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If a/b>c/d>1, which of the following could be true about the
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23 May 2020, 01:55
23 May 2020, 01:55
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If \(\frac{a}{b}>\frac{c}{d}>1\), which of the following could be true about the reciprocals of a,b,and c ?
A. \(\frac{1}{a}>\frac{1}{b}>0>\frac{1}{c}>\frac{1}{d}\)
B. \(\frac{1}{b}>\frac{1}{a}>0>\frac{1}{c}>\frac{1}{d}\)
C. \(\frac{1}{b}>\frac{1}{a}>0>\frac{1}{d}>\frac{1}{c}\)
D. \(\frac{1}{c}>\frac{1}{a}>0>\frac{1}{b}>\frac{1}{d}\)
E. \(\frac{1}{d}>\frac{1}{c}>0>\frac{1}{b}>\frac{1}{a}\)
A. \(\frac{1}{a}>\frac{1}{b}>0>\frac{1}{c}>\frac{1}{d}\)
B. \(\frac{1}{b}>\frac{1}{a}>0>\frac{1}{c}>\frac{1}{d}\)
C. \(\frac{1}{b}>\frac{1}{a}>0>\frac{1}{d}>\frac{1}{c}\)
D. \(\frac{1}{c}>\frac{1}{a}>0>\frac{1}{b}>\frac{1}{d}\)
E. \(\frac{1}{d}>\frac{1}{c}>0>\frac{1}{b}>\frac{1}{a}\)
Re: If a/b>c/d>1, which of the following could be true about the
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23 May 2020, 23:15
23 May 2020, 23:15
1
After viewing the options, we can see that c & d must be negative (first three option where c & d are less than 0).
Thus, select numbers accordingly. Let a = 15, b = 3, c = -12, d = -6. This satisfies the first condition as well.
Now on calculating the reciprocals, we can see that option B is perfect fit.
Thus, select numbers accordingly. Let a = 15, b = 3, c = -12, d = -6. This satisfies the first condition as well.
Now on calculating the reciprocals, we can see that option B is perfect fit.
Re: If a/b>c/d>1, which of the following could be true about the
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19 Aug 2020, 08:27
19 Aug 2020, 08:27
1
I have another suggestion, building on what sukrut was saying it might be clearer.
We could proceed by elimination.
The quickest thing we can say is that D isn't possible because if it were true, c and d (or a and b) would be of different signs, the ratio cannot be >1. --> eliminate D.
Then we look at answer A.
if \(\frac{1}{a}\) and \(\frac{1}{b}\) are positive this means that a,b are positive. But if a,b are positive and \(\frac{1}{a} > \frac{1}{b}\) this means \(\frac{b}{a}>1\) which is a contradiction --> eliminate A.
Answer B isn't ruled out so far.
Answer C is similar to B, so to see which one applies, we can check whether it's possible to have that \(0> \frac{1}{d} > \frac{1}{c}\).
If it were the case we would have \(d,c <0\) and thus \(0>c>d\) which would mean that \(\frac{c}{d} <1\) (take 0 > c=-2 > d=-4 to convince yourself). --> eliminate C.
And similarly we can also eliminate E because b is the analogous of d and a the analogous of c in this last assertion so the argument above still works. --> eliminate E.
So B could work and we could plug in the numbers by sukrut to complete the answer.
We could proceed by elimination.
The quickest thing we can say is that D isn't possible because if it were true, c and d (or a and b) would be of different signs, the ratio cannot be >1. --> eliminate D.
Then we look at answer A.
if \(\frac{1}{a}\) and \(\frac{1}{b}\) are positive this means that a,b are positive. But if a,b are positive and \(\frac{1}{a} > \frac{1}{b}\) this means \(\frac{b}{a}>1\) which is a contradiction --> eliminate A.
Answer B isn't ruled out so far.
Answer C is similar to B, so to see which one applies, we can check whether it's possible to have that \(0> \frac{1}{d} > \frac{1}{c}\).
If it were the case we would have \(d,c <0\) and thus \(0>c>d\) which would mean that \(\frac{c}{d} <1\) (take 0 > c=-2 > d=-4 to convince yourself). --> eliminate C.
And similarly we can also eliminate E because b is the analogous of d and a the analogous of c in this last assertion so the argument above still works. --> eliminate E.
So B could work and we could plug in the numbers by sukrut to complete the answer.
