{"id":59854,"date":"2024-04-10T13:05:25","date_gmt":"2024-04-10T07:35:25","guid":{"rendered":"https:\/\/apboardsolutions.com\/?p=59854"},"modified":"2024-04-10T13:05:25","modified_gmt":"2024-04-10T07:35:25","slug":"ts-10th-class-maths-model-paper-set-2","status":"publish","type":"post","link":"https:\/\/apboardsolutions.com\/ts-10th-class-maths-model-paper-set-2\/","title":{"rendered":"TS 10th Class Maths Model Paper Set 2 with Solutions"},"content":{"rendered":"
The strategic use of TS 10th Class Maths Model Papers<\/a> Set 2 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 \u2013 II (6 \u00d7 3 = 18 Marks)<\/span><\/p>\n Note :<\/p>\n Question 7. <\/p>\n Question 8. Question 9. Question 10. i) Prime numbers ii) Even numbers Question 11. In \u0394ABQ tan 45\u00b0 = \\(\\frac{\\mathrm{AB}}{\\mathrm{BQ}}\\) <\/p>\n Question 12. Section \u2013 III (6 \u00d7 5 = 30 Marks)<\/span><\/p>\n Note :<\/p>\n Question 13. B) Find the median for the following data. Question 14. (OR)<\/p>\n B) The angle of elevation of the top of a hill from the foot of a tower is 60\u00b0 and the angle of elevation of the top of the tower from the foot of the hill is 30\u00b0. If the tower is 50 m high. Find the height of the hill. <\/p>\n Question 15. (OR)<\/p>\n B) Question 16. (OR)<\/p>\n B) Sum of the areas of two squares is 850 m2<\/sup>. If the difference of their perimeters is 40 m, find the sides of the two squares. Question 17. (OR)<\/p>\n B) Solve the equations graphically 3x + 4y = 10 and 4x – 3y = 5. 4x – 3y = 5. ……………… (2)<\/p>\n Intersecting point is (2, 1) <\/p>\n Question 18. (OR)<\/p>\n B) Draw less than Ogive for the following frequency distribution. Find the median from obtained curve. Part \u2013 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. (Marks; 20 \u00d7 1 = 20)<\/span><\/p>\n Question 1. Question 2. <\/p>\n Question 3. Question 4. Question 5. Question 6. Question 7. Question 8. Question 9. Question 10. <\/p>\n 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 2 can significantly enhance a student’s problem-solving skills. TS SSC Maths Model Paper Set 2 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\/59854"}],"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=59854"}],"version-history":[{"count":6,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/59854\/revisions"}],"predecessor-version":[{"id":59910,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/59854\/revisions\/59910"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media?parent=59854"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/categories?post=59854"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/tags?post=59854"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}TS SSC Maths Model Paper Set 2 with Solutions<\/h2>\n
\nMaximum Marks: 80<\/p>\n\n
\nSection – I (6 \u00d7 2 = 12 Marks)<\/span><\/p>\n\n
\nExpand log a3<\/sup>b2<\/sup>c5<\/sup>.
\nSolution:
\nlog a3<\/sup>b2<\/sup>c5<\/sup> = log a3<\/sup> + log b2<\/sup> + log c5<\/sup>
\n= 3 log a + 2 log b + 5 log c<\/p>\n
\nIf p(x) = x2<\/sup> + 3x + 4, then find the values of p(0) and p(1).
\nSolution:
\np(x) = x2<\/sup> + 3x + 4
\np(0) = (0)2<\/sup> + 3(0) + 4
\n= 0 + 0 + 4 = 4
\n\u2234 p(0) = 4
\np(1) = (1)2<\/sup> + 3(1) + 4
\n= 1 + 3 + 4 = 8
\n\u2234 p(1) = 8<\/p>\n
\nFind the 10th<\/sup> term of the arithmetic progression 3, 5, 7, ……………
\nSolution:
\n3, 5, 7 ……………. A.P
\na = 3, d = 2, n = 10
\nan<\/sub> = a + (n – 1)d
\na10<\/sub> = 3 + (10 – 1)2
\n= 3 + (9 \u00d7 2)
\n= 3 + 18 = 21<\/p>\n
\nExpress tan \u03b8 in terms of sin \u03b8.
\nSolution:
\ntan \u03b8 = \\(\\frac{\\sin \\theta}{\\cos \\theta}\\)
\nsin2<\/sup> \u03b8 + cos2<\/sup> \u03b8 = 1
\ncos2<\/sup> \u03b8 = 1 – sin2<\/sup>\u03b8
\ncos \u03b8 = \\(\\sqrt{1-\\sin ^2 \\theta}\\)
\n\u2234 tan \u03b8 = \\(\\frac{\\sin \\theta}{\\sqrt{1-\\sin ^2 \\theta}}\\).<\/p>\n
\nIf a dice is rolled once, then find the probability of getting an odd number.
\nSolution:
\nTotal outcomes = {1, 2, 3, 4, 5, 6)
\nNumber of Total outcomes = 6
\nFavourable outcomes = {1, 3, 5}
\nNumber of favourable outcomes = 3
\nP (an odd number)
\n= \\(\\frac{\\text { Number of favourable outcomes }}{\\text { Number of Total outcomes }}\\)
\n= \\(\\frac{3}{6}\\) = \\(\\frac{1}{2}\\)<\/p>\n
\n“The top of a tower is observed at an angle of elevation 45\u00b0 and the foot of ‘ the tower is at a distance of 30 metres from the observer”. Draw a suitable diagram for this data.
\nSolution:
\n
\nC – Point of observation
\nAB – Tower
\nA – Top of the tower<\/p>\n\n
\nSolve : 2x + y = 5 and 5x + 3y = 11.
\nSolution:
\n2x + y = 5 ……………. (1)
\n5x + 3y = 11 …………… (2)
\n
\nSubstituting the value of x in equation (1)
\n\u21d2 2(4) + y = 5
\n\u21d2 8 + y = 5 \u21d2 y = 5 – 8 \u2234 y = -3
\n\u2234 x = 4, y = – 3 are the solution of the equations.<\/p>\n
\n5, 8, 11, 14, ………. is an arithmetic progression. Find the sum of first 20 terms of it.
\nSolution:
\n5, 8, 11, 14, ……………. A.P
\na = 5, d = 8 – 5 = 3, n = 20
\nSn<\/sub> = \\(\\frac{n}{2}\\)[2a + (n – 1) d]
\nS20<\/sub> = \\(\\frac{20}{2}\\)[(2 \u00d7 5) + (20 – 1)3]
\n= 10[10 + 57]
\n= 10 \u00d7 67 = 670<\/p>\n
\nWrite a Quadratic equation, whose roots are 3 + \u221a5 and 3 – \u221a5 .
\nSolution:
\nRoots are 3 + \u221a5 and 3 – \u221a5 .
\nSum of the roots = (3 + \u221a5) + (3 – \u221a5) = 6
\nProduct of roots = (3 + \u221a5) (3 – \u221a5)
\n= 9 – 5 = 4
\n\u2234 Required Quadratic equation = x2<\/sup> – x (sum of the roots) + Product of roots = 0
\n\u2234 x2<\/sup> – 6x + 4 = 0<\/p>\n
\nA box contains 20 cards which are numbered from 1 to 20. If one card is selected at random from the box, find the probability that it bears (i) a prime number, (ii) an even number.
\nSolution:
\nTotal possibilities
\n= {1, 2, 3, 4, 5, 6, 7, 8, 9, ……………,20}
\nNumber of Total possibilities = 20<\/p>\n
\n= {2, 3, 5, 7, 11, 13, 17, 19}
\nNumber of favourable outcomes = 8
\nP(Prime Number)
\n= \\(\\frac{\\text { Number of favourable outcomes }}{\\text { Total number of outcomes }}\\)
\n= \\(\\frac{8}{20}\\) = \\(\\frac{2}{5}\\)
\n\u2234 P(Prime Number) = \\(\\frac{2}{5}\\)<\/p>\n
\n= {2,4, 6, 8, 10,12, 14, 16, 18,20}
\nNumber of favourable outcomes = 10
\nP (Even Number)
\n\\(\\frac{\\text { Number of favourable outcomes }}{\\text { Total number of outcomes }}\\)
\n= \\(\\frac{10}{20}\\) = \\(\\frac{1}{2}\\)
\n\u2234 P(Prime Number) = \\(\\frac{1}{2}\\)<\/p>\n
\nIf two persons standing on either side of a tower of height 100 metres observes the top of it with angles of elevation of 60\u00b0 and 45\u00b0 respectively, then find the distance between the two persons.
\n[Note : Consider the two persons and the tower are on the same line.]
\nSolution:
\nTower height = 100 m
\nAngles of elevation = 60\u00b0 and 45\u00b0
\n
\nIn \u0394ABP tan 60\u00b0 = \\(\\frac{\\mathrm{AB}}{\\mathrm{PB}}\\)
\n\\(\\frac{\\sqrt{3}}{1}=\\frac{100}{x}\\)
\n\u221a3 = x = 100 \u21d2 x = \\(\\frac{100}{\\sqrt{3}}\\) m<\/p>\n
\n\\(\\frac{1}{1}=\\frac{100}{y}\\) \u21d2 y = 100 m
\n\u2234 The distance between Two persons = x + y
\n\\(\\frac{100}{\\sqrt{3}}\\) + 100 = \\(\\frac{100(\\sqrt{3}+1)}{\\sqrt{3}}\\) m<\/p>\n
\nIn a trapezium ABCD, AB || DC. If diagonals intersect each other at point ‘O’, then show that \\(\\frac{\\mathrm{AO}}{\\mathrm{BO}}\\) = \\(\\frac{\\mathrm{CO}}{\\mathrm{DO}}\\)
\nSolution:
\nGiven: In trapezium oABCD, AB \/\/ CD.
\nDiagonals AC, BD intersect at O.
\nR.T.P : \\(\\frac{\\mathrm{AO}}{\\mathrm{BO}}\\) = \\(\\frac{\\mathrm{CO}}{\\mathrm{DO}}\\)
\n
\nConstruction : Draw a line EF passing through the point \u2018O\u2019 and parallel to CD and AB.
\nProof: In \u0394ACD, EO\/\/CD
\n\u2234 \\(\\frac{\\mathrm{AO}}{\\mathrm{CO}}\\) = \\(\\frac{\\mathrm{AE}}{\\mathrm{DE}}\\)
\n[\u2235 line drawn parallel to one side of a triangle divides other two sides in the same ratio by Basic proportionality theorem]
\nIn \u0394ABD, EO \/\/AB
\nHence, \\(\\frac{\\mathrm{DE}}{\\mathrm{AE}}\\) = \\(\\frac{\\mathrm{DO}}{\\mathrm{BO}}\\)
\n[\u2235 Basic proportionality theorem]
\n\\(\\frac{\\mathrm{BO}}{\\mathrm{DO}}\\) = \\(\\frac{\\mathrm{AE}}{\\mathrm{ED}}\\) ………… (2)
\nFrom (1) and (2)
\n\\(\\frac{\\mathrm{AO}}{\\mathrm{CO}}\\) = \\(\\frac{\\mathrm{BO}}{\\mathrm{DO}}\\) [\u2235 Alternendo]<\/p>\n\n
\nA) If sec \u03b8 + tan \u03b8 = P, then prove that sin \u03b8 = \\(\\frac{\\mathbf{P}^2-1}{\\mathbf{P}^2+1}\\)
\nSolution:
\nIf sec \u03b8 + tan \u03b8 = P
\nThen sec \u03b8 – tan \u03b8 = \\(\\frac{1}{\\mathrm{P}}\\)
\n[\u2235 sec2<\/sup> \u03b8 – tan2<\/sup> \u03b8 = 1]
\nby adding (+) \u21d2 2 sec \u03b8 = P + \\(\\frac{1}{\\mathrm{P}}\\)
\nsec \u03b8 = \\(\\frac{\\mathrm{P}^2+1}{2 \\mathrm{P}}\\) ……………… (1)
\n
\n(OR)<\/p>\n
\n
\nSolution:
\n
\nMedian = l + (\\(\\frac{\\frac{N}{2}-c f}{f}\\)) \u00d7 h
\nl = 20, \\(\\frac{N}{2}\\) = \\(\\frac{44}{2}\\) = 22, cf = 16
\nf = 12, h = 10
\nMedian = 20 + (\\(\\frac{22-16}{12}\\)) \u00d7 10
\n20 + (\\(\\frac{1}{12}\\) \u00d7 10)
\n= 20 + (\\(\\frac{60}{12}\\)) = 20 + 5 = 25
\n\u2234 Median =25<\/p>\n
\nA) Prove that \u221a5 + \u221a7 Is an irrational number.
\nSolution:
\nSuppose \u221a5 + \u221a7 is not an irrational number.
\n\u221a5 + \u221a7 = \\(\\frac{p}{q}\\) ; p, q \u2208 Z, q \u2260 0
\nSquaring on both sides
\n(\u221a5 + \u221a7)2<\/sup> = (\\(\\frac{p}{q}\\))2<\/sup>
\n\u21d2 5 + 7 + 2\\(\\sqrt{35}\\) = \\(\\frac{p^2}{q^2}\\)
\n\u21d2 2\\(\\sqrt{35}\\) = \\(\\frac{p^2}{q^2}\\) – 12
\n\u21d2 \\(\\sqrt{35}\\) = \\(\\frac{p^2-12 q^2}{2 q^2}\\)
\n[p2<\/sup> – 12q2<\/sup>, 2q2<\/sup>\u2208 Z and 2q2<\/sup> \u2260 0]
\n\u2234 \\(\\frac{p^2-12 q^2}{2 q^2}\\) is a rational number.
\nBut \\(\\sqrt{35}\\) is an irrational number.
\nAn irrational number never be-comes equal to a rational number.
\nSo our supposition that \u221a5 + \u221a7 is not an irrational number is false.
\n\u2234 \u221a5 + \u221a7 is an irrational number.<\/p>\n
\nSolution:
\nGiven height of the tower = AB = 50 m
\nLet height of hill be CD = h m
\nand distance between their feet be AC = x m
\n\u2220ACB = 30\u00b0, \u2220CAD = 60\u00b0
\nFrom right angled \u0394 ABC, tan 30\u00b0 = \\(\\frac{\\mathrm{AB}}{\\mathrm{AC}}\\)
\n
\n\\(\\frac{1}{3}\\) = \\(\\frac{50}{x}\\) \u21d2 x = 50\u221a3 m
\nFrom right angled \u0394 ACD, tan 60\u00b0 = \\(\\frac{\\mathrm{CD}}{\\mathrm{AC}}\\)
\n\u221a3 = \\(\\frac{h}{x}\\) \u21d2 h = x\u221a3 m
\nh = 50\u221a3 .\u221a3 (\u2235 x = 50\u221a3m)
\nh = 50 \u00d7 3
\n\u21d2 h = 150 m.<\/p>\n
\nA) Show that the distance of the points (5, 12), (7, 24) and (35, 12) from the origin are arranged in ascending order forms an arithmetic progression. Find the common difference of the progression.
\nSolution:
\nThe distance of the point (x, y) from the origin = \\(\\sqrt{x^2+y^2}\\)
\nThe distance of the point (5, 12) from the origin = \\(\\sqrt{(5)^2+(12)^2}\\) = \\(\\sqrt{25+144}\\) = \\(\\sqrt{169}\\) = 13 units;
\nThe distance of the point (7, 24) from the origin = \\(\\sqrt{(7)^2+(24)^2}\\) = \\(\\sqrt{49+576}\\) = \\(\\sqrt{625}\\) = 25 units.
\nThe distance of the point (35, 12) from the origin = \\(\\sqrt{(35)^2+(12)^2}\\) = \\(\\sqrt{1225+144}\\) = \\(\\sqrt{1369}\\) = 37 units.,
\nThe ascending order of the distances is 13, 25, 37
\na1<\/sub> a2<\/sub> a3<\/sub>
\na2<\/sub> – a1<\/sub> = 25 – 13 = 12
\na3<\/sub> – a2<\/sub> = 37 – 25 = 12
\na2<\/sub> – a1<\/sub> = a3<\/sub> – a2<\/sub>
\na1<\/sub>, a2<\/sub>, a3<\/sub> are in A.R
\nCommon difference (d) = 12<\/p>\n
\nThe sum of the radius of base and height of a solid right circular cylinder is 37 cm. If it’s total surface area is 1628 square centimeters (cm2<\/sup>), then find the volume of the cylinder
\n(use \u03c0 = \\(\\frac{22}{7}\\))
\nSolution:
\n
\nIn the right circular cylinder
\nLet base radius = r
\nHeight = h
\nr + h = 37 cm
\nTotal surface area = 1628 sq.cm
\nVolume = ?
\nTotal surface area of the cylinder = 2\u03c0r (r + h)
\n2\u03c0r(r + h) = 1628
\n2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 r \u00d7 37 = 1628
\nr = \\(\\frac{1628 \\times 7}{2 \\times 22 \\times 37}\\) = 7 cm
\n7 + h = 37 \u21d2 h = 37 – 7 = 30 cm
\n\u2234 Volume = \u03c0r2<\/sup>h
\n= \\(\\frac{22}{7}\\) \u00d7 7 \u00d7 7 \u00d7 30 = 4620 cm3<\/sup><\/p>\n
\nA) From the given Venn diagram, write the sets A\u222aB, A\u2229B, A – B and B – A.
\n
\nSolution:
\nA = {1, 2, 3, 4, 5}, B = {2, 4, 6, 8}
\nA\u222aB = {1, 2, 3, 4, 5, 6, 8}
\nA\u2229B = (2, 4}
\nA- B = {1, 3, 5}
\nB – A = {6, 8}<\/p>\n
\nSolution:
\nLet the side of first square be ‘a’
\nthe side of second square be ‘b’
\nArea of first square = ‘a2<\/sup>‘ m2<\/sup> ,
\nArea of second square = ‘b2<\/sup>‘ m2<\/sup>
\nSum of the area of two squares = 850 m2<\/sup>
\na2<\/sup> + b2<\/sup> = 850 ………… (1)
\nPerimeter of first square = ‘4a’m
\nPerimeter of second square = ‘4b’m
\nDifference of their perimeters = 40 m
\n4a – 4b = 40
\na – b = 10 …………. (2)
\na = b + 10
\nsubstitute this in eqn (1)
\n(b + 10)2<\/sup> + b2<\/sup> = 850
\nb2<\/sup> + 20b + 100 + b2<\/sup> = 850
\n2b2<\/sup> + 20b – 750 = 0
\nb2<\/sup> + 10b – 375 = 0
\nb2<\/sup> + 25b – 15b – 375 = 0
\nb(b + 25) – 15(b + 25) = 0
\n(b – 15) (b + 25) = 0
\nb = 15, b = – 25
\nfrom (2) a – 15 = 10 \u21d2 a = 25
\n\u2234 The sides of two squares are 25m, 15m.<\/p>\n
\nA) Draw the graph of the polynomial p(x) x2<\/sup> + 2x – 3 and find the zeroes of the polynomial from the graph.
\nSolution:
\ny = p(x) = x2<\/sup> + 2x – 3
\n
\n
\nZeros of the polynomial are – 3, 1<\/p>\n
\nSolution:
\n3x + 4y = 10 ……………… (1)<\/p>\n\n\n
\n x<\/td>\n 2<\/td>\n -2<\/td>\n 6<\/td>\n<\/tr>\n \n y<\/td>\n 1<\/td>\n 4<\/td>\n -2<\/td>\n<\/tr>\n \n (x, y)<\/td>\n (2, 1)<\/td>\n (-2, 4)<\/td>\n (6, -2)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n \n\n
\n x<\/td>\n 2<\/td>\n -1<\/td>\n 5<\/td>\n<\/tr>\n \n y<\/td>\n 1<\/td>\n -3<\/td>\n 5<\/td>\n<\/tr>\n \n (x, y)<\/td>\n (2, 1)<\/td>\n (-1, -3)<\/td>\n (5, 5)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
\n\u2234 x = 2, y = 1
\n<\/p>\n
\nA) Draw a circle of radius 4 cm. from a point 9 cm away from it’s centre, construct a pair of tangents to the circle.
\nSolution:
\n
\nSteps of Construction :<\/p>\n\n
\n
\nSolution:
\n
\n
\nn = 103 \u21d2 \\(\\frac{n}{2}\\) = \\(\\frac{103}{2}\\) = 51.5
\nForm graph, median = 100<\/p>\n\n
\nIf A\u2282B, then A\u2229B is …………….
\nA) A
\nB) B
\nC) \u03bc
\nD) \u03a6
\nAnswer:
\nA) A<\/p>\n
\nThe coefficient of x3<\/sup> in the polynomial 2x4<\/sup> – 5x3<\/sup> + 6x2<\/sup> + 5 is ……………..
\nA) -5
\nB) 5
\nC) 6
\nD) 2
\nAnswer:
\nA) -5<\/p>\n
\nIf the slope of the line joining the points (2, 5) and (x, 3) is 2, then the value of ‘x’ is ……………
\nA) 0
\nB) 1
\nC) -1
\nD) 2
\nAnswer:
\nB) 1<\/p>\n
\nThe product of prime factors of 108 is ……………
\nA) 23<\/sup> \u00d7 32<\/sup>
\nB) 22<\/sup> \u00d7 32<\/sup>
\nC) 22<\/sup> \u00d7 33<\/sup>
\nD) 23<\/sup> \u00d7 33<\/sup>
\nAnswer:
\nC) 22<\/sup> \u00d7 33<\/sup><\/p>\n
\nThe number of solutions of the pair of linear equations
\n3x + 2y = 6 and 6x + 4y = 18 is ……………..
\nA) 0
\nB) 1
\nC) 2
\nD) infinite
\nAnswer:
\nA) 0<\/p>\n
\n“The total cost of 2 pens and 3 books is Rs. 110\u201d.
\nLinear equation representing this data is ……………..
\nA) x + y = 110
\nB) 5x = 110
\nC) x2<\/sup> + y3<\/sup> = 110
\nD) 2x + 3y = 110
\nAnswer:
\nD) 2x + 3y = 110<\/p>\n
\nIf the nth<\/sup> term of an arithmetic progression is 4n – 2, then its 10th<\/sup> term is ……………..
\nA) 38
\nB) 28
\nC) 42
\nD) 24
\nAnswer:
\nA) 38<\/p>\n
\nIf one root of the Quadratic equation x2<\/sup> – kx + 36 = 0 is 4, then the value of ‘k’ is ……………
\nA) 12
\nB) 17
\nC) 18
\nD) 13
\nAnswer:
\nD) 13<\/p>\n
\nThe nature of roots of the Quadratic equation x2<\/sup> + 6x + 9 = 0 is …………………
\nA) Real and distinct.
\nB) Real and equal.
\nC) No real roots.
\nD) One is positive and the other is negative.
\nAnswer:
\nB) Real and equal.<\/p>\n
\nThe sum of first 10 natural numbers is …………….
\nA) 45
\nB) 65
\nC) 55
\nD) 35
\nAnswer:
\nC) 55<\/p>\n
\nFrom the given Ogive curve, the value of the median of the data is ……………..
\n
\nA) 20
\nB) 25
\nC) 15
\nD) 30
\nAnswer:
\nD) 30<\/p>\n
\nIn the formula of volume of right circular cylinder V = \u03c0r2<\/sup>h, the letter ‘r’ represents ……………..
\nA) Diameter
\nB) Height
\nC) Volume
\nD) Radius
\nAnswer:
\nD) Radius<\/p>\n
\nIf E and \\(\\overline{\\mathbf{E}}\\) are complementary events in a random experiment and P(\\(\\overline{\\mathbf{E}}\\)) = 0.3, the value of P(E) is ……………….
\nA) 0.3
\nB) 0.7
\nC) 1
\nD) 0
\nAnswer:
\nB) 0.7<\/p>\n
\nIf one letter is selected randomly from the letters of the word “COVID”, then the probability of getting a vowel is ……………..
\nA) \\(\\frac{4}{5}\\)
\nB) \\(\\frac{3}{5}\\)
\nC) \\(\\frac{2}{5}\\)
\nD) \\(\\frac{1}{5}\\)
\nAnswer:
\nC) \\(\\frac{2}{5}\\)<\/p>\n
\n\u0394ABC ~ \u0394DEF, if \u2220A = 45\u00b0 and \u2220E = 75\u00b0, then \u2220C is ……………….
\nA) 90\u00b0
\nB) 120\u00b0
\nC) 30\u00b0
\nD) 60\u00b0
\nAnswer:
\nD) 60\u00b0<\/p>\n
\n\u0394ABC is a right triangle, right angled at B. If AB = 9 cm, BC = 12 cm, then AC is …………….
\nA) 13 cm
\nB) 14 cm
\nC) 15 cm
\nD) 16 cm
\nAnswer:
\nC) 15 cm<\/p>\n
\nIn the given figure OA and OB are radii. PA and PB are tangents to the circle at points A and B. If \u2220AOB = 130\u00b0, then \u2220APB = ?
\n
\nA) 40\u00b0
\nB) 50\u00b0
\nC) 60\u00b0
\nD) 70\u00b0
\nAnswer:
\nB) 50\u00b0<\/p>\n
\nIf sin \u03b8 = \\(\\frac{3}{5}\\), then the value of cos \u03b8 is (\u03b8 is acute angle) ……………..
\nA) \\(\\frac{1}{5}\\)
\nB) \\(\\frac{5}{3}\\)
\nC) \\(\\frac{4}{5}\\)
\nD) \\(\\frac{2}{5}\\)
\nAnswer:
\nC) \\(\\frac{4}{5}\\)<\/p>\n
\nThe maximum number of tangents can be drawn from an external point to a circle is ……………..
\nA) 1
\nB) 2
\nC) 3
\nD) 4
\nAnswer:
\nB) 2<\/p>\n
\nIf \u03b8 is acute angle, then sin \u03b8 \u00d7 sec \u03b8 = ……………….
\nA) tan \u03b8
\nB) cot \u03b8
\nC) 1
\nD) cosec \u03b8
\nAnswer:
\nA) tan \u03b8<\/p>\n","protected":false},"excerpt":{"rendered":"