QUANTITATIVE APTITUDE: NUMERICAL ABILITY - Practice Questions IT & Software Notes | EduRev

IT & Software : QUANTITATIVE APTITUDE: NUMERICAL ABILITY - Practice Questions IT & Software Notes | EduRev

 Page 1


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SYNDICATE PRACTICE SET  
QUANTITATIVE APTITUDE 
 
NUMERICAL ABILITY 
 
 Income Expenditure Profit% Profit 
2003 120 111.63 7.5 8.37 
2004 160 139.13 15 20.87 
2005 130 106.12 22.5 23.88 
2006 170 144.68 17.5 25.32 
2007 190 158.33 20 31.67 
2008 150 117.65 27.5 23.35 
Expenditure 
      
           
     
51. (e) Maximum Profit obtained in the year 2007  
52. (b) Total expenditure     
   
     
     
   
   
 
    
   
     
     
   
     
     
   
   
     
   
     
 
      
Average 
     
 
     
53. (c) 
       
    
             
54.  (c); Expenditure        
   
           
 
     
   
      
  
=140 (approx) 
55. (d); Expenditure     
   
   
     
56. (4):  
16
59.95 16.001 12.005 ? 60 80
12
? ? ? ? ? ? 
57. (3): 
5 162
183 850 ? 101 1377 1478
9 100
? ? ? ? ? ? ? 
58. (4): 
33 3
79509 7.002 ? 43 7 386 ? ? ? ? ? 
59. (2): 
4 5 161 20815 4 20
x 20815 x 72400
7 16 100 23
??
? ? ? ? ? ? ?
 
60. (3):  
? ? ?
? ? ? ? ? ? ? ?
28943.008 8762.950 x 20223
28943 8763 x 20223 x 20223 20180 43
 
61. (3):   17 × 1 = 17 + 
2
1 = 18 
18 × 1 = 18 + 
2
2 = 22 
22 × 1 = 22 + 
2
3 = 31 
31 × 1 = 31 + 
2
4 = 47 
47 × 1 = 47 + 
2
5 =69 
62. (3): The series of prime number. 
63. (5):   4 +2.5 = 6.5 
6.5 + 3.5 = 10.0 
10 + 4.5 = 14.5 
14.5 + 5.5 = 20.0 
20 + 6.5 = 26.5 
64. (5): All are multiples of 13 and 1 is added. ( 13 x 1 + 1 
= 14, 13 x 2 + 1 = 27, and so on….) 
 
65. (1):  5 * 1 + 1 =6 
6 * 2 + 2 = 14 
14 * 3 + 3 = 45 
45 * 4 + 4 = 184 
184 * 5 + 5 = 925 
66. (2) 
67. (3): either of them I & II alone are required to answer 
the question 
68. (5):  both I & II are required to answer the question 
69. (5): both I & II are required to answer the question 
70. (5): both I & II are required to answer the question 
71. (2): TROPHY 
6! = 6×5×4×3×2×1 = 720 
72. (2): let the numerator be x & denominator be y 
x 3x 3 4x 3 x 9
20x 9y
y 2y 5 3y 5 y 20
?
? ? ? ? ? ? ?
?
 
73. (3): Average speed = total distance /Total time = (36 + 
16 + 24)/(36/9 +16/8 +24/4) = 6.3  
74. (2): Required = [70 – (85 – 75)/150 ] =69.94 
75. (3):  10x + y – (10y +x ) = 45 
9x – 9y = 45 
x - y = 5 
(76-80): 
Total 2000 employees 
Departments Officer Clerk 
A 96 260 
B 144 216 
C 216 184 
D 184 156 
E 320 224 
Total 960 1040 
76.  (3): 216 
77.  (4): 184X100/156 = 117.94 
78.  (5): 224 
79.  (3): 96X100/2000 = 4.8 = 5% Approx 
80. (4): Total no. of clerks in department B & D = 216 + 
156 = 372 
(81-85): 
81. (4): 
11445
1635
7
? 
82. (3):  1750X100 /8700 = 20.11 % 
83. (3): 
2000
8:9
2250
? 
Page 2


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SYNDICATE PRACTICE SET  
QUANTITATIVE APTITUDE 
 
NUMERICAL ABILITY 
 
 Income Expenditure Profit% Profit 
2003 120 111.63 7.5 8.37 
2004 160 139.13 15 20.87 
2005 130 106.12 22.5 23.88 
2006 170 144.68 17.5 25.32 
2007 190 158.33 20 31.67 
2008 150 117.65 27.5 23.35 
Expenditure 
      
           
     
51. (e) Maximum Profit obtained in the year 2007  
52. (b) Total expenditure     
   
     
     
   
   
 
    
   
     
     
   
     
     
   
   
     
   
     
 
      
Average 
     
 
     
53. (c) 
       
    
             
54.  (c); Expenditure        
   
           
 
     
   
      
  
=140 (approx) 
55. (d); Expenditure     
   
   
     
56. (4):  
16
59.95 16.001 12.005 ? 60 80
12
? ? ? ? ? ? 
57. (3): 
5 162
183 850 ? 101 1377 1478
9 100
? ? ? ? ? ? ? 
58. (4): 
33 3
79509 7.002 ? 43 7 386 ? ? ? ? ? 
59. (2): 
4 5 161 20815 4 20
x 20815 x 72400
7 16 100 23
??
? ? ? ? ? ? ?
 
60. (3):  
? ? ?
? ? ? ? ? ? ? ?
28943.008 8762.950 x 20223
28943 8763 x 20223 x 20223 20180 43
 
61. (3):   17 × 1 = 17 + 
2
1 = 18 
18 × 1 = 18 + 
2
2 = 22 
22 × 1 = 22 + 
2
3 = 31 
31 × 1 = 31 + 
2
4 = 47 
47 × 1 = 47 + 
2
5 =69 
62. (3): The series of prime number. 
63. (5):   4 +2.5 = 6.5 
6.5 + 3.5 = 10.0 
10 + 4.5 = 14.5 
14.5 + 5.5 = 20.0 
20 + 6.5 = 26.5 
64. (5): All are multiples of 13 and 1 is added. ( 13 x 1 + 1 
= 14, 13 x 2 + 1 = 27, and so on….) 
 
65. (1):  5 * 1 + 1 =6 
6 * 2 + 2 = 14 
14 * 3 + 3 = 45 
45 * 4 + 4 = 184 
184 * 5 + 5 = 925 
66. (2) 
67. (3): either of them I & II alone are required to answer 
the question 
68. (5):  both I & II are required to answer the question 
69. (5): both I & II are required to answer the question 
70. (5): both I & II are required to answer the question 
71. (2): TROPHY 
6! = 6×5×4×3×2×1 = 720 
72. (2): let the numerator be x & denominator be y 
x 3x 3 4x 3 x 9
20x 9y
y 2y 5 3y 5 y 20
?
? ? ? ? ? ? ?
?
 
73. (3): Average speed = total distance /Total time = (36 + 
16 + 24)/(36/9 +16/8 +24/4) = 6.3  
74. (2): Required = [70 – (85 – 75)/150 ] =69.94 
75. (3):  10x + y – (10y +x ) = 45 
9x – 9y = 45 
x - y = 5 
(76-80): 
Total 2000 employees 
Departments Officer Clerk 
A 96 260 
B 144 216 
C 216 184 
D 184 156 
E 320 224 
Total 960 1040 
76.  (3): 216 
77.  (4): 184X100/156 = 117.94 
78.  (5): 224 
79.  (3): 96X100/2000 = 4.8 = 5% Approx 
80. (4): Total no. of clerks in department B & D = 216 + 
156 = 372 
(81-85): 
81. (4): 
11445
1635
7
? 
82. (3):  1750X100 /8700 = 20.11 % 
83. (3): 
2000
8:9
2250
? 
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84. (4): Average No. of players in Academy C = 
11005
1572
7
? 
Average No. of players in Academy D = 
10595
1514
7
? 
Difference = 1572 – 1514 = 58 
85. (5): Total No. of players in Academy B = 11,710 
Total No. of players in Academy D = 10,595 
Difference = 11710 – 10595 = 1,115 
 (86-90):  
86. 
(1): 
I. P
2
– 6p + 8 = 0 
P=  ( + 2,  + 4) 
II. q
2
 + 11q + 10 = 0 
q = (- 1, - 10) 
Clearly, p > q 
 
87. (2): I. p = 
2809
  
 P = 53 
 II. q
2
 = 2809 
 q = ± 53 
 p = q 
88. (3): 2p – 3q = – 3.5 
2p – 3q + 3.5 = 0 
II. 3p – 2q = – 6.5 
3 x (2p – 3q = – 3.5)….(1) 
2 x (3p – 2q = – 6.5………(2) 
From Equation (1) and (2) 
q = - 0.5 ,p = - 2.5 
 q  > p 
89. (5): I. p
2
 + 8p + 15 = 0   
p = ( – 3,  – 5) 
II. q
2
 + 10q + 24 = 0 
         q = (– 4  , – 6) 
Relationship cannot be determined. 
90. (4): I. p
2
 + 28p + 192 = 0  
p = ( – 12, – 16) 
II. q
2
 + 16q + 48 = 0 
q = (– 4, – 12)  
p = q 
91. (3) Let the number be X and Y 
Then 6y – x = 71 and 7x + y = 62 
 Solving the equation  
X = 7 , y = 13  
92. (4) Let the present ages of Yug, Vinod and karan be 
3x, 4x and 5x years respectively. 
Now, (3x + 4x + 5x)/3 = 28 ?12x = 84?x = 84/12 = 7 
So, required Sum = (3x + 4x + (5 + 5) years  
= (7x+ 10) years 
= (7 ×7 + 10) years 
=59 years  
93. (2) Area of the circle = 
??
22
22
(14) 616cm
7
 
Area of the rectangle 
2
1166 616 550cm ? ? ?
 
Breadth of the rectangle 
550
22cm
25
??
 
So, required sum = 
? ? ? ? ? ?
22
2 14 2(25 22) 182cm
7
  
94. (4) Let the length of the platform be x metres, Then, 
Speed of Metro train = 120 kmph = 
5 100
120 mps
18 3
??
 
100
320 x 24 x 800 320 480m
3
? ? ? ? ? ? ? ?
 
Required speed of woman 
480
2mps
4 60
??
?
 
95. (2) Number of questions attempted correctly = 
 70% of 10 + 50% of 30 + 60% of 45  = 7 + 15 + 27 = 
49 
Passing grade = (60/100 )* 85 = 51 
Reqd.  Ans = 51 - 49 = 2 
96. (1) First she faced loss of 12% 
Now she left with 88% amount then again she have 
gain of 12%  
Then amount she have = (88*112)/100 = 98.56% 
Loss% = 100 – 98.56 = 1.44% 
Loss = 46000*1.44% = 662.4 Rs. 
97. (5)  Let distance be D and speed of boat in still water = 
x kmph  
D = (x+3) 1 =x + 3 ……(1) 
D  =(x – 3) 3/2  …………(2) 
From eq (1) and (2)  
 X = 15 kmph  
98. (1).  Marked price = 130% 
Discount = 20%  
S.P. = (130*80)/100% = 104% 
Profit % = (104 – 100) = 4% 
 
99. (2);  
       
    
 = 
  
  
 
 x=7200 
100. (3); Speed downstream = 20+4=24 
 Distance = Speed Time 
= 
     
  
    km. 
 
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