{"id":6037,"date":"2023-11-04T13:44:05","date_gmt":"2023-11-04T08:14:05","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=6037"},"modified":"2023-11-06T11:36:57","modified_gmt":"2023-11-06T06:06:57","slug":"karnataka-sslc-maths-model-question-paper-1","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/karnataka-sslc-maths-model-question-paper-1\/","title":{"rendered":"Karnataka SSLC Maths Model Question Paper 1 with Answers"},"content":{"rendered":"

Students can Download Karnataka SSLC Maths Model Question Paper 1 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 1 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<\/p>\n

Question 2.
\nThe HCF and LCM of 12, 15 and 21 are ________ and ________ respectively
\na) 3,1
\nb) 420,3
\nc) 3, 420
\nd) 420,1
\nAnswer:
\nc) 3,420
\nSolution:
\nFactors of 12 = 2 \u00d7 2 \u00d7 3
\n= 3 \u00d7 5
\n= 7 \u00d7 3<\/p>\n

\"Karnataka<\/p>\n

Question 3.
\nThe value of \\(\\frac{\\sin 18}{\\cos 72}\\) is ________
\na) 1 \/ 2
\nb) 1
\nc) 0
\nd) \u221a3 \/ 2
\nAnswer:
\nb) 1
\nSolution:
\n\"Karnataka<\/p>\n

Question 4.
\nThe distance between the points p(2, 3) and Q(4, 1) is
\na) 3\u221a6
\nb) 2\u221a2
\nc) 2\u221a4
\nd) 3\u221a9
\nAnswer:
\nb) 2\u221a2
\nSolution:
\n\"Karnataka<\/p>\n

Question 5.
\nIn the given figure^ABC, DE||BC. If DE=5cm, BC=8cm and AD=3.5 cm, then the length of AB is
\na) 5.6cm
\nb) 4.8cm
\nc) 5.2cm
\nd) 6.4 cm
\nAnswer:
\na) 5.6cm
\nSolution:
\n\"Karnataka<\/p>\n

Question 6.
\n6. The 10th term of an A.P : 2, 7, 12,
\na) 1
\nb) 47
\nc) 9
\nd) 17
\nAnswer:
\nb) 47
\nSolution:
\na = 2, d= 7 – 2 = 5, ri= 10
\nan<\/sub> = a + (n – 1)d
\n= 2 + (10-1) (5)
\n= 2 + 45 = 47
\n\u2234 10th term of an AP is 47.<\/p>\n

\"Karnataka<\/p>\n

Question 7.
\nA box consists, of 4 red, 5 black and 6 white balls. One ball is drawn out at random, find the probability that the ball drawn is black.
\na) 1\/15
\nb) 1
\nc) 1\/4
\nd) 1\/3
\nAnswer:
\nd) 1\/3
\nSolution:
\n\"Karnataka<\/p>\n

Question 8.
\nIn the given fig, the angle of elevation \u03b8 is
\na) 30\u00b0
\nb) 45\u00b0
\nc) 90\u00b0
\nd) 60\u00b0
\nAnswer:
\nb) 45\u00b0
\nSolution:
\nTan \u03b8\u00a0 = \\(\\frac{A B}{B C}=\\frac{1}{1}\\)
\nTan \u03b8 = Tan 45\u00b0
\n\u03b8 = 45\u00b0
\n\"Karnataka<\/p>\n

II. Answer the following Questions : ( 1 \u00d7 8 = 8 )<\/span><\/p>\n

Question 9.
\n7 \u00d7 11 \u00d7 13 + 13 is a composite number. Why?
\nAnswer:
\n7 \u00d7 11 \u00d7 13 + 13 = 1014 is a number divisible by some other numbers other than itself and by one.
\n\u2234 1014 is a composite number.<\/p>\n

Question 10.
\nIf one of the roots of the quadratic equation 6x2<\/sup> – x – 2 = 0 is 2\/3. Find the other.
\nAnswer:
\nLet a and p are the roots.
\n\"Karnataka<\/p>\n

\"Karnataka<\/p>\n

Question 11.
\nFind the Value of cos90\u00b0 + tan 45\u00b0 Soln: cos90 + tam45 = 0+1 = 1
\nAnswer:
\ncos 90 + tan 45 = 0 + 1 = 1<\/p>\n

Question 12.
\nIf F(2, P) is the mid point of the Sine segment joining the points A(6, -5) and B(-2, 11). Find the value of P.
\nAnswer:
\nSince P is the mid point of AB,
\nCo – ordinates of P are
\n\"Karnataka
\n\"Karnataka
\nCo – ordinates of P are (2, 3)<\/p>\n

Question 13.
\nIf the probability of winning a game is 5\/11, what is the probability of losing it.
\nAnswer:
\nPE(E) + P(E\u0304) = 1
\n\\(\\frac{5}{11}\\) + P(E\u0304) = 1
\nP(E\u0304) = 1 – \\(\\frac{5}{11}\\) = 1 – \\(\\frac{5}{11}\\)
\n\u2234 P(E\u0304) = \\(\\frac{6}{11}\\)<\/p>\n

Question 14.
\nIf tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80\u00b0, then the what is the measure of \u2220POA = ?
\nAnswer:
\n\"Karnataka
\nIn \u2206 POA , \u2220OAP = 90\u00b0
\n\u2220APO = y = 40
\n\u2220POA = 90 – 40 = 50\u00b0
\n\u2234 \u2220POA = 50\u00b0<\/p>\n

Question 15.
\nWrite the formula to find the volume of a hemisphere.
\nAnswer:
\nVolume of hemisphere = 2\/3 \u03c0r3<\/sup><\/p>\n

Question 16.
\nWhat is the Area of the shaded region?
\nAnswer:
\n\"Karnataka
\nArea of the shaded region = Area of square – Area of circle
\n= (14)2<\/sup> \\(\\frac{22}{7}\\) \u00d7 7 \u00d7 7
\n= 196 – 154
\n= 42 sq.cms .<\/p>\n

III. Answer the following : ( 2 \u00d78 = 16 )<\/span><\/p>\n

Question 17.
\nP.T 5 – \u221a3 is irrational.
\nAnswer:
\nLet 5 – \u221a3 be rational no. in the form of p\/q
\nwhere p and q are integers and q \u2260 0.
\n5 – \u221a3 = p\/q
\n5 = \\(\\frac{p}{q}+\\sqrt{3}\\)
\n\u21d2 5 is rational number, \\(\\frac{p}{q}+\\sqrt{3}\\) is irrational which leads to the contradiction.
\nHence our assumption is wrong.
\n5 – \u221a3 is not a rational number ,
\n\u21d2 5 – \u221a3 is an irrational number.<\/p>\n

Question 18.
\nIn the figure DE || AC and DF || AE. BF BE
\nPrnvp that = \\(\\frac{\\mathbf{B F}}{\\mathbf{F E}}=\\frac{\\mathbf{B E}}{\\mathbf{E C}}\\)
\nAnswer:
\n\"Karnataka
\nBy Thales theorem,
\nIn \u2206 ABC, DE || AC
\n\"Karnataka<\/p>\n

\"Karnataka
\nOR<\/p>\n

In fig LM || CB and LN || CD Prove that \\(\\frac{A M}{A B}=\\frac{A N}{A D}\\)
\n\"Karnataka
\nAnswer:
\nIn \u2206 ABC
\nLM || CB
\nAccording to Thales theorem,
\n\\(\\frac{\\mathrm{AM}}{\\mathrm{AB}}=\\frac{\\mathrm{AL}}{\\mathrm{AC}}\\) ….. (1)
\nIn \u2206 ADC LN || CD
\nAccording to corollary of Thales
\n\\(\\frac{A N}{A D}=\\frac{A L}{A C}\\) ……. (2)
\nFrom (1) and (2)
\n\\(\\frac{\\mathrm{AM}}{\\mathrm{AB}}=\\frac{\\mathrm{AN}}{\\mathrm{AD}}\\)<\/p>\n

Question 19.
\nOn comparing the ratios \\(\\frac{\\mathbf{a}_{1}}{\\mathbf{a}_{2}}, \\frac{\\mathbf{b}_{1}}{\\mathbf{b}_{2}}\\) and \\(\\frac{\\mathbf{c}_{1}}{\\mathbf{c}_{2}}\\) find out whether the lines representing the following pair of linear equations interesect at a point, are parallel or coincident
\n5x – 4y + 8 = 0
\n7x + 6y – 9 = 0
\n5x – 4y + 8 = 0 is the line a1<\/sub>x + b1<\/sub>y + C1<\/sub> = 0
\n7x + 6y – 9 = 0 is the line a2<\/sub>x + b2<\/sub>y +c2<\/sub> =0
\na1<\/sub> = 5, b1<\/sub> = -4, c1<\/sub> = 8
\na2<\/sub> = 7, b2<\/sub> = 6, c2<\/sub> = -9
\n\\(\\frac{a_{1}}{a_{2}} \\neq \\frac{b_{1}}{b_{2}}\\) The two lines are intersecting lines.<\/p>\n

\"Karnataka<\/p>\n

Question 20.
\nDraw a circle of radius 4cm and construct a pair of tangents to the circle from a point 8cm away from the centre.
\nAnswer:
\n\"Karnataka<\/p>\n

Question 21.
\nFind a quadratic polynomial, the sum and product of whose zeroes are -3 and 2 respectively.
\nAnswer:
\nLet the quadratic polynomial be ax2<\/sup> + bx + c and its zeroes be \u03b1 and \u03b2
\n\u03b1 + \u03b2 = -b\/a = -3
\n\u03b1\u03b2 = c\/a = 2
\nEquation is x2<\/sup> – ( \u03b1 + \u03b2) x + \u03b1\u03b2
\nx2<\/sup> – (-3)x + 2
\nx2 + 3x + 2<\/p>\n

Question 22.
\nA kite is flying at a height of 60m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60\u00b0. Find the length of the string, assuming that is no slack in the string.
\nAnswer:
\n\"Karnataka<\/p>\n

OR<\/p>\n

Two solid right circular cones have same height and the radii of bases are r1<\/sub> and r2<\/sub>. Both the cones are metled and recasted to form a cylinder of same height. Show that the radius of the base of the cylinder is \\(\\sqrt{\\frac{\\mathbf{r}_{1}^{2}+\\mathbf{r}_{2}^{2}}{3}}\\)
\nAnswer:
\nVolume of the cylinder = Sum of the volumes 2 cones
\n\"Karnataka<\/p>\n

Question 23.
\nA bicycle wheel makes 5000 revolutions in moving 11 km. Find the diameter of the wheel.
\nAnswer:
\nDistance covered by the wheel in one revolution = Distance moved \/ Number of revolutions
\n= \\(\\frac{11}{5000}\\) cm
\n= \\(\\frac{11}{5000}\\) \u00d7 1000 \u00d7 100cm
\n= 220cm
\n\u2234 Circumference of the wheel = 220 cm
\n2\u03c0r = 220cm
\n= 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 r
\n= 220
\n=3- r = 35 cm
\n\u2234 Diameter = 2r = 2 \u00d7 35 = 70cm
\nHence, the diameter of the wheel = 70cm.<\/p>\n

Question 24.
\nA conical vessel whose internal radius is 5cm and height 24cm is full of water. The water is emptied into a cylindrical vessel with internal radius lOcms. Find the height to which the water rises.
\nAnswer:
\nr1<\/sub> = radius o the conical vessel = 5cm
\nh1<\/sub> = height of the conical vessel = 24cm
\nr2<\/sub> = radius of the cylindrical vessel = 10cm.
\n\u2234 Volume of water in conical vessel = Volume of water in cylindrical vessel.
\n\"Karnataka
\n\u2234 Height of the water in the cylindrical vessel = 2cm.<\/p>\n

III. Answer the following : ( 3 \u00d7 9 = 27 )<\/span><\/p>\n

Question 25.
\nIf 2 is added to numerator and denominator to a fraction, it becomes 9\/10 If 3 is subtracted from numerator and denominator, the fraction becomes 4\/5 Find the function.
\nAnswer:
\nLet the fraction be x \/ y
\nwhere x < y
\nIf 2 is added to Nr and Dr, we get 9\/10
\n\u2234 \\(\\frac{x+2}{y+2}=\\frac{9}{10}\\)
\n10(x + 2) = 9(y + 2)
\n10x + 20 = 9y + 18
\n10x – 9y + 20 – 18 = 0
\n10x – 9y + 2 = 0
\nl0x – 9y = -2 ..(1)
\nIf 3 is subtracted from same fraction, we
\n\\(\\frac{x-3}{y-3}=\\frac{4}{5}\\)
\n5(x – 3) = 4(y – 3)
\n5x – 15 = 4y – 12
\n5x – 4y = -12 + 15
\n5x – 4 = 3….. (2)
\nSolve (1) and (2)
\n(10x – 9y = -2) x 5
\n(5x – 4y = 3) x 10
\n\"Karnataka
\nSubstitute the value of y in (1)
\n10x – 9y = -2
\n10x – 9(8) = -2
\n10x – 72 = -2
\n10x = -2 + 72
\n10x = 70
\nx = 7
\n\u2234 Fraction = \\(\\frac{x}{y}=\\frac{7}{8}\\)<\/p>\n

OR<\/p>\n

A girl is twice as old as her sister. Four years, hence the product of their ages (in years) will be 160. Find their
\npresent ages.
\nAnswer:
\nLet Girls present age be x years.
\nSister\u2019s age = 2x years.
\nAfter 4 years, their respective ages will be (x + 4) years and (2x + 4) years.
\n(x + 4) (2x + 4) =160
\n2x2<\/sup> + 4x+ 8x + 16= 160
\n2x2<\/sup> + 12x+ 16= 360
\nx2<\/sup> + 6x + 8 = 80
\nx2<\/sup> + 6x + 8 – 80 = 0
\nx2<\/sup> + 6x – 72 = 0
\nx2<\/sup>+ 12x – 6x – 72 = 0
\nx(x + 12) – 6(x + 12) = 0
\n(x + 12) (x – 6) = 0
\nx + 12 = 0
\nx = -12 or x – 6 = 0, x = 6.
\nGirls present age = 6 years.
\nSister\u2019s age = 12 years.<\/p>\n

\"Karnataka<\/p>\n

Question 26.
\nThe difference of squares of two natural numbers is 84. The square of the larger number is 25 times the smaller number. Find the numbers.
\nAnswer:
\nLet the two natural numbers be x and y such that x > y.
\nx2<\/sup> – y2<\/sup> = 84 ……..(1)
\nx2<\/sup> = 25y …….(2)
\nBy eq.(1) and (2)
\n25y – y2<\/sup> = 84
\ny2<\/sup> – 25y + 84 = 0
\ny2<\/sup> – 21y – 4y + 84 = 0
\ny (y – 21) – 4(y – 21) = 0
\n(y – 21) (y – 4) = 0
\ny – 21 = 0,y = 4
\nSubstitute the value of y in (2)
\nx2<\/sup> = 25 x 21
\nx = \\(\\sqrt{25 \\times 21}=5 \\sqrt{21}\\)
\nx2<\/sup> = 25y
\nx2<\/sup> = 25 \u00d7 4= 100
\nx = \u221a10o = 1o
\n\u2234 Two numbers are 10 and 4.<\/p>\n

Question 27.
\nFind the value of K if the points A(2, 3) B (4, k) and C(6, -3) are collinear.
\nAnswer:
\nSince the given points are collinear, the area of the triangle formed by them must be O.
\n\\(\\frac{1}{2}\\)[2(k + 3) + 4(-3 -3) + 6(3 – k)] = 0 1
\n\\(\\frac{1}{2}\\)[2k + 6 + 4(-6) + 6(3 – k)] = 0
\n\\(\\frac{1}{2}\\)[2k + 6 – 24 + 18 – 6k] = 0
\n\\(\\frac{1}{2}\\)(-4k) = 0
\n-2k = 0 or k = 0.<\/p>\n

OR<\/p>\n

Find the area of a triangle whose vertices are (1, -1) (-4, 6) and (-3, -5)
\nAnswer:
\nArea of the triangle =
\n= \\(\\frac{1}{2}\\)[x1<\/sub>(y2<\/sub> – y3<\/sub>) + x2<\/sub> (y3<\/sub> – y2<\/sub>) + x3<\/sub> (y1<\/sub> – y2<\/sub>)]
\nx1<\/sub> = 1 x2<\/sub> = -4 x3<\/sub> = -3
\ny1<\/sub> = -1 y2<\/sub> = 6 y3<\/sub> = -5
\nArea = \\(\\frac{1}{2}\\)[1(6 + 5) + (-4) (-5 + 1) + (-3)(-l- 6)]
\n= \\(\\frac{1}{2}\\) [11 + (-4(-4) +(-3) (-7)]
\n= \\(\\frac{1}{2}\\) (11 + 16 + 21)
\n= \\(\\frac{1}{2}\\) (48) = 24 .
\n\u2234 Area of the triangle = 24 square units.<\/p>\n

Question 28.
\nProve that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
\n\"Karnataka
\nAnswer:
\nGiven
\ncentred \u2018O\u2019 at the point of contact \u2018P\u2019.
\nTo Prove that : OP \u22a5 AB
\nConstruction : Take any point Q, other than P, on the tangent AB, Join OQ, OQ cuts the circle at R.
\nProof : \u201cOf all the line segments joining the point \u2018O\u2019 to the tangent AB, perpendicular OP is the shortest distance from O to AB.
\n\u2234 OP = OR [ Radii of the same circle]
\nOQ = OR + RQ .
\nOQ > OR
\nOQ > OP (\u2235 OP = OR)
\nOP > OQ
\n\u2234 \u2018OP\u2019 is the shortest distance from point O to AB.
\nOP \u22a5 AB<\/p>\n

Question 29.
\nHow many spherical bullets can be made out of a solid cube of lead whose edge measures 44cm, each bullet being 4cm in diameter.
\nAnswer:
\nd = 4cm, r = 2cm, a = 44cm.
\nVolume of cube = a3<\/sup> = 443<\/sup>cm3<\/sup>.
\n\"Karnataka<\/p>\n

OR<\/p>\n

A circus tent is cylinderica! upto a height of 3m and conical above it. If the diameter of the base is 105m and the slant height of th conial part is 53m, find the total canvas used in making the tent.
\nAnswer:
\nd= 105m, r = 105\/2 = 52.5m, l = 53m,
\nh = 3m.
\nTotal canvas used = C.S.A of cylinder + C.S.A. of cone
\n\"Karnataka
\n\"Karnataka<\/p>\n

Question 30.
\nSolve the following system of equations graphically.
\nx + 3y = 6
\n2x – 3y = 12
\nAnswer:
\nx + 3y = 6
\n3y = 6- x
\ny = \\(\\frac{6-x}{3}\\)
\n\"Karnataka
\n2x – 3y = 12
\n-3y = 12 – 2x
\n3y = 2x – 12
\ny = \\(\\frac{2 x-12}{3}\\)<\/p>\n

\"Karnataka
\n\"Karnataka<\/p>\n

\"Karnataka<\/p>\n

Question 31.
\nDraw less than Ogive for the given data:
\n\"Karnataka
\nAnswer:<\/p>\n\n\n\n\n\n\n\n\n
C.I<\/td>\nF<\/td>\nC.f<\/td>\n<\/tr>\n
20- 30<\/td>\n5<\/td>\n5<\/td>\n<\/tr>\n
30-40<\/td>\n4<\/td>\n9<\/td>\n<\/tr>\n
40-50<\/td>\n3<\/td>\n12<\/td>\n<\/tr>\n
50-60<\/td>\n8<\/td>\n20<\/td>\n<\/tr>\n
60-70<\/td>\n7<\/td>\n27<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"Karnataka<\/p>\n

Question 32.
\nHow many terms of the series 54, 51, 48…. be taken. So that their sum is 513? Find the last term.
\nAnswer:
\nn = ? a = 54, d = 51 – 54 = -3, Sn = 513. S\u201e=
\nSn<\/sub>\\(\\frac{\\mathrm{n}}{2}\\) [2a + (n-1)d]
\n513 = \\(\\frac{\\mathrm{n}}{2}\\) [2(54) + (n – 1)(-3)]
\n513 = \\(\\frac{\\mathrm{n}}{2}\\) (l08-3n + 3)
\nn(111 -3n)= 1026
\n111n – 3n2<\/sup> – 1026 = 0
\n-3n2<\/sup> + 111n- 1026 = 0
\nDivide by -3.
\nn2<\/sup>-37n + 342 = 0
\nn2<\/sup> – 19n – 18n + 342 = 0
\nn(n – 19) – 18(n -19) = 0
\n(n – 19) (n – 18) = 0
\n\u2234 n = 19 or n = 18
\nT19<\/sub> = a + 18d
\n= 54-54 =54-51
\nT19<\/sub> = 0
\nOR
\nT18<\/sub> = a + 17d
\n= 54 + 18(-3)
\n= 54 + 17(-3)
\nT18<\/sub> = 3
\n\u2234 Sum of 18termsas well as that of 19 terms is 513.<\/p>\n

Question 33.
\nThe sum of two numbers is 15. If the sum of their reciprocals is 3\/10. Find the numbers.
\nAnswer:
\nLet the numbes be x and (15 – x)
\nBy data,
\n\"Karnataka
\n45x – 3x2<\/sup> = 150
\n3x2<\/sup> – 45x + 150 = 0
\nDivide by 3 .
\nx2<\/sup> – 45x + 150=0
\nDivide by 3
\nx2<\/sup> – 15x + 50 = 0
\nx2<\/sup> – 10x – 5x + 50 = 0
\nx(x – 10) – 5(x – 10) = 0
\n(x – 10) (x – 5) = 0
\nx -10 = 0
\nx = 10 OR
\nx – 5 = 0
\nx = 5
\nHence the two numbers are 10 and 5.<\/p>\n

OR<\/p>\n

Seven years ago Keshav\u2019s age was five times the square of Raghav\u2019s age. Three years hence Raghav\u2019s age will be two fifth of Keshav\u2019s age. Find their present ages.
\nAnswer:
\nSeven years ago, let Raghav\u2019s age be x years. Then, seven years ago Keshav\u2019s age was 5x2<\/sup> years.
\nRaghav\u2019s present age = (x + 7) years.
\nKeshav\u2019s present age = (5x2<\/sup> + 7) years.
\nThree years hence, we have R
\naghav\u2019s age = (x + 7 + 3) years = (x + 10) years.
\nKeshav\u2019s age = (5x2<\/sup> + 7 + 3) years = (5x2<\/sup> +10) years.
\nIt is given that three years hence Raghav\u2019s 2
\nage will be 2\/5 of Keshav\u2019s age.
\nx + 10 = \\(\\frac{2}{5}\\) (5x2<\/sup> +10)
\nx + 10 = 2x2<\/sup> + 4
\n2x2<\/sup> + 4-x – 10 = 0
\n2x2<\/sup> – 6 – x = 0
\n2x2<\/sup> – 4x + 3x – 6 = 0
\n2x(x – 2) + 3(x – 2) = 0 ,
\n(x – 2) (2x + 3) =0
\nx – 2 = 0, 2x + 3 = 0
\nx = 2, x = -3\/2
\nRaghav\u2019s present age = (2 + 7) years = 9 years.
\nKeshav\u2019s present age = (5 x 22<\/sup> + 7) = 27 years.<\/p>\n

\"Karnataka<\/p>\n

V. Answer the following : ( 4 \u00d7 4 = 16 )<\/span><\/p>\n

Question 34.
\nThe first and the last term of an A.P are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum? Soln: Let a be the first term and d be the common difference.
\nAnswer:
\nLet 1 be its last term. Then a = 17,1 = an<\/sub> =350 and d = 9
\n1 = an<\/sub> = 350 ”
\na + (n – 1)d = 350
\n9(n – 1) = 350 – 17 = 333
\nn – 1 = 333\/9 = 38
\nn = 37 + 1 = 38
\nPutting a = 17,1 = 350 and n = 38
\nSn<\/sub> = \\(\\frac{n}{2}\\) (a + 1)
\nS38<\/sub> = \\(\\frac{38}{2}\\) (17 + 350) .
\n= 19 \u00d7 367
\n= 6973
\nHence, there are 38 terms in A.P having their sum as 6973.<\/p>\n

OR<\/p>\n

Find the sum of first 51 terms of an A.P whose 2nd term and 3rd term are 14 and 18 respectively.
\nAnswer:
\nLet a be the first term and d be the common difference of given A.P. Then
\na2<\/sub>= 14 and a3<\/sub> = 18 a + d = 14 and a + 2d = 18
\nSolve
\n\"Karnataka
\nd= 4
\na + d= 14
\na + 4 = 14
\na = 14 – 4 = 10
\na = 10
\nPutting a = 10, d = 4 nd n = 51
\nSn = \\(\\frac{n}{2}\\) [2a + (n-l)d]
\nS5l= \\(\\frac{51}{2}\\)[2×10 + (51 – 1) \u00d7 4]
\n= \\(\\frac{51}{2}\\)[20 + 50 \u00d7 4]
\n= \\(\\frac{51}{2}\\)(20 + 200)
\n= \\(\\frac{51}{2}\\) \u00d7 220
\n= 51 \u00d7 110= 5610<\/p>\n

Question 35.
\nFind the mean for the following data
\n\"Karnataka
\nAnswer:
\n\"Karnataka<\/p>\n

\"Karnataka<\/p>\n

Question 36.
\nProve the Identity
\n\\(\\frac{\\tan ^{2} \\theta}{\\sin ^{2} \\theta-1}+\\frac{\\csc ^{2} \\theta}{\\sec ^{2} \\theta-\\csc ^{2} \\theta}=\\frac{1}{\\sin ^{2} \\theta-\\cos ^{2} \\theta}\\)
\nAnswer:
\n\"Karnataka
\n\"Karnataka<\/p>\n

Question 37.
\nConstruct a triangle with sides 5cm. 6cm and 7cm and then another triangle
\nwhose sides are 7\/5 of the corresponding sides of the first triangle.
\nAnswer:
\n\"Karnataka<\/p>\n

VI. Answer the following : ( 5 \u00d7 1 = 5 )<\/span><\/p>\n

Question 38.
\nAreas of similar triangles are proportional to the squares on the corresponding sides.
\nAnswer:
\nAreas of similar triangles are proportional to the squares on the corresponding sides.
\n\"Karnataka
\nData: \u2206 ABC \u2206DEF
\n\u2220A = \u2220D
\n\u2220B = \u2220F
\n\u2220C = \u2220F
\n\\(\\frac{\\mathrm{AB}}{\\mathrm{DE}}=\\frac{\\mathrm{AC}}{\\mathrm{DF}}=\\frac{\\mathrm{BC}}{\\mathrm{EF}}\\)
\n\"Karnataka<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Karnataka SSLC Maths Model Question Paper 1 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 1 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\/6037"}],"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=6037"}],"version-history":[{"count":1,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/6037\/revisions"}],"predecessor-version":[{"id":90879,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/6037\/revisions\/90879"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=6037"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=6037"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=6037"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}