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If cb < ab < 0, is |c - a| = |c| - |a|?

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If cb < ab < 0, is |c - a| = |c| - |a|?  [#permalink]

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01 Jul 2019, 08:00
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If $$cb < ab < 0$$, is $$|c - a| = |c| - |a|$$?

(1) $$c > a$$

(2) $$b < 0$$

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

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Re: If cb < ab < 0, is |c - a| = |c| - |a|?  [#permalink]

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19 Jul 2019, 02:01
Bunuel wrote:
If $$cb < ab < 0$$, is $$|c - a| = |c| - |a|$$?

(1) $$c > a$$

(2) $$b < 0$$

 This question was provided by Crack Verbal for the Game 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

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Re: If cb < ab < 0, is |c - a| = |c| - |a|?  [#permalink]

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01 Jul 2019, 10:36
6
1
cb<ab<0 -- given

so now 2 cases:
1. b>0
2. b<0

taking 1 : b>0:
cb-ab<0 ---> b(c-a)<0
b>0 so c-a should be <0. So c<a
also since ab<0 and b>0 so a<0. Similarly c<0
taking 2 : b<0
here c-a>0, so c>a
also since b<0 and ab<0, so a>0 and similarly c>0
To check if |c-a|=|c|-|a|

lets take statement 1:
c>a
means case 2 from above. This case gives a,c>0 and c>a. So for all values |c-a|=|c|-|a| is true. 1 is sufficient.

statement 2:
b<0:
again case 2. SO sufficient:

Hence Ans D
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If cb < ab < 0, is |c - a| = |c| - |a|?  [#permalink]

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Updated on: 02 Jul 2019, 10:16
4
1
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.

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

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|.

Originally posted by BelalHossain046 on 01 Jul 2019, 08:27.
Last edited by BelalHossain046 on 02 Jul 2019, 10:16, edited 2 times in total.
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If cb < ab < 0, is |c - a| = |c| - |a|?  [#permalink]

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01 Jul 2019, 08:29
1
We are told that cb<ab<0 i.e. both cb and ab are negative. So either 1) b is positive which would mean that a and c are negative; or 2) b is negative and thus a and c are positive.

for |c−a|to be equal to|c|−|a| we need to meet one of these two conditions: A) if c>a then both a and c have to be positive; or B) if c<a then both have to be negative. So if the statements individually or collectively confirm either of these two sets of conditions we would have our answer.

St-1: c>a: which leads us to condition A and both a and c have to be positive for this to work and b thus has to be negative (as bc<ac<0). But we do not know that. So insufficient.
St-2: b<0: this tells us that b<0 and thus both a and c are positive, but consider situation where a=5 and c=3, in this case LHS=2 but RHS=-2. Not sufficient.

Taken together we can clearly see that this is condition A where c>a>0 (and thus b<0). Sufficient.
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Re: If cb < ab < 0, is |c - a| = |c| - |a|?  [#permalink]

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01 Jul 2019, 08:30
1
as cb<ab, so c<a

and |c-a| would be equal to |c| - |a| only when c>a and result of RHS is positive.
So option A states exactly that, so option 1 is correct.

Option 2 is not required , so it is of no use.

Finally, Option A is correct as only option 1 is sufficient.
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Re: If cb < ab < 0, is |c - a| = |c| - |a|?  [#permalink]

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01 Jul 2019, 08:32
i) Insufficient, Not sure on the sign of a,c, and b.
ii)B<0, means c and a are positive for ab, bc to be less than zero; C > A as only then will CB<AB. Sufficient (gives yes to the stem)
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Re: If cb < ab < 0, is |c - a| = |c| - |a|?  [#permalink]

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01 Jul 2019, 08:34
6
It's given that cb<ab<0.
So both cb and ab are negative.

For this to be possible, there are only two scenarios.
C<a and b>0
Or
C>a and b< 0.

In the question, both the statements imply the same and point to scenario 2.

Which implies c>a and that implies the given statement will always never be true.

|C-a| is always positive whereas |c| - |a| will equal -(c-a).

Therefore both statements alone are sufficient and hence option D.

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Re: If cb < ab < 0, is |c - a| = |c| - |a|?  [#permalink]

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01 Jul 2019, 08:35
1
cb<ab<0.
Means the sign of the products cb and ab are opposite.

From S1:

c>a.
Let c and a be positive.
c = 5, a = 2.
And b is negative.
|c-a| = 3 = |c|-|a| = 3.
Satisfies.
If b is positive.
c and a are negative.
Then c = -2 and a = -4.
|c-a| = 2 = |c|-|a| = -2.
Not satisfies.
INSUFFICIENT.

From S2:

b is negative.
So c and a are positive.
And the |c-a| = |c|-|a|.
SUFFICIENT.

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If cb < ab < 0, is |c - a| = |c| - |a|?  [#permalink]

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Updated on: 01 Jul 2019, 08:40
2
(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.

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Originally posted by Rohan007 on 01 Jul 2019, 08:39.
Last edited by Rohan007 on 01 Jul 2019, 08:40, edited 1 time in total.
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If cb < ab < 0, is |c - a| = |c| - |a|?  [#permalink]

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Updated on: 02 Jul 2019, 08:13
1
If cb<ab<0cb<ab<0, is |c−a|=|c|−|a||c−a|=|c|−|a|?

(1) c>ac>a

(2) b<0

given cb<ab<0
so either of a,c or b has to be -ve
and |c−a|=|c|−|a| will hold true only when c & a are +ve
so #1
c>a means b is <0 -ve
c=2 and a=1 then |c−a|=|c|−|a| ; valid

#2
b<0
it means that a & c are both +ve so
|c−a|=|c|−|a| ; valid
IMO D

Originally posted by Archit3110 on 01 Jul 2019, 08:40.
Last edited by Archit3110 on 02 Jul 2019, 08:13, edited 1 time in total.
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If cb < ab < 0, is |c - a| = |c| - |a|?  [#permalink]

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Updated on: 02 Jul 2019, 10:44
This is a hard question.
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Re: If cb < ab < 0, is |c - a| = |c| - |a|?  [#permalink]

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01 Jul 2019, 08:48
2
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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Re: If cb < ab < 0, is |c - a| = |c| - |a|?  [#permalink]

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01 Jul 2019, 08:50
1
Let us rephrase the question

Are both c and an on same side of the number line?

Given inequality may have 2 cases based on the sign of b

First Case
If $$b>0$$
$$c<a<0$$

Second Case
If $$b <0$$
$$c>a>0$$

Statement 1: c>a

From the second case we know c>a only when b<0
Hence we know that $$c>a>0$$ . Clearly Sufficient.

Statement 2: $$b<0$$
This implies second case again.
So Clearly Sufficient.

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Re: If cb < ab < 0, is |c - a| = |c| - |a|?  [#permalink]

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01 Jul 2019, 08:52
1
Hi,
Given cb<ab<0 and Is |c-a| = |c| - |a|?

From looking at the question we get, either c <0 or b< 0 or a< 0.

1. c>a using this:
if c =2 and a = 1 then the Answer is Yes.
But if c = -1 and a = -2 then the answer is no.
Insufficient.

2. b<0
from this, we can say that c>0 and a>0, only this will hold the equality true as per the condition.
so, a> 0, b<0 , c> 0 is our deduction from this.
But again we don`t know about the relationship between a and c.
Insufficient.

3. Using both statements we can say:
b<0 , c>a and c>0 , a>0
That will give the answer Yes.

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Re: If cb < ab < 0, is |c - a| = |c| - |a|?  [#permalink]

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01 Jul 2019, 08:54
2
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

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Re: If cb < ab < 0, is |c - a| = |c| - |a|?  [#permalink]

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01 Jul 2019, 09:03
1
1. c>a => b<0 since cb<ab
b<0<a<c
c & a both are positive
|c| = c
& |a| = a
|c-a| = c-a =|c| - |a|
Statement 1 is sufficient.

2. b<0 => c>a since cb<ab
b<0<a<c
c & a both are positive
|c| = c
& |a| = a
|c-a| = c-a =|c| - |a|
Statement 2 is sufficient.

Since each statement alone is sufficient

IMO D
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Re: If cb < ab < 0, is |c - a| = |c| - |a|?  [#permalink]

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01 Jul 2019, 09:05
For cb to be less than ab, either the terms c and a play a major role or its the term b.
In case 1: c>a
For example, c= -1 and a= -3. Let b=3 [ Since both the terms have to be negative, b has to be positive]
|c-a| = 2 and |c|-|a|=-2. Hence not sufficient to answer.
In case 2: b<0. Therefore, c and a have to be positive for the terms cb and ab to be less than 0.
If b is negative, c has to be greater than a for it to satisfy cb<ab and positive.
Hence, statement 2 is enough.
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Re: If cb < ab < 0, is |c - a| = |c| - |a|?  [#permalink]

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01 Jul 2019, 09:08
1
A very good question indeed.

From cb < ab < 0,
we can tell that
(1) either b is negative & a,c are positive
or b is positive & a,c are negative.
(2) |cb| > |ab| ==> |c| > |a|
if both c and a are positive, |c−a|=|c|−|a|
but if both c and a are negative, |c−a|<>|c|−|a|
So, the question boils down to determine the sign for a/c.

If we know c > a, we can tell a/c are both positive. Sufficient.
If we know b < 0, we can tell a/c are both positive. Sufficient.

So, the right answer is (D).
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If cb < ab < 0, is |c - a| = |c| - |a|?  [#permalink]

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Updated on: 02 Jul 2019, 08:36
1
Given : cb<ab<0, so both cb and ab are negative. Since b is common, there can be 2 cases -

Case I - If b>0, then c < a < 0 (For example let b = 1, c = -3, a = -2)

Then |c - a| = 1 = |c| - |a|, True

Case II - If b<0, then c > a > 0 (For example let b = -1, c = 3, a = 2)

Then |c - a| = 1 = |c| - |a|, True

This is applicable for both statements 1 and 2. Hence D.

Originally posted by taransaurav on 01 Jul 2019, 09:11.
Last edited by taransaurav on 02 Jul 2019, 08:36, edited 1 time in total.
If cb < ab < 0, is |c - a| = |c| - |a|?   [#permalink] 01 Jul 2019, 09:11

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