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Data Sufficiency Competition--Prizes can be won

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 [#permalink] New post 28 Sep 2006, 23:59
ok, did a little more thinking and i am changing my answer to E

i totally forgot about a, which could be negative.

so to continue my previous thread, i found a yes scenario when combining both statements, now i need to find a no scenario to spoil answer C--

to do this, let's look at

(x+1)(-x+2)

we have -x^2+x+2, where b+c is positive and bc is also positive but this time we have one negative root and one positive root
x=-1 or x=2

therefore INSUFF together

so answer is E

Cicerone, is this right??
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 [#permalink] New post 29 Sep 2006, 00:20
keeeeeekse, not convinced........
Try it again...............
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 [#permalink] New post 29 Sep 2006, 01:32
Viperace wrote:
B

I) I just cant get
II) Sufficient

Question ask if both roots are positive
Only one condition for that is

b < -sqrt(b^2 - 4ac )
.
.
.
=> ac > 0

II) Product of two root > 0
which means ac>0 if u work it out

[-b + sqrt(b^2 - 4ac ) ]*[-b - sqrt(b^2 - 4ac ) ] >0
b^2 - (b^2 - 4ac )>0
=> ac >0


Woots, just found the mistake I made. However my answer doesnt changes


Question ask if both roots are positive
Only one condition for that is

b < -sqrt(b^2 - 4ac )
.
.
.
=> ac < 0
=> c <0 since given a>0

I) Sum of two root>0
This implies -b/a > 0
=> b <0
This is not sufficient to tell if c<0


II) Product of two root > 0
which means c/a>0 if u work it out

[-b + sqrt(b^2 - 4ac ) ]*[-b - sqrt(b^2 - 4ac ) ]/4a^2 >0
[b^2 - (b^2 - 4ac )]/4a^2>0
=> c/a >0
=> c > 0

Which is sufficient to answer the question. "both roots are positive?"NO
Condition from I) is not needed.
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 [#permalink] New post 29 Sep 2006, 01:45
In fact,
b < 0 is a redundant information.

By knowing that
i) a > 0
ii) both root are positive

We can already deduce b < 0
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 [#permalink] New post 29 Sep 2006, 05:34
From A we know that -(b/a)>0 , given a>0 then b must be NEG
In order both roots to be positive , discriminant must be >0
or b^2-4*a*c>0 .
A) is not suff

From B) we get (4a*c)/4*(a^2) or c/a>0 provided a>0 then c is >0
Given a+b+c>0 and b<0 then a+c>b
Substitute in equation of discriminant b with a+c and get
(a+c)^2-4ac>0 we get a^2-2ac+c^2>0
this is (a-c)^2>0 which is always positive BUT we do not know if A is not EQUAL TO C which would make (a-c)^2=0
SO E is the answer
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 [#permalink] New post 29 Sep 2006, 09:15
4. The roots of a quadratic equatin ax^2+bx+c=0 are integers and a+b+c>0 and a>0. Are both the roots of the equation positive?

(1) Sum of the roots is positive.
(2) Product of the roots is positive.
IMHO C it is
1st alone
roots may be -2 and 4 sum is +ve
roots may be 2 and 4 and so on insuff
2)both +ve and both -ve insuff
both together -->from 2 we know that throots must be both +ve or -ve
but in order to have positive sum both should be +ve only

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 [#permalink] New post 04 Oct 2006, 07:58
No correct answer yet...........
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 [#permalink] New post 04 Oct 2006, 16:06
cicerone wrote:
Here comes our next question

4. The roots of a quadratic equatin ax^2+bx+c=0 are integers and a+b+c>0 and a>0. Are both the roots of the equation positive?

(1) Sum of the roots is positive.
(2) Product of the roots is positive.

PRIZE: Vocabulary Practice Sotware (contains 3000 words)


ax^2+bx+c=0 => x^2 +(b/a)x +(c/a)=0 since a>0, a is not 0

If this quadratic equation has two integer roots m and n, then it can be written as
(x-m)(x-n)=0 where mn=c/a and -(m+n)=b/a

This is because (x-m)(x-n)=x^2-(m+n)x+mn

(1) We are told than m+n>0, which means that b/a<0 and so b<0
Also as m and n are integers m+n>=1, so |b/a|>=1 i.e |b|>=|a|=a

Summarizing, we know that a>0, b>0 and |b|=-b>=a

This means that b+a<=0 and since a+b+c>0, c>0 Thus ac>0 and so the roots have the same sign-

If the sum of the roots is positive, each must be >0 SUFFICIENT

(2) mn>0 means roots have the same sign. This doesn't tell me much.

Could they both be <0? Sure! (x+2)(x+1)=x^2+3x+2=0 has roots of -2 and -1 and a+b+c=6>0

Could they both be<0? Why not? (x-2)(x-6)=x^2-8x+12=0 has roots of 2 and 6 and a+b+c=5>0

NOT SUFFICIENT


My answer: A
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 [#permalink] New post 04 Oct 2006, 22:03
kevincan wrote:
cicerone wrote:
Here comes our next question

4. The roots of a quadratic equatin ax^2+bx+c=0 are integers and a+b+c>0 and a>0. Are both the roots of the equation positive?

(1) Sum of the roots is positive.
(2) Product of the roots is positive.

PRIZE: Vocabulary Practice Sotware (contains 3000 words)


ax^2+bx+c=0 => x^2 +(b/a)x +(c/a)=0 since a>0, a is not 0

If this quadratic equation has two integer roots m and n, then it can be written as
(x-m)(x-n)=0 where mn=c/a and -(m+n)=b/a

This is because (x-m)(x-n)=x^2-(m+n)x+mn

(1) We are told than m+n>0, which means that b/a<0 and so b<0
Also as m and n are integers m+n>=1, so |b/a|>=1 i.e |b|>=|a|=a

Summarizing, we know that a>0, b>0 and |b|=-b>=a

This means that b+a<=0 and since a+b+c>0, c>0 Thus ac>0 and so the roots have the same sign-

If the sum of the roots is positive, each must be >0 SUFFICIENT

(2) mn>0 means roots have the same sign. This doesn't tell me much.

Could they both be <0? Sure! (x+2)(x+1)=x^2+3x+2=0 has roots of -2 and -1 and a+b+c=6>0

Could they both be<0? Why not? (x-2)(x-6)=x^2-8x+12=0 has roots of 2 and 6 and a+b+c=5>0

NOT SUFFICIENT


My answer: A


Yes kevin, the answer is A.
I have a different approach...........

The roots of a quadratic equatin ax^2+bx+c=0 are integers and a+b+c>0 and a>0. Are both the roots of the equation positive?

Statement 1: Sum of the roots is positive.

ie -b/a = p where p>0
ie b= -(a x p).

Now consider the product of the roots.

Let c/a=q.

Clearly both p and q must be integers (since in the question it is given that both the roots are integers)

It is given that a+b+c>0
ie a-(axp)+(axq)>0
ie a(1-p+q)>0
It is given in the question that a>0
So (1-p+q)>0
ie q-p>-1
ie q-p>=0 (since p and q are integers)
ie q>=p
In statement 1 it is given that p>0
So clearly q>0

So from the first statement itself if sum is +ve we can conclude that product is also +ve.

Statement 2 alone is not sufficient..........

So A
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 [#permalink] New post 05 Oct 2006, 08:10
Here comes our next question

5. In triangle ABC angle A is the greatest angle. D is the foot of the
perpendicular dropped on to BC from A. Is triangle ABC right-angled?
1. AD^2= BD x DC.
2. AD/DC < BD/AD.

Prize: A file on logical ability to solve the critical reasoning questions from GMAT.
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 [#permalink] New post 06 Oct 2006, 02:46
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 [#permalink] New post 06 Oct 2006, 07:08
St 1 is sufficient.
triangle adb is similar to adc.
angle b = angle c
ab=ac
ad=ac
angle dac=45
thus angle bac =90

st 2 is not sufficient.

Answer is A.
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 [#permalink] New post 06 Oct 2006, 07:09
St 1 is sufficient.
triangle adb is similar to adc.
angle b = angle c
ab=ac
ad=ac
angle dac=45
thus angle bac =90

st 2 is not sufficient.

Answer is A.
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 [#permalink] New post 06 Oct 2006, 09:50
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 [#permalink] New post 06 Oct 2006, 13:28
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 [#permalink] New post 07 Oct 2006, 04:54
Changing my answer to d.
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 [#permalink] New post 07 Oct 2006, 11:29
cicerone wrote:
Here comes our next question

5. In triangle ABC angle A is the greatest angle. D is the foot of the
perpendicular dropped on to BC from A. Is triangle ABC right-angled?
1. AD^2= BD x DC.
2. AD/DC < BD/AD.

Prize: A file on logical ability to solve the critical reasoning questions from GMAT.


D.

If Angle A is right angle, then AD^2 = BD x DC
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 [#permalink] New post 08 Oct 2006, 04:56
Yes the answer must be D

In a right angled triangle

AD^2 = BDxDC

Clearly 1 says AD^2 = BD x DC
Sufficient

Clearly 2 says AD^2 !=BD x DC

Sufficient

So D

yogeshsheth, i have sent the link to the file.........

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 [#permalink] New post 08 Oct 2006, 11:54
Hey just relax with this question.................
No prize................
This is a simple one...........

6. How many sons does Mr. John have?

1. Mrs. John has 3 sons.
2. Peter is taller of the John's sons.

Answer this fast...........
Don't relax too much............

A lot more to come

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 [#permalink] New post 08 Oct 2006, 13:02
How many sons does Mr. John have?

1. Mrs. John has 3 sons.
2. Peter is taller of the John's sons

ONE INSUFF

TWO

THEY ARE TWO

(TALLER :lol: )
  [#permalink] 08 Oct 2006, 13:02
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