## Equations and Inequations Questions for IBPS SO

### Equations and Inequations Previous Year Questions for IBPS SO 2018 Exam

**In the following questions, two equations I and II are given. Solve the equations and give yur answer accordingly: (IBPS SO IT Officer – Pre 2018)**

**1.**

I, 5X^{2} + 28X + 15 = 0

II. 6Y^{2} + 35Y + 25 = 0

(a) If X > Y

(b) If Y > X

(c) If X ≥ Y

(d) If Y ≥ X

(e) If X = Y or a relation ship between X and Y cannot be established

**2.**

I. 12X^{2} + 82X + 140 = 0

II. 16Y^{2}+ 48Y + 32 = 0

(a) X > Y

(b) X ≥ Y

(c) X < Y

(d) X ≤ Y

(e) X = Y or the relationship cannot be established

**3.**

I. 4X^{2} – 48X + 143 = 0

II. 4Y^{2} – 52Y + 165 = 0

(a) If X > Y

(b) If Y > X

(c) If X ≥ Y

(d) If Y ≥ X

(e) If X = Y or a relation ship between X and Y cannot be established

**4.**

I. X^{2} + 4X = 21

II. Y^{2} – 6Y + 8 = 0

(a) X > Y

(b) Y > X

(c) X ≥ Y

(d) Y ≥ X

(e) X = Y or a relation ship between X and Y cannot be established

**5.**

I.

II.

(a) X > Y

(b) X ≥ Y

(c) X < Y

(d) X ≤ Y

(e) X = Y or the relationship cannot be established

**Equations and Inequations Previous Year Questions for IBPS SO 2018 Answers**

**1.** (e)

(X + 5) (5X + 3) = 0

⇒ X = -5, -3/5 (Y + 5) (6Y + 5) = 0

⇒ Y = -5, -5/6

No relation

**2.** (c)

I. 12X^{2} + 82X + 140 = 0

X = -7/2, -10/3

II. 16Y^{2} + 48Y + 32 = 0

Y = -1, -2

Y > X

**3.** (e)

(2X – 11) (2X – 13) = 0 ⇒ X = +11/2, +13/2

(2Y – 11) (2Y – 15) = 0 ⇒ Y = +11/2, +15/2

say X = 11/2 and Y = 15/2 ; Y > X

but if say X = 13/2 and Y = 11/2; then X > Y

Hence, No relation

**4.** (e)

(X – 3) (X + 7) = 0 ⇒ X = +3, -7

(Y – 4) (Y – 2) = 0 ⇒ Y = +4, +2

No relation

**5.** (e)

From I.

⇒ X = 14

From II.

⇒ Y^{5/2} = 14^{5/2}

⇒ Y = 14

So, X = Y

### Equations and Inequations Previous Year Questions for IBPS SO 2016 Exam

**In the given questions, two equations numbered I and /I are given. Solve both the equations and mark the appropriate answer. (IBPS SO 2016)**

(a) x > y

(b) x ≥ y

(c) x < Y

(d) Relationship between x and y cannot be determined

(e) x ≤ y

**1.**

I. 6x^{2} + 25x+24 = 0

II. 12y^{2} + 13y + 3 = 0

**2.**

I. 12x^{2}-x -1= 0

II. 20y^{2} – 41y + 20 = 0

**3.**

I. 10x^{2} + 33x +27 = 0

II. 5y^{2} + 19y +18 = 0

**4.**

I. 15x^{2} – 29x – 14 = 0

II. 6y^{2}– 5y – 25 = 0

**5.**

I. 3x^{2} – 22x + 7 = 0

II. y^{2} – 20y + 91 = 0

**Equations and Inequations Previous Year Questions for IBPS PO 2016 Answers**

**1.** (c)

I. 6x^{2} + 25x+24 = 0

II. 12y^{2} + 13y + 3 = 0

**2.** (c)

I. 12x^{2}-x -1= 0

II. 20y^{2} – 41y + 20 = 0

**3.** (b)

I. 10x^{2} + 33x +27 = 0

II. 5y^{2} + 19y +18 = 0

**4.** (d)

I. 15x^{2} – 29x – 14 = 0

II. 6y^{2}– 5y – 25 = 0

**5.** (b)

I. 3x^{2} – 22x + 7 = 0

3x^{2} – 21x – x + 7 = 0

x (3x – 1) – 7(3x – 1) = 0

(3x – 1) (x – 7) = 0

II. y^{2} – 20y + 91 = 0

y^{2} – 20y + 91 = 0

y^{2} – 13y – 7y + 91 = 0

y (y – 7) – 13 (y – 7) = 0

(y – 13) (y – 7) = 0

y = 13, 7 ⇒ y ≥ x

### Equations and Inequations Previous Year Questions for IBPS SO 2015 Exam

**In each of these questions two equations are given. You have to solve these equations and Give answer. (IBPS SO 2015)**

(a) if x < y (b) if x > y

(c) if x = Y

(d) if x ≥ y

(e) if x ≤ y

**1.**

I. x^{2} – 6x – 7

II. 2y^{2} + 13y +15 = 0

**2.**

I. 3 x^{2} – 7 x + 2 0

II. 2y^{2} – 11y +15 = 0

**3.**

I. 10x^{2} – 7 x + 1 = 0

II. 35y^{2} – 12y +1 = 0

**4.**

I. 4 x^{2} = 25

II. 2y^{2} – 13y + 21 = 0

**5.**

I. 3 x^{2} + 7 x = 6

II. 6(2y^{2} + 1) = 17y

**Equations and Inequations Previous Year Questions for IBPS SO 2015 Answers**

**1.** (b)

I. x^{2} – 6x = 7

or, x^{2} – 6x – 7 = 0

or, (x – 7) (x + 1) = 0

or, x = 7, – 1

II. 2y^{2} + 13y +15 = 0

or, 2y^{2} + 3y + 10y +15 = 0

or, (2y + 3) (y + 5) = 0 or,

y = -3/2, -5

Hence, x > y

**2.** (a)

I. 3x^{2} – 7x + 2 = 0

or, 3x^{2} – 6x – x + 2 = 0

or, (x – 2) (3x – 1) = 0

or, x = 2, 1/3

II. 2y^{2} – 11y +15 = 0

or, 2y^{2} -6y – 5y+15 = 0

or, (2y – 5) (y – 3) = 0

or, y = 5/2, 3

Hence, y > x

**3.** (d)

I. 10x^{2} -7 x +1 = 0

or, 10x^{2} -5x – 2x +1 = 0

or, (2x – 1) (5x – 1) = 0

or, x = 1/2, 1/5

II. 35y^{2} – 12y + 1= 0

or, 35y^{2} -7y – 5y +1 = 0

or, (5y – 1) (7y – 1) = 0

or, y = 1/5, 1/ 7

Hence, x ≥ y

**4.** (a)

I. 4x^{2} = 25

or, x^{2} = 25/4, or x = ± 5/2

II. 2y^{2} – 13y + 21 = 0

or, 2y^{2} – 6y – 7y + 21 = 0

or, (y – 3) (2y -7) = 0

or, y = 3, 7/2

Hence, y > x

**5.** (e)

I. 3x^{2} + 7 x – 6 = 0

or, 3x^{2} + 9 x – 2x – 6 = 0

or, (x + 3) (3x – 2) = 0

or, x = – 3, 2/3

II. 6(2/ + 1) = 17y

or, 12y^{2} + 6 – 17y = 0

or, 12y^{2} – 9y -8y + 6 = 0

or, (4y – 3) (3y – 2) = 0

or, y = 3/4, 2/3

Hence, y ≥ x

### Equations and Inequations Previous Year Questions for IBPS SO 2014 Exam

**1.** What is the value of m which satisfies 3m^{2} – 21m + 30 <0? **(IBPS SO 2014)**

(a) m < 2 or in > 5

(b) m> 2

(c) 2 < m < 5

(d) m < 5

(e) None of these

**2.** If one root of x^{2} + px + 12 = 0 is 4, while the equation x^{2} + px + q = 0 has equal roots, then the value of q is **(IBPS SO 2014)**

(a) 49/4

(b) 4/49

(c) 4

(d) 1/4

(e) None of these

**3.** Let p and q be the roots of the quadratic equation x^{2} – (α – 2)x – α -1 = 0 • What is the minimum possible value of p^{2}+q^{2} ? **(IBPS SO 2014)**

(a) 0

(b) 3

(c) 4

(d) 5

(e) None of these

**4.** If the roots, x_{1} and x_{2}, of the quadratic equation x^{2} – 2x + c = 0 also satisfy the equation 7x_{2} – 4x_{1}= 47, then which of the following is true? **(IBPS SO 2014)**

(a) c = – 15

(b) x_{1} = -5, x_{2} =3

(c) x_{1} = 4.5, x_{2} = -2.5

(d) c = 15

(e) None of these

**5.** If the sum of a number and its square is 182, what is the number? **(IBPS SO 2014)**

(a) 15

(b) 26

(c) 28

(d) 13

(e) None of these

**Equations and Inequations Previous Year Questions for IBPS SO 2014 Answers**

**1.** (c)

3m^{2} – 21m + 30 < 0

or m^{2} – 7m +10 < 0

or m^{2} – 5m – 2m +10 < 0

or m(m – 5) – 2(m – 5) < 0

or (m – 2)(m – 5) < 0

Case I : m – 2 > 0 and m – 5 < 0

⇒ m > 2 and m < 5 => 2

Case II: m – 2 < 0 and m – 5 > 0

⇒ m < 2 and m > 5

nothing common

Hence, 2 < m < 5

**2.** (a)

Given x^{2} + px +12 = 0

Since, x = 4 is the one root of the equation, therefore x = 4 will satisfy this equation

∴ 16 + 4p +12 = 0

⇒ p = -7

Other quadratic equation becomes x^{2} – 7x + q = 0

(By putting value of p)

Its roots are equal, so, b^{2} = 4ac

**3.** (d)

Given equation is x^{2} – (α – 2)x – α – 1 = 0

Sum of the roots, p + q = α – 2

Product of the roots pq = – α – 1

Now, p^{2} + q^{2} = (p+q)^{2} – 2pq

= (α – 2)^{2} + 2 (α +1)

= α^{4} + 4 – 4 α + 2α + 2 = (α – 1 )^{2} + 5

Hence, the minimum value of p^{2} + q^{2} will be 5

**4.** (a)

7x_{2} – 4x_{1} = 47

x_{1} + x_{2} = 2

Solving 11x_{2} = 55

x_{2} = 5 & x_{1} = -3

∴ c = -15

**5.** (d)

Let the number be x.

Then, x + x^{2} = 182 ⇒ x^{2} + x – 182 = 0

⇒ (x + 14) (x – 13) = 0 ⇒ x = 13.

### Equations and Inequations Previous Year Questions for IBPS PO 2013 Exam

**1.** Farah was married 8 years ago. Today her age is times to that at the time of marriage. At present her daughter’s age is of her age. What was her daughter’s age 3 years ago? (IBPS SO 2013)

(a) 6 years

(b) 7 years

(c) 3 years

(d) Cannot be determined

(e) None of these

**In each of the following questions two equations are given. Solve these equations and give answer:**

(a) if x > y, i.e., x is greater than or equal to y.

(b) if x > y, i.e., x is greater than y.

(c) if x < y, i.e., x is less than or equal to y.

(d) if x < y, i.e., x is less than y.

(e) x = y or no relation can be established between x and y** (IBPS SO 2013)**

**2.**

I. x^{2} + 5x+6 = 0

II. y^{2} + 7y +12 = 0

**3.**

I. x^{2} + 20 = 9x

II. y^{2} + 42 = 13y

**4.**

I. 2x + 3y = 14

II. 4x + 2y = 16

**5.**

I.

II.

**6.**

I. x^{2} + 4x + 4 = 0

II. y^{2} – 8y + 16 = 0

**Equations and Inequations Previous Year Questions for IBPS PO 2013 Answers**

**1.** (c)

Let Farah’s age 8 years ago be x years

Farah’s present age = (x + 8) years

⇒ 2x = 56

⇒ x = 28

Farah’s present age = 28 + 8 = 36 years

Her daughter’s age

Required age = 6 – 3 = 3 years

**2.** (a)

I. x^{2} + 2x + 3x + 6 = 0

⇒ x(x + 2) + 3(x + 2) = 0

⇒ (x + 3)(x + 2) = 0

⇒ x = -3 or – 2

II. y^{2} + 7y +12 = 0

⇒ y^{2} + 4y + 3y +12 = 0

⇒ y(y + 4) + 3(y + 4) = 0

⇒ (y + 3)(y + 4) = 0

⇒ y = -3 or = 4

On comparing the value of equ. (i) and equ. (ii)

x ≥ y

**3.** (d)

I. x^{2} – 9x + 20 = 0

⇒ x^{2} – 5x – 4x + 20 = 0

⇒ x(x – 5) – 4 (x – 5) = 0

= (x – 4) (x – 5) = 0

x = 4or 5

II. y^{2} -13y + 42 = 0

⇒ y^{2} – 7y – 6y + 42 = 0

⇒ y(y – 7) – 6(y – 7) = 0

⇒ (y – 6)(y – 7) = 0

⇒ y = 6 or 7

Here, y > x

**4.** (d)

2 x + 3 y =14 …(I)

4x+ 2y =16 …(II)

By equation (I) × 2 – equation II.

4 x + 6y – 4 x – 2 y = 28 – 16

⇒ 4 y =12 ⇒ y = 3

From equation I,

2x + 3 × 3 = 14

Here, y > x

**5.** (e)

I.

II. y

No relation can be established between x and y.

**6.** (d)

I. x^{2} + 4x +4 = 0

(x + 2)^{2} = 0 ⇒ x = – 2

II. y^{2} – 8 y +16 = 0

⇒( y – 4 )^{2} = 0

⇒ y = 4

Here, y > x

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