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Data Sufficiency Competition--Prizes can be won [#permalink]
21 Sep 2006, 09:34
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
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 _________________
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
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.
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......
Hey paddyboy, i want general answer please...... How can u proove that it holds for every example......
Never pass up a challenge
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.
Hey paddyboy, i want general answer please...... How can u proove that it holds for every example......
Never pass up a challenge
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..................
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.
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