{"id":89244,"date":"2021-07-06T09:06:36","date_gmt":"2021-07-06T03:36:36","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=89244"},"modified":"2021-07-06T09:06:36","modified_gmt":"2021-07-06T03:36:36","slug":"1st-puc-basic-maths-model-question-paper-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-basic-maths-model-question-paper-2\/","title":{"rendered":"1st PUC Basic Maths Model Question Paper 2 with Answers"},"content":{"rendered":"

Students can Download 1st PUC Basic Maths Model Question Paper 2 with Answers, 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 Model Question Paper 2 with Answers<\/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 4 – 5i
    \nAnswer:
    \n-5<\/p>\n

    Question 2.
    \nIf A = {1, 2, 3, 4} and B = {1, 2, 3, 4, 5, 6, 7} find A – B.
    \nAnswer:
    \nA – B = { }<\/p>\n

    Question 3.
    \nIf f: R \u2192 R is defined by f(x) = 3x + 5 then find f (-1)
    \nAnswer:
    \nf(-1) = 3(- 1) + 5 = -3 + 5 = 2<\/p>\n

    Question 4.
    \nSimplify (5\u00b0)2 + 5
    \nAnswer:
    \n(1)2 + 5′ = 1 +5 = 6<\/p>\n

    Question 5.
    \nFind the value of log10<\/sub>.01
    \nAnswer:
    \nlog100<\/sub>.01 = x \u21d2 10x = 0.01
    \n10x = \\(\\frac{1}{100}\\) \u21d2 10x = 10-2
    \nx = -2<\/p>\n

    \"KSEEB<\/p>\n

    Question 6.
    \nFind the 8th term of the progression -2, -4, -6……….
    \nAnswer:
    \na = -2, d = -4 + 2 = -2, n = 8, T8<\/sub> = ?
    \nTn = a + (n – l)d, T8<\/sub> = -2 + (8 – 1) (-2) = -2 + 7(-2) = -2 – 14 = -16.<\/p>\n

    Question 7.
    \nSolve for x : 2(7 + x) – 10 = 16 – 2 (x – 24).
    \nAnswer:
    \n14 + 2x – 10 = 16 – 2x + 48
    \n2x + 2x = 16 + 48 – 14 + 10
    \n4x = 60 \u21d2 x = 15<\/p>\n

    Question 8.
    \nConvert the ratio 3: 5 into percentages.
    \nAnswer:
    \n3: 5= \\(\\frac{3}{5}\\) \u00d7 100% = 3 \u00d7 20% = 60%.<\/p>\n

    Question 9.
    \nDefine perpetuity.
    \nAnswer:
    \nIf annuity payments are made for an infinite period then it is called a perpetuity.<\/p>\n

    Question 10.
    \nConvert 450\u00b0 into radians.
    \nAnswer:
    \n450\u00b0 = 450 \u00d7 \\(\\frac{\\pi}{180}=\\frac{45 \\pi}{18}=\\frac{5 \\pi}{2}\\)<\/p>\n

    \"KSEEB<\/p>\n

    Question 11.
    \nThe average age of 10 boys in a class is 13 years. What is the sum of their ages?
    \nAnswer:
    \nSum of their ages =10 \u00d7 13 = 130<\/p>\n

    Question 12.
    \nFind the slope of the line 2x + 5y – 11 = 0
    \nAnswer:
    \nSlope of the line 2x + 5y – 11 = 0 is \\(\\frac{-a}{b}=\\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 which when divided by 36,40 and 48 leaves the same remainder 5,
    \nAnswer:
    \nLet us find out the LCM of 36, 40, 48
    \n36 = 22<\/sup> \u00d7 32<\/sup>
    \n40 = 23<\/sup> \u00d7 5
    \n48 = 24<\/sup> \u00d7 3
    \nLCM = 24<\/sup> \u00d7 32<\/sup> \u00d7 51<\/sup>
    \n= 16 \u00d7 45
    \n= 720
    \nSince 5 has to be the remainder, we have to add 5 to 720 i.e. 720 + 5 = 725<\/p>\n

    Question 14.
    \nIf A = {1,23, 4) and B = (34, 5, 6) and U = {1,2, 3,4, 5,6, 7,8)
    \nVerify (A \u222a B)’ =A’ \u2229 B’
    \nAnswer:
    \nA \u222a B = {1,2,3,4,5,6}
    \n(A\u222aB)’ = (7,8)
    \nA’ = {5, 6,7,8) B’ = {1,2, 7,8)
    \nA’\u2229 B’ = {7, 8)
    \n\u2234 (A\u222aB)’ = A’ \u2229 B’<\/p>\n

    Question 15.
    \nFind the number of positive divisors and the sum of all positive divisors of 1200.
    \nAnswer:
    \n1200 = 24<\/sup> \u00d7 52<\/sup> \u00d7 31<\/sup> = P1<\/sub>\u03b11<\/sup> \u00d7 P2<\/sub>\u03b12<\/sup> \u00d7 P3<\/sub>\u03b13
    \n<\/sup>P1 <\/sub>= 2; P2<\/sub> = 5; P3<\/sub> = 3
    \n\u03b11 <\/sub>= 4; \u03b12 <\/sub>= 4; \u03b13 <\/sub>= 1
    \nT(n) = (1 + \u03b11<\/sub>)(1 + \u03b12<\/sub>)(1 + \u03b13<\/sub>) = (1 + 4)(1 + 2)(1 + 1) = (5) (3) (2) = 30<\/p>\n

    Sn<\/sub> = \\(\\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^{t}}{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)\\) = 31 \u00d7 124 = 3844
    \n\"1st<\/p>\n

    Question 16.
    \nSimplify \\(\\frac{2^{n+1}+2^{n-1}}{2^{n}+2^{n+2}}\\)
    \nAnswer:
    \n\"1st<\/p>\n

    \"KSEEB<\/p>\n

    Question 17.
    \nIf a, b, c are in G.P and ax<\/sup> = by<\/sup> = cz<\/sup> show that x, y and z are in H.P
    \nAnswer:
    \na,b,c are in GP
    \n\u2234 b2<\/sup> = ac
    \nax<\/sup> = by<\/sup> = cz<\/sup> = k
    \nax<\/sup> = k \u21d2 a = k1\/x<\/sup>
    \nby<\/sup> = k \u21d2 b = k1\/y<\/sup>
    \ncz<\/sup> = k \u21d2 c = k1\/z<\/sup><\/p>\n

    b2 = ac
    \n(k1\/y<\/sup>) = k1\/x+1\/z<\/sup>
    \n\\(\\frac{2}{y}=\\frac{1}{x}+\\frac{1}{z}\\)
    \n\\(\\frac{2}{y}=\\frac{x+z}{x z}\\)
    \ny = \\(\\frac{2 x z}{x+z}\\)
    \n\u2234 x, y, z are in HP.<\/p>\n

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

    Question 19.
    \nDetermine the principal which will amounts to \u20b9 15000 in 8 years at 41% per annum simple interest
    \nAnswer:
    \nH = \u20b9 15,000, R = 11, T = 8
    \nA = P + I = P + \\(\\frac{P+R}{100}\\)
    \n1500 = p(1 + \\(\\frac{\\mathrm{TR}}{100}\\)) = p(\\(\\frac{100+8 \\times 11}{100}\\)) = p\\(\\left(\\frac{188}{100}\\right)\\) \u21d2 p = \\(\\left(\\frac{188}{100}\\right)\\)
    \n\u2234 P = \u20b9 7978.72<\/p>\n

    Question 20.
    \nSolve 3x – 2 < 2x + 1 when x is an integer and x is a real number. Also represent on a number line.
    \nAnswer:
    \n3x – 2 < 2x + 1
    \n3x – 2x < 2 + 1
    \nx < 3
    \n(i) when x is an integer x = {………- 2, -1, 0, 1, 2}
    \n(ii) when x is a real number x \u2208 (-\u221e, 3)
    \n\"1st<\/p>\n

    Question 21.
    \nThe average score of 20 boys is 60% and the average score of 30 girls is 70%. Find the combined average.
    \nAnswer:
    \nX1<\/sub> = 60%, N1 = 20
    \nX2<\/sub> = 70%, N2 = 30
    \nX\u0304 = \\(\\frac{X_{1} N_{1}+X_{2} N_{2}}{N_{1}+N_{2}}=\\frac{(60)(20)+(70)(30)}{20+30}=\\frac{3300}{50}\\) = 66%<\/p>\n

    \"KSEEB<\/p>\n

    Question 22.
    \nIf the cost price of 10 articles is equal to the selling price of 9 articles, find the gain percent
    \nAnswer:
    \nLet the cost of each article be x.
    \nC.P of 10 articles = 10x
    \nS.P of each article = \\(\\frac{10 x}{9}\\)
    \nS.P of articles = \\(\\frac{10 x}{9}\\) \u00d7 10 = \\(\\frac{100 x}{9}\\)<\/p>\n

    Profit = SP – CP = \\(\\frac{100 x}{9}\\) – 10x = \\(\\frac{100 x-90 x}{9}=\\frac{10 x}{9}\\)
    \nProof% = \\(\\frac{\\text { Profit }}{\\text { C.P. }}\\) \u00d7 100 = \\(\\frac{\\frac{10 x}{9}}{10 x}\\) \u00d7 100 = \\(\\frac{10 x}{90 x}\\) \u00d7 100 = 11\\(\\frac{1}{9}\\)%<\/p>\n

    Question 23.
    \nFind the value of sin2<\/sup> \\(\\frac{\\pi}{6}\\)cos2<\/sup> \\(\\frac{\\pi}{3}\\) – tan2<\/sup> \\(\\frac{\\pi}{4}\\) + cot2<\/sup> \\(\\frac{\\pi}{3}\\)
    \nAnswer:
    \nsin2<\/sup> \\(\\frac{\\pi}{6}\\)cos2<\/sup> \\(\\frac{\\pi}{3}\\) – tan2<\/sup> \\(\\frac{\\pi}{4}\\) + cot2<\/sup> \\(\\frac{\\pi}{3}\\) = \\(\\left(\\frac{1}{2}\\right)^{2} \\cdot\\left(\\frac{1}{2}\\right)^{2}-(1)^{2}+\\left(\\frac{1}{\\sqrt{3}}\\right)^{2}\\)<\/p>\n

    \\(\\frac{1}{4} \\cdot \\frac{1}{4}-1+\\frac{1}{3}\\)
    \n\\(\\frac{1}{16}-1+\\frac{1}{3}=\\frac{3-48+16}{48}=\\frac{-29}{48}\\)<\/p>\n

    Question 24.
    \nProve that (SinA + Cos A)2<\/sup> + (Sin A – CosA)2<\/sup> = 2
    \nAnswer:
    \nLHS (sin A + cos A)2<\/sup> + (SinA-cos A)2<\/sup>
    \n\"1st<\/p>\n

    Question 25.
    \nFind the equation of the straight line passing through (2, 3) and (3, 4)
    \nAnswer:
    \n(x1<\/sub>, y1<\/sub>) = (2, 3) (x2, y2) =(3,4)
    \n\\(\\frac{y-y_{1}}{x-x_{1}}=\\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \\Rightarrow \\frac{y-3}{x-2}=\\frac{4-3}{3-2}\\)
    \n\u21d2 \\(\\frac{y-3}{x-2}=\\frac{1}{1}\\) \u21d2 y – 3 = x – 2
    \nx – y + 1 = 0<\/p>\n

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

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

    Question 26.
    \nIn a group of 600 people, 150 students were found to be taking tea, 225 like Coffee, 100 like both tea and coffee. Find out how many were taking neither tea nor coffee?
    \nAnswer:
    \nn(T) = 150
    \nn(C) =225
    \nn(C \u2229 T) = 100
    \nn(\u222a)=600
    \nn(C \u222a T) = n(C) + n(T) – n(C \u2229T) = 225 + 150 – 100 = 375 – 100 = 275
    \n\u2234 Number of students taking neither coffee nor tea = 600 – 275 = 325<\/p>\n

    \"KSEEB<\/p>\n

    Question 27.
    \nIf R-1<\/sup> = {(2, 4), (1, 2) (3, 1), (3, 2) Find R. Also find its domain and range.
    \nSol. R = {(4, 2), (2, 1), (1, 3), (2,3)}
    \nDomR= {1,2,4}
    \nRanR = {1, 2; 3}<\/p>\n

    Question 28.
    \nProve \\(\\sqrt{5}\\) is an irrational number.
    \nOR
    \nAn electronic device makes a beep after every 60 sec. Another device makes a beep after every 62 sec,. They beeped together at 10 am. Find the time when they will next take a beep together at the earliest.
    \nAnswer:
    \nIf possible let \\(\\sqrt{5}\\) is a rational number.
    \n\\(\\sqrt{5}\\) = \\(\\frac{\\mathrm{p}}{\\mathrm{q}}\\), p, q \u2208 z, q \u2260 0
    \nwe shall also assume that p and q do not have any common factor
    \n\u2234 5 = \\(\\frac{\\mathrm{p}}{\\mathrm{q}}\\)
    \n\u21d2 p2<\/sup> = 5q2<\/sup>
    \n\u2234 p2<\/sup> is a multiple of 5
    \n\u2234 5 divides p2<\/sup>
    \n\u2234 5 divides p
    \n\u2234 p = 5k where k \u2208 z, k \u2260 0
    \np2<\/sup> = 25k2<\/sup>
    \n5q2<\/sup> = 25k2<\/sup> \u2235 P2<\/sup>=5q2<\/sup>
    \nq2<\/sup> = 5k2<\/sup>
    \n\u2234 q2<\/sup> is a multiple of 5
    \n\u2234 5 divides q2<\/sup>
    \n\u2234 5 divides q
    \n\u2234 5 divides both p and q
    \nThis is a contradiction<\/p>\n

    \u2234 Our assumption is wrong
    \n\u2234 \\(\\sqrt{5}\\) is irrational.
    \nOR
    \n\"1st
    \nLCM = 22 \u00d7 31 \u00d7 51 \u00d7 31 = 1860 sec = 31 min.
    \nThey beeped together at 10am.
    \n\u2234 After 31 minutes i.e., at 10.31 AM they will next make a beep together at the earliest.<\/p>\n

    Question 29.
    \nProve that \\(\\frac{1}{\\log _{2} 4}+\\frac{1}{\\log _{8} 4}+\\frac{1}{\\log _{16} 4}\\) = 4
    \nAnswer:
    \nLHS = \\(\\frac{1}{\\log _{2} 4}+\\frac{1}{\\log _{8} 4}+\\frac{1}{\\log _{16} 4}\\) = log4<\/sub> 2 + log4<\/sub> 8 + log4<\/sub> 16 = log4<\/sub>(2 \u00d7 8 \u00d7 16) = log4<\/sub> 256 = log4<\/sub> 44 = 4 log4<\/sub> 4 = 4(1) = 4<\/p>\n

    Question 30.
    \nThe sum of three numbers in an A.P is 15 and their product is 105. Find the numbers.
    \nAnswer:
    \nLet the numbers be a – d, a, a + d
    \nsum = 15
    \na-d + a + a + d = 15
    \n3a = 15 \u21d2 a = 15
    \nProduct = 105
    \n(a-d) a(a + d) = 105
    \n(5 – d). 5. (5 + d) = 105
    \n(5-d)(5+d) = \\(\\frac{105}{5}\\) = 21
    \n52<\/sup> – d2<\/sup> = 21
    \n25 – 21 = d2<\/sup>
    \nd2<\/sup> = 4
    \nd = 2
    \n\u2234 The number are a – d, a, a + d,
    \n5-2, 5, 5 + 2
    \n3, 5, 7<\/p>\n

    \"KSEEB<\/p>\n

    Question 31.
    \nFind the difference between the compound interest and the simple interest on \u20b9 5000 invested for 4 years at 8% p.a.
    \nAnswer:
    \nP = \u20b9 5,000, T = 4yrs, R = 8%
    \nSI = \\(\\frac{\\mathrm{PTR}}{100}=\\frac{5000 \\times 4 \\times 8}{100}\\) = 50 \u00d7 4 \u00d7 8 = 1600
    \nA = P(1 + R)n = 5000(1 + 0.08)8 = 5000 (1.08)4 = 5000 \u00d7 1.3604 = 6802.44
    \n\u2234 CI = 6802.44 – 5000
    \n= 1802.44
    \nDifference between Cl and SI
    \n= CI – SI = 1802.44 – 1600 = \u20b9 202.44<\/p>\n

    Question 32.
    \nSolve the linear inequalities graphically:
    \n5x + 4y \u2265 40, x \u2265 2, y \u2265 3, x, y \u2265 0
    \nAnswer:
    \n5x + 4y = 40
    \ny = 0, 5x = 40 \u21d2 x = 8, A = (8, 0)
    \nx = 0, 4y = 40 \u21d2 y = 10, B= (0, 10)
    \n\"1st
    \nThe solution set of each equation is indicated by the arrow mark in the figure. The intersection of these solution sets is in the feasible region shaded as shown in the figure.<\/p>\n

    Question 33.
    \nSachin finds by getting out for a duck (0 runs) in the lith innings of his test matches his average of the previous innings is decreased by 5 runs. What is his average after the 11th innings?
    \nAnswer:
    \nLet the average after 10 inninngs = x
    \n\u2234 Total runs after 10 inninngs = 10x
    \nAverage after 11 inninngs = x – 5
    \n\u2234 Total runs after 11 inninngs = 11 (x – 5)
    \n11(x- 5) – 10x = 0
    \n11x- 55 – 10x = 0
    \nx- 55 = 0
    \nx = 55
    \n\u2234 Average after 11 inninngs = x – 5
    \n= 55 – 5 = 50<\/p>\n

    \"KSEEB<\/p>\n

    Question 34.
    \nFind the ratio in which the line joining the points (2,5) and (1,9) is divided by the x-axis. Also, find the point of division.
    \nAnswer:
    \nP(x, 0) A = (2, 5,) = (x1, y1)
    \nB = (1, 9) = (x2, y2)
    \nP = \\(\\left[\\frac{m x_{2}+n x_{1}}{m+n}, \\frac{m y_{2}+n y_{1}}{m+n}\\right]=\\left[\\frac{m(1)+n(2)}{m+n}, \\frac{m(9)+n(5)}{m+n}\\right]\\)
    \n(x, 0) = \\(\\left[\\frac{m+2 n}{m+n}, \\frac{9 m+5 n}{m+n}\\right]\\)
    \n\\(\\frac{9 m+5 n}{m+n}\\) = 0
    \n\\(\\frac{m}{n}=\\frac{-5}{9}\\) \u21d2 9m + 5n = 0 \u21d2 9m = – 5n
    \nm :n = 5:9 externally<\/p>\n

    Taking m = 5, n = 9 to find x use section formula for external division namely,
    \nx = \\(\\frac{m x_{2}-n x_{1}}{m-n}=\\frac{5(1)-9(2)}{5-9}=\\frac{5-18}{-4}=\\frac{-13}{-4}=\\frac{13}{4}\\)
    \nP = (\\(\\frac{13}{4}\\), 0) \u2192 Point of division.<\/p>\n

    Question 35.
    \nIf an article is sold at \u20b9 24 there is a profit of 4 and if it is sold at a loss of 10% find the selling price of the article.
    \nAnswer:
    \nSP of the article = \u20b9 24
    \nProfit = \u20b9 4
    \nCP = SP – Profit = 24 – 4 = \u20b9 20
    \nLoss = 10%
    \nWe have SP = \\(\\frac{100-\\operatorname{loss} \\%}{100}\\) \u00d7 CP = \\(\\frac{100-10}{100}\\) \u00d7 20 = \\(\\frac{90}{100}\\) \u00d7 20
    \n\u2234 SP = \u20b9 18<\/p>\n

    Question 36.
    \nIf Cot A = \\(\\frac{5}{12}\\) and A is acute, show that 2 cosec A – 4sec A = – \\(\\frac{247}{50}\\)
    \nAnswer:
    \n\"1st
    \nCosec A = \\(\\frac{13}{2}\\)
    \nSec A = \\(\\frac{13}{5}\\)
    \n2 cosec A – 4 sec A = 2\\(\\left(\\frac{13}{12}\\right)\\) – 4\\(\\left(\\frac{13}{5}\\right)\\) = – \\(\\frac{494}{60}=-\\frac{247}{30}\\)<\/p>\n

    Question 37.
    \nShow that the points A(2, 2) B(6, 3) and C(4, 11) form a right angled triangle.
    \nAnswer:
    \nAB = \\(\\sqrt{(6-2)^{2}+(3-2)^{2}}=\\sqrt{16+1}=\\sqrt{17}\\)
    \nBC = \\(\\sqrt{(4-6)^{2}+(11-3)^{2}}=\\sqrt{4+64}=\\sqrt{68}=\\sqrt{4 \\times 17}\\) = 2\\(\\sqrt{17}\\)
    \nCA= \\(\\sqrt{(4-2)^{2}+(11-2)^{2}}=\\sqrt{4+81}=\\sqrt{85}\\)
    \nAB2<\/sup> =17, BC2<\/sup> = 68, CA2<\/sup> = 85
    \n\u2234 AB2<\/sup> + BC2<\/sup> = CA2<\/sup>
    \n\u2234 ABC form a right angled triangle.<\/p>\n

    \"KSEEB<\/p>\n

    Question 38.
    \nIf a train travels 15 km \/ hr faster it would take 1 hour less to travel 180 km, find the original speed of the train.
    \nAnswer:
    \nLet s be the actual speed
    \nt be the actual time d = 180 km
    \ns = \\(\\frac{d}{t}=\\frac{180}{t}\\) \u21d2 t = \\(\\frac{180}{s}\\) \u21d2 st = 180
    \nIf speed is 15km\/hr more then speed becomes s + 15 and time taken will be 1 hour less it becomes t – 1.
    \n\"1st
    \n\u2234 Actual speed of the train = 45 km\/hr
    \nOR
    \nIf \u03b1 and \u03b2 are the roots of the equation 2x2<\/sup> + 5x + 5 = O then find the value of \\(\\frac{1}{\\alpha^{2}}+\\frac{1}{\\beta^{2}}\\)
    \nAnswer:
    \n\u03b1 + \u03b2 = \\(\\frac{-5}{2}\\) and \u03b1\u03b2 = \\(\\frac{5}{2}\\)
    \n\u2234 \\(\\frac{1}{\\alpha^{2}}+\\frac{1}{\\beta^{2}}=\\frac{\\alpha^{2}+\\beta^{2}}{\\alpha^{2} \\beta^{2}}=\\frac{(\\alpha+\\beta)^{2}-2 \\alpha \\beta}{(\\alpha \\beta)^{2}}=\\frac{\\left(\\frac{-5}{2}\\right)^{2}-2\\left(\\frac{5}{2}\\right)}{\\left(\\frac{5}{2}\\right)^{2}}=\\frac{\\frac{25}{4}-5}{\\frac{25}{4}}\\)
    \n\"1st<\/p>\n

    PART – D<\/span><\/p>\n

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

    Question 39.
    \nLet f = {(1, 1), (2, 3), (0, -1)} be a function from Z to Z defined by f(x) = ax + b some integers a and b. (i) Determine a & b. (ii) If f(x) = 2x + 1 g(x) = x2<\/sup> + 2x + 1 Find (a) fog(2)(b) gof (3)
    \nAnswer:
    \nf(x) = ax + b
    \nwhen x – 1 f(x) = 1
    \n\u2234 a + b = 1
    \nx = 0 f(x) = -1
    \na(0) + b = -1 \u21d2 b = -1
    \na + b = 1
    \na – 1 = 1 \u21d2 a = 1 + 1 = 2
    \n\u2234 a = 2
    \n\u2234 a = 2, b = -1
    \n(a) fog (2) = f [g(2)] = f [22 + 2(2) + 1]
    \n= f [4 + 4 + 1] = f [9] = 2(9) + 1 = 18 + 1 = 19<\/p>\n

    (b) gof (3) = g [f(3)] = g [2(3) + 1] – g(7)
    \n= 72 + 2(7) + 1 = 49 + 14 + 1 = 64<\/p>\n

    Question 40.
    \nFind the sum of all integers 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 – l)d
    \n390 = 65 + (n- 1)13
    \n390 = 65+ 13n-13
    \n13n= 338
    \n\u2234 n = 26
    \nSn<\/sub> = \\(\\frac{\\mathrm{n}}{2}\\)[a + l] = \\(\\frac{26}{2}\\)(65 + 390)= 13 (455) = 5,915
    \n\u2234 S = 5,9.15<\/p>\n

    \"KSEEB<\/p>\n

    Question 41.
    \nThe age of the father is 5 times that of the son’s 3 years if the age of the father was 8 times that of his son. Find their present ages.
    \nAnswer:
    \nLet the present age of the son = x
    \nPresent age of the father = 5x
    \nThree years-agey sons age = x – 3 and fathers age = -5x – 3
    \nGiven 5x – 3 = 8(x – 3)
    \n5x – 3 = 8x – 24
    \n8x – 5x = 24 – 3
    \n3x = 21
    \nx = 7
    \nSons age = x = 7 years
    \nFathers age 5x = 5(7) = 35 years<\/p>\n

    Question 42.
    \nA machine depreciates at 10% of its value at the beginning of a year. The cost and scrap value realized at the time of sale are 23240 and 9000 respectively. For how many years the machine was put to use?
    \nAnswer:
    \nr= 10% = 0.10 C =23,240 B=9,000 n=?
    \nB = C(1 – r)n<\/sup>
    \n9000 = 23240(1 – 0. 10)
    \n9000 = 23240 (0.90)n<\/sup>
    \n(0.9)n<\/sup> = \\(\\frac{9000}{23240}\\) =0.3872
    \nlog (0.9) = log 0.3 872
    \nn log 0.9 = log 0.3872
    \nn = \\(\\frac{\\log 0.3872}{\\log 0.9}=\\frac{1.5879}{1.9542}=\\frac{-1+0.5879}{-1+09542}=\\frac{-0.4121}{-0.0458}\\)
    \n= 8.99 \u2248 9 years<\/p>\n

    Question 43.
    \nCalculate the future value of the annuity immediate of \u20b9 1000 p.a for 12 years at 16% p.a compounded quarterly?
    \nAnswer:
    \nA = \u20b9 1,000 p.a, n = 12, R = 16%= 0.6, r = ?, q = 4
    \nr = \\(\\left(1+\\frac{\\mathrm{R}}{q}\\right)^{q}\\) – 1 = \\(\\left(1+\\frac{0.16}{4}\\right)^{4}\\) – 1 = 0.1698 = 16.98%
    \nF = \\(\\frac{\\mathrm{A}\\left[(1+r)^{n}-1\\right]}{r}=\\frac{1000\\left[(1+0.1698)^{12}-1\\right]}{0.1698}\\)
    \nF = 32,783.17<\/p>\n

    \"KSEEB<\/p>\n

    Question 44.
    \nFind the equation of the straight line passing through (-2,6) and the sum of the intercepts on the co-ordinate axes is 5.
    \nAnswer:
    \na + b = 5 \u21d2 b = 5 – a, Let the equation of the straight line be
    \n\\(\\frac{x}{a}+\\frac{y}{b}\\) = 1
    \n\\(\\frac{x}{a}+\\frac{y}{5-a}\\) = 1<\/p>\n

    It passes through (-2. 6) put x = -2, y = 6
    \n\\(\\frac{-2}{a}+\\frac{6}{5-a}\\) = 1 \u21d2 \\(\\frac{-10+2 a+6 a}{a(5-a)}\\) = 1
    \n\u21d2 – 10 + 8a = a(5 – a) \u21d2 – 10 + 8a = 5a – a2<\/sup>
    \n\u21d2 a2<\/sup>(a + 5) – 2(a + 5) = 0 \u21d2 (a + 5)(a + 2) = 0
    \n\u21d2 a = 2 \u2234 b = 5 – a = 5 – 2 = 3
    \nCase (i)
    \n\\(\\frac{x}{2}+\\frac{y}{3}\\) = 1
    \n3x + 2y = 6
    \n3x + 2y – 6 = 0
    \na = -5
    \n\u2234 b = 5 – a = 5 – (-5) = 10<\/p>\n

    Case (ii)
    \n\\(\\frac{x}{-5}+\\frac{y}{10}\\) = 1
    \n\\(\\frac{-2 x+y}{10}\\) = 1
    \n-2x + y – 10 = 0<\/p>\n

    \\(\\frac{-x}{5}+\\frac{y}{10}\\) = 1
    \n-2x + y = 10
    \n2x – y + 10 = 0<\/p>\n

    Question 45.
    \n(a) Form a quadratic equation whose roots are 2 + \\(\\sqrt{3}\\) & 2 – \\(\\sqrt{3}\\)
    \n(b) Evaluate using log. tables \\(\\frac{(0.5634)^{2} \\times(0.0635)}{(2.563)^{2}}\\)
    \nAnswer:
    \n(a) \u03b1 = 2 + \\(\\sqrt{3}\\)
    \n\u03b2 = 2 – \\(\\sqrt{3}\\)
    \n\u03b1 + \u03b2 = (2 + \\(\\sqrt{3}\\)) + (2 – \\(\\sqrt{3}\\)) =4
    \n\u03b1\u03b2 = (2 + \\(\\sqrt{3}\\))(2 – \\(\\sqrt{3}\\)) = 4 – 3 = 1
    \n\u2234 Quadratic equation’s
    \nx2<\/sup> – (\u03b1 + \u03b2)x + \u03b1\u03b2 = 0 \u21d2 x2 <\/sup>– 4x+l=0<\/p>\n

    (b) x = \\(\\frac{(0.5634)^{2} \\times(0.0635)}{(2.563)^{2}}\\)
    \nlog x = 2log 0.5634 + log 0.0635 – 2log 2.563
    \n= 2 (1.7508) + (2.8027)- 2(0.4087) = 2(-1 + 0.7508) + (-2+ 0.8027) – 2(0.4087)
    \n= -2 + 1.5016-2 + 0.8027-0.8174
    \n= 2.5131 =-2-0.5131
    \n= -2 – 1 + 1-0.5131
    \n= -3 + 0.4869
    \nlog x = 3.4869
    \nx = AL[34869] =0.00306<\/p>\n

    Question 46.
    \nFind the reflection of the point P(2, 1) in the line x + y = 5
    \n\"1st
    \nSlope of x + y – 5 = 0 is
    \nPQ – x + y – 5 =0
    \n\u2234 Slope of PQ = +1 =m<\/p>\n

    \u2234 Equation of PQ y – y1<\/sub> =m(x – x1<\/sub>)
    \ny – 1 = 1(x – 2)
    \ny – 1 = x – 2
    \nx – y – 1 = 0
    \nx + y – 5 =0
    \nAdd \u21d2 2x – 6 = 0
    \n\u21d2 x = 3<\/p>\n

    x – y – 1 = 0
    \nx + y – 5 =0
    \n3 + y – 5 = 0
    \nA = (3, 2) A is the mid point of PQ<\/p>\n

    \u2234 A = \\(\\left(\\frac{2+a}{2}, \\frac{1+b}{2}\\right)\\)
    \n(3, 2) = \\(\\left(\\frac{2+a}{2}, \\frac{1+b}{2}\\right)\\)
    \n\u21d2 \\(\\frac{2+a}{2}\\) = 3
    \n\u21d2 2 + a = 6 \u21d2 a = 6 – 2
    \na = 4<\/p>\n

    \u21d2 \\(\\frac{1+b}{2}\\) = 3
    \n\u21d2 1 + b = 4 \u21d2 b = 4 – 1
    \nb = 3
    \n\u2234 Q = (4, 3)<\/p>\n

    Question 47.
    \nIf x = r cos A cos B, y = r cos A sin B and z = r sin A Then P.T. x2 + y2 + z2 = r2
    \nAnswer:
    \nLHS = x2<\/sup> + y2<\/sup> + z2<\/sup> = r2<\/sup> cos2<\/sup> A cos2<\/sup> B + r2<\/sup> cos2<\/sup> A sin2<\/sup> B + r2<\/sup> sin2<\/sup> A = r2<\/sup> cos2<\/sup> A (cos2<\/sup> B + sin2<\/sup> B) + r2<\/sup> sin2<\/sup> A= r2<\/sup> cos2<\/sup> A(1) + r2 <\/sup>sin2<\/sup> A = r2<\/sup> (cos2<\/sup> A+ sin2<\/sup> A) = r2<\/sup> (1) = r2<\/sup> = RHS<\/p>\n

    \"KSEEB<\/p>\n

    Question 48.
    \nFind the locus of a point equidistant from (2, 0) and (-2, 0)
    \nAnswer:
    \nA = (2, 0), B = (-2, 0)
    \nLet P(x, y) be any point on the locus
    \n\u2234 PA = PB
    \nPA2<\/sup> = PB2<\/sup>
    \n\"1st
    \n– 4x – 4x + 4 – 4 = 0
    \n– 8x = 0
    \nx = 0<\/p>\n

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

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

    Question 49.
    \n(a) Find the equation of the straight line passing through the intersection of the line 2x + 3y = 5, 7x – y = 6 is perpendicular to the line 3x + 4y + 1 = 0.
    \n(b) If tan A + sin A = m and tanA – sin A = n show that m2<\/sup> – n2<\/sup>\u00a0= \u00b14\\(\\sqrt{\\mathrm{mn}}\\) .
    \n(c) Find the number of digits in 320?
    \nAnswer:
    \n(a) 2x + 3y = 5 \u00d7 1
    \n7x – y = 6 \u00d7 3
    \n\"1st
    \n23x = 23 \u21d2 x = 1
    \n7x – y = 6
    \n7(1) – y = 6
    \n7 – 6 = y
    \ny = 6
    \n\u2234 (x1, y1) = (1, 1)
    \nSlope of the given line = \\(\\frac{-a}{b}=\\frac{-3}{4}\\)
    \nRequired line is perpendicular to given line
    \n\u2234 slope of the required line = \\(\\frac{+4}{3}\\) = m
    \n\u2234 Equation of the required line
    \ny – y1<\/sub> = m(x – x1<\/sub>)
    \ny – 1 = \\(\\frac{4}{3}\\)(x – 1)
    \n3y – 3 = 4x – 4
    \n4x – 3y – 1 = 0<\/p>\n

    (b) m2<\/sup> – n2<\/sup> – (m – n) (m + n)
    \n= (tanA+ sin A- tan A + sin A) (tanA + sin A + tanA- sinA)
    \n= 2 sin A. 2 tan A = 4 sin A tan A = 4 sin A. tan A
    \n(m2<\/sup> – n2<\/sup>)2<\/sup> = 16 sin2<\/sup> A. tan2<\/sup> A …………..(1)
    \nmn = (tan A + sin A)(tan A – sin A) = tan2<\/sup> A – sin2<\/sup> A
    \n= \\(\\frac{\\sin ^{2} A}{\\cos ^{2} A}\\) – sin2<\/sup> A = sin2<\/sup>A(\\(\\frac{1}{\\cos ^{2} A}\\) – 1) = sin2<\/sup>A\\(\\frac{1-\\cos ^{2} A}{\\cos ^{2} A}=\\frac{\\sin ^{2} A}{\\cos ^{2} A}\\)sin2A
    \nmn = tan2<\/sup> A. sin2<\/sup> A ……….(2)
    \nsubstitue the value of mn from (2) in (1) we get
    \n(m2<\/sup> – n2<\/sup>)2<\/sup> = 16mn<\/p>\n

    (c) Let x = 320<\/sup>
    \nlog x = log 320<\/sup> = 20log3 = 20[0.4771]
    \nlog x = 9.5424
    \nAs the characteristic of log x is 9, the number of digits = 9 + 1 = 10.<\/p>\n

    \"KSEEB<\/p>\n

    Question 50.
    \n(a) Find the sum to n terms of the series: 5 + 55 + 555 +…. n term.
    \n(b) A confectioner makes and sells biscuits. Does he sell one pack of biscuits at \u20b9 80? His cost of manufacturing is \u20b9 40 per packet as variable cost and 3000 as a fixed cost. Find the
    \n(i) Revenue function
    \n(ii) Cost function
    \n(iii) Profit function
    \n(iv) If he limits his production to 100 packets can he make a profit?
    \n(v) What will be the number of boxes he must sell to make a profit so that he does not incur a loss?
    \n(e) If the product of two numbers is 216 and their LCM is 36. Find their HCF.
    \nAnswer:
    \n(a) Let Sn<\/sub> = 5 + 55 + 555 + …….n terms
    \nSn<\/sub> = 5(1 + 11 + 111 +…….. )
    \n\\(\\frac{9 \\mathrm{~S}_{\\mathrm{n}}}{5}\\) = 1 + 11 + 111 + ………
    \nMultiply both the sides by 9,
    \n\\(\\frac{9 \\mathrm{~S}_{\\mathrm{n}}}{5}\\) = 9 + 99 + 999 + …………
    \n\\(\\frac{9 \\mathrm{~S}_{\\mathrm{n}}}{5}\\) = (10 – 1) + (102<\/sup> – 1) + (103<\/sup> – 1) +
    \n\\(\\frac{9 \\mathrm{~S}_{\\mathrm{n}}}{5}\\) = (10 + 102<\/sup> + 103<\/sup> + ……….. ) – (1 + 1 + 1 + ………..)
    \n\\(\\frac{9 S_{n}}{5}=\\frac{10\\left(10^{n}-1\\right)}{10-1}\\) – n
    \n\\(\\frac{9 \\mathrm{~S}_{\\mathrm{n}}}{5}=\\frac{10\\left(10^{\\mathrm{n}}-1\\right)}{9}\\) – n
    \nSn<\/sub> = \\(\\frac{5}{9}\\left[\\frac{10\\left(10^{n}-1\\right)}{9}-n\\right]=\\frac{5}{9}\\left[\\frac{10}{9}\\left(10^{n}-1\\right)-n\\right]\\), n \u2208 N<\/p>\n

    (b) (a) Revenue function = R (x) = Selling price X quantity
    \n= 80 x (x = No. of pack of biscuit)
    \n(b) Cost function = C (x) = ax + b
    \n= 40x + 3000
    \n(c) Profit function = P (x)
    \n= R(x) – C(x)
    \n= 8Ox – (40x + 3000)
    \n= 80x – 40x – 3000
    \n= 40x – 3000<\/p>\n

    (d) If x = 100 packets
    \n\u2234 P(x) = 40x – 3000
    \nP(100) =40 (100)- 3000
    \n= \u20b9 1,000<\/p>\n

    (e) No. of packets required to ensure no Joss (at BEP)
    \nC (x) =R(x)
    \n40x + 3000 = 80 x
    \n3000 = 80x – 40x
    \n3000 = 40x
    \nx = \\(\\frac{3000}{40}\\)
    \nx =75 packet<\/p>\n

    (c) ab = 216 LCM = 36 HCF = ?
    \nHCF \u00d7 LCM = ab = 216
    \nHCF = \\(\\frac{216}{\\mathrm{LCM}}\\)
    \nHCF = \\(\\frac{216}{36}\\) = 6<\/p>\n","protected":false},"excerpt":{"rendered":"

    Students can Download 1st PUC Basic Maths Model Question Paper 2 with Answers, Karnataka 1st PUC Basic Maths Model Question Papers with Answers helps you to revise the complete syllabus. Karnataka 1st PUC Basic Maths Model Question Paper 2 with Answers Time: 3.15 Hours Max. Marks: 100 Instructions: The questions paper consists of five parts …<\/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\/89244"}],"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=89244"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/89244\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=89244"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=89244"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=89244"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}