{"id":89488,"date":"2021-07-12T11:23:12","date_gmt":"2021-07-12T05:53:12","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=89488"},"modified":"2021-07-16T14:55:56","modified_gmt":"2021-07-16T09:25:56","slug":"1st-puc-basic-maths-previous-year-question-paper-march-2018-north","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-basic-maths-previous-year-question-paper-march-2018-north\/","title":{"rendered":"1st PUC Basic Maths Previous Year Question Paper March 2018 (North)"},"content":{"rendered":"

Students can Download 1st PUC Basic Maths Previous Year Question Paper March 2018 (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 2018 (North)<\/h2>\n

Time: 3.15 Hours
\nMax. Marks: 100<\/p>\n

Instructions:<\/p>\n

    \n
  1. The questions paper consists of five parts A, B, C, D, and E.<\/li>\n
  2. Part – A carries 10 marks, Part – B carries 20 marks, Part – C curries 30 marks, Part – D carries 30, and Part – E carries 10 marks.<\/li>\n
  3. Write the question numbers properly as indicated in the questions paper<\/li>\n<\/ol>\n

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

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

    Question 1.
    \nWrite the imaginary part of the complex number -3i + 7
    \nAnswer:
    \n-3.<\/p>\n

    Question 2.
    \nIf (2x + 4, 0) = (8,0) find x.
    \nAnswer:
    \n2x + 4 = 8
    \n2x = 4
    \nx = 2<\/p>\n

    Question 3.
    \nSimplify: \\(\\left(\\frac{9}{4}\\right)^{-3 \/ 2}\\)
    \nAnswer:
    \n\\(\\left[\\left(\\frac{3}{2}\\right)^{2}\\right]^{-3 \/ 2}=\\left(\\frac{3}{2}\\right)^{-3}=\\left(\\frac{2}{3}\\right)^{3}=\\frac{8}{27}\\)<\/p>\n

    Question 4.
    \nExpress 34<\/sup> = 81 in logarithmic form.
    \nAnswer:
    \nlog3<\/sub>81 = 4<\/p>\n

    Question 5.
    \nFind the tenth term of HP \\(\\frac{1}{2}, \\frac{1}{4}, \\frac{1}{6}\\)………..
    \nAnswer:
    \nAP \u2192 2, 4, 6,…. T10<\/sub> = 2 + (n – 1)2 = 2 + 2n – 2 = 2n = 2(10) = 20
    \n\u2234 The tenth term of HP = \\(\\frac{1}{20}\\)<\/p>\n

    \"KSEEB<\/p>\n

    Question 6.
    \nForm the cubic equation whose roots are 3, 5, 7.
    \nAnswer:
    \n(x – 3)(x – 5) (x – 7) = 0
    \n(x – 3)(x2<\/sup> – 5x – 7x + 35) = 0
    \n(x – 3)(x2<\/sup> -12x + 35) = 0
    \nx3<\/sup> – 12x2<\/sup> + 35x – 3x2<\/sup> + 36x – 105 = 0
    \nx3<\/sup> – 15x2<\/sup> + 71x – 105 = 0<\/p>\n

    Question 7.
    \nWhat is the simple interest for 245 days for \u20b9 6000 at 8% p.a simple interest?
    \nAnswer:
    \nSI = 6000 \u00d7 \\(\\frac{245}{365} \\times \\frac{8}{100}\\) = 402.39<\/p>\n

    Question 8.
    \nWhat is the present value of a perpetuity of 5000 at 8% p.a
    \nAnswer:
    \nA = \\(\\frac{A}{r}=\\frac{5000}{0.08}\\) = 62,500<\/p>\n

    Question 9.
    \nIf the cost price of an item is 25000 and the profit is 10,000 find the profit percentage.
    \nAnswer:
    \nCP = 25,000 profit = \u20b9 10,000
    \nProfit% = \\(\\frac{\\text { Profit }}{\\mathrm{CP}}\\) \u00d7 100 = 40%<\/p>\n

    Question 10.
    \nExpress 144\u00b0 \u00a1n radian measure.
    \nAnswer:
    \n\\(\\frac{144 \\times \\pi}{180}=\\frac{4 \\pi}{5}\\)<\/p>\n

    \"KSEEB<\/p>\n

    Question 11.
    \nFind the value of cos 1500
    \nAnswer:
    \ncos150 = cos(180 – 30) = -cos 30 = \\(\\frac{\\sqrt{3}}{2}\\)<\/p>\n

    Question 12.
    \nFind k if the slope of the line joining the points (2,5) and (k, 13) is 20.
    \nAnswer:
    \n\\(\\frac{13-5}{k-2}\\) = 20 \u21d2 \\(\\frac{8}{k-2}\\) = 20 \u21d2 8 = 20k – 40 \u21d2 20k = 48 \u21d2 k = \\(\\frac{48}{20}=\\frac{12}{5}\\)<\/p>\n

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

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

    Question 13.
    \nFind the sum of all positive divisors of 768.
    \nAnswer:
    \n\"1st
    \n768 = 28<\/sup> \u00d7 31<\/sup>
    \n= p1<\/sub>\u03b11<\/sub><\/sup> \u00d7 p2<\/sub>\u03b12<\/sub><\/sup>
    \np1<\/sub> = 2 p2<\/sub> = 3
    \n\u03b11<\/sub> = 8, \u03b12<\/sub> = 1
    \n\"1st<\/p>\n

    Question 14.
    \nFind the least integer which when divided by 36,40 and 48 leaves the same remainder 5.
    \nAnswer:
    \nLCMof 36, 40, 48
    \n\"1st
    \nLCM = 24<\/sup> \u00d7 32<\/sup> \u00d7 51<\/sup> = 16 \u00d7 9 \u00d7 5 = 720
    \n\u2234 Required member = 720 + 5 = 725<\/p>\n

    Question 15.
    \nFind the domain and range of relation R = {(1, 2) (2, 5) (1, 6) (3, 6)}
    \nAnswer:
    \nDomain = {1, 2,3)
    \nRange = {2, 5, 6}<\/p>\n

    \"KSEEB<\/p>\n

    Question 16.
    \nIf ax<\/sup> = by<\/sup> = cz<\/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 17.
    \nIf the second term of the GP is 6 and fifth term is 162 then find the GP.
    \nAnswer:
    \nT2<\/sub> = 6, T5<\/sub> = ar4<\/sup> = 162 ar = 6 ar4<\/sup> = 162
    \nar.r3<\/sup> = 162
    \n6r3<\/sup> = 162
    \nr3<\/sup> = \\(\\frac{162}{6}\\) = 27
    \nr = 3<\/p>\n

    Question 18.
    \nThe sum of three consecutive numbers is 243. Find the three numbers.
    \nAnswer:
    \nx + x + 1 + x + 2 = 243
    \n3x + 3 = 2433x = 240
    \nx = 80
    \n80, 81, 82<\/p>\n

    Question 19.
    \nIf a and P are the roots of the equation 2x2<\/sup> + 4x + 1 = 0 find the value of \\(\\frac{1}{\\alpha}+\\frac{1}{\\beta}\\)
    \nAnswer:
    \n\u03b1 + \u03b2 = –\\(\\frac{4}{2}\\) = -2
    \n\\(\\frac{1}{\\alpha}+\\frac{1}{\\beta}=\\frac{\\alpha+\\beta}{\\alpha \\beta}\\)
    \n\u03b1\u03b2 = \\(\\frac{c}{a}=\\frac{1}{2}=\\frac{-2}{\\frac{1}{2}}\\) = -4<\/p>\n

    Question 20.
    \nSolve 7x + 3 < 5x + 9, x \u2208 R and represent on the number line.
    \nAnswer:<\/p>\n

    \"KSEEB<\/p>\n

    Question 21.
    \nFind the future value of an annuity of \u20b9 5000 at 12% p.a for 6 years.
    \nAnswer:
    \nA = Rs.5,000 r = 0.12 n = 6
    \nF = \\(\\frac{A\\left[(1+r)^{n}-1\\right]}{r}=\\frac{5000\\left[(1+0.12)^{6}-1\\right]}{0.12}=\\frac{4869.11}{0.12}\\) = 40,575.9<\/p>\n

    Question 22.
    \nProve that (1 + cotA)2<\/sup> + (1 – cotA)2<\/sup> = 2 cosec2<\/sup>A
    \nAnswer:
    \n\"1st<\/p>\n

    Question 23.
    \nFind the value of cos 60\u00b0 -sin 30\u00b0 – cot3<\/sup> 45\u00b0
    \nAnswer:
    \n\\(\\frac{1}{2}-\\frac{1}{2}\\) – (1)3<\/sup> = -1<\/p>\n

    Question 24.
    \nFind the distance of the centroid of the triangle formed by the points (7, 1) (1, 5) and (1, 6) from origin.
    \nAnswer:
    \nG = \\(\\left[\\frac{7+1+1}{3}, \\frac{1+5+6}{3}\\right]\\) = (3, 4)
    \nO = (0, 0)
    \nOG = \\(\\sqrt{(3)^{2}+(4)^{2}}=\\sqrt{9+16}=\\sqrt{25}\\) = 5<\/p>\n

    Question 25.
    \nFind the equation of the locus of the point which moves such that it is equidistant from (4,2) and the X – axis.
    \nAnswer:
    \nP = (x, y) distance from x – axis = y A = (4,2)
    \nPA = y
    \n\\(\\sqrt{(x-4)^{2}+(y-2)^{2}}\\) =y
    \n(x-4)2<\/sup> + (y-2)2<\/sup> = y2<\/sup>
    \nx2<\/sup> – 8x + 16 – 4y + 4 = 0<\/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 an irrational 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

    \"KSEEB<\/p>\n

    Question 27.
    \nA relation R is defined on the set of integers by
    \nR = {(X, Y): X-Y is a multiple of 5} Show that R is an equivalence relation on Z.
    \nAnswer:
    \nx – x = 0 is a multiple of 5
    \n\u2234 (x, x) \u2208 R \u2200 x
    \nR is reflexive<\/p>\n

    Let (x, y) \u2208 R
    \nx – y is a multiple of 5
    \ny – x is a multiple of 5 (y, x) \u2208 R
    \n\u2234 R is symmetric<\/p>\n

    Let (x, y) \u2208 R
    \n(y, z) \u2208 R
    \nx -y is a multiple of 5 and y – z is a multiple of 5
    \n(x – y) +(y – z) is a multiple of 5
    \nx-z is a multiple of 5 (x, z) \u2208 R<\/p>\n

    \u2234 R is transitive
    \nR is an equivalence relative<\/p>\n

    Question 28.
    \nSolve : 32x<\/sup> – 10.3x<\/sup> + 9 = 0
    \nAnswer:
    \nLet 3x<\/sup> = a \u21d2(3x<\/sup>)2<\/sup> = a2<\/sup>
    \n\u21d2 32x<\/sup> = a2<\/sup>
    \n\u2234 a2<\/sup> – 10a + 9 = 0
    \na2<\/sup> – 9a – a + 9 = 0
    \na(a – 9) -1 (a – 9) = 0
    \n(a – 9)(a – 1) = 0
    \na – = 0 a – 9 = 0
    \na = 1 a = 9
    \n3x<\/sup> = 30<\/sup> 3x<\/sup> = 32<\/sup>
    \nx = 0 x = 2<\/p>\n

    Question 29.
    \nIf x2<\/sup> + y2<\/sup> = 12xy, show that 2 log (x – y) = log2 + log 5 + log x + log y.
    \nAnswer:
    \nx2<\/sup> + y2<\/sup> = 12xy
    \nx2<\/sup> + y2<\/sup> – 2zy = 12xy – 2xy \u21d2 (x – y)2 =10xy
    \nlog(x-y)2<\/sup> = log10xy
    \n2 log (x – y) = log 10 + log x + log y
    \n2 log (x – y) = log (2 x 5)+log x + log y
    \n2 log (x – y) = log 2 + log 5 + log x + logy<\/p>\n

    Question 30.
    \nInsert 4 Arithmetic means between 14 and 34.
    \nAnswer:
    \n14, a1<\/sub>, a2<\/sub>, a3<\/sub>, 34
    \na = 14
    \nT =34
    \nn = 5
    \nTn<\/sub> = a + (n-1)d
    \n3 = 14 + (4 )d
    \n20 = 4 d
    \nd = 5
    \n14, 19, 24, 29, 34<\/p>\n

    Question 31.
    \nA number consists of 2 digits whose sum is 4, if 18 is added to the number, the digits get interchanged. Find the number.
    \nAnswer:
    \nLet the digit in the tens place be x and the digit in the unit place be y
    \nThe nunter is 10x + y
    \nGiven x + y = 4 …………(1)
    \nAlso given 10x + y + 18 = 10y + x
    \n10x – x + y – 10y + 18 = 0
    \n9x – 9y + 18 = 0
    \nx – y + 2 = 0
    \nx – y = -2……………..(2)
    \n\"1st
    \nx + y = 4
    \ny = 4 – x = 4 – 1 = 3
    \nThe numbcr is 10x + y
    \n10(1) + 3 = 10 + 3 = 13
    \nThe number is 13.<\/p>\n

    Question 32.
    \nSolve graphically: 2x + y \u2265 8 ; and x + y \u2265 110
    \nAnswer:
    \n\"1st<\/p>\n

    Question 33.
    \nPreitha bought a car for \u20b9 4,00,000. If it depreciates at the rate of 12% per year, how much will it be worth after 10 years?
    \nAnswer:
    \nC = Rs.4,00,000 r = 12% = 0.12 n = 10
    \nB = C(1 – r)n<\/sup>
    \n= 4,00,000 (1 – 0.12)10<\/sup> = 4,00,000 (0.88)10
    \nB = Rs. 1,11,400.39<\/p>\n

    Question 34.
    \nThe average age of 10 students in a class increases by 4.8 months, when a boy of 6 years is replaced by a new boy. What is the age of the new boy?
    \nAnswer:
    \nLet the average age of 10 students x
    \n\u2234 Sum of ages of 10 students = 10x
    \nA boy of 6 years is replaced by a new boy whose age is y
    \n\u2234 Total age of 10 students 10x + y – 6
    \n\u2234 Average age of 10 students = \\(\\frac{10 x+y-6}{10}\\)
    \n\u2234 \\(\\frac{10 x+y-6}{10}\\) = x + \\(\\frac{4.8}{12}\\) 4.8 months = \\(\\frac{4.8}{12}\\) years
    \nx by 10
    \n\u2234 10x + y – 6 = 10x + \\(\\frac{48}{12}\\)
    \n\"1st
    \ny = 6 + 4 = 10
    \n\u2234 Age of the new boy = 10 years<\/p>\n

    \"KSEEB<\/p>\n

    Question 35.
    \nIn a dance competition, 70% of the participants were girls. 35% of the boys and 65% of the girls are selected for the next round. If 49 girls were eliminated, And the number of boys who got selected.
    \nAnswer:
    \nLet the total number of participants – x
    \nIf 65% of girls got qualified then 35% of girls got eliminated
    \nTotal number of girls = \\(\\frac{70}{100}\\)x = 0.7x
    \nTotal number of boys = \\(\\frac{30}{100}\\) \u00d7 x = 0.3x
    \n\u2234 total girls eliminated = \\(\\frac{35}{100}\\) \u00d7 0.7x = 0.245x
    \nHence 0.245x = 49
    \nx = \\(\\frac{49}{0.245}\\)= 200
    \nHence total number of boys eliminated = \\(\\frac{35}{100}\\) \u00d7 0.3 \u00d7 200 = 21<\/p>\n

    Question 36.
    \nIf tan A = \\(\\frac{-12}{5}\\) and 270% < A < 360\u00b0 find the value of \\(\\frac{3 \\sin A-2 \\cos A}{9 \\cos A+4 \\sin A}\\)
    \nAnswer:
    \n\\(\\frac{3 \\sin A-2 \\cos A}{9 \\cos A+4 \\sin A}=\\frac{3\\left(-\\frac{12}{13}\\right)-2\\left(\\frac{5}{13}\\right)}{9\\left(\\frac{5}{13}\\right)+4\\left(\\frac{-12}{13}\\right)}=\\frac{-36-10}{45-48}=\\frac{-46}{-3}=\\frac{46}{3}\\)<\/p>\n

    Question 37.
    \nFind area of the triangle whose vertices are A(6, 3) B(-3, 5) C(4, -2)
    \nAnswer:
    \nA = (6,3) B = (-3,5) C = (4, -2)
    \n\u2206ABC = \\(\\frac{1}{2}\\)[x1(y2 – y3) + x3 (y3 – y) + X3 (y1 – y2)]
    \n= \\(\\frac{1}{2}\\) [6 (5 + 2) – 3 (-2 – 3) + 4 (3 – 5)] = \\(\\frac{1}{2}\\) [6(7) – 3(-5) + 4(-2)]
    \n= \\(\\frac{1}{2}\\) [42 + 15 – 8] = \\(\\frac{1}{2}\\) [49] squarts<\/p>\n

    \"KSEEB<\/p>\n

    Question 38.
    \nFind the equation of the line passing through (5,2) and cutting off intercepts that are equal in magnitude but opposite in sign.
    \nAnswer:
    \nb = -a \\(\\frac{x}{a}+\\frac{y}{-b}\\) = 1 put x = 5, y = 2\\(\\frac{5}{a}-\\frac{2}{a}\\) = 1
    \n3 = a, b = -3
    \n\u2234 \\(\\frac{x}{3}+\\frac{y}{-3}\\) = 1 \u21d2 \\(\\frac{x-y}{3}\\) = 1 \u21d2 x – y – 3 = 0<\/p>\n

    Part – D<\/span><\/p>\n

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

    Question 39.
    \nIn a class of 250 students if 75 students have taken statistics and 25 have taken both statistics and economics and 100 have taken neither statistics nor economics. Find how many have taken
    \n(a) economics (b) economics only Also represent in Venn diagram.
    \nAnswer:
    \n\"1st
    \nn(S\u222aE) = 250 – 100 = 150
    \nn(S) = 75, n(S \u2229 E) = 25
    \nn(S \u222a E) = n(S) + n(E) – n(S \u2229 E)
    \n150 = 75 + n(E) – 25
    \n150 = n(E)
    \nn(E)= 100 \u2192 Economics
    \nn(E – S) = n(E) – n[E \u2229 5) = 100 – 25 = 75 \u2192Economics only<\/p>\n

    Question 40.
    \nEvaluate using log \\(\\frac{0.5634 \\times 0.0635}{2.563 \\times 12.5}\\)
    \nAnswer:
    \nLet x = \\(\\frac{0.5634 \\times 0.0635}{2.563 \\times 12.5}\\)
    \nlog x = log[latex]\\frac{0.5634 \\times 0.0635}{2.563 \\times 12.5}[\/latex]
    \n= log 0.5634 + log 0.0635 – log 2.563 – log 12.5
    \n=1.7508 + 2.8028-0.4087-1.0969
    \n= 2.5536- 1.5056 = 2 + 0.5536-1.5056
    \n= -2 – 0.952 = -2.952
    \n= -3 + 3 -2.952 = – 3 + 0.048 = 3.048<\/p>\n

    \"KSEEB<\/p>\n

    Question 41.
    \nFind the sum of all integers between 200 and 500 which are divisible by 7.
    \nAnswer:
    \nSn<\/sub> = 203 + 210 + 217 +………….+ 497 a = 203, d = 7, T = 497 , n = ?
    \nTn<\/sub> = a + (n – 1)d
    \n497 = 203 + (n- 1)7
    \n497 = 203 + 7n-7
    \n497= 196 +7n
    \n7n = 301
    \nn = 43
    \nSn = \\(\\frac{n}{2}\\)(a + Tn<\/sub>) = \\(\\frac{43}{2}\\)(203 + 497) = 15,050<\/p>\n

    Question 42.
    \nFind an integral root between -3 and 3 by inspection and they using synthetic division solve the equation x3<\/sup> – 10x2<\/sup> + 29x – 20 = 0
    \nAnswer:
    \nPut x = 1
    \n1 – 10(1)2<\/sup> + 29(1) – 20
    \n1 – 10 + 29 – 20
    \n30 – 30 = 0
    \n\"1st
    \nQuotient = x2<\/sup> – 9x + 20
    \nRemainder = 0
    \nx3<\/sup> – 10x2<\/sup> + 29x – 20 = 0
    \n(x + 1)(x2<\/sup> – 9x + 20) = 0
    \nx – 1, x2<\/sup> – 9x + 200
    \nx = 4, x = 5<\/p>\n

    Question 43.
    \nA sum of money amounts to Rs. 19500 in 5 years and 22,200 after 8 years at the same rate of simple interest. Find the principal and rate of interest.
    \nAnswer:
    \nP = ?, A = Rs.19,500, T = 5years
    \nA = Rs.22200, T = 8years R = ?
    \nA = P + SI
    \nA = P + \\(\\frac{\\mathrm{PTR}}{100}\\)
    \n19,500= P(1 + \\(\\frac{5 \\mathrm{R}}{100}\\)) ………(1)
    \n22,220= P(i + \\(\\frac{8 \\mathrm{R}}{100}\\)) ………………(2)
    \n\"1st
    \n(100 + 5R)222 = (100 + 8R) 195,
    \n22200 + 1110R = 19500 + 1560 R
    \n22200 – 19500 = 1560 R – 1110
    \nR = \\(\\frac{2700}{450}\\) = 6%p.a. \u2234 R = 6%p.a.
    \nFrom (1)
    \n19,500 = P(1 + \\(\\frac{5 \\mathrm{R}}{100}\\))
    \n19500 = P(1 + \\(\\frac{5 \\cdot 6}{100}\\)) = P(1 + \\(\\frac{30}{100}\\)) = P(1 + \\(\\frac{3}{10}\\)) = P(\\(\\frac{10+3}{10}\\)) = \\(\\frac{13}{10}\\)P<\/p>\n

    P = \\(\\frac{19500 \\times 10}{13}\\) = 15,000
    \n\u2234 P = Rs. 15,000<\/p>\n

    Question 44.
    \nFind the present value of an annuity of 2000 payable at the beginning of each quarter for the next 3 years if the rate of interest is 4% p.a. compounded quarterly.
    \nAnswer:
    \nA = Rs.2000 n = 3 x 4 = 12 r = = 0.01
    \nP = \\(\\frac{A\\left[(1+r)^{n}-1\\right]}{r(1+r)^{n}}\\)(1 + r) = \\(\\frac{2000\\left[(1+0.01)^{12}-1\\right]}{0.01(1+0.01)^{12}}\\)(1 + 0.01)
    \n= \\(\\frac{2000\\left[(0.01)^{12}-1\\right]}{0.01(1.01)^{12}}\\)(1.01) = \\(\\frac{256.18}{0.011}\\) = 23,289.09<\/p>\n

    \"KSEEB<\/p>\n

    Question 45.
    \nA radio is sold at a profit of 25%. Cost price and selling price both are increased by 100. If the new profit is 20% find the original cost of the ratio.
    \nAnswer:
    \nLet x be the C.P. of the radio.
    \nthen SP = \\(\\frac{125 \\mathrm{x}}{100}\\)
    \nWhen CP changes to x + 100SP = \\(\\frac{125 \\mathrm{x}}{100}\\) + 100
    \ni.e., x + 1.00 +20% of(x + 100) = \\(\\frac{125 \\mathrm{x}}{100}\\) + 100
    \nx + 100 + \\(\\frac{20}{100}\\)(x + 100) = \\(\\frac{125 x+10000}{100}\\)
    \n100x + 10000 + 20x + 2000 = 125x + 10000
    \n5x = 2000
    \nx = Rs.400
    \nThe original cost of the radio = Rs.400<\/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= secA.cosecA + 1 = RHS<\/p>\n

    Question 47.
    \nShow that (-3 1), (-6, -7) (3, -9) (6, -1) taken in order are vertices of a parallelogram.
    \nAnswer:<\/p>\n

    Question 48.
    \nFind the image of the point (2, 4) on the line x + y – 10 = 10
    \nAnswer:
    \nLet P be the point (2,4) and Q(x,y) be its reflection on the line x + y – 10 = 0. Let PQ cut the line at R. Then R is the midpoint of PQ and PQ is perpendicular to the given line.
    \nNow R = \\(\\left(\\frac{x_{1}+2}{2}, \\frac{y_{1}+4}{2}\\right)\\) Since R lies on x + y – 100
    \nWe have \\(\\left(\\frac{x_{1}+2}{2}+\\frac{y_{1}+4}{2}\\right)\\) – 10 = 0 i.e. x1 + y1 – 14 = 0 ………………(1)<\/p>\n

    Slope ofPQ = \\(\\frac{y_{1}-4}{x_{1}-2}\\) Slope ofthegiven line x + y – 100 is -1.
    \nSince PQ is perpendicular to the line x + y – 10 = 0 m1m2 = -1
    \nSolving (1) and (2) we get the point of intersection as JC, = 6 and y] = 8 \u2234Image = Q (6,8)<\/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 = {1, 2, 3} B = {2, 3, 4}, C = {3, 4, 5} verify (A \u00d7 B) (A \u00d7 C) = A \u00d7 (B \u2229 C)
    \nAnswer:
    \nA \u00d7 B = {(1,2), (1,3), (1,4), (2,2), (2,3), (2,4), (3,2), (3,3), (3,4)}
    \nA \u00d7 C = {(1,3), (1,4), (1,5), (2,3), (2,4), (2,5), (3,3), (3,4), (3,5)}
    \n(A \u00d7 B)\u2229(A \u00d7 C) = {(1,3),(1,4),(2,3),(2,4),(3,3),(3,4)} B\u2229C = {3,4}
    \nA \u00d7 (B\u2229C) = {(1,3),(1,4),(2,3),(2,4),(3,3),(3,4)} \u2234 (A \u00d7 B)\u2229(A \u00d7 C) = A \u00d7 (B\u2229C)<\/p>\n

    (b) The daily cost of production C for x units is given by C(x) = 17.5 x + 7000.
    \n(i) If each unit is sold for 30, then determine the minimum units that should be produced and sold to ensure no loss.
    \n(ii) If it is known that 500 unit can be sold daily what price \/ unit should be charged to guarantee no loss.
    \nAnswer:
    \n(i) C(x) = 17.5x + 7000
    \nR(x) = 30x, R(x) = C(x)
    \n30x = 175x + 7000,
    \n12.5x = 7000
    \nx = \\(\\frac{7000}{12.5}\\) = 560 BEP<\/p>\n

    (ii) If the S.P. is reduced by Rs.3\/unit then R(x) = 27x
    \nFor BEP R(x) = C(x)
    \n27x = 17.5x + 7000,
    \n27x – 17.5x = 7000
    \n9.5x = 7000,
    \nx = \\(\\frac{7000}{9.5}\\) = 73684 = 737 units<\/p>\n

    (c) The average score of 20 boys is 60% and average score of 30 girls is 70% Find the combined averaged score.
    \nAnswer:
    \nx1<\/sub> = 60%
    \nx2<\/sub> = 70%
    \nN1<\/sub> = 20
    \nN2<\/sub> = 30<\/p>\n

    X\u0304 = \\(\\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{1200+2100}{50}=\\frac{3300}{50}\\) = 66%<\/p>\n

    \"KSEEB<\/p>\n

    Question 50.
    \n(a) Find the sum to n terms of the series 9 + 99 + 999 + ………………….
    \nAnswer:
    \nSn<\/sub> = 9 + 99 + 999 +…………. = (10 – 1) + (102 – 1) + (103 – 1) + ………………
    \n= (10 + 102 + 103 +………………. ) – (1 + 1 + 1 +……………. ) = \\(\\frac{10\\left(10^{n}-1\\right)}{10-1}\\) – n
    \nSn<\/sub> = \\(\\frac{10\\left(10^{n}-1\\right)}{9}\\) – n
    \nn \u2208 N<\/p>\n

    (b) Find k if line x – 2y + 1 = 0; 2x – 5y + 3 = 0; 5x – 9y + k = 0 are concurrent.
    \nAnswer:<\/p>\n

    (c) If the product of two numbers \u00a1s 216 and their LCM \u00a1s 36 find their HCF.
    \nAnswer:
    \nab = 216 LCM = 36 HCF = ?
    \nHCF \u00d7 LCM = ab = 216
    \nHCF = \\(\\frac{216}{\\mathrm{LCM}}\\) HCF = \\(\\frac{216}{36}\\) = 6<\/p>\n","protected":false},"excerpt":{"rendered":"

    Students can Download 1st PUC Basic Maths Previous Year Question Paper March 2018 (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 2018 (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\/89488"}],"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=89488"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/89488\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=89488"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=89488"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=89488"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}