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# If x is an integer and it satisfies the inequalities, x^2 – 6x - 7...

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If x is an integer and it satisfies the inequalities, x^2 – 6x - 7...  [#permalink]

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Updated on: 31 Oct 2018, 23:40
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35% (medium)

Question Stats:

76% (01:57) correct 24% (01:57) wrong based on 147 sessions

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If x is an integer and it satisfies the inequalities, $$x^2 – 6x - 7 < 0$$ and $$2x – x^2 + 3 < 0$$, then which of the following can be the value of x?

A. -1
B. 0
C. 1
D. 2
E. 5

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Originally posted by EgmatQuantExpert on 31 Oct 2018, 04:34.
Last edited by EgmatQuantExpert on 31 Oct 2018, 23:40, edited 1 time in total.
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Re: If x is an integer and it satisfies the inequalities, x^2 – 6x - 7...  [#permalink]

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31 Oct 2018, 05:59
EgmatQuantExpert wrote:
If x is an integer and it satisfies the inequalities, $$x^2 – 6x - 7 < 0$$ and $$2x – x^2 + 3 < 0$$, then which of the following can be the value of x?

A. -1
B. 0
C. 1
D. 2
E. 5

$$x^2 – 6x - 7 < 0$$
i.e. $$x^2 – 7x + x - 7 < 0$$

i.e.$$(x-7)*(x+1) < 0$$

Product of two things is less than zero i.e. one of them must be positive and other must be Negative. Also since (X-7) is smaller than (x+1) therefore (X-7) must be negative

i.e. $$(x-7) < 0$$ and $$(x+1) > 0$$

i.e. $$-1 < x < 7$$

Now, $$2x – x^2 + 3 < 0$$

i.e. $$x^2 - 2x - 3 > 0$$

i.e. $$x^2 - 3x +x - 3 > 0$$

i.e. $$(x-3)*(x+1) > 0$$

i.e. $$x > 3 or x < -1$$

Looking at both blue results we can infer that

$$3 < x < 7$$

i.e. x can only be 5 out of given option choices

Answer: Option E
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Re: If x is an integer and it satisfies the inequalities, x^2 – 6x - 7...  [#permalink]

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31 Oct 2018, 06:19
While I could solve the question by finding out the range of individual equations and then finding out the common value, I have a doubt.

Why can't I add the equations and still get the same result?

If we were to try add these equations ,we would get:

-4x-4<0
i.e x<-1

Why would this method not work?
I know there is a conceptual gap here because x<-1 is not a valid range.

Can some expert help out?
gmatbusters
EgmatQuantExpert
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Re: If x is an integer and it satisfies the inequalities, x^2 – 6x - 7...  [#permalink]

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31 Oct 2018, 07:30
1
nitesh50 wrote:
While I could solve the question by finding out the range of individual equations and then finding out the common value, I have a doubt.

Why can't I add the equations and still get the same result?

If we were to try add these equations ,we would get:

-4x-4<0
i.e x<-1

Why would this method not work?
I know there is a conceptual gap here because x<-1 is not a valid range.

Can some expert help out?
gmatbusters
EgmatQuantExpert
Gladiator59

nitesh50

Addition of inequalities is valid for Linear inequalities with same inequation signs.

Here we are dealing with quadratic inequations hence we can not simply add to get the solution

I hope this helps!!!
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Re: If x is an integer and it satisfies the inequalities, x^2 – 6x - 7...  [#permalink]

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31 Oct 2018, 07:50
Dear nitesh50

Adding the quadratic inequalities/equations might lead to loss of some roots, and you can get a larger Set as other conditions get lost with the loss of quadratic term.

Please note that there is a mistake in your solution which seems to show invalid range

Adding you get,
-4x-4<0
or, -4<4x
or x>-1

note that the actual answer x = 5 also satisfy this condition.

nitesh50 wrote:
While I could solve the question by finding out the range of individual equations and then finding out the common value, I have a doubt.

Why can't I add the equations and still get the same result?

If we were to try add these equations ,we would get:

-4x-4<0
i.e x<-1

Why would this method not work?
I know there is a conceptual gap here because x<-1 is not a valid range.

Can some expert help out?
gmatbusters
EgmatQuantExpert
Gladiator59

_________________
Manager
Joined: 04 Jun 2018
Posts: 157
GMAT 1: 610 Q48 V25
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Re: If x is an integer and it satisfies the inequalities, x^2 – 6x - 7...  [#permalink]

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31 Oct 2018, 08:38
gmatbusters wrote:
Dear nitesh50

Adding the quadratic inequalities/equations might lead to loss of some roots, and you can get a larger Set as other conditions get lost with the loss of quadratic term.

Please note that there is a mistake in your solution which seems to show invalid range

Adding you get,
-4x-4<0
or, -4<4x
or x>-1

note that the actual answer x = 5 also satisfy this condition.

nitesh50 wrote:
While I could solve the question by finding out the range of individual equations and then finding out the common value, I have a doubt.

Why can't I add the equations and still get the same result?

If we were to try add these equations ,we would get:

-4x-4<0
i.e x<-1

Why would this method not work?
I know there is a conceptual gap here because x<-1 is not a valid range.

Can some expert help out?
gmatbusters
EgmatQuantExpert
Gladiator59

Hi gmatbusters
Thank you for correcting my mistake.
this might be a stupid question, but
Why is that when we are adding 2 inequalities, it leads to loss of a certain range.
I guess What I am still looking for is a bit of a conceptual understanding to it.

Looking forward to you reply!

Regards
Nitesh
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Joined: 27 Oct 2017
Posts: 1225
Location: India
Concentration: International Business, General Management
GPA: 3.64
WE: Business Development (Energy and Utilities)
Re: If x is an integer and it satisfies the inequalities, x^2 – 6x - 7...  [#permalink]

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31 Oct 2018, 08:44
1
Hi
Since on adding you are loosing the quadratic term, leads to loss of roots/a certain range.

See this example

if x^2 = 2x
and we cancel out x from both sides ( it leads to loss of root, x = 0)
We get x = 2.

Whereas,
x^2 = 2x
or, x(x-2) = 0
or x = 0 or 2

By cancelling some terms you are loosing roots.

nitesh50 wrote:
gmatbusters wrote:
Dear nitesh50

Adding the quadratic inequalities/equations might lead to loss of some roots, and you can get a larger Set as other conditions get lost with the loss of quadratic term.

Please note that there is a mistake in your solution which seems to show invalid range

Adding you get,
-4x-4<0
or, -4<4x
or x>-1

note that the actual answer x = 5 also satisfy this condition.

nitesh50 wrote:
While I could solve the question by finding out the range of individual equations and then finding out the common value, I have a doubt.

Why can't I add the equations and still get the same result?

If we were to try add these equations ,we would get:

-4x-4<0
i.e x<-1

Why would this method not work?
I know there is a conceptual gap here because x<-1 is not a valid range.

Can some expert help out?
gmatbusters
EgmatQuantExpert
Gladiator59

Hi gmatbusters
Thank you for correcting my mistake.
this might be a stupid question, but
Why is that when we are adding 2 inequalities, it leads to loss of a certain range.
I guess What I am still looking for is a bit of a conceptual understanding to it.

Looking forward to you reply!

Regards
Nitesh

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Joined: 04 Jan 2015
Posts: 2888
If x is an integer and it satisfies the inequalities, x^2 – 6x - 7...  [#permalink]

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02 Nov 2018, 04:50

Solution

Given:
• We are given that x is an integer, and
• We are also given two inequalities,
o $$x^2 – 6x - 7 < 0$$
o $$2x – x^2 + 3 < 0$$

To find:
• We need to find out which among the given answer choices can be a possible value of x that satisfies the given information

Approach and Working:
• We need to solve both the quadratic inequalities to arrive at the answer.
• Let’s solve the first inequality, $$x^2 - 6x – 7 < 0$$
o Implies, (x - 7)(x + 1) < 0

Approach 1: Wavy-line method

• Thus, the expression is negative for -1 < x < 7
• The values of x that satisfy this inequality are, x = {0, 1, 2, 3, 4, 5, 6}

Now, let’s solve the second inequality, $$2x - x^2 - 3 < 0$$
o Multiplying by -1, we get, $$x^2 – 2x - 3 > 0$$
o Implies, (x - 3)(x + 1) > 0

• Thus, the expression is positive for x < -1 and x > 3

Combining both the results, -1 < x < 7, and (x < -1 or x > 3), we can say that the values of x that satisfy this inequality are, 3 < x < 7
• Since, x is an integer, x = {4 , 5 , 6}

Approach 2: Number-line method

Number- line for the expression (x + 1)(x - 7) < 0 is,

• Thus, the expression is negative for -1 < x < 7

Now, let’s draw the number-line for the expression, (x + 1)(x - 3) > 0

• Thus, the expression is positive for x < -1 or x > 3

Combining both these results, we get, 3 < x < 7
• Since, x is an integer, the values of x that satisfy the given expression are {4, 5, 6}

Hence, the correct answer is option E.

Answer: E

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If x is an integer and it satisfies the inequalities, x^2 – 6x - 7...  [#permalink]

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15 Jan 2019, 21:03
GMATinsight wrote:
EgmatQuantExpert wrote:
If x is an integer and it satisfies the inequalities, $$x^2 – 6x - 7 < 0$$ and $$2x – x^2 + 3 < 0$$, then which of the following can be the value of x?

A. -1
B. 0
C. 1
D. 2
E. 5

$$x^2 – 6x - 7 < 0$$
i.e. $$x^2 – 7x + x - 7 < 0$$

i.e.$$(x-7)*(x+1) < 0$$

Product of two things is less than zero i.e. one of them must be positive and other must be Negative. Also since (X-7) is smaller than (x+1) therefore (X-7) must be negative

i.e. $$(x-7) < 0$$ and $$(x+1) > 0$$

i.e. $$-1 < x < 7$$

Now, $$2x – x^2 + 3 < 0$$

i.e. $$x^2 - 2x - 3 > 0$$

i.e. $$x^2 - 3x +x - 3 > 0$$

i.e. $$(x-3)*(x+1) > 0$$

i.e. $$x > 3 or x < -1$$

Looking at both blue results we can infer that

$$3 < x < 7$$

i.e. x can only be 5 out of given option choices

Answer: Option E

Good night GMATinsight

I understood the whole method but I am struggling with on doubt:

i.e.$$(x-7)*(x+1) < 0$$

If I want to solve it algebraically, isn't the following correct?

x < 7
x < 1

What am I doing wrong?, I mean I understand the concept but why does the < sign change here if we are just adding/subtracting?

Kind regards!
If x is an integer and it satisfies the inequalities, x^2 – 6x - 7...   [#permalink] 15 Jan 2019, 21:03
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# If x is an integer and it satisfies the inequalities, x^2 – 6x - 7...

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