Paper:



Paper: |CBSE Sample paper 3 (Silver series) - X - Math -SA II | |

|Total marks of the paper: |90 |

|Total time of the paper: |3.5 hrs |

|General Instructions: |

| |

|1. All questions are compulsory. |

| |

|2. The question paper consists of 34 questions divided into four sections A, B, C, and D. Section – A comprises of 8 questions of |

|1 mark each, Section – B comprises of 6 questions of 2 marks each, Section – C comprises of 10 questions of 3 marks each and |

|Section – D comprises of 10 questions of 4 marks each. |

| |

|3. Question numbers 1 to 8 in Section – A are multiple choice questions where you are to select one correct option out of the |

|given four. |

| |

|4. There is no overall choice. However, internal choice has been provided in 1 question of two marks, 3 questions of three marks |

|each and 2 questions of four marks each. You have to attempt only one of the alternatives in all such questions. |

| |

|5. Use of calculator is not permitted. |

| |

|6. An additional 15 minutes has been allotted to read this question paper only. |

| |

|Questions: |

| |

|1] |The 9th term of an AP is 449 and 449th term is 9. The term |[Marks:1] |

| |which is equal to zero is | |

| |A. | |

| |459th | |

| | | |

| |B. | |

| |502th | |

| | | |

| |C. | |

| |501th | |

| | | |

| |D. | |

| |458th | |

| | | |

| | | |

|2] |Two circles touch each other externally at C and AB is a |[Marks:1] |

| |common tangent to the circles. Then, [pic]ACB= | |

| |A. | |

| |300 | |

| | | |

| |B. | |

| |450 | |

| | | |

| |C. | |

| |600 | |

| | | |

| |D. | |

| |90( | |

| | | |

| | | |

|3] |The length of an arc that subtends an angle of 24o at the |[Marks:1] |

| |centre of a circle with 5 cm radius is | |

| |A. | |

| |[pic]cm | |

| | | |

| |B. | |

| |[pic]cm | |

| | | |

| |C. | |

| |[pic]cm | |

| | | |

| |D. | |

| |[pic]cm | |

| | | |

| | | |

|4] |An observer 1.5 m tall is 28.5 m away from a tower. The |[Marks:1] |

| |angle of elevation of the top of the tower from his eyes is 450. | |

| |The height of the tower is | |

| |A. | |

| |40 m | |

| | | |

| |B. | |

| |20 m | |

| | | |

| |C. | |

| |10 m | |

| | | |

| |D. | |

| |30 m | |

| | | |

| | | |

|5] |The probability that a randomly chosen number from one to |[Marks:1] |

| |twelve is a divisor of twelve is | |

| |A. | |

| |[pic] | |

| | | |

| |B. | |

| |[pic] | |

| | | |

| |C. | |

| |[pic] | |

| | | |

| |D. | |

| |[pic] | |

| | | |

| | | |

|6] |The value of x, for which the points (x,-1), (2,1) and (4,5) lie |[Marks:1] |

| |on a line is | |

| |A. | |

| |3 | |

| | | |

| |B. | |

| |2 | |

| | | |

| |C. | |

| |0 | |

| | | |

| |D. | |

| |1 | |

| | | |

| | | |

|7] |The ratio in which the point R[pic] divides the join of |[Marks:1] |

| |P(-2,-2) and Q(2,-4) is | |

| |A. | |

| |1:2 | |

| | | |

| |B. | |

| |2:1 | |

| | | |

| |C. | |

| |4:3 | |

| | | |

| |D. | |

| |3:4 | |

| | | |

| | | |

|8] |A pendulum swings through an angle of 300 and describes an |[Marks:1] |

| |arc 8.8cm in length. The length of the pendulum is | |

| |A. | |

| |17 cm | |

| | | |

| |B. | |

| |15.8 cm | |

| | | |

| |C. | |

| |8.8 cm | |

| | | |

| |D. | |

| |16.8 cm | |

| | | |

| | | |

|9] |If the centroid of the triangle formed by points P(a, b),Q(b, c) |[Marks:2] |

| |and R(c, a) is at the origin, then find the value of a + b + c. | |

| |OR | |

| |Find the area of the quadrilateral ABCD whose vertices are | |

| |A(1,1), B(7,-3), C(12,2) and D(7,21) respectively. | |

| | | |

| | | |

|10] |An A.P. consists of 60 terms. If the first and the last term be 7 and 125 respectively, find the 32nd term. |[Marks:2] |

| | | |

| | | |

|11] |Find two consecutive positive integers, sum of whose squares is 25. |[Marks:2] |

| | | |

| | | |

|12] |Prove that the tangents at the extremities of any chord make equal angles with the chord. |[Marks:2] |

| | | |

| | | |

|13] |In given figure, find the area of the circle not included in the |[Marks:2] |

| |rectangle whose sides are 8cm and 6cm respectively. | |

| |[pic] | |

| | | |

| | | |

|14] |The perimeter of a sector of a circle of radius 5.2 cm is 16.4 |[Marks:2] |

| |cm, find the area of the sector. | |

| | | |

| | | |

|15] |If a [pic]b [pic]c, prove that the points (a,a2), (b,b2) and (c,c2) |[Marks:3] |

| |can never be collinear. | |

| | | |

| | | |

|16] |A body falls 8 metres in the first second of its motion, 24 metres in the second, 40 metres in the third |[Marks:3] |

| |second and so on. How long will it take to fall 2048 metres? | |

| | | |

| | | |

|17] |Draw a pair of tangents to a circle of any convenient radius, which are inclined to the line joining the |[Marks:3] |

| |centre of the circle and the point at which they intersect at an angle of 45°. Also write the steps of | |

| |construction. | |

| | | |

| | | |

|18] |From a balloon vertically above a straight road, the angles of depression of two cars at an instant are |[Marks:3] |

| |found to be 450 and 600. If the cars are 100m apart, find the height of the balloon. | |

| |OR | |

| |The angles of elevation of the top of a tower from two points at a distance a and b metres from the base and| |

| |in the same straight line with it are complementary. Prove that the height of the tower is [pic]metres. | |

| | | |

| | | |

|19] |A bag contains 12 balls out of which x are white. |[Marks:3] |

| |(i) If one ball is drawn at random, what is the probability that it will be a white ball? | |

| |(ii) If 6 more white balls are put in the bag, then the probability of drawing a white ball will be double | |

| |than that in (i). Find x. | |

| |OR | |

| |Two customers are visiting a particular shop in the same week (Monday to Saturday). Each is equally likely | |

| |to visit the shop on any day of the week. What is the probability that both will visit the shop on | |

| |(i) the same day? | |

| |(ii) two different days? | |

| |(iii) consecutive days? | |

| | | |

| | | |

|20] |In an equilateral triangle of side 24 cm, a circle is inscribed |[Marks:3] |

| |touching its sides. Find the area of the remaining portion of | |

| |the triangle. (take [pic]=1.732) | |

| |OR | |

| |A round table cover has six equal designs as shown in the | |

| |figure. | |

| |If the radius of the cover is 28 cm, find the cost of making the | |

| |design at the rate of Rs. 0.35 per cm2. | |

| |[pic] | |

| | | |

| | | |

|21] |A canal is 300cm wide and 120cm deep. The water in the canal is flowing with a speed of 20km/h. How much |[Marks:3] |

| |area will it irrigate in 20 minutes if 8cm of standing water is desired? | |

| | | |

| | | |

|22] |From a solid cylinder whose height is 2.4 cm and diameter 1.4cm, a conical cavity of the same height and |[Marks:3] |

| |same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2.(use | |

| |(=3.1416) | |

| | | |

| | | |

|23] |Solve for x, using the quadratic formula: |[Marks:3] |

| |abx2 _ (a2+b2)x + ab=0 | |

| | | |

| | | |

|24] |If D, E and F are the mid points of sides BC, CA and AB |[Marks:3] |

| |respectively of a ∆ABC, whose vertices are A(-4,1), B(6,7) and | |

| |C(2,-9), then prove that, | |

| |Ar. ∆DEF = [pic](Ar ∆ABC). | |

| | | |

| | | |

|25] |Find the value of the middle most term (s) of the A.P. : |[Marks:4] |

| |-11, -7, -3, ...., 49 | |

| |OR | |

| |The sum of the first three terms of an AP is 33. If the product of the first and the third term exceeds the | |

| |second term by 29, find the AP. | |

| | | |

| | | |

|26] |Two circles touch externally. The sum of their area is 130 ( sq. cm and the distance between their centres |[Marks:4] |

| |is 14 cm. Find the radii of the circles. | |

| | | |

| | | |

|27] |The angle of elevation of the top of a tower from a point on the same level as the foot of the tower is |[Marks:4] |

| |[pic]. On advancing p metres towards the foot of the tower, the angle of elevation becomes β. Show that the | |

| |height h of the tower is given by | |

| |[pic] | |

| | | |

| | | |

|28] |A passenger train takes 2 hour less for a journey of 300 km, if its speed is increased by 5km/hr from its |[Marks:4] |

| |usual speed. Find its usual speed. | |

| |OR | |

| |In a class test, the sum of the marks obtained by a student P in Mathematics and Science is 28. Had he got 3| |

| |more marks in Mathematics and 4 marks less in Science, the product of marks obtained in the two subjects | |

| |would have been 180. Find the marks obtained in the two subjects separately. | |

| | | |

| | | |

|29] |Prove that the lengths of tangents drawn from an external point to a circle are equal. |[Marks:4] |

| | | |

| | | |

|30] |In given figure, a circle is inscribed in a quadrilateral ABCD in which [pic]B=900. If AD=23 cm, AB=29 cm, |[Marks:4] |

| |and DS=5 cm, find the radius r of the circle. | |

| |[pic] | |

| | | |

| | | |

|31] |Kavita saved Rs 4 the first week of the year and then increased her weekly savings by Rs 1.75 each week. In |[Marks:4] |

| |what week will her weekly saving be Rs 19.75? | |

| | | |

| | | |

|32] |A solid consisting of a right circular cone, standing on a hemisphere, is placed upright, in a right |[Marks:4] |

| |circular cylinder, full of water, and touches the bottom. Find the volume of water left in the cylinder, | |

| |having been given that the radius of the cylinder is 3 cm and its height is 6 cm, the radius of the | |

| |hemisphere is 2 cm and the height of the cone is 4 cm. | |

| | | |

| | | |

|33] |A bag contains 5 white balls, 7 red balls, 4 black balls and 2 blue balls. One ball is drawn at random from |[Marks:4] |

| |the bag. What is the probability that the ball drawn is | |

| |(i) white or blue (ii) red or black (iii) not white | |

| |(iv) neither white nor black? | |

| | | |

| | | |

|34] |A beautiful park with flower plants is to be developed within the region joining the points A(0, -1), B(6, |[Marks:4] |

| |7), C(-2,3) and D(8, 3) as vertices of a quadrilateral such that AB and CD are diagonals. Show that AB and | |

| |CD bisects each other and AD2 + DB2 = AB2. Find the area of the park if all distances are in km. | |

| |As P.M. of your country, will you make a policy of creating green parks and gardens in every village and | |

| |town of you country? Give reasons. | |

| | | |

| | | |

[pic]

|Paper: |CBSE Sample paper 3 (Silver series) - X - Math -SA II |

|Total marks of the paper: |90 |

|Total time of the paper: |3.5 hrs |

|General Instructions: |

| |

|1. All questions are compulsory. |

| |

|2. The question paper consists of 34 questions divided into four sections A, B, C, and D. Section – A comprises of 8 questions of |

|1 mark each, Section – B comprises of 6 questions of 2 marks each, Section – C comprises of 10 questions of 3 marks each and |

|Section – D comprises of 10 questions of 4 marks each. |

| |

|3. Question numbers 1 to 8 in Section – A are multiple choice questions where you are to select one correct option out of the |

|given four. |

| |

|4. There is no overall choice. However, internal choice has been provided in 1 question of two marks, 3 questions of three marks |

|each and 2 questions of four marks each. You have to attempt only one of the alternatives in all such questions. |

| |

|5. Use of calculator is not permitted. |

| |

|6. An additional 15 minutes has been allotted to read this question paper only. |

| |

|Solutions: |

| |

|1] |a+8d = 449 and, | |

| |a+448d = 9 | |

| |on solving we get, 440d = -440 i.e. d = -1 | |

| |therefore, a-8 = 449 i.e. a = 457 | |

| |let its nth term be zero. | |

| |an = a+(n-1)d | |

| |0 = 457+(n-1)(-1) | |

| |457 = n-1 | |

| |or n = 458 | |

| | | | |

|2] |[pic] | |

| |Lengths of tangents drawn from an external point to a circle are equal. | |

| |PA=PC (from P) | |

| |Therefore, [pic]PAC= [pic]PCA = x (say) | |

| |Also, PC=PB (from P) | |

| |[pic]PBC= [pic]PCB=y | |

| |In ∆ABC, | |

| |[pic]ABC+ [pic]ACB+ [pic]BAC=180 | |

| |y + (x + y) + x = 180 | |

| |2(x + y) = 180 | |

| |x+ y = 90 | |

| |[pic]ACB = 900. | |

| | | | |

|3] |[pic] | |

| |The length of an arc that subtends an angle of 24o at thecentre of a circle with a 5 cm radius is | |

| |[pic][pic] | |

| |The length of the arc is [pic]cm. | |

| | | | |

|4] |[pic] | |

| |Let AC be the tower of height h metres and ED be the observer of height 1.5 m at a distance of DC=28.5 m from the tower | |

| |AC. | |

| |In right ∆AED, | |

| |Tan45 = [pic] | |

| |1= [pic] (as EB = DC = 28.5 m ) | |

| |AB = 28.5 m | |

| |Height of the tower = h = AB + BC = AB+DE | |

| |= 28.5 + 1.5 = 30m | |

| | | | |

|5] |Total number of outcomes = 12 | |

| |Numbers which are divisors of 12 are 1,2,3,4,6,12. | |

| |Number of favourable outcomes = 6 | |

| |P(divisor of 12) = [pic] | |

| | | | |

|6] |Area of a triangle = 0 | |

| |[x x1 + 2x5 + 4x(-1)] – [2x-1 + 4x1 + xx5] = 0 | |

| |(x+10-4) - (-2+4+5x) = 0 | |

| |(x+6)-(5x+2)=0 | |

| |-4x+4=0 | |

| |x=1 | |

| | | | |

|7] |Let R [pic]divides PQ in the ratio K:1 | |

| |[pic] | |

| |[pic]and [pic] | |

| |14K-14 = -2K-2 | |

| |16K = 12 | |

| |K= [pic] | |

| |Thus R divides PQ in the ratio 3:4. | |

| | | | |

|8] |Length of arc (l) = 2(r[pic] | |

| |8.8 [pic]360 [pic]7 = 30 [pic]2 [pic]22 [pic]r | |

| |r= [pic]cm = 16.8 cm | |

| | | | |

|9] |Let P(x1,y1)= P(a, b), Q(x2,y2) = Q(b, c)and R(x3,y3) = R(c, a) | |

| |be the vertices of ∆PQR. | |

| |We know that the coordinates of centroid of a triangle is given by [pic]i.e. | |

| |[pic] | |

| |Also, given that centroid is at origin i.e. its coordinates are (0,0). | |

| |So, [pic]. | |

| |OR | |

| |Area of quadrilateral ABCD | |

| |= area of triangle ABC + area of triangle ACD | |

| |Area of triangle ABC = [pic][1(-3-2) +7(2-1)+12(1+3)] | |

| |= [pic](-5+7+48) = 25 sq. units | |

| |Area of triangle ACD = [pic][1(2-21) + 12(21-1)+7(1-2)] | |

| |= [pic](-19+240-7) = 107 sq. units | |

| |Area of quadrilateral ABCD = 25 + 107 | |

| |= 132 sq. Units | |

| | | | |

|10] |Here, n = 60, a = 7 and [pic]= 125 | |

| |7+59d = 125 | |

| |d=2 | |

| |Therefore, 32nd term (t32) = a + 31d = 7 + 31 x 2 = 69 | |

| | | | |

|11] |let the required numbers be x and x+1 | |

| |Given x2 + (x+1)2 = 25 | |

| |2x2 +2x-24 =0 ( x2 + x – 12 = 0 | |

| |(x+4) (x-3) = 0 | |

| |X = -4 or x = 3 | |

| |Reject x = -4 | |

| |Therefore the given consecutive positive integers are | |

| |X = 3 and x + 1 = 3 + 1 = 4 | |

| | | | |

|12] |[pic] | |

| |Let AB be a chord of a circle with centre O, and let AP and BP be the tangents at A and B respectively. Suppose the | |

| |tangents meet at P. join OP. suppose OP meets A at C. | |

| |We have to prove that [pic]PAC= [pic]PBC | |

| |In triangles PCA and PCB, we have | |

| |PA = PB (tangents from an external point are equal) | |

| |[pic]APC = [pic]BPC (PA and PB are equally inclined to OP) | |

| |And, PC = PC (common) | |

| |∆PAC [pic]∆PBC ( by SAS ) | |

| |[pic][pic]PAC = [pic]PBC ( by cpct ) | |

| | | | |

|13] |Diameter of the circle = diagonal BD of the rectangle ABCD | |

| |In right triangle BCD, | |

| |BD2 = BC2+BD2 = 62+82 = 100 | |

| |BD2 = 100 i.e.BD = 10 cm | |

| |Radius of the circle = [pic]= 5cm | |

| |Area of the circle = (r2 = 3.14(5)2 = 78.50 cm2 | |

| |Area of rectangle ABCD = 6[pic]8 = 48 cm2 | |

| |Area of the circle not included in the rectangle | |

| |= area of the circle – area of rectangle ABCD | |

| |= 78.50 – 48 = 30.50 cm2 | |

| | | | |

|14] |[pic] | |

| |Let OAB be the given sector. Then, | |

| |Perimeter of sector OAB = 16.4 cm | |

| |OA+OB+ arc AB = 16.4 cm | |

| |5.2+5.2+arc AB =16.4 | |

| |Arc AB = 6 cm i.e. l =6cm | |

| | | |

| |Area of sector = [pic][pic]lr = [pic][pic]6 [pic]5.2 = 15.6 cm2 | |

| | | | |

|15] |If the area of the triangle formed by joining the given points is | |

| |zero, then only the points are collinear. | |

| |Area of triangle = [pic] | |

| |Here, x1 = a, y1 = a2; x2 = b, y2 = b2; x3 = c, y3 = c2 | |

| |Substituting the values in the formula for area of a triangle, we get | |

| |Area of triangle = [pic] | |

| |[pic] | |

| |[pic] | |

| |[pic] | |

| |[pic] | |

| |[pic] | |

| |It is given that a [pic]b [pic]c | |

| |[pic]Area of the triangle can never be 0 | |

| | | | |

|16] |The AP is 8, 24, 40, ….., and the sum is 2048. | |

| |We have to determine n. | |

| |Here a = 8, d = 16 | |

| |Since, Sn = 2048, we have | |

| |2048 = [pic][2x8+(n-1)16] | |

| |2048 = n[8+8n-8] | |

| |2048 = 8n2 | |

| |n2 = 256 i.e. n = 16 | |

| |(as n being number of terms, it can't be negative) | |

| |Hence, the body will take 16 seconds to fall 2048 metres. | |

| | | | |

|17] |The steps of construction are as follows: | |

| |(i) Draw a circle of any convenient radius with O as centre. | |

| |(ii) Take a point A on the circumference of the circle and join OA. Draw a perpendicular to OA at point A. | |

| |(iii) Draw a radius OB, making an angle of 90( with OA. | |

| |(iv) Draw a perpendicular to OB at point B. Let both the perpendiculars intersect at point P. | |

| |(v) Join OP. PA and PB are the required tangents, which make an angle of 45( with OP. | |

| | | |

| |[pic] | |

| | | | |

|18] |[pic] | |

| | | |

| |Let the height of the balloon at P be h metres. Let A and B be the two cars. Thus AB=100m. | |

| | | |

| |From right ∆PAQ, AQ = PQ = h (as tan 45 =1) | |

| |From right ∆PBQ, | |

| |[pic] = tan60 = [pic]or [pic]or h = [pic](h-100) | |

| |Therefore, h = [pic]= 50(3+[pic]) | |

| |i.e. the height of the balloon is 50(3+[pic]) m. | |

| |OR | |

| |[pic] | |

| |Let AB be the tower of height h metres and C and D be two points at a distance of a and b respectively from the base of | |

| |the tower. | |

| |AC = a m | |

| |AD = b m (a>b) | |

| |In right ∆CAB, | |

| |tanx = [pic]tanx = [pic]………………….(i) | |

| |In right ∆DAB, | |

| |Tan(90-x) = [pic]cotx = [pic]……………(ii) | |

| |From (i) and (ii), | |

| |tanx.cotx = [pic] | |

| |1 = [pic]( h2 = ab or h = [pic]metres. | |

| |Thus, the height of the tower is [pic]metres. | |

| | | | |

|19] |There are 12 balls in the bag. Out of these 12 balls, one ball | |

| |can be chosen in 12 ways. | |

| |Therefore, total number of outcomes = 12 | |

| |There are x white balls out of which one can be chosen in x | |

| |ways. | |

| |Therefore, favourable number of outcomes = x | |

| |Hence, p1=P(Getting a white ball) = [pic] | |

| |If 6 more white balls are put in the bag, then | |

| |Total number of balls in the bag = 12+6 = 18 | |

| |Number of white balls in the bags = x + 6 | |

| |Therefore p2=P(Getting a white ball) = [pic] | |

| |It is given that | |

| |p2 = 2p1 | |

| |[pic] | |

| |[pic] | |

| |6(x+6) = 18x | |

| |6x+36 = 18x | |

| |12x = 36 | |

| |x = [pic]= 3 | |

| |OR | |

| |Total number of equally likely outcomes for visiting shop in | |

| |the same week = 6x6=36 | |

| |(i) Same days are (M,M),(T,T),(W,W),(Th,Th),(F,F),(SAT,SAT) | |

| |Number of favourable outcomes = 6 | |

| |P(same day) = [pic] | |

| |(ii) Number of different days = 36-6 = 30 | |

| |Number of favourable outcomes = 30 | |

| |P(two different days) = [pic] | |

| |(iii) Favourable outcomes for (I customer,II customer) | |

| |= (M,T),(T,W),(W,Th),(Th,F),(F,SAT) and for | |

| |(II customer,I customer) | |

| |= (M,T),(T,W),(W,Th),(Th,F),(F,SAT) | |

| |Number of favourable outcomes = 10 | |

| |P(consecutive days) = [pic] | |

| | | | |

|20] | | |

| |Let ABC be an equilateral triangle of side 24cm, and let AD be an altitude from A on BC. Since the triangle is | |

| |equilateral, so AD bisects BC. | |

| |Therefore BD=CD=12 cm | |

| |The centre O of the inscribed circle will coincide with the centroid of ∆ABC | |

| |Therefore OD= [pic] | |

| |(as centroid divides each median in the ratio 2:1) | |

| |In right ∆ABD, we have | |

| |AB2 = AD2+BD2 [Using Pythagoras Theorem] | |

| |242 = AD2+122 | |

| |AD = 12 cm | |

| |Therefore OD = [pic] | |

| |Area of the incircle = ((OD)2 | |

| |= [pic] | |

| |Area of the triangle ABC = [pic](Side)2 | |

| |= [pic](24)2 = 249.4 cm2 Therefore, area of the remaining portion of the triangle | |

| |= (249.4-150.85)cm2 | |

| |= 98.55 cm2 | |

| |OR | |

| |Area of one design = [pic]– area of ∆OAB | |

| |= [pic][pic]- [pic] | |

| |= 282(0.5238-0.433) | |

| |=282x0.0908 | |

| |= 71.1872 | |

| |=71.19 cm2 (approx.) | |

| |Total cost = 6 [pic]71.19x 0.35 | |

| |= Rs149.50 | |

| | | | |

|21] |Volume of water flows in the canal in one hour = width of the canal x depth of the canal x speed of the canal water | |

| |= 3x1.2x20x1000m3 = 72000m3. | |

| | | |

| |In 20 minutes the volume of water = [pic]m3 | |

| |= 24000 m3. | |

| |Area irrigated in 20 minutes, if 8cm, i.e. 0.08 m standing water is required = [pic]= 300000 m2 = 30 hectares. | |

| | | | |

|22] |[pic] | |

| |Radius of the cylinder = 0.7cm | |

| |Height of the cylinder = 2.4 cm | |

| |Surface area of the cylinder = 2(rh+(r2 | |

| |= ([2[pic]0.7[pic]2.4+(0.7)2] | |

| |= 3.85 ( cm2 | |

| |Slant height of the cone is given by | |

| |l2 = (0.7)2+(2.4)2 = 6.25 (( l = 2.5 cm | |

| |surface area of interior conical portion = (rl = (x0.7[pic]2.5 | |

| |= 1.75( | |

| |Total surface area = 3.85( +1.75( = 5.60( cm2 | |

| |= 5.6 [pic]3.1416 = 17.59 cm2. | |

| | | | |

|23] |Here, abx2-(a2+b2)x + ab = 0 | |

| |X = [pic] | |

| |= [pic] | |

| |=[pic] | |

| |Therefore, x = [pic] | |

| | | | |

|24] |Let A = (x1,y1) = (-4,1), B = (x2,y2) = (6,7) and | |

| |C = (x3,y3)= (2,-9) | |

| |Area of ?ABC = [pic][x1 (y2 - y3 )+x2 ( y3- y1) +x3 (y1 - y2)] | |

| |Area of ?ABC = [pic][ -4(7+9) + 6(-9-1) + 2(1-7)] | |

| |Area of ?ABC = [pic][ -64-60-12] ? = 68 sq. Units. | |

| |Coordinates of D, E and F are | |

| |D [pic]and F [pic] | |

| |i.e. D(4,-1), E(-1,-4) and F(1,4) | |

| |Area of ?DEF =[pic] [ 4(-4-4) -1(4+1) + 1(-1+4)] | |

| |Area of ?DEF = ? [pic][ -32-5+3] ? = 17 sq. Units. | |

| |[pic](Area of ?ABC) = [pic](68) = 17 = Area of ?DEF | |

| | | | |

|25] |Here, a =-11, d =-7-(-11) =4,an = 49 | |

| |We have an = a+(n-1)d | |

| |So, 49 = -11+(n-1) [pic]4 | |

| |i.e., 60 = (n-1) [pic]4 | |

| |i.e., n = 16 | |

| |As n is an even number, there will be two middle terms which are | |

| |[pic] th and [pic]th | |

| |i.e., 8th term and the 9th term. | |

| |a8 = a+7d = -11+7[pic]4 =17 | |

| |a9 = a+8d = -11+8[pic]4 = 21 | |

| |So, the values of the two middle most terms are 17 and 21, respectively. | |

| |OR | |

| |Let the three terms in AP be d, a, a + d | |

| | | |

| |so, a-d+a+a+d=33 | |

| |or a = 11 | |

| |Also, (a-d)(a+ d) = a+29 | |

| |i.e., a2-d2 = a+29 | |

| | | |

| |i.e., 121-d2 = 11+29 | |

| |i.e., d2 = 81 | |

| |i.e., d = ± 9 | |

| |So there will be two APs and they are:2, 11, 20,... | |

| |And 20, 11, 2, ... | |

| | | | |

|26] |[pic] | |

| |Let r1, r2 be the radii of the given circles respectively. Since the circles touch externally/ | |

| |(( distance between their centres = (r1 + r2) | |

| |\\ r1 + r2 = 14 …(1) | |

| |Sum of their areas = (r12 = (r22 = ((r12 + r12) = 130 m (Given) | |

| |\\ r12 + r22 = 130 …(2) | |

| |Now, (r1 + r2)2 + (r1 – r2)2 = 2 (r12 + r22) | |

| |( (14)2 + (r1 – r2)2 = 2(130) = 260 | |

| |( (r1 – r2)2 = 260 – 196 = 64 ( r1 – r2 = 8 …(3) | |

| |(1) + (2) gives, | |

| |2r1 = 22 ( r1 = 11 (1) – (2) gives, 2r2 = 6 ( r2 = 3 | |

| |Thus, radii of the given circles are 11 cm and 3 cm. | |

| | | | |

|27] |[pic] | |

| |Let CQ be the tower of height h metres and A and B be the points of observation. AB = p metres and let BC = x metres. | |

| |In right ∆ACQ, | |

| |[pic] = tan[pic] ( h = (p+ x)tan[pic] or x = [pic]- p ………………(i) | |

| |In right ∆BCQ, | |

| |[pic] = tanβ ( h = xtanβ or x = [pic]…………………………(ii) | |

| |From (i) and (ii), | |

| |[pic] - p = [pic] | |

| |P = h [pic] | |

| |P =h [pic] | |

| |or h = [pic] | |

| | | | |

|28] |Let usual speed = x km / hr and | |

| |increased speed = (x+5) km/hr | |

| |According to question, | |

| |[pic] | |

| |(Because Distance/Speed = time) | |

| |[pic]2 | |

| |1500 = 2x(x + 5) | |

| |x2+5x = 750 | |

| |x2+5x-750 = 0 | |

| |(x+30) (x-25) = 0 | |

| |X = -30 or x = 25 | |

| |But x being speed can't be negative. | |

| |Therefore original speed = x = 25 km/hr | |

| | | |

| |OR | |

| |Let the marks obtained by P in Mathematics be x. | |

| |Therefore marks obtained by P in Science = 28 – x | |

| |New marks in Mathematics = x + 3 | |

| |New marks in science = 28-x-4 = 24-x | |

| |ATQ, | |

| |(x+3) (24-x) = 180 | |

| |24x+72-x2-3x = 180 | |

| |-x2+21x = 180-72 | |

| |x2-21x+108 = 0 | |

| |x2-12x-9x+108 = 0 | |

| |x(x-12)-9(x-12) = 0 | |

| |(x-12)(x-9) = 0 | |

| |x = 12,9 | |

| |Therefore marks in Mathematics = 12 | |

| |Marks in Science = 28-12 = 16 | |

| | | |

| |and marks in Mathematics = 9 | |

| |Marks in Science = 28-9 = 19 | |

| | | | |

|29] |Given: A circle with centre O; PA and PB are two tangents to the circle drawn from an external point P. | |

| |To prove: PA = PB | |

| |Construction: Join OA, OB, and OP. | |

| |[pic] | |

| | | |

| |It is known that a tangent at any point of a circle is perpendicular to the radius through the point of contact. | |

| |(( OA ( PA and OB ( PB ... (1) | |

| |In (OPA and (OPB: | |

| |(OAP = (OBP (Using (1)) | |

| |OA = OB (Radii of the same circle) | |

| |OP = PO (Common side) | |

| |Therefore, (OPA ( (OPB (RHS congruency criterion) | |

| |((PA = PB (Corresponding parts of congruent triangles are equal) | |

| |Thus, it is proved that the lengths of the two tangents drawn from an external point to a circle are equal. | |

| | | | |

|30] |Since tangents drawn from an exterior point to the circle are | |

| |equal. | |

| |Therefore, DR = DS = 5 cm | |

| |Now AR = AD-DR = 23-5 = 18 cm | |

| |But AR = AQ | |

| |Therefore AQ = 18 cm | |

| |Also BQ = AB-AQ = 29-18 = 11 cm | |

| |But BP = BQ | |

| |Therefore BP = 11 cm | |

| |Also [pic]OQB = [pic]OPB = 900 | |

| |[Because OQ and OP are perpendiculars to AB and BC | |

| |respectively.] | |

| |In quadrilateral BPOQ, | |

| |[pic]QOP+[pic]OPB+[pic]OQB+[pic]B = 3600 | |

| |[pic]QOP = 3600-(900+900+900)= 900 | |

| |Hence, OQBP is a square. | |

| |BQ = OQ = OP = BP = 11 cm | |

| |Hence, radius of circle is 11 cm. | |

| | | | |

|31] |The sequence of her saving (in Rs) is 4, 5.75, 7.5………… | |

| |Let in nth week her weekly saving be Rs 19.75 | |

| |Here, a = 4 and d = 5.75 – 4 = 1.75 and Tn = 19.75 346 | |

| |We know, Tn = a +(n – 1)d | |

| |[pic] 19.75 = 4 + (n – 1) [pic]1.75 | |

| |[pic] 15.75 = (n – 1) [pic]1.75 | |

| |[pic] [pic] | |

| |[pic] n = 10 | |

| |[pic] In 10th week her weekly saving be Rs 19.75. | |

| | | | |

|32] |Volume of water in cylinder when full = ((3)2 [pic]6 | |

| |= 54( cm3 | |

| |Volume of solid = [pic]((2)3 + [pic]( (2)2 [pic]4 | |

| |= [pic]cm3 | |

| |Volume of water in the cylinder when solid is immersed in it | |

| |= 54( - 32(/3 | |

| |=136.19 cm3 | |

| | | | |

|33] |Total number of balls = 5 + 7 + 4 + 2 = 18 | |

| |(i) Favourable outcomes = 5 + 2 = 7 | |

| |((Required probability = [pic] | |

| |(ii) Favourable outcomes = 7 + 4 = 11 | |

| |(( Required probability = [pic] | |

| |(iii) Number of balls which are not white = 18 – 5 = 13 | |

| |((Required probability = [pic] | |

| |(iv) Number of balls which are neither white nor black = Number of balls which are red or blue = 7 + 2 = 9 | |

| |(( Required probability =[pic]=[pic] | |

| | | | |

|34] | | |

| | | | |

[pic]

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