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Re: Data Sufficiency Competition--Prizes can be won [#permalink]
Thanks Ciceron........keep your ds coming
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Re: Data Sufficiency Competition--Prizes can be won [#permalink]
Here is our next question

2. If n>1, is n=2?

1. n has only two positive factors.
2. The difference between any two positive factors is odd.

The prize is a file containing 200 GMAT RC questions
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Re: Data Sufficiency Competition--Prizes can be won [#permalink]
cicerone wrote:
Here is our next question

2. If n>1, is n=2?

1. n has only two positive factors.
2. The difference between any two positive factors is odd.

The prize is a file containing 200 GMAT RC questions


SHould be B

1) 2,3,5 all primes fall here but we cannot say it is 2

2) 2 is the only number which is true here
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Re: Data Sufficiency Competition--Prizes can be won [#permalink]
Hey trivikram,

That's the right answer
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Re: Data Sufficiency Competition--Prizes can be won [#permalink]
The next one......

3. Is the five digited number ABCDE divisible by 13?

1. The number CD is divisible by 13.
2. 10A+B+4C+3D-E is divisible by 13.

Prize: VISA THINGS ONE NEED TO KNOW: An official book from consulate office.
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Re: Data Sufficiency Competition--Prizes can be won [#permalink]
cicerone wrote:
The next one......

3. Is the five digited number ABCDE divisible by 13?

1. The number CD is divisible by 13.
2. 10A+B+4C+3D-E is divisible by 13.

Prize: VISA THINGS ONE NEED TO KNOW: An official book from consulate office.


B
Case 1: Plugging in numbers.
It can be 22139 (Divisible) or 13135 (Indivisible). Hence Insuff.
Case 2: Plugging in numbers again. try 22139 or 14313 etc. Always evaluates to true. Hence Suff.
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Re: Data Sufficiency Competition--Prizes can be won [#permalink]
paddyboy wrote:
cicerone wrote:
The next one......

3. Is the five digited number ABCDE divisible by 13?

1. The number CD is divisible by 13.
2. 10A+B+4C+3D-E is divisible by 13.

Prize: VISA THINGS ONE NEED TO KNOW: An official book from consulate office.


B
Case 1: Plugging in numbers.
It can be 22139 (Divisible) or 13135 (Indivisible). Hence Insuff.
Case 2: Plugging in numbers again. try 22139 or 14313 etc. Always evaluates to true. Hence Suff.


Hey paddyboy, i want general answer please......
How can u proove that it holds for every example......
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Re: Data Sufficiency Competition--Prizes can be won [#permalink]
cicerone wrote:
Hey paddyboy, i want general answer please......
How can u proove that it holds for every example......


Never pass up a challenge :wink:

Here it is!

Number is 10000A + 1000B + 100C + 10D + E
= 10000A + 1000B + 4000C - 3900C + 3000D - 2990D - 1000E + 1001E
= 1000(10A + B + 4C + 3D - E) - (3900C + 2990D - 1001E)

Since all terms in 2nd set of parantheses are divisible, the number in the first set of parantheses also has to be divisible, if the number ABCDE is divisible by 13.

Now put C = 1 and D = 3. Self-explanatory?
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Re: Data Sufficiency Competition--Prizes can be won [#permalink]
paddyboy wrote:
cicerone wrote:
Hey paddyboy, i want general answer please......
How can u proove that it holds for every example......


Never pass up a challenge :wink:

Here it is!

Number is 10000A + 1000B + 100C + 10D + E
= 10000A + 1000B + 4000C - 3900C + 3000D - 2990D - 1000E + 1001E
= 1000(10A + B + 4C + 3D - E) - (3900C + 2990D - 1001E)

Since all terms in 2nd set of parantheses are divisible, the number in the first set of parantheses also has to be divisible, if the number ABCDE is divisible by 13.

Now put C = 1 and D = 3. Self-explanatory?


Hey paddyboy fine i agree with u.
But there's a simple way.

given number is 10000A+1000B+100C+10D+E
This when divided by 13 gives a remainder which will be of the form

3A+12B+9C+10D+E
The same value i can take it as
-10A-B-4C-3D+E ( when a number is divided by 13 if the remainder is 10 we can take it as -3 for making the calculations faster)

=> -(10A+B+4C+3D-E)

In the second statment it is clearly given that 10A+B+4C+3D-E is divisible by 13. Hence statement 2 ALONE is sufficient

So B.

Anyway, students come to know different approaches.............
fine i am sending u the link for the book paddyboy.........
Keep rocking..................
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Re: Data Sufficiency Competition--Prizes can be won [#permalink]
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)
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Re: Data Sufficiency Competition--Prizes can be won [#permalink]
statement 1

if sum of roots = -b/a is positive and a is >0 then b has to be nagative and one of the roots will necessary be negative

thus sufficent to say yes/no

statement 2 product = ac, a nd is positive thus c is positive since a is positive, but does not say anything about b
so insufficient

The answer is A since the signs are determined by b
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Re: Data Sufficiency Competition--Prizes can be won [#permalink]
jainan24 wrote:
statement 1

if sum of roots = -b/a is positive and a is >0 then b has to be nagative and one of the roots will necessary be negative

thus sufficent to say yes/no

statement 2 product = ac, a nd is positive thus c is positive since a is positive, but does not say anything about b
so insufficient

The answer is A since the signs are determined by b


Hey jainan,

When the sum of the roots is +ve we have two cases
1.Both the roots could be +ve
2.One root could be +ve and other root could be -ve.

So we can coclude that b has to be -ve since -b/a is +ve and a is +ve.
But from this alone how do we conclude whether
Both the roots are +ve or
One is +ve and the other is -ve.
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Re: Data Sufficiency Competition--Prizes can be won [#permalink]
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)


C

1 Sum of roots are positive .
Possible roots 1. + ve , -ve
2. -ve , +ve
3. + ve , +ve
InSuff

2. Product of the roots positive
Possible roots 1. + ve , +ve
2. -ve , -ve

InSuff


Combined

Only possible combination is +ve , +ve.
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Re: Data Sufficiency Competition--Prizes can be won [#permalink]
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
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Re: Data Sufficiency Competition--Prizes can be won [#permalink]
interesting thread 8-)
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Re: Data Sufficiency Competition--Prizes can be won [#permalink]
my answer is also C

from statement 1
sum of b and c is positive

therefore either b and c are both positive

or one is negative, one is positive (for example b=-2 and c=3 )

therefore INSUFFICIENT

from statement 2
product of b and c is positive

therefore b and c must either both be positive or both negative

INSUFF

to combine the 2 statements, we know that if the sum and product of b and c are both positive, b and c must both be positive. Therefore the roots are negative

SUFF

answer C
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Re: Data Sufficiency Competition--Prizes can be won [#permalink]
waiting for some more answers....................
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