Equations and Inequations Questions for SBI PO

Equations and Inequations Questions for SBI PO

Equations and Inequations Questions for SBI PO 2017 Exam

Given below are two equations in each question, which you have to solve and give answer
(a) If x < y
(b) If x > y
(c) If x ≤ y
(d) If x ≥ y
(e) If x = y or no relation can be established

Quantitative Aptitude

1. (I). x² – 3x + 2 = 0
(II). 2y² – 7y + 6 = 0

2. (I). 3x² + 4x + 1 = 0
(II). y² + 5y + 6 = 0

3. (I). 2x² + 5x + 2 = 0
II. y² + 9y + 20 = 0

4. (I). x² – 7x + 10 = 0
(II). y² – 12y + 35 = 0

5.
(I). (x-18)² = 0
(II). y² = 324

Equations and Inequations Questions for SBI PO 2017 Answers

1. (e) (I). x² – 2x – x + 2 = 0
⇒ x(x – 2) – 1 (x – 2) = 0
⇒ (x – 1)(x – 2) = 0
⇒ x = 1, 2
(II). 2y² – 7y + 6 = 0
2y² – 4y – 3y + 6 = 0
⇒ 2y(y – 2) – 3(y – 2) = 0
⇒ \(\mathrm { y } = 2 , \frac { 3 } { 2 }\)
∴ No relation established

2. (a) (I). 3x² + 3x + x + 1 = 0
⇒ 3x(x + 1) + 1(x + 1) = 0
⇒ x = – 1 , \(-\frac { 1 }{ 3 } \)
(II). y² + 3y + 2y + 6 = 0
⇒ y(y + 3) + 2(y + 3) = 0
⇒ y = -2, -3
∴ x > y

3. (a) (I). 2x² + 4x + x + 2 = 0
⇒  2x(x + 2) + 1(x + 2) = 0
⇒ \(\mathbf { x } = – 2 , \frac { – 1 } { 2 }\)
(II). y² + 5y + 4y + 20 = 0
⇒ y(y + 5) + 4(y + 5) = 0
⇒ y = -4, -5
∴ x > y

4. (d) (I). x² – 5x – 2x + 10 = 0
⇒ x(x – 5) -2(x – 5) =0
⇒ x = 2, 5
(II). y² – 7y – 5y + 35 = 0
⇒  y(y – 7) – 5(y – 7) = 0
⇒ y = 7, 5
∴ x ≤ y

5. (b) (I). (x – 18)(x – 18) = 0
⇒ x = 18
(II). y = ± 18
∴ x ≥ y

Equations and Inequations Questions for SBI PO 2016 Exam

In the following questions two equations numbered I and II are given. You have to solve both the equations and give answer.      (SBI PO 2016)

1.  (x + 2)(x + 1) = (x – 2)(x – 3)
(y + 3)(y + 2) = (y – 1)(y – 2)
(a) x > y
(b) x < y
(c) x ≥ y
(d) x ≤ y
(e) x = y or relation cannot be established

2. 12x² + 29x + 14 = 0
y² + 9y + 18 = 0
(a) x > y
(b) x < y
(c) x ≥ y
(d) x ≤ y
(e) x = y or relation cannot be established

3. 5x² – 18x + 9 = 0
3y² + 5y – 2 = 0
(a) x > y
(b) x < y
(c) x ≥ y
(d) x ≤ y
(e) x = y or relation cannot be established

4. 17²+ 144 ÷ 18 = x
26² – 18 * 21 = y
(a) x > y
(b) x < y
(c) x ≥ y
(d) x ≤ y
(e) x = y or relation cannot be established

5. 30x² + 11x + 1 = 0
42y²+ 13y + 1 = 0
(a) x > y
(b) x < y
(c) x ≥ y
(d) x ≤ y
(e) x = y or relation cannot be established

In you have to solve equation I and II, Give answer
(a) x > y
(b) x < y
(c) x ≤ y
(d) x ≥ y
(e) if x = y or relation cannot be established

6. (I). \(\frac { 4 } { \sqrt { x } } + \frac { 3 } { \sqrt { x } } = \sqrt { x }\)
(II). \(y ^ { 2 } – \left( 7 ^ { 5 / 2 } / \sqrt { y } \right) = 0\)

7. (I). 25x²+ 25x + 6= 0
(II). 5y² + 20y + 20 = 0

8. (I). 2x²+ 11x + 12= 0
(II). 2y²+ 19y + 45= 0

9. (I). 4x²- 19x + 12= 0
(II). 3y²+ 8y + 4 = 0

10. (I). 4x²- 13x – 12 = 0
(II).  y²- 7y – 60 = 0

11. \(\begin{array} { l l } { \mathrm { I. } } & { \mathrm { x } = ( 1024 ) ^ { 1 / 2 } } \\ { \mathrm { II. } } & { \mathrm { y } ^ { 2 } = 1024 } \end{array}\)

Equations and Inequations Questions for SBI PO 2016 Answers

1. (a)  (x + 2)(x + 1) = (x – 2)(x – 3)
\(\begin{array} { l } { \mathrm { x } = \frac { 1 } { 2 } = 0.5 } \\ { ( \mathrm { y } + 3 ) ( \mathrm { y } + 2 ) = ( \mathrm { y } – 1 ) ( \mathrm { y } – 2 ) } \\ { \mathrm { y } = – \frac { 1 } { 2 } = – 0.5 } \end{array}\)

2. (a) 12x² + 29x + 14 = 0
x = – 1.75, – 0.6
y² + 9y + 18 = 0
y = -6, -3

3. (a) 5x² – 18x + 9 = 0
x = 0.6, 3
3y² + 5y – 2 = 0
y = 0.33, -2

4. (b) 17²+ 144 ÷ 18 = x
x = 297
26² – 18 × 21 = y
y = 676 – 378 = 29

5. (d) 30x² + 11x + 1 = 0
30x² + 6x + 5x + 1 = 0
x = – 0.16, -0.19
42y² + 13y + 1 = 0
42y² + 6y + 7y + 1 = 0
y = -0.14, -0.16
Put on number line
-0.19, -0.16, -0.16, -0.14

6. (e)
\(\frac { 7 } { \sqrt { x } } = \sqrt { x }\)
x = 7
\(\left( y ^ { 35 / 2 } – 7 ^ { 5 / 2 } \right) / \sqrt { y } = 0\)
\(\mathrm { y } ^ { 5 / 2 } = 7 ^ { 5 / 2 }\)
y = 7; x =y

7. (a) (5x + 2)(5x + 3) = 0
\(x = – \frac { 2 } { 5 } , – \frac { 3 } { 5 } = – 0.4 , – 0.6\)
(5x + 10)(x + 2) = 0
\(x = – \frac { 10 } { 5 } , – 2 = – 2 , – 2\)
-0.4, -0.6, -2, -2
x > y

8. (a) (2x + 8) (2x + 3) = 0
\(\mathrm { X } = – \frac { 8 } { 2 } , – \frac { 3 } { 2 } = – 4 , – 1.5\)
(2y + 10)(2y + 9) = 0
\(Y = – \frac { 10 } { 2 } , – \frac { 9 } { 2 } = – 5 , – 4.5\)
-1.5, -4, -4.5, -5
⇒ x > y

9. (a)  (x – 4)(4x – 3) = 0
\(\mathrm { X } = 4 , \frac { 3 } { 4 }\)
(y + 2)(3y + 2) = 0
\(y = – 2 , – \frac { 2 } { 3 }\)
\(4 , – \frac { 3 } { 4 } , – \frac { 2 } { 3 } , – 2\)
x > y

10. (e)  (x – 4)(4x + 3) = 0
\(x = 4 , – 3 / 4\)
(x + 5)(x – 12) = 0
y = 12, -5

11. (c)  x = 32
y = ±32
x ≥ y

Equations and Inequations Questions for SBI PO 2015 Exam

In the following questions three equations numbered I, II and III are given. You have to solve all the equations either together or separately, or two together and one separately, or by any other method and Give answer If
\(\begin{array} { l l } { \text { (a) } } & { x < y = z \text { (b) } x \leq y < z } \\ { \text { (c) } } & { x < y > z \text { (d) } x = y > z } \\ { \text { (e) } } & { x = y = z \text { or if none of the above relationship is established } } \end{array}\)                  (SBI PO Main 2015)
1. (I). 7x + 6y + 4z = 122
(II). 4x +5y + 3z = 88
(III). 9x + 2y + z = 78

2. (I). 7x + 6y = 110
(II). 4x + 3y = 59
(III). x + z = 15

3. \(\begin{array} { l l } { \text { I. } } & { x = \sqrt { \left[ ( 36 ) ^ { 1 / 2 } \times ( 1296 ) ^ { 1 / 4 } \right] } } \\ { \text { II. } } & { 2 y + 3 z = 33 } \\ { \text { III. } } & { 6 y + 5 z = 71 } \end{array}\)

4. (I). 8x + 7y 135
(II). 5x + 6y 99
(III). 9y + 8z = 121

5. (I). \(( x + y ) ^ { 3 } = 1331\)
(II). x – y + z = 0
(III). xy = 28

Equations and Inequations Questions for SBI PO 2015 Answers

1. (a) 7x + 6y + 4z = 122 —— (i)
4x + 5y + 3z = 88 —— (ii)
9x + 2y + z = 78 —— (iii)
By equation (iii) × 3 – equation (ii),

By equation (iii) × 4 – equation (i),

By equation (iv) × 2 – equation (v),

⇒ x = 6
from equation (iv),
23 × 3 + y = 146
⇒ y = 146 – 138 = 8
From equation (iii),
9 × 6 + 2 × 8 + z = 78
⇒ 54 + 16 – z = 78
⇒ z = 78 – 70 = 8
Clearly, x < y = z

2. (c) By equation (II) × 2 – equation (I)
From equation (I),
7 × 8 + 6y = 110
⇒ 6y = 110 – 56 = 54
⇒ y = 9
From equation (iii),
8 + z = 15 ⇒ z = 7
Clearly, x < y > z

3. (b)
\(\quad \text { L } x = \sqrt { ( 36 ) \frac { 1 } { 2 } \times ( 1296 ) \frac { 1 } { 4 } } = \sqrt { 6 \times 6 } = \pm 6\)
By equation × 3 – equation I

From equation II,
2y + 3 × 7 = 33
⇒ 2y = 33 – 21 = 12
⇒ y = 6
x ≤ y < z

4. (d)  By equation 1 × 5 – II × 8

⇒ y = 9
From equation I,
8x + 7 × 9 = 135
⇒ 8x = 135 – 63 = 72
⇒ x = 9
From equation III,
9 × 9 + 8z = 121
⇒ 8z = 121 – 81 = 40
⇒ z = 5
Clearly, x = y > z

5. (e)
I. (x + y)³ = 1331
⇒ x + y = 11
⇒ y = 11 – x
From equation III,
x(11 – x) = 28
⇒ 11x – x² = 28
⇒  x² – 11x -+ 28 = 0
⇒  x² – 7x – 4x + 28 = 0
⇒ x(x – 7) – 4 (x – 7) = 0
⇒ (x – 7)(x – 4) = 0
⇒ x = 7, 4
From equation I,
y = 4, 7
From equation II,
7 – 4 + z = 0 ⇒ z = – 3
4 – 7 + z = 0 ⇒ z = 3

Equations and Inequations Questions for SBI PO 2011 Exam

1. The age of the father is 30 years more than the son’s age. Ten years hence, the father’s age will become three times the son’s age that time. What is the son’s present age in years?  (SBI PO 2011)
(a) Eight
(b) Seven
(c) Five
(d) Cannot be determined
(e) None of these

Equations and Inequations Questions for SBI PO 2011 Answers

1. (c)  Let the son’s present age be x years. Then the father’s present age is (x + 30) years.
Father’s age after 10 years = (x + 40) years
Son’s age after 10 years = (x + 10) years
(x + 40) = 3(x + 10)
x + 40 = 3x + 30
2x = 10
∴ x = 5