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lvillalon
Joined: 29 Jun 2025
Last visit: 25 Aug 2025
Posts: 88
Given Kudos: 14
Location: Chile
Concentration: Operations, Entrepreneurship
Schools: HEC Sept '26 INSEAD Aug '26 IMD'26 Rotterdam'26 Booth '26 Sloan '26 Kellogg '26 Anderson '26
GPA: 3,3
WE:Consulting (Consulting)
Schools: HEC Sept '26 INSEAD Aug '26 IMD'26 Rotterdam'26 Booth '26 Sloan '26 Kellogg '26 Anderson '26
Posts: 88
04 Jul 2025, 18:45
Kudos
Bookmarks
E. I, II, and III
The order of numbers change, if they have all the same even exponent, whilst being negative. Example: -4<-3 -> (-4)^2=16>(-3)^2=9
Since even exponents makes every number positive (a little like absolut value), you have to be cautious about numbers between -1 and 1. You have 2 possible options:
Option 1: {a,b,c} < 0 with |a|>|b|>|c|
Option 2: {a,b} < 0 with |a|>|b| and c>=0 with |b|>c
Now you try each option:
I. True with Option 2
II. True with Option 1
III. True with Option 2
->All could be True.
The order of numbers change, if they have all the same even exponent, whilst being negative. Example: -4<-3 -> (-4)^2=16>(-3)^2=9
Since even exponents makes every number positive (a little like absolut value), you have to be cautious about numbers between -1 and 1. You have 2 possible options:
Option 1: {a,b,c} < 0 with |a|>|b|>|c|
Option 2: {a,b} < 0 with |a|>|b| and c>=0 with |b|>c
Now you try each option:
I. True with Option 2
II. True with Option 1
III. True with Option 2
->All could be True.
Bunuel
04 Jul 2025, 19:09
Kudos
Bookmarks
If \(a < b < c\), and \(c^4 < b^4 < a^4\), which of the following could be true?
Case 1: Since \(a < b < c\) & \(c^4 < b^4 < a^4\), all a, b, c are negative numbers.
I. \(abc > 0\)
abc <0;
NOT TRUE
II. \(a^3b^5c^7 < 0\)
a^3<0
b^5<0
c^7<0
\(a^3b^5c^7 < 0\)
TRUE
III. \(a^2b^4c^6 = 0\)
\(a^2b^4c^6 > 0\)
NOT TRUE
Case 2: a = -3; b = -2; c = 0; c^4=0 < b^4 = 16 < a^4 = 81
I. \(abc > 0\)
abc =0;
NOT TRUE
II. \(a^3b^5c^7 < 0\)
a^3=0
b^5<0
c^7<0
\(a^3b^5c^7 = 0\)
NOT TRUE
III. \(a^2b^4c^6 = 0\)
\(a^2b^4c^6 = 0\)
TRUE
Case 3: a = -3; b = -2; c = 1; c^4=1 < b^4 = 16 < a^4 = 81
I. \(abc > 0\)
abc= 6 > 0;
TRUE
II. \(a^3b^5c^7 < 0\)
a^3>0
b^5<0
c^7<0
\(a^3b^5c^7 < 0\)
TRUE
III. \(a^2b^4c^6 = 0\)
\(a^2b^4c^6 > 0\)
NOT TRUE
It is observed that all 3 statement could be true in specific cases.
A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III
IMO E
Case 1: Since \(a < b < c\) & \(c^4 < b^4 < a^4\), all a, b, c are negative numbers.
I. \(abc > 0\)
abc <0;
NOT TRUE
II. \(a^3b^5c^7 < 0\)
a^3<0
b^5<0
c^7<0
\(a^3b^5c^7 < 0\)
TRUE
III. \(a^2b^4c^6 = 0\)
\(a^2b^4c^6 > 0\)
NOT TRUE
Case 2: a = -3; b = -2; c = 0; c^4=0 < b^4 = 16 < a^4 = 81
I. \(abc > 0\)
abc =0;
NOT TRUE
II. \(a^3b^5c^7 < 0\)
a^3=0
b^5<0
c^7<0
\(a^3b^5c^7 = 0\)
NOT TRUE
III. \(a^2b^4c^6 = 0\)
\(a^2b^4c^6 = 0\)
TRUE
Case 3: a = -3; b = -2; c = 1; c^4=1 < b^4 = 16 < a^4 = 81
I. \(abc > 0\)
abc= 6 > 0;
TRUE
II. \(a^3b^5c^7 < 0\)
a^3>0
b^5<0
c^7<0
\(a^3b^5c^7 < 0\)
TRUE
III. \(a^2b^4c^6 = 0\)
\(a^2b^4c^6 > 0\)
NOT TRUE
It is observed that all 3 statement could be true in specific cases.
A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III
IMO E
04 Jul 2025, 19:30
Kudos
Bookmarks
Correct Answer: (Option B) Only II
If a<b<c and c^4<b^4<a^4
then a, b and c should be in negative.
let's assume a= -4, b= -3, c= -2
Now both the cases will satisfy.
Let’s take
I. abc>0
If we do -2*-3*-4= -24
-24<0
This will be false as this will be less than 0, so eliminate this option.
II. a^3b^5c^7<0
Here all the powers are in odd, so it will definitively give negative number.
a^3b^5c^7<0
So this will be true as it will be less than 0.
III. a^2b^4c^6=0
Here we will get positive number but we have multiplication between these numbers, so definitely we will get a number which is greater than 0, so eliminate this option too.
Also we cannot put a, b and c value as 0 as because it will satisfy the seconds equation.
Hence our answer is Only II is correct.
If a<b<c and c^4<b^4<a^4
then a, b and c should be in negative.
let's assume a= -4, b= -3, c= -2
Now both the cases will satisfy.
Let’s take
I. abc>0
If we do -2*-3*-4= -24
-24<0
This will be false as this will be less than 0, so eliminate this option.
II. a^3b^5c^7<0
Here all the powers are in odd, so it will definitively give negative number.
a^3b^5c^7<0
So this will be true as it will be less than 0.
III. a^2b^4c^6=0
Here we will get positive number but we have multiplication between these numbers, so definitely we will get a number which is greater than 0, so eliminate this option too.
Also we cannot put a, b and c value as 0 as because it will satisfy the seconds equation.
Hence our answer is Only II is correct.
