Is |2a − 3b| < |a − b| + |a − 2b|? (1) b = 3 (2) a < b

That was a great explanation.
Mods is not a strong point for me.
It would be helpful if you can share some more tips and tricks on mods.

also please explain: If x and y have opposite signs, the inequality is strict.

Strict inequality means < , equality cannot hold.
For example |2+(-3)| < |2| + |-3| as 1 < 2 + 3.

I am confused at this point...For (1) and (2) together: x = a - b = a - 3 < 0, y =a - 2b = a - 6 < 0, so the given inequality doesn't hold (definite answer is NO), therefore sufficient.
Now b= 3 and a<b.
Consider case 1: a=4 a - 3 = 4-3=-1 and a - 6 = 4-6 = -2 i.e. -1 and -2 which is +ve
Consider case 1: a=2 a - 2 = 4-2= 2 and a - 6 = 2-6 = -4 i.e. 2 and -4 which is -ve
Hence the correct answer is E and not C..



jhe

Hi Eva

I am not sure on the solution mentioned above.

For statement 2 we have a< b, ie a-b <0 and similarly a-2b <0 . Therefore both X and Y are negative. In that case the inequality meets the condition. So shouldnt the answer be b

Is |2a − 3b| < |a − b| + |a − 2b|?

(1) b = 3

|2a − 3(3)| < |a − (3)| + |a − 2(3)|
|2a-9| < |a-3| + |a-6|

The checkpoints here are 4.5, 3, 6

The ranges to test are: x<3, 3<x<4.5, 4.5<x<6, x>6


a<3: -(2a-9) < -(a-3) + -(a-6) -2a+9 < -a+3 + -a+6 0 < 0 INVALID

3<a<4.5: -(2a-9) < (a-3) + -(a-6) -2a+9 < a-3 + -a+6 -2a < -6 a>3 VALID (a may fall within the range of 3<a<4.5)

4.5<a<6: (2a-9) < (a-3) + -(a-6) 2a-9 < a-3 + -a+6 2a < 12 a < 6 VALID 9 (a may fall within the range of 4.5<a<6)

a>6: (2a-9) < (a-3) + (a-6) 2a-9 < a-3 + a-6 0 < 0 INVALID

Some solutions are sufficient, some are not.
INSUFFICIENT

a<b

|2a − 3b| < |a − b| + |a − 2b|

If a<b then:

|2(2) - 3(3)| < |2-3| + |(2)-2(3)| |-5| < |-1| + |-6| 5<7 VALID
|2(-2)| - 3(3)| < |-2-3| + |-2 -2(3)| |-13| < |-5| + |-8| 13<13 INVALID
INSUFFICIENT

1+2) b=3 and a<b therefore a<3

Using the cases we found in #1, where a<3, the only solution where a<3 is invalid.
SUFFICIENT

(C)

(Is that correct reasoning I am using?)

Those two parts are not correct. Example for 3<a<4.5 you get a>3, so this could be a valid solution.

Just because \(a\) COULD fall in the range, this makes that given range a possible valid solution (for a=3.5 for example).
\(|2a - 3b| < |a - b| + |a - 2b|\) for a=3.5 and b=3 you get
\(|7-9|<|3.5-3|+|3.5-6|\) or \(2<3\) => valid

Hello all,

My approach to this question:

we are asked if |2a − 3b| < |a − b| + |a − 2b| ?
i.e. |(a-b) + (a-2b)| < |a − b| + |a − 2b|
i.e. |x + y| < |x| + |y|, where x = a-b and y = a-2b

this is possible only in 2 cases:
(i) x is +ve and y is -ve ; example check with (x,y) = (3,-4) or (4,-3)
(ii) x is -ve and y is +ve ; example check with (x,y) = (-3,4) or (-4,3)


now for case (1) when x is +ve and y is -ve
i.e. a-b > 0 and a-2b <0
i.e. a > b and a < 2b
i.e. b < a < 2b
clearly this is only possible when a and b both are positive (for b<2b to hold true) and a lies within the positive range {b,2b}

now for case (ii) when x is -ve and y is +ve
i.e. a-b < 0 and a -2b > 0
i.e. a < b and a > 2b
i.e. 2b < a < b
clearly this is only possible when both a and b are negaitve (for 2b<b to hold true) and a lies within the negative range {2b,b}

Now lets assess the choices:
(1) b = 3
this does not give us any idea about if a < b or a > b.
Hence insufficient

(2) a < b
this does not provide us information about if a and b are positive or negative real numbers

(1) & (2) together
b = 3 (positive real number) and a < b
From our derivation we know that, when a < b, both a and b should be negative (case ii mentioned above). However here, we have b as a positive number. hence both (1) and (2) taken together disproves the inequality, hence sufficient.

PS: if this question had provided us the 2 choices as
(1) a> b and
(2) b = 3,
than the choice (c) of finding sufficiency using both the options would still be insufficient.
because even if a>b and b is positive = 3, we still need an 'a' which lies within b (i.e. 3) and 2b (i.e. 6). And this information is absent.

Hope it helps..

unsure of this explanation. with b = 3 and a < b, how can a-b be negative but a-2b be positive? seems to me like both are negative and that would mean the inequality does not fail

We have that b = 3 and a < b, so a < 3. So, the question becomes is |2a − 9| < |a − 3| + |a − 6|?

Now, since a < 3, then 2a − 9 < 0, a − 3 < 0 and a − 6 <0. Now the question becomes, is -(2a − 9) < -(a − 3) - (a − 6)? --> is 9 < 9? The answer to this question is NO.

Answer is C.
This DS question expects an answer in Yes or No.

Consider the modulus expression |x - a|. As we know, Mod of any number gives out a positive value. But what is inside the Mod ("x - a" in this case) may have any sign or even have value zero.
Ex: "x - a" will be positive for x > a, negative for x < a and equal to 0 for x = a.

Two important things to note here. First, x = a is the point where sign of the expression gets reversed. Second, a negative value comes out of the Mod as positive, so when "x - a" is negative |x - a| = -(x - a), just as | -3 | = -(-3) = 3.

Now, coming to provided statements:
Statement 1:

Given b = 3.
So the given question becomes, Is |2a - 9| < |a - 3| + |a - 6| ?.
Now as mentioned above about the modulus expression, each of the three expression inside Mod will change sign based on value of a (in this case). The expression in question a has 3 sign reversal points: 9/2 , 3 and 6 respectively for "2a - 9", "a - 3" and "a - 6".

On number line:
<----------- 3 -------------- 9/2 -------------- 6 ----------------->

This is quite understandable that for each expression`s sign change, the complete equation will change.
For a < 3 (First case of Statement 1),
All the above "2a - 9", "a - 3" and "a - 6" will be negative: so, |2a - 9| = - (2a - 9),
|a - 3| = -(a - 3) and |a - 6| = -(a - 6). And the complete expression will be:
L.H.S = 9 - 2a
R.H.s = 3 - a + 6 - a = 9 - 2a.
Here we can see that for any a < 3, L.H.S = R.H.S. So according to First case of Statement 1, answer to asked question |2a - 9| < |a - 3| + |a - 6| ? is NO.

Again, for 3 < a < 9/2 (Second case of Statement 1),
Putting the values according to change signs (please try seeing which of the three have need to change sign),
L.H.S = 9 - 2a
R.H.S = 3.
Just put any value between 3 and 9/2 for a. L.H.S is always less than R.H.S. So according to Second case of Statement 1, answer to asked question |2a - 9| < |a - 3| + |a - 6| ? is YES.

We can stop here, since 2 different conditions for statement 1 gives contradicting results.
If this was not the case, we would have to check for all the ranges of values for a. For given situation statement 1 is clearly INSUFFICIENT.

Statement 2:
This also insufficient because we cannot decide for the signs of expressions inside Mod. INSUFFICIENT.

Statement 1 + 2:
b = 3 and a < b => a < 3.
From our previous analysis we can see that for a < 3, we have one undoubted answer (No) for question Is |2a - 9| < |a - 3| + |a - 6| ?
So, Statement 1 + 2 is sufficient to answer the question.
Hence C is correct.

**It looks lengthy process, but just because it is explained. With practice you can reduce time taken to solve below 2 mins.
Hope this helped.!!!!

Q: Is |2a-3b|<|a-b|+|a-2b|?
As rightly pointed out this is related to triangular inequality.

|x|=|a-b|; |y|=|a-2b|; |x+y|=|2a-3b|

question is asking whether x and y have opposite signs or not.

=> whether:
a-b<0 & a-2b>0 OR a-b>0 & a-2b<0

=>
a<b & a>2b (Only possible when both a and b are negative) OR a>b & a<2b
=> b<a<2b
Statement 1: b=3
b is positive. So, a<b and a>2b is not possible
since we do not know whether b<a<2b we cannot answer the question.
Not sufficient

Statement 2: a<b
we don't know whether a and b both are positive or negative.
We also don't know whether a<2b.
Not sufficient

Statement 1 + 2:
b=3 and a<b
=> neither a<b & a>2b (Only possible when both a and b are negative) NOR b<a<2b.
Hence we can definitely say |x+y| is not less than |x|+|y|.

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.

The first step of the VA (Variable Approach) method is to modify the original condition and the question, and then recheck the question.

We can modify the original condition and question using the following property.

\(|x+y| < |x| + |y| ⇔ xy < 0\)

The question asks if \((a-b)(a-2b) < 0\) since \(|2a-3b| = |(a-b)+(a-2b)| < |a-b| + |a-2b|\) by replacing \(x = a-b\) and \(y = a-2b\).


Since we have 2 variables (\(x\) and \(y\)) and 0 equations, C is most likely to be the answer and so we should consider 1) & 2) first.

Condition 1) & 2):

We have \(a < b < 2b\), since \(b = 3\).
\(a - b < 0\) and \(a - 2b < 0\).
Thus \((a-b)(a-2b) > 0\).

By CMT(Common Mistake Type) 1, since "No" is also an answer, both conditions 1) & 2) together are sufficient.

Since this question is an absolute valeu question (one of the key question areas), CMT 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1)
\(a = 1, b = 3\) : \((a-b)(a-2b) > 0\)
\(a = 4, b = 3\) : \((a-b)(a-2b) < 0\)
The condition 1) is not sufficient.

Condition 2)
\(a = 1, b = 3\) : \((a-b)(a-2b) > 0\)
\(a = -4, b = -3\) : \((a-b)(a-2b) < 0\)
The condition 2) is not sufficient.

Therefore, C is the answer.

Normally, in problems which require 2 or more additional equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.

|x-r| + |x-s|
Here, r and s are the CRITICAL POINTS.
Given this situation, we are generally concerned about two cases:
Case 1: x is BETWEEN the critical points
Case 2: x is NOT between the critical points



Statement 1 is clearly INSUFFICIENT.

Statement 2:
Let b=-2, with the result that a<-2.
Plugging b=-2 into |2a − 3b| < |a − b| + |a − 2b|, we get:
Is |2a − (-6)| < |a − (-2)| + |a − (-4)| ?
Here, the critical points on the right are -2 and -4.

Case 1: a=-3, with the result that \(a\) is BETWEEN the critical points
In this case, the answer to the question stem in blue is YES.
Case 2: a=-5, with the result that \(a\) is NOT between the critical points
In this case, the answer to the question stem in blue is NO.
INSUFFICIENT.

The two cases above indicate the following:
The answer will be YES if \(a\) is between the critical points.
The answer will be NO if \(a\) is not between the critical points.

Statements combined:
Plugging b=3 into |2a − 3b| < |a − b| + |a − 2b|, we get:
Is |2a − 9| < |a − 3| + |a − 6} ?
Here, the critical points on the right are 3 and 6.

Since a<b -- implying that a<3 -- we know that a is NOT between the critical points.
Thus, the answer to the question stem is NO.
SUFFICIENT.

Here is a visual explanation for this problem.
|x-y| = the distance between x and y



Question stem:
|2a − 3b| < |a − b| + |a − 2b|
2|a − 1.5b| < |a − b| + |a − 2b|

To understand the situation more clearly, test b=2.
Plugging b=2 into the question stem, we get:
2|a − 3| < |a − 2| + |a − 4| ?
In words:
Is twice the distance to 3 less than the sum of the distances to 2 and 4?

The answer will be YES if \(a\) BETWEEN 2 and 4, as in the following figure:

Here, the green bar (twice the distance to 3) is SHORTER than the blue bar and red bar combined (the sum of the distances to 2 and 4).

The answer will be NO if \(a\) is NOT between 2 and 4, as in the following figure:

Here, the green bar (twice the distance to 3) is EQUAL to the blue bar and red bar combined (the sum of the distances to 2 and 4).

In the case above:
The answer is YES if \(a\) is between 2 and 4.
We can generalize the question stem as follows:
2|a − 1.5b| < |a − b| + |a − 2b| if the value of \(a\) is BETWEEN the values of \(b\) and \(2b\).
When the statements are combined, it is not possible for \(a\) to be between \(b\) and \(2b\), so the answer to the question stem is NO.

I think the answer must be B by your method too.
if a<b, then \(a - b\) and \(a - 2b\) will always be of same sign.
i.e. If b is -ve then x & y are +ve
if b is +ve then x & y are - ve

Can any expert check this. For those who think C is the right answer, please share an a<b pair where the question stem is True.

If a = -4 and b = -3, then (|2a − 3b| = 1) < (|a − b| + |a − 2b| = 3) giving a YES answer to the question.

If a = 0 and b = 1, then (|2a − 3b| = 3) = (|a − b| + |a − 2b| = 3) giving a NO answer to the question.

In my initial post, I stated that the answer will be YES -- in other words, that the inequality will hold true -- when a=-3 and b=-2:
|2(-3) − 3(-2)| < |-3 − (-2)| + |-3 − 2(-2)|
|-6 + 6| > |-3 + 2| + |-3 + 4|
0 < 1 + 1
0 < 2

As noted in my second post, the inequality will hold true -- more generally -- if \(a\) is between \(b\) and \(2b\).
If a < b < 0, then it is possible for \(a\) to be between \(b\) and \(2b\):
2b < a < b < 0
-2 < -1.8 < -1 < 0
-4 < -3 < -2 < 0
-20 < -13 < -10 < 0
The inequality will hold true in all of these cases.