Answer-key & Solution

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1 nswer-key & Solution E (Mechanical) MOCK -(4) Date / 9/ 07. C. D. B 4. C. C D 0.. C 7. D. C 47. C. C 77. C D. B 8. C. D 48. C B B 4. B D 79. B 94. C 09. B D 5. B 50. C 5. C B 0.. B. C. B 5.. D 8. D 9. B. C 7. C. D 7. C C 8. D 97. B. D 8. D. 8. B 5. B C 98. D. D B C 9. C 84. C D 0. B 5. D 40. B 55. D 70. C C 5. B. C. B 4. B 5. B 7. B C.. 7. C B 7. C 87. D 0. D 7. D. 8. C 4. D 58. D 7. C 88. B C C 59. C 74. B D 9. C 5. D 0. B 45. C D 90. D D Note : If your opinion differ regarding any answer, please message the mock test and Question number to Note : If you face any problem regarding result or marks scored, please contact : 9777

2 SOLUTION (E Mechanical) MOCK TEST no. 4. (C) C.P. of 0 cups ` 0 8 ` 90 0 cups are broken S.P. of 00 cups ` (00 ) ` 00 Profit ` (00 90) ` 40 Profit percent % 90. (C) cos cosec 9 cosec 9 cos sec cosec 9 cosec (90 ) (B) Given equations are: x + 4y 5 x + y On solving (i) & (ii) We will get x and y x + 4y (i)...(ii) 4. () Let the first train meet the second in x hrs after the start, then 40x + (x ) 50 0 (the nd train takes (x )) hrs. as the train starts two hours later than the st) 90x x 0 90 hrs 7 hrs hrs hrs 0 min Two trains meet at (9a.m + hrs 0 min).0 a.m. 5. () x + x x S.I. at the rate of 5% for next 4 year P 5 4 0P S.I. at the rate of 8% for next year P 8 00 P 00 Total S.I. P P 00 + P P 00 ` P ` Required sum ` (D) Gain % Error True value Error % 9. () day's work of ( + B + C) part...(ii)...(iii) + B + C day's work 80 part Required number of days 80 9 days 0 x 4 4 x x x. (B) Let there be x men originally, then man will do the work in 50x days. lso, man does the work in (x + 8) 0 days. Now, 50x 0(x + 8) 0x 40 x 40 men 0 There were men originally. 7. (C) S.I. at the rate of % for year P 00 P 00...(i) 0. (B) (C) Total amount is 00% Percent of amount left (00 ( + 5) 5% So, mount left () S.P. ` 45 L% 8% C.P P 8% 00 ` 94400

3 S.P. 00 P % C.P. 00 ` 405. () sec a + 4a, cos 4a 4a tan sec 4 a 4a 4a 4a m 7. (D) sin + sin... sin sin 89 sin cos (sin 89 cos ) [ sin(90 ) cos ] It is a series of P 89 + (n ) n 45 sin 45 terms terms 4 a 4a 4a 4. (C) x 8 + ab a a + sin 45 + sin + sin + sin sin 85 + sin 87 + sin (C) Two angles and B and > B. + B 0 4 x (4 ) x (D) Strength of milk in the first. (D) mixture Strength of milk in the second mixture Required ratio 9 : B D C Required height 9 : 7 99 : 9 70 cot 0 cot radian + B radian...(i) lso, B On adding both equations, π radian 9. (B) rea of the floor 9 4 sq. m 00 sq. dm. rea of a square tile sq. dm Number of tiles (D) B P O Q C D...(ii) Centroid is the point where medians

4 intersect. Diagonals of parallelogram bisect each other. OP cm OQ cm PQ + cm. (C) Let required number of wickets x TQ, 0. x + (x + 5) (0. 0.) 0.x + (x + 5) 0 0.x + 0x x 0x 50 0.x 4. (D) x Required number of wickets 40 D 0 O E B OE 0 DO [Given] EOB OE OB C 5 crore. (B) Change poetry into poem as we are talking about one poem not a collection of them. 7. (C) Replace out of by of. 8. (C) If the subject is joined by with the verb will agree with the first subject. So we need to replace were by was. 9. () If of is used after one, the noun or pronoun that comes immediately after of will be plural in form. So we need to replace assistant by assistants -. (B) (i) nnoyed with someone. (ii) nnoyed at something. 9. () V of throw is threw. 4. (B) You stake a claim (the best way to express it) 4. () The sentence will be in present perfect continuous tense as the action started in the past and is still going on. 5.() E 0 N/m m E P.E P 0 7 N 5.(B) x y z 5MPa Dimension E 00 GPa 0. OEB OBE 0 0 CBE () Percentage increase % 9 v ( x + y + z ) net dv v ( x + y + z ) ( x + y + z ) ( x + y + z ) ( ) ( ) E 4. (B) Required verage Sale crore 00 9 % % crore 5. (D) Required Total Sale ( ) crore 5 ( 0.) 00 0 dv v dv v ( )mm dv 90 mm 5.(B) x 0 MPa y xy 40 MPa max

5 5 40 max 4.7 MPa 0 MPa cm 4.7 MPa cm Radius of circle 4.7 cm. 0.() R + R ( ) + ( ) 8T R R R R 8 R 4T S.F x + 4 ( < x < D) S.F (x 0) 4 S.F (x ) 0 S.F 0 (D < x < E) So S.F 0 at mid pt. 7.(C) (Wt) Body (F B ) Oil + (F B ) mer b.v.g o m b b 7. g / cm 70.(C) u x + y v 4xy (a x ) convective V V o g m g u v u u u w y z y ( x y ) ( x y ) ( 4 xy) ( x y ) ( x y )(4 x) 4 xy( y) 8x + 4xy 8xy 8x 4xy a x (, ) 8 () 4() () 8 a x (, ) 8 unit u 7.(C) U e o y/ s Condition of boundary layer flow u (y 0) 0 u (y - o) U 0 e 0 U 0 77.(C) So this profile doesn t satisfying the condition. N N H H N H H N H 0 m 79.(B) Work done 00 J 0 minutes s J 00 00s s W 80 kj No heat interaction, so. Q O 8.(D) W act 50 KW max max T T w max Q max w max 40 kw I w max w act I 0 kw 84.(C) LH 0.5 SH SH SHF SH LH SH 95.(B) SH 0.5SH.5 SHF 0.8 d dh 4 5

6 d 4 F F F F F d / 4 d / 4 dh d F d 4h 99.() I 9.8 kg m N 0 rpm E 9 J E I. () N 00 rpm 0.(C) mm g 0m/s w g w 00 rad s 07.(D) C KN N 00 rev/min L n 000 hrs 0 N 0 0 Lh N 0 L n 0 million of revolution L n 7 million revolution C and L n P million of revolution So 7 P P kn / 7 P 5.88 kn 8.() Normal time observed time rating factor 0.4. Normal time 0.48 min and standard time Normal time + llowance Standard time 0.58 min 9.(C) D springs/nnum C o 00 Rs/order C h 0 Rs/ EOQ DCo C h Piece nnum EOQ 0 EOQ 4000 units 0.(D) Total float on activity 0 0 T.F 8 0 T.F 5