{"id":89665,"date":"2021-07-16T16:28:37","date_gmt":"2021-07-16T10:58:37","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=89665"},"modified":"2021-07-16T16:28:37","modified_gmt":"2021-07-16T10:58:37","slug":"1st-puc-basic-maths-model-question-paper-8","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-basic-maths-model-question-paper-8\/","title":{"rendered":"1st PUC Basic Maths Model Question Paper 8 with Answers"},"content":{"rendered":"

Students can Download 1st PUC Basic Maths Model Question Paper 8 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 8 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 real part of 2 + 3i.
    \nAnswer:
    \n2.<\/p>\n

    Question 2.
    \nIf A = {a, b, c, d}, B = {a, b, c} find A n B.
    \nAnswer:
    \nA\u2229B = {a, b, c}.<\/p>\n

    Question 3.
    \nSimplify : (3\u00b0)2<\/sup> + (32<\/sup>)\u00b0.
    \nAnswer:
    \n(1)2<\/sup> + (9)\u00b0= 1 + 1= 2.<\/p>\n

    Question 4.
    \nFind the value of log10<\/sub> 1000
    \nAnswer:
    \nlog10<\/sub> 1000 = log10 <\/sub>103<\/sup> = 3 log10<\/sub> 10 = 3(1) = 3<\/p>\n

    Question 5.
    \nFind the 8th term of the A.P. 2, 4, 6………….
    \nAnswer:
    \na = 2, d = 2 ,n = 8, T8 = ?
    \nTn<\/sub> = a + (n – 1)d
    \n= 2 + (8 – 1) (2) = 2 + 7 (2) = 2 + 14 = 16<\/p>\n

    \"KSEEB<\/p>\n

    Question 6.
    \nSolve for x : 2 (- 8 + x) = 28 – 2x
    \nAnswer:
    \n– 16 + 2x = 28 – 2x
    \n2x + 2x = 28 + 16
    \n4x = 44 \u21d2 x = 11<\/p>\n

    Question 7.
    \nFind the S. I. on Rs. 500 at 5% for 5 years.
    \nAnswer:
    \nSI = \\(\\frac{500 \\times 5 \\times 5}{100}\\) = 25 \u00d7 5 = 125<\/p>\n

    Question 8.
    \nDefine perpetuity.
    \nAnswer:
    \nIf the installments (annuities) are made for an infinite period them it is called a perpetuity.<\/p>\n

    Question 9.
    \nConvert 0.25 into persentage.
    \nAnswer:
    \n0.25 = 0.25 \u00d7 100% = \\(\\frac{25}{100}\\) \u00d7 100% = 25%.<\/p>\n

    Question 10.
    \nConvent 45\u00b0 into radians.
    \nAnswer:
    \n45\u00b0 =45 \u00d7 \\(\\frac{\\pi}{180}=\\frac{\\pi}{4}\\)<\/p>\n

    Question 11.
    \nFind the value of Sin2<\/sup>60\u00b0 + Cos2<\/sup>60\u00b0.
    \nAnswer:
    \nSin2<\/sup> 60 +Cos2<\/sup> 60 = 1.<\/p>\n

    \"KSEEB<\/p>\n

    Question 12.
    \nFind the locus of a point that moves such that its distance x-axis is 5.
    \nAnswer:
    \ny = 5<\/p>\n

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

    II. Answer any TEN of the following questions: (10 x 2 = 20)<\/span><\/p>\n

    Question 13.
    \nFind the number of positive divisors of 360
    \nAnswer:
    \n360 = 2 \u00d7 32<\/sup> \u00d7 51<\/sup>
    \n= p1<\/sub>\u03b11<\/sub><\/sup> \u00d7 p2<\/sub>\u03b12<\/sub><\/sup> \u00d7 p3<\/sub>\u03b13<\/sub><\/sup>
    \nT(a) = (1 + \u03b11<\/sub>)(1 + \u03b12<\/sub>)(1 + \u03b13<\/sub>)
    \nT(360) =(1 + 3)(1 + 2)(1 + 1)
    \n= (4)(3)(2) = 12(2) = 24.
    \n\"1st<\/p>\n

    Question 14.
    \nFind the L.C.M. of 9, 15 and 24.
    \nAnswer:
    \n9 = 32
    \n15 = 3 \u00d7 5
    \n24 = 23<\/sup> \u00d7 31<\/sup>
    \nLCM = 23<\/sup> \u00d7 32<\/sup> \u00d7 51<\/sup> = 8 \u00d7 9 \u00d7 5 = 360.<\/p>\n

    Question 15.
    \nIf u= {1, 2, 3, ……….. 9}, A = {1, 2, 3, 4, 5}, B = {3, 4, 5, 6, 7} then verify (A \u222a B)’ = A’ \u2229 B’
    \nAnswer:
    \nA\u222aB = {1, 2, 3, 4, 5, 6,7} (A\u222aB)’ = {8,9}
    \nA’ = {6, 7, 8, 9} B’ = {0, 2, 8, 9} A’\u2229B’ = {8,9}
    \n\u2234 A (A\u222aB)’= A’\u2229B’<\/p>\n

    Question 16.
    \nIf K + 9, – 6 and 4 are in GP., then find the value of K.
    \nAnswer:
    \n(- 6)2 = (K + 9)4
    \n36 = 4K + 36
    \n\u21d2 4K = 0
    \nK = 0.<\/p>\n

    \"KSEEB<\/p>\n

    Question 17.
    \nIf \u03b1 and \u03b2 are the roots of the equation x2<\/sup> + 5x + 6 = 0. Then find the value of \\(\\frac{\\alpha}{\\beta}+\\frac{\\beta}{\\alpha}\\)
    \nAnswer:
    \n\u03b1 + \u03b2 = 5 \u03b1\u03b2 = 6
    \n\\(\\frac{\\alpha}{\\beta}+\\frac{\\beta}{\\alpha}=\\frac{\\alpha^{2}+\\beta^{2}}{\\alpha \\beta}=\\frac{(\\alpha+\\beta)^{2}-2 \\alpha \\beta}{\\alpha \\beta}=\\frac{25-12}{6}=\\frac{13}{6}\\)<\/p>\n

    Question 18.
    \nFind two numbers whose sum is 64 and the difference is 16.
    \nAnswer:
    \nx + y = 64
    \nx – y = 16
    \n2x + 80
    \n\u21d2 x = 40
    \nx + y – 64
    \n40 – y = 64
    \ny = 64 – 40 = 24
    \ny = 24<\/p>\n

    Question 19.
    \nSove the following linear inequality: 5x – 3 < 3x + 1, x \u2208 real numbers. Also represent it on a number line.
    \nAnswer:
    \n5x – 3 < 3x + 1
    \n5 x – 3x < 3 + 1
    \n2x < 4 \u21d2 x < 2 or x \u2208 (-\u221e, 2)
    \n\"1st<\/p>\n

    Question 20.
    \nThe average score of 40 boys is 60% and an average of 60 girls is 70%. Find the combined average of boys and girls.
    \nAnswer:
    \nN1<\/sub> = 40 X1<\/sub> = 60% N2<\/sub> = 60 X2<\/sub> = 70%
    \nX\u030412<\/sub> = \\(\\frac{\\mathrm{X}_{1} \\mathrm{~N}_{1}+\\mathrm{X}_{2} \\mathrm{~N}_{2}}{\\mathrm{~N}_{1}+\\mathrm{N}_{2}}=\\frac{60(40)+70(60)}{40+60}\\)
    \n\\(\\frac{2400+4200}{100}=\\frac{6600}{100}\\)
    \n= 66<\/p>\n

    Question 21.
    \nFind the value of: Sin2<\/sup> \\(\\frac{5 \\pi}{6}\\) + cos2<\/sup> \\(\\frac{5 \\pi}{6}\\) – tan2<\/sup>\\(\\frac{ \\pi}{4}\\)
    \nAnswer:
    \nSin 5\u03c0\/6 = Sin 150 = Sin (180-30) = Sin 30 = 1\/2
    \nCos 5\u03c0\/6 = Cos 150 = Cos (180-30) = -Cos 30 = –\\(\\)
    \ntan \u03c0\/4 = tan 45 = 1
    \nLHS = (1\/2)2<\/sup> – (-3\/2)2<\/sup> – 12<\/sup>
    \n= 1\/4 + 1\/3 – 1 = \\(\\frac{1+3-4}{4}=\\frac{0}{4}\\) = 0<\/p>\n

    Question 22.
    \nFind the value of x, if the distance between (x, 3) and (4, 5) is 5 units.
    \nAnswer:
    \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
    \n= \\(\\frac{8 \\pm \\sqrt{64-4(1)(-5)}}{2(1)}=\\frac{8 \\pm \\sqrt{84}}{2}\\)
    \n= \\(\\frac{8 \\pm 2 \\sqrt{21}}{2}\\)
    \nx = 4 \u00b1 \\(\\sqrt{2}\\).<\/p>\n

    \"KSEEB<\/p>\n

    Question 23.
    \nFind the locus of a point that moves such that it is equidistant from (5, 0) and (-5, 0).
    \nAnswer:
    \nP = (x, y) A = (5, 0) B = (- 5. 0)
    \nPA = PB
    \nPA2<\/sup> = PB2<\/sup>
    \n(x – 5)2<\/sup> + y2<\/sup> = (x + 5)2<\/sup> + y2<\/sup>
    \n\"1st
    \n20 x = 0 \u21d2 x = i.e., y – axis<\/p>\n

    Question 24.
    \nFind the equation of a straight line passing through (3, 4) and having a slope of 3.
    \nAnswer:
    \ny – 4 = 3 (x – 3)
    \ny – 4 = 3x – 9 \u21d2 3x – y – 5 = 0.<\/p>\n

    Question 25.
    \nDerive the equation of the straight line passing through two points (x1<\/sub>, y1<\/sub>) and (x2<\/sub>, y2<\/sub>).
    \nAnswer:
    \nLet P (x, y) be any point an the line joining points A (x1<\/sub>, y1<\/sub>) and B(x2<\/sub>, y2<\/sub>)
    \n\u2234 APB are collinear
    \nThe slope of AP = Slope of AB
    \n\"1st
    \n\\(\\frac{y-y_{1}}{x-x_{1}}=\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\) (Two Point form)<\/p>\n

    PART – C<\/span><\/p>\n

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

    Question 26.
    \nProve that \\(\\sqrt{2}\\) is not a rational number.
    \nAnswer:
    \nWe shall prove it by the method ofcontradiction.
    \nIf possible Let \\(\\sqrt{2}\\) be a rational number
    \nLet \\(\\sqrt{2}\\) = \\(\\frac{p}{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{p}{q}\\) \u21d2 \\(\\sqrt{2}\\)q = p
    \n\u21d2 2q2<\/sup> = p2<\/sup>
    \n\u21d2 2 divides p2<\/sup> \u21d2 2 divides p
    \n\u21d2 p is even
    \nLet p = 2k where k is an integer p2<\/sup> = 4k2<\/sup>
    \n2q2<\/sup> = 4k2<\/sup>
    \nq2<\/sup> = 2k2<\/sup> \u21d2 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
    \ncontradiction 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.
    \nIf R-1 = {(2, 4), (1, 2), (3, 1), (3, 2)}, find R. Also find its domain and range.
    \nAnswer:
    \nR = {(4, 2), (2, 1), (1, 3), (2, 3)}
    \nDomain of R= {1, 2, 4}
    \nRange of R= {1, 2, 3}.<\/p>\n

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

    Question 29.
    \nProve that \\(\\frac{1}{\\log _{2} 4}+\\frac{1}{\\log _{8} 4}+\\frac{1}{\\log _{16} 4}\\) = 4.
    \nAnswer:
    \nLHS = \\(\\frac{1}{\\log _{2} 4}+\\frac{1}{\\log _{8} 4}+\\frac{1}{\\log _{16} 4}\\) = log 42+ log4 8+ log4 16
    \n= log4 2. 8 .16 = log4 256 = log4 44 = 4 log4 4 = 4(1) = 4 = RHS.<\/p>\n

    \"KSEEB<\/p>\n

    Question 30.
    \nThe sum of three numbers in an A.P. is 15 and their product is 105. Find the numbers.
    \nAnswer:
    \nLet the numbers be a – b, a, a + d
    \n(a-d) + a + (a + d)= 15
    \n3a= 15
    \na = 15\/3 = 5
    \n(a – d) a (a + d) = 105 .
    \n(5 – d)5(5 + d) = 105
    \n(5 – d)(5 + d) = \\(\\frac{105}{5}\\)
    \n25 – d2<\/sup> = 21
    \n25 – 21 = d2 <\/sup>\u21d2 d2<\/sup> = 4
    \n\u21d2 d = \u00b12
    \nLet d = 2<\/p>\n

    \u2234 The numbers are
    \n5-2, 5, 5 + 2
    \n3, 5, 7.<\/p>\n

    Question 31.
    \nIf sum of the digits of a two digit numbers is 9 and if 9 is added to the number, then the digits get reversed.
    \nAnswer:
    \nLet the number be 10x + y, x + y = 9
    \n10x + y + 9 = 10y + x
    \n9x – 9
    \ny = – 9
    \n\"1st
    \nx = 4
    \ny = 5
    \n\u2234 The number in 10 x +y = 10 (4) + 5 = 40 + 5 = 45.<\/p>\n

    Question 32.
    \nIn what time, a sum of Rs. 500 will earn Rs. 975 at the rate of 6% p.a. if C. 1 is payable half yearly
    \nAnswer:
    \nP = 500 CI = 975
    \nA = P + CI = 500 + 975 = 1475
    \nR = 6% = 0. 06 (Nominal rate of interst)
    \n9 = 2
    \nr = \\(\\left(1+\\frac{R}{q}\\right)^{q}\\) – 1
    \n= \\(\\left(1+\\frac{0.06}{2}\\right)^{2}\\) – 1
    \n= 0. 0609 = 6. 09 %
    \nr= 0. 0609
    \nA + P (1 + r)n<\/sup>
    \nlog A= log P + n log (1 + r)<\/p>\n

    n = \\(\\frac{\\log \\mathrm{A}-\\log \\mathrm{P}}{\\log (1+r)}\\)
    \n= \\(\\frac{\\log 1475-\\log 500}{\\log 1.0609}\\)
    \n= \\(\\frac{3.1688-2.6990}{0.0253}\\)
    \nn = 18. 56 years.<\/p>\n

    Question 33.
    \nBy how much should the use of tea be decreased if the Price of tea is increased by 25% so that the expenditure remains unchanged.
    \nAnswer:
    \nLet the Price be 100 Let the quantity be 100
    \nTotal expenditure = 100 \u00d7 100 = 10,000
    \nNew Price = 100 + 25 = 125
    \nLet the new quantity be y
    \nTotal Expenditure = 125 y
    \n\u2234 125 y = 1000
    \ny = 80 \u2234 Decrease on quantity = 100 – 80 = 20
    \n% decrease = \\(\\frac{\\text { Decrease }}{\\text { Original Value }}\\) \u00d7 100%
    \n= \\(\\frac{20}{100}\\) x 100%
    \n= 20%.<\/p>\n

    Question 34.
    \nFind the B.E.P in units if total cost and the revenue functions are C(x) = 450 + 1. 5JC and R(x) = 3x. (x = level of out put).
    \nAnswer:
    \nAt Break Even point,
    \nTotal revenue = Total cost
    \n\u2234 R (x) = C(x)
    \n3x = 450 + 1.5x
    \n3x – 1.5x = 450
    \n1.5x = 450
    \n\u2234 x = \\(\\frac{450}{1.5}\\)
    \nx = 300 units.<\/p>\n

    Question 35.
    \nIf cot A = \\(\\frac{5}{12}\\) and A is acute show that 2 cosec A – 4SecA = – \\(\\frac{247}{30}\\)
    \nAnswer:
    \ncot A = \\(\\frac{5}{12}\\)
    \ncosec A = \\(\\frac{13}{12}\\)
    \nsec A = \\(\\frac{13}{5}\\)<\/p>\n

    2 cosec A – 4SecA = 2.\\(\\frac{13}{12}\\) – 4.\\(\\frac{13}{5}=\\frac{26}{12}-\\frac{52}{5}=\\frac{130-624}{60}=\\frac{-494}{60}=-\\frac{247}{30}\\)<\/p>\n

    \"KSEEB<\/p>\n

    Question 36.
    \nIf A, B, C are angles of a triangle then prove that.
    \n(i) Sin (A + B) = SinC
    \n(ii) Cos \\(\\left(\\frac{\\mathbf{A}+\\mathbf{B}}{2}\\right)\\) = Sin\\(\\frac{C}{2}\\)
    \nAnswer:
    \n(i) Sin (A + B) = Sin (180 – C) = Sinc
    \n(ii) Cos\\(\\left(\\frac{\\mathbf{A}+\\mathbf{B}}{2}\\right)\\) = Cos\\(\\left(\\frac{180-C}{2}\\right)\\) = Cos(90 – \\(\\frac{C}{2}\\))
    \n= Sin\\(\\frac{C}{2}\\)<\/p>\n

    Question 37.
    \nShow that the points A(2, 2), B(6, 3) and C (4, 11) form a right angled triangle.
    \nAnswer:
    \nAB = \\(\\sqrt{(6-2)^{2}+(3-2)^{2}}=\\sqrt{16+1}=\\sqrt{17}\\)
    \nBC = \\(\\sqrt{(4-6)^{2}+(11-3)^{2}}=\\sqrt{4+64}=\\sqrt{68}\\)
    \nCA = \\(\\sqrt{(4-2)^{2}+(11-2)^{2}}=\\sqrt{4+81}=\\sqrt{85}\\)
    \nAB2<\/sup> = 17 BC2<\/sup> =68 CA2<\/sup> =85
    \n\u2234 AB2<\/sup> + BC2<\/sup> = CA2<\/sup>
    \n\u2234 ABC form a right angled triangle<\/p>\n

    Question 38.
    \nFind the distance between the parallel lines : 3x + 4y + 5 = 0 and 6x + 8y + 20 = 0.
    \nAnswer:
    \n6x + 8y + 10 = 0
    \n6x + 8y + 20 = 0.
    \nd = \\(\\left|\\frac{\\mathrm{C}_{1}-\\mathrm{C}_{2}}{\\sqrt{a_{2}+6_{2}}}\\right|=\\left|\\frac{10-20}{\\sqrt{6^{2}+8^{2}}}\\right|=\\left|\\frac{-10}{10}\\right|\\) = |-1| = 1 units<\/p>\n

    PART – D<\/span><\/p>\n

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

    Question 39.
    \nOut of 50 people, 20 drink tea and 10 take both tea and coffee. How many take at least one of the two drinks. Also, represent the above problem by Venn diagram.
    \nAnswer:
    \n\"1st
    \nn(T\u222aC) = n(T) + n(C) – n(T\u2229C)
    \n50 = 20 + n(C) – 10
    \n40 = n(C)
    \nNumber of people taking atleast one of the two drinks 10 + 10 + 30 = 40 + 10 = 50.<\/p>\n

    Question 40.
    \nFind the value of the following using logarithmic table. \\(\\frac{213.781 \\times 7.434}{6.321}\\)
    \nAnswer:
    \nLet x = \\(\\frac{213.781 \\times 7.434}{6.321}\\)
    \nlogx = log\\(\\left[\\frac{213.781 \\times 7.434}{6.321}\\right]\\)
    \n= log213.781 + log 7.434 – log 6.321
    \n= 2.3298 + 0.8712 – 0.8008.
    \n= 2.4002
    \nx = Antilog 2.4002
    \nx = 251.3.<\/p>\n

    Question 41.
    \nFind the sum of all the numbers betw een 50 and 200 which are divisible by 11.
    \nAnswer:
    \nS = 55 + 66 + 77 +……………….+ 198
    \n55, 66, 77, …………….198
    \nHere a = 55, l = Tn<\/sub> = 198, d= 11
    \nNow, Tn<\/sub> = a + (n – 1) d
    \n198 = 55 + (n – 1) 11
    \n198 – 55 = (n – 1) 11
    \n2\\(\\frac{143}{11}\\) = n – 1
    \nn – 1 = 13
    \nn= 13 + 1
    \nn= 14
    \nNow consider:
    \nSn<\/sub> = \\(\\frac{n}{2}\\)(a + l)
    \nS14<\/sub> = \\(\\frac{14}{2}\\)(55 + 198)
    \n= 7(253)
    \n= 1,771.<\/p>\n

    Question 42.
    \nFind the integral roots between – 3 and + 3 by inspection and then using synthetic division, solve the equation.
    \nx3<\/sup> – 2x2<\/sup> – 5x + 6 = 0.
    \nAnswer:
    \nLet f (x) = x3<\/sup> – 2x2<\/sup> – 5x + 6
    \nf(1)= 1 – 2 – 5 + 6 = 7 – 7 = 0
    \n\u2234 x = 1 is a root of the given equation by using Synthetic division.
    \n\"1st
    \nThe resulting equation is x2<\/sup> – x – 6 = 0 is the quotient and remainder = 0
    \nx2<\/sup> – x – 6 = 0
    \n(x – 3) (x + 2) = 0
    \nx = 3 or – 2
    \nThus x = 1, -2, 3 are the roots of the given equation.<\/p>\n

    \"KSEEB<\/p>\n

    Question 43.
    \nFind the difference between S. 1. and C. 1. on Rs. 18000 invested for four years at 8% p.a. where C. 1. is compounded annually.
    \nAnswer:
    \nSimple Interest .
    \nP= 18,000
    \nT = 4
    \nR = 4
    \nSI = \\(\\frac{\\text { PTR }}{100}\\)
    \n= \\(\\frac{18,000 \\times 4 \\times 8}{100}\\)
    \n= \u20b9 5,760<\/p>\n

    Compound Interest
    \nP = 18,000
    \nn = 4
    \nI = 0.08
    \nA = P(1 + i)n<\/sup>
    \n= 18,000(1 +0.08)4<\/sup>
    \n= 24,488.8
    \nCI = A – P
    \n= 24,488.8 – 18,000
    \nCI = 6488.8<\/p>\n

    Difference between Cl and SI is
    \nCI – SI = 6488.8 – 5760
    \n= 728.8.<\/p>\n

    Question 44.
    \nFind the present value of an annuity of Rs. 2500 payable at the end of 6 months for 5 years, if money is worth 10% computed semi-annually.
    \nAnswer:
    \na = 2500
    \ni = \\(\\frac{0.10}{2}\\) = 0.05
    \nn = 5 x 2 = 10
    \nP = \\(\\frac{a\\left[(1+i)^{n}-1\\right]}{i(1+i)^{n}}\\)
    \n= 2500\\(\\frac{\\left[(1+0.05)^{10}-1\\right]}{0.05(1+0.05)^{10}}\\)
    \n= \\(\\frac{2500[1.6289-1]}{0.05 \\times 1.6289}\\)
    \n= \u20b9 19304.44.<\/p>\n

    Question 45.
    \nA watch is sold for Rs. 150 at a profit of 25%. At what price should it be sold in order to have 50% profit.
    \nAnswer:
    \n\"1st<\/p>\n

    Question 46.
    \nProve that: \\(\\frac{\\tan A}{1-\\cot A}+\\frac{\\cot A}{1-\\tan A}\\) = 1 + Sec A. Cosec A.
    \nAnswer:
    \n\"1st
    \n= sec A . cosec A + 1 = R.H.S.<\/p>\n

    Question 47.
    \nShow that the following points are the vertices of a square A (1, 1), B (4, 1),C(4, 4) and D (1, 4)
    \nAnswer:
    \nAB = \\(\\sqrt{(4-1)^{2}+(1-1)^{2}}=\\sqrt{9+0}\\) = 3
    \nBC = \\(\\sqrt{(4-4)^{2}+(4-1)^{2}}=\\sqrt{0+9}\\) = 3
    \nCD = \\(\\sqrt{(1-4)^{2}+(4-4)^{2}}=\\sqrt{9+0}\\) = 3
    \nDA = \\(\\sqrt{(1-1)^{2}+(4-1)^{2}}=\\sqrt{0+9}\\) = 3
    \nAB = BC = CD + AD
    \nSol.
    \nAC = \\(\\sqrt{(4-1)^{2}+(4-1)^{2}}=\\sqrt{9+9}=\\sqrt{18}\\)
    \nBD = \\(\\sqrt{(1-4)^{2}+(4-1)^{2}}=\\sqrt{9+9}=\\sqrt{18}\\)
    \n\u2234 AC = BD
    \nABCD \u2206 is a Square.<\/p>\n

    \"KSEEB<\/p>\n

    Question 48.
    \nFind the area of a quadrilateral whose vertices are A(1, 1), B (4, 1), C (4, 4), and D (1, 4).
    \nAnswer:
    \nThe required area ABCD = area of the triangle ABC + area of triangle ACD
    \nArea of the triangle ABC = \\(\\frac{1}{2}\\)[1(-3 – 2) + 7(2 -1) + 12(1 + 3)]
    \n= 25 sq. units
    \n\"1st
    \nSimilarly
    \nArea of the triangle ACD =107 sq.units
    \nArea of the triangle ABCD = 25 + 107 = 132 sq.units.<\/p>\n

    PART – E<\/span><\/p>\n

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

    Question 49.
    \n(a) If f(x) = x + 1 and g(x) = x2<\/sup> + 1.
    \nFind (i) fog (1), (ii) fog (2), (iii) gof (1), (iv) gof (2), (v) fog (3)
    \nAnswer:
    \nf(x) = x+ 1 and g (x) = x2<\/sup> + 1
    \n(i) fog (1) = f (g (1))
    \n= f(2)
    \n= 2 + 1 = 3<\/p>\n

    (ii) fog (2) = f (g(2))
    \n= f(5)
    \n= 5 + 1
    \n= 6<\/p>\n

    (iii) gof(1)
    \n= g(f(1))
    \n= g(2)
    \n= 4 + 1
    \n= 5<\/p>\n

    (iv) gof (2)
    \n= g(f(2))
    \n=g(3)
    \n= 9 + 1
    \n= 10<\/p>\n

    (v) fog (3) f(g(3))
    \nf (10) = 10 + 1
    \n= 11.<\/p>\n

    (b) Find the sum to n terms of series
    \n7 + 77 + 777 +………………n term.
    \nAnswer:
    \nLet S = 7 + 77 + 777 + ………….. n terms
    \n= 7 [1 + 11 + 111 +……………….to nterms] = \\(\\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}\\)[ [(10 + 102<\/sup> + 103<\/sup> + ……………… to n terms) – (1 + 1 + 1 ………………. to n terms] 9
    \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 in GP where a = 10, r = 10]
    \nSn<\/sub> = \\(\\frac{7}{9}\\left[\\frac{10\\left(10^{n}-1\\right)}{9}-n\\right]\\)<\/p>\n

    (c) Find the number of digits in 320.
    \nAnswer:
    \n(a) 320
    \nLet x = 320
    \nlog x = log 320 = 20 log 3 = 20 (0.4771) = 9.542
    \nAs the characterstic is 9, number of digits = 9 + 1 = 10.<\/p>\n

    \"KSEEB<\/p>\n

    Question 50.
    \n(a) Solve the following inequalities graphically
    \nx + 2y \u2264 8
    \n2x + y \u2264 8
    \nx \u2265 0
    \ny \u2265 0
    \nAnswer:
    \n\"1st<\/p>\n

    (b) Find the future value of an annuity of Rs. 500 at 5% p.a. payable for 5 years.
    \nAnswer:
    \nA = 500 r = 5% =0.05 n = 5
    \nF = \\(\\frac{\\mathrm{A}\\left[(1+r)^{n}-1\\right]}{r}\\)
    \n= \\(\\frac{500\\left[(1+0.005)^{5}-1\\right]}{0.05}\\)
    \n\\(\\frac{500\\left[(1 \\cdot 05)^{5}-1\\right]}{0 \\cdot 05}\\) = 2762.81<\/p>\n

    (c) Find the quadratic equation whose roots are 2 and 5.
    \nAnswer:
    \n(x – 2) (x – 5) = 0
    \nx2<\/sup> – 2x – 5x + 10 = 0
    \nx2<\/sup> – 7x + 10 = 0.<\/p>\n","protected":false},"excerpt":{"rendered":"

    Students can Download 1st PUC Basic Maths Model Question Paper 8 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 8 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\/89665"}],"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=89665"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/89665\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=89665"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=89665"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=89665"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}