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

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

(1) x² > 5
(2) x² + x < 5

Target question: Is x<5 ?

Statement 1: x² > 5
Let's TEST some values.
There are several values of x that satisfy statement 1. Here are two:
Case a: x = -10, in which case the answer to the target question is YES, x is less than 5
Case b: x = 10, in which case the answer to the target question is NO, x is not less than 5
Since we can’t answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: x² + x < 5
Subtract x² from both sides of the inequality to get: x < 5 - x²
Since x² is always greater than or equal to 0, the GREATEST possible value of 5 - x² is 5, and this occurs when x = 0
So, when x is 0, the expression 5 - x² = 5
For all other values of x, we know that 5 - x² < 5
In other words, 5 - x² ≤ 5
We can add this information to our existing inequality to get: x < 5 - x² ≤ 5
At this point, we can clearly see that x < 5
The answer to the target question is YES, x is less than 5, which means statement 2 is SUFFICIENT

Answer: B

Cheers,
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Re: Is x<5 ? (1) x^2 > 5 (2) x^2 + x < 5 [#permalink]
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Video solution from Quant Reasoning:
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Re: Is x<5 ? (1) x^2 > 5 (2) x^2 + x < 5 [#permalink]
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AbdurRakib wrote:
Is x<5 ?

(1) x^2 > 5

(2) x^2 + x < 5

Question: Is x<5 ?

Statement 1: x^2 > 5

ie.. x > √5 or x < -√5
√5 ≈ 2.2

ie.. x > 2.2 or x < -2.2
i.e. x may be smaller than 5 such as x=3 and x may be greater than as well such as x = 6 hence
NOT SUFFICIENT

Statement 2: x^2 + x < 5
i.e x*(x+1) < 5

Now, we need to think of some values separated by 1 whose product is less than 5
e.g. 2*3 = 6
i.e. x can NOT be 2 as x*(x+1) < 5
i..e x must be less than 2

We can observe similar result on negative side as well. the absolute value of x must be less than 2 in both positive and negative range hence we deduce
that -2 < x < 2
i.e. answer the question is DEFINITELY YES
SUFFICIENT

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

(1) x^2 > 5

(2) x^2 + x < 5

Solve the Official Questions more productively

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Answer: Option B

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

Here, we need to find whether x < 5 or not

(1) $$x^2 > 5$$
#Approach 1:

Assume that X =3, $$3^2 > 5$$
is X < 5 ? ==> Yes , 3 <5

Assume, X =10, $$10^2 > 5$$
is X < 5 ? ==> No , 10 > 5

Since we are not getting a definite YES or No for x <5 ?, Statement 1 alone is insufficient.

#Approach 2:
$$x^2 - 5 > 0$$
$$(x - \sqrt{5}) ( x + \sqrt{5}) > 0$$

Applying quadratic inequality,
we can conclude that$$x < - \sqrt{5}$$ or $$x > \sqrt{5}$$
$$\sqrt{5} ≈ 2.2$$

$$∴ x < -2.2$$ or $$x > 2.2$$

So here in this case we cannot confirm whether x < 5 or not.Hence its insufficient.

2)$$x^2 + x < 5$$
$$x < 5 - x^2$$
Since$$x^ 2$$ is non-negative, We can confirm that x <5.
Statement 2 alone is sufficient.

Option B is the answer

Thanks,
Clifin J Francis,
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Re: Is x<5 ? (1) x^2 > 5 (2) x^2 + x < 5 [#permalink]
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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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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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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]
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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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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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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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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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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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]
Hi Bunuel VeritasKarishma,

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]
Expert Reply
Shef08 wrote:
Hi Bunuel VeritasKarishma,

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]
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Bunuel VeritasKarishma

Can we use the following approach for statement 2?

statement 2 : x^2 + x < 5
As soon as i see this statement, my instinct tells me to break it in the form (x-a)(x-b)<0, which will give me a range for x.
But we can't do that easily because x^2 + x -5 doesn't yield integral roots.
Instead can we proceed as below ? :
x^2 + x < 5
=> x^2 + x < 6 (combining the 2 inequalities : x^2 + x < 5 and 5 < 6 )
=> (x-2)(x+3) < 0
=> -3 < x < 2 (Sufficient)

Do you see any risk in such an approach?
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