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I'm starting a new thread to collect some traps that are [#permalink]
12 Oct 2006, 19:12

1

This post received KUDOS

2

This post was BOOKMARKED

I'm starting a new thread to collect some traps that are commonly seen in GMAT. Please try to form your own answer to each question before you highlight the spoiler to see the official answer. You could also find more discussions regarding each question by searching the forum to find the original threads.

Please feel free to add to the thread when you discover a trap.

Hong

iced_tea wrote:

Is integer x positive ?

1) x > x^3 2) x < x^2

[spoiler]OA: A, trap: "integer"[/spoiler]

kevincan wrote:

A triangle has sides of length 12, 15 and k units. What is the length of the longest side?

(1) The perimeter of the triangle is a multiple of 9. (2) One of the interior angles of the triangle is equal to the sum of the other two interior angles.

[spoiler]OA: C, trap: special triangle[/spoiler]

haas_mba07 wrote:

If x is to be selected at random from set T, what is the probability that x/4 -5 < 0?

(1). T is a set of 8 integers (2). T is contained in the set o fintegers from 1 to 25, inclusive.

[spoiler]OA: E, trap: T is contained in a set[/spoiler]

gmatornot wrote:

If A is a prime number, what is the value of A?

(1) 0 < A <10>0 but A-2 can be 0[/color][/spoiler]

u2lover wrote:

Is 25/(a+1) an integer?

(1) a is a multiple of 25 (2) 0<=a<=25

[spoiler]OA: E, trap: 0 is a multiple of any umber[/spoiler]

n_sathya wrote:

In a class, 20 students speak spanish, 15 speak french and 15 speak chinese. how many students are in the class? 1. exactly 30% of the students in the class speak french 2. all of the students speak english and none speaks more than two languages.

[spoiler]OA: A, trap: You can't add students who speak Spanish, French and Chinese to get the number of students in the class. There might be students who speak Japanese and English, or any other languages. There could also be students who can only speak English.[/spoiler] _________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

Last edited by HongHu on 21 Dec 2006, 12:54, edited 5 times in total.

The first statement alone is sufficient .
Here is the break down

1. X>X^3
If u take postive value of X , then above value is false .
If u take x=-3 ,then x^3 = -27 , hence the given A is true and we know
X is not a positive integer.

2. X<X^2 - now its almost a common knowldege tht whenever u see ^2 or |X| ,its a trap .

For example x=2 < X^2 = 4 - So true
also if X=-2 < -2^2 = 4

For people who may not aware, I've been collecting different questions and updating the opening post. In other words there are more than one question there now. I think the best way to go is to compare your answer with the spoiler. If it is correct then no need to post. However if you have a question please post it here by all means.

I like Fig's suggestion too. I'll try to do something like that when we have a decent collection of the traps. _________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

The first statement alone is sufficient . Here is the break down

1. X>X^3 If u take postive value of X , then above value is false . If u take x=-3 ,then x^3 = -27 , hence the given A is true and we know X is not a positive integer.

2. X<X^2 - now its almost a common knowldege tht whenever u see ^2 or |X| ,its a trap .

For example x=2 < X^2 = 4 - So true also if X=-2 < -2^2 = 4

Hence insufficient .

So 1 alone is sufficient .

hope this helps

Thanks
I got trapped and for such silly lack of ratiocination. This is what happens when one is not fuuly attentive to the problem.
Thanks again.

1. x>x^3
=> x-x^3>0
=>x(1-x^2)>0
=>x(1-x)(1+x)>0
x cannot be 0, -1 or +1 or the inequality does not work.
So if x=+2, then 2*-1*3=-6 which is NOT > 0 (the inequality is not true)
if x=-2, then -2*3*-1 =+6 which is >0. so x is -ve
therefore x is negative.

2. x<x^2
=>x-x^2<0
=>x(1-x)<0
x cannot be o or 1
lets say x=+2, then inequality is true
lets say x=-2, then x(1-x)=-2*3=-6 <0 which is true.

In both cases (x being +ve and -ve), the inequality holds true and so this choice doesnot help determine x.

Re: Math trap collection [#permalink]
08 Nov 2006, 12:02

kevincan wrote:

A triangle has sides of length 12, 15 and k units. What is the length of the longest side?

(1) The perimeter of the triangle is a multiple of 9. (2) One of the interior angles of the triangle is equal to the sum of the other two interior angles.

From the stem and the triangle rules of 3 sides: 15-12<k<15+12 or 3<k<27.

(1) Perimeter = 12+15+k = 27+k. P is also a multiple of 9, thus P must be number such as 27,36,45,54 etc. However, since k is limited to the range of 3 to 27 exclusive, P must either be 36 or 45. That means k is either 9 or 18. Insuff.

(2) This means the triangle is a right triangle. However, 15 does not need to be the hypotenus of the triangle. There are 2 cases.
* 15 is the hypotenus. Then we have the special triangle 9-12-15.
* 15 is one of the leg. Then, using the theorem, the hypotenus k is about 19 (k^2 = 12^2 + 15^2= 369). Since k is still in the range provided in the stem. This is insuff to tell.

1+2: from 1: k = 9 or 18.
from 2: k = 9 or around 19. Thus, K must be 9 and the longest side is 15. Suff

but for the second it cannot be concluded that two angels of the triangle are equal. It can be that y=z+x (all angles are different) from this we can conclude that x+y+z = 180 i.e. y+y=180, y=90 so there is right triangle. But this is not enough because either side can have 90degrees angle so we need first statement: 12^2+15^=k^2, or 15^-12^2=k^2 and so on untill we will get the perimeter which will be the multiple of nine.
Am i right?

When to consider zero in DS questions ? [#permalink]
18 Nov 2006, 10:27

Is 25/(a+1) an integer?

(1) a is a multiple of 25
(2) 0<=a<=25

[spoiler]OA: E, trap: 0 is a multiple of any number[/spoiler]

Would "0" be considered a multiple ?

I consider "0" when calculating for a number, integer, even number, to the power of (any number to the power of zero is 1, note 0 to the power 0 does not exist ?).

I do not consider "0" as a divisor (any nymer divided by does not exist), sq. root (should I consider it ?? -- example - sq.root of x = x), and also as a multiple.(should it be considered as a multiple?)

but for the second it cannot be concluded that two angles of the triangle are equal. It can be that y=z+x (all angles are different) from this we can conclude that x+y+z = 180 i.e. y+y=180, y=90 so there is right triangle. But this is not enough because either side can have 90degrees angle so we need first statement: 12^2+15^=k^2, or 15^-12^2=k^2 and so on untill we will get the perimeter which will be the multiple of nine. Am i right?

I agree with you smiley - we cannot conclude that the two angels are the same but we do still need the first statement so I'm getting answer as C. However, when working through the problem I cant seem to get a suitable numerical answer.

We know that there is an angle of 90 degrees so if the triangle were an isosceles triangle then the sides would have to be in the ratio x: x: x*(sqrt2) which would not allow for sides of 12 and 15 or a perimeter with a multiple of 9.

From 1) we know the perimeter is a multiple of 9. The only configuration of sides with a 90 degree internal angle that would allow this is for the sides to be in the ratio 3:4:5 or 9:12:15 - giving a perimeter of 36.

C is the answer

Last edited by MBAlad on 05 Dec 2006, 14:35, edited 2 times in total.

Re: make a silly mistake - here ! [#permalink]
13 Dec 2006, 06:10

ItÂ´s A.

1): r<10 => r = 8 or 6 or 4, but as we also have p<q<r then r has to be 8, q = 6 and p = 4. ThereÂ´s no other set of numbers that fulfill the conditions stated so A is sufficient.

2): If p<6 then there are infinite values q can have, so B is not sufficient.

BLISSFUL wrote:

If p,q,r are even numbers such that 2<p<q<r what is the value of q ? 1) r<10 2)p<6

guys, take as less time as possible while trying this and let me know the answer you are getting..

Re: make a silly mistake - here ! [#permalink]
21 Dec 2006, 09:25

Andr359 wrote:

ItÂ´s A.

1): r<10 => r = 8 or 6 or 4, but as we also have p<q<r then r has to be 8, q = 6 and p = 4. ThereÂ´s no other set of numbers that fulfill the conditions stated so A is sufficient.

2): If p<6 then there are infinite values q can have, so B is not sufficient.

BLISSFUL wrote:

If p,q,r are even numbers such that 2<p<q<r what is the value of q ? 1) r<10 2)p<6

guys, take as less time as possible while trying this and let me know the answer you are getting..

Re: make a silly mistake - here ! [#permalink]
21 Dec 2006, 10:55

deep, I think you are thinking too deep -deeper than GMAT !??

I think even number means it refers to integer only.
Do we have even real numbers like 1.2,1.4 etc ??

whadya say folks ????
---------------------------------------------------------

deep wrote:

Andr359 wrote:

ItÂ´s A.

1): r<10 => r = 8 or 6 or 4, but as we also have p<q<r then r has to be 8, q = 6 and p = 4. ThereÂ´s no other set of numbers that fulfill the conditions stated so A is sufficient.

2): If p<6 then there are infinite values q can have, so B is not sufficient.

BLISSFUL wrote:

If p,q,r are even numbers such that 2<p<q<r what is the value of q ? 1) r<10 2)p<6

guys, take as less time as possible while trying this and let me know the answer you are getting..

LOL yes I agree that even numbers have to be integers.

Another trap:

n_sathya wrote:

In a class, 20 students speak spanish, 15 speak french and 15 speak chinese. how many students are in the class? 1. exactly 30% of the students in the class speak french 2. all of the students speak english and none speaks more than two languages.

[spoiler]OA: A, trap: You can't add students who speak Spanish, French and Chinese to get the number of students in the class. There might be students who speak Japanese and English, or any other languages. There could also be students who can only speak English.[/spoiler] _________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

LOL yes I agree that even numbers have to be integers.

Another trap:

n_sathya wrote:

In a class, 20 students speak spanish, 15 speak french and 15 speak chinese. how many students are in the class? 1. exactly 30% of the students in the class speak french 2. all of the students speak english and none speaks more than two languages.

[spoiler]OA: A, trap: You can't add students who speak Spanish, French and Chinese to get the number of students in the class. There might be students who speak Japanese and English, or any other languages. There could also be students who can only speak English.[/spoiler]

--------------------------------------------
Yes HongHu,
This was a classic trap that we had solved recently..
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