= 812 C.I. - S.I. = x 50. (E) = 20% of ? 1280 = = by solving it we get ? = = 5 = 10,000 P 51. (B) Part filled without leak in 1 hour = 54. (C) Area of rectangle = length × Breadth Part filled with leak in 1 hour = of given rectangle = 9 × 8 Area Part emptied by leak in 1 hour : = 72 cm2 As per the given data, area of square is twice the area of = rectangle x = 7  of square = 2 × 72 cm2 Area 52. (B) Average of 7 No. = 304 = 144 cm2 Sum of that seven no. = 304 × 7 Area of square = a2, where a = side of = 2128 the square Average of first three no. = 456 = a2 144 sum of that first three = a = No. = 456 × 3 1368  side of square = a = 12 cm Sum of last three No. = 234 × 3 for square, diameter of circumcircle = 702 = Diagonal of square = = cm Third Number = 2128 (1368 + Radius of circumcircle = cm  – 702) 55. (C) mixture of 90 liter of the spirit and = 58 water contains 20% water first mixture 53. (B) We know the difference b/w C.I & S.I for 3 is, = (20/100) × 90 = 18 litres
x = 812 50. (E) = 20% of ? = ? = = 5 51. (B) Part filled without leak in 1 hour = Part filled with leak in 1 hour = Part emptied by leak in 1 hour : =  x = 7 52. (B) Average of 7 No. = 304 Sum of that seven no. = 304 × 7 = 2128 Average of first three no. = 456 sum of that first three No. = 456 × 3  1368 Sum of last three No. = 234 × 3 = 702  Third Number = 2128 – (1368 + 702) = 58 53. (B) We know the difference b/w C.I & S.I for 3 is, C.I. - S.I. = 1280 = by solving it we get P = 10,000 54. (C) Area of rectangle = length × Breadth Area of given rectangle = 9 × 8 = 72 cm2 As per the given data, area of square is twice the area of rectangle Area of square = 2 × 72 cm2 = 144 cm2 Area of square = a2, where a = side of the square 144 = a2 = a = side of square = a = 12 cm for square, diameter of circumcircle = Diagonal of square = = cm Radius of circumcircle = cm 55. (C) mixture of 90 liter of the spirit and water contains 20% water first mixture = (20/100) × 90 = 18 litres