{"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 Time: 3 Hours General Instructions:<\/span><\/p>\n Part – A (60 Marks)<\/span> Note :<\/p>\n Question 1. Question 2. Question 3. <\/p>\n Question 4. Question 5. Question 6. Section – II (6 \u00d7 4 = 24 Marks)<\/span><\/p>\n Note : Question 7. <\/p>\n Question 8. Question 9. Question 10. Question 11. Question 12. <\/p>\n Section – III (4 \u00d7 6 = 24 Marks)<\/span><\/p>\n Note :<\/p>\n Question 13. Question 14. Question 15. Question 16. <\/p>\n Question 17. x + y + 1 = 0 …………… (2)<\/p>\n Question 18. Part – B (20 Marks)<\/span><\/p>\n Note :<\/p>\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. Question 2. <\/p>\n Question 3. Question 4. Question 5. Question 6. Question 7. Question 8. <\/p>\n Question 9. Question 10. Question 11. Question 12. Question 13. Question 14. Question 15. Question 16. <\/p>\n Question 17. Question 18. Question 19. Question 20. 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}]}}TS SSC Maths Model Paper Set 8 with Solutions<\/h2>\n
\nMaximum Marks: 80<\/p>\n\n
\nSection – I (6 \u00d7 2 = 12 Marks)<\/span><\/p>\n\n
\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
\nIn a rectangle ABCD,
\nAB = x + y, BC = x – y,
\nCD = 9 and AD = 3.
\n
\nFind the values of x and y.
\nAnswer:
\nx + y = 9
\n
\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
\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
\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
\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
\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
\n(i) Answer ALL the following questions.
\n(ii) Each question carries 4 marks.<\/p>\n
\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
\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
\nExpress 2016 as product of prime factors.
\nAnswer:
\n
\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
\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
\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
\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
\nDraw a circle with 5 cm radius and construct a pair of tangents to the circle.
\nAnswer:
\n<\/p>\n\n
\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
\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
\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
\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
\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
\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
\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<\/p>\n
\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 x<\/td>\n 0<\/td>\n 1<\/td>\n 2<\/td>\n<\/tr>\n \n 2x<\/td>\n 0<\/td>\n 2<\/td>\n 4<\/td>\n<\/tr>\n \n -4<\/td>\n -4<\/td>\n -4<\/td>\n -4<\/td>\n<\/tr>\n \n y<\/td>\n -4<\/td>\n -2<\/td>\n 0<\/td>\n<\/tr>\n \n (x, y)<\/td>\n (0, -4)<\/td>\n (1, -2)<\/td>\n (2, 0)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n \n\n
\n x<\/td>\n 0<\/td>\n 1<\/td>\n 2<\/td>\n<\/tr>\n \n -x<\/td>\n 0.<\/td>\n -1<\/td>\n -2.<\/td>\n<\/tr>\n \n -1<\/td>\n -1<\/td>\n -1<\/td>\n -1<\/td>\n<\/tr>\n \n y<\/td>\n -1<\/td>\n -2<\/td>\n -3<\/td>\n<\/tr>\n \n (x, y)<\/td>\n (0, -1)<\/td>\n (1, -2)<\/td>\n (2, -3)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
\nIntersecting point of equations (1) and (2) is (1, -2)
\nSo x = 1, y = -2
\n\u2234 The given equations are consistent.<\/p>\n
\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
\nSteps of Construction :<\/p>\n\n
\n
\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
\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
\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
\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
\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
\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
\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
\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
\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
\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
\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
\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
\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
\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
\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
\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
\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
\nExpress ‘x’ in terms of a, b and c in the following figure.
\n
\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
\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
\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":"