View Answer Correct Answer: 900
Required difference = \( \left(1440+1860\right) \left(900+500\right) \) = 2300  1400=900
 
2). 3/4th of the number of arts graduates in University A were female. If the number of female arts graduates in University A is less than that in University A by 175, what is the number of male arts graduates in University B?
View Answer Correct Answer: 600
The number of female arts graduates in university A = \( \frac{3}{4} \times 1300 = 975 \) The number of female arts graduates in university B = 975175 = 800 The number of male arts graduates in university B = 1400800 = 600
 
3). What is the respective ratio between the total number of graduates in engineering and commerce together in University A and that in the same courses together in University B ?
View Answer Correct Answer: 9:8
Required ratio = \( \left(1600+2000\right): \left(1400+1800\right) \) = 3600:3200 = 9:8
 
4). Number of science graduates in University B is what percent less than that in University A ?
View Answer Correct Answer: \(\large 16 \frac{2}{3} \)%
Required percentage = \( \frac{18001500}{1800} \times 100 \) = \( \frac{300}{1800} \times 100 \) = \(16 \frac{2}{3} \)%
 
5). Total number of graduates (in all the given six courses together) in University A, was what percent more than that in University B?
View Answer Correct Answer: 0.2
Required percentage = \( \frac{90007500}{7500} \times 100 \) = \( \frac{1500}{7500} \times 100 \) = 20%
 
6). Amit and Sujit together can complete an assignment of data entry in 5 days. Sujit's speed is 80% of Amit's speed and the total key depressions in the assignment are 5,76,000. What is Amit's speed in key depressions per hour if they work for 8 hours a day ?
View Answer Correct Answer: 8000 Ratio between work done by Amit and Sujit = 100:80 = 5:4 Total work done by Amit = \( 576000 \times \frac{5}{9} \) = 320000 Total working hours by Amit = \(8\times5 \)=40 hr Amit's speed = \( \frac{320000}{40}=8000 \)
 
7). Area of a rectangle is 96 m2 When the length of the same rectangle increased by 6 m and the breadth decreased by 3 m, then the area of the rectangle decreases by 30 m2. What is the perimeter of a square whose sides are equal to the length of rectangle ?
View Answer Correct Answer: 64 m Area of rectangle = 96 lb=96 ...... (i) According to question \( lb \left(l+6\right) \times \left(b3\right)=30 \) lblb+3l6b+18 = 30 3l6b = 12 l3b4 = 0 \( l = 2 \left(b+2\right) \) ...... (ii) By solving equations (i) and (ii), we get l = 16 m, b = 6m The side of square3 = l = 16m Perimeter of a square = \( 4 \times 16 = 64m \)
 
8). A vessel contains a mixture of milk and water in the respective ratio of 3 : 1. 32 litre of mixture was taken out and replaced with the same quantity of milk so that the resultant ratio between the quantities of milk and water in the mixture was 4 . : 1 respectively. If 10 litre of mixture is again taken out from the vessel, what is the resultant quantity of water in the mixture ? (in litre)
View Answer Correct Answer: 30 Let the total quantity of mixture = x litre According to question, \( \frac{4}{4+1}=\left[ \frac{x32}{x} \right] \) 4x=5x160 x = 160 Litrs The quantity of milk in mixture = \( 160 \times \frac{3}{4} \) = 120 Litre The quantity of water in misture = \( 160 \times \frac{1}{4} \) = 40 Litre 32 litre of mixture was taken out and replaced with the same quantity of milk. Then, the quantity of milk in mixture = \( 12032 \times \frac{3}{4}+32 \) = 12024+32 = 128 Litre After 10 litres of mixture is again taken out from the vessel, the resultant quantity of water in the mixture = \( 3210 \times \frac{1}{5} = 32  2 = 30 litre \)
 
9). A 476 m long moving train crosses a pole in 14 sec. The length of a platform is equal to the distance covered by the train in 20 sec. A man crosses the same platform in 7 m and 5 sec. What is the Speed of the man in m/s ?
View Answer Correct Answer: 1.6 m/s Speed of the traub = \( \frac{476}{14} = 34 m/s\) Length of the platform = \( 34 \times 20 = 680 m \) Speed of the man = \( \frac{680}{425} = 1.6 m/s \)
 
View Answer Correct Answer: \(\large 23 \frac{7}{11} \)% Total number of balls faced by O = 900 Total number of balls faced by Q = \( 900 \times \frac{3}{5} = 540 \) Total number of runs scored by O in the tournament = \( \frac{110 \times 900}{100} = 990 \) Total number of runs scored by Q in the tournament = \( \frac{140 \times 540}{100} = 756 \) Requireed percentage = \( \frac{990  756}{990} \times 100 \) = \( \frac{234}{99} \times 10 = 23\frac{7}{11} \)

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