/sup> + 4y2<\/sup> – 32x + 16y + 80
      \n9x2<\/sup> – 4x2<\/sup> + 9y2<\/sup> – 4y2<\/sup> \u2014 36x + 32x + 54y – 16y + 117 – 80 = 0
      \n5x2<\/sup> + 5y2<\/sup> – 4x + 38y + 37 = 0<\/p>\n

      Question 48.
      \nFind the equation of the line which passes through the point of intersection of the lines x – 2y + 4 = O and 4x – 3y + 1 = 0 and is inclined at an angle of 135\u00b0 with the x-axis.
      \nAnswer:
      \nx – 2y + 4 = 0 \u00d7 1
      \n4x – 3y + 1 = 0 \u00d7 1
      \n\"1st
      \n5y = 15
      \ny = 3
      \nx – 2y + 4 = 0
      \nx – 2 = 0<\/p>\n

      x – 6 + 4=O
      \nx = 2<\/p>\n

      \u2234 The point of intersection = (2, 3)
      \n\u03b8 = 135\u00b0
      \nm = tan\u03b8 = tan 135 = tan(180 – 45) = -tan45\u00b0= -1<\/p>\n

      \u2234 Equation of the required line \u00a1s
      \ny – y1<\/sub> = m(x – x1<\/sub>) \u21d2 y – 3 = -1(x – 2)
      \ny – 3 = -x + 2
      \nx + y – 5 = 0<\/p>\n

      PART – E<\/span><\/p>\n

      IV. Answer any One Question. (1 \u00d7 10 = 10)<\/span><\/p>\n

      Question 49.
      \n(a) Find the domain and Range of the function.
      \nf(x) = \\(\\frac{x^{2}+2 x+1}{x^{2}-8 x-12}\\) where x \u2208 R
      \n(b) What is the future value of 1000 deposited annually years for 12 gathering compound interest at 16%?
      \n(c) Form the cubic equation whose roots are 3, 5 and 7
      \nAnswer:
      \n(a) f(1) = \\(\\frac{4}{-19}=\\frac{4}{19}\\)
      \nf(3) = \\(\\frac{9+6+1}{9-24+12}\\)
      \n= \\(\\frac{16}{-3}\\)
      \nf(5) = \\(\\frac{5^{2}+10+1}{5^{2}-8(5)+12}\\)
      \n= \\(\\frac{36}{-3}\\) = – 12
      \nDomain = { 1, 3, 5 ………….}
      \nRange = {-4\/19, -16\/3, -12 …………..}<\/p>\n

      (b) A = 1000, n = 12, r = 0.16
      \nF = \\(\\frac{\\mathrm{A}\\left[(1+r)^{n}-\\right]}{r}=\\frac{100\\left[(1+0.16)^{12}-1\\right]}{0.16}=\\frac{1000\\left[(1.16)^{12}-1\\right]}{0.16}\\) = 30.850.16<\/p>\n

      (c) (x – 3)(x – 5)(x – 7) =0 \u21d2 (x – 3)(x2 <\/sup>– 5x – 7x + 35) =0
      \n\u21d2 (x – 3)(x2<\/sup> – 12x + 35) 0 \u21d2 x3<\/sup> – 12x2<\/sup> + 35x – 3x2<\/sup> + 36x – 105 =0
      \nx3<\/sup> – 15x2<\/sup> + 71x – 105 = 0<\/p>\n

      Question 50.
      \n(a) Find the value of x
      \nIf \\(\\frac{x \\cdot \\sin ^{2} 300 \\cdot \\sec ^{2} 240}{\\cos ^{2} 225 \\cdot {cosec}^{2} 240}\\) = cot2 315\u00b0. tan 300
      \n(b) Find the equation of the line perpendicular to 3x – 2y + 1 = O and passing through the point (1, -2)
      \n(c) Insert two geometric means between 1 and 1\/\/27
      \nAnswer:
      \nsin 300 = sin (360 – 60) = – sin 60 = \\(\\frac{\\sqrt{3}}{2}\\)
      \nsec 240 = sec(180 + 60)= – sec60\u00b0= – 2
      \ncos 225 = cos(180 + 45\u00b0) – cos45\u00b0= – \\(\\frac{1}{\\sqrt{2}}\\)
      \ncosec 240 = cosec (180 + 60\u00b0) = – cosec 60 = – \\(\\frac{2}{\\sqrt{3}}\\)
      \ncot 315\u00b0 = cot (360 – 45\u00b0) = – cot 45\u00b0 = – 1
      \ntan 300 = tan (360 – 60\u00b0) = – tan 600 = – \\(\\sqrt{3}\\)<\/p>\n

      \u2234 The given equation becomes:
      \n\\(\\frac{x \\cdot \\frac{3}{4} \\cdot 4}{\\frac{1}{2}, \\frac{4}{3}}\\) = 1.3
      \n\u21d2 x = \\(\\frac{6}{9}=\\frac{2}{3}\\)<\/p>\n

      (b) Slope of the given line 3x – 2y + 1 = 0 is \\(\\frac{3}{2}\\) = m1
      \nSince the two lines are perpendicular m1<\/sub> m2<\/sub> = – 1
      \n\u2234 m2<\/sub> = \\(\\frac{-1}{m_{1}}=\\frac{-1}{3 \/ 2}=\\frac{-2}{3}\\)<\/p>\n

      Thus the equation of the ime vith slope m = \\(\\) and passing through
      \n(x1<\/sub> y1<\/sub>) (1, -2) is given by
      \n\u21d2 y -y1<\/sub> = m(x – x1<\/sub>)
      \n\u21d2 y – 2 = (x – 1)
      \n\u21d2 3y – 6 = – 2x + 2
      \n\u21d2 2x + 3y + 4 = 0<\/p>\n

      (c) 1, g, g2<\/sub>, \\(\\frac{1}{27}\\)
      \na = 1, Tn<\/sub> = \\(\\frac{1}{27}\\), r =?, n = 4
      \nTn<\/sub> = arn-1<\/sup> \u21d2 \\(\\frac{1}{27}\\) = 1.r3<\/sup>
      \n\u21d2 \\(\\left(\\frac{1}{3}\\right)^{3}\\) = r3<\/sup>
      \n\u21d2 r = \\(\\frac{1}{3}\\)
      \n\u2234 1, 1\/3, 1\/9, 1\/21
      \n\u2234 Geometric means are 1\/3, 1\/9<\/p>\n","protected":false},"excerpt":{"rendered":"

      Students can Download 1st PUC Basic Maths Model Question Paper 1 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 1 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\/89174"}],"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=89174"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/89174\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=89174"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=89174"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=89174"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}