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Re: If k is an integer and x(x – k) = k + 1, what is the value of x? [#permalink]
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18 Jul 2016, 21:41
Bunuel wrote: If k is an integer and x(x – k) = k + 1, what is the value of x?
(1) x < k (2) x = 3 – k
Kudos for a correct solution. Given: x(xk) = k + 1 x^2  xk k  1 = 0 x^2  1 k(x+1) = 0 (x1)(x+1)  k (x+1) = 0 (x+1)(x1k) = 0  (i) Hence x =  1 or x = 1+k Required: x = ? Statement 1: x < k From this, we can say that ≠ k+1 Hence x can only take the value 1 SUFFICIENT Statement 2: x = 3k On substituting the value of k in the solutions for the equation, we get 3k =  1 and 3k = 1+k Hence k = 4 and k = 1 INSUFFICIENT Correct Option: A A very good question which can trap you in choosing the option C



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Re: If k is an integer and x(x – k) = k + 1, what is the value of x? [#permalink]
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18 Nov 2016, 09:48
Bunuel wrote: If k is an integer and x(x – k) = k + 1, what is the value of x?
(1) x < k (2) x = 3 – k
Kudos for a correct solution. I picked A... x<k so we have few options: 1) x negative k negative xk = negative > x*(xk) = positive. since k+1 must be positive, k must be either zero, or a noninteger. but since we assumed k is negative, then it doesn't work... 2) x positive k positive xk  negative negative * positive = negative it means that k+1 must be negative, meaning that k must be < 1. but we assumed k is positive. doesn't work... 3) we can't assume x positive and k negative, as we are told that x<k 4. x negative k positive xk  negative multiply by x (negative) = positive k+1 is positive. let's test some values: x=1 k=1 1(11) = 1*2 = 2 = 1+1  works x=2 k=1 2(21) = 6 but k=1. so doesn't work we can test more options, but it will not work unless x=1. A is sufficient. 2. x=3k x(x – k) = k + 1 k=3x x(x3+x) = 3x+1 x(2x3) = 4x 2x^2 3x = 4x 2x^2 2x 4 = 0  divide by 2 x^2 x 2 = 0 (x+1)(x2)=0 x=2; x=1 1st option x=2 x = 3 – k 2=3k k=1 x(x – k) = k + 1 = 2(21) = 1+1 2=2 2nd option: x=1 x = 3 – k 1 = 3k k=4 x(x – k) = k + 1 = 2(21) = 1+1 1(14) = 4+1 = 5=5 works 2 outcomes for x  not sufficient. answer is A.



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If k is an integer and x(x – k) = k + 1, what is the value of x? [#permalink]
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18 Nov 2016, 10:11
Bunuel wrote: If k is an integer and x(x – k) = k + 1, what is the value of x?
(1) x < k (2) x = 3 – k
Kudos for a correct solution. From statement 1> x(xk) = k+1 => x^2xkk1 = 0 => x = (k+(k^2+4k+2^2)^1/2)/2 or x = (k (k^2+4k+2^2)^1/2)/2 Since, k^2+4k+2^2 = (k+2)^2 By solving this we get either x= k+1 or x= 1; but x <k, therefore x can't be k+1 Hence, x = 1 => Sufficient From statement 2 > we can solve quadratic for k and get values 2 values for x and k each. Therefore it is insufficient Therefore, answer is A



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Re: If k is an integer and x(x – k) = k + 1, what is the value of x? [#permalink]
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19 Nov 2016, 00:42
(x1)*(x+1)k(x+1)=0
(x+1)*(x1k)=0
x+1=0, x=1 OR x1k=0, x=k+1
St.1 says that second scenario is impossible, so x=1
A



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If k is an integer and x(x – k) = k + 1, what is the value of x? [#permalink]
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01 Apr 2017, 00:27
Bunuel wrote: If k is an integer and x(x – k) = k + 1, what is the value of x?
(1) x < k (2) x = 3 – k
Kudos for a correct solution. correct answer is C statement 1: x(xk)=k+1 x^2xk=k+1 x^21=k+xk (x1)(x+1)(x+1)(k)=0 (x+1)[x1k]=0 this means x=1; or x=k+1 not valid since x<k(given) in sufficient as we cannot say anything about k when x=1 statement 2: putting x=3k given 2 values x=1;k=4 x=2;k=1 insufficient taking both together: x=1;k=4 solution therefore answer should be C give kudos if you liked this explanation



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If k is an integer and x(x – k) = k + 1, what is the value of x? [#permalink]
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01 Apr 2017, 00:35
Engr2012 wrote: Swaroopdev wrote: Hi Bunuel, None of the above solutions show A as the answer, is the OA wrong ? If not, could you please explain the solution ? Thanks. I think everyone is missing out on a very important information.Lets analyse the question: Given equation: x(xk) = k+1 > \(x^2kxk1=0\) >\(x = \frac {k\pm \sqrt{k^24(k1)}}{2}\) Thus you get 2 values of x as =\(\frac {k\pm(k+2)}{2}\) > x = k+1 or x=1 Per statement 1, x<k > x \(\neq\)k+1 > x = 1 . Thus this statement IS SUFFICIENTPer statement 2, x = 3k > susbtituting in the above given equation, you get, k = 4 or 1 > x = 1 or 2. Thus this statement is NOT sufficient. A is the correct answer.This question is a classic "CTrap" question. how can we conclude through statement A when x=1 x<k say x=1 and k=2 then also they satisfy the equation and when x=1 k=2 they also satisfy the equation then how can we find the values where k>x always???????? thus answer is C when x=1 k also takes unique value k=4



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Re: If k is an integer and x(x – k) = k + 1, what is the value of x? [#permalink]
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01 Apr 2017, 00:38
OptimusPrepJanielle wrote: Bunuel wrote: If k is an integer and x(x – k) = k + 1, what is the value of x?
(1) x < k (2) x = 3 – k
Kudos for a correct solution. Given: x(xk) = k + 1 x^2  xk k  1 = 0 x^2  1 k(x+1) = 0 (x1)(x+1)  k (x+1) = 0 (x+1)(x1k) = 0  (i) Hence x =  1 or x = 1+k Required: x = ? Statement 1: x < k From this, we can say that ≠ k+1 Hence x can only take the value 1 SUFFICIENT Statement 2: x = 3k On substituting the value of k in the solutions for the equation, we get 3k =  1 and 3k = 1+k Hence k = 4 and k = 1 INSUFFICIENT Correct Option: A A very good question which can trap you in choosing the option C how can we conclude through statement A when x=1 x<k say x=1 and k=2 then also they satisfy the equation and when x=1 k=2 they also satisfy the equation then how can we find the values where k>x always???????? thus answer is C when x=1 k also takes unique value k=4



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If k is an integer and x(x – k) = k + 1, what is the value of x? [#permalink]
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12 Apr 2017, 01:15
Bunuel, I was wondering about Statement 2. Using x = 3k, we could calculate as follows: (3k) (3kk) = k+1 ==> k^2  5k + 4 = 0 ; k= 4, 1 Using K=4 in the original equation x(xk) = k+1; we get: x^2  4x  5 =0 ==> x= 5 , 1 Using K=1 in the original equation x(xk) = k+1; we get: x^2  x  2 = 0 ==> x = 2 , 1 Since 1 is the only solution of x that satisfies both values of k in the original equation, can we not claim that statement 2 is sufficient?



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Re: If k is an integer and x(x – k) = k + 1, what is the value of x? [#permalink]
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12 Apr 2017, 02:13



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Re: If k is an integer and x(x – k) = k + 1, what is the value of x? [#permalink]
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12 Apr 2017, 03:13
Bunuel wrote: dg88074 wrote: Bunuel, I was wondering about Statement 2. Using x = 3k, we could calculate as follows: (3k) (3kk) = k+1 ==> k^2  5k + 4 = 0 ; k= 4, 1 Using K=4 in the original equation x(xk) = k+1; we get: x^2  4x  5 =0 ==> x= 5 , 1 Using K=1 in the original equation x(xk) = k+1; we get: x^2  x  2 = 0 ==> x = 2 , 1 Since 1 is the only solution of x that satisfies both values of k in the original equation, can we not claim that statement 2 is sufficient? We have: x(x – k) = k + 1 and x = 3 – k. k = 1 and x = 2 OR k = 4 and x = 1. Both sets satisfy the given inequalities. Oh it makes sense now. Thanks a lot Bunuel



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Re: If k is an integer and x(x – k) = k + 1, what is the value of x? [#permalink]
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21 Apr 2017, 16:18
Bunuel wrote: If k is an integer and x(x – k) = k + 1, what is the value of x?
(1) x < k (2) x = 3 – k
Kudos for a correct solution. st1: x^2xk=k+1 (x1)(x+1)=K(x+1) x+1= 0 x=1, k=0 x=k+1( irrelevant since we know k>x) st2: x=3k x=2, k=1 or x=1, k=4 (insufficient on its own) answer is A



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Re: If k is an integer and x(x – k) = k + 1, what is the value of x? [#permalink]
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07 Jul 2017, 06:34
smanujahrc wrote: OptimusPrepJanielle wrote: Bunuel wrote: If k is an integer and x(x – k) = k + 1, what is the value of x?
(1) x < k (2) x = 3 – k
Kudos for a correct solution. Given: x(xk) = k + 1 x^2  xk k  1 = 0 x^2  1 k(x+1) = 0 (x1)(x+1)  k (x+1) = 0 (x+1)(x1k) = 0  (i) Hence x =  1 or x = 1+k Required: x = ? Statement 1: x < k From this, we can say that ≠ k+1 Hence x can only take the value 1 SUFFICIENT Statement 2: x = 3k On substituting the value of k in the solutions for the equation, we get 3k =  1 and 3k = 1+k Hence k = 4 and k = 1 INSUFFICIENT Correct Option: A A very good question which can trap you in choosing the option C how can we conclude through statement A when x=1 x<k say x=1 and k=2 then also they satisfy the equation and when x=1 k=2 they also satisfy the equation then how can we find the values where k>x always???????? thus answer is C when x=1 k also takes unique value k=4 You are given that x=1 OR x=1+k, Statement 1 says that x<k, so how can x=1+k be true? x=1+k will ALWAYS make x>k, thus going against the information mentioned in statement 1. Thus, you MUST reject x=1+k, leaving you with x=1 as the only solution from statement 1. Hence this statement is sufficient.



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Re: If k is an integer and x(x – k) = k + 1, what is the value of x? [#permalink]
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16 Dec 2017, 06:20
Finally some sane question. x(xk) = k+1 => x^2  xk = k+1 => X^21  xk  k = 0 => (x1)(x+1)  k(x+1) = 0 => (x+1((x1k)=0 POssible values x=1, or, x=k+1
(1) x<k. Therefore, the only value is x=1. Sufficient (2) x=3k. Checking two possible values of x, x= k+1 => 3k =k+1 => k=1 & x=2. if x=1, => 3k=1, => k=4, x=1. Both values of x are possible. Ambiguous. Insufficient
Hence A



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Re: If k is an integer and x(x – k) = k + 1, what is the value of x? [#permalink]
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10 Apr 2018, 13:38
Hello, everybody!
I think one of quickest ways to solve it is by using a parabola method and Vieta's formulas ( x1*x2 = c/a; x1 + x2 = b/a).
Case 1:
1) So, we found x1*x2 = k1; and x1 + x2 = k.
Meaning, x1 = k +1, x2 = 1.
2) Our parabola has arms up and two possible xintercepts: x = k1; and x = 1.
If (k1) lays to the right of (1), then k1 > 1. So, k <0. And x<k.
Then, only possible solution is x2 = 1 (because whenever you add 1 to a negative number  it will become bigger, not smaller than it was; meaning x1 does not match the case). Sufficient.
If (k1) lays to the left of (1), then k1 < 1. So,k >0. And x < k.
Then, the only possible solution is x2 = 1 (becuse whenever you add 1 to a positive number  it will become bigger than it was; meaning x1 does not match the case). Sufficient.
Case 2: x = 3k. After soving the equation: k = 4; k = 1. So, 2 choices for x. Not sufficient.




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