ssc (tier-ii) - 2013 (mock test paper - 2) [solution] · pdf filessc (tier-ii) - 2013 (mock...

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MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR DILSHAD GARDEN ROHINIBADARPUR BORDER

SSC (Tier-II) - 2013 (Mock Test Paper - 2) [SOLUTION]

1. (C) 4847

4847

484747 4747

R 4847

4847

= 4894

R = 46

2. (D)Let the first odd integer = xSo, the second odd integer = x + 2Now,Difference between the squares of these twoconsecutive odd integers

= (x + 2)2 – x2

= x2 + 4 + 4x – x2

= 4( + 1)x

will be divisibleby 2 as ( + 1) will be even x

4(x + 1) will be divisible by4 × 2 i.e. by 8

3. (D)Let x = the larger numberand y = the smaller numberNow,

ATQ, x – y = 2395 ... (i)

and yx

; Quotient = 7 and Remainder = 25

x = 7y + 25 ... (ii)Now on putting value of y from (i) in (ii)We get,

x = 7(x – 2395) + 25or, x = 7x – 16765 + 25or, 6x = 16765 – 25

x = 616740

= 2790

4. (C)Let A, B, C and D are four consecutive primenumbers in descending order ( A > B > C > D)So, ATQ,

A × B × C = 2431 ... (i)and B × C × D = 1001 ... (ii)Now, both 2431 and 1001 are divisible by 11.i.e. 11 is a common factor of both 2431 and1001.and B & C are the common factors of 2431and 1001. Out of B & C, one must be 11.So, from (i),

Product of other two prime nos. = 112431

= 221and 221 = 17 × 13

Out of A, B and C, one is 17, one is 13and one is 11.

and A > B > C A = 17, B = 13 and C = 11So, the largest given prime no. = 17

5. (A) 182221164 3333

= 3 3 3( 2 2) (2 2 1) (1 1)

= 2323 )1()122()2(

= 23 )12(

So, square root of )1164( 33

i.e )1164( 33 = 123

6. (B) 0.5 = 21

105

2.1 = 1021

2.8 = 1028

= 514

So, LCM of 0.5, 2.1 & 2.8

= LCM of 1021,

21

& 514

= 5&10,2ofHCF14&21,1ofLCM

= 142

= 42

7. (C) Required multiplication= 2 × 4 × 6 × 8 × 10 × 12 × 14 × 16 × 18 × 20 × 5 × 10 × 15 × 20 × 25

Here, nos. (o) = no. of complete pairs of 2 & 5.= no. of 5 [ here n(5) < n(2)]= 7

8. (C) 42.03.03.07.07.037.03.03.03.03.07.07.07.0

= 0.7 0.7 0.7 0.3 0.3 0.3 3 0.7 0.3 1

0.7 0.7 0.3 0.3 2 0.7 0.3

= 3 2

2 2(0.7) (0.3) 0.3 0.7 0.3 (0.7 0.3)

(0.7) (0.3) 2 0.7 0.3

= 111

11

)3.07.0()3.07.0(

2

3

2

3

9. (D) each three distance is same.Let, S1, S2 and S3 respectively are the speedswith which three distances have beencovered.So,

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Average Speed = S S S

S S S S S S1 2 3

1 2 2 3 3 1

3

= 1540403030154030153

= 60012004504030153

= 540002250 = 24 km/hr

10. (B) A + C = 3722

part

B = 37221 part = 37

15part

and,

B + C = 3721

part

or, C15 2127 37 C = 37

63715

3721

part

So,

Wage of C = 9250376

= Rs. 1500

11. (C) Let,

(A + B) x days

So, A alone (x + 8) days

and, B alone (x + 18) daysIn such type of questions

The no. of days taken by A and B workingtogether to do the work = 188 days

= 144 days= 12 days

12. (*)Ratio of speedRatio of duration Ratio of distancecovered duringany same durationof time

== := :

A236

B111

( Distance = Speed × time )

13. (A) 600 km

570 km 30 kmB' BA

Distance covered by B before movement of A.i.e. Distance covered by B in 20 minutes

=

602090 km

= 30 km

When train from station A starts tomove; the another train will be B' anddistance between A & B'

= (600 – 30) km = 570 kmNow,

Relative speeds of trains= (100 + 90) km/hr= 190 km /hr

So, Time taken by each train to reach

each other =

190570

hr. = 3 hrs.

And in 3 hours, distance travelled by A= (100 × 3) km = 300 kms

Both train will cross each other at adistance 300 km & from A i.e. at the exactmiddle point of A and B.

14. (B) Let x = the required speed

Home Office

54 km/hr × hr.

x km/hr × hr.

1260

12 +660

601254 = 60

18x

x = 36 km/hr15. (C)

Ratio of speedsTime taken

As the distance is same.

(where x time taken by B to coverthe distance)

:

A3( + 8)hrx

B5x hr.

3(x + 8) = 5x 3x + 24 = 5x x = 12 hr.So,

Actual time taken by A to cover thedistance i.e. (x + 8) hr. = (12 + 8) hrs

= 20 hrs.16. (C) Sum of wrongly entered marks

= 73 + 78 + 80= 231

and Sum of correct marks= 63 + 70 + 82= 215

i.e. Sum of corrected marks <sum ofwrongly marks (by 16 marks)So,Actual average (after rectification)

= 40167240

= 402864

= 71.6 marks17. (A) n n + 1 n + 2 n + 3 n + 4

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The average of above 5 consecutive integers

= m = 24nn

m = n + 2

Now,n + 2 n + 3 n + 4 n + 5 n + 6 n + 7The average of above 6 consecutive integers

= 272 nn

= 292 n

= 2542 n

= 25)2(2 n

= 252 m

18. (D) Total money spent by all of them= (18 × 25) + {1 × (25 + 15)}= 450 + 40= 490

19. (C) Let x = no. of girlsSo,

600 × 11 yr 9 months= x × 11 yrs + (600 – x) × 12 yrs

7050 yr. = (11x + 7200 – 12x)yr. x = 7200 – 7050

= 15020. (A) Vessel '1' Vessel '2'

Water Water Milk Milk3 5 : : 2 3

Ratio of contents of vessel '1' to vessel '2' tomix with each other = 1 : 2 Ratio of water and milk in the resulting

mixture =

283

52

285

53

=

403016

405024

= 4674

21. (B) Initial Ratio = 4 : 6 : 9Initial nos. = 4x, 6x, 9xFinal Ratio = 7 : 9 : 12

(on increasing 12 students in each class)

126124

xx

= 97

or, 36x + 108 = 42x + 84 6x = 24 x = 4 Total no. of students in the threeclasses before the increase = 4x + 6x + 9x

= 19x = 7622. (B) ATQ

10% of 1st no. = 6

or, 101

of 1st no. 6 1st no. = 60

and, Ratio of 1st no. to 2nd no. = 4 : 3 Nos. are 60 & 45

Now, 20% of 2nd no. = 20% of 45 = 9& 5% of 1st no. = 5% of 60 = 3And, 9 is 3 times of 3. 20% of 2nd no. is 3 times of 5% of the

first no.23. (D) 1st no. : 2nd no. & 2nd no. : 3rd no.

= 3 : 2 = 3 : 2 1st no. : 2nd no. : 3rd no.

= 9 : 6 : 4So, Let the nos. are 9x, 6x & 4xATQ, (9x)2 + (6x)2 + (4x)2 + 532 133x2 = 532 x = 2So, the 2nd no.

i.e. 6x = 6 × 2 = 1224. (?) Table Chair

20% profit 25% profit

1 1

10

2 2

5

3 3

32

3 3

31

3 3

× profit13

23

× × 1 23 3

1300

= Rs. 433.33

1300

2 : :

:

:

1

1: 2

1 1

Cost price of table = Rs. 433.33Alternative method:-Let,

x = C.P. of table (1300 – x) = C.P. of chairNow, ATQ,

Profit on table + profit on chair= Total profit

or, 20% of x + 25% of (1300 – x)

= %3123 of 1300

or, )1300(10025

10020 xx = 1300

100370

or, 3910

441300

5

xx

or, 3253

91045

xx

or, 2054 xx

= 3975910

or, 20x

= 365

x = 32065

= Rs. 433.33

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25. (C) A's share = Rs. 12000

= 83

of (100% – 20%) of total

profit

= 83

of 80% of total profit

80% of total profit = Rs. 12000 × 38

So, Total profit (i.e. 100%)

= Rs. 12000 × 38

× 10080

= Rs. 4000026. (A) Let the third no. be 100.

So, A and B will be 80 and 72 respectively.

So, required % = AofvalueB&Aofdifference

× 100%

= %100808

= 10%

27. (B) Total no. of illiterate persons in the town= no. of literate men + no. of illiterate

women

=

311250

8340of%)24%100( +

311250

8343of%)8%100(

= 3112508343

10092311250

8340

10076

= 114000 + 148350= 262350

28. (B) Let x = maximum marksSo, ATQ,

30% of x + 50 = 320 + 30

or, x10030

= 300

x = 30100300

= 1000 marks

29. (D) Let x = man's initial incomeSo, Man's increased income = (x + 1200)Now, ATQ,

12% of x = 10% of (x + 1200) 12x = 10(x + 1200)or, 12x – 10x = 12000 x = 6000So, increased income;

i.e. x + 1200 = Rs. 6000 + Rs. 1200= Rs. 7200

30. (D) 20% profit on one and 20% loss on other net loss

and loss % = 100

)20( 2= 4%

Total SP = (100 – 4)% of total CP 2 × 4.5 lakh = 96% of total CPSo,

loss i.e. 4% of total CP

= 496

lakh5.42

= Rs. 37500 loss of Rs. 37500

31. (A) Required single discount= (1 – 0.9 × 0.8 × 0.75) × 100%= (1 – 0.54) × 100%= 0.46% × 100%= 46%

32. (D) SP per mango = Re. 101

= 140% of CP per mangoSo, CP per mango (i.e. 100%)

= Re. 140100

101

= Re. 141

14 mangoes for Re. 1.33. (A) Let x = CP of 2nd watch

So, (1120 – x) = CP of 1st watchATQ, Profit = Loss 15% of (1120 – x) = 10% of x 15(1120 – x) = 10x 25x = 15 × 1120

x = 25112015

= Rs. 672

34. (D)1

52

2

1710% (15 + 2.5)% %

=15% 1 : 2

2 : 1

=

gap = 5% gap = % 2.5%

50 pens (150 – 50) pens = 100 pens

35. (D) C.P. of 40 article = SP of 25 articlesIn such type of questions,

Gain percent = %10025

2540

= %1002515

= 60%

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36. (B)C.P. S.P.A

B

C

B

C

Rs. 600

=Rs.

= Rs.

Rs. x

Rs. x

Rs. x

Rs. + x

Rs. ×

Rs. –

Now,

Now,

x

120

36

6

6

1

36

36

6

36

x

x

×x

x

x

5

100

25

5

5

6

25

25

5

25

= Rs. x2536

65

Rs. 56

x

Rs. x56

= Rs. 600

So, CP of A i.e. Rs. x = Rs. 65600

= Rs. 50037. (*)

M.P. Price after 10% discount10% discount

6% discount(additional)

Rs.x

Again,

Rs.

Rs. Rs. 94

9

9 9x = Rs. 846×

x

x100

10

10 10

The original marked price i.e. x

= Rs. 99410100846

= Rs. 100038. (B) Single equivalent discount of 20% &

then %416 .

= %100

42520

%4

25%20

= 25% discount

C.P.Rs. 100(let)

S.P.Rs. 120

=75% of M.P.

M.P.20% gain 25% discount

So, 100% of MP = 10075

120

= Rs. 160 MP should be above CP by

= %100100

100160

= 60%39. (C) 20% discount on L.P. – 25% discount on

L.P. = Rs. 500 80% value of L.P. – 75% value of L.P.

= Rs. 500 5% value of L.P. = Rs. 500

L.P. (i.e. 100%) = Rs. 5100500

= Rs. 10000So, Tarun bought the TV at 80% of L.P.

= Rs. 1000010080

= Rs. 800040. (B) SI @ 3% p.a. for 4 yrs. = 12% of sum

SI @ 2% p.a. for 5 yrs. = 10% of sumATQ, (12% – 10%) of sum = Rs. 150

2% of sum = Rs. 150

sum = Rs. 2100150

= Rs. 750041. (C)

For 1st yearFor 2nd year

S.I.Rs. 135Rs. 135

C.I.Rs. 135

Now,if r = rate of interest per annum

r% of 135 = 27

r = 13510027

= 20

Also, 20% of the sum = 135

sum = 20100135

= Rs. 675

42. (C)For 4 yearsFor 5 years

C.I.Rs. 3840Rs. 3936 = Rs. 3840 + Rs. 96

Now,if r = rate of interest per annum

r% of 3840 = 96

r = 384010096

= 2.5%

43. (B) ATQ,where P Principal& t required no. of years

P + 20

t1 > 2P

100

or, 25011

t

or 256

t

Now,

256 1

, 256 2

, 256 3

but 256 4

Required least no. of complete yrs. = 4yrs.

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44. (D) 12M + 16B 10 days ... (i)18M + 50B 5 days

9M + 25B 10 days ... (ii)From (i) & (ii)

We have12M + 16B = 9M + 25B

3M = 9B 1M = 3BSo, Efficiency of a man to that of a boy = 3 : 1

45. (A)

1

1

10

12

Prem 10 days

12 days

part of work

part of work

in 1 day

in 1 day

PremAlso,

Raj

Raj

Now,Work done by Prem in initial 6 days

=

6101

= 53

part of work

Remaining part of the work

= 531

= 52

part of the work

Now,Amount of work done by Prem & Rajtogether in 1 day

=

121

101

= 6011

part of the work

So,No. of days taken by Prem & Raj togetherto do the remaining part of the work

= 1122

115602

601152

days

Raj actually did the work for 1122 days.

46. (*) A B6 hours 2 hours

Let the capacity of tank = 12 litres

Rate of filling of tank by pipe A

=

612

litre per hour

= 2 litre per hourand rate of filling of tank by pipe B

=

212

litre per hour

= 6 litre per hour Rate of filling of tank by pipe A & B together

= 8 litre per hourNow,

Pipe B was opened 21

hour earlier than

pipe A.

Volume of tank filled by pipe B in 21

hour

= 3 litre Remaining part of the tank which is

still empty = (12 – 3) litre= 9 litre

Time required to fill 9 litre of tank bypipe A & B together

= 89

hours

= 811 hours

= 1 hours 7.5 minutes Required point of time

= 2 : 30 pm + 1 hour 7.5 mins

= 3 : 37.5 pm or 3 : 37 21

pm

Alternative method

Part of tank filled by pipe B alone in 21

hour

= 21

× 21

= 41

part

Time taken by pipe A & B together to fillthe remaining part

= 89

6443

21

61

411

hours

Required point of time= 2 : 30 pm + 1 hour 7.5 min

= 3 : 37 21

pm

47. (B) Let x hour = time taken by pipe A alone to empty the pool

2x hr. = time taken by pipe B

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alone to empty the poolSo, Time taken by pipe A & B together to

empty the pool

= xxxx

22

hours

= xx

32 2

hours = x32

hours

Time taken by pipe C alone to empty

the pool =

232 x hours

= x34

hours

Part of the pool which will be emptywhen A, B & C work together

= x x x1 1 3

2 4

part

=

x4324

part

= x49

part

Total time taken by A, B & C workingtogether to empty the pool

= 94x

= 400 minutes

[6 hour 40 minutes = 400 min]

x = 49400

minutes

= 900 minutes = 15 hours48. (A) 12 km up + 18 km down in 3 hrs.

36 km up + 54 km down in 9 hrs...... (i)

Also, 36 km up + 24 km down in 216 hrs.

..... (ii)From (i) & (ii), we get,

30 km down in 212 hours

Sdown (i.e. SB + SC) = 12 km/hourAlso, Sup (i.e. SB + SC) = 8 km/hourSo,

Speed of current

i.e. SC = 2

SS updown

= 2812

km/hour

= 2 km/hour49. (C) –24, –20, –16, ..........

Let, n = required no. of termsNow,

Sn = 2n

{2a + (n – 1)d}

i.e. 180 = 2n

{2 × (–24) + (n –1)4}

or, 180 = 2n

(–48 + 4n – 4)

or, 360 = 4n2 – 52nor, 4n2 – 52n – 360 =0 n = 18

50. (D) 3 18 12 72 66 396 ?3 × 6 18 – 6 12 × 6 72 – 6 66 × 6 396 – 6

The missing no. = 396 – 6= 390

51. (A)21

xx = 3

xx 1 = 3

On cubing both the sides we get,

31

xx = 33

or, 331313

3

xx

xx

or, 333313

3 x

x 33 1

xx = 0

or, 3

6 1x

x = 0 x6 + 1 = 0 x6 = –1

Now,x72 + x66 + x54 + x36 + x24 + x6 + 1

= (x6)12 + (x6)11 + (x6)9 + (x6)6 + (x6)4 + x6 + 1= (–1)12 + (–1)11 + (–1)9 + (–1)6 + (–1)4 + (–1) + 1= 1 – 1 – 1 + 1 + 1 – 1 + 1= 1

52. (A) Let x = 3/13/1 )75627()75627(

x3 = 7562775627 +

1/3 1/33(27 756) (27 756)

)}75627()75627{(

x3 = 54 + 3(729 – 756)1/3 x= 54 + 3 × (–27) x

x3 = 54 – 9xx3 + 9x – 54 = 0(x – 3)(x2 + 3x + 18)

x = 353. (D) a2d2 + b2c2 – 2abcd + a2c2 + b2d2 + 2abcd

= a2(c2 + d2) + b2(c2 + d2)= (a2 + b2)(c2 + d2)= 2 × 1 = 2

54. (B) 347833

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= 2)32(833

= )32(833

= )38163(3

= 2)34(3

= 343

24

55. (A) x = 321

x – 1 = 3 2 11x 3 2

So, 11

x

x = 1 2 3 3 2

= 132

56. (B) n = 347

= 3434

= 322)3()2( 22

= 2)32(

So,

nn 1

= 32132

= 32

1347

= )32()32(4

= 4

57. (A) 12

31

)2()1(

K

K

or, –3(K – 1) = 2 – Kor, –3K + 3 = 2 – Kor, –3K + K = 2 – 3 –2K = –1

K = 21

58. (B) a + b + c = 6On squaring both sides, we get

(a + b + c)2 = 62

or, a2 + b2 + c2 + 2(ab + bc + ca) = 36or, 14 + 2(ab + bc + ca) = 36

ab + bc + ca = 36 14

2

i.e. ab + bc + ca = 11

Now,a3 + b3 + c2 – 3abc = (a + b + c) {a2 + b2 + c2

– (ab + bc + ca)}i.e. 36 – 3abc = 6(14 – 11) 3abc = 36 – 18 abc = 6

59. (D) (x2 + 5x + 10)-1 = )105(1

2 xx

)105(1

2 xx will be maximum when

x2 + 5x + 10 is minimum and maximum

value of x2 + 5x + 10 =

4101452

= 415

So, Required maximum value = 154

4151

60. (D) 2222 cosseccossin ec

22 cottan

= )tan(sec)cos(sin 2222

2222 cot)cot(costan ec

22 cottan

= )cot(tan2111 22

= }cottan2)cot{(tan23 2

= }2)cot{(tan23 2

= 2)cot(tan243

= 7)cot(tan27 2

0)cot(tan 2

61. (D) xxP

sin1sin1

P = xx

cossin1

and xxQ

cossin1

xx

xxR

sin1sin1

sin1cos

= x

xx2cos

)sin1(cos

R = 1 – sinx P = Q = R

62. (*)tan57º cot37º cot33º tan53ºtan33º cot53º tan33º cot53º

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=º53tan

)1º55tanº33(tanº33tan

)º33tanº53tan1(

º53tan1º33tan

º53º33tan

1

= º33tanº53tan

= º33cotº53tan

= º57tanº37cot

63. (D) 2 3sin sin sin 1

3 2sin sin cos

2 2) cossin (1 sin

22 cos)cos2(sin

222 coscos2(cos1

4222 cos]cos4cos4)[cos1(

6224 coscos4cos4cos4 44 coscos 04cos8cos4cos 246

4cos8cos4cos 246 64. (B) Let x = sin cos

x2 = cossin2cossin 22

x2 = 1 2sin cos

Now, 0 00 90

sin 1 and cos 1

sin cos 1

0 2sin cos 2

1 0 1 2sin cos 1 2

31 2 x

1 3x

1 sin cos 3

sin cos 1

65. (B) 0 0 0 0sin10 sin30 sin50 sin700 0 0 0 0 0 0sin(90 80 )sin(90 40 )sin(90 20 )sin30

1cos20 cos40 cos802

We know,

1cos cos2 cos4 cos34

1cos20cos40cos802

01 1cos(3 20 )

4 2

1 cos60º8

1 1 18 2 16

66. (B) 2sec 1 3 tan 3 1 0

21 tan – tan – 3 tan 3 – 1 0

2tan 3 tan tan 3 0

tan tan 3 1 tan 3 0

tan 3 tan 1 0

tan 3 0

tan 3

67. (C)

60º45º

B

D

C

A

xmtr.

5000mtr.

060ACB 045DCB AB = 5000 mtr.Let, AD= x mtr. From ABC ,

0tan60 ABBC

50003BC

mtr. BC= 5000

3 mtr.

From ,DBC

0 5000tan453

DB DB BCBC

Now, AD = AB - BD

50005000

3

15000 13

mtr.

68. (D)

90 – A A2 2

A

2hh

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2tan

2

h hA A

...(i)

2tan 90

2

hA

4cos hA

tan4Ah

...(ii)

From (i) & (ii)

24

h AA h

8h2 = A2

h2 = A2

8

22 Ah

2 2Ah

69. (D)2

2

12sin2x x

x

or, 212

sin22

xxx

a b a b ab2 2 2[ ( ) 2 ]

1

2sin21 2 x

xx

021

xx

1

2sin

= 0 Also,

2sin1 x

70. (D) DEAB

DEFPerABCPer

)()(

5.61.9

25)(

ABCPer

Per (ABC) = 35

71. (B)

B

D

A

C

ABCADB

ACAB

ABAD

AB2 = AD × AC72. (B) Let side of D = x

3243)2(

43 22 xx

324443 22 xxx

4x +4 =8 4x =4 x =1

73. (B)

B C

A

In ,ABCAB + BC = 12 cmBC + CA = 14 cm

& CA + AB = 18cm 2(AB + BC + CA) = 44 cm AB + BC + CA = 22 cm

r

Now,ATQ 22cm2 r π

cm227222

r72

cm

74. (D) Sum of all interior angles = 2× sum of all exterior angles º3602º1802n

or, º720º1802n (n – 2) = 4 n = 6 Required no. of sides of the polygon = 6

75. (B)

In an equilateral triangle,

side = 32 × inradius

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i. e. a = r32

= cm332

= cm36

76. (D)

B

A

D

5 cm 7 cm

CHere,AD = angle bisector of A DACBAD In such case,

7:575

ACAB

CDBD

77. (*) No. of diagonals of a polygon of n sides

= nnC 2

ATQ,nnC

2 = 54

nn2 2 = 54 + 1

n n n

n( 1) 2(2 1) 2 = 54 + n

n2 – n = 108 + 2n n2 – 3n – 108 = 0 n = 12, –9 n = 12

[ n cannot be negative.]

78. (B) A 35º B

P

Here, in ,AOPAO = OP 35APOPAO

)352(180AOP = 110º

º70º110º180POB

Also, In POB , BO = OP

º552

º70º180OPBPBO

º55ABP

79. (B) A

B C

I

º65ABC

º5.322

ABCIBC

Also, º55ACB

º5.272

ACBICB

)ICBIBC(º180IBC = 180º – 60º = 120º

80. (C) Ratio of no. of sides = 1: 2Then, Let the nos. of sides are n and 2nNow, Ratio of their interior angles = 2 : 3

nn

nn

2 4 90º 24 4 390º

2

nn

2 4 14 4 3

6n – 12 = 4n – 4 2n = 8 n = 4Respective nos. of sides of these polygons are 4 and 8.

81. (B)

:66

8a

34

a

2a

34

a3

3

3

3

82. (?) 4 r2 = 2 (r + h2) = r(l + r)or, 4r = 2(r + h2) = l + rNow, 4r = 2(r + h2)or, 4r = 2r + 2h2

or, 2r = 2h2

r : h2 = 1 : 1Again, 4r = l + r

or, 4r = 23

2 hr + r

or, 3r = 23

2 hr

9r2 = 23

2 hr

8r2 = 23h

r22 = h3

r : h3 = 1 : 22

r(h1) : h2 : h3 = 1 : 1 : 22

Centres at:==================================================================================

Ph: 011-27607854, (M) 8860-333-333 12


83. (C)

5x

12x

13x

Let the sides are 13x, 12x and 5x.ATQ,

13x + 12x + 5x = 450or, 30x = 450 x = 15 Sides are of length 195m, 180m and 75m.Now, Ratio of sides are 13 : 12 : 5

(Pythagorian triplet) Triangle is a right angled triangle.So,

Area of triangle = 21

× base × height

= 21

× 12x × 5x

= 21

× 180 × 75 m2

= 6750 sq. meter

84. (A)

r = 10 cm

h = 4 cm

Let radius is increased by x cm.

New volume of cylinder = 4)10( 2 x

Again,Let the height is increased by x cm.

New volume of cylinder = )4(102 x

4)10( 2 x = )4(102 x

(10 + x)2 × 4 = 100(4 + x) (10 + x)2 = 25(4 + x) 100 + x2 + 20x = 100 + 25x x2 – 5x = 0 x(x –5) = 0

x = 5 cm

85. (B)

10 cm

10 cm

44 cm r

Volume of the

cylinder = hr 2 r2 = 44 cm

= 1072 r = 222744

cm

= 1049722

r = 7 cm

= 1540 m3.86. (C) Volume of cube = a3 (where a = length of

a edge)When each edge is increased by 40%.

length of the new edge = 1.4a Volume of new cube = (1.4a)3

= 2.744 a3

Required % increase

= 3

33744.2a

aa × 100%

= (1.744 × 100)%= 174.4%

87. (B)

h = 24 cm

r

l

22 rhl +=22 724 +=

= 25 cm

Volume of cone = 1232 cm3

hr 2

31 = 1232 cm3

24722

31 2 r = 1232 cm3

r2 = 2422371232

r2 = 49 cm r = 7 cmSo, Curved surface area of cone = rl

= 257722

cm2

= 550 cm2

Centres at:==================================================================================

Ph: 011-27607854, (M) 8860-333-333 13


88. (D)18 cm

r

26 cm

Perimeter of the rectangle= 2 × (26 + 18) cm= 88 cm

= r2 r = 222788

= 14 cm

Area of the circle i.e. 2r

= 722

× 14 × 14

= 616 cm2

89. (A) A

B C

a16 cm 25 cm

28 cmO

a

a

Area of ABC

= Area of )( BOCAOCAOB

2

43 a = )282516(

21

a

a = 6923

4

cm

a = 346 cmSo,

Area of ABC = 2)346(43

= 3211643

= 31587 cm3

90. (C)emptyspace h

Volume of water needed to fill the emptyspace

= Volume of cylinder – Volume of cone

= hrhr 22

31

= hr 2

32

=

hr 2

312

= 272 cm3

= 54 cm3

91. (B)

7cm

7 cm

7cm AC

In right angled isosceles ABC,AB = BC

Also,AD = CD = BD = 7cm

= circum radius(where ACBD )

Now, Required area= Area of semicircle – Area of ABC

=

714

21

277722

cm2

= (77 – 49) cm2

= 28 cm2

92. (B) A

B CD

EF

G85

3

6 4

10

In the given ABC ,AD, BE and CF are medians and they cutone another at G.

GFCG

GEBG

GDAG

= 12

Here,AD = 9 cm, BE = 12 cm and CF = 15 cm

AG + GD = AD = 9 cm AG = 6 cm and GD = 3 cmAlso,

BG + GE = BE = 12 cm BG = 8 cm and GE = 4 cmAlso,

CG + GF = CF = 15 cm CG = 10 cm and GF = 5 cm

Area of AGB = 8621

= 24 cm2

So, Area of ABC = AGB3 = 2 × 24 cm2

= 72 cm2

Centres at:==================================================================================

Ph: 011-27607854, (M) 8860-333-333 14


93. (C) T.S.A. of prism = C.S.A. + 2 × Area of base

608 = Perimeter of base × height + 2 × Area of base

608 = 4x × 15 + 2x2 (where x = side of square)

x3 + 30x – 304 = 0 (x – 8) (x + 38) = 0 x = 8 Volume of prism = Area of base × height

= 8 × 8 × 15 = 960 cm3

94. (A) 22)1( rr = 22

)12( 22 rrr

r2 = 22

722

(2r + 1) = 22

2r + 1 = 7 2r = 6 r = 3 cm

95. (B) Volume of water due to 2 cm rain on asquare km land

= 1km × 1 km × 2 cm

= m100

2m1000m1000

= 20000 m3

50% of volume of rain drops= 10000m3

Now,Required level by which the water levelin the pool will be increased

= m10m100

m10000 3

= 10m96. (A) Total income = Rs. 150000

Required difference= (15% – 5%) of Rs. 150000= Rs. 15000

97. (C) Maximum expenditure of the familyother than on food, was on others.

98. (B) Saving = 15% = Housing99. (D) Required % = 10% + 12% + 5%

= 27%100. (D) Money spent on food

= 23% of Rs. 150000= Rs. 34500

Centres at:==================================================================================

Ph: 011-27607854, (M) 8860-333-333 15


SSC (Tier-II) - 2013 (Mock Test Paper - 2) (ANSWER SHEET)

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

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

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

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

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

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