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04 Jul 2025, 10:19
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For this one I tried some numbers.
For (I):
abc > 0?
Yes, it’s possible if a and b are negative while c is positive.
Example: a = -10, b = -5, c = 1
a < b < c
a^4 > b^4 > c^4
abc > 0
So, statement (I) is possible.
For (II):
a^3*b^5*c^7 < 0
Yes, it’s possible with all values lower than 0.
Example: a = -10, b = -5, c= -1
a < b < c
a^4 > b^4 > c^4
a^3*b^5*c^7 < 0
So, statement (II) is possible.
For (III):
a^2*b^4*c^6 = 0
In this case, we need at least one value equals 0.
Example: a = -10, b = -5, c = 0
a < b < c
a^4 > b^4 > c^4
a^2*b^4*c^6 = 0
So, statement (III) is possible.
Answer = E
For (I):
abc > 0?
Yes, it’s possible if a and b are negative while c is positive.
Example: a = -10, b = -5, c = 1
a < b < c
a^4 > b^4 > c^4
abc > 0
So, statement (I) is possible.
For (II):
a^3*b^5*c^7 < 0
Yes, it’s possible with all values lower than 0.
Example: a = -10, b = -5, c= -1
a < b < c
a^4 > b^4 > c^4
a^3*b^5*c^7 < 0
So, statement (II) is possible.
For (III):
a^2*b^4*c^6 = 0
In this case, we need at least one value equals 0.
Example: a = -10, b = -5, c = 0
a < b < c
a^4 > b^4 > c^4
a^2*b^4*c^6 = 0
So, statement (III) is possible.
Answer = E
04 Jul 2025, 10:29
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We are given that:
\(a<b<c\) and that, \(c^4<b^4<a^4\)
Firstly, take a square root of the latter equation; we can do this since, irrespective of whether a,b, or c are positive or negative, having a power of 4 or 2 is going to make them positive anyway. Since squares are easier to work with and this change has no impact on our answer, let's do this.
We now have,
\(a<b<c\) and \(c^2<b^2<a^2\)
Let's take cases where both of these statements are true:
Case 1:
\(a = -4, b= -3, c = -2\) ----> \(a<b<c\)
\(a = 16, b = 9, c = 4\) ----> \(c^2<b^2<a^2\)
So, having all three as negative numbers varying in magnitude works.
Case 2:
\(a = -4, b= -3, c = 2\) ----> \(a<b<c\)
\(a = 16, b = 9, c = 4\) ----> \(c^2<b^2<a^2\)
So, having both "a" and "b" as negative and "c" as positive works as well.
Case 3:
\(a = -4, b= -3, c = 0\) ----> \(a<b<c\)
\(a = 16, b = 9, c = 0\) ----> \(c^2<b^2<a^2\)
So, having both "a" and "b" as negative and "c" as 0 works as well.
Statement 1: This could be true if we have an even number of negatives among a,b, and c. This true basis our Case 2.
Statement 2: Since, all of them are raised to an odd power, the expression will be negative if we have odd number of negatives among a,b, and c. This is true basis our Case 1.
Statement 3: Since, the product is zero, this is only possible if at least one of a,b, or c is 0. This is true basis our Case 3.
All statements could be true.
So answer is E.
\(a<b<c\) and that, \(c^4<b^4<a^4\)
Firstly, take a square root of the latter equation; we can do this since, irrespective of whether a,b, or c are positive or negative, having a power of 4 or 2 is going to make them positive anyway. Since squares are easier to work with and this change has no impact on our answer, let's do this.
We now have,
\(a<b<c\) and \(c^2<b^2<a^2\)
Let's take cases where both of these statements are true:
Case 1:
\(a = -4, b= -3, c = -2\) ----> \(a<b<c\)
\(a = 16, b = 9, c = 4\) ----> \(c^2<b^2<a^2\)
So, having all three as negative numbers varying in magnitude works.
Case 2:
\(a = -4, b= -3, c = 2\) ----> \(a<b<c\)
\(a = 16, b = 9, c = 4\) ----> \(c^2<b^2<a^2\)
So, having both "a" and "b" as negative and "c" as positive works as well.
Case 3:
\(a = -4, b= -3, c = 0\) ----> \(a<b<c\)
\(a = 16, b = 9, c = 0\) ----> \(c^2<b^2<a^2\)
So, having both "a" and "b" as negative and "c" as 0 works as well.
Statement 1: This could be true if we have an even number of negatives among a,b, and c. This true basis our Case 2.
Statement 2: Since, all of them are raised to an odd power, the expression will be negative if we have odd number of negatives among a,b, and c. This is true basis our Case 1.
Statement 3: Since, the product is zero, this is only possible if at least one of a,b, or c is 0. This is true basis our Case 3.
All statements could be true.
So answer is E.
04 Jul 2025, 10:38
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Option E is the correct answer.
The catch here is "could be true". I fell for this bait and marked the wrong answer and realized it later.
prerequisites:
1. a < b < c
2. c^4 < b^4 < a^4
Case 1: a = -4 , b = -3, c =2
This satisfies the prerequisites.
Statement I : abc > 0, is true for these values.
Case 2: a=-4,b=-3,c=-2
This satisfies the prerequisites.
Statement II : a^3b^5c^7 < 0 , is true for these values.
Case3: a=-4,b=-3,c=0
This satisfies the prerequisites.
Statement III : a^2b^4c^6 = 0 , is true for these values.
So all I, II & III could be true.
The catch here is "could be true". I fell for this bait and marked the wrong answer and realized it later.
prerequisites:
1. a < b < c
2. c^4 < b^4 < a^4
Case 1: a = -4 , b = -3, c =2
This satisfies the prerequisites.
Statement I : abc > 0, is true for these values.
Case 2: a=-4,b=-3,c=-2
This satisfies the prerequisites.
Statement II : a^3b^5c^7 < 0 , is true for these values.
Case3: a=-4,b=-3,c=0
This satisfies the prerequisites.
Statement III : a^2b^4c^6 = 0 , is true for these values.
So all I, II & III could be true.
Bunuel
