   ## 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:
y4 = 2(9/2) / y(1/2)
(y4)( 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

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))]/y10 = - [ x(11/2) / sqrt(y20) ]
II. [y(11/2) - 2(5(11/2))]/x10 = - [ y(11/2) / sqrt(x20) ]

Options : a, b, c, d

Solution :

Since sqrt(y20) = y10 then the eqn1 becomes,

[x(11/2) - 2(5(11/2))]/y10 = - [ x(11/2) / y10 ]
[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 