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30 Sep 2021, 02:15
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
66% (01:38) correct
34%
(01:41)
wrong
based on 530
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If \(\frac{a}{b}>\frac{c}{d}\), and none of a, b, c, and d is equal to 0, which of the following must be true?
I. \(- \frac{a}{b }< \frac{c}{d}\)
II. \(| \frac{a}{b} | > | \frac{c}{d}|\)
III. \(- \frac{a}{b} < - \frac{c}{d}\)
A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III
I. \(- \frac{a}{b }< \frac{c}{d}\)
II. \(| \frac{a}{b} | > | \frac{c}{d}|\)
III. \(- \frac{a}{b} < - \frac{c}{d}\)
A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III
30 Sep 2021, 04:03
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This question might scare us a bit because of the number of variables and the fact that there are fractions. But in fact, we can just consider two variables: a/b (as, say, x) and c/d (as, say, y).
Translating: x > y and x and y are different from 0.
Let's test the statements while picking numbers for x and y. Don't forget to test for negative values in these types of questions (inequalities and/or absolute values). I usually do this right away since most of the statements are true with positive values but are quickly denied when testing negative values.
I. Translating to x and y: -x<y.
Say x=-2 and y=-3. This satisfies x>y but not -x<y (2 is not less than -3). Therefore I is not necessarily true.
II. Translating to x and y: |x|>|y|.
Again, let's test for x=-2 and y=-3. |x| = 2 and |y| = 3. This means II is not necessarily true.
III. Translating to x and y: -x<-y.
Again, let's test for x=-2 and y=-3. -x=2 and -y=3. It satisfies III.
Now let's test for x=4 and y=3. -x=-4 and -y=-3. It satisfies III.
Last combination: x=2 and y=-2. -x=-2 and -y=2. It satisfies III.
Only III is always true. Answer C
Translating: x > y and x and y are different from 0.
Let's test the statements while picking numbers for x and y. Don't forget to test for negative values in these types of questions (inequalities and/or absolute values). I usually do this right away since most of the statements are true with positive values but are quickly denied when testing negative values.
I. Translating to x and y: -x<y.
Say x=-2 and y=-3. This satisfies x>y but not -x<y (2 is not less than -3). Therefore I is not necessarily true.
II. Translating to x and y: |x|>|y|.
Again, let's test for x=-2 and y=-3. |x| = 2 and |y| = 3. This means II is not necessarily true.
III. Translating to x and y: -x<-y.
Again, let's test for x=-2 and y=-3. -x=2 and -y=3. It satisfies III.
Now let's test for x=4 and y=3. -x=-4 and -y=-3. It satisfies III.
Last combination: x=2 and y=-2. -x=-2 and -y=2. It satisfies III.
Only III is always true. Answer C
General Discussion
13 Oct 2024, 20:46
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Given that \(\frac{a}{b}>\frac{c}{d}\), and none of a, b, c, and d is equal to 0, And we need to find which of the following must be true?
To make it easier to understand the solution and to solve the problem lets assume that a is a multiple of b and c is a multiple of d
=> a = m*b (where m is an integer)
=> c = n*d (where n is an integer)
=> \(\frac{a}{b}\) = \(\frac{mb}{b}\) = m
=> \(\frac{c}{d}\) = \(\frac{nd}{d}\) = n
I. \(- \frac{a}{b }< \frac{c}{d}\)
=> -m > n
Now, this does not need to be true
Case 1: TRUE when BOTH are negative, m > n and m = -3, n = -4
=> -m = 3 > -4 (n)
Case 2: TRUE when m is positive and n is negative and |m| < |n|, m > n and m = 3, n = -4
=> -m = -3 > -4 (n)
Case 3: FALSE when BOTH are positive , m > n and m= 4, n = 3
=> -m = -4 < 3 (n)
Case 4: FALSE when m is positive and n is negative and |m| > |n|, m > n and m = 5, n = -4
=> -m = -5 < -4 (n)
=> This MUST NOT be TRUE
II. \(| \frac{a}{b} | > | \frac{c}{d}|\)
=> |m| > |n|
TRUE in case 3 and 4 above and FALSE in case 1 and 2
=> This [color=#c10300]MUST NOT be TRUE[/color]
III. \(- \frac{a}{b} < - \frac{c}{d}\)
=> -m < -n
Now, we know that m > n
Multiplying with -1 on both the sides (and changing sign of the inequality) we get
-m < -n
=> This MUST BE TRUE
So, Answer will be C
Hope it helps!
Watch the following video to MASTER Inequalities
To make it easier to understand the solution and to solve the problem lets assume that a is a multiple of b and c is a multiple of d
=> a = m*b (where m is an integer)
=> c = n*d (where n is an integer)
=> \(\frac{a}{b}\) = \(\frac{mb}{b}\) = m
=> \(\frac{c}{d}\) = \(\frac{nd}{d}\) = n
I. \(- \frac{a}{b }< \frac{c}{d}\)
=> -m > n
Now, this does not need to be true
Case 1: TRUE when BOTH are negative, m > n and m = -3, n = -4
=> -m = 3 > -4 (n)
Case 2: TRUE when m is positive and n is negative and |m| < |n|, m > n and m = 3, n = -4
=> -m = -3 > -4 (n)
Case 3: FALSE when BOTH are positive , m > n and m= 4, n = 3
=> -m = -4 < 3 (n)
Case 4: FALSE when m is positive and n is negative and |m| > |n|, m > n and m = 5, n = -4
=> -m = -5 < -4 (n)
=> This MUST NOT be TRUE
II. \(| \frac{a}{b} | > | \frac{c}{d}|\)
=> |m| > |n|
TRUE in case 3 and 4 above and FALSE in case 1 and 2
=> This [color=#c10300]MUST NOT be TRUE[/color]
III. \(- \frac{a}{b} < - \frac{c}{d}\)
=> -m < -n
Now, we know that m > n
Multiplying with -1 on both the sides (and changing sign of the inequality) we get
-m < -n
=> This MUST BE TRUE
So, Answer will be C
Hope it helps!
Watch the following video to MASTER Inequalities
