## Equations Based Bank Questions

Below are three problems of equations, you have to determine the relationship by solving the given equations.

**Directions to solve:**

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

a) x > y

b) x < y

C) x = y

d) relationship cannot be determined

**Question 1**

I. (25/x)^{(1/2)} - (16/x)^{(1/2)} = x^{(1/2)}

II. y^{(8/2)} = 2^{(9/2)} / y^{(1/2)}

Options : a, b, c, d

**Answer :** b)x < y

**Solution:**

Simplify eqn1 as follows:

[25^{(1/2)}]/[x^{(1/2)}] - [16^{(1/2)}]/[x^{(1/2)}] =x^{(1/2)}

[5/sqrt x ] - [4/sqrt x ] = sqrt x

5 - 4 = (sqrt x )(sqrt x) = x

x = 1.

Simplify eqn2 as follows:

y^{4} = 2^{(9/2)} / y^{(1/2)}

(y^{4})( y^{(1/2)})= 2^{(9/2)}

y^{(9/2)} = 2^{(9/2)}

y = 2 (cancelling the power on both sides)

Thus x = 1, y = 2

Hence x < y.

**Question 2**

I. 5x+y = 32

II. 7x-2y= 21

Options : a, b, c, d

**Answer :** b)x

**Solution:**

Multiply eqn 1 by 2, we have 10x+2y = 64

add above eqn to eqn2, we have 17x = 85 ==> x = 5

from eqn1, y = 32 - 25 = 7

Then obviously x < y

**Question 3**

I. [x^{(11/2)} - 2(5^{(11/2)})]/y^{10} = - [ x^{(11/2)} / sqrt(y^{20}) ]

II. [y^{(11/2)} - 2(5^{(11/2)})]/x^{10} = - [ y^{(11/2)} / sqrt(x^{20}) ]

Options : a, b, c, d

**Answer :** c)x=y

**Solution :**

Since sqrt(y^{20}) = y^{10} then the eqn1 becomes,

[x^{(11/2)} - 2(5^{(11/2)})]/y^{10} = - [ x^{(11/2)} / y^{10} ]

[x^{(11/2)} - 2(5^{(11/2)})] = - [x^{(11/2)} ] (by cancelling the denominator on both sides)

[x^{(11/2)} + x^{(11/2)}] = 2(5^{(11/2)})

2[x^{(11/2)}] = 2(5^{(11/2)})

[x^{(11/2)}] = 5^{(11/2)}

x = 5 (cancelling the power on both sides)

Similarly, from the 2nd eqn we have y=5 (since eqn2 has y instead of x and x instead of y)

Then x = y