Maths last 10 year papers

S ep-b -s ep e plana ion: In e clid's di ision algori hm, 7344 = 1260 5 + 1044 Al o, 1260 = 1044 1 + 216 1044 = 216 4 + 180 216 = 180 1 + 36 180 = 36 5 + 0 We ill follo he follo ing s eps, S ep 1 : di ide 7344 b 1260, Q o ien = 5 remainder = 1044, S ep 2: di ide di isor 1260 b remainder, Q o ien = 1 remainder = 216, S ep 3: Repea he abo e s eps n il e ge remainder 0. Th s, he las q o ien hich gi es he remainder ero is 36. OR An posi i e in eger is of he form 4q+1or4q+3 As per E clid s Di ision lemma. If a and b are o posi i e in egers, hen, a=bq+r Where 0 r<b. Le posi i e in egers be a.and b=4 Hence,a=bq+r Downloaded from www.padhle.in

Where, (0 r<4) R is an in eger grea er han or eq al o 0 and less han 4 Hence, r can be ei her 0,1,2and3 No , If r=1 Then, o r be eq a ion is becomes a=bq+r a=4q+1 This ill al a s be odd in eger. No , If r=3 Then, o r be eq a ion is becomes a=bq+r a=4q+3 This ill al a s be an odd in eger. Hence pro ed. Ans 8). ​The gi en A.P. i 3, 15, 27, 39, e m ill be i 31 e m. He e a = 3, d = 12 ​a21 = a + 20d = 3 + 20 12 = 3 + 240 = 243 No , an = a21 + 120 = 243 + 120 = 363 an =a + (n 1) d 363 = 3 + (n 1) 12 o 360 = (n 1) 12 o n 1 = 30 ​n = 31 Hence, he e m hich i 120 mo e han i 21 Downloaded from www.padhle.in

OR ​Gi en, Sn = 3n2 4n We kno ha ​an = Sn Sn-1 = 3n2 4n [3 (n 1)2 4 (n 1)] = 3n2 4n [3 (n2 2n + 1) 4n + 4] = 3n2 4n (3n2 6n + 3 4n + 4) = 3n2 4n 3n2 + 10n 7 = 6n 7 So, n h e m ill be 6n 7 An 9). Le he gi en oin be A (1, -3) and B (4, -5) and he line- egmen joining b he e oin i di ided b he -a i , o he co-o dina e of he oin of in e ec ion ill be P( , 0) Downloaded from www.padhle.in

A 10). ​The possible o comes of 3 imes ossing a coin 1s = HHH 2nd= HHT or HTH or THH 3rd= TTHor THTor HTT 4 h=TTT ​To al n mber of e en s=8 N​ o of fa o rable e en s= 6 ​P(E)= no. of f a o rable o come o al o come Downloaded from www.padhle.in

P(E)= 6/8 = 3/4 Ans ​ ​ Hence, he obabili of lo ing he game i 3/4. Ans 11). Possible n mbers of e en s on hro ing a dice = 6 N mbers on dice = 1,2,3,4,5 and 6 (i) Prime n mbers = 2, 3 and 5 Fa o rable n mber of e en s = 3 Probabili ha i ill be a prime n mber = no. of f a o rable o come =​ 3​ /6 = 1/2 o al o come (ii) N mbers l ing be een 2 and 6 = 3, 4 and 5 Fa o rable n mber of e en s = 3 Probabili ha a n mber be een 2 and 6 = 3/6 = Ans 12). Gi en S s em of eq a ions are -c + 3 + ( 3 - c ) = 0 and 12 - c - c = 0 To find: Val e of c hen s s em of eq a ion has infini el man sol ion Condi ion for infini el man sol ion is gi en b , From gi en s s em of eq a ion, P ing hese al e in condi ion e ge , c = c= c 12 c Downloaded from www.padhle.in

T​ o find al e of c consider firs eq ali .i.e., -c/12 = 3/-c c = 36 c= 6 ​Here c = -6 is rejec ed as i does no sa isf he 2nd eq ali . Therefore, c = 6 Ans 13). ​Le 2 i a a ional n mbe . So, 2 = he e a and b a e co ime in ege and b 0 o 2b=a ​S a ing on bo h ide , e ge 2b2 = a2 T​ he efo e, 2 di ide a2 o 2 di ide a (f om heo em) L​ e a = 2c, fo ome in ege c F om e a ion (i) 2b2 = (2c)2 o 2b2 = 4c2 o b2 = 2c2 Downloaded from www.padhle.in

​I mean ha 2 di ide b2 and o 2 di ide b The efo e a and b ha e a lea 2 a a common fac o . B​ hi con adic he fac ha a and b a e co- ime. Thi con adic ion i d e o o ong a m ion ha 2 i a ional. So, e concl de ha 2 i i a ional. Hence P o ed. An 14). - (k+6) + 2(2k-1) As s m of eros = -b/a = -(-(k+6))/1 = k+6 Prod c of eros = c/a = 2(2k-1)/1 = 2(2k-1) Acc di g he e i S m of eros =1/2 Prod c of eros Downloaded from www.padhle.in

-b/a = 1/2 c/a k + 6 = 1/2 2(2k -1) k + 6 = 2k -1 6 + 1 = 2k - k 7=k So Val e of k = 7 Ans 15). ​Le he e en age of fa he be ea and m of age of hi o child en be ea Acco ding o e ion = 3 (i) A​ f e 5 ea Fa he age = ( + 5) ea S m of age of o child en = ( + 5 + 5) ea = ( + 10) ea I​ n 2nd ca e a ion (i)) Acco ding o e ion + 5 = 2 ( + 10) o + 5 = 2 + 20 o 2 = 15 o 3 2 = 15 (U ing e = 15 No f om e a ion (i) = 3 (P = 15) Downloaded from www.padhle.in

o = 3 15 = 45 An ​ P e en age of fa he i 45 ea . OR Le he f ac ion be / Acco ding o e ion = 1/3 o 3( 2) = o 3 = 6 (i) again, Acco ding o e ion = 1/2 e ge o2 = 1 o 2 = -1 (ii) On ol ing e a ion (i) and (ii), = 7, = 15 An T​ he e i ed f ac ion i 7/15 An 16). We kno ha a oin on he -a i i of he fo m (0, ). So, le he oin P(0, ) be e idi an f om A (5, -2) and B (-3, 2) ​Then AP = BP ​o AP2 = BP2 ​o (5 0)2 + (-2 )2 = (-3 0)2 + (2 )2 o 25 + 4 + 2 + 4 = 9 + 4 + 2 4 8 = -16 = -2 So, he e i ed oin i (0, -2) Downloaded from www.padhle.in

OR ​The line egmen AB i i ec ed a he oin P and Q and P i nea e o A So, P di ide AB in he a io 1 : 2 P lie on he line 2 + k = 0 I ill a i f he e a ion. ​On ing = 3 and = -2 in he gi en e a ion, e ge 2(3) (-2) + k = 0 6+2+k=0 k = -8 Hence, k = -8 An 17). We kno he rigonome ric iden i ies , 1 ) sin A + cos A = 1 Downloaded from www.padhle.in

2 ) sec A = 1 + an A 3 ) cosec A = 1 + co A 4 ) sinA cosecA = 1 5 ) cosA secA = 1 No , LHS = (sinA+cosecA) + (cosA+secA) = sin A+cosec A + 2sinAcosecA+ cos A + sec A + 2cosAsecA = (Sin A + cos A ) + 2 + 2 + cosec A + Sec A = 1 + 2 + 2 + ( 1 + co A )+( 1 + an A ) = 7 + co A + an A Hence Pro ed. OR Downloaded from www.padhle.in

Ans 18). Le TR = Since OT i e endic la bi ec o of PQ. Downloaded from www.padhle.in

The efo e, PR=QR=4cm In igh iangle OTP and PTR, e ha e: ​TP2=TR2+PR2 Al o, OT2=TP2+OP2 OT2=(TR2+PR2) + OP2 ( +3)2= 2+16+25 (OR = 3, a OR2 = OP2 - PR2) ​6 =32 ​= ​TP2=TR2+PR2 ​TP2= +42 = +16 = ​TP= cm Ans 19). To pro e CD = BD AD In CAD, CA = CD + AD .... (1) Also in CDB, CB = CD + BD .... (2) (1) + (2) e ge : Downloaded from www.padhle.in

CA + CB = 2CD + AD + BD AB = 2CD + AD + BD AB - AD = BD + 2CD (AB + AD)(AB - AD) - BD = 2CD (AB + AD)BD - BD = 2CD BD(AB + AD - BD) = 2CD BD(AD + AD) = 2CD BD 2AD = 2CD CD = BD AD Hence pro ed. OR G​ i en, ABC i a igh -angled iangle in hich ∠C = 90 To o e : AQ2 + BP2 = AB2 + PQ2 Con c ion: Join AQ, PB and PQ P oof: In AQC, ∠C = 90 AQ2 = AC2 + CQ2 (i) (U ing P hago a heo em) In PBC, ∠C = 90 BP2 = BC2 + CP2 (ii) (U ing P hago a heo em) Adding e a ion (i) and (ii) AQ2 + BP2 = AC2 + CQ2 + BC2 + CP2 = AC2 + BC2 + CQ2 + CP2 Downloaded from www.padhle.in

o AQ2 + BP2 = AB2 + PQ2 Hence P o ed. Ans 20). G​ i en, ABCD i a ec angle i h ide AB = 8 cm and BC = 6 cm In ABC AC2 = 82` + 62 (B P hago a Theo em) ​ AC2 = 64 + 36 ​ AC2 = 100 ​ AC = 10 cm The diagonal of he ec angle ill be he diame e of he ci cle adi of he ci cle = = 5 cm A ea of haded o ion = A ea of ci cle A ea of Rec angle = 2 lb = 3.14 5 5 8 6 = 78.50 48 = 30.50 cm2 Hence, A ea of haded o ion = 30.5 cm2 An 21). ​Le b be he id h and h be he de h of he canal b = 6 m and h = 1.5 m Wa e i flo ing i h a eed = 10 km/h = 10,000 m/h L​ eng h of a e flo ing in 1 h = 10,000 m L​ eng h (l) of a e flo ing in h = 5,000 m Vol me of a e flo ing in 30 min. = l b h = 5000 6 1.5 m3 L​ e he a ea i iga ed in 30 min ( h ) be m2 Vol me of a e e i ed fo i iga ion = Vol me of a e flo ing in 30 min. Downloaded from www.padhle.in

​ = 5000 6 1.5 ​ = 562500 m2 = 56.25 hec a e . (∵ 1 hec a e = 104 m2) Hence, he canal ill i iga e 56.25 hec a e in 30 min. An 22). H​ e e, he ma im m cla f e enc i 16 M​ odal cla = 30-40 L​ o e limi (l) of modal cla = 30 C​ la i e (h) =10 F​ e enc (f1) of he modal cla = 16 ​F e enc (f0) of eceding cla = 10 ​F e enc (f2) of cceeding cla = 12 Mode = l + ( f1 f0 f2 )​ h 2f 1 f0 30 + ( 1 10 )​ 10 2 10 12 = 30 + 6/10 10 Ans Hence, Mode is 36. Ans 23). L​ e he a A i h longe diame e ake ho and he a B i h malle diame e ake ( + 2) ho o fill he ank. ​Po ion of ank filled b he a A in 1 h . = a​ nd Po ion of ank filled b he a B in 1 h . = Po ion of he ank filled b bo h a in 1 h . = + = T​ ime aken b bo h a o fill he ank = 1 7/8 h = h Downloaded from www.padhle.in

Po ion of he ank filled b bo h in 1 h . = 8/15 Acco ding o e ion, = ​= ​ 15 + 15 = 4 2 + 8 =3 ​ 4 2 12 + 5 15 = 0 ​ 4 ( 3) + 5 ( 3 ) = 0 ​ (4 + 5)( 3) =0 ​4 +5=0o 3=0 ​ = -5/4 Since, ime can no be nega i e hence, neglec ed hi al e i ; Hence, he ime aken i h longe diame e a = 3 ho and he ime aken i h a malle diame e a = 5 ho . OR Le he speed of he boa in s ill a er be km/h and he speed of he s ream be km/h. ​Speed ps ream= ( - ) km/h ​Speed do ns ream= ( + ) km/h O​ n s b rac ing, e ge , Downloaded from www.padhle.in

On sol ing, e ge , = 8 and =3 S​ peed of boa in s ill a er = 8 km/hr A​ nd, Speed of s ream = 3 km/hr Ans 24). GI en: S​4 =​ 40 S​14 ​=280 Sn=n/2[2a+(n-1)d] s=4/2[2a+(4-1)d] 40=2[2a+3d] 40/2=2a+3d 2a+3d=20 (1) similarl , S14=14/2[2a+(14-1)d] 280=7[2a+13d] 280/7=2a+13d 2a+13d=40. .(2) B s b rac ing eq(2) from(1) e ge , Downloaded from www.padhle.in

-10d =-20 10d = 20 d = 20/10 d=2 No s bs i e d=2 in eq(1) a=7 ​We ha e ,a=7 & d=2 Sn= n/2 [2a+(n-1)d] Sn= n/2 [2(7)+(n-1)2] Sn= n/2 [14+2n-2] Sn= n/2 [12+2n] Sn= n/2 [2(6+n)] Sn= n (6+n) Sn= 6n + n2​ Ans 25). Downloaded from www.padhle.in

Ans 26). Consider ABC Where AB is Heigh of To er BC is dis ance from boa o o er ​TanC = AB/BC = 3 No , as heigh of he o er is 100m, AB=100 100/BC = 3 BC = 57.73 N​ o le D be he poin he boa ra elled TanD = AB/BD = 1/ 3 = 100/BD = 173.2 ​Dis ance ra elled b boa CD=BD BC= 173.2 57.73 = 115.47m Downloaded from www.padhle.in

Speed of he boa = 115.47/2 = 57.73 m/min. Ans ​ S​ peed of he boa is 57.73 m/min. Ans 27). S e of Con c ion a e a follo : 1. D a AB = 5 cm ho n in he fig e. 2. A he oin , A d a ∠BAX = 45 3. F om AX c off AC = 6 cm 4. Join BC, ABC i fo med i h gi en da a. 5. D a AY making an ac e angle i h AB a 6. D a 5 a c P1, P2, P3, P4, and P5 i h e al in e al . 7. Join BP5. Downloaded from www.padhle.in

8. D a P3B P5B mee ing AB a B . 9. F om B , d a B C BC mee ing AC a C AB C ABC Hence AB C i he e i ed iangle. An 28). Gi en: ​Vol me of he b cke - 12308.8cm3 ​ 1 = 20cm and 2 = 12cm Downloaded from www.padhle.in

An H​ ence, heigh of he b cke = 15cm and A ea of he me al hee ed i 2160.32 .cm. An 29). Gi en: A righ angled ABC, righ angled a B To Pro e- AC =AB +BC Cons r c ion: Dra perpendic lar BD on o he side AC . Downloaded from www.padhle.in

Proof: We kno ha if a perpendic lar is dra n from he er e of a righ angle of a righ angled riangle o he h po en se, hen he riangles on bo h sides of he perpendic lar are similar o he hole riangle and o each o her. We ha e,​ ADB ABC. (b AA similari ) Therefore, AD/ AB=AB/AC (In similar Triangles corresponding sides are propor ional) AB =AD AC (1) Also, BDC ABC Therefore, CD/BC=BC/AC (in similar Triangles corresponding sides are propor ional) Or, BC =CD AC (2) Adding he eq a ions (1) and (2) e ge , Downloaded from www.padhle.in

AB +BC =AD AC+CD AC AB +BC =AC(AD+CD) ( From he fig re AD + CD = AC) AB +BC =AC . AC AC =AB +BC Ans Hence, Pro ed. Ans 30). Downloaded from www.padhle.in

Gi en: Median=32.5, hen 30 40 is class in er al. l = 30 , cf = 14+f​1 f=12 and h = 40 30 = 10. N = 40 and N/2 = 20 31+ f1 +f2 = 40. So, f1 + f2 =9 Downloaded from www.padhle.in

Median = N cf h 2 f 32.5 = 30 + 20 ( 1 + f 1) 10 12 2.5 = f1 ​ 5 15 = 30 5f​1 ​ 5f1​ ​​ = 15 ​ and on s bs i ing al es, f1 =3 and f2 = 6 OR T​ o d a a le han ogi e, e ma k he e -cla limi of he cla in e al on he -a i and hei c.f. on he -a i b aking a con enien cale. ​He e, n = 100 ⇒ n/2 = 50 Downloaded from www.padhle.in

T​ o ge median f om g a h F om 50, D​ a a e endic la , he oin he e hi e endic la mee on he -a i ill be he median. ​Median = 29 Downloaded from www.padhle.in

CBSE Ma hema ic 2018 Ge e a I c: . ()A e aec ( ) Th e aec f 30 e d ded f ec A, B, C a d D. ( ) Sec A c a 6 e f 1 a each. Sec Bc a 6 e f 2 a each, Sec C c a 10 e f3 a each. Sec D c a 8 e f 4 a each. ( ) The e e a ch ce. H e e , a e a ch ce ha bee ded f e f 3 a each a d 3 e f 4 a a e e f he a e a e a ch each. Y ha e e. ( ) U e f ca c a e ed. Downloaded from www.padhle.in

Qe i Sec i -A (1 Ma k Each) 1. If = 3 i e f he ad a ic e a i 2​ ​ 2k 6 = 0, he fi d he al e f k. 2. Wha i he HCF f malle ime mbe a d he malle c m i e mbe ? 3. Fi d he di a ce f a i P( , ) f m he igi . 4. I a AP, if he c mm diffe e ce (d) = 4 a d he e e h e m (a7) i 4, he fi d he fi e m. 5. Wha i he al e f (c 2​ 6​ 7 i ​2​ 23 ) ? 6. Gi e ​ A​ BC​ ​ ​ ​ ​ P​ QR, if AB = 1/3 , he fi d a ABC ? PQ a P QR Sec i -B (2 Ma k Each) 7. Gi e ha 2 i i a i al, e ha (5 + 3 2 ) i a i a i al mbe Downloaded from www.padhle.in

8. I fig. 1, ABCD i a ec a gle. Fi d he al e f a d . 9. Fi d he m f fi 8 m l i le f 3. 10. Fi d he a i i hich P (4, m) di ide he li e egme j i i g he i A (2, 3) a d B (6, 3). He ce fi d m. 11. T diffe e dice a e ed ge he . Fi d he babili : (i) f ge i g a d ble . (ii) f ge i g a dice. m 10, f he mbe he 12. A i ege i ch e a a d m be ee 1 a d 100. Fi d he babili ha i i : (i) di i ible b 8 (ii) di i ible b 8. Sec i - C (3 Ma k Each) 13. Fi d HCF a d LCM f 404 a d 96 a d e if ha HCF LCM = P d c f he gi e mbe . 14. Fi d all e e f he l mial (2 ​4​ 9 3​ ​ + 5 ​2​ + 3 1) if f i e e a e (2 + 3 ) a d (2 3 ). Downloaded from www.padhle.in

15. If A ( 2, 1), B (a, 0), C (4, b) a d D (1, 2) a e he e ice f a a allel g am ABCD, fi d he al e f a a d b. He ce fi d he le g h f i ide . OR If A ( 5, 7), B ( 4, 5), C ( 1, 6) a d D (4, 5) a e he e ice f a ad ila e al, fi d he a ea f he ad ila e al ABCD. 16. A la e lef 30 mi e la e ha i ched led ime a d i de each he de i a i 1500 km a a i ime, i had i c ea e i eed b 100 km/h f m he al eed. Fi d i al eed. 17. P e ha he a ea f a e ila e al ia gle de c ibed e ide f he a e i e al half he a ea f he e ila e al ia gle de c ibed e f i diag al. OR If he a ea f imila ia gle a e e al, e ha he a e c ge. 18. P e ha he le g h f a ge d a f m a e e al i a ci cle a e e al. 19. If 4 a ​​ = 3, e al ae 4 c1 4 c ( 1) OR If a 2A = c (A 18 ), he e 2A i a ac e a gle, fi d he al e f A. Downloaded from www.padhle.in

20. Fi d he a ea f he haded egi i Fig. 2, he e a c d a i h ce e A, B, C a d D i e ec i ai a mid- i P, Q, R a d S f he ide AB, BC, CD a d DA e ec i el f a 12 cm. [U e = 3.14] a e ABCD f ide 21. A de a icle a made b c i g a hemi he e f m each e d f a lid c li de , a h i he fig e. If he heigh f he c li de i 10 cm a d i ba e i f adi 3.5 cm. Fi d he al face a ea f he a icle. OR A hea f ice i i he f m f a c e f ba e diame e 24 m a d heigh 3.5 m. Fi d he l me f he ice. H m ch ca a cl h i e i ed j c e he hea ? 22. The able bel h he ala ie f 280 e : Sala ( i Th a d ​₹ ) N . f Pe 5-10 49 10-15 133 15-20 63 20-25 15 25-30 6 30-35 7 35-40 4 Downloaded from www.padhle.in

40-45 2 1 45-50 Calc la e he media ala f he da a. Sec i - D (4 Ma k Each) 23. A m b a h e eed i 18 km/h i ill a e ake 1 h eam he ame m e g 24 km eam ha ed . Fi d he eed f he eam. OR A ai a el a a ce ai a e age eed f a di a ce f 63 km a d he a el a a di a ce f 72 km a a a e age eed f 6 km/h m e ha i igi al eed. If i ake 3 h c m le e a al j e , ha i he igi al a e age eed ? 24. The m f f c ec i e mbe i a AP i 32 a d he a i f he d c f he fi a d he la e m he d c f middle e m i 7 : 15. Fi d he mbe . 25. I a e ila e al ABC, D i a i ide BC ch ha BD = 1/3 BC. P e ha 9(AD)2​ ​ = 7(AB)2​ ​ . OR P e ha , i a igh ia gle, he a e he h e e i e al he m f he a e he he ide . Downloaded from www.padhle.in

26. D a a ia gle ABC i h BC = 6 cm, AB = 5 cm a d ∠​ ​ ​ABC = 60 . The c c a ia gle h e ide a e 3/4 f he c e di g ide f he ABC. 27. P e ha : (Si A - 2Si 3​ ​A) / (2C 3​ ​A - C A) = Ta A 28. The diame e f he l e a d e e d f a b cke i he f m f a f m f a c e a e 10 cm a d 30 cm e ec i el . If i heigh i 24 cm, fi d : (i) The a ea f he me al hee ed make he b cke . (ii) Wh e h ld a id he b cke made b di a la ic ? [U e = 3.14] 29. A b e ed f m he f a 100 m high ligh h e f m he ea-le el, he a gle f de e i f hi a e 30 a d 45 . If e hi i e ac l behi d he he he ame ide f he ligh h e, fi d he di a ce be ee he hi . [U e 3 = 1.732] 30. The mea f he f ll i g di ib i i 18. Fi d he f e e c f f he cla 19 21. Cla 11 13 13-15 15-17 17-19 19-21 21-23 23-25 Fe e c 3 6 9 13 f 5 4 OR The f ll i g di ib i gi e he dail i c me f 50 ke f a fac : Downloaded from www.padhle.in

Dail I c me (i ​₹) 100 120 120 140 140 160 160 180 180 200 14 8 6 10 N mbe f ke 12 C e he di ib i ab e a le ha e c m la i e f e e c di ib i a d d a i gi e. Ae A 1). ad a ic e a i i : ​2​ 2k 6 = 0 ​ Gi e : =3i a f ab e e a i The , (3)​2 ​ 2k (3) 6 = 0 9 6k 6 = 0 3 6k = 0 3 = 6k k= A 2). ​ ​ Smalle ime mbe = 2 S​ malle c m i e mbe = 4 P​ ime fac i a i f 2 i 1 2 ​P ime fac i a i f 4 i 1 22​ A : HCF (2, 4) = 2 A 3). ​ ​ ​Di a ce f m P ( , ) igi Downloaded from www.padhle.in

P​ i f igi Q (0,0) i. ​B a l i g di a ce f m la ( 2 - 1) + ( 2 - 1) (0 - ) + (0 - ) + i. A ​ The di a ce f a i ( , ) f m he igi i ​ + A 4). Gi e : ​ ​c mm diffe e ce (d) = 4 ​ he e e h e m (a7) = 4 T fi d: The fi e m (a) S, d= 4 a7 = 4 a + 6d = 4 a + 6( 4) = 4 a = 28 A ​ The fi e m (a) i 28 A 5). Sl i : c ​2​ 67 i ​2 2​ 3 = c ​2 ​67 c ​2 (​ 90 23 ) Downloaded from www.padhle.in

We k , [ i (90 ​ )​ = c ​ ​ ] = c 2​ ​ 67 c ​2 ​67 = 0 A he al e f (c ​2​67 i ​2​23 ) i 0 A 6). Gi e , ABC PQR & AB = 1 PQ 3 S , a (AB C ) = a (P QR) =​ ( )​2 ​= 1/9 A a (AB C ) = 1/9 a (P QR) A 7). Gi e , 2 i i a i al mbe . Le 2 = m S e, 5 + 3 2 i a a i al mbe . S , 5 + 3 2 = a/b 3 2 = a/b - 5 3 2= a 5b b 2= a 5b 3b a 5b = m 3b Downloaded from www.padhle.in

B a 5b i ai al mbe a d ha be m, b hi 3b c adic he fac ha m = 2 hich i i a i al A He ce, 5 + 3 2 i al i a i al. A 8). ​Since ABCD is a rectangle. its opposite sides are equal.  A​ D = BC and AB = DC  x - y = 14  x + y = 30  S​ olving both the equations by elimination method,  Add the equations to get:   2x = 44  x = 22  ​Put the value of x in any equation, we get  x + y = 30  Downloaded from www.padhle.in

22 + y = 30    y = 8  Hence, x = 22 and y = 8    Ans 9).We know ​, m f e m f m 1 e m i   I​ he gi e e i , Fi e m = 3  C mm diffe e ce = 3 The , a l i g f m la f m ab e.  ​A mi g he al e f a 8 a d a a 3               Downloaded from www.padhle.in

Ans​ ​The sum of first 8 multiples of 3.    Ans 10).  Let P divides line segment AB in the ratio k : 1.     Le he a i be k : 1   B ec i f m la : 12 21 ​, 12 21 1 2 12   ( , ) = [ (k 2 + 1) / (k + 1) , (k 2 + 1) / (k + 1) ] (4, m) = [ k(6) + 2 / (k + 1) , k( 3) + 3 / (k + 1) ]   4 = (6k + 2) / (k + 1) a d m = ( 3k + 3) / (k + 1)   ​4k + 4 = 6k + 2   6​ k 4k = 4 2   ​ 2k = 2   ​k=1   N , b i i g k = 1 i m' al e   m = [ 3(1) + 3] / (1 + 1)   ​ 3+3 / 2   ​0/2   ​0 Downloaded from www.padhle.in

A 11). T al mbe a e 2, 3, 4, ........., 99 Le E be he e e f ge i g a mbe di i ible b 8. E = 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96 = 12 P(E) = Fa ab e O c e T aO c e 12 8 = 0​ .1224 Le E be he e e f ge i g a mbe di i ible b 8. The , P(E ) = 1 P(E) = 1 0.1224 = 0.8756 A 12). HCF f 404 a d 96 404 = 2*2*101 96 = 2*2*2*2*2*3 C mm Fac = 2*2 HCF f 404 a d 96 = 4 LCM f 404 a d 96 404 = 2*2*101 96 = 2*2*2*2*2*3 Downloaded from www.padhle.in

LCM = 2*2*2*2*2*3*101 = 9696 mbe LCM f 404 a d 96 i 9696 HCF LCM = P d c f he 4*9696 = 404*96 38784 = 38784 LHS = RHS A 14). Gi e : T e e f he l mial a e: (2 + 3 ) a d (2 3 ). Q ad a ic l mial i h e i gi e b : (2 + 3 ) . (2 - 3 ) ( 2 3 ) ( - (2 + 3 ) ( -2)2​ ​- ( 3 )​2 2​ -​ 4 + 4 - 3 2​ -​ 4 + 1 = g( ) N , g( ) ill be a fac f ( ) g( ) ill be di i ible b ( ) Downloaded from www.padhle.in

F he e e , 2 2​ ​ 1=0 2 ​2 ​ 2 + 1 = 0 2 ( 1) + 1 ( 1) = 0 ( 1) (2 + 1) = 0 1 = 0, 2 + 1 = 0 = 1, = -​ He ce, Ze e f ( ) = 2 4​ ​ 9 3​ ​ + 5 2​ ​ + 3 1 a e 1, - , 2 + 3 , 2 - 3 A 15). Downloaded from www.padhle.in

Gi e ide f a allel g am ​A(​ 2,1), B​ ​(​ ​,0), C​ (​ 4,​ ​), D​ ​(1,2) We k ha diag al f Pa allel g am bi ec each he . Mid- i le a ​ ​ f diag al ​AC​ i gi e b ​ = ​( 1+ 2 ) ​a d = ​( 1+ 2) 2 2 O(​ 2+4 , 1+ b ) 2 2 Mid- i le a P​ ​ f diag al ​BD​ i gi e b P​( a+1 , 0+2 ) 2 2 P i ​O​ a d P​ ​ a e ame E a i g he c e di g c di a e f b h mid i , e ge 2+4 a+1 2 2 a=1a d Downloaded from www.padhle.in

1+b 0+2 2 2 b=1a d N he Gi e c di a e f he a allel g am a e i e a A( 2,1), B(1,0), C(4,1), D(1,2) B di a ce f m la, ( 2 1)2 + ( 2 1)2 We ca fi d he le g h f each ide AB = ( 2 1 )2 + (1 0)2 AB = (3)2 + (1)2 = 10 AB = CD ( ai f i e ide f he a allel g am a e a allel a d e al) BC = (4 1)2 + (1 0)2 = 10 BC = (3)2 + (1)2 = 10 BC=AD ( ai f i e ide f he a allel g am a e a allel a d e al ) AB=BC=CD=AD = 10 ABCD i a Rh mb Downloaded from www.padhle.in

OR Gi e ABCD i ad ila e al. ad ila e al i di ided i B jiig i A a d C, he ia gle . N , A ea f ad. ABCD = A ea f ABC + A ea f ACD A ea f ABC A ea f ​ A​ BC = ​ [ 1 2 - 3 2 3 - 1 3 1- 2 ] = ​[ 5 5 6 4 6 7 17 5] = [ ​ 5(1) 4( 13) 1(12) ​] = (​ 5 52 12 )​ = (​ 35)​ ​ . i A ea f ​ ​ADC = ​ [ 1 2 - 3 2 3 - 1 3 1- 2 ] = ​ [ 5 5 6 4 6 7 ( 1) 7 5 ] = ​ [ 5 5 6 4 6 7 ( 1) 7 5 ] Downloaded from www.padhle.in

= [ ​ 5(11) 4( 13) 1(2) ​] = ​ ( 55 52 12) = 109/2 . i . S , A ea f ad ila e al ABCD = (​ 35) ​ ​(109) = 72 . i . A 16). Le he al eed f he la e be km/h . I c ea ed eed = 100 km/h Time ake c e 1500 km = 1500/ h . Time ake c e 1500 km i h i c ea ed eed = 1500/ + 100 h Acc di g he Q e i , 1500/ - 1500( + 100) = 30/60 + 100 - 300000 = 0 + 600 - 500 - 300000 = 0 ( + 600)( - 500) = 0 Downloaded from www.padhle.in

= -600 500 (Neglec i g ega i e ig a eed ca be ega i e) = 500 A He ce, he al eed f he la e i 500 km/h . A 17). I ​ ​ABC, AC2​ ​ = AB2​ +​ BC​2 a2​ +​ ​ ​a2​ ​ = 2a​2 med ide BC f a e ABCD) AC = 2 a​2 ​ = 2 a A ea f e ila e al BEC (f = 3 ( ide)​2 4 = 3 ​ a​2 4 Downloaded from www.padhle.in

A ea f e ila e al ACF (f med diag al AC f a e ABCD) = 3 ( 2 a ​)​2 4 = 3 2a 2 4 =2 3 a2​ 4 F m e . (i) a d (ii), a ​ ​ACF = 2 a ​ B​ CE O, a (​ B​ CE) = a (​ ​ACF) He ce P ed, a ea f ia gle de c ibed e ide f ae i half he a ea f ia gle de c ibed i diag al. OR Downloaded from www.padhle.in