{"id":89535,"date":"2021-07-13T11:19:45","date_gmt":"2021-07-13T05:49:45","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=89535"},"modified":"2021-07-13T11:19:45","modified_gmt":"2021-07-13T05:49:45","slug":"1st-puc-basic-maths-previous-year-question-paper-march-2017-north","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-basic-maths-previous-year-question-paper-march-2017-north\/","title":{"rendered":"1st PUC Basic Maths Previous Year Question Paper March 2017 (North)"},"content":{"rendered":"

Students can Download 1st PUC Basic Maths Previous Year Question Paper March 2017 (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 2017 (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.
    \nGive the canonical representation of 825.
    \nAnswer:
    \n825 = 31<\/sup> \u00d7 52<\/sup> \u00d7 11.<\/p>\n

    Question 2.
    \nConvert the set A = {a, e, i, o, u} from roster form into rule form.
    \nAnswer:
    \nA = {x\/x is a vowel in English alphabet}<\/p>\n

    Question 3.
    \nIf A = {a, b, c}; B = {c, d}. Find B \u00d7 A.
    \nAnswer:
    \nB \u00d7 A = {c, d) x {a, b, c}
    \n= c, a) (c, b) (c, c) (d, a) (d, b) (d, c)}.<\/p>\n

    Question 4.
    \nSimplify: \\(\\frac{a^{m+n} \\cdot a^{2 m-n}}{a^{m-n}}\\)
    \nAnswer:
    \n\"1st<\/p>\n

    Question 5.
    \nExpress Y = 27 in the .Lograrithmic Form.
    \nAnswer:
    \nlog3<\/sub> 27 = 3<\/p>\n

    \"KSEEB<\/p>\n

    Question 6.
    \nFind the 8th term of an A.P -2, -4, -6 ………..
    \nAnswer:
    \nT = a + (n – 1)d
    \nT8<\/sub> = -2 + (8 – 1)(-2)
    \nT8<\/sub> = -2 – 14 = -16<\/p>\n

    Question 7.
    \nSolve: \\(\\frac{x+2}{x-2}=\\frac{5}{2}\\)
    \nAnswer:
    \n2(x + 2) = 5(x – 1)
    \n2x + 4 = 5x – 5
    \n4 + 5 = 5x – 2x \u21d2 3x = 9 \u21d2x = 3.<\/p>\n

    Question 8.
    \nDefine perpetuity.
    \nAnswer:
    \nAn annuity that is payable forever is called a perpetuity.<\/p>\n

    Question 9.
    \nConvert 42% into a decimal.
    \nAnswer:
    \n42% = \\(\\frac{42}{100}\\) = 0.42.<\/p>\n

    Question 10.
    \nThe average score of 35 girls is 80 and the average score of 25 boys is 68. Find the average score of both boys and girls together.
    \nAnswer:
    \nX\u030412<\/sub> = \\(\\frac{35 \\times 80+25 \\times 68}{35+25}=\\frac{2800+1700}{60}\\) = 75<\/p>\n

    Question 11.
    \nConvent 75\u00b0 into radians.
    \nAnswer:
    \n\"1st<\/p>\n

    Question 12.
    \nIf the distance between the points (3, -2) and (-1, a) is 5 units. Find the values of
    \nAnswer:
    \n\\(\\sqrt{(3+1)^{2}+(-2-a)^{2}}\\) = 5
    \n42 + 4 + a2<\/sup> + 4a = 5
    \na2<\/sup> + 4a + 15 = 0
    \nUsing formula method we get the values of a.<\/p>\n

    Part – B<\/span><\/p>\n

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

    Question 13.
    \nFind the L.C.M of 12, 21 and 24.
    \nAnswer:
    \n\"1st
    \n\u2234 L.C.M = 23<\/sup> \u00d7 3 \u00d7 7 = 24 \u00d7 7 = 168<\/p>\n

    Question 14.
    \nFind the H.C.F of 18 and 24.
    \nAnswer:
    \n18 = 2 \u00d7 32<\/sup>, 24 = 23<\/sup> \u00d7 3
    \nH.C.F = 2 \u00d7 3 = 6<\/p>\n

    \"KSEEB<\/p>\n

    Question 15.
    \nIf A = {a, b,c, d}\\ B = {d, e, f, g, h, i}. Find A – B & B – A.
    \nAns.wer:
    \nA – B = {a, b, c}, B – A = {e, f g, h, i}<\/p>\n

    Question 16.
    \nFind the value of x so that the slope of the line joining the points (2, 5) and (x 13) is 20.
    \nAnswer:
    \nSlope = \\(\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\)
    \n\\(\\frac{13-5}{x-2}\\) = 20 \u21d2 \\(\\frac{8}{x-2}=\\frac{20}{1}\\)
    \n\u21d2 20x – 40 = 8 \u21d2 20x = 48
    \n\u21d2 x = \\(\\frac{48}{20}=\\frac{12}{5}\\) = 2.4<\/p>\n

    Question 17.
    \nFind the common difference of an A.P. whose first term is 6 and 12th term is 72.
    \nAnswer:
    \nGiven a = 6 and T12<\/sub> = a + 11d = 72
    \n6 + 11d = 72,
    \n11d = 72 – 6 = 66 \u21d2 d = 6.<\/p>\n

    Question 18.
    \nThe sum of two consecutive numbers is 23. Find them.
    \nAnswer:
    \nLet the 2 consecutive numbers be x and x + 1
    \nGiven sum = 23
    \nx + x + 1 = 23
    \n2x = 22 \u21d2 x = 11
    \n\u2234 The 2 consecutive numbers are 11 & 12.<\/p>\n

    Question 19.
    \nSolve 2x + 6 < 0; x \u2208 Z inequalities in one variable and represent the solution on the number line.
    \nAnswer:
    \nAnswer:
    \nGiven 2x + 6 < 0
    \n\u2234 2x \u2264 -6 \u21d2 x \u2264 -3
    \n\"1st
    \nSolution set is [-\u221e, -3].<\/p>\n

    \"KSEEB<\/p>\n

    Question 20.
    \nFind the interest on Rs. 1,500 at 4% pa. for 145 days.
    \nAnswer:
    \nS.I.= 1500 \u00d7 \\(\\frac{145}{365} \\times \\frac{4}{100}=\\frac{870000}{36500}\\) =23.836.<\/p>\n

    Question 21.
    \nIf x = a sec\u03b8; y = btan\u03b8 prove that \\(\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}\\) = 1
    \nAnswer:
    \n\"1st
    \n\u21d2 sec2<\/sup> \u03b8 – tan2<\/sup> \u03b8 = 1 = R.H.S<\/p>\n

    Question 22.
    \nThe weight of 6 men is 90 Kg, 70.5, 56 Kg, 45.5Kg, 85 Kg, and 78 Kg. Find average weight.
    \nAnswer:
    \nX\u0304 = \\(\\frac{90+70.5+56+45.5+85+78}{6}=\\frac{425}{6}\\) = 70.83<\/p>\n

    Question 23.
    \nKarthik received a scholarship of \u20b9 5,000 in 2011 and \u20b9 8,000 in 2012. Find the percentage increase.
    \nAnswer:
    \nDiff in the amount = 8000-5000 = 3000\u20b9
    \n\"1st<\/p>\n

    \"KSEEB<\/p>\n

    Question 24.
    \nProve that tan2<\/sup> A(1 – sin2<\/sup>A) = sin2<\/sup> A
    \nAnswer:
    \n\"1st<\/p>\n

    Question 25.
    \nFind the area of the triangle whose vertices are A(3, 4); B(2, -1), and C(4, -6).
    \nAnswer:
    \nArea of \u2206ABC = \\(\\frac{1}{2}\\)[x1<\/sub>(y2<\/sub> – y3<\/sub>) + x2<\/sub> (y3<\/sub> – y1<\/sub>) + x3<\/sub> (y1<\/sub> – y2<\/sub>)]
    \n= \\(\\frac{1}{2}\\)[3 (-1 + 6) + 2 (-6 – 4)+4 (4 – (-1)].
    \nA = \\(\\frac{1}{2}\\)[3(5) + 2(-10) + 4(5)]
    \n\"1st<\/p>\n

    Part – C<\/span><\/p>\n

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

    Question 26.
    \nIf the product of two numbers is 216 and their L.C.M is 36. Find their H.C.F.
    \nAnswer:
    \nGiven L.C.M of 36 and Product = 216, H.C.F = ?
    \nH.C.F = \\(\\frac{a \\times b}{L . C . M}=\\frac{216}{36}\\) = 6<\/p>\n

    Question 27.
    \nIf A= {1, 2}; B = {2, 3}; C = {3, 4}. Find (A \u00d7 B) \u2229 (A \u00d7 C).
    \nAnswer:
    \nA \u00d7 B = {1,2} \u00d7 {2,3}{(1,2) (1, 3) (2,2) (2, 3)}
    \nA \u00d7 C = {1, 2} \u00d7 {3, 4} – {(1,3) (1, 4) (2,3) (2,4)}
    \n(A \u00d7 B) \u2229 (A \u00d7 C) = {(1, 3) (2, 3)}.<\/p>\n

    Question 28.
    \nIf A = {2, 3, 4} write ail the proper subsets of ‘A’.
    \nAnswer:
    \nProper subsets of A
    \nX = P(A) = {{Q} (2) (3) (4) (2, 3) (3, 4) (4, 2)}.<\/p>\n

    \"KSEEB<\/p>\n

    Question 29.
    \nProve that (xb-c<\/sup>)a<\/sup> . (xc-a<\/sup>)b<\/sup> . (xa-b<\/sup>)c<\/sup> = 1.
    \nAnswer:
    \n\"1st<\/p>\n

    Question 30.
    \nIf the second term of the G.P is 6 and 5th term is 162. Then find the GP.
    \nAnswer:
    \nT2<\/sub> = ar = 6 ….(1), T5<\/sub> = ar4<\/sup> = 162 ….(2), G.P. = ?
    \n\"1st
    \nar = 6 \u21d2 3a = 6 \u21d2 a = 2
    \nTerms of G.P are a, ar ar2<\/sup>, ar3<\/sup>
    \n2, 2 \u00d7 3, 2 \u00d7 32<\/sup>, 2 \u00d7 33<\/sup>………….
    \n2, 6, 18…………..<\/p>\n

    Question 31.
    \nThe sum of 6 times a number and 5 times the number is 55. Which is that number?
    \nAnswer:
    \nLet the number be x
    \nGiven 6x + 5x = 55
    \n11x = 55 \u21d2 x = 5
    \n\u2234 The number is 5.<\/p>\n

    Question 32.
    \nSolve graphically:
    \n3x – 7 > 2(x – 6) and 6 – x > 11 – 2x; x\u2208R
    \nAnswer:
    \n3x – 7 > 2x – 12
    \n3x – 2x > -12 + 7
    \nx > -5<\/p>\n

    6 – x > 11 – 2x
    \n-x + 2x >11 – 6
    \nx > 5<\/p>\n

    Question 33.
    \nFind the present value of an annuity due of 8000 for 5 years at 5% p.a.
    \nAnswer:
    \nP = \\(\\frac{a\\left[(1+i)^{n}-1\\right]}{i(1+i)^{n}}\\)(1 + i)
    \nP = \\(\\frac{8000\\left[(1+0.05)^{5}-1\\right]}{0.05(1+0.05)^{5}}\\) \u00d7 (1 + 0.05)
    \n= \\(\\frac{8000\\left[(1.05)^{5}-1\\right](1.05)}{0.05 \\times(1.05)^{5}}=\\frac{8000[(1.2763-1)](1.05)}{0.05 \\times 1.2763}\\) = \u20b9 36369.50<\/p>\n

    Question 34.
    \nCalculate the arithmetic average mark from the following data.
    \n\"1st
    \nAnswer:
    \nX\u0304w<\/sub> = \\(\\frac{45 \\times 11+75 \\times 10+60 \\times 15+55 \\times 12+93 \\times 2}{11+10+15+12+2}\\)
    \n= \\(\\frac{495+750+900+660+186}{50}=\\frac{2991}{50}\\) = 59.82 \u2248 60<\/p>\n

    Question 35.
    \nIn an election, the winning candidate got 4,800 votes which are 80% of the total votes. Calculate the total number of votes.
    \nAnswer:
    \n\"1st
    \n\u2234 Total number of votes = 6000<\/p>\n

    Question 36.
    \nIf tan \u03b8 = \\(\\frac{a}{b}\\) show that \\(\\frac{a \\sin \\theta-b \\cos \\theta}{a \\sin \\theta+b \\cos \\theta}=\\frac{a^{2}-b^{2}}{a^{2}+b^{2}}\\)
    \nAnswer:
    \n\"1st<\/p>\n

    Question 37.
    \nA point ‘P’ moves such that PA2<\/sup> = 3PB2<\/sup>. If A = (5, 0) and B = (-5, 0). Find the equation of the Locus of ‘P’
    \nAnswer:
    \nLet P(x, y) be the point on the locus
    \nGiven PA2<\/sup> = 3PB2<\/sup> and A = (5, 0), B = (-5, 0)
    \n(x – 5)2<\/sup> + (y – 0)2<\/sup> = 3[(x + 5)2<\/sup> + (y – 0)2<\/sup>]
    \nx2<\/sup> + 25- 10x + y2<\/sup> = 3[x2<\/sup> + 25 + 10x + y2<\/sup>]
    \n3x2 <\/sup>+ 75 + 30x + 3y2<\/sup> = x2<\/sup> + y2<\/sup> – 10x + 25
    \n2x2<\/sup> + 2y2<\/sup> + 40x + 50 = 0 is the required equation.<\/p>\n

    \"KSEEB<\/p>\n

    Question 38.
    \nFind the point of intersection of the lines 3x = 2y – 5 = 0 and 4x – y – 3 = 0.
    \nAnswer:
    \n3x + 2y – 5 = 0
    \n4x – y – 3 = 0 \u2192 Multiplying by 2
    \n\"1st
    \n11x – 11 = 0 \u21d2 x = \\(\\frac{11}{11}\\) = 1
    \ny = 4x – 3 = 4 – 3 = 1.
    \n\u2234 Point of intersection is (1,1).<\/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 people drink tea, 10 take both tea and coffee. How many take at least one of the two drinks.
    \nAnswer:
    \nGiven n(T) = 20, n(T\u222aC) = 50, n(T\u2229C) = 10, n(C) = ?
    \n\"1st
    \nn(T\u222aC) = n(T) + n(C) – n(T\u2229C)
    \n50 = 20 + n(C)-10
    \nn(C) = 40
    \n\u2234 No. of people taking atleast one of the two drinks = 10 + 10 + 30 = 50.<\/p>\n

    Question 40.
    \nIf log\\(\\left(\\frac{a-b}{4}\\right)\\) = log\\(\\sqrt{a}\\) + log\\(\\sqrt{b}\\) show that (a + b)2 = 20ab.
    \nAnswer:
    \nGiven log\\(\\left(\\frac{a-b}{4}\\right)\\) = log\\(\\sqrt{a}\\) + log\\(\\sqrt{b}\\)
    \nlog\\(\\left(\\frac{a-b}{4}\\right)\\) = log\\(\\sqrt{ab}\\)
    \n\\(\\frac{a-b}{4}=\\sqrt{a b}\\) S.B.S
    \n(a – b)2<\/sup> = 16ab
    \na2<\/sup> + b2<\/sup> – 2ab = 16ab (Add 4ab both sides)
    \na2<\/sup> + b2<\/sup> + 2ab = 20ab,
    \n(a + b)2<\/sup> = 20ab.<\/p>\n

    Question 41.
    \nFind the sum of n terms of GP. 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

    Question 42.
    \nFind the integral root between -3 and 3 by inspection and then using synthetic division, solve the equation x3<\/sup> – 2x2<\/sup> – 5x + 6.
    \nAnswer:
    \nLet f(x) = x3<\/sup> – 2x2<\/sup> – 5x + 6,
    \nf(1)= 1 – 2 – 5 + 1 = 0
    \n\u2234 x = +1 is a root of the given equation. Let us remove this root by synthetic division.
    \n\"1st
    \n\u2234 The resulting equation is x2<\/sup> – x – 6 = 0 is the quotient and remainder is 0.
    \nx2<\/sup> – x – 6 = 0
    \n(x – 3) (x + 2) = 0 \u21d2 x = -2 or 3
    \nThus x = 1, -2, 3 are the roots of the given equation.<\/p>\n

    Question 43.
    \nAt what time a sum of \u20b91,200 will earn \u20b9573 as compound interest at the rate of 5% p.a. if the interest is added annually?
    \nAnswer:
    \nGiven P=1200, C.I = 573, A = 1200 + 573 = 1773
    \ni = \\(\\frac{5}{100}\\) = 0.05, n = ?
    \nA = P(1 + i)n<\/sup>
    \n1773 = 1200(1 +0.05)n<\/sup>
    \n\\(\\frac{1773}{1200}\\)1.4775 = (1.05)n<\/sup>
    \n(1.05)n = (1.05)8 \u21d2 n = 8.<\/p>\n

    \"KSEEB<\/p>\n

    Question 44.
    \nPreritha wants to buy a house after 5years when it is expected to cost 50 lakhs. How much should she save annually if her savings earn a compound interest of 12 percent?
    \nAnswer:
    \nGiven F = 50,00,000, n = 5, i = 0.12, a = ?
    \nF = \\(\\frac{a\\left[(1+i)^{n}-1\\right.}{i}\\) \u21d2 50,00,000 = \\(\\frac{a\\left[(1+0.12)^{5}-1\\right]}{0.12}\\)
    \n50,00,000 = 6.3528a \u21d2 a = \\(\\frac{50,00,000}{6.3528}\\) = \u20b9 787054.5<\/p>\n

    Question 45.
    \nIf \u03b8 = \\(\\frac{5}{2}\\) and \u03b8 is acute then prove that \\(\\frac{3 \\cos \\theta+2 \\sin \\theta}{3 \\cos \\theta-4 \\sin \\theta}=\\frac{19}{7}\\)
    \nAnswer:
    \nGiven cot\u03b8 = \\(\\frac{5}{2}=\\frac{a d j}{o p p}\\)
    \n\u2234 Hyp = \\(\\sqrt{5^{2}+2^{2}}=\\sqrt{29}\\)
    \ncos\u03b8 = \\(\\frac{5}{\\sqrt{29}}\\), sin\u03b8 = \\(\\frac{2}{\\sqrt{29}}\\)
    \nL.H.S = \\(\\frac{3 \\cos \\theta+2 \\sin \\theta}{3 \\cos \\theta-4 \\sin \\theta}\\)
    \n\"1st<\/p>\n

    Question 46.
    \nThe price of a pair of trousers was decreased by 22% to 390. Whast was the original price of the trousers?
    \nAnswer:
    \nLet the price of trouser = x
    \nIt was decreased by 22% to \u20b9 390
    \n\u2234 x – \\(\\frac{22 x}{100}\\) = 390
    \n\\(\\frac{100 x-22 x}{100}\\) = 390 \u21d2 78x = 390 \u00d7 100
    \nx = \\(\\frac{39,000}{78}\\) =500
    \n\u2234 Original price of the trouser = \u20b9500.<\/p>\n

    Question 47.
    \nFind the coordinates of the vertices of the triangle. Given the midpoints of the sides as (4, -1); (7, 9) & (4, 11).
    \nAnswer:
    \nLet A, B, C be the vertices of the A,e, and D, E, and F be the mid-points of the sides BC, CA, and AB respectively.
    \nD = Midpoint of BC
    \n(4, -1) = \\(\\left(\\frac{x_{2}+x_{3}}{2}, \\frac{y_{2}+y_{3}}{2}\\right)\\)<\/p>\n

    \u21d2 x2<\/sub> + x3<\/sub> = 8, y2<\/sub> + y3<\/sub> = 2 …………….(1)
    \nE = Mid point of CA
    \n(7, 9) = \\(\\left(\\frac{x_{3}+x_{1}}{2}, \\frac{y_{3}+y_{1}}{2}\\right)\\)
    \n\"1st
    \n\u21d2 x3<\/sub> + x1<\/sub> = 14, y3<\/sub> + y1<\/sub> = 18 ……………..(2)
    \nF = Mid point of AB
    \n(4,11) = \\(\\left(\\frac{x_{1}+x_{2}}{2}, \\frac{y_{1}+y_{2}}{2}\\right)\\)
    \n\u21d2 x1<\/sub> + x2<\/sub> = 8, y1<\/sub> + y2<\/sub> = 22 ………………..(3)<\/p>\n

    Solving equations 1,2 and 3 we get
    \nx1<\/sub> = 7, x2<\/sub> = 1, x3<\/sub> = 7
    \ny1<\/sub> =21, y2<\/sub> = 1, y3<\/sub> = -3
    \nThus A = (7, 21), B = (1, 1) and C = (7, -3).<\/p>\n

    \"KSEEB<\/p>\n

    Question 48.
    \nFind ‘a’ so that the lines x – 6y + a = 0; 2x + 3y + 4 = 0 and x + 4y + 1 = 0 concurrent.
    \nAnswer:
    \nGiven 2x + 3y + 4 = 0 ………….. (1)
    \nx + 4y + 1 = 0 …………. (2) \u00d7 2
    \nx – 6y + a = 0 …………. (3)
    \nSolving 1 and 2 we get
    \n\"1st
    \n= \\(\\frac{-8}{5}\\) – 1 = \\(\\frac{-8-5}{5}=\\frac{-13}{5}\\)<\/p>\n

    Put x = \\(\\frac{-13}{5}\\) and y = \\(\\frac{2}{5}\\) in equation 3
    \nWe get x – 6y + a = 0
    \na = 6y – x
    \na = 6.\\(\\frac{2}{5}-\\left(\\frac{-13}{5}\\right)\\)
    \n= \\(\\frac{12}{5}+\\frac{13}{5}=\\frac{25}{5}\\) = 5
    \n\u2234 a = 5<\/p>\n

    Part – E<\/span><\/p>\n

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

    Question 49.
    \n(a) If U = {1, 2, 3, 4, 5, 6, 7, 8, 9} A = {1, 2, 3, 4, 5} B = {3, 4, 5, 6, 7}
    \nShow that (A \u2229 B)’ =A’\u222aB’
    \nAnswer:
    \nA \u2229 B = {3,4, 5}
    \n(A \u2229 B)’ = {1,2, 6, 7, 8,9} ………….(1)
    \nA’\u222aB’ = {6,7, 8,9} \u222a {1,2, 8, 9}
    \n= {1,2, 6,7, 8, 9} ………………(2)
    \nFrom 1 and 2 we get A’\u222aB’ = (A \u2229 B)’.<\/p>\n

    (b) Insert G. Means [Geometric Mean] between \\(\\frac{1}{4}\\) and \\(\\frac{1}{64}\\).
    \nAnswer:
    \nG.M = \\(\\sqrt{\\frac{1}{4} \\times \\frac{1}{64}}=\\sqrt{\\frac{1}{4^{4}}}=\\frac{1}{4^{2}}=\\frac{1}{16}\\)<\/p>\n

    (c) If ax<\/sup> = b; & by<\/sup> = c; cz<\/sup> = a. Show that xyz = 1.
    \nAnswer:
    \nConsider ax<\/sup> = b’
    \n(cz<\/sup>)x<\/sup> = b
    \nczx<\/sup> = b
    \n(by<\/sup>)zx<\/sup> = b’ \u21d2 & bxyz = b’\u21d2 xyz = 1.<\/p>\n

    \"KSEEB<\/p>\n

    Question 50.
    \n(a) Find the equation of the locus of the point which moves such that its distance from 3x – 4y + 1 = 0 is equal to its distance from (1, -1).
    \nAnswer:
    \nLet A = (1, -1) and P(x, y) be any point on the locus, then PA = distance from the line 3x – 4y +1 = 0
    \n\\(\\sqrt{(x-1)^{2}+(y+1)^{2}}=\\left|\\frac{3 x-4 y+1}{\\sqrt{9^{2}+16^{2}}}\\right|\\) S.B.S<\/p>\n

    (x – 1)2<\/sup> + (y + 1)2<\/sup> = \\(\\frac{(3 x-4 y+1)^{2}}{25}\\)
    \n25(x2<\/sup> – 2x + 1 + y2<\/sup> + 2y + 1) = 9x2<\/sup> + 16y2<\/sup> + 1 – 24xy – 8y + 6x
    \n16x2<\/sup> + 24xy + 9y – 56x + 58y + 49 =0 is the required equation of the locus.<\/p>\n

    (b) A manufacturer produced and sells balloons at 8 per unit. His fixed cost is \u20b96,500 and the variable cost per balloon is \u20b93.50. Calculate
    \n(i) Revenue Function (ii) Cost Function
    \n(iii) Profi Function (iv) Break even point.
    \nAnswer:
    \ni.e. (i) R(x) = 8x<\/p>\n

    (ii) C(x) = 3.5x + 6500<\/p>\n

    (iii) P(x) = 4.5x – 6500
    \ni.e.,P(x) = 6500 = 0<\/p>\n

    (iv) At BEP P(A) = 0
    \n4.5x – 6500 = 0
    \n4.5x = 6500
    \nx = 144.4 units<\/p>\n

    (c) What is the present value of an income of 3000 to be received forever if the interest rate if 14% p.a.
    \nAnswer:
    \nP\u221e<\/sub> = \\(\\) = \u20b921428.5<\/p>\n","protected":false},"excerpt":{"rendered":"

    Students can Download 1st PUC Basic Maths Previous Year Question Paper March 2017 (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 2017 (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\/89535"}],"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=89535"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/89535\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=89535"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=89535"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=89535"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}