{"id":50419,"date":"2024-02-01T14:47:19","date_gmt":"2024-02-01T09:17:19","guid":{"rendered":"https:\/\/apboardsolutions.com\/?p=50419"},"modified":"2024-02-05T15:06:19","modified_gmt":"2024-02-05T09:36:19","slug":"ap-10th-class-maths-model-paper-set-2","status":"publish","type":"post","link":"https:\/\/apboardsolutions.com\/ap-10th-class-maths-model-paper-set-2\/","title":{"rendered":"AP 10th Class Maths Model Paper Set 2 with Solutions"},"content":{"rendered":"

Regularly solving\u00a0AP 10th Class Maths Model Papers<\/a> Set 2 contributes to the development of problem-solving skills.<\/p>\n

AP SSC Maths Model Paper Set 2 with Solutions<\/h2>\n

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

    \n
  1. In the duration of 3hours 15 minutes, 15 minutes of time is allotted to read the question paper.<\/li>\n
  2. All answers shall be written in the answer booklet only.<\/li>\n
  3. Question paper consists of 4 Sections and 33 questions.<\/li>\n
  4. Internal choice is available in section – IV only.<\/li>\n
  5. Answers shall be written neatly and legibly.<\/li>\n<\/ol>\n

    Section – I <\/span>
    \n(12 \u00d7 1 = 12M)<\/span><\/p>\n

    Note:<\/p>\n

      \n
    1. Answer all the questions in one word or phrase.<\/li>\n
    2. Each question carries 1 mark.<\/li>\n<\/ol>\n

      Question 1.
      \nIf A = {1, 2, 3, 4, 5} B = {4, 5, 6, 7}, then A – B is ?
      \nA) {1, 2, 3}
      \nB) {4, 5}
      \nC) {6, 7}
      \nD) {1, 2, 3, 4, 5}
      \nSolution:
      \nA) {1, 2, 3}<\/p>\n

      Explanation:
      \nA – B = {1, 2, 3, 4, 5} – {4, 5, 6, 7) = {1, 2, 3}<\/p>\n

      Question 2.
      \nThe number of parallel tangents drawn at the end points of a diameter is ………..
      \nSolution:
      \n2<\/p>\n

      \"AP<\/p>\n

      Question 3.
      \nWhen the line of sight is below the horizontal line, then the angle between the line of sight and horizontal line is called ……..
      \nSolution:
      \nAngle of depression<\/p>\n

      Question 4.
      \nFind the mode of the data : 5, 6, 9, 10, 6, 12, 3, 6, 11, 10, 4, 6, 7.
      \nSolution:
      \n6<\/p>\n

      Question 5.
      \nStatement \u2018p\u2019 : 2, 3, 5, 7, 8, 10, 15 …… are in G.P.
      \nStatement \u2018q\u2019 : -1, -3, -5, -7, ……. are in A.P.
      \nA) p\u2019 and \u2018q\u2019 both are true
      \nB) \u2018p\u2019 and \u2018q\u2019 both are false
      \nC) \u2018p\u2019 is true \u2018q\u2019 is false
      \nD) \u2018p\u2019 is false q\u2019 is true
      \nSolution:
      \nD) \u2018p\u2019 is false q\u2019 is true<\/p>\n

      Question 6.
      \nMatch the following:
      \n\"AP
      \nChoose the correct answer.
      \nA) a-i, b-ii, c-iii
      \nB) a – ii, b – iii, c – i
      \nC) a – iii, b – i, c – ii
      \nD) a – iii, b – ii, c – i
      \nSolution:
      \nC) a – iii, b – i, c – ii<\/p>\n

      Question 7.
      \nWhat is the last digit of 62023<\/sup> ?
      \nSolution:
      \n6<\/p>\n

      Question 8.
      \nWrite the general form of a linear equation in two variables.
      \nSolution:
      \nax + by + c = 0 (a2<\/sup> + b2<\/sup>, \u2260 0)<\/p>\n

      Question 9.
      \nIf \u03b8 is the angle made by the line with X-axis in positive direction, the slope m = ………
      \nA) Sin \u03b8
      \nB) Cos \u03b8
      \nC) Tan \u03b8
      \nD) Sec \u03b8
      \nSolution:
      \nC) Tan \u03b8<\/p>\n

      Question 10.
      \nPythagorean triplets are the sides of a …….
      \nA) acute angled triangle
      \nB) right angled triangle
      \nC) obtuse angled triangle
      \nD) All of these
      \nAnswer:
      \nB) right angled triangle<\/p>\n

      \"AP<\/p>\n

      Question 11.
      \nWrite the equation with roots \\(\\frac{1}{\\alpha}\\) and \\(\\frac{1}{\\beta}\\)?
      \nSolution:
      \ncx2<\/sup> + bx + a = 0<\/p>\n

      Question 12.
      \nThe graph of y = p(x) is given in the adjacent figure, then the number of zeroes of p (x) is…….
      \n\"AP
      \nA) 0
      \nB) 1
      \nC) 2
      \nD) 3
      \nSolution:
      \nD) 3<\/p>\n

      SECTION – II<\/span>
      \n(8 \u00d7 2 = 16 M)<\/span><\/p>\n

      Note:<\/p>\n

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

        Question 13.
        \nFind the value of Log2<\/sub>32.
        \nSolution:
        \nLet log2<\/sub> 32 be x \u21d2 log2<\/sub> 32 = x
        \n2x<\/sup> = 32
        \n2x<\/sup> = 25<\/sup>
        \n\u2234 x = 5
        \n\u2234 log2<\/sub>32 = 5<\/p>\n

        Question 14.
        \nFind the midpoint of the line segment joining the points (2, 7) and (12, – 7).
        \nSolution:
        \nThe midpoint of the line segment joining the points (2, 7) and (12, -7) is
        \n= \\(\\left(\\frac{x_1+x_2}{2}, \\frac{y_1+y_2}{2}\\right)\\) = \\(\\left(\\frac{2+12}{2}, \\frac{7+(-7)}{2}\\right)\\)
        \n= \\(\\left(\\frac{16}{2}, \\frac{0}{2}\\right)\\) = (8, 0)<\/p>\n

        Question 15.
        \nA ladder 5 m. long reaches a window of building 4 m. above the ground. Determine the distance of the foot of the ladder from the building.
        \nSolution:
        \nLength of a ladder = AC = 5 m
        \nHeight of a window = AB = 4 m
        \nLet the distance of the foot of the ladder from the building = BC = x m
        \nFrom \u2206ABC, \u2220B = 90\u00b0
        \nBy Pythagores theorem;
        \n52<\/sup> = 42<\/sup> + x2<\/sup>
        \n25 = 16 + x2<\/sup>
        \n\"AP
        \nx2<\/sup> = 25 – 16
        \nx2<\/sup> = 9
        \nx = \\(\\sqrt{9}\\) = 3 m
        \nThe distance of the foot of the ladder from the building = 3 m.<\/p>\n

        Question 16.
        \nEvaluate
        \n\"AP
        \nSolution:
        \ncosec 55\u00b0 cosec (90\u00b0 – 35\u00b0) = sec 35\u00b0
        \n[\u2235 cosec (90\u00b0 – \u03b8) = sec \u03b8]
        \n\u2234
        \n\"AP
        \n= \\(\\frac{\\sec 35^{\\circ}}{\\sec 35^{\\circ}}\\) = 1<\/p>\n

        Question 17.
        \nCan \\(\\frac{3}{2}\\) be the probability of an event? Explain.
        \nSolution:
        \nNo, \\(\\frac{3}{2}\\) cant be the probability of an event.
        \nThe probability of an event is always lies between 0 and 1
        \ni.e 0 \u2264 P(E) \u2264 1
        \nBut \\(\\frac{3}{2}\\) = 1.5, it is not lies between 0 and 1<\/p>\n

        Question 18.
        \nIn G. P. an<\/sub> = arn-1<\/sup>. Explain the terms ‘a’ and ‘r’.
        \nSolution:
        \nIn G.P. an = a.rn-1<\/sup>
        \na = first term of G.P
        \nr = common ratio
        \nn = number of terms of G.P<\/p>\n

        Question 19.
        \nFind the value of ‘k’ for which the pair of equations 2x – ky + 3 = 0, 4x + 6y – 5 = 0 represent parallel lines.
        \nSolution:
        \nGiven pair of equations
        \n2x – ky + 3 = 0 and 4x + 6y – 5 = 0
        \na1<\/sub> = 2 ; b1<\/sub> = -k ; c1<\/sub> = 3;
        \na2<\/sub> = 4 ; b2<\/sub> = 6 ; c2<\/sub> = -5
        \nGiven the pair of lines are parallel
        \n\u2234 \\(\\frac{a_1}{a_2}\\) = \\(\\frac{b_1}{b_2}\\) \u2260 \\(\\frac{c_1}{c_2}\\) ; \\(\\frac{a_1}{a_2}\\) = \\(\\frac{b_1}{b_2}\\) \u21d2 \\(\\frac{2}{4}\\) \\(\\frac{-k}{6}\\)
        \n\u21d2 -4k = 2 \u00d7 6 \u21d2 -4k = 12 \u21d2 k = \\(\\frac{12}{-4}\\) = -3<\/p>\n

        \"AP<\/p>\n

        Question 20.
        \nFrom the given figure, find A U B and A \u2229 B
        \n\"AP
        \nSolution:
        \nFrom venn diagram
        \nA \u222a B = {5, 6, 7, 8, 9, 10}; A \u2229 B = {7, 8}<\/p>\n

        Section – III<\/span>
        \n(8 \u00d7 4 = 32 M)<\/span><\/p>\n

        Note:<\/p>\n

          \n
        1. Answer all the questions.<\/li>\n
        2. Each question carries 4 marks.<\/li>\n<\/ol>\n

          Question 21.
          \nFind a quadratic polynomial whose zeroes are 2 and –\\(\\frac{1}{3}\\).
          \nSolution:
          \nLet the quadratic polynomial be ax2<\/sup> + bx + c, a \u2260 0 and its be \u03b1 and \u03b2. Here \u03b1 = 2, \u03b2 = \\(\\frac{-1}{3}\\)
          \nSum of the zeroes = (\u03b1 + \u03b2)
          \n= 2 + \\(\\left(\\frac{-1}{3}\\right)\\) = \\(\\frac{5}{3}\\)
          \nProduct of the zeroes = (\u03b1\u03b2)
          \n= \\(2\\left(\\frac{-1}{3}\\right)\\) = \\(\\frac{-2}{3}\\)
          \nTherefore the quadratic polynomial
          \nax2<\/sup> + bx + c is k[x2<\/sup> – (\u03b1 + \u03b2)x + \u03b1\u03b2], where k is a constant = k[x2<\/sup> – \\(\\frac{5}{3}\\)x – \\(\\frac{2}{3}\\)]
          \nWe can put different values of k.
          \nWhen k = 3, the quadratic polynomial will be 3x2<\/sup> – 5x – 2.<\/p>\n

          Question 22.
          \nWhich term of the A, P.: 3, 8, 13, 18, ……. is 78?
          \nSolution:
          \nHere a = 3; d = 8 – 3 = 5 and if an<\/sub> = 78 we have to find n.
          \nWe have an<\/sub> = a + (n – 1)d
          \n78 = 3 + (n – 1)5
          \n78 = 3 + 5n – 5
          \n78 = 5n – 2
          \n5n = 78 + 2
          \n5n = 80
          \nn = \\(\\frac{80}{5}\\) = 16
          \nTherefore the 16th<\/sup> term of the given AP is 78.<\/p>\n

          Question 23.
          \nA bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the proba-bility that she takes out
          \n(i) an orange flavoured candy?
          \n(ii) a lemon flavoured candy?
          \nSolution:
          \nThe bag contains only lemon flavoured candies.<\/p>\n

          i) So, the event that Malini takes an orange flavoured candy is an Impossible event. So, the corresponding probability is zero.
          \nii) The event that Malini takes a lemon flavoured candy is certain event. So, the corresponding probability is 1.<\/p>\n

          Question 24.
          \nState the reasons for the following :
          \ni) {1, 2, 3, ……, 10} \u2260 {x : x \u2208 N and 1< x < 10} ii) {2, 4 > 6, 8, 10} \u2260 {x : x = 2n + 1 and n \u2208 N]
          \niii) {5, 15, 30, 45} \u2260 {x : x is a multiple of 15}
          \niv) {2, 3, 5, 7, 9} \u2260 {x : x is a prime number}
          \nSolution:
          \ni) {1, 2, 3, ……., 10) \u2260 {2, 3, 4, 5, 6, 7, 8, 9}
          \nIn LHS set contains 1 and 10 are elements but in RHS they are not the elements of that set.<\/p>\n

          ii) {2, 4, 6, 8, 10} \u2260 {x : x = 2n + 1 and x \u2208 N)
          \n{2, 4, 6, 8, 10} \u2260 {3, 5, 7, 9 ……….}
          \nLHS is a set of even numbers less than 10. It is a finite set.
          \nBut RHS is a set of odd numbers. It is an infinite set.<\/p>\n

          iii) {15, 15, 30, 45) \u2260 {x : x is a multiple of 15}
          \n5 is not a multiple of 15.<\/p>\n

          iv) {2, 3, 5, 7,9) \u2260 {x : x is a prime number}
          \n9 is not a prime number.<\/p>\n

          Question 25.
          \nProve that
          \n\"AP
          \nSolution:
          \n\"AP<\/p>\n

          Question 26.
          \nWrite the formula to find the median of the grouped data and explain its terms.
          \nSolution:
          \nMedian = l + \\(\\frac{\\frac{n}{2}-c f}{f}\\) \u00d7 h
          \nWhere l = lower limit of the median class.
          \nh = length of the median class.
          \nf = frequency of the median class.
          \ncf = cumulative frequency of the class preceeding the median class.<\/p>\n

          Question 27.
          \nCalculate the length of a tangent from a point 15 cm away from the centre of a circle of radius 9 cm.
          \nSolution:
          \nIt is given that radius (r) = 9 cm
          \nOQ(d) = 15 cm
          \n\"AP<\/p>\n

          Question 28.
          \nFind the roots of
          \n(i) \\(\\frac{1}{x+4}\\) – \\(\\frac{1}{x-7}\\) = \\(\\frac{11}{30}\\); x \u2260 -4, 7
          \nSolution:
          \n\"AP
          \nwhich is a quadratic equation.
          \nHere a = 1 ; b = -3; c = 2;
          \nSo, b2<\/sup> – 4ac = (-3)2<\/sup> – 4(1)(2)
          \n= 9 – 8 = 1 > 0
          \nTherefore x = \\(\\frac{-(-3) \\pm \\sqrt{1}}{2 \\times 1}\\) = \\(\\frac{3 \\pm 1}{2}\\)
          \nx = \\(\\frac{3+1}{2}\\) = \\(\\frac{4}{2}\\) = 2
          \nx = \\(\\frac{3-1}{2}\\) = \\(\\frac{2}{2}\\) = 1
          \n\u2234 The roots are 1, 2<\/p>\n

          Section – IV<\/span>
          \n(5 \u00d7 8 = 40 M)<\/span><\/p>\n

          Note:<\/p>\n

            \n
          1. Answer all the questions,<\/li>\n
          2. Each question carries 8 marks,<\/li>\n
          3. Each question has internal choice.<\/li>\n<\/ol>\n

            Question 29.
            \na) If (2.3)x<\/sup> = (0.23)y<\/sup> = 1000, then find the value of \\(\\frac{1}{x}\\) – \\(\\frac{1}{y}\\)<\/p>\n

            (OR)<\/p>\n

            b) If A = {X : X is a natural number}
            \nB = {X : X is an even natural number}
            \nC = {X : X is a odd natural number) and
            \nD = {X : X is a prime number}<\/p>\n

            Find:
            \ni) A \u222a B
            \nii) A – C
            \niii) A – D
            \niv) B \u222a C
            \nSolution:
            \na) (2.3)x<\/sup> = (0.23)y<\/sup> = 1000
            \n(2.3)x<\/sup> = 1000;
            \n(2.3)x<\/sup> = 103<\/sup>
            \n2.3 = \\(10^{\\frac{3}{x}}\\);
            \n2.3 = \\(10^{\\frac{3}{x}}\\);
            \n2.3 = \\(10^{\\frac{3}{x}}\\);<\/p>\n

            (0.23)y<\/sup> = 1000
            \n(0.23)y<\/sup> = 103<\/sup>
            \n0.23 = \\(10^{\\frac{3}{y}}\\)
            \n\\(\\frac{2.3}{10}\\) = \\(10^{\\frac{3}{y}}\\)
            \n2.3 = \\(10^{\\frac{3}{y}}\\) \u00d7 101<\/sup>
            \n= \\(10^{\\frac{3}{y}+1}\\)<\/p>\n

            \u2234 \\(10^{\\frac{3}{x}}\\) = \\(10^{\\frac{3}{\\mathrm{y}}+1}\\)
            \n\\(\\frac{3}{x}\\) = \\(\\frac{3}{y}\\) + 1
            \n3(\\(\\frac{3}{x}\\)) = 3(\\(\\frac{1}{y}\\) + \\(\\frac{1}{3}\\))
            \n\\(\\frac{1}{x}\\) = \\(\\frac{1}{y}\\) + \\(\\frac{1}{3}\\)
            \n\u2234 \\(\\frac{1}{x}\\) – \\(\\frac{1}{y}\\) = \\(\\frac{1}{3}\\)<\/p>\n

            b)
            \nA {x : x is a natural number}
            \n= {1, 2, 3, 4, 5, 6, 7, 8……..}
            \nB = {x : x is an even natural number}
            \n= {2, 4, 6, 8…….}
            \nC = {x : x is an odd natural number)
            \n= {1, 3, 5, 7,…….. }
            \nD = {x : x is a prime number)
            \n= {2, 3, 5, 7,……..}<\/p>\n

            i) A \u2229 B
            \n{1, 2, 3, 4, 5, 6, 7, 8……} \u2229 {2, 4, 6, 8…….}
            \n= {2, 4, 6, 8……. }
            \n= {even natural numbers)<\/p>\n

            ii) A – C
            \nA – C = {1, 2, 3, 4, 5, 6, 7, 8……} – {1, 3, 5, 7…….}
            \n= {2, 4, 6, 8………..}
            \n= (even natural numbers)<\/p>\n

            iii) A – D
            \n= {1, 2, 3, 4, 5, 6, 7, 8…….} – {2, 3, 5, 7……..}
            \n= {1, 4, 6, 8……}<\/p>\n

            iv) B \u222a C
            \nB \u222a C = {2, 4, 6, 8……} \u222a {1, 3, 5, 7……}
            \n= {1, 2, 3, 4, 5, 6, 7, 8…..}
            \n= {natural numbers}<\/p>\n

            \"AP<\/p>\n

            Question 30.
            \na) Solve the equations by reducing them to a pair of linear equations. \\(\\frac{1}{3 x+y}\\) + \\(\\frac{1}{3 x-y}\\) = \\(\\frac{3}{4}\\)
            \n\\(\\frac{1}{2(3 x+y)}\\) + \\(\\frac{1}{2(3 x-y)}\\) = \\(\\frac{-1}{8}\\)<\/p>\n

            (OR)<\/p>\n

            b) One card is drawn from a well shuffled deck of 52 cards. Find the probability of getting:
            \ni) a king of red colour
            \nii) a queen of diamonds
            \niii) a jack of spade
            \niv) a black face card
            \nSolution:
            \na) \\(\\frac{1}{3 x+y}\\) + \\(\\frac{1}{3 x-y}\\) = \\(\\frac{3}{4}\\) …….. (1)
            \n\\(\\frac{1}{2}\\left(\\frac{1}{3 x+y}\\right)\\) – \\(\\frac{1}{2}\\left(\\frac{1}{3 x-y}\\right)\\) = \\(\\frac{-1}{8}\\) …… (2)
            \nIf we substitute \\(\\frac{1}{3 x+y}\\) = p and \\(\\frac{1}{3 x-y}\\) = p we get a pair of linear equations:
            \np + q = \\(\\frac{3}{4}\\) \u21d2 4p + 4q = 3 …….. (3)
            \n\\(\\frac{1}{2}\\)p – \\(\\frac{1}{2}\\)q = \\(\\frac{-1}{8}\\) \u21d2 4p – 4q = -1 …….. (4)
            \nElimination Method:
            \n\"AP
            \nSubstitute the value of x in (5)
            \n3(1) + y = 4
            \n3 + y = 4
            \n\u2234 y = 4 – 3 = 1
            \n\u2234 x = 1 ; y = 1<\/p>\n

            b) Total number of cards in a deck = 52
            \nTotal possible outcomes =52<\/p>\n

            i) A king of red colour
            \nLet E be the event of getting a king of red colour
            \nNumber of favourable outcomes to event
            \n(E) = 2
            \nProbability of an event P(E)
            \n\\(=\\frac{\\text { No. of favourable outcomes }}{\\text { Total possible outcomes }}\\)
            \n= \\(\\frac{2}{52}\\) = \\(\\frac{1}{26}\\)<\/p>\n

            ii) A queen of diamonds.
            \nLet \u2018F\u2019 be the event of getting a queen of diamonds
            \nNumber of favourable outcomes to event F = 1
            \nProbability of an event p(F) = \\(\\frac{1}{52}\\)<\/p>\n

            iii) A jack of spade
            \nLet \u2018G\u2019 be the event of getting a jack of spade.
            \nNumber of favourable outcomes to event G = 1
            \nProbability of an event P(G) = \\(\\frac{1}{52}\\)<\/p>\n

            iv) A black face card.
            \nLet \u2018H be the event of getting a black face card.
            \nNumber of favourable outcomes to getting event
            \nH = 6
            \nProbability of an event G = P(H) = \\(\\frac{6}{52}\\)
            \n= \\(\\frac{3}{26}\\)<\/p>\n

            Question 31.
            \na) Show that the points (1, 7), (4, 2), (-1, -1) and (-4, 4) are the vertices of a square.<\/p>\n

            (OR)<\/p>\n

            b) An iron pillar consists of a cylindrical portion of 2.8 m. height and 20 cm. in diameter and a cone of 42 cm. height surmounting it. Find the weight of the pillar if 1 cm3<\/sup> of iron weighs 7.5 g.
            \nSolution:
            \na) Let A (1, 7), B(4, 2), C(-1, -1) and D (-4, 4) be the given points.
            \nOne way of showing that ABCD is a square is to use the property that all its sides should be equal and both its digonals should also be equal. Now
            \n\"AP
            \n= \\(\\sqrt{4+64}\\) = \\(\\sqrt{68}\\) units
            \nBD = \\(\\sqrt{(4+4)^2+(2-4)^2}\\) = \\(\\sqrt{64+4}\\) = \\(\\sqrt{68}\\) units<\/p>\n

            Since AB = BC = CD = DA and AC = BD. So all the four sides of the quadrilateral ABCD are equal and its diagonals AC and BD are also equal.
            \nTherefore, ABCD is square.<\/p>\n

            b) Height of the cylinder portion = 2.8 m = 280 cm
            \nDiameter of the cylinder = 20 cm
            \nRadius of the cylinder = \\(\\frac{20}{2}\\) cm = 10 cm
            \nVolume of the cylinder = \u03c0r2<\/sup>h
            \n= \\(\\frac{22}{7}\\) \u00d7 10 \u00d7 10 \u00d7 280 cm3<\/sup>
            \n= 88000 cm3<\/sup>
            \nHeight of the cone (h) = 42 cm
            \nRadius of the cone (r) = 10 cm
            \nVolume of the cone (v) = \\(\\frac{1}{3}\\)\u03c0r2<\/sup>h
            \n= \\(\\frac{1}{3}\\) \u00d7 \\(\\frac{22}{7}\\) \u00d7 10 \u00d7 10 \u00d7 42 cm3<\/sup> = 4400 cm3<\/sup>
            \nVolume of the pillar
            \n= 88000 cm3<\/sup> + 4400 cm3<\/sup>
            \n= 92400 cm3<\/sup>
            \nWeight of 1 cm3<\/sup> of iron = 7.5 g
            \nWeight of the pillar = 7.5 \u00d7 92400 g
            \n= 693000 g
            \n= 693 kg<\/p>\n

            Question 32.
            \na) A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground by making 300 angle with the ground. The distance between the foot of the tree and the top of the tree on the ground is 6 m. Find the height of the tree before falling down.<\/p>\n

            (OR)<\/p>\n

            b) The following data gives the information on the observed life time (in hours) of 225 electrical components.<\/p>\n\n\n\n\n
            Life times
            \n(in hrs)<\/td>\n            		0-20<\/td>\n20-40<\/td>\n40-60<\/td>\n60-80<\/td>\n80-100<\/td>\n100-120<\/td>\n<\/tr>\n
            Frequency<\/td>\n10<\/td>\n35<\/td>\n52<\/td>\n61<\/td>\n38<\/td>\n29<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

            Determine the modal life time of the components.
            \nSolution:
            \na) Let Initial height of tree be AD when storm come, the tree which is broken at B and top touches the
            \nground at C.
            \n\u2234 height of tree = AB + BC
            \nAC = 6 m
            \n\"AP
            \nIn \u2206ABC, \u2220A = 90\u00b0, \u2220C = 30\u00b0
            \ntan C = tan 30\u00b0 = \\(\\frac{1}{\\sqrt{3}}\\)
            \n\"AP<\/p>\n

            b) 61 is the highest frequency corresponding to the class interval 60 – 80.
            \nSo, modal class : 60-80
            \nl = lower limit of the modal class = 60
            \nf1<\/sub> = frequency of the modal class = 61
            \nh = modal class length = 20
            \nf0<\/sub> = frequency of the preceeding class to 60 – 80 = 52
            \nf2<\/sub> = frequency of the suceeding class to 60 – 80 = 38
            \nf1<\/sub> – f0<\/sub> = 61 – 52 = 9
            \n2f1<\/sub> – f0<\/sub> – f2<\/sub> = 2 \u00d7 61 – 52 – 38 = 122 – 90 = 32
            \nSo,
            \n\"AP
            \nTherefore more electrical components have life time of 65.625 hours.<\/p>\n

            \"AP<\/p>\n

            Question 33.
            \na) Draw the graph of the polynomial p(x) = x2<\/sup> – x – 12 and find the zeroes.
            \n(OR)
            \nb) Construct a triangle of sides 4 cm, 5 cm and 6 cm. Then, construct a triangle similar to it, whose sides are \\(\\frac{2}{3}\\) of the corresponding sides of the first triangle.
            \nSolution:
            \na) p(x) = x2<\/sup> – x – 12; y = x2<\/sup> – x – 12
            \n\"AP<\/p>\n

            \"AP
            \n-3 and 4 are zeroes of the quadratic polynomial because (-3, 0) and (4, 0) are intersection points of X axis.<\/p>\n

            Justification:
            \nx2<\/sup> – x – 120 \u21d2 x2<\/sup> – 4x + 3x – 12.
            \n0 \u21d2 (x – 4) (x + 3) = 0
            \nx – 4 = 0 and x + 3 = 0
            \nx = 4, x = -3 zeroes of p(x) = -3, 4.<\/p>\n

            b)
            \n\"AP<\/p>\n

            Steps of construction:<\/p>\n

              \n
            1. Construct a triangle ABC with sides 4 cm, 5 cm and 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 from B2<\/sub> to C which is parallel to B3<\/sub>C and it is intersecting BC at C.<\/li>\n
            5. Draw a line through C parallel to AC to intersect AB at A’
              \nSo, \u2206A’BC’ is the required triangle.<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"

              Regularly solving\u00a0AP 10th Class Maths Model Papers Set 2 contributes to the development of problem-solving skills. AP SSC Maths Model Paper Set 2 with Solutions Instructions : In the duration of 3hours 15 minutes, 15 minutes of time is allotted to read the question paper. All answers shall be written in the answer booklet only. … Read more<\/a><\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/50419"}],"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\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/comments?post=50419"}],"version-history":[{"count":14,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/50419\/revisions"}],"predecessor-version":[{"id":50669,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/50419\/revisions\/50669"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media?parent=50419"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/categories?post=50419"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/tags?post=50419"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}