Solving DS questions with absolute value equations??

[Auzbee]

How do you solve such questions in less than 2 mins? Is there a quick strategy to solve DS questions that involve absolute value in equations?

If a and b are integers, and |a| > |b|, is a · |b| < a – b?

(1) a < 0

(2) ab 0

I would appreciate useful tips for such questions.

[camitava]

Auzbee,
There is a typo, man! I am not being able to understand about the second option - ab 0. What does it signify?
However, for a clarity I can suggest you a way to solve this kind of prob [Appears easier to me]. This type of problem really creates a confusion between positive and negatives. For worst, if a, b < 0, then a < b. And a . |b| is coming as less than a - b (As a & b are -ve, so the actual equation becomes -a + b). It is also true for b > 0 and a < 0.
If the second option says that a,b < 0 the above condition becomes true once again.
If the second option says that a, b > 0 then a . |b| becomes greater than a - b. But you can come to conclusion whether a . |b| < or > a - b. So I think OA is D [If and only if the second option is what I have assumed.].

[kajcha]

I would substitute values for a and b... I think that makes the task easy

[mayonnai5e]

It's not possible to solve this answer with the typo in (2) since it could be ab >0 or ab < 0 or ab = 0, each of which has implications on the problem.

But to answer your question, this is how I would approach this problem. Since the absolute value function is simple (e.g. |a| vs |3a+b|), I would categorize the variables into sets of possible values:

a --> {+, -, 0}
b --> {+, -, 0}
|a| --> {+, 0}
|b| --> {+, 0}
*remember your 0 properties and the stem did not state a and b are non-zero

a * |b| < a – b?
{+, -, 0} * {+, 0} < {+, -, 0} - {+, -, 0}

(1) a < 0
so a is not 0 and is negative, then the equation above looks like this:

a * |b| < a – b?
{-} * {+, 0} < {-} - {+, -, 0}

so the left hand side must be negative or 0

a * |b| < a – b?
{-} < {-} - {+, -} ?
0 < {-} - {0} ?

so now you can plug in different numbers knowing that you can try to find values that makes that last version of the equation sometimes true and sometimes false so that you can eliminate (1) as an answer choice, but you must remember that |a| > |b| as given by the stem.

let a = -2 then b = { -1, 0,1 } since |-2| = 2 and 2 > |b|:

a * |b| < a – b?
{-} < {-} - {-} ? becomes (-2)(1) < -2 - (-1) True
{-} < {-} - {+} ? becomes (-2)(1) < -2 - 1 False
0 < {-} - {0} ? becomes (-2)(0) < -2 - 0 False

So (1) is not sufficient.

You don't necessarily need to write out the formula with the different values, but I did it here to illustrate my point. You can just use the information given to constraint the values, then plug in; I wouldn't actually write out the formula so many times on a CAT because it's too time consuming. My suggestion is to take a methodical approach like the one above while practicing to get used to it.

[mayonnai5e]

Here's the faster way:

* a and b are integers
*|a| > |b|

a --> {+, -, 0}
b --> {+, -, 0}

(1) a < 0

constrain a:
a --> {-}

let a = -2

constrain b:
b = { -1, 0,1 } since |-2| = 2 and 2 > |b|

select values:
(-2)(1) < -2 - (-1) True
(-2)(1) < -2 - 1 False
(-2)(0) < -2 - 0 False

[Auzbee]

[quote="Auzbee"]
(2) ab 0
[/quote]

Folks sorry for not being clear on the second condition.
The second condition is:
(2) ab >= 0

[rtheman]

Folks sorry for not being clear on the second condition.
The second condition is:
(2) ab >= 0

---------

Now that we know Condition #2, here's my answer and let me know if this is what you all got:

for "ab >= 0", both a & b must have the same equality where either both a & b are Postive (+), or both a & b are Negative (-).

Using similar logic provide above (posts - which btw, they are awesome and neat trick to look at Absolute problems):

Postive: {0,+} * {0,+} < {0,+} - {0,+}; so for example:
if a=1 and b=2 --> 1*2 < 1 - 2 --> 2 < -2 --> FALSE
if a=2 and b=1 --> 2*1 < 2 - 1 --> 2 < 1 --> FALSE
if a=2 and b=0 --> 2*0 < 2 - 0 --> 2 < 0 --> FALSE

Negative: {0,-} * {0,+} < {0,-} - {0,-}; so for example:
if a=-2 and b=-1 --> -2*1 < -2 - (-1) --> -2 < -1 --> TRUE
if a=-1 and b=-2 --> -1*2 < -1 - (-2) --> -2 < 1 --> TRUE
if a=-2 and b=0 --> -2*0 < -2 - (0) --> 0 < -2 --> FALSE

So (2) is not sufficient due to inconsistence. And with So (1) being insufficient as well, that means the final result for this problem is E

p.s. I'd love to know if there's a short-cut because I feel like 6 examples for one DS problem is quite a bit, no?

[mayonnai5e]

Postive: {0,+} * {0,+} < {0,+} - {0,+}; so for example:
if a=1 and b=2 --> 1*2 < 1 - 2 --> 2 < -2 --> FALSE
if a=2 and b=1 --> 2*1 < 2 - 1 --> 2 < 1 --> FALSE
if a=2 and b=0 --> 2*0 < 2 - 0 --> 2 < 0 --> FALSE

Negative: {0,-} * {0,+} < {0,-} - {0,-}; so for example:
if a=-2 and b=-1 --> -2*1 < -2 - (-1) --> -2 < -1 --> TRUE
if a=-1 and b=-2 --> -1*2 < -1 - (-2) --> -2 < 1 --> TRUE
if a=-2 and b=0 --> -2*0 < -2 - (0) --> 0 < -2 --> FALSE

Just wanted to point out three things:
1) if a=1 and b=2 --> 1*2 < 1 - 2 --> 2 < -2 --> FALSE
if a=-1 and b=-2 --> -1*2 < -1 - (-2) --> -2 < 1 --> TRUE
==> these values do not work and should not be used because they violate the given condition: |a| > |b|

2) I would simply test one set of "picked numbers" for each set of possible values (e.g. do 1 set of numbers for both a, b positive and 1 set of numbers for both a,b negative) and see what the results are before doing more numbers. This saves time since had you just done one set of picked numbers for each, you would have seen pretty quickly that one is true while the other is false.

3) You forgot to check both (1) and (2) combined.