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01 Sep 2015, 22:51
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
18% (02:48) correct
82%
(02:46)
wrong
based on 1279
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If k is an integer and x(x – k) = k + 1, what is the value of x?
(1) x < k
(2) x = 3 – k
(1) x < k
(2) x = 3 – k
Updated on: 15 Apr 2018, 21:07
Originally posted by MathRevolution on 03 Sep 2015, 04:39.
Last edited by MathRevolution on 15 Apr 2018, 21:07, edited 2 times in total.
Last edited by MathRevolution on 15 Apr 2018, 21:07, edited 2 times in total.
Kudos
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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.
The original condition can be modified as follows.
x(x-k) = k+1
⇔ x^2 - xk = k + 1
⇔ x^2 - xk - k - 1 = 0
⇔ x^2 -k(x+1) - 1 = 0
⇔ (x+1)(x-1) -k(x+1) = 0
⇔ (x+1)(x-k-1) = 0
Since we have 2 variable (x and y) and 1 equation, D is most likely to be the answer. So, we should consider each of the conditions on their own first.
Condition 1)
Since x < k, x - k - 1 ≠ 0.
Thus we have x = -1.
Condition 1) is sufficient.
Condition 2)
We have two solutions.
x = -1, k = 4
x = 2, k = 1
Thus condition 2) is not sufficient.
Therefore, the answer is A.
The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.
The original condition can be modified as follows.
x(x-k) = k+1
⇔ x^2 - xk = k + 1
⇔ x^2 - xk - k - 1 = 0
⇔ x^2 -k(x+1) - 1 = 0
⇔ (x+1)(x-1) -k(x+1) = 0
⇔ (x+1)(x-k-1) = 0
Since we have 2 variable (x and y) and 1 equation, D is most likely to be the answer. So, we should consider each of the conditions on their own first.
Condition 1)
Since x < k, x - k - 1 ≠ 0.
Thus we have x = -1.
Condition 1) is sufficient.
Condition 2)
We have two solutions.
x = -1, k = 4
x = 2, k = 1
Thus condition 2) is not sufficient.
Therefore, the answer is A.
17 Jul 2016, 06:29
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Celestial09
I think the explanation as previously stated is the correct explanation for A.
x(x-k)=K+1
=> x^2-xk=k+1
=>
x^2-1=k(x+1)
or (x-1)(x+1)=k(x+1)
you cannot cancel x+1 on both sides directly.
Either we have k=(x-1) or x+1=0
=>
either x=k+1 or x=-1
Statement 1 says x<k => x <> k+1, thus x=-1, Hence it is sufficient.
Statement 2 says x=3-k => x+k=3. We cannot find out x here, as we don't know the value of K. Hence 2 is insufficient.
Thus, the answer is A.
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