Is |a-c| + |a| = |c|? (1) ab > bc (2) ab < 0

In this question,
|a-c| + |a| = |c|?
or |a-c| = |c| - |a|?
or |c-a| = |c| - |a|?
That is (ii) situation of my previous post.

Thus, the question is asking : IS a & c in same direction & is the absolute value of a is greater than that of C?

statement 1 & 2 don’t guarantee anything about same direction of a&c.
Only combination of of c says the both requirement (same direction & absolute value of c).
So I would go for C.

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Number picking (I think better approach).

Statement 1: Insufficient
Case 1: a=1, b=-2 & c=3 (yes to main question)
Case 2: a=-1, b=2, &c=-3 (yes to main question)
Case 3: a=1,b=2 & c=-3 (no to main question)

Statement 2: insufficient. Coz no info about c.

Combined:
Case 3 from statement 1 is out coz statement 2 says ab is negative.
So remaining case 1&2 say YES to main question.

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Absolute value concept.

Absolute always says about distance.
|a| or |a-0| is the distance between 0 and a.

|a-c| is same thing as |c-a| which means the distance between a and c.

|c| is the distance between 0 and c.

Thus the question says
Is the sum of (I) DISTANCE between (0&a) & (ii) DISTANCE between (a&c) is EQUAL to DISTANCE between (0 &c)?

This is only possible if
I) a&c are on the same side of zero and
(ii) a is closer to zero and c is distant from zero.
[this is possible in positive and negative sides, as shown in attached photo.]

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to have /a-c/+/a/=/c/
we have /a-c/=/c/-/a/
c and a must be the same sign

from both condition
b(a-c)>0
ab<0

case 1
b<0 and a<c
b<0 then a>0, this mean c also >0
this is good
case 2
b>0 and a>c
b>0 the a<0 , the mean c also < 0
this is good

answer C.

Small correction here.
so if b>0, then both a and c are negative and c>a in magnitude (|c| > |a|)
so if b<0, then both a and c are positive and c>a in magnitude (|c| > |a|)

This question follows the rules of the general form:
|x-y|=|x| - |y|
if:
a. y = 0; or
b. x and y have the same sign, and x>y

So reorganising the stem.
is |a-c| + |a| =|c|

is |a-c| =|c| - |a|
reshuffle by multiply by -1

is -|a-c| = |a| - |c|

For this to be true c must equal 0 or a and c must have the same sign and a must be greater than c
(1) ab > bc
++ >+(-) a and c could have different signs; or--> No
++>++ a and c could have the same signs; or -->Yes
++>+(0) c could be equal to zero -->Yes

Insufficient

(2) ab<0
could be +- or -+
either way we knowing about c
Insufficient

Combined (1+2)
ab<0
therefore
ab > bc
(-) > bc

Possible cases:
ab > bc
(-)+> +(-) a and c are both negative, but a must be greater than c to satisfy the constraint
+(-)>(-)+ a and c are both positive, but a must be greater than c to satisfy the constraint

In either case, a must be greater than c and must have the same sign as c, therefore combined statements are sufficient.

We can test numbers to validate the theory.
-|a-c| = |a| - |c|

condition 1: a and c are both negative and a > c
-|-2-(-3)| = |-2|-|-3|
-|1| = 2-3
-1 = -1

condition 2: a and c are both positive and a > c
-|3-2|=|3|-|2|
-|1| = |3|-|2|
-1 = 1 is not possible, therefore a and c must both be negative

Bunuel, sorry to pester but is my working out correct?

Is |a-c| + |a| = |c|?
Q. |c-a| = |c|-|a|

(1) ab > bc
ab - bc > 0
b (a-c) > 0
If b>0; a-c>0; a>c
If b<0; a-c<0; a<c
NOT SUFFICIENT

(2) ab < 0
If b>0; a<0
If b<0; a>0
No information provided for c
NOT SUFFICIENT

(1) + (2)
(1) ab > bc
ab - bc > 0
b (a-c) > 0
If b>0; a-c>0; a>c
If b<0; a-c<0; a<c
(2) ab < 0
If b>0; a<0
If b<0; a>0
If b>0; a-c>0; a>c; c<a<0; |c-a| = |c| - |a|
c------a-----------0
If b<0; a-c<0; a<c; c>a>0; |c-a| = |c| - |a|
0-------a------------c
SUFFICIENT

IMO C

Is |a-c| + |a| = |c|?

Is |a-c|=|c|-|a|

Square both side ,

Is a^2+c^2-2ac=c^2+a^2- 2|a|*|c|

Is ac=|a||c|

Now actually the question is asking whether a*c=>0 ? [ as |a|*|c|=>0]

(1) ab > bc

b(a-c)>0

either b>0 and a>c
or b<0 and a<c

Clearly not sufficient as we do not have any info on the sign of a and c

(2) ab < 0

this tell that a and b are of different sign , no info on c so not sufficient

(1)+ (2)

either b>0 and a>c ,as b>0 so a<0 (from s2)
now as a>c so if a<0 then c <0 so ac>0 (sufficient )

or b<0 and a<c, as b<0 so a>0(from s2)
now as a<c, so if a>0 then c>0 so ac>0 (sufficient )

On both the cases sufficient hence C

Statement 1: ab > bc
Case 1: b=1, with the result that a>c, allowing a=2 and c=1
If we plug a=2 and c=1 into |a-c| + |a| = |c|, we get:
|2-1| + |2| = |1|
3 = 1
NO

Case 2: b=-1, with the result that a<c, allowing a=1 and c=2
If we plug a=1 and c=2 into |a-c| + |a| = |c|, we get:
|1-2| + |1| = |2|
2 = 2
YES
INSUFFICIENT.

Statement 2: ab < 0
Case 2 also satisfies Statement 2.
In Case 2, the answer to the question stem is YES.

Case 3: b=-1, a=2 and c=1
If a=2 and c=1, the answer to the question stem is NO, as shown in Case 1 above.
INSUFFICIENT.

Statements combined:
Case 2 satisfies both statements.
In Case 2, the answer to the question stem is YES.

Case 4: b=1, with the result that a<0 in Statement 2 and a>c in Statement 1
If we plug a=-1 and c=-2 into |a-c| + |a| = |c|, we get:
|-1-(-2)| + |-1| = |-2|
2 = 2
YES

Cases 2 and 4 illustrate the only two scenarios that will satisfy both statements:
b<0, with the result that a>0 in Statement 2 and a<c in Statement 1
b>0, with the result that a<0 in Statement 2 and a>c in Statement 1
Since the answer is YES in both cases, SUFFICIENT.

Careful!
This line of reasoning is faulty.
If |a-c|=|c|-|a|, then we can conclude that ac=|a||c|.
But the reverse is not necessarily true.
If ac=|a||c|, we CANNOT conclude that |a-c|=|c|-|a|.
For example:
Plugging a=2 and c=1 into |a-c|=|c|-|a|, we get:
|2-1|=|1|-|2|
1 = -1
Thus, the rephrase in red -- is ac≥0? -- is invalid.

I would firstly create 4 cases from the question in order to rephrase the question.

1. A>0 C>0
2. A>0 C<0
3. A<0 C>0
4. A<0 C<0

Let’s apply each case to the question.

1. A - c + a = c
2a = 2c
A=c
Therefore, with the 1st case, we can answer that |a-c| + |a| = |c|.

2. A + c + a = -c
2a = -2c
A=-c
A and c are not same. So we can’t answer.

3. -a -c - a = c
-2a = 2c
-a = c
A and c are not the same. So we can’t answer.

4. It is the same as the 1st. So we can answer.

So if we rephrase the question, we know that we are asked to answer whether “A and C have the same sign”.

Now, look at the each statement.

(1) ab>bc
B(a-c) > 0
If b>0, then a>c. For a, even if A is negative we can still answer because c is always smaller than a. Therefore c is negative. So A and C have the same sign.
If b<0, then a<c, this one, a and c can have different sign. So this is not valid.
We have 2 different cases from statement 1, so statement 1 is not itself sufficient to answer the question.

(2) ab<0
A and b have different sign.
Obviously, we do not know what c is.

(1+2) together,
From statement 2, we know that,
A and b have different sign. Now, we see the statement 1, knowing that a and b have differenet sign.
Only the first case from the statement 1 is possible.
There, a and c have the same sign.

My answer is C.

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Note:
|x-y| = |y-x| = the distance between y and x
|x| = |x-0| = the distance between x and 0


Question stem, rephrased:
Is |c-a| + |a-0| = |c-0| ?
In words:
If the distance between c and a is added to the distance between a and 0, is the result equal to the distance between c and 0?

The answer will YES if a is between c and 0, as in the following cases:
c.....a.....0 --> (distance between c and a) + (distance between a and 0) = distance between c and 0
0.....a.....c --> (distance between c and a) + (distance between a and 0) = distance between c and 0

The answer will NO if a is NOT between c and 0, as in the following cases:
a.....c.....0 --> (distance between c and a) + (distance between a and 0) > distance between c and 0
0.....c.....a --> (distance between c and a) + (distance between a and 0) > distance between c and 0

Statement 1: ab>bc
If b=1, a=2 and c=1, then a is NOT between c and 0, so the answer to the question stem is NO.
If b=1, a=-1 and c=-2, then a IS between c and 0, so the answer to the question stem is YES.
INSUFFICIENT.

Statement 2: ab<0
No information about c.
INSUFFICIENT.

Statements combined: ab>bc and ab<0
Case 1: b>0, implying in Statement 1 that a>c and in Statement 2 that a<0
Linking together c<a and a<0, we get:
c<a<0
Since a is between c and 0, the answer to the question stem is YES.

Case 2: b<0, implying in Statement 1 that a<c and in Statement 2 that a>0
Linking together 0<a and a<c, we get:
0<a<c
Since a is between c and 0, the answer to the question stem is YES.

In both cases, the answer to the question stem is YES.
SUFFICIENT.

Hi Experts,

Since the question does not have "b" but answer options have "b", isn't the question wrong?

The question stem can be re-written as follows:
Is \(|a-c|=|c|-|a|\)?

Inferences:
1) \(|a-c|\) can't be negative.
2) \(|c| \ge |a|\)
<--------c-----a----0-----a------c------->

3) \(a\) & \(c\) should be on the same side of the number line.

Restating #3: is \(ac>0\)?

From S1: \(ab>bc\)
Insufficient

From S2: \(ab<0\)
Insufficient

S1 & S2:
\(0>ab>bc\)
Or, \(bc<ab<0\)

Inferences:
1) \(b\) & \(c\) are on the opposite sides of zero on the number line.
2) \(a\) & \(b\) are on the opposite sides of zero.
3) Hence, \(a\) & \(c\) are on the same side of the number line (both should have the same sign).
Restating #3: \(ac>0\)
Sufficient.
Hence C.

Video solution from Quant Reasoning:
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Statement A:
ab > bc
ab - bc > 0
b(a-c) > 0
=> Either b > 0 & (a - c) > 0,
or b < 0 & (a - c) < 0.
Depending on sign of (a - c), respective signs of a & c will differ. So we don't have a definite Yes or a definite No.
So, Not sufficient.

Statement B:
ab<0
=> Either a < 0 & b > 0,
or a >0 & b < 0.
Here we are not sure what sign c holds.
Not sufficient.

Statement C: [1] + [2]
ab>bc
ab < 0
=> bc is also < 0.
Now we have something.
bc < ab < 0
=>
Either a > 0, b < 0, c > 0, (One explanation for bc < 0 & ab < 0)
or a < 0, b > 0, c < 0, (Another explanation for bc < 0 & ab < 0)
Now we know that a & c share the same sign, and whatever it is, b has the opposite.

when b < 0, a-c <0 => a < c {from statement a} & a, c > 0 {from statement b}
=> |a-c| +|a| i.e (L.H.S of original statement that is to be proved)
=> |a-c| +|a| = c - a + a (Opening modulus w.r.t to signs)
=> |a-c| +|a| = c
=> |a-c| +|a| = c = |c| (since c > 0, |c| opens up with + ve sign, we did the reverse here.)
L.H.S = R.HS

when b > 0, a-c >0 => a > c {from statement a} & a, c < 0 {from statement b}
=> |a-c| +|a| i.e (L.H.S of original statement that is to be proved)
=> |a-c| +|a| = a - c - a (Opening modulus w.r.t to signs)
=> |a-c| +|a| = -c
=> |a-c| +|a| = -c = |c| (since c < 0, |c| opens up with - ve sign, we did the reverse here.)
L.H.S = R.HS
Hence C.