{"id":60298,"date":"2024-04-17T12:10:58","date_gmt":"2024-04-17T06:40:58","guid":{"rendered":"https:\/\/apboardsolutions.com\/?p=60298"},"modified":"2024-04-17T12:18:22","modified_gmt":"2024-04-17T06:48:22","slug":"ts-10th-class-maths-model-paper-set-8","status":"publish","type":"post","link":"https:\/\/apboardsolutions.com\/ts-10th-class-maths-model-paper-set-8\/","title":{"rendered":"TS 10th Class Maths Model Paper Set 8 with Solutions"},"content":{"rendered":"

The strategic use of TS 10th Class Maths Model Papers<\/a> Set 8 can significantly enhance a student’s problem-solving skills.<\/p>\n

TS SSC Maths Model Paper Set 8 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 – A (60 Marks)<\/span>
    \nSection – 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.
      \nFind the value of k, if 2 is one of the roots of the quadratic equation x2<\/sup> – kx + 6 = 0
      \nAnswer:
      \nx2<\/sup> – kx + 6 = 0
      \n(2)2<\/sup> – k(2) + 6 = 0
      \n4 + 6 – 2k = 0
      \n2k = 10 \u21d2 k = 5<\/p>\n

      Question 2.
      \nIn a rectangle ABCD,
      \nAB = x + y, BC = x – y,
      \nCD = 9 and AD = 3.
      \n\"TS
      \nFind the values of x and y.
      \nAnswer:
      \nx + y = 9
      \n\"TS
      \n{\u2235 two pairs of opposite sides are equal, in a rectangle.}
      \nx – y = 3.
      \nSolving the above equations, we get x = 6 and y = 3.<\/p>\n

      Question 3.
      \nIf the slope of the line passing through the two points (2, 5) and (5, 8) is represented by tan \u03b8; (where 0\u00b0 < \u03b8 < 90\u00b0) in trigonometry, then find angle ‘\u03b8’.
      \nAnswer:
      \nPoints on the given line are (2, 5) and (5, 8)
      \nSlope = Tan \u03b8 = \\(\\frac{y_2-y_1}{x_2-x_1}\\)
      \nTan \u03b8 = \\(\\frac{8-5}{5-2}\\) = \\(\\frac{1}{2}\\) = 1 \u21d2 Tan 45\u00b0
      \n\u2234 \u03b8 = 45\u00b0 ( \u2235 0\u00b0 < \u03b8 < 90\u00b0)<\/p>\n

      \"TS<\/p>\n

      Question 4.
      \nIt is given that \u0394ABC ~ \u0394DEF. Is it true to say that = \\(\\frac{\\mathrm{BC}}{\\mathrm{DE}}=\\frac{\\mathrm{AB}}{\\mathrm{EF}}\\) Justify your answer.
      \nAnswer:
      \nGiven that \u0394ABC ~ \u0394DEF
      \n\u2234 \\(\\frac{\\mathrm{AB}}{\\mathrm{DE}}\\) = \\(\\frac{\\mathrm{BC}}{\\mathrm{EF}}\\) = \\(\\frac{\\mathrm{AC}}{\\mathrm{DF}}\\)
      \n[\u2235 Ratio of corresponding sides of similar triangles are equal]
      \nBut \\(\\frac{\\mathrm{BC}}{\\mathrm{DE}}\\) = \u2018\\(\\frac{\\mathrm{AB}}{\\mathrm{EF}}\\) (given)
      \n\u2234 Given statement is wrong.<\/p>\n

      Question 5.
      \nEvaluate : \\(\\frac{{Sin} 58^{\\circ}}{{Cos} 32^{\\circ}}+\\frac{{Tan} 42^{\\circ}}{{Cot} 48^{\\circ}}\\).
      \nAnswer:
      \n\\(\\frac{{Sin} 58^{\\circ}}{{Cos} 32^{\\circ}}+\\frac{{Tan} 42^{\\circ}}{{Cot} 48^{\\circ}}\\) = \\(\\frac{{Sin} 58^{\\circ}}{{Cos}(90-58)^{\\circ}}+\\frac{{Tan} 42^{\\circ}}{{Cot}(90-42)^{\\circ}}\\)
      \n= \\(\\frac{{Sin} 58^{\\circ}}{{Sin} 58^{\\circ}}+\\frac{{Tan} 42^{\\circ}}{{Tan} 42^{\\circ}}\\) = 1 + 1 = 2<\/p>\n

      Question 6.
      \nWrite the formula to find the mean of a grouped data, using assumed mean method and explain each term.
      \nAnswer:
      \nMean = a + \\(\\frac{\\Sigma \\mathrm{f}_{\\mathrm{i}} \\mathrm{d}_{\\mathrm{i}}}{\\sum \\mathrm{f}_{\\mathrm{i}}}\\)
      \na – assumed mean
      \nf – frequency
      \nd – x – a
      \nx – class mark<\/p>\n

      Section – II (6 \u00d7 4 = 24 Marks)<\/span><\/p>\n

      Note :
      \n(i) Answer ALL the following questions.
      \n(ii) Each question carries 4 marks.<\/p>\n

      Question 7.
      \nShow that 2 and – \\(\\frac{1}{3}\\) are zeroes 04 the polynomial 3x2<\/sup> – 5x – 2.
      \nAnswer:
      \np(x) = 3×2 – 5x – 2
      \np(2) = 3(2)2 – 5(2) – 2
      \n= 12 – 10 – 2
      \n= 12 – 12 = 0
      \n= p(-\\(\\frac{1}{3}\\)) = 3(-\\(\\frac{1}{3}\\))2<\/sup> – 5(-\\(\\frac{1}{3}\\)) – 2
      \n= \\(\\frac{1}{3}\\) + \\(\\frac{5}{3}\\) – 2
      \n= 2 – 2 = 0
      \n\u2234 2 and –\\(\\frac{1}{3}\\) are zeroes of p(x)<\/p>\n

      \"TS<\/p>\n

      Question 8.
      \nIf the measure of angles of a triangle are x\u00b0, y\u00b0 and 40\u00b0, and difference between the two measures of angles x\u00b0 and y\u00b0 is 30\u00b0, then find the values of x\u00b0 and y\u00b0.
      \nAnswer:
      \nx + y + 40 = 180\u00b0
      \n(\u2235 sum of angles of a triangle is 180\u00b0)
      \nx + y = 140 ……………. (1)
      \nx – y = 30 …………… (2)
      \nSolving the above equations we obtain x = 85\u00b0, y = 55\u00b0<\/p>\n

      Question 9.
      \nExpress 2016 as product of prime factors.
      \nAnswer:
      \n\"TS
      \n2016 = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 7 = 25<\/sup> \u00d7 32<\/sup> \u00d7 7<\/p>\n

      Question 10.
      \nThere are 5 red balls, 4 green balls and 6 yellow balls in a box. If a ball is selected at random, what is the probability of not getting a yellow ball ?
      \nAnswer:
      \nTotal no. of balls in a bag = 15
      \nTotal no. of chances to select a ball from the box = 15
      \nFavourable outcomes to select not yellow ball = 9
      \nProbability of not getting a yellow ball
      \n= \\(\\frac{\\text { No. of favourable outcomes }}{\\text { Total no. of outcomes }}\\) = \\(\\frac{9}{15}\\) = \\(\\frac{3}{5}\\)<\/p>\n

      Question 11.
      \nA toy is in the form of a cone mounted on a hemisphere. The radius of the base and the height of the cone are 7 cm and 8 cm respectively. Find the surface area of the toy. (\u03c0 = \\(\\frac{22}{7}\\))
      \nAnswer:
      \n\"TS
      \nAccording to the data Radius of hemisphere = Radius of the base of cone = r
      \nr = 7 cm.
      \nHeight of cone = h = 8 cm
      \nSlant height of cone = l = \\(\\sqrt{\\mathrm{r}^2+\\mathrm{h}^2}\\) = \\(\\sqrt{7^2+8^2}\\) = \\(\\sqrt{49+64}\\) = \\(\\sqrt{113}\\)
      \nSurface area of toy = Curved surface area of cone. + Surface area of hemisphere
      \n= \u03c0 rl + 2\u03c0 r2<\/sup>
      \n= \\(\\frac{22}{7}\\) \u00d7 7 \u00d7 \\(\\sqrt{113}\\) + 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 72<\/sup>
      \n= 22\\(\\sqrt{113}\\) + 308 sq. cm.<\/p>\n

      Question 12.
      \nDraw a circle with 5 cm radius and construct a pair of tangents to the circle.
      \nAnswer:
      \n\"TS<\/p>\n

      \"TS<\/p>\n

      Section – III (4 \u00d7 6 = 24 Marks)<\/span><\/p>\n

      Note :<\/p>\n

        \n
      1. Answer any 4 of the following questions.<\/li>\n
      2. Each question carries 6 marks.<\/li>\n<\/ol>\n

        Question 13.
        \nA tree is broken without separating from the stem by the wind. The top touches the ground making an angle 30\u00b0 at a distance of 12 m from the foot of the tree. Find the height of the tree before breaking.
        \nAnswer:
        \nLet height of the tree before broken = (x + y)m
        \nAccording to the data Foot of the tree = A
        \nTree broken at B
        \n\"TS
        \nTop of the tree touches after broken at C
        \nGiven AC = 12 m. \u2220ACB – 30\u00b0
        \ntan 30\u00b0 = \\(\\frac{y}{12}=\\frac{1}{\\sqrt{3}}=\\frac{y}{12}\\)
        \ny = \\(\\frac{12}{\\sqrt{3}}\\) = 4\u221a3m
        \ncos 30\u00b0 = \\(\\frac{12}{x}\\)
        \n\\(\\frac{\\sqrt{3}}{2}=\\frac{12}{x}\\)
        \nx = \\(\\frac{12 \\times 2}{\\sqrt{3}}\\) = 8\u221a3mt
        \nHeight of the tree before broken = x + y
        \n= 8\u221a3 + 4\u221a3
        \n= 12\u221a3 m<\/p>\n

        Question 14.
        \nProve that \u221a3 + \u221a5 is an irrational number.
        \nAnswer:
        \n\u221a3 + \u221a5
        \nLet us suppose that \u221a3 + \u221a5 is rational
        \nLet \u221a3 + \u221a5 = \\(\\frac{a}{b}\\), where \\(\\frac{a}{b}\\) is rational, b \u2260 0
        \n\u2234 \u221a3 = \\(\\frac{a}{b}\\) – \u221a5
        \nSquaring on both sides, we get
        \n\"TS
        \nSince a,b are integers \\(\\frac{a^2+2 b^2}{2 a b}\\) is rational and so, \u221a5 is rational
        \nThis contradicts the fact that \u221a5 is irrational.
        \nHence \u221a3 + \u221a5 is irrational.<\/p>\n

        Question 15.
        \nIf the points P (-3, 9), Q (a, b) and R (4, -5) are collinear and a + b = 1, then find the values of a and b.
        \nAnswer:
        \nPoints P, Q, R are collinear
        \n\u21d2 area of \u0394PQR = 0
        \nP(-3,9) Q (a, b) R(4, 5)
        \nArea of triangle = \\(\\frac{1}{2}\\) |x1<\/sub> (y2<\/sub> – y3<\/sub>)
        \n+ x2<\/sub> (y3<\/sub> – y1<\/sub>) + x3<\/sub> (y1<\/sub> – y2<\/sub>)|
        \n0 = \\(\\frac{1}{2}\\) | -3 (b + 5) + 4 (-5 -9) + (9 – b) |
        \nAfter simplifications, we get
        \n2a + b = 3 …………… (1)
        \ngiven equation
        \na + b = 1 …………….. (2)
        \nSolving equations (1) and (2),
        \nwe obtain a = 2 and b = -1<\/p>\n

        Question 16.
        \nTwo boys on either side of their school building of 20 m height observe its top at the angles of elevation 30\u00b0 and 60\u00b0 respectively. Find the distance between two boys.
        \nAnswer:
        \n\"TS
        \nFrom the figure
        \nAD = height of the building
        \n= 20m
        \nBC = Distance between two boys.
        \n\u2220ABD = 60\u00b0, \u2220ACD = 30\u00b0
        \nIn \u0394ABD,
        \ncot 30\u00b0 = \\(\\frac{\\mathrm{BD}}{\\mathrm{AD}}\\)
        \n\"TS<\/p>\n

        \"TS<\/p>\n

        Question 17.
        \nDraw the graph for the equations 2x – y – 4 = 0 and x + y + 1 = 0 on the graph paper and check whether they are consistent or not.
        \nAnswer:
        \nB) 2x – y – 4 = 0 …………. (1)
        \ny = 2x – 4<\/p>\n\n\n\n\n\n\n\n
        x<\/td>\n0<\/td>\n1<\/td>\n2<\/td>\n<\/tr>\n
        2x<\/td>\n0<\/td>\n2<\/td>\n4<\/td>\n<\/tr>\n
        -4<\/td>\n-4<\/td>\n-4<\/td>\n-4<\/td>\n<\/tr>\n
        y<\/td>\n-4<\/td>\n-2<\/td>\n0<\/td>\n<\/tr>\n
        (x, y)<\/td>\n(0, -4)<\/td>\n(1, -2)<\/td>\n(2, 0)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

        x + y + 1 = 0 …………… (2)<\/p>\n\n\n\n\n\n\n\n
        x<\/td>\n0<\/td>\n1<\/td>\n2<\/td>\n<\/tr>\n
        -x<\/td>\n0.<\/td>\n-1<\/td>\n-2.<\/td>\n<\/tr>\n
        -1<\/td>\n-1<\/td>\n-1<\/td>\n-1<\/td>\n<\/tr>\n
        y<\/td>\n-1<\/td>\n-2<\/td>\n-3<\/td>\n<\/tr>\n
        (x, y)<\/td>\n(0, -1)<\/td>\n(1, -2)<\/td>\n(2, -3)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

        \"TS
        \nIntersecting point of equations (1) and (2) is (1, -2)
        \nSo x = 1, y = -2
        \n\u2234 The given equations are consistent.<\/p>\n

        Question 18.
        \nConstruct a triangle of sides 5 cm, 6 cm and 7 cm. Then construct a triangle similar to it, whose sides are IV2 times the corresponding sides of the first triangle.
        \nAnswer:
        \n\"TS
        \nSteps of Construction :<\/p>\n

          \n
        1. Construct a triangle ABC with sides 5 cm, 6 cm and 7 cm.<\/li>\n
        2. Draw a ray AX making an acute angle with AB on the side opposite to vertex C.<\/li>\n
        3. Locate 3 points A1<\/sub>, A2<\/sub>, A3<\/sub> on AX. So that AA1<\/sub> = A1<\/sub>A2<\/sub> = A2<\/sub>A3<\/sub>.<\/li>\n
        4. Join A3<\/sub>B’ and draw a line from A2<\/sub> to B. Which is parallel.to A3<\/sub>B’ and it is intersecting AB at B.<\/li>\n
        5. Draw a line through B parallel to B’C’ to intersect AC at AC’ So, AB’C is the required triangle.<\/li>\n<\/ol>\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 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.<\/span><\/p>\n

            Question 1.
            \nThe logarithmic form of ab<\/sup> = c is ……………….
            \nA) loga<\/sub>c = b
            \nB) logb<\/sub> c = a
            \nC) loga<\/sub>b = c
            \nD) logb<\/sub>a = c
            \nAnswer:
            \nA) loga<\/sub>c = b<\/p>\n

            Question 2.
            \nIf 3 log (x +3) = log 27,.then the value of x is ………………….
            \nA) 0
            \nB) 1
            \nC) 6
            \nD) 24
            \nAnswer:
            \nA) 0<\/p>\n

            \"TS<\/p>\n

            Question 3.
            \nIn the formula of nth<\/sup> term of a Geometric Progression, an<\/sub> = a. rn – 1<\/sup>, r denotes ……………..
            \nA) first term
            \nB) common ratio
            \nC) common difference
            \nD) number of terms
            \nAnswer:
            \nB) common ratio<\/p>\n

            Question 4.
            \nWhich one of the following rational numbers has terminating decimal expression?
            \nA) \\(\\frac{11}{7000}\\)
            \nB) \\(\\frac{91}{21000}\\)
            \nC) \\(\\frac{343}{2^3 \\times 5^3 \\times 7^3}\\)
            \nD) \\(\\frac{21}{9000}\\)
            \nAnswer:
            \nC) \\(\\frac{343}{2^3 \\times 5^3 \\times 7^3}\\)<\/p>\n

            Question 5.
            \nThe common difference of an Arithmetic Progression in which a25<\/sub> – a12<\/sub> = -52 is ……………..
            \nA) 4
            \nB) -4
            \nC) -3
            \nD) 3
            \nAnswer:
            \nB) -4<\/p>\n

            Question 6.
            \nWhich one of the following statements is false?
            \nA) Every set is subset of itself
            \nB) Empty set is subset of every set
            \nC) Intersection of two disjoint sets is empty set
            \nD) Cardinal number of an infinite set is zero
            \nAnswer:
            \nD) Cardinal number of an infinite set is zero<\/p>\n

            Question 7.
            \nIf the co-ordinates of the vertices of a rectangle are (0, 0), (4, 0), (4, 3) and (0, 3), then the length of its diagonal is
            \nA) 4
            \nB) 5
            \nC) 7
            \nD) 3
            \nAnswer:
            \nB) 5<\/p>\n

            Question 8.
            \nThe quadratic polynomial having \\(\\frac{1}{3}\\) and \\(\\frac{1}{2}\\) as its zeroes, is ……………….
            \nA) x2<\/sup> + \\(\\frac{5 x+1}{6}\\)
            \nB) -6x2<\/sup> – 5x + 1
            \nC) x2<\/sup> – \\(\\frac{5 x-1}{6}\\)
            \nD) 6x2<\/sup> – 5x – 1
            \nAnswer:
            \nC) x2<\/sup> – \\(\\frac{5 x-1}{6}\\)<\/p>\n

            \"TS<\/p>\n

            Question 9.
            \nSum of 10 terms of the progression log 2 + log 4 + log 8 + log 16 + ……………… is ……………
            \nA) 45 log 2
            \nB) 90 log 2
            \nC) 10 log 2
            \nD) 55 log 2
            \nAnswer:
            \nD) 55 log 2<\/p>\n

            Question 10.
            \nWhich term of the Arithmetic Progression 24, 21, 18, ……………. is the first negative term ?
            \nA) 8th<\/sup>
            \nB) 9th<\/sup>
            \nC) 10th<\/sup>
            \nD) 12th<\/sup>
            \nAnswer:
            \nC) 10th<\/sup><\/p>\n

            Question 11.
            \nThe value of Tan \u03b8 in terms of Cosec \u03b8 is ………………..
            \nA) \\(\\frac{1}{\\sqrt{{cosec}^2 \\theta-1}}\\)
            \nB) \\(\\frac{{cosec} \\theta}{\\sqrt{{cosec}^2 \\theta-1}}\\)
            \nC) \\(\\frac{2 {Cosec} \\theta}{\\sqrt{{cosec}^2 \\theta-1}}\\)
            \nD) \\(\\frac{2}{\\sqrt{{cosec}^2 \\theta-1}}\\)
            \nAnswer:
            \nA) \\(\\frac{1}{\\sqrt{{cosec}^2 \\theta-1}}\\)<\/p>\n

            Question 12.
            \nObserve the following:
            \nI) Sin2<\/sup> 20\u00b0 + Sin2<\/sup> 70\u00b0 = 1 II) Log2<\/sub> (Sin 90\u00b0)= 1 Which one is CORRECT ?
            \nA) (I) only
            \nB) (II) only
            \nC) Both (I) and (II)
            \nD) Neither (I) nor (II)
            \nAnswer:
            \nA) (I) only<\/p>\n

            Question 13.
            \nIn \u0394ABC, AC = 12 cm, AB = 5 cm and \u2220BAC = 30\u00b0, the area of \u0394ABC is
            \nA) 30 cm2<\/sup>
            \nB) 15 cm2<\/sup>
            \nC) 60 cm2<\/sup>
            \nD) 20 cm2<\/sup>
            \nAnswer:
            \nB) 15 cm2<\/sup><\/p>\n

            Question 14.
            \nWhich one of the following cannot be the probability of an event ?
            \nA) \\(\\frac{2}{3}\\)
            \nB) \\(\\frac{4}{5}\\)
            \nC) 0.7
            \nD) \\(\\frac{5}{4}\\)
            \nAnswer:
            \nD) \\(\\frac{5}{4}\\)<\/p>\n

            Question 15.
            \nThe X – coordinate of the point of intersection of the two ogives of grouped data is ……………….
            \nA) median of the data
            \nB) mode of the data
            \nC) mean of the data
            \nD) average of mid values of the data
            \nAnswer:
            \nA) median of the data<\/p>\n

            Question 16.
            \nVolumes of two spheres are in the ratio of 8 : 27, the ratio of their surface areas is ………………
            \nA) 2 : 3
            \nB) 4 : 3
            \nC) 2 : 9
            \nD) 4 : 9
            \nAnswer:
            \nD) 4 : 9<\/p>\n

            \"TS<\/p>\n

            Question 17.
            \nA solid ball is exactly fitted inside the cubical box of side ‘a’. The volume of the ball is
            \nA) \\(\\frac{1}{3}\\)\u03c0 a3<\/sup>
            \nB) \\(\\frac{1}{6}\\)\u03c0 a3<\/sup>
            \nC) \\(\\frac{4}{3}\\)\u03c0 a3<\/sup>
            \nD) \\(\\frac{8}{3}\\)\u03c0 a3<\/sup>
            \nAnswer:
            \nB) \\(\\frac{1}{6}\\)\u03c0 a3<\/sup><\/p>\n

            Question 18.
            \nExpress ‘x’ in terms of a, b and c in the following figure.
            \n\"TS
            \nA) x = \\(\\frac{a c}{b+c}\\)
            \nB) x = \\(\\frac{b c}{b+c}\\)
            \nC) x = \\(\\frac{b+c}{a c}\\)
            \nD) x = \\(\\frac{a b}{a+c}\\)
            \nAnswer:
            \nA) x = \\(\\frac{a c}{b+c}\\)<\/p>\n

            Question 19.
            \nIf the angle of elevation of sun increases from 0\u00b0 to 90\u00b0, then the length of shadow of the tower ………………
            \nA) no change
            \nB) increases
            \nC) decreases
            \nD) can’t be decided
            \nAnswer:
            \nC) decreases<\/p>\n

            Question 20.
            \nIn a right angled triangle with integral sides at least one of its measurements must be ……………..
            \nA) multiples of 3 and 2
            \nB) multiple of 9
            \nC) multiples of 8
            \nD) multiple of 7
            \nAnswer:
            \nA) multiples of 3 and 2<\/p>\n","protected":false},"excerpt":{"rendered":"

            The strategic use of TS 10th Class Maths Model Papers Set 8 can significantly enhance a student’s problem-solving skills. TS SSC Maths Model Paper Set 8 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\/60298"}],"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=60298"}],"version-history":[{"count":3,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/60298\/revisions"}],"predecessor-version":[{"id":60315,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/60298\/revisions\/60315"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media?parent=60298"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/categories?post=60298"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/tags?post=60298"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}