Cat Solved Paper. Fresh Paper. No. of Questions : 50 Time : 40 min

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1 t Solved Pper Fresh Pper No. of Questions : 50 Time : 0 min Note Ech wrong nswer crry rd negtive mrk. irections for question number to 5 : nswer the questions independently of ech other The infinite sum equls : () 7 (b) (c) 9 7 (d) onsider the sets Tn { n, n, n, n, n }, where n,,, 96. How mny of these sets contin 6 or ny integrl multiple there of (i. e., ny one of the numbers 6,, 8... )? () 80 (b) 8 (c) 8 (d) 8. Let EF be regulr hegon. Wht is the rtio of the tringle E to tht of the hegon EF? () (c) (b) (d) 5 6. The number of roots common between the two equtions 5 0 nd 7 0 : () 0 (b) (c) (d) 5. rel number stisfying, for every n n positive integer n, is best described by : () (b) (c) 0 (d) irections for question number 6, 7, nd 8 : nswer the questions on the bsis of the tbles given below. Two binry opertions nd * re defined over the set {, e, f, g, h} s per the following tbles : e f g h e f g h e e f g h f f g h e g g h e f h h e f g * e f g h e e f g h f f h e g g g e h f h h g f e Thus ccording to the first tble f g, while ccording to the second tble g h f, nd so on. lso, let f f f, g g g g, nd so on. 6. Wht is the smllest positive integer n such tht g n e? () (b) 5 (c) (d) 7. Upon simplifiction f [ f { f ( f f )} ] equls : () e (b) f (c) g (d) h Upon simplifiction { ( f g )} e equls : () e (c) g (b) f (d) h irections for question number of 9 nd 0 : nswer the questions on the bsis of the informtion given below. string of three English letters is formed s per the following rules : () The first letter is ny vowel. (b) The second letter is m, n or p. (c) If the second letter is m then the third letter is ny vowel which is different from the first letter. (d) If the second letter is n then the third letter is e or u. (e) If the second letter is p then the third letter is the sme s the first letter. 9. How mny strings of letters cn possibly be formed using the bove rules? () 0 (b) 5 (c) 0 (d) 5 0. How mny strings of letters cn possibly be formed using the bove rules such tht the third letter of the string is e? () 8 (b) 9 (c) 0 (d) irections for question number to 5 : nswer the questions independently of ech other.. Wht is the reminder when 96 is divided by 6? () 0 (b) (c) (d). If nd y re integers then the equlion 5 9y 6 hs : () no solution for 00 nd y 0 (b) no solution for 50 nd y 00 (c) solution for (d) solution for 59 y 56. If log M log N log , then : () M (c) M 9 9 N N (b) N (d) N 9 9 M 9 M

2 Solved Pper 00 (Fresh Pper). Using only, 5, 0, 5 nd 50 pise coins, wht will be the minimum number of coins required to py ectly 78 pise, 69 pise nd Re..0 to three different persons? () 9 (b) 0 (c) 7 (d) 8 5. The length of the circumference of circle equls the perimeter of tringle of equl sides, nd lso the perimeter of squre. The res covered by the circle, tringle, nd squre re c, t nd s respectively. Then : () s t c (b) c t s (c) c s t (d) s c t irection for question number 6 to 8 : nswer the questions on the bsis of the informtion given below. The seven bsic symbols in certin numerl system nd their respective vlues re s follows : T,V 5, X 0, L 50, 00, 500 nd M 000 In generl, the symbols in the numerl system re red from left to right, strting with the symbol representing the lrgest vlue; the sme symbol cnnot occur contiguously more thn three times; the vlue of the numerl is the sum of the vlues of the symbols. For emple, XXVII n eception to the left-to-right reding occurs when symbol is followed immeditely by symbol of greter vlue; then, the smller vlue is subtrcted from the lrger. For emple, XLVI ( 50 0) The vlue of the numerl MLXXXVII is : () 687 (b) 787 (c) 887 (d) The vlue of the numerl MMXIX is : () 999 (b) 899 (c) 989 (d) Which of the following cn represent the numerl for 995? () MMLXXV () MMXV () MV () MVM () only () nd () (b) only () nd () (c) only () nd () (d) only () irections for question number 9 to : nswer the questions independently of ech other. 9. There re towns grouped into four zones with three towns per zone. It is intended to connect the towns with telephone lines such tht every two towns re connected with three direct lines if they belong to the sme zone, nd with only one direct line otherwise. How mny direct telephone lines re required? () 7 (b) 90 (c) 96 (d) 0. In the figure (not drwn to scle) given below, P is point on such tht P : P :. PQ is prllel to nd Q is R S P Q prllel to P. In R, R 90, nd in PQS, PSQ 90. The length ofqs is 6 cms. Wht is rtio P : P? () 0 : (b) : (c) 7 : (d) 8 :. cr is being driven, in stright line nd t uniform speed, towrds the bse of verticl tower. The top of the tower is observed from the cr nd, in the process, it tkes 0 minutes for the ngle of elevtion to chnge from 5 to 60. fter how much time will this cr rech the bse of the tower? () 5( ) (b) 6 ( ) (c) 7 ( ) (d) 8 ( ). In the figure (not drwn to scle) given below, if nd E 96, how much is? () (b) 8 (c) 6 (d) cn t be determined. If both nd bbelong to the set{,,, }, then the number of equtions of the form b 0 hving rel roots is : () 0 (b) 7 (c) 6 (d). If log 0 log0 log 0, then possible vlue of is given by : () 0 (b) 00 (c) 000 (d) none of these 5. Wht is the sum of ll two digit numbers tht give reminder of when they re divided by 7? () 666 (b) 676 (c) 68 (d) In the figure given below (not drwn to scle), nd re three points on circle with centre O. The chord is etended to point T such tht T becomes tngent to the circle t point. If T 0 nd T 50, then the ngle O is : O 50 () 00 0 (b) 50 T (c) 80 (d) not possible to determine 7. Wht is the sum of n terms in the series : m m m log m log log log n n n n () log m m (c) log n ( n ) ( n ) ( n) ( n) n/ n/ m (b) log n m (d) log n m n/ n ( n ) ( n ) n/ E 96

3 Solved Pper 00 (Fresh Pper) 5 8. Let S be squre of side. nother squre S is formed by joining the mid points of the sides of S. The sme process is pplied to S to form yet nother squre S nd so on. If,, be the res nd P, P, P, be the perimeters,, respectively, then the rtio P P P of S S S equls : () ( ) (c) ( ) (b) ( ) (d) ( ) from points, nd respectively. Ech sprinter trverses her respective tringulr pth clockwise nd returns to her strting point. 9. If three positive rel numbers, y, z stisfy y z y nd yz, then wht is the minimum possible vlue of y? () / (b) / (c) / (d) / 0. n intelligence gency forms code of two distinct digits selected from 0,,, 9 such tht the first digit of the code is non zero. The code hndwritten on slip, cn however potentilly crete confusion when red upside down. For emple, the code 9 my pper s 6. How mny codes re there for which no such confusion cn rise? () 80 (b) 78 (c) 7 (d) 69. onsider two different cloth cutting processes. In the first one, n circulr cloth pieces re cut from squre cloth piece of side in the following steps : the originl squre of side is divided into n smller squres, not necessrily of the sme size, then circle of mimum possible re is cut from ech of the smller squres. In the second process, only one circle of mimum possible re is cut from the squre of side nd the process ends here the cloth pieces remining fter cutting the circles re scrpped in both the processes. The rtio of the totl re of scrp cloth generted in the former to tht in the ltter is : () : (b) : (c) n ( ) (d) n n n( ). In the figure below (not drwn to scle), rectngle is inscribed in the circle with centre t O. The length of side is greter thn tht of side. The rtio of the re of the circle to the E re of the rectngle is O :. The line segment E intersects t E such tht O E. Wht is the rtio E :? () : (b) : (c) : (d) : irections for question number to 5 : nswer the questions on the bsis of the informtion given below. onsider three circulr prks of equl size with centres t, nd respectively. The prks touch ech other t the edge s showm in the figure (not drwn to scle). There re three pths formed by the tringles, nd s shown. Three sprinters, nd begin running. Let the rdius of ech circulr prk be r, nd the distnces to be trversed by the sprinters, nd be, b nd c, respectively. Which of the following is true? () b c b r (b) b c b r c (c) b ( ) r (d) c b ( ) r. Sprinter trverse distnces,, t verge speeds of 0, 0 nd 5, respectively. trverses her entire pth t uniform speed of ( 0 0). trverses distnce 0, nd t verge speeds of ( ), 0 ( ) nd 0, respectively. ll speeds re in the sme unit. Where would nd be respectively when finishes her sprint? (), (b), (c) (d), somewhere between nd 5. Sprinters, nd trverse their respective pths t uniform speeds of u, v nd w respectively. It is known tht u : v : w is equl to re : re : re, where re, re nd re re the res of tringles, nd, respectively. Where would nd be when reches point? (), (b), (c), (d) Somewhere between nd, somewhere between nd irections for question number 6 to 8 : nswer the questions on the bsis of the informtion given below. onsider cyclinder of height h cms nd rdius r cms s shown in the figure (not drwn to scle). string of certin length when wound on its cylindricl surfce strting t point nd ending t point, gives mimum of n turns (in other words, the string s length is the minimum length required to wind n turns).

4 6 Solved Pper 00 (Fresh Pper) 6. Wht is the verticl spcing in cms between two consecutive turns? () h n (b) h n (c) h n (d) cnnot be determined with the given informtion 7. The sme string, when would on the eterior four wlls of cube of side n cms, strting t point nd ending t point, cn give ectly on turn (see figure, not drwn to scle). The length of the string, in cms, is : () n (b) 7n (c) n (d) n 8. In the setup of the previous two questions, how is h relted to n? () h n (b) h 7 n (c) h n (d) h n irections for question number 9 to 50 : nswer the questions independently of ech other. 9. Let nd y be positive integers such tht is prime nd y is composite. Then, () y cnnot be n even number (b) y cnnot be n even integer (c) ( y) cnnot be n even integer (d) none of the bvoe sttements is true 0. survey on smple of 5 new crs being sold t locl uto deler ws conducted to see which of the three populr options : ir conditioning, rdio nd power windows were lredy instlled. The survey found : 5 hd ir conditioning hd ir conditioning nd power windows but no rdios hd rdio 6 hd ir conditioning nd rdio but no power windows hd power windows hd rdio nd power windows hd ll three options Wht is the number of crs tht hd none of the options? () (b) (c) (d). If n is such tht 6 n 7, then n n( n ) 6 stisfies : n n () 0 5 (b) 58 (c) 5 6 (d) If z nd z 5y, then : () is necessrily less thn y (b) is necessrily greter thn y (c) is necessrily equl to y (d) none of the bove is necessrily true n. Let n ( ) be composite integer such tht n is not n integer. onsider the following sttements : ()n hs perfect integer vlued divisor which is greter thn nd less thn n () n hs perfect integer vlued divisor which is greter thn n but less thn n. Then : () both nd re flse (b) is true but is flse (c) is flse but is true (d) both nd re true. If b nd b, then which one of the following is necessrily true? () b 0 (b) b 0 (c) b 0 (d) b 0 5. piece of pper is in the shpe of right ngled tringle nd is cut long line tht is prllel to the hypotenuse, leving smller tringle. There ws 5% reduction in the length of the hypotenuse of the tringle. If the re of the originl tringle ws squre inches before the cut, wht is the re (in squre inches) of the smller tringle? () (b) (c) 5.65 (d) In costl villge, every yer floods destroy ectly hlf of the huts. fter the flood wter recedes, twice the number of huts destroyed re rebuilt. The floods occurred consecutively in the lst three yers nmely 00, 00 nd 00. If floods re given epected in 00, the number of huts epected to be destroyed is : () less thn the number of huts eisting t the begining of 00 (b) less then the totl number of huts destroyed by floods in 00 nd 00 (c) less thn the totl number of huts destroyed by floods in 00 nd 00 (d) more thn the totl number of huts built in 00 nd Let, b, c, d nd e be integers such tht 6b c nd b 9d e. Then which of the following pirs contins number tht is not n integer? b () 7, c (b) e 6, e bd c (c), (d) 8 6, d 8. If, nd re prime numbers, then the number of possible solutions for is : () one (b) two (c) three (d) more thn three 9. squre tin sheet of side inches is converted into bo with open top in the following steps : The sheet is plced horizontlly; then equl sized squres, ech of side inches, re cut from the four corners of the sheet; Finlly, the four resulting sides re bent verticlly upwrds in the shpe of bo. If is n integer, then wht vlue of mimizes the volume of the bo? () (b) (c) (d)

5 Solved Pper 00 (Fresh Pper) Two stright rods R nd R diverge from point t n ngle of 0. Rm strts wlking from point long R t uniform speed of km/hr. Shym strts wlking t the sme time from long R t uniform speed of km/h. They continue wlking for hours long their respective rods nd rech points nd on R nd R, respectively. There is stright line pth connecting nd. Then Rm returns to point fter wlking long the line segments nd. nswers Shym lso returns to fter wlking long line segments nd. Their speeds remin unchnged. The time intervl (in hours) between Rm s nd Shym s return to the point is : () (b) (c) 9 6 (d) 9 0 (c). (). (b). () 5. (c) 6. () 7. (d) 8. () 9. (d) 0. (c). (d). (c). (b). () 5. (c) 6. (b) 7. () 8. (c) 9. (b) 0. (c). (). (c). (b). (b) 5. (b) 6. () 7. (d) 8. (c) 9. (b) 0. (c). (). (). (). (c) 5. (b) 6. () 7. (b) 8. (c) 9. (d) 0. (d). (c). (d). (d). (b) 5. (d) 6. (c) 7. (d) 8. () 9. (d) 50. (b) Let S () S S () S 7S S 0 [( ) ( )] S S S 6S S Hints & Solutions 96 S S 9 7 {using sum of infinite GP} lterntively : Let S Putting, we get 7 S 5 5 S S S ( S S) 5 7 ( S S) { ( S S)} ( ) S ( ) S ; S ( ) (This cn be used s direct formul for solving this type of problem.) Now substituting, we get 7 S 9 7. Substituting the vlue of n,,, 96 in T n, we get the elements of T, T, T T96. e. g., T {,,,, 5} T {,,, 5, 6} T {,, 5, 6, 7} etc. y observtion we see tht for n, 7,, 9, 5,, 7, 9, there is no element int, T7, T T9 which contin 6 or ny integerl multiple of 6. Thus there re totl 6 sets ( T, T7, T T9) out of 96 sets which do not contin ny element 6 or its integrl multiple. Hence, there re only sets which contin 6 or ny integrl multiple.. EF is regulr hegon. Joining the centre O with vertices, E,, we get OE, O, EO. Since E EF F O EO O nd E EF EF O OE OE 0

6 8 Solved Pper 00 (Fresh Pper) Thus the O () E OE n n F Hence the best possible rnge of is described by 0. EF EO 6. g g g h Hence, the re of E O re of hegon EF E lterntively : In the given figure of hegon EF, there re smller congruent tringles with the ngles Out of these tringles 6 tringles re contined in the lrger tringle E. F E Hence, the rtio of re of E to re of hegon lterntively : E 0 E E 0 Let ech side of the hegon be, then E (using cosine rule or sine rule) re of E ( ) nd re of hegon 6 O 6 re of E re of hegon EF. Let be the common root. 5 0 () nd 7 0 () compring the two equtions, we get ( )( ) 0, No. of possible common roots between two equtions but when, re substituted in equtions () nd () none of them stisfies the equtions. Hence, the number of common roots 0 5. Since n is positive integer, therefore substituting the vlue of lest positive integer, we get 0 If n is very lrge i. e., n then 0 n F E () n g g g h g f g g g f g e 7. f [ f { f ( f f )} ] f [ f { f h} ] ( f f h) f [ f e] ( f h e) 8. 0 ( n or ) Similrly, e 8 e Now, f f f h f f ( f e f ) h ( f f h) f f f h f g f f f g f e 5 f f f e f f f f f f f h 0 lterntively : f f f f f f e Similrly, g 9 h h h h h h h h h e e h h h { ( f g )} e { ( h h)} e { g} e e e 9. There re eclusive cses : (i) when m is the second letter (ii) when n is the second letter (iii) when p is the second letter se (i) : First letter cn be selected in 5 wys out of 5 vowels. Since second letter is fied (i. e., m) therefore no. of wys of selection of second letter is. Third letter cn be selected in wys out of remining vowels (since the vowel which hs been used t first plce cn not be used t the plce of third letter. Totl no. of wys 5 0 se (ii) : First letter cn be selected in 5 wys. Second letter cn be selected in wy. Third letter cn be selected in wys out of e, u. Totl no. of wys 5 0 se (iii) : First letter cn be selected in 5 wys. Second letter cn be selected in wy. Third letter cn be selected in wy since third letter will be sme s the first letter. So if the first letter is selected, then there is no need to select the third letter. Totl no. of wys 5 5 Hence, sum of ll the possible no. of wys in which the string of letters cn be formed

7 Solved Pper 00 (Fresh Pper) 9 0. st Letter nd Letter rd Letter No. of wys, i, o, u m e, e, i, o, u n e 5 5 e p e. If is divided by 6, reminder is If is divided by 6, reminder is If is divided by 6, reminder is If is divided by 6, reminder is If 5 is divided by 6, reminder is If 96 is divided by 6, reminder is Totl no. of wys y 6 () 6 9y 5 Following tbles shows some vlues of nd y which stisfy the given eqution (). y y () () hoice () is wrong since for 00 nd y 0, there eists vrious solutions. hoice (b) is wrong since for 50 nd y 00 there eists vrious solutions e. g., for 56, y 6 nd 75, y 69 etc. hoice (d) is wrong since t y 58 nd y 57 there is no solution (see the ptter for y is tble ). hoice (c) is correct since there eists solution for 56, 75 etc.. LHS log M log log M / log / log ( M N ) N N RHS log log log 5 5 log5 5 log5 5 log5 5 LHS RHS / log ( M N ) M N / / (cubing both the sides) Vlue of MN 9 9 coins 9 9 mount N M required. In order to require minimum no. of coins we need to hve the coins of greter denomintions. 50 pise No. of oins of 5 pise 0 pise 5 pise pise Totl No. of oins 78 pise 7 69 pise 5 0 pise 7 Thus, the required no. of coins 9. Totl 9 5. For the given perimeter, the polygon with mimum no. of sides hs mimum re. Hence c s t. NOTE circle hs infinite no. of sides. lterntively : Let the circumference of the circle, perimeter of equilterl tringle nd perimeter of squre be sme nd equl to cm ( LM of,, ) Rdius of circle cm Ech side of cm nd ech side of squre cm re of circle r ( ) 86 cm 7 re of tringle ( ) cm nd re of squre ( ) 089 c s t lterntively : Let circumference of the circle Then = perimeter of tringle = perimeter of squre P rdius of circle P ech side of P ech side of Squre P re of circle P r re of tringle P 7 7 P 88 P P nd re of squre P P 6

8 0 Solved Pper 00 (Fresh Pper) 7 Since c s t 6. MLXXXVII MMXIX 000 ( ) ( 00 0) ( 0 ) () MMLXXV 000 ( ) (b) MMXV 000 ( ) ( 00 0) (c) MV 000 ( 500 5) 95 (d) MVM 000 ( 000 5) 995 Hence, (c) is the right choice. 9. Since ll the towns re connected with ech other with t lest one direct line. Totl no. of single direct lines mong towns 66 ut the towns which belong to the sme zone re connected by direct lines. So we need to connect towns of ech zone with more direct lines (since they re lredy connected with single direct line) No. of dditionl lines in one zone 6 Totl no. of dditionl lines in four zones 6 Required number of direct telephone lines lterntively : onsider prticulr zone hving towns. Ech town is directly connected with lines inside the sme zone. Therefore totl number of lines required for one zone 9 Hence, in four zones totl no. of required lines for internl E F Given tht P : P : Since ~ PQ ( PQ, is common) P : y : y ( Q P ) (s P divides in the rtio : so divides P in the sme rtio : ) ut P P y y 7 y y 7 we re required to find P P y y. Let T be the tower. Initilly cr is t where cr mkes 5 ngle nd it reches, where it mkes 60 with the top of the tower. 7 Let, y nd T h h In T, tn 5 y Q P 5 60 y h y h y () Q P h connections 9 6 Ech town in first zone cn be connected to towns in the second zone. Therefore between two prticulr zones totl no. of required lines 9 [, E, F;, E, F;, E, F] Therefore totl no. of lines required for connecting towns of different zones (Since out of zones ny two zones cn be selected in wys.) Hence, the totl no. of lines In T, tn 60 h y h y h y () From () nd () y y y ( ) Since the speed of cr is constnt, therefore the time tken by cr is directly proportionl to the distnce covered. y T T y ( ) (T time)

9 Solved Pper 00 (Fresh Pper) To cover m distnce cr tkes 0 min To cover y m distnce cr will tke 0 ( 0 ( ) ( ) ( ) 5( )min min ). Let ( ) (eterior ngle) ( ) Let y y 8 () nd in, 80 () y y y 80 () from equtions () nd (), we get, y 5 nd 6. b c 0 hs rel roots if b c 0 b 0 will hve rel roots if b c 0 or b (here c ) Vlue of orresponding Vlue of b for b No. of Wys,,, Totl 7 Hence totl no. of wys 7. log 0 log0 log 0 log 0 log0 log log 0 (log ) 0 0 log log 0 0 or 0 or Hence (b) is the correct option. 5. The required two digit number is of the form 7n. These numbers re s follows : 0, 7,,, 8, 9 The first number is obtined by substituting n nd lst number is obtined by substituting n, hence there re such numbers. Sum of these numbers S 676 n l n 6. T T nd T (ngle in lternte segment) 50 Hence, 80 ( ) 80 ( ) 50 Now, since O ( ) (using theorem) O 00 lterntively : O ( is n eterior ngle of T) ( O is the centrl ngle nd twice of ) gin OT 90 (T is tngent) O 0 O O 0 ( O O ) O 80 ( O O) O 00 O O O 60 O 60 ( 60 00) O 00 lterntively : O O 0 O O O O 0 O 80 ( 0 0 ) O 00 ( O O O 80 ) m m m 7. log m log log log n n n nth term m log n m log n ( n) ( ( n )) ( n ) ( n ) n/ n m m m m m log n n n n n n m log n 8. Let ech side of S be. O n( n ) ( n ) n 50 0 T s s 5 s s 6 s s

10 Solved Pper 00 (Fresh Pper) Now, since ech net squre i. e., S, S, S etc. is formed by joining the mid points of the sides of its previous squre i. e., S, S, S etc. Hence the ech side of the net squre will be times the side of the previous squre, respectively e.g., Side of S ( side of S ) Side of S Side of S ( side of S ) ( side of S ) etc. Thus if,, n be the sides of the squres S, S, S, S n respectively Then (sy) etc. P, P, P, P, nd,,,, 8 P P P ( ) ( ) nd 8 P P P ( ) ( ) ( ) ( ) 9. Since y z y, y nd z re in P Let d be the common difference of the P Therefore y d nd z y d yz ( y d) y( y d) y( y d ) y will be minimum when y y d will be mimum when d is 0. y( y d ) d will be mimum nd y( y 0) y y / (tking cube root of both sides) 0. Totl number of two digit numbers hving distinct digits There re four digits which crete confusion (ecept to 0) :, 6, 8, 9. From these four digits we cn form two digit codes which cn crete confusion ( digits re distinct) ut out of these numbers numbers re 69 nd 96 which do not crete confusion. Totl no. of two digit codes which crete confusion 0 Hence, the required no. of codes Let the side of originl squre be, then the re of squre nd re of lrgest possible circle in the squre re left fter cutting out the circle from squre re of scrpped cloth re of originl cloth, which is constnt. cm Since, the rtio of remining prt nd originl squre is constnt. cm Therefore we cn infer tht the percentge re of scrpped cloth is lwys constnt for ny size of squre shped cloth. Thus, we cn sy tht in ech cse (i. e., first process nd second process) the re of scrpped cloth is sme since the re of the originl piece of cloth is sme in ech cse. cm lterntively : Since it is given tht the originl squre of side is divided into n smller squres, not necessrily of the sme size. Therefore we cn ssume tht (for convenience) ech smller squre is of the sme size. Let the ech side of lrger (i. e., originl) squre be cm nd ssume tht it is being divided into four smller squres of side cm ech. Process I : re of ech of the E M smller squre ( ) cm re of mimum possible circle in ech of the squre ( ) cm re of scrpped (remining) cloth ( ) cm N O cm cm

11 Solved Pper 00 (Fresh Pper) NOTE Hence, totl re of scrpped cloth ( ) cm Process II : re of the originl squre ( ) 6 cm re of lrgest possible circle inside the squre ( ) re of scrpped cloth 6 ( ) cm Rtio of the scrpped cloth in both the process ( ) ( ) This cn lso be proved lgebriclly (i. e., by tking vribles) insted of ssuming some vlue of nd n s bove.. Recll tht the digonl of n inscribed rectngle bisect ech other t the centre of the circle. Now, join the mid points of nd (i. e., M nd N) pssing througho, the centre of the circle MN is prllel to nd both, where OM ON. Tringles E nd ON re similr where, E ON N E O Now go through options : Let us consider option () s from option () E ON N O (Using Pythgorus theorem) E ON N re of circle ( ) Now, length of rectngle N (rdius of the circle) nd bredth of rectngle MN ( ON ) re of rectngle ( ) ( ) re of circle re of rectngle Since, the given condition (tht the rtio of re s of circle to the rectngle is ) is stisfied, hence the ssumed option () is correct. lterntively : Since, re of circle re of rectngle ( O) O N ON ( N ) ( ON ) N ON N ON N ON ON N E P O Let ON ON N N ON N ON E N E ( NO ~ E ) lterntively : Produce O to nd drop perpendiculr from O on. OP ~ Now, go through options, Let E OP E P re of circle ( O) nd re of rectngle re of circle re of rectngle ( ) ( ( ) ), Hence chosen option () is correct. Solutions for question no.,, nd 5 : ech circulr prk be r. r In E, E 0 E E cos 0 r F E G E r E r H E EH H r r r H Let the rdius of

12 Solved Pper 00 (Fresh Pper) r( ) r( ) gin, In F, F 0 tn 0 F F r F F r G F FG G r r r r( ) r( ). ( r) 6r b c b c b r r r r. Time tken by [ r( )] r ( ) [ r( )] r ( ) r sec (sy) 0 istnce trversed by in r r sec ( 0 0) 0 0 r ( ) Since r ( ) is the distnce of the complete pth. Hence reches gin t. Time tken by to trverse Similrly, time tken by to trverse Hence covers distnce in r r r sec r ( ) r 0 ( ) 0 r 0 Thus reches t in r sec 0 Thus when reches, lso reches nd reches t. Hence, choice (c) is correct. 5. u : v : w re : re : re u : v : w ( ) : ( ) : ( ) u : v : w : : n n n n Since speeds re in the rtio of distce, the time tken will be the sme by ech sprinters. Thus ll of them will complete their sprint in sme time i. e.,,, will rech t,, t the sme time. Hence, choice (b) is correct. 6. Verticl spcing between ny two consecutive turns height of cylinder no. of turns h n 7.If ll the successive (or consecutive) fces re opened up then the following figure will represent the position of string. Then length of the string ( ) ( ) ( n) n n 7 8. The string of length n 7 is wound on cylinder of height h mking n rounds. Thus the totl circumference in n rounds n ctul circumference of cylinder n n If this phenomenon is represented in the geometricl form, we get the following figure. From the bove figure, ( n) ( h) ( n 7 ) h n 9. Let ( prime number) nd y 0 ( composite number) then choice () is wrong since y 0 8, which is n even integer. hoice (b) is wrong since y 0 60, which is n even integer hoice (c) is lso wrong. 0. length of string n 7 since n y 0 which is n even integer Hence, choice (d) is true. 5 6 P.W. Rtio 5 5 6, h (height of cylinder) 6 P.W. 6 5 P.W. Rdio Rdio

13 Solved Pper 00 (Fresh Pper) 5 From the bove venn-digrm it is cler tht no. of crs which hve tlest one option ( 5) ( 6 ) Hence the no. of crs which hve none of the options 5. Let us substitute n 6 in the given epression, we get, ( 6 ) 6 ( 6 ) 6 96 ( 0) gin if we put n 7, then we get, 8 ( 7) 7 ( 7 ) 6 59 (ppro.) 7 7 Since 8, hence choice (d) is eliminted nd for 59, choices () nd (b) re eliminted. Hence, choice (c) could be correct.. z nd z 5y ombining the two equtions, we get 7 0y Let us consider nd y, then we hve 7 0 ( ) 0 0 when( ) ( y ), the given inequlity is stisfied. Hence choice (b) nd (c) re eliminted. Now, we consider nd y, we hve 7 0 ( ) Hence, choice () is lso ruled out. Therefore, choice (d) is best nswer.. ny composite number ( n) which is not perfect squre hs tlest one fctor less thn n nd greter thn n, such tht their product is n. e. g., 6.5 then 6 hs two fctors nd ; (.5 ) 8.85 then 8 hs two fctors nd ; (.85 ) 0. then 0 hs two fctors, 5; (. ) The logic behind the question is tht if composite number is not perfect squre, then it must hve t lest two unequl fctors (since non perfect squre but composite) number cn not hve its squre root s n integer unlikely in perfect squre number two fctors re lwys equl. e. g., (i) 8 (ii) 0 5 (iii) 6 or (iv) 5 5 (v) 8 9 or 6 See the (i) emple GM of nd, which must lies between nd. Similrly, in (ii) emple 0 5. GM of nd 5, which must lies between nd 5. Thus we cn sy tht for non perfect squre composite number n is the geometric men of its fctors nd the GM of ny two or more numbers necessrily lies between them. So tlest one of the fctor lies below n nd other lies bove n. Hence, choice (d) is correct.. b b or b. Then we cn chrt out the following tble. b b 0 b ve b 0 0 b +ve b 0 0 b ve b 0 0 b +ve b 0 Hence, choice (b) is correct. lterntively : b nd b Now, b [ b] b b b 0 b 0 ( b is lwys positive nd b ) PQ (PQ is 5% less thn ) re of PQ PQ ( ) P re of ( ) re of PQ 0.5 re of sq. inch 6. Let us consider there re huts in the begining of the yer 00. Yer No. of Huts Rebuild No. of Huts in the egining of the Yer Huts estroyed Huts Remining in the End of the Yer Now, go through the options. Only option (c) stisfies the required condition. i. e., b c nd b 9d e : b : c : : 6 Q

14 6 Solved Pper 00 (Fresh Pper) nd b : d : e : : 9 or : b : c : : b : d : e 8 : : : b : c : d : e 08 : 8 : 9 : : or 08 k, b 8 k, c 9 k, d k nd e k; k I Now, go throug options : b hoice () is wrong, since gives integer. 7, e i.e., 08k 7 8k, ( k, 6k) k Similrly, choice (b) nd (c) re lso eliminted. choice (d) is correct c s,, 08 k 9k, b d 6 k (Since 9 is not n integer) 8. There is only one possible set of prime numbers viz.,, 5, 7. Hence, choice () is correct. Here you cn see some other sets where is prime number but either ( ) or ( ) is not prime number : 9. (5, 7, 9 ); (7, 9, ); (,, 5 ); (, 5, 7); (7, 9, ); (9,, ); (, 5, 7); (9,, ); (,, 5); (7, 9, ); etc Since the squre of side inch is being removed from ech corner, the length nd bredth of the bse of the sheet (i. e., the bse of the cuboid) reduces to ( ) inch. Therefore the re of the bse ( ) nd thus the height of the cuboid becomes inch. R the volume of the cuboid bse re height ( ) Now, going through options we find tht t, the volume of the cuboid mimizes. Hence, choice (d) is correct. 50. Since, the speed of Rm is km/h. Hence the distnce between nd is km. gin since the speed of Shym is km/h. Hence the distnce between nd is 8 km. km nd 8 km Using cosine rule, we get cos 6 cos or 9 km Time tken by Rm to return 9 8 hours Time tken by Shym to return 9 hours Time intervl 9 0 hours Hence, choice (b) is correct. 0 R