Equations and Inequations Questions for IBPS RRB

Equations and Inequations Questions for IBPS RRB

Equations and Inequations Questions for IBPS RRB 2017 Exam

In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer. (IBPS RRB Scale-I Prelim Exam 2017)

(a) if x > y
(b) if x ≥ y
(c) if x < y
(d) if x ≤ y
(e) if x = y or no relationship can be established

Quantitative Aptitude

1. (I). x² + 9x + 20 = 0
(II). y² = 16

2. (I). x² – 7x + 12 = 0
(II). 3y² – 11y + 10 = 0

3. (I). x² – 8x + 15 = 0
(II). y² – 12y + 36 = 0

4. (I.) 2x² + 9x + 7 = 0
(II). y² + 4y + 4 = 0

5. (I). 2x² + 9x + 7 = 0
(II). 2y² + 13y + 21 = 0

6. (I). 6x² + 25x + 24 = 0
(II). 12y² + 13y + 3 = 0

7. (I). 12x² – x – 1 = 0
(II). 20y² – 41y + 20 = 0

8. (I). 10x² + 33x + 27 = 0
(II). 5y² + 19y + 20 = 0

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

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

Equations and Inequations Questions for IBPS RRB 2017 Answers
1. (d)
(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 = 16
y = ± 4
∴ x ≤ y

2. (a)
(I). x² – 7x + 12 = 0
x² – 4x – 3x + 12 = 0
x(x – 4) – 3(x – 4) = 0
x = 3, 4
(II). 3y² – 11y + 10 = 0
3y² – 6y – 5y + 10 = 0
3y(y – 2) – 5(y – 2) = 0
(3y – 5)(y – 2) = 0
\(\mathrm { y } = 2 , \frac { 5 } { 3 }\)
∴ x > y

3. (c)
(I). x² – 8x + 15 = 0
x² – 3x – 5x + 15 = 0
x(x – 3) – 5(x – 3) = 0
(x – 3)(x – 5) = 0
x = 3, 5
(II). y² – 12y + 36 = 0
y² – 6y – 6y + 36 = 0
y(y – 6) -6(y – 6) = 0
(y – 6)(y – 6) = 0
y = 6
∴ x < y

4. (e)
(I). 2x² + 9x + 7 = 0
2x² + 7x + 2x + 7 = 0
x(2x + 7) + 1(2x + 7) = 0
(x + 1)(2x + 7) = 0
\(x = – 1 , – \frac { 7 } { 2 }\)
(II). y² + 4y + 4 = 0
y² + 2y + 2y + 4 = 0
y(y + 2) + 2(y + 2) = 0
(y + 2)(y + 2) = 0
y = – 2, – 2
∴ No relation.

5. (d)
(I). 2x² + 15x + 28 = 0
2x² + 8x + 7x + 28 = 0
2x(x + 4) + 7(x+ 4) = 0
(x + 4)(2x + 7) = 0
\(x = \left( – \frac { 7 } { 2 } \right) , – 4\)
(II). 2y² + 13y + 21 = 0
2y² + 7y + 6y + 21 = 0
y(2y + 7) + 3(2y + 7) = 0
(y + 3)(2y + 7) = 0
\(y = – 3 , \frac { – 7 } { 2 }\)
x ≤ y

6. (c)
(I). 6x² + 25x + 24 = 0
⇒ 6x² + 16x + 9x + 24 = 0
⇒ 2x(3x + 8x) + 3(3x + 8) = 0
⇒ (2x + 3)(3x + 8) = 0
\(\therefore \quad x = \frac { – 3 } { 2 } , \frac { – 8 } { 3 }\)
(II). 12y² + 13y + 3 = 0
⇒ 12y² + 9y + 4y + 3 = 0
⇒  3y(4y + 3) + 1(4y + 3)
⇒ (4y + 3)(3y + 1)
\(\therefore \quad y = \frac { – 3 } { 4 } \frac { – 1 } { 3 }\)
∴ x < y

7. (c)
(I). 12x² – x – 1 = 0
⇒ 12x² – 4x + 3x – 1
⇒ (4x + 1)(3x – 1) = 0
\(\Rightarrow \therefore \mathbf { x } = – 1 / 4,1 / 3\)
(II). 20y² – 41y + 20 = 0
⇒ 20y² – 25y – 16y + 20 = 0
⇒ 5y(4y -5) -4(4y – 5)
⇒ (5y -4)(4y – 5)
\(\therefore \quad y = \frac { 5 } { 4 } , \frac { 4 } { 5 } \Rightarrow x < y\)

8. (b)
(I). 10x² + 33x + 27 = 0
⇒ 10x² + 15x + 18x + 27 = 0
⇒ 5x(2x + 3) +9(2x + 3) = 0
⇒ (5x + 9)(2x + 3)
\(\therefore \quad x = – 9 / 5 , – 3 / 2\)
(II). 5y² + 19y + 18 = 0
5y² + 10y + 9y + 18 = 0
5y(y + 2) + 9(y + 2) = 0
(5y + 9)(y + 2) = 0
\(\therefore \quad y = – 9 / 5 , – 2\)
∴ x ≥  y

9. (d)
(I). 15x² – 29x – 14 = 0
⇒ 15x² – 35x +6x – 14 = 0
⇒ 5x(3x – 7) + 2(3x – 7)
⇒ (5x + 2)(3x – 7)
\(\therefore \quad x = \frac { – 2 } { 5 } , \frac { 7 } { 3 }\)
(II). 6y² – 5y – 25 = 0
⇒ 6y² – 15y + 10y – 25 = 0
⇒ 3y(2y – 5) + 5(2y – 5)
⇒ (3y + 5)(2y – 5)
\(\therefore \quad y = – 5 / 3,5 / 2\)
∴ So, relationship between x and y can’t be determined.

10. (e)
(I). 3x² – 22x + 7 = 0
3x² – 21x – x + 7 = 0
x(3x – 1) – 7(3x – 1) = 0
(3x – 1)(x – 7) = 0
\(x = \frac { 1 } { 3 } , 7\)
(II). y² – 20y + 91 = 0
y² – 13y -7y + 91 = 0
y(y – 13) – 7(y – 13) = 0
(y – 13)(y – 7) = 0
\(y = 13,7 \Rightarrow y \geq x\)

Equations and Inequations Questions for IBPS RRB 2015 Exam

In the following questions two equations numbered I and II are given. You have to solve both the questions and
Give answer (a) if x > y
Give answer (b) if x ≥ y
Give answer (c) if x ≥ y
Give answer (d) if x ≤ y
Give answer (d) if x = y or the relationship cannot be establlished betweeb ‘x’ and ‘y’.
(IBPS RRB 2015)

1. (I). x² + 5x + 6 = 0
(II). y² + 3y + 2 = 0

2. (I). x² –  10x + 24 = 0
(II). y² – 9y + 20 = 0

3. (I). (x² )² = 961
(II). y² = √961

4. (I). x² –  72 = x
(II). y² = 64

5. (I). x² –  463 = 321
(II). y² – 421 = 308

Equations and Inequations Questions for IBPS RRB 2015 Answers

1. (d)
(I). x² + 5x + 6 = 0
⇒ x² + 2x + 3x + 6 = 0
⇒ x(x + 2) + 3(x + 2) = 0
⇒ (x + 2)(x + 3) = 0
∴ x = – 3 or – 2
(II). y² + 3y + 2 = 0
⇒ y² + 2y + y + 2 = 0
⇒ y(y + 2) + 1(y + 2) = 0
⇒ (y + 2)(y + 1) = 0
∴ y = – 1 or – 2
∴ x ≤ y

2. (b)
(I). x² – 10x + 24 = 0
⇒ x² – 6x – 4x + 24 = 0
⇒ x(x – 6) – 4(x – 6) = 0
⇒  (x – 6)(x – 4) = 0
∴ x= 4 or 6
(II). y² – 9y + 20 = 0
⇒ y² – 5y – 4y+ 20 = 0
⇒ y(y – 5) – 4 (y – 5) = 0
⇒ (y – 5)(y – 4) = 0
∴ y = 5 or 4
∴ x ≥ y

3. (e)
(I). x² = 961
⇒ x = ±√961 = ± 31
(II). y = √961 = ± 31
∴ x = y

4. (b)
(I). x² – x – 72 = 0
⇒ x² – 9x + 8x – 72 = 0
⇒ x(x – 9) + 8 (x – 9) = 0
⇒ (x + 8)(x – 9) = 0
∴ x = – 8 or 9
(II). y² = 64
⇒ y = √64 = ± 8
∴ x ≥ y

5. (e)
(I). x² = 463 + 321 = 784
∴ x = √784 = ± 28
(II). y² = 308 + 421 = 729
∴ y = √729 = ± 27

Equations and Inequations Questions for IBPS RRB 2014 Exam

In the following questions two equations numbered I and II are given. You have to solve both the equations and Give Answers if
(a) x > y
(b) x ≥ y
(c) x < y
(d) x ≤ y
(e) x = y or the relationship cannot be established.  (IBPS RRB 2014)
1. (I). 20x² – x – 12 = 0
(II). 20y² + 27y + 9 = 0

2. (I). x² – 218 = 106
(II). y² – 37y + 342 = 0

3. (I). \(\frac { 7 } { \sqrt { x } } + \frac { 5 } { \sqrt { x } } = \sqrt { x }\)
(II). \(y ^ { 2 } – \frac { ( 12 ) ^ { 5 / 2 } } { \sqrt { y } } = 0\)

4. (I). \(\sqrt { 361 } x + \sqrt { 16 } = 0\)
(II). \(\sqrt { 441 } y + 4 = 0\)

5. In a family, a couple has a son and daughter. The age of the father is three times that of his daughter and the age of the son is half of his mother. The wife is nine years younger to her husband and the brother is seven years older than his sister. What is the age of the mother? (IBPS RRB 2014)
(a) 40 years
(b) 45 years
(c) 50 years
(d) 60 years
(e) 65 years

Equations and Inequations Questions for IBPS RRB 2014 Answers

1. (b) (I). 20x² – x – 12 = 0
⇒ 20x² – 16x – 15x – 12 = 0
⇒ 4x – (5x – 4) + 3(5x – 4) = 0
⇒ (5x – 4)(4x + 3) = 0
⇒ 5x – 4 = 0 or 4x + 3 = 0
\(\Rightarrow x = \frac { 4 } { 5 } \text { or } \frac { – 3 } { 4 }\)
(II). 20y² + 27y + 9 = 0
⇒ 20y² + 15y + 12y + 9 = 0
⇒ 5y(4y + 3) + 3(4y + 3) = 0
⇒ (4y + 3)(5y +3) = 0
\(\Rightarrow y = \frac { – 3 } { 5 } \text { or } \frac { – 3 } { 4 }\)
Clearly, x ≥ y

2. (d) (I).  x² = 106 + 218 = 324
∴ x = √324 = ± 18
(II). y² – 37y + 342 = 0
⇒ y² – 18y – 19y + 342 = 0
⇒ y(y – 18) 19(y – 18) = 0
⇒ (y – 18)(y – 19) = 0
⇒ y = 19 or 18

3. (e) (I). \(\frac { 7 } { \sqrt { x } } + \frac { 5 } { \sqrt { x } } = \sqrt { x }\)
⇒ 7 + 5 = √x × √x
⇒ x = 12
(II). \(y ^ { 2 } – \frac { ( 12 ) ^ { 5 / 2 } } { \sqrt { y } } = 0\)
\(\Rightarrow y ^ { 2 + \frac { 1 } { 2 } } – ( 12 ) ^ { ( 5 / 2 ) } = 0\)
\(\Rightarrow y ^ { 5 / 2 } = 12 ^ { 5 / 2 }\)
⇒ y = 12

4. (c) (I). 19x + 4 = 0
⇒ 19x = – 4
\(\Rightarrow x = \frac { – 4 } { 19 }\)
(II). 21y + 4 = 0
\(\Rightarrow y = \frac { – 4 } { 21 }\)

5. (d)
Let the mother’s age be y years.
∴ The age of father = (y + 9) years
The age of son \(= \frac { y } { 2 }\)years
The age of daughter \(= \left( \frac { y } { 2 } – 7 \right)\)years
Now according to the given condition,
\(( y + 9 ) = 3 \left( \frac { y } { 2 } – 7 \right)\)
\(\Rightarrow y + 9 = \frac { 3 y – 42 } { 2 }\)
⇒ 2y + 18 = 3y – 42
⇒ y = 60 years

Equations and Inequations Questions for IBPS RRB 2013 Exam

1. The present ages of Trisha and Shalini are in the ratio of 7 : 6 respectively. After 8 years the ratio of their ages will be 9 : 8. What is the difference in their ages  (IBPS RRB 2013)

(a) 4 years
(b) 8 years
(c) 10 years
(d) 12 years
(e) None of these

In the following questions two equations numbered I and II are given. You have to solve both equations and Give answer if
(a) x > y
(b) x ≥ y
(c) x < y
(d) x ≤ y
(e) x = y
or the relationship cannot be established   (IBPS RRBs 2013)

2. (I). x² – 7x + 10 = 0
(II). y² + 11y + 10 = 0

3. (I). x² + 28x + 192 = 0
(II).  y² + 16y + 48 = 0

4. (I). 2x – 3y = – 3.5
(II). 3x – 2y = – 6.5

5. (I). x² + 8x + 15 = 0
(II). y² + 11y + 30 = 0

6. (I). x = √3136
(II). y² = 3136

Equations and Inequations Questions for IBPS RRB 2013 Answers

1. (a)
Let Trisha’s and Shalini’s present ages be 7x and 6x years respectively.
After 8 years, \(\frac { 7 x + 8 } { 6 x + 8 } = \frac { 9 } { 8 }\)
⇒ 56x + 64 = 54x + 72
⇒ 2x = 72 – 64 = 8
⇒ x = 4
∴ Required difference = 7x – 6x ⇒ x = 4 years

2. (a) (I). x² – 7x + 10 = 0
⇒ 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 + 10) + (y + 1) = 0
⇒ y = – 1 or – 10

3. (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
Clearly, x ≤ y

4. (c)
2x – 3y = – 3.5 —— (i)
3x – 2y = – 6.5 —— (ii)
By equation (i) × 2 – equation (ii) × 3, we have

\(\Rightarrow \quad x = \frac { 12.5 } { – 5 } = – 2.5\)
From equation (i)
2 × (- 2.5) – 3y = – 3.5
⇒ 3y = – 5 + 3.5
\(\Rightarrow y = \frac { – 1.5 } { 3 } = – 0.5\)
Clearly, x < y

5. (b) (I). x² + 8x + 15 = 0
⇒ x² + 5x + 3x + 15 = 0
⇒ x(x + 5) + 3 (x + 15) = 0
⇒ (x + 15)(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 + 6)(y + 5) = 0
⇒ y = – 5 or – 6
Clearly, x > y

6. (e)
x = √3136 = ± 56
y² = 3136
⇒ y = √3136 = ± 56
Clearly, x = y