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

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Bunuel

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18 Dec 2025, 10:00

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Bunuel

Let's solve $x = 3\sqrt{x} + 4$.

Let $\sqrt{x}= t$, so $x = t^2$.

Substitute:

$t^2 = 3t + 4$
$t^2 - 3t - 4 = 0$

Factor:

$(t - 4)(t + 1) = 0$

So $t = 4$ or $t = -1$.

But $t = \sqrt{x}$ cannot be negative, so only $t = 4$ is valid.

Thus $\sqrt{x} = 4$, so $x = 16$.

Therefore $x ≠ 3\sqrt{x} + 4$ means $x ≠ 16$. That is the only restriction (along with x being a non-negative number).

Now test each statement to see whether it must be true when $x ≠ 16$.

I. (x - 1)(x - 16) ≠ 0

This expression is nonzero unless $x = 1$ or $x = 16$. We only know $x ≠ 16$, but x could be 1, which makes the expression 0. So Statement I is not guaranteed. Not always true.

II. |17 - x| ≠ 1

$|17 - x| = 1$ corresponds to $x = 16$ or $x = 18$. We only know $x ≠ 16$, but x could be 18, which would make $|17 - x| = 1$. So Statement II is not guaranteed. Not always true.

III. x is not a square of an integer

We only know $x ≠ 16$, but x could be 0, 1, 4, 9, 25, 36, etc. So Statement III is not guaranteed. Not always true.

Answer: E.

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paragw

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18 Dec 2025, 09:07

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x is non-negative and x not equals 3 sqrt(x) + 4

Check when equality would happen:
Let sqrt(x) = t >= 0
so x = t^2
then t^2 = 3t +4
t^2-3t-4 = 0
(t-4)(t+1) = 0
since t>= 0
t = 4, so x =16

I. x not equals 1 and 16
but x = 1 is allowed.

II. x =16 and x =18
But x =18 is allowed

III. x is not a square of an integer
but x =9 is allowed

None of the statements must be true

Answer E

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18 Dec 2025, 09:33

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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

for the given equation
sqrt x be a
a^2-3a-4=0
a= 4 , -1
sqrtx = 4
x= 16
from
#1
(x-1)(x-16) is not equal 0
x cannot be 1 or 16 not sufficient
#2
l17-xl is not =1
x can be 16 or 18
not valid option
#3
x is not square of an integer
this is not true

OPTION E none of above is correct

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18 Dec 2025, 09:35

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x != 3(x)^(1/2) + 4
x-4 != 3(x)^(1/2)
x^2-8x+16 != 9x
x^2-17x+16 != 0
(x-16)(x-1)!=0
x !=16 , x!=1
x could be 4,9,25,....

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18 Dec 2025, 09:47

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x is a non-negative number
x is not equal to 3(x^(1/2)) +4
We need to find excluded values of x
Let, x=t^2
t^2 =3t+4
Solving this equation we get (t-4)(t+1)=0
Ans since t is more than or equal to 0. Our only restrictions are x>=0 & x≠16

1. (x-1)(x-16)≠0
According to our restrictions, x can b equal to 0 .....so this statement must not be true.

2. |17-x| ≠1
If x=18 then this equation is not true.

3. x is not a square of an integer
But values like (0,1,4,9,25.....) are allowed and only 16 is excluded. This must mot be true.

None of the above must be true.

E

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18 Dec 2025, 09:49

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To find out quickly you can plug in values of x for which the conditions I II and III are not true.
The only restriction is that x≠3 sqrt x + 4 -> solving this we find that x≠16

I. works with x=1
II. works with x=18
III. works with x=4,9,25,ecc

None of the conditions MUST be true
Answer E

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18 Dec 2025, 10:02

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Bunuel

Lets substitute y = square_root(x)
then the equation becomes y^2 - 3y - 4 = 0;
The roots are y = approx 2.279 and approx 11.84

Case 1: x != 1 and x != 16, neither solves: 1 gives -6 and 16 gives +4.44.
So this is True

Case 2: |17 - x| != 1 means x != 16,18. After substituting we dont get 11.84
So this is True

Case 3: x is not a square, our solutions 2.27 and 11.84 aren't squares
So this is True

So all 1,2 and 3 are True
Option E

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18 Dec 2025, 10:14

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x is a non-negative number, e.g. x>=0
$x \neq 3\sqrt{x} + 4$

Let$ t = \sqrt{x}$
$x = t^2$

$t^2 \neq 3t + 4$
$t^2 - 3t - 4 \neq 0$
$(t-4)(t+1) \neq 0$

$t \neq 4$; $t \neq -1$; Not feasible since $t = \sqrt{x}$
$t \neq 4$;
$x = t^2 \neq 16$

I. $(x-1)(x-16) \neq 0$
At x = 1; (x-1)(x-16) = 0
Not necessarily true.

II. $|17-x| \neq 1$
At x = 18 ; |17-x| = 1
Not necessarily true

III. x is not a square of an integer
At x=9; $ 9 \neq 3\sqrt{9} + 4 = 13$
But x = 9 is a square of an integer
Not necessarily true

IMO E

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18 Dec 2025, 10:17

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x#3sqrtx+4
x2-17x+16#0
(x-1)(x-16)#0
Clearly if x=1, then condition is satisfied 1#7
if x=16, then condition 16=16, condition is not satisfied.

I So (x-1)(x-16)#0 is not correct
II |17-x|#1 again x#18, X#16, put x=18 condition is satisfied
So option II is also not correct
III x is not a square integer, put x=4 we get 4#10 condition is satisifed.
Option III is also not true.

So correct option is E.

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18 Dec 2025, 10:19

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This one is tricky question.
We need to solve equation with not equal sign.
Please check solutiion in image and any doubt, let me know.

Attachments

PS2.jpeg [ 104.16 KiB | Viewed 1717 times ]

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18 Dec 2025, 10:49

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Bunuel

Given , x>=0
and from inequality, x-3(x)ˆ1/2-4 not equal to 0
or [xˆ(1/2) +1].[xˆ(1/2)-4] not equals 0
or xˆ(1/2) not equals -1 or 4
or x not equals 4ˆ2 i.e. x not equals 16.

I. x can be 1, thus statement must be true - NO.
II. x can be 18, thus statement must be true - NO.
III. x can be 4, 9, etc. thus statement x is not a square of integer must be true - NO.

None of above

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18 Dec 2025, 10:52

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Given x>=0

Let’s assume x = 3*rt(x) +4

If rt (x)=y

y^2=3y+4

Solving for y, we get y=-1 or 4

Since rt(x) cannot be -1, it is equal to 4 and accordingly x is 16

In (1) If x=1, then the equation equals zero. Not a must be true

In (2) x can be 18 and the equation can be equal to 1. Not must be true.

In (3) x can be square of an integer such as 9 and still satisfy the constraints.

Therefore, Option E

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18 Dec 2025, 11:15

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$x\neq3\sqrt{x} +4$
$x-3\sqrt{x} - 4 \neq 0$
$(\sqrt{x} - 4) * (\sqrt{x} + 1) \neq 0$
$\sqrt{x} \neq 4,-1$
Since x is non negative, $\sqrt{x}$ will be positive
$x \neq 16$

Checking options
I Substituting x = 1, $3\sqrt{1} + 4 = 7$, which satisfies the equation and hence is allowed but it makes this statement in I 0.
II $17-x \neq 1$ => $x \neq 16 $ and $17 - x \neq -1$ => $18 \neq x$. 18 satisfies the constraint but doesn't satify the statement in II
III x can be square of an integer, say x = 1, works

None is the answer. E

Bunuel

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18 Dec 2025, 11:19

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b. II only

because x cannot be 16 given that 16 is the root of 4, so solving RHS, 3x4+4=16
while LHS is the value of x, ie, 16
given LHS is equal to RHS
Therefore, value of x cannot be 16

In statement II, it shows that x cannot be 16

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18 Dec 2025, 11:22

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Bunuel

$\neq$ functions just like = in this.

So simplifying the equation in the stem, we get:

$x^2-9x+16 \neq 0$

Now let's compare.

I. Simplifies to $x^2 - 17x +16 \neq 0$ Not the same as stem equation, so incorrect.

II. Simplifies to $x \neq 16$, the roots of our stem equation are $x \neq 16, x \neq 1$. If the absolute value is positive, we do get a potential answer of $1 \neq 1$ which is obviously not true, so this is incorrect.

III. We know that x is not the square of 4, but we don't know if it applies to all other integer squares. This need not be true, so incorrect.

Therefore answer is E.

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18 Dec 2025, 11:24

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$x not= 3\sqrt{x} +4$
$(x-4)^2 not= 9x$
$x^2 - 17x +16 not= 0$
$(x-16)(x-1) not= 0$....option A
if x=18, option B satisfies so false
if x=25, option C satiesfies so false

Ans A

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18 Dec 2025, 11:25

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Bunuel