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06 Oct 2022, 18:56
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5. If \(ab^2 > 0\) and \(ac < 0\), then which of the following must be true?
I. \(ab >0\)
II. \(b^2c < 0\)
III. \(a*c^2 > 0\)
A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III
Solution:
\(ab^2 > 0\) => \(a > 0\) & \(b\) could either be positive or negative
\(ac < 0\) => We already deduced \(a > 0\) from the previous inequality so here \(c < 0\)
So, we have \(a > 0, c < 0\) and \(b\) is inconclusive
I. \(ab >0\)
\(a > 0\) but \(b\) could either be + or - , so this CANNOT BE MUST BE TRUE
II. \(b^2c < 0\)
We already know that \(c < 0\) and \(b^2\) will always be positive so expression will always be < 0. MUST BE TRUE
III. \(a*c^2 > 0\)
We already know that \(a > 0, c < 0\) so \(c^2 > 0\), since both parts of the product are > 0, expression will always be > 0. MUST BE TRUE
II and III MUST BE TRUE
Answer: D
I. \(ab >0\)
II. \(b^2c < 0\)
III. \(a*c^2 > 0\)
A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III
Solution:
\(ab^2 > 0\) => \(a > 0\) & \(b\) could either be positive or negative
\(ac < 0\) => We already deduced \(a > 0\) from the previous inequality so here \(c < 0\)
So, we have \(a > 0, c < 0\) and \(b\) is inconclusive
I. \(ab >0\)
\(a > 0\) but \(b\) could either be + or - , so this CANNOT BE MUST BE TRUE
II. \(b^2c < 0\)
We already know that \(c < 0\) and \(b^2\) will always be positive so expression will always be < 0. MUST BE TRUE
III. \(a*c^2 > 0\)
We already know that \(a > 0, c < 0\) so \(c^2 > 0\), since both parts of the product are > 0, expression will always be > 0. MUST BE TRUE
II and III MUST BE TRUE
Answer: D
06 Oct 2022, 19:02
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6. What is the value of \(\sqrt{(49\%)} - \frac{7}{100}\) ?
A. 0
B. 0.63%
C. 0.063
D. 63%
E. 6.3
Solution:
\(\sqrt{(49\%)} - \frac{7}{100} = \frac{7}{10} - \frac{7}{100} = \frac{70-7}{100} = \frac{63}{100} = 63%\)
Answer: D
A. 0
B. 0.63%
C. 0.063
D. 63%
E. 6.3
Solution:
\(\sqrt{(49\%)} - \frac{7}{100} = \frac{7}{10} - \frac{7}{100} = \frac{70-7}{100} = \frac{63}{100} = 63%\)
Answer: D
06 Oct 2022, 19:08
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7. What is the value of \((12\%)^2\) ?
A. \(0.144\%\)
B. \(1.44\%\)
C. \(0.144\)
D. \(144\%\)
E. \(14.4\)
Solution:
\((12\%)^2 = \frac{12^2}{100^2} = \frac{144}{10000} =\frac{1.44}{100} = 1.44\%\)
Answer: B
A. \(0.144\%\)
B. \(1.44\%\)
C. \(0.144\)
D. \(144\%\)
E. \(14.4\)
Solution:
\((12\%)^2 = \frac{12^2}{100^2} = \frac{144}{10000} =\frac{1.44}{100} = 1.44\%\)
Answer: B
