## Equation Problems To Prepare For Bank Exams

Dear Reader,

Below are four problems on miscellaneous topics, you have to solve the given quadratic or linear equations.

Below are four problems on miscellaneous topics, you have to solve the given quadratic or linear equations.

Directions to solve:

In each of the following questions two equations numbered I and II are given. You have to solve both the equations and give answer if

(a) x = y

(b) x > y

(c) x < y

(d) x > = y

(e) x < = y

(a) x = y

(b) x > y

(c) x < y

(d) x > = y

(e) x < = y

**Question 1**

I. x - 7 = 0

II. 3y

II. 3y

^{2}- 10y + 7 = 0**Answer :**(b) x > y

Solution :

From I, x - 7 = 0

x = 7 ....(1)

x = 7 ....(1)

From II, 3y

3y

(3y - 7)(y - 1) = 0

y = 1 or y = 7/3 ....(2)

^{2}- 10y + 7 = 03y

^{2}- 3y - 7y + 7 = 0(3y - 7)(y - 1) = 0

y = 1 or y = 7/3 ....(2)

From (1) and (2), we have x > y.

Hence the answer is option b.

Hence the answer is option b.

**Question 2**

I.4y

II.x

^{2}+ 8y = 4y + 8II.x

^{2}+ 9p = 2p - 12
Answer : (c) x < y

Solution:

From I, 4y

y

y

(y - 1)(y + 2) = 0

y = 1 or y = -2 ...(1)

^{2}+ 8y = 4y + 8y

^{2}+ 2y = y + 2y

^{2}+ 2y - y - 2 = 0(y - 1)(y + 2) = 0

y = 1 or y = -2 ...(1)

From II, x

x

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

x = -4 or -3 ....(2)

^{2}+ 9p = 2p - 12x

^{2}+ 7p + 12 = 0(x + 4)(x + 3) = 0

x = -4 or -3 ....(2)

From (1) and (2), we have y > x.

Hence the answer is option c.

Hence the answer is option c.

**Question 3**

I. 2x

II. y

^{2}+ 40 = 18xII. y

^{2}= 13y + 42**Answer :**option c.

Solution :

From I, 2x^2 + 40 = 18x

2x

x

(x - 4)(x - 5) = 0

x = 4 or 5 ...(1)

2x

^{2}+ 40 - 18x = 0x

^{2}- 9x - 20 = 0(x - 4)(x - 5) = 0

x = 4 or 5 ...(1)

From II, y

y

(y - 7)(y - 6) = 0

y = 7 or y = 6 ...(2)

^{2}= 13y + 42y

^{2}- 13y - 42 = 0(y - 7)(y - 6) = 0

y = 7 or y = 6 ...(2)

From (1) and (2), we have y > x.

Hence the answer is option c.

Hence the answer is option c.

**Question 4**

I. 4x

II. y

^{2}= 16II. y

^{2}- 10y + 25 = 0**Answer :**option c.

Solution :

From I, 4x

x

x = sqrt4 = 2 or -2 ...(1)

^{2}= 16x

^{2}= 4x = sqrt4 = 2 or -2 ...(1)

From II, y

y

(y - 5)(y - 5) = 0

y = 5 ...(2)

^{2}- 10y + 25 = 0y

^{2}- 5y - 5y + 25 = 0(y - 5)(y - 5) = 0

y = 5 ...(2)

From (1) and (2), we have y > x.

Hence the answer is option c.

Hence the answer is option c.