If 2 < x < 4, what is the median of the numbers 0, 5, x, 1,

How do you get 1 and 5/2 for the 2nd statement? I am getting x=2 & x=5. Thanks.

Substitute 2 and 5 into 2x^2 - 7x + 5 = 0 to see that neither is the root of the equation, while 1 and 5/2 are.

Factoring Quadratics: https://www.purplemath.com/modules/factquad.htm

Solving Quadratic Equations: https://www.purplemath.com/modules/solvquad.htm

Hope this helps.

Here the 2(1, 2.5) values lead to 2 different medians so how is this sufficient?

I understand the fact that both values lie within the range specified, but it leads to 2 different answers- which is grounds for insufficiency. Can you pls explain how this explanation is wrong?

Hi TuringMachine,

You have to pay attention to ALL of the information that you've been given.

Notice at the beginning of the prompt, we were told that 2 < X < 4. That 'restriction' still applies.

With Fact 2, we have two potential values for X: 1 and 5/2, but ONLY 5/2 fits that initial range that we were given. Thus, 5/2 is the only possible value for X and we now have enough information answer to the question. Fact 2 is SUFFICIENT.

GMAT assassins aren't born, they're made,
Rich

the second eqn gives two values x =1 and 5/2 and when you insert you get two different median 1 and 5/2. SO that is why I choose a beause it gives only one value.

Question stem: 2<x<5; median of (0, 5, x, 1, 7, 3)

1: 2x-5 = 0 i.e. x = 2.5
This complies with question stem condition 2<x<5. Also, median can be easily determined.
Sufficient

2. 2x^2 - 7x + 5 = 0

i.e. 2x^2 - 5x - 2x + 5 = 0
(2x-5) (x-1) = 0
x = 5/2 or x = 1
As per the question stem, x cannot be 1. Therefore, x = 5/2.
Sufficient

Answer: D (Either statement are sufficient).

Hi alice7,

You have to pay attention to ALL of the information that you've been given.

Notice at the beginning of the prompt, we were told that 2 < X < 4. That 'restriction' still applies.

With Fact 2, we have two potential values for X: 1 and 5/2, but ONLY 5/2 fits that initial range that we were given. Thus, 5/2 is the only possible value for X and we now have enough information answer to the question. Fact 2 is SUFFICIENT.

GMAT assassins aren't born, they're made,
Rich

Thanks for clarification, I completely forgot initial x value provided.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If 2 < x < 4, what is the median of the numbers 0, 5, x, 1, 7, and 3?

(1) 2x−5=0

(2) 2x 2 −7x+5=0

There is one variable (X) and 2 equations from the 2 conditions, so (D) is our likely answer.
For condition 1, x=5/2. This is sufficient.
For condition 2, (2x-5)(x-1)=0, x=5/2, 1, but 1 is not possible, so x=5/2. This is sufficient as well, making the answer (D).

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.

(1)
2x - 5 = 0
2x = 5
x = \(\frac{5}{2}\) Sufficient

(2)
\(2x^2 -7x +5 = 0\)
(2x -5)(x - 1) = 0
2x = 5 , x = 1
\(x = \frac{5}{2}\) Sufficient, x = 1 <---- reject this root since x is between 2 and 4

Answer D

Great Official Question.
Here is what i did in this question -->

Data Set=>
0
1
3
5
7
x


Now #=6
Hence Median = 3rd term +4th term/2
Here x=>(2,4)

Case 1=> x--> (2,3)
Here median => x+3/2

Case 2=>x-->(3,4)
Here median = 3+x/2

Thus,
Irrespective of what value of x is => Median => x+3/2

So we actually just need the value of x.

Statement 1-->
x=5/2
Hence Sufficient
Statement 2-->
Its a Quadratic Equation.
Two solutions are --> 1,5/2
But x->(2,4)
Hence x=1 is not a solution
So x must be 5/2

Hence Sufficient

Hence D

Very important to remember the restriction 2 < x < 4, otherwise trap answer (A) is waiting.

write the numbers in order to get a clear idea: 0,1,3,5,7 and x will fit in this sequence somewhere

St 1: gives you x = 2.5, condition 2 < x < 4 fulfilled so , 0,1,2.5,3,5,7

Sufficient

St 2: gives you two solutions x = 1 & 2.5 as it is a quadratic equation.

One value (x = 2.5) lies inside the range and other value (x=1) outside the range. So sufficient as only valid value of

x is 2.5.

Sufficient

(D)

Hi, can some please explain me statement 2?

How do you get from 2x2−7x+5=0 to (2x -5)(x - 1) = 0

Can you explain through Foil process without skipping any detail. I really struggle with this statement 2.

Thanks!

Posted from my mobile device

Sometimes we can factorize quadratic expressions by splitting the middle term. Lets say a quadratic expression is:

ax^2 + bx + c

We have to split the middle term bx. We have to find two quantities whose sum is bx and whose product is ax^2*c .. this is the rule.. if we can split middle term in such a way then we can factorize the quadratic expression easily. Lets try this here on 2x^2 - 7x + 5 = 0

We have to find two quantities whose sum is '-7x' and whose product is 2x^2*5 = 10x^2. These two quantities are -2x and -5x. So we can write this quadratic equation in the following way:

2x^2 -2x -5x + 5 = 0. Now we can see that these are four terms. From first two terms, we can take 2x common, and from next two terms we can take -5 common. So our equation now becomes:

2x*(x-1) -5*(x-1) = 0. Now x-1 can be taken out common to get:
(x-1)*(2x-5) = 0

Hi Bunuel,

can you please help me figure out what am I missing here?

please see the attached picture. I didn't get (2x-5)(x-1), but (2x+5)(x+1)

thank!

(1) x=2.5, so there would be a defined value of median
Sufficient

(2) x=1,2.5
But in the question stem its given 2<x<4,
so x=2.5, and there would be ONE value of the median
Sufficient

Answer: D

Hi omavsp,

When factoring a Quadratic, the 'clues' that you will need will be found at a few specific spots in the original equation. In Fact 2, we have....

2X^2 - 7X + 5 = 0

To start, we know the 'first half' of each parentheses:
(X )(2X )

Next, since the equation ends in "+5", we know that the two 'numbers' in the parentheses are EITHER both positive (almost certainly +1 and +5) OR both negative (almost certainly -1 and -5).

The middle term is "-7X".... and we have to get to that NEGATIVE sum with either two positives or two negatives.... so the only option that makes sense is two negative numbers... With a little playing around, you can prove that the equation factors down into...

(X - 1)(2X - 5) = 0

GMAT assassins aren't born, they're made,
Rich