Quant Quiz

Directions (1-5): Study the following table carefully and answer the questions given below:
Percentage of students qualified w.r.t. those appeared from six states over the years in a competitive exam.

1. In the year 2010, a total of 22400 students appeared from state ‘D’ and ‘F’ together. How many students qualified from state ‘F’ in that year if the no. of appeared students from ‘F’ was 400 more than that from state ‘D’?
(1) 3520
(2) 2750
(3) 3280
(4) Can’t be determined
(5) None of these

2. If the no. of students qualified from state ‘B ‘ in 2005 was 50% more than that from state ‘A’ in 2006, what was the ratio of students appeared in state ‘B’ in 2005 to that in state ‘A’ in 2006?
(1) 1 : 1
(2) 3 : 4
(3) 4 : 3
(4) 2 : 3
(5) None of these

3. If the number of students appeared from each of the given states in the year 2007 was 2000, what was the total number of students qualified from all the states?
(1) 3920
(2) 3720
(3) 1960
(4) 1860
(5) None of these

4. If the numbers of students qualified from state ‘F’ in 2008 and 2009 were in the ratio of 2 : 1 respectively, what was the respective ratio of students appeared in these years from state ‘F’?
(1) 3 : 8
(2) 8 : 3
(3) 3 : 2
(4) Can’t be determined
(5) None of these

5. In the year 2008, as many as 12000 students appeared from state ‘C’. How many students were declared qualified from this state in the year 2009?
(1) 4080
(2) 3360
(3) 408
(4) Can’t be determined
(5) None of these

Directions (6-10): From the two given equations I and II give answer:
(1) If p is greater than q
(2) If p is smaller than q
(3) If p is equal to q
(4) If p is either equal to or greater than q
(5) If p is either equal to or smaller than q

6. I. 6p^2 + 5p + 1 = 0
II. 20q^2 + 9q = - 1

7. I. 3p^2 + 2p - 1 = 0
II. 2q^2 + 7q + 6 = 0

8. I. 3p^2 + 15p =  - 18
II. q^2 + 7q  + 12 = 0

9. I. p = √4/√9
II. 9q^2 - 12q + 4 = 0

10. I. p^2 + 13p + 42 = 0
II. q^2 = 36


ANSWERS
1. 5;
Total appeared students from 'D' and 'F' in 2010 = 22400 = x + (x + 400)
x = 11000
Appeared students from state 'F' = 11400
Qualified students = 25% of 11400 = 2850

2. 1;  
(Qualified from "B" in 2005)/ (Qualified from "A" in 2006)= 150/100 = 3 : 2
Reqd. ratio = 3/2 * 28/42 = 1 : 1

3. 2;
Total qualified students in 2007
= 2000{(32 + 35 + 30 + 22 + 32 + 35)/100} = 3720

4. 3; Reqd. ratio = 2/1 * 30/40 = 3 : 2

5. 4; Here, total appeared students from state 'C' in 2009 is not given.

Solutions (6 - 10):

6. (2)
I. 6p^2 + 5p + 1 = 0
or 6p^2 + 3p + 2p + 1 = 0
3p (2p + 1) + 1 (2p + 1) = 0
(3P + 1) (2p + 1) = 0
Hence p = -1/3, -1/2 

II. 20q^2 + 9q = - 1
or 20q^2 + 5q + 4q + 1= 0
or 5q (4q +1) +1 (4q + 1) =0
or (4q +1) (5q +1) = 0
Hence q = -1/5 , -1/4

Thus p < q

7. (1)
I. 3p^2 + 2p - 1 = 0
or 3p^2 + 2p - 1 = 0
(3p -1) (p+1) = 0
Therefore,
p = 1/3 , - 1

II. 2q^2 + 7q + 6 = 0
2q^2 + 4q +3q +  6 = 0
(2q + 3) (q+2) = 0
Therefore, q = -3/2 , - 2
Thus p >q

8. (4) 
I. 3p^2 + 15p =  - 18
3p^2 + 9p + 6p + 18 =  0
3p (p+3) + 6 (p+3) =0
p =  -6/3, - 3 = - 2, - 3

II. q^2 + 7q  + 12 = 0
q^2 + 4q + 3q  + 12 = 0
q = -3, - 4

Therefore p > q

9. (3)
I. p = √4/√9
p = 2/3

II. 9q^2 - 12q + 4 = 0

9q^2 - 6q – 6q +  4 = 0
(3q – 2) (3q -2) = 0
or q = 2/3
Therefore p = q

10. (5)
I. p^2 + 13p + 42 = 0
p^2 + 13p + 42 = 0

p^2 + 7p + 6p + 42 = 0
(p+6) (p+7) = 0
p = -6, - 7

II. q^2 = 36
q =  √36
q = +6, - 6
Therefore p <  q