{"id":6034,"date":"2023-11-03T13:56:09","date_gmt":"2023-11-03T08:26:09","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=6034"},"modified":"2023-11-06T11:40:09","modified_gmt":"2023-11-06T06:10:09","slug":"karnataka-sslc-maths-model-question-paper-5","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/karnataka-sslc-maths-model-question-paper-5\/","title":{"rendered":"Karnataka SSLC Maths Model Question Paper 5 with Answers"},"content":{"rendered":"

Students can Download Karnataka SSLC Maths Model Question Paper 5 with Answers, Karnataka SSLC Maths Model Question Papers with Answers<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/span><\/p>\n

Karnataka State Syllabus SSLC Maths Model Question Paper 5 with Answers<\/h2>\n

Time: 3 Hours
\nMax Marks: 80<\/p>\n

I. In the following questions, four choices are given for each question, choose and write the correct answer along with its alphabet: ( 1 \u00d7 8 = 8 )<\/span><\/p>\n

Question 1.
\nThe pair of linear equations 3a + 4b = k, 9a + 12b = 6 have infinitely many solutions when,.
\na) K = -2
\nb) K = 3
\nc) K = 2
\nd) K = -3
\nAnswer:
\nc) K = 2
\nSolution:
\n\"Karnataka<\/p>\n

Question 2.
\nn2<\/sup> – 1 is divisible by 8, if n is
\na) Prime numbers
\nb) Odd integer
\nc) Even integer
\nd) Natural number
\nAnswer:
\nb) Odd integer
\nSolution:
\nn2<\/sup> – 1
\nIf n is an odd. integer, 1,3,5,
\nEx: 12<\/sup>-1 = 1- 1= 0 divisible by 8.
\n32<\/sup> – 1=9-1=8 divisible by 8.
\n52<\/sup> – 1 = 25 – 1 = 24 divisible by 8.<\/p>\n

\"Karnataka<\/p>\n

Question 3.
\n\\(\\sqrt{1+\\tan ^{2} \\theta}\\) = , where 0<\u03b8< 90\u00b0
\na) sec\u03b8
\nb) cosec \u03b8
\nc) cos \u03b8
\nd) sin \u03b8
\nAnswer:
\na) sec\u03b8
\nSolution:
\n\\(\\sqrt{1+\\tan ^{2} \\theta}=\\sqrt{\\sec ^{2} \\theta}=\\sec \\theta\\)<\/p>\n

Question 4.
\nIf Q divides the line A(3, 5) and B (7,9) internally in the ratio 2:3, then the co ordinates of Q are.
\n\"Karnataka
\nAnswer:
\nc) \\(\\left(\\frac{23}{5}, \\frac{33}{5}\\right)\\)
\nSolution:
\n\"Karnataka<\/p>\n

Question 5.
\n\"Karnataka
\nIf Area of the sector\u00a0 OPRQ = \\(\\frac{5}{18}[ \/latex] of Area of circle. Then the value of x
\na) 25\u00b0
\nb) 50\u00b0
\nc) 75\u00b0
\nd) 100\u00b0
\nAnswer:
\nd) 100\u00b0
\nSolution:
\n\"Karnataka
\n\"Karnataka<\/p>\n

Question 6.
\nIf 1 + 2 + 3 + n terms = 28, then n is equal to
\na) 28
\nb) 7
\nc) 8
\nd) 56
\nAnswer:
\nb) 7
\nSolution:
\n[latex]\\frac{\\mathrm{n}(\\mathrm{n}+1)}{2}\\)
\nn(n + 1) = 28 x 2 = 56
\nn(n + 1) = 7 x 8
\n\u2234 n = 7<\/p>\n

Question 7.
\nIf E1<\/sub>E2<\/sub> E3<\/sub>…….E10<\/sub> are the possible elementary events of a random experiment, then P(E1<\/sub>) + P(E2<\/sub>) + P(E3<\/sub>) + ………P(E10<\/sub>) is equal to
\na) 0
\nb) 1
\nc) 2
\nd) 3
\nAnswer:
\nb) 1<\/p>\n

Question 8.
\nIf we express sec A in terms of sin A, then sec A is equal to
\n\"Karnataka
\nAnswer:
\n\\(\\frac{1}{\\sqrt{1-\\sin ^{2} A}}\\)
\nSolution:
\n\"Karnataka<\/p>\n

\"Karnataka<\/p>\n

II. Answer the following Questions : ( 1 x 8 = 8 )<\/span><\/p>\n

Question 9.
\nWhat is the \\(\\frac{p}{q}\\) form of 43.123456789 ?
\nAnswer:
\n\\(\\frac{43123456789}{999999999}\\)<\/p>\n

Question 10.
\nWrite the quadratic equation formed by the roots \u221a3 + \u221a5 and \u221a3 – \u221a5
\nAnswer:
\nx2<\/sup> -(\u03b1 + \u03b2)x + \u03b1\u03b2 = 0
\nX2<\/sup> – [3 + \u221a5 + 3 – \u221a5]x + (3 + \u221a5)(3 – \u221a5) = 0
\nx2<\/sup> – [6] x + (9 – 5) = 0
\nx2<\/sup> – 6 + 4 = 0<\/p>\n

Question 11.
\nFind the value of \\(\\frac{\\sin 26}{\\sec 64}+\\frac{\\cos 26}{\\csc 64}\\)
\nAnswer:
\n\"Karnataka<\/p>\n

Question 12.
\nWhat is the distance between the points p(cos0, sin0) and Q(sin0, -COS0)?
\nAnswer:
\n\"Karnataka<\/p>\n

Question 13.
\nIf a number \u201cx\u201d choosen at random from the numbers -2, -1, 0, 1, 2. What is the probability that x2<\/sup> < 3 ?
\nAnswer:
\nClearly \u201cx\u201d can take any of the five given values.
\n\u2234 n(x) = 5.
\nIf x2<\/sup> < 3, then x can take the values -1, 0, 1.
\n\u2234 This even A = ,{-1, 0, 1}
\nn(A) = 3.
\nP(A) = \\(\\frac{n(A)}{n(S)}=\\frac{3}{5}\\)<\/p>\n

Question 14.
\nWhat is the Area of a circle whose perimeter is 44 cms.
\nAnswer:
\n2\u03c0r = 44
\n2 x \\(\\frac{22}{7}\\) x r = 44
\nr = \\(\\frac{44 \\times 7}{22 \\times 2}\\)
\nr = 7
\nA = \u03c0r\u00b2 = \\(\\frac{22}{7}\\) x 7 x 7 = 154cm2<\/sup><\/p>\n

Question 15.
\nA toy was made by scooping out a hemisphere from each end of a solid cylinder. If the height of the cylinder is h cm and base radius is r cms, find the total surface area of the toy.
\nAnswer:
\nTotal surface Area of the toy = C.S.A of cylinder + 2 [Surface Area of Hemisphere]
\n= 2\u03c0rh + 2\u03c0r\u00b2
\n= 2 \u03c0r (h + r)
\n\"Karnataka<\/p>\n

Question 16.
\nThe circumference of a circle exceeds the diameter by 15cm. Find the raidus of the circle.
\nAnswer:
\nCircumference = 2r + 15
\n2\u03c0r = 2r + 15
\n2\u03c0r – 2r = 15
\n2 x \\(\\frac{22}{7}\\)r – 2r = 15
\nMultiply by 7
\n2 x 22r – 14r = 105 .
\n30r = 105
\nr = 3.5<\/p>\n

III. Answer the following : (2 x 8 = 16 )<\/span><\/p>\n

Question 17.
\nProve that \u221a5 + \u221a3 is an irrational number.
\nAnswer:
\n\"Karnataka
\n\u2234 \u221a3 is rational, is a contradiction
\n\u2234 \u221a5 + \u221a3 is irrational.<\/p>\n

\"Karnataka<\/p>\n

Question 18.
\nAX and BY are perpendiculars to segment XY. If AO = 5cm, BO = 7cm and Area of \u2206 AOX = 150 cm2<\/sup>, find the area of \u2206 BOY.
\nAnswer:
\n\"Karnataka
\nIn \u2206 OYB and \u2206 OXA,
\nWe have \u2220X = \u2220Y
\nSo, by AA – criterion of similarity, we have
\n\u2206 YOB \u2245 \u2206 XOA
\n\"Karnataka<\/p>\n

OR<\/p>\n

In the given figure, BD \u22a5 AC. Prove that AB2<\/sup> + CD2<\/sup> = AD2<\/sup> + BC2<\/sup>
\nAnswer:
\nSoln: In \u2206 BDC,
\n\u2220 BDC = 90\u00b0
\n\"Karnataka
\nBC2<\/sup> = BD2<\/sup> + DC2<\/sup> (By Pythagoras)
\nIn \u2206 BDA, \u2220 APB = 90\u00b0
\nAB2<\/sup> = AD2<\/sup> + BD2<\/sup> (By Pythagoras)
\nAB2<\/sup> – BC2<\/sup> = AD2<\/sup> + BD2<\/sup> – BD2<\/sup> – DC2<\/sup>
\nAB2<\/sup> + CD2<\/sup> = AD2<\/sup> + BC2<\/sup>
\nHence proved.<\/p>\n

Question 19.
\nIf zeroes of the polynomial p(y) = y3<\/sup> – 3y2<\/sup>+ y + 1 are a – b, a, a+ b. Find a and b.
\nAnswer:
\nSum of the zeroes of p(y) = y3<\/sup> – 3y2<\/sup>+ y + 1
\n\"Karnataka
\n(a-b)(a)(a + b)= \\(\\frac{-1}{1}\\) = -1
\n(a2<\/sup> – b2<\/sup>) a = -1
\n(1 – b2<\/sup>) (1) = -1
\n– b2<\/sup> = -1
\n-b2<\/sup> = – 1 – 1
\n-b2 = -2
\nb2 = 2
\nb = \u00b1\u221a2<\/p>\n

Question 20.
\nConstruct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length.
\nAnswer:
\n\"Karnataka<\/p>\n

Question 21.
\nFind the zeroes of the polynomial p(y) = y3<\/sup> – 5y2<\/sup> – 16y + 80. If zeroes are \u03b1, -\u03b1 and \u03b2.
\nAnswer:
\nSum of the zeroes
\n\"Karnataka
\n\u03b1 + (-\u03b1) + \u03b2 = 5
\n\u03b1 – \u03b1 + \u03b2 = 5
\n\u03b2 = 5
\n\"Karnataka
\n(\u03b1 )(-\u03b1 )(\u03b2) =-80
\n-\u03b12<\/sup>\u03b2 = -80
\n\u03b12<\/sup>\u03b2 = 80
\n\u03b12<\/sup>(5) = 80
\n\u03b1 = 80\/5 = 16
\n\u03b1 = \u221a16
\n= \u00b14<\/p>\n

Question 22.
\nTwo pillars of equal height and on either side of a road, which is 100m wide. The angles of elevation of the top of the pillars are 60 and 30 at a point on the road between the pillars. Find the position of the point between the pillars and height of each pillars.
\nAnswer:
\n\"Karnataka
\nIn \u2206 PAB,
\ntab 60 = \\(\\frac{\\mathrm{AP}}{\\mathrm{AB}}\\)
\n\u221a3 = h\/x
\nh = \u221a3x
\nIn \u2206 BCQ,
\ntan 30 = \\(\\frac{\\mathrm{CQ}}{\\mathrm{BC}}\\)
\n\\(\\frac{1}{\\sqrt{3}}=\\frac{h}{100-x}\\)
\nh\u221a3 -= 100 – x
\n(\u221a3x)(\u221a3x) = 100-x
\n4x = 100
\nx = 25
\n\u2234 h= \u221a3 X 25 = 25\u221a3
\n= 25 x 1.73 = 43.3<\/p>\n

OR<\/p>\n

The three metallic spheres of radii are in the ratio of 3 : 4 : 5 are melted to form a single solid sphere of radius 12cm. Find the radius of the three small metallic spheres.
\nAnswer:
\nVolume of the solid sphere = Sum of the volumes of each sphere
\n\\(\\frac{4}{3} \\pi r^{3}=\\frac{4}{3} \\pi\\left(r_{1}^{3}+r_{2}^{3}+r_{3}^{3}\\right)\\)
\n4 x 123<\/sup> = 4[(3x)3<\/sup> + (4x)3<\/sup> + (5x)3<\/sup>]
\n1728 = 27x3<\/sup> + 64x3<\/sup> + 125x3<\/sup> 1728 = 216 x3<\/sup>
\nx = \\(\\frac{1728}{216}\\)
\nx3<\/sup> = 8 x = 3\u221a8 = 2
\n\u2234 R1<\/sub> = 6, R2<\/sub> 8, R3<\/sub> = 10<\/p>\n

\"Karnataka<\/p>\n

Question 23.
\nIn the given fig. OACB is a quadrant of a circle with centre O and radius 3.5cm. If OD = 2cm, find the area of the shaded region
\nAnswer:
\n\"Karnataka
\n\"Karnataka<\/p>\n

Question 24.
\nFind the volume of the largest right circular cone that can be cut of a cube of edge 7cm.
\nAnswer:
\nGiven radius of the base of cone = \\(\\frac{7}{2}\\) cm
\nh = height of cone = 7cm
\nVolume of the cone = \\(\\frac{1}{3}\\)\u03c0r\u00b2h
\nVolume = \\(\\frac{1}{3} \\times \\frac{22}{7} \\times \\frac{7}{2} \\times \\frac{7}{2} \\times 7\\)
\n= \\(\\frac{22 \\times 7 \\times 7}{12}\\)
\n= 89.83cm3<\/sup><\/p>\n

IV. Answer the following : ( 3 x 9 = 27 )<\/span><\/p>\n

Question 25.
\nThe sum of a two digit number and the number obtained by reversing the order of its digits is 165. If the digits differ by 3, find the number.
\nAnswer:
\nLet the digits at units and tens place of the given number be y and x respectively, then
\nGiven number = 10x + y
\nNumber obtained by reversing the order of the digits = 10y + x
\n(10x + y) + (10y + x) = 165
\n10x + y+ 10y + x = 165
\n(11x+11y=165) = 11 \u2019
\nx + y = 15 ……(1)
\nDigits differ by 3, if x > y
\nx – y = 3 ……..(2)
\nSolve (1) and (2)
\n\"Karnataka
\nConsider x + y =15
\n9 + y = 15
\ny = 15 – 9
\ny = 6
\n\u2234 The number is 96, if y >x, number is 69.<\/p>\n

OR<\/p>\n

Ten years ago Sudhir was twelve times as old as his son Raghav and ten years, hence, he will be twice as old as his son will be. Find their present ages.
\nAnswer:
\nLet the present ages of Sudhir and Raghav be x years and y years respectively. Ten years ago, Father \u2019s age = (x -10) years Son\u2019s age = (y – 10) years,
\nx- 10= 12(y – 10)
\n=> x – 10 = 12y – 120
\nx – 12y = 120+ 10
\nx – i2y = – 110
\nx – 12y + 110 = 0. …….(1)
\nTen years later, Father\u2019s age = (x + 10) years
\nSon\u2019s age = (y + 10)
\nx + 10 = 2(y + 10)
\n\u21d2 x+ 10 = 2y + 20
\nx – 2y = 20 – 10
\nx – 2y = 10
\nx – 2y – 10 = 0 ……(2)
\nSolve (1) and (2)
\n\"Karnataka
\ny = 12
\nx – 12y = 110
\nx – 12(12) = -110
\nx – 144 = -110
\nx = -110 + 144 = 34
\n\u2234 Present age of Sudhir = 34 years and present age of Raghav = 12 years.<\/p>\n

Question 26.
\nThe altitude of a right triangle is 7cm less than its base. If the hypotenuse is 13cm, find the other two sides.
\nAnswer:
\nLet the base BC be = x cm
\nThen its height AB = (x – 7) cm
\nHypotenuse AC = 13 cm
\nIn \u2206ABC, AC2<\/sup> = AB2<\/sup> + BC2<\/sup>
\n132 = (x – If + (x)2<\/sup>
\n169 = x2<\/sup> + 49 – 14x + x2<\/sup>
\n169 = 2x2<\/sup> – 14x + 49
\n2x2<\/sup> – 14x + 49 – 169 = 0
\n2x2<\/sup> – 14x-120 = 0
\nx2<\/sup> – 7x – 60 = 0
\nx2<\/sup>– 12x + 5x-60 = 0
\nx(x- 12) + 5(x – 12) = 0
\n(x-12) (x + 5) = 0
\n(x – 12) = 0 or x + 5 = 0
\nx = 12 or x = -5
\nThe base of the triangle = 12 cm Length of the altitude = (12 – 7)cm = 5 cm.
\n\"Karnataka<\/p>\n

\"Karnataka<\/p>\n

Question 27.
\nX(2, 5), Y (-1, 2) and Z(5, 8) are the co – ordinates of the vertices of A XYZ. The point A and B lie on XY and XZ respectively such that XA : AY = XB: BZ =1:2. Calculate the co – ordinates of X and Y.
\nAnswer:
\n\"Karnataka
\n\"Karnataka
\nB(X2<\/sub> y2<\/sub>)= [3, 6]
\n\u2234 Co – ordinates of A = (1,4)
\nCo – ordinates of B = (3, 6)<\/p>\n

OR<\/p>\n

Find the Area of the triangle formed by joining the mid points of the sides of the triangle whose vertices are (0, -1) (2, 1) and
\nAnswer:
\n\"Karnataka
\nCo – ordinates of A
\n\\(\\frac{0+2}{2}, \\frac{-1+1}{2}\\)
\n= [1,0]
\nCo – ordinates of B
\n\\(\\frac{0+0}{2}, \\frac{-1+3}{2}\\)
\n= [0,1]
\nCo – ordinates of B
\n\\(\\frac{2+0}{2}, \\frac{1+3}{2}\\)
\n= [1,2]
\n\u2234 Area of A ABC = \\(\\frac{1}{2}\\) \u03a3 x1<\/sub> (y2<\/sub> – y3<\/sub>)
\n\"Karnataka
\n= \\(\\frac{1}{2}\\) [1(1-2) + 0(2 – 0) + 1(0-1)]
\n= \\(\\frac{1}{2}\\) [-1 + 0 + 1]
\n\u2234 Area of \u2206 ABC = 1 sq. unit<\/p>\n

Question 28.
\nProve that the tangent at any point of a circle is perpendicular to the radius through the point of contacts.
\nAnswer:
\nData : AB is the tangent drawn to a circle centered at O
\n\u2018C\u2019 is the point of contact.
\n\"Karnataka
\nTo prove that :
\nOC \u22a5AB
\nConstruction : Mark D .on AB. Join OD.
\nLet it cut the circle at E
\nProof: OE < OD
\nOE is the radius of the circle.
\nOE = OC (radii of the same circle)
\n\u21d2 OC < OD
\n\u21d2 OC is the shortest distance between centre of the circle and tangent AB
\n\u2234 OC \u22a5 AB
\nHence proved.<\/p>\n

Question 29.
\nA hollow sphere of internal and external radii are 6cm and Scm respectively is melted and recast into small cones of base radius 2cm and height 8cm. Find the number of cones.
\nAnswer:
\nInner radius of a hollow sphere (r) = 6cm
\nand outer radius (R) = 8cms.
\n\"Karnataka
\nRadius of one small cone (r,) 2cm and
\nheigh (h) = 8 cm.
\n\"Karnataka<\/p>\n

OR<\/p>\n

A medicine capsule is in the shape of a cylinder with two hemispheres stuch to each of its ends. The length of the entire capsule is 14mm and the diameter of the capsule is 5 mm. Find its surface area.
\n\"Karnataka
\nAnswer:
\nLet r mm be the radius and h mm be the height of the cylinder.
\nr = \\(\\frac{5}{2}\\) mm = 2.5mm
\nh = (14 – 2 x \\(\\frac{5}{2}\\) mm
\n= (14-5) mm
\n= 9 mm
\nThe radius of hemisphere = r = \\(\\frac{5}{2}\\) mm
\nNow, the surface area of the capsule. = Curved surface Area of the cylinder surface Area of two hemispheres.
\n= 2\u03c0rh + 2 x 2 \u03c0r\u00b2
\n= 2\u03c0r\u00b2 (h + 2r)
\n\"Karnataka<\/p>\n

\"Karnataka<\/p>\n

Question 30.
\nSolve graphically
\ny = \\(\\frac{1}{2}\\) x and 3x + 4y – 20 = 0
\ny = \\(\\frac{1}{2}\\) x
\n\"Karnataka
\n3x + 4y – 20 = 0
\n3x + 4y = 20
\n4y = 20 – 3x
\n\"Karnataka
\n\"Karnataka<\/p>\n

Question 31.
\nThe following distribution gives the daily income of 65 workers of a factory. Convert this into less than type cumulative frequency distribution and draw its ogive.
\n\"Karnataka
\nAnswer:
\n\"Karnataka
\n\"Karnataka<\/p>\n

Question 32.
\nThe fourth term of an AP is 11, and 8th term exceeds twice the fourth term by 5. Find AP and find the sum of first 100 terms.
\nAnswer:
\nGiven : T4<\/sub> = -11 and T8<\/sub> = 2T4<\/sub> + 5
\nT8<\/sub> = 2(11) + 5
\na + 7d = 22 + 5 = 27
\n\"Karnataka
\nd = 4
\na + 7d = 27
\na + 7(4) = 27
\na+ 28 = 27
\na = 27-28 = -1
\na = -1
\nA.P = -1,3, 7, …..
\nSn<\/sub> = \\(\\frac{n}{2}\\) [2a + (n-1)d] = \\(\\frac{100}{2}\\) [2(-1) + (99)4]
\n= 50 [-2 + 396]
\n= 50 [394]
\n= 19700<\/p>\n

Question 33.
\nA person on tour has \u20b9 360 for his expenses. If he extends his tour for 4 days, he has to cut down his daily expenses by \u20b9 3. Find the original duration of tour.
\nAnswer:
\nLet the original duration of the tour be x days.
\nTotal expenditure on four = 360 \u20b9
\nExpenditure per day = \\(\\frac{360}{x}\\)
\nDuration of the extended tour = (x +4) days
\n\u2234 Expenditure per day according to new schedule = \\(\\frac{360}{x+4}\\)
\nIt is given that daily expenses are cut down by 3\u20b9
\n\\(\\frac{360}{x}-\\frac{360}{x+4}\\) = 3
\n360 (x + 4) – 360x = 3(x) (x + 4)
\n360x + 1440 – 360x = 3x (x + 4)
\n1440 = 3x2<\/sup>+ 12x
\n3x2<\/sup> + 12x – 1440 = 0
\nx2<\/sup> + 4x – 480 = 0
\nx2<\/sup> + 24x – 20x – 480 = 0
\nx(x + 24) – 20(x + 24) = 0
\n(x + 24) (x – 20) = 0
\nx + 24 = 0 and x – 20 = 0
\nx = -24 and x = 20
\n\u2234 Duration of tour = 20 days.<\/p>\n

OR<\/p>\n

Two pipes running together can fill a cistern in 3\\(\\frac{1}{13}\\) minutes. If one pipe
\ntakes 3 minutes more than the other to fill it, find the time in which each pipe would fill the cistern.
\nAnswer:
\nLet the faster pipe takes x minutes.
\nLet the smaller pipe take (x + 3) minutes to fill.
\nPortion of the cistern filled by the faster 1
\npipe in one minute = \\(\\frac{1}{x}\\)
\n\u21d2 Portion of the cisterin filled by the faster pipe in \\(\\frac{40}{13}\\) minutes
\n\"Karnataka
\n|||ly portion of the cistern filled by the slower pipe in \\(\\frac{40}{13}\\) minutes.
\n\"Karnataka
\nIt is given that the cistern is filled in \\(\\frac{40}{13}\\) min.
\n\"Karnataka
\n[40(x + 3) + 40x] = [(x) (x +3)] 13
\n40x + 120 + 40x = (x2<\/sup> + 3x) 13
\n80x + 120= 13x2<\/sup> + 39x
\n13x2<\/sup> + 39x – 80x – 120 = 0
\n13x2<\/sup> – 41x-120 = 0
\n13x2<\/sup> – 65x + 24x – 120 = 0
\n13x (x – 5) + 24(x – 5) = 0
\nx – 5 = 0, 13x + 24 = 0
\nx = 5, 13x = -24, x = \\(\\frac{-24}{13}\\)
\nFaster pipe fills the cistern in 5 minutes and the slower pipe takes 8 min to fill the cistern.<\/p>\n

V. Answer the following : ( 4 x 4 = 16 )<\/span><\/p>\n

Question 34.
\nIn an AP whose first term is 2, the sum of first five terms is one fourth the sum of the next five terms. Show that T20 = -112. Find S20<\/sub>.
\nAnswer:
\na = 2, T1<\/sub> + T2<\/sub> + T3<\/sub> + T4<\/sub>+ T5<\/sub>
\n= \\(\\frac{1}{4}\\) (T6<\/sub> + T7<\/sub> + T8<\/sub> + T9<\/sub> + T10<\/sub>)
\na + a + d + a + 2d + a + 3d + a + 4d
\n= \\(\\frac{1}{4}\\)(a + 5d + a + 6d + a + 7d + a + 8d + a + 9d)
\n5a+ 10d = \\(\\frac{1}{4}\\) (5a + 35d)
\n5(a + 2d) = \\(\\frac{1}{4}\\) x 5( a + 7d)
\na + 2d = \\(\\frac{1}{4}\\) (a + 7d)
\n4(a+2d) = a+7d
\n4a + 8d – a – 7d = 0
\n3a + d = 0
\n3(2) + d = 0
\nd = -6
\nT20<\/sub> = a + 19d
\n= 2 +19(-6)
\n= 2 – 114
\nT20<\/sub> = -112
\nSn<\/sub> = \\(\\frac{n}{2}\\) [2a + (n-1)d]
\nSn<\/sub> = \\(\\frac{20}{2}\\) [2(2) + (20-1)(-6)]
\n= 10 [4- 114]
\n= (10) (-110)
\nS20<\/sub> = -1100.<\/p>\n

OR<\/p>\n

A man repays a loan of \u20b9 3250 by paying 7 20 in the first month and then increases the payment by \u20b9 15 every month. How long will it take to clear his loan?
\nAnswer:
\nHere, a = 20, c.d = 15, n = ?
\nSn<\/sub> = \\(\\frac{n}{2}\\)[2a + (n-1)d]
\n3250 = \\(\\frac{n}{2}\\) [2 x 20 + (n-1)15]
\n3250 x 2 = n(40 + 15n- 15)
\n6500 = n (25 +15n)
\n6500 = 25n + 15n2
\n(15n2<\/sup> + 25n – 6500 = 0)\u00f7 5
\n3n2<\/sup> + 5n – 1300 = 0
\n3n2<\/sup>-60n + 65n- 1300 = 0
\n3n(n – 20) + 65 (n – 20) = 0
\n(n – 20) (3n + 65) = 0
\n(n – 20) = 0, 3n + 65 = 0
\nn = 20, 3n = -65
\nn = \\(\\frac{-65}{3}\\)
\n\u2234 Total amount will be paid in 20 months.<\/p>\n

\"Karnataka<\/p>\n

Question 35.
\nFind the mean, median and mode for * the following frequency distribution.
\n\"Karnataka
\nAnswer:
\n\"Karnataka
\n\"Karnataka<\/p>\n

Question 36.
\nP.T \\(\\frac{\\tan \\theta}{1-\\cot \\theta}+\\frac{\\cot \\theta}{1-\\tan \\theta}\\) = 1+ sec\u03b8 cosec\u03b8
\nAnswer:
\n\"Karnataka
\n\"Karnataka<\/p>\n

Question 37.
\nDraw a right triangle in which the sides (other than hypotenuse) arc of lengths 8cm and 6cm, then construct another triangle whose sides are 5\/3 times the corresponding sides of given triangle.
\nAnswer:
\n\"Karnataka
\nVerify
\nBC = 8cm
\nBC’= \\(\\frac{5}{3} \\times 8=\\frac{40}{3}\\)
\nAB =6cms
\nA’B= \\(\\frac{5}{3} \\times 6\\)
\n=10cms.<\/p>\n

VI. Answer the following : (5 x 1 = 5 )<\/span><\/p>\n

Question 38.
\nState and prove pythagoras theorem.
\nIn a right angled triangle, square on the hypotenuse is equal to sum of A the squares on the other sides.
\nAnswer:
\n\"Karnataka
\nData : \u2206 ABC, \u2220B = 90\u00b0
\nT.P.T : AC2<\/sup> = AB2<\/sup> + BC2<\/sup>
\nConstruction : Draw BD \u22a5 AC
\nProof: In \u2206 ABC and \u2206 ABD
\n\u2220 ABC = \u2220ADB = 90 [\u2235 Data & construction]
\n\u2220A = \u2220A (\u2235 common Angle)
\n\u2220ACB =\u2220 ABD [ \u2235 Re maining angle]
\n\u2234 \u2206 ABC and \u2206 ABD are equiangular
\n=> \u2206 ABC ~ \u2206ABD
\n\"Karnataka
\nAB2<\/sup>= AC x AD ……… (1)
\nIn \u2206ABC and \u2206 BDC
\n\u2220 ABC = \u2220 BDC = 90 [\u2235 Data & construction]
\n\u2220C = \u2220C (\u2235 common Angle)
\n\u2220BAC =\u2220 DBC [ \u2235 Re maining angle]
\n\u2234 \u2206 ABC and \u2206 DBC are equiangular
\n\u2206ABC ~ \u2206BDC
\n\"Karnataka
\nAB2<\/sup> + BC2<\/sup> = AC (AD + DC)
\nAB2<\/sup> + BC2<\/sup> = AC(AC)
\nAB2<\/sup> + BC2<\/sup>=AC2<\/sup>
\nAC2<\/sup> = AB2<\/sup>+ BC2<\/sup><\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Karnataka SSLC Maths Model Question Paper 5 with Answers, Karnataka SSLC Maths Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka State Syllabus SSLC Maths Model Question Paper 5 with Answers Time: 3 Hours Max Marks: 80 I. …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/6034"}],"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\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=6034"}],"version-history":[{"count":1,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/6034\/revisions"}],"predecessor-version":[{"id":90889,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/6034\/revisions\/90889"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=6034"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=6034"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=6034"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}