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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If a = b, and c = d, is |a| = |c|?

(1) |b| = |d|
(2) b = -d


When you modify the original condition and the question, they become |a|=|c|?--> |b|=|d|?--> b=-d or d?, which makes 1)=2). Also, each of them is yes, which is sufficient.
Therefore, the answer is D.


 Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.
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From the question it is clear that we have to check if numerical values of a and c are equal (signs will not matter)
Statement 1 says |b| = |d| . It clearly shows that |a| = |c| as a = b and c = d so sufficient
Statement 2 says b = -d. It clearly tells that numerical value of b and d is same which in turn says |a| = |c| so sufficient
Correct answer - D
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chetan2u Bunuel VeritasPrepKarishma niks18

I am really confused about opening modulus in both sides of equality
in spite of going through chetan2u signature

I could get away by simply substituting a and c for b and d resp to establish suff for St 1
but St 2 was confusing.

Can you please add your two cents?
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chetan2u Bunuel VeritasPrepKarishma niks18

I am really confused about opening modulus in both sides of equality
in spite of going through chetan2u signature

I could get away by simply substituting a and c for b and d resp to establish suff for St 1
but St 2 was confusing.

Can you please add your two cents?

Hi adkikani

mod is always positive, so if \(x<0\) i.e negative then \(|x| =-x\)

for eg. if \(x=-3\) then \(|-3|=-(-3)=3\)

so for Statement 2: \(b=-d\) is simply \(|b|=|-d|\), apply mod on both sides

which in turn is \(|b|=|d|\), now substitute the value of \(b\) & \(d\) to get \(|a|=|c|\)

Alternatively, we are asked is \(|a|=|c|\), square both sides to remove mod. so the question becomes Is \(a^2=c^2\)?

Statement 2: \(b=-d\), substitute the value of \(b\) & \(d\) to get \(a=-c\). Now square both sides to get \(a^2=c^2\). Sufficient
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Hi niks18

Quote:

mod is always positive, so if \(x<0\) i.e negative then \(|x| =-x\)

for eg. if \(x=-3\) then \(|-3|=-(-3)=3\)

how did you use above knowledge to infer below :

Quote:
\(|b|=|-d|\), apply mod on both sides

which in turn is \(|b|=|d|\),

To be more specific, \(|b|=|d|\), note that we do not know whether b or d is less than zero
but all we know is that b and d have opposite signs. Let me know if I missed anything.
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Hi niks18

Quote:

mod is always positive, so if \(x<0\) i.e negative then \(|x| =-x\)

for eg. if \(x=-3\) then \(|-3|=-(-3)=3\)

how did you use above knowledge to infer below :

Quote:
\(|b|=|-d|\), apply mod on both sides

which in turn is \(|b|=|d|\),

To be more specific, \(|b|=|d|\), note that we do not know whether b or d is less than zero
but all we know is that b and d have opposite signs. Let me know if I missed anything.

Hi adkikani

As I mentioned mod is always positive, because it represents the distance from the 0-point in the number line and distance cannot be negative.
so irrespective of the value of b or d; |-d|=|d|

let b=2 then b=-d will result in d=-2

So |2|=2 and |-2|=-(-2)=2=|2| so we have |b|=|-d| =>|b|=|d|
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niks18

Crystal clear now. +1 Kudos :-)
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chetan2u Bunuel VeritasPrepKarishma niks18

I am really confused about opening modulus in both sides of equality
in spite of going through chetan2u signature

I could get away by simply substituting a and c for b and d resp to establish suff for St 1
but St 2 was confusing.

Can you please add your two cents?

Hi adkikani

mod is always positive, so if \(x<0\) i.e negative then \(|x| =-x\)

for eg. if \(x=-3\) then \(|-3|=-(-3)=3\)

so for Statement 2: \(b=-d\) is simply \(|b|=|-d|\), apply mod on both sides

which in turn is \(|b|=|d|\), now substitute the value of \(b\) & \(d\) to get \(|a|=|c|\)

Alternatively, we are asked is \(|a|=|c|\), square both sides to remove mod. so the question becomes Is \(a^2=c^2\)?

Statement 2: \(b=-d\), substitute the value of \(b\) & \(d\) to get \(a=-c\). Now square both sides to get \(a^2=c^2\). Sufficient

Hey niks18,

Red part marked above is inconsistent as to my understanding absolute values are always non-negative rather than always positive.

Please clarify.

QZ
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niks18
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chetan2u Bunuel VeritasPrepKarishma niks18

I am really confused about opening modulus in both sides of equality
in spite of going through chetan2u signature

I could get away by simply substituting a and c for b and d resp to establish suff for St 1
but St 2 was confusing.

Can you please add your two cents?

Hi adkikani

mod is always positive, so if \(x<0\) i.e negative then \(|x| =-x\)

for eg. if \(x=-3\) then \(|-3|=-(-3)=3\)

so for Statement 2: \(b=-d\) is simply \(|b|=|-d|\), apply mod on both sides

which in turn is \(|b|=|d|\), now substitute the value of \(b\) & \(d\) to get \(|a|=|c|\)

Alternatively, we are asked is \(|a|=|c|\), square both sides to remove mod. so the question becomes Is \(a^2=c^2\)?

Statement 2: \(b=-d\), substitute the value of \(b\) & \(d\) to get \(a=-c\). Now square both sides to get \(a^2=c^2\). Sufficient

Hey niks18,

Red part marked above is inconsistent as to my understanding absolute values are always non-negative rather than always positive.

Please clarify.

QZ

Hi QZ

You are right that mod is "Always Non negative" i.e. either positive or 0. I think either I missed mentioning 0 or I might had wrote that statement in the context of the question. Anyways thanks for highlighting.
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From statement 1 we get that even if b and d are of opposite signs the absolute value of a would be equal to the absolute value of c. Sufficient.
From statement 1 we get that b and d are of opposite signs, the absolute value of a would be equal to the absolute value of c even then. Sufficient.
IMO D

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As we our taking absolute values, this seems obvious that irrespective of positive or negative numbers, we will get sufficient statements.

=> |b| = |d|

sufficient.

=> b = -d

Sufficient. D

We can also plug numbers.
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Bunuel
If a = b, and c = d, is |a| = |c|?

(1) |b| = |d|
(2) b = -d


Kudos for correct solution.


#1
|b| = |d|
irrespective of b & d sign ; value of a & c will be equal since they are in modulus
sufficient
#2
b=-d
again same logic as #1; sufficient
IMO D
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