Quant Quiz

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 ?

1) 148 

2) 158 

3) 128 

4) 138 

5) None of these

3.What is the difference between the number of boys who are passed and number of girls who are failed ?

1) 102 

2) 202 

3) 52 

4) 302 

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 ?

1) 60 % 

2) 120 % 

3) 150 % 

4) 180 % 

5) 270 %

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 ?

1) 54/145

2) 66/455

3) 46/105

4) 31/455

5) None of these

7.What is the probability that the balls drawn contains no brown ball ?

I) 11/65

2) 31/91

3) 24/91

4) 31/455

5) None of these

8. What is the probability that the balls drawn are not of the same colour ?

1) 34/455

2) 31/105

3) 74/105

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 ?

1) 31/105

2) 74/105

3) 34/455

4) 421/455

5) None of these

10.What is the probability that the three balls drawn are of different colour ?

1) 21/91

2) 31/105

3) 24/105

4) 24/91

5) None of these

Answers:

1. 3

Ratio: 90/108 = 5/6

2. 4

Sum = 108 + 30 = 138

3. 1

Diff = 162 - 60 = 102

4. 2

Ratio: 192/168 = 8/7

5. 4

Req% = 162/90 * 100 = 180%

6. 2

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

7. 3

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

8. 4

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

9. 1

n(S) = 15C2 = 105

n(E) = 4C2 + 5C2C + 6C2 = 6 + 10 + 15 = 31

P(E) = 31/105

10. 4

n(S) = 15C3 = 455

n(E) = 4C1 + 5C1 + 6C1 = 4 * 5 * 6 = 120

P(E) = 120/455 = 24/91