= 90 18 = 72 litres > y spirit x – After adding some water to the 60 (D) x = 15, 24 mixture water became 50% of the y = 25, 24 new mixture > x y = litres of in new mixture 7 61. (C) No. of A level employees in II = 50% of new mixture = 0.14 × 25500 = 3570 litres of new mixture total No. of D level employees in I = = 144 litres = 0.33 × 2050 = 6765 So water to be added = 144 90 – % = Required = 54 litres = 52.77 53% 56-60 Total No. of F level employees in I 62. (E) 56. (B) x = = 0.13 × y = = 2665 x > y total no. of E level employees in II 57. (A) Multiple (i) by 7 and multiply (ii) = 0.18 × 2550 by 3 and subtract (ii) from (i) = 4590 y = No. = 4590 + 2665 Required x = = 7255 x > y 58. (E) x = 7, 63. (B) No. of B level employees in I – y = 6, = 0.05 × 20500 – 59. (A) x = = 1025 y = No. of B level employees II
spirit = 90 – 18 = 72 litres After adding some water to the mixture water became 50% of the new mixture 7 = litres of in new mixture = 50% of new mixture total litres of new mixture = = 144 litres So water to be added = 144 – 90 = 54 litres 56-60 56. (B) x = y = x > y 57. (A) Multiple (i) by 7 and multiply (ii) by 3 and subtract (ii) from (i) y = x = x > y 58. (E) x = – 7, y = – 6, 59. (A) x = y = x > y 60 (D) x = 15, 24 y = 25, 24 y > x 61. (C) No. of A level employees in II = 0.14 × 25500 = 3570 No. of D level employees in I = 0.33 × 2050 = 6765 Required % = = 52.77 53% 62. (E) Total No. of F level employees in I = 0.13 × = 2665 total no. of E level employees in II = 0.18 × 2550 = 4590 Required No. = 4590 + 2665 = 7255 63. (B) No. of B level employees in I = 0.05 × 20500 = 1025 No. of B level employees II