Time 3 Hours Maximum Marks : 90. x 2 = 75 3 = 25. x = ± 5. QR = 7.6 cm 1. = 1 2 [x 1 (y 2 y 3 ) + x 2 (y 3 y 1 ) + x 3 (y 1 y 2 )]

Transcription

1 SOLUTIONS SMPLE QUESTION PPER - 6 Self ssessment Time Hours Maximum Marks : 90. SECTION x x x x 75 5 x ± 5 ( mark each) \ Positive root 5. QP.8 QP PT (Length of tangents from external points are equal) PT.8 cm PR PT.8 cm [CSE Marking Scheme, 05] QR 7.6 cm. rea of triangle [CSE Marking Scheme, 0] [x (y y ) + x (y y ) + x (y y )] [0(0 5) + 6(5 0) + 0(0 0)] [6 5] 5 sq units [CSE Marking Scheme, 05] MTHEMTICS Oswaal CSE (CCE) Class -0, S- Examination Sample Question Paper

2 OSWL CSE (CCE), Mathematics, Class 0 4. Volume of reduced cylinder Volume of original cylinder r π h πrh 4 : 4 [CSE Marking Scheme, 0] SECTION ( marks each) 5. C 0º O D Join OC. C 90 (ngle in a semi-circle) OC 80 ( ) (ngle sum property) 60 O OC (Radii of same circle) OC OC 60º (ngles opposite to equal sides) CD OC CD + DC (ext. angle prop.) DC DC 0 CD DC C D Hence, Proved (CSE Marking Scheme, 04, 5TYPEZ) 6. In right DC, C sin x 90 cm x x C

3 Solutions (S-) x Hence length of string 0 9 m. [CSE Marking Scheme, 0] 7. (a) P (a vowel is selected) 5 6 (b) P (a consonant is selected) 6 (CSE Marking Scheme, 04, 5TYPEZ) 8. Given points are (8, ), (, k) and (k, 5) are collinear. rea of triangle formed 0 [8( k + 5) + ( 5 ) + k( + k)] 0 k 5k + 0 (k ) (k ) 0 k, [CSE Marking Scheme, 05] 9. Given : rea of the sector radius r 6 cm length l cm (length of the corresponding arc) radius i.e., l r 6 0. Volume of cylinder π(5) 4 cm 6 cm. (CSE Marking Scheme, 04, 5TYPEZ) 00π cm Volume of cone 8 π 4π Required ratio 00π : 4π 5 : 6. [CSE Marking Scheme, 0] SECTION C ( marks each). (x + y ) (a + b ) (ax + by) a x + b x + a y + b y a x + b y + abxy b x + a y abxy 0 (bx ay) 0 bx ay x a y b Hence Proved (CSE Marking Scheme, 04, 5TYPEZ)

4 4 OSWL CSE (CCE), Mathematics, Class 0. Let first term be a Common difference be d a 4 a + d 5...() a 9 a + 8d 0...() From () and () d, a 6 s 7 7 [ + 6 ] 7 ( 60) 50 Or Detailed nswer : Let first term of an P is a and common difference is d. Given, a 4 5 and a 9 0 We know that, T n a + (n )d a 4 a + d 5...() a 9 a + 8d 0...() Subtract eq () from eq (), 5d 5 d 5 5 Put the value of d in eq. (), n Now, S n [ a+ ( n ) d ] a + ( ) 5 a 9 5 a S 7 7 [( 6) + (7 )( )] 7 [ 48] 50 (CSE Marking Scheme, 04, 5TYPEZ). Given : circle of radius cm with centre O and a point P at a distance of 7 cm from O. Construction : We have to construct the two tangents from P to the circle. Q P M O R Step of construction :. Draw a line segment PO 7 cm.. From the point O, draw a circle of radius cm

5 Solutions (S-) 5. Draw a perpendicular bisector of PO. Let M be the mid-point of PO. 4. Taking M as centre and OM as radius draw a circle. 5. Let this circle intersect the given circle at the point Q and R 6. Join PQ and PR. Thus PQ and PR are the required two tangents. Length of the tangent PQ PR PQ PR OP OQ (7) () cm. [CSE Marking Scheme, 05] 4. P θ O is the common point of the radius O and tangent P. OP 90 Let PO q and O r OP r In right triangle OP, Opposite side sin q Hypotenuse sin q O OP sinq r r Join. sin q q 0 PO 0 P In DP; P P P P Using angle sum property of DP [ngle opp. to equal sides are equal] P + P + P 80 P [Q P P] P Here each angle of DP is 60. Hence, P is an equilateral D. (CSE Marking Scheme, 04, 5TYPEZ)

6 6 OSWL CSE (CCE), Mathematics, Class m C x Let be a point and and C be two objects. Let C x In DCO, O tan 45 CO lso in DO, 00 CO O CO 00 m O tan 60 O 00 O C CO O O m 6. C (CSE Marking Scheme, 04, 5TYPEZ) 500 Let, In DL 60 0 x L L x, L x M tan 60 L CM 500 In DCM 500 x x 500 m. CM M tan 0, CM L + LM LM 500 () 4500 LM 000 m.

7 Solutions (S-) 7 Speed distance Time 000 m 5s 00m./s km/hour 70 km/hr. [CSE Marking Scheme, 0] 7. S {HH, HT, TH, TT} (a) P (No heads) 4 E {TT} (b) P (t most one tail) 4 E {HH, HT, TH} (c) P (One tail) 4 E {HT, TH} (d) P (One head and one tail) 4 E 4 {HT, TH} (CSE Marking Scheme, 04, 5TYPEZ) 8. Total number of outcomes 8 (i) P (three heads) 8 E {HHH} (ii) P (atleast two tails) 4 E {HTT, THT, TTH, TTT} 8 9. Given : [CSE Marking Scheme, 05] r cm h 0. cm Volume of a plate pr h p (0.75) 0. Radius of cylinder R cm Height of cylinder H 0 cm Volume of cylinder pr H p (.5) 0 Volume of cylinder No. of plates Volume of plate π (.5) 0 π (0.75) (CSE Marking Scheme, 04, 5TYPEZ) 0. C O

8 8 OSWL CSE (CCE), Mathematics, Class 0 s DC is a triangle in a semicircle, so C 90 rea of semicircle πr rea of DC OC Now, rea of shaded region area of semicircle area of DC πr OC cm. (CSE Marking Scheme, 04, 5TYPEZ) SECTION D (4 marks each). Let original Number of persons x ccording to st condition ccording to nd condition ccording to the question Each person s share 9000 x Each person s share 9000 ( x + 0) 9000 x 9000 ( x + 0) + 60 x + 0x 5 0 x + 45x 5x 5 0 (x + 45) (x 5) 0 (x 5) 0 or x x 5 or x 45 (which is not possible) Hence original number of persons 5 (CSE Marking Scheme, 04, 5TYPEZ). S n n + 4n put n, S a a put n, S a + a a S S a 0 7 a + d 7 + d d 6 P becomes : 7,, 9,... Now a n a + (n )d 7 + (n ) n 6 6n + a n 6n + (CSE Marking Scheme, 04, 5TYPEZ)

9 Solutions (S-) 9. b 4ac 0 For x + kx k 56 0 k 6... () For x 8x + k 0, 64 4k 0, 64 4k 4k 64 k 6...() From () & () k 6 (CSE Marking Scheme, 04, 5TYPEZ) 4. t (60 ) Sum of last 0 terms (t 5 + t t 60 ) t Sum of last 0 terms 0 [08 + 6] 70 [CSE Marking Scheme, 05] 5. Given : tangent is the circle with centre O at the point of contact P. To Prove : OP O P Q Construction : Let Q be any point other than P on the tangent. Draw line segment OQ. Proof : Since Q is any point other than P on the tangent. So Q will be outside the circle. [If Q is inside the circle, then the line passing through P and Q again will intersect the circle at some point consequently is may not remain a tangent] OQ > OP, that is OP < OQ Since the shortest line segment that can be drawn from centre O to be line, is perpendicular to. Therefore OP Hence Proved. [CSE Marking Scheme, 05] 6. P S Q R

10 0 OSWL CSE (CCE), Mathematics, Class 0 Given : PQ, PR and are tangents, To find : perimeter of DP PQ 8 cm Solution : PQ PR 8 cm Q S S R [Length of tangents drawn from an external point to a circle are equal]. P + Q P + R P + S P + S 8 cm Perimeter of DP P + S + S + P cm. (CSE Marking Scheme, 04, 5TYPEZ) 7. The vertices of a DC are (6, ), (, 5) and C (, ) (6 ) + ( 5) () + () C ( + ) + (5 + ) (6) + (6) C (6 + ) + ( + ) (9) + () Since + C C Hence, by converse of Pythagoras theorem DC is a right angled triangle, right angled at. (CSE Marking Scheme, 04, 5TYPEZ) 8. X' ( 4, 6) (, ) (, 5) 4 C 6 Y' 6 4 From the graph, we observe that the C is scalene. The coordinates of the vertices of C are (, ), ( 4, 6) and C (, 5) respectively Then, ( + 4) + ( 6) X

11 Solutions (S-) C ( 4 + ) + (6 + 5) + C ( + ) + ( + 5) since C C C is scalene Now, area of C [(6 + 5) + ( 4) ( 5 + ) + ( )( 6)] [ ] 4 sq. units. (CSE Marking Scheme, 04, 5TYPEZ) 9. rea of circular portion rea of circle rea of sector rea of equilateral triangle cm 4 6 cm rea of shaded portion cm Value Environmental concern 4 (CSE Marking Scheme, 04, 5TYPEZ) Detailed nswer : OR Given, radius of the circle 6 cm Side of the equilateral triangle cm Now area of circle pr Hence, rea of circular portion πr θ rea of sector 60 area of circle area of sector pr πr θ 60 pr 60 60

12 OSWL CSE (CCE), Mathematics, Class 0 rea of equilateral triangle So, Value : Environmental Concern cm 4 (side) 4 () 6 cm rea of shaded portion cm (CSE Marking Scheme, 04, 5TYPEZ) 0. O D h r N l C L H Now \ R M DONC ~ DOM l r h L R H C.S.. of frustum 5 6 C.S.. of cone OCD 6 C.S..of cone OCD C.S..of cone O 6 πrl π RL 6 r l R L 6 h h H H 6 h H 4 [C.S.. of cone O] [C.S.. of cone O] h 4 H ON 4 H

13 Solutions (S-) and MN 4 H ON : MN :. (CSE Marking Scheme, 04, 5TYPEZ). Since a cone, a sphere and a cylinder are on equal bases. Let the radius of each of them be r and height also be equal to r. Then volume of cone V πrh π rr πr 4 Volume of sphere V πr Volume of cylinder V pr h pr. V pr Ratio in their volumes 4 V : V : V : πr : πr : πr : 4 : [CSE Marking Scheme] qqq

14 SOLUTIONS MTHEMTICS Oswaal CSE (CCE) Class -0, S- Examination Sample Question Paper SMPLE QUESTION PPER - 7 Self ssessment Time Hours Maximum Marks : 90. 5x. 0 6 x x 6x + 0 x 6 x 6x + 0 x( x ) ( x ) 0 ( x )( x ) 0 x 0 x 0 x P SECTION 0 0, cm Here, P and P are two tangents to circle with centre O. O P OP 0 In OP, we have P cot 0 O P O ( mark each) P cm. P P cm.. Let (, ), (, ), C (x, y) be the points of equivalent triangle. ( + ) + ( ) 6 OR 5x 0 6x x 6x + 0 x b ± b 4ac a 6 ± ( 6) 4()(). 6 6

15 Solutions (S-) 5 C (x ) + (y ) 6... () C (x + ) + (y ) 6... () From equation () and () (x + ) (x ) 0 x 0, x 0 x 0, y Surface area of sphere 4pr 66 4 r 7 66 r So, radius of sphere 49 7 cm SECTION ( marks each) 5. O P OP 70 5 OP 90 OP 80 ( ) 55 O 5 P m C Let C be the point where the ball is In DC, 60 C 60 (alt. angles) tan 60 C 0 x x m. 7. Total cards 0 Number divisible by,, 6, 9,, 5, 8,, 4, 7, 0 Total number 0 P (number divisible by ) 0 0 P (Not divisible by )

16 6 OSWL CSE (CCE), Mathematics, Class 0 8. (, a) lies on x y 5 0 So, a 5 0 a x y 5 0 cuts the x-axis at (x, 0). a x 5 0 x 5 5 Point is, Surface area of two hemispheres cm 7 Surface area of cylinder cm Total surface area of article cm. Vol. of liquid in hemisphere π 0. Number of bottles Vol. of one cylindrical bottle R π rh (9) 4 54 bottles. SECTION C ( marks each). Let total number of camels x ccording to question, x + 4 x + 5 x x 8 x 60 0 Let x y, then y 8y 60 0 y 8y + 0y 60 0 y(y 6) + 0(y 6) 0 (y + 0)(y 6) 0 0 y 6 or y y 0 (not possible) So, y 6 y 6 6 Hence, number of camels 6. Series of two digits numbers which is divisible by 6 is :, 8, 4,..., 96 Here a, d 8 6, Tn 96

17 Solutions (S-) 7 T n a + (n )d 96 + (n ) 6 84 (n ) 6 4 n n 5 n S n [a + (n )d] 5 [ + (5 ) 6]. 5 [ + 4 ] 5[ + 4] cm O r 6cm C rea of C cm 4. C cm ar ( C) ar (DOC) + ar (DOC) + ar (DO) 6 r + 0 r + 8 r r r 4 r cm. ' 5cm O' 60 6cm O C' C 4 Following steps will be followed to draw a C whose sides are /4 of corresponding sides of C. X

18 8 OSWL CSE (CCE), Mathematics, Class Draw a line segment C of 6 cm. Draw an arc of any radius while taking as centre. Let it intersect line C at point O. Now taking O as centre draw another arc to cut the previous arc at point O. Join O which is the ray making 60 with line C.. Now draw an arc of 5 cm radius, while taking as centre, intersecting extended line segment O at point. Join C. C is having 5 cm, C 6 cm and C 60.. Draw a ray X making an acute angle with C on opposite side of vertex. 4. Locate 4 points (as 4 is greater in and 4),,, 4 on line segment X. 5. Join 4 C and draw a line through parallel to 4 C intersecting C at C. 6. Draw a line through C parallel to C intersecting at. C is the required triangle. (60 h ) 60º 0º 60 m E x 0º D h m x Let is a building 60 m high and a tower h m high. ngle of depressions of top and bottom are given 0º and 60º respectively. DC E h m and let C x E (60 h) m In ED, In C, 60º h m 60 h ED tan 0 60 h C C (60 h ) x...(i) 60 x tan Putting the value of x from equation (i) in equation (ii), we get Height of tower is 40 m. 60 x...(ii) 60 (60 h) 60 (60 h) 0 60 h h 40 m C D h h 60º 0º x 800 m E

19 Solutions (S-) 9 In 600 sec distance travelled by plane m In 0 sec distance travelled by plane m In C, h tan 60º x h x h x...(i) In ED, h x tan 0º h x h x (ii) From equations (i) and (ii), we get x x x x x 800 x 900 m h x m Height of jet m. 7. Number of male fish 5 Number of female fish 8 Total number of fish in the tank Now, probability of getting male fish 5 8. ll possible outcomes are 6 (, ) (, ) (, ) (, 4) (, 5) (, 6) (, ) (, ) (, ) (, 4) (, 5) (, 6) (, ) (, ) (, ) (, 4) (, 5) (, 6) (4, ) (4, ) (4, ) (4, 4) (4, 5) (4, 6) (5, ) (5, ) (5, ) (5, 4) (5, 5) (5, 6) (6, ) (6, ) (6, ) (6, 4) (6, 5) (6, 6) E {(4, 5), (5, 4), (, 6), (6, )} (a) Sum of two numbers appearing on top of dice is E {(, ), (, ), (, ), (4, 4), (5, 5), (6, 6)} (b) Outcomes appearing on top of dices are same E Equal to {(6, 6)} Number of out comes, less then 6 5 P(Number of Outcomes sum < ) 5 6

20 0 OSWL CSE (CCE), Mathematics, Class 0 9. rea of shaded region p[r r θ ] 60 7 [7 (.5) ] cm For cylinder, 08 r 8 cm, h cm For hemisphere, r 8 cm C.S.. of solid C.S.. of cylinder + (C.S.. of hemisphere) Cost of polishing ` 0 70 per sq. cm prh + 4pr prh + 4pr pr (h + r) (7 + 6) ` SECTION D x x x x 5 ( x )( x 5 ) + ( x 4 )( x ) ( x )( x 5) 0 [(x 5x x + 0) + (x x 4x + )] 0(x 5x x + 5) (4 marks each) [x 7x x 7x + ] 0[x 8x + 5] [x 4x + ] 0(x 8x + 5) 6x 4x x + 80x x + 8x 84 0 x 9x x x 7x x (x 6) 7(x 6) 0 x 7 0 or x 6 0

21 Solutions (S-) x 7/ or x 6 Hence, x 7,6. Let the original list price be ` x. No. of books bought for ` x Reduced list price of the book ` (x 5) No. of books bought for ` x 5 ccording to the question, x 5 x 5 x 5x 00 0 (x 0) (x + 5) 0 x 0 or 5 (not possible) Original list price ` 0. Let first term of an.p. is a and common difference is d then, ccording to the first condition, T 5 + T 9 0 a + 4d + a + 8d 0 a + d 0 a + 6d 5 () ccording to the second condition T 5 T 8 a + 4d (a + 7d) a + 4d a + d a d 0 a d () Put the value of a in eq. () d + 6d 5 d + d 0 5d 0 d 0 5 So a d Hence,.P. is a, a + d, a + d.., 5, 7, 9, ccording to the question, Class-I will plant tree Class-II will plant trees Class-III will plant trees Class-XII will plant trees

22 OSWL CSE (CCE), Mathematics, Class 0 lso each class has sections. So, total number of plants planted by students of Class I trees Total number of plants planted by students of class II 4 trees Total number of plants planted by students of Class III 6 trees Similarly total number of plants planted by students of class XII 4 trees Thus, we get series.p. i.e., 4, 6,, 4 Here, a, d 4, n So, S n [a + (n )d] [4 + ()] 5. 6[4 + ] Hence total 56 trees will be planted by the students. Value : Social, Environmental. P T Given : PQ 6 cm and radius OP 0 cm OR PQ, bisects PQ R Q O In right triangle ORP, PR 6 8 cm OP OR + PR (0) OR + (8) OR (0) (8) OR 6 6 cm TPR + RPO 90 [Q Radius is to the tangent at the point of contact] lso, TPR + PTR 90 RPO PTR DTRP ~ DPRO [] TP PO RP RO TP TP cm

23 Solutions (S-) 6. O R P Q Given : circle C (O, r) and a tangent at a point P. To prove : OP Construction : Take any point Q, other than P on the tangent. Join OQ. OQ meets the circle at R. We know that OP is the shorter distance. OP OR OQ OR + RQ OQ > OR OQ > OP, [Q OP OR] Thus, OP is shorter than any other segment joining O to any point on. OP. ngle in semi-circle is y-axis divides the line segment joining the point ( 4, 6) and (0, ), then x coordinate of this point of division is 0. mx + nx x m+ n 0 0m n 4 m n ( m/ n) 0 + ( 4) ( m/ n) m : n : 5 lso, my + ny y m+ n y ( m/ ny ) + y ( m/ n) + 8. The co-ordinates of the point of division (, 4) (/5) + ( 6) (/5) , (, ) E D( x, y) C(-6,)

24 4 OSWL CSE (CCE), Mathematics, Class 0 Diagonals of a parallelogram bisect each other. Using mid-point formula, [E is mid-point of C] gain, 6 4+ E, E, x y / E, E, x x + x nd, y y y Fourth vertex D is (, ). Therefore, rea of C [x (y y ) + x (y y ) + x (y y )] [( ) ( + 4) 6( 4 + )] [ ] ( 5) 5 5 square unit 9. rea of parallelogram CD rea of C 5 5 square unit. O C 60 D Given : R 4 cm r cm O 60 rea of region CD rea of sector O rea of sector COD π (4) π () 6 π [(4) () ]

25 Solutions (S-) 5 6 π [764 44] cm rea of circular ring π R πr (4) () cm. 7 Hence, rea of shaded region cm O h C N r D h h M r DOND ~ DOM ON OM ND M h h r V Volume of cone O, V V Volume of frustum V V π rh π r h π rh rh V V πrh π( rh rh ) r h r h r h 4r h r h r h rh 4 i.e.,. r r r r 4 r r 4 r r 4 i.e., h h h 4 h 4/

26 6 OSWL CSE (CCE), Mathematics, Class 0. h h 4 h h h 4 5/ mm 5/ mm 4 mm Total Length 4 mm Height of cylinder 4 (.5) mm Radius of cylinder.5 mm Radius of hemisphere.5 mm Volume of capsule Volume of two hemispheres + Volume of cylinder π r +πrh π +π π +π π π mm. qqq

27 SOLUTIONS SMPLE QUESTION PPER - 8 Self ssessment Time Hours Maximum Marks : 90 SECTION. Given : x, x So, the equation is (x ) (x + ) 0 x + x x 0 x + x 0 ut equation is given x + ax + b 0 On comparing a ( mark each) and b. Since OP QP So, In DOPQ OQ QP + OP OP OQ QP () () OP 5 cm. So, radius of circle is 5 cm,. Let the point P be (0, y) P (y 5) + ( 6) nd, P (y ) + (4) Since, P P (y 5) + ( 6) (y ) + (4) y 9 Point is (0, 9). 4. Radius cm Perimeter of a sector 66 cm Length of arc + r 66 Length of arc cm. MTHEMTICS Oswaal CSE (CCE) Class -0, S- Examination Sample Question Paper

28 8 OSWL CSE (CCE), Mathematics, Class 0 SECTION ( marks each) 5. O L In LO and LO, O O, (radii of bigger circle) OL OL, (common (from figure) LO OL 90º, (tangent is to radius at the point of contact) LO LO, (y RHS) L L is bisected at the point of contact. 6. Tower In C, tan 60º C 60 0 C 40 m C...(i) In D, tan 0º C + 40 D C C + 40 ; [using (i)] C 0 m 0 m. Height of tower 0 m. 7. (a) Total possible outcomes P (lack ball) 7 4 (b) Not green balls P (not green ball)

29 Solutions (S-) 9 8. P (, 5) k (, 5) (4, 9) Let P divides in k :. ( k 4) + ( ) k + 4 k k + 4k k + k 5 k 5 or P : P 5 : 9. rea of rectangle sq. cm rea of one circle pr.4 sq. cm. rea of 6 circles sq. cm rea to be painted sq. cm Volume pr h h 660 m 7 h Curved surface area m m Cost 880 ` SECTION C ( marks each) 5x cm ( x )cm C rea of triangle 5x (x ) ccording to questions, 5x 5x 0 x x 4 0 x 9x + 8x 4 0 x (x ) + 8 (x ) 0 (x + 8) (x ) 0 8 x, x

30 0 OSWL CSE (CCE), Mathematics, Class 0 Length can t be negative, so x 5 cm, C 9 8 cm C cm.. Given, a 9, d Q S n 6 n [a + (n )d] 6. n [(9) + (n ) ( )] 6 n[8 n + ] 4 n + n or n n 4 0 or n 7n 44 0 n 6n + 9n 44 0 n (n 6) + 9 (n 6) 0 (n 6) (n + 9) 0 n 9 or n 6 n 9 (rejected) n 6. X Y Z C X Y () (Tangents from an X Z () external point to a CZ CY () circle are equal) C, (Given) X + X Y + YC X YC Z CZ Z is the mid-point of C and Z is the point of contact. C is bisected at the point of contact. 4. C' C S R Q ' P 4 5 X

31 Solutions (S-) 5. Given that sides other than hypotenuse are 4 cm and cm. Clearly these will be perpendicular to each other. Following steps will be followed to draw the required triangle :. Draw a line segment 4 cm, draw a ray S making 90 with it.. Draw an arc of cm radius while taking as its centre to intersect S at C. Join C. C is required triangle.. Draw a ray X making an acute angle with, opposite to vertex C. 4. Locate 5 points (as 5 is greater in 5 and ),,, 4, 5 on line segment X. Such that Join. Draw a line through 5 parallel to intersecting extended line segment at. 6. Through draw a line parallel to C intersecting extended line segment C at C. C is required triangle. h 6. Let the height be h m and breadth be b m. In C, 60º 0º b C 0 m D h b tan 60º h b In D, h b + 0 h b tan 0º b + 0 b + 0 b b + 0 b 0 b 0 m h b m Height of tree is 7 m and breadth of river is 0 m. C 0 m P 0º 45º is the building of height 0 m. C is the flag pole. In P, PC 90º

32 OSWL CSE (CCE), Mathematics, Class 0 In CP, tan 0º P 0 P, P 0 7. tan 45º C P C P P C C 7. Height of flag pole m. 7. (a) Total number of coins Total number of possible outcomes of a coin will fall out 80 Number of 50p coins 00 Number of favourable outcomes relating to fall out of a 50 p coin 00 Now, P (of getting a 50 p coin) Number of favourable outcomes Total number of possible outcomes (b) P(not a Rs. 5 coin) P (Rs. 5 coin) (a) Total number of points 8 (i.e.,,,, 4, 5, 6, 7, 8) Total number of possible outcomes in which an arrow comes to rest pointing at one of the number 8 Number of favourable outcomes in which an arrow will point at 8 P (arrow will point at 8) Number of favourable outcomes Total number of possible outcomes 8 (b) Number of odd number points 4 i.e., (,, 5, 7) Number of favourable outcomes in which an arrow will point at odd number 4 Now, P (arrow will point at odd number) Number of favourable outcomes Total number of possible outcomes 4 8 (c) Number of points greater than 6 i.e., (, 4, 5, 6, 7, 8) Number of favourable outcomes in which an arrow will point at a number greater than 6 Now, P (arrow will point at a number greater than )

33 Solutions (S-) Number of favourable outcomes Total number of possible outcomes D O C D C 90 (ngle in a semi-circle) rea of DD D D cm lso rea of DC 60 cm rea of shaded region rea of circle (rea of DD + rea of DC) rea of circle (0 cm ), 7 Q cm radius cm rea of shaded region cm (6.87 0) cm cm. 60 cm r 8cm 0 cm Rcm Radius of lower cylinder R cm Radius of upper cylinder r 8 cm Height of upper cylinder h 60 cm Height of lower cylinder H 0 cm Volume of solid iron pole pr H + pr h

34 4 OSWL CSE (CCE), Mathematics, Class 0.4 () (8) cm Mass of the pole g kg. SECTION D (4 marks each). Let the number of books be x and cost of each book be ` y. xy 80 lso (x + 4)(y ) 80 xy x + 4y x x 80 0 x x x + 0 4x 0 x + 4x 0 0 x + 0x 6x x(x + 0) 6(x + 0) 0 (x 6) (x + 0) 0 x 6 or x 0 (Not possible) Number of books bought 6 +. Let numerator be x. denominator is x +. fraction ccording to question, x x+ + x+ x x x (x + x + 4x + 4) 4(x + x) 0x + 60x x + 68x 4x + 8x 60 0 x + x 5 0 x + 5x x 5 0 x(x + 5) (x + 5) 0 (x + 5) (x ) 0 x or x 5 ( Not possible) Fraction 5..P. is 6,, 5... Here, a 6 d Given, S n 5 6 +

35 Solutions (S-) 5 S n n [a + (n )d] 5 n ( n ) + 4 ( ) 50 + n n 00 n[n 5] n 5n (n 0) (n 5) 0 n 0 or 5 S 0 S 5 Two answers Q a is negative and d is positive and the sum of the terms from 6th to 0 th is zero. 4. (i)let P be the principle, R rate of interest and In be the interest at the end of n years. 5. PRN We know that Interest 00 Here, we have P ` 000 R 8% per annum n In ` 00 ` 80n Putting n,,... In ` 80, I ` 60, I ` 40 ans. so on. Since In is a linear expression in n. Therefore, the sequence of interest forms and.p. with common difference 80. (ii) Interest at the end of 0 years I 0 ` (80 0) ` 400 (iii) rithmetic Progression. (iv) Slow and steady wins the race. a O 60 P In DOP OP 90 PO 60 0 tan 0 O P a P P a Hence Proved

36 6 OSWL CSE (CCE), Mathematics, Class F x C x E 8 x x D 0 C 8 cm 0 cm C cm Let, CF x CF EC x F 8 x D E x D ut, D + D 0 8 x + x 0 0 x 0 x 0 x 5 D cm E 7 cm and CF 5 cm. 7. P(a, b) is mid-point of, where (0, 6) and (k, 4) k P (a, b), a k + 0, b a b 8 a + 8 a 6 k + 0 a k k P(a, b) (6, ) ( 0) (4 6) unit. 8. P(x, y) divides in the ratio m : m. (8, 9), (, ) m + 8m x m + m m + 8m m + m + m 9m m + m y 5 m 9m m + m 4m + 6m + 6m m 5m + 5m m 6m m m 6 8

37 Solutions (S-) 7 9. m : m 8 : 8+ 8 x 8+ 8 y P(x, y), 9. P rea of shaded region ar (DC) + ar (semi-circle with diameter C) ar (quadrant PC) C ( ) cm 98 cm Calculation of C 4 cm. 0. Let r be the radius of base and h be the height of the cylinder Total surface area pr(r + h) cm and Curved surface area prh cm ccording to question, prh [pr(r + h)] h r Curved surface area 46 cm 54 cm r π r 54 r r 7 cm Volume of cylinder pr h 7 (7) 7 59 cm.

38 Surface area of double cone prl + prl πr(l + l ) 8 OSWL CSE (CCE), Mathematics, Class 0. 4 x D E 5cm C y revolving right triangle about longest side double cone is generated. Let radius of double cone x cm. In DE and DC, ED DC 90 DE DC (common angle) DDE DDC (by ) y (i) and (ii), we get x C D DC DE D x 4 5 DE (i) (ii) (iii) x 5.4 cm DE cm 7.4 ( + 4) cm. qqq

39 SOLUTIONS SMPLE QUESTION PPER - 9 Self ssessment Time Hours Maximum Marks : 90. x ax a 0 x ax + ax a 0 x(x a) + a (x a) 0 (x a) (x + a) 0 x a 0 or x + a 0 Hence, SECTION x a or x a x a, a. Here PT.8 cm PT PR PQ [length of tangents from external points are equal] PR.8 QR PR + PQ If the points (x, ), (, 4) and C (7, 5) are collinear then, rea of C 0 [x (y y ) + x (y y ) + x (y y )] 0 x ( 4 + 5) + ( ) ( 5 ) + 7( + 4) 0 x ( mark each) 7.6 cm x Height of the cylinder h 45 cm Radius of the cylinder r cm Radius of sphere R 6 cm MTHEMTICS Oswaal CSE (CCE) Class -0, S- Examination Sample Question Paper

40 90 m 40 OSWL CSE (CCE), Mathematics, Class 0 Number of solid spheres Volume of cylinder Volume of sphere πrh 4 πr SECTION ( marks each) 5. Let PQ and RS be two parallel tangents to a circle with centre O. P Q C O 6. R Join O and O. Draw OC PQ Now, P CO PO + CO 80 [Q Sum of the angles on the same side of a transversal is 80 ] 90 + CO 80 [Q PO angle between a tangent and radius 90 ] CO 90 Similarly, CO 90 CO + CO Hence, O is a straight line passing through O. S x In right DC, C 60 sin 60 C 90 x x 80 x 60 m 60.7 Hence, length of the string 0.9 m. 7. If two coins are tossed simultaneously then Total number of events {HH, TH, HT, TT} 4

41 Solutions (S-) 4 t least one tail O is obtained, if any one of the following elementary events happens. TH, HT, TT So, the required probability 4 8. (6, 4), (5, ), C (7, ) C + ( x x ) ( y y ) ( x x ) ( y y ) (5 7) 0 + C ( x x ) + ( y y ) C Q Two sides of triangle are equal in length. Therefore, triangle is an isosceles triangle. Q C O P Here O side of square 0 cm So, area of square (0) 400 Now, radius of circle diagonal of square O + (0) + (0) 0 Now area of shaded region area of quadrant OPQ area of square OC 90 ( 0 ) 60 π 400 π cm

42 4 OSWL CSE (CCE), Mathematics, Class cm.4 cm Volume of remaining solid Volume of cylinder Volume of cone πrh πrh πrh cm SECTION C ( marks each). 4 x 5 x + 4 x x 5 x + (4 x) (x + ) 5x 8x + 6x 9x 5x 6x x + 5x 6x 6x + 0 x + x 0 x + x x 0 x(x + ) (x + ) 0 (x + ) (x ) 0 x 0 or x + 0 x or x Hence, x,. Let the first term of an P is a and common difference is d then Given : T 7 9 & T 9 7 Since, T n a + (n )d T 7 a + 6d 9 () T 9 a + 8d 7 () Subtracting eq. () from eq. (), we get d 9 7

43 Solutions (S-) 4 6 d 6. From eq. (), 6 a a Hence, T 6 a + (6 )d P 6 6. O Q Given : Two tangents from with centre O. To prove : PQ OQP Construction : Draw a line joining PQ. Proof : In DPOQ, PO OQ [radii of same circle] DPOQ is an isosceles triangle. OPQ OQP xº OP 90º PQ 90º xº OQ 90º PQ 90º xº In PQ, PQ + 90º xº + 90º xº 80º PQ x 0 PQ x PQ OPQ. Q P 5 cm 4. Consider the situation. If a circle is drawn through, D and C, C will be its diameter as DC is of 90. The centre E of this circle will be the mid-point of C. D 4cm E C

44 44 OSWL CSE (CCE), Mathematics, Class 0 Following steps will be followed for constructing the required tangents on given circle :. Join E and bisect it. Let F be the mid-point of E.. Now taking F as centre and FE as its radius draw a circle which will intersect the circle at point and G. Join G. and G are the required tangents. F D G E C 5. Let P is the position of light house. Then PM 00 m and and be the position of ship. P X' 45º 60º X 00 m Let In DMP, 45º y M M x, and M y tan 60 PM M 00 x 60º x x 00 () In DMP, we have tan 45 PM M 00 y y 00 () Now Distance between two ships x + y

45 h 0 0 m Solutions (S-) º 00(4.7) 5.46 m h 0º C x 60º 0º P In right CP, tan 60º h x x h h x...(i) In right CP, tan 0º h 0 x h 0, [from (i)] h/ h h 0 h 0 m. Height of the tower is 0 m. 7. Remaining cards 5 49 (a) P (heart) 49 (b) P (king) 49 (c) P (club) Number of playing cards remaining 5 49 (a )P (diamond) 0 49 (b) P (jack) (c) P (heart) C 90 (ngle in a semi-circle) In C, by Pythagoras theorem, C + C 9 + 6

46 46 OSWL CSE (CCE), Mathematics, Class cm Radius 5.5 cm rea of shaded region rea of semi-circle area ( C) πr C C Speed of water flowing through the pipe 5 km/hr 5000 m/hr Volume of water flowing in hr pr H Volume of water in the tank when the depth is cm.8 cm m m lbh m 46 m Time taken to fill 46 m 46 hrs. SECTION D (4 marks each). + + a+ b+ x a b x + a+ b+ x x a b x a b x b x(a+ b+ x) + a ab a+ b ( ) + + a + b ab ( ) ax bx x x + (a + b)x ab x + ax + bx + ab 0 x(x + a) + b(x + a) 0 (x + a)(x + b) 0 x + a 0 x a (OR) x + b 0 x b

47 Solutions (S-) 47. Let the speed of train x km/hr x x+ 5 x + 5x (x + 0)(x 5) 0 x 0 or x 5 Since, speed cannot be negative. Hence x 0, x 5 km/hr. Speed of train 5 km/hr.. 4th term of an.p., 0th term of the.p., ccording to given question, 7nd term of the.p., 5th term of the.p., T 4 a + (4 )d a + d...(i) T 0 a + (0 )d T 0 a + 9d...(ii) a + d (a + 9d) a 5d T 7 a + 7d or T 7 76d, [by (iii)]...(iii)...(iv) T 5 a + 4d or T 5 9d, [by (iii)]...(v) T 7 4 T 5 [by (iv) and (v)] Hence proved. 4. (i) Since, distance between the first potato and the bucket 5 m and also there are 0 potatoes which are m apart. Distance covered by the competitor in first pick 5 0 m Distance covered by the competitor in second pick (5 + ) 8 6 m Distance covered by the competitor in third pick (5 + ) (5 + 6) m Similarly, distance covered by the competitor in 0 th pick (5 + 9) (5 + 7) 64 m Therefore, the sequence becomes, 0, 6,,..., 64. This is an.p. Let S be the total distance covered by the competitor. i.e., S Here, a 0, d 6 0 6, n 0, l 64 Now, S n n (a + l) S 0 0 (0 + 64) S 0 5(74) 70 m Hence, the total distance covered by the competitor 70 m.

48 48 OSWL CSE (CCE), Mathematics, Class 0 5. In the given figure, S C R D P Q P DP...(i) length of tangents Q Q...(ii) drawn from an CR R...(iii) external point SC DS...(iv) are equal. 6. On adding (i), (ii), (iii) and (iv), we get P + Q + CR + SC DP + Q + R + DS PQ + SR PS + QR. 5 D 4 R C S O Q Let (, ), (5, ), C (, ) P 7 ORD OSD 90, (Tangent to radius) D 90 DR DS (Tangents from ext. point) P Q 7 cm C Q + QC CQ C Q 8 7 RC CQ DC DR + RC DR DC RC 5 4 DR OS 4 cm OS r 4 cm. (5 ) + ( + ) units C ( 5) + ( + ) ( ) + ( ) units

49 Solutions (S-) 49 C ( ) ( ) () 8. Since 0+ 4 units C D is isosceles. and + C C D is isosceles right triangle. C ( ) + (6 ) 0 units (5 ) + (7 6) 0 units D (4, 4) C (5, 7) CD D (4 5) + (4 7) 0 units (4 ) + (4 ) 0 units (, ) (, 6) C CD D 0 units C (5 ) + (7 ) 4 units D (4 ) + (4 6) units 9. C D Hence, CD is a rhombus. C Let radius of each circle be r cm, them C C r cm ar DC 70.5 ( ) r r , rea of each sector π π r r 60 6 rea of shaded region rea of DC (rea of sector) 70.5 π r , cm.

50 50 OSWL CSE (CCE), Mathematics, Class 0 0. ccording to question, h 7 cm R 4 cm r 0 cm V ( ) π h R + r + Rr. Cost of oil at ` 50 per litre 7( ) 7 44 cm.44 litre ` 57.0 HOME 8 m Road HOME HOME Option (i) Diameter of semi-circle 8 m Radius (r) 4 m rea of parking πr Option m. (ii) Option (iii) We suggest option as it will leave space on road for smooth traffic. (iv) Social responsibility and understanding. qqq

51 SOLUTIONS SMPLE QUESTION PPER - 0 Self ssessment Time Hours Maximum Marks : 90 SECTION. 7, 8, 6... Here, a 7, d Next term is a + d ( mark each) 4 7. Given : x, C 7 cm, CR cm, S 5 cm, S P 5 cm nd CR CQ cm Q Q C CQ 7 4 cm Hence, P Q 4 cm Now, x P + P cm. Here, (0,4) (0, 0) C (, 0) C C (0 0) (4 0) + 4 ( 0) (0 0) + ( 0) + (0 4) Hence perimeter of a triangle unit MTHEMTICS Oswaal CSE (CCE) Class -0, S- Examination Sample Question Paper

52 5 OSWL CSE (CCE), Mathematics, Class 0 4. Radius of a wheel m Distance covered in 500 revolutions 500 πr Total balls m SECTION ( marks each) (a) P(red ball ) 8 (b) P(not red ball) P( x, y) ccording to question, ( a + b, b a) P P [ x ( a+ b)] + [ y ( b a)] (, + ) a b a b [ x ( a b)] + [ y ( a+ b)] Squaring, both sides, we get [x (a + b)] + [y (b a)] [x (a b)] + [y (a + b)] [x (a + b)] [x (a b)] [y (a + b)] [y (b a)] [On applying formula a b (a + b)(a b)] [x a b] [x a + b] [y a b] [y b + a] (x a b + x a + b) (x a b x + a b) (y a b + y b + a) (y a b y + b a) (x a) ( b) (y b) ( a) (x a)b (y b)a bx ay. Proved 7. Circumference of a circle pr r 9 cm and r 9 cm Sum of the two circumferences of two circles pr + pr p 9 + p 9 56p cm Let the radius of required circle is R Then pr 56p R 56 π π 8 cm.

53 Solutions (S-) 5 8. Let the height of water raised measured h cm. Volume of water displaced in cylinder p(0) h Volume of cube cm p(0) h h cm. 9. PO OP 0 TP TPO + OP PT PT 80 (0 + 0 ) 0 0. Rope m 0 C Let be the vertical pole of the height m. So, sin 0 C C C 4m Hence, the distance covered by the circus artist is 4 m. SECTION C. x + px 5 0 Since 5 is the one root of this equation, So ( 5) + p( 5) p p 0 ( marks each) p gain, p (x + x) + k 0 7(x + x) + k 0 7x + 7x + k 0 Re-Equation has equal roots b 4ac 7 4(7)(k) 4k 7 k (i). Here a 8, d T n T n a + (n )d 5

54 54 OSWL CSE (CCE), Mathematics, Class (n ) ( ) n 5 5 ( ) n 7 n l n 8 S n [ + ( ) ] 8 8 (8 ) ( 6) 44 C D O E F Let be a diameter of a given circle and let CD and EF be the tangent lines drawn to the circle at and respectively. CD and EF C 90 and F 90 C F C and F are alternate int. angles. CD EF 4. '.5 cm C' O 7.5 cm C 4 5 Steps of construction :. Draw a line C 7.5 cm.. Draw a perpendicular bisector of C which cut the line C at O. 6 7

55 Solutions (S-) Cut the line O.5 cm. 4. Join to and C. 5. C is the required triangle. 6. Draw a ray X making an acute angle. 7. Locate 7 points,,..., 7 on line segment X. Such that Join 7 C. Draw a parallel line through 4 to 7 C intersecting extended line segment C at C. 9. Through C draw a line parallel to C intersecting x the segment at. 0. Hence, D C is the required triangle. 0 Let CD be the light house of height h. Let and be the two ships. Given : 00 m CD 0 and CD 45 In CD tan 0 0 D h In CD tan 45 C h D h D h D D h Since 00 D + D 00 h+ h 00 h ( + ) h ( ) ( ) 50(.7 ) m. E 60 m D 0 60 h C 60 x Let EC h and distance between tower & building is x. Then

56 56 OSWL CSE (CCE), Mathematics, Class 0 In DCE tan 0 h DC or h x () In D tan x x 60 0 m From (), h 0 h 0 0 m i.e., difference between their heights h 0 m. 7. Total numbers 0 (a) Multiples of 7 7, 4,, 8 P (a multiple of 7) (b) P (even number) 5 0 (c) Prime numbers,, 5, 7,,, 7, 9,, 9, P (a prime number) Total number of possible outcomes 6 (a) For a sum 8, the outcomes are (, 5) (5, ), (4, 4), (, 6), (6, ) i.e., 5 P(a sum of 8) 5 6 (b) For a sum > 0, outcomes are (5, 6), (6, 5), (6, 6) P (sum >0) 6 9. rea of bigger semi-circle πr sq cm πr rea of one smaller semi-circle sq cm Required area sq cm.

57 Solutions (S-) Slant height of cone, l h + r Total surface area of the toy Surface area of hemisphere + curved surface area of cone pr + prl cm. SECTION D (4 marks each). Suppose alone can finish the work in x days and alone takes (x 6) days. ccording to the question, + x x 6 4 x 4x x x x x (x ) (x ) 0 (x ) (x ) 0 x or x ut x cannot be less than 6. So x can finish the work in days.. Let the original price of the book x Number of books bought for ` x Reduced list price of the book x 5 Number of books bought for ` x 5 It is given that, x 5 x 00 x 00x+ 500 xx ( 5) x 5x 5 x 5x 00 0 x 0x + 5x 00 0 x(x 0) + 5(x 0) 0 (x 0) (x + 5) 0 x 0 0 or x x 0 or x 5, which is not possible Hence, x 0 So, the list price of the book ` 0.. Let a be the first term and d be the common difference. S [ + ( ) ] n a + n d S [ ( ) ]

58 58 OSWL CSE (CCE), Mathematics, Class 0 n a + n d S [ ( ) ] n n Now (S S ) { a+ ( n ) d} { a+ ( n ) d} n { 4a+ ( n ) d} { a+ ( n ) d} n { 4a + 4 nd d a nd + d} n { a + nd d} n { a+ ( n ) d} S Hence Proved. 4. a + a 7 6; a a 7 8 a + 8d 6; (a + d)(a + 6d) 8. a + 4d a 4d. ( 4d + d) ( 4d + 6d) 8 ( + d) ( d) 8 9 4d 8, d 4 d ± Case (i) d a ; S 0 5 Case (ii) d a 5; S P T R O Q In D s TPO and TQO, we have TPO TQO [each equal to 90 ] So, by RHS congruence criterion, We have, DTPO DTQO PTO QTO lso, DPTR DQTR [by SS Congruence criterion] PR QR and TRP TRQ ut, TRP + TRQ 80 TRP TRQ 90 Hence, PQ and OT are right bisectors of each other.

59 Solutions (S-) S P O 4 5 R C Q D Let CD be a quadrilateral circumsribing a circle centered at O such that it touches the circle at points P, Q, R, S. Let us join the vertices of the quadrilateral CD to the centre of the circle. Consider DOP and DOS. P S (Tangents from the same point) OP OS (Radii of the same circle) O O (Common side) DOP DOS (SSS congruence criterion) Therefore,, P S, O O nd thus, PO OS 8 Similarly, ( + 8) + ( + ) + ( 4 + 5) + ( 6 + 7) ( + ) + ( 5 + 6) 60 ( l + ) + ( 5 + 6) 80 O + COD 80 Similarly, we can prove that OC + DO 80 Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angle at the centre of the circle. 7. x + y 90 (, ) k (, x y) (, 7) Let the line x + y 9 0 divide the line at (x, y) in the ratio k : So, x k +, y 7 k + k + k + (x, y) lies on x + y 9 0 so, k 7 k + k+ k+ 9 0

60 60 OSWL CSE (CCE), Mathematics, Class 0 6k + + 7k + 9k 9 0 4k 0 k 4 8. So, line divides the given line segment in the ratio of : 4. (, ) (, 6) O D( x, y) C (5, 0) CD is a parallelogram. Diagonal of a gm bisect each other. If C and D are diagonals, O is the mid point of C and D. x + x Co-ordinates of O in C y + y, O is also the mid-point of D, hence + x 6, 8 [, 4], 6 + y 4 x, y Fourth vertex is (, ) 9. rea of the rectangle cm In O, O + O 400 O 90 rea of segment E rea of sector EO area of triangle O cm rea of the shaded region cm. 0. (a) Radius of the Gold scoring area 0.5 cm rea of the Gold scoring region [ pr ] 46.5 cm Radius of combined Gold and Red region cm rea of Red scoring region rea of combined Gold and Red regions rea of the Gold region cm

61 Solutions (S-) 6 Radius of combined Gold, Red and lue regions cm rea of lue scoring region rea of combined Gold, Red and lue region rea of combined Gold and Red region cm Radius of combined Gold, Red, lue and lack region cm rea of black scoring regions rea of combined Gold, Red, lue and lack regions rea of combined Gold, Red and lue regions cm Radius of combined Gold, Red, lue, lack and White regions cm rea of white scoring region rea of combined Gold, Red, lue, lack and White regions rea of combined Gold, Red, lue and lack regions cm (b) reas of related to circle. (c) Logical proportion is required every where.. Radius of the sphere cm Volume of the sphere 4 π () 6p cm Radius of the cylindrical vessel 6 cm Suppose water level rises by h cm in the cylindrical vessel then, Volume of the cylinder of height h cm and radius 6 cm (p 6 h) cm 6πh cm Clearly, volume of water displaced by the sphere is equal to the volume of the sphere 6p cm 6ph cm h cm Hence, water level rises by cm. qqq