CBSE Continuous and Comprehensive Evaluation (CCE) QUESTION. Term 2 (October to March 2015) Mathematics

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1 CSE Continuous and Comprehensive Evaluation (CCE) SMPLE QUESTION PPERS Solutions Term (October to March 05) Mathematics *Solutions for SQP 6-0 can be downloaded from Class0 Published by : OSWL OOKS Oswaal House /, Sahitya Kunj, M.G. Road, GR-800 Ph.: , 5778, Fax : , contact@oswaalbooks.com, website :

2 SOLUTIONS SMPLE QUESTION PPER - 6 Solved Time 3 Hours Maximum Marks : 90. SECTION 0x 3 x 0x 3x 0x 3x 0 (5x + ) (x ) 0 x or x 5.. We know that, the tangents drawn from an external point to the circle are equal in length. L N 4 cm CL 6 cm Now, C L + CL cm. 3. (5, ) : P (4, ) mx + nx my + ny Point of division, m+ n m+ n. y + ( ) + (, ) y y Let the radius of the protractor be r cm. Given, perimeter 08 ( π r ) + r 08 pr + r 08 r+ r 08 7 MTHEMTICS Oswaal CSE Class -0, S- Examination Sample Question Paper

3 OSWL CSE (CCE), Mathematics Class r 08 7 r cm diameter of the protractor 4 cm. T SECTION P O Join PO and produce to D. OP TP and TP DP 90 OD D D DDP DP, (SS) P P, (CPCT) Hence, P is an isosceles triangle 6. D D x 45 0 m P 30 From figure, Tan 30 P P 0 3 i.e., In DPD 5 7. (i) 6 (ii) 6 8. Distance formula pplying it, we get distance of the building m tan 45 D P 0 + x 0 3 x 0( 3 ) 7.3 m ( x x ) + ( y y ) 50, C 0 and C 50. Since C, then triangle is isosceles.

4 Sample Question Papers (S-) 3 9. Shaded portion indicates the area which the horse can graze. Clearly, shaded area is the area of a quadrant of a circle of radius r m. Required area pr Required area cm cm 0. So, πr + πr + πr π R πr(6) + πr(8) + πr(0) R cm 3 π 3 4 ( ) 3 π π R 78 R 3 cm R radius of resulting sphere cm. SECTION C. Marks obtained in maths x Marks obtained in science 8 x s per question, (x + 3) (8 x 4) 80 (x + 3) (4 x) 80 4x x + 7 3x 80 x x x ( + 9)x x x 9x x(x ) 9 (x ) 0 (x 9) (x ) 0 x 9 or x. Case I : When x 9 then Marks obtained in maths 9 Marks obtained in science 9

5 4 OSWL CSE (CCE), Mathematics Class 0 Case II : When x then Marks obtained in maths Marks obtained in science 6.. Let the first term of an P is a and common difference is d, then ccording to first condition a 4 + a 8 4 a + 3d + a + 7d 4 a + 0 d 4 a + 5d...(i) ccording to second condition a 6 + a 0 44 a + 3d + a + 9d 44 a + 4d 44 a + 7d a + 5d a + 7d d 0 d 5 Putting d 5 in (i), we get a + 5 a 5 a 3. Hence P is a, a + d, a + d,... 3, 8, 3, P. O R PQ and RS are two parallel tangents to a circle with centre O. is tangent to circle at C, intersecting and PQ and RS at respectively. Since P RS and is transversal , 3 4 (by congruancy) In DO, by angle sum property of a triangle, O 80 ( + 3) O Following steps will be followed for constructing the tangents on the given circles :. Draw a line segment of 8 cm. Taking and as centre draw two circles of 4 cm and 3 cm radius.. isect the line. Let mid-point of is C. Taking C as centre draw a circle of radius C which will intersect the circles at point P, Q, R and S. Join P, Q, S and R These are required tangents. 4 3 C Q S

6 Sample Question Papers (S-) 5 P O S C Q O' R 5. C H m x D H x E H + 00 F In DDC, tan 30 H 00 x 3 H 00 x or x 3 (H 00) m In DDF, tan 60 3 H + 00 x H + 00 x 3 H ( H 00) 3 (H 00) H H H H 800 So, required height H 400 m. 6. l D l h C 45º 0 m

7 6 OSWL CSE (CCE), Mathematics Class 0 Let is the height of a tree which is broken at the point D and its top touches the gound at point C. h 0 tan 45 h 0 h 0 m lso, h l 0 l sin 45 l 0 Height of tree (a) (a) (b) P(sum is 8) 5 36 (b) ( + ) m. S {HH, HT, TH, TT} (c) 4 9. rea of semi-circle with diameter cm, r, π r rea of semi-circle with diameter M rea with diameter MN rea with diameter N radius 7 cm rea π r cm cm rea of shaded region rea on (rea on MN + rea on M + rea on N) as diameter cm.

8 Sample Question Papers (S-) 7 0. Height of the cone 4 cm Given, pr 8 R 8 π R 9 π cm pr 6 Curved surface area of frustum pl(r + r) r 6 π 3 π cm 9 3 π 4 + π π 4π π 48 cm. SECTION D. Let the usual speed be x km/h, then ccording to question, x x 0 3 x+ 0 x 360 xx ( + 0) (x + 0x) x + 0x 00 x + 0x 00 0 x + 40x 30x 00 0 x(x + 40) 30(x + 40) 0 (x 30) (x + 40) 0 x 30; x 40 (neglected) Usual speed of the train is 30 km/h.. (a) Suppose rjun had x arrows. x Number of arrows used to cut arrows of heeshm Number of arrows used to kill the rath driver 6 Number of other arrows used 3 Remaining arrows 4 x + So, x x + x

9 8 OSWL CSE (CCE), Mathematics Class 0 x x x x x Put x y, then y 0 + 8y y 8y 0 0 y 0y + y 0 0 y(y 0) + (y 0) 0 (y 0) (y + ) 0 y 0 or y 5 0, [y can not negative] x y x 00 Hence, the number of arrows which rjun had 00 (b) Quadratic equation. (c) Friendship. 3., 9, 6,, is the given.p., then, d 9 ( ) 3, T n T n a + (n )d + (n )3 + 3(n ) 33 3(n ) (n ) n + Now, sum of all terms of an.p. S n n [a + (n ) d] S [ ( ) + ( ) 3] 6 [ ] Let r, r, be the radii of semi-circle and l, l,.. be the lengths of circumferences of semi-circles, then l pr p() p cm l pr p() p cm l 3 3p cm : : l p cm Total length of spiral l + l l l p + p + 3p p l p( ) π l cm. l

10 Sample Question Papers (S-) 9 5. D F 6. E C F D E D, (tangents from external points) CE CF C D + D F + FC D FC, ( D F) E EC, ( D E, CE CF) E bisects C. x cm x cm 6cm F 4cm I 4cm 4cm E 8cm D C 6cm 8cm C (x + 8) cm, (x + 6) cm S semi-perimeter x + 4 rea of DC s( s a)( s b)( s c) rea of DC 48x 67 + x...(i) ar (DIC) cm...(ii) ar (DI) (x + 6) 4 cm...(iii) ar (DIC) (x + 8) 4 cm...(iv) ar (DC) ar (DIC) + ar (DI) + ar (DIC) 48x 67 + x 8 + (x + 6) 4 + (x + 8) 4 4x x + 67x 6 (x + 4) x 7 C x cm x cm.

11 0 OSWL CSE (CCE), Mathematics Class 0 7. (3, 4) (, 3) Here D is the mid-point of C, then ( D 3, 9 ( C (5, 6) D + 5, , Length of D unit rea of DD rea of DCD (4 3) square unit (4 6) square unit Hence, area of DD rea of DCD.

12 Sample Question Papers (S-) 8. (, ) (3, ) C (, 4) Here P, Q and R are the mid-point of, C and C, then Co-ordinates of rea of DPQR rea of DC P Q + 3 +, 3 + 4, (. 0) (, 3) + 4 R, (0, ) [(3 ) + ( - 0) + 0(0 3)] [ ] square unit [( 4) + 3(4 + ) ( )] [ ] 0 0 square unit m E 06 cm C O O H D F We have O O C 30 m CD 0 m O O D (30 + 0) m 40 m. ar (track) ar (rectangle CD) + ar (rectangle EFGH) + (ar semi-circle of radius 40) ar (semi-circle of radius 30) G [(0 06) + (0 06)] + (40) (30) 7 7 (0 + 00) m 430 m.

13 OSWL CSE (CCE), Mathematics Class Volume of cylinder Volume of cone p(8) (3) π ( r ) (4) 3 r r 36 cm Slant height of cone l r + h 3. Water flows in hr 0 km (36) + (4) cm. Water flows in hr 0 5 km 5000 m Now volume of water flows in ccording to question, hr lbh m m 3. Volume of water in hr area of field 8 00 m rea 8 rea hectares.

14 SOLUTIONS SMPLE QUESTION PPER - 7 Solved Time 3 Hours Maximum Marks : 90 SECTION. 5x 0 6x x 6x + 0 3x 6x 6x + 0 3x ( 3x ) ( 3x ) 0 ( 3x )( 3x ) x, 3 3 P cm Here, P and P are two tangents to circle with centre O. O P OP 30 In OP, we have P cot 30 O P 3 3 P 3 3 cm. 3. Let (3, ), ( 3, ), C (x, y) be the points of equivalent triangle. (x 3) + (y ) 36 (x + 3) + (y ) 36 O MTHEMTICS Oswaal CSE Class -0, S- Examination Sample Question Paper

15 4 OSWL CSE (CCE), Mathematics Class 0 from equation () and () x 0, y Surface area of sphere 4pr 66 4 r 66 7 r So, radius of sphere 49 7 cm SECTION 5. PO OP 30 TP TPO + OP PT PT 80 ( ) Rope m C 30 Let be the vertical pole of the height m. So, sin 30 C C C 4m Hence, the distance covered by the circus artist is 4 m. 7. Total ball 8 (a) P(red ball ) 3 8 (b) P(not red ball) P( x, y) Q ( a + b, b a) R ( a b, a + b)

16 Sample Question Papers (S-) 5 ccording to question, PQ PR [ x ( a b)] [ y ( b a)] [ 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 (a + b)] (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 0 a) ( b) (y b) ( a) (x a)b (y b)a bx ay. Proved 9. 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 0. 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 cm. h cm. SECTION C x x+ + 4x + 4x+ x x+ x 5x + x+ x x 5x + x + 4x x x + 4x (x + )(x + ) 0 x,

17 6 OSWL CSE (CCE), Mathematics Class 0. Let first term of an P is a and common difference is d then ccording to question, 3. and a a + 6d a + 6d 3 a + d 7 a 3 a + d 7 4d 6 d 4 a a Now, a 0 a + 9d nd, a n a + (n )d + 4n 4 4n 5 P R Q C O C, (Tangents drawn from an ext. points are equal) P PR QC QR P + P P + PR C Q + QR + C P + PR + Q + QR P + PQ + Q Perimeter of DPQ (Perimeter of DPQ) 4. Steps of Construction :. Draw a circle of radius 4 cm with centre O.. Draw another circle of radius 6 cm with same centre O. 3. Take a point P on second circle and join OP. 4. Draw perpendicular bisector of OP which intersect OP at O. 5. Draw a circle with centre O which intersect inner circle at points and. 6. Join P and P. P and P are required tangents.

18 Sample Question Papers (S-) 7 5. Let PO x and QR h then, R h Q In triangle DPOQ P 45 30º tan 30 OQ PO x 3 0 x O 0 m x 0 3 m...(i) 6. Now, in DPOR, From equation (i) and (ii) tan 45 OR PO h + 0 x x h + 0 h h (ii) 0( 3 ) 0(.73 ) 0(0.73) 7.3 m The length of the flagstaff is 7.3 m and the distance of the building from the point P is 0 3 m (or 7.3 m.) In DO, tan 60 O 3 h x Now, in DDCO, h 3x tan 30 DC CO

19 8 OSWL CSE (CCE), Mathematics Class 0 3 h 80 x 3 3 x 80 x 80 x 3x 3x + x 80 4x 80 x 0 m Height of the poles is 0 3 m. Distance from first pole is O 0 m. Distance from second pole is OC m. 7. Favourable event Some a (H), (H), (H3), (H4), (H5), (H5) (T), (T), (T3), (T4), (T5), (T6) (a) P(number on die is prime) 6 6 (b) P(head) (c) P(head and even number) (a) P(extremely patient) 3 4 (b) P(extremely kind or honest) Extremely Honest. 9. 7cm D 7 cm O 7cm 7cm C rea of circle on DO as diameter rea of semi-circle on as diameter r π R sq. cm sq. cm. 7 rea of C sq. cm. rea of shaded region rea of circle on DO + rea of semi-circle on rea of DC sq. cm.

20 Sample Question Papers (S-) 9 0. (i) C C 5 cm ar ( C) ar ( C) C C D D E D C C D D cm Volume of double cone Volume of upper cone + Volume of lower cone 3 π(d) D + 3 π(d) CD 3 π(d) {D + CD} 3 π(d) (C) cm. Surface area C.S.. of upper cone + C.S.. of lower cone π()(0) + π()(5) π{0 + 5} cm. SECTION D. 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 x x x x 0 x + 4x 30 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 +

21 0 OSWL CSE (CCE), Mathematics Class 0. Let numerator be x. denominator is x +. x fraction x + ccording to question, x x x+ x 5 5(x + x + 4x + 4) 34(x + x) 30x + 60x x + 68x 4x + 8x 60 0 x + x 5 0 x + 5x 3x 5 0 x(x + 5) 3(x + 5) 0 (x + 5) (x 3) 0 x 3 or x 5 (Not possible) Fraction P. is 6,, 5... Here, a 6 Given, S n 5 d + 6 S n n [a + (n )d] 5 n ( n ) ( n ) 50 n 00 n[n 5] n 5n (n 0) (n 5) 0 n 0 or 5 S 0 S 5 Two answers 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 I n be the interest at the end of n years. We know that Interest Here, we have P ` 000 R 8% per annum I n ` ` 80n Putting n,, 3... I n ` 80, I ` 60, I 3 ` 40 ans. so on.

22 Sample Question Papers (S-) Since I n is a linear expression in n. Therefore, the sequence of interest forms and.p. with common difference 80. (ii) Interest at the end of 30 years I 30 ` (80 30) ` 400 (iii) rithmetic Progression. (iv) Slow and steady wins the race. 5. a O 60 P In DOP OP 90 PO tan 30 O p 3 a P P a 3 Hence Proved C E F Let, D C 8 cm b 0 cm C cm CF x CF EC x F 8 x D E x D 8 x + x 0 0 x 0 x 0 x 5 D 3 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), k + 0 a, b

23 OSWL CSE (CCE), Mathematics Class 0 a b 8 a + 8 a 6 k + 0 a k k P(a, b) (6, ) ( 0) + (4 + 6) 6 unit. 8. P(x, y) divides in the ratio m : m. (8, 9) : (, ) 9. m 8 + m m 9m 3 m + m + m + m m + 8m x m + m m 9m y m + m 5 0 m 6m 0 i.e., m : m 8 : x y P(x, y) , 3 9. rea of shaded region ar (DC) + ar (semi-circle with diameter C) ar (quadrant PC) ( ) cm 98 cm Calculation of C 4 cm. 30. 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)] 3

24 Sample Question Papers (S-) 3 3. h r Curved surface area 46 3 cm 54 cm r πr 54 r r 7 cm Volume of cylinder pr h 7 (7) cm 3. 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 DE 3 (i) (ii) (iii) x 5.4 cm DE cm Surface area of double cone prl + prl πr(l + l ) 7.4 (3 + 4) cm.

25 SOLUTIONS MTHEMTICS Oswaal CSE Class -0, S- Examination Sample Question Paper SMPLE QUESTION PPER - 8 Solved Time 3 Hours Maximum Marks : 90 SECTION. Let the x th term of an.p be zero, then a n 0 a + (n ) d (n ) ( 4) 0 4n 4 n The first negative term of the.p. is (n + ) th term 3 th term. DC C 50 In DCD, CD 80 ( DC + DC ) 80 ( ) x + x + 3. Mid-point of 5 x 6 nd, y + + y 3 4. Radius of circle (r) cm ngle of sector (θ) 50 θ rea of sector πr 360 y cm.

26 Sample Question Papers (S-) 5 SECTION 5. Let O x and C h C h 3000 h O x Given, C 3000 m, O 45º, OC 60º In O, O 3000 h x tan 45 x 3000 h...(i) C 3000 In OC, tan 60 3 O x From (i) and (ii), we get x (ii) x 73 m h h h 68 m. 6. O P O 5 cm, OP 3 cm OP P + OP O P cm 4 8 cm. 7. There are seven different kind of leap year calenders. n(s) 7 out of these 7 calenders, two calenders will have 53 Wednesdays

27 6 OSWL CSE (CCE), Mathematics Class 0 P (53 Wednesdays in a leap year) Mid-point of C Mid-point of D + 0, +, 3 +,, Mid-point of C Mid-point of D Diagonal bisect each other. CD is a parallelogram. 9. Given, r cm q 0 rea of segment ( ) 0 sin 80 Proved. 44 ( ) 08.7 cm. 0. Diameter of hemisphere Side of cubical box R 7 R 7 Surface area of solid surface area of the cube area of base of hemisphere + curved surface area of hemisphere 6l pr + pr cm SECTION C. Let the age of father x year and son y year ccording to Questions, x + y 35 and xy 50 y 35 x Putting the value of y, x (35 x) 50 x 35x (x 30) (x 5) 0 x 30; x 5 (rejected) y 5 The age of father 30 years and age of son 5 years.. Two digit odd positive numbers are, 3, 5, 7, Here, a d 3 T n 99 T n a + (n )d

28 Sample Question Papers (S-) (n ) n 90 n Now, S n x [a + (n )d] [ + (45 ) ] 45 [ + 44 ] 45 [55] 475. Ȯ Q P xº T Let PTQ xº, then by angle sum property of a triangle, TQP + TPQ 80º x...(i) ut TP TQ TQP TPQ (80 x ) x 90 OPQ ( OPT TPQ) x PTQ OPQ PTQ 4. Let us assume that C is right-angled at, with base C 5 cm and C + 0 cm. C whose sides are times of C can be drawn as follows :. Draw a line segment C of length 5 cm.. t, draw XC 90. Taking as centre and radius as 0 cm, draw an arc that cuts the ray X at Y. 3. Join CY and draw its perpendicular bisector to intersect Y at. Join C. 4. Draw a ray Z making an acute angle with line segment C on the opposite side of vertex. 5. Locate 7 points,, 3, 4, 5, 6 and 7 on Z such that Join C 5 and draw a line C 7 parallel to C 5 to intersect extended line segment C at point C. 7. Draw a line through C parallel to C intersecting the ray X at.

29 8 OSWL CSE (CCE), Mathematics Class 0 C is the required triangle. X Y 0 cm ' 5cm C C' Justification : The construction can be justified by proving that : Z 7 5, C 7 5 C, C 7 5 C In C and C, C C C C C ~ C ( similarity criterion) ' C C ' C C ' ' In 5 C and 7 C, 5 C 7 C 5 C 7 C (Corresponding angles) 5 C ~ 7 C ( similarity criterion) C C ' 5 C C ' 5 7 On comparing equations (i) and (ii), we obtain ' C C ' C C ' ' , C 7 5 C, C 7 5 C 7 (Common) (Corresponding angles) (i) (Common)...(ii) This justifies the construction.

30 Sample Question Papers (S-) 9 5. P Q O M L L' 88. M Let P be the position of the balloon when its angle of elevation from the eyes of the girl is 60 and Q be the position when angle of elevation is 30. M M' In DOLP, tan 60 PL QL OL OL 87 3 PL' LL' OL 88.. OL In DOMQ, tan 30 QM OM QM ' MM ' OM OM OM 87 3 Distance travelled by the ballon, PQ LM OM OL m m 74 3 m 58 3 m.

31 30 OSWL CSE (CCE), Mathematics Class m P O 60º 30º H 0 P R H H ' In OP, tan 30º In DOP, tan 60º From (i) and (ii), we get 3 H 0 OP 3 H 0 OP H (i) OP ( ) OP H + 0 H + 0 OP H + 0 OP H ( H 0)...(ii) 3 So, height of cloud H 40 m. 7. Total number of defective pens Total number of good pens 3 Total number of pens Probability that the pen taken out is good The possible out comes of a coin is tossed 3 times. S { (HHH), (TTT), (HTT), (THT), (TTH), (THH), (HTH), (HHT)} n(s) 8 (a) Let E Event of getting all heads {HHH} n(e ) ne ( ) P(E ) ns () 8 (b) Let E Event of getting at least heads { (HHT) (HTH) (THH) (HHH)} n(e ) 4 ne ( ) P(E ) 4 ns () 8

32 Sample Question Papers (S-) 3 9. C 54 cm, C 0 cm 44 cm radius of small circle cm r (shaded region) r (big circle) r (small circle) p(7 7) p( ) p[(7) () ] p[79 484] cm. Volume of godown Volume of cuboid + Volume of cylinder L H πr h. a+ b+ x a+ b+ x x x a b x x(a+ b+ x) ( a+ b) ax + bx + x m 3. SECTION D + + a b x + a b b+ a a+ b ( a + b) ab x + (a + b)x ab x + ax + bx + ab 0

33 3 OSWL CSE (CCE), Mathematics Class 0 x(x + a) + b(x + a) 0 (x + a)(x + b) 0 x + a 0 x a x + b 0 b x. Let the speed of train x km/hr x x + 5 x + 5x (x + 30)(x 5) 0 x 30 or x 5 Since, speed cannot be negative. Hence x 30, x 5 km/hr. Speed of train 5 km/hr th term of an.p., T 4 a + (4 )d a + 3d...(i) 0 th term of n.p., T 0 a + (0 )d a 9d...(ii) ccording to given question, a + 3d (a + 9d) a 5d...(iii) 7 nd term of.p., T 7 a + 7d or T 7 76d, [by (iii)]...(iv) 5 th term of.p., T 5 a + 4d or T 5 9d, [by (iii)]...(v) T 7 4 T 5 Now, by (iv) and (v), we get T 7 4 9d 76d Hence proved. 4. (i) Since, distance between the first potato and the bucket 5 m and also there are 0 potatoes which are 3 m apart. Distance covered by the competitor in first pick 5 0 m Distance covered by the competitor in second pick (5 + 3) 8 6 m Distance covered by the competitor in third pick (5 + 3) (5 + 6) m Similarly, distance covered by the competitor in 0 th pick ( ) (5 + 7) 64 m Therefore, the sequence becomes, 0, 6,,..., 64 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)

34 Sample Question Papers (S-) 33 S 0 0 (0 + 64) S 0 5(74) 370 m Hence, the total distance covered by the competitor 370 m. 5. In the given figure, 6. P DP Q Q CR R SC DS...(i)...(ii)...(iii)...(iv) (length of tangents drawn from an external point are equal.) 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 38 7 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 38 7 RC CQ DC DR + RC DR DC RC 5 4 DR OS 4 cm OS r 4 cm. 7. Let (3, ), (5, ), C (3, 3) (5 3) + ( + ) units C (3 5) + ( 3 + )

35 34 OSWL CSE (CCE), Mathematics Class 0 C ( ) + ( ) units (3 3) + ( + 3) 0 + () 0+ 4 units Since C D is isosceles. and + C C D is isosceles right triangle. 8. ( ) + (6 3) 0 unit C CD (5 ) + (7 6) 0 unit (4 5) + (4 7) 0 unit D (4 ) + (4 3) 0 unit C CD D 0 unit C (5 ) + (7 3) 4 unit D (4 ) + (4 6) 4 unit Hence, CD is a rhombus. 9. Let radius of each circle be r cm, them C C r cm ar DC ( ) r r rea of each sector 60 πr 360 rea of shaded region rea of DC 3 (rea of sector) π r 6 π 6 r cm.

36 Sample Question Papers (S-) ccording to question, h 7 cm R 4 cm r 0 cm V 3 h + r + Rr) 7( ) cm 3.44 litre Cost of oil at ` 50 per litre ` (i) Diameter of semi-circle 8 m Radius (r) 4 m rea of parking πr m. (ii) Option (iii) We suggest option as it will leave space on road for smooth traffic. (iv) Social responsibility and understanding.

37 SOLUTIONS MTHEMTICS Oswaal CSE Class -0, S- Examination Sample Question Paper SMPLE QUESTION PPER - 9 Solved Time 3 Hours Maximum Marks : 90 SECTION. Since roots are equal so D 0 b 4ac 0 4K 4K 0 K O + COD COD 80 COD y 6 7 x Hence, the coordinates of (, ) 4:3 P (, x y) 4. Perimeter of circle pr Perimeter of square 4a ccording to question, pr 4a r a π (, 3) P 6 5, 7 7. Ratio of their areas 4πr a 4 π 4 7 π a a π 4.

38 Sample Question Papers (S-) 37 SECTION 5. O P OP OP 90º OP 80 ( ) 55º O 35º P 90º 35º 55º m C Let C be the point where the ball is C 60 (alt. angles) In DC, tan 60 C x x m. 7. Total cards 30 Number divisible by 3 3, 6, 9,, 5, 8,, 4, 7, 30 Total number 0 P (number divisible by 3) P (Not divisible by 3) (3, a) lies on x 3y 5 0 So, 3 3a 5 0 3a a 3 x 3y 5 0 cuts the x-axis at (x, 0).

39 38 OSWL CSE (CCE), Mathematics Class 0 x 5 0 x 5 Point is 5,0 9. Surface area of two hemispheres cm 7 Surface area of cylinder cm Total surface area of article cm. 0. Number of bottles Vol. of liquid in hemisphere Vol. of one cylindrical bottle (9) π 3 R π 3 rh 54 bottles. SECTION C. Let total number of camels x ccording to question, x + 4 x + 5 x 3x 8 x 60 0 Let x y, then 3y 8y y 8y + 0y y(y 6) + 0(y 6) 0 (3y + 0)(y 6) 0 y 6 or y y (not possible) So, y 6 y 6 36 Hence, number of camels 36. Series of two digits numbers which is divisible by 6 is :, 8, 4,..., 96 Here a, d 8 6, Tn 96 T n a + (n )d 96 + (n ) 6 84 (n ) 6 4 n n 5

40 Sample Question Papers (S-) 39 S n n [a + (n )d] 5 [ + (5 ) 6] 3. 5 [ + 4 3] 5[ + 4] cm r O 6cm C rea of C cm C cm ar ( C) ar (DOC) + ar (DOC) + ar (DO) r r 4 r cm. 4. 5cm O' ' 60 6cm O C' C 3 4 Following steps will be followed to draw a C whose sides are 3/4 of corresponding sides of C.. 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 0. X

41 40 OSWL CSE (CCE), Mathematics Class Draw a ray X making an acute angle with C on opposite side of vertex. 4. Locate 4 points (as 4 is greater in 3 and 4),, 3, 4 on line segment X. 5. Join 4 C and draw a line through 3 parallel to 4 C intersecting C at C. 6. Draw a line through C parallel to C intersecting at. C is the required triangle. Let is a building 60 m high and x is a tower h m high. ngle of depressions of top and bottom are given 30º and 60º respectively. DC E h m and let C x E (60 h) m In ED, In C, 60 h ED tan h C 3(60 h) x...(i) 60 x tan 60 Putting the value of x from equation (i) in equation (ii), we get 60 3x...(ii) 6. Height of tower is 40 m (60 h) 60 3 (60 h) 0 60 h h 40 m C D h h 60º 30º x 800 m In 3600 sec distance travelled by plane m E

42 Sample Question Papers (S-) 4 In 0 sec distance travelled by plane In C, In C, From equations (i) and (ii), we get h x h x h x tan 60º h x m h x 3...(i) h x 3 tan 30º 3 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 ll possible outcomes are 36 (, ) (, ) (, 3) (, 4) (, 5) (, 6) (, ) (, ) (, 3) (, 4) (, 5) (, 6) (3, ) (3, ) (3, 3) (3, 4) (3, 5) (3, 6) (4, ) (4, ) (4, 3) (4, 4) (4, 5) (4, 6) (5, ) (5, ) (5, 3) (5, 4) (5, 5) (5, 6) (6, ) (6, ) (6, 3) (6, 4) (6, 5) (6, 6) (a) Sum of two numbers appearing on top of dice is (b) Outcomes appearing on top of dices are same (c) (ii)

43 4 OSWL CSE (CCE), Mathematics Class 0 9. rea of shaded region p[r r θ ] [7 (3.5) ] cm For cylinder, r 8 cm, h cm For hemisphere, r 8 cm C.S.. of solid C.S.. of cylinder + (C.S.. of hemisphere) prh + 4pr prh + 4pr pr (h + r) 8(7 + 36) 7 Cost of polishing ` 0 70 per sq. cm ` SECTION D. Let marks in Mathematics x Let marks in Science 3 x or (3 x ) (x + 4) 53 or (30 x) (x + 4) 53 6x x or x 6x or x 9x 7x x(x 9) 7(x 9) 0 x 7 or 5 9 If x 7, then marks in Mathematics 7 marks in Science 5 and if x 9, then marks in Mathematics 9 marks in Science 3.. (b c)x + (c a)x + (a b) 0 D 4C 0 Equal roots (c a) 4(b c)(a b) 0 c + a ac 4(ab ac b + bc) 0

44 Sample Question Papers (S-) 43 c + a ac 4ab + 4ac + 4b 4bc 0 c + a + ac 4ab 4bc + 4b 0 4b + c + a 4ab + ac 0 (b c a) 0 b c a 0 b a + c Hence proved. 3. S n 300 a 0, d S n n [a + (n )d] 300 n (0) ( n ) n 40 n n[0 n + ] n n 800 n n n 6n n 36n 5n n(n 36) 5(n 36) 0 (n 36) (n 5) 0 n 36, 5. i.e., the sum of all terms from 5 th to 36 th term is zero. 4. Let, first term a Common difference d a + d...(i) and a + 9d (a + 4d) + a + d...(ii) Solving (i) and (ii), we get a 3, d 4 S 30 [6 + (9) 4] P C Q R We know that, tangents from an external point to a circle are equal in length. PQ P (tangents from P) QR RC (tangents from R) C (tangents from ) Now, perimeter of DPR P + PR + R

45 44 OSWL CSE (CCE), Mathematics Class 0 6. P + PQ + QR + R (P + P) + (RC + R) ( PQ P and QR RC) + C + ( C ) 0 40 cm. Given : PT is a tangent drawn from an external point P and a line segment P is drawn to a circle with centre O. ON. To Prove : (i) P. P PN N (ii) PN N OP OT (iii) P. P PT Proof : (i) P. P (PN N) (PN + N) (PN N) (PN + N) PN N...(i) (ii) PN N (OP) (ON) N OP ON N OP (ON + N ) OP O OP OT...(ii) (iii) From eqn (i) and (ii), we get P. P OP OT PT [DOTP, PT OP OT ] Proved 7. (x, y ), (x, y ), C (x 3, y 3 ) (, 3) (4, p) (6, 3) Since the points are collinear rea 0 [x (y y 3 ) + x (y 3 y ) + x 3 (y y )] 0 [(p + 3) + 4( 3, 3) + 6(3 p)] 0 [p p] 0 [ 4 p] 0 4p 0, p 0 8. (7, 0), (, 5) and C (3, 4) are the vertices of a triangle. C ( 7) + (5 0) (3 + ) + ( 4 5)

46 Sample Question Papers (S-) 45 C (3 7) + ( 4 0) Here, C + C and C Therefore, and C are the vertices of an isosceles right triangle. 9. Let OR OQ OP r (radii) RQ QP r (sides of a rhombus are equal) Since diagonals of a rhombus bisect each other at right angles. OS (OQ) r In right OSR, r RP (RS) x (say RS x) OSR 90 (as OQ RP) r (Pythogoras Theorem) x x 3r 4 x 3 r rea of rhombus 3 3 (OQ PR) 3 3 (x r) 3 3 3r 3 3 r 64 r 8 cm Radius 8 cm. 30. rea of square (4) cm 96 cm rea of internal circle 7 7 cm 7 77 cm

47 46 OSWL CSE (CCE), Mathematics Class 0 rea of semi-circle with 4 cm diameter rea of two quarter circles of radius 7 cm 7 cm 7 77 cm cm Shaded area cm. 3. (i) Diameter 5 cm radius.5 cm height 0 cm Volume of glass of type pr h cm 3 Volume of hemisphere r cm 3 Volume of glass of type cm 3 Volume of cone pr h cm 3 Volume of glass of type C cm 3 (ii) The glass of type has the minimum capacity of cm 3. (iii) Volume of solid figure (Mensuration) (iv) Honesty.

48 SOLUTIONS SMPLE QUESTION PPER - 0 Solved Time 3 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 and b. Since OP QP So, In DOPQ OQ QP + OP OP OQ QP (3) () OP 5 cm. So, radius of circle is 5 cm, 3. Let the point P be (0, y) P (y 5) + ( 6) nd, P (y 3) + (4) Since, P P (y 5) + ( 6) (y 3) + (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 Class -0, S- Examination Sample Question Paper

49 48 OSWL CSE (CCE), Mathematics Class 0 5. SECTION 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 C 40 m 3C...(i) In D, tan 30º C m. D 3 3 C ; [using (i)] C + 40 C 0 m Height of tower 0 3 m. 7. (a) Total possible outcomes P (lack ball) 7 4 (b) Not green balls P (not green ball)

50 Sample Question Papers (S-) Let P divides in k :. ( 3, 5) P k (, 5) (4, 9) 4k 3 k + 4k 3 k + k 5 k 5 or 5 : 9. rea of rectangle sq. cm rea of one circle pr 3.4 sq. cm. rea of 6 circles sq. cm rea to be painted sq. cm 0. Volume pr h h 660 m h m Curved surface area prh m Cost of painting ` 30. SECTION C 5x cm (3 x )cm C rea of triangle 5x (3x ) ccording to questions, 5x 5x 0 3x x 4 0 3x 9x + 8x 4 0 3x (x 3) + 8 (x 3) 0 x 3, x 8 3 Length can t be negative, so x 3

51 50 OSWL CSE (CCE), Mathematics Class 0 5 cm, C 9 8 cm C cm.. Given, a 9, d 3 S n 6 n [a + (n )d] 6 n [(9) + (n ) ( 3)] 6 n[8 3n + 3] 43 3n + n or 3n n 43 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 X Y () X Z () (Tangents from an external point to a CZ CY (3) 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 X

52 Sample Question Papers (S-) 5 5. Given that sides other than hypotenuse are 4 cm and 3 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 3 cm radius while taking as its centre to intersect S at C. Join C. C is required triangle. 3. Draw a ray X making an acute angle with, opposite to vertex C. 4. Locate 5 points (as 5 is greater in 5 and 3),, 3, 4, 5 on line segment X. 5. Join 3. Draw a line through 5 parallel to 3 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. 60º 30º b C 0 m D Let the height be h m and breadth be b m. In C, h tan 60º 3 b h b 3 In D, h b + 0 h b 3 tan 30º b b + 0 3b b + 0 b 0 b 0 m h b m Height of tree is 7 3 m and breadth of river is 0 m. 3 3 C 0 m P 30º 45º is the building of height 0 m. C is the flag pole. In P, PC 90º

53 5 OSWL CSE (CCE), Mathematics Class 0 In CP, tan 30º P 3 0 P tan 45º C P C P P P C C 7.3 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) (b) P(not a Rs. 5 coin) P (Rs. 5 coin) Number of favourable outcomes Total number of possible outcomes (a) Total number of points 8 (i.e.,,, 3, 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., (, 3, 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., (3, 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 )

54 Sample Question Papers (S-) 53 Number of favourable outcomes Total number of possible outcomes D O C D C 90 (ngle in a semi-circle) DD or DC D D cm rea of DC 60 cm rea of shaded region rea of circle (rea of DD + rea of DC) rea of circle (0 cm ), cm radius cm 7 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 3.4 () (8) 60

55 54 OSWL CSE (CCE), Mathematics Class 0. Let the rate of my walking is x km/hr ccording to question, 53 8 cm 3 Mass of the pole g kg. SECTION D Time taken at this speed x hr Increased speed (x + ) km/hr Time taken at this speed x x ( x + ) hr or x + x 6 (x + x) or x + x or x + x 0 x + 4x 3x 0 x(x + 4) 3(x + 4) 0 (x + 4) (x 3) 0 x 4 or x 3 x 4 (rejected) Rate of walking is 3 km/hr.. N 50 km W O E Let and station is at O. S speed of two trains x, (x + 5) km/hr Distance Time Speed Distance travelled by first train O (x + 5) km Distance travelled by second train O x km 50 km, (given) O + O 50 4(x + 5) + (x) x + 5x x 0, 5 x 5, as x can t be negative x speed of second train 5 km/hr x + 5 speed of first train 0 km/hr.

56 Sample Question Papers (S-) The number of rose plants in the st, nd,... are 3,, 9,..., 5. a 3, d, a n 5 a n a + (n )d (n ) ( ) n 0 Total number of rose plants in the flower bed, 4. n S n n [a + (n )d] Middle most term + S 0 5(46 8) 40. th 3 middle most terms same 0 th, th, th a 0 + a + a 9 a + 9d + a + 0d + a + d 9 3a + 30d 9 a + 0d 43...(i) a 9 + a 0 + a 37 a + 8d + a + 9d + a + 0d 37 3a + 57d 37 a + 9d 79...(ii) y solving equations (i) and (ii), we get 9d 36 d 4 a Hence, the required.p. is 3, 7,, 5, O C P 60 (a) Given, P P 8 m and we know that tangents drawn from an external point to the circle are equal in length. So, P and P are tangents Now, draw OP which bisects P and perpendicular to the chord PC PC 30 and CP CP 90 In CP, PC + CP + PC PC 80 PC Similarly PC 60 Thus, DP is an equilateral triangle P P 8 m

57 56 OSWL CSE (CCE), Mathematics Class 0 (b) Given, OC 3 m We know that, If a perpendicular drawn from the centre of the circle to the chord of that circle, then it bisect the chord. C C 4 m In DCO, O C + OC (4) + (3) 5 O 5m Which is the radius of a circle (c) Pythagoras theorem. (d) Creating awareness and taking initiative. 6. P D Q O C 7. (a) We know that, the lengths of two tangents drawn from an external point to a circle are equal. PD P...(i) & QD QC...(ii) Now perimeter of the triangular garden PQ P + PQ + Q p + (PD + DQ) + Q (P + P) + (QC + Q) + C cm (b) Given, radius of circle O cm Here O O 90 In DO O O + () + (5) O 69 3 cm. Let P (x, y ), Q (x, y ) divides into 3 equal parts. P divides in the ratio of : x y

58 Sample Question Papers (S-) 57 Co-ordinates of Q is mid-points of P. P,0 3. x y ( 3) 3 5 Co-ordinates of Q, If points are collinear, then area of the triangle formed by these points is 0. [x (y y 3 ) + x (y 3 y ) + x 3 (y y )] 0 9. So, [k(k 6 + k) + ( k + ) (6 k + k) + ( 4 k) ( k k)] 0 4k 6k 4k ( 4 k) ( 4k 0 4k 0k k k + 4k 0 8k + 4k 4 0 k + k k + k k 0 k (k + ) (k + ) 0 (k ) (k + ) k,. Let us mark the unshaded portion as I, II, III, IV. rea of (I + III) r of CD r of () semi-circles each of radius cm 7 7 p cm Similarly, area of II and IV 0 5 cm rea of shaded portion rea of CD area of unshaded portions cm.

59 58 OSWL CSE (CCE), Mathematics Class r.4 cm 7 0 cm Volume of marble Number of marbles Volume due to rise of water Volume of marble πr H or 0..4 cm.4 cm.4 cm.4 cm.4 cm.4 cm. cm.4 cm.4 cm (i) Radius of cylindrical portion and hemispherical portion of a gulab jamun.8.4 cm Length of cylindrical portion cm. Now, Volume of one gulab jamun Volume of cylindrical part + Volume of hemispherical part Volume of 45 gulab jamun p(.4). + 3 p(.4) (.4) cm cm 7.8 cm 3 Volume of syrup in 45 gulab jamuns 30% of cm cm 3 (approx).