{"id":89320,"date":"2021-07-06T17:03:11","date_gmt":"2021-07-06T11:33:11","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=89320"},"modified":"2021-07-06T17:09:19","modified_gmt":"2021-07-06T11:39:19","slug":"1st-puc-basic-maths-model-question-paper-5","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-basic-maths-model-question-paper-5\/","title":{"rendered":"1st PUC Basic Maths Model Question Paper 5 with Answers"},"content":{"rendered":"

Students can Download 1st PUC Basic Maths Model Question Paper 5 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 5 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 canonical representation of 60.
    \nAnswer:
    \n\"1st<\/p>\n

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

    Question 3.
    \nIf f: R \u2192 R is defined by\/(x) = 2x + 3 then find f(1\/2).
    \nAnswer:
    \nf(1\/2) = 2 \u00d7 \\(\\frac{1}{2}\\) + 3 =1 + 3 = 4<\/p>\n

    Question 4.
    \nSimplify : (3\u00b0)2 + (32)\u00b0
    \nAnswer:
    \n= 12 + 9\u00b0 = 1 + 1 = 2<\/p>\n

    Question 5.
    \nEvaluate : log\\(\\sqrt{x}\\) 4 = 2
    \nAnswer:
    \n\u21d2 (\\(\\sqrt{2}\\))2 = 4 \u21d2 x = 4<\/p>\n

    \"KSEEB<\/p>\n

    Question 6.
    \nFind the 11th term of the A.P. 3, 5, 7, 9…………
    \nAnswer:
    \nHere a = 3, d = 5 – 3 = 2, n = 11
    \nTn<\/sub> = a + (n – 1 )d
    \nT11<\/sub> = 3 + (11 – 1)2 = 3 + (10) = 3 + 20 = 23<\/p>\n

    Question 7.
    \nThe sum of 6 times a number and 5 times the number is 55. Which is that number?
    \nAnswer:
    \nLet the no. be x.
    \nGiven 6x + 5x =55
    \n11x =55
    \nx = \\(\\frac{55}{11}\\) = 5
    \nHence the number is 5.<\/p>\n

    Question 8.
    \nConvert the ratio 2: 5 into percentages.
    \nAnswer:
    \n2:5 = \\(\\frac{2}{5}\\) \u00d7 100% = 40%<\/p>\n

    Question 9.
    \nWrite the formula for the future value of annuity immediate.
    \nAnswer:
    \nF = \\(\\frac{\\mathrm{A}\\left[(1+r)^{n}-1\\right]}{r}\\)<\/p>\n

    Question 10.
    \nConvert 225\u00b0 into radians.
    \nAnswer:
    \n225\u00b0 = 225 \u00d7 \\(\\frac{\\pi}{180}=\\frac{5 \\pi}{4}\\)<\/p>\n

    \"KSEEB<\/p>\n

    Question 11.
    \nThe average age of 10 girls \u00a1n a class in 17 years. What is the sum of their ages?
    \nAnswer:
    \nSum of their ages 10 \u00d7 17 = 170 years.<\/p>\n

    Question 12.
    \nFind the slope of the line 3x + 4y – 29 = O
    \nAnswer:
    \nSlope = \\(\\frac{-a}{b}=\\frac{-3}{4}\\)<\/p>\n

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

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

    Question 13.
    \nFind the real part of \\(\\frac{1+2 i}{3-4 i}\\)
    \nAnswer:
    \n\\(\\frac{1+2 i}{(3-4 i)} \\times \\frac{(3+4 i)}{(3+4 i)}=\\frac{(1+2 i)(3+4 i)}{9+16}=\\frac{(3-8)+i(10)}{25}=\\frac{-5+10}{25}=-\\frac{1}{5}+\\frac{2 i}{5}\\)
    \nReal part = –\\(\\frac{1}{5}\\) Imaginary part = \\(\\frac{2}{5}\\)<\/p>\n

    Question 14.
    \nLet f = {(1, 2), (2,3), (3, 4)} be a function from 2 to 2 where z is the set of integers defined by\/W = ax + b v some integers a and b. Determine a and b.
    \nAnswer:
    \n2 = a + b …(i) 3 = 2a + b …………. (ii)
    \nequation (1) and (2)
    \n-1 = -a
    \n\u2234 a = 1 \u2234 b = 1<\/p>\n

    Question 15.
    \nSolve : 32x – 10. 31 + 9 = 0.
    \nAnswer:
    \n32x – 10. 3x + 9 = 0
    \nLet 3x = a \u21d2 (3x)2<\/sup> = a2<\/sup>
    \n\u21d2 32x<\/sup> = a2<\/sup>
    \n\u2234 a2<\/sup> – 10a + 9 = 0
    \na2 <\/sup>– 9a – a + 9 = 0
    \na(a – 9) -1 (a – 9) = 0
    \n(a – 9) (a – 1) = 0
    \na – 1 = 0
    \na – 9 = 0
    \na = 1 a = 9
    \n3x<\/sup> = 3\u00b0
    \n3x<\/sup> = 32<\/sup>
    \nx = 0 x = 2<\/p>\n

    Question 16.
    \nProve that loga<\/sub> mn = loga<\/sub>\u00a0m + loga<\/sub> n
    \nAnswer:
    \nLet loga<\/sub> m=x \u21d4 ax<\/sup> = m
    \nloga<\/sub> n = y \u21d4 ay<\/sup> = n
    \nloga<\/sub> mn = z \u21d4 az<\/sup> = mn
    \nConsider az<\/sup> = mn \u21d4 az<\/sup> = ax<\/sup>.ay<\/sup> \u21d2 az<\/sup> = ax+y<\/sup> \u21d2 z = x + y \u21d2 loga<\/sub> mn = loga<\/sub> m + loga<\/sub>\u00a0n<\/p>\n

    Question 17.
    \nIf \\(\\frac{3}{5}\\), K, \\(\\frac{13}{5}\\) are inA.P., then find the value of K.
    \nAnswer:
    \nk – \\(\\\\frac{3}{5}=\\frac{13}{5}\\) – k
    \nk + k = \\(\\frac{13}{5}+\\frac{3}{5}=\\frac{16}{5}\\)
    \n2k = \\(\\frac{16}{5}\\) \u21d2 k = \\(\\frac{8}{5}\\)<\/p>\n

    Question 18.
    \nForm cubic equation whose roots are 3, 5, 7
    \nAnswer:
    \n3, 5, 7 are the roots
    \n\u2234 x – 3, x – 5, x – 7 are the factors
    \nEquation is (x – 3) (x – 5) (x – 7) = 0
    \n\u21d2 x3<\/sup> – 15x2<\/sup> + 71x – 105 = 0<\/p>\n

    \"KSEEB<\/p>\n

    Question 19.
    \nSolve 3x – 4 > 7 – 2x (x\u2208R)
    \nAnswer:
    \n3x – 4 > 7 – 2x
    \n3x + 2x > 7 + 4
    \n5x > 11
    \nx > \\(\\frac{11}{5}\\) or x \u2208 (\\(\\frac{11}{5}\\), \u221e)<\/p>\n

    Question 20.
    \nFind the simple interest on Rs. 800 at 8% p.a. for one year one month and 10 days.
    \nAnswer:
    \nP = 800. R = 8 T = 1yr, I month and 10 days
    \nT = (1 + \\(\\frac{1}{12}+\\frac{10}{365}\\))years (1 + 0.08 + 0.02)(1.1)year
    \nSI = \\(\\frac{\\mathrm{PTQ}}{100}=\\frac{800 \\times 8 \\times 1.1}{100}\\) = 71.11<\/p>\n

    Question 21.
    \nFind the effective rate of interest when a sum lent at 12% is computed half yearly.
    \nAnswer:
    \nR = Nominal rate of interest = 12% = \\(\\frac{12}{100}\\) = 0.12
    \nq = number of times interest is compounded in one year = 2
    \nr = effective rate of interest
    \n= \\(\\left(1+\\frac{R}{q}\\right)^{q}\\) – 1 = \\(\\left(1+\\frac{0.12}{2}\\right)^{2}\\) – 1 = (1 + 0.6)2<\/sup> – 1 = (1.06)2<\/sup> – 1 = 1.1236 – 1
    \n= 0.1236 = 12.36%
    \n\u2234 Effective rate of interest = 12.36%<\/p>\n

    Question 22.
    \nFind the present value of a perpetuity of Rs.3000 to be received forever at 4% p.a.
    \nAnswer:
    \nA = Rs.3000 r = 0.04
    \nP\u221e = \\(\\frac{\\mathrm{A}}{\\mathrm{r}}=\\frac{3000}{0.04}\\) = 75,000<\/p>\n

    Question 23.
    \nThe average mark of is students in a class is 45. A student who has scored 50 marks leaves the classroom. Find the average marks of the remaining 14 students.
    \nAnswer:
    \nn= 15, x\u0304 =45, \u03a3x = nx\u0304 = 15(45) = 675
    \nTotal marks of 14 students = 675 – 50 = 625
    \nAverage marks of 14 students = \\(\\frac{625}{14}\\) =44.6<\/p>\n

    Question 24.
    \nThe angles of a triangle are in the ratio 3:4:5. Find them in radians and in degrees.
    \nAnswer:
    \n3:4:5 or 3x, 4x, 5x
    \n3x + 4x + 5x = 180\u00b0, 12x = 180
    \nx = \\(\\frac{180}{12}\\) = 15\u00b0
    \nAngles are 3(15\u00b0), 4(15\u00b0), 5(15\u00b0)
    \n45\u00b0, 60\u00b0, 75\u00b0,
    \n45\u00b0 = 45 \u00d7 \\(\\frac{\\pi}{180}=\\frac{\\pi}{4}\\)
    \n60\u00b0 = 60 \u00d7 \\(\\frac{\\pi}{180}=\\frac{\\pi}{3}\\)
    \n75\u00b0 = 75 \u00d7 \\(\\frac{\\pi}{180}=\\frac{5 \\pi}{12}\\)<\/p>\n

    \"KSEEB<\/p>\n

    Question 25.
    \nIf the distance between (2, a) and (-1, 1) is \\(\\sqrt{13}\\) find the value of a
    \nAnswer:
    \nLet A = (2, a) B = (-1, 1)
    \n\u2234 AB = \\(\\sqrt{(2+1)^{2}+(a-1)^{2}}\\)
    \ni.e \\(\\sqrt{13}=\\sqrt{9+(a-1)^{2}}\\)
    \nsquaring
    \n\u2234 13 = 9 + (a – 1)2<\/sup>
    \n\u2234 (a – 1)2<\/sup> = 4
    \n\u2234 (a – 1) = \u00b12
    \n\u2234 a – 1 = 2 or a – 1 = -2
    \ni.e., a = 3 or a = -1<\/p>\n

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

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

    Question 26.
    \nProve that 3 + \\(\\sqrt{5}\\) is an irrational number.
    \nAnswer:
    \nIf possible let 3 + \\(\\sqrt{5}\\) be a rational number
    \n\u2234 3 + \\(\\sqrt{5}\\) = \\(\\frac{p}{q}\\) where p, q \u2208 z, q \u2260 0
    \n\u2234 \\(\\sqrt{5}\\) = \\(\\frac{p}{q}\\) – 3 = rational – rational rational number
    \n\u2234 \\(\\sqrt{5}\\) is a rational number which is a contradiction.
    \n\u2234 our assumption is wrong
    \n\u2234 3 + \\(\\sqrt{5}\\) is irratjona1.<\/p>\n

    Question 27.
    \nIn a class of 150 students, It was found that 95 like burgers and 79 like pizzas. Assuming every student like at least one of the above, find the number of students who like both burgers and pizzas.
    \nAnswer:
    \nn(B\u222aP) = n(B) + n(P) – n(B \u2229 P)
    \n150 = 95 + 79 – n(B \u2229 P)
    \n\u2234 n(B \u2229 P) – 95 + 79 – 150 = 24
    \nn(B \u2229 P) = 24<\/p>\n

    Question 28.
    \nIf 451\/x<\/sup> = 31\/y<\/sup> = 51\/z<\/sup>, prove that x = 2y + z.
    \nAnswer:
    \nLet 451\/x<\/sup> = 31\/y<\/sup> = 51\/z<\/sup> = k
    \n(45)1\/x <\/sup>= k \u21d2 45 = kx<\/sup>
    \n31\/y<\/sup> = k \u21d2 3 = ky<\/sup>
    \n51\/z<\/sup> = k \u21d2 5 = kz<\/sup>
    \nConsider kx = 45 = 9 \u00d7 5 = 32<\/sup> \u00d7 5 = (ky<\/sup>)2<\/sup> kz<\/sup> = k2<\/sup>y kz<\/sup> = k2y-z<\/sup>
    \n\u2234 x = 2y + z<\/p>\n

    Question 29.
    \nIf x = log2a<\/sub> a, y = log3a<\/sub> 2a, z = log4a<\/sub> 3a. Show that xyz + 1 = 2yz.
    \nAnswer:
    \nxyz +1 = log2a<\/sub> a. log3a<\/sub> 2a. log4a<\/sub> 3a + 1
    \n= log3a<\/sub> a. log4a<\/sub> 3a. +1 (\u2235 logb<\/sub> a. logc<\/sub> b = logc<\/sub> a)
    \n= log4a<\/sub> a +1
    \n= log4a<\/sub>a + log4a <\/sub>4a (\u2235 loga<\/sub> a = 1)
    \n= log4a<\/sub> 4a2<\/sup>
    \n= log4a<\/sub> (23)2<\/sup>
    \n= 2 log4a<\/sub> 2a (\u2235 log an = m log a)
    \n= 2log3a<\/sub> 2a. log4a<\/sub> 3a = 2 yz = RHS<\/p>\n

    Question 30.
    \nThe sum of three numbers in APis -18 and sum of their squares is 140. Find the numbers.
    \nAnswer:
    \n(a – d) + a + (a + d) = -18
    \n3a = -18 \u21d2 a = -6
    \n(a – d) + a2<\/sup> + (a + d)2<\/sup> = 140
    \n\"1st
    \n3a2<\/sup> + 2d2<\/sup> = 140
    \n3(-6)2<\/sup> + 2d2<\/sup> = 140
    \n108 + 2d2<\/sup> = 140
    \n2d2<\/sup> = 140- 108
    \n2d2<\/sup> = 32
    \nd2<\/sup> = 16
    \nd = 4
    \n\u2234 The numbers are
    \n-6 – 4, -6, -6 + 4
    \n-10, -6, -2<\/p>\n

    \"KSEEB<\/p>\n

    Question 31.
    \nFind an integral root between -3 and 3 by inspersion and then using synthetic division solve the equation x3<\/sup> + 2x2<\/sup> – 11x -12 = 0.
    \nAnswer:
    \nPut x = – 1
    \n(-1)3<\/sup> + 2(- 1)2<\/sup> – 11 (- 1) – 12
    \n-1 + 2 + 11 – 12 = 13 – 13 = 0
    \nx = -1 is a root, x + 1 is a factor
    \n\"1st
    \nQuotient = x2<\/sup> + x – 12
    \nRemainder = 0
    \nx3<\/sup> + 2x2<\/sup> – 11x – 12 = 0
    \n(x + 1)(x2<\/sup> + x – 12) = 0
    \nx + 1 = 0, x2<\/sup> + x – 12 = 0
    \nx = 3, -4 x = -1
    \nThe roots are -1. 3, -4<\/p>\n

    Question 32.
    \nPradeep invested equal amounts one at 6% SI and the other at 5% CI. If the former earns Rs.437.5 more as interest at the end of two years, find the total amount invested.
    \nAnswer:
    \nLet the amount invested = x
    \nP = x r = 6% T = 2years
    \nSI = \\(\\frac{P \\times T \\times R}{100}=\\frac{x \\times 2 \\times 6}{100}=\\frac{12 x}{100}\\) = 0.12x
    \nP = x r = 5% = 0.05 n = 2
    \nA =P(1 + r)n<\/sup> = x(1 + 0.05)2<\/sup> = x(1.05)2<\/sup> = 1.1025x
    \nCI = A – P = 1.1025 x – x = 0.1025 x
    \nGiven 0.12x – 0.1025x = 437.5
    \n0.0175x = 437.5
    \nAmount invested at 6% SI = 25,000
    \nAmount invested at 5% Cl = 25,000
    \n\u2234 Total amount invested = 25,000 + 25,000
    \n= Rs. 50,000<\/p>\n

    Question 33.
    \nA school runs in the morning and afternoon shifts and employs 40 teachers. The average salary of 25 teachers working in the morning shift is Rs.2800\/- and the average salary of teachers working the afternoon shift is Rs.3000\/- find
    \n(i) the average salary of the teachers in the school.
    \n(ii) the average salary is 5 teachers shifted from morning to afternoon shift,
    \nAnswer:
    \n(i) N1<\/sub> = 25 X1<\/sub> = 2800
    \nN2<\/sub> = 15 X2<\/sub> = 3000
    \nX\u0304 = \\(\\frac{\\mathrm{X}_{1} \\mathrm{~N}_{1}+\\mathrm{X}_{2} \\mathrm{~N}_{2}}{\\mathrm{~N}_{1}+\\mathrm{N}_{2}}=\\frac{2800(25)+3000(15)}{25+15}=\\frac{70000+45000}{40}\\) = Rs. 2875<\/p>\n

    (ii) X1<\/sub> = 2800 N1<\/sub> = 20
    \nX2<\/sub> = 3000 N2<\/sub> = 20
    \nX\u0304 = \\(\\frac{X_{1} N_{1}+X_{2} N_{2}}{N_{1}+N_{2}}\\)
    \n= \\(\\frac{2800(20)+3000(20)}{20+20}=\\frac{56000+60000}{40}\\) = Rs. 2900<\/p>\n

    Question 34.
    \nA wholesale dealer sold a machine to a shopkeeper at 20% profit. The shopkeeper sold it to a customer so as to get 25% profit for himself. The difference between the selling price of the shopkeeper and that of the wholesale dealer was found to be Rs.129. Find the initial price of the machine.
    \nAnswer:
    \nLet x be C.P. of the machine
    \nSP of wholesale dealer = x + 20% of x = \\(\\frac{120 x}{100}\\)<\/p>\n

    SP of the shopkeeper = 25% of \\(\\frac{120 x}{100}+\\frac{120 x}{100}\\)
    \n\\(\\frac{120 x}{100}+\\frac{25}{100}\\left(\\frac{120 x}{100}\\right)-\\frac{120 x}{100}\\) = 129
    \n\\(\\frac{25}{100} \\times \\frac{120 x}{100}\\) = 129
    \n30x = 129 \u00d7 100
    \nx = \\(\\frac{12900}{30}\\) = Rs. 430<\/p>\n

    Question 35.
    \nFind the orthocentre of the triangle formed by the vertices A (5, -2), B (-1, 2) C (1, 4).
    \nAnswer:
    \nLet AD, BE be the altitudes.
    \nNow slope of BC =\\(\\frac{4-2}{1+1}\\) = 1
    \n\u2234 Slope of AD = -1 since AD perpendicular to BC.
    \n\u2234 Equation to the altitude AD is y + 2 = -1 (x – 5)
    \nor x + y – 3 = 0
    \nAgain slope of AC = \\(\\frac{4+2}{1-5}=-\\frac{3}{2}\\)
    \n\u2234 Slope of altitudes BE = \\(-\\frac{1}{3}=\\frac{2}{3}\\)
    \n\u2234 Equation to the altitude BE is
    \ny – 2 = \\(\\frac{2}{3}\\)(x+ 1)
    \nor 2x – 3y + 8 = 0 (2)
    \nSolving (1)and (2), we get x = \\(\\frac{1}{5}\\), y = \\(\\frac{14}{5}\\)
    \n\u2234 The orthocentre is \\(\\left(\\frac{1}{5}, \\frac{14}{5}\\right)\\).<\/p>\n

    \"KSEEB<\/p>\n

    Question 36.
    \nProve that tan 120\u00b0 +3 sin2 3000_ 2 cosec2 240\u00b0 – \\(\\frac{3}{4}\\) cot2 60\u00b0 = \\(\\frac{10}{3}\\)
    \nAnswer:
    \ntan 120 = tan(180 – 60) = -tan60\u00b0 = \\(\\sqrt{3}\\)
    \nsin 300 = sin (360 – 60) = -sin 60 = \\(-\\frac{\\sqrt{3}}{2}\\)
    \ncosec 240 = cosec (180 + 60) = \u2014-cosec 60 = \\(-\\frac{2}{\\sqrt{3}}\\)
    \ncot 60 = \\(\\frac{1}{\\sqrt{3}}\\)
    \nLHS = \\(\\frac{4}{3}(-\\sqrt{3})^{2}+3\\left(-\\frac{\\sqrt{3}}{2}\\right)^{2}-2\\left(\\frac{-2}{\\sqrt{3}}\\right)^{2}-\\frac{3}{4}\\left(\\frac{1}{\\sqrt{3}}\\right)^{2}\\)
    \n= \\(\\frac{4}{3}(3)+3\\left(\\frac{3}{4}\\right)-2\\left(\\frac{4}{3}\\right)-\\frac{3}{4}\\left(\\frac{1}{3}\\right)\\) = 4 + \\(\\frac{9}{4}-\\frac{8}{3}-\\frac{1}{4}=\\frac{48+27-32-3}{12}=\\frac{40}{12}=\\frac{10}{3}\\) = RHS<\/p>\n

    Question 37.
    \nFind a point that is equidistant from the points (1, 2), (5, -6), and (3, -4).
    \nAnswer:
    \nP = (x, y) A(1, 2) B(5, -6)
    \nPA = PB = PC
    \nPA = PB
    \n\\(\\sqrt{(x-1)^{2}+(y-2)^{2}}=\\sqrt{(x-5)^{2}+(y+6)^{2}}\\)
    \n(x – 1)2<\/sup> + (y – 2)2<\/sup> = (x – 5)2<\/sup> + (y + 6)2<\/sup>
    \n\"1st
    \n-2x + 10x – 4y – 12y + 5 – 61 = 0, 8x – 16y -56 = 0
    \nx – 2y -7 = 0 (1)
    \nPB = PC
    \n\\(\\sqrt{(x-5)^{2}+(y+6)^{2}}=\\sqrt{(x-3)^{2}+(y+4)^{2}}\\)
    \n(x – 5)2<\/sup> + (y + 6)2<\/sup> = (x – 3)2<\/sup> + (y +4)2<\/sup>
    \n\"1st
    \n-10x + 6x + 12y – 8y + 61 – 25 = 0
    \n-4x + 4y + 36 = 0
    \nx – y – 9 = 0 (2)
    \n\u00f7 -4
    \n\"1st
    \ny = 2
    \nx – y – 9 = 0
    \nx – 2 – 9 = 0
    \nx – 11 = 0
    \nx = 11
    \nRequired Point = P = (11, 2)
    \nWhich is nothing but the circumcenter of the \u2206le ABC.<\/p>\n

    Question 38.
    \nFind the lengths of the altitudes of the triangle whose vertices are (5, 2) (3, -3) and (-4, 3).
    \nAnswer:
    \nA = (5,2) B = (3, -3) C = (-4, 3)
    \nArea of \u2206ABC = \\(\\frac{1}{2}\\) [x1<\/sub>(y2<\/sub> – y3<\/sub>) + x2<\/sub> (y3<\/sub> – y1<\/sub>) + x3<\/sub> (y1<\/sub> – y2<\/sub>)]
    \n= \\(\\frac{1}{2}\\)[5 (-3 – 3) + 3 (3 – 2) -4 (2 + 3)] = \\(\\frac{1}{2}\\)[5 (-6) + 3 (1) – 4 (5)]
    \n= \\(\\frac{1}{2}\\) [-30 + 3 – 20] =\\(\\frac{1}{2}\\)[-47] = \\(\\left|\\frac{-47}{2}\\right|=\\frac{47}{2}\\) = Square units
    \nArea of the \u2206ABC = \\(\\frac{1}{2}\\) \u00d7 BC \u00d7 AD
    \nAD = \\(\\frac{2 \\Delta \\mathrm{ABC}}{\\mathrm{BC}}\\)
    \nBC = \\(\\sqrt{(-4-3)^{2}+(3+3)^{2}}=\\sqrt{49+36}=\\sqrt{85}\\)
    \n\"1st
    \nAC = \\(\\sqrt{(-4-5)^{2}+(3-2)^{2}}=\\sqrt{81+1}=\\sqrt{82}\\)
    \n\u2206ABC = \\(\\frac{1}{2}\\) \u00d7 AC \u00d7BE
    \nBE = \\(\\frac{2 \\times \\Delta \\mathrm{ABC}}{\\mathrm{AC}}=\\frac{2 \\times 47}{2 \\times \\sqrt{82}}\\)
    \nBE = \\(\\frac{47}{\\sqrt{82}}\\)
    \nAB = \\(\\sqrt{(3-5)^{2}+(-3-2)^{2}}=\\sqrt{4+25}=\\sqrt{29}\\)
    \n\u2206ABC = \\(\\frac{1}{2}\\) \u00d7 AB \u00d7CF
    \n\"1st<\/p>\n

    PART – D<\/span><\/p>\n

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

    Question 39.
    \nIf x5\/3<\/sup> – y1\/3<\/sup> – z = 0, then show that (x5<\/sup> – y – z3<\/sup>)3<\/sup> = 27 x5<\/sup> yz3<\/sup>
    \nAnswer:
    \nx5\/3<\/sup> – y1\/3<\/sup> – z = 0 \u21d2 x5\/3<\/sup> – y1\/3<\/sup> = z
    \nCubing on both sides
    \n(x5\/3<\/sup> – y1\/3<\/sup>) = z3<\/sup>
    \n(x5\/3<\/sup>)3<\/sup> – (y1\/3<\/sup>)3<\/sup> – 3x5\/3<\/sup>y1\/3<\/sup>(x5\/3<\/sup> – y1\/3<\/sup>) = z3<\/sup>
    \nx5<\/sup> – y – 3x5\/3<\/sup> y1\/3<\/sup>(z) = z3<\/sup> \u21d2 x5<\/sup> – y – z3<\/sup> = 3x5\/3<\/sup> y1\/3<\/sup> z
    \nCubing on both sides
    \n(x5<\/sup> – y – z3)3<\/sup> = 27x5<\/sup>yz3<\/sup><\/p>\n

    Question 40.
    \nUsing log tables find the value of \\(\\frac{\\sqrt{14.5} \\times \\sqrt[3]{8.571}}{(16.751)^{2 \/ 3}}\\)
    \nAnswer:
    \nlogx = log \\(\\frac{\\sqrt{14.5} \\times \\sqrt[3]{8.571}}{(16.751)^{2 \/ 3}}\\)
    \n= log\\(\\sqrt{14.5}\\) + log\\(\\sqrt[3]{8.571}\\) – log(16.751)2\/3
    \n= \\(\\frac{1}{2}\\)log14.54 + \\(\\frac{1}{3}\\)log8.571 – \\(\\frac{2}{3}\\)log(l6.751) .
    \n= \\(\\frac{1}{2}\\)(l.l614) + \\(\\frac{1}{3}\\)(0.9331) – \\(\\frac{2}{3}\\)(1.2240)
    \n= 0.5807 + 0.3110 – 0.816 = 0.0757
    \nx = antilog 0.0757= 1.191<\/p>\n

    Question 41.
    \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 = 390
    \nTn<\/sub> = a + (n – 1)d
    \n390 = 65 + (n – 1)13
    \n390 = 65 + 13n- 13
    \n13n = 338 n = 26
    \nSn<\/sub> = \\(\\frac{n}{2}\\) [a + l] = \\(\\frac{26}{2}\\)(65 +390) = 13 (455) = 5,915
    \n\u2234 S = 5,915<\/p>\n

    Question 42.
    \nTwo brothers have their annual income in the ratio 8: 5, while their annual expenditures are in the ratio 5 : 3, if they save \u20b9 1200\/- and \u20b9 1000\/- per annum. Find their incomes.
    \nAnswer:
    \nLet the income be x and expenditure be y.
    \nSo the income of two brothers would be 8x and 5x and expenditure would be 5y and 3y.
    \nWe knew that Income – Expenditure = Saving
    \n\u2234 we get:
    \n8x – 5y = 1200 …………..(1)
    \n5x – 3y = 1000 …………..(2)
    \nSolving (1) and (2), we get: x = 1400 and y – 2000
    \n\u2234 Annual incomes of two brothers are 8x and 5x
    \n= 8 \u00d7 1400 & 5 \u00d7 1400 = \u20b9 11200 and \u20b9 7000.<\/p>\n

    \"KSEEB<\/p>\n

    Question 43.
    \nIf the interest on 800 is more than the interest on \u20b9 400 by Rs 40 in 2 years. Find the rate of interest.
    \nAnswer:
    \nP = 800, T = 2, R = R
    \nSI = \\(\\frac{800 \\times 2 \\times R}{100}\\)
    \nI1<\/sub> = 16R
    \nP = 400, T = 2 R = R
    \nSI = \\(\\frac{400 \\times 2 \\times \\mathrm{R}}{100}\\)
    \nI2<\/sub> = 8R
    \nI1<\/sub> = I2<\/sub> = 40
    \n16R – 8R = 40
    \nR = \\(\\frac{40}{8}\\) = 5%<\/p>\n

    Question 44.
    \nIf the cost function C (x) of producing ‘x’ unit of a product is given by C(x) = 500x2<\/sup> + 2500x + 5000 and if each unit of the product is sold at \u20b9 6000. then find BEP.
    \nAnswer:
    \nGiven C(x) = 500 x2<\/sup> + 2500x + 5000 (x = output)
    \nTotal Revenue = R(x) = Selling price X quantity = 6000 x
    \nFor BEP : C (x) = R (x)
    \n500x2<\/sup> + 250Qx + 5000 = 6000x
    \n500x2<\/sup> + 2500x – 6000x + 5000 = 0
    \n500x2<\/sup> – 3500x – 5000x = 0
    \n\u00f7 500
    \nx2<\/sup> – 7x + 10 = 0
    \n(x – 5) (x – 2) = 0
    \nx = 5 or 2 units<\/p>\n

    Question 45.
    \n\\(\\frac{\\mathrm{x} \\cdot{cosec}^{2} 30^{\\circ} \\cdot \\sec ^{2} 45^{\\circ}}{8 \\cos 45^{\\circ} \\cdot \\sin 60^{\\circ}}\\) = tan2 60\u00b0 – tan2 30\u00b0
    \nAnswer:
    \n\"1st<\/p>\n

    Question 46.
    \nFind the ratio in which the co-ordinate axes, divide the line joining the points (2, 5) and (1, 9). Find also the coordinates of the points of division.
    \nAnswer:
    \nThe co-ordinate of the point dividing the join of A (2, 5) and B (1, 9) in the ratio
    \nK: 1 are \\(\\left(\\frac{K+2}{K+1}, \\frac{9 K+5}{K+1}\\right)\\) (1)
    \n(i) If the point lies on x axis then its y co-ordinate is zero
    \ni.e. \\(\\frac{9 \\mathrm{~K}+5}{\\mathrm{~K}=1}\\) = 0
    \n\u21d2 K = \\(\\frac{-5}{9}\\)
    \n\u2234 x axis divides AB externally in the ratio 5 : 9
    \nSubstituting K = \\(\\frac{-5}{9}\\) in (1) the co-ordinates of the point of division is \\(\\left(\\frac{K+2}{K+1}, \\frac{9 K+5}{K+1}\\right)=\\left(\\frac{\\frac{-5}{9}+2}{\\frac{-5}{9}+1}, 0\\right)=\\left(\\frac{13}{4}, 0\\right)\\)<\/p>\n

    (ii) If the point lies on y axis then its abscissa (x coordinate) = 0
    \ni.e \\(\\frac{\\mathrm{K}+2}{\\mathrm{~K}+1}\\) = 0
    \n\u21d2 K = -2
    \n\u2234 y axis divides AB externally in the ratio 2 : 1
    \nCoordinates of the point of division
    \n= \\(\\left(\\frac{K+2}{K+1}, \\frac{9 K+5}{K+1}\\right)=\\left(0, \\frac{9(-2)+5}{-2+1}\\right)\\) = (0, 13)<\/p>\n

    \"KSEEB<\/p>\n

    Question 47.
    \nFind the ratio in which the line joining (1, 2) and (4, 3) is divided by the line joining the points (2, 3) and (4, 1)
    \nAnswer:
    \nLet A = (2, 3) and B (4, 1). The equation of the line AB is given by
    \n\\(\\frac{y-3}{x-2}=\\frac{1-3}{4-2}\\) \u21d2 y – 3 = x + 2 \u21d2 x + y = 5
    \nLet the line joining the pins C (1, 2) and D (4, 3) cut the line AB at P (x, y) in the ratio r : 1. Then
    \nP = \\(\\left(\\frac{4 r+1}{r+1}, \\frac{3 r+2}{r+1}\\right)\\)
    \nThis lies on the line x + y = 5
    \nThus the co-ordinates of P must satisfy this equation.
    \n\u2234 \\(\\frac{4 r+1}{r+1}+\\frac{3 r+2}{r+1}\\) = 5
    \n4r + 1 + 3r + 2 = 5 (r + 1)
    \n7r + 3 = 5r + 5 \u21d2 r = 1
    \nThus the required ratio is 1 : 1<\/p>\n

    Question 48.
    \nShow that the following points are the vertices of a rectangle. (1, 6), (-1, -2), (4,1) (-4, 3)
    \nAnswer:
    \nLet A = (1, 6), B = (-1, -2), C = (4, 1), D = (-4, 3)
    \nAB = \\(\\sqrt{(1+1)^{2}+(6+2)^{2}}=\\sqrt{4+64}=\\sqrt{68}\\)
    \nBC = \\(\\sqrt{(-1-4)^{2}+(-2-1)^{2}}=\\sqrt{25+9}=\\sqrt{34}\\)
    \nCD = \\(\\sqrt{(4+4)^{2}+(1-3)^{2}}=\\sqrt{64+4}=\\sqrt{68}\\)
    \nDA = \\(\\sqrt{(1+4)^{2}+(6-3)^{2}}=\\sqrt{25+9}=\\sqrt{34}\\)
    \n\u2234 AB = CD and BC = AD
    \ni.e., opposite sides are equal.
    \nFurther
    \nAC = \\(\\sqrt{(1+4)^{2}+(6-1)^{2}}=\\sqrt{9+25}=\\sqrt{34}\\)
    \nBD = \\(\\sqrt{(-1+4)^{2}+(-2-3)^{2}}=\\sqrt{9+25}=\\sqrt{34}\\)
    \n\u2234 The diagonals AC and BD are equal.
    \n\u2234 ABCD is a parallelogram with diagonals equal.
    \n\u2234 ABCD is a rectangle.<\/p>\n

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

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

    Question 49.
    \n(a) Derive the equation of the straight line is the form \\(\\frac{x}{a}+\\frac{y}{b}\\) = 1
    \nAnswer:
    \nLet a line cut the x-axis at A and the y-axis at B. Let OA = a and OB = b. Then the length OA = a is called the x-intercept of the line and the length OB = b is called the y-intercept of the line. If ‘ the line passes the origin. Then both the x-intercept and y-intercepts are zero.<\/p>\n

    Consider the XOY plane. Let the line cut the x-axis at A and the y-axis at B. By data OA = a and OB = b. Thus A = (0, 0) and B = (0, b). Hence the required line is the line joining A and B, Its equation can be obtained by using a two-point form of the equation of a line.
    \n\"1st
    \nThen the equation of AB is
    \n\\(\\frac{y-0}{x-a}=\\frac{0-b}{a-0}\\)
    \n\u21d2 \\(\\frac{y}{x-a}=\\frac{-b}{a}\\) \u21d2 ay = -bx + ab \u21d2 bx + ay = ab
    \nDividing throughout by ‘ab’ we have
    \n\\(\\frac{b x}{a b}+\\frac{a y}{a b}=\\frac{a b}{a b}\\)
    \nThis form of equation of the line is called Intercept form.
    \n\u21d2 \\(\\frac{x}{a}+\\frac{y}{b}\\) = 1<\/p>\n

    (b) The angles of a triangle are in A.P. and the ratio of a number of degrees in the least to the number of radians in the greatest is 60: \u03c0. Find the angles of the triangle in radians.
    \nAnswer:
    \nA – D, A, A + D
    \nA – D + A + A + D= 180
    \n3A = 180
    \nA = 60\u00b0<\/p>\n

    A- D, A, A+ D Angles in degrees
    \n(A – D)\\(\\frac{\\pi}{180}\\), A, \\(\\frac{\\pi}{180}\\)(A+D)\\(\\frac{\\pi}{180}\\) Angles in radians
    \nGiven
    \n\\(\\frac{A-D}{(A+1) \\frac{\\pi}{180}}=\\frac{60}{\\pi}\\)
    \n\"1st
    \n(A – D) 180 = 60(A + D)
    \n(A – D)3 = A + D
    \n3A – 3D = A + D
    \n3A – A = 3D + D
    \n2A = 4D
    \n2(60) = 4D
    \n4D = 120
    \nD = 30\u00b0
    \nAngles in degrees are
    \n60 – 30, 60, 60 + 30
    \nA – D, A, A + D
    \n300, 60\u00b0, 90\u00b0
    \nor Angles in radians are
    \n\\(\\frac{\\pi}{6}, \\frac{\\pi}{3}, \\frac{\\pi}{2}\\)<\/p>\n

    (c) Find the number of digits in 2S0. (log 2 = 0,3010)
    \nAnswer:
    \nLet x = 250 log x = log 250 = 50 log 2 = 50 \u00d7 0.3010 = 15.05
    \nAs the characteristic of log x is 15, no. of digits are 15 + 1 = 16.<\/p>\n

    \"KSEEB<\/p>\n

    Question 50.
    \n(a) Shanmukh buys every year Bank’s cash certificate of value exceeding the last year’s purchase by Rs. 500. After 15 years, he finds that the total value of the certificates purchased by him is Rs.82,500. Find the value of the certificates purchased by him
    \n(a) in the first year and b in the 10lh year.
    \nAnswer:
    \nd = Rs.500 n=15 Sn<\/sub> = 82,500 a = ? T10<\/sub> = ?
    \nSn<\/sub> = \\(\\frac{\\mathrm{n}}{2}\\)[2a + (n – 1)d]
    \n82500 = \\(\\frac{15}{2}\\) [2a+ 14(500)]
    \n\\(\\frac{82500 \\times 2}{15}\\) = 2a + 7000
    \n11000 = 2a + 7000
    \n2a = 11000 – 7000 = 4,000
    \na = 2,000
    \na = Rs. 2,000
    \nT10<\/sub> = a + (10 – 1) d = 2000 + 9 (500) = 2000 + 4500 = 6500
    \nT10<\/sub> = 6,500<\/p>\n

    (b) x + 3y \u2265 3,2x + y \u2265 2, x \u2265 0, y \u2265 0
    \n\"1st<\/p>\n

    (c) Find the numbers which when divided by 36,40 and 48 leaves 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 = 720
    \nSince 5 has to be the remainder, we have to add 5 to 720
    \ni.e. 720 + 5 = 725<\/p>\n","protected":false},"excerpt":{"rendered":"

    Students can Download 1st PUC Basic Maths Model Question Paper 5 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 5 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\/89320"}],"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=89320"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/89320\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=89320"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=89320"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=89320"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}