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GMAT Club Olympics 2025 (Day 5): If a < b < c, and c^4 < b^4 < a^4

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AviNFC

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1. consider -3,-2,1..satisfies.. abc>0
2. consider -4,-3,-2..satisfies.. the product will be negative.
3. consider -3,-2,0..satisfies.. product is 0

All satisfy. Ans E

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04 Jul 2025, 08:44

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given condition
If $a 0$
possible when two values are -ve and one is + or else all are +ve
the given condition is not possible at all +ve
but when c=1 , b= -2 , a = -3
a^4 = 81 , b^4 = 16 , c^4 =1
true
sufficient
#2
$a^3b^5c^7 < 0$
this can stand true
all are -ve values
a=-3, b=-2 , c =-1
sufficient
#3
$a^2b^4c^6 = 0$
this will be true only when either of a,b,c or all are 0
it cannot be true for given condition to stand when all are +ve
$c^4 < b^4 < a^4$
but when c=0, b=-2 , c=-3
c^4= 81 , b^4=16 , a^4 = 0 this will not match condition

OPTION D is correct D. I and II only

Bunuel

If $a 0$

II. $a^3b^5c^7 < 0$

III. $a^2b^4c^6 = 0$

A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III

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04 Jul 2025, 08:47

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Checking with cases
Let a= -4, b=-3, c=2.
It satisfies $a 0
So, I is possible

Let a= -4, b=-3, c=-2.
It satisfies $a < b < c$, and $c^4 < b^4 < a^4$ as 16<81<256
abc<0
Since II is using the odd powers to determine the sign, we can use abc instead and it is -ve. So, II is possible

Let a= -4, b=-3, c=0.
It satisfies $a $a^2b^4c^6 = 0$
So, III is possible

Answer E

Bunuel

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Bunuel

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If a, b and c is less than -1

Point II can hold true

If a, b and c are between 0 and 1

Point I can hold true

If c is equal to 0, point III can hold true.

Option E

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Hi,

The following solution is what I consider the fastest way to solve and not the best explanation, give Kudos if this makes sense and you can use it to solve these type of questions faster. Cheers.

The fastest way to solve would be:
Given : a<b<c -> c^even < b^even < a^even
now this can clearly happen when a,b,c are negative : Even powers making them positive and more negative they are larger they become with even power and since powers are equal so we don't have to worry about them getting bigger disproportionately.
Looking at options (II) becomes could be true. We can reject A and C.

So answer is only from B,D,E. Next I would check (III) as it can lead me to one option.
(III) says a*b*c with some powers = 0 that means one of them has to be 0, now this can again happen pulling from earlier analysis a < b < c=0, then when even powers are given to each then c=0 < b^even < a^even.
[didn't even had to check (I), and while marking I knew this can be true for numbers which are between 0 and 1, even if you dont know this this can be solved under 1.5 mins]
Hence the answer is E

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04 Jul 2025, 08:57

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We are given:

a < b < c
c^4 <b^4 <a^4

All three statements can be true

1) a=−3, b=−2, c=1 => satisfyies both conditions

2) a=−3, b=−2, c=−1 => satisfies both conditions

3) a=−3, b=−2, c=0 => both conditions

(E)

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04 Jul 2025, 09:03

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The question is "could be true", so let's take 1 case for each condition.
All cases should abide by these rules: a<b<c and a^4>b^4>c^4

1. abc > 0

Assume a=-4, b=-3, c=1. Could be true

2. a^3b^5c^7 < 0

Assume a=-4, b=-3, c=-2. Could be true

3. a^2b^4c^6 = 0

Assume a=-2, b=-1, c=0. Could be true.

Correct answer is E

Bunuel

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Bunuel

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so through the question statement we can conclude that in order to satisfy the above statements a,b,c all three are negative integers .

only II statement satisfies the conditions since all powers are odd so together it is less than 0 , so option B

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Consider the below values satisfying the equations:
a,b,c = -4,-3,-2
a,b,c = -4,-3,+2
among the both cases, the one with all negatives satisfies option 1 and 2 (as this said it could be true so yes there is a possibility)

now the 3rd option, is possible if one of a,b,c is zero: (c=0)can work as order of a,b,c is getting reversed when
like -4,-3,0

Hence all I , II and III are possible

Bunuel

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Given a<b<c and c^4 < b^4 < a^4
From c^4 < b^4 < a^4
, since even powers are involved, this implies ∣c∣<∣b∣<∣a∣.

We need to find if there are values of a,b,c that satisfy both conditions for each statement.

Possible Scenarios for a,b,c:

All negative: a<b<c<0.
Example: a=−3,b=−2,c=−1.
Check conditions:
a<b<c⟹−3<−2<−1 (True)
∣c∣<∣b∣<∣a∣⟹∣−1∣<∣−2∣<∣−3∣⟹1<2<3 (True)
c^4 < b^4 < a^4
⟹(−1)^4 < (−2)^4 < (−3)^4
⟹1<16<81 (True)
This scenario is possible.

a,b negative, c positive: a<b<0<c.
However, we also need ∣c∣<∣b∣. This means c must be a positive number smaller than the absolute value of b.
Example: a=−3,b=−2,c=1.
Check conditions:
a<b<c⟹−3<−2<1 (True)
∣c∣<∣b∣<∣a∣⟹∣1∣<∣−2∣<∣−3∣⟹1<2<3 (True)
c^4 < b^4 < a^4
⟹(1)^4 < (−2)^4 < (−3)^4
⟹1<16<81 (True)
This scenario is possible.

c=0: If c=0, then a<b<0.
Check conditions:
a<b<c⟹a<b<0 (e.g., a=−2,b=−1,c=0) (True)
∣c∣<∣b∣<∣a∣⟹∣0∣<∣−1∣<∣−2∣⟹0<1<2 (True)
c^4 < b^4 <a^4
⟹0^4 <(−1)^4 <(−2)^4
⟹0<1<16 (True)
This scenario is possible.

Now let's evaluate each statement:

I. abc>0

This would be true if there's an even number of negative values.

In Scenario 2 (a=−3,b=−2,c=1), abc=(−3)(−2)(1)=6>0.

Therefore, I could be true.

II. a^3 * b^5 * c^7 <0

The sign of an odd power is the same as the sign of the base. So, this statement has the same sign as abc.

This means a^3 * b^5 * c^7 <0 if abc<0.

In Scenario 1 (a=−3,b=−2,c=−1), abc=(−3)(−2)(−1)=−6<0.

Therefore, a^3 * b^5 * c^7 = (−3)^3 * (−2)^5 * (−1)^7 =(−27)(−32)(−1)=−864<0.

Therefore, II could be true.

III. a^2 * b^4 * c^6 =0

For this expression to be 0, at least one of a,b,c must be 0.

In Scenario 3 (a=−2,b=−1,c=0), a^2 * b^4 * c^6 = (−2)^2 * (−1)^4 *(0)^6 =4×1×0=0.

Therefore, III could be true.

Since I, II, and III all have possible scenarios where they could be true, all of them could be true.

Answer is E

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[ltr]a<b<c[/ltr]
, and
[ltr]c4<b4<a4
[/ltr]
which of the following could be true?

I.
[ltr]abc>0abc>0

abc>0,[/ltr]
[ltr]Solve by substituting values of a, b and c which satisfies the equation
Let a=-4, b=-2 and c=1
abc>0 hence could be true.[/ltr]

II.
[ltr]a3b5c7<0a3b5c7<0
Let a=-4, b=-2 and c=-1
a3b5c7 all three will be negative values and will be less than 0, could be true.
[/ltr]

III.
[ltr]a2b4c6=0a2b4c6=0
Let a=-4, b=-2 and c=0
a2b4c6=0
Could be true.

Thus answer is E[/ltr]

A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III

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i chose values that could work

a = -4
b = -2
c = 0

these satisfy the condition c^4 < b^4 < a ^4
with these values only C works

C

Bunuel

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Identify that this question is asking, " which of the following could be true? " , which means even a single satisfying solution makes the statement right.

Now,
I. $abc > 0$, If $a=-3, b=-2, c=1$, then $ c^4 < b^4 < a^4 $ is satisfied and also, the statement is satisfied. So select I,
II. $a^3b^5c^7 < 0$. Now replace $ c = -1 $, now all the above conditions are satisfied and also the Statement is satisfied. So select II also.
III. $a^2b^4c^6 = 0$, Now replace $c=0$, again all the above conditions are satisfied along with the statement. So select III also.

So answer is E.

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04 Jul 2025, 09:32

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Lets falsify the statements as per given conditions:
Condition 1. $a 0$

all 3 numbers are negative
So not true

II. $a^3b^5c^7 < 0$
This satisfies the conditions and is true

III. $a^2b^4c^6 = 0$
Clearly not true

Hence B is correct

Bunuel

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04 Jul 2025, 09:33

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Option E
Imagine a, b, c on number line

We are given:
a |b| > |c|, in this case statement-1 will be true

Statement-2 can be written as (a^2)*(b^4)*(c^6)*(abc), we know that even powers are positive so we have to show abc < 0
It is possible that a, b, c are all negative but modulus is |a| > |b| > |c|, we can also say that statement-2 is true

Statement-3 is also true
It is possible that a, b negative and c is zero but modulus is |a| > |b| > |c|
so (a^2)*(b^4)*(c^6) = 0

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I. abc>0
a = -0.4; b = -0.3; c = 0.1 satisfy

II. a^3.b^5.c^7<0
a = -0.4; b = -0.3; c = -0.2 satisfy

III. a^2.b^4.c^6=0
a = -0.4; b = -0.3; c = 0 satisfy

Answer: E

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Solution:

The question asks us could be true even if one case satisfies the condition, then it is considered to be true. Let's solve the question by considering three cases:

I. $abc > 0$

Case 1 a=-3,b=-2,c=1

$(-3)(-2)(1) >0$ True

Case 2 a=-3,b=-2,c=-1

$(-3)(-2)(-1)>0$ Not True

Case 3 a=-2,b=--1,c=0

$(-2)(-1)(0)>0 $Not True

2. $a^3b^5c^7 < 0$

Case 1 a=-3,b=-2,c=1

$(-3)^3(-2)^5(1)^7<0$ Not True

Case 2 a=-3,b=-2,c=-1

$(-3)^3(-2)^5(-1)^7<0$ True

Case 3 a=-2,b=--1,c=0

$(-2)^3(-1)^5(0)^7<0$ Not True

3. $a^2b^4c^6 = 0$

Case 1 a=-3,b=-2,c=1

$(-3)^2(-2)^4(1)^6=0$ Not True

Case 2 a=-3,b=-2,c=-1

$(-3)^2(-2)^4(-1)^6=0$ Not True

Case 3 a=-2,b=--1,c=0

$(-2)^2(-1)^4(0)^6=0$ True

Hence, Option E is the Correct one

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Given a<b<c & c^4 < b^4 < a^4

Above is be possible when all are negative
Also possible when it is in +ve decimal form, and raising it to the power of 4 will decrease it.
Also possible when a is negative but highest absolute number and B and C is positive number and b<c

Since the question asks the following could be true, so if we get any case for which it is true, then that option is correct

Let's evaluate

(1) ABC>0 : It is possible when all are in positive decimal form . Eg c=0.5, b=0.3, a =0.2 so a<b<c and c^4 < b^4 < a^4 also it's product is more than 0 so it is true

(2) a^3b^5c^7 < 0: It is possible when all numbers are negative and odd power keeps the sign as it is and Eg a= -8, b=-6, c=-5 and it satisfies the given condition in the question and the value of a^3b^5c^7 is less than 0

(3)a^2b^4c^6 =0: It is possible when we take a and b value negative but absolute value of a>absolute value of b and c =0 eg: a=-8 , b =-6, c=0
This is also true

So the answer is E

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The given inequalities are `a < b < c` and `c^4 < b^4 < a^4`. The first inequality establishes the order of the numbers on the number line. The second inequality involves an even exponent, which means it provides information about the absolute values of the variables. Taking the fourth root of the second inequality yields `|c| < |b| < |a|`. This means `c` is the closest to zero, and `a` is the farthest from zero. For `a` to be the smallest number (`a < c`) while also being the farthest from zero (`|a| > |c|`), `a` must be negative. Similarly, for `c` to be greater than `b` (`b < c`) while being closer to zero (`|c| < |b|`), `b` must also be negative. Therefore, the conditions require that `a < b < 0`. The variable `c` must satisfy `b < c` and `|c| < |b|`, which allows `c` to be negative, zero, or positive.

Each of the three statements can be proven true by finding a single valid example that satisfies the initial conditions.

Statement I, `abc > 0`, could be true. This requires an even number of negative factors. Since `a` and `b` must be negative, `c` must be positive. The set of values `a = -3, b = -2, c = 1` satisfies the conditions: `-3 < -2 < 1` is true, and `1^4 < (-2)^4 < (-3)^4` (or `1 < 16 < 81`) is true. For these values, `abc = 6`, which is greater than 0.

Statement II, `a^3 * b^5 * c^7 < 0`, could be true. The odd exponents preserve the sign of the base, so this condition is equivalent to `abc < 0`. This requires an odd number of negative factors. Since `a` and `b` are negative, `c` must also be negative. The set of values `a = -3, b = -2, c = -1` satisfies the conditions: `-3 < -2 < -1` is true, and `(-1)^4 < (-2)^4 < (-3)^4` (or `1 < 16 < 81`) is true. For these values, the product is negative.

Statement III, `a^2 * b^4 * c^6 = 0`, could be true. This requires at least one variable to be zero. The condition `c^4 < b^4 < a^4` means `a` and `b` cannot be zero. However, `c` can be zero. The set of values `a = -2, b = -1, c = 0` satisfies the conditions: `-2 < -1 < 0` is true, and `0^4 < (-1)^4 < (-2)^4` (or `0 < 1 < 16`) is true. If `c = 0`, the expression `a^2 * b^4 * c^6` equals 0.

Since valid examples exist for all three statements, all three could be true. The correct answer is (E).

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Imagine three numbers sitting in a row so that the first is the smallest, the second is in the middle, and the third is the largest. Now supose that, when you compare how big they each are in absolute value, the first is farthest from zero, the second is in between, and the third is closest to zero. The only way both of those facts can be true is if the first two numbers are negative (and quite large in magnitude) while the third sits nearer to zero (it could even be positive).
In that situation, if you multiply all three toether, the two negatives make a positive, and multiplying by the third keeps it positive so it’s perfectly possible for their product to be positive. Likewise, if you gave each number an odd power before multiplying, you’d end up with the same sign as the simple product, and by choosing a version where all three remain ngative you could make that odd‐powered product negative. Finally, because none of the three numbers is actually zero, there’s no way any combination of powers could give exactly zero. Thus only the first two claims could possibly hold.

Bunuel

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