Arithmetic Progression - Nayaks TutorialsArithmetic Progression 1 . Determine the 10th term from the...

AlgebraAlgebraAlgebraAlgebra (Std.(Std.(Std.(Std. X)X)X)X)

ArithmeticArithmeticArithmeticArithmetic ProgressionProgressionProgressionProgression1111.... DetermineDetermineDetermineDetermine thethethethe 10th10th10th10th termtermtermterm fromfromfromfrom thethethethe endendendend ofofofof thethethethe A.P.A.P.A.P.A.P. 4,9,14,4,9,14,4,9,14,4,9,14, ………… ,,,, 254.254.254.254.Solution:Solution:Solution:Solution: We h ave ,

l = Las t te rm = 2 5 4 an d , d = Com m on d iffe ren ce = 5 . 1 0 th term from th e en d = l (1 0 1 ) d = l 9 d = 2 5 4 9 5 = 2 0 9 .

2222.... IsIsIsIs 184184184184 aaaa termtermtermterm ofofofof thethethethe sequencesequencesequencesequence 3,3,3,3, 7,7,7,7, 11,11,11,11, ………… ,,,, ????Solution:Solution:Solution:Solution: Clearly, th e seq u en ce is an A.P. with firs t te rm a (= 3 ) an d com m on d iffe ren ce d (= 4 ).

Let th e n th te rm of th e g iven seq u en ce b e 1 8 4 . Th en ,a n = 1 8 4

a + (n 1 ) d = 1 8 4 3 + (n 1 ) 4 = 1 8 4 4 n = 1 8 5

n = 4 641

Sin ce n is n o t a n a tu ra l n u m b er. So , 1 8 4 is n o t a te rm of th e g iven seq u en ce .

3333....ForForForFor whatwhatwhatwhat valuevaluevaluevalue ofofofof nnnn isisisis thethethethe nnnnthththth termtermtermterm ofofofof thethethethe followingfollowingfollowingfollowing twotwotwotwo A.PA.PA.PA.P’’’’ssss thethethethe same?same?same?same?i)i)i)i) 1,7,13,19,1,7,13,19,1,7,13,19,1,7,13,19, ………… ii)ii)ii)ii) 69,68,67,69,68,67,69,68,67,69,68,67,…………

Solution:Solution:Solution:Solution: Clearly, 1 ,7 ,1 3 ,1 9 ,… form s an A.P. with firs t te rm 1 an d com m on d iffe ren ce 6 .Th ere fore , its n th te rm is g iven b y

a n = 1 + (n 1 ) 6 = 6 n 5Als o , 6 9 , 6 8 , 6 7 , 6 6 , … form s an A.P. with firs t te rm 6 9 an d com m on d iffe ren ce 1 .So , its n th te rm is g iven b y

a n = 6 9 + (n 1 ) ( 1 ) = n + 7 0Th e two A.Ps will h ave id en tica l n th te rm s , if

a n = a n 6 n 5 = n + 7 0 7 n = 7 5

n =7

7 5 , wh ich is n o t a n a tu ra l n u m b er.

Hen ce , th ere is n o valu e of n for wh ich th e two A.Ps will h ave id en tica l te rm s .

4444....IfIfIfIf thethethethe 8888thththth termtermtermterm ofofofof anananan A.P.A.P.A.P.A.P. isisisis 31313131 andandandand thethethethe 15151515thththth termtermtermterm isisisis 16161616 moremoremoremore thanthanthanthan thethethethe 11111111thththth term,term,term,term, findfindfindfind thethethethe A.P.A.P.A.P.A.P.Solution:Solution:Solution:Solution: Let a b e th e firs t te rm an d d b e th e com m on d iffe ren ce of th e A.P.

We h ave ,a 8 = 3 1 an a 1 5 = 1 6 + a 1 1

a + 7 d = 3 1 an d a + 1 4 d = 1 6 + a + 1 0 d a + 7 d = 3 1 an d 4 d = 1 6 a + 7 d = 3 1 an d d = 4 a + 7 4 = 3 1 a + 2 8 = 3 1 a = 3

Hen ce , th e A.P. is a , a + d , a + 2 d , a + 3 d , … i.e ., 3 ,7 ,1 1 ,1 5 ,1 9 ,…

5555....WhichWhichWhichWhich termtermtermterm ofofofof thethethethe arithmeticarithmeticarithmeticarithmetic progressionprogressionprogressionprogression 5,15,25,5,15,25,5,15,25,5,15,25,………… willwillwillwill bebebebe 130130130130 moremoremoremore thanthanthanthan itsitsitsits 31313131stststst tem?tem?tem?tem?Solution:Solution:Solution:Solution: We h ave , a = 5 an d d = 1 0 a 3 1 = a + 3 0 d = 5 + 3 0 1 0 = 3 0 5

Let n th te rm of th e g iven A.P. b e 1 3 0 m ore th an its 3 1 s t te rm . Th en ,a n = 1 3 0 + a 3 1

a + (n 1 ) d = 1 3 0 + 3 0 5 5 + 1 0 (n 1 ) = 4 3 5 1 0 (n 1 ) = 4 3 0 n 1 = 4 3 n = 4 4

Hen ce , 4 4 th te rm of th e g iven A.P. is 1 3 0 m ore th an its 3 1 s t te rm .

AlgebraAlgebraAlgebraAlgebra (Std.(Std.(Std.(Std. X)X)X)X)2222

6666....IfIfIfIf fivefivefivefive timestimestimestimes thethethethe fifthfifthfifthfifth termtermtermterm ofofofof anananan A.P.A.P.A.P.A.P. isisisis equalequalequalequal totototo 8888 timestimestimestimes itsitsitsits eightheightheightheighth tem,tem,tem,tem, showshowshowshow thatthatthatthat itsitsitsits 13131313thththth termtermtermterm isisisiszero.zero.zero.zero.

Solution:Solution:Solution:Solution: Let a 1, a 2, a 3, … , a n, … b e th e A.P. with its firs t te rm a an d com m on d iffe ren ce d .It is g iven th a t

5 a 5 = 8 a 8

5 (a + 4 d ) = 8 (a + 7 d ) 5 a + 2 0 d = 8 a + 5 6 d 3 a + 3 6 d = 0 3 (a + 1 2 d ) = 0 a + 1 2 d = 0 a + (1 3 1 ) d = 0 a 1 3 = 0 .

Hen ce , 1 3 th te rm is z ero .

7777....IfIfIfIf 2x,2x,2x,2x, xxxx ++++ 10,10,10,10, 3x3x3x3x ++++ 2222 areareareare inininin A.P.,A.P.,A.P.,A.P., findfindfindfind thethethethe valuevaluevaluevalue ofofofof x.x.x.x.Solution:Solution:Solution:Solution: Sin ce , 2 x , x + 1 0 , 3 x + 2 are in A.P. 2 (x + 1 0 ) = 2 x + (3 x + 2 ) 2 x + 2 0 = 5 x + 2 3 x = 1 8 x = 6 .

8888....IfIfIfIf thethethethe sumsumsumsum ofofofof mmmm termtermtermterm ofofofof anananan A.P.A.P.A.P.A.P. isisisis thethethethe samesamesamesame asasasas thethethethe sumsumsumsum ofofofof itsitsitsits nnnn terms,terms,terms,terms, showshowshowshow thatthatthatthat thethethethe sumsumsumsum ofofofof itsitsitsits (m(m(m(m ++++n)n)n)n) termstermstermsterms isisisis zero.zero.zero.zero.

Solution:Solution:Solution:Solution: Let a b e th e firs t te rm an d d b e th e com m on d iffe ren ce of th e g iven A.P. th en ,Sm = Sn

2m {2 a + (m 1 ) d } =

2n {2 a + (n 1 ) d }

2 a (m n )+ {m (m 1 ) n (n 1 )} d = 0 2 a(m n ) + {(m 2 n 2)(m n )} d = 0 (m n ) {2 a + (m n 1 ) d } = 0 2 a + (m + n 1 ) d = 0 [Qm n 0 ] …….(i)

Sm + n =2

nm {2 a + (m + n 1 ) d }

Sm + n =2

nm 0 = 0 [Us in g (i)]

9999....TheTheTheThe sumssumssumssums ofofofof nnnn termstermstermsterms ofofofof threethreethreethree arithmethicalarithmethicalarithmethicalarithmethical progrssionsprogrssionsprogrssionsprogrssions areareareare SSSS1111,,,, SSSS2222 andandandand SSSS3333.... TheTheTheThe firstfirstfirstfirst termtermtermterm ofofofof eacheacheacheach isisisisunityunityunityunity andandandand thethethethe commoncommoncommoncommon differencedifferencedifferencedifference areareareare 1,21,21,21,2 andandandand 3333 respectively.respectively.respectively.respectively. ProveProveProveProve thatthatthatthat SSSS1111 ++++ SSSS3333 ==== 2S2S2S2S2222....

Solution:Solution:Solution:Solution: We h ave ,S1 = Su m of n term of an A.P. with firs t te rm 1 an d com m on d iffe ren ce 1

S1 =2n {2 1 + (n 1 ) 1 } =

2n (n + 1 )

S2 = Su m of n term s of an A.P. with firs t te rm 1 an d com m on d iffe ren ce 2

S2 =2n {2 1 + (n 1 ) 2 } = n 2

S3 = Su m of n term s of an A.P. with firs t te rm 1 an d com m on d iffe ren ce 3

S3 =2n {2 1 + (n 1 ) 3 }

2n (3 n 1 )

Now, S1 + S3 =2n (n + 1 ) +

2n (3 n 1 ) = 2 n 2 an d S2 = n 2

Hen ce , S1 + S3 = 2 S2.


10101010.... TheTheTheThe sumsumsumsum ofofofof firstfirstfirstfirst sixsixsixsix termtermtermterm ofofofof anananan arithmeticarithmeticarithmeticarithmetic progressionprogressionprogressionprogression isisisis 42.42.42.42. TheTheTheThe ratioratioratioratio ofofofof itsitsitsits 10th10th10th10th termtermtermterm totototo itsitsitsits 30th30th30th30thtermtermtermterm isisisis 1:3.1:3.1:3.1:3. CalculateCalculateCalculateCalculate thethethethe firstfirstfirstfirst andandandand thethethethe thirteenththirteenththirteenththirteenth termtermtermterm ofofofof thethethethe A.P.A.P.A.P.A.P.

Solution:Solution:Solution:Solution: Let a b e th e firs t te rm an d d b e th e com m on d iffe ren ce of th e g iven A.P. Th en ,

S6 = 4 2 26 {2 a + (6 1 ) d } = 4 2 2 a + 5 d = 1 4 ……….(i)

It is g iven th a ta 1 0 : a 3 0 = 1 :3

d2 9a

d9a =

31

3 a + 2 7 d = a + 2 9 d 2 a 2 d = 0 a = d ………..(ii)

So lvin g (i) an d (ii), we g et a = d = 2 a 1 3 = a + 1 2 d = 2 + 2 1 2 = 2 6

Hen ce , firs t te rm = 2 an d th irteen th te rm = 2 6 .

11111111.... DivideDivideDivideDivide 32323232 intointointointo fourfourfourfour partspartspartsparts whichwhichwhichwhich areareareare inininin A.P.A.P.A.P.A.P. suchsuchsuchsuch thatthatthatthat thethethethe productproductproductproduct ofofofof extremesextremesextremesextremes isisisis totototo thethethethe productproductproductproduct ofofofofmeansmeansmeansmeans isisisis 7:15.7:15.7:15.7:15.

Solution:Solution:Solution:Solution: Let th e fou r p arts b e (a 3 d ),(a d ), (a + d ) an d (a + 3 d ). Th en ,Su m of th e n u m b ers = 3 2

(a 3 d ) + (a d ) + (a + d ) + (a + 3 d ) = 3 2 4 a = 3 2 a = 8It is g iven th a t dada

d3ad3a =

1 57

22

22

dad9a

=

1 57

2

2

d6 4d96 4

=

1 57

1 2 8 d 2 = 5 1 2 d 2 = 4 d = 2

Th u s , th e fou r p arts are a 3 d , a d , a + d an d a + 3 d i.e . 2 ,6 ,1 0 ,1 4 .

11112222.... IfIfIfIf ppppthththth,,,, qqqqthththth andandandand rrrrthththth termstermstermsterms ofofofof anananan A.P.A.P.A.P.A.P. areareareare a,a,a,a, b,b,b,b, cccc respectively,respectively,respectively,respectively, thenthenthenthen showshowshowshow thatthatthatthati)i)i)i) a(qa(qa(qa(q r)r)r)r) ++++ b(rb(rb(rb(r p)p)p)p) ++++ cccc (p(p(p(p q)q)q)q) ==== 0000ii)ii)ii)ii) (a(a(a(a b)b)b)b) rrrr ++++ (b(b(b(b c)c)c)c) pppp ++++ (c(c(c(c a)qa)qa)qa)q ==== 0000Solution:Solution:Solution:Solution: (i) Le t A b e th e firs t te rm an d D b e th e com m on d iffe ren ce of th e g iven A.P. Th en ,

a = P th term a = A + (p 1 )D …(i)b = q th term b = A + (q 1 )D …(ii)c = r th te rm c = A + (r 1 )D …(iii)We h ave ,a (q r) + b (r p ) + c (p q )

= {A + (p 1 )D} (q r) + {A + (q 1 )d } (r p ) + {A + (r 1 )D} (p q ) [Us in g (i), (ii),& (iii)]= A{(q - r) + (r - p ) + (p - q )} + D {(p - 1 )(q - r) + (q - 1 )(r - p ) + (r - 1 )(p - q )}= A 0 + D{p (q r) + q (r –p )+ r(p –q ) –(q – r)- (r – p )- (p – q )}= A 0 + D 0 = 0

(ii) On su b tract in g eq u atio n (ii) from eq u ation (i), eq u a tio n (iii) from eq u atio n (ii) an d eq u atio n (i) fromeq u ation (iii), we g eta b = (p - q )D, (b c)= (q r)D an d c a = (r p )D

(a b )r + (b c) p + (c a) q= (p q ) D r + (q r) D p + (r p ) D q= D{(p q )r + (q r)p + (r p )q }= D 0 = 0 .


11113.3.3.3. TheTheTheThe sumsumsumsum ofofofof thethethethe thirdthirdthirdthird andandandand seventhseventhseventhseventh termstermstermsterms ofofofof anananan A.P.A.P.A.P.A.P. isisisis 6666 andandandand theirtheirtheirtheir productproductproductproduct isisisis 8.8.8.8. FindFindFindFind thethethethe sumsumsumsum ofofofoffirstfirstfirstfirst sixteensixteensixteensixteen ofofofof thethethethe A.P.A.P.A.P.A.P.

Solution:Solution:Solution:Solution: Let a b e th e firs t te rm an d d b e th e com m on d iffe ren ce of th e A.P.We h ave ,a 3 + a 7 = 6 an d a 3a 7 = 8

(a + 2 d ) + (a + 6 d )= 6 an d (a + 2 d )(a + 6 d ) = 8 2 a + 8 d = 6 an d (a + 2 d )(a + 6 d ) = 8 a + 4 d = 3 an d (a + 2 d )(a + 6 d ) = 8 a = 3 4 d an d (a + 2 d )(a + 6 d ) = 8 (3 4 d + 2 d )(3 4 d + 6 d ) = 8 (3 2 d )(3 + 2 d ) = 8

9 4 d 2 = 8 4 d 2 = 1 d 2 =41

d =21

Case I Wh en d =21 :

Pu tt in g d =21 in a = 3 4 d , we g et

a = 3 4 21 = 3 2 = 1

S1 6 =2

1 6 {2 a + (1 6 1 )d } = 8 {2 1 + 1 5 21 } = 8

21 9 = 7 6

Case II Wh en d = -21 :

Pu tt in g d = -21 in a = 3 4 d , we g et a = 3 + 2 = 5

S1 6 =2

1 6 {2 a + (1 6 1 ) d }

S1 6 = 8 {1 0 + 1 5 -21 } = 8

25 = 2 0 .

11114.4.4.4. IfIfIfIf inininin anananan A.P.A.P.A.P.A.P. thethethethe sumsumsumsum ofofofof mmmm termstermstermsterms isisisis equalequalequalequal totototo nnnn andandandand thethethethe sumsumsumsum ofofofof nnnn termtermtermterm isisisis equalequalequalequal totototo m,m,m,m, thenthenthenthen proveproveproveprove thatthatthatthatthethethethe sumsumsumsum ofofofof (m(m(m(m ++++ n)n)n)n) termstermstermsterms isisisis (m(m(m(m ++++ n).n).n).n).

Solution:Solution:Solution:Solution: Let a b e th e firs t te rm an d b e d b e th e com m on d iffe ren ce of th e g iven A.P. Th em ,Sm = n

2m {2 a + (m - 1 )d } = n

2 am + m (m - 1 ) d = 2 n ….(i)An d , Sn = m

2n {2 a + (n – 1 ) d } = m

2 am + n (n 1 ) d = 2 m ….(ii)Su b tract in g eq u ation (ii)from eq u atio n (i). we g et2 a (m –n ) + {m (m - 1 ) – n (n - 1 )}d = 2 n 2 m

2 a (m –n )+ {m 2 n 2) (m –n )}d = - 2 (m - n ) 2 a + (m + n 1 ) d = - 2 [On d ivid in g b oth s id es b y (m - n ) ……(iii)

Now,

Sm + n =2

nm {2 a + (m + n 1 ) d }

Sm + n = 2

nm (- 2 ) [Us in g (iii)]

Sm + n = (m + n ).


15151515.... TwoTwoTwoTwo carscarscarscars startstartstartstart togethertogethertogethertogether inininin thethethethe samesamesamesame directiondirectiondirectiondirection fromfromfromfrom thethethethe samesamesamesame place.place.place.place. TheTheTheThe firstfirstfirstfirst goesgoesgoesgoes withwithwithwith uniformuniformuniformuniformspeedspeedspeedspeed ofofofof 10101010 km/h.km/h.km/h.km/h. TheTheTheThe secondsecondsecondsecond goesgoesgoesgoes atatatat aaaa speedspeedspeedspeed ofofofof 8888 km/hkm/hkm/hkm/h inininin thethethethe firstfirstfirstfirst hourhourhourhour andandandand increasesincreasesincreasesincreases thethethethespeedspeedspeedspeed bybybyby ½½½½ kmkmkmkm inininin eacheacheacheach succeedingsucceedingsucceedingsucceeding hour.hour.hour.hour. AfterAfterAfterAfter howhowhowhow manymanymanymany hourshourshourshours willwillwillwill thethethethe secondsecondsecondsecond carcarcarcar overtakeovertakeovertakeovertake thethethethefirstfirstfirstfirst carcarcarcar ifififif bothbothbothboth carscarscarscars gogogogo non-stop?non-stop?non-stop?non-stop?

Solution:Solution:Solution:Solution: Su p p os e th e secon d car overtakes th e firs t car afte r t h ou rs . Th en , th e two cars trave l th e sam ed is tan ce in t h ou rs .Dis tan ce trave lled b y th e firs t car in t h ou rs .Dis tan ce trave lled b y th e secon d car in t h ou rs .= Su m of t te rm s of an A.P. with firs t te rm 8 an d com m on d iffe ren ce ½.

=2t

211t82 =

43 1tt

Wh en th e secon d car will overtake th e firs t car, we h ave

1 0 t = 4

3 1tt 4 0 t = t 2 + 3 1 t t 2 3 1 t t 2 9 t = 0 t (t 9 ) = 0 t = 9 [Q t

0 ]Th u s , th e secon d car overtake th e firs t car in 9 h ou rs .

16161616.... 150150150150 workersworkersworkersworkers werewerewerewere engagedengagedengagedengaged totototo finishfinishfinishfinish aaaa piecepiecepiecepiece ofofofof workworkworkwork inininin aaaa certaincertaincertaincertain numbernumbernumbernumber ofofofof days.days.days.days. FourFourFourFour workersworkersworkersworkersdroppeddroppeddroppeddropped thethethethe secondsecondsecondsecond day,day,day,day, fourfourfourfour moremoremoremore workersworkersworkersworkers droppeddroppeddroppeddropped thethethethe thirdthirdthirdthird daydaydayday andandandand sosososo on.on.on.on. ItItItIt takestakestakestakes 8888 moremoremoremoredaysdaysdaysdays totototo finishfinishfinishfinish thethethethe workworkworkwork now.now.now.now. FindFindFindFind thethethethe numbernumbernumbernumber ofofofof daysdaysdaysdays inininin whichwhichwhichwhich thethethethe workworkworkwork waswaswaswas completed.completed.completed.completed.

Solution:Solution:Solution:Solution: Su p p ose th e work is com p le ted in n d ays wh en th e workers s ta rted d rop p in g . Sin ce 4 workers ared rop p ed on every d ay excep t th e firs t d ay. Th erefore , th e to ta l n u m b er of workers wh o worked all th e nd ays is th e su m of n term s of an A.P. with firs t te rm 1 5 0 an d com m on d iffe ren ce 4

i.e .,2n {2 1 5 0 + (n - 1 ) 4 } = n (1 5 2 2 n )

Had th e workers n o t d rop p ed th en th e work wou ld h ave fin ish ed it in (n – 8 ) d ays with 1 5 0 workersworkin g on each d ay. Th ere fore , th e to ta l n u m b er of workers wh o wou ld h ave worked all th e n d ays is1 5 0 (n – 8 ).

n (1 5 2 2 n ) = 1 5 0 (n – 8 ) 1 5 2 n – 2 n 2 = 1 5 0 n – 1 2 0 0 2 n 2 2 n 1 2 0 0 = 0 n 2 n 6 0 0 = 0 (n 2 5 ) (n + 2 4 ) = 0 [Q n > 0 ] n = 2 5

Hen ce , th e work is com p le ted in 2 5 d ays .

17.17.17.17. HowHowHowHow manymanymanymany twotwotwotwo digitdigitdigitdigit numbersnumbersnumbersnumbers leaveleaveleaveleave thethethethe remainderremainderremainderremainder 1111 whenwhenwhenwhen divideddivideddivideddivided bybybyby 5555 ????Solution:Solution:Solution:Solution:

Th e firs t 2 d ig it n o . wh ich is d ivis ib le b y 5 is 1 0 & th e las t 2 d ig it n o . d ivis ib le b y 5 is 9 5 . Th e two d ig it n o . th a t leave th e rem ain d e r 1 wh en d ivid ed b y 5 , are as fo llows .

1 1 , 1 6 , 2 1 , …., 9 6 .Wh ich is an A.P. wh erea = 1 1 , d = 1 1 – 6 = 5 & t n = 9 6We kn ow,t n= a + (n – 1 )d

9 6 = 1 1 + (n – 1 )5 9 6 = 1 1 + 5 n – 5 9 6 = 5 n + 6 9 6 – 6 = 5 n 5 n = 9 0

n =5

9 0

n = 1 8 No.No.No.No. ofofofof twotwotwotwo digitdigitdigitdigit numbersnumbersnumbersnumbers thatthatthatthat leaveleaveleaveleave thethethethe remainderremainderremainderremainder 1111 whenwhenwhenwhen divideddivideddivideddivided bybybyby 5555 isisisis 18181818....


18.18.18.18. HowHowHowHow manymanymanymany termstermstermsterms ofofofof thethethethe A.A.A.A. P.P.P.P. 16,16,16,16, 14,14,14,14, 12,12,12,12, ………….... areareareare neededneededneededneeded totototo givegivegivegive thethethethe sumsumsumsum 60606060 ???? ExplainExplainExplainExplain whywhywhywhy wewewewe getgetgetgetdifferentdifferentdifferentdifferent answersanswersanswersanswers ????

Solution:Solution:Solution:Solution:Th e g iven n u m b ers in A.P. are 1 6 , 1 4 , 1 2 , ….Wh ere a = 1 6 , d = 1 4 – 1 6 = –2 ,Sn = 6 0We kn ow,

Sn=n2

[2 a + (n – 1 )d ]

6 0 = n2

[2 × 1 6 + (n – 1 )(–2 )]

1 2 0 = n (3 2 – 2 n + 2 ) 1 2 0 = n (3 4 – 2 n ) 1 2 0 = 3 4 n – 2 n 2

2 n 2 – 3 4 n + 1 2 0 = 0Divid in g b y 2 on b oth s id es

n 2 – 1 7 n + 6 0 = 0 (n – 1 2 )(n – 5 ) = 0 n – 1 2 = 0 Or n – 5 = 0 n = 1 2 Or n = 5

For n = 5S5 = 6 0

i.e SumSumSumSum ofofofof 16,16,16,16, 14,14,14,14, 12,12,12,12, 10,10,10,10, 8888 isisisis 60.60.60.60.For n = 1 2S1 2 = 6 0

i.e SumSumSumSum ofofofof 16,16,16,16, 14,14,14,14, 12,12,12,12, 10,10,10,10, 8,8,8,8, 6,6,6,6, 4,4,4,4, 2,2,2,2, 0,0,0,0, -2,-2,-2,-2, -4,-4,-4,-4, -6-6-6-6 isisisis 60.60.60.60.

19.19.19.19. IfIfIfIf thethethethe 9999thththth termtermtermterm ofofofof anananan A.A.A.A. P.P.P.P. isisisis zerozerozerozero thenthenthenthen proveproveproveprove thatthatthatthat thethethethe 29292929thththth termtermtermterm isisisis doubledoubledoubledouble thethethethe 19191919thththth term.term.term.term.Solution:Solution:Solution:Solution:

ToToToTo proveproveproveprove ::::t 2 9 = 2 × t 1 9

For th e A.P.,We kn ow,t n = a + (n – 1 )d

For t 9, p u t n = 9 , t 9 = a + (9 – 1 )d t 9 = a + 8 d a + 8 d = 0 a = –8 d ….(ii)

For t 1 9 , p u t n = 1 9 , t 1 9= a + (1 9 – 1 )d t 1 9= a + 1 8 d t 1 9= –8 + 1 8 d [From (ii)] t 1 9 = 1 0 d …..(iii)

For t 2 9 , Pu t n = 2 9 t 2 9 = a + (2 9 – 1 )d

= a + 2 8 d= –8 d + 2 8 d ….(ii)

t 2 9 = 2 0 d ….(iv)From (iii) & (iv),

t 2 9 = 2 × t 1 9

29292929thththth termtermtermterm ofofofof anananan A.P.A.P.A.P.A.P. IsIsIsIs doubledoubledoubledouble ofofofof 19191919thththth term.term.term.term.


20.20.20.20. AAAA manmanmanman arrangesarrangesarrangesarranges totototo paypaypaypay aaaa debtdebtdebtdebt ofofofof Rs.3600Rs.3600Rs.3600Rs.3600 inininin 40404040 monthlymonthlymonthlymonthly installmentsinstallmentsinstallmentsinstallments whichwhichwhichwhich areareareare inininin aaaa AP.AP.AP.AP.WhenWhenWhenWhen 30303030 installmentsinstallmentsinstallmentsinstallments areareareare paidpaidpaidpaid hehehehe stopsstopsstopsstops payingpayingpayingpaying thethethethe installmentsinstallmentsinstallmentsinstallments leavingleavingleavingleaving oneoneoneone thirdthirdthirdthird ofofofofthethethethe debtdebtdebtdebt unpaid.unpaid.unpaid.unpaid. FindFindFindFind thethethethe valuevaluevaluevalue ofofofof thethethethe firstfirstfirstfirst instalinstalinstalinstalllllmentmentmentment.

Solution:Solution:Solution:Solution:Tota l Deb t = ` 3 6 0 0No. of in s ta llm en ts = 4 0No. of in s ta llm en ts p a id = 3 0

Rem ain in g am o u n t =31

x 3 6 0 0

= ` 1 2 0 0 Am t. p a id in 3 0 in s ta llm en ts = 3 6 0 0 - 1 2 0 0 = ` 2 4 0 0

Let th e firs t in s ta llm en t b e 'a ' an d th e com m on d iffe ren ce 'd '.

We kn ow,

d)1n(a22nsn

Case I:Pu t n = 4 0 , Sn = 3 6 0 0

3 6 0 0 = d)140(a2240

3 6 0 0 = 2 0 [2 a + 3 9 d ] 1 8 0 = 2 a + 3 9 d - - - - (I)

Cas e II:Pu t n = 3 0 , Sn = 2 4 0 0

2 4 0 0 = d)130(a2230

2 4 0 0 = 1 5 [2 a + 2 9 d ] 1 6 0 = 2 a + 2 9 d - - - - (II)

So lve eq u ation s I an d II, to g e t 'a '.

��


QuadraticQuadraticQuadraticQuadratic EquationsEquationsEquationsEquations1111.... WriteWriteWriteWrite thethethethe valuevaluevaluevalue ofofofof forforforfor whichwhichwhichwhich xxxx2222 ++++ 4x4x4x4x ++++ isisisis aaaa perfectperfectperfectperfect square.square.square.square.Solution:Solution:Solution:Solution: For (x 2 + 4 x + ) to b e a p erfect sq u are ,

Th ird te rm () = (21

coefficien t of x)2

= (21 4 )2

= (2 )2

= 4Hen ce , fo r = 4 , x 2 + 4 x + is a p erfect sq u are .

2222.... IfIfIfIf oneoneoneone rootrootrootroot ofofofof thethethethe equationequationequationequation 4x4x4x4x2222 2x2x2x2x ++++ (((( 4)4)4)4) ==== 0000 bebebebe thethethethe reciprocalreciprocalreciprocalreciprocal ofofofof thethethethe other,other,other,other, thenthenthenthen findfindfindfind ....Solution:Solution:Solution:Solution: Com p arin g th e eq u atio n 4 x 2 2 x + ( 4 ) = 0

With ax 2 + b x + c = 0 ,we g et a = 4 , b = 2 , c = ( - 4 )s in ce , th e two roo ts are recip roca ls of each o th er, th e ir p rod u ct wou ld b e 1 .

Prod u ct of roo ts =ac

4

4 = 1

4 = 4 = 8 .

3333....DivideDivideDivideDivide 16161616 intointointointo twotwotwotwo partspartspartsparts suchsuchsuchsuch thatthatthatthat twicetwicetwicetwice thethethethe squaresquaresquaresquare ofofofof thethethethe largerlargerlargerlarger partpartpartpart exceedsexceedsexceedsexceeds thethethethe squaresquaresquaresquare ofofofof thethethethesmallersmallersmallersmaller partpartpartpart bybybyby 164.164.164.164.

Solution:Solution:Solution:Solution: Let th e la rg er p art b e x . Th en , th e sm alle r p art = 1 6 x .By h yp oth es is , we h ave

2 x 2 = (1 6 x)2 + 1 6 4 2 x 2 (1 6 – x)2 1 6 4 = 0 x 2 + 3 2 x 4 2 0 = 0 (x + 4 2 ) (x 1 0 ) = 0 x = 4 2 or, x = 1 0 x = 1 0 [Q x> 0 x = 4 2 is n o t p oss ib le ]

Hen ce , th e req u ired p arts are 1 0 an d 6 .

4444....SolveSolveSolveSolve thethethethe followingfollowingfollowingfollowing quadraticquadraticquadraticquadratic bybybyby factorizationfactorizationfactorizationfactorization method:method:method:method:

3xx2

++++3x2

1

++++ 3x23x9x3

==== 0000

Solution:Solution:Solution:Solution: Clearly, th e g iven eq u atio n is va lid if x – 3 0 an d 2 x + 3 0 i.e . wh en x 23 ,3 .

Now,

3xx2

+3x2

1

+ 3x23x9x3

= 0

2 x(2 x + 3 )+ (x – 3 ) + 3 x + 9 = 0 4 x 2 + 6 x + x 3 + 3 x + 9 = 0 4 x 2 + 1 0 x + 6 = 0 2 x 2 + 5 x + 3 = 0 2 x 2 + 2 x + 3 x + 3 = 0 2 x (x + 1 ) + 3 (x + 1 ) = 0 (2 x + 3 ) (x + 1 ) = 0 x + 1 = 0 x = - 1 [Q 2 x + 3 0 ]

Hen ce , x = 1 is th e on ly so lu t ion of th e g iven eq u atio n .


5555....SolveSolveSolveSolve thethethethe followingfollowingfollowingfollowing quadraticquadraticquadraticquadratic equationsequationsequationsequations bybybyby factorizationfactorizationfactorizationfactorization method:method:method:method:

xba1

====a1++++

b1++++

x1,,,, aaaa ++++ bbbb 0000

Solution:Solution:Solution:Solution: We h ave ,

xba1

-x1 =

a1 +

b1

xbax

xbax =

abba

)xba(x

)ba(

=ab

ba

ab (a + b )= (a + b ) x (a + b + x) (a + b ) {x(a + b + x) + ab } = 0 x (a + b + x) + ab = 0 [Q a + b 0 ] x 2 + ax + b x + ab = 0 x (x + a) + b (x + a) = 0 (x + a) (x + b ) = 0 x + a = 0 or x + b = 0 x = - a or x = b .

6666....ProveProveProveProve thatthatthatthat thethethethe equationequationequationequation xxxx2222(a(a(a(a2222 ++++ bbbb2222)))) ++++ 2x2x2x2x (ac(ac(ac(ac ++++ bd)bd)bd)bd) ++++ (c(c(c(c2222 ++++ dddd2222)))) ==== 0000 hashashashas nononono realrealrealreal root,root,root,root, ifififif adadadad bc.bc.bc.bc.SolutionSolutionSolutionSolution :::: Th e d is crim in an t of th e g iven eq u ation is g iven b y

DDDD = 4 (ac + b d )2 4 (a 2 + b 2) (c2 + d 2) = 4 [(ac + b d )2 – (a 2 + b 2) (c2 + d 2)] = 4 [a 2c2 + b 2d 2 + 2 ac.b d a 2c2 a 2d 2 b 2c2 b 2 d 2] = 4 [2 ac.b d a 2c2 b 2c2] = 4 [a 2d 2 + b 2c2 2 ad .b c]

= 4 (ad b c)2

We h ave , ad b c ad b c 0 (ad b c)2 > 0 4 (ad – b c)2 < 0 D < 0

Hen ce , th e g iven eq u ation h as n o rea l roo ts .

7777....TheTheTheThe areaareaareaarea ofofofof anananan isoscelesisoscelesisoscelesisosceles triangletriangletriangletriangle isisisis 60606060 cmcmcmcm2222 andandandand thethethethe lengthlengthlengthlength ofofofof eacheacheacheach oneoneoneone ofofofof itsitsitsits equalequalequalequal sidessidessidessides isisisis 13131313 cm.cm.cm.cm.FindFindFindFind itsitsitsits base.base.base.base.

Solution:Solution:Solution:Solution: Let ABC b e th e g iven isos ce les trian g le in wh ich AB = AC = 1 3 cm . Draw AD p erp en d icu lar from Aon BC.Let BC = 2 x cm . Th en , BD = DC = x cm .In ABD, we h aveAB2 = AD2 + BD2 [By Pyth ag oras th eor.]

1 3 2 = AD2 + x 2

AD = 22 x1 3 = 2x1 6 9

Now, Area = 6 0 cm 2

21 (BC AC) = 6 0

21 {2 x 2x1 6 9 } = 6 0

x 2x1 6 9 = 6 0 x 2 (1 6 9 x 2) = 3 6 0 0 x 4 1 6 9 x 2 + 3 6 0 0 = 0 (x 2 – 1 4 4 ) (x 2 – 2 5 ) = 0 x 2 = 1 4 4 or, x 2 = 2 5 x = 1 2 or, x = 5

Hen ce , Base = 2 x = 2 4 cm , or 1 0 cm .


8888.... AAAA swimmingswimmingswimmingswimming poolpoolpoolpool isisisis filledfilledfilledfilled withwithwithwith threethreethreethree pipespipespipespipes withwithwithwith uniformuniformuniformuniform flow.flow.flow.flow. TheTheTheThe firstfirstfirstfirst twotwotwotwo pipespipespipespipes operatingoperatingoperatingoperatingsimultaneously,simultaneously,simultaneously,simultaneously, fillfillfillfill thethethethe poolpoolpoolpool inininin thethethethe samesamesamesame timetimetimetime duringduringduringduring whichwhichwhichwhich thethethethe poolpoolpoolpool isisisis filledfilledfilledfilled bybybyby thethethethe thirdthirdthirdthird pipepipepipepipealone.alone.alone.alone. TheTheTheThe secondsecondsecondsecond pipepipepipepipe fillsfillsfillsfills thethethethe poolpoolpoolpool fivefivefivefive hourshourshourshours fasterfasterfasterfaster thanthanthanthan thethethethe firstfirstfirstfirst pipespipespipespipes andandandand fourfourfourfour hourshourshourshours slowerslowerslowerslowerthanthanthanthan thethethethe thirdthirdthirdthird pipe.pipe.pipe.pipe. FindFindFindFind thethethethe timetimetimetime requiredrequiredrequiredrequired bybybyby eacheacheacheach pipepipepipepipe totototo fillfillfillfill thethethethe poolpoolpoolpool separately.separately.separately.separately.

Solution:Solution:Solution:Solution: Let V b e th e volu m e of th e p ool an d x th e n u m b er of h ou rs req u ired b y th e secon d p ip e a lon e tofill th e p ool . Th en , th e firs t p ip e takes (x + 5 ) h ou rs , wh ich th e th ird p ip e takes (x- 4 ) h ou rs to fill th ep ool. So , th e p arts of th e p ool filled b y th e firs t , secon d an d th ird p ip es in on e h ou r res p ective ly

5xV

,xV an d

4xV

Let th e t im e taken b y th e firs t an d secon d p ip es to fill th e p ool s im u ltan eou s ly b e t h ou rs .Th en , th e th ird p ip e a lso takes th e sam e tim e to fill th e p ool.

xV

5xV t = Volu m e of th e p ool.

Als o ,4x

V

t = Volu m e of th e p ool.

xV

5xV t =

4xV

t

5x

1

+51 =

4x1

(2 x + 5 ) (x – 4 ) = x 2 + 5 x x 2 – 8 x – 2 0 = 0 x 2 - 1 0 x + 2 x – 2 0 = 0 (x – 1 0 )(x + 2 ) = 0 x = 1 0 or, x = 2

Bu t, x can not b e n eg ative . So , x = 1 0Hen ce , th e t im in g s b y firs t , secon d an d th ird p ip es to fill b e p ool in d ivid u a lly are 1 5 are , 1 0 h ou rs an d6 h ou rs res p ective ly.

9.9.9.9. AAAA peacockpeacockpeacockpeacock isisisis sittingsittingsittingsitting onononon thethethethe toptoptoptop ofofofof aaaa pillar,pillar,pillar,pillar, whichwhichwhichwhich isisisis 9999 mmmm high.high.high.high. FindFindFindFind aaaa pointpointpointpoint 27272727 mmmm awayawayawayaway fromfromfromfrom thethethethebottombottombottombottom ofofofof thethethethe pillar,pillar,pillar,pillar, aaaa snakesnakesnakesnake isisisis comingcomingcomingcoming totototo itsitsitsits holeholeholehole atatatat thethethethe basebasebasebase ofofofof thethethethe pillar.pillar.pillar.pillar. SeeingSeeingSeeingSeeing thethethethe snakesnakesnakesnake thethethethepeacockpeacockpeacockpeacock pouncespouncespouncespounces onononon it.it.it.it. IfIfIfIf theirtheirtheirtheir speedsspeedsspeedsspeeds areareareare equal,equal,equal,equal, atatatat whatwhatwhatwhat distancedistancedistancedistance fromfromfromfrom thethethethe wholewholewholewhole isisisis thethethethe snakesnakesnakesnakecaught?caught?caught?caught?

Solution:Solution:Solution:Solution: Le t PQ b e th e p o le an d th e p eaco ck is s it t in g a t th e top P of th e p o le . Le t th e h o le b e a t Q. In it ia lly,th e sn ake is a t S wh en th e p eacock n otices th e sn ake su ch th a t p eacock ca tch es th e sn ake afte r tsecon d s a t p o in t T. Clearly, d is tan ce trave lled b y th e sn ake in t secon d s is sam e as th e d is tan ce flownb y p eacock.

PT = ST = x (say)Th u s , in rig h t trian g le PQT, we h ave

QT = 2 7 x , PT = x an d PQ = 9Usin g Pyth ag oras th eorem , we h ave

PT2= PQ2 + QT2

x 2 = 9 2 + (2 7 x)2

x 2 = 8 1 + 7 2 9 x 2

0 = 8 1 0 5 4 x 5 4 x = 8 1 0 x = 1 5

Q2 7 m

P

TS

9 m


QT = SQ ST = (2 7 1 5 )m = 1 2 mHen ce , th e sn ake is cau g h t a t a d is tan ce of 1 2 m from th e h ole .

10.10.10.10. One-fourthOne-fourthOne-fourthOne-fourth ofofofof aaaa herdherdherdherd ofofofof camelscamelscamelscamels waswaswaswas seenseenseenseen inininin thethethethe forest.forest.forest.forest. TwiceTwiceTwiceTwice thethethethe squaresquaresquaresquare rootrootrootroot ofofofof thethethethe herdherdherdherd hadhadhadhadgonegonegonegone totototo mountainsmountainsmountainsmountains andandandand thethethethe remainingremainingremainingremaining 15151515 camelscamelscamelscamels werewerewerewere seenseenseenseen onononon thethethethe bankbankbankbank ofofofof aaaa river.river.river.river. FindFindFindFind thethethethe totaltotaltotaltotalnumbernumbernumbernumber ofofofof camels.camels.camels.camels.

Solution:Solution:Solution:Solution: Let th e to ta l n u m b er of cam els b e x . Th en ,Nu m b er of cam els seen in th e fores t =

4x

Nu m b er of cam els of g on e to m ou n ta in s = 2 xNu m b er of cam els on th e b an k of river = 1 5

Tota l n u m b er of cam els =4x + 2 x + 1 5

By Hyp oth es is , we h ave

4x + 2 x + 1 5 = x

3 x 8 x 6 0 = 0 3 y2 8 y – 6 0 = 0 3 y2 8 y 6 0 = 0 , wh ere x = y2

3 y2 1 8 y + 1 0 y – 6 0 = 0 3 y(y 6 ) + 1 0 (y 6 ) = 0 (3 y + 1 0 )(y 6 ) = 9

y = 6 or, y = -3

1 0

Now, y = -3

1 0 x =

2

31 0

=

91 0 0 [Qx = y2]

Bu t, th e n u m b er of cam els can not b e a fract ion . y = 6 x = 6 2 = 3 6 [Q x = y2]

Hen ce , th e n u m b er of cam els = 3 6 .

11.11.11.11. TwoTwoTwoTwo pipespipespipespipes runningrunningrunningrunning togethertogethertogethertogether cancancancan fillfillfillfill aaaa cisterncisterncisterncistern inininin 333313131313

1111minutes.minutes.minutes.minutes. IfIfIfIf oneoneoneone pipepipepipepipe takestakestakestakes 3333 minutesminutesminutesminutes moremoremoremore

thanthanthanthan thethethethe otherotherotherother totototo fillfillfillfill it,it,it,it, findfindfindfind thethethethe timetimetimetime inininin whichwhichwhichwhich eacheacheacheach pipepipepipepipe wouldwouldwouldwould fillfillfillfill thethethethe cistern.cistern.cistern.cistern.Solution:Solution:Solution:Solution: Su p p os e th e fas te r p ip e takes x m in u tes to fill th e cis te rn . Th ere fore , th e s lower p ip e will take

(x+ 3 ) m in u tes to fill th e cis te rn .Sin ce th e fas te r p ip e takes x m in u tes to fill th e cis te rn .

Port ion of th e filled b y th e fas te r p ip e in on e m in u te =x1

Port ion of th e cis te rn filled b y th e fas te r p ip e in1 34 0 m in u tes =

x1

1 34 0 =

x1 34 0

Sim ila rly,

Port ion of th e cis te rn filled b y th e s lower p ip e in1 34 0 m in u tes

=3x

1

1 34 0 = 3x1 3

4 0

It is g iven th a t th e cis te rn is filled in1 34 0 m in u tes .

x1 3

4 0 + 3x1 34 0

= 1

x1 +

3x1

=4 01 3

3xxx3x

=

4 01 3


4 0 (2 x + 3 ) = 1 3 x (x + 3 ) 8 0 x + 1 2 0 = 1 3 x 2 + 3 9 x 1 3 x 2 4 1 x 1 2 0 = 0 1 3 x 2 6 5 x + 2 4 x 1 2 0 = 0 1 3 x 2(x 5 ) + 2 4 (x – 5 ) = 0 (x - 5 )(1 3 x + 2 4 ) = 0 x 5 = 0 or, 1 3 x + 2 4 = 0

x = 5 or, x =1 32 4

x = 5 [Qx> 0 ]

Hen ce , th e fas te r p ip e th e cis te rn in 5 m in u tes an d th e s lower p ip e takes 8 m in u tes to fill th e cis te rn .

12.12.12.12. OutOutOutOut ofofofof aaaa groupgroupgroupgroup ofofofof swans,swans,swans,swans,2222

7777timestimestimestimes thethethethe squaresquaresquaresquare rootrootrootroot ofofofof thethethethe numbernumbernumbernumber areareareare playingplayingplayingplaying onononon thethethethe shoreshoreshoreshore ofofofof aaaa

tank.tank.tank.tank. TheTheTheThe twotwotwotwo remainingremainingremainingremaining onesonesonesones areareareare playing,playing,playing,playing, withwithwithwith amorousamorousamorousamorous fight,fight,fight,fight, inininin thethethethe water.water.water.water. WhatWhatWhatWhat isisisis thethethethe totaltotaltotaltotalnumbernumbernumbernumber ofofofof swans?swans?swans?swans?

Solution:Solution:Solution:Solution: Let th e to ta l n u m b er swan s b e x . Th en ,

Nu m b er of swan s p layin g on th e sh ore of th e tan k =27 x

It is g iven th a t th ere are two rem ain in g swan s .

x =27 x + 2

x 27 x - 2 = 0

y2 -27 y – 2 = 0 , wh ere y2 = x

2 y2 7 y 4 = 0 2 y2 8 y + y 4 = 0 2 y(y - 4 ) + (y – 4 ) = 0 (y – 4 ) (2 y + 1 ) = 0

y = 4 or, y =21

y = 4 [Q y =21

is n o t p os s ib le ]

x = y2 = 4 2 = 1 6Hen ce , th e to ta l n um b er of swan s is 1 6 .

13.13.13.13. MadhulikaMadhulikaMadhulikaMadhulika andandandand ritaritaritarita ranranranran aaaa 2222 kmkmkmkm raceraceracerace twice.twice.twice.twice. InInInIn thethethethe firstfirstfirstfirst roundroundroundround ritaritaritarita finishesfinishesfinishesfinishes 2222 minminminmin earlierearlierearlierearlier thanthanthanthanmadhulika.madhulika.madhulika.madhulika. InInInIn thethethethe secondsecondsecondsecond round,round,round,round, madhulikamadhulikamadhulikamadhulika increasedincreasedincreasedincreased herherherher speedspeedspeedspeed bybybyby 2222 km/hrkm/hrkm/hrkm/hr andandandand ritaritaritarita reducedreducedreducedreduced herherherherspeedspeedspeedspeed bybybyby 2222 km/hr,km/hr,km/hr,km/hr, inininin thethethethe secondsecondsecondsecond round,round,round,round, madhulikamadhulikamadhulikamadhulika finishesfinishesfinishesfinishes 2222 minminminmin earlierearlierearlierearlier thanthanthanthan rita.rita.rita.rita. FindFindFindFind thethethethespeedsspeedsspeedsspeeds ofofofof theirtheirtheirtheir runningrunningrunningrunning thethethethe firstfirstfirstfirst round.round.round.round.

Solution:Solution:Solution:Solution: Let say Mad h u lika Sp eed in Firs t Rou n d = x km / Hr

Let say Rita Sp eed in firs t Rou n d = y km / Hr

Dis tan ce = 2 km

Tim e taken b y Mad h u lika in firs t rou n d = 2 / x Hr

Tim e taken b y Ritu in firs t rou n d = 2 / y Hr

2 / x - 2 / y = 2 / 6 0 ( rita fin ish es two m in s earlie r 2 m in = 2 / 6 0 Hr)

= > 1 / x - 1 / y = 1 / 6 0

= > 6 0 y - 6 0 x = xy - eq 1


in Secon d rou n d

Mad h u lika 's Sp eed = x+ 2 km / Hr

Rita 's sp eed = y- 2 Km / Hr

Tim e taken b y Mad h u lika = 2 / (x+ 2 ) h r

Tim e Taken b y Rita = 2 / (y- 2 ) h r

2 / (y- 2 ) - 2 / (x+ 2 ) = 2 / 6 0 ( m ad h u lika fin ish es two m in s earlie r in secon d rou n d )

= > 1 / (y- 2 ) - 1 / (x+ 2 ) = 1 / 6 0

= > 6 0 x + 1 2 0 - 6 0 y + 1 2 0 = xy - 2 x + 2 y - 4

= > 6 2 x - 6 2 y = xy - 2 4 4

= > 6 2 y - 6 2 x = 2 4 4 - xy - eq 2

Mu lt ip lyin g eq 1 b y 3 1 & eq 2 b y 3 0 & eq u atin g b oth

3 1 xy = 3 0 (2 4 4 - xy)

= > 3 1 xy = 3 0 *2 4 4 - 3 0 xy

= > 6 1 xy = 3 0 *2 4 4

= > xy = 1 2 0

Pu tt in g in eq 1

6 0 y - 6 0 x = 1 2 0

= > y - x = 2

= > y - (1 2 0 / y) = 2

= > y² - 1 2 0 = 2 y

= > y² - 2 y - 1 2 0 = 0

= > y² - 1 2 y + 1 0 y - 1 2 0 = 0

= > y(y- 1 2 ) + 1 0 (y- 1 2 ) = 0

= > (y+ 1 0 )(y- 1 2 ) = 0

y = 1 2

Rita 's sp eed in firs t rou n d = 1 2 km / Hr

y - x = 2

= > x = 1 0

Mad h u lika 's sp eed in firs t rou n d = 1 0 km / Hr


11114444.... IfIfIfIf thethethethe rootsrootsrootsroots ofofofof thethethethe quadraticquadraticquadraticquadratic equationequationequationequation aaaaxxxx2222 ++++ ccccxxxx ++++ cccc ==== 0000 areareareare inininin thethethethe ratioratioratioratio pppp :::: qqqq

showshowshowshow thatthatthatthatqp++++

pq++++

ac==== 0000

Solution:Solution:Solution:Solution:Th e g iven eq u atio n is ax 2 + cx + c = 0 ,

Com p arin g with ax 2 + b x + c = 0a = a , b = c, c = c

Let , b e th e roo ts of g iven eq u atio n

+ =ac ……(i)

& =ac ……(ii)

Als o , =

qp ……Given ……(iii)

LHS =qp +

pq +

ac

= +

+

ac …(from iii)

=

+

+

ac

=

+ac

=

ac

a/c +ac …(from (i) & (ii))

=a/c

a/c +ac

=c

aac

+

ac

=ac

caac

=cacc

= 0= RHS

qp++++

pq++++

ac==== 0000

11115.5.5.5. IfIfIfIf thethethethe sumsumsumsum ofofofof thethethethe rootsrootsrootsroots ofofofof thethethethe quadraticquadraticquadraticquadratic equationequationequationequation axaxaxax2222 ++++ bxbxbxbx ++++ cccc ==== 0000 isisisis equalequalequalequal totototo thethethethe sumsumsumsum ofofofof thethethethesquaressquaressquaressquares ofofofof theirtheirtheirtheir reciprocalsreciprocalsreciprocalsreciprocals thenthenthenthen proveproveproveprove thatthatthatthat 2a2a2a2a2222cccc ==== cccc2222bbbb ++++ bbbb2222aaaa

Solution:Solution:Solution:Solution: For ax 2 + b x + c = 0 ,

+ =2

1

+

21

..... Given

+ =2 2

2 2 +

( + )=2

2( + ) – 2

( )

... (i)


[Q 2 + 2 = ( + )2 – 2]We kn ow,

+ = –ba

& = ca

–ba

=

2

2

–b c – 2a a

ca

–ba

=

2

2

2

2 2

acac

ab

–ba

=

2

2

2

2

b – 2 a caca

–ba

=2

2b – 2 a c

c –b c2 = a(b 2 – 2 ac) –b c2 = ab 2 – 2 a 2C 2a2a2a2a2222cccc ==== cccc2222bbbb ++++ bbbb2222aaaa

11116.6.6.6. FormFormFormForm thethethethe quadraticquadraticquadraticquadratic equationsequationsequationsequations whosewhosewhosewhose rootsrootsrootsroots areareareare thethethethe squaressquaressquaressquares ofofofof thethethethe sumsumsumsum ofofofof rootsrootsrootsroots andandandand squaresquaresquaresquare ofofofof thethethethedifferencedifferencedifferencedifference ofofofof rootsrootsrootsroots ofofofof thethethethe equationequationequationequation 2x2x2x2x2222 ++++ 2(m2(m2(m2(m ++++ n)xn)xn)xn)x ++++ mmmm2222 ++++ nnnn2222 ==== 0.0.0.0.

Solution:Solution:Solution:Solution:Th e g iven q u ad ra t ic eq u ation is 2 x 2 + 2 (m + n )x + (m 2 + n 2) = 0Let & b e th e roo ts of th is eq n.

We kn ow, + = –ba

& = ca

Com p arin g th e g iven eq n with ax 2 + b x + c = 0We g et ,

a = 2 , b = 2 (m + n ), c = (m 2 + n 2)

+ = –2(m + n)2

+ = –(m + n ) ….. (i)

An d =2 2m + n

2….. (ii)

Le t th e roo ts of th e req u ired q u ad ra t ic eq u ation b e r & r.As p er g iven con d it ion ,r = ( + )2 & ….. (III)r = ( – )2 ….. (IV)

Now,( – )2 = ( + )2 – 4

= [–(m + n )]2 – 4 ×2 2(m + n )

2= m 2 + 2 m n + n 2 – 2 (m 2 – n 2)= m 2 + 2 m n + n 2 – 2 m 2 – 2 n 2

( – )2 = –m 2 + 2 m n – n 2

We kn own ,Th e req u ired eq u atio n is of th e formx 2 – (r + r)x + rr = 0


x 2 – [( + )2 + ( – )2] x + ( + )2 ( – )2 = 0x 2 – [–(m + n )2 + (–m 2 + 2 m n – n 2)] x + [–(m + n )2 × (–m 2 + 2 m n – n 2)] = 0

x 2 – (m 2 + 2 m n + n 2 – m 2 + 2 m n – n 2) x [–(m + n )2 × – (m – n )2] = 0 x 2 – 4 m n x – (m 2 – n 2)2 = 0

11117777.... FindFindFindFind thethethethe valuevaluevaluevalue ofofofof P,P,P,P, ifififif thethethethe equationsequationsequationsequations3x3x3x3x2222 2x2x2x2x ++++ pppp ==== 0000 andandandand 6x6x6x6x2222 –––– 17x17x17x17x ++++ 12121212 ==== 0000 havehavehavehave aaaa commoncommoncommoncommon root.root.root.root.

Solution:Solution:Solution:Solution:We kn own 6 x 2 1 7 x + 1 2 = 0

6 x 2 9 x 8 x + 1 2 = 0 3 x(2 x 3 ) 4 (2 x 3 ) = 0 (3 x 4 ) (2 x 3 ) = 0 3 x 4 = 0 or 2 x 3 = 0 3 x = 4 or 2 x = 3

x =34 or x =

23

Pu t x =34 in eq n . 3 x 2 2 x + p = 0

32

34

2 34 + p = 0

3 33

1 6

38 + p = 0

38

31 6

+ p = 0

38 + p = 0

p =38

p =38

or p u t x =23 in eq n . 3 x 2 2 x + p = 0

32

23

2 23 + p = 0

3 49

26 + p = 0

4

2 7 3 + p = 0

4

1 22 7 + p = 0

4

1 5 + p = 0

p =41 5

1 8 . SameSameSameSame asasasas 15151515


19.19.19.19. IfIfIfIf thethethethe rootsrootsrootsroots ofofofof aaaa quadraticquadraticquadraticquadratic equationequationequationequation (m(m(m(m –––– nnnn )))) xxxx2222 ++++ (n(n(n(n ---- l)l)l)l) xxxx ++++ (l(l(l(l –––– mmmm )))) ==== 0000 areareareare eeeequal,qual,qual,qual, showshowshowshow thatthatthatthat l,l,l,l, mmmm ,,,, nnnnareareareare inininin A.P.A.P.A.P.A.P.

Solution:Solution:Solution:Solution:(m(m(m(m –––– nnnn )))) xxxx2222 ++++ (n(n(n(n ---- l)l)l)l) xxxx ++++ (l(l(l(l –––– mmmm )))) ==== 0000Com p arin g with ax 2 + b x + c = 0 ,a = (m - n ), b = (n - l), c = (l - m )Roots of th is eq u ation are eq u al.

b 2 - 4 ac = 0 ((((n - l)2 - 4 ((((m - n )((((l - m ) = 0 n 2 - 2 n l + l2 - 4 (m l - m 2 - n l + m n ) = 0 n 2 - 2 n l + l2 - 4 m l + 4 m 2 + 4 n l - 4 m n = 0 n 2 + 2 n l + l2 - 4 m l + 4 m 2 - 4 m n = 0 ((((n + l)2 - 4 m ((((l + m ) + 4 m 2 = 0 ((((n + l)2 - 2 x 2 m ((((n + l) + ((((2222m )2 = 0 (n + l - 2 m )2 = 0 (n + l - 2 m ) = 0 n + l = 2 m n + l = m + m n - m = m - l l, m an d n are in AP

2 0 . A p art o f m on th ly exp en s es of a fam ily is con s tan t an d th e rem ain in g varies with th ep rice of wh eat . Wh en th e ra te of wh eat is 2 5 0 a q u in ta l, th e to ta l m on th ly exp en s es ofth e fam ily are 1 0 0 0 an d wh en it is 2 4 0 a q u in ta l, th e to ta l m on th ly exp en ses are

9 8 0 . Fin d th e to ta l m on th ly exp en ses of th e fam ily wh en th e cos t o f wh eat is 3 5 0 aQu in ta l.

Solution:Solution:Solution:Solution:Let th e con s tan t exp en d itu re b e ₹.x

Let th e con su m p tion of rice b y th e fam ily b e y q u in ta ls .As p er firs t con d it ion ,x+ 2 5 0 y= 1 0 0 0 - - - - - - - - - - - - - - - - - - (1 )

As p er secon d con d it ion ,x+ 2 4 0 y= 9 8 0 - - - - - - - - - - - - - - - - - - - (2 )

Su b tract in g eq (1 )an d (2 )

x + 2 5 0 y = 1 0 0 0

x + 2 4 0 y = 9 8 0

(- ) (- ) (- )

- - - - - - - - - - - - - - - - - - - - - - - - - - -

1 0 y = 2 0

y= 2 ………….(3 )

Pu t th e va lu e of y in eq 1x+ 2 5 0 y= 1 0 0 0x + 2 5 0 (2 ) = 1 0 0 0x+ 5 0 0 = 1 0 0 0

x= 5 0 0 ………….(4 )

Tota l m on th ly exp en d itu re wh en th e cos t o f rice ₹3 0 0 = con s tan t exp en d itu re + con su m p t ion of rice a t₹ 3 0 0 p er q u in ta l.

AlgebraAlgebraAlgebraAlgebra (Std.(Std.(Std.(Std. X)X)X)X)1 3

13131313

= x+ 3 0 0 y= 5 0 0 + 3 0 0 × 2 ( From eq 3 & 4 )= 5 0 0 + 6 0 0= ₹ 1 1 0 0Hen ce , th e Tota l m on th ly exp en d itu re wh en th e cos t o f rice ₹3 0 0 is ₹ 1 1 0 0

��

LinearLinearLinearLinear EquationsEquationsEquationsEquations InInInIn TwoTwoTwoTwo VariablesVariablesVariablesVariables1)1)1)1) ForForForFor whatwhatwhatwhat valuevaluevaluevalue ofofofof kkkk willwillwillwill bebebebe thethethethe equationsequationsequationsequations xxxx ++++ 2y2y2y2y ++++ 7777 ==== 0,0,0,0, 2x2x2x2x ++++ kykykyky ++++ 14141414 ==== 0000 representrepresentrepresentrepresent coincidentcoincidentcoincidentcoincident

lines?lines?lines?lines?Solution:Solution:Solution:Solution:

Th e g iven eq u ation s are of th e forma 1x + b 1y + c1 = 0 an d a 2x + b 2y + c2 = 0 ,

Wh erea 1 = 1 , b 1= 2 , c1 = 7 an d a 2 = 2 , b 2 = k, c2 = 1 4

Th e g iven eq u atio n s will rep re sen t co in cid en t lin es if th ey h ave in fin ite ly m an y so lu t ion s . Th e con d it ionfor wh ich is

2

1

aa =

2

1

bb =

2

1

cc

21 =

k2 =

1 47

k = 4Hen ce , th e g iven sys tem of eq u ation s will rep resen t co in cid en t lin es , If k = 4 .

2.2.2.2.Solve:Solve:Solve:Solve: axaxaxax ++++ bybybyby ==== ccccbxbxbxbx ++++ ayayayay ==== 1111 ++++ cccc

Solution:Solution:Solution:Solution: Th e g iven sys tem of eq u atio n s m ay b e writ ten asax + b y c = 0b x + ay (1 + c) = 0By cross - m u lt ip lica t ion , we h ave

cac1bx

= cbc1a

y

=bbaa

1

bb cac

x

=ab cac

y

= 22 ba1

bbacx

= abac

y

= baba1

x = baba

bbac an d y =

babaabac

x =ba

c

babab

an d y =

bac

+ babaa

Hen ce , th e so lu t ion of th e g iven sys tem of eq u ation s is

x =ba

c

22 bab

, y =ba

c

+ 22 baa

.

3.3.3.3. DetermineDetermineDetermineDetermine thethethethe valuesvaluesvaluesvalues ofofofof mmmm andandandand nnnn sosososo thatthatthatthat thethethethe followingfollowingfollowingfollowing systemsystemsystemsystem ofofofof linearlinearlinearlinear equationsequationsequationsequations havehavehavehave infiniteinfiniteinfiniteinfinitenumbernumbernumbernumber ofofofof solutions:solutions:solutions:solutions:(2m(2m(2m(2m 1)1)1)1) xxxx ++++ 3y3y3y3y 5555 ==== 00003x3x3x3x ++++ (n(n(n(n 1)1)1)1) yyyy 2222 ==== 0000

Solution:Solution:Solution:Solution: Th e g iven sys tem of eq u atio n s will h ave in fin ite n u m b er of so lu t ion s , if


31m2 ====

1n3

====25

3

1m2 =1n

3

=25

3

1m2 =25 an d

1n3

=25

4 m 2 = 1 5 an d 6 = 5 n 5 4 m = 1 7 an d 5 n = 1 1

m =4

1 7 an d n =5

1 1

Hen ce , th e g iven sys tem of eq u ation s will h ave in fin ite n u m b er of so lu t ion s , if m =4

1 7 an d n =5

1 1 .

4.4.4.4.DetermineDetermineDetermineDetermine thethethethe valuevaluevaluevalue ofofofof kkkk sosososo thatthatthatthat thethethethe followingfollowingfollowingfollowing linearlinearlinearlinear equationsequationsequationsequations havehavehavehave nononono solution:solution:solution:solution:(3k(3k(3k(3k ++++ 1)1)1)1) xxxx ++++ 3y3y3y3y 2222 ==== 0000(k(k(k(k2222 ++++ 1)1)1)1) xxxx ++++ (k(k(k(k 2)2)2)2) yyyy 5555 ==== 0000

Solution:Solution:Solution:Solution: Th e g iven sys tem of eq u atio n s will h ave n o so lu t ion , if

1k1k3

2

====2k

3

52

1k1k3

2 =

2k3

an d2k

3

52

Now,1k1k3

2 =

2k3

(3 k + 1 ) (k 2 ) = 3 (k 2 + 1 ) 3 k 2 5 k 2 = 3 k 2 + 3 5 k 2 = 3 5 k = 5 k = - 1

Clearly,2k

3

52 for k = 1 .

Hen ce , th e g iven sys tem of eq u ation s will h ave n o so lu t ion for k = 1 .

5.5.5.5. IfIfIfIf threethreethreethree timestimestimestimes thethethethe largerlargerlargerlarger ofofofof thethethethe twotwotwotwo numbersnumbersnumbersnumbers isisisis divideddivideddivideddivided bybybyby thethethethe smallersmallersmallersmaller one,one,one,one, wewewewe getgetgetget 4444 asasasas quotientquotientquotientquotientandandandand 3333 asasasas thethethethe remainder.remainder.remainder.remainder. Also,Also,Also,Also, ifififif sevensevensevenseven timestimestimestimes thethethethe smallersmallersmallersmaller numbernumbernumbernumber isisisis divideddivideddivideddivided bybybyby thethethethe largerlargerlargerlarger one,one,one,one, wewewewegetgetgetget 5555 asasasas quotientquotientquotientquotient andandandand 1111 asasasas remainder.remainder.remainder.remainder. FindFindFindFind thethethethe numbers.numbers.numbers.numbers.

Solution:Solution:Solution:Solution: Let th e la rg er n u m b er b e x an d sm alle r on e b e y. we kn ow th atDivid en d = (Divis or Qu otien t) + Rem ain d erWh en 3 x is d ivid ed b y y, we g et 4 as q u otien t an d 3 as rem ain d e r. Th ere fore , b y u s in g (i) as , we g et3 x = 4 y + 3 3 x 4 y 3 = 0 ….(ii)Wh en 7 y is d ivid ed b y x , we g et 5 as q u otien t an d 1 as rem ain d e r. Th ere fore , b y u s in g (i), w g et7 y = 5 x + 1 5 x 7 y + 1 = 0 ….(iii)So lvin g eq u atio n s (ii) an d (iii), b y cross - m u lt ip lica t ion , we g et

2 14x

=1 53y

=

2 02 11

x = 2 5 an d y = 1 8

Hen ce , th e req u ired n u m b ers are 2 5 an d 1 8 .

6666.... AAAA andandandand BBBB areareareare friendsfriendsfriendsfriends andandandand theirtheirtheirtheir agesagesagesages differdifferdifferdiffer bybybyby 2222 years.years.years.years. AAAA’’’’ssss fatherfatherfatherfather DDDD isisisis twicetwicetwicetwice asasasas oldoldoldold asasasas AAAA andandandand BBBB isisisistwicetwicetwicetwice asasasas oldoldoldold asasasas hishishishis sistersistersistersister C.C.C.C. TheTheTheThe ageageageage ofofofof DDDD andandandand CCCC differdifferdifferdiffer bybybyby 40404040 years.years.years.years. FindFindFindFind thethethethe agesagesagesages ofofofof AAAA andandandand B.B.B.B.

Solution:Solution:Solution:Solution: Let th e ag es of A an d B b e x an d y years resp ective ly. Th en ,x y = 2 [Given ]

D’s ag e = 2 x years an d , C’s ag e =2y years .

Clearly, D is o ld er th an C

2 x 2y = 4 0 4 x y = 8 0


15151515

Th u s , we h ave th e fo llowin g two sys tem s of lin ear eq u ation sx y = 2 ….(i)

an d , 4 x y = 8 0 ….(ii)x y = 2 …..(iii)4 x y = 8 0 ….(iv)

Su b tract in g eq u ation (i) from eq u atio n (ii), we g et3 x = 7 8 x = 2 6

Pu tt in g x = 2 6 in eq u ation (i), we g et

3 x = 8 2 x =3

8 2 = 2 731

Pu tt in g x =3

8 2 in eq u ation (iii), we g et

Y =3

8 2 + 2 =3

8 8 = 2 931

Hen ce , A’s ag e = 2 6 years an d B’s ag e = 2 4 yearsOr,

A’s ag e = 2 731 years an d B’s ag e = 2 9

31 years .

7777.... StudentsStudentsStudentsStudents ofofofof aaaa classclassclassclass areareareare mademademademade totototo standstandstandstand inininin rows.rows.rows.rows. IfIfIfIf oneoneoneone studentstudentstudentstudent isisisis extraextraextraextra inininin aaaa row,row,row,row, theretheretherethere wouldwouldwouldwould bebebebe 2222rowsrowsrowsrows less.less.less.less. IFIFIFIF oneoneoneone studentsstudentsstudentsstudents isisisis lesslesslessless wouldwouldwouldwould bebebebe 3333 rowsrowsrowsrows more.more.more.more. FindFindFindFind thethethethe numbernumbernumbernumber ofofofof studentsstudentsstudentsstudents inininin thethethethe class.class.class.class.

SolutionSolutionSolutionSolution :::: Let th e n u m b er of s tu d en ts b e x an d th e n u m b er of rows b e y. Th en ,Nu m b er of s tu d en ts in each row = x/ y

Wh en on e s tu d en t is ex tra in each row, th ere are 2 rows less i.e ., wh en each row h as

1

yx s tu d en ts ,

th e n u m b er of rows is (y 2 ). Tota l n u m b er of s tu d en ts = No. of rows No. of s tu d en ts in each row

x =

1

yx (y 2 )

x = x yx2 + y 2

yx2 + y 2 = 0 …..(i)

If on e s tu d en ts is les s in each row, th en th ere are 3 rows m ore i.e ., wh en each row h as

1

yx s tu d en ts ,

th e n u m b er of rows is (y + 3 ) Tota l n u m b er of s tu d en ts = No. of rows No. of s tu d en ts in each row

x =

1

yx (y + 3 )

x = x +yx3 y 3

yx3

y 3 = 0

….(ii)

Pu tt in gyx = u in (i) an d (ii), we g et

2 u + y 2 = 0 …..(iii) 3 u y 3 = 0 …..(iv)

Ad d in g (iii) an d (iv), we g et u 5 = 0 u = 5

Pu tt in g u = 5 in (iii), we g et y = 1 2

Now, u = 5 yx = 5

1 2x = 5 x = 6 0

Hen ce , th e n u m b er of s tu d en ts in th e class is 6 0 .


8888.... 8888 menmenmenmen andandandand 12121212 boysboysboysboys cancancancan finishfinishfinishfinish aaaa piecepiecepiecepiece ofofofof workworkworkwork inininin 10101010 daysdaysdaysdays whilewhilewhilewhile 6666 menmenmenmen andandandand 8888 boysboysboysboys cancancancan finishfinishfinishfinish itititit inininin14141414 days.days.days.days. FindFindFindFind thethethethe timetimetimetime takentakentakentaken bybybyby oneoneoneone manmanmanman alonealonealonealone andandandand thatthatthatthat bybybyby oneoneoneone boyboyboyboy alonealonealonealone totototo finishfinishfinishfinish thethethethe work.work.work.work.

Solution:Solution:Solution:Solution: Su p p os e th a t on e m an alon e can fin ish th e work in x d ays an d on e b oy alon e can fin ish it in yd ays . Th en ,

On e m an ’s on e d ay’s work =x1

On e b oy’s on e d ay’s work =y1

Eig h t m en ’s on e d ay’s work =x8

1 2 b oy’s on e d ay’s work =y

1 2

Sin ce 8 m en an d 1 2 b oys can fin ish th e work in 1 0 d ays

1 0

y1 2

x8 = 1

x8 0 +

y1 2 0 = 1

Ag ain , 6 m en an d 8 b oy’s can fin ish th e work in 1 4 d ays .

1 4

y8

x6 = 1

x8 4 +

y1 1 2 = 1

Pu tt in gx1 = u an d

y1 = v in eq u ation s (i) an d (ii), we g et

8 0 u + 1 2 0 v 1 = 08 4 u + 1 1 2 v 1 = 0

By u s in g cross - m u lt ip lica t ion , we h ave

1 1 21 2 0u

=8 48 0

v

=8 41 2 01 1 28 0

1

8

u

=4

v

=1 1 2 0

1

u =1 1 2 0

8 =

1 4 01 an d v =

1 1 2 04

=

2 8 01

Now, u =1 4 0

1

x1 =

1 4 01

x = 1 4 0

an d , v =2 8 0

1

y1 =

2 8 01

y = 2 8 0 .

Hen ce , on e m en alon e can fin ish th e work in 1 4 0 d ays an d on e b oy alon e can fin ish th e work in 2 8 0d ays .

9999.... OnOnOnOn sellingsellingsellingselling aaaa tea-settea-settea-settea-set atatatat 5%5%5%5% losslosslossloss andandandand aaaa lemon-setlemon-setlemon-setlemon-set atatatat 15%15%15%15% gain,gain,gain,gain, aaaa crockerycrockerycrockerycrockery sellersellersellerseller gainsgainsgainsgains RsRsRsRs 7.7.7.7. IfIfIfIf thethethethesellssellssellssells thethethethe tea-settea-settea-settea-set atatatat 5%5%5%5% gaingaingaingain andandandand thethethethe lemon-setlemon-setlemon-setlemon-set atatatat 10%10%10%10% gain,gain,gain,gain, hehehehe gainsgainsgainsgains RsRsRsRs 13.13.13.13. FindFindFindFind thethethethe actualactualactualactual pricepricepriceprice ofofofofthethethethe tea-settea-settea-settea-set andandandand thethethethe lemon-set.lemon-set.lemon-set.lemon-set.

Solution:Solution:Solution:Solution: Let th e cos t p rice of th e tea- s e t an d th e lem o n - se t b e Rs x an d Rs y resp ective ly.Case Wh en tea se t is so ld a t 5 % loss an d lem on - se t a t 1 5 % g ain .

Loss on tea- se t = Rs1 0 0

x5 = Rs2 0x

Gain on lem on - se t= Rs1 0 0

y1 5 = Rs2 0

y3

Net g a in = Rs2 0

y3

2 0x

2 0

y3

2 0x = 7 3 y x = 1 4 0 x 3 y + 1 4 0 = 0 …..(i)

Case Wh en tea- s e t is so ld a t 5 % g ain an d th e lem on - se t a t 1 0 % g ain .

Gain on tea- s e t = Rs1 0 0

x5 = Rs2 0x


17171717

Gain on lem on - se t = Rs1 0 0

y1 0 = Rs1 0y

Tota l g a in = Rs2 0x +

1 0y

2 0x +

1 0y = 1 3 x + 2 y = 2 6 0 x + 2 y 2 6 0 = 0 ……(ii)

Su b tract in g eq u ation (ii) from eq u ation (i), we g et 5 y + 4 0 0 = 0 y = 8 0

Pu tt in g y = 8 0 in eq u atio n (i), we g etx 2 4 0 + 1 4 0 = 0 x = 1 0 0

Hen ce , cos t p rices of tea- se t an d lem on - se t a re Rs 1 0 0 an d Rs8 0 resp ective ly.

10101010.... TheTheTheThe denominatordenominatordenominatordenominator ofofofof aaaa fractionfractionfractionfraction isisisis 4444 moremoremoremore thanthanthanthan twicetwicetwicetwice thethethethe numerator.numerator.numerator.numerator. WhenWhenWhenWhen bothbothbothboth thethethethe numeratornumeratornumeratornumerator andandandanddenominatordenominatordenominatordenominator areareareare decreaseddecreaseddecreaseddecreased bybybyby 6,6,6,6, thenthenthenthen thethethethe denominatordenominatordenominatordenominator becomesbecomesbecomesbecomes 12121212 timestimestimestimes thethethethe numerator.numerator.numerator.numerator.DetermineDetermineDetermineDetermine thethethethe fraction.fraction.fraction.fraction.

Solution:Solution:Solution:Solution: Let th e n u m era to r an d d en om in ator of th e fract ion b e x an d y resp ective ly. Th en ,

Fract ion =yx

It is g iven th a tDen om in ator = 2 (Nu m era to r) + 4

y = 2 x + 4 2 x y + 4 = 0

Accord in g to th e g iven con d it ion , we h avey 6 = 1 2 (x 6 )

y 6 = 1 2 x 7 2 1 2 x y 6 6 = 0

Th u s , we h ave th e fo llowin g sys tem of eq u atio n s2 x y + 4 = 0 …..(i)1 2 x y 6 6 = 0 …..(ii)

Su b tract in g eq u ation (i)from eq u atio n (ii), we g et1 0 x 7 0 = 0

x = 7Pu tt in g x = 7 in eq u atio n (i), we g et

1 4 y + 4 = 0 y = 1 8

Hen ce , req u ired fract ion =1 87 .

11111111.... RoohiRoohiRoohiRoohi travelstravelstravelstravels 300300300300 kmkmkmkm totototo herherherher homehomehomehome partlypartlypartlypartly bybybyby traintraintraintrain andandandand partlypartlypartlypartly bybybyby bus.bus.bus.bus. SheSheSheShe takestakestakestakes 4444 hours,hours,hours,hours, ifififif sheshesheshetravelstravelstravelstravels 60606060 kmkmkmkm bybybyby traintraintraintrain andandandand thethethethe remainingremainingremainingremaining bybybyby bus.bus.bus.bus. IfIfIfIf sheshesheshe travelstravelstravelstravels 100100100100 kmkmkmkm bybybyby traintraintraintrain andandandand thethethethe remainingremainingremainingremainingbybybyby bus,bus,bus,bus, sheshesheshe takentakentakentaken 10101010 minutesminutesminutesminutes longer.longer.longer.longer. FindFindFindFind thethethethe speedspeedspeedspeed ofofofof thethethethe traintraintraintrain andandandand thethethethe busbusbusbus separately.separately.separately.separately.

Solution:Solution:Solution:Solution: Let th e sp eed of tra in b e x km / h an d sp eed of b u s b e y km / h

Th en ,x

6 0 +y

2 4 0 = 4 …..(i)

an d ,x

1 0 0 +y

2 0 0 = 4 +6 01 0 =

62 5 ….(ii)

Le tx1 = u an d

y1 = v. Th en eq u atio n s (i) an d (ii) b ecom es

6 0 u + 2 4 0 v = 4 …..(iii)

1 0 0 u + 2 0 0 v =6

2 5 …..(iv)

Mu lt ip lyin g (iii) b y 5 an d (iv) b y 6 , we g et3 0 0 u + 1 2 0 0 v = 2 06 0 0 u + 1 2 0 0 v = 2 5

3 0 0 u = 5


u =3 0 0

5 =

6 01

From (iii),

6 0 6 01 + 2 4 0 v = 4

v =2 4 0

3 =8 01

So, x =u1 = 6 0 an d y =

v1 = 8 0

Hen ce , sp eed of tra in is 60606060 km/hkm/hkm/hkm/h an d sp eed of b u s is 80808080 km/hkm/hkm/hkm/h.

11112.2.2.2. AAAA partpartpartpart ofofofof monthlymonthlymonthlymonthly hostelhostelhostelhostel chargeschargeschargescharges isisisis fixedfixedfixedfixed andandandand remainingremainingremainingremaining dependsdependsdependsdepends onononon thethethethe numbernumbernumbernumber ofofofof daysdaysdaysdays oneoneoneone hashashashastakentakentakentaken foodfoodfoodfood inininin thethethethe mess.mess.mess.mess. WhenWhenWhenWhen aaaa student,student,student,student, AAAA takestakestakestakes foodfoodfoodfood forforforfor 20202020 days,days,days,days, AAAA hashashashas 1000as1000as1000as1000as hostelhostelhostelhostel charges,charges,charges,charges,whereaswhereaswhereaswhereas aaaa studentsstudentsstudentsstudents B,B,B,B, whowhowhowho takestakestakestakes foodfoodfoodfood forforforfor 26262626 daysdaysdaysdays payspayspayspays Rs1180Rs1180Rs1180Rs1180 aaaa hostelhostelhostelhostel charges.charges.charges.charges. FindFindFindFind thethethethe fixedfixedfixedfixedchargechargechargecharge andandandand thethethethe costcostcostcost ofofofof foodfoodfoodfood perperperper day.day.day.day.

Solution:Solution:Solution:Solution: Let th e fixed h os te l ch arg e b e Rs . xAn d th e cos t o f food p er d ay b e Rs . yAccord in g to th e q u es t io n ,A’s h os te l ch arg es areFixed Hos te l ch arg es + Cos t of food for 2 0 d ays = Rs . 1 0 0 0

x + 2 0 y = 1 0 0 0 ……(i)B’s h os te l ch arg es areFixed Hos te l ch arg es + Cos t of food for 2 6 d ays = Rs . 1 1 8 0

x + 2 6 y = 1 1 8 0 …..(ii)Su b tract in g eq u ation (i) from eq u atio n (ii), we g et

6 y = 1 8 0 y = 3 0Pu tt in g y = 3 0 in eq u atio n (i), we h ave

x + 2 0 (3 0 ) = 1 0 0 0 x + 6 0 0 = 1 0 0 0 x = 1 0 0 0 6 0 0 = 4 0 0

Hen ce , fixed ch arg e = Rs.Rs.Rs.Rs. 400400400400 an d cos t o f food p er d ay = Rs.Rs.Rs.Rs. 30303030.

11113.3.3.3. RRRRssss.... 9,0009,0009,0009,000 werewerewerewere divideddivideddivideddivided equallyequallyequallyequally amongamongamongamong aaaa certaincertaincertaincertain numbernumbernumbernumber ofofofof persons.persons.persons.persons. HadHadHadHad theretheretherethere beenbeenbeenbeen 20202020 moremoremoremorepersons,persons,persons,persons, eacheacheacheach wouldwouldwouldwould havehavehavehave gotgotgotgot Rs.Rs.Rs.Rs. 160160160160 less.less.less.less. FindFindFindFind thethethethe originaloriginaloriginaloriginal numbernumbernumbernumber ofofofof persons.persons.persons.persons.

Solution:Solution:Solution:Solution: Let orig in a l n u m b er of p ers on s= x

Th e in creased n u m b er of p ers on s = yAccord in g to th e g iven con d it ion ,

y = x + 2 0 …..(i)To ta l am o u n t = Rs . 9 0 0 0Accord in g to th e q u es t io n .Orig in a l sh are of each p ers on sh are of each of th e in creased p ers on s = Rs . 1 6 0

x

9 0 0 0

y9 0 0 0 = 1 6 0

p ersonsof.NoAm ountTotalgotEach

Divid in g b oth s id es b y 4 0 , we g et

x2 2 5

y

2 2 5 = 4

Pu tt in g y = x + 2 0 from eq u atio n (i) in eq u atio n (ii), we g et

x2 2 5

y

2 2 5 = 4

2 0xxx2 2 54 5 0 0x2 2 5

= 4 2 0xx

4 5 0 0

= 4

4 x 2 + 8 0 x = 4 5 0 0 x 2 + 2 0 x = 1 1 2 5 x 2 + 2 0 x 1 1 2 5 = 0 x 2+ 4 5 x 2 5 x – 1 1 2 5 = 0 x (x + 4 5 ) 2 5 (x + 4 5 )= 0 (x 2 5 ) (x + 4 5 ) = 0 x = 2 5 or x = 4 5


19191919

Bu t, n u m b er of p ers on s can’t b e n eg ative .Hen ce , th e orig in a l n u m b er of p ers on are is 2 5 .

11114.4.4.4. TwoTwoTwoTwo pipespipespipespipes runningrunningrunningrunning togethertogethertogethertogether cancancancan fillfillfillfill aaaa cisterncisterncisterncistern inininin 33331 31

minutes.minutes.minutes.minutes. IfIfIfIf oneoneoneone pipepipepipepipe takestakestakestakes 3333 minutesminutesminutesminutes moremoremoremore

thanthanthanthan thethethethe otherotherotherother totototo fillfillfillfill thethethethe cistern,cistern,cistern,cistern, findfindfindfind thethethethe timetimetimetime inininin whichwhichwhichwhich eacheacheacheach pipepipepipepipe wouldwouldwouldwould fillfillfillfill thethethethe cistern.cistern.cistern.cistern.Solution:Solution:Solution:Solution: Let th e two p ip es b e A an d B.

Let th e t im e taken b y p ip e A = x m in u tesAn d tim e taken b y p ip e B to fill = y m in u tes .As p er th e q u es t ion ,

y = x + 3 …….(i)

In 1 m in u te p ip e A fillsx1 cis te rn .

In 1 m in u te p ip e B fillsy1 cis te rn .

In 1 m in u te b o th p ip es A an d B will fill

y1

x1 cis te rn . …….(ii)

In 31 31 m in u te b o th p ip es fill 1 cis te rn

In 1 m in u te b o th p ip es fill4 01 3 cis te rn .

……(iii)Accord in g to th e q u es t io n ,

x1 +

y1 =

4 01 3 ……(iv)

Pu tt in g y = x + 3 from eq u ation (i) in eq u ation (iv), we g et

x1 +

3x1

=4 01 3

4 0 (x + 3 ) + 4 0 x = 1 3 x(x + 3 ) 4 0 x + 1 2 0 + 4 0 x = 1 3 x 2 + 3 9 x 1 3 x 2 + 3 9 x 4 0 x 1 2 0 4 0 x = 0 1 3 x 2 4 1 x 1 2 0 = 0

x =

2 61 2 01 344 14 1 2

x =2 6

6 2 4 01 6 8 14 1

x =2 6

7 9 2 14 1 x =

2 68 94 1

x =2 6

1 3 0 or2 64 8

x = 5 or 1 32 4

Sin ce tim e can n ot b e n eg ative , th erefore x = 5Hen ce , Tim e taken b y p ip e A = 5555 minutesminutesminutesminutes an d tim e taken b y p ip e B = 5 + 3 = 8888 minutesminutesminutesminutes.

11115.5.5.5. TwoTwoTwoTwo placesplacesplacesplaces AAAA andandandand BBBB areareareare 120120120120 kmkmkmkm apartapartapartapart fromfromfromfrom eacheacheacheach otherotherotherother onononon aaaa highway.highway.highway.highway. OneOneOneOne carcarcarcar startsstartsstartsstarts fromfromfromfrom AAAA andandandandanotheranotheranotheranother fromfromfromfrom BBBB atatatat thethethethe samesamesamesame time.time.time.time. IfIfIfIf theytheytheythey movemovemovemove inininin thethethethe samesamesamesame direction,direction,direction,direction, theytheytheythey meetmeetmeetmeet inininin 6666 hourshourshourshours andandandand ififififtheytheytheythey movemovemovemove inininin oppositeoppositeoppositeopposite directions,directions,directions,directions, theytheytheythey meetmeetmeetmeet inininin 1111 hourhourhourhour andandandand 12121212 minutes.minutes.minutes.minutes. FindFindFindFind thethethethe speedsspeedsspeedsspeeds ofofofof thethethethecars.cars.cars.cars.

Solution:Solution:Solution:Solution: Let th e sp eed of car s ta rt in g from A = x km / h r.An d th e sp eed of car s ta rt in g from B = y km / h r.Wh ile m ovin g in th e sam e- d irect ion le t th ey m eet a t C.Dis tan ce trave lled b y firs t car in 6 h ou rs = AC = 6 xDis tan ce trave lled b y secon d car in 6 h ou rs = BC = 6 yAccord in g to firs t con d it ion ,

AC = AB + BC 6 x = 1 2 0 + 6 y (Q Dis tan ce =

Sp eed Tim e) x = 2 0 + y ……(i)

Accord in g to th e secon d con d it ion ,

Dis tan ce trave lled b y firs t car in56 h ou rs = AD =

56 x


Dis tan ce trave lled b y secon d car in56 h ou rs = BD =

56 y

Wh ile m ovin g in th e op p o s ite d irect ion le t th ey m eet a t D.AD + DB = AB

56 x +

56 y = 1 2 0

[1 h ou r 1 2 m in u tes =56 h ou rs ]

x + y = 1 2 0 65

x + y = 1 0 0Su b s titu t ion x = 2 0 + y from eq u atio n (i) in eq u atio n (ii), we g et(2 0 + y) + y = 1 0 0 2 y = 8 0 y = 4 0 km / h ou rPu tt in g y = 4 0 in eq u ation (i), we h ave x = 2 0 + 4 0 = 6 0 Q/ h ou rHen ce , th e sp eed of firs t car = 60606060 km/hour.km/hour.km/hour.km/hour.An d th e sp eed of secon d car = 40404040 km/hour.km/hour.km/hour.km/hour.

16161616.... SolveSolveSolveSolve thethethethe followingfollowingfollowingfollowing simultaneoussimultaneoussimultaneoussimultaneous equations:equations:equations:equations:axaxaxax ++++ bybybyby==== (a(a(a(a –––– b);b);b);b);bxbxbxbx ==== ayayayay ++++ aaaa ++++ bbbb WhereWhereWhereWhere aaaa &&&& bbbb areareareare constantconstantconstantconstant....

Solution:Solution:Solution:Solution:ax + b y= (a – b ) …(I)

b x = ay + a + b b x – ay = (a + b ) …(III)

Wh ere a & b are con s tan tHere

D =a bb –a

D = –a 2 – b 2

Dx =a – b ba + b –a

= –a(a – b ) –b (a + b )= –a 2 + ab – ab – b 2

Dx = –a 2 – b 2

Dy =a a – bb a + b

= a(a + b ) – b (a – b )= a 2 + ab – ab + b 2

Dy = a 2 + b 2

By Cram er’s Ru le ,

x = DxD

& y = DyD

x =2 2

2 2(–a – b )

(–a – b ) x = 1

y =2 2

2 2a + b

–a – b

=2 2

2 2a + b

–(a – b ) y = –1 x = 1 & y = –1 is th e so lu t ion of g iven eq u atio n s


21212121

17.17.17.17. SolveSolveSolveSolve thethethethe followingfollowingfollowingfollowing simultaneoussimultaneoussimultaneoussimultaneous equations:equations:equations:equations:m(xm(xm(xm(x ++++ y)y)y)y) ++++ n(xn(xn(xn(x y)y)y)y)(m(m(m(m2222 ++++ mnmnmnmn ++++ nnnn2222)))) ==== 0000m(xm(xm(xm(x ++++ y)y)y)y) ++++ n(xn(xn(xn(x y)y)y)y)(m(m(m(m2222 mnmnmnmn ++++ nnnn2222)))) ==== 0000

Solution:Solution:Solution:Solution:m (x + y) + n (x y)(m 2 + m n + n 2) = 0

m (x + y) + n (x y) = m 2 + m n + n 2

m x + m y + n x n y = m 2 + m n + n 2

m x + n x + m y n y = m 2 + m n + n 2

x(m + n ) + y(m n ) = m 2 + m n + n 2 …(i)m (x + y) + n (x y) (m 2 m n + n 2) = 0

m (x + y) + n (x y) = m 2 m n + n 2

m x + m y + n x n y = m 2 m n + n 2

m x + n x + m y n y = m 2 m n + n 2

x(m + n ) + y(m n ) = m 2 m n + n 2 …(ii)Com p arin g eq n. (i) an d eq n. (ii) witha 1x + b 1y + c1 = 0 an d a 2x + b 2y + c2 = 0a 1 = m + n an d a 2 = m + nb 1 = m n b 2 = m nc1 = m 2 + m n + n 2 c2 = m 2 m n + n 2

as2

1

aa

=2

1

bb

2

1

cc

th e s im u ltan eou s eq u ation s m (x + y) + n (x y) (m 2 + m n + n 2) = 0 an d

m (x + y) + n (x y) (m 2 m n + n 2) = 0 h ave n o so lu t ion

18181818.... TheTheTheThe foreforeforefore wheelwheelwheelwheel ofofofof aaaa carriagecarriagecarriagecarriage makesmakesmakesmakes 6666 revolutionsrevolutionsrevolutionsrevolutions moremoremoremore thanthanthanthan thethethethe rearrearrearrear wheelwheelwheelwheel inininin goinggoinggoinggoing 120120120120 m.m.m.m. IfIfIfIf thethethethediameterdiameterdiameterdiameter ofofofof thethethethe foreforeforefore wheelwheelwheelwheel bebebebe increasedincreasedincreasedincreased ¼¼¼¼ itsitsitsits presentpresentpresentpresent diameterdiameterdiameterdiameter andandandand thethethethe diameterdiameterdiameterdiameter ofofofof thethethethe rearrearrearrearwheelwheelwheelwheel bebebebe increasedincreasedincreasedincreased bybybyby one-fifthone-fifthone-fifthone-fifth ofofofof itsitsitsits presentpresentpresentpresent diameter,diameter,diameter,diameter, thenthenthenthen thethethethe foreforeforefore wheelwheelwheelwheel makesmakesmakesmakes 4444 revolutionsrevolutionsrevolutionsrevolutionsmoremoremoremore thanthanthanthan thethethethe rearrearrearrear wheelwheelwheelwheel inininin goinggoinggoinggoing thethethethe samesamesamesame distance.distance.distance.distance. FindFindFindFind thethethethe circumferencecircumferencecircumferencecircumference ofofofof eacheacheacheach wheelwheelwheelwheel ofofofof thethethethecarriage.carriage.carriage.carriage.

Solution:Solution:Solution:Solution:Let th e d iam e ters of th e fore wh eel & rear wh eel x m & y m resp ective ly.

Circu m feren ce of fore wh eel = x m …(i)An d Circu m feren ce of rear wh eel = y m …(ii)

Now,For a wh eel,Dis tan ce = circu m feren ce x n o . of revo lu t io n s

For a fore wh eel1 2 0 = x × No. of revo lu t io n s b y a fore wh eel

No. of revo lu t io n s b y a fore wh eel =

120x

For a rear wh eel1 2 0 = y × No. of revo lu t ion b y a rear wh eel

No. of revo lu t io n s b y a rear wh eel =

1 2 0y

As p er 1 s t con d it ion ,

120

x=

1 2 0

y+ 6

1 2 0x

–

1 2 0y

= 6 (iii)

Now,

Diam ete r of freewh ee l is in creased 14

its p res en t d iam e ter.


Now Diam ete r of fo rewh ee l= x + x4

= 5x4

m

New circu m feren ce of a forewh eel = 5 x4

m

Diam ete r of rear wh eel is in creased 15

t im es its p res en t d iam ete r.

New d iam ete r of rear wh eel= y + y5

= 6 y5

m

New circu m feren ce of a rear wh eel = 6 y5

m

Now, for fore wh eel

1 2 0 = 5 x4

× No. of revo lu t io n b y a fore wh eel

No. of revo lu t io n b y a fore wh eel= 1 2 0 ×4

5 x=

9 6

xFor a rear wh eel

1 2 0 = 6 y5

× No. of revo lu t io n b y a rear wh eel

No. of revo lu t ion s b y a rear wh eel = 1 2 0 ×5

6 y=

1 0 0

y As p er 2 n d con d it ion ,

9 6

x=

1 0 0

y+ 4

96

x–

1 0 0

y= 4 ..(IV)

Su b s titu t in g1x

= a &1y

= b in eq u ation (III) & (IV)

1 2 0 a – 1 2 0 b = 6 …(V)9 6 a – 1 0 0 b = 4 …(VI)

Mu lt ip lyin g eq u ation V b y 1 0 , 1 2 0 0 a – 1 2 0 0 b = 4 8 …(viii)

Su b s t itu t in g eq u ation viii from vii 1 2 0 0 a – 1 2 0 0 b = 6 0

– 1 1 5 2 a – 1 2 0 0 b = 4 8– + –

4 8 a = 1 2

a = 1 24 8

a = 14

Su b s titu t in g a = 14

is …(v)

1 2 0 × 14

– 1 2 0 b = 6

3 0 – 1 2 0 b = 6 –1 2 0 b = –2 4

b = 2 41 2 0


23232323

b = 15

a =1x

& b =1y

We g et

1x

= 14

&1y

= 15

x = 4 y = 5 Circu m feren ce of fore wh eel = 4 m

Circu m feren ce of rear wh eel = 5 m

19.19.19.19. DrawDrawDrawDraw thethethethe graphsgraphsgraphsgraphs representingrepresentingrepresentingrepresenting thethethethe equationsequationsequationsequations 2x2x2x2x ==== yyyy ++++ 2222 andandandand 4x4x4x4x ++++ 3y3y3y3y ==== 24242424 nnnn thethethethe samesamesamesame graphgraphgraphgraph papers.papers.papers.papers.FindFindFindFind thethethethe areaareaareaarea ofofofof thethethethe triangletriangletriangletriangle formedformedformedformed bybybyby thesethesethesethese lineslineslineslines andandandand thethethethe XXXX –––– axis.axis.axis.axis.

20202020.... WhenWhenWhenWhen thethethethe sonsonsonson willwillwillwill bebebebe asasasas oldoldoldold asasasas hishishishis fatherfatherfatherfather today,today,today,today, thethethethe sumsumsumsum ofofofof theirtheirtheirtheir agesagesagesages thenthenthenthen willwillwillwill bebebebe 126.126.126.126. WhenWhenWhenWhen thethethethefatherfatherfatherfather waswaswaswas asasasas oldoldoldold asasasas hishishishis sonsonsonson isisisis today,today,today,today, thethethethe sumsumsumsum ofofofof theirtheirtheirtheir agesagesagesages thenthenthenthen waswaswaswas 38.38.38.38. FindFindFindFind theirtheirtheirtheir presentpresentpresentpresent ages.ages.ages.ages.

Solution:Solution:Solution:Solution:Let th e p res en t ag es of son b e x years & th e d iffe ren ce b etween th e ag es of son & fa th er b e y years

p res en t ag e of fa th er = (x + y)yearsNow,After y years ,

Ag e of son = (x + y)yearsAg e of fa th er = x + y + y

= (x + 2 y)years As p er 1 s t con d it ion ,

x + y + x + 2 y = 1 2 6 2 x + 3 y = 1 2 6 …(i)An d

y years ag o ,Ag e of son = (x – y)years

Ag e of fa th er = x years As p er 2 n d Con d it ion ,


x – y + x = 3 8 2 x – y = 3 8 …(ii)

Su b tract in g eq u ation (ii) from (i) 2 x + 3 y = 1 2 8

– 2 x – y = 3 8– + –

4 y = 8 8

y = 8 84

y = 2 2Su b s titu t in g y = 2 2 in eq n (ii),

2 x – 2 2 = 3 8 2 x = 3 8 + 2 2 2 x = 6 0 x = 3 0

& x + y = 3 0 + 2 2= 5 2

p res en t ag e of Son = 3 0 years& p res en t ag e of fa th er = 5 2 years .

FinFinFinFinaaaancialncialncialncial PlanningPlanningPlanningPlanning

4.4.4.4. TheTheTheThe marketmarketmarketmarket valuevaluevaluevalue ofofofof aaaa mutualmutualmutualmutual fundfundfundfund isisisis 400400400400 crorecrorecrorecrore rupees.rupees.rupees.rupees. WhichWhichWhichWhich isisisis divideddivideddivideddivided intointointointo 8888 crorecrorecrorecrore units?units?units?units?a.a.a.a. SupposeSupposeSupposeSuppose youyouyouyou investinvestinvestinvest Rs.Rs.Rs.Rs. 10,00010,00010,00010,000 inininin thethethethe units,units,units,units, howhowhowhow manymanymanymany unitsunitsunitsunits willwillwillwill youyouyouyou get?get?get?get?b.b.b.b. WhileWhileWhileWhile sellingsellingsellingselling thethethethe unitsunitsunitsunits ifififif theirtheirtheirtheir marketmarketmarketmarket valuevaluevaluevalue isisisis increasedincreasedincreasedincreased bybybyby 10%,10%,10%,10%, howhowhowhow muchmuchmuchmuch amountamountamountamount willwillwillwill youyouyouyou getgetgetget bybybybysellingsellingsellingselling them?them?them?them?Solution:Solution:Solution:Solution:

The price of one unit =crores8crores40

= Rs. 50.

a.No. of units by investing Rs. 10,000 =50

10000= 200

b.If the market value is increased by 10% by selling one unit, the profit will be50 x 0.1= Rs. 5By selling 200 units, the profit will be 200 x 5 = Rs. 1000

5.5.5.5. AAAA personpersonpersonperson buysbuysbuysbuys 120120120120 sharessharessharesshares atatatat nominalnominalnominalnominal valuevaluevaluevalue ofofofof 40404040 each,each,each,each, whichwhichwhichwhich hehehehe sellssellssellssells outoutoutout atatatat 45454545 each.each.each.each. FindFindFindFind hishishishisprofitprofitprofitprofit andandandand profitprofitprofitprofit perperperper cent.cent.cent.cent.Gain in 1 sh are = (4 5 – 4 0 ) = 5Gain in 1 2 0 sh ares = 1 2 0 × 5 = 6 0 0

Gain % = 1 2 .5 %

6.6.6.6. WhichWhichWhichWhich isisisis aaaa betterbetterbetterbetter investment:investment:investment:investment: 12%12%12%12% 100100100100 sharessharessharesshares atatatat 120120120120 orororor 8%8%8%8% atatatat 90?90?90?90?

ForForForFor IstIstIstIst casecasecasecaseIn com e on 1 2 0 = 1 2 % of 1 0 0 = 1 2In com e on 1 = 0 .1 0ForForForFor IIndIIndIIndIInd casecasecasecaseIn com e on 9 0 = 8 % of 1 0 0 = 8In com e on 1 = 0 .0 9 Firs t in ves tm en t is b e t te r.

7.7.7.7. AAAA companycompanycompanycompany withwithwithwith 10,00010,00010,00010,000 sharessharessharesshares ofofofof 100100100100 eacheacheacheach declaresdeclaresdeclaresdeclares anananan annualannualannualannual dividenddividenddividenddividend ofofofof 5%.5%.5%.5%.((((iiii)))) WhatWhatWhatWhat isisisis thethethethe totaltotaltotaltotal amountamountamountamount ofofofof dividenddividenddividenddividend paidpaidpaidpaid bybybyby thethethethe company?company?company?company?((((iiiiiiii)))) WhatWhatWhatWhat wouldwouldwouldwould bebebebe thethethethe annualannualannualannual incomeincomeincomeincome ofofofof aaaa man,man,man,man, whowhowhowho hashashashas 72727272 sharessharessharesshares inininin thethethethe company?company?company?company?((((iiiiiiiiiiii)))) IfIfIfIf hehehehe receivedreceivedreceivedreceived onlyonlyonlyonly 4%4%4%4% onononon hishishishis investment,investment,investment,investment, findfindfindfind thethethethe pricepricepriceprice hehehehe paidpaidpaidpaid forforforfor eacheacheacheach share.share.share.share.


25252525

(i) Tota l face va lu e = 1 0 0 × 1 0 ,0 0 0 = 1 0 ,0 0 ,0 0 0Tota l am o u n t of d ivid en d = 5 % of 1 0 ,0 0 ,0 0 0 = 5 0 ,0 0 0(ii) An n u al in com e on 7 2 sh ares = 5 % of (7 2 × 1 0 0 ) = 3 6 0iii) Price p a id for each sh are = 1 2 5

8.8.8.8. Mr.Mr.Mr.Mr. SharmaSharmaSharmaSharma buysbuysbuysbuys 60606060 sharessharessharesshares ofofofof nominalnominalnominalnominal valuevaluevaluevalue ofofofof 100100100100 andandandand hehehehe decidesdecidesdecidesdecides totototo sellsellsellsell themthemthemthem whenwhenwhenwhen theytheytheythey areareareare atatatataaaa premiumpremiumpremiumpremium ofofofof 60%.60%.60%.60%. HeHeHeHe investsinvestsinvestsinvests thethethethe proceedsproceedsproceedsproceeds inininin sharessharessharesshares ofofofof nominalnominalnominalnominal valuevaluevaluevalue ofofofof 50,50,50,50, quotedquotedquotedquoted atatatat 4%4%4%4% discount,discount,discount,discount,payingpayingpayingpaying 18%18%18%18% dividenddividenddividenddividend annually.annually.annually.annually. Calculate:Calculate:Calculate:Calculate:((((iiii)))) thethethethe salesalesalesale proceeds,proceeds,proceeds,proceeds,((((iiiiiiii)))) thethethethe numbernumbernumbernumber ofofofof sharessharessharesshares hehehehe buysbuysbuysbuys andandandand((((iiiiiiiiiiii)))) thethethethe annualannualannualannual dividenddividenddividenddividend fromfromfromfrom thesethesethesethese shares.shares.shares.shares.i) Nom in al va lu e of 6 0 sh ares = 1 0 0 × 6 0 = 6 ,0 0 0 Th e sa le p roceed s = 6 0 0 0 + 6 0 % of 6 ,0 0 0= 6 0 0 0 + 3 6 0 0 = 9 6 0 0(ii) Discou n t on each sh are = 4 % of 5 0 = 2 m arke t va lu e of each sh are = 4 8So, n u m b er of sh ares b ou g h t = 2 0 0(iii) Tota l face va lu e = 5 0 × 2 0 0 = 1 0 ,0 0 0An n u al d ivid en d = 1 8 % of 1 0 ,0 0 0 = 1 8 0 0

9.9.9.9. AAAA manmanmanman investsinvestsinvestsinvests 9600960096009600 onononon 100100100100 sharessharessharesshares atatatat 80.80.80.80. IfIfIfIf thethethethe companycompanycompanycompany payspayspayspays himhimhimhim 18%18%18%18% dividend,dividend,dividend,dividend, find:(find:(find:(find:(iiii)))) thethethethenumbernumbernumbernumber ofofofof sharessharessharesshares hehehehe buys.buys.buys.buys.((((iiiiiiii)))) hishishishis totaltotaltotaltotal dividend.dividend.dividend.dividend.((((iiiiiiiiiiii)))) hishishishis percentagepercentagepercentagepercentage returnreturnreturnreturn onononon thethethethe shares.shares.shares.shares.

(i) Nu m b er of sh ares b ou g h t = 1 2 0(ii) Divid en d = 1 8 % of to ta l face va lu e = 2 1 6 0(iii) Percen tag e re tu rn = 2 2 .5 %

10.10.10.10. AmitAmitAmitAmit KumarKumarKumarKumar investsinvestsinvestsinvests 36,00036,00036,00036,000 inininin buyingbuyingbuyingbuying 100100100100 sharessharessharesshares atatatat 20202020 premium.premium.premium.premium. TheTheTheThe dividenddividenddividenddividend isisisis 15%15%15%15%perperperper annum.annum.annum.annum. Find:Find:Find:Find:((((iiii)))) thethethethe numbernumbernumbernumber ofofofof sharessharessharesshares hehehehe buysbuysbuysbuys((((iiiiiiii)))) hishishishis yearlyyearlyyearlyyearly dividenddividenddividenddividend((((iiiiiiiiiiii)))) thethethethe percentagepercentagepercentagepercentage returnreturnreturnreturn onononon hishishishis investment.investment.investment.investment. GiveGiveGiveGive youryouryouryour answeransweransweranswer correctcorrectcorrectcorrect totototo thethethethe nearestnearestnearestnearest wholewholewholewhole number.number.number.number.

In ves tm en t = 3 6 ,0 0 0Face valu e of each sh are = 1 0 0Marke t va lu e of each sh are = 1 0 0 + 2 0 % of 1 0 0 = 1 2 0

(i) No. of sh ares p u rch ased = = 3 0 0(ii) Tota l face va lu e of 3 0 0 sh ares = 3 0 0 × 1 0 0 = 3 0 0 0 0

Divid en d = 1 5 % of 3 0 0 0 0 = × 3 0 0 0 0 = 4 5 0 0

(iii) Percen tag e re tu rn = × 1 0 0 % = 1 2 .5 % 1 3 %

11.11.11.11. RakeshRakeshRakeshRakesh investedinvestedinvestedinvested 43,92043,92043,92043,920 totototo buybuybuybuy sharessharessharesshares ofofofof aaaa companycompanycompanycompany whosewhosewhosewhose marketmarketmarketmarket valuevaluevaluevalue isisisis 122122122122 each.each.each.each. TheTheTheThefacefacefaceface valuevaluevaluevalue ofofofof aaaa shareshareshareshare isisisis 100.100.100.100. FindFindFindFind thethethethe numbernumbernumbernumber ofofofof sharessharessharesshares purchased.purchased.purchased.purchased.In ves tm en t = 4 3 ,9 2 0

Marke t va lu e of each sh are = 1 2 2

No. of sh ares p u rch ased = = 3 6 0


12.12.12.12. SalmanSalmanSalmanSalman investsinvestsinvestsinvests aaaa sumsumsumsum ofofofof moneymoneymoneymoney inininin 50505050 shares,shares,shares,shares, payingpayingpayingpaying 15%15%15%15% dividenddividenddividenddividend quotedquotedquotedquoted atatatat 20%20%20%20% premium.premium.premium.premium. IfIfIfIfhishishishis annualannualannualannual dividenddividenddividenddividend isisisis 600,600,600,600, calculate:calculate:calculate:calculate:((((iiii)))) thethethethe numbernumbernumbernumber ofofofof sharessharessharesshares hehehehe bought.bought.bought.bought.((((iiiiiiii)))) hishishishis totaltotaltotaltotal investment.investment.investment.investment.((((iiiiiiiiiiii)))) thethethethe raterateraterate ofofofof returnreturnreturnreturn onononon hishishishis investment.investment.investment.investment.

Face valu e of each sh are = 5 0

Prem iu m = 2 0 % of 5 0 = × 5 0 = 1 0 Marke t va lu e of each sh are = (5 0 + 1 0 ) = 6 0Let th e n u m b er of sh ares b e p u rch ased = xTh en face valu e of x sh ares = 5 0x(i) Now d ivid en d = 6 0 0 1 5 % of 5 0x = 6 0 0

× 5 0x = 6 0 0

x = = 8 0Hen ce , n u m b er of sh ares b ou g h t = 8 0(ii) Tota l in ves tm en t = 6 0 × 8 0 = 4 8 0 0

(iii) Rate of re tu rn = × 1 0 0 % = 1 2 .5 %13.13.13.13. WhatWhatWhatWhat sumsumsumsum shouldshouldshouldshould AshokAshokAshokAshok investinvestinvestinvest inininin 25252525 shares,shares,shares,shares, sellingsellingsellingselling atatatat 36363636 totototo obtainobtainobtainobtain anananan incomeincomeincomeincome ofofofof 720720720720 ifififif thethethethedividenddividenddividenddividend declareddeclareddeclareddeclared isisisis 12%?12%?12%?12%? AlsoAlsoAlsoAlso find:find:find:find:((((iiii)))) thethethethe numbernumbernumbernumber ofofofof sharessharessharesshares boughtboughtboughtbought bybybyby Ashok,Ashok,Ashok,Ashok,((((iiiiiiii)))) thethethethe percentagepercentagepercentagepercentage returnreturnreturnreturn onononon hishishishis investment.investment.investment.investment.Face valu e of each sh are = 2 5

Marke t va lu e of each sh are = 3 6Let n o . of sh ares b ou g h t = xTota l face va lu e of x sh ares = 2 5xDivid en d = 1 2 % of 2 5x

× 2 5x = 7 2 0

x = = 2 4 0(i) No. of sh ares p u rch ased b y Ash ok = 2 4 0(ii) Tota l in ves tm en t (m arke t va lu e) = 3 6 × 2 4 0 = 8 6 4 0

Percen tag e re tu rn = × 1 0 0 % = 8 .3 3

14.14.14.14. RohitRohitRohitRohit investedinvestedinvestedinvested 9,6009,6009,6009,600 onononon 100100100100 sharessharessharesshares atatatat 20202020 premiumpremiumpremiumpremium payingpayingpayingpaying 8%8%8%8% dividend.dividend.dividend.dividend. RohitRohitRohitRohit soldsoldsoldsold thethethethesharessharessharesshares whenwhenwhenwhen thethethethe pricepricepriceprice roseroseroserose totototo 160.160.160.160. HeHeHeHe investedinvestedinvestedinvested thethethethe proceedsproceedsproceedsproceeds (excluding(excluding(excluding(excluding dividend)dividend)dividend)dividend) inininin 10%10%10%10% 50505050sharessharessharesshares atatatat 40.40.40.40. FindFindFindFind the:the:the:the:((((iiii)))) originaloriginaloriginaloriginal numbernumbernumbernumber ofofofof shares.shares.shares.shares.((((iiiiiiii)))) salesalesalesale proceeds.proceeds.proceeds.proceeds.((((iiiiiiiiiiii)))) newnewnewnew numbernumbernumbernumber ofofofof shares.shares.shares.shares.((((iviviviv)))) changechangechangechange inininin thethethethe twotwotwotwo dividends.dividends.dividends.dividends.

(i) In ves tm en t = 9 6 0 0Face valu e of each sh are = 1 0 0Marke t va lu e of each sh are = 1 0 0 + 2 0 = 1 2 0


27272727

Orig in a l n u m b er of sh ares = = 8 0Tota l face va lu e of 8 0 sh ares = 1 0 0 × 8 0 = 8 0 0 0

Divid en d rece ived = 8 % of 8 0 0 0 = × 8 0 0 0 = 6 4 0(ii) Marke t va lu e of each sh are to b e so ld = 1 6 0 Sa le p roceed s = 1 6 0 × 8 0 = 1 2 8 0 0

(iii) New n u m b er of sh ares = = 3 2 0Tota l face va lu e = 5 0 × 3 2 0 = 1 6 0 0 0

Divid en d = 1 0 % of 1 6 0 0 0 = × 1 6 0 0 0 = 1 6 0 0(iv) Ch an g e in two d ivid en d = (1 6 0 0 – 6 4 0 ) = 9 6 0

15.15.15.15. AshokAshokAshokAshok investedinvestedinvestedinvested 26,40026,40026,40026,400 onononon 12%,12%,12%,12%, 25252525 sharessharessharesshares ofofofof aaaa company.company.company.company. IfIfIfIf hehehehe receivesreceivesreceivesreceives aaaa dividenddividenddividenddividend ofofofof 2,475,2,475,2,475,2,475,findfindfindfind the:the:the:the:((((iiii)))) numbernumbernumbernumber ofofofof sharessharessharesshares hehehehe boughtboughtboughtbought((((iiiiiiii)))) marketmarketmarketmarket valuevaluevaluevalue ofofofof eacheacheacheach shareshareshareshare

(i) Let th e n u m b er of sh ares p u rch ased b y Ash ok b e xFace valu e of each sh are = 2 5Tota l face va lu e of x sh ares = 2 5xDivid en d = 2 4 7 5 1 2 % of 2 5x = 2 4 7 5

× 2 5x = 2 4 7 5

x = = 8 2 5 Nu m b er of sh ares p u rch ased b y Ash ok = 8 2 5

(ii) Marke t va lu e of each sh are = = 3 2

ProbabilityProbabilityProbabilityProbability1)1)1)1) TheTheTheThe probabilityprobabilityprobabilityprobability ofofofof guessingguessingguessingguessing thethethethe correctcorrectcorrectcorrect answeransweransweranswer totototo aaaa certaincertaincertaincertain questionquestionquestionquestion isisisis

1 2x.... IfIfIfIf thethethethe probabilityprobabilityprobabilityprobability ofofofof

notnotnotnot guessingguessingguessingguessing thethethethe correctcorrectcorrectcorrect answeransweransweranswer totototo thisthisthisthis questionquestionquestionquestion isisisis32,,,, thenthenthenthen findfindfindfind x.x.x.x.

Solution:Solution:Solution:Solution:Let A b e th e even t of g u ess in g th e correct an swer.

P(A) =1 2x P(A) =

32

We kn ow,P(A) + P(A) = 1

1 2x +

32 = 1

1 2x = 1 -

32

1 2x =

31

xxxx ==== 4.4.4.4.

2222.... IfIfIfIf aaaa numbernumbernumbernumber xxxx isisisis chosenchosenchosenchosen atatatat randomrandomrandomrandom fromfromfromfrom thethethethe numbersnumbersnumbersnumbers 2,2,2,2, 1,1,1,1, 0,0,0,0, 1,1,1,1, 2.2.2.2. WhatWhatWhatWhat isisisis thethethethe probabilityprobabilityprobabilityprobability thatthatthatthatxxxx2222 <<<< 2222 ????

Solution:Solution:Solution:Solution:


Clearly, n u m b er x can take an y on e of th e five g iven valu es .So , to ta l n u m b er of e lem en tary even ts = 5We ob se rve th a t x 2 < 2 wh en x takes an y on e of th e fo llowin g th ree va lu es 1 , 0 an d 1 .So , favou rab le n u m b er of e lem en tary even ts = 3 .

Hen ce , P (x 2 < 2 ) =53 .

3333.... ItItItIt isisisis knowknowknowknow thatthatthatthat aaaa boxboxboxbox ofofofof 600600600600 electricelectricelectricelectric bulbsbulbsbulbsbulbs containscontainscontainscontains 12121212 defectivedefectivedefectivedefective bulbs.bulbs.bulbs.bulbs. OneOneOneOne bulbbulbbulbbulb isisisis takentakentakentaken outoutoutout atatatatrandomrandomrandomrandom fromfromfromfrom thisthisthisthis box.box.box.box. WhatWhatWhatWhat isisisis thethethethe probabilityprobabilityprobabilityprobability thatthatthatthat itititit isisisis aaaa no-defectiveno-defectiveno-defectiveno-defective bulbbulbbulbbulb ????

Solution:Solution:Solution:Solution:Ou t of 6 0 0 electric on e b u lb can b e ch osen in 6 0 0 ways .

Tota l n u m b er of e lem en tary even ts = 6 0 0Th ere are 5 8 8 (= 6 0 0 1 2 ) n on - d efect ive b u lb s ou t of wh ich on e b u lb can b e ch osen in 5 8 8 ways .

Favou rab le n u m b er of e lem en tary even ts = 5 8 8

Hen ce , P(Gett in g a n on - d efect ive b u lb ) =6 0 05 8 8 =

5 04 9 = 0 .9 8 .

4444.... AAAA gamegamegamegame consistsconsistsconsistsconsists ofofofof tossingtossingtossingtossing aaaa oneoneoneone rupeerupeerupeerupee coincoincoincoin 3333 timestimestimestimes andandandand notingnotingnotingnoting itsitsitsits outcomeoutcomeoutcomeoutcome eacheacheacheach time.time.time.time. HanifHanifHanifHanifwinswinswinswins ifififif allallallall thethethethe tossestossestossestosses givegivegivegive thethethethe samesamesamesame resultresultresultresult i.e.i.e.i.e.i.e. threethreethreethree headsheadsheadsheads orororor threethreethreethree tails,tails,tails,tails, andandandand losesloseslosesloses otherwise.otherwise.otherwise.otherwise.CalculateCalculateCalculateCalculate thethethethe probabilityprobabilityprobabilityprobability thatthatthatthat HanifHanifHanifHanif willwillwillwill loseloseloselose thethethethe game.game.game.game.

Solution:Solution:Solution:Solution:Wh en a co in is toss ed th ree t im es , p os s ib le ou tcom es areHHH, HHT, HTH, THH, HTT, THT, TTH, TTT.

Tota l n u m b er of e lem en tary even ts = 8Han if will lose th e g am e if a ll th e tosse s d o n ot g ive th e sam e resu lt i.e . a ll h ead s or a ll ta ils . So ,favou rab le ou tcom es areHHT, HTH, THH, TTH, HTT, THT,

Favou rab le n u m b er of e lem en tary even ts = 6

Hen ce , P(Han if will lose th e g am e) =86 =

43 .

5555.... AAAA jarjarjarjar containscontainscontainscontains 24242424 marblesmarblesmarblesmarbles somesomesomesome areareareare greengreengreengreen areareareare othersothersothersothers areareareare blue.blue.blue.blue. IfIfIfIf aaaa marblemarblemarblemarble isisisis drawndrawndrawndrawn atatatat randomrandomrandomrandomfromfromfromfrom thethethethe jar,jar,jar,jar, thethethethe probabilityprobabilityprobabilityprobability thatthatthatthat itititit isisisis greengreengreengreen isisisis 2/3.2/3.2/3.2/3. FindFindFindFind thethethethe numbernumbernumbernumber ofofofof blueblueblueblue marblesmarblesmarblesmarbles inininin thethethethe jar.jar.jar.jar.

Solution:Solution:Solution:Solution:Let th ere b e b b lu e m arb les an d g g reen m arb les in th e ja r. Th en ,Tota l n u m b er of m arb les in th e ja r= b + gIt is g iven th a t th ere are 2 4 m arb les in th e ja r.

b + g = 2 4 ……..(i)

P (Gett in g a g reen m arb le from th e jar) =2 4g

32 =

2 4g

g = 2 4 32 = 1 6

Pu tt in g g = 1 6 in (i), we g et b = 8Hen ce , th ere are 8 b lu e m arb les in th e ja r.

6666.... AAAA numbernumbernumbernumber xxxx isisisis selectedselectedselectedselected fromfromfromfrom thethethethe numbersnumbersnumbersnumbers 1,1,1,1, 2,2,2,2, 3333 andandandand thenthenthenthen aaaa secondsecondsecondsecond numbernumbernumbernumber yyyy isisisis randomlyrandomlyrandomlyrandomlyselectedselectedselectedselected fromfromfromfrom thethethethe numbersnumbersnumbersnumbers 1,1,1,1, 4,4,4,4, 9.9.9.9. WhatWhatWhatWhat isisisis thethethethe probabilityprobabilityprobabilityprobability thatthatthatthat thethethethe productproductproductproduct xyxyxyxy ofofofof thethethethe twotwotwotwo numbersnumbersnumbersnumberswillwillwillwill bebebebe lesslesslessless thanthanthanthan 9999 ????

Solution:Solution:Solution:Solution:Nu m b er x can b e se lected in th ree ways an d corresp on d in g to each su ch way th ere are th ree ways ofse lect in g n u m b er y. Th ere fore , two n u m b ers can b e se lected in 9 ways as lis ted b elo w:(1 ,1 ), (1 ,4 ), (1 ,9 ), (2 ,1 ), (2 ,4 ), (2 ,9 ), (3 ,1 ), (3 ,4 ), (3 ,9 )So , to ta l n u m b er of e lem en tary even ts = 9Th e p rod u ct xy will b e less th an 9 , if x an d y are ch osen in on e of th e fo llowin g ways :(1 ,1 ), (1 ,4 ), (2 ,1 ), (2 ,4 ), (3 ,1 )


Hen ce , req u ired p rob ab ility =95 .


29292929

0

7777.... InInInIn aaaa musicalmusicalmusicalmusical chairchairchairchair game,game,game,game, thethethethe personpersonpersonperson playingplayingplayingplaying thethethethe musicmusicmusicmusic hashashashas beenbeenbeenbeen advisedadvisedadvisedadvised totototo stopstopstopstop playingplayingplayingplaying thethethethe musicmusicmusicmusicatatatat anyanyanyany timetimetimetime withinwithinwithinwithin 2222 minutesminutesminutesminutes afterafterafterafter sheshesheshe startsstartsstartsstarts playing.playing.playing.playing. WhatWhatWhatWhat isisisis thethethethe probabilityprobabilityprobabilityprobability thatthatthatthat thethethethe musicmusicmusicmusic willwillwillwillstopstopstopstop withinwithinwithinwithin thethethethe firstfirstfirstfirst halfhalfhalfhalf minuteminuteminuteminute afterafterafterafter starting?starting?starting?starting?

Solution:Solution:Solution:Solution:Here th e p os s ib le ou tcom es are a ll th e n u m b ers b e tween 0 an d 2 . Th is is th e p ort ion of th e n u m b er lin efrom 0 to 2 as sh own in th e fig u re .Let A b e th e even t th a t ‘th e m u s ic is s top p ed with in th e firs t h a lf m in u te ’. Th en , Ou tcom es favou rab le

to th e even t A are a ll p o in ts on th e n u m b er lin e from O to Q i.e . from 0 to21 .

Th e to ta l n u m b er of ou tcom es are th e p o in ts on th e n u m b er lin e from O to P i.e . from 0 to 2 .

P(A) =OPLen g th

OQLen g th =2

2/1 =41 .

8888.... 1000100010001000 ticketsticketsticketstickets ofofofof aaaa lotterylotterylotterylottery werewerewerewere soldsoldsoldsold andandandand theretheretherethere areareareare 5555 prizesprizesprizesprizes onononon thesethesethesethese tickets.tickets.tickets.tickets. IfIfIfIf SaketSaketSaketSaket hashashashas purchasedpurchasedpurchasedpurchasedoneoneoneone lotterylotterylotterylottery ticket,ticket,ticket,ticket, whatwhatwhatwhat isisisis thethethethe probabilityprobabilityprobabilityprobability ofofofof winningwinningwinningwinning aaaa prize?prize?prize?prize?

Solution:Solution:Solution:Solution:Ou t of 1 0 0 0 lo tte ry t icke ts on e can b e ch osen in 1 0 0 0 ways .

Tota l n u m b er of e lem en tary even ts = 1 0 0 0It is g iven th a t th ere are 5 p riz es on th ese 1 0 0 0 ticke t can b e ch osen in 1 0 0 0 ways . Th ere fore ,Nu m b er of ways of se lect in g a p riz e t icke t = 5


Hen ce , P (Win n in g a p riz e) =1 0 0 0

5 = 0 .0 0 5 .

9999.... InInInIn fig.,fig.,fig.,fig., aaaa dartdartdartdart isisisis thrownthrownthrownthrown andandandand landslandslandslands inininin thethethethe interiorinteriorinteriorinterior ofofofof thethethethe circle.circle.circle.circle. WhatWhatWhatWhat isisisis thethethethe probabilityprobabilityprobabilityprobability thatthatthatthat thethethethe dartdartdartdartwillwillwillwill landlandlandland inininin thethethethe shadedshadedshadedshaded region?region?region?region?

Solution:Solution:Solution:Solution:We h ave ,AB = CD = 8 an d AD = BC = 6

Usin g Pyth ag oras th eorem in ABC, we h aveAC2 = AB2 + BC2

AC2 = 8 2 + 6 2

AC = 1 0 OA = OC = 5 [Q O is th e m id - p o in t o f AC] Area of th e circle = (OA)2 = 2 5 sq . u n its [Q Area = r2]

Area of rectan g le ABCD = AB BC= 8 6= 4 8 sq . u n its

21

21

D C

A B

O

8

6


Area of sh ad ed reg ion = Area of th e circle Area of rectan g le ABCD Area of sh ad ed reg ion = 2 5 4 8 sq . u n its

Hen ce ,

P (Dart lan d s in th e sh ad ed reg ion ) =circleofArea

reg ionsh ad edofArea =

2 54 82 5 .

10101010.... AAAA piggypiggypiggypiggy bankbankbankbank containscontainscontainscontains hundredhundredhundredhundred 50505050 paisapaisapaisapaisa coins,coins,coins,coins, fiftyfiftyfiftyfifty Rs.Rs.Rs.Rs. 1111 coins,coins,coins,coins, twentytwentytwentytwenty Rs.Rs.Rs.Rs. 2222 coinscoinscoinscoins andandandand tentententen Rs.Rs.Rs.Rs. 5555coins.coins.coins.coins. IfIfIfIf itititit isisisis equallyequallyequallyequally likelylikelylikelylikely thatthatthatthat oneoneoneone ofofofof thethethethe coinscoinscoinscoins willwillwillwill falloutfalloutfalloutfallout whenwhenwhenwhen thethethethe bankbankbankbank isisisis turnedturnedturnedturned upupupup sidesidesideside down,down,down,down,WhatWhatWhatWhat isisisis thethethethe probabilityprobabilityprobabilityprobability thatthatthatthat thethethethe coincoincoincoin (i)(i)(i)(i) willwillwillwill bebebebe aaaa 50505050 paisapaisapaisapaisa coin?coin?coin?coin? (ii)(ii)(ii)(ii) willwillwillwill notnotnotnot bebebebe aaaa Rs.Rs.Rs.Rs. 5555 coins?coins?coins?coins?

SolutionSolutionSolutionSolution ::::We h ave ,

Tota l n u m b er of co in s = 1 0 0 + 5 0 + 2 0 + 1 0 = 1 8 0So, on e co in can b e ch osen ou t of 1 8 0 co in s in 1 8 0 ways .

Tota l n u m b er of e lem en tary even ts = 1 8 0(i)Th ere are 1 0 0 fifty p a is a co in s ou t of wh ich on e co in can b e ch osen in 1 0 0 ways .

Prob ab ility th a t a 5 0 p ais a co in will fa ll =1 8 01 0 0 =

95 .

(ii) Oth e r th an Rs . 5 co in s th ere are 1 7 0 co in s .

Prob ab ility th a t co in fa llen ou t is n o t a Rs . 5 co in =1 8 01 7 0 =

1 81 7 .

11111111.... ThereThereThereThere areareareare 40404040 studentsstudentsstudentsstudents inininin classclassclassclass XXXX ofofofof schoolschoolschoolschool ofofofof whomwhomwhomwhom 25252525 areareareare girlsgirlsgirlsgirls andandandand 15151515 areareareare boys.boys.boys.boys. TheTheTheThe classclassclassclassteacherteacherteacherteacher hashashashas totototo selectselectselectselect oneoneoneone studentstudentstudentstudent asasasas aaaa classclassclassclass representative.representative.representative.representative. HeHeHeHe writeswriteswriteswrites thethethethe namenamenamename ofofofof eacheacheacheach studentstudentstudentstudentonononon aaaa separateseparateseparateseparate card,card,card,card, thethethethe cardscardscardscards beingbeingbeingbeing identical.identical.identical.identical. ThenThenThenThen sheshesheshe putsputsputsputs cardscardscardscards inininin aaaa bagbagbagbag andandandand stirsstirsstirsstirs themthemthemthemthoroughly.thoroughly.thoroughly.thoroughly. SheSheSheShe thenthenthenthen drawsdrawsdrawsdraws oneoneoneone cardcardcardcard fromfromfromfrom thethethethe bag.bag.bag.bag. WhatWhatWhatWhat isisisis thethethethe probabilityprobabilityprobabilityprobability thatthatthatthat thethethethe namenamenamename writtenwrittenwrittenwrittenonononon thethethethe cardcardcardcard isisisis thethethethe namenamenamename ofofofof (i)(i)(i)(i) aaaa girl?girl?girl?girl? (ii)(ii)(ii)(ii) aaaa boy?boy?boy?boy?

Solution:Solution:Solution:Solution:Sin ce th ere are 4 0 s tu d en ts an d th ere is on e card for each s tu d en t .So , on e card can b e ch ose ou t of 4 0 card s in 4 0 ways .

Tota l n u m b er of e lem en tary even ts = 4 0(i)Th ere are 2 5 g irls an d corresp o n d in g to each g irl th ere is card of h er n am e.

Th ere fore , a card with th e n am e of a g irl can b e ch osen in 2 5 ways . Favou reab le n u m b er of e lem en tary even ts = 2 5 .

Hen ce , P(Gett in g a card with th e n am e of a g irl) =4 02 5 =

85

(ii) We h ave ,P(Gett in g a card with n am e of a b oy) = 1 P(Gett in g a card with n am e of a g irl)

= 1 85 =

83 .

12121212.... AAAA cartoncartoncartoncarton consistsconsistsconsistsconsists ofofofof 100100100100 shirtsshirtsshirtsshirts ofofofof whichwhichwhichwhich 88888888 areareareare good,good,good,good, 8888 havehavehavehave minorminorminorminor defectsdefectsdefectsdefects andandandand 4444 havehavehavehave majormajormajormajordefectsdefectsdefectsdefects Jimmy,Jimmy,Jimmy,Jimmy, aaaa trader,trader,trader,trader, willwillwillwill onlyonlyonlyonly acceptacceptacceptaccept thethethethe shirtsshirtsshirtsshirts whichwhichwhichwhich areareareare good,good,good,good, butbutbutbut Sujatha,Sujatha,Sujatha,Sujatha, anotheranotheranotheranother trader,trader,trader,trader,willwillwillwill onlyonlyonlyonly rejectrejectrejectreject thethethethe shirtsshirtsshirtsshirts whichwhichwhichwhich havehavehavehave majormajormajormajor defects.defects.defects.defects. OneOneOneOne shirtshirtshirtshirt isisisis drawndrawndrawndrawn atatatat randomrandomrandomrandom thethethethe carton.carton.carton.carton.WhatWhatWhatWhat isisisis thethethethe probabilityprobabilityprobabilityprobability thatthatthatthat itititit isisisis acceptableacceptableacceptableacceptable totototo (i)(i)(i)(i) Jimmy?Jimmy?Jimmy?Jimmy? (ii)(ii)(ii)(ii) Sujatha?Sujatha?Sujatha?Sujatha?

Solution:Solution:Solution:Solution:On e sh irt is d rawn at ran d om from th e carton of 1 0 0 sh irts . Th is can b e d on e in 1 0 0 ways

Tota l n u m b er of e lem en tary even ts = 1 0 0 .(i) Sin ce Jim m y accep ts on ly g ood sh irts an d th e n u m b er of g ood sh irts is 8 8 Nu m b er of e lem en tary even ts favou rab le to Jim m y = 8 8

So, p rob ab ility th a t a sh irt is accep tab le to Jim m y =1 0 08 8 = 0 .8 8 .

(ii) Su ja th a accep ts g ood as well as sh irts h avin g m in or d efects . Th e n u m b er of su ch sh irt is 8 8 + 8 = 9 6 . Nu m b er of e lem en tary even ts favou rab le to an even t of se lect in g a g ood sh irt o r a sh irt with m in or

d efect is 9 6 .

Hen ce , p rob ab ility th a t a sh irt is accep tab le to Su ja th a =1 0 09 6 = 0 .9 6 .

13131313.... AAAA bagbagbagbag containscontainscontainscontains 12121212 ballsballsballsballs outoutoutout ofofofof whichwhichwhichwhich xxxx areareareare white.white.white.white.


31313131

(i)(i)(i)(i) IfIfIfIf oneoneoneone ballballballball isisisis drawndrawndrawndrawn atatatat random,random,random,random, whatwhatwhatwhat isisisis thethethethe probabilityprobabilityprobabilityprobability thatthatthatthat itititit willwillwillwill bebebebe aaaa whitewhitewhitewhite ball?ball?ball?ball?(ii)(ii)(ii)(ii) IfIfIfIf 6666 moremoremoremore whitewhitewhitewhite ballsballsballsballs areareareare putputputput inininin thethethethe bag,bag,bag,bag, thethethethe probabilityprobabilityprobabilityprobability ofofofof drawingdrawingdrawingdrawing aaaa whitewhitewhitewhite ballballballball willwillwillwill bebebebe doubledoubledoubledoublethanthanthanthan thatthatthatthat inininin (i).(i).(i).(i). FindFindFindFind x.x.x.x.

Solution:Solution:Solution:Solution:(i)Th ere are 1 2 b alls in th e b ag . Ou t of th ese 1 2 b alls on e can b e chose in 1 2 ways . Tota l n u m b er of e lem en tary even ts = 1 2

Th ere are x wh ite b a lls ou t of wh ich on e can b e ch osen in x ways . Favorab le n u m b er of e lem en tary even ts = x

Hen ce , p 1 = P (Gett in g a wh ite b a ll) =1 2x

(ii) If 6 m ore wh ite b a lls a re p u t in th e b ag , th enTota l n u m b er of b a lls in th e b ag = 1 2 + 6 = 1 8Nu m b er of wh ite b a lls in th e b ag s = x + 6

p 2 = P (Gett in g a wh ite b a ll) =1 8

6x

It is g iven th a tP2 = 2 p 1

1 8

6x = 2 1 2x

1 8

6x =6x

6 (x + 6 ) = 1 8 x 6 x + 3 6 = 1 8 x 1 2 x = 3 6 x =1 23 6 = 3 .

14141414.... TheTheTheThe king,king,king,king, queenqueenqueenqueen andandandand jackjackjackjack ofofofof clubsclubsclubsclubs areareareare removedremovedremovedremoved fromfromfromfrom aaaa deckdeckdeckdeck ofofofof 52525252 playingplayingplayingplaying cardscardscardscards andandandand thethethethe wellwellwellwellshuffled.shuffled.shuffled.shuffled. OneOneOneOne cardcardcardcard isisisis selectedselectedselectedselected fromfromfromfrom thethethethe remainingremainingremainingremaining cards.cards.cards.cards. FindFindFindFind thethethethe probabilityprobabilityprobabilityprobability ofofofof getting.getting.getting.getting.(i)(i)(i)(i) aaaa heartheartheartheart(ii)(ii)(ii)(ii) aaaa kingkingkingking(iii)(iii)(iii)(iii) aaaa clubclubclubclub(iv)(iv)(iv)(iv) thethethethe ‘‘‘‘10101010’’’’ ofofofof hearts.hearts.hearts.hearts.

Solution:Solution:Solution:Solution:After rem ovin g kin g , q u een an d jack of clu b s from a d eck of 5 2 p layin g card s th ere are 4 9 card s le ft inth e d eck. Ou t of th ese 4 9 card s on e card can b e ch osen in 4 9 ways .

Tota l n u m b er of e lem en tary even ts = 4 9 .(i)Th ere are 1 3 h eart card s in th e d eck con ta in ing 4 9 card s ou t of wh ich on e h eart card can b e chose in

1 3 ways . Favou rab le n u m b er of e lem en tary even ts = 1 3

Hen ce , P(Gett in g a h eart) =4 91 3 .

(ii) Th ere are 3 kin g s in th e d eck con ta in in g 4 9 card s . Ou t of th ese th ree kin g s on e kin g can b e ch osen in3 ways .


Hen ce , P(Gett in g a kin g ) =4 93 .

(iii) Afte r rem o vin g kin g , q u een an d jack of clu b s on ly 1 0 clu b card s are le ft in th e d eck. Ou t of th ese 1 0clu b card s on e clu b card is chosen in 1 0 ways .

Favou rab le n u m b er of e lem en tary even ts = 1 0

Hen ce , P(Gett in g a clu b ) =4 91 0 .

(iv) Th ere is on ly on e ‘1 0 ’ of h earts . Favou rab le n u m b er of e lem en tary even ts = 1

Hen ce ,

P(Gett in g th e ‘1 0 ’ to h earts ) =4 91 .

15151515.... FindFindFindFind thethethethe probabilityprobabilityprobabilityprobability ofofofof aaaa fourfourfourfour turningturningturningturning upupupup atatatat leastleastleastleast onceonceonceonce inininin twotwotwotwo tossestossestossestosses ofofofof aaaa fairfairfairfair die.die.die.die.Solution:Solution:Solution:Solution:


Let S b e th e sam p le sp ace S =

{(1 ,1 ), (1 ,2 ), (1 ,3 ), (1 ,4 ), (1 ,5 ), (1 ,6 ),(2 ,1 ), (2 ,2 ), (2 ,3 ), (2 ,4 ), (2 ,5 ), (2 ,6 ),(3 ,1 ), (3 ,2 ), (3 ,3 ), (3 ,4 ), (3 ,5 ), (3 ,6 ),(4 ,1 ), (4 ,2 ), (4 ,3 ), (4 ,4 ), (4 ,5 ), (4 ,6 ),(5 ,1 ), (5 ,2 ), (5 ,3 ), (5 ,4 ), (5 ,5 ), (5 ,6 ),(6 ,1 ), (6 ,2 ), (6 ,3 ), (6 ,4 ), (6 ,5 ), (6 ,6 )}

n (S) = 3 6Let ‘A’ b e th e even t of a fou r tu rn in g u p at leas t on ce .

A ={(1 ,4 ) (2 ,4 ) (4 ,1 ) (4 ,2 ) (4 ,3 ) (4 ,4 )(4 , 5 ) (4 , 6 ) (5 , 4 ) (6 , 4 )}

n (A) = 1 0

P(A) =)S(n)A(n =

3 61 0 =

1 85

16161616.... WhatWhatWhatWhat isisisis thethethethe probabilityprobabilityprobabilityprobability thatthatthatthat anananan ordinaryordinaryordinaryordinary yearyearyearyear andandandand aaaa leapleapleapleap yearyearyearyear hashashashas 53535353 SundaysSundaysSundaysSundays ????Solution:Solution:Solution:Solution:

(i)(i)(i)(i) OrdinaryOrdinaryOrdinaryOrdinary yearyearyearyear1 year = 3 6 5 d ays

=7

3 6 5

= 5 2 weeks + 1 d ay= 5 2 Su n d ays + 1 d ay

Let ‘S’ b e th e sam p le sp ace of th is 1 d ay b e ‘S’. S = {Su n , Mon , Tu e , Wed , Th u , Fri, Sa t} n (S) = 7

Let ‘A’ b e th e even t th a t th e 1 d ay is a Su n d ay. A = {Su n } n (A) = 1

P(A) =71

)S(n)A(n

(ii)(ii)(ii)(ii) LeapLeapLeapLeap yearyearyearyear

1 leap year = 3 6 6 d ays

=7

3 6 6

= 5 2 weeks + 2 d ays= 5 2 Su n d ays + 2 d ays

Let ’s b e th e sam p le sp ace of th ese 2 d ays S = {SM, MT, TW, WTh , Th F, FSa , SaS} n (S) = 7

Let A b e th e even t th a t 1 d ay from it is a Su n d ay. A = {SM, SaS} n (A) = 2

P(A) =)S(n)A(n =

72

17171717.... TheTheTheThe probabilityprobabilityprobabilityprobability thatthatthatthat atatatat leastleastleastleast oneoneoneone ofofofof thethethethe eventseventseventsevents AAAA andandandand BBBB occursoccursoccursoccurs isisisis 0.6.0.6.0.6.0.6. IfIfIfIf AAAA andandandand BBBB occuroccuroccuroccursimultaneouslysimultaneouslysimultaneouslysimultaneously withwithwithwith probabilityprobabilityprobabilityprobability 0.2,0.2,0.2,0.2, evaluateevaluateevaluateevaluate P(A)P(A)P(A)P(A) ++++ P(B)P(B)P(B)P(B)

Solution:Solution:Solution:Solution:P(A B) = 0 .6P(A B) = 0 .2P(A B) = P(A) + P(B) P (A B)

0 .6 = P(A) + P(B) 0 .2 0 .6 + 0 .2 = P(A) + P(B) 0 .8 = P(A) + P(B)


33333333

P(A) + P(B) = 0 .8

18181818.... ThreeThreeThreeThree horseshorseshorseshorses A,A,A,A, BBBB andandandand CCCC areareareare inininin aaaa race.race.race.race. AAAA isisisis twicetwicetwicetwice asasasas likelylikelylikelylikely totototo winwinwinwin asasasas BBBB andandandand BBBB isisisis twicetwicetwicetwice asasasas likelylikelylikelylikely totototowinwinwinwin asasasas C.C.C.C. WhatWhatWhatWhat areareareare theirtheirtheirtheir probabilitiesprobabilitiesprobabilitiesprobabilities ofofofof winningwinningwinningwinning ????

Solution:Solution:Solution:Solution:P(A) = 2 P(B)P(B) = 2 P(C)

P(A) = 2 2 P(C)= 4 P(C)P(A) + P(B) + P(C) = 1

4 P(C) + 2 P(C) + P(C) = 1 7 P(C) = 1

P(C) =71

Hen ce , th e even ts are m u tu a lly exclu s ive . P(A) = 4 P(C)

=74

P(B) = 2 P(C)

=72

P(C) =71

19191919.... SavitaSavitaSavitaSavita andandandand HamidaHamidaHamidaHamida areareareare friends,friends,friends,friends, whatwhatwhatwhat isisisis thethethethe probabilityprobabilityprobabilityprobability thatthatthatthat bothbothbothboth withwithwithwith havehavehavehavei)i)i)i) differentdifferentdifferentdifferent birthdays.birthdays.birthdays.birthdays.ii)ii)ii)ii) thethethethe samesamesamesame birthdaybirthdaybirthdaybirthday (ignoring(ignoring(ignoring(ignoring aaaa leapleapleapleap year)year)year)year)

Solution:Solution:Solution:Solution:Sin ce th ere are 3 6 5 d ays in a year an d th ere are two b irth d ays of Savita an d Ham id a .

Th e n u m b er of ou tcom es for th e sam p le sp ace for th e ir b irth d ays= 3 6 5 3 6 5

n (S) = 3 6 5 3 6 5i) Let ‘A’ b e th e even t th a t Savita an d Ham id a h ave d iffe ren t b irth d ays .If th ey h ave th e sam e b irth d ay, th en th ere is on ly on e ou tcom e .Bu t if th ey h ave d iffe ren t b irth d ays , n u m b er of ou tcom es in th is even t = (3 6 5 1 ) 3 6 5

= 3 6 4 3 6 5 n (A) = 3 6 4 3 6 5

P(A) =)S(n)A(n

=3 6 53 6 53 6 53 6 4

=

3 6 53 6 4

ii) Let B b e th e even t th a t th e ir b irth d ays are on th e sam e d ay.Hen ce ‘A’ an d ‘B’ are com p lem en tary even ts .

P(B) = 1 P(A)

= 1 3 6 53 6 4

====3 6 5

1

22220000.... EachEachEachEach coefficientcoefficientcoefficientcoefficient inininin equationequationequationequation axaxaxax2222 ++++ bxbxbxbx ++++ cccc ==== 0000 isisisis obtainedobtainedobtainedobtained bybybyby throwingthrowingthrowingthrowing aaaa fairfairfairfair die.die.die.die. FindFindFindFind thethethetheprobabilityprobabilityprobabilityprobability thatthatthatthat thethethethe equationsequationsequationsequations havehavehavehave realrealrealreal andandandand equalequalequalequal roots.roots.roots.roots.

Solution:Solution:Solution:Solution:ax 2 + b x + c = 0 is th e q u ad ra t ic eq u atio n .Its roo ts will b e rea l an d eq u al if b 2 4 ac = 0 or b 2 = 4 acLet ‘S’ b e th e even t of toss in g a d ie th rice to g e t th e va lu er of a , b an d c.

S = {(1 , 1 , 1 ) (1 , 1 , 2 ) (1 , 1 , 3 )…..(6 , 6 , 5 )}For a = 1 , th ere will b e 3 6 ou tcom es for va lu es of b an d c.


Hen ce to ta l n u m b er of ou tcom es n (S) = 6 3 6= 2 1 6

Let ‘A’ b e th e even t th a t th e va lu es of a , b an d c g ive rea l an d eq u al roo ts .A ==== {(1 , 2 , 1 ), (2 , 4 , 2 ), (1 , 4 , 4 ), (4 , 4 , 1 ), (3 , 6 , 3 )}

n (A) = 5Th e p rob ab ility th a t th e q u ad ra t ic eq u atio n h as rea l roo ts .

2165

)S(n)A(n)A(P

��

StatisticsStatisticsStatisticsStatistics1111.... FindFindFindFind thethethethe meanmeanmeanmean ofofofof 1,1,1,1, 2,2,2,2, 3,3,3,3, 4,4,4,4, ……………………,,,, n.n.n.n.Solution:Solution:Solution:Solution:

Mean =n

n...4321

(1 + 2 + 3 + 4 + … + n ) is in APa = 1 d = 1 t n= n .

Sn =2n [2 a + (n 1 )d ]

=2n [2 + (n 1 )]

Sn =2n (n + 1 )

Mean =

n2

1nn

Mean =n

1n .

2222.... TheTheTheThe meanmeanmeanmean ofofofof nnnn observationobservationobservationobservation isisisis x .... IfIfIfIf everyeveryeveryevery observationobservationobservationobservation isisisis increasedincreasedincreasedincreased bybybyby 5,5,5,5, whatwhatwhatwhat wouldwouldwouldwould bebebebe thethethethe mean?mean?mean?mean?Solution:Solution:Solution:Solution:

Mean = ( x + 5 )

3333.... FindFindFindFind thethethethe missingmissingmissingmissing frequenciesfrequenciesfrequenciesfrequencies inininin thethethethe followingfollowingfollowingfollowing frequencyfrequencyfrequencyfrequency distributiondistributiondistributiondistribution ifififif itititit isisisis knownknownknownknown thatthatthatthat thethethethe meanmeanmeanmeanofofofof thethethethe distributiondistributiondistributiondistribution isisisis 1.1.1.1. 46.46.46.46.

N0 . of accid en ts (x ): 0 1 2 3 4 5 to ta lFreq u en cy (f): 4 6 ? ? 2 5 1 0 5 2 0 0 .Solution:Solution:Solution:Solution:

Let th e m is s in g freq u en cies b e f1 an d f2 .Com p u ta t ion of Arith m etic Mean

x i fi fi x i

0 4 6 01 f1 f1

2 f2 2 f2

3 2 5 7 54 1 0 4 05 5 2 5


35353535

N = 8 6 + f1

+ f2

fix i = 1 4 0 +f1 + 2 f2

We h ave ,N = 2 0 0 2 0 0 = 8 6 + f1 + f2 f1 + f2 = 1 1 4 …..(i)Als o ,Mean = 1 .4 6

1 .4 6 =N

xf ii

1 .4 6 =2 0 0

f2f1 4 0 21

2 9 2 = 1 4 0 + f1 + 2 f2 f1 + 2 f2 = 1 5 2…. (ii)

So lvin g eq u atio n (i) an d (ii), we g etf1 = 7 6 an d f2 = 3 8 .

4444.... TheTheTheThe meanmeanmeanmean ofofofof thethethethe followingfollowingfollowingfollowing frequencyfrequencyfrequencyfrequency tabletabletabletable 50.50.50.50. ButButButBut thethethethe frequenciesfrequenciesfrequenciesfrequencies ffff1111 andandandand ffff2222 inininin classclassclassclass 20-4020-4020-4020-40 andandandand 60-60-60-60-80808080 areareareare missing.missing.missing.missing. FindFindFindFind thethethethe missingmissingmissingmissing frequencies.frequencies.frequencies.frequencies.

Clas s : 0 - 2 0 2 0 - 4 0 4 0 - 6 0 6 0 - 8 0 8 0 - 1 0 0F. : 1 7 , fi , 3 2 , fi , 1 9

Tota l1 2 0

Solution:Solution:Solution:Solution:

We h ave ,N = fi = 1 2 0 [Given ]

6 8 + f1 + f2 = 1 2 0 f1 + f2 = 5 2 ….(i)

Now,Mean = 5 0

A + h

iiuf

N1 = 5 0

5 0 + 2 0

1 2 0ff4 21 = 5 0

5 0 +6

ff4 21 = 5 0

6

ff4 21 = 0

4 f1 + f2 = 0 f1 f2 = 4 ….(ii)

So lvin g eq u atio n s (i) an d (ii), we g et f1 = 2 8 an d f2 = 2 4 .

5555.... IfIfIfIf thethethethe medianmedianmedianmedian ofofofof thethethethe distributiondistributiondistributiondistribution givengivengivengiven belowbelowbelowbelow isisisis28.5,28.5,28.5,28.5, findfindfindfind thethethethe valuevaluevaluevalue ofofofof xxxx andandandand y.y.y.y.

Clas s in te rva l: 0 - 1 0 ,1 0 - 2 0 ,2 0 - 3 0 ,3 0 - 4 0 , 4 0 - 5 0 , 5 0 - 6 0 , to ta l.No .of s tu d en ts : 5 , x , 2 0 , 1 5 , y , 5 , 6 0Solution:Solution:Solution:Solution:

Com p u ta t io n of Med ian

Classin tervals

Freq u en cy (f) Cu m ula tiveFr. (Cf)

0 - 1 0 5 5

1 0 - 2 0 X 5 + x2 0 - 3 0 2 0 2 5 + x3 0 - 4 0 1 5 4 0 + x

Clas s Freq u en cyfi

Mid -va lu es

x i

u i=

hAx i

fiui

0 - 2 0 1 7 1 0 2 3 4

2 0 -4 0

fi 3 0 - 1 - fi

4 0 -6 0

3 2 5 0 0 0

6 0 -8 0

fi 7 0 1 f2

8 0 -1 0 0

1 9 9 0 2 3 8

N = fi = 6 8 + f1 + f2 fiu i = 4 f1 + f2


4 0 - 5 0 Y 4 0 + x + y5 0 - 6 0 5 4 5 + x + y

f1 = 6 0

We h ave ,Med ian = 2 8 .5

Clearly, it lies in th e class in te rva l 2 0 - 3 0 . So , 2 0 - 3 0 is th e m ed ian class . l = 2 0 , h = 1 0 , f = 2 0 , F= 5 + x an d N = 6 0

Now,

Med ian = l +f

F2N

h

2 8 .5 = 2 0 + 2 0

x53 0 1 0

2 8 .5 = 2 0 +2

x2 5

8 .5 =2

x2 5 2 5 x = 1 7

x = 8We h ave ,

N = 6 0 4 5 + x + y = 6 0 x + y = 1 5

Pu tt in g x = 8 in x + y = 1 5 , we g et y = 7Hen ce , x = 8 an d y = 7 .

6666.... AAAA lifelifelifelife insuranceinsuranceinsuranceinsurance agentagentagentagent foundfoundfoundfound thethethethe followingfollowingfollowingfollowing datadatadatadata forforforfor distributiondistributiondistributiondistribution ofofofof agesagesagesages ofofofof 100100100100 policypolicypolicypolicy holders.holders.holders.holders.CalculateCalculateCalculateCalculate thethethethe medianmedianmedianmedian age,age,age,age, ifififif policiespoliciespoliciespolicies areareareare onlyonlyonlyonly givengivengivengiven totototo personpersonpersonperson havinghavinghavinghaving ageageageage 18181818 yearsyearsyearsyears onwardsonwardsonwardsonwards butbutbutbutlesslesslessless thanthanthanthan 60606060 year.year.year.year.

Ag e (in years ) No. of p licy h o ld ersBelow 2 0 2Below 2 5 6Below 3 0 2 4Below 3 5 4 5Below 4 0 7 8Below 4 5 8 9Below 5 0 9 2Below 5 5 9 8Below 6 0 1 0 0

Solution:Solution:Solution:Solution: Let u s firs t con s tru ct a freq u en cy d is t rib u tion tab le .

Class in terva l Freq u en cy C.F

1 5 - 2 0 2 22 0 - 2 5 4 62 5 - 3 0 1 8 2 43 0 - 3 5 2 1 4 53 5 - 4 0 3 3 7 84 0 - 4 5 1 1 8 94 5 - 5 0 3 9 25 0 - 5 5 6 9 85 5 - 6 0 2 1 0 0

Here n = 1 0 0 .

So ,2n =

21 0 0 = 5 0


37373737

Th is ob s erva tion lies in th e class 3 5 - 4 0l = 3 5 , n / 2 = 5 0 , f = 3 3 , c.f = 4 5Ap p lyin g th e form u la Med ian =

l +f

f.c2n

h . , we g et

Med ian = 3 5 +3 3

4 55 0 5 = 3 5 +

3 32 5

Hen ce th e m ed ian ag e = 3 5 .7 6 year.

7777.... TheTheTheThe medianmedianmedianmedian andandandand modemodemodemode ofofofof thethethethe followingfollowingfollowingfollowing wagewagewagewage distributiondistributiondistributiondistribution areareareare knownknownknownknown totototo bebebebe Rs.Rs.Rs.Rs. 33.533.533.533.5 andandandand Rs.Rs.Rs.Rs. 34343434respectively.respectively.respectively.respectively. ThreeThreeThreeThree frequencyfrequencyfrequencyfrequency valuesvaluesvaluesvalues fromfromfromfrom thethethethe tabletabletabletable areareareare howeverhoweverhoweverhowever missing.missing.missing.missing. YouYouYouYou requiredrequiredrequiredrequired totototo findfindfindfindoutoutoutout thesethesethesethese values.values.values.values.

Wag es in Rs . No. of p erson s0 - 1 0 41 0 - 2 0 1 62 0 - 3 0 ?3 0 - 4 0 ?4 0 - 5 0 ?5 0 - 6 0 66 0 - 7 0 4Tota l 2 3 0

Solution:Solution:Solution:Solution:Let th e m is s in g freq u en cies of th e class in te rva ls 2 0 - 3 0 , 3 0 - 4 0 an d 4 0 - 5 0 b e f1, f2 an d f3 res p ective ly.Sin ce ,fi = 2 3 0

f1 + f2 + f3 = fi (4 + 1 6 + 6 + 4 ) = 2 3 0 3 0 = 2 0 0As m ed ian Rs . 3 3 .5 , so 3 0 - 4 0 is th e m ed ian class

Med ian = l +f

f.c2n

h = 3 0 +2

1

f)f1 64(1 1 5 1 0

3 3 .5 = 3 0 +2

1

ff9 5

1 0

3 .5 f2 = 9 5 0 1 0 f1 1 0 f1 + 3 .5 f2 = 9 5 0 …..(ii)Ag ain m od e is 3 4 , th erefore 3 0 - 4 0 is th e m od al class an d th erefore

Mod e = l +201

01

fff2ff

h

3 4 = 3 0 + 312

12

fff2ff

1 0

4 = 312

12

fff2ff1 0

2 (f2 f1 f3) = 5 f2 5 f1

3 f1 f2 2 f3 = 0 …….(iii)Now from (i), we h ave f3 = 2 0 0 – f1- f2

Su b s titu t in g th is va lu e of f3 in (iii), we g et .

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Arithmetic Progression - Nayaks TutorialsArithmetic Progression 1 . Determine the 10th term from the...

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Transcript of Arithmetic Progression - Nayaks TutorialsArithmetic Progression 1 . Determine the 10th term from the...