QUANTITATIVE APTITUDE - QUADRATIC EQUATIONS  FOR UPCOMING EXAMINATION 

Directions (Q. 1-5): For the two given equations I and II.

Give answer:

a)   If p is greater than q.

b)   If p is smaller than q.

c)   If p is equal to q.

d)   If p is either equal to or greater than q.

e)   If p is either equal to or smaller than q.

1) I. 6p²+5p+1=0

II. 20q²+9q=-1

2)  I. 3p²+2p-1=0

II. 2q²+7q+6=0

3)  I. 3p²+15p=-18

II. q²+7q+12=0

4) I. p=√4/√9

II. 9q²-12q+4=0

5) I. p²+13p+42=0

II. q²=36

Direction (Q. 6-10) : 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 the relationship can't be established.

6). I.  x = √1369

II. y = ₃ √29791

7). I.  8x - 3y = 31

II.  5x + 4y = 84

8). I.  20x² - 79x + 77 = 0

II. 4y² + 9y - 28 = 0

9). I.  6x² + 29x + 28 = 0

II. 6y² + 11y + 4 = 0

10). I. x² + 3x - 54 = 0

II.  y²  + 4y - 77 = 0

ANSWER 

Solution:

1).B)

I. 6p²+5p+1=0

(3p+1)(2p+1)=0 

p=-1/3,-1/2 

II. 20q²+9q+1=0

(4q+1)(5q+1)=0

q=-1/4,-1/5

∴p<q

2). A)

I. 3p²+2p-1=0

(3p-1)(p+1)=0

p=1/3,-1 

II. 2q²+7q+6=0

(2q+3)(q+2)=0

q=-3/2,-2

3). D)

I. 3p²+15p+18=0

(3p+6)(p+3)=0

p=-2,-3

II. q²+7q+12=0

(q+4)(q+3)=0

q=-3,-4

∴p≥q

4). C)

I. p=√4/√9=2/3

II. 9q²-12q+4=0

(3q-2)²=0

q=2/3

∴p=q

5). E)

p²+13p+42=0

(p+7)(p+6)=0

p=-6,-7

II. q²=36

q=±6

∴p≤q

6). A)

 x =  √1369  = 37 .................... (I)

y = ₃√29791 = 31 ................... (II)

x > y

7). C)

  equn. (I)  ×4 +  equn (II) ×3

32 x – 12 y = 124

15 x + 12 y = 252

_____________

 47 x           =    376

x = 8 and from this y = 11

x < y

8). B)

20 x² – 35 x – 44 x + 77 = 0

5 x (4 x – 7) – 11 (4 x – 7) = 0

(4x – 7) (5 x – 11) = 0

x = (7/4), (11/5)

4 y²  + 16 y – 7 y – 28 = 0

4 y (y + 4) – 7 (y + 4) = 0

(4 y – 7) (y + 4) = 0

y = – 4, (7/4)
x ≥ y

9). D)

6 x²  + 8 x + 21 x + 28 = 0

2 x (3 x + 4) + 7 (3x + 4) = 0

(3x + 4) (2 x + 7) = 0

x = (-4/3), (-7/2)

6 y² + 3 y + 8 y + 4 = 0

3 y (2 y + 1) + 4 (2 y + 1) = 0

(3 y + 4) (2 y + 1) = 0

y = (-4/3), (-1/2)

x  ≤ y

10). E)

 x²  + 9 x – 6 x - 54 = 0

x (x + 9) –6 ( x + 9) = 0

x = 6, –9

y²  + 11 y – 7 y – 77 = 0

y (y + 11) – 7 (y + 11) = 0

(y – 7) (y + 11) = 0

y = 7, –11

i.e. No relation between x & y