Time Allowed : 3 hours<\/strong> General Instructions<\/strong><\/span><\/p>\n SECTION-A<\/strong><\/p>\n Question 1.<\/strong><\/span> Question 2.<\/strong><\/span> Question 3.<\/strong><\/span> Question 4.<\/strong><\/span> Question 5.<\/strong><\/span> Question 6.<\/strong><\/span> SECTION-B<\/strong><\/p>\n Question 7.<\/strong><\/span> Question 8.<\/strong><\/span> Question 9.<\/strong><\/span> Question 10.<\/strong><\/span> Question 11.<\/strong><\/span> Question 12.<\/strong><\/span> SECTION-C<\/strong><\/p>\n Question 13.<\/strong><\/span> Question 14.<\/strong><\/span> OR<\/strong><\/p>\n If the points A( 1, – 2), B(2, 3), C(a, 2) and D(- 4, – 3) form a parallelogram, find the value of a and height of the parallelogram taking AB as base.<\/p>\n Question 15.<\/strong><\/span> Question 16.<\/strong><\/span> Question 17.<\/strong><\/span> OR<\/strong> Question 18.<\/strong><\/span> OR<\/strong><\/p>\n A wall 24 m long, 0.4 m thick and 6 m high is constructed with the bricks each of dimensions 25 cm x 16 cm x 10 cm. If the mortar occupies 1\/10 of the volume of the wall, then find the number of bricks used in constructing the wall.<\/p>\n Question 19.<\/strong><\/span> OR<\/strong><\/p>\n Find the sum of first 17 terms of an A.P. whose 4th and 9th terms are – 15 and – 30 respectively.<\/p>\n Question 20.<\/strong><\/span> Question 21.<\/strong><\/span> Question 22.<\/strong><\/span> SECTION-D<\/strong><\/p>\n Question 23.<\/strong><\/span> OR<\/strong><\/p>\n A boy is standing on the ground and flying a kite with 100 m of string at an elevation of 30\u00b0. Another boy is standing on the roof of a 10 m high building and is flying his kite at an elevation of 45\u00b0. Both the boys are on opposite sides of both the kites. Find the length of the string that the second boy must have so that the two kites meet.<\/p>\n Question 24.<\/strong><\/span> Question 25.<\/strong><\/span> Question 26.<\/strong><\/span> OR<\/strong><\/p>\n A gulab jamun, contains sugar syrup about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm.<\/p>\n Question 27.<\/strong><\/span> OR<\/strong><\/p>\n Solve graphically x – 2y = 1; 2x + y = 7. Also find the coordinates of points where the lines meet X-axis.<\/p>\n Question 28.<\/strong><\/span> Question 29.<\/strong><\/span> Question 30.<\/strong><\/span> SOLUTIONS<\/strong><\/span><\/p>\n SECTION-A<\/strong><\/p>\n Solution 1.<\/strong><\/span> Solution 2.<\/strong><\/span> Solution 3.<\/strong><\/span> Solution 4.<\/strong><\/span> Solution 5.<\/strong><\/span> Solution 6.<\/strong><\/span> SECTION-B<\/strong><\/p>\n Solution 7:<\/strong><\/span> Solution 8:<\/strong><\/span> Solution 9:<\/strong><\/span> Solution 10:<\/strong><\/span> Solution 11:<\/strong><\/span> Solution 12:<\/strong><\/span> SECTION-D<\/strong><\/p>\n Solution 13:<\/strong><\/span> Solution 14:<\/strong><\/span> Solution 15:<\/strong><\/span> Solution 16:<\/strong><\/span> Solution 17:<\/strong><\/span> Solution 18:<\/strong><\/span> Solution 19:<\/strong><\/span> Solution 20:<\/strong><\/span> Solution 21:<\/strong><\/span> Solution 22:<\/strong><\/span> SECTION-D<\/strong><\/p>\n Solution 23:<\/strong><\/span> Solution 24:<\/strong><\/span>
\nMaximum Marks : 80<\/strong><\/p>\n\n
\nIf two integers a and b are written as a = x3<\/sup>y2<\/sup> and b = xy4<\/sup>; x, y are prime numbers, then find H.C.F. (a, b).<\/p>\n
\nFind the perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0).<\/p>\n
\nIf tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80\u00b0, then find the measure of \u2220POA.<\/p>\n
\nDetermine the pair of linear equations from 2x + 3y = 5, y + \\(\\\\ \\frac { 2 }{ 3 } \\) x = 5, 4x = 6y + 10, which has infinite solution.<\/p>\n
\nThe first term of an A.P. is p and its common difference is q. Find its 10th term.<\/p>\n
\nIf 9 sec A = 41, find cos A and cot A.<\/p>\n
\nFind a point which is equidistant from the points A(- 5, 4) and B(- 1, 6). How many such points are there ?<\/p>\n
\nWhich term of A.P. 129, 125, 121, … is its first negative term ?<\/p>\n
\nWrite the denominator of the rational number \\(\\\\ \\frac { 257 }{ 5000 } \\) in the form 2m<\/sup> x 5n<\/sup>, where m, n are non – negatives. Hence, write its decimal expansion without actual division.<\/p>\n
\nIt is known that a box of 600 electric bulbs contains 12 defective bulbs. One bulb is taken out at random from this box. What is the probability that it is a non-defective bulb ?<\/p>\n
\nIn a lottery of 50 tickets numbered 1 to 50, one ticket is drawn. Find the probability that the drawn ticket bears a prime number.<\/p>\n
\nFor what value of k the quadratic equation (2k + 3) x2<\/sup> + 2x – 5 = 0 has equal roots ?<\/p>\n
\nFind the smallest number which when divided by 161, 207 and 184 leaves remainder 21 in each case.<\/p>\n
\nFind the values of k if the points A(k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear.<\/p>\n
\nO is the point of intersection of the diagonals AC and BD of a trapezium ABCD with AB || DC. Through O, a line segment PQ is drawn parallel to AB meeting AD in P and BC in Q. Prove that PO = QO.<\/p>\n
\nTwo concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.<\/p>\n
\nThe diameter of coin is 1 cm (fig.). If four such coins be placed on a table so that the rim of each touches that of the other two, find the area of the shaded region (Take \u03c0 = 3.1416).
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1909\/43319737960_9c3a07b93a_o.png?resize=153%2C145&ssl=1\")
<\/p>\n
\nArea of a sector of a circle of radius 36 cm is 54\u03c0 cm2<\/sup>. Find the length of the corresponding arc of the sector.<\/p>\n
\nA hemispherical tank of diameter 3 m, is full of water. It is being emptied by a pipe at the rate of \\(3 \\frac { 4 }{ 7 } \\) litres per second. How much time will it take to make the tank half empty ?<\/p>\n
\nThe first term of an A.P. is 2 and the last term is 50. The sum of all these terms is 442. Find the common difference.<\/p>\n
\nSolve for x :
\n\\({ \\left( \\frac { 2x+1 }{ x+1 } \\right) }^{ 4 }-13{ \\left( \\frac { 2x+1 }{ x+1 } \\right) }^{ 2 }+36=0\\)<\/p>\n
\nIf cosec A + cot A = p, then prove that sec A = \\(\\frac { { p }^{ 2 }+1 }{ { p }^{ 2 }-1 } \\)<\/p>\n
\nFind mode of the following data :
\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1960\/43319737840_2fbe1c756c_o.png?resize=275%2C175&ssl=1\")
<\/p>\n
\nFrom the top of a lighthouse, the angles of depression of two ships on the opposite sides of it are observed to be a and p. If the height of the lighthouse is h metres and the line joining the ships passes through the foot of the lighthouse, show that the distance between the ships is \\(\\frac { h\\left( tan\\alpha +tan\\beta \\right) }{ tan\\alpha \\quad tan\\beta } \\) meters<\/p>\n
\nConstruct a right triangle in which sides (other than the hypotenuse) are of lengths 8 cm and 6 cm. Then construct another triangle whose sides are 3\/4 times the corresponding sides of the first triangle.<\/p>\n
\nState and prove the converse of Pythagoras theorem.<\/p>\n
\nA hemispherical bowl of internal radius 9 cm is full of liquid. The liquid is to be filled into cylindrical shaped bottles each of radius 1.5 cm and height 4 cm. How many bottles are needed to empty the bowl ?
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1943\/45132881671_65e781dd82_o.png?resize=108%2C147&ssl=1\")
<\/p>\n
\nA boat goes 12 km downstream and 26 km upstream in 8 hours. It can go 16 km upstream and 32 km downstream in same time. Find the speed of boat in still water and the speed of stream.<\/p>\n
\nA sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each prize.<\/p>\n
\nShow that :
\n\\(\\frac { sinA-cosA+1 }{ sinA+cosA-1 } =\\frac { 1 }{ secA-tanA } \\)<\/p>\n
\nThe median of the following data is 20.75. Find the missing frequencies ‘x’ and \u2018y’, if the total frequency is 100.
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1911\/43319737640_4f34f107fb_o.png?resize=257%2C253&ssl=1\")
<\/p>\n
\nGiven : a = x3<\/sup>y2<\/sup>
\nb = xy4<\/sup>
\nH.C.F.(a, b) = xy2<\/sup>.<\/p>\n
\nGiven vertices are A(3, 0), B(0, 4) and 0(0, 0).
\nOA = 3 units
\nOB = 4 units
\nIn \u2206AOB, \u2220O = 90\u00b0
\n\u2234From Pythagoras theorem,
\nAB2<\/sup> = OA2<\/sup> + OB2<\/sup> = 32<\/sup> + 42<\/sup>
\nAB = \u221a9+16 = 5 unit
\nPerimeter of \u2206OAB = OA + OB + AB
\n= 3 + 4 + 5 = 12 unit.
\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1903\/44412510984_1a2fd05e0f_o.png?resize=190%2C206&ssl=1\")
<\/p>\n
\nGiven :
\n\u2220APB = 80\u00b0
\nIn \u2206OAP and \u2206OBP,
\n\u2220B = \u2220A [each 90\u00b0]
\nOA = OB [radii]
\nOP = OP [common]
\nBy RHS congruency rule
\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1924\/44412510734_9da1b00e9e_o.png?resize=219%2C130&ssl=1\")
\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1922\/44412510894_51a6a86db8_o.png?resize=400%2C207&ssl=1\")
<\/p>\n
\nGiven equations are,
\n2x + 3y – 5 = 0
\ny + \\(\\\\ \\frac { 2 }{ 3 } \\)x – 5=0
\nand 4x + 6y = 10
\n=> 2x + 3y – 5 = 0
\n2x + 3y – 15 = 0
\nand 4x + 6y – 10 =0
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1942\/44412510654_5e6d86a6f1_o.png?resize=782%2C152&ssl=1\")
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1948\/44412510634_48c2c9ab72_o.png?resize=453%2C93&ssl=1\")
<\/p>\n
\nGiven : The first term of A.P. = a
\nand common difference = q
\nwe know, an<\/sub> = a + (n – 1)d
\n\u2234 a10<\/sub> = p + (10 – 1)d
\n=> a10<\/sub> = P + 9d.<\/p>\n
\nGiven : 9 sec A = 41
\n=> sec A = \\(\\\\ \\frac { 41 }{ 9 } \\)
\ncos A = \\(\\\\ \\frac { 1 }{ sec A } \\)
\ncos A = \\(\\\\ \\frac { 9 }{ 41 } \\)
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1957\/44412510494_7c16069f5b_o.png?resize=763%2C269&ssl=1\")
<\/p>\n
\nLet the point P(x, y) be equidistant from the points A (- 5, 4) and B(- 1, 6).
\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1967\/45133494301_f1c02d1e5d_o.png?resize=165%2C195&ssl=1\")
\n\\(\\left| \\sqrt { { (x+5) }^{ 2 }+{ (y-4) }^{ 2 } } \\right| =\\left| \\sqrt { { (x+1) }^{ 2 }+{ (y-6) }^{ 2 } } \\right| \\)
\nSquaring both sides
\n(x + 5)2<\/sup> + (y – 4)2<\/sup> = (x + 1)2<\/sup> + (y – 6)2<\/sup>
\nx2<\/sup> + 10x + 25 + y2<\/sup> – 8y + 16 = x2<\/sup> + 2x + 1 + y2<\/sup> – 12y + 36
\n10x – 2x – 8y + 12y + 41 – 37 = 0
\n8x + 4y + 4 = 0
\n2x + y + 1 =0
\nThe points lying on 2x + y + 1 = 0 are equidistant from A and B.<\/p>\n
\nGiven A.P. is 129, 125, 121,…
\nHere, a = 129, d = 125 – 129 = 121 – 125 = – 4
\nLet nth term be the first negative term.
\nThen Tn<\/sub> < 0.
\nWe know, nth term, Tn<\/sub> = a + (n – 1)d
\n= 129 + (n -1) (- 4)
\n= 129 – 4n + 4
\n= 133 – 4n
\n\u2234Tn<\/sub>< 0
\n\u2235133 – 4n < 0
\n=> 133 < 4n
\n=> 4n > 133
\n=> n =\\(\\\\ \\frac { 133 }{ 4 } \\)
\n= \\(33 \\frac { 1 }{ 4 } \\)
\nHence, 34th term will be the first negative term.<\/p>\n
\nWe have,
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1943\/45133494111_4d084548d7_o.png?resize=239%2C215&ssl=1\")
<\/p>\n
\nTotal bulbs = 600
\nNumber of defective bulbs = 12
\nNumber of non-defective bulbs = 600 – 12 = 588
\nprobability of a non defective bulb = \\(\\frac { Number\\quad of\\quad non\\quad defective\\quad bulbs }{ total\\quad bulbs } \\)
\n= \\(\\\\ \\frac { 588 }{ 600 } \\)
\n= \\(\\\\ \\frac { 147 }{ 150 } \\)<\/p>\n
\nTotal tickets 1, 2, 3, 4, …, 50
\n=> n(S) = 50
\nPrime number tickets = 2, 3, 5, 7,11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.
\nn(E) = 15
\nProbability that the ticket bears a prime number
\nP(E) = \\(\\\\ \\frac { n(E) }{ n(S) } \\)
\n= \\(\\\\ \\frac { 15 }{ 50 } \\)
\n= \\(\\\\ \\frac { 3 }{ 10 } \\)
\n= 0.3<\/p>\n
\nGiven quadratic equation is,
\n(2k + 3)x2<\/sup> + 2x – 5 = 0
\nOn comparing the equation with ax2<\/sup> + bx + c = 0, we get
\na = 2k + 3, b = 2, c = – 5
\nFor equal roots,
\nb2<\/sup> – 4ac = 0
\n(2)2<\/sup> – 4(2k + 3) ( – 5) = 0
\n4 + 20(2k + 3) = 0
\n4 + 40k + 60 = 0
\n40k = – 64
\nk = \\(\\\\ \\frac { -64 }{ 40 } \\)
\nk = \\(\\\\ \\frac { -8 }{ 5 } \\)<\/p>\n
\nGiven numbers are 161, 207 and 184.
\n161 = 7 x 23
\n207 = 3 x 3 x 23
\n\u2234 184 = 2 x 2 x 2 x 23
\nL.C.M. (161, 207, 1841) = 2 x 2 x 2 x 3 x 3 x 23 x 7
\n= 11592
\nSmallest number which when divided by 161, 207 and 184, leaves remainder 21 in each case
\n= 11592 + 21
\n= 11613. Ans.<\/p>\n
\nGiven points are A(k + 1, 2k), B(3k, 2k + 3), C(5k – 1, 5k)
\nIf the points A, B and C are collinear, then
\nar (\u2206ABC) = 0
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1957\/45194369511_b7f2741a44_o.png?resize=478%2C389&ssl=1\")
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1948\/43381005090_cb4d72dba3_o.png?resize=772%2C739&ssl=1\")
\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1910\/45194369771_f2fc3c4fe9_o.png?resize=537%2C415&ssl=1\")
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1972\/43381005490_52c098a0b2_o.png?resize=519%2C203&ssl=1\")
<\/p>\n
\nGiven, ABCD is trapezium in which
\nAB || CD || PQ
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1915\/45194369361_77e6865daa_o.png?resize=777%2C715&ssl=1\")
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1971\/43381004750_3b572fe5ea_o.png?resize=761%2C213&ssl=1\")
<\/p>\n
\nLet O be the common centre of two concentric circles and let AB be a chord of large circle touching the smaller circle at P.
\nConstruction : Join OP
\nSince OP is the radius of the smaller circle and AB is tangent to this circle at P,
\n\u2234 OP \u22a5 AB
\nWe know that the perpendicualr drawn from the centre of a circle to any chord of the circle bisects the chord.
\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1939\/43381004480_6e02fb1acf_o.png?resize=791%2C305&ssl=1\")
<\/p>\n
\nGiven : Diameter of coin = 1 cm
\nRadius of coin, r = 0.5 cm
\nPQ = QR = RS = PS
\n= 2 x 0.5 = 1 cm
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1908\/45194368851_9731a9605d_o.png?resize=166%2C172&ssl=1\")
\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1957\/45194369161_cd9bb5c0ff_o.png?resize=777%2C562&ssl=1\")
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1979\/43381004240_68c3b6c7a8_o.png?resize=607%2C196&ssl=1\")
<\/p>\n
\nGiven : Diameter of hemispherical tank = 3 cm
\nRadius, r = \\(\\\\ \\frac { 3 }{ 2 } \\)
\nNow, Volume of hemispherical tank = \\(\\frac { 2 }{ 3 } { \\pi r }^{ 3 } \\)
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1928\/43381003820_602241fea3_o.png?resize=775%2C496&ssl=1\")
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1945\/43381004050_453aa57800_o.png?resize=577%2C338&ssl=1\")
\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1973\/45194368721_26425d91f6_o.png?resize=605%2C243&ssl=1\")
<\/p>\n
\nGiven, a = 2, an<\/sub> = 50, Sn<\/sub> = 442
\nWe know that
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1933\/44283156085_04c500ef11_o.png?resize=444%2C402&ssl=1\")
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1917\/43381003750_f56113f6de_o.png?resize=780%2C492&ssl=1\")
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1939\/43381003530_72418dd84f_o.png?resize=441%2C118&ssl=1\")
<\/p>\n
\nWe have
\n\\({ \\left( \\frac { 2x+1 }{ x+1 } \\right) }^{ 4 }-13{ \\left( \\frac { 2x+1 }{ x+1 } \\right) }^{ 2 }+36=0\\)
\nlet
\n\\({ \\left( \\frac { 2x+1 }{ x+1 } \\right) }^{ 2 }=a\\)
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1971\/44283155805_f80219c7dc_o.png?resize=739%2C538&ssl=1\")
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1964\/43381003200_7aab25058f_o.png?resize=776%2C330&ssl=1\")
<\/p>\n
\nGiven : cosec A + cot A = p ….(i)
\nWe know,
\ncosec\u00b2 A – cot\u00b2 A = 1
\n=> (cosec A + cot A) (cosec A – cot A) = 1 [\u2235 a\u00b2 – b\u00b2 = (a+b)(a-b)]
\n=> (p) (cosec A – cot A) = 1
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1964\/43381003070_fea1ecfc37_o.png?resize=784%2C713&ssl=1\")
<\/p>\n
\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1909\/44283155505_56f3b7e418_o.png?resize=372%2C356&ssl=1\")
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1957\/43381002940_8dda11ba95_o.png?resize=268%2C146&ssl=1\")
<\/p>\n
\nLet PO be the lighthouse and A, B be the position of the two ships.
\nPO = h m
\nAngle of depression of ship at the point A = a\u00b0.
\nAngle of depression of ship at the point B = P\u00b0.
\nIn \u2206OPA, \u2220P = 90\u00b0
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1967\/31320923398_ff3f7dd684_o.png?resize=213%2C198&ssl=1\")
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1965\/31320923688_97325f02bf_o.png?resize=806%2C517&ssl=1\")
\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1916\/31320923808_26a29297c9_o.png?resize=574%2C350&ssl=1\")
\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1957\/31320924028_0e31e093c8_o.png?resize=791%2C381&ssl=1\")
<\/p>\n
\nSteps of construction :<\/strong>
\n(1) Draw a line segment BC = 8 cm and make a right angle (\u2220DBC) at the point B.
\n(2) Taking B as centre and radius 6 cm, draw an arc which intersect BD at the point A.
\n(3) Join AC. Thus, \u2206ABC is obtained.
\n(4) Draw a ray BX such that \u2220CBX is an acute angle.
\n(5) Make four points B1<\/sub> B2<\/sub>, B3<\/sub> and B4<\/sub> such that
\nBB1<\/sub> = B1<\/sub>B2<\/sub> = B2<\/sub>B3<\/sub> = B3<\/sub>B4<\/sub>
\n(6) Join B4<\/sub>C.