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From statement 1 : x>-1 so x could be -0.5,-0.8,0 1,5 ,7 .... hence insufficient from statement 2 x<-1 (sufficient) explanation : we have 3 cases to check : (x>2),(-4<x<2) and (x<-4) when x>2 x+4-x+2>0 - no solution when -4<x<2 x+4-(-(x-2))<0 => x<-1 when x<-4 -x-4-(-(x-2))<0 again no solution Hence B is the answer

(1) First statement is obviously insufficient because x+1 > 0 tells us that x > -1. So x could lie between 0 and -1 or it could be >0.

(2) |x+4| > |x-2| One way of looking at this statement is as follows:

|x-(-4)| < |x-2|

This tells us that 'Distance between x and -4' on the number line is lesser than 'Distance between x and 2'. Now which number lies exactly in the middle of -4 and 2 ? That number is -1 (3 steps away from -4 and 3 steps away from 2). So, all numbers to the right of -1 will have the property that their distance from -4 will be greater than their distance from 2. So they wont satisfy the condition given in this statement. But, all the numbers to the left of -1 will have the property that their distance from -4 will be less than their distance from 2. All these numbers are the only numbers which will satisfy the given inequality in this statement.

Thus on the basis of second statement alone we can conclude that x lies to the left of -1 on the number line. So x must be negative and hence cannot be >0. We get NO as the answer.

From statement 1 : x>-1 so x could be -0.5,-0.8,0 1,5 ,7 .... hence insufficient from statement 2 x<-1 (sufficient) explanation : we have 3 cases to check : (x>2),(-4<x<2) and (x<-4) when x>2 x+4-x+2>0 - no solution when -4<x<2 x+4-(-(x-2))<0 => x<-1 when x<-4 -x-4-(-(x-2))<0 again no solution Hence B is the answer

Good solution Raksat. May I point out though that in the last step we get -2 < 0, which is TRUE. This means all the values of x which are less than -4 will satisfy the given inequality. So x < -1, YES, and it also includes all values which are less than -4.

I think you already know this, I just thought writing 'no solution' probably doesnt fit in here. Hence pointing it out

From statement 1 : x>-1 so x could be -0.5,-0.8,0 1,5 ,7 .... hence insufficient from statement 2 x<-1 (sufficient) explanation : we have 3 cases to check : (x>2),(-4<x<2) and (x<-4) when x>2 x+4-x+2>0 - no solution when -4<x<2 x+4-(-(x-2))<0 => x<-1 when x<-4 -x-4-(-(x-2))<0 again no solution Hence B is the answer

Good solution Raksat. May I point out though that in the last step we get -2 < 0, which is TRUE. This means all the values of x which are less than -4 will satisfy the given inequality. So x < -1, YES, and it also includes all values which are less than -4.

I think you already know this, I just thought writing 'no solution' probably doesnt fit in here. Hence pointing it out

yes... any value less than -4 will satisfy the given inequality and in turn answer the question stem.

Can there be any number to which we add 4 and it becomes less than the number that we get after subtracting 2 from it? NOT POSSIBLE. This is only possible when x is negative and we compare absolute values. Thus, x has to be negative or x < 0.

Its good to solve the inequality and arrive at the conclusion methodically. But even if we don't do that, we can address the sufficiency question.

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 1 variable (x) and 0 equations, D is most likely to be the answer.

Condition 1) \(x + 1 > 0\) \(x > -1\) Since the range of the question does not include the range of the condition, this is not sufficient.

Condition 2) \(|x+4| < |x-2| ⟺|x+4|^2 < |x-2|^2 ⟺(x+4)^2 < (x-2)^2 ⟺x^2 + 8x + 16 < x^2 -4x + 4 ⟺8x + 16 < -4x + 4 ⟺12x < -12\) \(⟺x < -1\) The answer is ‘no’. By CMT (Common Mistake Type) 1, ‘no’ is also an answer. This is sufficient.

Therefore, the answer is B.

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.

chetan2uMathRevolution My query is : Can we always square both the sides in an inequality involving modulus on both the sides as it is in this question. And why will the sign remain same after squaring ?

chetan2uMathRevolution My query is : Can we always square both the sides in an inequality involving modulus on both the sides as it is in this question. And why will the sign remain same after squaring ?

Hi Raksat

I think we can square here because both sides are under modulus. Modulus of anything can never be negative. Lets say you are given that a non negative quantity a is less than a non negative quantity b. Since both are non-negative, we can safely say that square of a will be less than square of b. (you can take any two non negative quantities whether integers or fractions and check) Now |x+4| is < |x-2|. Since both |x+4| and |x-2| are non negative (since they are absolute values) we can safely square both sides.

chetan2uMathRevolution My query is : Can we always square both the sides in an inequality involving modulus on both the sides as it is in this question. And why will the sign remain same after squaring ?

1. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality). For example: \(2<4\) --> we can square both sides and write: \(2^2<4^2\); \(0\leq{x}<{y}\) --> we can square both sides and write: \(x^2<y^2\);

But if either of side is negative then raising to even power doesn't always work. For example: \(1>-2\) if we square we'll get \(1>4\) which is not right. So if given that \(x>y\) then we cannot square both sides and write \(x^2>y^2\) if we are not certain that both \(x\) and \(y\) are non-negative.

2. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality). For example: \(-2<-1\) --> we can raise both sides to third power and write: \(-2^3=-8<-1=-1^3\) or \(-5<1\) --> \(-5^3=-125<1=1^3\); \(x<y\) --> we can raise both sides to third power and write: \(x^3<y^3\).

I have a different approach to view Statement 2. Please see and confirm my understanding is correct.

Bunuel wrote:

Raksat wrote:

chetan2uMathRevolution My query is : Can we always square both the sides in an inequality involving modulus on both the sides as it is in this question. And why will the sign remain same after squaring ?

1. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality). For example: \(2<4\) --> we can square both sides and write: \(2^2<4^2\); \(0\leq{x}<{y}\) --> we can square both sides and write: \(x^2<y^2\);

But if either of side is negative then raising to even power doesn't always work. For example: \(1>-2\) if we square we'll get \(1>4\) which is not right. So if given that \(x>y\) then we cannot square both sides and write \(x^2>y^2\) if we are not certain that both \(x\) and \(y\) are non-negative.

2. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality). For example: \(-2<-1\) --> we can raise both sides to third power and write: \(-2^3=-8<-1=-1^3\) or \(-5<1\) --> \(-5^3=-125<1=1^3\); \(x<y\) --> we can raise both sides to third power and write: \(x^3<y^3\).

Dear Bunuel, MathRevolution As per the Statement 1: x>-1 ahile Statement 2 ,x< -1. Doesnt it makes the question inconsistent.

Regards

MathRevolution wrote:

=>

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 1 variable (x) and 0 equations, D is most likely to be the answer.

Condition 1) \(x + 1 > 0\) \(x > -1\) Since the range of the question does not include the range of the condition, this is not sufficient.

Condition 2) \(|x+4| < |x-2| ⟺|x+4|^2 < |x-2|^2 ⟺(x+4)^2 < (x-2)^2 ⟺x^2 + 8x + 16 < x^2 -4x + 4 ⟺8x + 16 < -4x + 4 ⟺12x < -12\) \(⟺x < -1\) The answer is ‘no’. By CMT (Common Mistake Type) 1, ‘no’ is also an answer. This is sufficient.

Therefore, the answer is B.

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.

Answer: B

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Dear Bunuel, MathRevolution As per the Statement 1: x>-1 ahile Statement 2 ,x< -1. Doesnt it makes the question inconsistent.

Regards

MathRevolution wrote:

=>

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 1 variable (x) and 0 equations, D is most likely to be the answer.

Condition 1) \(x + 1 > 0\) \(x > -1\) Since the range of the question does not include the range of the condition, this is not sufficient.

Condition 2) \(|x+4| < |x-2| ⟺|x+4|^2 < |x-2|^2 ⟺(x+4)^2 < (x-2)^2 ⟺x^2 + 8x + 16 < x^2 -4x + 4 ⟺8x + 16 < -4x + 4 ⟺12x < -12\) \(⟺x < -1\) The answer is ‘no’. By CMT (Common Mistake Type) 1, ‘no’ is also an answer. This is sufficient.

Therefore, the answer is B.

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.

Answer: B

I am no expert in quants but as far as I could understand.. The question stem is

*IS X<0* and we need a definite answer to this question.

Statement A does not give us a definite answer.... Whereas Statement B gives us a definite answer....

I have a different approach to view Statement 2. Please see and confirm my understanding is correct.

Bunuel wrote:

Raksat wrote:

chetan2uMathRevolution My query is : Can we always square both the sides in an inequality involving modulus on both the sides as it is in this question. And why will the sign remain same after squaring ?

1. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality). For example: \(2<4\) --> we can square both sides and write: \(2^2<4^2\); \(0\leq{x}<{y}\) --> we can square both sides and write: \(x^2<y^2\);

But if either of side is negative then raising to even power doesn't always work. For example: \(1>-2\) if we square we'll get \(1>4\) which is not right. So if given that \(x>y\) then we cannot square both sides and write \(x^2>y^2\) if we are not certain that both \(x\) and \(y\) are non-negative.

2. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality). For example: \(-2<-1\) --> we can raise both sides to third power and write: \(-2^3=-8<-1=-1^3\) or \(-5<1\) --> \(-5^3=-125<1=1^3\); \(x<y\) --> we can raise both sides to third power and write: \(x^3<y^3\).