{"id":89426,"date":"2021-07-08T16:38:48","date_gmt":"2021-07-08T11:08:48","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=89426"},"modified":"2021-07-09T15:59:33","modified_gmt":"2021-07-09T10:29:33","slug":"1st-puc-basic-maths-previous-year-question-paper-march-2019-south","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-basic-maths-previous-year-question-paper-march-2019-south\/","title":{"rendered":"1st PUC Basic Maths Previous Year Question Paper March 2019 (South)"},"content":{"rendered":"

Students can Download 1st PUC Basic Maths Previous Year Question Paper March 2019 (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 2019 (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.
    \nDefine a prime number
    \nAnswer:
    \nNumbers which are divisible by one and the number itself is called prime number. {2,3,5,7,11,…………}<\/p>\n

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

    Question 3.
    \nSimplify \\(\\left\\{\\frac{5 x^{3}}{y}\\right\\}^{2}\\)
    \nAnswer:
    \n\\(\\frac{25 x^{6}}{y^{2}}\\)<\/p>\n

    Question 4.
    \nFind x if logx<\/sub>625 = 4.
    \nAnswer:
    \nx4<\/sup> = 625 \u21d2 x4<\/sup> = 54<\/sup> \u21d2 x = 5.<\/p>\n

    Question 5.
    \nFind the 11\u201c’ term of the AP 3, 5, 7, 9
    \nAnswer:
    \nT11<\/sub> = 3 + (11 – 1)2 \u21d2 Tn<\/sub> = 3 + 20 = 23.<\/p>\n

    \"KSEEB<\/p>\n

    Question 6.
    \nSolve for x if (x + 2) (x + 3) = (x – 2) (x – 4).
    \nAnswer:
    \nx2<\/sup> + 6 + 5x = x2<\/sup> + 8 – 6x
    \n11x = 2 \u21d2 x = \\(\\frac{2}{11}\\)<\/p>\n

    Question 7.
    \nFind the simple interest on \u20b9 600 for 3 years at 4% p.a.
    \nAnswer:
    \nSI = \\(\\frac{600 \\times 3 \\times 4}{100}\\) = \u20b9 72<\/p>\n

    Question 8.
    \nFind the present value of a perpetuity of \u20b9 3000 to be received forever at 4% p.a.
    \nAnswer:
    \nA = Rs, 3000
    \nr = 0.04
    \nP\u221e = \\(\\frac{\\mathrm{A}}{\\mathrm{r}}=\\frac{3000}{0.04}\\) = 75000<\/p>\n

    Question 9.
    \nConvert \\(\\frac{1}{4}\\) into percentage.
    \nAnswer:
    \n\\(\\frac{1}{4}\\) \u00d7 100 = 25%<\/p>\n

    \"KSEEB<\/p>\n

    Question 10.
    \nExpress \\(\\frac{3 \\pi}{2}\\) in degrees.
    \nAnswer:
    \n\\(\\frac{3 \\pi}{5}=\\frac{3 \\times 180}{5}\\) = 108\u00b0<\/p>\n

    Question 11.
    \nFind the value of sin2<\/sup>120 + cos2<\/sup>120.
    \nAnswer:
    \nsin2<\/sup>120 + cos2<\/sup>120 = 1<\/p>\n

    Question 12.
    \nFind the slope of line 2x + 3y – 11 = 0.
    \nAnswer:
    \n5y = -2x + 11
    \ny = \\(\\frac{-2}{5}\\)x + \\(\\frac{11}{5}\\) compare with y = mx + c
    \n\u2234 m = slope = \\(\\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 divisors of 825.
    \nAnswer:
    \n825 = 31<\/sup> x 52<\/sup> x 111<\/sup>
    \nP1<\/sub> = 3, \u03b11<\/sub> = 1,P2<\/sub> = 5, \u03b12<\/sub> = 2, P3<\/sub> = 11, \u03b13<\/sub> = 11
    \nT (n) = (1 + \u03b11<\/sub>) (1 + \u03b12<\/sub>) (1 + \u03b13<\/sub>)
    \n= (1+ 1)(1 +2)(1 + 1)
    \n= (2)(3)(2)
    \n= 12
    \n\"1st
    \nS(n) = \\(\\left(\\frac{p_{1}^{\\alpha_{1}+1}-1}{p_{1}-1}\\right)\\left(\\frac{p_{2}^{a_{i}+1}-1}{p_{2}-1}\\right)\\left(\\frac{p_{1}^{a_{1}+1}-1}{p_{3}-1}\\right)\\)
    \nS(825) = \\(\\frac{3^{2}-1}{3-1} \\times \\frac{5^{3}-1}{5-1} \\times \\frac{11^{2}-1}{11-1}\\)
    \n= \\(\\frac{8}{2} \\times \\frac{124}{4} \\times \\frac{120}{10}\\)
    \n= 4 \u00d7 31 \u00d7 12 = 1488<\/p>\n

    Question 14.
    \nFind the number which, when divided by 16, 20 and 40 leaves the same remainder 4.
    \nAnswer:
    \n16 = 24<\/sup>, 20 = 22<\/sup> \u00d7 51<\/sup>, 40 = 23<\/sup> \u00d7 51<\/sup>
    \nLCM = 24<\/sup> \u00d7 51<\/sup> = 16 \u00d7 5 = 80
    \n\u2234 Required number = 80 + 4 = 84.<\/p>\n

    \"KSEEB<\/p>\n

    Question 15.
    \nIf A = {1, 2, 3, 4}, B = {3, 4, 5, 6} and U = {1, 2-, 3, 4, 5, 6, 7, 8} verify (A\u222aB)’ = A’\u2229B’
    \nAnswer:
    \n\"1st
    \n1200 = 24<\/sup> \u00d7 52<\/sup> \u00d7 31<\/sup> = p1<\/sub>\u03b11<\/sub><\/sup>\u00d7 p2<\/sub>\u03b12<\/sub><\/sup> \u00d7 p3<\/sub>\u03b13<\/sub><\/sup>
    \nP1<\/sub> = 2 P2<\/sub> = 5; P3<\/sub> = 3 \u03b11<\/sub> = 4; \u03b12<\/sub> = 2; \u03b13<\/sub> = 1
    \nT(n) = (1 + \u03b11<\/sub>)(1 + \u03b12<\/sub>)(1 + \u03b13<\/sub>) = (1 + 4)(1 + 2)(1 + 1) = (5)(3)(2) = 30
    \nS(n) = \\(\\left(\\frac{\\mathrm{P}_{1}^{\\alpha_{1}+1}-1}{\\mathrm{P}_{1}-1}\\right)\\left(\\frac{\\mathrm{P}_{2}^{\\alpha_{2}+1}-1}{\\mathrm{P}_{2}-1}\\right)\\left(\\frac{\\mathrm{P}_{3}^{\\alpha_{3}+1}-1}{\\mathrm{P}_{3}-1}\\right)\\)
    \n= \\(\\left(\\frac{2^{5}-1^{\\circ}}{2-1}\\right)\\left(\\frac{5^{3}-1}{5-1}\\right)\\left(\\frac{3^{2}-1}{3-1}\\right)=\\left(\\frac{31}{1}\\right)\\left(\\frac{124}{4}\\right)\\left(\\frac{8}{2}\\right)\\)
    \n= 31 \u00d7 124 = 3844<\/p>\n

    Question 16.
    \nSimplify \\(\\left[\\frac{x^{a}}{x^{b}}\\right]^{a+b}\\left[\\frac{x^{b}}{x^{c}}\\right]^{b+c}\\left[\\frac{x^{c}}{x^{a}}\\right]^{c+a}\\)
    \nAnswer:
    \n\"1st<\/p>\n

    Question 17.
    \nWhich clement of the GP 5, 10, 20………. is 80.
    \nAnswer:
    \nTn<\/sub> = arn-1<\/sup>
    \n80 = 5.2x-1<\/sup>
    \n2n-1<\/sup> = 16 \u21d2 2n-1<\/sup> = 24<\/sup> \u21d2 n-1 = 4\u21d2 n = 5<\/p>\n

    Question 18.
    \nThe sum of the two numbers is 107 and their difference is 17. Find the numbers.
    \nAnswer:
    \n\"1st
    \n\u2234 a = 62
    \na + b = 107
    \nb = 107 – a = 107 – 62 = 45
    \n\u2234 b = 45<\/p>\n

    Question 19.
    \nSolve by formula method: 5x2<\/sup> – 7x – 12 = 0.
    \nAnswer:
    \nx = \\(\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}=\\frac{-(-7) \\pm \\sqrt{(-7)^{2}-4(5)(-12)}}{(2)(5)}\\) \u21d2 x = \\(\\frac{24}{10}\\) and x = -1<\/p>\n

    \"KSEEB<\/p>\n

    Question 20.
    \nSolve : 3x – 2 < 2x + 1 ; x\u2208 R and represent on the number line.
    \nAnswer:
    \n3x – 2 < 2x + 1
    \n3x – 2x < 2+ 1
    \nx < 3<\/p>\n

    Question 21.
    \nFind the future value of an annuity due of \u20b9 800 for 3 years at 5% p.a.
    \nAnswer:
    \nF = \\(\\frac{a\\left[(1+i)^{n}-1\\right]}{i}\\)(1 + i) = \\(\\frac{800\\left[(1+0.05)^{3}-1\\right]}{(0.05)}\\)(1 + 0.05)
    \nF = \u20b9 2287.14<\/p>\n

    Question 22.
    \nProve that tan2<\/sup>A(1 – sin2A) = sin2A.
    \nAnswer:
    \nLHS = tan2<\/sup> A cos2<\/sup> A = \\(\\frac{\\sin ^{2} A}{\\cos ^{2} A}\\)cos2<\/sup> A- = sin2<\/sup> A = RHS<\/p>\n

    Question 23.
    \nFind the value of cot2<\/sup>60\u00b0 + sin2<\/sup>45\u00b0 + sin2<\/sup>30\u00b0 + cos2<\/sup>90\u00b0.
    \nAnswer:
    \n\\(\\left(\\frac{1}{\\sqrt{3}}\\right)^{2}+\\left(\\frac{1}{\\sqrt{2}}\\right)^{2}+\\left(\\frac{1}{2}\\right)^{2}\\) + 0 = \\(\\frac{1}{3}+\\frac{1}{2}+\\frac{1}{4}=\\frac{13}{12}\\)<\/p>\n

    Question 24.
    \nA point P moves such that PA2<\/sup> = 3PB2<\/sup>. If A = (5, 0), 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 (\u20145, 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 25.
    \nIf the distance between the points (3, -2) and (-1, a) is 5 units find the values of a.
    \nAnswer:
    \nAB = \\(\\sqrt{(-1-3)^{2}+(a+2)^{2}}\\) =5
    \nSquaring and simplifying we get
    \na = 1 and a = -5<\/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

    Question 27.
    \nIf f(x) = x + 1 and g(x) = x2<\/sup> + 1. Find (1) fog(1) (2) fog(2) (3) gof(2)
    \nAnswer:
    \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 = 6<\/p>\n

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

    Question 28.
    \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 Equatlig powers on both sides, we get.
    \n\\(\\frac{2}{y}=\\frac{1}{x}+\\frac{1}{z}\\)<\/p>\n

    \"KSEEB<\/p>\n

    Question 29.
    \nSolve log x + log (x – 4) – log (x – 6) = 0
    \nAnswer:
    \nlogx(x-4) – log(x-6) = 0
    \n\u21d2 log\\(\\) = 0 \u21d2 \\(\\) = 1 \u21d2 x2<\/sup> – 5x + 6 = 0
    \nx = 2 or 3<\/p>\n

    Question 30.
    \nFind the three numbers in GP whose sum is 39 and their product is 729.
    \nAnswer:
    \nLet the three numbers be \\(\\frac{a}{r}\\),a,ar
    \nProduct = 729
    \n\u21d2 a3<\/sup> = 729 \u21d2 a3<\/sup> = 93<\/sup> \u21d2a = 9
    \nSum = 39
    \n\\(\\frac{a}{r}\\) + a + ar = 39
    \n\\(\\frac{9}{r}\\) + 9 + 9r = 39
    \n\\(\\frac{9}{r}\\) + 9r = 39 – 9
    \n\\(\\frac{9}{r}\\) + 9r = 30
    \n\u21d2 9 + 9r2<\/sup> = 30
    \nr \u21d2 9r2<\/sup> – 30r + 90
    \n\u21d2 9r2<\/sup> – 27r – 3r + 90
    \n9r(r – 3)-3 (r-3) = 0
    \n(r – 3)(9r – 3) = 0
    \nr = 3 or 9r = 3
    \nr = 3 or r = \\(\\frac{3}{9}=\\frac{1}{3}\\)
    \nr = 3 or r = \\(\\frac{1}{3}\\)<\/p>\n

    The numbers are \\(\\frac{a}{3}\\),a,ar
    \n\\(\\frac{9}{3}\\), 9, 9(3)
    \n3,9,27.<\/p>\n

    Question 31.
    \nFind the quotient and remainder when x4<\/sup> + 10x3<\/sup> + 39x2<\/sup> + 76x + 65 is divided by x +4.
    \nAnswer:
    \n\"1st
    \nQuotient = x3<\/sup> + 6x2<\/sup> + 15x + 16
    \nRemainder = 1<\/p>\n

    \"KSEEB<\/p>\n

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

    Question 33.
    \nFind the compound interest on \u20b9 7000 at 5% p.a for 4 years?
    \nAnswer:
    \nA = P(l + i)n<\/sup> \u21d2 A = 7000(1 + 0.05)4<\/sup> = 8508.54
    \nCI – A – P = 8508.54 – 7000 \u21d2 Cl = \u20b9 1508.54<\/p>\n

    Question 34.
    \nRamya purchased 3 varieties of cooking oil 4 kg of oil at \u20b9 50\/kg. 5 kg of oil at \u20b9 60\/ kg and 9 kg of oil at \u20b9 70\/kg. What is the average price of oil\/kg?
    \nAnswer:
    \nX\u0304w = \\(\\frac{x_{1} w_{1}+x_{2} w_{2}+x_{3} w_{3}}{w_{1}+w_{2}+w_{3}}=\\frac{50 \\times 4+60 \\times 5+70 \\times 9}{4+5+9}=\\frac{1130}{18}\\) = \u20b9 63.2 \/kg<\/p>\n

    Question 35.
    \nThe price of a pair of trousers was decreased by 22% of \u20b9 390. What was the original price of the trousers.
    \nAnswer:
    \nLet the original price be x. .
    \nThen (x – \\(\\frac{22 x}{100}\\)) = 390 \u21d2 x(1 – \\(\\frac{22}{100}\\)) = 390 \u21d2 x\\(\\frac{78}{100}\\) = 390
    \nx = \\(\\frac{390 \\times 100}{78}\\) = \u20b9 500<\/p>\n

    Question 36.
    \nFind the value of \\(\\frac{\\sin \\frac{\\pi}{2} \\cos ^{2} \\frac{\\pi}{6} \\sec ^{2} \\frac{\\pi}{4}}{\\tan \\frac{\\pi}{3}+\\cot \\frac{\\pi}{3}}\\)
    \nAnswer:
    \n\\(\\frac{\\sin \\frac{\\pi}{2} \\cos ^{2} \\frac{\\pi}{6} \\sec ^{2} \\frac{\\pi}{4}}{\\tan \\frac{\\pi}{3}+\\cot \\frac{\\pi}{3}}=\\frac{1 \\times\\left(\\frac{\\sqrt{3}}{2}\\right)^{2} \\cdot(\\sqrt{2})^{2}}{\\sqrt{3}+\\frac{1}{\\sqrt{3}}}=\\frac{\\frac{\\sqrt{3}}{4} \\times 2}{\\frac{(\\sqrt{3})^{2}+1}{\\sqrt{3}}}=\\frac{\\sqrt{3}}{2} \\times \\frac{\\sqrt{3}}{4}=\\frac{3}{8}\\)<\/p>\n

    Question 37.
    \nFind the third vertex of a triangle if two of its vertices are at (-2, 4) and (7, -3) and the centroid at (3, -2).
    \nAnswer:
    \nCentroid G(x, y) = \\(\\left(\\frac{x_{1}+x_{2}+x_{3}}{3}, \\frac{y_{1}+y_{2}+y_{3}}{3}\\right)\\)
    \n(3, 2) = \\(\\left(\\frac{-2+7+x_{3}}{3}, \\frac{4-3+y_{3}}{3}\\right)\\) \u21d2 x3<\/sub> = 4, y3<\/sub> = -7
    \n\u2234 Third vertex (x3<\/sub>, y3<\/sub>) = (4, -7)<\/p>\n

    \"KSEEB<\/p>\n

    Question 38.
    \nFind the distance between the two lines 2x – 3y + 4 = 0 and 4x – 6y – 5 = 0.
    \nAnswer:
    \nDistance between parallel lines = \\(\\left|\\frac{c_{1}-c_{2}}{\\sqrt{a^{2}+b^{2}}}\\right|=\\left|\\frac{8-(-5)}{\\sqrt{4^{2}+(-6)^{2}}}\\right|=\\frac{13}{\\sqrt{52}}=\\frac{\\sqrt{13}}{2}\\) units<\/p>\n

    Part – D<\/span><\/p>\n

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

    Question 39.
    \nIn a group of 50 people, 35 speak Kannada and 25 speak both English and Kannada and all the people speak at least one of the two languages. How many speak English? How many speak only English and not Kannada? How many speak only Kannada?
    \nAnswer:
    \nn(K \u222a E) = n[K) + n(E) – n(K \u2229 E)
    \n50 = 35 + n(E) – 25 \u21d2 K(E) = 40
    \n(i) n(E-K) = n(E)-n(E\u2229K) = 40-25 = 15
    \n(ii) n(K-E) = n(K) – n[K \u2229 E) = 35 -25 = 10<\/p>\n

    Question 40.
    \nEvaluate \\(\\frac{213.78 \\times 7.434}{6.321}\\) using logarithmic tables.
    \nAnswer:
    \nLet x = \\(\\frac{213.781 \\times 7.434}{6.321}\\)
    \nlogx = \\(\\left[\\frac{213.781 \\times 7.434}{6.321}\\right]\\)
    \n= log213.781 + logl.434 – log6.321
    \n= 2.3298 + 0.87 12 –0.8008.
    \n= 2.4002
    \nx = Antilog 2.4002
    \nx = 25 1.3.<\/p>\n

    \"KSEEB<\/p>\n

    Question 41.
    \nFind the sum of all numbers between 60 and 400 which are divisible by 13.
    \nAnswer:
    \nSn<\/sub> = 65 + 78 + 91 +…………..+390
    \na = 5 d = 13 n =? Tn<\/sub> = 390
    \nTn<\/sub> = a + (n – 1)d
    \n390 = 65 + (n – 1)13
    \n390 = 65 + 13n – 13
    \n13n = 338
    \n\u2234 n=26
    \nSn<\/sub> = \\(\\frac{n}{2}\\)[a + l] = (65+390) = 13 (455)= 5,915
    \n\u2234 S = 5,915<\/p>\n

    Question 42.
    \nA mother is 32 years older than her son. After 4 years the mother’s age will be 8 years more than twice that of her son. Find their present ages.
    \nAnswer:
    \nLet the present age of the son = x yrs.
    \nand the present age of the mother = x + 32 years
    \nAfter 4 years age of the mother = x + 36
    \nGiven that x + 36 = 8 + 2 (x + 4) = 8 + 2x + 8
    \n36 – 16 = 2x – x
    \nx = 20 years
    \n\u2234 son’s present age is x = 20 yrs
    \nmother’s present age is 20 + 32 = 52 years.<\/p>\n

    Question 43.
    \nFind the difference between simple interest and compound interest on \u20b9 18,000 invested for 4 years at 8% p.a. where compound interest is compounded annually.
    \nAnswer:
    \nSimple Interest
    \nP = 18,000 T = 4 R = 4
    \nSI = \\(\\frac{\\mathrm{PTR}}{100}\\)
    \n= \\(\\frac{18,000 \\times 4 \\times 8}{100}\\)
    \n= \u20b95,760<\/p>\n

    Compound Interest
    \nP = 18,000 n = 4 I = 008
    \nA = P(1 + i)n
    \n= 18,000 (1 + 0.08)4
    \n= 24,488.8
    \nCI = A – P
    \n= 24,488.8 – 18,000
    \nCI = 6488.8<\/p>\n

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

    Question 44.
    \nIf you want to have \u20b9 80,000 after 5 years, how much should you deposit every year if the bank offers 12% p.a. interest compounded quarterly.
    \nAnswer:
    \nF = 80,000, n = 5yrs, r 12% 0.12, i = \\(\\left(1+\\frac{0.12}{4}\\right)^{4}\\) – 1 = 0.125
    \nF = \\(\\frac{a\\left[(1+i)^{n}-1\\right]}{i}\\)
    \na = \\(\\frac{\\mathrm{F} \\times i}{(1+i)^{n}-1}=\\frac{80,000 \\times 10.125}{(1.125)^{5}-1}=\\frac{80,000}{6.416}\\) = \u20b912,468.8<\/p>\n

    \"KSEEB<\/p>\n

    Question 45.
    \nA businessman sells an article for \u20b9 720 and earns a profit of 20% Find the (a) cost price (b) profit percentage at the selling price.
    \nAnswer:
    \nS.P = 720 = \\(\\frac{120}{100}\\)x \u21d2 x = \\(\\frac{720 \\times 100}{120}\\) = 600
    \n\u2234 Profit = 720 – 600 = 120 \u21d2 profit% = \\(\\frac{\\text { profit in Rs. }}{\\text { S.P. }}\\) \u00d7 100 = \\(\\frac{120}{720}\\) \u00d7 100 = 16.66%<\/p>\n

    Question 46.
    \nIf cot \\(\\frac{\\theta}{2}=\\frac{5}{2}\\) q is acute. Show that \\(\\frac{3 \\cos \\theta+2 \\sin \\theta}{3 \\cos \\theta-4 \\sin \\theta}=\\frac{19}{7}\\)
    \nAnswer:
    \ncot\u03b8 = \\(\\frac{5}{2}\\) \u2234 Hyp = \\(\\sqrt{29}\\)
    \ncos\u03b8 = \\(\\frac{5}{\\sqrt{29}}\\), sin\u03b8 = \\(\\frac{2}{\\sqrt{29}}\\)<\/p>\n

    Substituting and simplifying we get
    \nLHS = \\(\\frac{3 \\cos \\theta+2 \\sin \\theta}{3 \\cos \\theta-4 \\sin \\theta}=\\frac{3 \\cdot \\frac{5}{\\sqrt{29}}+2 \\cdot \\frac{2}{\\sqrt{29}}}{3 \\cdot \\frac{5}{\\sqrt{29}}-4 \\cdot \\frac{2}{\\sqrt{29}}}=\\frac{15+4}{15-8}=\\frac{19}{7}\\)<\/p>\n

    Question 47.
    \nFind the ratio in which the line joining the points (3, 5) and (-7, 9) \u00a1s divided by the point (\\(\\frac{1}{2}\\),6).
    \nAnswer:<\/p>\n

    \"KSEEB<\/p>\n

    Question 48.
    \nFind the equation of a line that passes through (-4, 1), and a portion of it between the axes is divided by the point in the ratio 1:2.
    \nAnswer:
    \nA straight line passes through the point (-4, 7) and the portion of the line intercepted between the axes is divided at this point in the ratio 3: 2. Find the equation of the line.<\/p>\n

    Let the required equation of the line be
    \n\\(\\frac{\\mathrm{x}}{\\mathrm{a}}+\\frac{\\mathrm{y}}{\\mathrm{b}}\\) = 1 …………(1)
    \nThen it meets the axes at the points A (a, 0) and B (0, b). The coordinates of the point which divides the line AB in the ratio 3: 2 are
    \n\\(\\left(\\frac{2 \\cdot a+3 \\cdot 0}{2+3}, \\frac{2 \\cdot 0+3 \\cdot b}{2+3}\\right)=\\left(\\frac{2 a}{5}, \\frac{3 b}{5}\\right)\\) …….(2)
    \nThis is the same as (-4, 7) ……………(3)
    \nComparing (2) and (3), \\(\\frac{2 a}{5}\\) =-4, \\(\\frac{3b}{5}\\) = 7 or a = -10, b = \\(\\frac{35}{3}\\)
    \nSubstituting these values of a, b in (1) we get
    \n\\(\\frac{x}{-10}+\\frac{y}{\\frac{35}{3}}\\) = 1<\/p>\n

    Part – E<\/span><\/p>\n

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

    Question 49.
    \n(a) Find the sum to n terms of the series 5 + 55 + 555 +………..n terms.
    \nAnswer:
    \nLet S = 5 + 55 + 555 + ………. to n terms
    \n\\(\\frac{S}{2}\\) = 1 + 11 + 111 +……….to n terms
    \n\\(\\frac{9S}{2}\\) = 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 + to n terms)
    \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

    (b) A manufacture produced and sells balloons at \u20b9 8 per unit. His fixed cost is X 6500 and the variable cost per balloon is \u20b9 3.50. Calculate
    \n(i) Revenue function (ii) Cost function (iii) Profit function (iv) Break even point
    \nAnswer:
    \nManufacture produce\u00e7i and sells balloons at \u20b9 8 per unit. His fixed cost is \u20b9 6500 and the variable cost per Balloon is \u20b9 3.50 calculate,<\/p>\n

      \n
    1. Revenue Function<\/li>\n
    2. Cost Function<\/li>\n
    3. Profit Function<\/li>\n
    4. Break-Even point.<\/li>\n<\/ol>\n

      i.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) The weight of 6 men are 90 kg, 70.5 kg, 56 kg, 45.5 kg, 85 kg and 78 kg. Find the average weight.
      \nAnswer:
      \nAverage weight of 6 men = \\(\\frac{90+70.5+56+45.5+85+78}{6}=\\frac{425}{6}\\) = 70.83 kg<\/p>\n

      \"KSEEB<\/p>\n

      Question 50.
      \n(a) Find the equation of the straight line which passes through the point of intersection of 2x – 3y = 4 and 2x + 2y = 1 and perpendicular to the line x + 4y = 8.
      \nAnswer:<\/p>\n

      (b) If A = (1, 3, 5) B = {5} C = {7} verify A \u00d7 (B – C) = (A \u00d7 B) – (A \u00d7 C)
      \nAnswer:
      \nB – C = {5}
      \nA \u00d7 (B -C) = {1,3,5} \u00d7 {5} ={(1,5), (3,5), (5,5)}…………………. (1)
      \nA \u00d7 B= {(1,5), (3,5), (5,5)}
      \nA \u00d7 C = {(1,7), (3,7), (5,7)}
      \n(A \u00d7 B) – (A \u00d7 C) = {(1,5), (3,5), (5,5)} ………..(2)
      \nFrom (1) and (2), A – (B – C) = (A \u00d7 B) – (A \u00d7 C)<\/p>\n

      (c) If the HCF of two numbers is 42 and their product is 52920, find their LCM.
      \nAnswer:
      \nH.C.F \u00d7 L.C.M = A \u00d7 B
      \n42 \u00d7 LCM = 52920 \u21d2 LCM = \\(\\frac{52920}{42}\\) = 1260
      \n\u2234 LCM = 1260<\/p>\n","protected":false},"excerpt":{"rendered":"

      Students can Download 1st PUC Basic Maths Previous Year Question Paper March 2019 (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 2019 (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\/89426"}],"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=89426"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/89426\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=89426"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=89426"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=89426"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}