Directions (Q. 1-5): From a class total 360 students appeared in an exam. Three fourth of these are boys. 40 % boys failed in the exam, two third of the girls are failed in the exam and remaining passed :
1.What is the ratio of the total number of girls to the number of boys who failed in the exam ?
1) 4 : 9
2) 9 : 5
3) 5 : 6
4) 6 : 3
5) None of these
2.What is the sum of the number of boys who failed and number of girls who passed in the exam ?
5) None of these
3.What is the difference between the number of boys who are passed and number of girls who are failed ?
5) None of these
4.What is the ratio of the total number of boys and girls who are passed to the number of boys and girls who are failed ?
1) 7 : 8
2) 8 : 7
3) 9 : 17
4) 5 : 7
5) None of these
5. Total number of boys passed is approximately what percent of total number of girls in the class ?
2) 120 %
3) 150 %
4) 180 %
Direction (Q. 6 – 10): A bag contains 4 red, 5 brown and 6 white balls. Three balls are drawn randomly :
6.What is the probability that balls drawn contains exactly two red balls ?
5) None of these
7.What is the probability that the balls drawn contains no brown ball ?
5) None of these
8. What is the probability that the balls drawn are not of the same colour ?
4) 421/455
5) None of these
9. If two balls are drawn randomly then what is the probability that both the balls are of same colour ?
4) 421/455
5) None of these
10.What is the probability that the three balls drawn are of different colour ?
5) None of these
Answers:
Ratio: 90/108 = 5/6
Sum = 108 + 30 = 138
Diff = 162 - 60 = 102
Ratio: 192/168 = 8/7
Req% = 162/90 * 100 = 180%
n(S) = 15C3 = (15 * 14 * 13)/6 = 455
2 red balls can be selected in 4C2 = 6 ways
And remaining one ball can be selected in 11C1 = 11 ways
P(E) = (6*11)/455 = 66/455
n(S) = 15C3 = 455
Three balls have to be selected from 4 red and 6 white balls
No. of ways = 10C3 = 120
P(E) = 120/455 = 24/91
n(S) = 15C3 =455
If all three balls are of same colour, then no. of ways = 4C3 + 5C3 + 6C3 = 4 + 10 + 20 = 34
P(E) = 34/455
For being different colours P(E) = 1 - 34/455 = 421/455
n(S) = 15C2 = 105
n(E) = 4C2 + 5C2C + 6C2 = 6 + 10 + 15 = 31
P(E) = 31/105
n(S) = 15C3 = 455
n(E) = 4C1 + 5C1 + 6C1 = 4 * 5 * 6 = 120
P(E) = 120/455 = 24/91
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