GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 19 Jun 2018, 14:32

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Two students were asked to solve the equation

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Expert Post
1 KUDOS received
Math Revolution GMAT Instructor
User avatar
D
Joined: 16 Aug 2015
Posts: 5577
GMAT 1: 800 Q59 V59
GPA: 3.82
Two students were asked to solve the equation [#permalink]

Show Tags

New post 17 Jan 2018, 01:07
1
1
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

82% (02:08) correct 18% (00:49) wrong based on 57 sessions

HideShow timer Statistics

[GMAT math practice question]

Two students were asked to solve the equation \(x^2+px+q=0\). Alice’s answer was \(x=1\) and \(5\). This answer is incorrect for the original equation, but correct for the equation \(x^2+px+r=0.\) Bob’s answer was \(x=-2\) and \(-4\). This answer is incorrect for the original equation, but correct for \(x^2+sx+q=0\). What are the solutions of the original equation?

\(A. 1, 8\)
\(B. -1,-8\)
\(C. 2,4\)
\(D. -1,-5\)
\(E. 1,-5\)

_________________

MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
"Only $99 for 3 month Online Course"
"Free Resources-30 day online access & Diagnostic Test"
"Unlimited Access to over 120 free video lessons - try it yourself"

Senior Manager
Senior Manager
User avatar
G
Joined: 17 Oct 2016
Posts: 326
Location: India
Concentration: Operations, Strategy
GPA: 3.73
WE: Design (Real Estate)
GMAT ToolKit User Premium Member CAT Tests
Re: Two students were asked to solve the equation [#permalink]

Show Tags

New post 17 Jan 2018, 01:29
1
sum of roots of x^2+px+r=0 is -p=1+5 (Alice's eqn)

p=-6

product of roots of x^2+sx+q=0 is q=-2*(-4) (Bob's eqn)

q=8

now the original equation is

(x^2)-6x+8=0

whose roots are 2 and 4 which is option C

But OA is B. Can someone explain?
_________________

Help with kudos if u found the post useful. Thanks

Intern
Intern
avatar
Joined: 15 Jan 2018
Posts: 1
Re: Two students were asked to solve the equation [#permalink]

Show Tags

New post 17 Jan 2018, 11:15
I also think it is Answer C:

x^2+px+r with x1=1 , x2=5 correspons to (x-1)*(x-5) = x^2-6x+5
x^2+sx+q with x1=-2 , x2=-4 correspons to (x+2)*(x+4) = x^2+6x+8

so x^2+px+q = x^2-6x+8 = 0
if you use pq-formula you will get x1=2 , x2=4
Manager
Manager
avatar
B
Joined: 22 May 2017
Posts: 118
Re: Two students were asked to solve the equation [#permalink]

Show Tags

New post 17 Jan 2018, 11:25
I think there is some error in OA
solved it individually as well

(x-1)(x-5) = x2-6x+5
(x+2)(x+4) = x2+6x+8

therefore , x2-6x+8 correct equation
roots are (x-2)(x-4)
_________________

Kudos please if explanation helped
-------------------------------------------------------------------------------------------------
Don't stop when you are tired , stop when you are DONE .

Expert Post
Math Revolution GMAT Instructor
User avatar
D
Joined: 16 Aug 2015
Posts: 5577
GMAT 1: 800 Q59 V59
GPA: 3.82
Two students were asked to solve the equation [#permalink]

Show Tags

New post Updated on: 21 Jan 2018, 09:41
=>

Using the factor theorem with Alice’s solution yields
\((x-1)(x-5) = x^2 – 6x + 5 = x^2+px+r.\)
So, \(p = -6\).

Using the factor theorem with Bob’s solution yields
\((x+2)(x+4) = x^2 + 6x + 8 = x^2 +sx +q.\)
So, \(q = 8.\)

Thus, the original equation is \(x^2-6x+8 = (x-2)(x-4) = 0\). Its roots are \(2\) and \(4\).

Therefore, the answer is C.
_________________

MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
"Only $99 for 3 month Online Course"
"Free Resources-30 day online access & Diagnostic Test"
"Unlimited Access to over 120 free video lessons - try it yourself"


Originally posted by MathRevolution on 19 Jan 2018, 01:19.
Last edited by MathRevolution on 21 Jan 2018, 09:41, edited 1 time in total.
Senior Manager
Senior Manager
User avatar
G
Joined: 17 Oct 2016
Posts: 326
Location: India
Concentration: Operations, Strategy
GPA: 3.73
WE: Design (Real Estate)
GMAT ToolKit User Premium Member CAT Tests
Re: Two students were asked to solve the equation [#permalink]

Show Tags

New post 19 Jan 2018, 03:03
MathRevolution wrote:
=>

Using the factor theorem with Alice’s solution yields
\((x-1)(x-5) = x^2 – 6x + 5 = x^2+px+r.\)
So, \(p = -6\).

Using the factor theorem with Bob’s solution yields
\((x+2)(x+4) = x^2 + 6x + 8 = x^2 +sx +q.\)
So, \(q = 8.\)

Thus, the original equation is \(x^2-6x+8 = (x-2)(x-4) = 0\). Its roots are \(2\) and \(4\).

Therefore, the answer is C.
Answer: B


Please clarify what is the OA? It has to be C. in that case you have to edit the OA
_________________

Help with kudos if u found the post useful. Thanks

Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 46167
Re: Two students were asked to solve the equation [#permalink]

Show Tags

New post 19 Jan 2018, 03:53
Sasindran wrote:
MathRevolution wrote:
=>

Using the factor theorem with Alice’s solution yields
\((x-1)(x-5) = x^2 – 6x + 5 = x^2+px+r.\)
So, \(p = -6\).

Using the factor theorem with Bob’s solution yields
\((x+2)(x+4) = x^2 + 6x + 8 = x^2 +sx +q.\)
So, \(q = 8.\)

Thus, the original equation is \(x^2-6x+8 = (x-2)(x-4) = 0\). Its roots are \(2\) and \(4\).

Therefore, the answer is C.
Answer: B


Please clarify what is the OA? It has to be C. in that case you have to edit the OA


The OA is C. Edited. Thank you.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Re: Two students were asked to solve the equation   [#permalink] 19 Jan 2018, 03:53
Display posts from previous: Sort by

Two students were asked to solve the equation

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.