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If 2 < x < 4, what is the median of the numbers 0, 5, x, 1,
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30 Nov 2013, 10:54
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If 2 < x < 4, what is the median of the numbers 0, 5, x, 1, 7, and 3? (1) \(2x  5 = 0\) (2) \(2x^2  7x + 5 = 0\)
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Re: If 2 < x < 4, what is the median of the numbers 0, 5, x, 1,
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05 Jan 2014, 19:53
What's important to remember when faced with a Data Sufficiency Question is that you don't not need to solve. You need to figure out if you have enough information to solve.
First look at the prompt to figure out exactly what you need for sufficiency. If we know the value of x we can find the median. So when you evaluate the statements, ask yourself  "Can I find the value of x?"
Statement1: We have a linear equation with one variable, x. We can solve for x  Sufficient
Statement 2: Here we have a quadratic equation which typically has two solutions. Remember though that we have a range for x in the prompt, so it is possible only one of the solutions will fit in the range and lead us to sufficiency. 2x^2 7x + 5 = 0 factors into (2x 5)(x1) = 0 Leading us to the solutions x = 2.5 and x =1. Only x= 2.5 fits in the range 2<x<4, so we have one value for x and sufficiency.
The answer is D




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Re: If 2 < x < 4, what is the median of the numbers 0, 5, x, 1,
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30 Nov 2013, 11:05
(1) gives you a single solution: 5/2. (2) gives you two solutions: 1 and 5/2. Each equation only gives you a single solution in the given range (2 < x < 4). Thus, you know the answer is
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Re: If 2 < x < 4, what is the median of the numbers 0, 5, x, 1,
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01 Jan 2014, 17:18
How do you get 1 and 5/2 for the 2nd statement? I am getting x=2 & x=5. Thanks.



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Re: If 2 < x < 4, what is the median of the numbers 0, 5, x, 1,
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02 Jan 2014, 04:25



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Re: If 2 < x < 4, what is the median of the numbers 0, 5, x, 1,
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11 Apr 2015, 06:01
Bunuel wrote: 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.
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?



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Re: If 2 < x < 4, what is the median of the numbers 0, 5, x, 1,
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11 Apr 2015, 15:26
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
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Re: If 2 < x < 4, what is the median of the numbers 0, 5, x, 1,
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09 Nov 2015, 07:46
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.



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Re: If 2 < x < 4, what is the median of the numbers 0, 5, x, 1,
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09 Nov 2015, 08:58
Question stem: 2<x<5; median of (0, 5, x, 1, 7, 3)
1: 2x5 = 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 (2x5) (x1) = 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).



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Re: If 2 < x < 4, what is the median of the numbers 0, 5, x, 1,
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09 Nov 2015, 12:03
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
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Re: If 2 < x < 4, what is the median of the numbers 0, 5, x, 1,
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09 Nov 2015, 12:47
Thanks for clarification, I completely forgot initial x value provided.



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Re: If 2 < x < 4, what is the median of the numbers 0, 5, x, 1,
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10 Nov 2015, 10:56
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, (2x5)(x1)=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.
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Re: If 2 < x < 4, what is the median of the numbers 0, 5, x, 1,
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15 Jul 2016, 13:28
bgribble wrote: If 2 < x < 4, what is the median of the numbers 0, 5, x, 1, 7, and 3?
(1) \(2x  5 = 0\)
(2) \(2x^2  7x + 5 = 0\) (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
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Re: If 2 < x < 4, what is the median of the numbers 0, 5, x, 1,
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19 Dec 2016, 19:01
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
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If 2 < x < 4, what is the median of the numbers 0, 5, x, 1,
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30 Dec 2017, 08:59
bgribble wrote: If 2 < x < 4, what is the median of the numbers 0, 5, x, 1, 7, and 3?
(1) \(2x  5 = 0\)
(2) \(2x^2  7x + 5 = 0\) 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)
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Re: If 2 < x < 4, what is the median of the numbers 0, 5, x, 1,
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12 Jun 2018, 06:34
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!
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Re: If 2 < x < 4, what is the median of the numbers 0, 5, x, 1,
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12 Jun 2018, 21:19
Rebaz wrote: 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*(x1) 5*(x1) = 0. Now x1 can be taken out common to get: (x1)*(2x5) = 0




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