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

1 2 3 4 5 6

UfuomaOh

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05 Jul 2025, 00:56

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

The second expression is raised to power 4 which means that C to the power 4 and B to the power 4 as well as C to the power 4 are all positive.

However when raised to the power of 1 the order of magnitude reverses. So it can only mean that a is less than 0 and C is closest to zero than either a or b. In other words a, b and c are negative.

Option I abc> 0

It means negative x negative x negative is positive. However, this is not true. So this leaves this option as incorrect

Option II a3b5c7<0

given that a, b and c are negative and the power of a, b and c is odd. It means
a3, b5 and c7 are all negative. So negative x negative x negative is negative which is less than 0. so this option is true

Option III
a2b4c6= 0

For this to be true, it means that either a , b or c is 0. But since a<b<c and
c4<b4<a4. It means a cannot be zero, as the order of magnitude reverses for negative numbers.

Lets say a= -16 b=-4 and c=-2

Then of course c4<b4<a4. a cannot be zero because then a would be greater than b and c as we have established that a, b and c are negative numbers. This is also apparent for b. However c is the largest of the value and is closer to zero. so if we assume a=-16, b=-4 and c =0. Then a<b<c and C4<b4<a4. And it is possible that a2b4c6= 0 because -16*-4*0=0

We do not have option II and III. and given that option II is more likely, the answer is thus option II. which is B

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the answer is B. The other options are wrong . only the second option holds true

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Given: a < b < c and c4 < b4 < a4
All three numbers must be negative!
Because if they were positive, a < b < c would mean a4 < b4 < c4, but we're told c4 < b4 < a4.

Example: a = -6, b = -4, c = -2

Check: -6 < -4 < -2
Check: (-2)4 < (-4)4 < (-6)4 = 16 < 256 < 1296

Now test each option:
I. abc > 0?
negative × negative × negative = negative So abc < 0.
FALSE

II. a3b5c7 < 0?
negative × negative × negative = negative < 0.
TRUE

III. a2b4c6 = 0?
positive × positive × positive = positive ≠ 0.
FALSE

Only II works.
Answer: B

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If a<b<c
�
<
�
<
�
, and c4<b4<a4

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

Since it is a "could be true" question, we can check with only 1 solution if the options come out to be true.

I. $abc > 0$

Let a=(-3), b=(-2), c=1
-3<-2<1 & 1^4<(-2)^4<(-3)^4

Further abc = (-3)*(-2)*(1) = 6 > 1. Hence, this could be true.

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

Let a=(-3), b=(-2), c=(-1) [a<b<c & c^4<b^4<a^4]

a^3*b^5*c^7 = (-3)^3*(-2)^5*(-1)^7 = -64 < 0. Hence, this could be true.

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

Let a=(-3), b=(-2), c=0 [a<b<c & c^4<b^4<a^4]

a^2b^4c^6 = (-3)^2*(-2)^4*(0)^6 = 0. Hence, this could be true.

Answer: E. I, II, and III.

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We have a |b| > |c|
So a and b have to be negative (they are the smallest numbers but have the biggest sizes). c can be -ve, 0, or +ve.
Let's try some values
• a = –3, b = –2, c = 1. Then abc = (–3)(–2)(1) = 6 > 0, so I can happen.
• a = –3, b = –2, c = –1. Then (a^3*b^5*c^7) = (–3)^3 * (–2)^5 * (–1)^7 = –864 < 0, so II can happen.
• a = –3, b = –2, c = 0. Then (a^2*b^4*c^6) = (–3)^2 * (–2)^4 * (0)^6 = 0, so III can happen.
All three statements are possible, so the answer is E.

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Constraints : $a 0$

Suppose, $a= -3, b = -2, c = 1$, then $abc = -3*-2*1= 6 > 0$.

then $-3 <-2 < 1 $
and $(-3)^4 = 81, (-2)^4 = 16, 1^4 = 1 ---> 1^4 < (-2)^4 < (-3)^4 $

This satisfies the constraints given to us. So we have an example where $abc>0$ and the constraints are followed.

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

Suppose,$ a= -3, b = -2, c = -1$, then $a^3b^5c^7 < 0$ will be less than 0 since we have three negative numbers raised to odd powers so -ve * -ve*-ve will give a -ve number ehich is <0
also $-3 <-2 < -1 $
and $(-3)^4 = 81, (-2)^4 = 16, (-1)^4 = 1 ---> (-1)^4 < (-2)^4 < (-3)^4 $

This satisfies the constraints given to us. So we have an example where $a^3b^5c^7 < 0$ and the constraints are followed.

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

Suppose, $a= -3, b = -2, c = 0$, then $a^2b^4c^6= (-3)^2 * (-2)^4 * (0)^6 = 0 $

also $-3 <-2 < 0 $
and $(-3)^4 = 81, (-2)^4 = 16, 0^4 = 0 ---> [m] 0^4 < (-2)^4 < (-3)^4 $

This satisfies the constraints given to us. So we have an example where $a^2b^4c^6 = 0$ and the constraints are followed.

So all three statements can be satisfied using certain examples..hence answer is E , I, II, and III

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05 Jul 2025, 04:31

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Try to draw a number line, keep a, b < 0 and keep the c flexible to check different options.

I. abc>0 : Let's take a=-4, b=-2, c=1 , it satisfies.

II. a3b5c7<0 : Let's take a=-4, b=-2, c=-1 , it satisfies.

III. a2b4c6=0 : Let's take a=-4, b=-2, c=0, it satisfies.

E

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If a,b & c are three numbers & if we assume c to be zero & a & b to be negative numbers then a^2*b^4*c^6 can be equal to 0.

A is equal to -6, B is equal to -5 & C is equal to -1. Hence a^3*b^5*c^7 will be negative

A is equal to -6 & B is equal to -5 & C is equal to 1. Hence a*b*c >0.

Hence all three conditions are possible & hence the correct answer is option (E)

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The question asks which of these could be true. Here the relationship is reversed totally if exponents become even, which happens when a no. is negative. Keeping that in mind, statement 1- cant be true, 3 negative no.s cant result into something postive.
Statement 2- can be true , all powers are odd.
statement3- cant be true, as can never be equal to 0. Hence, B

Bunuel

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Let's go statement by statement and find if a set of values for a,b,c satisfy it.

i) abc > 0

if a=-10, b=-5, c=1
a<b<c & c^4 < b^4 < a^4
then abc=50, which is greater than zero

ii) a^3 * b^5 * c^7 < 0
if all a,b,c are less than -1, then this is true

iii) a^2 * b^4 * c^6 = 0
if a and b are negative integers, and c is zero, this is true

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Bunuel

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For could be true, we need only one case to satisfy the given equation,

a<b<c (1st inequality) => c^4<b^4<a^4 (2nd inequality)

1. abc>0,
either a,b,c all should be positive ==> multiple set of numbers will satisfy the 1st inequality but not second one.

or any 2 out of a,b,c can be positive.. --> case 1 - a= -3, b=-2, c=1 --> 1<16<81 (satisfying second equation too )

- Could be true.

2. a^3b^5c^7 < 0,
either all are negative or one of 3 is negative.

case 1. - a=-3, b=-2, c=-1 --> 1<16<81 -- could be true

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

any one is 0 --> a=0 not possible a^4 can't be highest

b=0 --> a=-4 , b=0 , c= 4 (can't be less than or equal to b)

c=0 ---> a=-4, b= -1, c=0 ---> 0<1<16 --could be true.
Hence E is the answer.

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05 Jul 2025, 05:44

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We are given : a b means c is positive.
So only valid sign combination: a < 0 0; only

Statement II could be true.

Answer: B

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This is a "could be" question.
The easiest way is to find examples that meet the requirements.

(a, b, c) = (-10, -5, 2) for I.
(a, b, c) = (-10, -5, -2) for II.
(a, b, c) = (-10, -5, 0) for III.

All the three options are possible.

The right answer is E

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Given:
a < b < c and c^4 < b^4 < a^4

there are three cases where this is true
1. All a, b and c are negative.
2. a and b are negative and c is positive, such that |c| < |b| < |a|
3. a and b are negative and c is 0.

I. abc > 0

Case 2 checks out for this. If c=+ve, a=-ve, b=-ve
(-ve)*(-ve)*(+ve)=(+ve) which is more than 0.
Can be true.

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

Case 1 checks out for this.
We know sign of variable with odd power remains as is.
So, (-ve)*(-ve)*(-ve)=(-ve) which is less than 0.
Can be true.

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

Case 3 checks out for this. If c=0
Zero raise to power of anything is 0. And anything multiplied by 0 is 0.
Can be true.

Answer: E

Bunuel

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For could be true questions, try to assume one set of variables that satisfies the given condition and can prove the statement to be true.

For St.1, if a= -4, b= -2 and c= 1

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

abc = -4*-2*1 = 8>0

St. 1 could be true

St.2, if a= -4, b= -3, c=-1

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

a^3b^5c^7<0
(-4)^3*(-3)^5*(-1)^7

You don't need to calculate here since all the values have odd power which means a^3b^5c^7<0

St.2 could be true

St.3 if a= -2, b= -1, c= 0

a^2=4, b^4=1, c^6=0

a^2b^4c^6=0

St.3 could be true

All statements could be true.

Option E

Bunuel

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05 Jul 2025, 06:52

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Let us consider a=-3, b=-2, c=1. Then c^4 < b^4 < a^4.

Case I is possible.
For case III, consider a=-2, b=-1 and c=0. This would satisfy case III
At this point, we can directly choose option E since that is the only option which has both Case I and Case III.
But just to be sure, let us simulate case II as well.

a= -3, b=-2,c =-1
This will make case II possible.

Correct answer is option E

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The only way $a<b<c $and $c^4<b^4<a^4$ can be true at the same time is when all three of the variables are negative.

(1). $abc > 0$ -> This can never be true. Product of three negatives cannot be greater than$ 0$.

(2). $a^3b^5c^7 < 0$ -> This has to be true. All variables are raised to the odd power, meaning they retain their sign (which is negative). Product of three negatives is negative.

(3). $a^2b^4c^6 = 0$ -> $a$, $b$, and $c$ are different values. Their product can never be $0$.

Answer - B

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05 Jul 2025, 07:27

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If $a<b<c$ but $c^4<b^4<a^4$ then there could be multiple possible cases. Let's try solving this problem by plugging in numbers:
i) $a=-2$, $b=-1$, $c=0$ then $a^4=16$, $b^4=1$, $c^4=0$
$a^2b^4c^6 =(-2)^2(-1)^4(0)^6= 0$ ; possible

ii) $a=-3$, $b=-2$, $c=1$ then $a^4=81$, $b^4=16$, $c^4=1$
$abc$=$(-3)(-2)(1)=6 > 0$ ; possible

iii) $a=-3$, $b=-2$, $c=-1$
$a^3b^5c^7$ = $(-3)^3(-2)^5(-1)^7$ = $(-27)(-32)(-1)$ = neg*neg*neg = neg < 0 ; possible

Since all three cases could be true, the answer is E.

Bunuel

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I. If a=-30, b=-20 and c=10 then c^4<b^4<a^4 and abc>0
II. If a=-30, b=-20 and c=-10 then c^4<b^4<a^4 and a^3b^5c^7<0
III. If a=-30, b=-20 and c=0 then c^4<b^4<a^4 and a^2b^4c^6=0

I, II, and III could be true.

Correct answer is E