Solutions / CBSE 10th Mathematics Sample Paper

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  • 1. Solutions / CBSE 10th Mathematics Sample PaperSection A 1.We haveax + by = a b ... ( i )bx ay = a + b...( ii) Multiplying (i) by a and (ii) by b, we get a2 x + aby = a2 ab... ( iii )2 b x aby = ab + b 2 ... ( iv )(1 mark) Adding (iii) and (iv), we get ( a 2 + b2 ) x = a2 + b2 x = 1 ( mark) Substituting x = 1 in (i), we geta 1 + by = a b by = b y = 1( mark) ORLet the fathers age be x years and sons age be y years. Now according to problemx = 3y + 3 x 3y = 3(i) ( mark) and x + 3 = 2 ( y + 3 ) + 10 x 2y = 13 (ii) ( mark) Subtracting (i) from (ii), we gety = 10( mark) Substituting y = 10 in (i), we getx 3 10 = 3 x = 33 ( mark) Hence, fathers age = 33 years, sons age = 10 years2.We have HCF = x(x + a) , LCM = 12x 2 (x + a)(x 2 a2 ) and P(x) = 4x(x + a)2 . We Know P(x) Q(x) = HCF LCMHCF LCM Q(x) = ( mark)P(x)x(x + a) 12x 2 (x + a)(x 2 a2 ) Q(x) = (1 mark)4x(x + a)2 Q(x) = 3x 2 (x 2 a2 ) ( mark)2.We have1 1 1 1= + + ; a 0 , b 0, x 0a+b+x a b x11 1 1 = +a+b+x x a b

2. x ab x b+a= ( mark)x ( a + b + x) ab1 1 x 2 + ( a + b ) x = ab x + ( a + b ) x + ab = 02 ( mark) ( x + a) ( x + b) = 0 ( mark) x = a or x = b( mark) 3.Let a be the first term and d be the common difference of an A.P., then a7 = a + ( 7 1) d a7 = a + 6d [Using an = a + (n 1)d] and a1 1 = a + ( 11 1) d a11 = a + 10d( mark) According to the problem7a 7 = 11a 1 1 7 ( a + 6d) = 11( a + 10d ) ( mark) 4 ( a + 17d) = 0 a + ( 18 1) d = 0( mark) a1 8 = 0 ( mark)5.Cash price of the electric iron = Rs. 440 Cash down payment = Rs. 200 Balance to be paid = Rs. (440 200) = Rs. 240( mark) Instalment paid = Rs. 244 Total price charged under the instalment plan = Rs. (200 + 244) = Rs. 444 Therefore, interest charged = Rs. 444 Rs. 440 = Rs. 4 ( mark) 1 Therefore, Rs. 4 is the interest charged on Rs. 240 for 1 month i.e.year12Int. 100 rate % = Pr incipal Time ( mark) 4 100=1240 124 100= = 20 20 ( mark) Hence, rate = 20% per annum 6. ABC D In ABC and DAC , we have 3. BAC = ADC(Given)And C = C[Common] ABC : DAC [By AA criterion of similarity] ( mark)AB BC AC==( mark)DA AC DC CB CA =(mark) CA DC CA 2 = CB DC ( mark) ORS R40 65 P Q O 50M QRP = 90 [Angle in semi circle]In triangle QRP,QRP + QPR + PQR = 180 90 + QPR + 65 = 180 QPR = 25Q PQRS is a cyclic quadratic. QRS + QPS = 180 90 + PRS + 25 + 40 = 180 PRS = 25 (1 mark)Q PMQR is also a cyclic quadratic. MQR + RPM = 180 50 + 65 + 25 + QPM = 180 QPM = 40 (1 mark) 7. Total number of outcomes = 17(i) Odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17Number of favourable outcomes = 9P(an odd number) = 9/17(1 mark)(ii) Numbers divisible by 3 or 5 are 3, 5, 6, 10, 12, 15Number of favourable outcomes = 6P(a number divisible by 3 or 5) =6/17(1 mark)Section B8. We have y = x 4. x0 14y0 14 y = 2x x0 12y0 24 and x + y = 6 y=6xx 013 y 653(1 mark) y 2x= x + y = 6y x ) y = ,64 ) (0 2,),5( (1 B4) (4 ,,3 ) ) A (3 ,2 (1x( 1 ,1 )x O (0 ,0 ) y(1 mark) From figure, the three straight lines intersect at O(0, 0), A(3, 3) and B(2, 4). Vertices of the required triangle OAB are (0, 0), (3, 3) and (2, 4). ( mark) 11 x2 y2 1 9. We have 2 x y x + y x y xy xy 2 x + y x + y ( x + y) ( x y) x y = (1 mark) ( x y) ( x + y) xy ( x y )1 2y ( x + y) ( x y) x y= (1 mark)( x y) ( x + y)xy ( x y ) 12=(1 mark)x 5. 10. Let x be the larger number, then square of the smaller number will be 4x x 2 4x = 45 (1 mark) x 2 4x 45 = 0 ( x 9) ( x +5) = 0( mark) x = 9 or x = 5( mark) Q The numbers are natural numbers x=9 ( mark) Square of smaller number = 4 9 = 36 Smaller number = 36 = 6( mark) Hence, larger number = 9 and smaller number = 611. Let a be the first term and d be the common difference of the A.P. a3 = 7 and a7 = 3a3 + 2( mark) a + 2d = 7 and a + 6d = 3 ( a + 2d ) + 2 Q Tn = a + ( n 1) d ( mark) a + 2d = 7 and a = 1 a = 1 and d = 4 ( mark)22n S22 = 2 1 + ( 22 1) 4 Q Sn = 2a + ( n 1) d (1 mark) 2 2 S22 = 11[ 2 + 84 ]( mark) S22 = 902 12.Let each instalment be Rs. x. r = Rate of interest = 5% per annum CI. n = Time = 2 years For first instalment: A = Rs.x, time = 1 year, r = 5% per annum and P = P11 r n 5 Using A = P 1 + x = P1 1 + 100 100 ( mark) 20 P1 = Rs. x 21 For second instalment: A = Rs. x, time = 2 years, r = 5% per annum and P = P22 5 x = P2 1 + ( mark) 100 2 20 P2 = Rs. x 21 Total sum borrowed = Rs. 4100[Given] Therefore, P1 + P2 = Rs.4100220 20 x+ x = Rs.4100(1mark)21 21 20 20 x1 + 21 = Rs. 410021 6. 20 41 x = Rs.4100 ( mark) 21 2121 21 x = Rs.4100 20 41 x = Rs. 2,205 ( mark) Hence, annual instalment is Rs. 2,205. 22 sin15 cos75 13. cos75 + sin15 2 tan1 tan89 + 3 cos35 cos ec 55 2 2 sin(90 75 ) cos(90 15 ) = cos75 + sin15 2 tan(90 89 )tan89 + 3 cos35 cosec ( 90 35 ) (1 mark) 2 2 cos75 sin15 = + 2 cot 89 tan89 + 3 cos35 s ec35 (1 mark) cos75 sin15 1 1 Q tan = cot and sec = cos =1+12+3(1 mark) =3 OR 2p 1 ( sec + tan ) 12LHS = =[ Given p = sec + tan ] ( mark) 2p +1 ( sec + tan ) 2 + 1 sec 2 + tan2 + 2 tan sec 1 = ( mark) sec 2 + tan2 + 2 tan sec + 1= ( sec 2 ) 1 + tan2 + 2 tan sec ( mark) ( ) sec + 1 + tan2 + 2 tan sec 2 tan2 + tan2 + 2 tan sec = ( mark) sec 2 + sec 2 + 2 tan sec 2 tan2 + 2 tan sec = ( mark) 2 sec 2 + 2 tan sec 2 tan ( tan + sec ) tan sin cos = == ( mark) 2 sec ( sec + tan ) sec cos 1 = sin = RHS21 14. Radius of solid metallic sphere =cm = 10.5 cm 24 Volume of the sphere = (10.5)3 cm3 = 1543.5 cm3 (1 mark)3 3.5 Radius of the base of the cone =cm = 1.75 cm21 Volume of the cone = ( 1.75 ) 3 cm3 = 3.0625 cm32 (1 mark)3 Let n be the number of cones so formed. 7. n volume of one cone = volume of sphere n 3.0625 = 1543.5 1543.5 n= = 504 (1 mark)3.062515.A cm3 c 4Q I PmB 5 cmL C (2 marks) Steps of construction:(1 mark) 1. Construct the ABC in which BC = 5 cm, CA = 3 cm and AB = 4 cm. 2. Bisectand B and C . Let bisectors of these angles be BP and CQ respectively. Let these intersect at I. 3. Draw the perpendicular IL on BC. 4. Taking I as centre and IL as radius, draw the circle. This is the required incircle of the ABC. Radius IL = 1 cm 16.1 2A (3 , 4) P (p , 2) 5B (1, 2 )Q ,q3 Let AB be the line-segment. Clearly P divides AB in the ratio of 1 : 2 internally. ( mark) 1 1 + 2 3 p = [Using section formula]1+ 2 7 p= (1 mark) 3 Now Q is mid-point of PB( mark) 2 + 2q = [Using mid-point formula] 2 q=0(1 mark) ORA (x1, y1) 2 1B (x2, y2)C ( x 3, y 3)D 8. Let D be the mid-point of BC, x + x3 y 2 + y3 D 2 , ( mark) 22 Let G (x, y) be the centroid of the triangle. Since, G divides AD in the ratio 2:1.( mark) x 2 + x3 2 + 1 x1 x + x + x x =2 = 123 (1 mark) 2 +13 y + y3 2 2+ 1 y1 y + y + y and y =2= 1 2 3 (1 mark)2 +13 Hence,the co-ordinatesof the centroid of thetriangle are x1 + x 2 + x 3 y1 + y 2 + y 3 , . 33 17. T A BCDQ AT is a tangent and TB is a chord of the circle. ATB = TCB ... (i) [Angles in the alternate segments] ( mark) Q TD is the bisector of BTC. BTD = CTD... (ii) ( mark) Adding (i) and (ii), we get ATB + BTD = TCB + CTD ATD = TCB + CTD ... (iii) ( mark) As TDA is the exterior angle of TDC TDA = TCB + CTD ... (iv)( mark) From (iii) and (iv), we get ATD = TDA Hence, DAT is an isosceles triangle(1 marks)18. Let the co-ordinates of point on the Y-axis is P(0, y). ( mark) Also let A and B denote the points ( 5, 2) and (3, 2) . Since AP = BP, AP2 = BP2( mark) ( 0 + 5) + ( y + 2) = ( 0 3) + ( y 2)2 222 ( mark) 25 + 4 + y 2 + 4y = 9 + 4 + y 2 4y ( mark) 8y = 16 ( mark) y = 2 ( mark) Hence, the point on Y-axis is (0, -2). . 9. 19. The following table gives the share of each item as a component of 360.ItemsExpenditure (in percent) Share as a component of 36040 Food 40 360 0 = 144 100 Entertainment2525 360 0 = 90 100Other2020 expenditure 360 0 = 72 100Savings15 15 360 0 = 54 100 Total100360 (1 mark) Pie Chart: OtherSavingsexpenditure 5472 Entertainm ent90F ood 144 (1 mark) Section C20. 10. C lo u dx PH QOhy RP(1 mark) According to figure, the cloud is at P and the image of the cloud P is at with respect to the lake. PR = P' R Let OP = x, OQ = y and PR = PR = H In right triangle OPQ, Hh tan =...(i)(1 mark)y In right triangle OP 'Q , H+h tan = (ii)(1 mark)y Subtracting (ii) and (iii), we get 2htan tan = (iii)( mark)y Again, in right triangle OPQ, xsec = (iv) ( mark) y Dividing (v) and (iv), we getxsec y= ( mark)tan tan 2hysec x 2hsec =x=( mark) tan tan 2h tan tan OR 11. PQ4 ,0 0 0 m 60 45Ox A(1 mark) Let P and Q be the positions of the two aeroplanes. In right-angled OAP , AP4000 tan60 = 3=OA x 4000 x=... (i) (1 mark)3 In right-angled OAQ , AQ AQ tan45 = 1=OA x4000 AQ = x =[From (i)] (1 mark)3Distance between two aeroplanes 4000 3 1 = PQ = AP AQ = 4000 = 4000 33 = 4000 0.244 3 = 976 3 m (1 mark) 21. Part I PAB D CQ SR Given: ABC : PQRarea ( ABC ) AB2BC2AC2 To prove: ===area ( PQR ) PQ2QR2PR2 Construction: Draw AD BC and PS QR( mark)1 area ( ABC ) BC AD Proof: = 2 area ( PQR ) 1QR PS2BCAD =... (i) ( mark) QR PS In ADB and PSQ B = Q [Q ABC : PQR ] 12. ADB = PSQ = 90 ADB : PSQADAB =... (ii) (1 mark)PSPQ ABBCCA But = = [Q ABC : PQR] PQQRRP ADBC =[From (ii)] ... (iii)( mark) PSQR area ( ABC ) BC BC BC2 ==[From (i) and (iii)] ( mark) area ( PQR ) QR QR QR2 area ( ABC )AB2 BC2 CA 2 Hence,= = =( mark) area ( PQR )PQ2 QR2 RP2Part II area ( ABC )BC2 =area ( PQR ) QR2 64 BC2 =( mark) 121( 15.4 ) 2 8 BC = ( mark) 11 15.4 15.4 8 BC = = 11.2 cm ( mark)1122. Part IMPB O Q S A TGiven: ST is a tangent to the circle with centre O. P and Q are points onmajor and minor arc AB respectively. To prove: BAT = APB and BAS = AQB Construction: Draw a diameter AM and join MB.( mark) Proof: ABM = 90 [Angle in a semicircle] MAT = 90 In AMB , AMB + MAB = 90 ... (i) Also MAB + BAT = 90 ... (ii) (1 mark) From (i) and (ii), AMB + MAB = MAB + BAT 13. AMB = BAT ( mark) Also AMB = AP