Quadratic Equation Questions for SBI Clerk Exam

The Quadratic Equation section is a crucial part of the Quantitative Aptitude section in the SBI Clerk exam, with 5 questions typically included out of the total 35 questions. Given its significance in the exam, mastering quadratic equations is essential for achieving a high score in the quant section. To aid in your preparation, Adda247 offers a variety of Quadratic Equation Questions for SBI Clerk Exam. Regular practice of these questions will help you solve them efficiently and accurately, boosting your overall performance in the exam.

Quadratic Equation Questions for SBI Clerk Exam

Quadratic equation questions in the SBI Clerk exam 2025 are designed to assess the candidates’ ability to solve quadratic equations efficiently. These questions usually involve finding the roots of quadratic equations using various methods such as factorization, completing the square or applying the quadratic formula. To successfully solve these questions, candidates need to have a solid understanding of algebraic concepts and the ability to quickly identify the best method to solve the equation. Regular practice with these problems will increase problem-solving speed and accuracy.

Solutions

S1. Ans.(d)
Sol.
I. 2x² – 17x+ 36 = 0
2x² – 8x – 9x + 36 = 0
2x (x – 4) – 9 (x – 4) = 0
(2x – 9) (x- 4) = 0
x=9/2, 4
II. 2y² – 19y + 45 = 0
2y² – 10y – 9y + 45 = 0
2y (y- 5) – 9 (y- 5) = 0
(2y- 9) (y- 5) = 0
y=9/2,5
∴ y ≥ x

S2. Ans.(e)
Sol.
I. x² – 25x + 154 = 0
x² – 14x – 11x + 154 = 0
x (x – 14) – 11 (x- 14) = 0
(x – 11) (x- 14) = 0
x = 11, 14
II. y² – 28y + 195 = 0
y² – 13y – 15y + 195 = 0
y (y- 13) – 15 (y -13) =0
(y- 13) (y – 15) = 0
y = 13, 15
∴ no relation

S3. Ans.(a)
Sol.
I. 10/x-24/x² =1
Multiplying by x² on both side
10x – 24 = x²
x² – 10x + 24 = 0
x² – 6x -4x+ 24 = 0
x(x – 6) – 4 (x- 6) = 0
(x – 4) (x- 6) = 0
x= 4, 6
II. 5/y-6/y² =1
Multiplying by y² on both side
5y – 6 = y²
y² – 5y + 6 = 0
y² – 3y – 2y + 6 = 0
y (y- 3) – 2 (y- 3) = 0
(y – 2) (y- 3) = 0
y = 2, 3
∴ x > y

S4. Ans.(d)
Sol.
I. 3x² – 10x – 8 = 0
3x² – 12x + 2x – 8 = 0
3x (x – 4) + 2 (x- 4) = 0
(3x+ 2) (x- 4) = 0
x= -2/3,4
II. 2y²-23y+60=0
2y² – 8y- 15y + 60 = 0
2y (y- 4) -15(y-4) = 0
(y- 4) (2y- 15) = 0
y=4,15/2
∴ y ≥ x

S5. Ans.(a)
Sol.
I. 12x – 16y +16 = 0
3x – 4y + 4 = 0 …(i)
II. 17y- 13x = 12 …(ii)
By multiplying equation (i) by 13 & equation (ii) by 3
39x – 52y = -52
-39x + 51y = 36
y = 16 & x = 20
∴ x > y

S6. Ans (c)
Sol. I. 4x²+4x+1=0
(2x+1)²=0
x=-1/2
II. 9y²+6y+1=0
(3y+1)²=0
y=-1/3
∴ x<y

S7. Ans (e)
Sol. I. (x-2)²=4
x-2=±2
x=0,4
II. y²-2y+1=0
(y-1)²=0
y=1
∴ no relation can be obtained.

S8. Ans (e)
Sol. I. 3x+2y=5
II. 4x+6y=10
Applying 2 × I and equate with II
x = y = 1
∴ x=y

S9. Ans (d)
Sol. I. x²-4x-21=0
x²-7x+3x-21=0
x(x-7)+3(x-7)=0
(x+3)(x-7)=0
x=-3,7
II. y²-16y+63=0
y²-7y-9y+63=0
y(y-7)-9(y-7)=0
(y-7)(y-9)=0
y=7,9
So, y≥x

S10. Ans (a)
Sol. I. 2x=3y-1
II. x+y=7
Applying 2×II-I
2x+2y-2x=14-3y+1
5y=15
y=3
And x=4
∴ x>y

S11. Ans(d)
Sol. I. 4x²+6x-2x-3=0
2x(2x+3)-1(2x+3)=0
(2x-1)(2x+3)=0
So, x=1/2,-3/2
II. 4y²-6y-2y+3=0
2y(2y-3)-1(2y-3)=0
(2y-1)(2y-3)=0
y=1/2,3/2
So, y≥x

S12. Ans(c)
Sol. I. 11x-13y+48=0
II. 13y+11x=290
Adding I and II
22x+48=290
x=242/22
x=11
Put x = 11 in I
121-13y+48=0
13y=169
y=13
So, y > x

S13. Ans(a)
Sol. I. 2x+3xy=207
II. 15x=945/y
From II
xy=63
So, 3xy=189
Put value of 3xy in I
2x+189=207
x=18/2
x=9
y=7
So, x > y

S14. Ans(e)
Sol. I. x²-14x+33=0
x²-11x-3x+33=0
x(x-11)-3(x-11)=0
(x-11)(x-3)=0
x=3,11
II. y²-15y+44=0
y²-11y-4y+44=0
y(y-11)-4(y-11)=0
(y-4)(y-11)=0
y=4,11
So, no relation can be obtained between x and y.

S15. Ans(e)
Sol. I. 8x²+28x-6x-21=0
4x(2x+7)-3(2x+7)=0
(4x-3)(2x+7)=0
x=-7/2,3/4
II. 18y²+42y-15y-35=0
6y(3y+7)-5(3y+7)=0
(6y-5)(3y+7)=0
y=-7/3,5/6
So, no relation can be obtained between x and y.

S16. Ans.(e)
Sol. I. x²+9x-22=0
⇒ x² + 11x – 2x – 22 = 0
⇒ (x + 11) (x – 2) = 0
⇒ x = – 11, 2
II. 2y² – 7y + 6 = 0
⇒ 2y² – 4y – 3y + 6 = 0
⇒ 2y(y–2)–3(y–2) =0
⇒ (y–2) (2y–3) = 0
⇒ y = 2, 3/2
No relation

S17. Ans.(e)
Sol. I. 2y² – 13y – 34 = 0
⇒ 2y² – 17y + 4y – 34 = 0
⇒ y(2y–17) + 2(2y–17) = 0
⇒ (2y–17) (y+2) = 0
⇒ y = 17/2,–2
II. 3x² – 11x – 20 = 0
⇒ 3x² – 15x + 4x – 20 = 0
⇒ 3x (x–5) + 4(x–5) =0
⇒ (x – 5) (3x + 4) = 0
⇒ x = 5, (-4)/3
No relation

S18. Ans.(c)
Sol. I. x⁴ = 256
⇒ x = ± 4
II. y² – 16y + 64 = 0
⇒ (y – 8) ² = 0
⇒ y = 8
y > x

S19. Ans.(b)
Sol. I. 9x² – 54x + 77 = 0
9x² – 21x – 33x + 77 = 0
3x (3x – 7) – 11(3x – 7) = 0
(3x – 7)(3x – 11) = 0
x = 7/3,11/3
II. 12y² – 55y + 63 = 0
12y² – 28y – 27y + 63 = 0
4y (3y – 7) –9 (3y – 7) = 0
(4y – 9) (3y – 7) = 0
y = 9/4,7/3
So, x ≥ y

S20. Ans.(d)
Sol. I. (x – 1)² = 121
x – 1 = ± 11
x = 12, – 10
II. y² –24y + 144 = 0
(y – 12)² = 0
y = 12
So, y ≥ x