adwaitkasbekar wrote:If x > y^2 > z^4, which of the following could be true?
x > y > z
z > y > x
x > z > y
a. I
b. I and II
c. I and III
d. II and III
e. I, II, and III
This problem states "could be scenario". Kindly anyone suggest how to tackle such problem.
OA is [spoiler]E
[/spoiler]
Solution:
The first thing to notice in this problem is that we are being asked which of the following
could be true. This means that we can create different scenarios for the values of x, y, and z, to determine whether each Roman numeral could be true. Another thing to consider is that any time we are presented with a problem with inequalities and exponents, there is a high likelihood that we are being tested on the way in which different types of numbers react to being raised to positive integer exponents. For example, when a number greater than 1 is raised to an integer exponent greater than 1, the resulting value increases (e.g., 32 > 3). When a number between 0 and 1 (i.e., a positive proper fraction) is raised to an integer exponent greater than 1, the resulting value decreases (e.g., (1/3)^2 < 1/3). Let's keep that in mind as we test convenient numbers for each Roman numeral. Remember, too, that whatever convenient numbers we choose to use must fulfill the statement:
x > y² > z�.
I. x > y > z
Notice that the order of arrangement of x, y, and z in the inequality x > y > z is the same as the order of arrangement of x, y^2, and z^4 in the inequality x > y^2 > z^4, so we want to test positive integers in this case.
x = 10
y = 3
z = 1
Notice that 10 > 3 > 1 AND 10 > 9 > 1
We see that I could be true.
II. z > y > x
Notice that the order of arrangement of x, y, and z in the inequality z > y > x differs from the order of arrangement of x, y^2, and z^4 in the inequality x > y^2 > z^4, so we want to test positive proper fractions in this case. This is because we need to decrease the value of y and z to make it work within the given inequality.
x = 1/5
y = 1/3
z = ½
Notice that ½ > 1/3 > 1/5 AND 1/5 > 1/9 > 1/16.
We see that II could be true.
III. x > z > y
Notice that the order of arrangement of y and z in the inequality x > z > y differs from the order of arrangement of y^2 and z^4 in the inequality x > y^2 > z^4, so we once again want to test positive proper fractions in this case. This is because we need to decrease the value of z to make it work within the given inequality (that is, we want to swap the order of z^4 and y^2 even if z > y).
x = 1/2
y = 1/4
z = 1/3
Notice that ½ > 1/3 > 1/4 AND ½ > 1/16 > 1/81.
We see that III could be true.
The answer is
E
