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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.