Aptitude Questions (Quadratic Equations) for IBPS Clerk Mains

Aptitude Questions (Quadratic Equations) for IBPS Clerk Mains:
Dear Readers, Important Practice Aptitude Questions for Upcoming IBPS Clerk V Mains Exam was given here with Explanations. Aspirants those who are preparing for the examination can use this.

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Directions (Q.1- 5): In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and given answer:
a)   If x< y
b)   If x ≤ y
c)   If x = y, or no relation can be established between x and y.
d)   If x> y
e)   If x ≥ y
1). I.9x2 = 1                    II.4y2 + 11y – 3 =0

2). I.3x 2 + 5x – 2 =0       II.2y2 – 7y + 5 =0

3).I.6x 2 + 13x + 5 =0    II.3y2+ 11y + 10 =0

4).I.7x – 4Y =29            II.5x + 3y – 50 =0

5).I.x2 – 5 =0               II.4y2 – 24y + 35 =0

Directions (Q.6- 10): In each of these questions, two equations are provided. On the basis of these you have to find out the relation between p and q. Give answer

a)   If p = q, or no relation can be established between p and q.
b)   If p> q
c)   If q> p
d)   If p ≥ q
e)   If q ≥ p

6).I. 3p + 2q = 19              II.p + q = 8

7).I.10q2 + 19q + 9 =0     II. 13p2 – 2p – 11 = 0

8).I. 2p2 + 3p – 5 =0         II.2q2 + 11q + 15 =0

9).I.2401p2 = p-2             II.7(p + q) = 2

10). I.p2 -3p + 2 =0        II. q2 – 7q +10 = 0

Solutions:
1). I. 9x2 = 1
:. x2 = 1/9
:. x = ± 1/3
II.4y2 +11y – 3 =0
Or, 4y2 + 12y – y – 3= 0
Or, 4y(y + 3) – 1 (y + 3) = 0
   :.y = 1/ 4, – 3
Hence, there is no relation between x and y.
Answer: c)

2). I. 3x2 + 5x – 2 =0
Or, 3x2 + 6x – x -2= 0
Or, 3x(x + 2) – 1(x + 2) =0
Or, (3x -1) (x +2) = 0
:. X = -2, 1/3
II. 2y2 – 7y + 5 =0
2y2 – 2y – 5y + 5 = 0
Or, 2y(y – 1) – 5 (y – 1) =0
:. y = 1, 5/2
Hence, x< y
Answer: a)

3). I. 6x2 + 13x +5 =0
Or, 6x2 + 3x +10x + 5=0
Or, 3x (2x +1)+ 5(2x + 1)= 0
Or, (3x + 5) (2x +1 ) = 0
:. X = – 5/3, -1/2
II.3y2+11y+10 = 0
Or, 3y+ 6y + 5y+10 = 0
Or, 3y(y + 2) + 5(y + 2) = 0
Or, (3y + 5) (y + 2) = 0
:. y= – 5/3, -2
Hence, x ≥ y
Answer: e)

4). I. 7x – 4y = 29
II. 5x + 3y = 50
 (I)×3 + (II)×4
21x – 12y = 87
20x + 12y = 200
  41x         = 287
:. x = 7
Putting the value of x in (I), we get
y = 5
Hence, x> y
Answer: d)

5). I. x2 = 5
:. x = +=√5 ≈ ± 2 .236
II.4y2 – 24y + 35 =0
Or, 4y2 – 14y – 10y + 35 = 0
Or, 2y (2y – 7) – 5(2y – 7)= 0
Or,(2y – 5) (2y – 7) = 0
:. y = 5/2, 7/2 = 2.5, 3.5
Hence, x< y
Answer: a)

6). II. p + q = 8
Or, p = 8 – q ……(i)
I. 3p + 2q = 19
Or, 3(8 – q) + 2q =19
Or, q = 5
:.p =8 -5 = 3
Hence, q> p
Answer: c)
7). I.10q2 + 19 q + 9= 0
Or, 10q2 + 10q + 9q + 9= 0
Or, 10q (q + 1) + 9(q + 1) =0
Or, (q + 1)(10q + 9) =0
Or, q = -9/10, – 1
II. 13p2 – 2p – 11 =0
Or, 13p2 – 13p + 11p – 11 = 0
Or, 13p (p – 1) + 11(p – 1) = 0
Or,(13p + 11) (p – 1) =0
Or, p = – 11/13, 1
Hence, p> q
Answer: b)

8). I. 2p2 + 3p – 5 =0
Or, p = 1, – 5/2
II. 2q2 + 11q + 15 = 0
Or, 2q2 + 6q + 5q+ 15 = 0
Or, 2q(q + 3)+ 5(q + 3) = 0
Or,(q + 3) (2q + 5) = 0
Or, q = -3, – 5/2
Hence, p ≥ q.
Answer: d)

9). I. 2401 p2= p-2
  Or, p4 = 1/ 2401
Or, p4 = (1 / 7)4 or, p= 1/ 7
II. 7p + 7q = 2
Or, 7q + 1 =2
Or, 7q =1
Or, q = 1/ 7
Hence, p = q
Answer: a)

10).  I. p2 – 3p + 2 =0
Or, (p – 2) (p -1) = 0
Or, p =1,2
II. q2 – 7q + 10 = 0
Or, (q – 2) (q – 5) =0
Or, q = 2, 5
Hence, q ≥ p.
Answer: e)

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