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If A < 0, 10 < B < 30, and 50 < C < 80, what is the relative
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20 Nov 2017, 11:06
20 Nov 2017, 11:06
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65% (00:55) correct
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If \(A < 0\), \(10 < B < 30\), and \(50 < C < 80\), what is the relative order of the reciprocals \(\frac{1}{A}\), \(\frac{1}{B}\), and \(\frac{1}{C}\) ?
A. \(\frac{1}{A} < \frac{1}{B} < \frac{1}{C}\)
B. \(\frac{1}{A} < \frac{1}{C} < \frac{1}{B}\)
C. \(\frac{1}{C} < \frac{1}{A} < \frac{1}{B}\)
D. \(\frac{1}{C} < \frac{1}{B} < \frac{1}{A}\)
E. \(\frac{1}{B} < \frac{1}{C}< \frac{1}{A}\)
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A. \(\frac{1}{A} < \frac{1}{B} < \frac{1}{C}\)
B. \(\frac{1}{A} < \frac{1}{C} < \frac{1}{B}\)
C. \(\frac{1}{C} < \frac{1}{A} < \frac{1}{B}\)
D. \(\frac{1}{C} < \frac{1}{B} < \frac{1}{A}\)
E. \(\frac{1}{B} < \frac{1}{C}< \frac{1}{A}\)
Kudos for correct solution.
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Re: If A < 0, 10 < B < 30, and 50 < C < 80, what is the relative
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21 Nov 2017, 13:40
21 Nov 2017, 13:40
2
let us consider A=-2, B=20 and C=60.
Then 1/A= -1/2 => -0.5,
1/B=1/20 => .05,
1/C=1/60 => .016
Therefore, the order will be 1/A < 1/C < 1/B . Option (B) is the answer.
Then 1/A= -1/2 => -0.5,
1/B=1/20 => .05,
1/C=1/60 => .016
Therefore, the order will be 1/A < 1/C < 1/B . Option (B) is the answer.
Re: If A < 0, 10 < B < 30, and 50 < C < 80, what is the relative
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01 Dec 2017, 21:16
01 Dec 2017, 21:16
2
Bunuel wrote:
If A < 0, 10 < B < 30, and 50 < C < 80, what is the relative order of the reciprocals 1/A, 1/B, and 1/C?
A. 1/A < 1/B < 1/C
B. 1/A < 1/C < 1/B
C. 1/C < 1/A < 1/B
D. 1/C < 1/B < 1/A
E. 1/B < 1/C < 1/A
A. 1/A < 1/B < 1/C
B. 1/A < 1/C < 1/B
C. 1/C < 1/A < 1/B
D. 1/C < 1/B < 1/A
E. 1/B < 1/C < 1/A
Here a quick thought - In fraction, the larger the denominator the smaller the value
Here A = will always be negative so is the smallest.
For B = it will range from 11 to 29 ,
whereas for C= it will range from 51 to 79.
Hence \(\frac{1}{A} < \frac{1}{C} < \frac{1}{B}\)
