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Bunuel
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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.
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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(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 \(\neq\)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.
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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
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abhimahna
Celestial09
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

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.

-- Hit Kudos if you get the answer.


Why CANT I cancel
x+1 on both sides?
thanks
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Celestial09
abhimahna
Celestial09
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

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.

-- Hit Kudos if you get the answer.


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

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.
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Bunuel
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(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
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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?
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dg88074
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?

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.
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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.
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rick0917
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.

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.
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