Frequency<\/strong><\/td>\n| 18<\/td>\n | 4<\/td>\n | 6<\/td>\n | 4<\/td>\n | 8<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n SECTION-D<\/strong><\/p>\nQuestion 23.<\/strong>
\nConstruct a triangle ABC in which AC = AB = 4.5 cm and \u2220A = 90\u00b0, then constuct a triangle similar to \u2206ABC with its sides equal to (\\(\\frac { 5 }{ 4 } \\))th of the corresponding sides of \u2206ABC.\u00a0\u00a0[4]<\/strong><\/p>\nQuestion 24.<\/strong>
\nThe angle of elevation of the top of a tower from a point A on the ground is 30\u00b0. On moving a distance of 20 metre towards the foot of the tower to a point B, the angle of elevation increases to 60\u00b0. Find the height of the tower and the distance of the tower from the
\npoint A.\u00a0\u00a0[4]<\/strong>
\nOR<\/strong>
\nAs observed from the top of a lighthouse, 100 m above sea level, the angle of depression of a ship, sailing directly towards it, changes from 30\u00b0 to 45\u00b0. Determine the distance travelled by the ship during the period of observation.\u00a0[4]<\/strong><\/p>\nQuestion 25.<\/strong>
\nIn figure, P is the mid-point of BC and Q is the mid-point of AP. If BQ when produced meets AC at R, prove that RA = \\(\\frac { 1 }{ 3 } \\)CA\u00a0\u00a0[4]<\/strong>
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1949\/43299084810_5ba6e857a3_o.png?resize=140%2C166&ssl=1\") <\/p>\nQuestion 26.<\/strong>
\nA milk container of height 16 cm is made of metal sheet in the form of a frustum of a cone with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find the cost of milk at the rate of ? 22 per litre which the container can hold.\u00a0\u00a0[4]<\/strong>
\nOR<\/strong>
\nFrom a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2<\/sup>.<\/p>\nQuestion 27.<\/strong>
\nIn a rectangle if length is increased and breadth is reduced by 2 metre, the area is reduced by 28 sq. m. If length is reduced by 1 metre and breadth is increased by 2 metre, the area increased by 33 sq. m. Find the length and breadth of the rectangle.\u00a0\u00a0[4]<\/strong>
\nOR<\/strong>
\nOut of group of swans, \\(\\frac { 7 }{ 2 } \\) times the square root of the total number are playing on the shore of a pond. The remaining two are swimming in water. Find the total number of swans.\u00a0\u00a0[4]<\/strong><\/p>\nQuestion 28.<\/strong>
\nThe students of a school decided to beautify the school on the Annual Day by fixing colourful flags on the straight passage of the school. They have 27 flags to be fixed at intervals of every 2 m. The flags are stored at the position of the middle most flag. Ruchi was given the responsibility of placing the flags. Ruchi kept her books where the flags were stored. She could carry only one flag at a time. How much distance did she cover in completing this job and returning back to collect her books ?\u00a0\u00a0[4]<\/strong><\/p>\nQuestion 29.\u00a0\u00a0<\/strong>
\nShow \\(\\frac { { cot }^{ 2 }A }{ 1+cosecA } +1=cosec A\\)\u00a0\u00a0[4]<\/strong><\/p>\nQuestion 30.<\/strong>
\nFind the value of p, if the mean of the following distribution is 7.5.\u00a0\u00a0[4]<\/strong><\/p>\n\n\n\nX<\/strong><\/td>\n| 3<\/td>\n | 5<\/td>\n | 7<\/td>\n | 9<\/td>\n | 11<\/td>\n | 13<\/td>\n<\/tr>\n | \nf<\/strong><\/td>\n| 6<\/td>\n | 8<\/td>\n | 15<\/td>\n | p<\/td>\n | 8<\/td>\n | 4<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n SOLUTIONS<\/strong><\/span>
\nSection – A<\/strong><\/p>\nAnswer 1.<\/strong>
\nHere A(2,3) B(5,8) and C(2,1).
\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1965\/45129095841_11b0589bd7_o.png?resize=686%2C264&ssl=1\") <\/p>\nAnswer 2.<\/strong>
\nLet, radius of circle OP = r
\n\u2235 Tangent to a circle is perpendicular to the radius
\n\u2234 In \u2206POQ, using Pythagoras theorem
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1911\/45080099812_d0c6f485e1_o.png?resize=252%2C155&ssl=1\")
\nOQ\u00b2 = PQ\u00b2 + OP\u00b2
\n25\u00b2 = 24\u00b2 + r\u00b2
\n\u21d2 r\u00b2 = 25\u00b2 – 24\u00b2
\n\u21d2 r\u00b2 = (25 + 24)(25 – 24)
\n\u21d2 r\u00b2 = 49
\n\u21d2 r = 7 cm<\/p>\nAnswer 3.<\/strong>
\nWe know that (2 + \u221a3) and (2 – \u221a3) are irrational numbers
\n(2 + \u221a3) + (2 – \u221a3)
\n2 + \u221a3 + 2 – \u221a3 = 4<\/p>\nAnswer 4.<\/strong>
\nGiven system of equations is,
\n3x – y – 8 =0
\nand 4x – 12y + 16 = 0
\nFor infinite solution
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1966\/44217210065_f519b8a6cc_o.png?resize=463%2C321&ssl=1\") <\/p>\nAnswer 5.<\/strong>
\nWe have
\n3x\u00b2 – 2x – 1 = 0
\nIf x = 1,
\nThen,
\n3 \u00d7 (1)2<\/sup> – 2 \u00d7 1 – 1 = 0
\n3 – 2 – 1 = 0
\n3 – 3 = 0
\n0 = 0
\nL.H.S = R.H.S
\nHence, x = 1 is a solution of the given equation.<\/p>\nAnswer 6<\/strong>.
\nWe have,
\ntan R =\\(\\frac { 24 }{ 7 } \\)
\nFrom \u2206PQR, tan R = \\(\\frac { PQ }{ RQ } \\)
\nIn \u2206PQR, PR\u00b2 = PQ\u00b2 + RQ\u00b2
\nPR\u00b2 = 24\u00b2 + 7\u00b2
\nPR\u00b2 = 576+ 49
\nPR\u00b2 = 625
\nPR = \u221a625
\nPR 25
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1947\/45080099662_0ea4b859d1_o.png?resize=132%2C179&ssl=1\")
\nsec R = \\(\\frac { PR }{ RQ } =\\frac { 25 }{ 7 }\\)
\nsin R = \\(\\frac { PQ }{ PR } =\\frac { 24 }{ 25 } \\)<\/p>\nSECTION-B<\/strong><\/p>\nAnswer 7.<\/strong>
\nWe have, A( 1, 2), O(0, 0) and C(a, b) are collinear.
\nSince, the given points are collinear, therefore, area of triangle formed by them is zero.
\ni.e., ar (\u2206AOB) = 0
\n\u21d2 \\(\\frac { 1 }{ 2 } \\) x1(y2 – y3) + x2(y3 – yx) + x3{yx – y2) 1=0
\n\u21d2 \\(\\frac { 1 }{ 2 } \\) 1(0 – b) + 0(b – 2) + a(2 – 0) i =0
\n\u21d2 – b + 2a = 0
\n\u21d2 b = 2a.<\/p>\nAnswer 8.<\/strong>
\nGiven, Sania takes 24 minutes for one round and Ravi takes 16 minutes for one round. It means that Ravi takes lesser time than Sonia to drive one round of field.
\nNow,
\nRequired time = L.C.M. of the time taken by Sonia and Ravi for completing one round
\n= L.C.M. of 24 and 16
\n\u2235 24 = 23<\/sup> x 3 and 16 = 24<\/sup>
\n\u2234 L.C.M. = 24<\/sup> x 3 = 16 x 3 = 48
\nHence, Ravi and Sonia will meet each other at the starting point after, 48 minutes.<\/p>\nAnswer 9.<\/strong>
\nNumbers are 1, 2, 3,…., 25.
\n\u2234 Total numbers n(S) = 25
\nPrime numbers between 1 to 25 are 2, 3, 5, 7, 11, 13, 17, 19, 23
\n\u21d2 Total prime numbers = n(E) = 9
\n\u2234 P(Prime Number) = \\(\\frac { n(E) }{ n(S) } \\quad \\frac { 9 }{ 25 }\\)<\/p>\nAnswer 10.<\/strong>
\nGiven numbers are -2,-1, 0, 1, 2
\n\u21d2 Total numbers = 5
\nHere, (-2)2<\/sup> = 4, (-1)2<\/sup> = 1, 02<\/sup> = 0, 12<\/sup> = 1, 22<\/sup> = 4
\nFor x = – 1, 0, 1\u00a0 x2\u00a0<\/sup>< 2
\n\u21d2 Favourable outcomes = 3
\np(x\u00b2 < 2) = \\(\\frac { Favourable\\quad outcomes }{ Total\\quad outcomes } = \\frac { 3 }{ 5 } \\)<\/p>\nAnswer 11.<\/strong>
\nGiven equation is,
\nx\u00b2 – 2x + k = 0
\nHere a = 1, b = -2, c = k
\nWe know, D = b2<\/sup> – 4ac
\n\u21d2 D = (-2)2<\/sup> – 4 \u00d7 1 \u00d7 k
\n\u21d2 D = 4 – 4k<\/p>\nFor real roots,
\nD =0
\n\u21d2 4 – 4 k =0
\n\u21d2 – 4k = – 4
\n\u21d2 k = \\(\\frac { -4 }{ -4 } \\)
\n\u21d2 k = 1<\/p>\n Answer 12.<\/strong>
\nGiven A.P. is, 4, 9, 14, 19, 24
\nHere, a = 4
\nd = 9 – 4=5
\nWe know
\nan<\/sub>\u00a0= a + (n – 1) d
\na30<\/sub><\/span>\u00a0= 4 + (30 – 1) 5
\n\u21d2 a30<\/sub> = 4 + 29 x 5
\n\u21d2 a30<\/sub> = 4 + 145
\n\u21d2 a30<\/sub> = 149
\nAnd an<\/sub> = a + (10 – 1) d = 4 + (30 -1 )5
\n= 4 + (9 x 5) = 4 + 45
\n\u21d2 a30<\/sub> = 49
\n\u2234 a30<\/sub> – a10<\/sub> = 149 – 49
\n= 100<\/p>\nSECTION-C<\/strong><\/p>\nAnswer 13.<\/strong>
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1917\/44217209855_7612dff356_o.png?resize=148%2C203&ssl=1\")
\nAP =AQ
\n\u21d2 AP2<\/sup> = AQ2<\/sup>
\nLet (x1<\/sub>, y1<\/sub>) = (2, – 4)
\n(x2<\/sub>, y2<\/sub>) = (3, 8)
\n(x3<\/sub>, y3<\/sub>) = (-10, y)
\nThen,
\n(x2<\/sub> – x1<\/sub>)2<\/sup> + (y2<\/sub> – y1<\/sub>)2<\/sup> = (x3<\/sub> – x1<\/sub> )2<\/sup> + (y3<\/sub> – x1<\/sub>)2<\/sup>
\n(3 – 2)2<\/sup> + (8 + 4)2<\/sup> = (-10 – 2)2<\/sup> + (y + 4)2<\/sup>
\n(1)2<\/sup> + (12)2<\/sup> = (12)2<\/sup> + (y + 4)2<\/sup>
\n1 + 144 = 144 + (y + 4)2<\/sup>
\n1 = (y + 4)2<\/sup>
\n\u00b11 = y+4
\n\u21d2 y = -1 – 4 or y = 1 – 4
\n\u21d2 y=-5 or y = -3
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1959\/43315809580_56f433cca1_o.png?resize=604%2C214&ssl=1\") <\/p>\nAnswer 14.<\/strong>
\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1962\/44217209765_3447711d51_o.png?resize=810%2C464&ssl=1\") <\/p>\nAnswer 15.<\/strong>
\nGiven : A circle with centre at O. AB and CD are tangents drawn at the end of diameter.
\nTo prove : AB || CD
\nProof : We know that a tangent at any point of a circle is perpendicular to the radius throught the point of contact.
\n\u2234 OM \u22a5 AB and ON \u22a5 CD
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1940\/45080099172_2477825cc3_o.png?resize=213%2C191&ssl=1\")
\n\u21d2 \u2220OMB = 90\u00b0
\n\u2220OMB = 90\u00b0
\n\u2220OND = 90\u00b0
\n\u2220ONC = 90\u00b0
\n\u2220OMA = \u2220OND
\n\u2220OMB = \u2220ONC
\nThese are the pair of alternate interior angles.
\nSince, alternate angles are equal, the line AB and CD are parallel to each other.
\ni.e, AB || CD<\/p>\nAnswer 16.<\/strong>
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1942\/44217209475_ca7b54aabd_o.png?resize=300%2C171&ssl=1\")
\nPrime factorisation of the given numbers are,
\n396 = 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 11 = 2\u00b2 \u00d7 3\u00b2 \u00d7 11
\n429 = 3 \u00d7 143
\n561 = 3 \u00d7 187
\n\u2234 LCM (396, 429, 561) = 2\u00b2 \u00d7 3\u00b2 \u00d7 143 \u00d7 187 \u00d7 11 = 10589436<\/p>\nAnswer 17.<\/strong>
\nGiven, Diameter of front wheel = 80 cm
\n\u2234 Radius, r = \\(\\frac { 80 }{ 2 }\\) = 40 cm
\nand Diameter of rear wheel = 2 m
\n\u2234 Radius, R = \\(\\frac { 2 }{ 2 }\\) = 1 m = 100 cm
\nNumber of revolutions covered by front wheel = 140.
\nLet the number of revolutions covered by rear wheel be n.
\nAccording to the question,
\nDistance covered by real wheel in n revolutions = Distance covered by front wheel in 140 revolutions
\n\u21d2 n x 2\u03c0R = 140 x 2\u03c0r
\nn \u00d7 R = 140 x r
\nn x 100 = 140 x 40
\nn = \\(\\frac { 140\\times 40 }{ 100 }\\) = 56
\nThus, the number of revolutions covered by rear wheel is 56.
\nOR<\/strong>
\nGiven : Radius, OA = 3.5 cm
\nand OD = 2 cm
\nArea of shaded region = Area of quadrant OACB – Area of \u2206DOB
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1904\/45080098962_858cc06c40_o.png?resize=317%2C215&ssl=1\")
\nHence, area of shaded region is 6-125 cm2<\/sup>.<\/p>\nAnswer 18.<\/strong>
\nGiven : Length of cuboid, l = 6 m
\nBreadth of cuboid, b = 11 m
\nHeight of cuboid, h = 5 m
\nRadius of cylinder, r = \\(\\frac { 3.5 }{ 10 } \\) m = \\(\\frac { 7 }{ 2 } \\) m
\nLet height of water level in the cylindrical tank be H m
\nSince, water in rectangular tank is transferred to cylindrical tank.
\n\u2234 Volume of cuboid = Volume of cylinder of height H
\nl \u00d7b \u00d7 h =\u03c0r\u00b2H
\n6 x 11 x 5 = \\(\\frac { 22 }{ 7 } \\times \\frac { 7 }{ 2 } \\times \\frac { 7 }{ 2 }\\) \u00d7 H
\n\\(\\frac { 30\\times 2 }{ 7 } \\) = H
\nH = \\(\\frac { 60 }{ 7 }\\) m = 8.57 m.
\nOR<\/strong>
\nGiven : Diameter of the copper rod = 1 cm
\nSo, Radius (r) of the copper rod = 1\/2 cm
\nLength of copper rod, h = 8 cm
\nLength of wire, H = 18 m = 18 x 100 cm
\nLet R be the radius of the cross-section (circular) of the wire.
\nWe know that volume of cylinder = \u03c0r\u00b2h Now, according to the question
\nNow, according to the question,
\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1946\/43315646920_21e9c31904_o.png?resize=662%2C446&ssl=1\") <\/p>\nAnswer 19.<\/strong>
\nLet p(x) = x\u00b2 + 13x + 30
\n= x\u00b2 + 10x + 3x + 30
\n= x(x + 10) + 3(x + 10)
\n=(x + 3) (x + 10)
\n\u21d2 x + 3 = 0; x + 10 = 0
\n\u21d2 x = – 3, x = – 10
\nLet \u03b1 = – 3
\nand \u03b2 = – 10
\nNow, \u03b1 + \u03b2 = \\(-\\frac { b }{ a }\\)
\n– 3 – 10 = \\(-\\frac { (-13) }{ 1 } \\)
\n– 13 = – 13
\nand \u03b1\u03b2 = \\(\\frac { c }{ a } \\)
\n(-3) (-10) = \\(\\frac { 30 }{ 1 }\\)
\n30 = 30
\nThus, the releationship between the zeroes and coefficients is verified.<\/p>\nAnswer 20.<\/strong>
\nGiven : First term of A.P., a = 17
\nLast term, l = 350
\nCommon difference, d = 9
\nWe know,
\nan<\/sub> = a + (n – 1) d
\n\u2235 350 is the nth term of given A.P.,
\n\u2235 350 = 17 + (n – 1) 9
\n350 – 17 = (n – 1) 9
\n333 = (n – 1) 9
\n\\(\\frac { 333 }{ 9 } \\) = n – 1
\n37 = n – 1
\n37 + 1 =n
\n38 = n
\nWe know that sum of n terms of A.P is given as,
\nSn<\/sub> = \\(\\frac { n }{ 2 } \\)[a + l]
\n\u21d2 S38<\/sub> = \\(\\frac { 38 }{ 2 }\\)[17 + 350]
\n= 19 [367] = 6973
\nn = 38 and Sn<\/sub> = 6973.
\nOR<\/strong>
\nThe first number greater than 10 which when divided by 4 leaves a remainder 3 is 11.
\nSo, the next number will be 11 + 4 = 15
\nThe other numbers will be 15 + 4 = 19, 19 + 4 = 23….
\nThus, A.P. = 11, 15, 19, 23 …..
\nThe last term of this A.P. will be 299.
\nWe have to find n. i.e., number of terms in the A.P.
\nWe know,
\na + (n – 1 )d = an<\/sub>
\n\u21d2 11 + (n – 1)4 =299
\n\u21d2 11 + 4n – 4 = 299
\n\u21d2 7 + 4n =299
\n\u21d2 4n = 299 – 7
\n\u21d2 4n = 292
\n\u21d2 n = \\(\\frac { 292 }{ 4 } \\)
\n\u21d2 n = 73
\n\u2234 numbers lie between 10 and 300 which divided by 4 leaves a remainder 3.<\/p>\nAnswer 21.<\/strong>
\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1917\/44217209335_e0396f8107_o.png?resize=740%2C160&ssl=1\") <\/p>\nAnswer 22.<\/strong><\/p>\n | | | | ![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1935\/45112813821_d9fb97bda5_o.png?resize=177%2C165&ssl=1\")
<\/p>\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1942\/45112813611_6940ba6370_o.png?resize=326%2C63&ssl=1\")
<\/p>\n