Quadratic Equation Questions for SBI PO Pre – Download PDF

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We are sharing Data Interpretation Questions which are very important for  IBPS Clerk, IBPS PO, SBI Clerk, SBI PO, LIC AAO, Assitant, Bank Exams, SSC and Railway. Attempt these questions. 5 questions are asked in every bank exams. Previous year Question, Data Interpretation Questions

Quadratic Equation Questions

Quadratic Equation Questions for SBI PO Pre – Download PDF



Direction (1-10): In the following questions two equations numbered I and II are given. You have to solve both equations and give answer

1. I. x2 – 7x + 10 = 0   

II. y2 + lly + 10 = 0

(a) x > y

(b) x ≥ y

(c) x < y

(d) x ≤ y

(e) x = y or the relationship cannot be established

2. I x2 + 28x + 192 = 0   

II. y2 + 16y + 48 = 0

(a) x > y

(b) x ≥ y

(c) x < y

(d) x ≤ y

(e) x = y or the relationship cannot be established

3. I. 2x – 3y = – 3.5   

II. 3x – 2y = – 6.5

(a) x > y

(b) x ≥ y

(c) x < y

(d) x ≤ y

(e) x = y or the relationship cannot be established

4. I. x2 + 8x + 15 = 0

II. y2 + 11y + 30 = 0

(a) x > y

(b) x ≥ y

(c) x < y

(d) x ≤ y

(e) x = y or the relationship cannot be established

5. I. x = √3136

II. y2 = 3136

(a) x > y

(b) x ≥ y

(c) x < y

(d) x ≤ y

(e) x = y or the relationship cannot be established

6. I. x2 + 5x + 6 = 0

II. y2 + 7y + 12 = 0

(a) x > y

(b) x ≥ y

(c) x < y

(d) x ≤ y

(e) x = y or the relationship cannot be established

7. I. x2 + 20 = 9x

II. y2 + 42 = 13y

(a) x > y

(b) x ≥ y

(c) x < y

(d) x ≤ y

(e) x = y or the relationship cannot be established

8. I. 2x + 3y = 14

II. 4x + 2y = 16

(a) x > y

(b) x ≥ y

(c) x < y

(d) x ≤ y

(e) x = y or the relationship cannot be established

9. I. x = √625

II. x = √676

(a) x > y

(b) x ≥ y

(c) x < y

(d) x ≤ y

(e) x = y or the relationship cannot be established

10. I. x2 + 4x + 4 = 0

II. y2 – 8y + 16 = 0

(a) x > y

(b) x ≥ y

(c) x < y

(d) x ≤ y

(e) x = y or the relationship cannot be established

Answer 1. (a)

I. x2 – 7x + 10 = 0

x2 – 5x – 2x + 10 = 0

x(x – 5) – 2(x – 5) = 0

(x – 2) (x – 5) = 0

x = 2 or 5

II. y2 + 11y + 10 = 0

y2 + 10y + y + 10 = 0

y(y + 10) + 1(y + 10) = 0

(y + 1) (y + 10) = 0

y = – 1 or – 10

Here, x > y

Answer 2. (d)

I. x2 + 28x + 192 = 0

x2 + 16x + 12x + 192 = 0

x(x + 16) + 12(x + 16) = 0

(x + 12) (x + 16) = 0

x = – 12 or –16

II. y2 + 16y + 48 = 0

y2 + 12y + 4y + 48 = 0

y (y + 12) + 4 (y + 12) = 0

(y + 12) (y + 4) = 0

y = – 12 or – 4

Here, x ≤ y

Answer 3. (c)

2x – 3y = – 3.5 … (i)

3x – 2y = – 6.5 … (ii)

By equation (i) × 2 – equation (ii) × 3, we have

-5x = 12.5

x = -12.5/5 = –2.5

From equation (i)

2 × (–2.5) – 3y = – 3.5

3y = – 5 + 3.5

3y = – 1.5

y = -0.5

Here, x < y

Answer 4. (b)

I. x2 + 8x + 15 = 0

x2 + 5x + 3x + 15 = 0

x(x + 5) + 3(x + 5) = 0

(x + 5) (x + 3) = 0

x = –5 or –3

II. y2 + 11y + 30 = 0

y2 + 6y + 5y + 30 = 0

y (y + 6) + 5 (y + 6) = 0

(y + 5) (y + 6) = 0

y = – 5 or – 6

Here, x ≥ y

Answer 5. (b)

I. x = √3136

x = 56

II. y2 = 3136

y = √3136

x = ± 56

Here, x ≥ y

Answer 6. (b)

I. x2 + 5x + 6 = 0

x2 + 2x + 3x + 6 = 0

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

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

x = -2, -3

II. y2 + 7y + 12 = 0

y2 + 4y + 3y + 12 = 0

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

(y + 4)(x + 3) = 0

x = -3, -4

Here, x ≥ y

Answer 7. (c)

I. x2 – 9x + 20 = 0

x2 – 5x – 4x + 20 = 0

x(x – 5) – 4(x – 5) = 0

(x – 4)(x – 5) = 0

x = 4, 5

II. y2 – 13y + 42 = 0

y2 – 6y – 7y + 42 = 0

y(y – 6) -7(y – 6) = 0

(y – 6)(y – 7) = 0

y = 6, 7

Here, x < y

Answer 8. (c)

2 x + 3 y = 14 …(i)

4 x + 2 y =16 …(ii)

By equation (I) × 2 – equation II.

4 x + 6y – 4 x – 2 y = 28 – 16

4y =12

y = 3

From equation I,

2 x + 3× 3 = 14

2 x = 14 – 9 = 5

x = 5/2

Here, x < y

Answer 9. (c)

I. x = √625 = 25

II. y = √676 = 26

Here, x < y

Answer 10. (c)

I. x2 + 4x + 4 = 0

(x + 2)2 = 0

x = – 2

II. y2 – 8y + 16 = 0

(y – 4)2 = 0

y = 4

Here, x < y



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