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ans A.. 1) by looking at the statement 1 itself , one can find that x=0.. also by opening the mod, we can find that x gives only one value that is 0.. sufficient 2) 7 and 3 straight way gives one value of x as 0 and by opening mod, we have x can take multiple values 2 ,-2,3,4etc ... so insufficient
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When both 7 - x and 3 + x are positive (so when -3 <= x <= 7), then |7 - x| + |3 + x| = 10 expands as 7 - x + 3 + x = 10 --> 10 = 10, which is true. This means that any x where -3 <= x <= 7 satisfies the equation. Not sufficient.

Can anyone help in how do we open the modules sign for 1 and 2?

this is how i solved it but its time consuming 1. |x-1| - |x+1| = 0 -(x-1) - (x+1) = 0 therefore x= 0 (x-1) + (x+1) = 0 therefore x = 0

any alternate technique to solve this problem

Make sure to format your question in such a way that puts your analysis under "spoiler".

As for your question,

Both statements can be solved in a similar way,

Statement 1, |1-x|-|x+1|=0 , now as |a-x|=|x-a| , you get |x-1|-|x+1|=0 ---> |x-1|=|x+1| ----> \(\pm (x-1) = \pm (x+1)\) , you only get 2 possible cases giving you x=0 for both cases. Hence this statement is sufficient.

Statement 2, |7 - x| + |3 + x| = 10, using the same principle as shown in statement 1, you get the equation as \(\pm (x-7) \pm (x+3) = 10\), giving you 2 possible cases again but giving you x=-3 and x=7. Thus this statement is not sufficient.

In statement 2, you will notice that there is a 7 and 3 on the left side and 10 on the right. We also know 7+3=10, and if we choose x=0, we see the equation is indeed satisfied, and if we change x to 1, then also it is satisfied. So just by observation we can come up with two values for x, and say that statement 2 alone is not sufficient.

For statement 1, one could quickly graph the functions |1-x| and |x+1| and show that they only intersect at x=0. Actually, being able to graph these basic types of absolute value equations is important for the GMAT.

And finally, the statement 1 is based on the following official GMAT question from official guide 10th edition, data sufficiency question#81. See the attached image.

I didn't square, since I didn't know how to tackle this question rather than try the possible options, and although I selected A as the answer I later check and found something interesting: both positive both negative 1st positive second negative 1st negative second positive:

1. |1 - x| - |x + 1| = 0 ++ 1-x-x+1=0=> 0=2x=>x=0 -- -1+x+x+1=>2x=0 => x=0 -+ -1+x-x+1 => 0=0 - doesn't help us much, since there is no way to solve it. or x can take any value. +- 1-x+x+1 => 2=0 - this is not a solution.

x 2. |7 - x| + |3 + x| = 10

only 2 options tested, which yielded 2 different values for x, thus I dismissed B and D.

For such questions the most straightforward method to solve is to realize that |x-a| with a = constant is the distance of x from 'a' and |a-x| = |x-a|

Thus |1-x| = |x-1| --> distance of x from 1,

Similarly for |x+1|, |x-7| and |x+3|.

Do remember that x = integer (given information).

Per statement 1, |1 - x| - |x + 1| = 0 ---> |x-1| - |x + 1| = 0 ---> |x-1| = |x + 1| ---> distance of integer x is same from -1 and 1 ---> x can only take 0 as the possible value. Unique value of x ---> sufficient statement.

Per statement 2, |7 - x| + |3 + x| = 10 --> |x-7| + |3 + x| = 10 ---> distance of x from 7 + distance of x from -3 = 10 units. This is satisfied by all integer values of x between -3 and 7. Consider x = 4 , distance from 7 = 3 , distance of x from -3 = 7 , total = 10.

But if x = 0, distance from 7 = 7 and distance of x from -3 = 3 , total = 10 units.

Thus this statement provides multiple values of x ---> not sufficient.

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.

When it comes to absolute value, it indicates a distance between two dots. That is, it is a distance between |x-y|=x and y in this question. In the original condition, there is 1 variable(x), which should match with the number of equations. So you need 1 equation. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer. For 1), from |1-x|=|x-(-1)|, since a distance from x to 1 is as same as a distance from x to -1, x=0 is derived, which is unique and sufficient. For 2), from |7-x|+|x-(-3)|=10, the sum of a distance between x and 7, and a distance between x and -3 is 10. That is, all values are possible in -3<=x<=7, which is not unique and not sufficient. Therefore, the answer is A.

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Bunuel: How do I deal with second case - |7-x| + |3+x| = 10 Now four possibilities can exist Case 1: 7-x>0 and 3+x>0 -3 < x < 7 On solving 10=10

Case 2: 7-x<0 and 3+x<0 4=10 Dump this!

Case 3: 7-x>0 and 3+x<0 x<7 and x<-3 or x<-3 On solving we get ; x=-3

Case 4: 7-x<0 and 3+x>0 x > -3 and x >7 or x>7 On solving x = 7

Bunuel: Can you please suggest how do I conclude this.

\(|3 + x| + |7 - x| = 10\)

2 key points at -3 and 7.

When \(x < -3\), then \(3 + x\) is negative and \(7 - x\) is positive, thus \(|3 + x| + |7 - x| = 10\) becomes \(-(3 + x) + (7 - x) = 10\) --> \(x=-3\). Discard since -3 is not in the range;

When \(-3 \leq x \leq 7\), then \(3 + x\) is positive and \(7 - x\) is positive, thus \(|3 + x| + |7 - x| = 10\) becomes \((3 + x) + (7 - x) = 10\) --> \(10=10\). Since this is true, then this means that ANY x from the given range satisfies the equation;

When \(x > 7\), then \(3 + x\) is positive and \(7 - x\) is negative, thus \(|3 + x| + |7 - x| = 10\) becomes \((3 + x) - (7 - x) = 10\) --> \(x=-3\). Discard since 7 is not in the range.

Therefore, \(|3 + x| + |7 - x| = 10\) is true for any x where \(-3 \leq x \leq 7\).

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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