[Math Revolution GMAT math practice question]
If x is integer and 3 | x |+x < 4, what is the value of x?
1) x < 0
2) x > -2
If x is integer and 3 | x |+x < 4, what is the value of x
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- Max@Math Revolution
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$$3\left| x \right| + x < 4\,\,\,\,\,,\,\,\,\,x\,\,{\mathop{\rm int}} $$Max@Math Revolution wrote:[Math Revolution GMAT math practice question]
If x is integer and 3 | x |+x < 4, what is the value of x?
1) x < 0
2) x > -2
$$? = x$$
$$\left( 1 \right)\,\,\,x < 0\,\,\,\,\, \Rightarrow \,\,\,\,3\left( { - x} \right) + x < 4\,\,\,\,\, \Rightarrow \,\,\,\, - 2x < 4\,\,\,\,\,\mathop \Rightarrow \limits^{:\,\,\left( { - 2} \right)} \,\,\,\,\,x > - 2$$
$$x > - 2\,\,\,\,\,\,\mathop \Rightarrow \limits^{x\,\,{\mathop{\rm int}} } \,\,\,\,\,\,x = - 1\,\,\,\,\,\left( {x < 0} \right)\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{SUFF}}.$$
$$\left( 2 \right)\,\,\,x > - 2\,\,\,\left\{ \matrix{
\,\,{\rm{Take}}\,\,x = - 1\,\,\,\,\,\left[ {\,\,3\left| { - 1} \right| + \left( { - 1} \right) < 4\,\,\,} \right]\,\,\,\,\,\,\, \Rightarrow \,\,\,? = - 1\,\, \hfill \cr
\,\,{\rm{Take}}\,\,x = 0\,\,\,\,\,\left[ {\,\,3\left| 0 \right| + \left( 0 \right) < 4\,\,\,} \right]\,\,\,\,\,\,\, \Rightarrow \,\,\,? = 0\,\, \hfill \cr} \right.\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{INSUFF}}.$$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
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3|x| < 4-xMax@Math Revolution wrote:[Math Revolution GMAT math practice question]
If x is integer and 3 | x |+x < 4, what is the value of x?
1) x < 0
2) x > -2
Case 1: Signs unchanged
3x < -x
4x < 4
x < 1
Case 2: Signs changed in the absolute value
3(-x) < 4-x
-2x < 4
x > -2
Question stem, rephrased:
If x is an integer such that -2 < x < 1, what is the value of x?
Statement 1: x < 0
Here, x must be an integer such that -2 < x < 0.
Thus, x = -1.
SUFFICIENT.
Statement 2: x > -2
Statement 2 offers no new information: the prompt itself indicates that - 2 < x < 1,
Since it's possible that x=-1 or that x=0, INSUFFICIENT.
The correct answer is A.
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=>
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. We then recheck the question.
Modifying the original condition:
There are two cases to consider.
Case 1) x ≥ 0:
3|x|+x < 4
=> 3x + x < 4
=> 4x < 4
=> x < 1
=> 0 ≤ x < 1
Case 2) x < 0:
-3x+x < 4
=> -2x < 4
=> x > -2
=> -2 < x <0
Therefore, x is an integer with -2 < x < 1. Thus, the original condition tells us that x = -1 or 0.
Since we have 1 variable (x) and 0 equations, D is most likely to be the answer. So, we should consider each condition on its own first.
Condition 1)
Since x < 0, we must have x = -1 as the original condition tells us that x = 0 or x = -1.
Condition 1) is sufficient, because it yields a unique solution.
Condition 2)
Both x = 0 and x = -1 satisfy condition 2).
Since it does not yield a unique solution, condition 2) is not sufficient.
Therefore, A is the answer.
Answer: A
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.
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. We then recheck the question.
Modifying the original condition:
There are two cases to consider.
Case 1) x ≥ 0:
3|x|+x < 4
=> 3x + x < 4
=> 4x < 4
=> x < 1
=> 0 ≤ x < 1
Case 2) x < 0:
-3x+x < 4
=> -2x < 4
=> x > -2
=> -2 < x <0
Therefore, x is an integer with -2 < x < 1. Thus, the original condition tells us that x = -1 or 0.
Since we have 1 variable (x) and 0 equations, D is most likely to be the answer. So, we should consider each condition on its own first.
Condition 1)
Since x < 0, we must have x = -1 as the original condition tells us that x = 0 or x = -1.
Condition 1) is sufficient, because it yields a unique solution.
Condition 2)
Both x = 0 and x = -1 satisfy condition 2).
Since it does not yield a unique solution, condition 2) is not sufficient.
Therefore, A is the answer.
Answer: A
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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