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.
For any real number x, the operator & is defined as
This topic has expert replies
-
- Legendary Member
- Posts: 1622
- Joined: Thu Mar 01, 2018 7:22 am
- Followed by:2 members
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
-
- Legendary Member
- Posts: 2898
- Joined: Thu Sep 07, 2017 2:49 pm
- Thanked: 6 times
- Followed by:5 members
Hello Gmat_mission.Gmat_mission wrote: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.
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.
GMAT/MBA Expert
- Brent@GMATPrepNow
- GMAT Instructor
- Posts: 16207
- Joined: Mon Dec 08, 2008 6:26 pm
- Location: Vancouver, BC
- Thanked: 5254 times
- Followed by:1268 members
- GMAT Score:770
Let's look at a few examples of this operator (&) in action.Gmat_mission wrote: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
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
GMAT/MBA Expert
- Jeff@TargetTestPrep
- GMAT Instructor
- Posts: 1462
- Joined: Thu Apr 09, 2015 9:34 am
- Location: New York, NY
- Thanked: 39 times
- Followed by:22 members
We can create the equation:Gmat_mission wrote: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
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
Jeffrey Miller
Head of GMAT Instruction
[email protected]
See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews
GMAT/MBA Expert
- ceilidh.erickson
- GMAT Instructor
- Posts: 2095
- Joined: Tue Dec 04, 2012 3:22 pm
- Thanked: 1443 times
- Followed by:247 members
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:Gmat_mission wrote: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.
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.
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
EdM in Mind, Brain, and Education
Harvard Graduate School of Education