Is t a non integer?
(1) |t-3|<1
(2) t(does not not = )3
I understand that it's not (2)
but I am confused on (1) only because this is the solution:
|t-3|<1
-1<t<-3 can someone explain this concept to me
2<t<4 and then this part, where i got lost as well.
shouldnt it be
|t-3|<1
t<4???
any help appreciated
thanks
danny
silly inequality question :)
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dzelkas wrote:Is t a non integer?
(1) |t-3|<1
(2) t(does not not = )3
I understand that it's not (2)
but I am confused on (1) only because this is the solution:
|t-3|<1
-1<t<-3 can someone explain this concept to me
2<t<4 and then this part, where i got lost as well.
shouldnt it be
|t-3|<1
t<4???
any help appreciated
thanks
danny
Let me rewrite it to make things clearer...
Is t a non integer?
(1) |t - 3| < 1.
(2) t ≠3.
I. If t is an integer, then so is t - 3; and if |t - 3| < 1, then it's possible only if the integer t is 3. Reason being that the absolute value of any integer is non-negative, that's why by means of t being an integer in the company of |t - 3| < 1 would lead us to believe that |t - 3| = 0 or t = 3. All we can decide is either t is 3 or a non integer. Insufficient
II. If t is not 3, then t may or may not be an integer. Insufficient
Taking I and II together, we can [spoiler]definitely say that t is a non integer. Sufficient
Take C[/spoiler]
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- Whitney Garner
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Hi Danny!dzelkas wrote:but I am confused on (1) only because this is the solution:
|t-3|<1
-1<t<-3 can someone explain this concept to me
2<t<4 and then this part, where i got lost as well.
shouldnt it be
|t-3|<1
t<4???
Let's first attack the problem of solving an inequality when there is an absolute value. This should help the confusion you mentioned:
If we need to solve the expression |t-3| < 1, we first need to "jailbreak" that (t-3) from the absolute value symbols. To do so, we need to consider that 1 of 2 things may be true.
(A) (t-3) might be positive; therefore the absolute value symbols aren't really doing anything.
(B) (t-3) is actually negative; therefore the absolute value symbols are working to change the sign of (t-3).
Okay, so let's take (A) since it's easier. If the absolute value symbols aren't doing anything, we can just remove them and solve:
(t-3) < 1
t < 4
Now, we need to address scenario (B). If t-3 is actually negative, then if we throw away the absolute value symbol, WE need to manually change the sign on (t-3). We can accomplish that by just multiplying it times a -1.
-1*(t-3) < 1
-t + 3 < 1
2 < t
So, if we put A and B together, we get the proper range of values for |t-3|<1 = 2 < t < 4 (test some values to check...if I plug in 2 and 4, I see that it equals 1, and if I pick stuff in between (2.5, 3, 3.5, etc) I see that it works!
Now, let's get back to the original problem:
Statement (1) only tells us that 2<t<4, but doesn't guarantee we get an integer! NOT SufficientOriginal Question wrote:Is t a non integer?
(1) |t - 3| < 1.
(2) t ≠3.
Statement (2) only tells us that t≠3; it could therefore be any other possible integer or non-integer in the world! NOT Sufficient
Statement (1+2) = now we know that t must be a value between (but not including) 2 and 4. Well, the only integer in that range is 3, but statement 2 says it cannot be 2...Therefore we know that it MUST be a NON-integer! Sufficient
Hope this clears that up a bit!
Whit
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Math is a lot like love - a simple idea that can easily get complicated
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Whitney Garner wrote:Hi Danny!dzelkas wrote:but I am confused on (1) only because this is the solution:
|t-3|<1
-1<t<-3 can someone explain this concept to me
2<t<4 and then this part, where i got lost as well.
shouldnt it be
|t-3|<1
t<4???
Let's first attack the problem of solving an inequality when there is an absolute value. This should help the confusion you mentioned:
If we need to solve the expression |t-3| < 1, we first need to "jailbreak" that (t-3) from the absolute value symbols. To do so, we need to consider that 1 of 2 things may be true.
(A) (t-3) might be positive; therefore the absolute value symbols aren't really doing anything.
(B) (t-3) is actually negative; therefore the absolute value symbols are working to change the sign of (t-3).
Okay, so let's take (A) since it's easier. If the absolute value symbols aren't doing anything, we can just remove them and solve:
(t-3) < 1
t < 4
Now, we need to address scenario (B). If t-3 is actually negative, then if we throw away the absolute value symbol, WE need to manually change the sign on (t-3). We can accomplish that by just multiplying it times a -1.
-1*(t-3) < 1
-t + 3 < 1
2 < t
So, if we put A and B together, we get the proper range of values for |t-3|<1 = 2 < t < 4 (test some values to check...if I plug in 2 and 4, I see that it equals 1, and if I pick stuff in between (2.5, 3, 3.5, etc) I see that it works!
Now, let's get back to the original problem:
Statement (1) only tells us that 2<t<4, but doesn't guarantee we get an integer! NOT SufficientOriginal Question wrote:Is t a non integer?
(1) |t - 3| < 1.
(2) t ≠3.
Statement (2) only tells us that t≠3; it could therefore be any other possible integer or non-integer in the world! NOT Sufficient
Statement (1+2) = now we know that t must be a value between (but not including) 2 and 4. Well, the only integer in that range is 3, but statement 2 says it cannot be 2...Therefore we know that it MUST be a NON-integer! Sufficient
Hope this clears that up a bit!
Whit
coooolll the same as i did... its because you taught me about the inequallity before, i remember that i got mistake with this in another question and you corrected it... cool thank you