XAT 2021 Question 20

Question 20

Find z, if it is known that:
a: $$-y^2 + x^2 = 20$$
b: $$y^3 - 2x^2 - 4z \geq -12$$ and
c: x, y and z are all positive integers

Solution

Since $$x^2-y^2=20$$ and x,y,z are positive integers, 

(x+y)*(x-y) = 20, Hence x-y, x+y are factors of 20.

Since x, y are positive integers, x+y is always positive, and for the product of (x+y)*(x-y)  to be positive x-y must be positive.

x, y are positive integers and x-y is positive x must be greater than y.

The possible cases are : (x+y = 10, x-y = 2), (x+y = 5, x-y = 4).

The second case fails because we get x =9/2, y = 1/2 but x, y are integral values

For case one x = 6, y = 4.

$$y^3 - 2x^2 - 4z \geq -12$$

Substituting  the values of x and y, we have :

64 - 72 - 4*z $$\ge\ -12$$

 -8 - 4*z $$\ge\ -12$$

z$$\le\ 1$$

Since x, y, z are positive integers, the only possible value for z is 1.


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