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3. (*)

100 [(0.003)2  (0.21)2  (0.0065)2 ] [(0.003)2  (0.21)2  (0.0065)2 ]

! 10

P(x) = 2×3×x2 (x – y) (x2+ x y + y ) Q(x) = 2×3×3x ×y 2 (x–y) (x2 +y2 –2xy) R(x) = 2×2×3×x ×y (x–y)3 H.C.F. = 6x (x – y) 5. (A) A – 10 = B + 10 A – B = 20 .... (i) Ÿ Ÿ and A + 20 = 2(B – 20) A – 2B = –60 .... (ii) Ÿ From Eqs. (i) and (ii), A = 100, B = 80 6. (C) A number is divisible by 8 if the number formed by the last three digits is divisible by 8, i.e. 58N is divisible by 8. Hence, N = 4 Again a number is divisible by 11 if the difference between the sum of digits at even place and sum of digits at the odd places is either 0 or divisible by 11, i.e., (M + 9 + 4 + 4 + 8) – (3 + 0 + 8 + 5 + N) = M + 25 – (16 + N) = M – N + 9 (M – N) + 9 must be zero or it must be divisible by 11. ( 584 is divisible by 8. ? N = 4) i.e. M – N = 2 M=2+4=6 Ÿ M = 6, N = 4 ? 7. (D) Height of pole = 15 metre. Speed of climbing = 5 metre/min Speed of sliding = 3 metre/min @ Distance climbed by monkey in 11 min. = 5 × 6 – 3 × 5 = 15 m. @ required time = 11 min. 8. (A) The four prime numbers are 5, 7, 11, 13 as 5 × 7 × 11 = 385 7 × 11 × 13 = 1001 Hence, the first prime number is 5. 9. (B) Let the present ages be x and y yrs. Then, x – y = 20 ..... (i)

gl

4. (B)

.in

2. (A)

14

5 7 9 1001 1001 x x x .. = 3 5 7 999 3 3600 = 4 × 9 × 100 = 22 × 32 × 52 × 22 = 24 × 3 2 × 5 2 3240 = 810 × 4 = 32 × 32 × 2 × 5 × 22 = 34 × 23 × 5 Third number = 22 × 35 × 72 NOTE : READ (0.021) 2 in place of (0.21) 2 in the denominator of question then the solution is

=

and (x – 5) = 5(y – 5) ..... (ii) From Eqs. (i) and (ii) 20 + y – 5 = 5y – 25 y = 10 yrs and x = 30 yrs Ÿ 10. (A) Let no. of persons buying the tickets on the three days are 2x, 5x, 13x respectively. No. of total tickets bought = 20x @ then from question, Total cost of tickets = 1 5 × 2 x +7 . 5× 5 x +2. 5× 13 x = (30 + 37.5 + 32.5)x = (100.0)x = ` 100x average cost of ticket per person @ = 100 x/20 x = ` 5 11. (D) Let the age's of three children be x1, x2 and x3 yrs. Then,

20

1. (C)

ª ª 1 ºª 3 ºª 5º 997 º « «2 – 3 »« »«2 – 5 »« »«2 – 7 » ».......... « «2 – 999 » » ¬ ¼¬ ¼¬ ¼ ¬ ¼

(SOLUTION)

=

20 100

§ 26  x 3 · ¨ ¸ 2 © ¹

x1  x 2  x 3 26  x 3 = ... (i) 3 10 Also, M + x1 = 39 ... (ii) From Eqs. (i) and (ii), we cannot determine the value of x2. 12. (D) Let x kg of good quality wheat is added in 150 kg of wheat. ATQ, 95% of (150 + x) = 135 + x 150 × 95 + 95x = 5x

-c

sc

.s

w

w

w

x1  x 2  x 3 3

Ÿ

750 =x 5 x = 150 kg Ÿ 13. (D) Given, b + c + d + g = 23

.... (i)

a + b + g + e = 15 .... (ii) e+f+g+d = 18 .... (iii) and a + b + c + d + e + f + g = 50 .... (iv) Solving Eqs. (i), (ii),. (iii) and (iv) b = 3, f = 6, d = 6, c = 9 and g = 5 14. (C) Let the price of sugar be ` x per kg. ? Initial expenditure = ` 30x New expenditure = ` 33x New monthly consumption ?

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=

33x =25 kg 1.32x

For more free Video / Audio Tutorials & Study Material visit www.ssc-cgl2014.in Facebook Page - www.facebook.com/cgl.ssc2014 15. (B) 21. (B) S. I. =

px rx t 100

S. I. x100 620 x100 ! pxt 2000 x 5 = 6.2% if r = 6.2 + 3 = 9.2%

r=

? Required number of musician

pxr xt 92x 5 ! 2000 x = ` 920 100 10 x100 @ Amount = ` 2000 + ` 920 = ` 2920 22.(B) Given P = ` 3000 r = 10% p.a. n = 3 year Let the total amount given by man = A Now, by formula

= 120 – (40 + 6 + 30) = 44 16.(*) Let weight of diamond = x From question, @ Initial cost of diamond = kx 2 where k = constant Let the weights of 4 pieces be y, 2y, 3y, 4y respectively, x = y + 2y + 3y + 4y @ x = 10y................(i) again, from question, ky 2 + k (2y) 2 +k (3y) 2+k (4y) 2 = 140000 30 k y2 = 140000

.in

[using (i)]

140000x100 30 Kx2 = ` 4.7 lakh (approx.) Initial cost of diamond = ` 4.7 Lakh @ 17. (A) Let the number of male and female participants at the start of seminar be 3x and x respectively.

­ª r ºn1 ª r ºn2½ ®«1 ¾ » ¾ ®« 100»»  ««1100» ¼ ¬ ¼ ¿¾ ¯®¬

­ª 10 º2 ª 10 º1½ ª 10 º3 ¾ » »» – 1000 ®®««1 »  ««1 100» »¾ = 3000 ««1  100 ¬ ¼ ¬ ¼ ® ¾¿ ¬ 100 ¼ ¯ = 3993 – 1210 – 1100 = ` 1683 23. (A) Amount remaining after

-c

gl

kx 2 !

14

x2 !140000 100

ª r ºn Amount ! P ««1 »» – A ¬ 100¼

20

30k

then, S. I. =

sc

3x  16 2 = x 6 1 3x – 16 = 2x + 12 Ÿ x = 28 Ÿ ? Total number of participants at the start of seminar = 3x + x = 4 × 28 = 112 18. (B) (17 + 19) = 36% of the cost price = ` 162

w

w

.s

Then,

162 × 100 36 = ` 450 19. (A) Original price of 250 chairs. = 250 × 50 = ` 12500 Price after discount

w

? Cost price

=

80 85 95 = 12500 × × × 100 100 100 = ` 8075 20.(D) SP = ` 17,940, Discount = 8% ?



17940 = ` 19500 0.92 Gain = 19.6% (given)

MP

=

17940 CP = = ` 15000 ? 1.196 New SP without discount = ` 19500 Gain = (19500 – 15000) = ` 4500 ? Gain percent =

4500 × 100 = 30% 15000

7.5 · § ¸ – 1500 = ` 2800 1 yr = 4000 ¨1  100 © ¹

7.5 · § ¸ – 1500 = ` 1510 2 yr = 2800 ¨1  © 100 ¹

7.5 · § ¸ – 1500 = ` 123.25 3 yr = 1510 ¨1  100 © ¹ 24. (B) Let the work be finished in x days. Then,

x (x  1) 2 + + 8 16 24

1

=

Ÿ

11 12

=

2x  x  1 16

Ÿ

3x – 1

=

16 u 11 12

=

47 = 5 days 9

Ÿ

x

25. (A) Anu's 1 day work

=

1 part 10

Manu's 1 day work =

125 1 1 × = part 100 10 8

Sonu's 1 day work

=

160 1 1 × = part 100 8 5

Total work

=

1 1 1 17 + + = 10 8 5 40

Total days

www.ssc-cgl2014.in

=

40 6 =2 days 17 17

For more free Video / Audio Tutorials & Study Material visit www.ssc-cgl2014.in Facebook Page - www.facebook.com/cgl.ssc2014 26. (C) In 600 m race, ratio of distances A : B 600 : 540 10 : 9 In 500 m race, ratio of distances B : C 500 : 475 20 : 19

As both trains are travelling in opposite direction.

Ÿ

Now,

40x = (48)2 – x2 x2 + 40x – 2304 = 0 (x + 72)(x – 32) = 0 x = 32 m/min 33. (C) Let the first instalment be 'a' and the common difference between any two consecutive instalments be 'd'. Using the formula for the sum of an AP. Ÿ Ÿ Ÿ Ÿ

gl

= 1125 ×

Sn

9 × 370 = ` 90 37 29. (B) X's investment = (700 × 3) ? A's share =

w

w

= (7320  7200 ) × 726 = ` 366 30.(C) Let two train meet at a distance x from Delhi then, x = 60 × t1 (Mumbai express) also, x = 80 × (t1 – 2) (Rajdhani express) Ÿ 60 t1 = 80 t1 – 160 or, 20 t1 = 160 or, t1 = 8h @ Required distance x = 60 × t1 = 60 × 8 = 480 km 31. (A) Let the length of platform be x m, length of first train be y m and length of second train be

y m. 2

n [2a + (n – 1)d] 2

40 [2a + (40 – 1)d] 2 = 20(2a + 39d) = 2a + 39d ... (i)

3600 = Ÿ

180

30 [2a + (30 – 1)d] 2 160 = 2a + 29d ... (ii) Solving Eqs. (i) and (ii), 20 = 10d d =2 Ÿ Therefore, 180 = 2a + 39 × 2 2a = 102 Ÿ a = 51 Ÿ Value of 8th instalment = 51 + (8 – 1) × 2 = 51 + 14 = Rs. 65 34. (C) Let n be the number of members in the club.

Again, 2400 =

5 3· § · § + ¨ 700 u u 3 ¸ + ¨ 500  200 u ¸ × 6 7 5¹ © ¹ © = ` 7320 Y's investment = 600 × 12 = ` 7200 ? X's share from profit 7320

=

We have,

3u3 9 B 3u4 12 = = : = = 4u3 12 C 4u4 16 A : B : C = 9 : 12 : 16

w

A 28. (D) B ?

2 = 50 km/h 45 Speed of train on return journey = 40 km/h

x

sc

?

1125 1 + x 2

.s

Ÿ

=

200 200 – (48  x ) = 10 (48  x )

Ÿ

-c

Ÿ 23

500 1 500 + + x 2 0.8x

.in

71u 400 200 = 342m A can beat C by 400 – 342 = 58 m 27. (B) Let the speed of train on onward journey be x km/h. Then, the speed of train on return journey = 0.8 x km/h.

When A runs 400 m o C runs

14

171 m 200

5 × 45 = 600 18 x = 600 – 200 = 400 m Ÿ 32. (D) Let the speed of current be x m/min. Then, speed with current = (48 + x) m/min & Speed against the current = (48 – x) km/h ATQ,

y + x = 48 ×

20

= 200 : 180 : 171 So, when A runs 200 m o C runs 171 m

Total time =

3 y = 300 2 y = 200 m

Ÿ

? A : B : C= (10 × 20) : (9 × 20) : (19 ×9)

When A runs 1 m o C runs

y 5 = (48 + 42) × × 12 2 18

So, y +

Then,

250 =

nª 3º 2 u 7  (n  1) » 2 «¬ 12 ¼

Ÿ

250 =

nª 1 1º 14   2 «¬ 4n 4 »¼

Ÿ

250 = 7n +

Ÿ

n = 25

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n n2 – 8 8

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2.5 5 (x – 4000) = x + 600 100 100 2.5x = 30000 Ÿ x = ` 12000 Ÿ 42.(A) Quantity of alcohol in 1 l mixture of first

2 1 ×1= l 10 5 As second bottle does not contains alcohol.

bottle =

.in

1 1 1 × = l 3 5 15 43. (C) Let the man purchased x pairs of brown socks. Price of black socks and brown socks be ` 2a and ` a per pair respectively. ATQ,

So, required fraction =

sc

-c

2 § 12 u 48 · 2 ¸ h = 10 h × ¨ 3 © 48  12 ¹ 3 So, total time to fill the tank

2 2 = 14 h. 3 3 38. (C) Work done by both the pipes in 4 min

.s

= 4 + 10

w

1 · 2 § 1  ¸ = work. = 4¨ 15 10 © ¹ 3 When all the pipes working together.

w

1 1 1 1   = 15 10 5 30

=

w

Work done =

1 part of th tank is emptied in 1 min 30

2 2 u 30 of the tank can be emptied in 3 3 = 20 min ?

39. (A) Required average

10 u 4.5  30 u 3.5 = 40

=

45  105 40

?

3 (4 × 2a + x × a) = x × 2a + 4 × a 2

Ÿ

12a +

Ÿ

3 xa = 2xa + 4a 2

12 +

3 x = 2x + 4 2

x =8 2 x = 16

Ÿ

Ÿ

4 1 = 16 4 44.(B) Given, C. P. of the goods = ` 840 ?

Required ratio =

1 x 840 = ` 210 4 & C. P. of the remaining quantity of goods = ` 840 – ` 210 = ` 630 Let x % be the profit at which the remaining quantity of goods be sold. then, from question,

@

C.P. of 1/4 of the goods =

210 x

80 (100  x ) 120  630 x ! 840 x 100 100 100

168 

63 (100  x ) ! 1008 10

1 % 3 45. (A) Let he bought x number of CDs. (x – 1) × 6 = 114 ? x – 1 = 19 Ÿ x = 20 Ÿ 46. (*) Ratio of the efficiencies of A, B, C & D

x = 33

150 15 = 40 4 40.(A) Let the quantity of haematite mined be x kg. ATQ, Pure Iron = 8000 kg

=

=

1000 +

gl

§ 20 u 30 · ¨ ¸ h = 12h © 20  30 ¹ One third tank can be filled in 4 h. Now, there is a leak which can empty the tank in (12 × 4)h = 48h So, two-third tank can be filled in

x

14

95 106 + (2200 – x) × = 2200 100 100 95x + 233200 – 106x = 220000 Ÿ 11x = 13200 Ÿ x = `1200 Ÿ and 2200 – x = ` 1000 36. (A) Let P and M denote Pintu and Mintu respectively. Case 1:P + n = 4(M – n) P – 4M = –5n ... (i) Ÿ Case 2:(P – n) = 3(M + n) P – 3M = 4n ... (ii) Ÿ Solving Eqs. (i) and (ii), we get M = 9n and P = 31n Put n = 1, we get P = 31 37. (A) Together both pipes can fill the tank in

x×

80000 u 100 u 100 80 u 25 = 400000 kg 41. (D) Let his sales were x. Then, ?

20

35. (C) Let the cost price of one table be ` x. Then, Cost price of other table will be ` (2200 – x) ATQ,

=

80 25 xu u = 80000 100 100

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1 1 1 1 : : : 32 20 30 24

= 15 : 24 : 16 : 20

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d h ? Usual time (t) = s

t–

..... (i)

1 d = 2 s4

2n [2a + (2n – 1)d] 2

and S3n =

3n [2a + (3n – 1)d] 2

w

Ÿ

d

?

Sn =

=

and S3n =

20

gl

2n n [2a + (2n – 1)d] =3 [2a + (n – 1)d] 2 2 4a + (4n – 2)d = 6a + (3n – 3)d d(4n – 2 – 3n + 3) = 2a

w

Ÿ Ÿ

w

Given S2n = 3Sn

­° sin2 D  sin2 E ½° ¾ AB2 = h2 ® °¯ sin2 D sin2 E °¿

.s

sc

S2n =

-c

n [2a + (n – 1)d] 2

h OB Ÿ OB = h cot E Now:In ' OAB:OB2 = OA2 + AB2 Ÿ AB2 = OB2 – OA2 AB2 = h2 cot2 E – h2 cot2 D AB2 = h2[cot2 E – cot2 D ] AB2 = h2[(cosec2 E – 1) – (cosec2 D – 1] = h2[cosec2 E – cosec2 D ]

tan E =

t+

Then, Sn =

?

h OA Ÿ OA = h cot D In ' OBP:-

tan D =

..... (ii)

1 d = ..... (iii) 3 s4 Solving Eqs. (i), (ii) and (iii), d = 60 km, s = 20 km/h and t = 3 h 49. (B) Let 'a' be first term and d be the common difference.

and

In ' OAP:-

14

?

51. (A)

.in

40 × 25 99 = ` 10.10 47. (C) Let the number of computers required = N Tasks done by the computers 6h = 30 tasks 1h = 5 tasks 3h = 15 tasks So, 15 N = 80 N = 5.33 | 6 Ÿ 48. (A) Let the distance be 'd' km and usual speed of the Toy train be 's' km/h ? C's share

2a n 1

AB sin D sin E ?

h

=

sin2 D  sin2 E

52. (B) sec T + tan T = P (sec T + tan T )2 = P2

(on squaring)

2

§ 1  sin T · ¸ = P2 Ÿ¨ © cos T ¹

(1  sin T)2

2an 2 n 1

Ÿ

12an 2 n 1

1  sin T = P2 1  sin T Applying C & D :-

1  sin2 T

= P2

Ÿ

?

Sn n 1 1 12an 2 = × = S 3n 6 12an 2 n 1

1  sin T  1  sin T P 2 1 Ÿ 1  sin T  (1  sin T) = 2 P 1

Ÿ

S 3n =6 Sn

Ÿ

50. (C) a = –3, d = 3 ? T10 = a + (10 – 1)d T10 = –3 + 9 × 3 = 24

P 2 1 2 = 2 2 sin T P 1

Ÿ sin T =

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P 2 1 P 2 1

For more free Video / Audio Tutorials & Study Material visit www.ssc-cgl2014.in Facebook Page - www.facebook.com/cgl.ssc2014 53. (A)

tan D =

cot D  cot2 E

54. (A) sin T + cos T =

Ÿ

20

150 u 49 = 294 cm2 25 60. (B)  PQ||BC & also P is the mid point of AC. Ÿ Q is the mid point of AB. Now, PB2 = PQ2 + QB2 ? ar ('QOA) =

-c

2

Ÿ sin T = ( 2 – 1) cos T

1 a3

1 b3

1

1

1 c3

2 1 2 1

=

w

55. (B)

2 1

=

2 1

.s

1

Ÿ cot T =

=0

w 3

2

1 [BC2 + AB2] 4

=

1 AC2 4

ª « PQ ¬

1 º BC» 2 ¼

1 AC Ans. 2

61. (B) 3

w

1· § 1 § 1· ¨a 3  b 3 ¸ ¨c 3 ¸ Ÿ ¨ ¸ = ¨ ¸ © ¹ © ¹

=

? PB =

1

Ÿ a3  b3 = c 3

2

§1 · §1 · = ¨ BC ¸ + ¨ AB ¸ ©2 ¹ ©2 ¹

sc

2 cos T

...... (B)

150 52 = ar ('QOA ) 72

gl

=

...... (A)

ar ('POB ) PO 2 Ÿ ar ('QOA ) = QO 2

d

h

...... (ii)

.... (i)

h tan E = OB .... (ii) Ÿ OB = h cot E Since ' OAB is a right angle triangle. ? AB2 = OA2 + OB2 d2 = h2cot2 D + h2cot2 E ?

131x + 217y = 827 Adding equation (i) and (ii):348x + 348y = 1740 x+ y =5 Subtracting (ii) from (i):86x – 86y = 86 x – y =1 Now:from (A) and (B):2x = 6 x=3 and, x + y = 5 ?y=2 59. (A)  ' POB ~ ' QOA

14

h OA Ÿ OA = h cot D In ' OBP:-

...... (i)

.in

In ' OAP:-

58. (B) 217x + 131y = 313

1 1§ 1 1· Ÿ a  b  3a 3 .b 3 ¨a 3  b 3 ¸ = – c ¨ ¸ © ¹ 1

1

1

Ÿ a + b + c = 3a 3 .b 3 .c 3 Ÿ (a + b + c)3 = 27abc

 OD A BE & AE A BE Ÿ OD || AE, also 'O' is the mid point of AB.

1 Ÿ ? OD = AE 56. (B) Volume of the cubiod = 3x2 – 27 2 = 3(x2 – 9) 1 = 3(x – 3)(x + 3) 8 = AE Ÿ 2 ? possible dimensions are 3, (x – 3), (x + 3) AE = 16 cm Ans. In ' AED, 57(C)  x2 – 1 is a factor of the polynomial AD2 = AE2 + DE2 P(x) = ax4 + bx3 + cx2 + dx + e = 162 + 105 ? P(–1) = 0 or P(1) = 0 4 3 2 = 256 + 105 Ÿ a(–1) + b(–1) + c(–1) + d(–1) + e = 0 = 361 or, a – b + c – d + e = 0 AD = 19cm Ÿ Ÿ a + c + e = b + d Ans. www.ssc-cgl2014.in

For more free Video / Audio Tutorials & Study Material visit www.ssc-cgl2014.in Facebook Page - www.facebook.com/cgl.ssc2014 65. (A)

62. (C) DE || BC

(  ' ADE ~ ' ABC)

A1A2 = 1000 m = 1 km and ‘ A2A1B1 = 30º ‘ TA2B2 = 60º In ' A1A2Q

3 u 5.6 = 2.1 cm 8 63. (D) r1 & r2 be the outer & inner radii of the cylinderical pipe. ? AE =

= 44

2S(r1  r2 ) u 14

=

44 u 7 1 = 2 u 22 u 14 2

.... (i)

Ÿ A1Q =

gl

.s

w

=

3 km 2

......(ii)

TB1 x = A1B1 A1B1

......(iii)

In ' TA2B2:TB2 TB1  B 2B1 TB1  B 2B1 tan 60º= A B = = AB AQ QB 2 2 1 1 1 1

sc

-c

tan 45º =

9 1 5 ÷ = .... (ii) 4 2 2 On solving (i) & (ii) r1 = 2.5 cm, r2 = 2 cm

r1 + r2

3 ×1= 2

In ' TB1B2:-

99 u 7 9 = = 22 u 14 4

9 (r1 + r2) (r1 – r2) = 4

......(i)

A1Q cos 30º = A A 1 2

= 99

22 2 (r1  r22 ) u 14 = 99 7  r22

1 1 ×1= km = B2B1 2 2

20

Sr12h  Sr22h

1 2 3 = 3 x 2 x

Ÿ

[From (i), (ii) and (iii)

w

Ÿ x = 1.366 km

66. (A)

w

64. (A)

? A2Q =

14

1 r1 – r2 = 2 Again

r12

A 2Q sin 30º = A A 1 2

.in

2Sr1h  2Sr2h

Volume of water in the reserviour = 0.075 × 0.055 × 18 × 30 × 60 m3 = 133.65 m3 Now:Volume = Area × height or, 133.65 = 10.3 × 3.75 × h ?

Given that:Height of pryamid = 21 cm [square] Base = 40 cm Now:Height of triangle =

20 2  212

=

400  441

= 841 = 29 cm Slant Surface Area of Pyramid = 4 × {Area of triangle} ª1 º = 4 × « u 40 u 29» 2 ¬ ¼ = 4 × 20 × 29 = 2320 cm2.

h

=

133.65 = 3.3 m 10.8 u 3.75

67. (A) Let:sec T + tan T = K Then:sec2 T – tan2 T = 1 K(sec T – tan T ) = 1 1 sec T – tan T = K Adding (i) and (ii):2sec T = K +

1 K

1 º ª 1 2 «x  =K+ » 4x ¼ K ¬

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..... (i)

..... (ii)

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1 1 =K+ 2x K K = 2x

2

or, 2x 

Ÿ [sin 3 T]3  [cos 3 T] 3

Ÿ

Ÿ sin2 T + cos2 T = 1

71. (A)

1 1 =– K 2x

Ÿ Ÿ Ÿ Ÿ Ÿ Ÿ Ÿ Ÿ Ÿ

69. (A) ?

sin T + sin2 T + sin3 T = 1 sin T + sin3 T = 1 – sin2 T sin T [1 + sin2 T ] = cos2 T sin2 T [1 + sin2 T ]2 = cos4 T [on squarring] (1 – cos2 T )[1 + (1 – cos2 T )]2 = cos4 T (1 – cos2 T )[2 – cos2 T ]2 = cos4 T (1 – cos2 T )[4 – 4cos2 T + cos4 T ] = cos4 T 4 – 4cos2 T + cos4 T – 4cos2 T + 4cos4 T – cos6 T = cos4 T –cos6 T + 4cos4 T – 8cos2 T + 4 = 0 cos6 T – 4cos4 T + 8cos2 T = 4 3x = 2x + x cot (3x) = cot (2x + x)

gl

cot 2x cot x  1 cot 2x  cot x or, cot 3x.cot 2x + cot 3x.cot x = cot 2x.cot x –1 ? cot 2x.cot x – cot 3x.cot 2x – cot 3x.cot x = 1 70. (A) cosec T – sin T = m

A = (4, 3) B = (x, 5) Origin at (2, 3) Now:OA2 = OB2 or, (2 – x)2 + (3 – 3)2 = (2 – 4)2 + (3 – 3)2 or, 4 + x2 – 2x + 4 = 4 or, (2 – x)2 = 0 ? x=2 72. (B)

.in

68. (A)

1 2x

14

? sec T + tan T = K = 2x or –

20

Ÿ

-c

or, cot 3x =

sc

1 – sin T = m sin T

Area of ' AOB =

1 ×6×3 2 = 9 sq. unit. 73. (A) From the figure

w

cos 2 T =m sin T m = cot T . cos T sec T – cos T = n

5 1 = 10 2 ? T = 60º Now, reflex ‘ AOB = 360º – 120º = 240º Now, length of the belt

w

w

cos T =

1 – cos T = n cos T sin2 T =n cos T n = tan T . sin T 2 mn2 3

Ÿ

 m2n

2 3

2 (cot T. cos T).(tan T. sin T)2 3

SrT 180º

=

S u 5 u 240º 180º

20S cm 3 74. (A)Reflex ‘ AOC = 360º – 40º = 320º

+

2

Required area = 2

Ÿ (cot T. tan T. sin2 ) 3 + (cot T. sin T. cos 2 ) 3 2

=

=

(cot T. cos T)2.(tan T. sin T) 3 2

1 ×b×h 2

=

.s

1  sin2 T =m sin T

Y = |x| Y=3 Now:-

= 2

sinT ª º3 ª cos T º3 .sin2 T» + « .sin T. cos 2 T» Ÿ «cosT. cos T ¬ ¼ ¬ sin T ¼

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Sr 2 u 320 º Sr12 u 320 º – 2 360 º 360 º

22 320 2 2 u r  r2 7 360 1

22 320 u [142  72 ] 7 360 = 410.67 cm2.

=

For more free Video / Audio Tutorials & Study Material visit www.ssc-cgl2014.in Facebook Page - www.facebook.com/cgl.ssc2014 75. (A) Volume of regular tetrahedron:-

?

a a

a

=

3

6 2

= 6 2 × 2.45 = 2.75 m

2 = 2.75 × 0.816 3 = 2.245 m 76. (B)Volume of the cube = a3 cubic unit Volume of the cylinder

height = a u

?

§a · ©2¹

2

= Su ¨ ¸ ua =

a2  b2 ab

or,

or, a2 + b2 = –ab 2 or, (a + b2 + ab) = 0 Now:(a3 – b3) = (a – b) (a2 + b2 + ab) = (a – b) × 0 = 0 =0

S 3 a cubic unit 4

Volume of the cone

80. (D) Multiplying with

2

1 §a · S 3 S¨ ¸ u a = a cubic unit 3 ©2¹ 12

3 5

S 3 S 3 Required Ratio = a3 : a : a 4 12

22 × r = 28 cm 7

28 u 7 2 u 22 r = 4.4545 cm Now, height of cup will be h cm. ?

=

=

l2  r 2

=

142  (4.4545 )2

=

196  19 .8429

w

w

h

w

= 176 .1570 = 13.12 cm (approx.) 78. (D) Let the number of camels = x According to question:1 x + 2 x + 15 = x 4 Replace x and by y2 :1 2 y + 2y + 15 = y2 4 1 2 or, y2 – y – 2y – 15 = 0 4 or, 3y2 – 8y – 60 = 0  8 r 82  4 u 3 u 60 y= 2u 3

y=

8 r 784 6

y = 6,

 10 3

20

sc

r

.s

?

1 =3 5a Squaring both sides:-

gl

2×

1 · § 3 ¨ 5a  ¸ =5× 3 a 5 © ¹

3a 

-c

S S =1: : 4 12 = 12 : 3 S : S Ans. 77. (D) Base of the ice cream cup:2 S r = 28 cm

3 5

14

=

= –1

.in

V

Now:y2 o x ? x = (6)2 = 36 ? No. of camels = 36 m a b 79. (D) + = –1 b a

3

9a2 +

Ÿ 9a2 +

1 25a 2

1 25a

2

+ 2 × 3a ×

=9– =

1 =9 5a

6 5

45  6 39 = 5 5

81.(D)  ' AOB ~ ' COD OB OA = ? OD OC OB × OC = OA × OD (x – 3)(x – 5) = (3x – 19) × 3 x2 – 8x + 15 = 9x – 57 x2 – 17x + 72 = 0 (x – 9) (x – 8) = 0 x = 9, 8 82. (D) Given,

Area of PQRS = P Area of PQUT = R Area of PSQ = T 1 Area of 2 parallelogram PQRS. Area of parallelogram = area of rectangle (constructed on the same base and between the same parallels) Hence, options (a), (b) and (c) are correct.

Now, area of triangle =

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For more free Video / Audio Tutorials & Study Material visit www.ssc-cgl2014.in Facebook Page - www.facebook.com/cgl.ssc2014 83. (C) Let O be the starting point.

Now:§d · a2  ¨ 1 ¸ © 2¹

d2 = 2

d2 = (32.5)2  (16 .5)2 2 ? d2 = 56 m Now:-

Area of base =

d1d2 33 u 56 = = 924 m2 2 2

Now:Volume of prism = Area × height = 924 × 2.6 = 2402.40 m3 88. (D) To construct a triangle sum of any two sides is greater than the third side. Hence, option (d) is the correct answer.

14

.in

OQ is the desired shortest distance. ? (OQ)2 = (50)2 + (120)2 = 2500 + 14400 ? = 16900 OQ = 130 m ? 84. (B) Let r1, r2 and r3 be the radii of three circles, then

2

sc

.s

w

w

Ÿ l2 + b2 + h2 = 121 cm2 ? Area of cuboid

w

= 2{lb + bh + lh} = (l + b + h)2 – {l2 + b2 + h2} = 192 – 121 = 361– 121 = 240 cm2 The base is isosceles triangle.

? Area of base =

90. (C) tan T =

=

a – (BD + CD) 2

=

a b – –c 2 2

=

1 (a – b) – c 2

1  sin T  cos T 1  sin T  cos T Dividing by :- cos T 1  tan T  1 cos T 1  tan T  1 cos T

40 4 u (25 )2  402 4 = 300 m2

=

Volume 2400 = = 8m ? Height = Area 300 87. (A) The base of a rhombus whose perimeter = 4a = 130

20 21

=

b 4a 2  b 2 4

130 = 32.5 m 4 One diagonal, d1 = 33 m

b and CD = c 2

Then, AB = AC – BC

l 2  b 2  h 2 = 11 cm

86. (A)

a 2

gl

BD =

-c

r1 + r2 = 2.2 ...... (i) r2 + r3 = 3.2 ...... (ii) r1 + r3 = 4.0 ...... (iii) Adding Eqs. (i), (ii) and (iii) 2(r1 + r2 + r3) = 9.6 r1 + r2 + r3 = 4.8 Ÿ r3 = (4.8 – 2.2) = 2.6 ? r1 = (4.8 – 3.4) = 1.4 r2 = (4.8 – 4.0) = 0.8 ? The diameter are 2 × 1.4 = 2.8, 2 × 0.8 = 1.6 and 2 × 2.6 = 5.2 85. (A) l + b + h = 19 cm length of diagonal = 11 cm Now:-

20

89. (D) Given, AC =

tan T = ?

h=

20 P = 21 b 212  202

= 441  400 h = 29

?a =

cos T =

www.ssc-cgl2014.in

b 21 = h 29

For more free Video / Audio Tutorials & Study Material visit www.ssc-cgl2014.in Facebook Page - www.facebook.com/cgl.ssc2014 9 20  1 29  20  21 30 3 = = = Ÿ 21 21 9 20 29  20  21 70 7  1 21 21

25 = 9.75 tonnes 100 Total revenue earned = 9.75 × 1000 kg × 20 = 1.95 lakh 99. (C) When total fruit production is increased by 12% in 2003, then

= 39 ×

91. (D) Cannot be determined. 92. (D) Cannot be determined. 93. (B)

112 = 196 tonnes 100 Production of mangoes

= 175 ×

26 = 50.96 tonnes 100 100. (C) Total fruit production in 1998 = 100 tonnes Grapes production in 1998

.in

= 196 ×

14 = 14 tonnes 100

14

= 100 ×

14 2 = 7 tonnes

Price tonnes

sc

-c

gl

Only agriculture and commercial increases by more than 50% during the same period. 94. (C) 95. (B) Industrial + Agriculture will be more will be more than 50%. 96. (A) 20%. 97. (B) In 1998, the production of total fruits = 100 The Guava production in 1999

20

Half of grapes exported =

15 = 15 tonnes 100 In 1996 = 10% lower than 1998

w

.s

= 100 ×

90 = 13.5 tonnes 100 98. (A) Mangoes produced in 2001

w

= 15 ×

w

26 = 39 tonnes 100 25% exported earned

= 150 ×

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1.4 u 100000 7 = ` 20000

=

For more free Video / Audio Tutorials & Study Material visit www.ssc-cgl2014.in Facebook Page - www.facebook.com/cgl.ssc2014 SSC MAINS (MATHS) MOCK TEST -1 (ANSWER SHEET) 61. 62. 63. 64. 65. 66. 67. 68. 69. 70. 71. 72. 73. 74. 75. 76. 77. 78. 79. 80.

B C D A A A A A A A A B A A A B D D D D

81. 82. 83. 84. 85. 86. 87. 88. 89. 90. 91. 92. 93. 94. 95. 96. 97. 98. 99. 100.

sc .s w w www.ssc-cgl2014.in

D D C B A A A D D C C D B C B A B A C C

.in

D A C B A * C A B C A B A A B B C B A B

14

41. 42. 43. 44. 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56. 57. 58. 59. 60.

20

B B A B A C B D B C A D C C C A A C A A

gl

21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40.

-c

C A * B A C D A B A D D D C B * A B A D

w

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20.