SSC MOCK TEST- Question with Answer (Solution, Exam Preparation Notes | EduRev

: SSC MOCK TEST- Question with Answer (Solution, Exam Preparation Notes | EduRev

 Page 1


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=============================================================
1
1. (B) Given +  >

– 
×  =
I  <
L 
 
Given expression A | B × C  A < B = C
because option (B) follows the given
expression here because C - B + A means
C 

 B > A, one of which meaning is  A<B= C
2. (C)2 M C 3 K
(2 M = N) (2 N  3 K)
               (2 N  3 K)
2 × 2 M  3 K (since N = 2 M)
4 M  3 K
4 M is not more than 3 K, then 2 M is less
than 3 K.
Hence 2 M C 3 K is correct.
3. (B) (15 +12) ÷ 9 = 3 and (44 + 28) ÷ 9 = 8
Similarly,
(64 + 53) ÷ 9 = 13
4. (B)
5. (C)
6. (D)54 – 30 = 24 and 112 – 42 = 70
Similarly,
x – 28 = 38
 x = 38 + 28 = 66
7. (D)216 – 7 = 209  209 – 7 = 202 and
522 – 7 = 515  515 – 7 = 508
Similarly,
633 – 7 = 626  626 – 7 = 619
8. (B)
9. (B)
10. (A)
11. (A)
12. (A)a a b b/abba/a abb/a b ba
13. (A) AZ B Y , A Z B Y, A Z B Y, A Z B Y
14. (C) 
15. (D) (A) 
(B) 
(C) 
(D) 
16. (B)
17. (A)
18. (B)(A) 7 × 11 – 3 = 74
(B) 9 × 11 – 3 = 96 

(C) 4 × 11 – 3 = 41
(D) 6 × 11 – 6 = 63
19. (B) Given set = 
Similarly, in option (B)
 20. (A)
21. (B)
22. (D)
23. (B)
24.  (D)
SSC MOCK TEST 29 (SOLUTION)
Page 2


 Centres at:     MUKHERJEE NAGAR     MUNIRKA    UTTAM NAGAR    DILSHAD GARDEN    ROHINI     BADARPUR BORDER
=============================================================
1
1. (B) Given +  >

– 
×  =
I  <
L 
 
Given expression A | B × C  A < B = C
because option (B) follows the given
expression here because C - B + A means
C 

 B > A, one of which meaning is  A<B= C
2. (C)2 M C 3 K
(2 M = N) (2 N  3 K)
               (2 N  3 K)
2 × 2 M  3 K (since N = 2 M)
4 M  3 K
4 M is not more than 3 K, then 2 M is less
than 3 K.
Hence 2 M C 3 K is correct.
3. (B) (15 +12) ÷ 9 = 3 and (44 + 28) ÷ 9 = 8
Similarly,
(64 + 53) ÷ 9 = 13
4. (B)
5. (C)
6. (D)54 – 30 = 24 and 112 – 42 = 70
Similarly,
x – 28 = 38
 x = 38 + 28 = 66
7. (D)216 – 7 = 209  209 – 7 = 202 and
522 – 7 = 515  515 – 7 = 508
Similarly,
633 – 7 = 626  626 – 7 = 619
8. (B)
9. (B)
10. (A)
11. (A)
12. (A)a a b b/abba/a abb/a b ba
13. (A) AZ B Y , A Z B Y, A Z B Y, A Z B Y
14. (C) 
15. (D) (A) 
(B) 
(C) 
(D) 
16. (B)
17. (A)
18. (B)(A) 7 × 11 – 3 = 74
(B) 9 × 11 – 3 = 96 

(C) 4 × 11 – 3 = 41
(D) 6 × 11 – 6 = 63
19. (B) Given set = 
Similarly, in option (B)
 20. (A)
21. (B)
22. (D)
23. (B)
24.  (D)
SSC MOCK TEST 29 (SOLUTION)
 Centres at:     MUKHERJEE NAGAR     MUNIRKA    UTTAM NAGAR    DILSHAD GARDEN    ROHINI     BADARPUR BORDER
=============================================================
2
25. (B) Total age of father and son = 22 × 2
    = 44 years
Father : Son
10           1
 Son's age = 
44
1 10
1


= 4 years
26.  (B) Veni > Smith > Raju > Salim
27.  (B) G E N E R A T E
28.  (C) Q U A I N T
29.  (B) According to question, the odd numerical
value of HOTEL will be –
HOTEL = 15 + 29 + 39 + 9 + 23
                      = 115
30.  (A) 
Similarly,
31.  (A) 29 × 48   2 × 9 × 4 × 8 = 576
    35 × 16  3 × 5 × 1 × 6 = 90,
    22 × 46  2 × 2 × 4 × 6 = 96 and
  Similarly,
42 × 17  4 × 2 × 1 × 7 = 56
32.  (C) 12 P 6 M 15 T 16 B 4
 12 × 6 + 15 – 16 ÷ 4
 12 × 6 + 15 – 4
 72 + 15 – 4
 87 – 4 = 83
33.  (B)  
     Meaningful word
34.  (C) 
Similarly,
35.  (A)
(Here = DA = CB =  2 km)
Required distance AE = DE – DA
   = 3 – 2
   = 1 km
36. (C) 
 According to given direction the faces of
Rani and Sarita will be East and South
direction with respect to x.
37. (D)
38. (B)
Conclusion I — 
II — ×
39. (B) 40. (B) 41. (A)
42. (C)
43. (B)
44. (A)
45. (A) 46. (B) 47. (C) 48. (A)
49. (D)
50. (D) The numerical groups of PLAY will be –
P – 15, 43
L – 36, 65
A – 42, 46, 62
Y – 45
51. (B)Straight line 4x + 3y = 12 passes
through 1st, 2nd & 4th quadrant.
52. (B)
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=============================================================
1
1. (B) Given +  >

– 
×  =
I  <
L 
 
Given expression A | B × C  A < B = C
because option (B) follows the given
expression here because C - B + A means
C 

 B > A, one of which meaning is  A<B= C
2. (C)2 M C 3 K
(2 M = N) (2 N  3 K)
               (2 N  3 K)
2 × 2 M  3 K (since N = 2 M)
4 M  3 K
4 M is not more than 3 K, then 2 M is less
than 3 K.
Hence 2 M C 3 K is correct.
3. (B) (15 +12) ÷ 9 = 3 and (44 + 28) ÷ 9 = 8
Similarly,
(64 + 53) ÷ 9 = 13
4. (B)
5. (C)
6. (D)54 – 30 = 24 and 112 – 42 = 70
Similarly,
x – 28 = 38
 x = 38 + 28 = 66
7. (D)216 – 7 = 209  209 – 7 = 202 and
522 – 7 = 515  515 – 7 = 508
Similarly,
633 – 7 = 626  626 – 7 = 619
8. (B)
9. (B)
10. (A)
11. (A)
12. (A)a a b b/abba/a abb/a b ba
13. (A) AZ B Y , A Z B Y, A Z B Y, A Z B Y
14. (C) 
15. (D) (A) 
(B) 
(C) 
(D) 
16. (B)
17. (A)
18. (B)(A) 7 × 11 – 3 = 74
(B) 9 × 11 – 3 = 96 

(C) 4 × 11 – 3 = 41
(D) 6 × 11 – 6 = 63
19. (B) Given set = 
Similarly, in option (B)
 20. (A)
21. (B)
22. (D)
23. (B)
24.  (D)
SSC MOCK TEST 29 (SOLUTION)
 Centres at:     MUKHERJEE NAGAR     MUNIRKA    UTTAM NAGAR    DILSHAD GARDEN    ROHINI     BADARPUR BORDER
=============================================================
2
25. (B) Total age of father and son = 22 × 2
    = 44 years
Father : Son
10           1
 Son's age = 
44
1 10
1


= 4 years
26.  (B) Veni > Smith > Raju > Salim
27.  (B) G E N E R A T E
28.  (C) Q U A I N T
29.  (B) According to question, the odd numerical
value of HOTEL will be –
HOTEL = 15 + 29 + 39 + 9 + 23
                      = 115
30.  (A) 
Similarly,
31.  (A) 29 × 48   2 × 9 × 4 × 8 = 576
    35 × 16  3 × 5 × 1 × 6 = 90,
    22 × 46  2 × 2 × 4 × 6 = 96 and
  Similarly,
42 × 17  4 × 2 × 1 × 7 = 56
32.  (C) 12 P 6 M 15 T 16 B 4
 12 × 6 + 15 – 16 ÷ 4
 12 × 6 + 15 – 4
 72 + 15 – 4
 87 – 4 = 83
33.  (B)  
     Meaningful word
34.  (C) 
Similarly,
35.  (A)
(Here = DA = CB =  2 km)
Required distance AE = DE – DA
   = 3 – 2
   = 1 km
36. (C) 
 According to given direction the faces of
Rani and Sarita will be East and South
direction with respect to x.
37. (D)
38. (B)
Conclusion I — 
II — ×
39. (B) 40. (B) 41. (A)
42. (C)
43. (B)
44. (A)
45. (A) 46. (B) 47. (C) 48. (A)
49. (D)
50. (D) The numerical groups of PLAY will be –
P – 15, 43
L – 36, 65
A – 42, 46, 62
Y – 45
51. (B)Straight line 4x + 3y = 12 passes
through 1st, 2nd & 4th quadrant.
52. (B)
 Centres at:     MUKHERJEE NAGAR     MUNIRKA    UTTAM NAGAR    DILSHAD GARDEN    ROHINI     BADARPUR BORDER
=============================================================
3
 In ADC  ,
AC is the hypotenuse, which is the
longest side of triangle.
 AC > AD
Similarly, AC > CF
AB > AD
AB > BE
BC > CF
and BC > BE
On adding above inequalities, we have
2(AB + BC + CA) > 2(AD + BE + CF)
  (AB + BC + CA) > (AD + BE + CF)
53. (B) ) ( ) ( C B B A 	  	  	  	 = 65º + 140º
B C B A 	  	  	  	 ) ( = 205º
180º + B 	 = 205º
B 	
= 205º – 180º
= 25º
54. (A)
OT = 4 cm
PO = 5 cm
PO
2
= PT
2
 + TO
2
  5
2
= PT
2
 + 4
2
 PT = 3 cm
55. (B)








x
x
2
3
= 
x
3
3x – 2 = 
x
3
3x – 
x
3
= 2
x – 
x
1
= 
3
2
2
2
1
x
x 
= 
2
1
2
 







x
x
= 
2
3
2
2
 






= 
1
2
9
4

 = 
9
22
= 
9
4
2
56. (A) 16a
2
 – 12a
= (4a)
2
 – 2.4a × 
2
2
3
2
3








= 
2
2
3
4 






 a
4
9
 must be added in 16a
2
 – 12a to make it a
perfect square.
57. (C)
Let AB & CD are two parallel chords of
C(0, 10 cm).
AB = 12 cm  AM = 
2
1
× 12 = 6 cm
[perp. from centre to any chord,
 bisects the chord]
Similarly CN = 
2
1
× 16 = 8 cm
AM
2
 + OM
2
= OA
2
 6
2
   + OM
2
=10
2
 OM = 
36 100
= 
64
= 8 cm
Again,
CN
2
 + ON
2
= OC
2
  8
2
  + ON
2
= 10
2
 ON = 
36 100
= 
36
 = 6 cm
 The distance between the chords
= MN = OM + ON
=    8  + 6
= 14 cm
58. (D)
BD = AD – AD
= 10 – 4 = 6 cm
BD : DA = BE : EC
 6 : 4= BE : EC  BE : CE = 3 : 2
59. (B) 
In AOB  ,
 OA = OB
 º 45  	  	 OBA OAB
In 
AOC 
,
OA = OC
 º 35  	  	 OCA OAC
 OAC OAB BAC 	  	  	
º 80 º 35 º 45   
Page 4


 Centres at:     MUKHERJEE NAGAR     MUNIRKA    UTTAM NAGAR    DILSHAD GARDEN    ROHINI     BADARPUR BORDER
=============================================================
1
1. (B) Given +  >

– 
×  =
I  <
L 
 
Given expression A | B × C  A < B = C
because option (B) follows the given
expression here because C - B + A means
C 

 B > A, one of which meaning is  A<B= C
2. (C)2 M C 3 K
(2 M = N) (2 N  3 K)
               (2 N  3 K)
2 × 2 M  3 K (since N = 2 M)
4 M  3 K
4 M is not more than 3 K, then 2 M is less
than 3 K.
Hence 2 M C 3 K is correct.
3. (B) (15 +12) ÷ 9 = 3 and (44 + 28) ÷ 9 = 8
Similarly,
(64 + 53) ÷ 9 = 13
4. (B)
5. (C)
6. (D)54 – 30 = 24 and 112 – 42 = 70
Similarly,
x – 28 = 38
 x = 38 + 28 = 66
7. (D)216 – 7 = 209  209 – 7 = 202 and
522 – 7 = 515  515 – 7 = 508
Similarly,
633 – 7 = 626  626 – 7 = 619
8. (B)
9. (B)
10. (A)
11. (A)
12. (A)a a b b/abba/a abb/a b ba
13. (A) AZ B Y , A Z B Y, A Z B Y, A Z B Y
14. (C) 
15. (D) (A) 
(B) 
(C) 
(D) 
16. (B)
17. (A)
18. (B)(A) 7 × 11 – 3 = 74
(B) 9 × 11 – 3 = 96 

(C) 4 × 11 – 3 = 41
(D) 6 × 11 – 6 = 63
19. (B) Given set = 
Similarly, in option (B)
 20. (A)
21. (B)
22. (D)
23. (B)
24.  (D)
SSC MOCK TEST 29 (SOLUTION)
 Centres at:     MUKHERJEE NAGAR     MUNIRKA    UTTAM NAGAR    DILSHAD GARDEN    ROHINI     BADARPUR BORDER
=============================================================
2
25. (B) Total age of father and son = 22 × 2
    = 44 years
Father : Son
10           1
 Son's age = 
44
1 10
1


= 4 years
26.  (B) Veni > Smith > Raju > Salim
27.  (B) G E N E R A T E
28.  (C) Q U A I N T
29.  (B) According to question, the odd numerical
value of HOTEL will be –
HOTEL = 15 + 29 + 39 + 9 + 23
                      = 115
30.  (A) 
Similarly,
31.  (A) 29 × 48   2 × 9 × 4 × 8 = 576
    35 × 16  3 × 5 × 1 × 6 = 90,
    22 × 46  2 × 2 × 4 × 6 = 96 and
  Similarly,
42 × 17  4 × 2 × 1 × 7 = 56
32.  (C) 12 P 6 M 15 T 16 B 4
 12 × 6 + 15 – 16 ÷ 4
 12 × 6 + 15 – 4
 72 + 15 – 4
 87 – 4 = 83
33.  (B)  
     Meaningful word
34.  (C) 
Similarly,
35.  (A)
(Here = DA = CB =  2 km)
Required distance AE = DE – DA
   = 3 – 2
   = 1 km
36. (C) 
 According to given direction the faces of
Rani and Sarita will be East and South
direction with respect to x.
37. (D)
38. (B)
Conclusion I — 
II — ×
39. (B) 40. (B) 41. (A)
42. (C)
43. (B)
44. (A)
45. (A) 46. (B) 47. (C) 48. (A)
49. (D)
50. (D) The numerical groups of PLAY will be –
P – 15, 43
L – 36, 65
A – 42, 46, 62
Y – 45
51. (B)Straight line 4x + 3y = 12 passes
through 1st, 2nd & 4th quadrant.
52. (B)
 Centres at:     MUKHERJEE NAGAR     MUNIRKA    UTTAM NAGAR    DILSHAD GARDEN    ROHINI     BADARPUR BORDER
=============================================================
3
 In ADC  ,
AC is the hypotenuse, which is the
longest side of triangle.
 AC > AD
Similarly, AC > CF
AB > AD
AB > BE
BC > CF
and BC > BE
On adding above inequalities, we have
2(AB + BC + CA) > 2(AD + BE + CF)
  (AB + BC + CA) > (AD + BE + CF)
53. (B) ) ( ) ( C B B A 	  	  	  	 = 65º + 140º
B C B A 	  	  	  	 ) ( = 205º
180º + B 	 = 205º
B 	
= 205º – 180º
= 25º
54. (A)
OT = 4 cm
PO = 5 cm
PO
2
= PT
2
 + TO
2
  5
2
= PT
2
 + 4
2
 PT = 3 cm
55. (B)








x
x
2
3
= 
x
3
3x – 2 = 
x
3
3x – 
x
3
= 2
x – 
x
1
= 
3
2
2
2
1
x
x 
= 
2
1
2
 







x
x
= 
2
3
2
2
 






= 
1
2
9
4

 = 
9
22
= 
9
4
2
56. (A) 16a
2
 – 12a
= (4a)
2
 – 2.4a × 
2
2
3
2
3








= 
2
2
3
4 






 a
4
9
 must be added in 16a
2
 – 12a to make it a
perfect square.
57. (C)
Let AB & CD are two parallel chords of
C(0, 10 cm).
AB = 12 cm  AM = 
2
1
× 12 = 6 cm
[perp. from centre to any chord,
 bisects the chord]
Similarly CN = 
2
1
× 16 = 8 cm
AM
2
 + OM
2
= OA
2
 6
2
   + OM
2
=10
2
 OM = 
36 100
= 
64
= 8 cm
Again,
CN
2
 + ON
2
= OC
2
  8
2
  + ON
2
= 10
2
 ON = 
36 100
= 
36
 = 6 cm
 The distance between the chords
= MN = OM + ON
=    8  + 6
= 14 cm
58. (D)
BD = AD – AD
= 10 – 4 = 6 cm
BD : DA = BE : EC
 6 : 4= BE : EC  BE : CE = 3 : 2
59. (B) 
In AOB  ,
 OA = OB
 º 45  	  	 OBA OAB
In 
AOC 
,
OA = OC
 º 35  	  	 OCA OAC
 OAC OAB BAC 	  	  	
º 80 º 35 º 45   
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=============================================================
4
60. (A)
Let h m be the height of the ballon.
In CBD  ,
tan60º = 
BD
CB
3
  = 
BD
h
 BD     = 
3
h
cm
In CBA 
tan30º = 
BA
CB
3
1
   = 
DA BD
h

 = 
1000
3

h
h
 1000
3

h
 = 
3 h
3
3
h
h 
  = 1000









 
3
1 3
h
  = 1000
h  = 
2
3 1000
  = 500 3
  = 
1000
3 500
km = 3
2
1
km
61. (C)
CBD CAD 	  	
 (Angles in the same
 segment of a circle)
   	CBD
Now, COB AOB    (by RHS)
 CBO ABO 	  	 (by CPCT)
  	  	 CBD ABO
 ABC 	 =   	  	 2 CBO ABO
62. (C)
Let ABC  is equilateral & AD is its height.
Let 'a' unit is the side of the ABC  .
 AB = a  & BD = 
2
a
height AD = 
2 2
BD AB 
      15 = 
4
2
2
a
a 
      15 = a
2
3
 a = 
3
2 15
cm
Hence BC = 
3
30
 = 
3
3 3 10 
 = 
3 10
cm
Area of 
ABC 
= 
2
4
3
a 
= 3 10 3 10
4
3
 
= 
3 75
cm
2
.
63. (A)Work done by Ram & Shyam in 1 day
= 
12
1
Work done by Shyam & Hari in 1 day
= 
15
1
Work done by Hari & Ram in 1 day
= 
20
1
Work done by 2[Ram + Shyam + Hari]
= 
20
1
15
1
12
1
 
= 
60
3 4 5  
 = 
60
12
= 
5
1
Work done by (Ram + Shyam + Hari)
in 1 day = 
10
1
Work done by Ram alone in 1 day
= 
10
1
5
1

= 
30
2 3
= 
30
1
 Ram can do the whole work in 30 days.
Page 5


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=============================================================
1
1. (B) Given +  >

– 
×  =
I  <
L 
 
Given expression A | B × C  A < B = C
because option (B) follows the given
expression here because C - B + A means
C 

 B > A, one of which meaning is  A<B= C
2. (C)2 M C 3 K
(2 M = N) (2 N  3 K)
               (2 N  3 K)
2 × 2 M  3 K (since N = 2 M)
4 M  3 K
4 M is not more than 3 K, then 2 M is less
than 3 K.
Hence 2 M C 3 K is correct.
3. (B) (15 +12) ÷ 9 = 3 and (44 + 28) ÷ 9 = 8
Similarly,
(64 + 53) ÷ 9 = 13
4. (B)
5. (C)
6. (D)54 – 30 = 24 and 112 – 42 = 70
Similarly,
x – 28 = 38
 x = 38 + 28 = 66
7. (D)216 – 7 = 209  209 – 7 = 202 and
522 – 7 = 515  515 – 7 = 508
Similarly,
633 – 7 = 626  626 – 7 = 619
8. (B)
9. (B)
10. (A)
11. (A)
12. (A)a a b b/abba/a abb/a b ba
13. (A) AZ B Y , A Z B Y, A Z B Y, A Z B Y
14. (C) 
15. (D) (A) 
(B) 
(C) 
(D) 
16. (B)
17. (A)
18. (B)(A) 7 × 11 – 3 = 74
(B) 9 × 11 – 3 = 96 

(C) 4 × 11 – 3 = 41
(D) 6 × 11 – 6 = 63
19. (B) Given set = 
Similarly, in option (B)
 20. (A)
21. (B)
22. (D)
23. (B)
24.  (D)
SSC MOCK TEST 29 (SOLUTION)
 Centres at:     MUKHERJEE NAGAR     MUNIRKA    UTTAM NAGAR    DILSHAD GARDEN    ROHINI     BADARPUR BORDER
=============================================================
2
25. (B) Total age of father and son = 22 × 2
    = 44 years
Father : Son
10           1
 Son's age = 
44
1 10
1


= 4 years
26.  (B) Veni > Smith > Raju > Salim
27.  (B) G E N E R A T E
28.  (C) Q U A I N T
29.  (B) According to question, the odd numerical
value of HOTEL will be –
HOTEL = 15 + 29 + 39 + 9 + 23
                      = 115
30.  (A) 
Similarly,
31.  (A) 29 × 48   2 × 9 × 4 × 8 = 576
    35 × 16  3 × 5 × 1 × 6 = 90,
    22 × 46  2 × 2 × 4 × 6 = 96 and
  Similarly,
42 × 17  4 × 2 × 1 × 7 = 56
32.  (C) 12 P 6 M 15 T 16 B 4
 12 × 6 + 15 – 16 ÷ 4
 12 × 6 + 15 – 4
 72 + 15 – 4
 87 – 4 = 83
33.  (B)  
     Meaningful word
34.  (C) 
Similarly,
35.  (A)
(Here = DA = CB =  2 km)
Required distance AE = DE – DA
   = 3 – 2
   = 1 km
36. (C) 
 According to given direction the faces of
Rani and Sarita will be East and South
direction with respect to x.
37. (D)
38. (B)
Conclusion I — 
II — ×
39. (B) 40. (B) 41. (A)
42. (C)
43. (B)
44. (A)
45. (A) 46. (B) 47. (C) 48. (A)
49. (D)
50. (D) The numerical groups of PLAY will be –
P – 15, 43
L – 36, 65
A – 42, 46, 62
Y – 45
51. (B)Straight line 4x + 3y = 12 passes
through 1st, 2nd & 4th quadrant.
52. (B)
 Centres at:     MUKHERJEE NAGAR     MUNIRKA    UTTAM NAGAR    DILSHAD GARDEN    ROHINI     BADARPUR BORDER
=============================================================
3
 In ADC  ,
AC is the hypotenuse, which is the
longest side of triangle.
 AC > AD
Similarly, AC > CF
AB > AD
AB > BE
BC > CF
and BC > BE
On adding above inequalities, we have
2(AB + BC + CA) > 2(AD + BE + CF)
  (AB + BC + CA) > (AD + BE + CF)
53. (B) ) ( ) ( C B B A 	  	  	  	 = 65º + 140º
B C B A 	  	  	  	 ) ( = 205º
180º + B 	 = 205º
B 	
= 205º – 180º
= 25º
54. (A)
OT = 4 cm
PO = 5 cm
PO
2
= PT
2
 + TO
2
  5
2
= PT
2
 + 4
2
 PT = 3 cm
55. (B)








x
x
2
3
= 
x
3
3x – 2 = 
x
3
3x – 
x
3
= 2
x – 
x
1
= 
3
2
2
2
1
x
x 
= 
2
1
2
 







x
x
= 
2
3
2
2
 






= 
1
2
9
4

 = 
9
22
= 
9
4
2
56. (A) 16a
2
 – 12a
= (4a)
2
 – 2.4a × 
2
2
3
2
3








= 
2
2
3
4 






 a
4
9
 must be added in 16a
2
 – 12a to make it a
perfect square.
57. (C)
Let AB & CD are two parallel chords of
C(0, 10 cm).
AB = 12 cm  AM = 
2
1
× 12 = 6 cm
[perp. from centre to any chord,
 bisects the chord]
Similarly CN = 
2
1
× 16 = 8 cm
AM
2
 + OM
2
= OA
2
 6
2
   + OM
2
=10
2
 OM = 
36 100
= 
64
= 8 cm
Again,
CN
2
 + ON
2
= OC
2
  8
2
  + ON
2
= 10
2
 ON = 
36 100
= 
36
 = 6 cm
 The distance between the chords
= MN = OM + ON
=    8  + 6
= 14 cm
58. (D)
BD = AD – AD
= 10 – 4 = 6 cm
BD : DA = BE : EC
 6 : 4= BE : EC  BE : CE = 3 : 2
59. (B) 
In AOB  ,
 OA = OB
 º 45  	  	 OBA OAB
In 
AOC 
,
OA = OC
 º 35  	  	 OCA OAC
 OAC OAB BAC 	  	  	
º 80 º 35 º 45   
 Centres at:     MUKHERJEE NAGAR     MUNIRKA    UTTAM NAGAR    DILSHAD GARDEN    ROHINI     BADARPUR BORDER
=============================================================
4
60. (A)
Let h m be the height of the ballon.
In CBD  ,
tan60º = 
BD
CB
3
  = 
BD
h
 BD     = 
3
h
cm
In CBA 
tan30º = 
BA
CB
3
1
   = 
DA BD
h

 = 
1000
3

h
h
 1000
3

h
 = 
3 h
3
3
h
h 
  = 1000









 
3
1 3
h
  = 1000
h  = 
2
3 1000
  = 500 3
  = 
1000
3 500
km = 3
2
1
km
61. (C)
CBD CAD 	  	
 (Angles in the same
 segment of a circle)
   	CBD
Now, COB AOB    (by RHS)
 CBO ABO 	  	 (by CPCT)
  	  	 CBD ABO
 ABC 	 =   	  	 2 CBO ABO
62. (C)
Let ABC  is equilateral & AD is its height.
Let 'a' unit is the side of the ABC  .
 AB = a  & BD = 
2
a
height AD = 
2 2
BD AB 
      15 = 
4
2
2
a
a 
      15 = a
2
3
 a = 
3
2 15
cm
Hence BC = 
3
30
 = 
3
3 3 10 
 = 
3 10
cm
Area of 
ABC 
= 
2
4
3
a 
= 3 10 3 10
4
3
 
= 
3 75
cm
2
.
63. (A)Work done by Ram & Shyam in 1 day
= 
12
1
Work done by Shyam & Hari in 1 day
= 
15
1
Work done by Hari & Ram in 1 day
= 
20
1
Work done by 2[Ram + Shyam + Hari]
= 
20
1
15
1
12
1
 
= 
60
3 4 5  
 = 
60
12
= 
5
1
Work done by (Ram + Shyam + Hari)
in 1 day = 
10
1
Work done by Ram alone in 1 day
= 
10
1
5
1

= 
30
2 3
= 
30
1
 Ram can do the whole work in 30 days.
 Centres at:     MUKHERJEE NAGAR     MUNIRKA    UTTAM NAGAR    DILSHAD GARDEN    ROHINI     BADARPUR BORDER
=============================================================
5
64. (A) 3 men = 5 women
 1 man = 
3
5
women
 6 men = 
3
5
 × 6 = 10 women
6 men + 5 women = (10 + 5) women
 5 women complete a work in 15 days.
 1 woman completes a work in 15 × 5
 15 women complete a work in 
15
5 15
= 5 days
65. (C) Work done by (A + B) in 1 day = 
12
1
Work done by (B + C) in 1 day = 
15
1
Work done by (C + A) in 1 day = 
20
1
Work done by 2(A + B + C) in 1 day
= 
20
1
15
1
12
1
 
= 
60
3 4 5  
= 
60
12
 = 
5
1
Work done by (A + B + C) in 1 day
= 
2 5
1

= 
10
1
Work done by A alone in 1 day
= 
15 10
1 

 = 
30
2 3
= 
30
1
Hence A can complete the work alone
in 30 days.
66. (A)Perimeter – Diameter = X
r r 2 2   = X
) 1 ( 2   r = X
2r = 
1  
X
67. (D)
r  2 = a
r = 
 2
a
Volume = V
h r
2
 = V
h
a
² 4
²

 
= V
 h  = 
2
4
a
V 
68. (D) 4
th
 root of 24010000
= 
4
1
) 24010000 (
= 
4
1
) 10 10 10 10 7 7 7 7 (       
= 7 × 10 = 70
69. (C) Greatest 4-digit number = 9999
Greatest 4-digit perfect square number
= 9999 – 198
= 9801
70. (A)MP = Rs. x (say)
SP = 90% of x = Rs. 
10
9x
% profit = 12%
        CP =
% 12 100
100 SP


=
112
100
10
9

x
=
112
90x
Now,
MP
CP
=
x
x
112
90
= 
56
45
71. (D) MP = Rs. 160
SP after two successive discounts
= Rs. 122.40
   Ist discount = 10%
ATQ,






 






 
100
100
100
10 100
160
y
  = Rs. 122.40
100
100
100
90
160
y 
 
    = Rs. 122.40
    (100 – y) = 
160 90
100 100 40 . 122

 
= 85
      y = 100 – 85 = 15
2nd discount = 15%
72. (B)Let  'g' stands for the no. of girls
 & 'b' stands for the no. of boys
10% of g = 
20
1
of b

100
10
 × g = 
20
1
 × b

g
b
= 
100
20 10
= 
1
2
 b : g = 2 : 1
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