If k is an integer and x(x k) = k + 1, what is the value of x?

Question

If k is an integer and x(x – k) = k + 1, what is the value of x?

(1) x < k
(2) x = 3 – k

MathRevolution · Accepted Answer

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.

abhimahna · Answer

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 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. -- Hit Kudos if you get the answer.

Swaroopdev · Answer

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.

ENGRTOMBA2018 · Answer

I think everyone is missing out on a very important information.

Lets analyse the question:

Given equation: x(x-k) = k+1 ---->  x^2-kx-k-1=0  ---> x = \frac {k\pm \sqrt{k^2-4(-k-1)}}{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  
eq k+1 ---> x = -1 is the only solution left. Thus this statement  IS SUFFICIENT

Per statement 2, x = 3-k ---> substituting 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 "C-Trap" question.

Celestial09 · Answer

Hi!
I too picked C.

bunuel , chetan4u, empowergmat richi, magoosh mike, veritas..
plz help us on this 
1) correct answer and method
2) why answer c is wrong

thanks

Celestial09 · Answer

Why CANT I cancel 
 x+1 on both sides?
thanks

Bunuel · Answer

Because x + 1 can be 0 and we cannot divide by 0.

Never reduce  equation  by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We cannot divide by zero.

Never multiply (or reduce)  inequality  by variable (or expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero.

OptimusPrepJanielle · Answer

Given: x(x-k) = k + 1 
x^2 - xk -k - 1 = 0 
x^2 - 1 -k(x+1) = 0 
(x-1)(x+1) - k (x+1) = 0 
(x+1)(x-1-k) = 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 = 3-k 
On substituting the value of k in the solutions for the equation, we get 
3-k = - 1 and 3-k = 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

dg88074 · Answer

Bunuel , I was wondering about Statement 2. Using x = 3-k, we could calculate as follows:

(3-k) (3-k-k) = k+1 ==> k^2 - 5k + 4 = 0 ; k= 4, 1

Using K=4 in the original equation x(x-k) = k+1; we get:

x^2 - 4x - 5 =0 ==> x= 5 , -1

Using K=1 in the original equation x(x-k) = 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?

Bunuel · Answer

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.

rick0917 · Answer

Hi  Bunuel ,

The solution says A.

But if the value of k is less than -1 then it cannot be true. 
x<k and x=-1 
but if k<-1 then there are no solutions.

If this explanation is correct then I think we need to change the answer to E.

Bunuel · Answer

The OA is correct.

The point is that from x(x – k) = k + 1 and x < k it follows that k > -1, so your example is not valid: k cannot be less than or equal to -1 IF both x(x – k) = k + 1 and x < k are true.

bumpbot · Answer

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If k is an integer and x(x k) = k + 1, what is the value of x?

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If k is an integer and x(x k) = k + 1, what is the value of x?

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