nitin6305 wrote:
If x and y are negative numbers, is x<y?
(1) 3x+4<2y+3
(2) 2x−3<3y−4
Bunuel, can you please explain the graphical method to solve this question?
Thanks
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.
Since we have 2 variables and 0 equation, C is most likely to be the answer.
Condition 1) & 2)
\(3x + 4 < 2y + 3\)
\(2x - 3 < 3y - 4\)
If we add both inequalities, we have \(5x + 1 < 5y - 1\).
\(5x + 1 < 5y - 1 < 5y + 1 ⇔ 5x < 5y or x < y\).
Both conditions together are sufficient.
Since this question is one of key questions, we need to check A or B by CMT(Common Mistake Type) 4 (A).
Condition 1)
\(3x + 4 < 2y + 3\)
\(x = -2\), \(y = -1\) : Yes
\(x = -20\), \(y = -21\) : No.
This is not sufficient.
Condition 2)
\(2x - 3 < 3y - 4 ⇒ 3x - 4 < 3x - 3 < 2x - 3 < 3y - 4\) since \(x < 0\).
We have \(3x < 3y\) or\(x < y\).
This is sufficient.
Therefore, the answer is B.
Normally for cases where we need 2 more equations, such as original conditions with 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore C has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using 1) and 2) together. (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
for first statement 1, I tested values and concluded that st 1 is enough.
I checked extreme values (x=-100 y= -1) and close values (x= -4 and y =-3) both cases gave me a YES. therefore I concluded with st 1 as sufficient.