Equations and Inequations Questions for IBPS SO

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, 5X² + 28X + 15 = 0
II. 6Y² + 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

Quantitative Aptitude

2.
I. 12X² + 82X + 140 = 0
II. 16Y²+ 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. 4X² – 48X + 143 = 0
II. 4Y² – 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. X² + 4X = 21
II. Y² – 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. \(\frac { 8 } { \sqrt { X } } + \frac { 6 } { \sqrt { X } } = \sqrt { X }\)
II. \(\mathrm { Y } ^ { 2 } – \left( 14 ^ { 5 / 2 } / \mathrm { Y } ^ { 1 / 2 } \right) = 0\)
(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. 12X² + 82X + 140 = 0
X = -7/2, -10/3
II. 16Y² + 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.
\(\frac { 8 } { \sqrt { X } } + \frac { 6 } { \sqrt { X } } = \sqrt { X }\)
⇒ X = 14
From II.
\(\mathrm { Y } ^ { 2 } – \left( 14 ^ { 5 / 2 } / \mathrm { Y } ^ { 1 / 2 } \right) = 0\)
⇒ 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. 6x² + 25x+24 = 0
II. 12y² + 13y + 3 = 0

2.
I. 12x²-x -1= 0
II. 20y² – 41y + 20 = 0

3.
I. 10x² + 33x +27 = 0
II. 5y² + 19y +18 = 0

4.
I. 15x² – 29x – 14 = 0
II. 6y²– 5y – 25 = 0

5.
I. 3x² – 22x + 7 = 0
II. y² – 20y + 91 = 0

Equations and Inequations Previous Year Questions for IBPS PO 2016 Answers

1. (c)
I. 6x² + 25x+24 = 0
\(\mathrm { D } = \sqrt { \mathrm { b } ^ { 2 } – 4 \mathrm { ac } }\)
\(\mathrm { D } = \sqrt { 625 – 4 \times 24 \times 6 }\)
\(= \sqrt { 49 } = 7\)
\(x _ { 1 } = \frac { – b + 7 } { 12 } = \frac { – 25 + 7 } { 12 } = \frac { – 18 } { 12 } = – \frac { 3 } { 2 }\)
\(x _ { 2 } = \frac { – b – 7 } { 12 } = \frac { – 25 – 7 } { 12 } = \frac { – 32 } { 12 } = – \frac { 8 } { 3 }\)
\(\mathrm { x } = \frac { – 3 } { 2 } , \frac { – 8 } { 3 }\)
II. 12y² + 13y + 3 = 0
\(\mathrm { y } _ { 1 } = \frac { – 13 + \sqrt { 169 – 144 } } { 24 }\)
\(= \frac { – 13 + 5 } { 24 } = \frac { – 8 } { 24 } = \frac { – 1 } { 3 }\)
\(\mathrm { y } _ { 2 } = \frac { – 13 – \sqrt { 169 – 144 } } { 24 } = \frac { – 18 } { 24 } = \frac { – 3 } { 4 }\)
\(\mathrm { y } = \frac { – 1 } { 3 } , \frac { – 3 } { 4 } \Rightarrow \mathrm { x } < \mathrm { y }\)

2. (c)
I. 12x²-x -1= 0
\(x _ { 1 } = \frac { – b + \sqrt { D } } { 2 a } = \frac { 1 + \sqrt { 1 – 4 \times 12 \times 1 } } { 24 }\)
\(= \frac { 1 + 7 } { 24 } = \frac { 8 } { 24 } = \frac { 1 } { 3 }\)
\(\mathrm { x } _ { 2 } = \frac { – \mathrm { b } – \sqrt { \mathrm { D } } } { 2 \mathrm { a } }\)
\(\mathrm { x } _ { 2 } = \frac { 1 – 7 } { 24 } = \frac { – 6 } { 24 } = \frac { – 1 } { 4 }\)
\(x = \frac { 1 } { 3 } , – \frac { 1 } { 4 }\)
II. 20y² – 41y + 20 = 0
\(\mathrm { y } _ { 1 } = \frac { 41 – \sqrt { 1681 – 1600 } } { 40 }\)
\(\mathrm { y } _ { 2 } = \frac { 41 – \sqrt { 1681 – 1600 } } { 40 }\)
\(\mathrm { y } _ { 1 } = \frac { 41 + 9 } { 40 } = \frac { 50 } { 40 } , \mathrm { y } _ { 2 } = \frac { 32 } { 40 }\)
\(\mathrm { y } = \frac { 5 } { 4 } , \frac { 4 } { 5 } \quad \Rightarrow \quad \mathrm { x } < \mathrm { y }\)

3. (b)
I. 10x² + 33x +27 = 0
\(x _ { 1 } = \frac { – 33 + \sqrt { b ^ { 2 } – 4 a c } } { 2 a } = \frac { – 33 + \sqrt { 1089 – 4 \times 10 \times 27 } } { 20 }\)
\(\mathrm { x } _ { 2 } = \frac { – 33 – \sqrt { \mathrm { b } ^ { 2 } – 4 \mathrm { ac } } } { 2 \mathrm { a } }\)
\(\mathrm { x } _ { 2 } = \frac { – 33 – \sqrt { 1089 – 1080 } } { 20 }\)
\(\mathrm { x } _ { 1 } = \frac { – 33 + 3 } { 20 } , \mathrm { x } _ { 2 } = \frac { – 33 – 3 } { 20 }\)
\(\mathrm { x } _ { 1 } = \frac { – 30 } { 20 } , \mathrm { x } _ { 2 } = \frac { – 36 } { 20 } = \frac { – 9 } { 5 } , \mathrm { x } = \frac { – 3 } { 2 } , \frac { – 9 } { 5 }\)
II. 5y² + 19y +18 = 0
\(\mathrm { y } _ { 1 } = \frac { – 19 – \sqrt { 361 – 4 \times 18 \times 5 } } { 10 }\)
\(\mathrm { y } _ { 2 } = \frac { – 19 – \sqrt { 361 – 360 } } { 10 }\)
\(\mathrm { y } _ { 1 } = \frac { – 19 + 1 } { 10 }\)
\(\mathrm { y } _ { 2 } = \frac { – 19 – 1 } { 10 } = \frac { – 18 } { 10 } = \frac { – 9 } { 5 } = \frac { – 20 } { 10 } = – 2\)
\(\mathrm { y } = \frac { – 9 } { 5 } , – 2 \quad \Rightarrow \quad \mathrm { x } \geq \mathrm { y }\)

4. (d)
I. 15x² – 29x – 14 = 0
\(\mathrm { x } _ { 1 } = \frac { 29 + \sqrt { 841 + 60 \times 14 } } { 30 }\)
\(= \frac { 29 + 41 } { 30 } = \frac { 70 } { 30 }\)
\(\mathrm { x } _ { 2 } = \frac { 29 – \sqrt { 1681 } } { 30 }\)
\(\mathrm { x } _ { 2 } = \frac { 29 – 41 } { 30 } = \frac { – 12 } { 30 }\)
\(\mathrm { x } = \frac { 7 } { 3 } \cdot \frac { – 2 } { 5 }\)
II. 6y²– 5y – 25 = 0
\(\mathrm { y } _ { 1 } = \frac { 5 + \sqrt { 25 – 4 \times 6 \times – 25 } } { 12 } = \frac { 5 + \sqrt { 625 } } { 12 } = \frac { 30 } { 12 }\)
\(\mathrm { y } _ { 2 } = \frac { 5 – \sqrt { 25 – 4 \times 6 \times – 25 } } { 12 } \Rightarrow \mathrm { y } _ { 2 } = \frac { 5 – \sqrt { 625 } } { 12 } = \frac { – 20 } { 12 }\)
\(\mathrm { y } = \frac { 5 } { 2 } , \frac { – 5 } { 3 }\)

5. (b)
I. 3x² – 22x + 7 = 0
3x² – 21x – x + 7 = 0
x (3x – 1) – 7(3x – 1) = 0
(3x – 1) (x – 7) = 0
\(\mathrm { x } = \frac { 1 } { 3 } , 7\)
II. y² – 20y + 91 = 0
y² – 20y + 91 = 0
y² – 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. x² – 6x – 7
II. 2y² + 13y +15 = 0

2.
I. 3 x² – 7 x + 2 0
II. 2y² – 11y +15 = 0

3.
I. 10x² – 7 x + 1 = 0
II. 35y² – 12y +1 = 0

4.
I. 4 x² = 25
II. 2y² – 13y + 21 = 0

5.
I. 3 x² + 7 x = 6
II. 6(2y² + 1) = 17y

Equations and Inequations Previous Year Questions for IBPS SO 2015 Answers

1. (b)
I. x² – 6x = 7
or, x² – 6x – 7 = 0
or, (x – 7) (x + 1) = 0
or, x = 7, – 1
II. 2y² + 13y +15 = 0
or, 2y² + 3y + 10y +15 = 0
or, (2y + 3) (y + 5) = 0 or,
y = -3/2, -5
Hence, x > y

2. (a)
I. 3x² – 7x + 2 = 0
or, 3x² – 6x – x + 2 = 0
or, (x – 2) (3x – 1) = 0
or, x = 2, 1/3
II. 2y² – 11y +15 = 0
or, 2y² -6y – 5y+15 = 0
or, (2y – 5) (y – 3) = 0
or, y = 5/2, 3
Hence, y > x

3. (d)
I. 10x² -7 x +1 = 0
or, 10x² -5x – 2x +1 = 0
or, (2x – 1) (5x – 1) = 0
or, x = 1/2, 1/5
II. 35y² – 12y + 1= 0
or, 35y² -7y – 5y +1 = 0
or, (5y – 1) (7y – 1) = 0
or, y = 1/5, 1/ 7
Hence, x ≥ y

4. (a)
I. 4x² = 25
or, x² = 25/4, or x = ± 5/2
II. 2y² – 13y + 21 = 0
or, 2y² – 6y – 7y + 21 = 0
or, (y – 3) (2y -7) = 0
or, y = 3, 7/2
Hence, y > x

5. (e)
I. 3x² + 7 x – 6 = 0
or, 3x² + 9 x – 2x – 6 = 0
or, (x + 3) (3x – 2) = 0
or, x = – 3, 2/3
II. 6(2/ + 1) = 17y
or, 12y² + 6 – 17y = 0
or, 12y² – 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 3m² – 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 x² + px + 12 = 0 is 4, while the equation x² + 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 x² – (α – 2)x – α -1 = 0 • What is the minimum possible value of p²+q² ? (IBPS SO 2014)
(a) 0
(b) 3
(c) 4
(d) 5
(e) None of these

4. If the roots, x₁ and x₂, of the quadratic equation x² – 2x + c = 0 also satisfy the equation 7x₂ – 4x₁= 47, then which of the following is true? (IBPS SO 2014)
(a) c = – 15
(b) x₁ = -5, x₂ =3
(c) x₁ = 4.5, x₂ = -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)
3m² – 21m + 30 < 0
or m² – 7m +10 < 0
or m² – 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 x² + 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 x² – 7x + q = 0
(By putting value of p)
Its roots are equal, so, b² = 4ac
\(\Rightarrow 49 = 4 q \text { or } q = \frac { 49 } { 4 }\)

3. (d)
Given equation is x² – (α – 2)x – α – 1 = 0
Sum of the roots, p + q = α – 2
Product of the roots pq = – α – 1
Now, p² + q² = (p+q)² – 2pq
= (α – 2)² + 2 (α +1)
= α⁴ + 4 – 4 α + 2α + 2 = (α – 1 )² + 5
Hence, the minimum value of p² + q² will be 5

4. (a)
7x₂ – 4x₁ = 47
x₁ + x₂ = 2
Solving 11x₂ = 55
x₂ = 5 & x₁ = -3
∴ c = -15

5. (d)
Let the number be x.
Then, x + x² = 182 ⇒ x² + 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 \(1 \frac { 2 } { 7 }\) times to that at the time of marriage. At present her daughter’s age is \(\frac { 1 } { 6 } \mathrm { th }\) 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. x² + 5x+6 = 0
II. y² + 7y +12 = 0

3.
I. x² + 20 = 9x
II. y² + 42 = 13y

4.
I. 2x + 3y = 14
II. 4x + 2y = 16

5.
I. \(x = \sqrt { 625 }\)
II. \(x = \sqrt { 676 }\)

6.
I. x² + 4x + 4 = 0
II. y² – 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
∴ \(x + 8 = \frac { 9 x } { 7 } \Rightarrow 7 x + 56 = 9 x\)
⇒ 2x = 56
⇒ x = 28
Farah’s present age = 28 + 8 = 36 years
Her daughter’s age
\(= 36 \times \frac { 1 } { 6 } = 6\) Required age = 6 – 3 = 3 years

2. (a)
I. x² + 2x + 3x + 6 = 0
⇒ x(x + 2) + 3(x + 2) = 0
⇒ (x + 3)(x + 2) = 0
⇒ x = -3 or – 2
II. y² + 7y +12 = 0
⇒ y² + 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. x² – 9x + 20 = 0
⇒ x² – 5x – 4x + 20 = 0
⇒ x(x – 5) – 4 (x – 5) = 0
= (x – 4) (x – 5) = 0
x = 4or 5
II. y² -13y + 42 = 0
⇒ y² – 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
\(\Rightarrow 2 x = 14 – 9 = 5 \Rightarrow x = \frac { 5 } { 2 }\)
Here, y > x

5. (e)
I. \(x = \sqrt { 625 } = \pm 25\)
II. y\(y = \sqrt { 676 } = \pm 26\)
No relation can be established between x and y.

6. (d)
I. x² + 4x +4 = 0
(x + 2)² = 0 ⇒ x = – 2
II. y² – 8 y +16 = 0
⇒( y – 4 )² = 0
⇒ y = 4
Here, y > x