{"id":60196,"date":"2024-04-16T11:56:45","date_gmt":"2024-04-16T06:26:45","guid":{"rendered":"https:\/\/apboardsolutions.com\/?p=60196"},"modified":"2024-04-18T09:44:29","modified_gmt":"2024-04-18T04:14:29","slug":"ts-10th-class-maths-model-paper-set-6","status":"publish","type":"post","link":"https:\/\/apboardsolutions.com\/ts-10th-class-maths-model-paper-set-6\/","title":{"rendered":"TS 10th Class Maths Model Paper Set 6 with Solutions"},"content":{"rendered":"

The strategic use of TS 10th Class Maths Model Papers<\/a> Set 6 can significantly enhance a student’s problem-solving skills.<\/p>\n

TS SSC Maths Model Paper Set 6 with Solutions<\/h2>\n

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

General Instructions:<\/span><\/p>\n

    \n
  1. Answer all the questions under Part – A on a separate answer book.<\/li>\n
  2. Write the answers to the questions under Part – B on the question paper itself and attach it to the answer book of Part – A.<\/li>\n<\/ol>\n

    Part \u2013 A (60 Marks)<\/span>
    \nSection \u2013 I (6 \u00d7 2 = 12 Marks)<\/span><\/p>\n

    Note :<\/p>\n

      \n
    1. Answer ALL the following questions.<\/li>\n
    2. Each question carries 2 marks.<\/li>\n<\/ol>\n

      Question 1.
      \nIf A = {x : x is a factor of 24}, then find n(A).
      \nSolution:
      \nA = (1, 2, 3, 4, 6, 8, 12, 24} ; n(A) = 8
      \n24 = 1 \u00d7 24
      \n= 2 \u00d7 12 = 3 \u00d7 8 = 4 \u00d7 6<\/p>\n

      Question 2.
      \nFind the roots of the Quadratic equation x2<\/sup> + 2x – 3 = 0.
      \nSolution:
      \nx2<\/sup> + 2x – 3 = 0
      \nx2<\/sup> + 3x – x – 3 = 0
      \nx(x + 3) – 1 (x + 3) = 0
      \n(x + 3) (x – 1) = 0
      \nx = – 3 (or) x = 1
      \nRoots of Quadratic Equation are – 3, 1.<\/p>\n

      Question 3.
      \nFor what value of ‘t’ the following pair of linear equations has a no solution ?
      \n2x – ty = 5 and 3x + 2y = 11.
      \nSolution:
      \nPair of linear equations has no solution
      \nIf \\(\\frac{a_1}{a_2}\\) = \\(\\frac{b_1}{b_2}\\) \u2260 \\(\\frac{c_1}{c_2}\\) \u21d2 \\(\\frac{2}{3}\\) = \\(\\frac{-t}{2}\\) \u2260 \\(\\frac{-5}{-11}\\)
      \n– 3t = 4 \u21d2 t = – \\(\\frac{4}{3}\\)<\/p>\n

      \"TS<\/p>\n

      Question 4.
      \nIn a right triangle ABC, right angled at ‘C’ in which AB =13 cm, BC = 5cm, determine the value of cos2<\/sup> B + sin2<\/sup> A.
      \nSolution:
      \nWe have, Cos B = \\(\\frac{\\mathrm{BC}}{\\mathrm{AB}}\\) = \\(\\frac{5}{13}\\)
      \nSin A = \\(\\frac{\\mathrm{BC}}{\\mathrm{AB}}\\) = \\(\\frac{5}{13}\\)
      \n\"TS
      \nCos2<\/sup> B + Sin2<\/sup> A
      \n= \\(\\frac{25}{169}\\) +\\(\\frac{25}{169}\\) = \\(\\frac{25+25}{169}\\) = \\(\\frac{50}{169}\\)<\/p>\n

      Question 5.
      \nA point P is 25 cm from the centre O of the circle. The length of the tangent drawn from P to the circle is 24 cm. Find the radius of the circle.
      \nSolution:
      \nFrom right angled \u0394 AOP
      \n\"TS
      \nOP2<\/sup> = OA2<\/sup> + AP2<\/sup>
      \n(25)2<\/sup> = OA2<\/sup> + (24)2<\/sup>
      \n625 = OA2<\/sup> + 576
      \nOA2<\/sup> = 625 – 576 = 49 = 72<\/sup>
      \nOA = 7 cm
      \n\u2234 The radius of the circle is 7 cm.<\/p>\n

      Question 6.
      \nIn a hemispherical bowl of 2.1 cm radius ice-cream is there. Find the volume of the bowl.
      \nSolution:
      \nRadius (r) = 2.1 cm
      \nVolume (V) = \\(\\frac{1}{2}\\)\u03c0r3<\/sup>
      \n= \\(\\frac{2}{3}\\) \u00d7 \\(\\frac{22}{7}\\) \u00d7 (2.1)3<\/sup>
      \n= 19.404 cm3<\/sup><\/p>\n

      Section – II (6 \u00d7 3 = 18 Marks)<\/span><\/p>\n

      Note :<\/p>\n

        \n
      1. Answer ALL the following questions.<\/li>\n
      2. Each question carries 3 marks.<\/li>\n<\/ol>\n

        Question 7.
        \nIf x2<\/sup> + y2<\/sup> = 7xy, then prove that log(\\(\\frac{x+y}{3}\\)) = \\(\\frac{1}{2}\\)
        \nGiven x2<\/sup> + y2<\/sup> = 7xy, now add (2xy) on both sides. We get
        \nx2<\/sup> + y2<\/sup> + 2xy = 7xy + 2xy = 9xy
        \nConsidering square root on both sides we get
        \n\\(\\sqrt{x^2+y^2+2 x y}\\) = \\(\\sqrt{9 x y}\\)
        \nx + y = 3\\(\\sqrt{x y}\\)
        \n\u21d2 \\(\\frac{x+y}{3}\\) = \\(\\sqrt{x y}=(x y)^{\\frac{1}{2}}\\) considering logarithm on both sides, we get
        \nlog(\\(\\frac{x+y}{3}\\)) = log\\((x y)^{\\frac{1}{2}}\\) = \\(\\frac{1}{2}\\) log(xy)
        \nlog(\\(\\frac{x+y}{3}\\)) = \\(\\frac{1}{2}\\) (log x + log y) (\u2235 log mn = log m + log n)<\/p>\n

        \"TS<\/p>\n

        Question 8.
        \n“If we multiply or divide both sides of a linear equation by a non-zero number, then the roots of that linear equation will remain the same.”
        \nIs it true ? If so, justify with an example.
        \nSolution:
        \nGiven statement : “If we multiply or divide both sides of a linear equation by a non-zero number, then the roots of that linear equation will remain the same”.
        \nConsider a linear equation
        \nx + y = 5 ………….. (1)
        \nOne of the solutions for this equation (1) is (2, 3)
        \nNow multiply equation (1) by ‘3’.
        \n3 (x + y) = 3 \u00d7 5
        \n3x + 3y = 15 ………….. (2)
        \nNow substitute the point (2, 3) in the equation (2).
        \n\u2234 3(2)+ 3 (3) = 15
        \n16 + 9= 15 \u21d2 15 = 15
        \n\u2234 (2, 3) is also a solution for the equation (2).
        \nSo the given statement is true.<\/p>\n

        Question 9.
        \nMeasures of sides of a triangle are in Arithmetic Progression. Its perimeter is 30cm, and the difference between the longest and shortest side is 4cm; then find the measures of the sides.
        \nSolution:
        \nLet the 3 sides of given triangle = a – d, a, a + d
        \nThen its perimeter
        \n= a – d + a + a + d = 30 cm.
        \n3a = 30 cm \u21d2 a = \\(\\frac{30}{3}\\) = 10 cm.
        \nThe larger side = a + d
        \nThe shorter side = a – d
        \nThe difference between the above two
        \n= (a + d) – (a – d) = 4 cm.
        \na + d – a + d = 4 cm.
        \n2d = 40 \u21d2 d = \\(\\frac{4}{2}\\) = 2 cm.
        \na = 10 cm, d = 2 cm
        \nSo the sides a – d = 10 – 2 = 8 cm
        \nand a + d=10 + 2 = 12 cm.
        \nSo 8, 10, 12 cm are the sides of the triangle.<\/p>\n

        Question 10.
        \nA bag contains balls which are numbered from 1 to 50. A ball is drawn at random from the bag, the probability that it bears a two digit number multiple of 7.
        \nSolution:
        \nNumber of possible outcomes = 50 Number of required outcomes = 6 {14,21,28,35,42,49}
        \n\u2234 Probability of getting two digit number which is a multiple of 7
        \n= \\(\\frac{\\text { Number of favourable outcomes }}{\\text { Number of total outcomes }}\\)
        \n= \\(\\frac{6}{50}\\) = \\(\\frac{3}{25}\\)<\/p>\n

        Question 11.
        \nIf cot \u03b8 = \\(\\frac{7}{8}\\) then,
        \nEvaluate :
        \ni) \\(\\frac{(1+\\sin \\theta)(1-\\sin \\theta)}{(1+\\cos \\theta)(1-\\cos \\theta)}\\)
        \nii) \\(\\frac{1+\\cos \\theta}{\\sin \\theta}\\)
        \nSolution:
        \ni) \\(\\frac{(1+\\sin \\theta)(1-\\sin \\theta)}{(1+\\cos \\theta)(1-\\cos \\theta)}\\)
        \n= \\(\\frac{1-\\sin ^2 \\theta}{1-\\cos ^2 \\theta}=\\frac{\\cos ^2 \\theta}{\\sin ^2 \\theta}\\) = cot2<\/sup> \u03b8
        \n= (\\(\\frac{7}{8}\\))2<\/sup> = \\(\\frac{49}{64}\\) ………… (1)
        \n(\u2235 sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8 = 1)<\/p>\n

        ii) \\(\\frac{1+\\cos \\theta}{\\sin \\theta}\\) = \\(\\frac{1}{\\sin \\theta}+\\frac{\\cos \\theta}{\\sin \\theta}\\) = cosec \u03b8 + cot \u03b8
        \nNow given cot \u03b8 = \\(\\frac{7}{8}\\)) then we have
        \n(1 + cot2<\/sup> \u03b8) = cosec2<\/sup> \u03b8
        \n\u21d2 1 + (\\(\\frac{7}{8}\\))2<\/sup>
        \n= 1 + \\(\\frac{49}{64}\\) + \\(\\frac{64+49}{64}\\) = \\(\\frac{113}{64}\\)
        \n\u2234 cosec \u03b8 = \\(\\sqrt{\\frac{113}{64}}\\)
        \n\u2234 \\(\\frac{1+\\cos \\theta}{\\sin \\theta}\\) = cosec \u03b8 + cot \u03b8
        \n= \\(\\frac{\\sqrt{113}}{8}\\) + \\(\\frac{7}{8}\\) = \\(\\frac{7+\\sqrt{113}}{8}\\)<\/p>\n

        Question 12.
        \nA right circular cylinder has radius 3.5 cm and height 14 cm. Find curved surface area.
        \nSolution:
        \nradius (r) = 3.5cm; height (h) = 14cm Curved surface area of the right circular cylinder
        \n= 2\u03c0rh = 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 3.5 \u00d7 14
        \n= 44 \u00d7 7 = 308 cm2<\/sup><\/p>\n

        \"TS<\/p>\n

        Section – III (6 \u00d7 5 = 30 Marks)<\/span><\/p>\n

        Note :<\/p>\n

          \n
        1. Answer all the following questions.<\/li>\n
        2. In this section, every question has internal choice. Answer any one alternative.<\/li>\n
        3. Each question carries 5 marks.<\/li>\n<\/ol>\n

          Question 13.
          \nA) If two dice are thrown at the same time, find the probability of getting sum of the dots cm top is prime.
          \nSolution:
          \nTotal possible outcomes when two dice are rolled = 36
          \nLet E be an event to get sum of the tops is prime
          \nTotal favourable outcomes : (1, 1) (1, 2), (2, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1, 6), (2, 5), (3, 4), (4, 3) (5, 2), (6, 1), (5, 6), (6, 5)
          \nNumber of total favourable outcomes = 15
          \nP(E) = \\(\\frac{\\text { No. of favourable outcomes }}{\\text { No. of total possible outcomes }}\\)
          \n\\(\\frac{15}{36}\\) =\\(\\frac{5}{12}\\)<\/p>\n

          (OR)<\/p>\n

          (B) Show that \\(\\frac{\\cos \\theta}{1-\\sin \\theta}+\\frac{1-\\sin \\theta}{\\cos \\theta}\\) = 2 sec \u03b8
          \nSolution:
          \n\"TS<\/p>\n

          Question 14.
          \nA) Find the ratio in which X-axis divides the line segment joining the points (2, – 3) and (5, 6). Then find the intersecting point on X-axis.
          \nSolution:
          \nLet X – axis divides the line segment joining points (2, – 3) and (5, 6) in the ratio m1<\/sub> : m2<\/sub>
          \nx1<\/sub> = 2 x2<\/sub> = 5
          \ny1<\/sub> = -3 y2<\/sub> = 6
          \nCo-ordinate of point
          \nP = (\\(\\frac{m_1 x_2+m_2 x_1}{m_1+m_2}, \\frac{m_1 y_2+m_2 y_1}{m_1+m_2}\\))
          \n= (\\(\\frac{m_1(5)+m_2(2)}{m_1+m_2}, \\frac{m_1(6)+m_2(-3)}{m_1+m_2}\\))
          \nBut this is a point on X-axis, so its y co-ordinate is zero.
          \n\\(\\frac{m_1(6)+m_2(-3)}{m_1+m_2}\\) = 0
          \n6m1<\/sub> – 3m2<\/sub> = 0
          \n6m1<\/sub> = 3m2<\/sub>
          \n\\(\\frac{\\mathrm{m}_1}{\\mathrm{~m}_2}\\) = \\(\\frac{3}{6}\\) = \\(\\frac{1}{2}\\)
          \n\u2234 Required ratio = 1 : 2
          \n\"TS<\/p>\n

          (OR)<\/p>\n

          B) DWACRA is supplied cuboidal shaped wax block with measurements 88 cm \u00d7 42 cm \u00d7 35 cm. From this how many number of cylinderical candles of 2.8 cm diametre and 8 cm of height can be prepared ?
          \nSolution:
          \nShape of wax block = cuboid
          \nIts length (l) = 88 cm ;
          \nbreath (b) = 42 cm ; height (h) = 35 cm
          \nThen the volume of wax present in block = lbh
          \n= 88 \u00d7 42 \u00d7 35 cm3<\/sup> …………… (1)
          \nShape of candle = cylinder
          \nDiameter of candle = (d) = 2.8 cm 2.8
          \n\u21d2 radius = r = \\(\\frac{2.8}{2}\\) = 1.4 cm
          \nheight (h) = 8 cm
          \nVolume of wax required to make one candle = V = \u03c0r2<\/sup>h
          \n= \\(\\frac{22}{7}\\) \u00d7 1.4 \u00d7 1.4 \u00d7 8 cm3<\/sup>
          \n\u2234 Total number of candles that can be
          \nmade = \\(\\frac{\\text { Total volume of block }}{\\text { Volume of each candle }}\\)
          \n= \\(\\frac{88 \\times 42 \\times 35}{\\frac{22}{7} \\times 1.4 \\times 1.4 \\times 8}\\) = 2625
          \nSo 2625 candles can be made with given measurements.<\/p>\n

          Question 15.
          \nA) Two trains leave a railway station at the same time. The first train travels towards west and the second train towards north. The first train travels 5 km\/hr faster than the second train. If after two hours they are 50 km. apart, find the average speed of each train.
          \nSolution:
          \nLet the speed of the slower train = x kmph
          \nthen speed of the faster train = x + 5 kmph.
          \n\"TS
          \nDistance = Speed \u00d7 Time
          \n\u2234 Distance travelled by the first train = 2(x + 5) = 2x+ 10
          \nDistance travelled by the second train = 2.x = 2x
          \nBy Pythagoras Theorem
          \n(hypotenuse)2<\/sup> = (side)2<\/sup> + (side)2<\/sup>
          \n\u21d2 (2x)2<\/sup> + (2x + 10)2<\/sup> = 502<\/sup>
          \n\u21d2 4x2<\/sup> + (4x2<\/sup> + 40x + 100) = 2500
          \n\u21d2 4x2<\/sup> + 4x2<\/sup> + 40x + 100 = 2500
          \n\u21d2 8x2<\/sup> + 40x – 2400 = 0
          \n= x2<\/sup> + 5x – 300 = 0
          \n\u21d2 x2<\/sup> + 20x – 15x – 300 = 0
          \n\u21d2 x(x + 20) – 15(x + 20) = 0
          \n\u21d2 (x + 20) (x – 15) = 0
          \n\u2234 x – 15 = 0 (or) x + 20 = 0
          \n\u21d2 x = 15 (or) -20
          \nBut x can’t be negative.
          \n\u2234 Speed of the slower train x = 15kmph.
          \nSpeed of the faster train
          \nx + 5 = 15 + 5 = 20 kmph.<\/p>\n

          (OR)<\/p>\n

          B) In a right angle triangle, the hypotennse is 10 cm more than the shortest side. If third side is 6 cm less than the hypotenuse, find the sides of the right angle triangle.
          \nSolution:
          \nLet the shortest side of right angled triangle be x cm,
          \nHypotenuse = (x + 10) cm (given)
          \nThird side = (x + 10 – 6) cm
          \n= (x + 4) cm (given)
          \nAC = (x + 10) cm, AB = x cm,
          \nBC = (x + 4) cm
          \nIn a right angled triangle ABC
          \n\"TS
          \nAC2<\/sup> = AB2<\/sup> + BC2<\/sup>
          \n(x + 10)2<\/sup> = x2<\/sup> + (x + 4)2<\/sup>
          \nx2<\/sup> + 20x + 100 = x2<\/sup> + x2<\/sup> + 8x + 16
          \n\u21d2 x2<\/sup> – 12x – 84 = 0
          \n\"TS
          \n\u21d2 x = 6 + 2\\(\\sqrt{30}\\) cm (\u2235 x < 0)
          \nx + 4 = 10 + 2\\(\\sqrt{30}\\) cm
          \nx + 10 = 16 + 2\\(\\sqrt{30}\\) cm<\/p>\n

          \"TS<\/p>\n

          Question 16.
          \nA) Show that \u221a5 – \u221a3 is an irrational number.
          \nSolution:
          \nSuppose \u221a5 – \u221a3 is not an irrational number. \u221a5 – \u221a3 is a rational number.
          \nLet \u221a5 – \u221a3 = \\(\\frac{p}{q}\\) where q \u2260 0 and p, q e \u2208 Z squaring on both sides
          \n5 + 3 – \\(2 \\sqrt{15}=\\frac{p^2}{q^2}\\)
          \n\\(\\sqrt{15}=\\frac{8 q^2-p^2}{2 q^2}\\)
          \n\u2234 p, q \u2208 Z & q \u2260 0
          \n8q2<\/sup> – p2<\/sup> & 2q2<\/sup> \u2208 Z and also 2q2<\/sup> \u2260 0.
          \nSo \\(\\frac{8 q^2-p^2}{2 q^2}\\) JL is a rational number.
          \nbut \\(\\sqrt{15}\\) is an irrational number.
          \nAn irrational number never be equal to a rational number.
          \nSo our supposition that \u221a5 – \u221a3 is not an irrational number is false.
          \n\u2234 \u221a5 – \u221a3 is an irrational number.<\/p>\n

          (OR)<\/p>\n

          (B) Prove that :
          \n(1 + tan2<\/sup> \u03b8) + ( 1 + \\(\\frac{1}{\\tan ^2 \\theta}\\)) = \\(\\frac{1}{\\sin ^2 \\theta-\\sin ^4 \\theta}\\)
          \nSolution:
          \n\"TS<\/p>\n

          Question 17.
          \nA) Draw the graph of the polynomial p(x) = x2<\/sup> – 7x + 12, then find its zeroes from the graph.
          \nSolution:
          \np(x) = x2<\/sup> – 7x + 12
          \n\"TS
          \n\"TS
          \nzeroes of the polynomial is 3, 4<\/p>\n

          (OR)<\/p>\n

          B) Find the solution of x + 2y = 10 and 2x + 4y = 8 graphically.
          \nSolution:
          \nx + 2y = 10 ………… (1)<\/p>\n\n\n\n\n\n
          x<\/td>\n0<\/td>\n2<\/td>\n4<\/td>\n6<\/td>\n<\/tr>\n
          y<\/td>\n5<\/td>\n4<\/td>\n3<\/td>\n2<\/td>\n<\/tr>\n
          (x, y)<\/td>\n(0, 5)<\/td>\n(2, 4)<\/td>\n(4, 3)<\/td>\n(6, 2)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

          2x + 4y = 8 ……….. (2)<\/p>\n\n\n\n\n\n
          x<\/td>\n0<\/td>\n2<\/td>\n4<\/td>\n6<\/td>\n<\/tr>\n
          y<\/td>\n2<\/td>\n1<\/td>\n0<\/td>\n-1<\/td>\n<\/tr>\n
          (x, y)<\/td>\n(0, 2)<\/td>\n(2, 1)<\/td>\n(4, 0)<\/td>\n(6, -1)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

          \"TS
          \nThe lines are parallel
          \n\u2234 No solution for the given pair of equations.<\/p>\n

          Question 18.
          \nA) Construct a triangle of sides 4.2 cm, 5.1 cm and 6 cm. Then construct a triangle similar to it, whose sides are 2\/3 of corresponding sides of the first triangle.
          \nSolution:
          \n\"TS
          \nConstruction Steps :<\/p>\n

            \n
          1. Draw a triangle ABC, with sides AB = 4.2 cm, BC = 5.1 cm, CA = 6 cm.<\/li>\n
          2. Draw a ray BX making an acute angle with BC on the side opposite to vertex A.<\/li>\n
          3. Locate 3 points B1<\/sub>, B2<\/sub>, B3<\/sub> on BX so that BB1<\/sub> = B1<\/sub>B2<\/sub> = B2<\/sub>B3<\/sub>.<\/li>\n
          4. Join B3<\/sub>, C and draw a line through B2<\/sub> parallel to B3<\/sub>C intersecting BC at C’.<\/li>\n
          5. Draw a line through C’ parallel to CA intersect AB at A’.<\/li>\n
          6. \u0394A’BC’ is required triangle.<\/li>\n<\/ol>\n

            (OR)<\/p>\n

            B) Draw a circle of radius 6 cm and construct two tangents to the circle so that angle between the tangents is 60\u00b0.
            \nSolution:
            \nTo begin let us consider a circle with centre ‘O’ and radius 6 cm.
            \nLet PA and PB are two tangents draw from a point ‘P’ outside the circle and the angle between them is 60\u00b0.
            \nIn this \u2220APB = 60\u00b0. Join OP.
            \n\"TS
            \nAs we know, OP is the bisector of \u2220APB, then
            \n\u2220OAP = \u2220OPB = \\(\\frac{60^{\\circ}}{2}\\) = 30\u00b0 (\u2235 \u0394OAP\u2245\u0394OBP)
            \nNow in \u0394OAP
            \nsin 30\u00b0 = \\(\\frac{\\text { Opposite side }}{\\text { Hypotenuse }}=\\frac{\\mathrm{OA}}{\\mathrm{OP}}\\)
            \n\\(\\frac{1}{2}=\\frac{6}{\\mathrm{OP}}\\) (From trigonometric ratio)
            \nOP = 12 cm
            \nNow we can draw a circle of radius 6 cm with centre ‘O’. We then mark a point at a distance of 12 cm from the centre of the circle. Join OP and complete the construction.
            \n\"TS
            \nHence PA and PB are the required pair of tangents to the given circle.<\/p>\n

            \"TS<\/p>\n

            Part – B (20 Marks)<\/span><\/p>\n

            Note :<\/p>\n

              \n
            1. Answer all the questions.<\/li>\n
            2. Each question carries 1 mark.<\/li>\n
            3. Answers are to be written in the Question paper only.<\/li>\n
            4. Marks will not be awarded in any case of over writing, rewriting or erased answers.<\/li>\n<\/ol>\n

              Note : Write the capital letters (A, B, C, D) showing the correct answer for the following questions in the brackets provided against them. (Marks: 20 \u00d7 1 = 20)<\/span><\/p>\n

              Question 1.
              \nIf both roots are common to the Quadratic equations
              \nx2<\/sup> – 4 = 0 and x2<\/sup> + px – 4 = 0, then p =
              \nA) 2
              \nB) 0
              \nC) 4
              \nD) 1
              \nAnswer:
              \nB) 0<\/p>\n

              Question 2.
              \nThe distance of the point p(x, y) from Y-axis is
              \nA) |x|
              \nB) |y|
              \nC) |x + y|
              \nD) |x – y|
              \nAnswer:
              \nA) |x|<\/p>\n

              Question 3.
              \nThe exponential form of loga<\/sub>\\(\\sqrt{x^4}\\) = y is
              \nA) ay<\/sup> = x4<\/sup>
              \nB) ya<\/sup> = 4
              \nC) ay<\/sup> = x2<\/sup>
              \nD) xy<\/sup> = a2<\/sup>
              \nAnswer:
              \nC) ay<\/sup> = x2<\/sup><\/p>\n

              Question 4.
              \nIf A \u2282 B, then A – B =
              \nA) A
              \nB) B
              \nC) B – A
              \nD) \u03a6
              \nAnswer:
              \nD) \u03a6<\/p>\n

              Question 5.
              \nIf (a, b), (b, c) and (c, a) art the vertices of a triangle and the centroid of triangle is origin, then a3<\/sup> + b3<\/sup> + c3<\/sup> =
              \nA) abc
              \nB) a + b + c
              \nC) 3abc
              \nD) 0
              \nAnswer:
              \nC) 3abc<\/p>\n

              \"TS<\/p>\n

              Question 6.
              \nIf \\(\\frac{5}{x-1}+\\frac{1}{y-2}\\) = 2, \\(\\frac{6}{x-1}+\\frac{-3}{y-2}\\) = 1 then x = …………..
              \nA) 4
              \nB) 7
              \nC) -1
              \nD) 3
              \nAnswer:
              \nA) 4<\/p>\n

              Question 7.
              \nThe last (unit place) digit of 62019<\/sup> in its standard form
              \nA) 6
              \nB) 4
              \nC) 9
              \nD) 19
              \nAnswer:
              \nA) 6<\/p>\n

              Question 8.
              \nIf \u03b1, \u03b2 are the zeroes of the polynomial x2<\/sup> + 5x + k and \u03b1 – \u03b2 = 3, then the value of k.
              \nA) 6
              \nB) 9
              \nC) 5
              \nD) 4
              \nAnswer:
              \nD) 4<\/p>\n

              Question 9.
              \nThe 25th<\/sup> term of -300, -290, -280, is ………
              \nA) -60
              \nB) -80
              \nC) 60
              \nD) 80
              \nAnswer:
              \nA) -60<\/p>\n

              Question 10.
              \nIf A {x : x2<\/sup> – 16 = 0, x \u2208 R} and
              \nB – {x : x2<\/sup> – 5x + 6 = 0, x\u2208 R}, then A\u222aB is a ………..
              \nA) Singleton set
              \nB) Infinite set
              \nC) Null set
              \nD) Finite set
              \nAnswer:
              \nD) Finite set<\/p>\n

              Question 11.
              \nWhat change will be observed in the angle of elevation as we move away from the object?
              \nA) increase
              \nB) decrease
              \nC) can’t be determined
              \nD) no change
              \nAnswer:
              \nA) increase<\/p>\n

              \"TS<\/p>\n

              Question 12.
              \nLet x1<\/sub>, x2<\/sub>, x3<\/sub>, x4<\/sub>,……. xn<\/sub> be the n observations and \\(\\overline{\\mathbf{x}}\\) be the mean of n observations, then \\(\\sum_{i=1}^n\\left(x_1-\\bar{x}\\right)\\) = ……..
              \nA) 0
              \nB) n\\(\\overline{\\mathbf{x}}\\)
              \nC) \\(\\frac{\\bar{x}}{n}\\)
              \nD) \\(\\frac{2 \\bar{x}}{n}\\)
              \nAnswer:
              \nA) 0<\/p>\n

              Question 13.
              \nA 20 m long ladder is placed on a pole of 10 m height Shaking ‘\u03b1’ angle with the ground, then \u03b1 =
              \nA) 60\u00b0
              \nB) 45\u00b0
              \nC) 30\u00b0
              \nD) 0\u00b0
              \nAnswer:
              \nC) 30\u00b0<\/p>\n

              Question 14.
              \nTangents PA and PB inclined at an angle 60\u00b0 as shown in the figure, the ratio of lengths of OA, OP and AP is
              \n\"TS
              \nA) 1 : 2 : 3
              \nB) 3 : 2 : 1
              \nC) \u221a3 : 2 : 1
              \nD) 1 : 2 : \u221a3
              \nAnswer:
              \nD) 1 : 2 : \u221a3 (Add Score)<\/p>\n

              Question 15.
              \nABC is a right angle triangle and \u2220C = 90\u00b0, Let BC = a, CA = b, AB = c and p be the length of the perpendicular from C on AB, then ………………
              \nA) \\(\\frac{1}{\\mathrm{p}^2}=\\frac{1}{\\mathrm{a}^2}-\\frac{1}{\\mathrm{~b}^2}\\)
              \nB) \\(\\frac{1}{\\mathrm{p}^2}=\\frac{1}{\\mathrm{~b}^2}-\\frac{1}{\\mathrm{a}^2}\\)
              \nC) \\(\\frac{1}{\\mathrm{p}^2}=\\frac{1}{\\mathrm{a}^2}+\\frac{1}{\\mathrm{~b}^2}\\)
              \nD) \\(\\frac{2}{p^2}=\\frac{1}{a^2}+\\frac{1}{b^2}\\)
              \nAnswer:
              \nC) \\(\\frac{1}{\\mathrm{p}^2}=\\frac{1}{\\mathrm{a}^2}+\\frac{1}{\\mathrm{~b}^2}\\)<\/p>\n

              Question 16.
              \nLet r, h had l be the radius, height and slant height of a cone respectively, then express l in terms of r and h is ……………
              \nA) \\(\\sqrt{h^2-r^2}\\)
              \nB) \\(\\sqrt{\\mathrm{r}^2+\\mathrm{h}^2}\\)
              \nC) \\(\\sqrt{\\mathrm{r}^2-\\mathrm{h}^2}\\)
              \nD) \\(\\sqrt{4 r^2+h^2}\\)
              \nAnswer:
              \nB) \\(\\sqrt{\\mathrm{r}^2+\\mathrm{h}^2}\\)<\/p>\n

              \"TS<\/p>\n

              Question 17.
              \nNumber of circles passing through 3 collinear points in a plane is
              \nA) 1
              \nB) 0
              \nC) 9
              \nD) 12
              \nAnswer:
              \nB) 0<\/p>\n

              Question 18.
              \nChoose the correct figure for which sin A = \\(\\frac{5}{13}\\)
              \n\"TS
              \nAnswer:
              \n\"TS<\/p>\n

              Question 19.
              \nA letter is choosen from the word “BAHUBALI\u201d, the probability that it was not a vowel is
              \nA) \\(\\frac{1}{2}\\)
              \nB) \\(\\frac{3}{2}\\)
              \nC) \\(\\frac{4}{3}\\)
              \nD) \\(\\frac{3}{4}\\)
              \nAnswer:
              \nA) \\(\\frac{1}{2}\\)<\/p>\n

              Question 20.
              \nWhich of the following statement is true ?
              \nA) All acute angle triangles are similar.
              \nB) All obtuse angle triangles are similar.
              \nC) All right angle triangles are similar.
              \nD) All isosceles right triangles are similar.
              \nAnswer:
              \nD) All isosceles right triangles are similar.<\/p>\n","protected":false},"excerpt":{"rendered":"

              The strategic use of TS 10th Class Maths Model Papers Set 6 can significantly enhance a student’s problem-solving skills. TS SSC Maths Model Paper Set 6 with Solutions Time: 3 Hours Maximum Marks: 80 General Instructions: Answer all the questions under Part – A on a separate answer book. Write the answers to the questions … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[25],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/60196"}],"collection":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/comments?post=60196"}],"version-history":[{"count":7,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/60196\/revisions"}],"predecessor-version":[{"id":60352,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/60196\/revisions\/60352"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media?parent=60196"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/categories?post=60196"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/tags?post=60196"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}