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e-GMAT Representative V
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If x is a positive integer, how many values of x satisfy the..........  [#permalink]

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Question Stats: 35% (02:03) correct 65% (02:20) wrong based on 281 sessions

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If x is a positive integer, how many values of x satisfy the inequality, $$\frac{(x - 2)}{(x^2 – 13x + 36)} ≤ 0$$?

A. 4
B. 5
C. 6
D. 7
E. Cannot be determined

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Re: If x is a positive integer, how many values of x satisfy the..........  [#permalink]

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EgmatQuantExpert wrote:
If x is a positive integer, how many values of x satisfy the inequality, $$\frac{(x - 2)}{(x^2 – 13x + 36)} ≤ 0$$?

A. 4
B. 5
C. 6
D. 7
E. Cannot be determined

$$\frac{(x - 2)}{(x^2 – 13x + 36)} ≤ 0$$

Factorizing denominator,

Or, $$\frac{x-2}{\left(x-4\right)\left(x-9\right)}\le \:0$$

Now, applying Wavy-curve method, we have the following ranges of x:-

$$x\leq{2}$$ or 4<x<9

As 'x' is a positive integer, hence the posssible value of x:- 1,2,5,6,7,8

Therefore, SIX nos. of values of 'x' satisfy the given inequality.

Ans. (C)
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How did you get 1 2 5 6 7 and 8 from those 2 intervals?

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Re: If x is a positive integer, how many values of x satisfy the..........  [#permalink]

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rahulkashyap wrote:
How did you get 1 2 5 6 7 and 8 from those 2 intervals?

Posted from my mobile device

Hi rahulkashyap,
a) x≤2 and x is a +ve integer. hence x could be: 1, 2
b) 4<x<9 and x is a +ve integer. hence x could be: 5,6,7,8
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e-GMAT Representative V
Joined: 04 Jan 2015
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If x is a positive integer, how many values of x satisfy the..........  [#permalink]

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Solution

Given:
• We are given that x is a positive integer
• And, we are given an inequality, $$\frac{(x - 2)}{(x^2 – 13x + 36)} ≤ 0$$

To find:
• We need to find out the number of values of x that satisfies the given inequality

Approach and Working:
• If we observe the given inequality, we can see that the denominator is a quadratic expression.
• So, let’s try to factorise the quadratic expression, $$x^2 – 13x + 36$$
o $$x^2 – 13x + 36 = (x – 9) * (x – 4)$$

• Thus, the inequality can be written as, $$\frac{(x - 2)}{(x – 9)(x - 4)} ≤ 0$$
• Now, we need to multiply the numerator and denominator of LHS with (x – 9)*(x – 4)
o $$\frac{(x - 2) (x – 9)(x - 4)}{[(x – 9)(x - 4)]^2} ≤ 0$$
o The denominator of the above inequality is always greater than 0 and x ≠ 4 or 9, since the denominator cannot be equal to 0.
o So, the numerator must be ≤ 0, for $$\frac{(x - 2)(x – 9)(x - 4)}{[(x – 9)(x - 4)]^2} ≤ 0$$.
o We need to find the values of x, for which (x - 2) (x – 4)(x - 9) ≤ 0

Approach 1: Wavy-line method

• The zero points of the inequality are {2, 4, 9} and the wavy-line will be as follows: • So, the expression will be equal to zero for x = {2, 4, 9}
o But, as we already know that, x cannot take the values 4 and 9
o Thus, x = {2}
• The expression will be negative in the regions, 4 < x < 9 and x < 2
o But we know that x is a positive integer, so, the expression will be negative for x = {1, 5, 6, 7, 8}
• Therefore, the given inequality satisfies for x = {1, 2, 5, 6, 7, 8}

Approach 2: Number-line method

The zero points of the inequality are {2, 4, 9} and highlighting these points on a number line, we get: • Thus, (x - 2) (x – 4)(x - 9) will be ≤ 0 in two regions, x ≤ 2 and 4 ≤ x ≤ 9, but we already inferred that x ≠ {4 , 9}
• And we know that x is a positive integer
• Therefore, the given inequality satisfies for x = {1, 2, 5, 6, 7, 8}

Hence, the correct answer is option C.

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If x is a positive integer, how many values of x satisfy the..........  [#permalink]

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EgmatQuantExpert wrote:

Solution

Given:
• We are given that x is a positive integer
• And, we are given an inequality, $$\frac{(x - 2)}{(x^2 – 13x + 36)} ≤ 0$$

To find:
• We need to find out the number of values of x that satisfies the given inequality

Approach and Working:
• If we observe the given inequality, we can see that the denominator is a quadratic expression.
• So, let’s try to factorise the quadratic expression, $$x^2 – 13x + 36$$
o $$x^2 – 13x + 36 = (x – 9) * (x – 4)$$

• Thus, the inequality can be written as, $$\frac{(x - 2)}{(x – 9)(x - 4)} ≤ 0$$
Now, we need to multiply the numerator and denominator of LHS with (x – 9)*(x – 4)
o $$\frac{(x - 2) (x – 9)(x - 4)}{[(x – 9)(x - 4)]^2} ≤ 0$$
o The denominator of the above inequality is always greater than 0 and x ≠ 4 or 9, since the denominator cannot be equal to 0.
o So, the numerator must be ≤ 0, for $$\frac{(x - 2)(x – 9)(x - 4)}{[(x – 9)(x - 4)]^2} ≤ 0$$.
o We need to find the values of x, for which (x - 2) (x – 4)(x - 9) ≤ 0

Approach 1: Wavy-line method

• The zero points of the inequality are {2, 4, 9} and the wavy-line will be as follows: • So, the expression will be equal to zero for x = {2, 4, 9}
o But, as we already know that, x cannot take the values 4 and 9
o Thus, x = {2}
• The expression will be negative in the regions, 4 < x < 9 and x < 2
o But we know that x is a positive integer, so, the expression will be negative for x = {1, 5, 6, 7, 8}
• Therefore, the given inequality satisfies for x = {1, 2, 5, 6, 7, 8}

Approach 2: Number-line method

The zero points of the inequality are {2, 4, 9} and highlighting these points on a number line, we get: • Thus, (x - 2) (x – 4)(x - 9) will be ≤ 0 in two regions, x ≤ 2 and 4 ≤ x ≤ 9, but we already inferred that x ≠ {4 , 9}
• And we know that x is a positive integer
• Therefore, the given inequality satisfies for x = {1, 2, 5, 6, 7, 8}

Hence, the correct answer is option C.

i would appreciate you could take time to answrr my questions below thank you! Question 1. Why do we need to multiply the numerator and denominator of LHS with (x – 9)*(x – 4) in which cases do we need to do so ?

Question 2. if the denominator of the above inequality is always greater than 0 and x ≠ 4 or 9, since the denominator cannot be equal to 0. WHY do we remove whole denominator in the end, leave only this (x - 2) (x – 4)(x - 9) ≤ 0 ???

Question 3. And the last question is it gramatically paralel structure to say "Approach and Working" may be "Aprroach and Work" is better or "Approach first and then start Working" how about "approach while working and by the time you approach it will be solved" Senior PS Moderator V
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dave13 wrote:

i would appreciate you could take time to answrr my questions below thank you! Question 1. Why do we need to multiply the numerator and denominator of LHS with (x – 9)*(x – 4) in which cases do we need to do so ?

Question 2. if the denominator of the above inequality is always greater than 0 and x ≠ 4 or 9, since the denominator cannot be equal to 0. WHY do we remove whole denominator in the end, leave only this (x - 2) (x – 4)(x - 9) ≤ 0 ???

Question 3. And the last question is it gramatically paralel structure to say "Approach and Working" may be "Aprroach and Work" is better or "Approach first and then start Working" how about "approach while working and by the time you approach it will be solved" Hey dave13

The answer to question 1 is that they do so to make the denominator a square.
Remember a square of a number/expression can never be negative. Now, that
we have that in order, for the overall expression to be neagtive or equal to zero,
the numerator must be negative or have a value of zero at maximum.

That's the reason why they remove the denominator and keep only the numerator
Hope that answers your second question. As for your thid question, I am sure generis will have an answer. Unfortunately,
I am not qualified to make that judgement  _________________
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Senior PS Moderator D
Status: It always seems impossible until it's done.
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GMAT 1: 740 Q50 V40 GMAT 2: 770 Q51 V42 Re: If x is a positive integer, how many values of x satisfy the..........  [#permalink]

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dave13,

I'd like to add that this is a common trick in inequality questions, to make perfect squares and then "ignore them" as the perfect squares of real numbers are greater than or equal to zero.

However, as an advanced step you can go ahead and skip that part. Treat the numerator and denominator the same and use the wavy approach. ( While multiplying by same factor in both num and den we are doing this itself!)

To add a little to q2 raised by you above, -ve * +ve is -ve and +ve * +ve is +ve... Which means multiplying by positive entity does not change the sign... Hence we can ignore the perfect square part in denominator.

Q3: no clue what you're talking about! Let me know if you wanna followup.

Best,

Posted from my mobile device
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Re: If x is a positive integer, how many values of x satisfy the..........  [#permalink]

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EgmatQuantExpert wrote:
If x is a positive integer, how many values of x satisfy the inequality, $$\frac{(x - 2)}{(x^2 – 13x + 36)} ≤ 0$$?

A. 4
B. 5
C. 6
D. 7
E. Cannot be determined

Asked: If x is a positive integer, how many values of x satisfy the inequality, $$\frac{(x - 2)}{(x^2 – 13x + 36)} ≤ 0$$?

(x-2)/(x-4)(x-9) <=0

Possible values of x = {1,2,5,6,7,8}

IMO C
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Re: If x is a positive integer, how many values of x satisfy the..........  [#permalink]

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Why can't 0 be one of the integer.

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Re: If x is a positive integer, how many values of x satisfy the..........  [#permalink]

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rahulkashyap wrote:
How did you get 1 2 5 6 7 and 8 from those 2 intervals?

Posted from my mobile device

3 is also satisfying the equation
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Re: If x is a positive integer, how many values of x satisfy the..........  [#permalink]

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EgmatQuantExpert wrote:
If x is a positive integer, how many values of x satisfy the inequality, $$\frac{(x - 2)}{(x^2 – 13x + 36)} ≤ 0$$?

A. 4
B. 5
C. 6
D. 7
E. Cannot be determined

Attachments

File comment: You can use this method of plotting transition points on the number line.

This is fast method and with practice one can get used to it. 6226da61-516a-4a8a-b3d5-c724a9240d3f.jpg [ 81.43 KiB | Viewed 536 times ]

_________________ Re: If x is a positive integer, how many values of x satisfy the..........   [#permalink] 25 Apr 2020, 06:57

# If x is a positive integer, how many values of x satisfy the..........  