Hello All,
I would like to listen to some comments and find alternate way(s) to approach the following question. If this question has been explained already in the forum, I apologize for the re-post.
OG Quant Review 2ED - PS - Question 103
If a, b, and c are constants, a > b > c, and (x^3 - x) = (x-a)(x-b)(x-c) for all numbers x, what is the value of b ?
I tried to expand x^3-x trying to find a way to solve but ended up nowhere. Hence I switched to this route.
The question stem states the Left Hand Side(LHS) = Right Hand Side(RHS) and also mentions "for all numbers x".
So I set x=2 (some random number) and arrived here:
8-2 = (2-a) (2-b) (2-c)
6 = (2-a) (2-b) (2-c)
From here - three numbers when multiplied together gives 6 (3x2x1) and a>b>c.
I needed a 3, 2 and 1 and picked numbers for a,b and c.
6 = (2-1) (2-0) (2-(-1))
I believe my approach does not follow a set rule, might take extra time and break in a few scenarios(that I couldn't think of now). Please explain how to solve this question in a systematic manner.
Thank you!
Best Regards,
Arvind.
OG Quant Review - PS - Q 103
This topic has expert replies
- Uva@90
- Master | Next Rank: 500 Posts
- Posts: 490
- Joined: Thu Jul 04, 2013 7:30 am
- Location: Chennai, India
- Thanked: 83 times
- Followed by:5 members
Hi Arvind,
so,(x-1)(x)(x+1) = (x-a)(x-b)(x-c)
since a>b>c , there fore value of (x-b) will be between (x-a) and (x-c)
in LHS of the above equation we can see that (x+1)>x>(x-1)
so x = x-b
hence [spoiler]b=0[/spoiler]
Hope it helps you.
Regards,
Uva.
x^3 - x can be re-written as x(x^2-1) => x(x+1)(x-1)If a, b, and c are constants, a > b > c, and (x^3 - x) = (x-a)(x-b)(x-c) for all numbers x, what is the value of b ?
so,(x-1)(x)(x+1) = (x-a)(x-b)(x-c)
since a>b>c , there fore value of (x-b) will be between (x-a) and (x-c)
in LHS of the above equation we can see that (x+1)>x>(x-1)
so x = x-b
hence [spoiler]b=0[/spoiler]
Hope it helps you.
Regards,
Uva.
Known is a drop Unknown is an Ocean
- GMATGuruNY
- GMAT Instructor
- Posts: 15539
- Joined: Tue May 25, 2010 12:04 pm
- Location: New York, NY
- Thanked: 13060 times
- Followed by:1906 members
- GMAT Score:790
Since a>b>c, (x-a)(x-b)(x-c) represents 3 factors in ASCENDING order, where (x-b) is the value of the MIDDLE factor.arvysri wrote: If a, b, and c are constants, a > b > c, and (x^3 - x) = (x-a)(x-b)(x-c) for all numbers x, what is the value of b ?
-3
-1
0
1
3
Your approach -- to plug in a value for x and solve for b -- could lead to issues here.
Let x=3.
Then x³ - x = 3³ - 3 = 24.
Here, the middle factor = 3-b.
Two options for the 3 factors in ascending order:
Case 1: 1*4*6
Here, 3-b=4, implying that b=-1 (answer choice B).
Case 2: 2*3*4
Here, 10-b=3, implying that b=0 (answer choice C).
A test-taker who considers only Case 1 would be led to select answer choice B, which is incorrect.
For this reason, algebra is safer here:
x³ - x = x(x²-1) = x(x+1)(x-1).
Rearranging the factors in ascending order, we get:
(x-1)(x)(x+1).
Since the middle factor is x:
x-b = x
b=0.
The correct answer is C.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
- ygcrowanhand
- Senior | Next Rank: 100 Posts
- Posts: 30
- Joined: Mon Nov 24, 2014 1:47 pm
- Location: London
Is Your GMAT Score Stuck in the 600s? This FREE 8-Video, 20-Page Guide Can Help.
https://privategmattutor.london/move-yo ... -the-700s/
PS have you seen the new GMAT Work and Rates guide? Comes with a free 8-video course.
https://yourgmatcoach.podia.com/courses ... s-problems
Learn more about Private GMAT Tutoring at: https://privategmattutor.london
https://privategmattutor.london/move-yo ... -the-700s/
PS have you seen the new GMAT Work and Rates guide? Comes with a free 8-video course.
https://yourgmatcoach.podia.com/courses ... s-problems
Learn more about Private GMAT Tutoring at: https://privategmattutor.london
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
Let's simplify the given equation:arvysri wrote: If a, b, and c are constants, a > b > c, and (x^3 - x) = (x-a)(x-b)(x-c) for all numbers x, what is the value of b ?
A. -3
B. -1
C. 0
D. 1
E. 3
x^3 - x = (x-a)(x-b)(x-c)
x(x^2 - 1) = (x-a)(x-b)(x-c)
x(x - 1)(x + 1) = (x-a)(x-b)(x-c)
(x - 1)x(x + 1) = (x-a)(x-b)(x-c)
Since a > b > c:
(x - b) = x
b = 0
Alternate Solution:
If we let x = a in the equation, we get a^3 - a = 0; i.e., a^3 = a. Similarly, letting x = b and x = c, we get b^3 = b and c^3 = c. We know that the only three numbers whose cube is equal to the number itself are 1, 0, and -1. We have a > b > c; therefore, we must have a = 1, b = 0, and c = -1.
Answer: C
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
It helps to recognize that x^3 - x is GMAT code for the product of 3 consecutive integers. As Jeff pointed out, we can factor it as (x - 1)(x)(x + 1). This structure (or variations on it) have shown up in a number of official questions testing the properties of consecutive products.arvysri wrote:Hello All,
I would like to listen to some comments and find alternate way(s) to approach the following question. If this question has been explained already in the forum, I apologize for the re-post.
OG Quant Review 2ED - PS - Question 103
If a, b, and c are constants, a > b > c, and (x^3 - x) = (x-a)(x-b)(x-c) for all numbers x, what is the value of b ?
I tried to expand x^3-x trying to find a way to solve but ended up nowhere. Hence I switched to this route.
The question stem states the Left Hand Side(LHS) = Right Hand Side(RHS) and also mentions "for all numbers x".
So I set x=2 (some random number) and arrived here:
8-2 = (2-a) (2-b) (2-c)
6 = (2-a) (2-b) (2-c)
From here - three numbers when multiplied together gives 6 (3x2x1) and a>b>c.
I needed a 3, 2 and 1 and picked numbers for a,b and c.
6 = (2-1) (2-0) (2-(-1))
I believe my approach does not follow a set rule, might take extra time and break in a few scenarios(that I couldn't think of now). Please explain how to solve this question in a systematic manner.
Thank you!
Best Regards,
Arvind.
For example, look at problem DS #170 in OG12:
Here are a few other similar examples:If n is a positive integer, is n^3 - n divisible by 4?
(1) n = 2k + 1, where k is an integer
(2) n2 + n is divisible by 6
https://www.beatthegmat.com/x-x-1-x-k-t ... tml#770020
https://www.beatthegmat.com/totaly-lost ... tml#716315
https://www.beatthegmat.com/is-x-x-2-x- ... tml#718646
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
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
Given: x³ - x = (x - a)(x - b)(x - c)If a, b, and c are constants, a > b > c, and x^3 - x = (x - a)(x - b)(x - c) for all numbers x, what is the value of b ?
(A) - 3
(B) -1
(C) 0
(D) 1
(E) 3
Factor x from left side: x(x² - 1) = (x - a)(x - b)(x - c)
x² - 1 is a difference of squares, so we can factor that: x(x + 1)(x - 1) = (x - a)(x - b)(x - c)
Rewrite first 2 terms as: (x - 0)(x - -1)(x - 1) = (x - a)(x - b)(x - c)
Rearrange as: (x - 1)(x - 0)(x - -1) = (x - a)(x - b)(x - c)
Since, we're told that c < b < a, we can see that a = 1, b = 0 and c = -1
Answer: C
Cheers,
Brent