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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.

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.

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 ?

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!

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. _________________

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.

What if b=0 and d=-1?

In that situation, wouldn't the 2nd equation become:

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.

What if b=0 and d=-1?

In that situation, wouldn't the 2nd equation become:

a(0) > (0)c – d

If b = 0 and d = -1, then ab^2 – b = 0 and b^2c – d = 1. Anyway, what are you trying to say? _________________

Harvard asks you to write a post interview reflection (PIR) within 24 hours of your interview. Many have said that there is little you can do in this...