Quantitative Aptitude 1 Algebraic Identities Answers and Explanations

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1 Quntittive ptitude lgebric Identities nswers nd Eplntions P- (S) d 5 6 b 7 c 8 b 9 c 0 d d 5 d 6 d 7 b 8 9 c 0 c c b c 9 d 0 b d b b 5 6 b 7 c 8 b 9 c 0 d c c b c 5 d 6 c 7 d 8 b 9 c 50 b. Q + y + z ( )( )( ) + y + z yz ( ) Given 997, y 998, z 999 Q + y y yz z [ y z y yz z] + + [( y) + (y z) + (z ) ] [ ] d Q + b + c 8 ( ) + (b ) + (c ) 0 If + y + z 0, then + y + z yz 0 ( ) + (b ) + (( ) ( )( )(b ) (c ) Given + nd y + y nd y Q + y y ( y ) [( + y)( y)] b If + b + c 0, then + b + c bc b c + + bc c b + b + c bc bc bc. 7. c + b + c bc 0 ( + b + c)( + b + c b bc c) 0 ( b c)[( b) (b c) + (c ) ] 0 Q + b + c 0 Then, ( b) + (b c) + (c ) 0 b b c c 0 b 0 b b c 0 b c c 0 c b c. 8. b Q + b b + () b + b + b c 5 yz + z + y y z 7 yz 7 5 yz + z + y Q ( + y+ z) + y + z + ( y+ yz+ z) ( ) ( ) + y + z + y+ z y+ yz+ z + y + z y + z ( b), (b c) 5, (c ) Q + b + c bc ( + b + c)[ + b + c b bc c] ( + b + c) [ + b + c b bc c] ( + b + c) [( b) + (b c) + (c ) ] Q / Eercise - EX-50/PS/7 / Pge

2 + b + c bc + b+ c 6. d. d [( b) + (b c) + (c ) ] ( ) + ( 5 ) + ( ) [ ]. b c + + b c b c b c + + b + b c c b c + +. b c. d ( ) + (b + ) + (c + ) 0 0 ; b + 0 b c + 0 c b + 7c ( ) + 7 ( ) y y 0 ( + ) + y nd y 0 + y 5.. ( + ) + (b ) + (c ) b 0 b c 0 c + b + c d Similrly, b 5 + ; + c > > < b < c. b c. 7. b + b + c + + b + c + + b b + + c c + 0 ( ) + (b ) + (c ) 0 0 ; b 0 b nd, c 0 c + b + c. 8. y + y 7 y k y k + y 7k ( + y) ( y) 9k k y 8k 6k 8k k y k. 9. c + b + c 0 + b c; b + c, c + b + b + b + c + c + c b 0. b c + + b+ c c + + b Epression ( ) ( ) 9. + b + b b b b...(i) gin, b + c c b c b On multiplying (f) & (ii) bc b. b...(ii) EX-50/PS/7 / Pge Q / Eercise -

3 . c + b 5b + b 5 b. c. b + b 5 b On squring both sides, b + b 5 b b b + 5. b p+ p + k ( ) k d c p + p. + k b k ± 8. 8 c d + b b b c + d + b + b c d + b + b b (y componendo nd dividendo) c d b (c d) b c d (c + d) (c d) b.. + b b 0... (i) We know, 6 + b 6 ( ) + (b ) ( + b ) ( b + b ) ( + b ) 0 0 [From eqution (i)] ubing both sides, ( ) Now, ( 6 + ) + ( 6 + ) ( + ) ( 6 + ) 0. + Multiplying both sides by + ubing both sides, y + + y y y ( + y) y + y + y y + y + y 0 y ( y) ( + y + y ) c y ( + y) + y y ubing both sides, + y + y ( + y) y + y + y y y 9. d y y Q / Eercise - EX-50/PS/7 / Pge

4 ubing both sides, y z + y + ( z) 0 + y z + yz + y + ( z).y ( z) 0.. b + On cubing both sides, (i) ( ) d + (Given) +...(i) Epression ( + ) 6 + ( + ). 6 + ( ) [ ].. b +. b On cubing both sides ( + ) On squring ( + ) + + ( + ) ( + ) + ( + )... (i) gin, cubing ( + ) + ( + ) ( + ) + + ( ) + + ( + ) ( + ) 8 ( + ) + ( + ) 8 ( + ) + ( + ) ( + ) + ( + ) 5 ( + ) ( + ) ( + ) 5 ( + ) 5 ( + ) + 5 ( + ). EX-50/PS/7 / Pge Q / Eercise -

5 5. b b b b b ( b) ( b) b b + b b b + b 0 + b ( + b) ( b + b) b ( + b + c) + b + c + (b + bc + c) 6 + (b + bc + c) b + bc + c (6 ) b + bc + c + b + c bc ( + b + c) ( + b + c b bc c) 6 bc 6( ) (y (i)] 6 bc bc bc c + y + z yz ( + y + z) [( y) + (y z) + (z ). c Epression ( + y + z)( + y z) y z Putting, y, z (+ 9 )( + ). 7. c + b + c bc will be minimum if b, c Lest vlue b On squring both side, On squring gin, + + ( + + )(0 + + ) b p q On cubing both sides, (p q) 6 p 8q + p. q p. q 6 p 8q + pq 6p q 6 p 8q 6pq (p q) 6 p 8q 6pq 6 p 8q pq c ( ) 0 0. d On cubing, +. ( ) c d + y + + y y 0 ( ) + y 0 0 nd y 0 + y + 0. Q / Eercise - EX-50/PS/7 / Pge 5

6 6. c p 99 (Given) p(p + p + ) p + p + p p + p + p + (p + ) (99 + ) (00) d 997 y 998 z 999 y y z z y + z y yz z ( + y + z y yz z) ( y y y z yz z z) [( y) (y z) (z ) ] + + [( ) ( ) () ] + + (+ + ) c t t + 0 t + t t + t t + t On cubing both sides, t + t t t t t 6 t t t t 50. b ( b) b + b + k ( ).. + k k (). 8. b p + p + p 7 p + p + p (p + ) () p + p p + p +. EX-50/PS/7 / Pge 6 Q / Eercise -

7 Quntittive ptitude Logrithm Identities nd Progressions nswers nd Eplntions P- (S) b c c d 5 b 6 c c 0 c c b c b 5 c 6 c 7 8 d 9 0 b b c d 8 9 b 0 c b d d c 5 d 6 b 7 d 8 d 9 b 0 c b d b 5 b 6 b 7 8 d 9 c 50 d. b logloglog6 log (log (log6)) loglog ( ) log ( log) log. ( Qlog6 log log ) 5. b + y Q log (log + logy) 5 ( + y) log log(.y) ( + y) log log(.y) y + y + y (y) y 5 5. c log0 79 log log0 + y + y y 5 5 y y log0 log log0 6log0 log log0 y. y + log c logloglog 0 0 loglog log 8.. d Q log (b) 6. c y log( + y) log y + y 5y 5 y log log5 y log logy log5. log + log b + log b log b log log + + b b + logb + logbb 7. log + log + log 6 log + log + log log n [ Q logm nlogm] 8. log b. b Q / Eercise - EX-50/PS/7 / Pge

8 Let 7log + 5log + log 5 80 m Q log logm logn n 7(log6 log5) + 5(log5 log) + (log8 log80) 7(log log( 5)) + 5(log5 log(8 ) + (log log(6 5)) 7(log log log5) + 5(log5 log log) + (log log log5) log. 9. c Let logy logz.y logz log.zlog logy logy logz log y z.. logz log logy y z logy logz logz y.. logz log y logz y. 0. c Let log b log b c log c k k b, b k c, c k k.b k.c k (bc) k b c.. c Q log 0 log 0 log0 log0 log0 00 log0 log logb log ( Q b ). b Q + b c c b...(i). c Given series is n P. S n n [ + (n )d] n 5,, d 5 S n [ + (5 )] 5 [ + 8] 5.. b Given series is GP.,r nd T n Q T n r n 56 n n 7 n 8. 8 n c Lst term + (n )d + (50 ) c Let its n th term be. Q 68, d 68 + (n ) ( ) 6 n + n 8 n th term + 6d 8...(i) th term + d 66...(ii) y using equtions (i) nd (ii), we get 0, d, Then, its 0 th term is ( ) d Let the nth term be Let + log c+ b log (c ) b log b (c + ) + log b (c ) log b (c + )(c ) log b (c ) log b b log b b. (from (i)) Q 5 nd r, then n n n 7 n 7 n EX-50/PS/7 / Pge Q / Eercise -

9 9. Let the rithmetic men be, nd, then,,,,, re in P d b, here b,, n n+ d 7 + So, the required rithmetic mens re + 7, + 7 8, Hence, three.m s re, 8, Q, b nd c re in GP b c Tking logrithm on both sides, we get log b log (c) log b log + log c log b log + log c log b is the rithmetic men of log nd log c. Therefore, log, log b nd log c re in P.. b,,,,... re in P. nd + (n )d 5.5 +( )d (Since n ) d b The students will follow the rule 75, 70, 65,..., (n P series) where n + (n )d Here n 75 nd d (n ) ( 50) n 5 n 6. Hence, number of rounds c Let the numbers be ( d),, ( + d). Then, ( d) + + ( + d) 7 ( d) ( + d) 5 ( d ) 5 7(9 d ) 5 9 d 5 d d± If d, then numbers re 5, 7, 9 If d, then numbers re 9, 7, 5.. lerly, the numbers between 00 nd 00 which re divisible by re 0, 0, 07,..., 99. Let n be the number of terms in P. Then, n (n ) 98 n n 0 n 67 n 67 + n Sum ( ) ( 0 99) , 9 6, then r (i) r (ii) On dividing eqution (ii) by (i), we get 8 r 6 5 r 8 r 8 r Substituting the vlue of r in eqution (i), we get The GP is,,, Given sequence is,,,, The reciprocl of the sequence is in P i.e., 0 8,,, It cn be written s,,,,... Here in P,, d 8th term, 8 + (n )d + (8 ) Here 8th term of HP is. 7. d We hve; b lerly, R.H.S. is geometric series with first term nd common rtio. b b b b b + Hence, the vlue of b b + is. 8. Given, r, lst term l 9 nd sum S 8. Let be the first term. lr S, 8 9 r b Here 6, common rtio r 0 r (0 ) r 9, then 0th term 9 ( 6) ( 6) 5. Q / Eercise - EX-50/PS/7 / Pge

10 0. c Here,, r 89 Sum of si terms, S6 6. ( ) 6. b Let be the first term nd d be the common difference of the given P. Then, p times p th term q times q th term. p [ + (p ) d] q [ + (q ) d] p [ + (p ) d] q [ + (q ) d] 0 (p q) + [p (p )] q (q )] d 0 (p q) + [(p q ) (p q)] d 0 (p q) + (p q) (p + q ) d 0 (p q) [ + (p + q ) d] 0 + (p + q ) d 0 p+q 0 The (p + q)th term of the given P is zero.. d Sum 9 ( + r + r +... ) 9 r 9 Sum of the squres 7 ( + r + r +... ) 7 (i) 7 (ii) r On dividing the squre of (i), by (ii), we get ( r) ( 9) r + r r. 7 r Putting r in (i), we get 9 9 The GP is 9, 9, 9, d Let nd d be the first term nd common difference of given P. Then, S sum of n terms n S [ + (n )d] (i) S sum of n terms n S [ + (n )d] (ii) nd S sum of n terms n S [ + (n )d] (iii) n Now, S S [ + (n )d] n [ + (n )d] n [ + (n ) d] n (S S ) [ + (n ) d] S.. c Here, 7, d n 0 n + (n )d (n ) 9 7 9(n ) n 8. n d lerly, repetition tkes plce for ech set of four terms. Hence, 507 th term will be 507, when divided by, gives s reminder nd rd term is. 6. b Sum of n terms of GP n (r ) (when r > ) r 8 ( ) 6560 (656 ) d, b, b is n P with first term nd common difference b Now, t 0 + (0 ) ( b) 0 9b (i) t 0 + (0 ) ( b) 0 9b (ii) From eqution (i) (ii), 0 0 9b + 9b 0b 0 b From eqution (i), 0 9b 9 t 9 + ( ) ( ) d Epression n(n+ ) n(n+ ) n n + n+ n n + n + n +. EX-50/PS/7 / Pge Q / Eercise -

11 9. b First term, n ommon difference d n + n + n n n nth term + (n ) d + (n ). n + n n n n+. n n 0. c Epression to n terms ( to n terms) ( to n terms) 9 9 [(0 ) + (00 ) + (000 ) +... to n terms] 9 [( to n terms) n] [Q hs been dded n times] 9 [ 0 ( to n terms) n] n ( 0 ) 0. n [ Q to n terms 0 n 0 n. 8 9 ( ). b We know tht n(n + ) n (000 + ) n 0 ] n to n terms is Geometric series whose first term is nd the common rtio (r) is. n ( r ) Sn r n n n. n n. n n.. d ( ) Number of terms 99 Middle term th term 00. Second Method It is n rithmetic series., n 98, d common difference Number of terms n n + ( n )d 98 + (n ) (n ) n Middle term th term 50 + (50 ) b Q ( + b) ( b) b (0 + 5) (0 5) n n. 5. b The lrgest number will be 6. For n (n ).n(n + ) 6, for n, (n ) (n) (n + ) etc. 6. b Epression, n n(n + )(n + ) Q / Eercise - EX-50/PS/7 / Pge 5

12 8. d S S ( ) ( ) We know tht sum of squres of first n odd nturl numbers n(n ) Sum of squres of first n even nturl numbers n(n + )(n + ) Hence, S 5( 5 5 ) 5(5 + )( 5 + ) S c n n(n + )(n + ) ( ) ( ) d Using formul n n(n + ) we hve, (55) EX-50/PS/7 / Pge 6 Q / Eercise -

13 Quntittive ptitude Mensurtion- (Solid Figures) nswers nd Eplntions P- (S) d b b d 5 b 6 d 7 b 8 9 d 0 b d c 5 6 b 7 b 8 d 9 c 0 b b c d 5 b 6 7 d 8 b 9 0 d d d 5 b 6 d 7 d 8 b 9 d 0 b d c c 5 b 6 c 7 b 8 9 d 50. d Thickness of the wood.5 cm cm Internl length of bo cm. Internl bredth of bo cm. Internl height of bo cm. Volume of the bo 5 7 cm. 7 Volume of one brick 6 cm. Let the required number of bricks be n. Volume of thebo 5 7 n Volume of one brick 7 5. b Let the dimensions of the rectngulr bo be 5, 6 nd. Totl surfce re (lb + bh + hl) ( ) Volume of the bo m. 6. d Volume of wter to be filled in tnk m Volume of wter flow into tnk per hour m/hour Time required to fill 8 m wter level in tnk hours b Totl re of room to be pinted [ + ] + 5 m cm bricks. 7. b Q hectre 0,000 m re of the ground m Volume of rin wter collected m b Rte of flow of wter m/min 60 Fll of wter into the se in one minute m 0,00,000 litre.. d The dimensions of the new room re ll twice the originl one. Hence, the re shll be four times the originl one. ost shll be four times the originl cost of covering the wlls with pper. Hence, cost of covering the wlls of new room with pper 80 `,90. 0 cm 0 cm 0 cm 5 cm 5 cm Volume of the bo length bredth height cm. 5 cm Q / Eercise - EX-50/PS/7 / Pge

14 9. d Let sides of the cuboid be l, b nd h. Then, lb, bh 0, hl 5 lb bh hl 0 5 (lbh) 600 lbh 60 cm. 0. Volume of wter to be filled Wter flowing per hour out of pipe π (7 0 ) 5 0 Totl required time hours b cm cm 8 cm Length of bo ( ) 8 cm Width of bo 8 cm Height of bo cm Surfce re of bo 8 + ( + 8) cm.. d Volume of the bo (56 5 8) cm Volume of sop cke (8 5 ) cm Number of sop ckes c Let r be the rdius of the hole. Thickness t + r 0 r 0 t V r 000 Internl volume π ( ) 000π( 0 t) Volume of solid cylinder V π( 0) ( 000) ( ) π V ( 0 t) V ( 0 ) Thickness of the shell is 5 cm.. Number of turns 0 t t 5 0 Length of cylinder 0 imeter of wire 0. 9 Length of wire in ech turn 5 cm 7 Length of totl wire 5 0 cm 5.6 m. 5. re of curved surfce πrh π b Volume of ech mrble Q π 880 cm. 7 r π. π Volume of wter tht rises in the cylinder πr h 7 π. (5.6) 7 7 π 5.6 Totl number of mrbles... π b Volume of wter through circulr pipe Volume of wter through cistern Time πrh volume of cistern volume of wter/sec through pipe sec hour 0 minutes. 600 πrh πrh π r h 8. d Volume of wter column n volume of sphericl mrble (where, n is the totl number of sphericl mrbles submerged).. π (9). n π 9 9. π 9 9 n... π c Volume of erth dug out π () 0 m Volume of erth dug out QRise in field level re of rec tngulr field 0 0 cm EX-50/PS/7 / Pge Q / Eercise -

15 0. b π r r 7 Volume of cylinder 7 r πr h cm 7. b If the height of incresed wter level be h cm, then π rh πr h h cm. c Initil re of the cylinder Volume of the new cylinder. πr h πr h π (.r).h Increse in re (. ) πr h 0. πr h Percentge increse 0.πr h 00.% πrh. re of the curved surfce 6 5 sq. cm. π rh + π r π r 6 π r r 9 π r 9 7 cm. d Volume of wter drwn from cylinder 5 5 π rh h h 000 cm h cm b ircumference of the bse of the cone π r m r 7 m, height 9 m Volume of ir contined in the conicl tent πr²h m R h r V π R h V π (r) (r) π (9r r) 9πr. 7. d Let the rdius of their bse be r nd the height be h. Slnt height of the cone Then, πrh πrl rh r h r π π + h h r + h h + r r h i.e. :. l h r + 8. b We hve r R h H or R r, V πr²h, V π R²H V RH V r h 7 V cm dditionl wter cm. 9. Volume of cone rh π π 6 6 Volume of sphere Hence, r 6 cm. r π r π π 6 6 Q / Eercise - EX-50/PS/7 / Pge

16 0. Volume of cone π rh 7 7 cu.cm. Volume of cubicl block 0 5 cm. 00 cm. Wstge of wood 00 cm cm % Wstge % 00. d Let height nd rdius both of cylinder chnge by %, then volume chnges by + + % % ( )% 7.8%. Volume of the cone π rh, new height h + 00%h Percentge increse in volume 00%. d Rdius of lrger sphere R units Its volume R π cu. units Volume of smller cone R π cubic units Volume of smller sphere R π π r πr R R r r Surfce re of smller sphere Surfce re of lrger sphere πr :πr r :R R :R : ( ) : ( ) :. d re of the curved surfce πrl where, l r h () (60) 6 68 cm + + re of the curved surfce π rl 60 7 Totl cost of pinting 5 60 `.9 ppro b Volume of the copper sphere Volume of the wire Let l be the length of the wire. Then, π π (0.) l π 6 l,800 cm 8 m. π d Surfce re of sphere πr Volume of sphere r π π r πr r unit. Volume π () 6π unit Now, let the rdii of the spheres be,, nd. Totl volume r π ( ) π + () + () + () (8 ) π 7 6π π 8 8 Therefore, rdius of lrgest sphere unit. EX-50/PS/7 / Pge Q / Eercise -

17 7. d Volume of hemisphere r π So, π r 6. r.6 π r 6 r 6 6. Let the volume be 8 nd 7 Their rdii re nd The rtio of their surfce re : 9 : 9. c Originl surfce re of sphere πr Surfce re of sphere π(r) 6πr πr (originl surfce re). c Surfce re of sphere r π r r 6 i.e. :. r r b curved surfce re of hemisphere Rtio curved surfce re of cone r πr π r r + h :. [ Qr h] r cm 5. b Lterl surfce re of prism Perimeter of bse height cm. 9. d Volume of bigger sphere (8) π Let n be the number of smller spheres. Then, n π () π(8) n b Volume of sphere Volume of wire π (r) l π (0.) 600 Let rdius of sphere be r. Then, r 6 r cm. π π. d Volume of the sphere or 88 π r () r π 6. c Volume of pyrmid 6 bsere 9 se re cm bse re height Q se re (where is one side of the bse of the pyrmid) 8 cm Perimeter of the bse of the pyrmid cm. or r () 7 7 or r The curved surfce of the sphere πr 6 cm. 7 Q / Eercise - EX-50/PS/7 / Pge 5

18 7. b E 9. d Totl surfce re Perimeter of bse height + re of bse (5 + ) cm. F 50. re of the tetrhedron re of bse height Height of the tringle cm re of the lterl surfce of pyrmid re of tringle bse height cm. re of the bse 9 cm (side) Now, length of the perpendiculr In the equilterl tringle 9 9 cm 8. re of the bse of prism cm. Volume re of bse height 8 9 height 8 height 9 cm. 9 Height 7 6 cm Required re cm. EX-50/PS/7 / Pge 6 Q / Eercise -

19 Quntittive ptitude Geometry - Qudrilterls, ircles & oordinte Geometry nswers nd Eplntions P- (S) c d b b 5 d 6 c 7 b 8 b 9 d 0 c b c c 0 d c c 5 b 6 7 b 8 c 9 b 0 c d b b 5 b 6 d 7 c 8 c 9 d 0 d c b d d 5 b 6 c 7 b 8 9 c 50 c. c 0º 0º 0º 80º 60º E 0º 60º 0º In E, E 80º nd E 60º. E 0º, 0º (ngles in the sme segment). lso, 0º (given). bisects. nd subtend 0º t the circumference.. subtends 0º t (... E 80º, E 0º) subtends 60º t ( ).. 0º nd 0º. bisects.. d 5 cm E 9 cm re of trpezium ( ) 5 ( ) F 5 7 F F 5 cm F + F. b EF Now, F nd EF 5 cm 9 5 F cm F + F cm. P O S Q cm PQR 0 PQO 60 In POQ, OPQ 80º 90º 60º 0º sin OPQ OQ PQ OQ PQ sin 0º 6 QS (OQ) 6 cm. {Q OS OQ}. b Sum of the ngles of regulr polygon of n sides (n ) 90º (n ) 90º 080º n n + 6 n 8. R Q / Eercise - EX-50/PS/7 / Pge

20 5. d If the number of sides of regulr polygon be n, then (n ) n (n ) 5n 6n 5n n. 0. P 6. c Sum of interior ngles of regulr polygon (n ) 90º Where n number of sides (n ) 90º 0º n n 6 n n b Q Q { is rhombus} P In s PQ nd Q, P Q; Q: Q Q PQ nd Q re similr. + Q Q + P Q + Q E Point is the mid-point of E. Q is the mid-point of E, In s Q nd QE, Q QE Q QE Q QE oth tringles Q nd QE re similr. Q QE Q Q :.. c Q Q :. H O G F 8. b O E The length of the digonl of is. y midpoint theorem, the length of ech side of squre 9. d 90º (ngle of semi-circle) 0º 80º 90º 0º 60º gin, + 80º 80º 60º 0º EFGH is.. Let O be the centre of the circles. If touches the inner circle t, then O is perpendiculr to nd is the midpoint of. O O 5 cm. cm. The sum of opposite ngles of concylic qudrilterl 80º º O 5 cm cm EX-50/PS/7 / Pge Q / Eercise -

21 . b We hve, TQR similr to PTR, ( R is common QTR TPQ, lternte segment) TQ PT PT QR 6 TR.5 cm. QR TR TQ 7.. c r cm L M O 6 cm 8 cm r E is the bisector of. Let r be the rdius of the circle. Then, In MO, 8 + r... (i) In LO, 6 + ( + ) r... (ii) Solving (i) nd (ii), we get 6 cm nd r 0 cm imeter 0 0 cm. lternte: r 6 r 8 r 0 cm cm P ~ P P 5. P cm P 5. is tngent to circle t. is chord through. Hence, E 6º (ngle between the chord nd tngent is equl to the ngle in the lternte segment.) E + E + E 80º (Sum of the ngles of tringle) E 80 {6 + 6 } N P P 90º ; N P 8 cm ; P 6 cm P + P cm Now, PN PN 90 {Q N is the foot of the perpendiculr drwn from the point P on } P N + NP nd P N + NP P P N N 6 6 (0 ) cm c O P PO OP 90º nd O 0º P 60º {Q Sum of ll ngles of qudrilterl is 60º} lso, line PO bisects P eqully in two prts. PO 0 P 60 :. PO 0 0. S T Q O P R OQ OP 90º QOP + QP 80º QOP 80º º 8º QOP SOR RTS {Q ngle mde by the rc on the circumcircle is hlf the ngle mde t the centre} RTS 8 7. Q / Eercise - EX-50/PS/7 / Pge

22 . M nd M re secnts to the circle from the sme point M. M M M M (M + ) 5 M (6 + ) 5 M 0 5 M M 8 cm M M 8 5 cm.. d X P Q T R S Join XT perpendiculr to line l. QT RT...(i) PT ST...(ii) (Perpendiculr from the centre of circle to chord bisects the chord.) Subtrcting (i) from (ii), we get PT QT ST RT PQ SR Hence, SR cm.. c Let ech side of the tringle be cm. E R Then, r G cm l. c Required re of the tringle 56 cm. P R O Q S OP OR 5 cm, PQ 6 cm nd RS 8 cm. Since O PQ nd O RS, then P Q cm nd R S cm. In OP, right-ngled t, OP O + P 5 O + O O cm Similrly, in OR, OR O + R 5 O + O O cm Now O O cm. 5. b OQ OR OP (Rdii of the sme circle) ORQ OQR 0º Similrly, OPQ OQP 0º PQR OQP + OQR 0º + 0º 70º Since POR PQR 70º 0º (ngle t the centre is twice the ngle t circumference). G cm Let rdius of circumcircle be R cm nd rdius of incircle be r cm. Then, R G R nd cm. r G E πr πr cm 6. O Let nd be the centres of the two circles.. 5 O O nd O O Since 5 cm EX-50/PS/7 / Pge Q / Eercise -

23 O O cm re of the portion bounded by the rcs re of sq. (re of qudrnt) π () ( π) sq. cm. ommon chord (O) 5 cm. 0. c Trnsverse common tngent 7. b P S 5 cm cm R Q O (r + r ) () (5 + ) Let be the chord of length 6 cm. 5º O 5º { Q O 90 } cm cos O O cos 5º O O cm 8. c O cm. Length of the direct common tngent 5 cm O cm O is right-ngled tringle, right-ngled t. So, + O O O 5 cm, O cm Hence, cm. () 6 cm. 9. b Let be squre nd cm P P Q Q R R S S cm cm S cm cm P cm cm Q cm cm R cm. d (distnce between two centres) (r r ) () (7 ) 69 5 cm. E lso [ngle between the chord nd tngent is equl to the ngle mde by the chord in the lternte segment] In 7 cm. E Q / Eercise - EX-50/PS/7 / Pge 5

24 . b PR PQ + PR + 5 PR 5 5 cm O P r re of tringle Semiperimeter of tringle. b O 0º OP 90º OP 90º P + O + OP + OP 60º P + 0º + 90º + 90º 60º P 80º 0º 70º + 5 (Pythogoren Triplet] is right ngled tringle. 90º ngle t the circumference Since, imeter of circle 5 cm ircum rdius.5 cm. 5. b Simply, check through options & find Pythgoren Triplet 7. c 8. c 6 cm [Or, in the bove figure, QR r + + r r ] O P Join. O + O ( + ) In P P + P P eterior ngle sum of interior opposite ngles P O (50 + 0) 5. P O 6. d P Sides 8,, 0 r r r r + r P O 0º O 5º In P P P + P 5º (Eterior ngle) 9. d P r Q r r + r R Q O QOR 0º OPR 5º QPR 0º 55º R EX-50/PS/7 / Pge 6 Q / Eercise -

25 OR OP OPR PRO 5º. d y y d 70 OQR ORQ 5 PRQ 5º + 5º 60º E H G F E H E F G F G H E + E + G + G H + F + F + H cm (0, ) (0, ) (, 0) (, y ) (, 0) + y Required ltitude Since point is the intersection of lines y + 0 nd + y Solving the two equtions, we get nd y units.. d Y Q(5, ). c Y P(, 0) X (0, ) y PQ (5 ) ( 0) + PQ re of circle with rdius PQ π(pq) π(5) 5π sq. units.. b re of tringle Y O (0, ) (, 0) ()() (, 0) X + y ()() 6 sq. units. O 90º dimeter {if ngle mde by chord on the circle is 90º, then the chord is the dimeter of the circle} + 5 rdius.5. X 5. b Eqution of line is y 7. Points (, b) nd ( +, b + k) re lying on the line. Then, b 7 b 7 nd + (b + k) 7 + b k b k 0 k 0 {Q b + 7 0} k. 6. c (0,) O y (0,) (,0) (,0) y 0 is the eqution of y-is. y 0 is the eqution of -is. Putting 0 in + y, y Putting y 0 in + y, Putting 0 in + y 6 y 6 y Q / Eercise - EX-50/PS/7 / Pge 7

26 Putting y 0 in + y 6 6 O ; O O ; O Required re O O 9. c Y X O P Q X sq. units. 7. b + 0 nd y 9 0 y, 8. + y 8... (i) y... (ii) Eqution (i) + (ii) gives, 6 + y y y 9 y Putting y in (ii) y (p, q) (, ) nd hence, p + q 7 Y Eqution of stright line prllel to -is ; y Here, Eqution is : y 50. c + y 0... (i) + y 0 y From eqution (i), y + y 0 0y 0 y 0 0 (, b) (, ) + b + EX-50/PS/7 / Pge 8 Q / Eercise -

27 Quntittive ptitude 5 Trigonometric Rtios nd Identities nswers nd Eplntions P- (S) b c c 5 c 6 c 7 c 8 d 9 d 0 b c c b 9 b 0 c c b c 5 b 6 b 7 8 c 9 c 0 b d d 5 6 b 7 c 8 c 9 d 0 d b b 5 6 d 7 d 8 b 9 c 50 d. b sin6 sec cos8 cosec8 ( ) ( ) cos 90 6 cosec 90 cos8 cosec8 cos8 cosec 8 (). cos8 cosec 8 [Q sin (90º θ ) cos θ cosec (90º θ ) sec θ ]. c tn5 tn5 tn0 tn5 tn65 tn85 ( tn5 tn85 )( tn5 tn65 ) tn0 [Q tn 5 ] ( ) ( ) tn5 tn 90 5 tn5 tn 90 5 tn0 ( tn5 cot5 )( tn5 cot 5 ) tn0.. tn tn tn...tn89 ( ) ( ) ( ) tn tn tn tn87 tn88 tn89 cot89 cot88 cot87...tn87 tn88 tn89 ( cot89 tn89 )( cot88 tn88 )( cot87 tn87 )...(cot tn )tn c ( cos 60 + sin 0 ) ( tn 60 + cot 5 ) + sec 0 ( ) () c 6. c + ( + ) sinθ + sinθ + sinθ sinθ + sinθ sin θ + sinθ + sinθ sinθ + cos θ cos θ cos θ cos θ sin cot cos lternte method: cos sin ( cos ) + sin ( + cos )( cos ) cos sin ( cos ) + sin cos cos + cos cosec. sin sec θ+ tn θ. Q / Eercise - 5 EX-505/PS/7 / Pge

28 7. c tnθ sec θ + tn θ cosθ sec θ 5 cosθ cosθ b cos(0º + ) sin0º cos(0º + ) sin(90º 60º) cos(0º + ) cos60º 0º + 60º 0º. cot + cot cot + cot. c ( ) [cosec ] + (cosec ) ( Q + cot cosec ) cos ec cos ec + + cos ec cosec cosec. 8. d 9. d 5sin 0 cos 5 tn sin0 cos60 + tn5 5 ( ) π π< < eists in III qudrnt. Hence, sin is negtive nd tn is positive. sin ± cos sin sin, cos ec. Using sinα sinβ cos( α β) cos( α+β ), we get sin sin cos( ) cos( + ) cos cos.. c Using + sin sin sin cos 5. + nd cos cos sin sin + sin cos sin sin We get, cos cos + sin sin + cot. ( sinθcosθ) (sin θ) sin θcos θ ( + cos θ) cos θ ( cos θ) sin nd tn cos tn cosec sin( ) sin( ) sin0º 0º... (i) nd cos( + ) cos( + ) cos60º + 60º...(ii) From (i) nd (ii), we get 5º nd 5º. EX-505/PS/7 / Pge 6. sin θ tn θ. cos θ sin θ (cosθ ) sin θ (cosθ ) sin θ+ (cosθ ) sin θ (cosθ ) sin θ+ cos θ cosθ+ sin θ(cosθ ) sin θ cos θ+ cosθ cosθ+ sinθ sinθcos cos θ+ cosθ ( cos θ ) + sin θ( cos θ) cos θ( cos θ) ( + sin θ) + sin θ. cosθ cosθ Q / Eercise - 5

29 7. Given sin α cos α 6 cot 6 α cot cos α α cot α 6 sin α sin α cot α 6 sin α cot α sin α cosec α cot α. 8. b ( + tnθ + sec θ)( + cot θ cosec θ) 9. b sinθ cosθ cosθ cosθ sinθ sinθ cosθ+ sinθ+ sinθ+ cosθ cosθ sinθ (sinθ+ cos θ) sinθcosθ + sinθcosθ. sinθcosθ sin sin cos + cos cos sin cos sin tn. + cos cos 0. c We hve, cos θ cot θ sin θ cot θ+ cosec θ cos θ sin θ sin θ cos θ sin θ sin θ cos θ. ( Qcos cos sin ). c We hve, sin(60 +θ) sin(60 θ ) cos60 sinθ [ Q cos.sin sin( + ) sin( )] sinθ sin θ tn( + ) π π tn( + ) tn tn Q π ' ' ' ' c c 599 π 8π c u n cos n α + sin n α u 6 cos 6 α + sin 6 α (cos α) + (sin α) (cos α + sin α) cos α sin α(sin α + cos α) [Q + b ( + b) b( + b)] cos α sin α u cos α + sin α. (cos α) + (sin α) (cos α + sin α) cos α. sin α cos αsin α u 6 u + ( sin α cos α) ( sin α cos α) b ( + sec 0º + cot 70º)( cosec 0º + tn 70º) ( + sec 0º + tn 0º)( cosec 0º + cot 0º) [Q tn (90º θ) cot θ; cot(90º θ) tn θ ] sin0 cos cos0 cos0 sin0 sin0 + cos0 + sin0 cos0 ( ) cos0 + sin0 sin0 cos0 sin0 + cos0 sin0. b tn + tn tn( + ) tn tn We hve, 7 tn nd 8 tn 5 cos 0 + sin 0 + sin0 cos0 sin0 cos0. [Q sin θ + cos θ ] Q / Eercise - 5 EX-505/PS/7 / Pge

30 6. b sin α + 5 cos α 7 sin α + 5( sin α) 7 sin α sin α 7 5 sin α sin α sin α sin α + 0 sin α 8 0 sin α (5 sin α ) + (5 sin α ) 0 ( sin α + )(5 sin α ) 0 sin α 5 cos α cosα cot α sinα {becuse sin α sin θ sin 0º θ 0º θ 5º cos(75º θ) cos(75º 5º) cos 60º. } + sin θ + sinθ sin θ sin θ + sin θ 0 sin θ + sin θ 0 sinθ or sin θ.. cos θ sin θ cos θ ividing by cos θ, sec θ tn θ ( + tn θ) tn θ tn θ tn θ + 0 tn θ or tn θ. ut tnθ { Qsinθ cos θ} tn θ.. d sin θ + cos θ cos(90 θ ) ividing both sides by sin θ, + cotθ cot θ. sinθ 8. c P sinθ P cosθ On squring nd dding, P sin θ + P cos θ + P (sin θ + cos θ) P P. 9. c nd re complementry ngles. So, + 90º 90º sin sin (90º ) cos cos cos (90º ) sin tn tn (90º ) cot cot cot (90º ) tn Given, sin cos + cos sin tn tn + sec cot sin + cos cot tn + sec tn b sec θ + tn θ sinθ + cosθ cosθ + sin θ cosθ + sin θ + sinθ cos θ. cos + cos cos cos sin Squring both sides, cos sin sin + sin cos + cos.. d.m. G.M. 5. sec α+ 9cos α sec α 9cos α sec α+ 9cos α 6sec αcosα Lest vlue of sec α+ 9cos α. sin + cos sin cos cos sin cos + sin + cos sin cos sin + + cos sin( cos ) + sin cos + + sin cos ( + cos ) sin ( + sin) cos + sin cos + sin. 6. b sin θ + sin θ ( sin θ) (cos θ) cos θ EX-505/PS/7 / Pge Q / Eercise - 5

31 7. c sin θ 0.7 cos θ sin θ (0.7) c.sin 60º tn 0º sec 60º cot 5º 9. d + sinθ + sinθ ( + sin60 + sin60 ) + + ( + + ) ( + + ) ( + ) + ( ) ( + + ) cos0 θ cos 0. d tn θ e sec θ + tn θ cosec θ sec θ + tn θ tn θ cosec θ sec θ + tn θ sin θ cosθ sinθ sec θ + tn θ sec θ ( + tn θ) ( ) + tn θ (+ tn θ) e ( ) ( + tn θ + e ) ( ). sin 60º tn 0º tn 5º cosec 60º cot 0º sec 5º. ( ) 0.. b Epression cos 60 + sec 0 tn 5 sin 0 + cos 0 + [Q sin θ + cos θ ] sin θ + 5 cos θ 5...(i) Let 5 sin θ cos θ...(ii) On squring nd dding both equtions ( sin θ + 5 cos θ) + (5 sin θ cos θ) sin θ + 5 cos θ + 0 sin θ cos θ + 5 sin θ + 9 cos θ 0 sin θ cos θ sin θ + 9 cos θ + 5 cos θ + 5 sin θ (sin θ + cos θ) + 5 (cos θ + sin θ) ±.. b tn sin 5º cos 5º + sin 0º + + tn tn 5º 5º 5. sin cos ( 6º) cos (90º ) cos ( 6º) 90º 6º 90º + 6º + 6º 6 9 Q / Eercise - 5 EX-505/PS/7 / Pge 5

32 6. d Epression + θ θ sin + sin sinθ + sinθ + θ + θ θ θ ( sin )( sin ) + ( sin )( sin ) ( sin θ )(+ sin θ ) (+ sin θ)( sin θ) (+ sin θ) ( sin θ) + sin θ sin θ (+ sin θ) ( sin θ) + cos θ cos θ + sinθ sinθ + cosθ cosθ + sinθ+ sinθ sec θ. cosθ cosθ 9. c tn θ sin θ sinθ sin θ cosθ cosθ cosθ sin θ cos θ sin θ cos θ 7. d tn + cot sec, cosec tn + cot ( + tn ) ( + cot ) tn + cot ( + tn + cot + cot tn tn + cot tn cot cot tn [tn cot ] 8. b cos θ cos θ + 0 ( cos θ ) 0 cos θ 0 cos θ cos θ cos 60 θ 60º tn (θ 5º) tn (60º 5º) tn 5º 50. d 60 90º 60º 80º 90º 60º 0º cos cos 0º :. EX-505/PS/7 / Pge 6 Q / Eercise - 5

33 Quntittive ptitude 6 Trigonometric Identities (ont.), Heights nd istnces nswers nd Eplntions P- (S) b b c c 5 b 6 d 7 c 8 b 9 d 0 b b b b 5 d 6 c 7 8 b 9 c 0 b c b 5 b 6 d 7 8 c 9 b 0 b b d b 5 b 6 d 7 d 8 b 9 0 c d c b b 5 b 6 b c. b Mimum vlue of sin θ + b cos θ + b Mimum vlue of sin θ + cos θ tnθ 5 5tnθ b tn θ cot θ Q cosec θ cot θ cosec θ + cot θ c cosecθ cotθ 7...(i) cosec θ cot θ (cosecθ + cotθ) (cosecθ cotθ) cosecθ + cotθ cosec θ cot θ 7...(ii) On dding both equtions. 7 cosecθ cosecθ 5 8. c 5 tn θ tnθ 5 5sinθ cosθ 5sinθ+ cosθ ividing numertor nd denomintor by cos θ. sinθ cosθ 5 cos θ cos θ sinθ cosθ 5 + cos θ cos θ 5. b cosec º cot 67º sin º sin 67º cot 67º.cosec (90º 67º) cot 67º sin º sin 90º º) cot 67º sec 67º cot 67º sin º cos º cot 67º cosec 67º cot 67º cosec 67º sec º Q sec(90 θ ) cosec θ; sin(90 θ ) cos θ& cosec (67 ) cot 67º 6. d Epression cosec 8º cot 7 cosec 8º tn 7º [Q tn θ. cot θ ] cosec 8º tn (90º 8º) cosec 8º cot 8º [Q tn (90º θ cotθ; cosec θ cot θ ] 7. c tn θ + 9 cot θ ( tnθ cotθ) + Minimum vlue [Q ( tnθ cotθ) 0]. 8. b sin 7 cos sin (90º ) 7 90º 8 90º 5º tn 9 + cot 9 tn 5º + cot 5º + Q / Eercise - 6 EX-506/PS/7 / Pge

34 9. d cos º, cos º, cosº,... cos 90º... cos 00º 0 [cos90º 0] 0. b sec cosecy cos siny π sin siny 7. sinθ cosθ cosθ+ sinθ ividing numertor nd denomintor by sin θ. cotθ cot θ+ cot θ cot θ + cot θ cot θ y π + y π sin ( + y) sin π. b cos θ + cos θ cos θ cos θ sin θ tn θ cos θ tn θ + cos θ cos θ + cos θ. sec θ + tn θ 7 + tn θ + tn θ 7 tn θ 7 6 tn θ tnθ θ 60º.. b sinθ + cosecθ sinθ + sinθ sin θ sinθ + 0 (sinθ ) 0 sinθ cosecθ sin 00 θ + cosec 00 θ +.. b cos α + cos β sin α + sinβ sin α + sin β 0 sin α 0 & sin β 0 α β 0 tn α + sin 5 β d cosθ + secθ cosθ + cosθ cos θ cosθ + 0 (cosθ ) 0 cosθ secθ cos 6 θ + sec 6 θ + 6. c tn 7θ tn θ tn 7θ cot θ tnθ tn 7θ tn (90º θ) 7θ 90º θ 9θ 90º θ 0º tn θ tn 0º EX-506/PS/7 / Pge 8. b sin θ + cosec θ sin θ + sinθ sin θ sin θ + 0 (sin θ ) 0 sin θ cosec θ sin 9 θ + cosec 9 θ + 9. c sinθ cosθ y k k sin θ; y k cos θ + y k (sin θ + cos θ) k k + y sin θ cos θ y y k k k y + y 0. b Epression cosecθ cot θ sinθ cosec θ cot θ cosec θ cosec θ cot θ cosec θ + cot θ cosec θ cot θ [cosec θ cot θ & cosec θ] sinθ Method-: cosθ sinθ sinθ sinθ sinθ sin θ + cosθ cosθ sinθ sin θ( cos θ) cos θ + cosθ sin θ( cos θ) cos θ ( cos θ+ ) cos θ cot θ. sin( cos θ) sinθ. c tn θ cot θ 0 tn θ cot θ tn (90º θ) θ 90º θ θ 90º θ 5º sin θ + cos θ sin 5º + cos 5º + Q / Eercise - 6

35 . tn θ + cot θ tn θ + tnθ tn θ+ tnθ tn θ + tn θ tn θ tn θ + 0 (tn θ ) 0 tn θ 0 tn θ tn 5º θ 5º. 6. d 8 m Pole In, Rope 0 8 sin0 6 m. 7.. b h k, k (k) + k 5k 5k sin + cot + k k + 5k k Epression sin cos9 + 8cos 60 cos7 sin7 sin cos9 + 8 cos(90 ) sin(90 9) sin + cos9 8 sin cos9 [sin (90º θ) cos θ; cos (90º θ) sin θ] b Epression cot 9º cot 7º cot 6º cot 8º cot 9º cot 7º cot (90º 7º) cot (90º 9º) cot 9º cot 7º tn 7º tn 9º [tn (90º θ) cot θ; cot (90º θ) tn θ] cot 9º tn 9º cot 7º tn 7º [tn (90º θ) (tn θ, cot θ ] 0 60 O 00 m Let h be the height of tower. In, h h tn60...(i) In O, h h tn0...(ii) From (i) nd (ii), we get h 00 m. 8. c m 5 0 Let be light horse nd nd be two points on the opposite bnks of river In, 60 tn 5 60 tn 5 60 m Similrly, in, 60 tn tn 0 60 m Width of the river ( + ) m. Q / Eercise - 6 EX-506/PS/7 / Pge

36 9. b F E 0 5 E 90 m h 7 m 7 m 60 Let be building nd be tower. In E, tn5 E E E In, tn m Height of the tower m.. b 60 Let be cliff nd be tower of height h. In, 90 tn 60º 90 0 m. In E, E 90 h tn 0º E 0 90 h 0 h m. R 0. b 0 m h 90 α α 9 ft 6 ft Let h be the height of the tower. Then, h tnα 6 tn (90º α) cot α h 9 h h tn α cot α 6 9 h h feet. 90 θ θ S P Q 0 m Let RS be building. In RSQ, 0 tn θ + 0 In RSP, tn (90º θ) 0...(i) cot θ 0...(ii) From (i) nd (ii), or 0 ( cn not be negtive) 0 m. EX-506/PS/7 / Pge Q / Eercise - 6

37 . d 60 6 m Let the height of the wll () be h m. h tn 60 6 h 6 tn 60º h 6 h 6.7 h m.. b h 6. d From (i) nd (ii), we get ( ) ( + ) m. h uilding 56m In, tn tn 60 0 E Pole m. θ 60 Let be the flg stff nd let θ be the ngle between the sun rys nd the ground t the time of longer shdow. In, tn 60º h In, h tnθ h h tnθ tnθ θ b 7. d ird ird m Let the height of the tower be m. In, tn In, tn 60 tn 60 h...(i)...(ii) 60 0 S N oy Let distnce covered in min be. 50 cot0º + 50 cot 60º 00 m Speed of bird km/hr m/min 50 m Q / Eercise - 6 EX-506/PS/7 / Pge 5

38 8. b 0. c m Let be the broken prt of the tree. tn 0º sin Totl height m h E. d Q 0 0 m P 5 d Let the height of tower be h m nd distnce of foot of tower from point P be d m. In PO, h tn5 h d d In QO, h h tn0 0 + d 0 + h h 0 + h 0 + h h h + h O ( ) Height of tower 5( + ) m. 5( + ) h 60 d Let the height of tower be h m nd distnce between the building nd the tower be d m. In, 60 tn60 d 60 d 60 d In E, 60 h 60 h tn0 d h 60 h 5 m 0 60 nd position of plnes 5 m Let metre In, tn 60º 5 + In, tn 0º h h 0 m Height of tower 0 m (5) metre. EX-506/PS/7 / Pge 6 Q / Eercise - 6

39 . c. b m Tower h metre 0º 5º 60 metre metre From, tn 5º Tower units Shdow units tn ( ) 0º tn 0º 6 5. b h (ssuming h be the height of flg pole) h 75 feet 6. b P Q h h. b From, tn 0º h + 60 h h+ 60 h h + 60 h h 60 h( ) ( + ) h ( )( + ) ( + ) 0 metre 500 m 0 60 P & Q re the positions of the plne P 60º; Q 0º P 500 metre. In P, tn 60º P metre In Q, tn 0º Q metre PQ metre 000 m trvelled in 5 sec. Speed of plne m/sec Post 5 feet The post breks t point. feet (5 ) feet 0º From, sin 0º feet. θ 90 θ Q P Tower h units Let, Q θ P 90º θ P ; Q b From Q, tnθ Q h Q / Eercise - 6 EX-506/PS/7 / Pge 7

40 tn θ h b From P tn(90º θ) h P EX-506/PS/7 / Pge 8... (i) cot θ h... (ii) y multiplying (i) nd (ii) h h tnθ. cotθ b h b h b m River 00 metre Height of plne h metre metre (let) (00 ) metre From, tn 60º h h metre h metre... (i) From, tn 0º h 00 h 00 h 00 [From eqution (i)] h + h 00 h + h 00 h 00 h metre metre 9. h metre, h metre h 90 θ O θ metre h 50. c O O metre From O, tn θ h h... (i) From O, tn (90º θ) O cotθ h h... (ii) Multiplying both equtions, h h tn θ. cot θ 8h [Q tnθ cotθ ] h 5000 m 8 h 60 60º 5º 5000 metre metre From, tn 60º metre metre From, tn 5º m. Q / Eercise - 6