For any real number x, the operator & is defined as

[Gmat_mission]

For any real number x, the operator & is defined as: $$\&(x)=x(1−x)\ .$$ If p + 1 = &(p + 1), then p =

A. −1
B. 0
C. 1
D. 2
E. 3

[spoiler]OA=A[/spoiler].

How can I solve this PS question? Should I try option by option? Or there is an algebraic way? Thanks in advanced.

[Vincen]

For any real number x, the operator & is defined as: $$\&(x)=x(1−x)\ .$$ If p + 1 = &(p + 1), then p =

A. −1
B. 0
C. 1
D. 2
E. 3

[spoiler]OA=A[/spoiler].

How can I solve this PS question? Should I try option by option? Or there is an algebraic way? Thanks in advanced.

Hello Gmat_mission.

This is how I'd solve this question.

First, let's calculate the value expression &(p + 1) as follows: $$\&(p+1)=\left(p+1\right)\left(1-\left(p+1\right)\right)=\left(p+1\right)\left(-p\right)=-p\left(p+1\right).$$ Now, we set the given equation and solve for p. $$p+1=\&(p+1)$$ $$p+1=-p\left(p+1\right)$$ $$p+1+p\left(p+1\right)=0$$ $$\left(p+1\right)\left(1+p\right)=0$$ $$\left(p+1\right)^2=0$$ $$p+1=0$$ $$p=-1.$$ Therefore, the correct answer is the option A.

I hope this answer may help you.

[Brent@GMATPrepNow]

For any real number x, the operator & is defined as: $$\&(x)=x(1−x)\ .$$ If p + 1 = &(p + 1), then p =

A. −1
B. 0
C. 1
D. 2
E. 3

Let's look at a few examples of this operator (&) in action.
If &(x) = x(1 − x), then...
&(3) = 3(1 − 3) = 3(-2) = -6
&(7) = 7(1 − 7) = 7(-6) = -42
&(-5) = -5(1 − -5) = (-5)(6) = -30
And now.....
&(p+1) = (p+1)[1 − (p+1)] = (p + 1)(-p)

Now onto the question.....
We're told that p + 1 = &(p + 1)
We can now rewrite the right side of the equation as: p + 1 = (p + 1)(-p)
We need to solve this equation for p
Expand the right side to get: p + 1 = -p² - p
Add p² to both sides: p² + p + 1 = -p
Add p to both sides: p² + 2p + 1 = 0
Factor: (p +1)(p +1) = 0
So, p = -1

Answer: A

Cheers,
Brent

[Jeff@TargetTestPrep]

For any real number x, the operator & is defined as: $$\&(x)=x(1−x)\ .$$ If p + 1 = &(p + 1), then p =

A. −1
B. 0
C. 1
D. 2
E. 3

We can create the equation:

p + 1 = (p + 1)(1 - (p + 1))

p + 1 = (p + 1)(-p)

p + 1 = -p^2 - p

p^2 + 2p + 1 = 0

(p + 1)(p + 1) = 0

p = -1

Answer: A

[swerve]

$$\&(x)=x(1−x)$$
$$\&(p+1)=(p+1)(1-p-1)=-p(p+1)$$
We are given that
$$p+1=\&(p+1)$$
Therefore
$$-p(p+1)=(p+1)$$
Or
$$(p+1)+p(p+1)=0$$
$$(p+1)^2=0$$
$$p=-1$$
Regards!

[ceilidh.erickson]

For any real number x, the operator & is defined as: $$\&(x)=x(1−x)\ .$$ If p + 1 = &(p + 1), then p =

A. −1
B. 0
C. 1
D. 2
E. 3

[spoiler]OA=A[/spoiler].

How can I solve this PS question? Should I try option by option? Or there is an algebraic way? Thanks in advanced.

Quite often on this type of problem, testing the answer choices is a good option. The other posters have all given you the (identical) algebraic solution, so here's what answer testing would look like:

A. p = −1
p + 1 = -1 + 1 = 0
&(p + 1) = (p + 1)(1 - (p + 1)) --> (-1 + 1)(1 - (-1 + 1)) --> (0)(1) = 0
p + 1 = &(p + 1) ?
Yes!

B. p = 0
p + 1 = 0 + 1 = 1
&(p + 1) = (p + 1)(1 - (p + 1)) --> (0 + 1)(1 - (0 + 1)) --> (1)(0) = 0
p + 1 = &(p + 1) ?
No.

C. p = 1
p + 1 = 1 + 1 = 2
&(p + 1) = (p + 1)(1 - (p + 1)) --> (1 + 1)(1 - (1 + 1)) --> (2)(-1) = -2
p + 1 = &(p + 1) ?
No.

D. p = 2
p + 1 = 2 + 1 = 3
&(p + 1) = (p + 1)(1 - (p + 1)) --> (2 + 1)(1 - (2 + 1)) --> (3)(-2) = -6
p + 1 = &(p + 1) ?
No.

E. p = 3
p + 1 = 3 + 1 = 4
&(p + 1) = (p + 1)(1 - (p + 1)) --> (3 + 1)(1 - (3 + 1)) --> (4)(-3) = -12
p + 1 = &(p + 1) ?
No.

The answer is A.

Doing all of the arithmetic on each answer choice took slightly longer than the algebra on this problem, but of course we could have stopped after answer choice A and not tested the others. The question structure implies that there is only one value of p for which this will be true, so we can stop as soon as we hit that answer (I just wrote out the others to demonstrate the process).

As you're practicing, trying solving the problem BOTH ways, to develop the instinct of whether number testing or algebra will be faster on a given problem.