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amanvermagmat
Is it possible to form a triangle with its three sides as 'a', 'b' and 'c' units respectively?

(1) a + b < c

(2) a - b < c

Hi amanvermagmat

In the question, its not mentioned that a,b,c >0 or not. Hence, E. Not sure if I am missing anything.
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amanvermagmat
Is it possible to form a triangle with its three sides as 'a', 'b' and 'c' units respectively?

(1) a + b < c

(2) a - b < c

Hi amanvermagmat

In the question, its not mentioned that a,b,c >0 or not. Hence, E. Not sure if I am missing anything.

Hello Rahul

Thanks for pointing that out. I have now edited the question to mention that a, b, c are all positive numbers.
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amanvermagmat
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amanvermagmat
Is it possible to form a triangle with its three sides as 'a', 'b' and 'c' units respectively?

(1) a + b < c

(2) a - b < c

Hi amanvermagmat

In the question, its not mentioned that a,b,c >0 or not. Hence, E. Not sure if I am missing anything.

Hello Rahul

Thanks for pointing that out. I have now edited the question to mention that a, b, c are all positive numbers.

Thanks amanvermagmat

As a,b,c > 0, by combining the two (i.e. by subtracting the two equations) we get b < 0. But as per question b > 0. I see a contradiction here, can you help me to point out the flaw or the way forward. As I understand, the statements shouldn't contradict the question statement.
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As a,b,c > 0, by combining the two (i.e. by subtracting the two equations) we get b < 0. But as per question b > 0. I see a contradiction here, can you help me to point out the flaw or the way forward. As I understand, the statements shouldn't contradict the question statement.[/quote]

Hello

As per my understanding, we cannot subtract inequalities which are in the same direction. We can only add them.
So on adding the two inequations, we get a+b+a-b < c+c OR 2a < 2c OR a < c
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amanvermagmat
As a,b,c > 0, by combining the two (i.e. by subtracting the two equations) we get b < 0. But as per question b > 0. I see a contradiction here, can you help me to point out the flaw or the way forward. As I understand, the statements shouldn't contradict the question statement.

Hello

As per my understanding, we cannot subtract inequalities which are in the same direction. We can only add them.
So on adding the two inequations, we get a+b+a-b < c+c OR 2a < 2c OR a < c[/quote]

Thanks amanvermagmat. The answer should be A.

Statement I: As per property of triangle, sum of two sides must be greater than third side. So, not possible.

Statement II: Let \((a,b,c) = (0.5,1,0.5)\) --> The Traingle cannot be formed as it violates property of Triangle.
Let \((a,b,c) = (2,1,4)\) --> The traingle can be formed as its in accordance with Property of Triangle.

Hence, A.
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IMO A.

Statement 1: It is clear . C>a+b. Not possible to form a triangle. Hence a clear no
Statement 2: c+b>a. but what about other combinations. a+b / a+c. Hence not clear.

Please clarify if I am wrong
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amanvermagmat
Is it possible to form a triangle with its three sides as 'a', 'b' and 'c' units respectively, such that a, b and c are all positive numbers?

(1) a + b < c

(2) a - b < c

amanvermagmat I think it was clear in the previous version also. The question stated " triangle with its three sides as 'a', 'b' and 'c' units".
Length of the sides can never be of non-positive unit length !
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amanvermagmat
Is it possible to form a triangle with its three sides as 'a', 'b' and 'c' units respectively, such that a, b and c are all positive numbers?

(1) a + b < c

(2) a - b < c

amanvermagmat I think it was clear in the previous version also. The question stated " triangle with its three sides as 'a', 'b' and 'c' units".
Length of the sides can never be of non-positive unit length !


Hello Rumanshu

Yes I think you are correct. Question was alright in its original form also, thanks.
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First thing to mention:
In DS the statements don't contradict each other, If you arrive that such triangle doesn't exist with first statement and such triangle exists with second statement, the first thing you should do is to check the statements once again and detect the trap

1st statement: I dont see any problem, there is a rule and according to this rule such Triangle cannot exist-- Suffic
2nd statement: Here we have a trap, the rule says: The Positive difference of any two sides of triangle must be less than the third side. we dont know a-b is a positive difference or not. a=5 b=7 c=8 Such triangle has a right to exist but a-b<c as a-b is a negative number and always will be less than positive third side. -- Insuff.

IMO
Ans: A
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I dont understand why Statement 2 alone is not sufficient.

We're given a-b<c, which can be re-arranged as a<b+c. This means that the sum of 2 sides is greater than the 3rd side, which is always true for a triangle. Thus, it's possible to form a triangle with sides a,b,c.
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