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Bunuel
If \(cb < ab < 0\), is \(|c - a| = |c| - |a|\)?


(1) \(c > a\)

(2) \(b < 0\)


 

This question was provided by Crack Verbal
for the Heroes of Timers Competition

 


OFFICIAL EXPLANATION: FROM CRACK VERBAL:



Statement I is sufficient
If c > a and cb < ab then b is negative and c > a > 0
Since c and a both are positive and c > a then |c - a| will be equal to c - a and |c| - |a| will also be equal to c - a

Statement II is sufficient
If b is negative then c > a > 0
Since c and a both are positive and c > a then |c - a| will be equal to c - a and |c| - |a| will also be equal to c – a

Answer: D
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Given,
Case 1: If b is positive, a & c both negative and c<a. Or
Case 2: If b is negative, both a& c positive & c>a.
Question says: are a&c in SAME DIRECTION?
Statement 1: c>a that means second case & a&c both positive.
Sufficient.

Statement 2: b is negative that means second case & a&c both positive.
Sufficient.

My answer:D

Posted from my mobile device

======================================================================

DECODING THE MAIN QUESTION:is |c - a| = |c| - |a|?
In my opinion,
Such type of ABSOLUTE VALUE question (mentioned in main question) ask u a simple thing I.e. the two variables are in same directions or not.

TWO TYPES of SUCH QUESTION:
(i) {|c|+|a|} is either EQUAL or GREATER than {|c+a|}.
When the two numbers (c & a) are in SAME DIRECTION (both positive or both negatives or both zero), both expressions [(|c|+|a|) & |c+a|] are EQUAL.

When the two numbers (c & a) are in OPPOSITE DIRECTION (one positive and another one negative), (|c|+|a|) is GREATER than |c+a|.

(ii) {|c|-|a|} is either EQUAL or LESS than {|c-a|}.
When the two numbers (c & a) are in SAME DIRECTION (both positive or both negatives or both zero) & c has greater ABSOLUTE value, both expressions [(|c|-|a|) & |c-a|] are EQUAL.

When the two numbers (c & a) are in OPPOSITE DIRECTION (one positive and another one negative), (|c|-|a|) is LESS than |c-a|.
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(1) b can take +ve or -ve values. but only -ve value of b can satisfy both cb<ab & c>a .

So both c & a are positive, Which can answer the asked question. (if both c,a are of the same sign then only given equality will hold)

(2) b<0
here directly we can say that c>0, a>0. (2) can also answer the asked question.

Answer: D

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Either is independently sufficient.

#1 c>a
Let c=5 and a=2
Statement only possible when b<0
5*(-1) < 2*(-1),

For c=-2 and a=-5
Statement not possible.

So |c−a|=|c|−|a|

#2
For b<0, statement is only possible iff a and c are +ve.

So, |c−a|=|c|−|a| for and c are positives.

D IMO.
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Given, cb<ab<0 , is |c−a|=|c|−|a||?

Deduce:
1. ab, cb is -v
2. If b is -ve, then c>a (a,c are +ve)
3. If b is +ve, then a>c (a,c are -ve)

(1) c>a : b is -ve, c=5, a=2 --> |5-2| = |5|-|2| YES . so, AD/BCE
(2) b<0 : OR b is -ve. This is same as (1). Hence we know this statement is also sufficient. So, AD

The answer is D
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IMO D

cb<ab<0

Case 1: b<0 & c>a => c>0 & a>0

Case 2: b>0 & c<a => c<0 & a <0

S1) c>a i.e Case 1 . c>0 & a>0 so sufficient
S2) b<0 i.e Case 1 . c>0 & a>0 so sufficient

Therefore all bandhulog ... answer is D

Regards,
Anjanaputra
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cb<ab<0 tells us that c and b; a and b should have different signs;

1) c>a
Let us take random numbers to make the problem easier:
First case: a = 2; b = -5; c = 10
Then, cb<ab<0
10*-5<2*-5<0

a and c cannot be negative and b cannot be positive, because c>a and cb<ab will not work together
So, it's sufficient to answer the question

2) b<0
The same case as considered above

Also sufficient to answer the question

Answer: D
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Please find attached the video solution.

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My 2 cents on this:

It feels like the question is really just asking about the relationship between c and a in the given equation. How I did it was as follows:

St 1: c>a which means c-a>0 which means C is positive for sure.

If we take the given equation, it says cb<ab<0 --> b(c-a) < 0 suggesting that b<0 since c-a>0

by that logic you can take values of c and a and then solve easily. Sufficient

St 2: Same logic except we are already given b<0 and so same numbers used from St1 can apply. Sufficient

Hence, D
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