Ssc Mains

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�MUKHERJEE NAGAR �MUNIRKA �UTTAM NAGAR� DILSHAD GARDEN �ROHINI�BADARPUR BORDER

SSC (Tier-II) - 2013 (Mock Test Paper - 1) [SOLUTION]

1. (A) ∵ Ratio of diameters of the cylinders

= 3 : 2

⇒ Ratio of radii of the cylinders

= 3 : 2

So, Let the radii of the two cylinders are 3rand 2r

and, Let the heights of the two cylinders are h

1 and h

2.

Now,

Volume of first cylinder

= volume of second cylinder

i.e. 12)3( hrπ = 2

2)2( hrπ

⇒ 2

1

h

h= 2

2

9

4

r

r

×π

×π=

9

4⇒ 4 : 9

2. (B) Difference in votes of candidates

= (100% – 46%) – 46% of the total votes polled

= 8% of the total votes polled

= 3680 votes

So,

Total votes polled (i.e. 100%)

= 1008

3680× = 46000

3. (C)Let C.P. = x

⇒ M.P. = 1.4x

⇒ S.P. = 0.75 × 1.4x = 1.05 x

⇒ % Profit = %100)05.1(×

−

x

xx

= 5%

4. (D)To keep the expenditure (Rs. 608) constant,

If % discount in initial cost of sugar = 5%

⇒ % increases in initial consumption

= %1005100

5×

−

= %19

100= 2 kg

⇒ Initial consumption (i.e. 100%)

= (2 × 19) kg

= 38 kg

So, Inititial S.P. of sugar

= kg38

608.Rs = Rs. 16/kg

5. (B) Ratio of the two numbers = 3 : 4

[L.C.M. = 12]

L.C.M. of the two numbers = 48

= 412

48

So, the two numbers are (3 × 4) & (4 × 4)

(∵ 3 & 4 are co-prime numbers)

i.e. 12 & 16

and so, sum of the two numbers

= 12 + 16

= 28

6. (C) 2A = 3B = 4C = k (let)

So, A : B : C

=2

k:

3

k:

4

k

= 122×

k: 12

3×

k: 12

4×

k

= 6k : 4k : 3k

= 6 : 4 : 3

7. (A) 43.5 : 25

= (22)3.5 : 25= 27 : 25

= = 22 : 1

= 4 : 1

8. (D) Required average price

= )812(

)408()3012(

+

×+×

= Rs.34/ kg

9. (A)

15cm

30cm

∵ Perimeter of square = 120 cm

⇒ Each side of the square

= 4

120cm = 30 cm

⇒ Radius of the inscribed greatest possible

circle = 2

30cm = 15 cm

⇒ Area of the circle = 22cm)15(×π

= 22 cm)15(

7

22×

10. (D)Now,

Required ratio =

T.S.A. of one small cubes : T.S.A. of the big cube

= 6 × (1)2 : 6 × (5)2

= 1 : 25

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11. (D) From both the conditions, we have relation

2.5 km/hr × (t + 6) min

= 3.5 km/hr × (t – 6) min

(where t = actual time in minute)

⇒5.2

5.3

6t

6t=

−

+ ⇒ t = 36 minutes

So,

Required distance

= 2.5 km/hr × (36 + 6) minutes

[or, = 3.5 km/hr × (36 – 6) minutes]

= 4

31 km

12. (C) As distance is the same,

So, the required distance (from one side)

= 2

timetotalSpeedverage ×Α

= 2

hr60

348hr/km

425

4252

×

+

××

= 2

40km = 20 km

13. (D) Volume of cone = hr 2

3

1π

After increment,

New vol. = )h2.1()r2.1(3

1 2π

= 728.13

1 2 ×π hr

So, Required % increase

= %1001

1728.1×

−

= 0.728 × 100%

= 72.8%

14. (C) % (Fail in Hindi U Fail in English)

= % (Fail in Hindi + % (Fail in English – % (Fail in both Hindi & English)

= 52% + 42% – 17%

= 77%

So,

% (Passed in both the subjects)

= 100% – 77%

= 23%

15. (A) 5mainderRe:9

6709=

⇒ The required least number = 5

16. (C) 13.103.103.10

13.103.103.10

+−×

+××

r + 20%

1.2hh + 20%

1.2r

= 22

33

1)13.10()3.10(

1)3.10(

+×−

+

+=

+−

+∴ ba

baba

ba22

33

= (10.3 +1)

= 11.3

17. (D) Required no. = H.C.F. of (122 – 2) & (243 –3)

= H.C.F. of 120 + 240

= 120

18. (B) Money left = 100% – (80% + 6% of 20%)

= 100% – 81.2%

= 18.8% of total pocket money

And,

ATQ.

18.8% of total pocket money

= 47 paise

= Rs. 100

47

So, Total pocket money

(i.e. 100%)= Rs. 8.18

100

100

47×

= Rs. 2.5

19. (C) Let, r = radius of the circular field

Land portion of the circular field

= Total area of the circular field – Area of

the rectangular tank

⇒ 40000m2 = 22 m)120180(r ×−π

⇒ 2rπ = 61600 m2

⇒ r = 22

761600×m

= 19600 m

= 140 m

20. (B) Total C.I. in 2 years @ 12.5% p.a.

= (12.5%) + (12.5% + 12.5% of 12.5%) of sum

= sumof%8

5.12%25 +

= sumof%8

5.212

= Rs. 510

So,

Total S.I. in 2 years @ 12.5% p.a.

= (2 × 12.5%) of the sum

= 25% of the sum

= 255.212

8510×

×

= 480

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21. (C)

1st condition

2nd condition

x 20% 1oss 0.8x

x 5% profit 0.8 x + 100= 1.05x

Let initial C.P. of the article.x =

C.P. S.P.

⇒ 1.05 x = 0.8x + 100

⇒ x = 25.0

100= 400

22. (D) Total S.I. = (3 ×12)% of the principal amount

= 36% of the principal amount

= Rs. 5400

So, The principle amount (i.e. 100%)

= Rs. 10036

5400×

= Rs. 15000

23. (C) S.I. in 4 years – S.I. in 2.5 years

= S.I. in 1.5 years

⇒ S.I. in 1.5 yrs = Rs. (1067.20 – 1020)

= Rs. 55.2

⇒ S.I. per year = Rs. 5.1

2.55= Rs. 36.8

So,

Principle money = Rs. 1067.2 – Rs. (36.8 × 4)= Rs. 1067.2 – Rs. 147.2

= Rs. 920

So,

Rate of interest per annum

= %100920

8.36×

= 4%

24. (C) 5012832 +−

= 252642162 ×+×−×

= 252824 +−

= 2 = 1.414

25. (C) Required distance

= 2

TimeTotalSpeedAverage ×

= 2

1)15()15(

)15)(15(2×

−++

−+×

km

= 2

8.4km = 2.4 km

26. (A) Total % discount on M.P.

= %404

120

4

10

2

1

×+×+×

= (0 + 5 + 10)%

= 15%

Now,

If C.P. = x

then, M.P. = 20% above C.P.

= 1.2x

⇒ M.P. after discount = 8.5% of 1.2x

= 0.85 × 1.2x

= 1.02x

So,

Total gain % = %10002.1

×−

x

xx

⇒ = 0.02 × 100%

= 2%

27. (?) Money spent on article

= 25% of total amount

Money spent on cloths

= 10% of remaining (75%) amount

= 7.5% of total amount

⇒ (25% + 7.5%) of total amount + Rs. 531.25

= Total amount – Rs. 8000

⇒ Total (100%) amount – 32.5% of totalamount = Rs. 8000 + 531.25

= Rs. 8531.25

⇒ 67.5% of the total amount = Rs. 8531.25

So,

Money spent on clothes

i.e. 7.5% of the total amount

= 5.75.67

25.8531×

= Rs. 948

28. (D) Required time = 4060

4060

−

× minutes

= 20

2400minutes

= 120 minutes

29. (A) Average Speed= 6040

60402

+

××km/hr

= 100

4800 km/hr

= 48 km/hr

30. (B)

Total distance covered

= 12 km + 12 km

= 24 km12 km

down warddown ward

up ward

12 km

Total time taken = 3 hours

⇒ Average Speed = 3

24= 8 km/hr

Now,

⇒updown

updown

SS

SS2

+

××

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⇒)3()3(

)3)(3(2

−++

−+×

BB

BB

SS

SS

⇒ SB = 9 km/hr

(we can find it out from options also)

31. (B) B

C

A

B

A = 1.6B ...(i)

B = 0.8C ...(ii)

+60%

–20%

From (i) & (ii),

A = 1.6 B = 1.6 × 0.8C

⇒ A = 1.28C

⇒ A : C

1.28 : 1

128 : 100

32 : 25

32. (B) ATQ,

1352 = x × (1.04)2 (where x → sum)

⇒ x = 2)04.1(

1352

= 0816.1

1352

= Rs. 1250

33. (A) 33

22

yx

xyyx

−

++=

yx −

1

[ ))(()( 2233 xyyxyxyx ++−=−∴ ]

= 11

1

1819

1==

−

34. (C) Sum of first 25 natural numbers

i.e. 1 + 2 + 3 + 4 + ...... + 25

=2

2625×

= 25 × 13

⇒ Required factor = 13

35. (C)

2

2

12

+ =

2

122

2

1)2(

2

2 ××+

+

= 22

12 ++

= 2

14

36. (B) i.e. 22 + 24 + 26 + ....... + 50

= no. of terms × average

=

+×

+

−

2

50221

2

2250

= 15 × 36 = 540

37. (D) 3.276.5 −

= 9

32

99

765 − =

9

21

99

571−

= 99

231571−=

99

340= 43.3

38. (A) (0.5 × 5 + 0.25 × 0.5 + 0.5 × 4 + 0.5 × 0.75)

= 2.5 + 0.125 + 2 + 0.375

= 5

39. (D) A : B : C

1 : 2 : 3 [Average = 3

321 ++=2]

∵ Average = 600 ⇒ 2 ≅ 600

So,

A : B : C

1 : 2 : 3

300 600 900

Now,

A

B

Average

+10%300 + 30 = 330 (new value of A)

600 – 120 = 480 (new value of B)

600 + 30 = 630 (new average)

–20%

+5%

Now,

3

Cofvaluenew480330 ++= 630

⇒ new value of C (630 × 3) – (330 + 480) = 1080

⇒ Increase in C = 1080 – 900 = 180

40. (C) Weight of new person = (80 – 6 × 3) kg

= (80 – 18) kg

= 62 kg

41. (A) D – A = 3(5000 – 4000)

or, D – A = 3 × 1000 = Rs. 3000

⇒ D – 2750 = Rs. 3000

⇒ D = Rs. (3000 + 2750)

= Rs. 5750

42. (B) Required % decrease = %10010100

10×

+

= %100110

10×

= %11

19

43. (D)

S.P.

0.8 = Rs. 24x

List pricex

+20% discount

(–20%)

S.P. after 20% discount i.e. 0.8x = 24

So,

S.P. after 30% discount i.e. 0.7x = 7.08.0

24×

= Rs. 21

44. (B) Required single discount

= (1 – 0.7 × 0.8 × 0.9) × 100%

= (1 – 0.504) × 100%

= 0.496 × 100%

= 49.60%

45. (A) Let x = original number of workers

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So, ATQ, 22x = 24(x – 3)

or, 22x = 24x – 72

⇒ x = 36

46. (B) The number of days taken by A, B and C tocomplete the work while working together

=

×

++

2

36

.M.C.L

24

.M.C.L

18

.M.C.L

36&24,18of.M.C.L days

=

×

++

2

36

72

24

72

18

72

72days

=

×++

2)234(

72days

= 9

272× days

= 16 days

47. (B)

16 m

20 m 20 m

12 m

In the given figure,

width of the street

= (x + y)

= 2222 12201620 −+−

= 256144 +

= 12 + 16

= 28 m

48. (B) Cost price of each marble @ 20 per rupee

= Rs. 20

1

Cost price of each marble @ 30 per rupee

= Rs. 30

1

⇒ Average cost price of each marble

= Rs. 2

30

1

20

1+

[∴ no. of marbles are equal]

= Rs. 24

1

ATQ,

Selling price of each marble @ 25 per rupee

= Rs. 25

1

⇒<< loss.P.C.P.Sin

24

1

25

1∵

So, % loss = 100.P.C

.P.C.P.S×

−

= %100

24

124

1

25

1

×−

= %100600

24)2524(×

×−

= – 4%

⇒ 4% loss

49. (B) Loss of 20% on one and gain of 20% on other.

⇒ there will be a loss

and loss % = %100

)20( 2

= 4% loss (on total C.P.)

⇒ C.P.(100%)

S.P.–4%

(96%)=12,000 ×2 = 24000

S.P. (i.e. 96%) = 24000

So, loss in transaction (i.e. 4%)

= 496

24000× = 1000

⇒ loss of Rs. 1000

50. (D)

116 – 92 = 24 Let x = profit when S.P. = 92i.e. 3 x – x = 24 So, 3 x = profit when S.P. = 116 ⇒ x = 12

⇒ When S.P. = 92, Profit = Rs. 12

So, C.P. = 92 – 12 = Rs. 80

51. (C) C.P. of each article = Rs. 73

5110= Rs. 70

S.P. of each article = Rs. 89

5607= Rs. 63

Here,

S.P. < C.P. ⇒ loss

and % loss = %100.P.C

.P.C.P.S×

−

= %10070

7063×

−

= –10%

= 10% loss

52. (D) C.P.(100%)

S.P.–5%

(95%) = Rs. 665

S.P. at 5% loss i.e. 95% of C.P. = Rs. 665

So,

SP at 12% profit i.e. 112% of C.P.

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= Rs. 11295

665×

= Rs. 784

53.(C) Less by %100

)10( 2

= i.e. less by 1%

54. (B) 32

32

−

+=

32

32

32

32

+

+×

−

+= 22

2

)3(2

)32(

−

+

= 34

3434

−

++=

1

732.147 ×+

= 7 + 6.928 = 13.928

55. (C) 10272 −+

= )2)(5(2)2()5(2 22 −++

= 2)25(2 −+

= 252 −+ = 5

56. (B) Required number + 0.01 = 1.1

⇒ Required number = 1.1 – 0.01 = 1.09

57. (B) n7

1n

2

1n

3

2n +++ = 97

or,

+++

42

6212842n = 97

⇒ n = 97

4297×

⇒ n = 42

58. (D)

h b

h s

rb

rs

Volume of small cone = 27

conebigof.vol

i.e. )h()r(3

1s

2sπ =

27

)h()r(3

1b

2bπ

or, (rs)2(h

s) =

27

)h()r( b2

b

⇒

2b b

2s s

(r ) (h )

(r ) (h ) = 27

or,111

333

××

××=

××

××

sss

bbb

hrr

hrr

or,1

3

h

h

s

b = ⇒ hs =

3

hb=

3

30= 10 cm

⇒ The required height above the base

= (30 – 10) cm

= 20 cm

59. (A) (8.5% + 6.25%) of C.P. = Rs. 590

i.e. 14.75% of C.P. = Rs. 590

So, C.P. (i.e. 100%) = Rs. 10075.14

590×

= Rs. 4000

60. (A) Let C.P. = x

So, M.P. = 25% of x

= x100

125

and S.P. = Price after discount of 12.5% on M.P.

= x100

125

100

5.87×

= x4

5

8

7× = x

32

35

So, % profit = %10032

35

×−

x

xx

= %8

75 = %

8

39

61. (C)

C.P.100

S.P. M.P.+20%+20

–4% discount

120 = 96% of MP

96% of M.P. = 120

So, M.P. (i.e. 100%) = 10096

120× = 125

62. (B)

3m

Circumference of wheel = dπ

=

×3

7

22m

⇒ Distance covered in 1 minute

= m37

2228 ××

= 264 m

So, Time taken by wheel to cover a distanceof 5.280 km (or, 5280 m)

= 264

5280minute = 20 minutes

=

s

b

s

b

h

h

r

r∵

here,

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63. (C)

B

A

CE

F

D

E

4cm

2cm2cm

3cm

2cm

1.5cm1.5cm

1.5cm

2.5cm

2.5cm

5cm

Sides are 3, 4 and 5 cm

⇒ Triangle ABC is a right angled triangle

where º90B =∠ .

Now, D, E and F are mid points of the sides AB, BC and CA respectively.

Here,

FE || AB and DF ||BC

Also, In º90F,DEF =∠∆

⇒ DEF∆ is a right angled triangle.

So, also, from mid point theorem,

FE = 2

1AB = 1.5 cm

& DF = 2

1 BC = 2 cm

So, Area of 5.122

1DEF ××=∆ =

2

3cm2

64. (C)

Externally

Internally

l

3.3 m330 m

320 m

b

2.6 m260 cm

250 cm

h

1.1 m110 cm

Volume

9438000 cm3

8000000 cm3

⇒ Internal height = 2

3

cm)250320(

cm8000000

×

= 2

2

cm80000

cm8000000

= 100 cm

⇒ thickeness of the bottom

= (110 – 100) cm

= 10 cm

= 1 dm

65. (A) 50412833228 +−+

= 25246423162242 ×+×−×+×

= 2202242822 +−+

= 26 = 6 × 1.414

= 8.484

66. (C)

11.2cm

21cmVolume of metal

= External vol. of cylindrical tube – Internal vol. of cylindrical tube

=

})(){()()( 2222inexinex rrhhrhr −π=π−π

=

−

π

22

2

2.11

2

12h

= )36.3136(217

22−××

= 22 × 3 × 4.64

= 306.24 cm3.

67. (B) 1 ÷ [1 + 1 ÷ {1 + 1 ÷ (1 + 1 ÷ 2)}]

=

+÷+÷+÷

2

1111111

=

×+÷+÷

3

211111

=

÷+÷3

5111

=

+÷

5

311

=5

81÷

=8

5

8

51 =×

68. (C) x – y = 3

x2 + y2 = 369

(y + 3)2 + y2 = 369

y2 + 6y + 9 + y2 = 369

2y2 + 6y – 360 = 0

y2 + 3y – 180 = 0

y2 + 15y – 12y – 180 = 0

(y + 15)(y – 12) = 0

y = 12, – 15

x = 15

Sum of numbers = 12 + 15 = 27

69. (B) L.C.M. for 4, 6, 10 & 15 = 60

N will be in form of N = 60n + 2

Now,

least six digit number of form 60 n

(i.e. divisible by 60)

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= 100020

So,

⇒ least six digit number of form N

= 100020 + 2

= 100022

⇒ Sum of digits of N = 1 + 0 + 0 + 0 + 2 + 2

= 5

70. (D) Given fractions are proper.

Diff. beetween N & D of every fraction = 1

∴ The fraction with greatest N or D islargest.

Ans : 8

7

71. (?) 4

3of

4

3

7

5

4

1

3

1

8

7

6

51

4

32

÷+

+×÷

= 4

3

4

3

7

5

12

7

8

7

6

114

11

×÷+×÷

= 9

16

7

5

12

7

7

8

4

6×+××

= 63

143

63

801 =+

72. (A) x = 13

13

−

+=

2

)13( 2+=

2

3213 ++= 32 +

y = 13

)13( 2

−

−=

2

3213 −+= 32 −

x2 + y2 = 22 )32()32( −++

= 14]34[2])3(2[2 22 =+=+

73. (C) H.C.F. × L.C.M. = 7 × 140

= 980

= 1st no. × 2nd no.

(between 20 & 45)

= 28 × 35

So,

Sum of the numbers = 28 + 35

= 63

74. (A) Present age of son = x yrs.

Present age of father = 3x + 3 yrs

After 3 years, son = x + 3 yrs

Father = 3x + 3 + 3

= 3x + 6 yrs

ATQ,

3x + 6 = 2(x + 3) + 10

3x + 6 = 2x + 6 + 10

3x – 2x = 10

x = 10

Father's present age = 3 × 10 + 3

= 33 yrs

75. (C) A B C Let the three co-prime numbers

ATQ, A × B = 551 and B × C = 1073

And ∵ 19 × 29 = 551 and 29 × 37 = 1073

⇒ A = 19, B = 29 and C = 37

⇒ Sum of three nos. = 19 + 29 + 37 = 85

76. (A) No. of boys = x

Constribution by each= Rs. 4

x

Total contribution = 4

xx ×

4

xx × = 400

x2 = 1600

x = 40

77. (C) 21 22 rr π−π = 132

(r1 – r

2) =

π2

132

= 222

7132

×

× = 21 m

Width of the path = (r1 – r

2) = 21 m

78. (B) Side of the rhombus = 10 m

Height = 5 m

Area = b × h = 105 = 50 sq. m.

79. (C) D. of the circle = 21 cm

Rad. = 2

21cm

Area of the biggest circle =

2

2

1

2×π

2

21

2

21

7

22×× =

4

1386 = 346.5 sq. cm

80. (?) )(2 bl

l

+=

16

5

10l + 10b = 16l

10b = 6l

b

l=

6

10 =

3

5

l : b = 5 : 3

81. (C)

Alloy 'A'

Gold : Copper5 : 3

Total (5 + 3 = 8)or, (5 × 2) + (3 × 2) = 16

Alloy 'B'

Gold : Copper5 : 11

Total (5 + 1 = 16)

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to equalise

[L.C.M. of 8 & 16 = 16 unit of each alloy]

The ratio of gold and copper in the alloy C

= 11)23(

5)25(

+×

+×=

17

15

⇒ 15 : 17

82. (C) A = B + 120 = C – 120

A = x (say)

B = x – 120

C = x + 120

A + B + C = 561

x + (x – 120) + (x + 120) = 561

3x = 561

x = 3

561= 187

B's share = 187 – 120 = 67

83. (A)

In 17 kg 500 gm of Alloy

Copper Zinc5

× 17500 gm,12500 gm, 5000 gm

× 17500 gm

: 25 2

7 7

Now,

When 1.250 kg (or 1250 gm) of zinc is

mixed in 17 kg 500 gm of alloy.

then, in the new mixed,

amount of zinc = (5000 + 1250) gm

= 6250 gm

So, In new mixture, ratio of copper & zinc

= 12500 gm : 6250 gm

= 2 : 1

84. (A)

Respective values

So, Respective no. of coins

Re 1 50 p 25 p

13 (13 × 1)

: 11 (11 × 2): 7

(7 × 4)

Total no. of coins in 1 set = 13 + 22 + 28

= 63 coins

Total no. of sets of coins = 63

378coins

= 6 set

⇒ No. of 50 paise coins = 22 × 6

= 132 coins

85. (C)

Rate of InterestS.I. after 4 years

Rs. 500 Required Sum

12%480

10%480

S.I. is same

⇒ SumquiredRe

500.Rs =

%12

%10

⇒SumquiredRe

500.Rs =

6

5

⇒ Required sum = Rs. 65

500× = Rs. 600

86. (C) Decrease in rate of interest

= 8% – 4

37 %

= 8% – 4

31% =

4

1%

Now, ATQ,

4

1% of capital = Rs. 61.50

⇒ Total capital (i.e. 100%)

= Rs. 61.50 × 4 × 100

= Rs. 24600

87. (B) Total ages of 40 students = 40 × 15

= 600 yrs

Let the average age of 10 new students = x yrs.

50

10600 x+= 15.2

600 + 10x = 15.2 × 50

600 + 10x = 760

x = 10

600760 −

= 10

160 = 16 yrs.

88. (B) T1 =

6

24= 4 hrs, T

2 =

8

24 = 3 hrs,

T3 =

12

24 = 2 hrs.

Average speed = 234

242424

++

++

= 9

72 = 8 km/h

89. (C) Let the average of 8 innings = x

Total runs = 8x

9

1008 +x= x + 9

9x + 81 = 8x + 100

x = 19

Centres at:

==================================================================================

Ph: 011-27607854, (M) 8860-333-333 10


New average = x + 9 = 19 + 9 = 28

90. (B) Radius of the shot put ball = 7 cm

Height of the cylinder = 3

7cm

Volume of the shot put = Volume of the

cylinder

373

4×π =

3

72 ××π R

R2 = π

××π7

37

3

4 3

R = 274× = 2 × 7

= 14 cm

d = 2R = 2 × 14

= 28 cm

91. (B) Let the initial area of the rectangle = xy sq. unit

New area = yx100

110

100

90×

= xy100

99 sq. unit

Change in area = xyxy100

99−

= 100

xy

% change in area = 100100 ×xy

xy

= 1%

92. (D) h = 100

251+ m =

4

5m

Total area of the wet surface

= Area of the cistern without top

= 2[lb + bh + lh] – lb

= 244

56

4

54462 −

×+×+×

= 244

305242 −

++

= 244

304292 −

+×

= 244

1462 −×

= 73 – 24

= 49 m2.

93. (D) Total no. of late arrivals of trains

= 114 + 31 + 5

= 150

94. (C) Total no. of late departures of trains

= 82 + 5 + 3

= 90

95. (C) Required % = %1001400

)531114(×

++

= %1001400

150× = 10.7%

96. (B) Required punctuality = %10014901400

14001250×

+

+

= %1002890

2650×

= 91.7%

97. (C) x – y = w + z + 6 x + y = w – z – 3

2 x = 2 w + 3

2x – 2w = 3

x – w = 2

3= 1.5

98. (A)

−

−

−

−

n

11....

4

11

3

11

2

11

n

n 1......

4

3

3

2

2

1 −×××

= n

1

99. (C) })5.2(75.8)5.3){(5.25.3( 2332333 +−+

= 3333 )5.2()5.3( +

[by using (a + b)(a2 – ab + b2) = a3 + b3]

= 3.5 + 2.5

= 6

100. (A) Total CP = CP + repairing charge

= 1200 + 200

= Rs. 1400

SP = Rs. 1680

% of profit = CP

100)14001680( ×−

= 1400

100280× = 20%

Centres at:

==================================================================================

Ph: 011-27607854, (M) 8860-333-333 11


SSC (Tier-II) - 2013 (Mock Test Paper - 1) (ANSWER SHEET)

1. (A)2. (B)3. (C)4. (D)5. (B)6. (C)7. (A)8. (D)9. (A)10. (D)11. (D)12. (C)13. (D)14. (C)15. (A)16. (C)17. (D)18. (B)19. (C)20. (B)

21. (C)22. (D)23. (C)24. (C)25. (C)26. (A)27. *28. (D)29. (A)30. (B)31. (B)32. (B)33. (A)34. (C)35. (C)36. (B)37. (D)38. (A)39. (D)40. (C)

41. (A)42. (B)43. (D)44. (B)45. (A)46. (B)47. (B)48. (B)49. (B)50. (D)51. (C)52. (D)53. (C)54. (B)55. (C)56. (B)57. (B)58. (D)59. (A)60. (A)

61. (C)62. (B)63. (C)64. (C)65. (A)66. (C)67. (B)68. (C)69. (B)70. (D)71. (*)72. (A)73. (C)74. (A)75. (C)76. (A)77. (C)78. (B)79. (C)80. (D)

81. (C)82. (C)83. (A)84. (A)85. (C)86. (C)87. (B)88. (B)89. (C)90. (B)91. (B)92. (D)93. (D)94. (C)95. (C)96. (B)97. (C)98. (A)99. (C)100. (A)

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