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# if x is a positive integer x = ? (1) {(x-5)(x+4)}^1/2 >10

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Senior Manager
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if x is a positive integer x = ? (1) {(x-5)(x+4)}^1/2 >10 [#permalink]  10 Jun 2008, 21:44
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if x is a positive integer x = ?

(1) {(x-5)(x+4)}^1/2 >10
(2) {(x-5)(x+4)}^1/2 <11

Analysing statement 1

oh k if we solve the inequality

{(x-5)(x+4)}^1/2 >10

squaring both sides we get

(x-5)(x+4) >100

which means x-5>100 or x+4 >100
which translates to x>105 or x>96
which obviously means x>105

therefore x can be any number greater than 105. so this statement 1 is insufficent

similarly analysing statement 2 we get

{(x-5)(x+4)}^1/2 <11

squaring both sides we get

(x-5)(x+4) <121

which means x-5<121 or x+4 <121
which translates to x<126 or x<117
which obviously means x<126

therefore x can be any number less than 126. so this statement 2 is insufficent

put together both statements we get to know that x can be any value greater than 105 and less than 126

which means there are a number of possible values for x

therefore i deduce that E is the answer

which however, is not the OA

Can you please tell me where i am going wrong or is the OA for this one is wrong

wud appreciate u r help on this one guys
_________________

The world is continuous, but the mind is discrete

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Re: DS question [#permalink]  10 Jun 2008, 22:01
1)
if we simplify this , we get
x*(x-1) >120 ....not suff

2) if we simplyfy this ,we get

x*(x-1) < 141 ....again not suff

combining both the only value possible is x*(x-1) = 132 , so x=12 ...ANS C
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Re: DS question [#permalink]  11 Jun 2008, 06:10
Hi, vdhawan1!

I agree with the solution that the previous author has posted, but still decided to post since it might be helpful for you to know where the flaw in your reasoning was.

This is how you approached the inequality:

Quote:
(x-5)(x+4) >100

which means x-5>100 or x+4 >100.

This is where the mistake is. In general, a*b>c is not equivalent to a>c or b>c. In fact, the latter does not even follow from the former. Consider a*b>6, for example. a=2, b=4 satisfy this inequality, while neither is greater than 6.

Then, you wrote

Quote:
(x-5)(x+4) <121

which means x-5<121 or x+4 <121

Here you applied the same method. While in this case x-5<121 or x+4 <121 indeed follow from the initial inequality, this two inequalities are not equivalent to the initial one, since the range of possible solutions for the system of these two inequalities is larger than that of the initial inequality. Consider x=100. It is one of the solutions of the system {x-5<121 or x+4 <121}, but not of the initial inequality.

In general, I think that it might be useful to review the properties of inequalities of various types, the equivalent ways to transform them etc.
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Re: DS question [#permalink]  11 Jun 2008, 22:09
greenoak wrote:
Hi, vdhawan1!

I agree with the solution that the previous author has posted, but still decided to post since it might be helpful for you to know where the flaw in your reasoning was.

This is how you approached the inequality:

Quote:
(x-5)(x+4) >100

which means x-5>100 or x+4 >100.

This is where the mistake is. In general, a*b>c is not equivalent to a>c or b>c. In fact, the latter does not even follow from the former. Consider a*b>6, for example. a=2, b=4 satisfy this inequality, while neither is greater than 6.

Then, you wrote

Quote:
(x-5)(x+4) <121

which means x-5<121 or x+4 <121

Here you applied the same method. While in this case x-5<121 or x+4 <121 indeed follow from the initial inequality, this two inequalities are not equivalent to the initial one, since the range of possible solutions for the system of these two inequalities is larger than that of the initial inequality. Consider x=100. It is one of the solutions of the system {x-5<121 or x+4 <121}, but not of the initial inequality.

In general, I think that it might be useful to review the properties of inequalities of various types, the equivalent ways to transform them etc.

ok i see the mistake now and thanks a lot for pointing it out

i have reviewed the wiki guide (http://gmatclub.com/wiki/Algebra) on this and i see the following

consider the following example

x^2>x
x^2-x>0
x(x-1) >0
now either both x>0 and x-1>0 which translates to x>0 and x>1
or
both x<0 and x-1<0 which translates to x<0 and x<1
so the solution can be
x>1 or x<0

now coming back to the question
(x-5)(x+4) >100
therefore this means
(x-5)>100 or (x-5)<100 which translates to x>105 or x<105

(x+4) >100 or (x+4)<100 which translates to x>96 or x<96
combining both statements together we get the solution
as
x>96 or x<96

similarly
now coming back to the question
(x-5)(x+4) <121
therefore this means
(x-5)<121 or (x-5)>121 which translates to x<126 or x>126

so statement 1 does not help

(x+4) <121 or (x+4)>121 which translates to x<117 or x>125
combining both statements together we get the solution
as
x<117 or x>125
so statement 2 does not help
combining 2 statements together we get
x<117 or x>125

from this again it can be any value greater than 125 or less than 117

so shdnt E be the answer

wud appreciate u r help , if u can tell me where exactly i am going wrong
Many thanks
_________________

The world is continuous, but the mind is discrete

GMAT Instructor
Joined: 04 Jul 2006
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Re: DS question [#permalink]  12 Jun 2008, 04:11
If x is a positive integer x = ?

(1) {(x-5)(x+4)}^1/2 >10
(2) {(x-5)(x+4)}^1/2 <11

Together, we see that 100 < (x - 5)(x + 4) < 121, where x is a positive integer

Note that x - 5 and x + 4 differ by 9

15 x 6 < 100
100 < 16 x 7 < 121
17 x 8 > 121

Thus x + 4 must be 16, i.e. x = 12
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Re: DS question [#permalink]  12 Jun 2008, 04:25
2
KUDOS
Hello again, vdhawan1!

Quote:
now coming back to the question
(x-5)(x+4) >100
therefore this means
(x-5)>100 or (x-5)<100 which translates to x>105 or x<105

If we have zero on the right side, this reasoning would be correct. Otherwise, it is not applicable. So, in this case, in order to apply this logic, we should transform the equation so that it will have zero.

Thus, we may write:
(x-5)(x+4) = x^2-x-20 and thus we obtain
x^2-x-20>100 and consequently
x^2-x-120>0

This last one we could have solved by factorizing (theoretically), as you suggested, only it is not practical in the given example, because the roots of this quadratic equation are not integers, and we are looking for an integer x. Moreover, from the form of the last equation, it is clear that we have infinite possibilities for x, so clearly it is not suff.

That’s why it’s better to left the equation in form:
x^2-x>120 =>
x(x-1)>120

and look at the second one.

The equation from the second condition, written as x^2-x-141<0, again, leaves more than one possibility for x and it’s impractical to solve the q. equation, again.

So, let’s leave it in the form x(x-1)<141

And, finally, we have
120< x(x-1)< 141

and at this stage it’s time to pick numbers . Luckily, only one possibility for x, namely x=12, satisfies the equation.

***
So, in general, if you have an eq. of the kind (x-a)(x-b)>c (or < c), you may apply the reasoning you described only if c=0. Otherwise, it won’t work, and we need to search for another approach.
P.S. I've PM you.
Senior Manager
Joined: 29 Aug 2005
Posts: 283
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Kudos [?]: 30 [0], given: 0

Re: DS question [#permalink]  12 Jun 2008, 05:45
greenoak wrote:
Hello again, vdhawan1!

Quote:
now coming back to the question
(x-5)(x+4) >100
therefore this means
(x-5)>100 or (x-5)<100 which translates to x>105 or x<105

If we have zero on the right side, this reasoning would be correct. Otherwise, it is not applicable. So, in this case, in order to apply this logic, we should transform the equation so that it will have zero.

Thus, we may write:
(x-5)(x+4) = x^2-x-20 and thus we obtain
x^2-x-20>100 and consequently
x^2-x-120>0

This last one we could have solved by factorizing (theoretically), as you suggested, only it is not practical in the given example, because the roots of this quadratic equation are not integers, and we are looking for an integer x. Moreover, from the form of the last equation, it is clear that we have infinite possibilities for x, so clearly it is not suff.

That’s why it’s better to left the equation in form:
x^2-x>120 =>
x(x-1)>120

and look at the second one.

The equation from the second condition, written as x^2-x-141<0, again, leaves more than one possibility for x and it’s impractical to solve the q. equation, again.

So, let’s leave it in the form x(x-1)<141

And, finally, we have
120< x(x-1)< 141

and at this stage it’s time to pick numbers . Luckily, only one possibility for x, namely x=12, satisfies the equation.

***
So, in general, if you have an eq. of the kind (x-a)(x-b)>c (or < c), you may apply the reasoning you described only if c=0. Otherwise, it won’t work, and we need to search for another approach.
P.S. I've PM you.

Greenoak

Buddy cant thank u enough for this help

now i see where i m going wrong

Kudos to u for this
_________________

The world is continuous, but the mind is discrete

Re: DS question   [#permalink] 12 Jun 2008, 05:45
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# if x is a positive integer x = ? (1) {(x-5)(x+4)}^1/2 >10

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