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# Is x<5 ? (1) x^2 > 5 (2) x^2 + x < 5

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Is x<5 ? (1) x^2 > 5 (2) x^2 + x < 5  [#permalink]

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24 Jun 2017, 03:15
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Is x<5 ?

(1) x^2 > 5

(2) x^2 + x < 5

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Md. Abdur Rakib

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Re: Is x<5 ? (1) x^2 > 5 (2) x^2 + x < 5  [#permalink]

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17 Nov 2017, 12:38
8
4
AbdurRakib wrote:
Is x<5 ?

(1) x^2 > 5

(2) x^2 + x < 5

We need to determine whether x < 5.

Statement One Alone:

x^2 > 5

If x = 3, then x is less than 5. However, if x = 6, then x is not less than 5. Statement one alone is not sufficient to answer the question.

Statement Two Alone:

x^2 + x < 5

Thus, we have x < 5 - x^2. Since x^2 is nonnegative, we have 5 - x^2 ≤ 5. Since x < 5 - x^2 and 5 - x^2 ≤ 5, we have x < 5.

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Re: Is x<5 ? (1) x^2 > 5 (2) x^2 + x < 5  [#permalink]

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24 Jun 2017, 03:37
12
5
AbdurRakib wrote:
Is x<5 ?

(1) $$x^2$$>5

(2) $$x^2$$+x<5

(1) $$x^2>5$$

If $$x = 6$$ or$$-6$$. $$x^2$$ will be greater than 5 in both cases.

However 6 is greater than 5 and -6 is less than 5.

(1) has multiple values. Hence I is Not Sufficient.

(2) $$x^2+x<5$$

$$x(x+1) < 5$$

$$x<5$$ or

$$x+1<5 = x < 4$$

(2) has x less than 5 or 4. Therefore x is less than 5. II is Sufficient. Answer (B)...
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Re: Is x<5 ? (1) x^2 > 5 (2) x^2 + x < 5  [#permalink]

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24 Jun 2017, 03:44
1
AbdurRakib wrote:
Is x<5 ?

(1) $$x^2$$>5

(2) $$x^2$$+x<5

Solution:

Statement 1: x can be 3,4,5 or 10. Insufficient.

Statement 2: The greatest positive value that satisfies the equation is 1. Therefore, x<5 . Sufficient.

Answer is Option B.
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Re: Is x<5 ? (1) x^2 > 5 (2) x^2 + x < 5  [#permalink]

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26 Jun 2017, 14:55
2
1
Is x<5 ?

$$(1) x^2 > 5$$

This means x can have Positive and Negative values , lets check

$$x = 10 = x^2 = 100$$

$$x = - 10 = x^2 = 100$$

As we are getting answer as YES & NO

Eq. (1) =====> is NOT SUFFICIENT

$$(2) x^2 + x < 5$$

$$x^2 + x < 5$$

$$x(x + 1) < 5$$

$$x < 5$$ or

$$x < 4$$

As both these values are < 5

(2) =====> is SUFFICIENT

Hence, Answer is B
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Re: Is x<5 ? (1) x^2 > 5 (2) x^2 + x < 5  [#permalink]

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18 Nov 2017, 16:39
AbdurRakib wrote:
Is x<5 ?

(1) x^2 > 5

(2) x^2 + x < 5

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 1 variables and 0 equations, D is most likely to be the answer and so we should consider each of conditions first.

Condition 1)
$$x = 10$$ : Yes
$$x = -10$$ : No
This is not sufficient.

Condition 2)
$$x^2 + x < 5$$
$$x < 5 - x^2 < 5$$
This is sufficient.

Therefore, the answer is B.

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.
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Re: Is x<5 ? (1) x^2 > 5 (2) x^2 + x < 5  [#permalink]

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06 Apr 2018, 21:34
ydmuley wrote:
Is x<5 ?

$$(2) x^2 + x < 5$$

$$x^2 + x < 5$$

$$x(x + 1) < 5$$

$$x < 5$$ or

$$x < 4$$

As both these values are < 5

(2) =====> is SUFFICIENT

Hence, Answer is B

can anyone please solve the second statement in detail ?
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Is x<5 ? (1) x^2 > 5 (2) x^2 + x < 5  [#permalink]

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06 Apr 2018, 23:27
1
pranavpal ,

St(2): $$x^{2} + x < 5$$

There are a couple of ways, we can solve the second statement.

Method1: Use Quadratic equation root formula:

For a given Quadratic equation $$ax^{2} + bx + c = 0$$ (where $$a \neq 0$$), roots can be obtained using following frmula:

$$\frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$$

Lets first solve the equality case of st(2) i.e. : $$x^{2} + x -5 = 0$$ . Here we have, a = 1, b = 1, c = -5. Put all the values in the above formula, we get:

Roots = $$\frac{-1 - \sqrt{21}}{2}, ~~~~ \frac{-1 + \sqrt{21}}{2}$$ => approximately roots are -3 and +2.

In order to solve the inequality, put the root's value on the number line and check whether the inequality gets satisfied or not in each section.

----------------Not satisfy ----------(-3) --------Satisfy--------(2)--------Not satisfy------------

It is very clear that the st(2) will satisfy only when -3 < x < 2. => x is definitely less than 5. Hence, Sufficient.

I hope this helps.

Thanks.
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Re: Is x<5 ? (1) x^2 > 5 (2) x^2 + x < 5  [#permalink]

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07 Apr 2018, 17:40
3
Although most of the folks got it right, I want to mention something which most of the folks got wrong..

x(x+1)<5

DOES NOT MEAN - x < 5 or (x+1) < 5.

and you easily check it by substitution

See bunuel's post - https://gmatclub.com/forum/inequalities ... 06653.html
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Re: Is x<5 ? (1) x^2 > 5 (2) x^2 + x < 5  [#permalink]

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14 Sep 2018, 22:22
pranavpal wrote:
ydmuley wrote:
Is x<5 ?

$$(2) x^2 + x < 5$$

$$x^2 + x < 5$$

$$x(x + 1) < 5$$

$$x < 5$$ or

$$x < 4$$

As both these values are < 5

(2) =====> is SUFFICIENT

Hence, Answer is B

can anyone please solve the second statement in detail ?

Hi,

The second statemnt if we read it along we can understand if this would be sufficient.

We are told that square of a number added to the number itself is less than 5. that means at least x< 5 and also that at least $$x^2< 5$$

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Re: Is x<5 ? (1) x^2 > 5 (2) x^2 + x < 5  [#permalink]

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16 Oct 2018, 05:55
Hi all,

I understand that plugging in values is a very good approach for this question. Still, I am wondering how to solve Statement 1 algebraically.

My way:

(1) x^2 > 5
|x| > √5 (correct?)
x > √5 or x < -√5 --> Therefore x can be bigger than 5, insufficient

Thanks a lot!
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Re: Is x<5 ? (1) x^2 > 5 (2) x^2 + x < 5  [#permalink]

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26 Mar 2019, 01:03
Can someone please show the exact steps from: "Since x < 5 - x^2 and 5 - x^2 ≤ 5, we have x < 5."
Why is x ≤ 5 not possible? Or how do I know which sign I should take when adding inequalities?

ScottTargetTestPrep Bunuel chetan2u

Thank you in advance!
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Is x<5 ? (1) x^2 > 5 (2) x^2 + x < 5  [#permalink]

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26 Mar 2019, 02:07
1
lstsch wrote:
Can someone please show the exact steps from: "Since x < 5 - x^2 and 5 - x^2 ≤ 5, we have x < 5."
Why is x ≤ 5 not possible? Or how do I know which sign I should take when adding inequalities?

ScottTargetTestPrep Bunuel chetan2u

Thank you in advance!

$$x < 5 - x^2$$ and $$5 - x^2 ≤ 5$$
When we combine the two, we get $$x< 5 - x^2 ≤ 5$$... This, x< 5 - x^2 ≤ 5 gives us x<5
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Re: Is x<5 ? (1) x^2 > 5 (2) x^2 + x < 5  [#permalink]

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27 Mar 2019, 19:05
2
First of all, we are NOT adding the two inequalities to arrive to the conclusion that x < 5. Let’s be clear on that. It’s a transitive property in inequalities.

For example, if a < b and b < c, then a < c. Another example is: if a ≤ b and b ≤ c, then a ≤ c. Of course, here we have if a < b and b ≤ c, then a < c. You can see that the premise actually can be combined into a double inequality, that is, for short, we can say: a < b < c implies that a < c; a ≤ b ≤ c implies that a ≤ c; and last but not least, a < b ≤ c implies a < c.

So your question is, if a double inequality have both < and ≤ signs, why we take the < sign, instead of the ≤ sign?

The reason is we always take the sign of the < (“strictly less than”) sign when a double inequality have both. That is because in a < b ≤ c, it says b is no more than c, so if a is strictly less than b, it will be also strictly less than c, hence the conclusion inequality a < c. The other way around is also true.

That is, a ≤ b < c also implies a < c. Here, it means a is no more than b, but if b is strictly less than c, so does a.
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Is x<5 ? (1) x^2 > 5 (2) x^2 + x < 5  [#permalink]

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11 Apr 2019, 23:32

I need ur expert opinion in this..

I got the answer correct but need to know whether my process is correct or no..

S2: x^2+x<5
X^2+x-5<0
(X+1)(x-5)<0

X<-1 or x<5

Posted from my mobile device
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Re: Is x<5 ? (1) x^2 > 5 (2) x^2 + x < 5  [#permalink]

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11 Apr 2019, 23:48
Shef08 wrote:

I need ur expert opinion in this..

I got the answer correct but need to know whether my process is correct or no..

S2: x^2+x<5
X^2+x-5<0
(X+1)(x-5)<0

X<-1 or x<5

Posted from my mobile device

(x + 1)*(x - 5) is not the same as (x^2 + x - 5). It is same as (x^2 - 4x -5).

(x^2 + x - 5) does not have integer roots but we can guess the roots using the formula as done in this comment above: https://gmatclub.com/forum/is-x-5-1-x-2 ... l#p2042359

If the roots are approximately -3 and 2, then (x^2 + x - 5) = (x + 3)(x - 2) < 0
which gives us -3 < x < 2
So in any case, x is less than 5.
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Re: Is x<5 ? (1) x^2 > 5 (2) x^2 + x < 5   [#permalink] 11 Apr 2019, 23:48
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