Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 06 May 2016, 16:09

### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

Author Message
TAGS:

### Hide Tags

Manager
Status: Still Struggling
Joined: 03 Nov 2010
Posts: 137
Location: India
GMAT Date: 10-15-2011
GPA: 3.71
WE: Information Technology (Computer Software)
Followers: 5

Kudos [?]: 57 [0], given: 8

### Show Tags

20 Dec 2010, 08:20
6
This post was
BOOKMARKED
00:00

Difficulty:

15% (low)

Question Stats:

83% (02:28) correct 17% (01:47) wrong based on 207 sessions

### HideShow timer Statictics

John and Jane started solving a quadratic equation. John made a mistake while copying the constant term and got the roots as 5 and 9. Jane made a mistake in the coefficient of x and she got the roots as 12 and 4.What is the equation?

A. x^2 + 4x + 14 = 0
B. 2x^2 + 7x -24 = 0
C. x^2 -14x + 48 = 0
D. 3x^2 -17x + 52 = 0
E. 2x^2 + 4x + 14 = 0
[Reveal] Spoiler: OA

_________________

Knewton Free Test 10/03 - 710 (49/37)
Princeton Free Test 10/08 - 610 (44/31)
Kaplan Test 1- 10/10 - 630
Veritas Prep- 10/11 - 630 (42/37)
MGMAT 1 - 10/12 - 680 (45/34)

Last edited by Bunuel on 09 Jul 2013, 10:02, edited 1 time in total.
Edited the question.
Math Expert
Joined: 02 Sep 2009
Posts: 32664
Followers: 5662

Kudos [?]: 68843 [0], given: 9819

Re: Problem in the roots [#permalink]

### Show Tags

20 Dec 2010, 08:41
Expert's post
7
This post was
BOOKMARKED
krishnasty wrote:
John and Jane started solving a quadratic equation. John made a mistake while copying the constant term and got the roots as 5 and 9. Jane made a mistake in the coefficient of x and she got the roots as 12 and 4.What is the equation?

a. x2 + 4x +14 =0
b. 2x2 +7x -24 =0
c. x2 -14x +48 =0
d. 3x2 -17x +52 =0
e. 2x2 + 4x +14 =0

Kindly explain..!!

Viete's formula for the roots $$x_1$$ and $$x_2$$ of equation $$ax^2+bx+c=0$$: $$x_1+x_2=-\frac{b}{a}$$ AND $$x_1*x_2=\frac{c}{a}$$.

John made a mistake while copying the constant term and got the roots as 5 and 9 so he copied coefficient of x and x^2 correctly: $$5+9=14=-\frac{b}{a}$$;

Jane made a mistake in the coefficient of x and she got the roots as 12 and 4 so she copied the constant term and the coefficient of x^2 correctly: $$12*4=48=\frac{c}{a}$$;

Only option C satisfies the above conditions $$x^2-14x+48=0$$: $$-\frac{b}{a}=-\frac{-14}{1}=14$$ and $$\frac{c}{a}=\frac{48}{1}=48$$ (any of the condition would be sufficient).

_________________
Intern
Joined: 09 Nov 2010
Posts: 9
Followers: 0

Kudos [?]: 7 [0], given: 1

Re: Problem in the roots [#permalink]

### Show Tags

21 Dec 2010, 04:56
Bunuel wrote:
krishnasty wrote:
John and Jane started solving a quadratic equation. John made a mistake while copying the constant term and got the roots as 5 and 9. Jane made a mistake in the coefficient of x and she got the roots as 12 and 4.What is the equation?

a. x2 + 4x +14 =0
b. 2x2 +7x -24 =0
c. x2 -14x +48 =0
d. 3x2 -17x +52 =0
e. 2x2 + 4x +14 =0

Kindly explain..!!

Viete's formula for the roots $$x_1$$ and $$x_2$$ of equation $$ax^2+bx+c=0$$: $$x_1+x_2=-\frac{b}{a}$$ AND $$x_1*x_2=\frac{c}{a}$$.

John made a mistake while copying the constant term and got the roots as 5 and 9 so he copied coefficient of x and x^2 correctly: $$5+9=14=-\frac{b}{a}$$;

Jane made a mistake in the coefficient of x and she got the roots as 12 and 4 so she copied the constant term and the coefficient of x^2 correctly: $$12*4=48=\frac{c}{a}$$;

Only option C satisfies the above conditions $$x^2-14x+48=0$$: $$-\frac{b}{a}=-\frac{-14}{1}=14$$ and $$\frac{c}{a}=\frac{48}{1}=48$$ (any of the condition would be sufficient).

I just solve it with (x-y)(x-y) aproach not sure if I was just lucky and landed on the right answer or if it does always work

So: John got values 9 and 5 for x; (x-5)(x-9)=0 --> x^2-14x+45=0 (his mistake 45)
Jane got values 12 and 4 for x; (x-12)(x-4)=0 --> x^2-16x+48=0 (her mistake 16x)

Eliminate mistakes and plug in the right values each picked: X^2-14x(from Jonhs equ)+48(from janes)=0

Guess that this might only work when X^2 does not have a multiplier...

Will be adding to my "Blackbook" Viete's formula for the roots.
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 6493
Location: Pune, India
Followers: 1763

Kudos [?]: 10528 [0], given: 207

Re: Problem in the roots [#permalink]

### Show Tags

21 Dec 2010, 13:43
Expert's post
itnas wrote:
John and Jane started solving a quadratic equation. John made a mistake while copying the constant term and got the roots as 5 and 9. Jane made a mistake in the coefficient of x and she got the roots as 12 and 4.What is the equation?

a. x2 + 4x +14 =0
b. 2x2 +7x -24 =0
c. x2 -14x +48 =0
d. 3x2 -17x +52 =0
e. 2x2 + 4x +14 =0

I just solve it with (x-y)(x-y) aproach not sure if I was just lucky and landed on the right answer or if it does always work

So: John got values 9 and 5 for x; (x-5)(x-9)=0 --> x^2-14x+45=0 (his mistake 45)
Jane got values 12 and 4 for x; (x-12)(x-4)=0 --> x^2-16x+48=0 (her mistake 16x)

Eliminate mistakes and plug in the right values each picked: X^2-14x(from Jonhs equ)+48(from janes)=0

Guess that this might only work when X^2 does not have a multiplier...

Will be adding to my "Blackbook" Viete's formula for the roots.

This approach isn't incorrect. It is perfectly fine and Viete's formula is good to know (I assume it is discussed in detail in high school in most curriculums).

It will work even if the equation has a co-efficient other than 1 for x^2. Lets say, rather than x2 -14x +48 =0, the given equation in options is: 2x^2 - 28x +96 =0. It doesn't matter since 2 is common to all terms and will be taken out and eliminated. So, in essence, the equation is still x2 -14x +48 =0.
Also if the equation is something like 4x^2 -28x + 45 =0, where nothing is common, the roots you obtain will reflect the co-efficient of x^2. i.e. roots of this equation are 5/2 and 9/2 and when you put it in the (x-a)(x-b) = 0 form, you get:
(x - 5/2)(x - 9/2) = 0
x^2 - 14/2 x + 45/4 = 0
4x^2 - 28x + 45 = 0

So according to the given roots, you will always get the accurate equation. It might have something common in all the terms but that equation will still be the same.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Math Expert Joined: 02 Sep 2009 Posts: 32664 Followers: 5662 Kudos [?]: 68843 [0], given: 9819 Re: John and Jane started solving a quadratic equation. John mad [#permalink] ### Show Tags 09 Jul 2013, 10:03 Expert's post Bumping for review and further discussion. _________________ GMAT Club Legend Joined: 09 Sep 2013 Posts: 9317 Followers: 456 Kudos [?]: 115 [0], given: 0 Re: John and Jane started solving a quadratic equation. John mad [#permalink] ### Show Tags 09 Oct 2014, 05:59 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Manager Joined: 21 Sep 2012 Posts: 219 Location: United States Concentration: Finance, Economics Schools: CBS '17 GPA: 4 WE: General Management (Consumer Products) Followers: 3 Kudos [?]: 138 [0], given: 31 Re: John and Jane started solving a quadratic equation. John mad [#permalink] ### Show Tags 09 Oct 2014, 07:30 John made a mistake while copying the constant term and got the roots as 5 and 9. (x-5)(x-9)=0 x^2-14x+45=0 ....... I Jane made a mistake in the coefficient of x and she got the roots as 12 and 4. (x-12)(x-4)=0 x^2-16x+48=0 ....... II John made a mistake while copying constant term, so ignore constant term and pick first term of his solution x^2 Jane made a mistake in the coefficient of x, so ignore that term and pick the last term of her solution +48 Therefore correct equation must contain x^2 as its first term and +48 as its last term. only C satisfies the condition. Ans= C Intern Joined: 23 Dec 2014 Posts: 44 Followers: 0 Kudos [?]: 2 [0], given: 50 Re: John and Jane started solving a quadratic equation. John mad [#permalink] ### Show Tags 23 Dec 2014, 15:15 I found this very hard. Can you tell me what is the level of this question? _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6493 Location: Pune, India Followers: 1763 Kudos [?]: 10528 [0], given: 207 Re: John and Jane started solving a quadratic equation. John mad [#permalink] ### Show Tags 23 Dec 2014, 21:35 Expert's post Salvetor wrote: I found this very hard. Can you tell me what is the level of this question? The question isn't very hard. You can easily get the values for the co-efficient of x and the constant term from the two equations as shown by itnas above. I would say this is around 600 level. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 9317
Followers: 456

Kudos [?]: 115 [0], given: 0

### Show Tags

13 Jan 2016, 09:30
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Similar topics Replies Last post
Similar
Topics:
1 Solving for x using the quadratic equations by using the formula 1 20 Dec 2015, 06:56
7 John and Amanda stand at opposite ends of a straight road and start 7 05 Jun 2015, 06:21
7 John and Jane went out for a dinner and they ordered the 4 26 Jul 2014, 11:39
1 One hour before John started walking from P to Q, a distance 7 15 Apr 2014, 15:28
24 John can complete a given task in 20 days. Jane will take 15 20 Jul 2008, 05:23
Display posts from previous: Sort by