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Bunuel
On a number line, 0 < A < B, is |20 - A| < |20 -B| ?

(1) A + 20 < 2B
(2) B = 3A

Now here, the crux is that for the question to hold true A should be greater than 20 and B should be greater than 20 and yes also greater than A.
A>20 & B>20>A

Reasoning- if A=1, B=2, |20-A|=19 & |20-B|=18 therefore in no case |20 - A| < |20 -B| till A< 20.

Now, Statement (1) says the same thing,
(1) A + 20 < 2B
A is greater than 20 therefore LHS is a value greater than 40. therefore B> 20( 40 divided by 2)
Not Sufficient
Now,
(2) B = 3A
values such as A=1, B=3 condition falls apart.(19 is not less than 17)
whereas when A=22, B=66, condition agrees.(2<46).

Combining it gives us no more info than we already have.

Therefore, (E).

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Bunuel
On a number line, 0 < A < B, is |20 - A| < |20 -B| ?

(1) A + 20 < 2B
(2) B = 3A

The question simple asks about if distance between A & 20 is less than distance between 20 & B

(1) A + 20 < 2B

Let A = 5 & B = 15 .........|20 - 5| < |20 -15|...........Answer is NO

Let A = 15 & B = 45 .......|20 - 15| < |20 -45|...........Answer is Yes

Insufficient

(2) B = 3A

Let A = 5 & B = 15 .........|20 - 5| < |20 -15|...........Answer is NO

Let A = 15 & B = 45 .......|20 - 15| < |20 -45|...........Answer is Yes

Insufficient

Combine 1 & 2
same examples used without clear cut

Answer: E

Another algebraic approach for combing 1 & 2

From 2: B =3A

Substitute in 1: A +20 < 6A.............5A > 20......A > 4

Then our examples in Statement 1 & 2 stratify the conclusion.
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Bunuel
On a number line, 0 < A < B, is |20 - A| < |20 -B| ?

(1) A + 20 < 2B
(2) B = 3A


Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 2 variables (A and B) and 1 equation, D is most likely to be the answer. So, we should consider each of the conditions on their own first.

Condition 1)
If A = 20 and B = 60, the answer is "yes".
If A = 10 and B = 30, the answer is "no".
Since we don't have a unique solution, condition 1) is not sufficient.

Condition 2)
If A = 20 and B = 60, the answer is "yes".
If A = 10 and B = 30, the answer is "no".
Since we don't have a unique solution, condition 1) is not sufficient.


Conditions 1) & 2)
If A = 20 and B = 60, the answer is "yes".
If A = 10 and B = 30, the answer is "no".
Since we don't have a unique solution, condition 1) is not sufficient.

Therefore, the answer is E.

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.
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Bunuel
On a number line, 0 < A < B, is |20 - A| < |20 -B| ?

(1) A + 20 < 2B
(2) B = 3A

Let's break down this question to understand it better..

Number line..
0.....A......B
Now |20-A|<|20-B|
This happen only when A is closer to 20 than B is.

Now when does this happen??
When 20 lies LEFT of middle of A and B..
0.....A..|...B
So \(20<\frac{A+B}{2}....40<A+B\)

Let's see what each statement tells us..
1) A+20<3B...
Say A is 4, so B>8....
If B is 10, ans is NO, but if B is 50, ans is YES
Insufficient

2)B=3A..
A=5, B=15 ans NO
A=20, B=60.. Ans is YES
Insufficient

Combined
A+20<3*3A.....7A>20..
The values of statement II still stand.
Insufficient

E
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