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

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Jeremy Dean
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1 SET - GSE tch : 0th Std. Eg. Medium MHESH TUTILS SUJET : Mths(0) First Prelimiry Exm Model swer Pper PRT -.. () like does ot exist s biomil surd. () (c) touches () - d () -4 7 (c) (d) ifiite (d) (d) x 64. (c) (c) (c) (c) - 8. (c) ( +) (d) 00 :: () 7 (d) (d) (, 0) () right gled (c) - (c) , (c) cos. ().. t (c) (d) > : 9 (c) (d) l r (c) % r 0 7 (c) 0 0 (c) 47 (d) 4 (d) πrθ 80 Dte: Mrks : 00 Time: Hrs.

2 PRT - SETION - Solve the followig sums : [ mrks]. Let 9 4 Now, x + y x. x + y 9, xy 7 d y Here, + - d -4 b - b b b k, k 0 b k d k Now, -4 let c -4 c (-4) c (-4) k c -4 k p(x) x + bx + c kx + kx - 4k k(x + x - 4), k 0. x - y (i) x-y (ii) From equtio (i), we get x y y x Substitutig y x i equtio (ii), we get x - x x x -6 6 x Substitutig x i equtio (i), we get () - y 0 9-y 0 9 y y 9 {(x, y)} {(, 9)} is the solutio of the give pir of equtios. 4. x Here, d 0 T 00 + ( ) d + ( ) +,?

3 00 0 S 0 [ 9 ] 0 [0 + 9] 00 S0 S0 S0 4. [ + ( - )d] S S0 S0 l x 0 00 Tm T We kow, d m Here T9 T + So, T9 T T9 T 9 d d T 7 7 T + 4d The.P. is,, 7,, 7... I, : Stircse : Wll : distce of bse of stircse from the wll. I, m (y Pythgors Theorem) 6m (6) + (6.) () (.). The distce of bse of stircse from the wll is.m (x, y) (, 7) d (x, y) 6.m? (4, ) re give poits. P Let P(x, y) be the poit o such tht. P m m P(x, y) divides from i the rtio. x mx + x m+, y my + y m+ x 4 + +, y x, y 6 + 4

4 0 0, y x, y 4 The co-ordites of the poit which divides i the rtio : from re (, 4). x 7. I, let m 90 d m θ. cosec θ I k k, sy k d k, by Pythgors theorem + (k) + (k) 69k + k 69k - k 44k k 44k k k, k, k 7. t θ k k cot θ k k L.H.S. ( + cot - cosec ) ( + t + sec ) 8. cos si si si cos cos si cos cos si si cos [(si cos ) ] [(si cos ) ] si cos (si cos ) () si cos si cos si cos si cos si cos si cos si cos si cos R.H.S. Here, mximum clss frequecy is 8. Modl clss 8 [(x - y)(x + y) x - y]

5 Lower limit l 8 lss size c 7 Frequecy of modl clss f 8 Frequecy of the clss precedig modl clss f0 Frequecy of the clss succeedig modl clss f Now, Mode Z f f l + f f f c Hece, mode of the give dt is SETION - Solve the followig sums : [ mrks] Let the price of petrol be Rs x per litre. The qutity of petrol obtied for Rs 0 0 litres. x If price of petrol is icresed by Rs per litre the ew price Rs (x + ) per litre 0 The the qutity of petrol obtied t Rs. 0 (x + ) litres. If the price of petrol is icresed, oe gets litres less petrol. 0 0 x x Multiplyig the equtio by x ( x + ), 0 x 0 (x + ) x (x + ) 0 x 0x x 0x x + 0x Dividig the equtio by, x + x 0 0 x + 60x x 00 0 x (x + 60) (x + 60) 0 (x + 60) (x ) 0 x or x 0 x 60 or x Price of petrol cot be egtive. x 60 x Origil price of petrol is Rs per litre. Hece icresed price of petrol is Rs (x + ) Rs ( + ) Rs 60 per litre 0. Here, is the height of the pedestl. is the height of the sttue. D is the poit o the groud. The gle of elevtio of the top of the sttue from poit D is

6 60. m D 60 The gle of elevtio of the bottom of the sttue from poit D is 4. m D 4. I D, m 90, m D 4. t D D t 4 D.46 m t D D t 60 D? D [From (i) d.46m] 46 7 m Height of the pedestl is m.. (i) (iii) Here selectio of oe crd from well-shuffled pck of crd is eqully likely outcomes. There re fce crd (4 kigs, 4 quees, 4 jcks). Let be the evet tht the selected crd is bse crd. So, the umber of outcomes fvourble to the evet is. There re dimod crd. Let be the evet tht the selected crd is of dimod. Therefore, umber of outcomes fvourble to is. P() 4 (ii) D D...(i) I D, m 90, m D 60. P() Let be the evet tht the selected crd is ot ce.the the evet is the selected crd is ce. the the evet hs 4 elemets. ut P( ) 4 P () - P ( ) -

greter vlue th 88 lies i the clss 0-0 l 0, cf 07, f 4, c 0 M 88 07 0 + 0 4 7 0 4 0 +.. Hece medi of give dt is.

7 (iv) Let D be the evet tht the selected crd is ce of blck colour. So the umber of outcomes fvourble to D is (i.e. ce of spde d club). P (D) 6. Deposit (Rs i thousd) Number of depositors (fi) Here 776 umultive frequecy (cf) Nerest d greter vlue th 88 lies i the clss 0-0 l 0, cf 07, f 4, c 0 M Hece medi of give dt is... cf l+ f c 0 + lss frequecy (fi) umultive frequecy (cf) b b b b b b b Here, 400 (give) b b (i) Medi is 8 which lies i the clss 0-40 l 0, 00, cf 80, cf M l+ f c f, c 0

8 Plcig vlue of i equtio (i) 0 + b 98 b 48 Hece vlue of is 0 d b is 48.. SETION - Solve the followig sums : [4 mrks] Dt : circle with cetre O touches ll the sides D of D. is the lrgest side. O To Prove : D is the smllest side of D. Proof : ircle touches ll the sides of D. + D + D (i) Suppose, D > D + D + D ( ddig D o both the sides) D + D > + D ( From (i)) D > Which is costrditio with the give, becuse is the lrgest side of qudrilterl. D < (ii) Similrity, if D > D the D + > + D (From (i)) > Which cotrdicts the give, becuse is the lrgest side. D < D (iii) lso, D < (Give) (iv) Thus, from the results (ii), (iii) d (iv), D is the smllest side i D. Rdius of circle r re of coloured regio 0.cm. re of sector O - re of OD... (i) re of sector O re of OD r cm bse ltitude OD O O 6 D 0. 4.

6 0..cm Substitutig vlues i equtio (i) re of coloured regio 86.6 -..cm re of coloured regio is.cm. Sphere r cm?

9 6 0..cm Substitutig vlues i equtio (i) re of coloured regio cm re of coloured regio is.cm. Sphere r cm? volume of sphere l ube 44cm volume of cube 4 r 4 7 l The volume of the coe The volume of the cylider r h cm r H cm The totl volume of the model volume of coe + volume of cylider cm The volume of ir i the model is 4.8 cm Hece, 4 sphericl blls re mde. Here rdius of the coe d cylider is r 4 cm d the height of coe is h cm. The height of the cylider H - () 7 cm s show i the figure, the totl volume is divided ito three prts, mely two coicl prts d oe cylidricl prt.. SETION - D swer the followig questios : [ mrks] Dt : ostruct with m 90, 4 cm d cm. 4 To ostruct : costruct XY similr to where the scle fctor is.

10 M Y c m X 4 cm 4 i. ii. N Steps of costructio : ostruct with m 90, 4 cm d cm. Drw N mkig cute gle with such tht N d re i differet hlf ples of. iii. Select y rdius d drw rc with cetre which itersects N i. iv. Similrly with N cetre d the sme rdius, drw rc itersectig N t such tht. Similrly cotiue the procedure uptil poit 4. v. vi. vii. viii. ix. 7. Drw. Drw 4 X prllel to itersectig i X. Drw XY prllel to itersectig M i Y. XY is the required trigle of desired mesures. If lie prllel to oe of the sides of trigle itersects the other two sides i distict poits, the the segmets of the other two sides i oe hlfple re proportiol to the segmets i other hlfple. N M Q P P l N Q l M Give : I the ple of, lie l d l itersects d t poits P d Q repectively.

11 To prove : Q P Q P Proof : Let QM, d PN. ostruct Q d P. re of trigle bse ltitude re of PQ P QM re of PQ P QM re of PQ re of PQ P QM P QM P P lso re of PQ Q PN re of PQ Q PN re of PQ re of PQ Q PN Q Q Q PN (i) (ii) PQ d PQ re hvig commo bse PQ d they re lyig betwee two prllel lies PQ d. re of PQ re of PQ From (i), (ii) d (iii) 7. Give : PQR, m Q M QR, N PQ To Prove : PM + RN (iii) Q P. Q P 90., PR + MN Proof : I PQR, m Q 90, M QR, N PQ. PQR, PQM, NQR, NQM re right gled trigles i which Q is right gle. y Pythgors Theorem. P I PQR, PR PQ + QR (i) I NQM, MN NQ + QM (ii) I PQM, PM PQ + QM (iii) I NQR, RN NQ + QR (iv) Now, PM + RN PQ + QM + NQ + QR (from (iii) d (iv) ) N Rerrgig the terms PQ + QR + QM + NQ PR + MN Q M R PM + RN PR + MN. **** est of Luck ****

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