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

 It is currently 24 May 2015, 09:30

### 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

# y = x^2 + 6x 7, x= 1). x is positive integer 2). y^2=y

Author Message
TAGS:
Manager
Joined: 03 Jul 2006
Posts: 178
Followers: 1

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

y = x^2 + 6x 7, x= 1). x is positive integer 2). y^2=y [#permalink]  12 Sep 2006, 18:50
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
y = x^2 + 6x – 7, x=

1). x is positive integer

2). y^2=y
Manager
Joined: 03 Jul 2006
Posts: 178
Followers: 1

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

why not 'A'

from the stem, x = -7 or 1

and A says x has to be +ve then x must be 1 ?
VP
Joined: 28 Mar 2006
Posts: 1384
Followers: 2

Kudos [?]: 20 [0], given: 0

rkatl wrote:
why not 'A'

from the stem, x = -7 or 1

and A says x has to be +ve then x must be 1 ?

here the eq is not =0 but equal to y.

How can you deduce anything from 1 only?

either it has to be C or E
CEO
Joined: 20 Nov 2005
Posts: 2913
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
Followers: 18

Kudos [?]: 127 [0], given: 0

C

St1: clearly INSUFF

St2: y = 0 or y = 1.
then there are two equations
x^2 + 6x-7 = 0 gives us x = 1 or -7 and
x^2+6x-8 = 0 gives us two non-integer roots.: INSUFF

Together:
x = 1.: SUFF
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

SVP
Joined: 05 Jul 2006
Posts: 1518
Followers: 5

Kudos [?]: 118 [0], given: 39

the key here is statment b

y^2 =y thus y is either = {-1,1,0}....insuff

both together

x is positive and y = x^2 + 6x â€“ 7

if y = -1 thus x^2 + 6x-6 = 0....this equation has two non intiger answers one +ve and one -ve ( test with (b^2 - 4ac)....insuff

y = 1 thus x^2 + 6x-8 = 0 this equation has non intiger answers one +ve and one -ve....insuff

if y = 0 the equation has two intiger roots and as it is mentioned in one that x is positive intiger

Similar topics Replies Last post
Similar
Topics:
If x and y are positive integers and 5^x - 5^y = 2^(y-1) 0 24 Sep 2013, 13:48
1 if x is NOT = 0, is (x^2 +1)/x > Y? (1) x = y (2) y>0 4 06 Oct 2011, 12:06
1 If x!=0, is (x^2 + 1)/x > y? (1) x = y (2) y > 0 What 9 14 Jun 2011, 05:23
Set A is {x,2x+y,2y,y+z,x+z}. If x,y, and z are positive 2 11 Mar 2010, 05:40
Let *x* = x^2-1 and $y$ = 6y^2 - (*y*) for all integers x 3 15 May 2008, 11:49
Display posts from previous: Sort by