GMAT Club World Cup 2022 (DAY 10): If a data set A consists of all x

(x2−5x+5)^(x2+9x)=1
then(x2−5x+5)=1 => (x2−5x+4=0
=> x=1 or 4, then the median is 2.5

Our answer is E

\((x^2 - 5x + 5)^{(x^2 + 9x)} = 1\), it means either {(x^2 + 9x)} = 0 or (x^2 - 5x + 5)= 1;

By Solving {(x^2 + 9x)} = 0 , we will get x= 0, -9; & by solving (x^2 - 5x + 5)= 1, we will get x= 4,1.

A={-9,0,1,4}, Median = 0.5

Option B is the correct answer.

Asked: If a data set A consists of all x such that \((x^2 - 5x + 5)^{(x^2 + 9x)} = 1\), then what is the median of A ?

Case 1: x^2 - 5x +5 = 1
or
Case 2: x^2 + 9x = 0
or
Case 3: x^2 - 5x +5 = -1 and x^2 + 9x is even

Case 1: x^2 - 5x +5 = 1
x^2 - 5x + 4 = 0
(x-4)(x-1) = 0
x = 1 or x = 4

Case 2: x^2 + 9x = 0
x(x+9) = 0
x = 0 or x =-9

Case 3: x^2 - 5x +5 = -1 and x^2 + 9x is even
x^2 - 5x + 6 = 0
(x-2)(x-3) = 0
x = 2 ; x^2 + 9x = 4 + 18 = 22
x =3; x^2 + 9x = 9 + 27 = 36
x = 2 or x = 3

A = {-9,0,1,2,3,4}
Median of set A= (1+2)/2 =1.5

IMO D

I can get X = -9, -3, -2, 0, etc...
Guess it's B

for given expression \((x^2 - 5x + 5)^{(x^2 + 8x)} = 1\)
this would hold true at x = 1 , 4
median of set A ; ( 1+4)/2 ; 2.5
OPTION E

Correct answer : Choice B

For the given statement to be true ,
either x square + 8 x = 0 or x square -5x + 5 = 1

if x squre + 8x =0, then x can be 0 or -8
if x square -5x + 5 =1 , then x can be 1 or 4

so our set is {-8,0,1,4}
the median is (0 + 1)/2 = 0.5

Correct answer : Choice B

Answer B for me.

Explanation in the picture.


Posted from my mobile device

Clearly we see that
x^2 - 5x +5 = 1
solving we get x=1,4
therefore set A = (1,4)
median of set A = (1+4)/2 = 2.5
Answer E

\((x^2−5x+5)^(x^2+8x)\)=1 ; to find median of A

If \((x^2−5x+5) = 1\) then \((x^2−5x+5)^(x^2+8x)\)=1 for all values of x
\(=> x^2−5x+5 -1 = 0\\
=> x^2−5x+4 = 0\\
=> x^2−4x - x+4 = 0\\
=> (x-4)(x-1) = 0\)
\(=> x=4\) or \(x=1\) .......... (1)

If \((x^2−5x+5)\) = -1 then \((x^2−5x+5)^(x^2+8x)\)=1 for all even values of x
\(=> x^2−5x+5 +1 = 0\\
=> x^2−5x+6 = 0\\
=> x^2−2x-3x+6 = 0\\
=> (x-2)(x-3) = 0\\
=> x=2 or x=3\)
but x has to be even for this case. Hence, \(x=2\)......(2)

if \((x^2+8x)=0\) then \((x^2−5x+5)^(x^2+8x)=1\) for all values of x
\(=> x(x+8) = 0\)
\(=> x=0\) or \(x= -8 \) ..........(3)

Hence, all possible values of x are: 4,1,2,0,-8
Arranging in increasing order to find median {-8,0,1,2,4). Hence, median of all possible values of x = 1.

Answer: C

If (x^2 - 5x + 5)^(x^2+8x) = 1,

three cases are possible:
a) x^2+8x = 0, i.e., x=-8
b) x^2-5x+5 = 1, solving which, we get x = 1, 4
c) x^2-5x+5 = -1, where x^2 is even, so x is even - solving which, we get x = 2

So, the required "set" of values = {-8, 1, 2, 4}

Median of this set = average of the 2nd and 3rd term, i.e., (1+2)/2 = 1.5

The answer is option D.

OA = B

If a data set A consists of all x such that \((x2−5x+5)^{x2+8x}\)=1

x2−5x+5 = 1 or x2+8x = 0

solving both equation x = -8 , 0 . 1 , 4

so median = (0+1)/2 = 0.5

1 & 0 are the only numbers that belong to this set. So the median of 1 and 0 is 0.5
Ans 0.5

IMO A

only x = 0,1,2 & 4 satisfies above equation

median = 1+2/2 = 3 / 2 = 1.5

Answer D

X can be 0 or 1 os the mwdian will be •5,

Posted from my mobile device

IMO, Option C is the answer.

After manipulating the expression,
x can be -8,0,1,2 and 4. The median is 1.