{"id":89628,"date":"2021-07-15T16:16:01","date_gmt":"2021-07-15T10:46:01","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=89628"},"modified":"2021-07-15T16:16:01","modified_gmt":"2021-07-15T10:46:01","slug":"1st-puc-basic-maths-previous-year-question-paper-march-2015-north","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-basic-maths-previous-year-question-paper-march-2015-north\/","title":{"rendered":"1st PUC Basic Maths Previous Year Question Paper March 2015 (North)"},"content":{"rendered":"

Students can Download 1st PUC Basic Maths Previous Year Question Paper March 2015 (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 2015 (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 canonical representation of 156.
    \nAnswer:
    \n22<\/sup> \u00d7 31<\/sup> \u00d7 131<\/sup><\/p>\n

    Question 2.
    \nIf A = {1, 2, 3, 4, 5} and relation R on A is defined as R = {(x, y) : x = y}, find R.
    \nAnswer:
    \nR = {(1, 1), (2,2), (3, 3), (4,4), (5, 5)}<\/p>\n

    Question 3.
    \nIf f is a function defined by f(x) = 3x + 5, find f(3).
    \nAnswer:
    \nf(3) = 3 \u00d7 3 + 5 = 14<\/p>\n

    Question 4.
    \nSimplify : ax+y<\/sup> a2x-y<\/sup>
    \nAnswer:
    \nax+y<\/sup>.a2x-y<\/sup> = ax+y+2x-y<\/sup> = a3x<\/sup><\/p>\n

    \"KSEEB<\/p>\n

    Question 5.
    \nExpress 3-2<\/sup> = \\(\\frac{1}{9}\\) in logarithmic form.
    \nAnswer:
    \n-2 = log3<\/sub>(\\(\\frac{1}{9}\\))<\/p>\n

    Question 6.
    \nFind the 30th term of the A.P. 2, 5, 8,………………….
    \nAnswer:
    \nT30 <\/sub>= 2 + 29 \u00d7 3 = 2 + 87 = 89<\/p>\n

    Question 7.
    \nSolve for x : 2(7 + x) – 10 = 16 – 2(x – 24).
    \nAnswer:
    \n14 + 2x – 10 = 16 – 2x + 48 = 4x = 68 \u21d2 x = 17<\/p>\n

    Question 8.
    \nFind the present value of a perpetuity of \u20b9 3,000 to be received forever at 4% p.a.
    \nAnswer:
    \nP80<\/sub> = \\(\\frac{A}{r}=\\frac{3000}{0.04}\\) = \u20b9 75,000<\/p>\n

    Question 9.
    \nThe average score of 20 boys is 60% and the average score of 30 girls is 70%. Find their combined average score.
    \nAnswer:
    \nX\u030412<\/sub> = \\(\\frac{20 \\times 60+30 \\times 70}{20+30}\\) = 66%<\/p>\n

    Question 10.
    \nConvert 0.32 into percentages.
    \nAnswer:
    \n0.32 \u00d7 100 = 32.00 = 32%<\/p>\n

    \"KSEEB<\/p>\n

    Question 11.
    \nExpress \\(\\frac{3 \\pi}{5}\\) in degree measure.
    \nAnswer:
    \n\\(\\frac{3 \\pi}{5}=\\frac{3 \\times 180}{5}\\) = 108<\/p>\n

    Question 12.
    \nFind the distance of the point (5,4) from the origin.
    \nAnswer:
    \nDistance = \\(\\sqrt{5^{2}+4^{2}}=\\sqrt{25+16}=\\sqrt{41}\\)<\/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 34<\/sup> \u00d7 53<\/sup> \u00d7 72<\/sup>.
    \nAnswer:
    \nSum of all positive divisors = T(n) \u2234 T(n) = (4+ 1)(3 + 1)(2+ 1) = 60<\/p>\n

    Question 14.
    \nFind the number which when divided by 16, 20, and 40 leaves the same remainder 4.
    \nAnswer:
    \nL.C.M of the given numbers 16,20 & 40 is 80 \u2234 the required number = 80 + 4 = 84<\/p>\n

    Question 15.
    \nIf A = (1, 3, 5}, B{5}, C = {7}, find A \u00d7 (B – C)
    \nAnswer:
    \nB – C = {5}
    \nA \u00d7 (B – C) = {1, 3, 5} \u00d7 {5} = {(1,5) (3,5) (5,5)}<\/p>\n

    Question 16.
    \nSimplify: \\(\\frac{2^{\\left(2^{0}\\right)}+\\left(2^{0}\\right)^{3}}{\\left(2^{3}\\right)^{2}+\\left(2^{2}\\right)^{0}}\\)
    \nAnswer:
    \n\\(\\frac{2^{\\left(2^{0}\\right)}+\\left(2^{0}\\right)^{3}}{\\left(2^{3}\\right)^{2}+\\left(2^{2}\\right)^{0}}=\\frac{2^{1}+1^{3}}{8^{2}+1}=\\frac{2+1}{64+1}=\\frac{3}{65}\\)<\/p>\n

    \"KSEEB<\/p>\n

    Question 17.
    \nInsert 3 geometric means between -4 and 64.
    \nAnswer:
    \nLet g1<\/sub>, g2<\/sub>, g3<\/sub> are G.Ms. \u2234 -4. g1<\/sub>, g2<\/sub>, g3<\/sub>, -64 are in G.P.
    \nHere a = -4, ar4<\/sup> = -64 \u21d2 r4<\/sup> = \\(\\frac{-64}{-4}\\) = 16 = 24 \u21d2 r = 2
    \nthe 3 G.Ms are g1<\/sub>\u00a0= ar = -8; g2<\/sub> = ar2<\/sup> = -16; g3<\/sub> = ar3<\/sup> = -32<\/p>\n

    Question 18.
    \nSolve by formula method : 5x2<\/sup> – 7x – 12 = 0
    \nAnswer:
    \nx = \\(\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}\\)
    \na = 5, b = -7, c = -12.
    \nx = \\(\\frac{7 \\pm \\sqrt{(-7)^{2}-4(5)(-12)}}{10}=\\frac{7 \\pm \\sqrt{49+240}}{10}=\\frac{7 \\pm \\sqrt{289}}{10}=\\frac{7 \\pm \\sqrt{17^{2}}}{10}=\\frac{7 \\pm 17}{10}\\)
    \n\u2234 x = \\(\\frac{12}{5}\\), -1<\/p>\n

    Question 19.
    \nSolve 3x – 2 < 2x + 1, x \u2208 R and represent on the number line.
    \nAnswer:
    \n3x – 2 < 2x + 1
    \n3x – 2x < 1 + 2
    \n\"1st
    \nx < 3<\/p>\n

    Question 20.
    \nWhat sum will amount to \u20b9 46,000 in 7 years at 12% p.a. SI?
    \nAnswer:
    \nA = P(1 + \\(\\frac{\\mathrm{TR}}{100}\\))
    \nP = \\(\\frac{A}{\\left(1+\\frac{T R}{100}\\right)}=\\frac{46,000}{\\left(1+\\frac{7 \\times 12}{100}\\right)}\\) = \u20b9 25,000<\/p>\n

    Question 21.
    \nThe average height of a group of people is 6 ft. 10 more people with an average height of 5 ft. join the group. Find the average height of the new group of 60 people.
    \nAnswer:
    \nTotal height of 50 people = 50 \u00d7 6 = 3 00
    \nTotal height of 10 people = 10 \u00d7 5 = 50
    \nTotal height of 60 people = 300 + 50 = 350
    \nHence new average = \\(\\frac{350}{60}\\) = 5.83 ft.<\/p>\n

    Question 22.
    \nWhile taking measurement, a tailor writes 34 instead of 24. What is the percentage error?
    \nAnswer:
    \nError = 34 – 24 = 10
    \n\u2234 Error percent = \\(\\frac{10}{24}\\) \u00d7 100 = 41.6%<\/p>\n

    Question 23.
    \nFind (1 + tan2<\/sup>\u03b8)(1 – sin2<\/sup>\u03b8) = 1
    \nAnswer:
    \nLHS = (1 + tan2<\/sup>\u03b8)(1 – sin2<\/sup>\u03b8) = sec2<\/sup>\u03b8.cos2<\/sup>\u03b8 = \\(\\frac{1}{\\cos ^{2} \\theta}\\) .cos2<\/sup>\u03b8 = 1 = RHS<\/p>\n

    Question 24.
    \nFind the small value of : cos60\u00b0- sin30\u00b0 – cot3<\/sup>345\u00b0
    \nAnswer:
    \ncos60 – sin30 – cot3<\/sup>(45\u00b0) = \\(\\frac{1}{2}-\\frac{1}{2}\\) – (1)3<\/sup> = -1<\/p>\n

    \"KSEEB<\/p>\n

    Question 25.
    \nFind k if the lines 2x – 3y + 4 = 0 and x + ky = 3 are perpendicular.
    \nAnswer:
    \nm1<\/sub> = \\(\\frac{-a}{b}=\\frac{2}{3}\\)
    \nm2<\/sub> = \\(\\frac{-1}{K}\\).
    \n\u2235 lines are perpendicular .. m1 \u00d7 m, = -1
    \n\u21d2 \\(\\frac{2}{3} \\times \\frac{-1}{\\mathrm{k}}\\) = -1
    \n\u21d2 \\(\\frac{2}{3} \\times \\frac{-1}{k}\\) = -1
    \n\u21d2 k = \\(\\frac{2}{3}\\)<\/p>\n

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

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

    Question 26.
    \nFind the HCF of \\(\\frac{8}{9}, \\frac{32}{81}, \\frac{16}{27}\\)
    \nAnswer:
    \nH.C.F of 8, 32, 16 = 8
    \nL.C.M of 9, 81, 27 = 81
    \n\u2234 Required HCF = \\(\\frac{\\text { HCF of N.R }}{\\text { LCM of DNR }}=\\frac{8}{81}\\)<\/p>\n

    Question 27.
    \nIf f(x) = x2<\/sup> g(x) = x + 1, find the value of
    \n(i) fog(1) (ii) gof(1) and (iii) fof(2)
    \nAnswer:
    \n(i) fog(1) = f(g(1)) = f(1 + 1) = f(2) = 22<\/sup> = 4.
    \n(ii) gof(1) = g(f(1)) = g(12<\/sup>) = g(1) = 1 + 1 = 2
    \n(iii) fof(2) = f(f(2)) f(22<\/sup>) = f(4) = 42<\/sup> = 16<\/p>\n

    Question 28.
    \nIf A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {4, 5, 6, 7, 8}, then verify Au(BnC) = (Au B)n(AuC)
    \nAnswer:
    \nA\u222a B = {1, 2, 3,4, 5, 6}
    \nA\u222a C = {1, 2,3,4, 5, 6, 7, 8}
    \nB\u2229 C= {4,5, 6}
    \nA\u222a(B\u2229C) = [1,2,3,4,5,6] = LHS
    \n(A\u222aB)\u2229(A\u222aC) = {1,2,3,4,5,6} = RHS
    \n\u2234 LHS = RHS<\/p>\n

    Question 29.
    \nIf x = log, 9, y = log, 7, z = log, 4 then show that xyz = 2.
    \nAnswer:
    \nLHS = xyz = \\(\\frac{\\log 9}{\\log 2} \\times \\frac{\\log 7}{\\log 9} \\times \\frac{\\log 4}{\\log 7}=\\frac{\\log 2^{2}}{\\log 2}=2 \\frac{\\log 2}{\\log 2}\\) = 2 = RHS<\/p>\n

    Question 30.
    \nThe sum of 3 numbers in an AP is 15 and their product is 105. Find the numbers.
    \nAnswer:
    \nLet the 3 numbers be (a – b), a, (a + d)
    \nGiven sum =15
    \n\u21d2 a – d + a + a + d = 15
    \n\u21d2 3a = 15
    \n\u21d2 a = 5
    \nProduct =105
    \n\u21d2 (a-d)(a)(a + d) = 105
    \n\u21d2 d = \u00b12
    \n\u2234 The three numbers are 3, 5, 7<\/p>\n

    \"KSEEB<\/p>\n

    Question 31.
    \nSolve x3<\/sup> + 6x2<\/sup> + 9x + 4 = 0 using synthetic division, given that it has an integral root between -3 and 3.
    \nAnswer:
    \nx = -1 is a root of the equation x3<\/sup>\u00a0+ 6x2<\/sup> + 9x + 4 = 0
    \n\"1st
    \nUsing synthetic division and getting the reduced equation as x2<\/sup> + 5x + 4 = 0 solving we get
    \nx = -1 & x = -4<\/p>\n

    Question 32.
    \nSolve the system of linear inequalities graphically : x + y < 6 and x + y > 4.
    \nAnswer:
    \nConsider x + y = 6, x + y = 4
    \nx = 0, y = 6 \u21d2 (0,6) (0,4)
    \ny = 0, x = 6 \u21d2 (6,0) (4,0)
    \n\"1st<\/p>\n

    Question 33.
    \nThe cost of a refrigerator is \u20b9 27,000. If it depreciates at the rate of 8% p.a., find its value after 4 years.
    \nAnswer:
    \nA = 27,000(1 -0.08)4 = 27,000(0.02)4= \u20b919,342.61
    \nRefrigerators value after 4 years = 19,342.61<\/p>\n

    Question 34.
    \n10 years ago, the average age of a family of 4 was 24 years. 2 children have been born and the average age remains the same. What are the present ages of the 2 children if their ages differ by 2 years?
    \nAnswer:
    \nLet the total age of 4 members = 96 = 4 \u00d7 24
    \nAfter 10 years total age of 4 = 96 + 40 = 136
    \nLet Ages of 2 children x & x + 2.
    \n\u2234 Total age of 6 = 136 + x + x + 2 = 138 + 2x
    \nGiven average age = 24 \u21d2 \\(\\frac{138+2 x}{6}\\) = 24 \u21d2 x = 3
    \n\u2234 Ages of 2 children = 3 & 5 years<\/p>\n

    Question 35.
    \nRitu\u2019s salary was increased by 10% and then again by 5%. If the present salary is \u20b9 9,240, then what was her previous salary?
    \nAnswer:
    \nLet the original salary – x
    \nWith 10% increase, salary = x + 0.1x = 1.1x
    \nwith 5% increase, salary = 1.155x
    \n1.155x = 9240
    \n\u21d2 x = \\(\\frac{9240}{1.155}\\) = \u20b9 8,000<\/p>\n

    \"KSEEB<\/p>\n

    Question 36.
    \nIf cosec\u03b8 = and 90\u00b0 < \u03b8 < 1800, then prove that \\(\\frac{4 \\sin \\theta-7 \\cos \\theta}{3 \\sin \\theta+2 \\cos \\theta}\\) = 40
    \nAnswer:
    \nsin\u03b8 = \\(\\frac{3}{5}\\) cos\u03b8 = \\(\\frac{-4}{5}\\)
    \n(\u2235 in II quadrant cos\u03b8 is negative)
    \nLHS = \\(\\frac{4 \\sin \\theta-7 \\cos \\theta}{3 \\sin \\theta+2 \\cos \\theta}=\\frac{4 \\cdot \\frac{3}{5}-7\\left(-\\frac{4}{5}\\right)}{3 \\cdot \\frac{3}{5}+2 \\cdot\\left(\\frac{-4}{5}\\right)}=\\frac{\\frac{12}{2}+\\frac{28}{5}}{\\frac{9}{5}-\\frac{8}{5}}=\\frac{\\frac{40}{5}}{\\frac{1}{5}}\\) = 40 = RHS<\/p>\n

    Question 37.
    \n3 consecutive vertices of a parallelogram are A (3, 0), B(5, 2), and C (-2, 6). Find the fourth vertex.
    \nAnswer:
    \nLet D(x, y) be the fourth vertex
    \nw.k. that in a parallelogram diagonal bisect each other
    \n\u2234 midpoints are equal.
    \n\"1st
    \n\\(\\left(\\frac{x+5}{2,}, \\frac{y+2}{2}\\right)=\\left(\\frac{3-2}{2}, \\frac{0+6}{2}\\right)\\)
    \n\u21d2 x + 5 = 1, y + 2 = 6
    \n\u21d2 x = -4, y = 4
    \n\u2234 the fourth vertex is D(-4, 4)<\/p>\n

    Question 38.
    \nDerive the slope-intercept form of a straight line.
    \nAnswer:
    \nRequired equation is y = mx + c<\/p>\n

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

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

    Question 39.
    \nIn a class of 50 students, 15 do not participate in any games, 25 play cricket, and 20 play football. Find the number of students who play both. Represent the results using the Venn diagram.
    \nAnswer:
    \n\"1st
    \nn(A\u222aB) = 50-15 = 35
    \nn(A \u2229 B) = n(A) + n(B) – n(A \u222a B)
    \nn(A\u2229B) = 25 + 20-35 = 10<\/p>\n

    Question 40.
    \nEvaluate using log tables : \\(\\frac{5.634 \\times 25.64}{12.72}\\)
    \nAnswer:
    \nlogx = log\\(\\left(\\frac{5.634 \\times 25.64}{12.72}\\right)\\)
    \n= log 5.634 + log 25.64 – log12.72
    \nlogx = 0.7508+ 1.4089-1.1044
    \nx = A.L (1.0553)
    \nx = 11.36<\/p>\n

    Question 41.
    \nFind the sum of all integers between 60 and 400 which are divisble by 13.
    \nAnswer:
    \nS = 65, 78, 91 ……………………390
    \na = 65, d = 13, Tn<\/sub> = 390, n = ?, S =?
    \nTn<\/sub> = a + (n – 1)d
    \n390 = 65 + (n- 1)13 \u21d2 n = 26
    \nSn<\/sub> = \\(\\frac{n}{2}\\)(a + l) = \\(\\frac{26}{2}\\)(65 + 390) \u21d2 Sn<\/sub> = 5915<\/p>\n

    \"KSEEB<\/p>\n

    Question 42.
    \nA certain two-digit number is 4 times the sum of the digits. If 18 is added to the number, the digits get interchanged. Find the number.
    \nAnswer:
    \nLet x be in 10th place, y in units place
    \n\u2234 the number = 10x + y
    \nGiven 10x + y = 4 (x + y)
    \n\u21d2 y = 2x…………………(1)
    \nAlso 10x + y + 18 = 10y + x
    \n\u21d2 x + 2 = y……………….(2)
    \nSolving (1) and (2) we get
    \n2x = x + 2 \u21d2 x = 2
    \n\u2234 x = 2 & y = 4
    \nHence required number = 24.<\/p>\n

    Question 43.
    \nFind the Cl on \u20b9 7500 at 14% for 14 years while interest is calculated half-yearly.
    \nAnswer:
    \nEffective rate of interest = r = \\(\\left(1+\\frac{0.14}{2}\\right)^{2}\\) – 1 = 15%
    \n\u2234 C.I = A-P = 7500(1 + 0.15)4 – 7500 = \u20b95617.5<\/p>\n

    Question 44.
    \nFind the future value of an annuity of \u20b9 5000 12% p.a for 6 years.
    \nAnswer:
    \nFuture value of an annuity = F = \\(\\frac{A\\left((1+i)^{n}-1\\right)}{i}\\)
    \nF = \\(\\frac{5000\\left((1+0.12)^{6}-1\\right)}{0.12}\\) = \u20b9 40,575.94<\/p>\n

    Question 45.
    \nA radio is sold at a profit of 25% cost price and selling price are both increased by 100. If the new profit is at the rate of 20%, find the original cost of the radio, i.
    \nAnswer:
    \nLet the original cost price be x
    \nI: Profit = 25% SP = 1.25x
    \nII: C.P. = 100 + x then SP= 1.25x + 100 ………(1)
    \nProfit = 20%
    \nS.P = (100 + x) + 0.2(100 + x) ……………..(2)
    \nFrom 1 & 2 we get
    \n1.25x + 100 = 100 + x + 20 + 0.2x
    \nSolving we get x = \u20b9 400 = original cost of the radio.<\/p>\n

    Question 46.
    \nProve that : \\(\\frac{\\tan A}{1-\\cot A}+\\frac{\\cot A}{1-\\tan A}\\) = 1 + secA.cosecA
    \nAnswer:
    \n\\(\\frac{\\tan A}{1-\\cot A}+\\frac{\\cot A}{1-\\tan A}=\\frac{\\tan A}{1-\\frac{1}{\\tan A}}+\\frac{\\frac{1}{\\tan A}}{1-\\tan A}\\)
    \n\"1st
    \n= RHS<\/p>\n

    \"KSEEB<\/p>\n

    Question 47.
    \nFind the area of a quadrilateral whose vertices are (1,2), (6,2), (5,3), and (3,4) taken in as order.
    \nAnswer:
    \nArea of a quadrilateral = Area of \u2206ABC + Area of \u2206ACD
    \n\"1st
    \nArea of \u2206ABC = \\(\\frac{1}{2}\\)\u03a3x1<\/sub> (y2<\/sub> – y3<\/sub>) = \\(\\frac{5}{2}\\) sq. units
    \nArea of \u2206ACD = 3 units
    \n\u2234 Area of \u2206BCD = \\(\\frac{5}{2}\\) + 3 = \\(\\frac{11}{2}\\)sq. units<\/p>\n

    Question 48.
    \nFind the equation of a line through he intersection of the lines x – y + 2 = 0, 3x + 2y = 9 and parallel to the line 2x + y – 3.
    \nAnswer:
    \nThe point of intersection of the lines x -y + 2 = 0, 3x + 2y = 9 is (1, 3)
    \nAny line parallel to the given line 2x + y = 3 is of the form 2x + y + k – 0
    \nBut this line passes through the point (1, 3)
    \n2(1) + 3 + k = 0
    \n\u2234 k = -5
    \nRequired equation of the line is 2x + y – 5 = 0<\/p>\n

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

    V. Answer any ONE question: (6 \u00d7 5 = 30)<\/span><\/p>\n

    Question 49.
    \n(a) Find the equation to the locus of a point moving in a plane such that it is equidistant from (-1, 1) and (4, 2). 4
    \nAnswer:
    \nLet P(x, y) be the point on the locus
    \nGiven PA = PB where A = (-1, 1) B = (4, 2)
    \nPA2<\/sup> = PB2<\/sup>
    \n(x + 1)2<\/sup> + (y – 1)2<\/sup> = (x – 4)2<\/sup> + (y – 2)2<\/sup>
    \nx2<\/sup> + 2x +1 + y2<\/sup> – 2y + 1 = x2<\/sup> +16 – 8x + y2<\/sup> + 4 – 4y
    \n10x – 6y = 18 or 5x – 3y = 9 is the required equation of the locus.<\/p>\n

    (b) Solve for x if:
    \nx = sec30\u00b0, tan60\u00b0 + sin45\u00b0cosec45\u00b0 + cos30\u00b0.cot60\u00b0.
    \nx = \\(\\frac{2}{\\sqrt{3}} \\cdot \\sqrt{3}+\\frac{1}{\\sqrt{2}} \\sqrt{2}+\\frac{\\sqrt{3}}{2} \\cdot \\frac{1}{\\sqrt{3}}\\)
    \nx = 2 + 1 + \\(\\frac{1}{2}\\) = 3 + \\(\\frac{1}{2}\\)
    \nx = \\(\\frac{6+1}{2}=\\frac{7}{2}\\) \u21d2 x = \\(\\frac{7}{2}\\)<\/p>\n

    (c) Find the number of digits in the integral part of 320<\/sup>. (2M)
    \nAnswer:
    \nConsider log320<\/sup>\u00a0= 20 log 3 = 20 \u00d7 0.4771 = 9.542
    \nSince the characteristic is 9, there are 10 digits in 320<\/sup>.<\/p>\n

    \"KSEEB<\/p>\n

    Question 50.
    \n(a) A firm finds that the production cost directly attributed to each product is 25 and fixed costs are 10,000. If each product can be sold for 35, find the
    \n(i) cost function (ii) revenue function
    \n(iii) profit function (iv) break-even level of output.
    \nAnswer:
    \nV.C. = \u20b9 125 , F.C. = \u20b9 10,000, price = \u20b9 35
    \n(i) cost function = C(x) = 10,000 + 25x
    \n(ii) Revenue function = R(x) = P \u00d7 x = 35x
    \n(iii) Profit function = TR – TC = 35x – 10,000 – 25 = 10x – 10,000.<\/p>\n

    At B.E. level of output T.R. = T.C.
    \n35x = 10,000 + 25x
    \n10x = 10,000
    \n\u21d2 x = 10,000 units<\/p>\n

    (b) Find the sum to n terms of the series 5 + 55 + 555 + ……………………
    \nAnswer:
    \nLet S = 5 + 55 + 555 +…………… to n terms
    \n\\(\\frac{\\mathrm{S}}{5}\\) = 1+ 11 + 111+………..to n terms
    \n\\(\\frac{9 \\mathrm{~S}}{5}\\) = 9 + 99 + 999 +…………… to n terms
    \n=(10 – 1) + (100 – 1) + (1000 – 1) +……….. to n terms
    \n= (10 + 100 + 1000 +………… to n terms) – (1 + 1 + 1………. to n ternis)
    \n= \\(\\frac{10\\left(10^{n}-1\\right)}{9}\\) – 1 \u00d7 n
    \nS = \\(\\frac{5}{9}\\left[\\frac{10\\left(10^{n}-1\\right)}{9}-n\\right]\\)<\/p>\n

    (c) The HCF of two numbers is 16 and their LCM is 160. If each of the numbers is 64, then find the other number.
    \nAnswer:
    \nGiven H = 16, L = 160, a = 64, b = ?
    \nH \u00d7 L = a \u00d7 b
    \nb = \\(\\frac{H \\times L}{a}=\\frac{16 \\times 160}{64}\\) =40
    \n\u2234 the other number is 40.<\/p>\n","protected":false},"excerpt":{"rendered":"

    Students can Download 1st PUC Basic Maths Previous Year Question Paper March 2015 (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 2015 (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\/89628"}],"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=89628"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/89628\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=89628"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=89628"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=89628"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}