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YES, definitely, you can add two inequalities that have the same direction.

In the above DS question, obviously neither (1) nor (2) alone is sufficient.

(1) and (2) together: Adding the two inequalities side-by-side we obtain \(ab^2-b+b>b^2c-d+d\) or \(b^2(a-c)>0\), which means necessarily \(a-c>0\) or \(a>c.\) Sufficient.

Answer C _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

You can only add inequalities when their signs are in the same direction:

If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) --> \(a+c>b+d\). Example: \(3<4\) and \(2<5\) --> \(3+2<4+5\).

You can only apply subtraction when their signs are in the opposite directions:

If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) --> \(a-c>b-d\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) --> \(3-5<4-1\).

You can only add inequalities when their signs are in the same direction:

If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) --> \(a+c>b+d\). Example: \(3<4\) and \(2<5\) --> \(3+2<4+5\).

You can only apply subtraction when their signs are in the opposite directions:

If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) --> \(a-c>b-d\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) --> \(3-5<4-1\).

Hope it helps.

When we subtract two inequalities with their signs in opposite directions, we are in fact using addition of two inequalities in the same direction:

\(a>b\)

\(C<D\) -> this can be rewritten as

\(-C>-D\)

Now we can add the first and the third inequality, because they have the same direction and get \(a-C>b-D.\) _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: Can we add inequalities [#permalink]
12 Aug 2013, 07:05

EvaJager wrote:

rphardu wrote:

Is a > c?

(1) b > d

(2) ab2 – b > b2c – d

(1) and (2) together: Adding the two inequalities side-by-side we obtain \(ab^2-b+b>b^2c-d+d\) or \(b^2(a-c)>0\), which means necessarily \(a-c>0\) or \(a>c.\) Sufficient.

Answer C

I don't understand the solution beyond this part...\(b^2(a-c)>0\) as per me by dividing both sides of equation by \(b^2\) we are assuming value of B is not equal 0, else it will be 0/0 which is not defined. so how is this correct? _________________

Life is very similar to a boxing ring. Defeat is not final when you fall down… It is final when you refuse to get up and fight back!

Re: Can we add inequalities [#permalink]
12 Aug 2013, 09:22

1

This post received KUDOS

nikhil007 wrote:

EvaJager wrote:

rphardu wrote:

Is a > c?

(1) b > d

(2) ab2 – b > b2c – d

(1) and (2) together: Adding the two inequalities side-by-side we obtain \(ab^2-b+b>b^2c-d+d\) or \(b^2(a-c)>0\), which means necessarily \(a-c>0\) or \(a>c.\) Sufficient.

Answer C

I don't understand the solution beyond this part...\(b^2(a-c)>0\) as per me by dividing both sides of equation by \(b^2\) we are assuming value of B is not equal 0, else it will be 0/0 which is not defined. so how is this correct?

From (1) and (2) we see \(b^2(a-c)>0\)

Since LHS >0 we must have \(b =! 0\) and\(a > c\) as if \(b = 0\) then LHS = 0 . _________________

The question is not can you rise up to iconic! The real question is will you ?

YES, definitely, you can add two inequalities that have the same direction.

In the above DS question, obviously neither (1) nor (2) alone is sufficient.

(1) and (2) together: Adding the two inequalities side-by-side we obtain \(ab^2-b+b>b^2c-d+d\) or \(b^2(a-c)>0\), which means necessarily \(a-c>0\) or \(a>c.\) Sufficient.

Answer C

I understood till this point - b^2 (a-c) > 0 can someone explain after this step please. _________________

YES, definitely, you can add two inequalities that have the same direction.

In the above DS question, obviously neither (1) nor (2) alone is sufficient.

(1) and (2) together: Adding the two inequalities side-by-side we obtain \(ab^2-b+b>b^2c-d+d\) or \(b^2(a-c)>0\), which means necessarily \(a-c>0\) or \(a>c.\) Sufficient.

Answer C

I understood till this point - b^2 (a-c) > 0 can someone explain after this step please.

We have \(b^2(a-c)>0\) (\(b\neq{0}\)). Now, since \(b^2>0\), then the other multiple must also be greater than 0 --> \(a-c>0\) --> \(a>c.\).

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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