Solution:x²y³ = 200
Observe that 200 can be factorised as:
#1. 2³ * 5² where x = 5 and y = 2 ... (i)
However, we can also say that:
#2. \(200 = [2^\frac{1}{3}]³ * 10² \), where x = \(10\) and y = \([2^\frac{1}{3}]\) ...(ii)
Or
#3. \(200 = [200^\frac{1}{2}]² * 1³\), where x = \([200^\frac{1}{2}]\) and y = \(1\) ...(iii)
#4. \(200 = [50^\frac{1}{2}]² * [4^\frac{1}{3}]³ \), where x = \([50^\frac{1}{2}]\) and y = \([4^\frac{1}{3}]\) ...(iv)
Thus, there can be multiple ways of expressing 200.
Let's now look at the 2 statements:
St1.
y is an integerBoth (i) and (iii) are valid - Insufficient
St2.
x is an integerBoth (i) and (ii) are valid - Insufficient
Combining:
Since both x and y are integers apparently it seems that only (i) is correct.
However, the trick here is that it doesn't say x and y are
positive integers.
Thus, another possible solution is:
2³ * (-5)² where x = -5 and y = 2
Thus, even after combining the statements, we have:
xy = 10 or -10
Hence, the [banswer is E[/b]
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