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Re: What is the value of (21*x^2*y)(5*y^2*x)? [#permalink]
shasadou wrote:
What is the value of \(\frac{(21*x^2*y)}{(5*y^2*x)}\)?

(1) x^2*y^2=16

(2) x+y=5


What is \(\frac{21x}{5y}\)?

(1) \(x^2y^2 = 16\)
\(xy = 4\)

We can't determine x or y individually; INSUFFICIENT.

(2) \(x+y = 5\)

Clearly insufficient.

(1&2) Combined, we have \(xy = 4\) and \(x + y = 5\). INSUFFICIENT.

Answer is E.
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Re: What is the value of (21*x^2*y)(5*y^2*x)? [#permalink]
The given fraction reduces to:
\(\frac{21x}{5y}\)
We need to find a value of x/y.

1) \(x^2y^2=16\)
Many possible cases.
Consider x=4, y=1 or the inverse case of y=4, x=1. Both yield different values.

2) similar case here. Both have many combinations

3)
knowing that x+y=5
x=5-y
\((5-y)^2y^2=16\)
A possible solution is y=1, x=4
However, because we do not know which of the values of x or y is larger, we cannot exclude the inverse case of x=1, y=4.

INS.
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Re: What is the value of (21*x^2*y)(5*y^2*x)? [#permalink]
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