Page 1 www.bankersadda.com | www.sscadda.com | www.careerpower.in | www.careeradda.co.in Page 1 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 onâ€¦.) 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 â€“ (85 â€“ 75)/150 ] =69.94 75. (3): 10x + y â€“ (10y +x ) = 45 9x â€“ 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 www.bankersadda.com | www.sscadda.com | www.careerpower.in | www.careeradda.co.in Page 1 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 onâ€¦.) 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 â€“ (85 â€“ 75)/150 ] =69.94 75. (3): 10x + y â€“ (10y +x ) = 45 9x â€“ 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 ? www.bankersadda.com | www.sscadda.com | www.careerpower.in | www.careeradda.co.in Page 2 84. (4): Average No. of players in Academy C = 11005 1572 7 ? Average No. of players in Academy D = 10595 1514 7 ? Difference = 1572 â€“ 1514 = 58 85. (5): Total No. of players in Academy B = 11,710 Total No. of players in Academy D = 10,595 Difference = 11710 â€“ 10595 = 1,115 (86-90): 86. (1): I. P 2 â€“ 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 â€“ 3q = â€“ 3.5 2p â€“ 3q + 3.5 = 0 II. 3p â€“ 2q = â€“ 6.5 3 x (2p â€“ 3q = â€“ 3.5)â€¦.(1) 2 x (3p â€“ 2q = â€“ 6.5â€¦â€¦â€¦(2) From Equation (1) and (2) q = - 0.5 ,p = - 2.5 q > p 89. (5): I. p 2 + 8p + 15 = 0 p = ( â€“ 3, â€“ 5) II. q 2 + 10q + 24 = 0 q = (â€“ 4 , â€“ 6) Relationship cannot be determined. 90. (4): I. p 2 + 28p + 192 = 0 p = ( â€“ 12, â€“ 16) II. q 2 + 16q + 48 = 0 q = (â€“ 4, â€“ 12) p = q 91. (3) Let the number be X and Y Then 6y â€“ 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 â€“ 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 â€¦â€¦(1) D =(x â€“ 3) 3/2 â€¦â€¦â€¦â€¦(2) From eq (1) and (2) X = 15 kmph 98. (1). Marked price = 130% Discount = 20% S.P. = (130*80)/100% = 104% Profit % = (104 â€“ 100) = 4% 99. (2); = x=7200 100. (3); Speed downstream = 20+4=24 Distance = Speed Time = km.Read More

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