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

 It is currently 11 Dec 2018, 16:59

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 December
PrevNext
SuMoTuWeThFrSa
2526272829301
2345678
9101112131415
16171819202122
23242526272829
303112345
Open Detailed Calendar
• Free GMAT Prep Hour

December 11, 2018

December 11, 2018

09:00 PM EST

10:00 PM EST

Strategies and techniques for approaching featured GMAT topics. December 11 at 9 PM EST.
• The winning strategy for 700+ on the GMAT

December 13, 2018

December 13, 2018

08:00 AM PST

09:00 AM PST

What people who reach the high 700's do differently? We're going to share insights, tips and strategies from data we collected on over 50,000 students who used examPAL.

In the xy plane, at what points does the graph of y=

Author Message
TAGS:

Hide Tags

Intern
Joined: 26 Nov 2009
Posts: 10
In the xy plane, at what points does the graph of y=  [#permalink]

Show Tags

Updated on: 26 Feb 2013, 04:23
3
35
00:00

Difficulty:

35% (medium)

Question Stats:

71% (01:51) correct 29% (01:59) wrong based on 814 sessions

HideShow timer Statistics

In the xy plane, at what points does the graph of y=(x+a)(x+b) intersect the x-axis?

(1) a + b = -1
(2) The graph intersects the y axis at (0,-6)

Originally posted by JimmyWorld on 24 Dec 2009, 09:00.
Last edited by Bunuel on 26 Feb 2013, 04:23, edited 2 times in total.
Edited the question and added the OA
Math Expert
Joined: 02 Sep 2009
Posts: 51100
Re: GMAT Prep Question, Any Shortcuts?  [#permalink]

Show Tags

24 Dec 2009, 09:16
11
21
JimmyWorld wrote:
This question took me much longer then 2 minutes to figure out and sucked up a lot of time for me. Are there any shortcuts that would make this problem easier to solve?

In the xy plane, at what points does the graph of y=(x+a)(x+b) intersect the x-axis?

(1) a + b = -1
(2) the graph intersects the y axis at (0,-6)

OA

Thanks.

X-intercepts of the function $$f(x)$$ or in our case the function (graph) $$y=(x+a)(x+b)$$ is the value(s) of $$x$$ for $$y=0$$. So basically the question asks to find the roots of quadratic equation $$(x+a)(x+b)=0$$.

$$(x+a)(x+b)=0$$ --> $$x^2+bx+ax+ab=0$$ --> $$x^2+(a+b)x+ab=0$$.

Statement (1) gives the value of $$a+b$$, but we don't know the value of $$ab$$ to solve the equation.

Statement (2) tells us the point of y-intercept, or the value of $$y$$ when $$x=0$$ --> $$y=(x+a)(x+b)=(0+a)(0+b)=ab=-6$$. We know the value of $$ab$$ but we don't know the value of $$a+b$$ to solve the equation.

Together we know the values of both $$a+b$$ and $$ab$$, hence we can solve the quadratic equation, which will be the x-intercepts of the given graph.

For more on this topic check Coordinate Geometry chapter of Math Book (link in my signature).

Hope it helps.
_________________
General Discussion
Intern
Joined: 17 Dec 2009
Posts: 38
Re: GMAT Prep Question, Any Shortcuts?  [#permalink]

Show Tags

24 Dec 2009, 09:37
1
y=(x+a)(x+b) when y=0
To solve this one, what do we need to know?
Obviously a or b, which are not stated in the information (1) & (2)
So one rule advise by MGMAT Book, always expand when the information given is factorized or Factorized when the information given is expended.
We know from a quadratic expression, the x axis intersect when y=0
So let's expand, and we have x^2 + (a + b)x + ab = 0
So now, we need to know ab and (a+b) to solve this equation
Therefore the correct answer is C, since only both information taken together permit to answer the question.
Retired Moderator
Joined: 02 Sep 2010
Posts: 766
Location: London
Re: DS question : need help  [#permalink]

Show Tags

15 Oct 2010, 00:05
satishreddy wrote:
need help on DS question

Right, so when the graph intersects the x-axis, y is 0. Those points are clearly x=-a and x=-b

(1) a+b=1. Not sufficient to know a or b
(2) We have point (0,-6). Plus it into the equation to get -6=ab. Again not sufficient to know either number

(1+2). Now we have 2 equations, sufficient. a-6/a=1; a^2-a-6=0; a=3 or a=-2 Hence b=-2 or b=3 .. Either way the points are 3 and -2

_________________
Manager
Joined: 06 Aug 2010
Posts: 173
Location: Boston
Re: GMAT Prep Question, Any Shortcuts?  [#permalink]

Show Tags

02 Nov 2010, 06:47
JimmyWorld wrote:
This question took me much longer then 2 minutes to figure out and sucked up a lot of time for me. Are there any shortcuts that would make this problem easier to solve?

In the xy plane, at what points does the graph of y=(x+a)(x+b) intersect the x-axis?

(1) a + b = -1
(2) the graph intersects the y axis at (0,-6)

OA

Thanks.

We know that the graph will intersect the x-axis when y=0, which occurs when x = -a and x = -b.

1) This doesn't tell us what a and b are. For example, a could be 5 and b could be -6, or a could be -4 and b could be 3. Insufficient.

2) This tells us that either a or b is 6, but we don't know the other point. Insufficient.

Together, we know that one of the values is 6, and that a + b = -1. So we can get the other value. Sufficient.
VP
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
Posts: 1033

Show Tags

03 May 2011, 09:19
y = x^2 + (a+b)x + ab is a parabola.Intersecting x axis at two points.
hence we need to find the roots of this equation.

a. a+b = -1 means a= -1 | b = -1 | a=b= -0.5 each etc. Hence not sufficient.

b y = -6 for x = 0

means ab = -6 (substituting in the original quadratic equation.

a= -3,b=1 | a = -6, b = 1 etc. hence not sufficient.

together 1+2

y = x^2 -x -6 = (x-3) * (x + 2) hence two points for X axis.

thus C.
Senior Manager
Joined: 24 Mar 2011
Posts: 364
Location: Texas

Show Tags

03 May 2011, 09:21
1
y=(x+a)(x+b) = x^2+(a+b)x+ab
this is parabola equation. intersects x-axis means y=0

st 1--> a+b = -1
x^2-x+ab=0 cannot solve the equation; not sufficient

st 2--> graph intersects y-axis at -6
i.e. ab = -6
x^2+(a+b)-6=0 cannot solve the eq, not sufficient.

Both together, x^2-x-6=0
Solving gives x = -2 or x=3

Hence C.
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8659
Location: Pune, India

Show Tags

03 May 2011, 17:33
4
5
udaymathapati wrote:
Attachment:
M-Q19.JPG

At what two points does the graph of y = (x+a)(x+b) intersect the x axis?

You don't need to worry what the equation represents. Just think, what does 'intersection with x axis' imply? It means the y co-ordinate is 0.

0 = (x+a)(x+b)
or x = -a or -b
Hence the graph must intersect the x axis at points (-a, 0) and (-b, 0). We need the values of a and b now.

Statement 1: a + b = -1
Two variables, only one equation. Not sufficient.

Statement 2: Graph intersects the y axis at (0, -6).
At y axis, x = 0. This means when x = 0, y co-ordinate is -6.
Put these values in y = (x+a)(x+b) to get -6 = ab.
Again, two variables, one equation. Not sufficient alone.

Using both statements, we have two variables and two different equations so we will be able to find the values of a and b. It doesn't matter which is 'a' and which is 'b'. We find that the two of them are -3 and 2. Since we need the points (-a, 0) and (-b, 0), the required points are (3, 0) and (-2, 0). Sufficient.

_________________

[b]Karishma
Veritas Prep GMAT Instructor

GMAT Tutor
Joined: 24 Jun 2008
Posts: 1325

Show Tags

25 May 2011, 02:26
1
1
To find x-intercepts, you set y=0 and solve. The question asks for the x-intercepts of y = (x+a)(x+b). Setting y=0, the question is asking when (x+a)(x+b) = 0. Since one of the factors on the left side must be 0, we find we have intercepts when x = -a and x = -b.

Statement 1 tells us the sum of a and b, which doesn't let us find their values alone.

Statement 2 gives us a point on the curve; if we plug that point into our equation, what we get must be true. So -6 = (0+a)(0+b), and ab = -6. So Statement 2 gives us the product of a and b, which isn't enough to find their individual values.

Together, we know that a+b = -1 and ab = -6. That is, we know the sum and product of two numbers. Notice this is what happens every time you factor a quadratic: you look for two numbers with a certain product and a certain sum. For example, if you were to factor the quadratic x^2 - x - 6, you'd look for two numbers which multiply to -6 and add to -1. So if a+b = -1 and ab = -6, then our solutions for a and b must be -3 and 2, in some order. Since our intercepts are at -a and -b, the intercepts are at x=3 and x=-2, so the information is sufficient.

Notice we can't figure out which value is equal to a and which is equal to b, but we don't need to do that to answer the question; we just need to know where the two points are, and not which point is a and which point is b.
_________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

Retired Moderator
Joined: 20 Dec 2010
Posts: 1820

Show Tags

24 Jul 2011, 15:37
1
1
tt2011 wrote:
In xy plane, at what 2 points does the graph of y=(x+a)(x+b) intersect the x-axis?

(1) a+b= -1

(2) The graph intersects the y-axis at (0,-6)

1) y=(x+a)(x+b)

Question is asking for the solution of x
(x+a)(x+b)=0
x^2+ax+bx+ab=0
x^2+(a+b)x+ab=0

1) a+b=-1
We got a+b. Yet, we don't know about "ab"
Not Sufficient.

2) Intersect y at (0,-6)
x=0; y=-6
-6=(0+a)(0+b)
ab=-6
We got ab. We don't know a+b.

Together;
Sufficient.

Ans: "C"
_________________
Intern
Joined: 15 Mar 2010
Posts: 8

Show Tags

24 Jul 2011, 20:22
We need to find the values of x when x = a or when x =b

Stmt 1: a+b = -1 gives numerous values of a and b . Insuff

Stmt 2: plugging in (0, -6) in the equation:

ab= -6 still cannot determine values of x

Combining 1 and 2 gives values of x as a quadratic eqn forms:

x^2-x-6=0. Hence C
Manager
Joined: 20 Aug 2011
Posts: 128

Show Tags

Updated on: 26 Oct 2011, 06:17
2
1
kotela wrote:
In the XY plane what two points does the graph of y=(x+a)(x+b) intersect the x -axis?

1. a+b=-1

2. The graph intersects the y-axis at (0,-6)

can anyone plz explain?

Intersect the x axis means y is 0.
the x coordinates will be -a and -b respectively.

1. a+b=-1
Insufficient

2.
-6=(0+a)(0+b)
ab=-6
Insufficient.

1+2
a+b=-1
ab=-6
a=-3, b=2
a=2, b=-3
Sufficient as we want two points we can say (-3,0) and (-2,0) without knowing the exact values of a & b.
_________________

Hit kudos if my post helps you.
You may send me a PM if you have any doubts about my solution or GMAT problems in general.

Originally posted by blink005 on 26 Oct 2011, 02:04.
Last edited by blink005 on 26 Oct 2011, 06:17, edited 2 times in total.
Intern
Joined: 11 Jan 2012
Posts: 2
Re: Intersection at X axis?  [#permalink]

Show Tags

16 Feb 2012, 17:41
Here is my attempt:

In the xy–plane, at what two points does the graph of y = (x+a)(x+b) intersect the x–axis?
(1) a + b = –1
(2) The graph intersects the y–axis at (0, –6).

Essentially they are asking us to solve for x when y=0:

y = (x+a)(x+b) -> x^2 + xb + xa + ab = 0 (What is X?)

1) a = -b - 1 -> This is insufficient because it still leaves us with one unknown: b

2) -6 = ab -> This allows us to solve for ab in the line equation, but we still don't know what a and b are individually, which we need because we have an xb and xa term.

1) + 2) -> Knowing a = -6/b, we can plug into 1) and find b and a separately. Then the only unknown is x which we can then solve for.

Senior Manager
Joined: 08 Jun 2010
Posts: 311
Location: United States
Concentration: General Management, Finance
GMAT 1: 680 Q50 V32
Re: In the xy plane, at what points does the graph of y=  [#permalink]

Show Tags

26 Feb 2012, 22:50
I tried plugging in values in this one.

Started from option 2:
ab = -6
Possible values are (2,-3); (3,-2); (-6,1); (1,-6) --> INSUFFICIENT.

Option 1:
substitute for x and y in equation we get
a+b = -1
Several possible values such as (-3,2); (-8,7) and so on. --> INSUFFICIENT.

Combining both --> find from option 2 which possible value leads to a+b =-1, only one of the four choices does that (2,-3). Hence SUFFICIENT. Answer choice C.
Math Expert
Joined: 02 Sep 2009
Posts: 51100
Re: In the xy plane, at what points does the graph of y=  [#permalink]

Show Tags

26 Feb 2012, 23:03
mourinhogmat1 wrote:
I tried plugging in values in this one.

Started from option 2:
ab = -6
Possible values are (2,-3); (3,-2); (-6,1); (1,-6) --> INSUFFICIENT.

Option 1:
substitute for x and y in equation we get
a+b = -1
Several possible values such as (-3,2); (-8,7) and so on. --> INSUFFICIENT.

Combining both --> find from option 2 which possible value leads to a+b =-1, only one of the four choices does that (2,-3). Hence SUFFICIENT. Answer choice C.

This is not a good idea to plug the numbers for this problem, it's better to understand the concept and you won't need any math at all (certainly you won't need to solve a+b=-1 and ab=-6).

Next, there are infinitely many values of a and b possible to satisfy ab=-6, not just four: notice that we are not told that a and b are integers only, so for example a=1/2 and b=-12 is also a solution.
_________________
Intern
Joined: 22 Jan 2012
Posts: 31
Re: In the xy plane, at what points does the graph of y=  [#permalink]

Show Tags

26 Apr 2012, 20:33
1
y= x^2+ax+bx+ab
y= x^2+x(a+b)+ab

the question asks us to find the value of two points where the graph intersects x axis. So y=0

x^2+x(a+b)+ab=0 =?

1) a+b=-1 , but we dot have any value of ab, hence not sufficient

2) putting the value x=0 , y=-6
y= x^2+x(a+b)+ab
ab=-6
but we dot have any value of a+b , hence not sufficient

1&2) x^2+x(a+b)+ab=0
now we have a+b=-1
ab =-6
hence sufficient to find 2 values of x.

Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4489

Show Tags

30 Apr 2012, 11:17
Stiv wrote:
In the xy-plane, at what two points does the graph of $$y= (x + a) (x + b)$$ intersect the x - axis?
1) $$a + b = -1$$
2) The graph intersects the y-axis at (0, -6)

So, it's important to know --- when the quadratic is given in factored form --- y= (x + a) (x + b) --- then we know the two roots, x = -a, and x = -b. Roots are the x-intercepts, the places where the graph intersects the x-axis. Basically, the prompt is asking us to find the values of a & b.

Statement #1: a + b = -1

One equation for two unknowns. Not enough to solve. Not sufficient.

Statement #2: The graph intersects the y-axis at (0, -6)

Plugging in x = 0 (the condition of the y-axis), we get y = (0+a)(0+b) = ab = -6

Again, one equation for two unknowns. Not enough to solve. Not sufficient.

Combined statements:
a + b = -1
ab = -6

Two equations with two unknowns ---> we can solve for the values of a & b, which will answer the question. Sufficient.

Here's another practice question on quadratics for practice.
http://gmat.magoosh.com/questions/120
When you submit your answer to that question, the next page will have a full video explanation.

Let me know if you have any further questions.

Mike
_________________

Mike McGarry
Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

Intern
Joined: 28 Aug 2012
Posts: 43
Location: Austria
GMAT 1: 770 Q51 V42
Re: Coordinate DS; y = (x+a)(x+b)  [#permalink]

Show Tags

02 Sep 2012, 10:50
Intersecting the x-axis meens that we have to find the roots of the equation (when will y=0?).
y = (x+a) * (x+b) = x^2 + a*x + b*x + a*b = x^2 + x*(a+b) + a*b

Statement (1) is not sufficient. We don't know specific values of a and b, nor can we substitute all variables (a and b).
Statement (2) just gives us one point.

Combining both statements:
0 = (-6)^2 + (-6)*(-1) + a*b
0 = 36 + 6 + a*b
-42 = a*b

Substituting a*b (and (a+b)):
x^2 + x*(-1) + (-42) = 0

You don't have to calculate any further. It's easy to see now that we have sufficient information to calculate the other point.
Manager
Joined: 28 Feb 2012
Posts: 111
GPA: 3.9
WE: Marketing (Other)
Re: In the xy plane, at what points does the graph of y=  [#permalink]

Show Tags

04 Sep 2012, 00:38
1 statement is clearly insufficient, since we have nothing about x or y
2 statement tells us that the ab=-6, but nothing about a or b individually

Combining statements we have: a=-6/b, -6/b+b=-1, b^2+b-6=0, a=-3, b=2, so both statements together are sufficient (C)
_________________

If you found my post useful and/or interesting - you are welcome to give kudos!

VP
Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1063
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8

Show Tags

23 Apr 2013, 00:44
kabilank87 wrote:
In the xy plane at what 2 points does the graph of y = (x+a)(x+b) intersect the X - axis ?

1.a+b=-1
2.The graph intersects the y axis at (0, -6 )

The function is a parabola $$y=x^2+x(a+b)+ab$$

1)Does $$y=x^2-x+ab$$ intersec x? It depend on ab.Not sufficient

2) Tells us that point (0,6) is on the parabol, so $$ab=-6$$. Does $$y=x^2+x(a+b)-6$$ intersect x? It depends on a+b. Not sufficient

1+2) The equation of the parabola is complete $$y=x^2-x-6$$, we can tell if it intersect x. Sufficient
C
_________________

It is beyond a doubt that all our knowledge that begins with experience.

Kant , Critique of Pure Reason

Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant

Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b]

Re: Coordinate geometry &nbs [#permalink] 23 Apr 2013, 00:44

Go to page    1   2    Next  [ 28 posts ]

Display posts from previous: Sort by