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12 Days of Christmas GMAT Competition 2025 - Day 4: If x is a

1 2 3

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Bunuel

If x is a non-negative number and $x ≠ 3\sqrt{x} + 4$, which of the following must be true?

I. (x - 1)(x - 16) ≠ 0

II. |17 - x| ≠ 1

III. x is not a square of an integer

A. I only
B. II only
C. III only
D. I, and II only
E. None of the above

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Given that x is a non negative number. x >=0.

x is not equal to 3* sqrt (x)+4

Let’s take sqrt x = a, then x = a^2

a^2 =3*a + 4

Solving it we get a =4 and a =-1.

Sqrt x = 4, then x = 16

Sqrt x = -1, this is not possible.

So, x is not equal to 16.

Which of the following is MUST true.

I) (x-1)*(x-16) is not equal to 0.

so, x cannot take the values 1 or 16.

x being non negative. We know that x is not equal to 16. But, x can take the value 1.

So, must true is not possible.

II) mod ( 17-x) is not equal to 1.

Removing the mod, we get (17-x) = 1. OR. (17-x) = -1

x is not equal to 16, is already derived.

x is not equal to 18, cannot be said exactly. Since, x is a non negative.

So, must true is not possible.

III) x is not the square of the integer.

x cannot be equal to 16.

But x being a non negative number, can take values 1, 4,9,25 etc.

So, must true is not possible.

Option E

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non -ve numbers -> 0, 1, 2, 3, 4, 5
x != 3sqRoot(x) + 4 // not sure y given, might be some equation condition -->
solving by substituting sqRoot(x) = y -> y^2 != 3y + 4 => (y-4)(y+1) != 0 -> y = -1 can't be true, y = 4 -> x = 16
therefore x -> 0, 1, 2, 3, 4, 5.....15,17,19
evaluating options:
I. (x-1)(x-16) != 0 => x can be 1 hence can be zero
II. we can put x = 18 will give value 1 hence cannot be must be true
III. what if x = 49 possible from our list -> hence not always true

therefore, E -> none of the above

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Answer E none of the above

x greater or equal to zero

let square root of x be m
therefore

m^2 = 3m +4
m^2 -3m -4 = 0
(m-4)(m+1) = 0

m equals 4, x equals 16

1. x can equal 1 incorrect
2. x can equal 16 incorrct
3. x can equal 0,1,4, 9 ... incorrect

Bunuel

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Bunuel

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x-4=3[square_root][/x]squaring both sides and solving we get
(x-1)(x-16)=/ 0
x=/1 and x=/ 16
But x=/1 satisfies the contraints and the equation
x>0 and x=/3[square_root][/x] +4
so ,1- must not be true since x=1 satisfies the equation
2- on simplifying modulus ,x=/ 16 and x=/ 18 , but x= 18 satisfies the equation, so (2 ) must not be true
#- x= 9 is a possible solution, so x can be a square of an integer
so ans is =E

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X is non-negative.
So X in (0,+ve )
Also X<>3*X^1/2+4

Now I am going to check the options :
I. (X-1)(X-16)<>0 ------- If I take X= 4 then equation is justified but if X=16 then it's not justified . It' some-time true but not must be true --So, This is not correct option.
II. |X-17|<>1 -------- If X= 15 the the equation is justified but if X=16 then it's not justified . So it's also some-time true but not must be true -- So, this is not correct option .
III. X is not a square of any an integer . --- So X can't be 16 but X can all other integer . So this is not must be true .

SO CORRECT option is E.

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Must be true ques (=/ means not equals to)

1) (x-1) (x-16) =/ 1
Let x = 1, then equation will be equal to 1

we will check if we can have x= 1 by putting in the main equation 1 =/ 7, hence x can be one, not true

2) |17 - x| =/ 1

to prove this false, we can either have x = 16 or x= 18, checking both in our main equation

for x = 16, gives 16 both sides, hence this wont work
for x = 18, gives 3 root18 + 4 = 9 root 2 + 4 on one side & 18 on another, hence works

Not true

3) x is not a square of an integer

let x = 4, putting in our main equation gives 4 on one side & 10 on other side

hence x can be a sq of an integer and statement is not true

Ans E

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Bunuel

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solving for x in the equation
we get x cannot be 16

so x can be any number > 0 and not 16
1) for x=1, this equation won't hold true so it is false
2) for x= 18, this equation does not hold true
3) for x = 4, this does not hold true to answer is

E none of above

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Given equation: x != 3 * square_root of x + 4

(1) Suppose (x-1)(x-16) = 0, this gives two solutions x=1 and x=16
If we substitute this in given equation, 16 will be satisfied and 1 will not.
This means that as per given equation, x cannot be 16, but it can be 1.
And the (1) equation will be satisfied when x=1
Therefore, (1) will not be true.

(2) Solving for x, x = 16 or 18
Similarly as above, 18 could be x as per given equation but cannot be as per (2) equation.
Therefore, (2) will not be true.

(3) Since for x = 16, which is a square of an integer, the given equation is satisfied.
Therefore, (3) will not be true.

The answer is (E)

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I. (x - 1)(x - 16) not equal to 0
We know that if x = 1 OR x = 16, the equation will become 0.
x not equal to 3_/x + 4
Putting x = 1 in this,
1 not equal to 3_/1 + 4
1 not equal to 7, which is valid and means that x can be one.
In this case (x - 1) will become 0 and the statement will not be valid. Hence, option 1 does not satisfy.

II. |17 - x| not equal to 1
For this to be equal, x must be 16 or 18
Putting x = 16 in the given equation
16 not equal to 3_/16 + 4
16 not equal to 12 + 4, which is not true. Hence, x cannot be 16.
Similarly we can find that x can be 18.

In that case the statement will be equal and hence not true.

III. x can be 1 which is a square of 1. Hence not true

None of the statements must be true. Hence, Option E is the answer

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I solved for the equation as if it were true by assigning variables. I then got x =16. I used the answer choices to test the conditions and got 'E', none of the above.

Bunuel

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Let's solve the equation first ,

x not equal to 3root(x) +4
x-4 not equal to 3root(x)
squaring both sides ,
(x-4)^2 not equal to 9x
which simplified to
(x-1)(x-16) not equal to 0.

SO from we are sure that x not equal to 16. But lets have a look for x having 1 as value. x=1 satisfies original equation but here x=1 is not satisfying that. so (x-1)(x-16) must not be true

|17-x| not equal to 1 also must not be true since x=18 satisfies original equation but is not satisfying this one.

x is not a square i=of an integer must not be true since there are so many values of x like x=4 ,25 that satisfies original equation and they are square of one integer.

SO final answer is E None of the above.

Bunuel

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From given information we have x >= 0 and x # 16

I. x = 1 => (x-1)(x-16) = 0 (eliminate)
II. x = 18 => |17-x| = 1 (eliminate)
III. x can be 4, 9, 25... => eliminate

Answer: E

Bunuel

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let t=sqrt(x) so t^2 =x
now solving for t^2=3t+4 we get t=-1 not possible since x>=0 and t=4
for t=4 x=16
from I x!=1 and x!=16 so only x!=16 is satisifed from x!=3sqrt(x)+4 but not x!=1 so only partially correct since
if we put 1!=3(1)+4 =7 so (x-1)(x-16) could equal 0
now in II
We know x!=16
17-x=1 or 17-x=-1 or x=18
18= 3sqrT(18)+4=16 point something satisfies the first equation
so this is false
in III
if we put x= 0
we get 0!=4 which is a square
so not true
hence none of the statements are correct

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Given x ≥ 0
x ≠ 3√x + 4
Then values which will be x = 3√x +4
let √x= p , then equation becomes p^2 = 3p +4 , p^2-3p-4 , simplifying this ;
(p-4)(p+1)=0 , p=4 , x=16 , t=-1 (Not allowed since √x≥0
Only excluded value x ≠ 16
ST 1 : (x-1) (x-16) ≠ 0 ❌
means x≠1 , x ≠16 , But x = 1 is allowed only 16 is not allowed.

ST 2 : |17-x| ≠1
If |17-x| = 1 , x = 16 or 18
but we know x ≠ 16 , only 18 can be allowed.❌

ST 3
x= 0,1 4,9,25 ,36..all allowed only 16 is excluded. Other's still work . So even this is eliminated ❌

Therefore Option E ✅

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It is asking for something that must be true.
x >= 0
Also, assuming $\sqrt{x} = t$ in the given condition, it simplifies to a quadratic equation and gives
x is not equal to 16, so x can take any number >=0 except for 16.

I. x can be 1, so not always true.

II. x can be 18, not always true

III. x can be 4, 9, etc.

Option E.

Bunuel

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This looks a bit scary, but I'll try to explain it nonetheless.

First to understand what the stem is saying:

x is non-negative. This means it is 0 and above. It can be 0.00001 as well, or a gazillion if need be.

Also, x cannot be equal to 3*(root x) + 4.

Now, to get a sense of what this means, we take x to be 16, and then this comes out to be, 3*(root 16) + 4, that's 3*4 + 4 = 16. This cannot be the case. (I couldn't really find any others that can be the case here, so we can just take that x cannot be 16 but it can be literally any other number).

Now, I: (x - 1) (x - 16) is not equal to 0.

Here, substituting the 16, we immediately get (17 - 1) (16 - 16) = 16*0 = 0, which contradicts the statement and makes it untrue, but then again we cannot use x = 16, so we cannot consider this.

But we can take x = 1. Then we get (1 - 1) (1 - 16) or 0*(-15) = 0. Hence, I is not a must.

Now, II: |17 - x| is not equal to 1.

Again, places x = 16 here contradicts the statement, as 17 - 16 = 1, but note that if 17 - x can be (-1), as we're given the absolute value; so if we go for 17 - 18 = |(-1)| = 1, then that nullifies the statement as well. Hence, II is not a must.

Now, III:

We, we know that x is the square of an integer if we take x to be equal to 16. But we also know that we can take any other value of x that is not 16 and also the square of an integer, like 25, 36 or 49, then it is the square of an integer.

Hence, III is also not necessary. Also, for MAYBE THE THIRD TIME IN THIS SET OF 9 QUESTIONS, I have realized I went wrong while solving this. I can't seem to recall why I made these *brilliant* decisions, I'm guessing in this case because I know 16, which is the square of an integer, isn't the option, I somehow applied that to a 25, 36, etc., as well, when I even used those numbers to test the viability of the first two options. Anyway, good learning curve.

The answer is: E - None of the above.

Bunuel

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Bunuel

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root(x) = y

y^2 = 3y + 4

(y-4)(y+1) = 0

y = 4
y = -1

root(x) cannot be -1

Therefore root(x) = 4

x is a non negative number and the value of x cannot be 16. All other values of x is allowed.

I) If x = 1, this is not true

II) If x = 18, this is not true

III) If x = 4, this is not true

Option E

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Firstly, the only way for $X=3\sqrt{x}+4$ is when $x=16$. Therefore, we need to find the option which must be true so that X is not equal to 16.

I. Here, X can be equal to neither 16 or 1.

II. Here, X can be equal to neither 16 or 18.

III. Here, X cannot be any full square, including 16, but also 9, 49, 225 etc.

Interestingly, while each option says that X isn't 16, each of them also points out that there're other constraints. For instance, if the first option is not true, then yes, X can be 16 - but it can also be 1. Same for option two, if it's not observed and x is actually 16, it can also very easily be 18.

Therefore, 'breaking' any of the options and making them untrue wouldn't necessarily (only possibly) turn X into 16. Thus, the answer is E.

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Lets say root of x = a
we get a^2 = 3a + 4
a^2 - 3a -4 = 0
(a-4)(a+1) = 0
from this we can say a = 4 or a = -1. Discard a =-1 which is not possible in this case.
so a = 4 = root of x
a^2 = 16.
If x not equal to 3(x)^(1/2) + 4 then x is no equal to 16.

Condition 1: has roots x = 1 and x =16 which cannot be the case.

Condition 2: |17 - x| not equal to 1
17 - x = 1 when |17-x| > 0
x = 16

And for |17-x| < 1
17-x = -1
x =18

X cannot be 16 but x can be 18. Hence 2 is not true either.

Condition 3: X is not a square of an integer:
Only restriction is that x cannot be 16 which is 4^2. But can be 25, 49, etc.
Hence condition 3 is also not true/

Ans is E. None of the above.

Bunuel

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We have two conditions given
x is greater than or equal to zero
x should not satisfy the given equation.
Lets solve the given equation to understand which value of x cannot be taken.
Let [square_root][/square_root x] = u
then we can write the equation as
$u^2$ - 3u - 4 = 0
solving this equation gives
u = 4, u = -1
Since u cannot be negative (root of x cannot be negative), u = 4
Therefore x =$ u^2$
= 16

This gives us the conditions as
x greater than or equal to zero
x not equal to 16

Analysing options
I
This equation gives
x not equal to 1
x not equal to 16
But this equation excludes 1 also along with 16. So this need not be true

II
This equation gives
x = 16
x = 18
But this equation excludes 18 also apart from 16. So this need not be true

III
x not equal to 1, 4, 9, 16 ...
This excludes many squares. This need not be true

Hence E is the answer

Bunuel