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Data Sufficiency Competition--Prizes can be won

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Senior Manager
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Data Sufficiency Competition--Prizes can be won [#permalink] New post 21 Sep 2006, 09:34
Hi Folks,

I will be posting Data Sufficiency Questios on a regular basis in this thread. The first person to answer them can win a prize. The prizes are nothing but I'll give them a link to download some softwares as well as e-books and also GMAT material related stuff.........

EDITING THE POST IS NOT ALLOWED

Here is the first question...

1. The sum of the first N-1 terms of an Arithmetic Progression is zero or positve and sum of the first N terms is negative. What is the value of N?

1. The common difference is -4 and the 7th term is the last positive term.
2. The first term is 25 and there are 7 positive terms and all are integers.

The prize is file that contains 205 GMAT Critical Reasoning questions
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Averages Accelerated:Guide to solve Averages Quickly(with 10 practice problems)


Last edited by cicerone on 25 Sep 2008, 00:47, edited 1 time in total.
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 [#permalink] New post 21 Sep 2006, 10:06
The sum of the first N-1 terms of an Arithmetic Progression is zero or positve and sum of the first N terms is negative. What is the value of N?

1. The common difference is -4 and the 7th term is the last positive term.
2. The first term is 25 and there are 7 positive terms and all are integers.

from one

if the difference is -4 and the 7th term = a-24 is the last positive

thus 24<a<28 thus a could be anything in this interval .

if a = 25 sum = 91 , a = 26 thus sum = 98 if we assumed a = 27 thus sum = 105
for the sum to be -ve the sum of terms after the last positive term (7) must be greater than , either

-91 or -98 or -105...

the sum of terms from 8 to 14 if we assumed a = 25 is 105

thus n-1 = 14 thus n= 15

the sum of terms from 8 to 14 if we assumed a = 26 is -98

thus n-1 = 14 thus n = 13
insuff

we need not to try fractions because either ways it is insuff

two...suff B is my answer
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 [#permalink] New post 22 Sep 2006, 06:06
YEs, Yezz, the right answer is B.

I have sent a personal message to u containing the link to the file
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 [#permalink] New post 22 Sep 2006, 07:24
Thanks Ciceron........keep your ds coming
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 [#permalink] New post 22 Sep 2006, 18:17
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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 [#permalink] New post 22 Sep 2006, 18:21
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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 [#permalink] New post 22 Sep 2006, 20:00
Hey trivikram,

That's the right answer
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 [#permalink] New post 22 Sep 2006, 21:35
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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 [#permalink] New post 22 Sep 2006, 22:36
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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 [#permalink] New post 23 Sep 2006, 01:20
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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 [#permalink] New post 23 Sep 2006, 04:43
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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 [#permalink] New post 25 Sep 2006, 19:46
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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 [#permalink] New post 27 Sep 2006, 20:12
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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 [#permalink] New post 27 Sep 2006, 20:58
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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 [#permalink] New post 27 Sep 2006, 21:13
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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 [#permalink] New post 27 Sep 2006, 21:35
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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 [#permalink] New post 27 Sep 2006, 23:25
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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 [#permalink] New post 27 Sep 2006, 23:57
interesting thread 8-)
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 [#permalink] New post 28 Sep 2006, 01:00
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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 [#permalink] New post 28 Sep 2006, 22:57
waiting for some more answers....................
  [#permalink] 28 Sep 2006, 22:57
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