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SAMPLE QUESTION PAPER-Y C Explanations) Cobb, Titel pumice tof -owkeomes mids total numbew of Lying between 2 and 6 =3 therefore, . 7 Probobilit, = 2 = j_ a Vv 6 ak, (a) length ef 66 =0D and Oc= on i So, AOBD and 4OAC are iSosceles. A Te 7 Now, In AAOC and ABOD { OA = 0B Cradius of civcle) \ 0 } oc= 00 (4 a) ~ ° - — Laoc = L800 (verticod opp: L's) ‘. AOC = BOD CSAS) _ |) one rook is a So put x= 44+ Qb+\a =o Ueb= 46 aba-8 — Now, vooks sf equation x*+bx+q=0 are esual,=> gale. 5: |Cd> We need to convert the angle to yadtus ' O=lto r= QK (Qo 3 The formula for the length of on orc bs. $= ve “Substituting the given values a a Re ve KR Dd %=3 cm. - 3 ol 6-| (a) According to question we hove, 22(x-+10) = ROC B+ 2% 2D ke teto = SH+R > 3x =1e st > m= G6 = _ A Co) ton 60°= 60 5» DE= bo = 2ofZ IN - an {3 \ ton 30 = AB wABH RofZxl =20 8] 2 ols 3 Now, as AD= AB+B0> 60= 20+8D 5 BO =4om. (d)\ This is a’ national number. It.can be expressed aso, which t& the ratio of tuso integess expressed in the ! fosm P where oo. a L : 2. | @) The gquadvatte equatten Yor+3oct 2 =o can be written fn the standard Form as ax*t+bo+c=0, where a=y ,b=3 and c=. The aum of the coefficients of oc4 c& and the conatant term iS given by atb+tc. Substituting. the values, of a,b: eer & se get atot+c =44+3+2=9..... \ vd ; 18: |Cc) The given equation is : 6x2-Fx-22o. heve a=6 ,b=~%,c=~ 220. Disewiminant O0= b’-Yac . ; = Cay 4x6 K6 X (- 2205 a = 81+ 5280 = S36] oS Po O= S3bI>0o, khexefore, the “roots « S Ore > real a and cunequol. — a \4- | Cb) Given equations are a _ M+ky +t =o atx ky tly =O For coincident Lines _ - _ Qio= br = Cy a2 br Cc, _ (2 8. = F 2 k 14 > { = 2 > e=4. o2 k IS. | Cad tt volume of a cone =! mv, he @ ; ! 3 | 7 volume of cylinder = nvyho - | According to queation, : a mH = 2 & hi = Y TT Ve 3 he =) j | Se, ratio = Volume of cone i 7 Volume of cylinder i - = Vern @/3sYxy » ) yy x4 3S 16 1135 7 R 3 3° 9 °~«OS 19 |€> Assertion CA) is true but Reason CQ is False. [ Since the product of zeroes ig |o = lo ) For axt+ bx2+cac+d ,pvoduct of voots =-d a | 2o- | CO Assertion CA) t& trve but Reason(e) is False. An tnteger {s a vational number, ao the asaewtion is krve. Whereas yoot of any ftntegery can't be said as iyvationel . _| Vib) main A+ncesA = Pp i(d) CmsinaA+ neosA) = p? Caquesing beth aides). mainAt nzco?A +&mnsina-cosA =p? _ _@ (meos A- nsinAY = g2 ‘mteos?A +n? sin? A — @mnecosA:sian = ee) Adding @ ond @ . m? Csin2?A + Cos’ A) +n? CSin?a +cos’A) +0 = P+a2 m?+n2 = pe+9 2 Hence proved. ?+562x4+650 SD 4+ Mxe4+3x+6 =90 2.2. oc(x+2) +3 C2+3) =0 _ Cx+2 (x43) =0 x =-2,-3 Xe-2, B=-3 - Y¥+B=-b SB -2-3 =-S 3-S=-S fs lenort } ails ve E22G3)= 6 os b=6. | ] 83. Given | . _ ae In AABC, O is midpoint of AB and OF ; is parallel to &c- ; : Ade os AOo= De. a . ( To proove i AE = EC \ Proof: Since, DE II BC : \ a. By Basic proportionality theovem,. ® ce AD = AE [ 086 Ec AD=D6, Thevefore, AE =| » AEX=EC. Ec Noy, 26. Cb) In QABC We have LM AB AL = 8m _ [ey thales Theorem]: tc mc > AL = 86™ Ac-AL 6c- 6M > x-3 = “ee ox Cx-3) Cdxct+3> -(x-25 > x3 = x-#& X4+3 M45 > Ce-B>CxK+S) = Cor-29 C43) > we42xrc-1S = mH 6 > x= 4. 7 27-(b)| ton A= ntanB > tonB= 1 tanA> cotB= n tana Zin A= msinB > aing =) sina ScosecBa™ | m sinA substituting the values of cot & and cosec 8. In cose? — cot’ R=] uy, 1 We get, > _m —~ n? =) aintA kan? A > m*__ =I ainz A ton? A > me-rcos*a_ =| Bin? A 3 m2-n?cos? A = dsin?A > me-en’cosa =\-cos’A > m-l=n?cos*A -costA E) m?-| = cost*A Cn*1) > m1 = cos*A. n-} 24: Given, T= 3-G em, h=locm TSA of the article = 2x cSA of the hemispherical part . +CS5A of the cylindrical past. = AXVWnxVA*+ Qash = &xnr(rth) = 2K ¥3S CRAB SHO) F = 22 43-S (FAlod 3S = 22A%)F = 344 cm? Thua, the totel auyface avea of sing axticle is 344 cm? To proove: AE =aAc = cE L 3o- = eF BD €D A Proof : In A AEP and ABFP, , Qllm CGivend 7 b Zr =d2 @ £324 [Atkernaote interior angle] eh ee , _ DAEP ~ aBeP, CBy AA similarity) iE + AE = AP - EP ~- -- —(Iy é BF ep FP - | In Acer & BDEP, Litm (Given) 27=28 & (5=26 (Alternote interior angles} J a. AceP~ AvP [By AA similavity] > CE 2ceP EP . ~~ Gi) oF oe ¢P In OAcP ond ABDF- Lilm CGivend 4o=22 & @s=6 CALteynate intevior angles] vw. DACP» DBP (By AR similarity) Ac . AP scp ...... Cit) 8D BP oP From ¢iy Gi) and Gil) AP = AC = cP = ce =~ EP = AE SAC =AE ~- ce PB BD be DE FR BE BD BF DF 3.2. Cb) length of minute hand = yadivs of the clock. Radius (+) of the civle =)4em Civen) Angle swept by minute hand fn 60 minutes =3 60° d So, the angle asuwept by the minute hand in S minute = 360° xs = 30° 60° We know, Avea of sector = © yay? 360 Now area of the sector making an angle of 30° =3B0° XY 22 KX IYXIY | - » B6O® F. ~ _ : = 1 x kexexy Ix a = IS64Y cm, 3 —— 7 Hence; the ‘requived area Awept by the minute hand in S minutes is 164 cm’. Marks = Fs ar. 20-30 Pe - e BS= i \s is+P Yo-So_ - 85 4yorP cl so-60 | go 60+ 6-o | 4 cee'd Fo-80 é 6O+P +L 80-Fo | lo Fetety | We ore given medion =S0 Thus gum of atl the Frequencies = Jo > PHISH+2S +204+4494!) 0 = FH4PHIU=Jo > P+Q=12 Also, es median = L+ (%e-c&) xh F Se = Yot C4S- (ISHE) yo S a5 =30-p foo 2S - -—p =-S ovr p=S,hevre PTT= 12 5qr12-p > q=7 “. P=S14=F fe 2 ¥9S-1S-20. mode occurs in (Yo-Se) 03 25 is highest Frequenry, mode= 10+[/25— 5 = 46-61- 3S- Jacks ond Oveens of black colour ave wemoved > 2 each is removed = 4 cazds. kings and Aces of ved colour are vemoved. Tv > 2 each card of Cking& Ace) removed =4 cards. @) A black kina: Total number oF black king = 2. and totol cards ave $2-@ =44. | | w. PCblack king) = 2 _ | | G4 22 (b) A ved colour cand: |_No. of wed colour cavd= 26-Y= va. | Plved covd) = 72 =| | | 44 2 | | (€) No- of black coloured jacks in a deck ~ o | 7 PCblack jack) = © =o | ay @) No- of face cards In the deck = 12-6=6 P (face card) = & _ 3 uy RD 36. dy Money daniel on Tt day = Bs. 27-5) uy : Sehaj Incveases ‘his: Savings by a Fixed ‘amount of Rs. 2-5. “. His saving from an AP. with a 2 2hs and d= 2-5 v Money saved .on- toh day .: Gn = Atdd = 27-5 4+9(2-5) ES 2ES +295 HL = Ra. So Gi) Money aavedion I day = 2&- 24-5 money aaved on 35h dow ea where a=27:5 and d= 8-5 (O95 = O484d > Tet SAIUS) = Rs a7-5 Gii) Tetal amount saved by Sehoj_in 30 dye S305 Be [axis re 925) = we(ss+29(e-s s)] = 1912-5. xO 
