5. Given, a first term 1 and d common difference 4 ( 1) Let the nth term of the given AP be 63. Then, a n 63

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1 Sample Questio Paper (Detailed Solutios) Mathematics lass 0th 5 9. We have, Now, 04 Thus, it is the product of prime factors. Hece, 7 is a composite umber.. Let the legth of the shadow be B x m 8 m B Legth of pole 8 m B I right agled B, (/) B B ta 0 Perpedicular ta Base 8 x 8 m ta 0 x Hece, the legth of the shadow is 8 m. (/). Whe we take a third poit (say) o a semi-cir cle ad joi by the ed poits of di am e ter (say ad B), the 90 [ agle i a semi-circle] D B r rea of B Base ltitude B D r r r sq uits 4. Give equatio is x 4 x 4 0. O comparig it with ax bx c 0, we get a, b 4 ad c 4 D b 4ac ( 4 ) 4( )( 4) r x 0º which shows that the give quadratic equatio has real ad equal roots. 5. Give, a first term ad d commo differece 4 ( ) 4 5 Let the th term of the give P be 6. The, a 6 a ( ) d 6 [ a a ( ) d ] ( )( 5) Number of terms caot be a fractio. Thus, 6 is ot a term of the P, 4, 9, 4,.... (/) 6. Mea of th term ad ( th ) term is the required media. i.e. (th ( ) th) observatio. 7. Let x. 46 x (i) O multiplyig Eq. (i) by 000, we get 000 x (ii) O subtractig Eq. (i) from Eq. (ii), we get x 40 x 999 Hece, which is the required form of p q. 8. Give, PQ ad PR are tagets at Q ad R, respectively ad SQR 8 Here, OQP 90 Now, OQP OQR PQR [agle betwee taget ad radius] 90 8PQR [ OQR SQR 8 ] PQR 5 s, PQ PR [ tagets draw form a exteral poit to the circle are equal i legth] So, PRQ PQR 5 [ agles opposite to the equal sides are equal] Now, applyig agle sum property i PQR, we get PQR PRQ QPR 80

2 6 m 0 5 5QPR 80 QPR QPR 76 gai, QOR 80 QPR [ sum of opposite agles of quadrilateral OQPR 80 ] QOR Further, OSR ORS [ OS OR, radii of same circle] OSR QOR [exterior agle theorem] QSR 04 [ OSR QSR] 04 QSR 5 Thus, QSR 5, PRQ 5, QPR 76 ad PQR 5 9. Let BD be a rect a gu lar field of di me sios 0 m 6 m. Suppose, a cow is tied at a poit ad legth of the rope, E 4 m r [say] D G 4 m B F 0 m rea of the field i which the cow graze r ( ) rea of sector FEG [ 90, i.e. shape of the field is rectagular] m 0. Give, B cm, 4 cm I B, B 90 [agle i a semi-circle] B B [ by Pythagoras theorem] B 4 B 6 9 B 5 B 5 5 cm [takig positive square root both sides] lso give, B ~ PO B OP 5 OP P 4 P OP i.e. P 5 4. Odd i te gers be twee ad 000, which are di vis i ble by, are, 9, 5,, 999, which form a P. Here, first term ( a), commo differece ( d ) 9 6 ad last term, a 999 a ( ) d 999 [ a a ( ) d ] ( ) E Now, sum of 67 terms a a ( ) ( 999) Hece proved.. Let the X-axis di vides the lie seg met joi ig the poits (, ) ad ( 5, 6 ) i the ra tio k :. The, by usig sectio formula, 5k 6k oordiates of itersectio poit, k k But it is a poit o the X-axis. So, its y-coordiate 0. 6k k 0 6k 0 k Hece, required ratio : :.. Sice, each fried is left with 5 toys. Number of toys distributed by isha (/) ad umber of toys distributed by Suchi (/) Now, each child gets equal umber of toys. Number of toys received by child HF (5, 56) 7 (/) Now, um ber of chil dre whom isha distributed the 5 toys 5 (/) 7 ad umber of childre whom Suchi distributed the 56 toys 8 (/) 7 Total umber of childre 5 8 (/) 4. Let the preset age of Salim be x yr ad that of his daughter be y yr. The accordig to the questio, we have x ( y ) x y 4... (i) ad x 6 ( y 6) 4 x 6 y 4 x y 0... (ii) Now, from Eq. (i), we have x y 4 O substitutig this value of x i Eq. (ii), we get y 4 y 0 y 4 0 y 4 O puttig y 4 i x y 4, we get x ( 4) Hece, the preset age of Salim is 8 yr ad preset age of his daughter is 4 yr.

3 Let the orig i al list price of the book be ` x. The, umber of books bought for ` x Reduced list price of the book `( x 5 ) 00 Number of books bought for ` 00 x 5 ccordig to the questio, we have x 5 x 00x 00x x ( x 5) x 5x x 5x 00 x 5x 00 0 x 0x 5x 00 0 [by splittig the middle term] x( x 0) 5( x 0) 0 ( x 0)( x 5 ) 0 x 0 0 or x 5 0 x 0 or x 5 But x 5 is ot possible, because price caot be egative. x 0 Hece, the origial list price of the book is ` Let woma fi ish the work i x days ad ma fi ish the work i y days. The, work of woma i day x ad work of ma i day y Work of wome ad 5 me i oe day 5 x y 5x y The umber of days required for complete work 5x y ad work of wome ad 6 me i oe day 6 x y 6x y Number of days required for complete work 6x y ccordig to the questio, 4 ad 5x y The, O puttig x 5x y ad ad y x 6x y u ad v, we get y 6x y 8 9 y x 0v 8u (i) ad 8v 9u (ii) O multiplyig Eq. (i) by 9 ad Eq. (ii) by 8 ad the subtractig Eq. (ii) from Eq. (i), we get 80v 44v 9 8 6v v 6 O substitutig v i Eq. (ii), we get 6 8 9u 6 9u u 8 Now, u ad v 8 6 ad x 8 y 6 x 8 ad y 6 Hece, sigle woma fiish the work i 8 days ad sigle ma fiish the work i 6 days. 6. Let the speed of the stream be x km/h. The, speed of motor boat i dowstream ( 5 x ) km/h ad speed of motor boat i upstream ( 5 x ) km/h Distace travelled i dowstream 0 km ad distace travelled i upstream 0 km Time take by motor boat i dowstream ad 0 upstream are 5 x h ad 0 h, respectively. 5 x time distace speed Total time take 4 h 0 mi 4 h 9 h x 5 x x 5 x 0 ( 5 x) 0 ( 5 x) ( 5 x)( 5 x) 0 ( 5 x 5 x) ( 5 x)( 5 x) 0 0 ( 5 x)( 5 x ) 00 5 x 5 00 x 5 x [ ( a b)( a b) a b ] x 5 [takig square root] s, speed caot be egative. So, x 5 is eglected. Hece, speed of the stream is 5 km/h.

4 7. Let a be the first term ad d the commo differece of P. a a [give] p q a ( p ) d [ a ( q ) d ] [ a a ( ) d ] a pd a qd a ( p q) d... (i) Now, a p a ( p ) d ( p q) d pd [usig a ( p q) d] ( p q p) d ( 4p q) d ( p q) d... (ii) gai ap q a ( p q ) d ( p q) d ( p q) d [ from Eq. (i), a ( p q) d] ( p q p q) d ( p q) d... (iii) From Eqs. (ii) ad (iii), we get a a Hece proved. p p q Let be the umber of sides of the polygo. Sice, iterior agles of polygo are i P. So, smallest agle first term i.e. a 0 ad commo differece, d 5 Now, sum of iterior agles, S [(0 ) ( ) 5 ] [ S [ a ( ) d ]] ( ) ( 5 5 ) (i) s, we kow that, sum of iterior agles of a polygo of sides ( 4) 90 From Eq. (i), we get (5 + 5) ( ) 5 5 ( 4) [by factorisatio] ( 9)( 6) 0 9, or 6 If 9, the greatest agle, i.e. a 9 0 ( 9 ) If 6, the greatest agle, i.e. a (6 )5 95 which is ot possible, sice o iterior agle of a polygo ca be more tha 80. Hece, the umber of sides of the polygo is Give B is a equi lat eral tri a gle ad D is a poit o B such that BD B. To prove 9D 7 B ostructio Let us draw E B. Proof I right agled EB, B E BE [by Pythagoras theorem] E B BE B 4 B E B B D E 4 [ B is a equilateral, so B B ] E B (i) 4 gai i right agled ED, D E DE [ by Pythagoras theorem] E ( BE BD) E B B B E 6 E B 6 E B 6 give, BD B [ B is a equilateral triagle, so B B ] B B [usig Eq.(i)] B B 8 B 7 B Hece, 9D 7 B Hece proved. 9. We have, si 0 cos 45 4ta 0 si 90 cos ( ) ( 0) 4 si 0, cos 45,ta 0, si 90 ad cos 90 0] () 4 4 (i) We have, si 9 cos 5 si( 90 5 ) cos 5 (/) cos 5 cos 5 0 [ si ( 90 ) cos ] (ii) We have, cosec 5 sec 65 cosec ( ) sec 65 (/) sec 65 sec 65 [ cosec ( 90 ) sec ] 0 0. Let the side of a equi lat eral tri a gle be a. The, area of a equilateral B 4 ( a)

5 But give, area of equilateral B cm ( a ) a a cm 00 Radius of circle 00 cm (/) [ radius of circle half the legth of the side of the B] Sice, B is a equilateral triagle. B 60 rea of sector of a circle 60 r (00) 400 cm 6 (/) rea of three equal sectors cm (/) 6 Now, required area rea of B rea of three sectors cm (/). Ta ble for cu mu la tive fre quecy is give be low lass iterval Frequecy umulative frequecy Total N = 66 Here, N 66, which lies i the cumulative frequecy 7. So, media lies i the class iterval 0-5. () Highest frequecy is 0, so mode lies i the class iterval 5-0. Sum of the lower limit of media ad the modal class Here assumed mea a 09. 5ad class iterval h 0. lass No. of Mid-poit x marks frequecy ( x i ) u i i f i u i Total k u f i u i i x a hu ( 0. 06) Hece, the average erolmet per H.S. School is (i) Let the mea of x, x,, x be x. x x x i.e. x (i) (/) Now, ew observatios are ax, ax,, ax. a( x x x ) x ax [from Eq. (i)] (ii) Give, mea of 0 observatios 9. If a ew observatio is added, the the mea is reduced by 0.5. (/) So, ew observatio ( ) 9. 0 (/) (/). (i) Let the umber of pats be x ad the umber of skirts be y. Now, accordig to the questio, y x (i) ad y 4x 4 (ii) From Eqs. (i) ad (ii), we get 4 x 4 x 4x x 4 x x O puttig the value of x i Eq. (i), we get y 0 Hece, the umber of pats, she purchased is ad she did ot buy ay skirt. (ii) Pair of liear equatios i two variables. (/) (iii) Friedly ature ad logical way of talkig are idicated here. (/) 4. Give two vertices of a equilateral triagle are (, 4) ad B(, ). Let third vertex of equilateral triagle be ( x, y). Sice, the triagle is equilateral. B B B B [ i equilateral triagle, all sides are equal] Now, B B ( ) ( 4) ( x ) ( y ) [ distace betwee two poits 5 x 4x 4 y 6y 9 ] ( x x ) ( y y )

6 4 x y 4x 6y 0... (i) gai B ( x ) ( y 4) ( ) ( 4) x y 6x 8y 5 6 x y 6x 8y 0... (ii) From Eqs. (i) ad (ii), we get x y 4x 6y x y 6x 8y 0x y 0 5x y 6 0 y 6 5x O substitutig y 6 5 x i Eq. (i), we get x ( 6 5x) 4x 6( 6 5x) 0 6x 6x 0 x x 0 x 4 8 b b 4ac x 4 a For x ; y For x ; y Hece, the required poits are 7 5, ad 7 5,. Let the two ver ti ces of a BP be ( 7, 6 ) ad B ( 8, 5) ad let the third ver tex be P ( x, y). Sice, ce troid of the tri a gle is give as (, ). ( 7) 8 x 6 5 y, (, ) cetroid of triagle havig vertices ( x, y ), ( x, y ), ) x x x y y y, O comparig the coodiates both sides, we get ( 7) 8 x 6 5 y ad x ad y 9 x ad y 9 x ad y Thus, the coordiates of the third vertex are (, ). Now, let three vertices of B be ( 7, 6 ), B ( 8, 5) ad (, 4 ), respectively ad let the cetroid of B be G ( x, y). The, x ( 7) ad y x ad y 5 x ad y 5 Hece, the coordiates of cetroid are (, 5). 5. Give, D ad E are altitudes which itersectio each other at the poit P. (i) I EP ad DP, EP DP [each 90 ] ad PE PD [vertically opposite agles] EP ~ DP [by similarity criterio] (ii) I BD ad BE, DB EB [each 90 ] ad BD BE [commo agle] BD ~ BE [by similarity criterio] (iii) I EP ad DB, EP DB [each 90 ] ad PE BD [commo agle] EP ~ DB (iv) I PD ad BE [by similarity criterio] PD BE [each 90 ] ad PD BE [commo agle] PD ~ BE [by similarity criterio] 6. Give Two circles meet at poit, tagets draw at meet the circle at B ad. To prove P is the circumcetre of the B. ostructio Draw perpedicular lies from cetre O ad Q to the poit. Proof Q B [ a taget to a circle is perpedicular to the radius through the poit of cotact] Q OP OP B [ opposite sides of a gm are parallel] [ Q OP ad Q B, so OP B] Let OP itersects B at M. B O OM B M BM P M [ perpedicular draw from the cetre of a circle to a chord bisects the chord] OM ad OP is the perpedicular bisector of B. Similarly, PQ is perpedicular bisector of. Q

7 5 Now i B, OP is the perpedicular bisector of side B. P PB [ ay poit o the perpedicular bisector is equidistat from the fixed poits] Similarly, P P P PB P P is equidistat from three vertices of B The circles with P as cetre ad its distace from ay vertex as radius passes through the three vertices of B ad the poit P is the circumcetre of the B. Hece proved. learly, OPT 90 [ radius is perpedicular to the taget at the poit of cotact] Now, from OPT, we have OT OP PT [by Pythagoras theorem] PT ( ) ( 5) [ OT= cm ad OP 5 cm] PT PT cm Sice, tagets draw from a exteral poit to a circle are equal i leghts, therefore we have P E x cm (say) (i) T PT P x (ii) Now, as OE is radius ad B is taget to the circle at E. OE B ET BET 90 Now, i ET, T E TE [by Pythagoras theorem] ( x) x ( 5) [usig Eqs. (i) ad (ii)] 44 x 4x x x 64 4x x 4 cm E 0 cm Similarly, BE 0 cm Hece, B E BE We have, (sec ta )(sec B ta B)(sec ta ) 0 cm (sec ta )(sec B ta B)(sec ta ) O multiplyig both sides by (sec ta ) (sec B ta B)(sec ta ), we get, (sec ta )(sec B ta B)(sec ta ) (sec ta )(sec B ta B)(sec ta ) (sec ta ) (sec B ta B) (sec ta ) (sec ta )(sec B ta B) (sec ta ) [ ( a b)( a b) a b ] (sec ta ) (sec B ta B) (sec ta ) [(sec ta )(sec B ta B) (sec ta )] [ sec ta ] (/) (sec ta )(sec B ta B) (sec ta ) (/) gai, multiplyig both sides by (sec ta )(sec B ta B)(sec ta ), we get (sec ta )(sec B ta B)(sec ta ) Hece proved. 8. Let B be the buildig of height 60 m, D be the tower of height xm ad distace betwee buildig ad tower be ym, i.e. BD ym. Draw E BD. The E BD y lso, DB XD 60 [alterate agles] ad E X 0 [alterate agles] (/) Now, i right agled BD, ta 60 B DB X x m D y 0 y m 0 60 ta perpedicular base [ ta 60 ] y y 0 m (i) ad i right agled E, ta 0 E E (/) 60 x 0 [ ta 0 ad from Eq. (i)] 60 x 0 x 0 60 x 40 Hece, the height of the tower is 40 m. 9. Height of the cylider ( H) 40 cm Radius of the cylider ( r ) 7 cm Radius of the hemisphere ( r ) 7cm E 60 m B

8 6 Radius of the coe ( r ) 7 cm Height of the coe ( h) 4 cm Volume of the article Volume of the cylider Volume of the coe Volume of the hemisphere r H r h r r H h r r 40 ( 4) ( 7) cm Slat height of the coe ( l ) h r ( 4) ( 7) cm 5 cm Total surface area of the article urved surface area of the cylider urved surface area of the coe Surface area of the hemisphere rh rl r r[ H l r ] (/) 7 [ ] 7 [ ] 9 68 cm 0. coi is tossed three times. Therefore, sample space is S {HHT, HTH, THH, HTT, THT, TTH, HHH, TTT} There are 8 possible outcomes, ( S ) 8 The game cosists of toosig a coi times. If oe or two heads show, Sweta gets her etry fee back. If she throws heads, she receives double the etry fees. If she gets TTT, she loses the etry fees (i) loses the etry Out of 8 possible outcomes, oly oe (TTT) is favourable. P (she loses etry fee) 8 (ii) gets double etry fee Out of 8 possible outcomes, oly oe (HHH) is favourable. P (she get double etry fees) 8 (iii) Let E be evet that she just gets her etry fee. The E {HHT, HTH, THH, HTT, THT, TTH} ( E) 6 ( E) 6 P (she just gets her etry fee) ( S ) 8 4 Total umber of discs 80 (i) Number of discs with two-digit umbers Required probability 7 80 (ii) Number of discs with perfect square umbers 8 (i.e., 4, 9, 6, 5, 6, 49 ad 64) 8 Required probability 80 0 (½) (iii) Number of discs with umbers divisible by 5 6 (i.e. 5, 0, 5, 0, 5, 0, 5, 40, 45, 50, 55, 60, 65, 70, 75 ad 80) 6 Required probability (½) 80 5

MAHESH TUTORIALS SUBJECT : Maths(012) First Preliminary Exam Model Answer Paper

MAHESH TUTORIALS SUBJECT : Maths(012) First Preliminary Exam Model Answer Paper SET - GSE atch : 0th Std. Eg. Medium MHESH TUTORILS SUJET : Maths(0) First Prelimiar Exam Model swer Paper PRT -. ulike.... 6. 7.. 9. 0...... - i two distict poits (d) - ad - (d) 0 x + 0 6 does ot exist