SSC TIER II (MATHS) MOCK TEST - 31 (SOLUTION)

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1 007, OUTRM LINES, ST FLOOR, OOSITE MUKHERJEE NGR OLIE STTION, DELHI-0009 SS TIER II (MTHS) MOK TEST - (SOLUTION). () We know tht x + y + z xyz (x + y + z) (x + y + z xy yz zx) (x + y + z)[(x + y + z) (xy + xy + zx)] b + b c + c b c b c b c b c b c b c On putting the vlues, we get b + b c 9(8 ). () 6 0 O M 8 + c 9 [9 ] Hee, O M O M M Dimete dius cm. (D) Let the distnce of the cicul tck be 80 m (LM of 6 nd 0) Velocity of Vipul 80 6 m/s nd, eltive velocity of Vipul nd Sumit m/s Velocity of Sumit 9 m/s Time tken by Sumit to complete. () one ound 80 sec. OL h: , E D F In figue, FDQ Q nd, ED FD nd, DE e of the EF (D) + (DEF) 80 + DE FD cm. () Tin tkes ( 8) 6 seconds to tvel distnce of 90 m (length of pltfom) So, speed of tin m/s s tin tkes 8 sec. to coss the men So, length of the tin 8 8 m 6. (D) S R K In the figue, O O is the cente of cicle. nd, RL L Q Using pythgos, we get Q

2 007, OUTRM LINES, ST FLOOR, OOSITE MUKHERJEE NGR OLIE STTION, DELHI-0009 We know tht KL is equl to the side of sque. So, KO + OL KL + Squing both sides, we get + 8 Rdius of the cicle units 8 7. () Q R 00 T Tking points, Q, R nd T concyclic, we get RT 80 tking points, T, S nd R concylic, we get S RT In RT, RT 80º ( RT + RT) 80º ( + 80 ) 8. (D) Let the length of ech cndle units 9. () We know tht, inteio ngle + exteio ngle 80 9 exteio ngle 80 exteio ngle 0 n 60º exteio ngle 60º 0º 8 Numbe of sides of polygon 8 0. (D) The pecentge incese in the numbe of membes of the club % Using lligtion Method, +% +% +% 7 Men Women ( + 7) units 0 units 0 unit diffeence 7 units 60. () Income 7 Expenditue 6 8 sves of his income Rte of buning of fist cndle units/hou i.e. expenditue 0 6 units 0 Rte of buning of second cndle units/hou let the equied time tke t hous ccoding to the question, t t t t t 8t hous minute units 9 On multiplying expenditue by nd income by 9, the new tio becomes Income 6 08 Expenditue Sving The equied tio : : h: ,

3 007, OUTRM LINES, ST FLOOR, OOSITE MUKHERJEE NGR OLIE STTION, DELHI () o 7 E D 8 Hee, O is the cente of the sphee with dius Using pythgos, we get 8 7 cm EO D O EO D cm Rdius of sphee.8 cm. (D) Fo e to be mximum S SR RQ nd, OS SOR QOR s we know tht totl ngle t the cente 60 OS + 60 OS 7 cos O + OQ Q O.OQ e of qudiltel sin + sin7 +. (c) x, y. () h: , S R T E D Q Dw EQD, S nd RE in D, EQD nd E is the mid point of So, by m.p.t. Q will be the mid point of D. Since D divides line in the tio :. D DQ Similly, by m.p.t in EQ, we get E gin in D, TED nd E is the mid point of. So, by m.p.t T TD in RE, SRE, is the mid point of E. So, S SR Similly in D, T D Hee, we get, T TD nd T D We know tht, D T + TD TD D D D Subtct fom both sides, D D D D D 6. () Given, D D : x x + y y... (i) nd, x y + y x 0 xy x y 0... (ii) x y x x + y y + x y x y

4 007, OUTRM LINES, ST FLOOR, OOSITE MUKHERJEE NGR OLIE STTION, DELHI-0009 On putting the vlues, we get x y + 0 x y 8... (iii) On putting the vlue of eqution (iii) in (ii), we get xy 0 8 Squing eqution (iii), we get x + y + xy 6 x + y + 6 x + y (D) S J 6 Wok done by Stypeet in 0 dys units Remining wok units time tken by Jspeet to complete the wok 90 8 dys Requied time 8 dys 8. () Let the numbe of boys in the clss be x nd x espectively. (x ) (x ) x 9x x 9x 8x On solving, we get x 6 Totl numbe of students x + x () S R L M q q K Q Let LQR y symmety We find QS nd, KLQ QM KQ... (i) cos MQ LQ sin... (ii) nd, LQ 6 cos... (iii) Fom eqution (i), (ii) nd (iii), we get 6 cos sin KQ 6 sin cos tn KQ 6 tn let Q KQ On putting the vlues, we get S 6 Q S Q ( 08) Q 8 units 0. (D) We know tht oduct of sides nd ltitudes emins equl s e of the tingle emins constnt. Let, b, c be the sides of the tingle 8. b. c On simplificition,we get 60 b 8c LM of (60,, 8) b c : : 7 0 : b : c 7 : 0 :. (D) f(x) x x + x +... (i) s,, nd e the oots of the eqution f(x) (x ) (x ) (x ) put x f( ) ( ) ( )( ) nd, put x in eqution (i), we get f( ) ( ) + ( )( )( ) (+ )(+ )(+ ) h: ,

5 . (). (D) 007, OUTRM LINES, ST FLOOR, OOSITE MUKHERJEE NGR OLIE STTION, DELHI-0009 S D R O Q cicle cn be inscibed in qudiltel only if O + OR OQ + OS OS O + OR OQ OS + 6 OS OS cm Given, 0 D 8 se of tingle 8 cm D cm nd, D 0 cm Using pythgos, we get 0 6cm eimete cm. () s we know tht tn sin tn. () To chnge din into degee, we multiply by 80º c ' 6. () Let the ngles of the tingle be (60 d), 60 nd (60 + d) lest ngle in degee (60 d) getest ngle in din (60 + d) 80º 60 d 60 d 80º d 60 d 0 d 60 + d d 60 d 0 Getest ngle () (cosec sin)(sec cos) (tn + cot ) sin cos sin cos + sin cos cos sin sin sin cos cos sin + cos cos. sin cos sin sin. cos sin.cos 8. (D) (cos sin + ) sec [(cos sin ) (cos + sin ) + ].sec [cos sin + ].sec [cos + cos ].sec cos. sec 9. (D) Given, tn + tn This eqution is stified t sec sec sec sec nd given tht, sin II method:- sec sec sec (sec ) On comping,we get ( + tn ) (tn ) tn tn + tn h: ,

6 0. () 007, OUTRM LINES, ST FLOOR, OOSITE MUKHERJEE NGR OLIE STTION, DELHI-0009 O x 8 Q 0 D Hee, O is the dius of the cicle Let O be x Now, 8 nd, QD 0 cm 0 cm Using pythgos, we get x +... (i) nd, (x + 8) (ii) Subtct eqution (i) fom eqution (ii), we get, (x + 8) + 0 x 0 6x x 7 On putting the vlue of x in eqution (i) we get cm Dimete of the cicle 0 cm. () We know tht, tn. tn(60º ). tn (60º + ) tn tn6º. tn(60º 6º). tn (60º + 6º) tn( 6º) tn6º. tnº. tn66º tn8º tn6º. tn66º tn8º tnº... (i) tn8º.tn(60º 8º).tn(60º + 8º) tn( 8º) tn8º. tnº. tn78º tn º tnº. tn78º tnº tn8º... (ii). () Discount 0% 0 ofit 8 % 6 M S We get M 90 nd units `800 M 90 units `7 List pice of the cycle `7. () Mke the goup of the numbe s (0, 99) (, 98) (, 97)... Thee will be totl of 0 pis nd, One pi gives sum Sum of 0 pis Hee, we missed the numbe 00 So, totl sum (D) Reduction in the pice of te 0% Incesed Quntity 7 units kg unit 9 kg (oiginl te) Num. Den Num fte incesed quntity, quntity of te kg educed pice 0 70 ` pe kg Multiply eqution (i) nd (ii), we get 6. () Use the fomul tn6º. tnº. tn66º.tn78º xy x + y +. () onside ( ) 00 ( + 7 ) + ( + 6 ) x n + y n is lwys divided by (x + y) if n is 0 n odd numbe. hnge in e So, whole fction is divided by. Tht's why, it will be divisible by Reminde 0 Decese in e 6.% h: ,

7 007, OUTRM LINES, ST FLOOR, OOSITE MUKHERJEE NGR OLIE STTION, DELHI () We know tht, Digonl of cube 9 cm 9 cm volume of cube 8 cm 8. () Diffeence between compound inteest nd simple inteest fo yes p ` () e of the field tht cn be gzed 60 [sum of the ngle of tingle 80 ] m 0. () Efficiency time tken 7 7 will tke dys to finish the wok.. () onside x + y 6x + 0y 0 (x ) + (y + ) 0 x 0 nd y + 0 x nd y x y ( ) 8. () efficiency of ( + ) ipe cn empty the tnk in 80 minutes pcity of tnk gllon. () Let vege scoe of innings be x. x (x + ) x + 9 6x + 8 x 7 vege fte 6th inning ltentive Method: vege fte 6th inning 9. () Let... y On squing both sides, we get... y y y y ( ) x x x. () Let the numbe of the coins be x. volume of n coins must be equl to the volume of ight cicul cylinde. i.e., n h h n n () Let the dius of the sphee be cm. [( + ] ] 06 [( + ) ( + + )] 06 ( + ) Rdius of the oiginl sphee 6 cm 7. (D) 7 l cm 9 0 h: , D l b l b l b 0... (i) nd, l b 7 l + b (ii) Using eqution (i) nd (ii), we get l + b (iii) nd, l b (iv) using eqution (iii) nd (iv), we get 7

8 007, OUTRM LINES, ST FLOOR, OOSITE MUKHERJEE NGR OLIE STTION, DELHI () Inteest on `66 fo one ye `.. () Rte of inteest % 9. (D) ` efficiency of efficiency of on comping, we get Efficiency of time tken by to complete the wok 0 60 dys 0. () If the diffeence is sme then numbe with smlle digits will be the getest.. () Hee, is the lgest numbe. 6 D 0 Using ppollonius theoem, + (D + D ) On putting the vlues, we get 6 + (D + 0 ) (00 + D ) D 6 Hee, Rdius of cylinde Height of cylinde tio of thei volumes volume of cylinde volume of hemisphee :. () : units `00 unit `00 pofit of 00 `6000. (D) Let the two numbes be x nd y. xy 0... (i) nd, x y 8 ut x 8 nd y in eqution (i) x nd y 6 0 Sum of the numbes () (07 + ) To mke the numbe pefect sque 07 hs to be subtcted. 6. () 6 6 Wok done by in hous 6 units nd, Wok done by in hou units wte in cisten units t 7 pm. effective efficiency + 6 i.e., D 79 cisten will be emptied in 8 8 hous Length of medin D 79 cm Requied time 7pm + 8 hous.m. h: ,

9 007, OUTRM LINES, ST FLOOR, OOSITE MUKHERJEE NGR OLIE STTION, DELHI (D) Let S.. of one pple be ` S.. of 60 pple `60 ofit S. of 0 pple `0.. S.. pofit 60 0 `0 pofit pecentge pofit 00% 0% 8. () onside the equtions x + b y + c 0 x + b y + c 0 Fo lines to be pllel b b c c k k k + 9 k k 0 k 9. () onside x 7 + x ( ) + + ( + ) x +... (i) nd, x... (ii) dding eqution (i) nd (ii), we get x + x + + sque oot of x x 6. () 6. () x 9 x x x 9 0 x 6x + x 9 0 x (x ) +(x ) 0 x nd x x + x ut x + x + x D 9 In pllelogm, + D [ + ] 6 + D [6 + 9 ] + D [6 + 8] D D 90 0 cm 6. () M of the ticle `00 Fist discount `0 ice 00 0 ` () Let the speed of the bot be x km/h nd the speed of the stem be y km/ h 0 x y + 6 x y nd, 0 x y + 0 x y 8... (i)... (ii) On solving the equtions, we get x + y nd, x y 6 speed of the bot 6 9 km/h x% x 6. () Requied tio h: ,

10 007, OUTRM LINES, ST FLOOR, OOSITE MUKHERJEE NGR OLIE STTION, DELHI % Rte of inteest 6 % 6. (D) sec tn sin cos cos sin cos cos sin cos sin cos sin cos sincos + sin cos sin cos + sin Divide numeto nd Denominto by cos sec tn tn + tn tn tn tn tn.tn tn[ ] 66. () 6 % pofit S Using lligtion method Requied tio : 0 : 67. () Let x + 6 x 6 x + x We know tht cm 7 h: , Subtct 6 fom both sides 6 6 ut 6 x x x 6 6 [ ] 68. () 80 qo q 7 Q R 9 Using the popety R SR QR 7 SR 9 SR cm S 7 cm S Rdius of the cicle (OQ) cm e of QR e of OQR + e of OQ 9 + sin (80º ) + 7 sin [OR (OS + SR) + ] cm e of QR 97. cm 69. () Sufce e of sphee Sufce e of two hemisphees 6 incesed sufce e 6 0

11 70. () 007, OUTRM LINES, ST FLOOR, OOSITE MUKHERJEE NGR OLIE STTION, DELHI D Using the popety D 6 (6 + ) cm 7. (D) Weight of teche kg gm kg + 7 kg 68 kg 7. () Fe of km fe of km 87 ` Given, chge of 0 km `60 km (fixed) + 6 km (dditionl) `60 km (fixed) + 6 `60 fixed chge 60 0 `0 7. () Digonls of hombus bisect ech othe of 90. So, O is the mid point of nd slope of line D slope of line 7 oodintes of O, (, 6) nd, slope of line 7 6 Slope of line D eqution of line D y 6 x y 6 x + y y x x m x y () Let the speed of Rjdhni tin be x km/h. nd, tht of expess tin be y km/h. 78 y 78 x nd, 78 x 78 y 8... (i)... (ii) 7. () we hve to find... (i) 0 [ + ] Totl inteest obtined fte yes ` (D) time tken by nd to do this 0 wok 6 dys 0 nd, emining wok time tken by to do this wok dys Totl time tken dys 77. () Rte of Inteest % Using lligtion 8 6 On solving the eqution, we get x 9 km/h 7 0 Speed of Rjdhni tin 9 km/h Requied quntity 0 kg 7 h: , : 6 ( + 8) units units 0 unit 0 units 0 ` 60 nd, 8 units 0 8 ` ().T.Q. 6 6% 00 0 Initil Now ice 0 7 Quntity 7 0

12 007, OUTRM LINES, ST FLOOR, OOSITE MUKHERJEE NGR OLIE STTION, DELHI ().T.Q mn womn child 0 0 Wok done by Mn nd Womn in one dy + 7 emining wok 0 7 Requied numbe of boys 80. () onside x y 8 z 8. () Time Speed x y z On comping, we get x y z x : y : z 6 : : let x 6, y nd z x + y + z vlue of z [/ less mens is chnged into ] [time nd speed emins in evese tio] incesed speed 00 % 8. () % 8 inteest on `600 till July ` 00 nd, net mount of July ` 00 inteest t the end of ye 00 8 ` 8. (D) Time fte which they will meet gin LM of (0, 0, 0) 0 sec. 87. min 8. () Let the pice of onge, pple nd bnn be x, y nd z espectively. x + y + z 6... (i) x + y + z... (ii) Multiply eqution (i) by nd eqution (ii) by nd on solving, we get z y 8 multiply eqution (i) by nd on solving, we get x 7 y the pice of onge, pple nd 7 bnns x + y + 7z (7 y) + y + 7 (y 8) 0 8y + y + y 6 ` 8 8. () Speed of moto bike fo Ist km 0 km/h Speed of moto bike fo IInd km 0 km/h nd, Speed of moto bike fo next hlf km 0 km/h vege speed km/h () We know tht, Distnce Totl inteest 00 + ` lite h: , Totl Distnce Totl time oduct of speeds Diffeence of speeds time 0 D 0 60 D km Distnce between his house nd office km 87. () : : 7 0 [Mke the quntity equl in both the cses.] New tio: 6 : : 7 liquid tken out 6 totl quntity 0 lite units 0 units Next time quntity of liquid 6 units

13 007, OUTRM LINES, ST FLOOR, OOSITE MUKHERJEE NGR OLIE STTION, DELHI () onside the eqution mx + nx + x 0 sum of the oots ( + ) n m nd, tio of gils to boys in school who scoed between 80 nd 90 pecent 7 : numbe of gils nd, poduct of the oots ( ) n m Given, p q p q n q p m totl numbe of students of school of F who scoed moe thn 90 pecent nd, tio of gils to boys : 7 + n n m m 89. () S 90. (D) 00 0% % +7% pofit pecentge % : : : : : 600 : 7 : ( ) units ` Lon ecieved by 7 ` () Totl numbe of students of school who scoed between 80 nd 90 pecent numbe of gils equied pecentge % () Numbe of boys of diffeent school who scoed 90 pecent e bove e D E F Totl students vege numbe of students () Numbe of gils of school E who scoed 90% nd bove nd, numbe of boys who scoed between 80 90% fom school h: ,

14 007, OUTRM LINES, ST FLOOR, OOSITE MUKHERJEE NGR OLIE STTION, DELHI equied diffeence (D) Numbe of boys of school D who scoed who scoed 90% nd bove 00 nd, numbe of boys of school D who scoed between 80 90% 00 p Requied tio 00 : 86 6 : % 9. () Numbe of boys of school D nd F who scoed 90% nd bove () 70 nd, numbe of gils of school nd E who scoed 80 90% R Q Hee, c + b + c b + c (i) R RQ nd, Q Requied pecentge % RQ 90 b b c c Then RQ + c using eqution (i) we get + b 90 b 96. () LM of the expessions (x + 7x + ) (x ) (x + )(x + ) (x ), HF of the expessions x, nd, One expession x + x (x ) (x + ) We know tht, Ist numbe nd numbe LM HF (x ) (x + ) nd expession (x + )(x + )(x )(x ) nd expession (x + )(x ) x +x 97. () () 0,, h: , Diffeence 8 When diffeence becomes mximum we get the lest numbe. So, 00. () onside is the smllest numbe ( ) + (6 ) + (0 9 ) + ( ) ( ) ( + ) + (6 ) (6 + ) +(0 9) (0 + 9) + ( ) ( + )

15 007, OUTRM LINES, ST FLOOR, OOSITE MUKHERJEE NGR OLIE STTION, DELHI-0009 SS TIER II (MTHS) MOK TEST - (NSWER KEY). (). (). (D). (). () 6. (D) 7. () 8. (D) 9. () 0. (D). (). (). (D). (). () 6. () 7. (D) 8. () 9. () 0. (D). (D). (). (D). (). () 6. () 7. () 8. (D) 9. (D) 0. (). (). (). (). (). (D) 6. () 7. () 8. () 9. () 0. (). (). (). (). (). () 6. () 7. (D) 8. () 9. (D) 0. (). (). (). (). (D). () 6. () 7. (D) 8. () 9. () 60. () 6. () 6. () 6. () 6. () 6. (D) 66. () 67. () 68. () 69. () 70. () 7. (D) 7. () 7. () 7. () 7. () 76. (D) 77. () 78. () 79. () 80. () 8. () 8. () 8. (D) 8. () 8. () 86. () 87. () 88. () 89. () 90. (D) 9. () 9. () 9. () 9. (D) 9. () 96. () 97. () 98. () 99. () 00.() Note:- If you opinion diffes egding ny nswe, plese messge the mock test nd question numbe to Note:- Whtspp with Mock Test No. nd Question No. t fo ny of the doubts. Join the goup nd you my lso she you suggestions nd expeience of Sundy Mock Note:- If you fce ny poblem egding esult o mks scoed, plese contct 9777 h: ,

SSC [PRE+MAINS] Mock Test 131 [Answer with Solution]

SSC [PRE+MAINS] Mock Test 131 [Answer with Solution] SS [PRE+MINS] Mock Test [nswe with Solution]. () Put 0 in the given epession we get, LHS 0 0. () Given. () Putting nd b in b + bc + c 0 we get, + c 0 c /, b, c / o,, b, c. () bc b c c b 0. b b b b nd hee,