Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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.

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

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 _________________

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.

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.