Quantitative Aptitude Quiz - 27 | Bank Exams

Directions (Q.1 - 5): In the following questions, two equations numbered I and II are given. You have to solve both the equations and 

Give answer -

(a) if X > Y 
(b) if X < Y 
(c) if X >= y
(d) if x =< y 
(e) if x = y or no relation can be established between X and Y

1. I. 2x2 - 11x + 15 = 0 
   II. 21y2 - 23y + 6 = 0

2. I. 5x2 - 16x + 11= 0 
   II. 5y2 - 3y - 2 = 0

3. I. x2 + 11x + 28 = 0 
   II. 2y2 + 13y + 20 = 0

4. I. 6x2 + 29x + 35 = 0 
   II. 3y2 + 19y + 30 = 0

5. I. 2x + 5y = 6 
   II. 5x + 11y = 9

Directions (Q.6 - 10): Study the following table carefully to answer the questions that follow.

6. What is the total number of students from all the states who passed in 2006 examination?
a) 3692
b) 4050
c) 4588
d) 3898
a) None of these

7. What is the difference between the number of students passed from State P in 2009 and number of students passed from State R in the same year?
a) 254
b) 268
c) 277
d) 289
e) None of these

8. What is the respective ratio of students failed from state Q in 2007 to the students passed from same state and year?
a) 19 : 81
b) 17 : 77
c) 81 : 17
d) 77 : 19
e) None of these

9. In the given years, what is the average number of students appeared from State S?
a) 4850
b) 5090
c) 5170
d) 4900
e) None of these

10. What is the difference between the students failed from State R in 2005 and number of students passed from State S in the same year?
a) 2724
b) 2624
c) 2735
d) 2727
e) None of these

Answers:
1. A; I. 2x2 - 6x - 5x + 15 = 0
or,2x(x - 3) - 5(x - 3) = 0
or, (2x - 5) (x - 3) = 0
x = 3,5/2

II. 21y2 - 14y - 9y + 6 = 0
or,7y(3y - 2) - 3 (3y - 2) = 0
or,(7y - 3)(3y - 2) = 0
y = 3/7, 2/3
So, x > y

2. C; I. 5x2 - 5x - 11x + 11 = 0
or,5x(x - 1) - 11(x - 1) = 0
or,(x - 1) (5x - 11) = 0
Then, x = 1, 11/5. 

II. 5y2 - 5y + 2y - 2 = 0
or, 5y (y - 1) + 2(y - 1) = 0
or, (5y + 2)(y - 1) = 0
Then, y = 1, - 2/5. So, x >= y

3. D; I. x2 + 4x + 7x + 28 = 0
or, x(x + 4) +7(x + 7) = 0
or, (x + 4) (x + 7) = 0
Then, x = - 4, - 7

II. 2y2 + 8y + 5y + 20 = 0
or, 2y(y + 4) + 5(y + 4) = 0
or, (y + 4) (2y + 5) = 0
Then, y = -4, -5/2. So, x =< y

4. A; I. 6x2 + 15x + 14x + 35 = 0
or, 3x(2x + 5) + 7(2x + 5) = 0
or,(3x + 7) (2x + 5) = 0
Then, x = -7/3, -5/2

II. 3y2 + 9y + 10y + 30 = 0
or, 3y(y + 3) +10(y + 3) = 0
or,(3y + 10) (y + 3) = 0
Then, y = -3, -10/3. So, x > y

5. B; After Solving equ:  x = -7 and y = 4. So, y > x.

6. C; Total students passed in 2006 = 5600 * 16/100 + 6250 * 18/100 + 5000 * 30/100 + 4850 * 22/100
= 896 + 1125 + 1500 + 1067 = 4588.

7. C; Number of passed students from State P in year 2009 = 7100 * 13/100 = 923
Number of passed students from state R in year 2009 = 7500 * 16/100 = 1200
Difference = 1200 - 923 = 277.

8. E; Students failed from State Q in 2007 = 5900 * (100-19)/100 = 4779
Students passed from State Q in 2007 = 5900 * 19/100 = 1121
Ratio = (4779/59) / (1121/59) = 81/19 = 81 : 19.

9. C; (3200+4850+4800+6000+7000)/5 = 25850/5 = 5170.


10. A; 2724.