{"id":89564,"date":"2021-07-14T11:50:58","date_gmt":"2021-07-14T06:20:58","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=89564"},"modified":"2021-07-14T11:50:58","modified_gmt":"2021-07-14T06:20:58","slug":"1st-puc-basic-maths-previous-year-question-paper-march-2016-south","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-basic-maths-previous-year-question-paper-march-2016-south\/","title":{"rendered":"1st PUC Basic Maths Previous Year Question Paper March 2016 (South)"},"content":{"rendered":"

Students can Download 1st PUC Basic Maths Previous Year Question Paper March 2016 (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 2016 (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.
    \nGive the canonical representation of 385.
    \nAnswer:
    \n385 = 5′ \u00d7 71<\/sup> \u00d7 111<\/sup><\/p>\n

    Question 2.
    \nIf R = {(1, 2), (2, 3), (3, 3} a relation defined on the set A = {1, 2, 3} find R-1<\/sup>.
    \nAnswer:
    \nR-1<\/sup> = {(2, 1), (3,2), (3, 3)}<\/p>\n

    Question 3.
    \nIf A = (3, 5, 7}, B = {7, 9, 11} find B-A.
    \nAnswer:
    \nB – A = {9, 11}<\/p>\n

    Question 4.
    \nSimplify \\(\\sqrt[4]{x^{-4 \/ 3}}\\)
    \nAnswer:
    \n(x-4\/3<\/sup>)1\/4<\/sup> = x-1\/3<\/sup> = x1\/1\/3<\/sup> = \\(\\frac{1}{\\sqrt[3]{x}}\\)<\/p>\n

    Question 5.
    \nFind the value of x if log\\(\\sqrt{2}\\) x = 4
    \nAnswer:
    \n(\\(\\sqrt{2}\\))4 = x \u21d2 x = 22 \u21d2 x = 4<\/p>\n

    \"KSEEB<\/p>\n

    Question 6.
    \nFind the 8th term of the sequence 1, \\(\\frac{1}{2}\\), \\(\\frac{1}{3}\\)
    \nAnswer:
    \nT8<\/sub> = \\(\\frac{1}{8}\\)<\/p>\n

    Question 7.
    \nForm a quadratic equation with roots \u2018O\u2019 and \u2018-1′
    \nAnswer:
    \n(x – 0)(x + 1) = 0 \u21d2 x2 + x= 0<\/p>\n

    Question 8.
    \nWhat is the simple interest on Rs. 1000 for half a year at 10% p.a.?
    \nAnswer:
    \n\"1st<\/p>\n

    Question 9.
    \nThe sum of the ages of 10 girls in a class is 120 years. If one more girl of age 1 year is added to the group, find the average of 11 girls.
    \nAnswer:
    \nX\u030411<\/sub> = \\(\\frac{120+1}{11}=\\frac{121}{11}\\) = 11yrs<\/p>\n

    Question 10.
    \nConvert 15% to the ratio form.
    \nAnswer:
    \n3:20<\/p>\n

    Question 11.
    \nWrite the value of cosec2<\/sup> 10\u00b0 – cot2<\/sup>10\u00b0.
    \nAnswer:
    \ncosec2<\/sup>10 – cot2<\/sup>10 = 1<\/p>\n

    Question 12.
    \nFind the distance from the origin to the line 3x – 4y + 5 = 0.
    \nAnswer:
    \n\\(\\frac{5}{\\sqrt{3^{2}+4^{2}}}=\\frac{5}{\\sqrt{25}}=\\frac{5}{5}\\) = 1<\/p>\n

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

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

    Question 13.
    \nFind the total number of positive divisors of 180.
    \nAnswer:
    \n180 = 22<\/sup> \u00d7 32<\/sup> \u00d7 5
    \nT(180) = (2 + 1) (2 + 1) (1 + 1) = 3 \u00d7 3 \u00d7 2 = 18<\/p>\n

    Question 14.
    \nIf f(x) = x2<\/sup>, g(x) = x – 1, find go f(x) and gof(a).
    \nAnswer:
    \ngof (x) = gf(x) = g(x2<\/sup>) = x2<\/sup> – 1
    \ngof(a) = a2<\/sup> – 1<\/p>\n

    \"KSEEB<\/p>\n

    Question 15.
    \nSimplify: \\(\\frac{2^{7 a-2 b} \\cdot 8^{2 a-b}}{16^{a+b}}\\)
    \nAnswer:
    \n\\(\\frac{2^{7 a-2 b} \\cdot\\left(2^{3}\\right)^{2 a-b}}{\\left(2^{4}\\right)^{a+b}}=\\frac{2^{7 a-2 b} \\cdot 2^{6 a-3 b}}{2^{4 a+4 b}}\\) = 27a-2b+a-3b-4a-4b<\/sup> = 29a-9b<\/sup><\/p>\n

    Question 16.
    \nProve that log \\(\\sqrt{a \/ b}\\) + log \\(\\sqrt{b \/ c}+\\) + log \\(\\sqrt{c \/ a}+\\) = 0
    \nAnswer:
    \nLHS = \\(\\frac{1}{2}\\)log\\(\\frac{a}{b}\\) + \\(\\frac{1}{2}\\)log\\(\\frac{b}{c}\\) + \\(\\frac{1}{2}\\)log\\(\\frac{c}{a}\\)
    \n\"1st
    \n= \\(\\frac{1}{2}\\) \u00d7 0 = 0 = R.H.S<\/p>\n

    Question 17.
    \nFind the sum to infinity of GP. 3, -1, \\(\\frac{1}{3}\\), –\\(\\frac{1}{9}\\) ………. if it exists.
    \nAnswer:
    \nS\u221e<\/sub> = \\(\\frac{a}{1-r}\\) Given a = 3, r = –\\(\\frac{1}{3}\\)
    \nS\u221e<\/sub> = \\(\\frac{3}{1+\\frac{1}{3}}=\\frac{9}{4}\\)<\/p>\n

    Question 18.
    \nFind two numbers whose sum is 64 and the difference is 16.
    \nAnswer:
    \nLet 2 numbers be x and y
    \n\u2234 Given x + y = 64 & x – y = 16
    \nSolving the above equations we get
    \nx = 40 & y = 24
    \n\u2234 The two numbers are 24 and 40<\/p>\n

    Question 19.
    \nSolve the unequalities 5x – 3 > 3x + 1, x\u2208R and represent on the number line.
    \nAnswer:
    \n5x – 3 > 3x + 1 & x \u2208 R
    \n\"1st
    \n5x – 3x > 1 + 3
    \n2x > 4 \u21d2 x > 2<\/p>\n

    Question 20.
    \nA certain sum of money amounts to Rs.24,200 in 2 years
    \nAnswer:
    \nGiven F = 24,200, n = 2yrs r = 10%, P = ?
    \nF = P(i + r)n<\/sup>
    \nP = \\(\\frac{\\mathrm{F}}{(1+r)^{n}}=\\frac{24,200}{(1+0.1)^{2}}=\\frac{24,200}{(1.1)^{2}}\\) = \u20b920,000
    \n\u2234 The Sum = \u20b920,000<\/p>\n

    Question 21.
    \nRam and Shyam went up a hill at a speed of 20kmph. Both of them came tumbling down the same distance at a speed of 30 km ph. Find the average speed for the round trip.
    \nAnswer:
    \nAverage speed = \\(\\frac{2 d}{\\frac{d}{20}+\\frac{d}{30}}=\\frac{2 d}{\\frac{5 d}{60}}=\\frac{120}{5}\\) = 24 kmph<\/p>\n

    Question 22.
    \nWhile taking measurement a tailor writes 34 instead of 24. What is the percentage error?
    \nAnswer:
    \nFor 24 \u2192 10 Errors
    \n\u2234 100 \u2192 ?
    \n\\(\\frac{10 \\times 100}{24}\\) = 41.66
    \nHence percentage error = 41.66<\/p>\n

    Question 23.
    \nIf tan A = 5\/12 acute find the value of \\(\\frac{\\sin A+\\cos A}{\\sin A-\\cos A}\\)
    \nAnswer:
    \nGiven tan A = \\(\\frac{5}{12}=\\frac{\\mathrm{opp}}{\\mathrm{adj}}\\)
    \n\u2234 HYP = \\(\\sqrt{12^{2}+5^{2}}=\\sqrt{169}\\) = 13
    \n\u2234 sinA = \\(\\frac{5}{13}\\), & cos A = \\(\\frac{12}{13}\\)
    \n\"1st<\/p>\n

    Question 24.
    \nShow that the points A(4, -2), B(2, -4), and C(7, 1) are collinear using the slope method.
    \nAnswer:
    \nA(4, -2) B(2, -4) C(7, 1)
    \nSlope AB = \\(\\frac{y_{2}-y_{1}}{y_{2}-x_{1}}=\\frac{-4-(-2)}{2-4}=\\frac{-2}{-2}\\) = 1
    \nSlope of BC = \\(\\frac{1-(-4)}{7-2}=\\frac{5}{5}\\) = 1
    \nSlopes are equal \u2234 the points are collirear<\/p>\n

    \"KSEEB<\/p>\n

    Question 25.
    \nFind the equation of the line through (1, 2) and making an angle of 60\u00b0 with the positive direction of x-axis.
    \nAnswer:
    \nGiven = (1, 2) and \u03b8 = 60\u00b0
    \nm = tan \u03b8 = tan 60\u00b0 = \\(\\sqrt{3}\\)
    \nRequired equation is y – y1<\/sub> = m(x – x1<\/sub>)
    \ny – 2 = \\(\\sqrt{3}\\)(x – 1)
    \ny – 2 = \\(\\sqrt{3}\\)x – \\(\\sqrt{3}\\)
    \n\u21d2 \\(\\sqrt{3}\\)x – y + 2 – \\(\\sqrt{3}\\) = 0<\/p>\n

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

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

    Question 26.
    \nFind the greatest number which divides 98, 120, and 153 leaving the same remainder.
    \nAnswer:
    \nWe have to find the H.C.F of 120 – 98, 153 – 120, 153-98
    \nRequired is H.C.F of 22, 33, 55
    \n\u2234 H.C.F is 11
    \nHence the greatest number is 11.<\/p>\n

    Question 27.
    \nShow that the relation \u201cCongruent to\u201d on the set of all triangles is an equivalence relation.
    \nAnswer:
    \nEvery triangle is congruent to itself
    \n\u2234 the relation is reflexive
    \n\u2206T1<\/sub> is congruent to \u2206T2<\/sub> \u21d2 \u2206T2<\/sub> is congruent to \u2206T1<\/sub> \u21d2 R is symmetric
    \nIf triangle T, is congruent to T2<\/sub>, \u2206T2<\/sub> is congruent to \u2206T3<\/sub> then \u2206T1<\/sub> is congruent to \u2206T3<\/sub>
    \n\u2234 the relative is transitive Hence the relation \u201ccongruent to\u201d is an equivalence relation.<\/p>\n

    Question 28.
    \nIf A = {1, 3, 5}, B = {5, 7}, C = {7} verify that A \u222a (B \u2229 C) – (A \u222a B) \u2229 (A \u222a C)
    \nAnswer:
    \nA \u222a (B \u2229 C) = {1, 3, 5} \u222a {7} = {1, 3, 5, 7}
    \n(A \u2229 B) \u2229 (A \u222a C) = {1, 3, 5, 7} \u2229 {1, 3, 5, 7} = {1, 3, 5, 7}<\/p>\n

    Question 29.
    \nIf px<\/sup> = qy<\/sup> = rz<\/sup> = sw<\/sup> and pq = rs then prove that \\(\\frac{1}{x}+\\frac{1}{y}=\\frac{1}{z}+\\frac{1}{w}\\)
    \nAnswer:
    \nLet px<\/sup> = qy<\/sup> = rz<\/sup> = sw<\/sup> = k
    \n\u21d2 p = k1\/w<\/sup>, q = k1\/y<\/sup>, r = k1\/z<\/sup>, s = k1\/w<\/sup>
    \nAlso given pq = rs
    \n\u21d2 k1\/x<\/sup>k1\/y<\/sup>= k1\/z<\/sup>k1\/w<\/sup>
    \n\u21d2 k1\/x+1\/y<\/sup>\u00a0= k1\/z+1\/w<\/sup>
    \n\u21d2 \\(\\frac{1}{x}+\\frac{1}{y}=\\frac{1}{z}+\\frac{1}{w}\\)<\/p>\n

    Question 30.
    \nFind the nature of roots of the equation 3x2<\/sup> + 2x + 1 = 0 without solving. If \u03b1 and \u03b2 are the roots of 3x2<\/sup> + 2x + 1 = 0 find the value of \\(\\).
    \nAnswer:
    \n\u0394 = b2<\/sup> – 4ac = 4 – 12 = -8 < 0 \u2234 the roots are imaginary
    \n\u03b1 + \u03b2 = \\(\\frac{-b}{a}=\\frac{-2}{3}\\) & \u03b1.\u03b2 = \\(\\frac{c}{a}=\\frac{1}{3}\\)
    \n\"1st<\/p>\n

    \"KSEEB<\/p>\n

    Question 31.
    \nThe sum of four numbers which are in A.P. is 28 and 10 times the least numbers is 4 times the greatest. Find the numbers.
    \nAnswer:
    \nLet the 4 numbers are a – 3d, a-d, a+d, a+3d
    \nGiven sum = 28
    \n\"1st
    \n4a – 28 \u21d2 a = 1
    \nAlso given 10(a – 3d) – 4(a + 3d)
    \n10a – 30d = 4a + 12d
    \n6a = 42 d
    \n42 = 42d \u21d2 d = 1
    \n\u2234 the 4 numbers are 4, 6,8,10<\/p>\n

    Question 32.
    \nSolve the linear inequalities x + 3y \u2264 3 2x + y, 2, x,y \u2265 0 graphically.
    \nAnswer:
    \nConsider x + 3y = 3 .
    \nIf x = 0, then y = 1 (0,1)
    \nIf y = 0 then x = 3 (3, 0)<\/p>\n

    2x + y – 2
    \nIf x = 0, y = 2 (0, 2)
    \nIf y = 0 then x = 1, (1, 0)
    \nPlot the two lines on the graph there
    \n\"1st<\/p>\n

    Question 33.
    \nXYZ suppliers buy a machine for Rs. 20,000 the rate of depreciation is 10%. Find the depreciated value of the machine after 3 years. Also, find the amount of depreciation.
    \nAnswer:
    \nB = C (1 – r)n<\/sup>
    \nGiven C = 20,000, r = 0.1, n = 3 B = ?
    \nB = 20,000 (1~0.1)3<\/sup> = \u20b9 14,580
    \nthe amount of depreciation = 14,580 \u20b9<\/p>\n

    \"KSEEB<\/p>\n

    Question 34.
    \nThe average age of A and B is 18 years, the average age of B and C is 17 years and that of C and A is 20 years. Find A\u2019s age.
    \nAnswer:
    \nGiven a + b = 36 – 0 \u2192 (1)
    \nb + c = 34 (2) \u2192 (2)
    \nc + a = 40 \u2192 (3)
    \nAdding we get 2{a + 6 + c) = 110 \u21d2 a + b + c = 55 \u2192 (4)
    \nEqu 4 – Equ 2 gives a = 55 – 34 = 21
    \nWhere A\u2019s age = a = 21 yrs<\/p>\n

    Question 35.
    \nA certain amount invested at 4% p.a. compounded semi-annually amount to Rs. 78030 at the end of one year. Find the sum.
    \nAnswer:
    \nF = 78,030, r = 4% = 0.04, n = 1 yr
    \nF = P\\(\\left(1+\\frac{r}{2}\\right)^{n \\times 2}\\)
    \n78,030 = P\\(\\left(1+\\frac{0.02}{2}\\right)^{2}\\)
    \nP = \\(\\frac{78.030}{(1.01)^{2}}\\) = \u20b975,000<\/p>\n

    Question 36.
    \nProve that \\(\\sqrt{\\frac{\\sec \\theta+1}{\\sec \\theta-1}}=\\frac{1+\\cos \\theta}{\\sin \\theta}\\)
    \nAnswer:
    \nL.H.S = \\(\\sqrt{\\frac{\\sec \\theta+1}{\\sec \\theta-1}} \\times \\frac{\\sqrt{\\sec \\theta+1}}{\\sqrt{\\sec \\theta+1}}\\)
    \n\\(\\frac{(\\sqrt{\\sec \\theta+1})^{2}}{\\sqrt{\\sec ^{2} \\theta-1}}=\\frac{\\sec \\theta+1}{\\tan \\theta}=\\frac{\\frac{1}{\\cos \\theta}+1}{\\frac{\\sin \\theta}{\\cos \\theta}}=\\frac{1+\\cos \\theta}{\\sin \\theta}\\) = R.H.S<\/p>\n

    Question 37.
    \nDerive slope-intercept form of line y = mx + c. Also, write the equation of passing through origin with slope m.
    \nAnswer:
    \nLet the line passess thr y-axis at A(0, C) and x-axis at B, and m = tan\u03b8 is slope of the line by using one point formula we get y – y1<\/sub> = m(x – x1<\/sub>)
    \ny – c = m(x – 0)
    \n\u21d2 y – c – mx
    \n\u21d2 y = mx + c is the required equation of the line.
    \nAlso equation passing thr (0, 0) is y = mx
    \n\"1st<\/p>\n

    Question 38.
    \nFind the equation of the line which passes through the point (-4, 5) and whose intercepts are equal in magnitude but opposite in sign.
    \nAnswer:
    \nIntercept form is \\(\\frac{x}{a}+\\frac{y}{b}\\) = 1
    \nGiven b = -a
    \n\u2234 \\(\\frac{x}{a}+\\frac{y}{-a}\\) – 1
    \nof x – y = a
    \nThis equn passess through (-4, 5)
    \n-4 – 5 = a \u21d2 a = – 9
    \n\u2234 The required equation is x – y = 9<\/p>\n

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

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

    Question 39.
    \nIn a college, 2\/5th of the students play basketball, and 3\/4th play volleyball. If 50 students play none of these two games and 125 play both find the numbers of students who play.
    \n(a) at least one of the two games.
    \n(b) exactly one
    \nRepresent the results using the Venn diagram.
    \nAnswer:
    \nLet total students = x,
    \nLet A = basket ball and B = volley ball
    \n\u2234 n(A) = \\(\\frac{2}{5}\\)x
    \nn(B) = \\(\\frac{3}{4}\\)x<\/p>\n

    Given n (AUB)’ = 50 n(AnB) = 125
    \nn(AUB) = x – x(AUB)’ = x – 50<\/p>\n

    (a) n(AuB) = n(A) + n(B) – n(A n B)
    \nx – 50 = 0.4x + 0.75x – 125
    \n\u2234 75 = 1.15x – 1x = 0.15x
    \nx = \\(\\frac{75}{0.15}\\) = 500
    \nn( A u B) = x – 50 = 500 – 50 = 450<\/p>\n

    (b) n(only A) + n(only B)
    \nn(A) = \\(\\frac{2}{5}\\) \u00d7 500 = 200
    \n= 200 – 125 + 375 – 125
    \n\"1st
    \n= 325<\/p>\n

    Question 40.
    \nEvaluate \\(\\frac{1.234 \\times .0 .8921}{43.43 \\times 0.0092}\\) using logarithm table.
    \nAnswer:
    \nLet x = \\(\\frac{1.234 \\times .0 .8921}{43.43 \\times 0.0092}\\)
    \nTaking logm both sides we get
    \nlogx = log 1.234 + log 0.8921 – log43.43 – log 0.0092
    \n= 0.931 + 1.9504 – 1.6378 – 3.9638
    \nlogx = 0.4401
    \nx = A.L(0.440l)
    \nx = 2.755.<\/p>\n

    Question 41.
    \nMs. Sneha buys a second-hand bike for Rs. 18,000 she pays Rs. 12,000 cash and agrees to pay the balance in 12 annual installments of Rs. 500 each plus 10% interest on the unpaid amount. How much will the bike cost her?
    \nAnswer:
    \nBalance amount 18,000 – 12,000 = 6,000 Rs.
    \nNo. of instalment = 12, Each unit amt = Rs. 500
    \n1st instalment = 500 + 6000 \u00d7 0.1 = 1,100
    \n2nd instalment = 500 + 5000 \u00d7 0.l = 1050
    \n3rd instalment = 500 + 5000 \u00d7 o. 1 = 1000
    \nCost of instalments = 1,100 + 1050 + 1000 +……………. 12 instalment
    \nThe terms are in A.P with d = -50, a = 1,100
    \n\u2234 Sm<\/sub> = \\(\\frac{12}{2}\\)[1,100 + (12-1) \u00d7 -50] = 6(1,100 – 550) = 6 \u00d7 550 = \u20b93300
    \n\u2234 Total cost= 12,000 +3,300 = \u20b915,300<\/p>\n

    \"KSEEB<\/p>\n

    Question 42.
    \nObtain a root of the equation x3<\/sup> – 2x2<\/sup> -2x + 3- 0 by inspection and then solve using synthetic division method.
    \nAnswer:
    \nf(x) = x3<\/sup> – 2x2<\/sup> – 2x + 3
    \nf(1) = 0 \u2234 1 is a root
    \n\u2234 by sythetic division we have
    \n\"1st
    \n\u2234 Reduced equation is x2<\/sup> – x – 3 = 0
    \n\u2234 Roots are x = \\(\\frac{1 \\pm \\sqrt{13}}{2}\\)
    \nHence the 3 roots are 1, \\(\\frac{1+\\sqrt{13}}{2}\\), \\(\\frac{1-\\sqrt{13}}{2}\\)<\/p>\n

    Question 43.
    \nA person spent 30% of his wealth and thereafter \u20b9 20,000 and a further 10% of the remainder. If 29,250 is still remaining, what was his total wealth amount?
    \nAnswer:
    \nLet total wealth be Rs. x
    \n\"1st
    \n0.63x = 18,000
    \nx = \u20b9 75,000<\/p>\n

    Question 44.
    \nA father wants to send his child for higher studies after 15 years. He expects the cost of higher studies to be Rs. 1,00,000. How much should he share annually to have one lakh after 15 years if the interest rate \u00a1s 12% C.I. per annum?
    \nAnswer:
    \nGiven data F = 1,00000, a = ?, n = 15, r = 0.12
    \nF = \\(\\frac{a\\left[(1+i)^{n}-1\\right]}{i}\\)
    \n1,00,000 = a\\(\\frac{\\left[(1+0.12)^{15}-1\\right]}{0.12}\\)
    \na = \\(\\frac{1,00000}{(1.12)^{75}-1}\\) = \u20b9 2682.47<\/p>\n

    Question 45.
    \nFind x if \\(\\frac{x \\sin ^{2} 300^{\\circ} \\cdot \\sec 240^{\\circ}}{\\cos ^{2} 225^{\\circ} \\cdot {cosec} 240^{\\circ}}\\) = cot 135\u00b0 tan 315\u00b0
    \nAnswer:
    \nsin 300 = sin (360 – 60) = – sin 60 = \\(\\frac{\\sqrt{3}}{2}\\)
    \nsec 240 = sec(180 + 60)= – sec60\u00b0= – 2
    \ncos 225 = cos(180 + 45\u00b0) – cos45\u00b0= – \\(\\frac{1}{\\sqrt{2}}\\)
    \ncosec 240 = cosec (180 + 60\u00b0) = – cosec 60 = – \\(\\frac{2}{\\sqrt{3}}\\)
    \ncot 315\u00b0 = cot (360 – 45\u00b0) = – cot 45\u00b0 = – 1
    \ntan 300 = tan (360 – 60\u00b0) = – tan 600 = – \\(\\sqrt{3}\\)<\/p>\n

    \u2234 The given equation becomes:
    \n\\(\\frac{x \\cdot \\frac{3}{4} \\cdot 4}{\\frac{1}{2}, \\frac{4}{3}}\\) = 1.3
    \n\u21d2 x = \\(\\frac{6}{9}=\\frac{2}{3}\\)<\/p>\n

    \"KSEEB<\/p>\n

    Question 46.
    \nIn what ratio is the line segment joining the points (4, 5) and (1, 2) is divided by the x-axis. Also, find the coordinates of the point of division.
    \nAnswer:
    \nPoint on x-axis is y = 0
    \nLet P(x, y) divide AB in
    \nthe ratio k: 1 ratio
    \n\u2234 P(x, y) = \\(\\left(\\frac{4 k+1}{k+1}, \\frac{5 k+2}{k+1}\\right)\\)
    \n(x, 0) = \\(\\left(\\frac{4 k+1}{k+1}, \\frac{5 k+2}{k+1}\\right)\\)
    \n\u21d2 \\(\\frac{5 k+2}{k+1}\\) = 0 \u21d2 k = \\(\\frac{-2}{5}\\)
    \nratio is 2: 5 externaily
    \n\u2234 P = (-1, 0)<\/p>\n

    Question 47.
    \nFind the equation for the locus of a point that moves such that the sum of its distances from (0, 3) and (0, -3) is 8 units.
    \nAnswer:
    \nLet P(x, y)be a moving point
    \nGiven |PA| + |PB| = 8
    \n|PA| = \\(\\sqrt{x^{2}+(y-3)^{2}}\\) & |PB| = \\(\\sqrt{x^{2}+(y+3)^{2}}\\)
    \n|PA| = 8 – |PB|
    \nS.B.S We get
    \nx2<\/sup> + (y – 3)2<\/sup> = \\(\\left(8-\\sqrt{x^{4}+(y+3)^{2}}\\right)^{2}\\)
    \nSimplifying we get the required equation of the locus 16x2<\/sup> + 7y2<\/sup> = 112.<\/p>\n

    Question 48.
    \nShow that the points (2, -2), (8, 4), (5, 7), and (-1, 1) are taken murder the vertices of a rectangle.
    \nAnswer:
    \nAB = \\(\\)
    \nSimilarly BC = \\(\\sqrt{18}\\) , CD = \\(\\sqrt{72}\\) , DA = \\(\\sqrt{18}\\) , AC = BD = \\(\\sqrt{90}\\)
    \nOpposite sides are equal and diagonals are equal<\/p>\n

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

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

    Question 49.
    \n(a) If x = a r sin A cos B, y = br sin A sin B, z = cr cos A, prove that \\(\\)
    \nAnswer:
    \n\"1st
    \n= r2<\/sup>[sin2<\/sup> A(cos2<\/sup>B + sin2<\/sup>B) + cos2<\/sup>A] = r2<\/sup> (sin2<\/sup> A + cos2<\/sup> A)
    \n= r2<\/sup> = RHS.<\/p>\n

    (b) Prove that the lines x + y + 4 = 0, 2x = 3y + 7 and 3x + y = -6 are concurrent. Also find the point of concerrency.
    \nAnswer:
    \nSolving equations 1 and 2 we get
    \nx + y + 4 = 0 …(1)
    \n2x – 3y – 7 = 0 …(2)
    \nx = -1 & y = -3
    \nPut x = -1 and y = -3 in eqn (3) We get
    \n3x + y = -6
    \n3(-1) + -3 = -6 – 6 = – 6
    \nHence the 3 lines are concurrent and the point of concurrency (-1, -3)<\/p>\n

    (c) Find the present value of a perpetuity of Rs. 3000 to be received forever at 4 p.a.
    \nAnswer:
    \nGiven A = 3000, r = 0.04, P\u221e = ?
    \nP\u221e = \\(\\frac{A}{r}=\\frac{3000}{0.04}\\) = 75,000
    \n\u2234 Present value of perpetuity = 75,000<\/p>\n

    \"KSEEB<\/p>\n

    Question 50.
    \n(a) Find the sura of all numbers between 50 and 200 which are divisible by 11.
    \nAnswer:
    \nNumbers are 55, 66, 77, ……………….. 198
    \nwhich are in A.P
    \nHere a = 55, Tn<\/sub>\u00a0= 198, d= 11
    \n198 = 55 + (n – 1)11
    \nn = 14
    \n\u2234 Sn<\/sub> = \\(\\frac{14}{2}\\)(55 + 198) = 1771<\/p>\n

    (b) A company sells \u2018x\u2019 tins of talcum powder per day at Rs. 10\/tin. The cost of manufacturing a tin is Rs. 6 and distributor charge per tin is Rs. 1. Besides, the daily overhead cost is Rs. 600. Determine
    \n(i) Cost function
    \n(ii) Revenue function
    \n(iii) Profit function
    \n(iv) Break even level of production.
    \nAnswer:
    \n(i) C(x) =V.C + E.C
    \n= 6x + x + 600
    \n\u2234 C(x) = 7x + 600<\/p>\n

    (ii) R(x) = Price x Quantity
    \nR(x) = 10x<\/p>\n

    (iii) P(x) = R(x) – C(x)
    \n= 10x – 7x – 600
    \nP(x) = 3x – 600<\/p>\n

    (iv) At BEP:P(x) = 0 or R(x) = C(x)
    \n3x – 600 = 0
    \nx = 200 units<\/p>\n

    \u2234 Break even level of production 200 units<\/p>\n

    (c) Form the LCM of 2\/3, 4\/5, 6\/7
    \nAnswer:
    \nLCM of fractions = \\(\\frac{\\text { LCM of NR }}{\\text { HCF of DNR }}=\\frac{\\text { LCM of } 2,4,6}{\\text { HCF of } 3,5,7}=\\frac{12}{1}\\) = 12<\/p>\n","protected":false},"excerpt":{"rendered":"

    Students can Download 1st PUC Basic Maths Previous Year Question Paper March 2016 (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 2016 (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\/89564"}],"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=89564"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/89564\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=89564"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=89564"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=89564"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}