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20 Oct 2022, 13:13
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6. If \(x + y ≠ 0\), is \(\frac{1}{(x + y)} < 2\)?
(1) \(x^3 = y^3\)
(2) \(\frac{1}{x} < 2\)
Solution:
(1) \(x^3 = y^3\)
=> x = y
If x = y = 2: YES
If x = y = 1/10: NO
Not sufficient
(2) \(\frac{1}{x} < 2\)
This does not give us any information about y
Not sufficient
TOGETHER: Both statements combined:
\(x = y\) and \(\frac{1}{x} < 2\) => \(\frac{1}{y} < 2\)
=> \( x = y > \frac{1}{2}\) or \(x = y < 0\)
In any possible case, \(\frac{1}{(x + y)} < 2\) ALWAYS
SUFFICIENT
Answer: C
(1) \(x^3 = y^3\)
(2) \(\frac{1}{x} < 2\)
Solution:
(1) \(x^3 = y^3\)
=> x = y
If x = y = 2: YES
If x = y = 1/10: NO
Not sufficient
(2) \(\frac{1}{x} < 2\)
This does not give us any information about y
Not sufficient
TOGETHER: Both statements combined:
\(x = y\) and \(\frac{1}{x} < 2\) => \(\frac{1}{y} < 2\)
=> \( x = y > \frac{1}{2}\) or \(x = y < 0\)
In any possible case, \(\frac{1}{(x + y)} < 2\) ALWAYS
SUFFICIENT
Answer: C
20 Oct 2022, 13:15
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7. Is \(x = 0\) ?
(1) \(x^2*y^3 > 0\)
(2) \(x^3*y^2 < 0\)
Solution:
(1) \(x^2*y^3 > 0\)
If the expression is not equal to 0, it means that NONE of the terms can be equal to 0 => x NOT EQUAL TO 0
SUFFICIENT
(2) \(x^3*y^2 < 0\)
If the expression is not equal to 0, it means that NONE of the terms can be equal to 0 => x NOT EQUAL TO 0
SUFFICIENT
Answer: D
(1) \(x^2*y^3 > 0\)
(2) \(x^3*y^2 < 0\)
Solution:
(1) \(x^2*y^3 > 0\)
If the expression is not equal to 0, it means that NONE of the terms can be equal to 0 => x NOT EQUAL TO 0
SUFFICIENT
(2) \(x^3*y^2 < 0\)
If the expression is not equal to 0, it means that NONE of the terms can be equal to 0 => x NOT EQUAL TO 0
SUFFICIENT
Answer: D
20 Oct 2022, 13:26
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8. Is \(x > y\)?
(1) \(x + y > 0\)
(2) \(y^2 > x^2\)
Solution:
(1) \(x + y > 0\)
\(x > -y\)
If \(y = 3: x > -3\): If x = -2 NO, If x = 4 YES
Not sufficient
(2) \(y^2 > x^2\)
\(x^2 - y^2 < 0\)
\((x - y)(x + y) < 0\)
Either \(-y < x < y\) OR \(x > y & x < -y\)
In one case, x > y, in other it is not
Not Sufficient
Together: Both statements combined
From statement 2: Either \(-y < x < y\) OR \(x > y & x < -y\)
From statement 1: \(x > -y\)
=> \(-y < x < y\)
SUFFICIENT
Answer: C
(1) \(x + y > 0\)
(2) \(y^2 > x^2\)
Solution:
(1) \(x + y > 0\)
\(x > -y\)
If \(y = 3: x > -3\): If x = -2 NO, If x = 4 YES
Not sufficient
(2) \(y^2 > x^2\)
\(x^2 - y^2 < 0\)
\((x - y)(x + y) < 0\)
Either \(-y < x < y\) OR \(x > y & x < -y\)
In one case, x > y, in other it is not
Not Sufficient
Together: Both statements combined
From statement 2: Either \(-y < x < y\) OR \(x > y & x < -y\)
From statement 1: \(x > -y\)
=> \(-y < x < y\)
SUFFICIENT
Answer: C
are there explanations on how to solve these? 
