{"id":89510,"date":"2021-07-12T16:54:50","date_gmt":"2021-07-12T11:24:50","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=89510"},"modified":"2021-07-12T16:54:50","modified_gmt":"2021-07-12T11:24:50","slug":"1st-puc-basic-maths-previous-year-question-paper-march-2017-south","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-basic-maths-previous-year-question-paper-march-2017-south\/","title":{"rendered":"1st PUC Basic Maths Previous Year Question Paper March 2017 (South)"},"content":{"rendered":"

Students can Download 1st PUC Basic Maths Previous Year Question Paper March 2017 (South), 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 (South)<\/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 3 – Imaginary part of 3 – i is -1.
    \nAnswer:
    \nThe imaginary part of 3 – i is -1<\/p>\n

    Question 2.
    \nIf A has 4 elements, how many elements, how many elements will P(A) have.
    \nAnswer:
    \nP(A) will have 16 Elements.<\/p>\n

    Question 3.
    \nIf R-1 = {(2, 4) (1, 2) (3, 1) (3, 2)} find R.
    \nAnswer:
    \nR= {(4, 2) (2,1) (1,3) (2, 3)}.<\/p>\n

    Question 4.
    \nSimplify: (22<\/sup>)2<\/sup> – 2(3\u00b0)<\/sup>
    \nAnswer:
    \n= 1 – 2′ = 1 – 2 = -1<\/p>\n

    \"KSEEB<\/p>\n

    Question 5.
    \nSolve for x if log7<\/sub> x = 2.
    \nAnswer:
    \nx = 72<\/sup> = 49.<\/p>\n

    Question 6.
    \nFind the sum to infinity of the GP 3,1, \\(\\frac{1}{3}\\), ……………
    \nAnswer:
    \nSn<\/sub> = \\(\\frac{a}{1-r}=\\frac{3}{1-\\frac{1}{3}}=\\frac{9}{3-1}=\\frac{9}{2}\\)<\/p>\n

    Question 7.
    \nForm the quadratic equation whose roots are 2, 3.
    \nAnswer:
    \n(x – 2)(x – 3) = 0 \u21d2 x2<\/sup> – 5x + 6 = 0.<\/p>\n

    Question 8.
    \nWhat is the simple interest on 650 for 14 weeks at 6% p.a?
    \nAnswer:
    \nS.I = 650 \u00d7 \\(\\frac{14}{52} \\times \\frac{6}{100}\\) = 10.5<\/p>\n

    \"KSEEB<\/p>\n

    Question 9.
    \nThe average score of 35 girls is 80 and the average score of 25 boys \u00a1s 68. Find the average score of both boys and girls together.
    \nAnswer:
    \n\"1st<\/p>\n

    Question 10.
    \nBy selling a book at’ 250 the profit made is \u20b9 50. What is the cost price of the book?
    \nAnswer:
    \nC.P = S.P – Profit = 250 – 50 = 200\u20b9.<\/p>\n

    Question 11.
    \nConvent 315\u00b0 to radians.
    \nAnswer:
    \n\"1st<\/p>\n

    Question 12.
    \nFind the slope of the line 2x + 5y – 11 = 0.
    \nAnswer:
    \nSlope = –\\(\\frac{2}{5}\\)<\/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 360.
    \nAnswer:
    \n360 = 23<\/sup> \u00d7 32<\/sup> \u00d7 51<\/sup>
    \nn = P1<\/sub>\u03b11<\/sub><\/sup>. P2<\/sub>\u03b12<\/sub><\/sup>. P3<\/sub>\u03b13<\/sub><\/sup>
    \nT(n) = T(360) = (1 + \u03b11<\/sub>) (1 + \u03b12<\/sub>) (1 + \u03b13<\/sub>) = (3 + 1) (2 + 1) (1 + 1) = 24.
    \n\u2234 If has 24 positive divisors.<\/p>\n

    \"KSEEB<\/p>\n

    Question 14.
    \nFind the HCF of two numbers if their LCM is 1260 and product is 52920.
    \nAnswer:
    \nH.C.F = \\(\\frac{a \\times b}{\\mathrm{LCM}}=\\frac{52920}{1260}=\\frac{5292}{126}\\) = 42<\/p>\n

    Question 15.
    \nIf f(x) = x, g(x) = x3<\/sup> + 1 find (a) fog (2) (b) gof (1)
    \nAnswer:
    \nfog (2) = f(g (2)) = f( 23<\/sup> + 1) = f( 9) = 9
    \ngof(1) – g (f(1)) = g (1) = 13<\/sup> + 1 = 1 + 1 = 2.<\/p>\n

    Question 16.
    \nFind the domain and range of the relation R = {(x, y): y = x2<\/sup>, x is a positive prime number less than 10}
    \nAnswer:
    \nR= {(2,4)'(3, 9) (5,25) (7, 49)}
    \nDomain – {2, 31 5, 7}, Range = {4,9,25, 40}.<\/p>\n

    Question 17.
    \nIf a = 3x<\/sup> b = 3y<\/sup> c = 3z and ab – c2<\/sup>. P.T. x + y = 2z.
    \nAnswer:
    \nGiven ab = c2<\/sup>
    \n3x<\/sup>.3y<\/sup> = (32z<\/sup>) \u21d2 3x+y<\/sup> = 32z<\/sup>
    \nx + y = 2z.<\/p>\n

    \"KSEEB<\/p>\n

    Question 18.
    \nInsert 3 Arithmetic means between -2 and -10.
    \nAnswer:
    \nLet A1<\/sub>, A2<\/sub>, A3<\/sub> are 3 A.m between -2 and -10
    \n\u2234 -2, A1<\/sub>, A2<\/sub>, A3<\/sub>………. -10 are in A.P .
    \na a + d a + 2d a + 3d a + Ad are in A.P
    \n\u2234 a = – 2 and a + 4d = – 10 \u21d2 4d = -10 + 2 = -8
    \nd = – 2 .
    \nHence A1<\/sub> = -2 – 2 = -4,
    \nA2<\/sub> = a + 2d = – 2 – 4 = – 6,
    \nA3<\/sub> = a + 3d = -2 – 6 = -8
    \nHence 3 Ams are -4, -6, and -8.<\/p>\n

    Question 19.
    \nIf a and b are the roots of the equation 2x2<\/sup> + 5x + 5 = 0 the find the value of \\(\\frac{1}{\\alpha}+\\frac{1}{\\beta}\\)
    \nAnswer:
    \nGiven 2x2<\/sup> + 5x + 5 = 0
    \n\u03b1 + \u03b2 = \\(\\frac{-b}{a}=\\frac{-5}{2}\\)
    \n\u03b1\u03b2 = \\(\\frac{c}{a}=\\frac{5}{2}\\)
    \n\\(\\frac{1}{\\alpha}+\\frac{1}{\\beta}=\\frac{\\alpha+\\beta}{\\alpha \\beta}=\\frac{-5 \/ 2}{5 \/ 2}\\) = -1<\/p>\n

    Question 20.
    \nSolve 7x + 3 < 5x + 9, x\u2208R and represent on number line.
    \nAnswer:
    \n7x + 3 < 5x + 9
    \n\"1st
    \n7x – 5x <9 – 3
    \n2x < 6
    \nx < 3
    \n\u2234 Solution set is (- \u221e, 3)<\/p>\n

    Question 21.
    \nFind the effective rate of interest when a sum lent at 12% is computed half yearly.
    \nAnswer:
    \nr = \\(\\left(1+\\frac{i}{q}\\right)^{q}\\) – 1 = \\(\\left(1+\\frac{0.12}{2}\\right)^{2}\\) – 1
    \n= (1 + 0.06)2 – 1 =(1.06)2 – 1 = 12.36.<\/p>\n

    Question 22.
    \nThe average age of 12 boys is 8 years. Another boy of 21 years joins the group. Find the average age of the new group.
    \nAnswer:
    \nAve age of 12 boys = 8 yrs.
    \n\u2234 Total age of boys = 8 x 12 = 96 yrs.
    \nWhen a boy of 21 yrs joins the group
    \n\u2234 Total age of the 13 boys = 96 + 21 = 117 yrs.
    \n\u2234 Average age of the new group = \\(\\frac{117}{13}\\) = 9 yrs.<\/p>\n

    Question 23.
    \nFind the value of sin (480\u00b0) + tan 135\u00b0.
    \nAnswer:
    \nsin (480\u00b0) + tan (135\u00b0) = sin (360 + 120) + tan (180 – 45)
    \n= sin 120-tan 45
    \n= sin (180 – 60) – 1 = sin 60 – 1 = \\(\\frac{\\sqrt{3}}{2}\\) – 1<\/p>\n

    \"KSEEB<\/p>\n

    Question 24.
    \nThe CEO the triangle ABC is the point (2, 3). The coordinates of A are (5, 6) and B(-l, 4). Find the coordinates of C.
    \nAnswer:
    \nG(x, y) = \\(\\left(\\frac{x_{1}+x_{2}+x_{3}}{3}, \\frac{y_{1}+y_{2}+y_{3}}{3}\\right)\\), C(x3<\/sub>, y3<\/sub>) = ?
    \n(2, 3) = \\(\\left(\\frac{5-1+x_{3}}{3}, \\frac{6+4+y_{3}}{3}\\right)\\)
    \n(2, 3) = \\(\\left(\\frac{4+x_{3}}{3}, \\frac{10+y_{3}}{3}\\right)\\) \u21d2 6 = 4 + x3<\/sub>,
    \n10 = y3<\/sub> = 9
    \n\u21d2 x3<\/sub> – 6 – 4, y3<\/sub> = 9 – 10
    \n\u21d2 x3<\/sub> = 2, y3<\/sub> = -1
    \nHence co-ordinates of C(x3<\/sub>, y3<\/sub>) = C(2, – 1)<\/p>\n

    Question 25.
    \nFind equation of line passing through (0, -4) and making an angle of 30\u00b0 with the x-axis.
    \nAnswer:
    \nRequired equation is y – y1<\/sub> = m(x – x1<\/sub>)
    \ni.e., (y – (-4) = tan 30\u00b0 (x = 0)
    \ny + 4 = \\(\\)(x)
    \nx = \\(\\sqrt{3}\\)y + 4\\(\\sqrt{3}\\) = 0<\/p>\n

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

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

    Question 26.
    \nFind the number which when divided by 36, 40 and 48 leaves the same remainder 5.
    \nAnswer:
    \nLet us find out the L.C.M of 36,40 and 48
    \n36 = 22<\/sup> \u00d7 32<\/sup>, 40 = 23<\/sup> \u00d7 51<\/sup>, 48 = 24<\/sup> \u00d7 3
    \n\u2234 LCM = 24<\/sup> \u00d7 32<\/sup> \u00d7 51<\/sup>
    \n= 16 \u00d7 9 \u00d7 5 = 720
    \nSince 5 has to be the remainder. We have to add 5 to 720 i.e., 720 + 5 = 725<\/p>\n

    Question 27.
    \nIf A = {1, 2, 3} and R = {(1, 1) (1, 2) (2,1) (2, 2) (3, 3)} Prove that R is an equivalence relation on A.
    \nAnswer:
    \nR = { (1, 1) (1, 2) (2, 1) (2, 2) (3, 3)}
    \nR is reflexive \u2235 (1, 1) (2, 2) (3, 3) \u2208 R
    \nR is Symmetric \u2235 (1, 2) \u2208 R => (2, 1) \u2208 R
    \nR is transitive \u2235 (1, 2) (2, 1) \u2208 R => (1, 1) \u2208 R<\/p>\n

    Question 28.
    \nIf a1\/3<\/sup> + b1\/3<\/sup> = c1\/3<\/sup> Prove that (a + b + c)3<\/sup> = 27 abc.
    \nAnswer:
    \nGiven a1\/3<\/sup> + b1\/3<\/sup> = -c1\/3<\/sup>
    \nCubing both sides we get
    \n(a1\/3<\/sup> + b1\/3<\/sup>)3<\/sup> = -c
    \na + b + 3a1\/3<\/sup>b1\/3<\/sup>(a1\/3<\/sup> + b1\/3<\/sup>)3<\/sup> = -c
    \na + b + c = -3a1\/3<\/sup>, b1\/3<\/sup>, c1\/3<\/sup>.
    \nCubing both sides we get
    \n{a + b + c)3<\/sup> = 27 abc.<\/p>\n

    Question 29.
    \nSolve: log2<\/sub> x + log4<\/sub> x = 3.
    \nAnswer:
    \n\\(\\frac{\\log x}{\\log 2}+\\frac{\\log x}{2 \\log 2}\\) = 3
    \n\\(\\frac{2 \\log x+\\log x}{2 \\log 2}\\) = 3
    \n\u21d2 3 log x = 6 log 2 \u21d2 log x3<\/sup> = log 26<\/sup> \u21d2 x3<\/sup> =(22<\/sup>)3<\/sup> \u21d2 x = 4<\/p>\n

    Question 30.
    \nThe sum of 3 numbers in A.P. is 15 and their product is 105. Find the three numbers.
    \nAnswer:
    \nLet the 3 numbers be a – d, a, a + d
    \nGiven
    \n\"1st
    \n3a = 15 \u21d2 a = 5
    \nAlso (a – d) (a) (a + d) = 105
    \n5(25 – d2<\/sup>) = 105
    \n25 – d2<\/sup> = 21 \u21d2 25 – 21 = d2<\/sup> \u21d2 d2<\/sup> = 4 \u21d2 d = 2
    \nThe 3 numbers are 5 – 2, 5, 5 + 2 = 3, 5, 7.<\/p>\n

    Question 31.
    \nThe age of a father is 5 times that of a son. 3 years ago the age of the father was 8 times that of his son. Find their present ages.
    \nAnswer:
    \nLet the present age of the son be x yrs.
    \nLet the present age of father be 5x yrs.
    \n3yr ago, the age of son is x – 3 and that of father is 5x – 3
    \nGiven 5x – 3 = 8 (x – 3)
    \n5x – 3 = 8x – 24
    \n24 – 3 = 8x – 5x = 3x
    \nx = 21 \u21d2 x = 7
    \n\u2234 The present age of son = 7 yrs and father is 35 yrs.<\/p>\n

    \"KSEEB<\/p>\n

    Question 32.
    \nSolve graphically:
    \nx + 3y \u2265 3; 2x + y \u2264 2; x \u2265 0, y \u2265 0.
    \nAnswer:
    \nLet x + 3y = 3,
    \nPut x = 0, y= 1, (0,1)
    \nPut y = 0 x = 0, x = 3, (3,0)<\/p>\n

    2x + y = 2,
    \nPut x = 0, y = 2, (0,2)
    \nPut y = 0, x = 1, (1, 0)
    \nPlot the two lines in the graph we get
    \n\"1st<\/p>\n

    Question 33.
    \nFind the future value of an annuity of 5000 at 12% p.a. for 6 years.
    \nAnswer:
    \nF = \\(\\frac{a\\left[(1+i)^{n}-1\\right]}{i}\\) = 500\\(\\frac{\\left[(1+0.12)^{6}-1\\right]}{0.12}\\) = 500\\(\\frac{\\left[(1.12)^{6}-1\\right]}{0.12}\\) = 40575.9 Rs
    \nFuture value = 4.575.9<\/p>\n

    Question 34.
    \nRitu\u2019s salary was increased by 10% and then again by 5%. If the present salary is \u20b9 9,240. What was Ritu\u2019s previous salary?
    \nAnswer:
    \nLet his salary be x
    \n\"1st
    \n\\(\\frac{220 x+11 x}{200}\\) = 9240 \u21d2 231x = 9240 \u00d7 200
    \nx = \\(\\frac{9240 \\times 200}{231}\\) = 8000<\/p>\n

    Question 35.
    \nProve that \\(\\frac{1}{1+\\cos A}+\\frac{1}{1-\\cos A}\\) = 2cosec2<\/sup>A
    \nAnswer:
    \n\\(\\frac{1}{1+\\cos A}+\\frac{1}{1-\\cos A}\\)
    \n\"1st<\/p>\n

    Question 36.
    \nFind area of the quadrilateral whose vertices are A(1, 2) B(6, 2) C(5, 3) and D(3, 4) in order.
    \nAnswer:
    \nArea of the Quadrilateral ABCD
    \n= Area of triangle \u2206 ABC + Area of \u2206ACD
    \n= \\(\\frac{5}{2}\\) + 3 = \\(\\frac{11}{2}\\) sq units
    \nBut Area of \u2206ABC = \\(\\frac{1}{2}\\)\u03a3x1<\/sub> (y2<\/sub> – y3<\/sub>)
    \n\"1st
    \n= | Area of \u2206ABC| + |Area of \u2206ACD| = \\(\\left|\\frac{5}{2}\\right|\\) + |3| = \\(\\frac{5}{2}\\) + 3 = \\(\\frac{5+6}{2}=\\frac{11}{2}\\) sq.units<\/p>\n

    \"KSEEB<\/p>\n

    Question 37.
    \nFind the equation of the locus of the point which moves such that it \u00a1s equidistant from (4, 2) and x-axis.
    \nAnswer:
    \nLet P(x, y) be the point on the locus and A = (4,2)
    \n\u2234 PA = Y = (PA)2<\/sup> = y2<\/sup>
    \n\u21d2 (x – y)2<\/sup> + (y – 2)2<\/sup> = y2<\/sup>
    \n\"1st
    \n\u21d2 x2<\/sup> – 8x – 4y + 20 = 0 is the required equation.<\/p>\n

    Question 38.
    \nFind k so that the distance from (2, 3) to the line 8x + 15y + k = 0. may be equal to 4 units.
    \nAnswer:
    \n\\(\\left|\\frac{8(2)+15(3)+k}{\\sqrt{8^{2}+15^{2}}}\\right|\\) = 4
    \n\\(\\left|\\frac{16+45+k}{\\sqrt{64+225}}\\right|\\) = 4
    \n16 + k = 4\\(\\sqrt{289}\\) = 4 \u00d7 17 = 68
    \nk = 4(17) – 61 = 68 – 61 = 7. \u2234 k = 1.<\/p>\n

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

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

    Question 39.
    \nIn a survey of 100 persons, it was found that 28 read magazine A, 30 read magazine B, 42 read magazine C, 8 need magazines A and B, 10 read magazines A and C, 5 read magazines B and C while 3 read all the three magazines.
    \nFind: (a) How many read none of the three magazines
    \n(b) How many read-only magazines
    \n(c) How many read exactly one magazine only.
    \nAnswer:
    \nGiven U = 100, n(A) = 28, n(B) = 30, n(C) = 42
    \nn(A\u2229B) = 8, n(A\u2229C) = 10, n(B\u2229C) = 5, n(A\u2229B\u2229C) = 3
    \n\"1st
    \n(i) Number of people who read none of the magazines
    \n= 100 – (13 + 5 + 20 + 7 + 3 + 2 + 30)
    \n= 100 – 80 = 20<\/p>\n

    (ii) No. of people who read only magazines C = 30<\/p>\n

    (iii) No. of people who read exactly one magazines only = 13 + 20 + 30 = 63<\/p>\n

    Question 40.
    \nUsing log table find the value of \\(\\frac{21.54 \\times 72.6}{6.5}\\).
    \nAnswer:
    \nLet log = log\\(\\left(\\frac{21.54 \\times 72.6}{6.5}\\right)\\)
    \nlogx = log 21.54 + log 72.6 – log 6.5
    \n= 1.3332+ 1.8609-0.8129
    \nx = A.L (2.3812) = 240.58.<\/p>\n

    Question 41.
    \nFind the sum of all even integers from 40 to 160.
    \nAnswer:
    \nSn<\/sub>\u00a0= 40, 42, 44……………160
    \na = 40, d = 2, Tn<\/sub> = 160
    \nTn<\/sub> = a + (n – 1)d
    \n160 = 40 + (n – 1)2
    \n120 = 2n – 2
    \n122 = 2n \u21d2 n = 61
    \nConsider Sn = \\(\\frac{n}{2}\\)(a + l) = \\(\\frac{61}{2}\\)(40+160)= \\(\\frac{61 \\times 200}{2}\\) =6100
    \n\u2234 Sum of all even integers = 6100.<\/p>\n

    Question 42.
    \nSolve x3<\/sup> – 10x2<\/sup> + 29x – 20 = 0. Using synthetic division given that it has an integral root between -3 and 3.
    \nAnswer:
    \nf(x) = x3<\/sup> – 10x2<\/sup> + 29x – 20
    \nf(1)= 1 – 10 + 29 – 20 = 0
    \nx = 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> – 9x + 20 = 0 is the quotient and remainder is 0.
    \nx2<\/sup> – 9x + 20 = 0 (x – 5) (x – 4) = 0
    \nx = 4 or 5
    \nThus x = 1, 4, 5 are the roots of the given equation.<\/p>\n

    Question 43.
    \nThe difference between simple interest and compound interest on a certain sum of money invested for 3 years at 6% p.a. is f 110.16. Find the sum.
    \nAnswer:
    \nLet the sum = x
    \nS.I = \\(\\frac{x \\times 3 \\times 6}{100}\\) = 0.18x
    \nC.I. = A – P where A = P(1 + i)n<\/sup>
    \n= x(1 +0.06)3<\/sup> = 1.191016x
    \nC.I.= A – P = 1.191016x – 1x = 0.191016x
    \nGiven C.I. – S.I = 110.16
    \n0.191016x – 0.18x = 110.16
    \n0.011016x = 110.16
    \nx = \\(\\frac{110.16}{0.011016}\\) = 10,000
    \n\u2234 The sum= 10,000.<\/p>\n

    Question 44.
    \nFind x if x3<\/sup> sin45\u00b0cos 60\u00b0 = \\(\\frac{\\tan ^{2} 60^{\\circ} \\cdot \\cos e c 30^{\\circ}}{\\sec 45^{\\circ} \\cdot \\cot ^{2} 30^{\\circ}}\\)
    \nAnswer:
    \nx3<\/sup>. \\(\\frac{1}{\\sqrt{2}} \\cdot\\left(\\frac{1}{2}\\right)^{2}=\\frac{(\\sqrt{3})^{2} \\cdot 2}{\\sqrt{2} \\cdot\\left(\\frac{1}{\\sqrt{3}}\\right)^{2}}\\)
    \nx3<\/sup>.\\(\\frac{1}{\\sqrt{2}} \\cdot \\frac{1}{4}=\\frac{3.2}{\\sqrt{2} \\cdot \\frac{1}{3}}\\)
    \n\"1st
    \nx3<\/sup>. \\(\\frac{1}{4 \\sqrt{2}}=\\frac{18}{\\sqrt{2}}\\) \u21d2 x3 = \\(\\frac{72 \\sqrt{2}}{2}\\) \u21d2 x = 3\\(\\sqrt{72}\\)<\/p>\n

    Question 45.
    \nRaj wants to invest a lump-sum amount in the bank so that he can get an annual income of Rs.15,000 every year for the next 10 years. If the bank offers 16% p.a. compound interest, what is the amount he should invest today?
    \nAnswer:
    \nGiven, n = 10yrs, a = 15,000, i = 0.16, P = ?
    \nP = \\(\\frac{a\\left[(1+i)^{n}-1\\right]}{i(1+i)^{n}}\\)
    \n= \\(\\frac{15,000\\left[(1+0.16)^{10}-1\\right]}{0.16(1+0.16)^{10}}=\\frac{15,000\\left[(1.16)^{10}-1\\right]}{0.16 \\times(1.16)^{10}}\\) = 72498.2
    \n\u2234 He should invert today 7249.8<\/p>\n

    Question 46.
    \nA person spent 30% of his wealth and thereafter Rs. 20,000 and further 10% of the remainder. If Rs. 29,250 is still remaining, what was his total wealth?
    \nAnswer:
    \nLet his wealth be x Rs
    \n\"1st
    \nAgain he spent 20,000 \\(\\frac{7 x}{10}\\) – 20,000 \\(\\frac{7 x-2,00,000}{10}\\)
    \nAgain he spent 10% = \\(\\left(\\frac{7 x-2,00,000}{10}\\right)-\\frac{7 x-2,00,000}{10} \\times \\frac{1}{10}\\)x
    \n\\(\\left(\\frac{7 x-2 ; 00,000}{10}\\right)\\left(1-\\frac{x}{10}\\right)=\\frac{(7 x-2,00,000)(10-x)}{100}\\)
    \n\\(\\frac{(7 x-2,00,000)(10-x)}{100}\\) = 29,250 =x 75,000
    \n\u2234 Total wealth = 75,000.<\/p>\n

    \"KSEEB<\/p>\n

    Question 47.
    \nProve that A(-3, 6) B(-2,11) C(3,12) and D(2, 7) are the vertices of a rhombus. Also And its area.
    \nAnswer:
    \nA(-3, 6) B(-2, 11) C(3, 12) and D(2, 7)
    \nAB = \\(\\sqrt{(-2+3)^{2}+(11-6)^{2}}=\\sqrt{1^{2}+5^{2}}=\\sqrt{1+25}=\\sqrt{26}\\)
    \nBC = \\(\\sqrt{(3+2)^{2}+(12-11)^{2}}=\\sqrt{5^{2}+1^{2}}=\\sqrt{25+1}=\\sqrt{26}\\)
    \nCD = \\(\\sqrt{(2-3)^{2}+(7-12)^{2}}=\\sqrt{(-1)^{2}+(-5)^{2}}=\\sqrt{1+25}=\\sqrt{26}\\)
    \nDA = \\(\\sqrt{(2+3)^{2}+(7-6)^{2}}=\\sqrt{5^{2}+1^{2}}=\\sqrt{25+1}=\\sqrt{26}\\)
    \nd1<\/sub> = AC\\(\\sqrt{(3+3)^{2}+(12-6)^{2}}=\\sqrt{(6)^{2}+(6)^{2}}=\\sqrt{36+36}=\\sqrt{72}\\) = 6\\(\\sqrt{2}\\)
    \nd2<\/sub>\u00a0= BD\\(\\sqrt{(3+3)^{2}+(12-6)^{2}}=\\sqrt{(6)^{2}+(6)^{2}}=\\sqrt{36+36}=\\sqrt{72}\\) = 4\\(\\sqrt{2}\\)
    \nAll sides are equal = \\(\\sqrt{26}\\) \u2234 ABCD forms Rhombus.
    \nArea = \\(\\frac{1}{2}\\) d1<\/sub>d2<\/sub> = \\(\\frac{1}{2}\\).6\\(\\sqrt{2}\\).4\\(\\sqrt{2}\\) = 24 sq. units<\/p>\n

    Question 48.
    \nFind the equation of a straight line passing through the point (2, 2) such that the sum of its intercepts on the axes is 9.
    \nAnswer:
    \nIntercept equation is \\(\\frac{x}{a}+\\frac{y}{b}\\) = 1
    \nAlso given a + b = 9
    \nb = 9 – a
    \n\u2234 Equation becomes \\(\\frac{x}{a}+\\frac{y}{9-a}\\) = 1. But this equation passes through (2,2)
    \n\\(\\frac{2}{a}+\\frac{2}{9-a}\\) = 1
    \n\"1st
    \n18 = 9a – a2<\/sup>
    \na2<\/sup> – 9a + 18 = 0
    \n(a-6) (a-3) = 0
    \na = 6 or 3,
    \nb = 3 or 6
    \nThe required equations are \\(\\frac{x}{6}+\\frac{y}{3}\\) = 1 or \\(\\frac{x}{3}+\\frac{y}{6}\\) = 1<\/p>\n

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

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

    (a) If A = {3, 5, 7, 8} B = {4, 5, 7, 9} C = {1, 2, 3, 4, 5}
    \nFind (i) (A – B) \u00d7 (B – C)
    \n(ii) (A \u2229 B) \u00d7 (B \u2229 C)
    \nAnswer:
    \n(A – B)= {3, 8}, B – C = {7, 9}
    \n(i) (A – B) \u00d7 (B – C) = {(3, 7) (3, 9) (8, 7) (8, 9)}
    \n(ii) (A \u2229 B) \u00d7 (B \u2229 C) = {5, 7} \u00d7 {4, 5}
    \n= {5, 4) (5, 5) (7, 4) (7, 5)}.<\/p>\n

    (b) A manufactures sells his product at Rs. 8.35 per unit, he is able to sell his entire production. His fixed cost is Rs. 2116 and his variable cost per unit is \u20b9 7.20 Find
    \n(i) Level of production at which he can make a profit of \u20b9 4, 600
    \n(ii) Level of output at which he will incur a loss of \u20b91150
    \n(iii) Break-even level of production.
    \nAnswer:
    \nC(x)= 7.20x + 2116,
    \nR(x)= 8.35x
    \nP(x) = R(x) – C(x)<\/p>\n

    (i) P(x) = 4,600,
    \n4600 = 8.35x – [7.20x + 2116] = 1.15x – 2116
    \n1.15x = 4600 + 2116 = 6716 = 5840 .<\/p>\n

    (ii) C(x) – R(x) = 1150,
    \n7.20x + 2116 – 8.35x = 1150
    \n115x = 966,
    \nx = 840 units<\/p>\n

    (iii) C(x) = R(x)
    \n720x + 2116 = 8.35x,
    \n2116 = 8.35x – 7.20x
    \n2116 = 1.1 5x
    \nx = 1840 units BEP<\/p>\n

    (c) Using log find the number of digits in 350.
    \nAnswer:
    \nConsider log320<\/sup> =20 log 3 = 20 \u00d7 0.477 1 = 9.542
    \nSince the characteristic is 9, there are 10 digits in 320<\/sup>.<\/p>\n

    \"KSEEB<\/p>\n

    Question 50.
    \n(a) If the lines 2x – y = 5,
    \nKx-y = 6 and 4x – y = 1 are concurrent find k.
    \nAnswer:
    \nSolving equations 2x – y = 5, & 4x – y = 7 we get
    \n\"1st
    \n\u2234 y = 2x – 5 = 2 – 5 = -3<\/p>\n

    Since lines are concurrent put x = 1 and y = -3 in kx – y = 6
    \nKx – y = 6 \u21d2 K(1) – (-3) = 6
    \n\u21d2 K + 3 = 6
    \n\u21d2 K = 6 – 3
    \n\u21d2 K = 3<\/p>\n

    (b) Find sum of 7 + 77 + 777 + ……… n terms.
    \nAnswer:
    \nLet Sn<\/sub>\u00a0= 7 + 77 + 777 +…………….n terms
    \n= 7 [1 + 11 + 111 + –\u2014–\u2014 to n terms] = \\(\\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]
    \n= \\(\\frac{7}{9}\\left[\\frac{10\\left(10^{n}-1\\right)}{10-1}-n\\right]\\)[… 10 + 102<\/sup> + 103<\/sup>……….. to n terms is 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) A scholarship of Rs. 2000 every year has to be instituted. How much should be invested today \u00a1f the rate of interest, is 8% p.a. .
    \nAnswer:
    \nP = \\(\\frac{a}{i}=\\frac{2000}{0.08}\\) = 25,000<\/p>\n","protected":false},"excerpt":{"rendered":"

    Students can Download 1st PUC Basic Maths Previous Year Question Paper March 2017 (South), 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 (South) 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\/89510"}],"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=89510"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/89510\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=89510"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=89510"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=89510"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}