If ac/(db) < 0, is d – b < 0 ?

Could you please explain why

|a − c| = |a| − |c|
This statement holds only if both a and c have the same sign

Thanks.

Hi,

If both a and c are of opposite sign....
|a-c| will result in a higher value than |a|-|c|, as a-c will result in numeric value of a and c getting added up, and then MODULUS value will make it a positive value..

lets check with example..
let a = 5, and c = -10...... a-c = 5-(-10)=15 and |15| = 15....... but |a| = |5| and |c| = |-10|, so |a|-|c| = |5|--10|=5-10 =-5....

But if a and b are of same sign...
a= 5 and c=10...
\(|a-c| = |5-10| = |-5| = 5........\) and\(|a|-|c| = |5|-|10| = -5....\) But \(5\neq{-5}\)..
What does it tell us?
That Not only both a and c should be of same sign BUt also numeric value of a should be mor than that of c...... Numeric value is value without sign..

If ac/(db) < 0, is d – b < 0 ?

(1) |a − c| = |a| − |c|
(2) d – b > d + b


From the question stem, we can infer that either one or three of the variables (a through d) is negative and all of them are non-zero.
(1) It means that a > c and both are of same sign. |a| - |c| is a positive quantity (modulus on left side of the eqn). so |a|>|c|.
(2) this implies b < 0 .

so either a/c = +/+ or -/- giving ac = always +ve
Each stmt is insufficient. Combining two, we know that d > 0.


(+ve) - (-ve) is always > 0

So answer is C

Hi, can you kindly explain how you arrived at 'd' as >0?
That is the missing link for me to understand the solution to this problem.

Thanks!

Bunuel chetan2u VeritasKarishma

Could you please help me in explaining where I am going wrong. Because I am getting answer B because of following reasons

given \(\frac{ac}{db} <0\)
This implies all 3 have same sign, only one has a different sign

(1) \(|a − c|\) = \(|a| − |c|\)
This says \(a\) and \(c\) are positive but it does not give any info on \(b\) and \(d\)

(2) \(d – b > d + b\)
This says that \(b<0\) and \(d>0.\)
Now that we see that \(b\) and \(d\) have 2 different signs, it hardly matters what is \(a\) and \(c\). \(a\) & \(c\) can be both positive or negative because \(\frac{ac}{db} <0\)
As \(b\) is -ve and \(d\) is +ve, hence b<d.

What am I missing here? Why do we also need 1st statement to find the solution?

d-b>d+b is equal to 2b<0 or b<0.
Nothing about d as d cancels out. It could be 0, <0 or >0.

Why cant the answer be just B. The stem clearly says and AC/DB <0.

So there is no chance of any of the variables being = 0.