Ssc Mains (Maths) Mock Test-2 (Solution)

8/10/2019 Ssc Mains (Maths) Mock Test-2 (Solution)

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MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR

DILSHAD GARDEN ROHINI

BADARPUR BORDER

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

1. (C)4847

4847

484747 4747

R 48

47

48

47 =

48

94 R = 46

2. (D)Let the first odd integer = x

So, 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 evenx

4(x+ 1) will be divisible by

4 2 i.e. by 83. (D)Let x= the larger number

and y= the smaller numberNow,

ATQ, x y= 2395 ... (i)

andy

x; 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) + 25

or, x= 7x 16765 + 25or, 6x= 16765 25

x=6

16740= 2790

4. (C) Let A, B, C and D are four consecutive prime

numbers 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.

andB & 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. =11

2431

= 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 = 11

So, 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 =2

1

10

5

2.1 =10

21

2.8 =10

28=

5

14

So, LCM of 0.5, 2.1 & 2.8

= LCM of10

21,

2

1&

5

14

=5&10,2ofHCF

14&21,1ofLCM=

1

42= 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)42.03.03.07.07.0

37.03.03.03.03.07.07.07.0

=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

= 11

1

1

1

)3.07.0(

)3.07.0(2

3

2

3

9. (D)each three distance is same.Let, S

1, S

2and S

3respectively are the speeds

with which three distances have beencovered.

So,


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Average Speed =S S S

S S S S S S 1 2 3

1 2 2 3 3 1

3

=154040303015

4030153

=6001200450

4030153

=54000

2250= 24 km/hr

10. (B) A + C =37

22part

B =37

221 part =

37

15part

and,

B + C = 37

21

part

or, C15 21

27 37 C =

37

6

37

15

37

21 part

So,

Wage of C = 925037

6 = Rs. 1500

11. (C) Let,

(A + B) xdays

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 working

together to do the work = 188 days

= 144 days

= 12 days

12. (*)Ratio of speedRatio of durationRatio of distancecovered during

any same durationof time

== := :

A

236

B

111

( Distance = Speed time )

13. (A) 600 km

570 km 30 km

B' BA

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

=

60

2090 km

= 30 km

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

= (600 30) km = 570 km

Now,

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

= 190 km /hr

So, Time taken by each train to reach

each other =

190

570hr. = 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.

xkm/hr hr.

1260

12 +660

60

1254 =

60

18x

x= 36 km/hr

15. (C)

Ratio of speedsTime taken

As the distance is same.

(where x time taken by B to coverthe distance)

:

A

3( + 8)hrx

B

5x 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

= 231and Sum of correct marks

= 63 + 70 + 82

= 215

i.e. Sum of corrected marks


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The average of above 5 consecutive integers

= m=2

4 nn m = n+ 2

Now,

n + 2 n + 3 n + 4 n + 5 n + 6 n + 7

The average of above 6 consecutive integers

=2

72 nn=

2

92 n

=2

542 n=

2

5)2(2 n=

2

52 m

18. (D) Total money spent by all of them

= (18 25) + {1 (25 + 15)}

= 450 + 40

= 490

19. (C) Let x= no. of girls

So,600 11 yr 9 months

= x 11 yrs + (600 x) 12 yrs

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

x = 7200 7050

= 150

20. (A) Vessel '1' Vessel '2'

Water WaterMilk 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 =

28

3

5

2

28

5

5

3

=

40

301640

5024

=46

74

21. (B) Initial Ratio = 4 : 6 : 9

Initial nos. = 4x, 6x, 9x

Final Ratio = 7 : 9 : 12

(on increasing 12 students in each class)

126

124

x

x

= 9

7

or, 36x+ 108 = 42x+ 84

6x = 24 x= 4

Total no . of students in the threeclasses before the increase = 4x+ 6x+ 9x

= 19x= 76

22. (B) ATQ

10% of 1st no. = 6

or,10

1of 1stno. 6 1stno. = 60

and, Ratio of 1stno. to 2ndno. = 4 : 3

Nos. are 60 & 45

Now, 20% of 2ndno. = 20% of 45 = 9

& 5% of 1stno. = 5% of 60 = 3

And, 9 is 3 times of 3.

20% of 2ndno. is 3 times of 5% of thefirst no.

23. (D) 1st

no. : 2nd

no. &2nd

no. : 3rd

no.= 3 : 2 = 3 : 2

1stno. : 2ndno. : 3rdno.

= 9 : 6 : 4

So, Let the nos. are 9x, 6x& 4x

ATQ, (9x)2+ (6x)2+ (4x)2+ 532

133x2= 532

x= 2

So, the 2ndno.

i.e. 6x= 6 2 = 12

24. (?) Table Chair

20% profit 25% profit

1 1

10

2 2

5

3 3

3

2

3 3

3

1

3 3

profit1

323

1 23 3

1300

= Rs. 433.33

1300

2 : :

:

:

1

1: 2

1 1

Cost price of table = Rs. 433.33

Alternative method:-

Let,

x = C.P. of table

(1300 x) = C.P. of chair

Now, ATQ,

Profit on table + profit on chair

= Total profit

or, 20% of x+ 25% of (1300 x)

= %3

123 of 1300

or, )1300(100

25

100

20xx = 1300

1003

70

or,3

910

44

1300

5

xx

or, 3253

910

45

xx

or,20

54 xx=

3

975910

or,20

x =

3

65

x= 3

2065

= Rs. 433.33


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

=8

3of (100% 20%) of total

profit

=83 of 80% of total profit

80% of total profit = Rs. 12000 3

8

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

= Rs. 12000 3

8

100

80

= Rs. 40000

26. (A) Let the third no. be 100.

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

So, required % =Aofvalue

B&Aofdifference 100%

= %10080

8 = 10%

27. (B) Total no. of illiterate persons in the town

= no. of literate men + no. of illiteratewomen

=

31125083

40of%)24%100( +

31125083

43

of%)8%100(

= 31125083

43

100

92311250

83

40

100

76

= 114000 + 148350

= 262350

28. (B) Let x= maximum marks

So, ATQ,

30% of x+ 50 = 320 + 30

or, x100

30= 300

x =30

100300 = 1000 marks

29. (D) Let x= man's initial income

So, Man's increased income = (x+ 1200)

Now, ATQ,

12% of x = 10% of (x+ 1200)

12x = 10(x + 1200)

or, 12x 10x = 12000

x = 6000

So, 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 CP

So,

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.10

1

= 140% of CP per mango

So, CP per mango (i.e. 100%)

= Re.140

100

10

1 = Re.

14

1

14 mangoes for Re. 1.

33. (A) Let x= CP of 2ndwatch

So, (1120 x) = CP of 1stwatch

ATQ, Profit = Loss

15% of (1120 x) = 10% of x

15(1120 x) = 10x

25x = 15 1120

x =25

112015 = Rs. 672

34. (D)1

5

2

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 articles

In such type of questions,

Gain percent = %10025

2540

= %10025

15

= 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. x25

36

6

5Rs.

5

6x

Rs. x5

6= Rs. 600

So, CP of A i.e. Rs. x = Rs.6

5600

= Rs. 500

37. (*)M.P. Price after 10% discount

10% 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.994

10100846

= Rs. 1000

38. (B) Single equivalent discount of 20% &

then %4

16 .

= %100

4

2520

%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.5

100500

= Rs. 10000

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

= Rs. 10000100

80

= Rs. 8000

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

SI @ 2% p.a. for 5 yrs. = 10% of sum

ATQ,

(12% 10%) of sum = Rs. 150

2% of sum = Rs. 150

sum = Rs.2

100150

= Rs. 7500

41. (C)

For 1st year

For 2nd year

S.I.

Rs. 135Rs. 135

C.I.

Rs. 135

Now,

if r= rate of interest per annum

r% of 135 = 27

r=135

10027 = 20

Also,

20% of the sum = 135

sum =20

100135 = Rs. 675

42. (C)

For 4 years

For 5 years

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

Now,

if r = rate of interest per annum

r% of 3840 = 96

r = 3840

10096

= 2.5%

43. (B) ATQ,

where P Principal

& t required no. of years

P +20

t

1 > 2P100

or, 250

11

t

or 25

6

t

Now,

25

61

, 2

5

62

, 2

5

63

but 2

5

64

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 have

12M + 16B = 9M + 25B 3M = 9B

1M = 3B

So,

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

Prem

Also,

Raj

Raj

Now,

Work done by Prem in initial 6 days

=

6

10

1

=5

3part of work

Remaining part of the work

=5

31

=5

2part of the work

Now,

Amount of work done by Prem & Raj

together in 1 day

=

12

1

10

1

=60

11part of the work

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

=11

22

115

602

60

115

2

days

Raj actually did the work for11

22 days.

46. (*) A B

6 hours 2 hours

Let the capacity of tank = 12 litres

Rate of filling of tank by pipe A

=

6

12litre per hour

= 2 litre per hour

and rate of filling of tank by pipe B

=

2

12litre per hour

= 6 litre per hour

Rate of filling of tank by pipe A & B together= 8 litre per hour

Now,

Pipe B was opened2

1hour earlier than

pipe A.

Volume of tank filled by pipe B in2

1hour

= 3 litre

Remaining part of the tank which isstill empty = (12 3) litre

= 9 litre

Time required to fill 9 litre of tank by

pipe A & B together

=8

9hours

=8

11 hours

= 1 hours 7.5 minutes Required point of time

= 2 : 30 pm + 1 hour 7.5 mins

= 3 : 37.5 pm or 3 : 372

1pm

Alternative method

Part of tank filled by pipe B alone in2

1hour

=2

1

2

1=

4

1part

Time taken by pipe A & B together to fill

the remaining part

=8

9

6

44

3

2

1

6

14

11

hours

Required point of time

= 2 : 30 pm + 1 hour 7.5 min

= 3 : 372

1pm

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

2xhr. = 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

=xx

xx

2

2

hours

=x

x

3

2 2hours = x

32

hours

Time taken by pipe C alone to empty

the pool =

2

3

2x hours

= x3

4hours

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

=x x x

1 1 3

2 4

part

=

x4

324part

=x4

9part

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

=9

4x= 400 minutes

[6 hour 40 minutes = 400 min]

x =4

9400 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 in2

16 hrs.

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

30 km down in2

12 hours

Sdown

(i.e. SB+ S

C) = 12 km/hour

Also, S

up

(i.e. SB

+ SC

) = 8 km/hourSo,

Speed of current

i.e. SC

=2

SS updown

=2

812 km/hour

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

Let, n= required no. of terms

Now,

Sn = 2

n

{2a+ (n 1)d}

i.e.180 =2

n{2 (24) + (n1)4}

or, 180 =2

n(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

x

x1

= 3

On cubing both the sides we get,

31

xx = 33

or, 331

313

3

xx

xx

or, 333313

3 x

x 33 1

xx = 0

or, 3

6 1

x

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)1/3x

= 54 + 3 (27) x

x3 = 54 9x

x3+ 9x 54 = 0

(x 3)(x2+ 3x+ 18)

x= 3

53. (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 1

1

x 3 2

So,1

1

xx = 1 2 3 3 2

= 132

56. (B) n = 347

= 3434

= 322)3()2( 22

= 2)32(

So,

nn

1

= 32

132

=32

1347

=)32(

)32(4

= 4

57. (A)1

2

3

1

)2(

)1(

K

K

or, 3(K 1) = 2 K

or, 3K + 3 = 2 K

or, 3K + K = 2 3 2K = 1

K =2

1

58. (B) a+ b+ c= 6

On squaring both sides, we get

(a + b + c)2= 62

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

or, 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(

12 xx

)105(

12 xx

will be maximum when

x2+ 5x+ 10 is minimum and maximum

value of x2+ 5x+ 10 =

4

101452

=4

15

So, Required maximum value =15

4

415

1

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(tan2

61. (D)x

xP

sin1

sin1

P=

x

x

cos

sin1

andx

xQ

cos

sin1

x

x

x

xR

sin1

sin1

sin1

cos

=x

xx2cos

)sin1(cos

R= 1 sinx

P= Q= R

62. (*)

tan57 cot37 cot33 tan53

tan33 cot53 tan33 cot53


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=

53tan

)155tan33(tan33tan

)33tan53tan1(

53tan

133tan

5333tan

1

= 33tan

53tan= 33cot53tan

= 57tan37cot

63. (D) 2 3sin sin sin 1

3 2sin sin cos

2 2cossin (1 sin

22 cos)cos2(sin

222 coscos2(cos1

4222cos]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 sin70

0 0 0 0 0 0 0sin(90 80 )sin(90 40 )sin(90 20 )sin30

1cos20 cos40 cos80

2

We know,

1

cos cos2 cos4 cos34

1

cos20cos40cos802

01 1cos(3 20 )

4 2

1

cos60

8

1 1 1

8 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)

60

45

B

D

C

A

xmtr.

5000mtr.

060ACB

045DCB AB = 5000 mtr.

Let, AD= xmtr. From ABC ,

0

tan60AB

BC

5000

3BC

mtr. BC=5000

3mtr.

From ,DBC

0 5000

tan453

DBDB BC

BC

Now, AD = AB - BD

5000

5000 3

1

5000 13

mtr.

68. (D)

90 A A

2 2

A

2hh


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2tan

2

h h

A A ...(i)

2

tan 90

2

h

A

4cos

h

A

tan4

A

h ...(ii)

From (i) & (ii)

2

4

h A

A h

8h2= A2

h2= A2

8

22 Ah

2 2

Ah

69. (D)2

2

12sin

2

xx

x

or, 21

2sin2

2

xx

x

a b a b ab 2 2 2

[ ( ) 2 ]

1

2sin2

12

x

xx

021

xx

1

2sin

= 0 Also,

2sin1

x

70. (D)DE

AB

DEFPer

ABCPer

)(

)(

5.6

1.9

25

)(

ABCPer

Per (ABC) = 35

71. (B)

B

D

A

C

ABCADB

AC

AB

AB

AD

AB2= AD AC

72. (B) Let side of D = x

3243)2(

43 22 xx

32444

3 22 xxx

4x +4 =8

4x =4

x =1

73. (B)

B C

A

In ,ABC

AB + BC = 12 cm

BC + CA = 14 cm

& CA + AB = 18cm

2(AB + BC + CA) = 44 cm

AB + BC + CA = 22 cm

r

Now,

ATQ

22cm2 r

cm227

222

r7

2 cm

74. (D) Sum of all interior angles

= 2 sum of all exterior angles

36021802n

or, 7201802n (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 cm7 cm

C

Here,

AD = angle bisector of A

DACBAD

In such case,

7:5

7

5

AC

AB

CD

BD

77. (*) No. of diagonals of a polygon of nsides

= nnC 2

ATQ,

nnC 2 = 54

n

n2 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

[ncannot be negative.]

78. (B) A35

B

P

Here, in ,AOP

AO = OP

35APOPAO

)352(180AOP

= 110

70110180POB

Also, In POB ,

BO = OP

552

70180OPBPBO

55ABP

79. (B) A

B C

I

65ABC

5.322

ABCIBC

Also, 55ACB

5.272

ACB

ICB

)ICBIBC(180IBC

= 180 60

= 12080. (C) Ratio of no. of sides = 1: 2

Then, Let the nos. of sides are nand 2n

Now, Ratio of their interior angles = 2 : 3

n

nn

n

2 490 2

4 4 390

2

n

n

2 4 1

4 4 3

6n 12 = 4n 4 2n= 8 n= 4

Respective nos. of sides of these polygons are 4 and 8.

81. (B)

:66

8

a

3

4

a

2

a

3

4

a3

3

3

3

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

or, 4r = 2(r + h2) = l+ r

Now, 4r = 2(r + h2)

or, 4r = 2r + 2h2

or, 2r = 2h2

r : h2= 1 : 1

Again, 4r = l+ r

or, 4r = 232 hr + r

or, 3r = 232 hr

9r2 = 232 hr

8r2 = 23h

r22 = h3

r: h3= 1 : 22

r(h1) : h

2: h

3= 1 : 1 : 22


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83. (C)

5x

12x

13x

Let the sides are 13x, 12xand 5x.

ATQ,

13x+ 12x+ 5x= 450

or, 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 =2

1 base height

=2

1 12x 5x

=2

1 180 75 m2

= 6750 sq. meter

84. (A)

r = 10 cm

h = 4 cm

Let radius is increased by xcm.

New volume of cylinder = 4)10( 2 x

Again,

Let the height is increased by xcm.

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(x5) = 0

x = 5 cm

85. (B)

10 cm

10 cm

44 cm r

Volume of the

cylinder = hr2 r2 = 44 cm

= 1072 r= 222

744

cm

= 10497

22 r= 7 cm

= 1540 m3.

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

33744.2

a

aa 100%

= (1.744 100)%

= 174.4%

87. (B)

h = 24cm

r

l

22 rhl +=22 724 +=

= 25 cm

Volume of cone = 1232 cm3

hr2

3

1 = 1232 cm3

247

22

3

1 2 r = 1232 cm3

r2 =2422

371232

r2 = 49 cm

r = 7 cm

So, Curved surface area of cone = rl

= 2577

22 cm2

= 550 cm2


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88. (D)

18 cmr

26 cm

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

= 88 cm

= r2 r =222

788

= 14 cm

Area of the circle i.e. 2r

=7

22 14 14

= 616 cm2

89. (A) A

B C

a

16 cm 25 cm

28 cmO

a

a

Area of ABC

= Area of )( BOCAOCAOB

2

4

3a = )282516(

2

1a

a= 6923

4

cm

a= 346 cm

So,

Area of ABC =2)346(

4

3

= 321164

3

= 31587 cm3

90. (C)emptyspace

h

Volume of water needed to fill the emptyspace

= Volume of cylinder Volume of cone

= hrhr22

3

1

= hr2

3

2

=

hr2

3

12

= 272 cm3

= 54 cm3

91. (B)

7cm

7 cm

7cmAC

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

2

1

27

7722cm2

= (77 49) cm2

= 28 cm2

92. (B) A

B CD

EF

G8

53

6 4

10

In the given ABC ,

AD, BE and CF are medians and they cut

one another at G.

GF

CG

GE

BG

GD

AG =

1

2

Here,

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

AG + GD = AD = 9 cm

AG = 6 cm and GD = 3 cm

Also,BG + GE = BE = 12 cm

BG = 8 cm and GE = 4 cm

Also,

CG + GF = CF = 15 cm

CG = 10 cm and GF = 5 cm

Area of AGB = 862

1 = 24 cm2

So, Area of ABC = AGB3

= 2 24 cm2

= 72 cm2


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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

7

22(2r+ 1) = 22

2r+ 1 = 7

2r= 6

r= 3 cm

95. (B) Volume of water due to 2 cm rain on a

square 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 level

in the pool will be increased

=m10m100

m10000 3

= 10m

96. (A) Total income = Rs. 150000

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

= Rs. 15000

97. (C) Maximum expenditure of the family

other than on food, was on others.

98. (B) Saving = 15% = Housing

99. (D) Required % = 10% + 12% + 5%

= 27%

100. (D) Money spent on food

= 23% of Rs. 150000

= Rs. 34500


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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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Ssc Mains (Maths) Mock Test-2 (Solution)

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