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Statement 1) y>0 Not suff, X could be anything larger or smaller than X. Statement 2) x=1-y x+y=1 Let x=3 and y=-2 then y-x < 0. But if x=1/4 and y=3/4 then y-x >0 Not suff.

1 and 2 together) From the example above we have: if x=1/4 and y=3/4 then y-x >0 but if we flip it around: if x=3/4 and y=1/4 then y-x <0 not suff.

ANS = E

I selected that both stmts were sufficient together because the examples I chose worked both ways so if x=-2 y=3 you get 3-(-2)=5 or if y=2 than x = (-1) so 2-(-1)=3. Can the insufficiency only be seen with fractional numbers? Thanks.

Statement 1) y>0 Not suff, X could be anything larger or smaller than X. Statement 2) x=1-y x+y=1 Let x=3 and y=-2 then y-x < 0. But if x=1/4 and y=3/4 then y-x >0 Not suff.

1 and 2 together) From the example above we have: if x=1/4 and y=3/4 then y-x >0 but if we flip it around: if x=3/4 and y=1/4 then y-x <0 not suff.

ANS = E

I selected that both stmts were sufficient together because the examples I chose worked both ways so if x=-2 y=3 you get 3-(-2)=5 or if y=2 than x = (-1) so 2-(-1)=3. Can the insufficiency only be seen with fractional numbers? Thanks.

Yes, in order to get NO answer you should choose values of x and y from (0, 1), note that this range can give you the YES answer as well.

Is \(y-x>0\)? --> is \(y>x\)?

(1) y > 0, not sufficient as no info about \(x\). (2) x = 1 - y --> \(x+y=1\) --> the sum of 2 numbers equal to 1 --> we can not say which one is greater. Not sufficient.

(1)+(2) \(x+y=1\) and \(y>0\), still can not determine which one is greater: if \(y=0.1>0\) and \(x=0.9\) then \(y<x\) but if \(y=0.9>0\) and \(x=0.1\) then \(y>x\). Not sufficient.

For question # 8, please explain how this is true :

either 4y=32 y=8 x=-3, xy=-24 OR -4y=32

Regards, Subhash

If x and y are non-zero integers and |x| + |y| = 32, what is xy?

(1) \(-4x-12y=0\) --> \(x=-3y\) --> \(x\) and \(y\) have opposite signs.

So either: \(|x|=x\) and \(|y|=-y\) --> in this case \(|x|+|y|=x-y=-3y-y=-4y=32\): \(y=-8\), \(x=24\), \(xy=-24*8\);

OR: \(|x|=-x\) and \(|y|=y\) --> \(|x|+|y|=-x+y=3y+y=4y=32\) --> \(y=8\) and \(x=-24\) --> \(xy=-24*8\), the same answer.

Sufficient.

(2) \(|x| - |y| = 16\). Sum this one with th equations given in the stem --> \(2|x|=48\) --> \(|x|=24\), \(|y|=8\). \(xy=-24*8\) (x and y have opposite sign) or \(xy=24*8\) (x and y have the same sign). Multiple choices. Not sufficient.

1st number = 10w+x and 2nd number = 10c+x so three digit number = 100*(wc)+10*(w+c)+x^2

In 1, We are given, wc=x*(w+c)=x^2

or x*(w+c-x) = 0 implying either x=0 or w+c-x = 0 Since it is given that x is non-zero, we have w+c-x = 0 so 1 is sufficient w+c and x being odd just says that w+c-x is even number, so insufficient Answer A

Re: The sum of n consecutive positive integers is 45 [#permalink]

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20 Apr 2012, 21:16

ENAFEX wrote:

Bunuel wrote:

Please find below new set of DS problems:

TIP: many of these problems act in GMAT zone, so beware of ZIP trap.

1. The sum of n consecutive positive integers is 45. What is the value of n? (1) n is even (2) n < 9

Bunuel, Is this approach correct?

sum of n consecutive +ve integers = \(\frac{n(n+1)}{2}\)

so, \(\frac{n(n+1)}{2}\) = 45 n(n+1) = 90 \(n^{2}+ n - 90 = 0\) solving this equation gives n= -10 or 9

Both A and B doesn't help. Hence E.

Bunuel, Ignore my request. I found out the mistake. The question does not say sum of first consecutive integers, so, I cant use this formula.
_________________

Hi Bunuel/Karishma, Request you to clarify the below colored part. What difference does this makes in the question? Thanks H

Bunuel wrote:

ANSWERS:

7. If b, c, and d are constants and x^2 + bx + c = (x + d)^2 for all values of x, what is the value of c? (1) d = 3 (2) b = 6

Note this part: "for all values of x" So, it must be true for x=0 --> c=d^2 --> b=2d (1) d = 3 --> c=9 Sufficient (2) b = 6 --> b=2d, d=3 --> c=9 Sufficient

Re: The sum of n consecutive positive integers is 45 [#permalink]

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29 Jul 2012, 04:31

1. The sum of n consecutive positive integers is 45. What is the value of n? (1) n is even (2) n < 9

Cant we use the formula for sum of n consecutive +ve integers (n)*(n+1)/2 = 45 n^2 +n -90 = 0 n=9 or -10
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Re: The sum of n consecutive positive integers is 45 [#permalink]

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29 Jul 2012, 05:02

dvinoth86 wrote:

1. The sum of n consecutive positive integers is 45. What is the value of n? (1) n is even (2) n < 9

Cant we use the formula for sum of n consecutive +ve integers (n)*(n+1)/2 = 45.

[/color] n^2 +n -90 = 0 n=9 or -10

Cant we use the formula for sum of n consecutive +ve integers (n)*(n+1)/2 = 45.

This formula gives the sum of the first n consecutive positive integers: 1,2 3, ..., n. Nowhere is stated that our sequence starts with 1.
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PhD in Applied Mathematics Love GMAT Quant questions and running.

1. The sum of n consecutive positive integers is 45. What is the value of n? (1) n is even (2) n < 9

Cant we use the formula for sum of n consecutive +ve integers (n)*(n+1)/2 = 45 n^2 +n -90 = 0 n=9 or -10

As stated in the previous post \(\frac{n(n+1)}{2}\) gives the sum of first \(n\) positive integers: \(1+2+3+...+n=\frac{n(n+1)}{2}\) and we cannot use that formula since we are not told that we have this case. Check this for more: math-number-theory-88376.html (Evenly spaced set chapter).

Solution of this problem is as follows:

The sum of n consecutive positive integers is 45. What is the value of n?

(1) n is even --> n can be 2: 22+23=45. But it also can be 6 --> x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)=45 --> x=5. At least two values of n are possible. Not sufficient.

(2) n<9 --> the above example is also valid for this statement, hence not sufficient.

(1)+(2) Still at least two values of n are possible. Not sufficient.

Re: The sum of n consecutive positive integers is 45 [#permalink]

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29 Jul 2012, 05:32

Q1

The sum of n consecutive integers is either a multiple of the middle term, in case n is odd, or a multiple of the sum of the two middle terms, in case n is even.

(1) Since n is even and 45 is odd, we must have an odd number of pairs in our sequence, such that the sum of each pair is a factor of 45, and it is the same as the sum of the two middle terms. For example, we can have just one pair (22, 23) - or three pairs, each sum being 45/3 = 15 - 5, 6, 7, 8, 9, 10, with 5+10=6+9=7+8=15. So, (1) is not sufficient.

(2) From what we have seen above, (2) is not sufficient either. Just as an example, if n is odd and less than 9, we can have the sequence 14, 15, 16.

(1) and (2) taken together is obviously not sufficient.

Answer E
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Re: The sum of n consecutive positive integers is 45 [#permalink]

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29 Jul 2012, 05:49

Q2. Is a product of three integers XYZ a prime? (1) X=-Y (2) Z=1

(1) We can take X=1, then Y=-1, and taking Z any non-negative integer will get a non-prime, as the product will be either negative or 0. But if we take X=1, Y=-1 and for example Z=-2, than XYZ=2, which is a prime. Therefore, (1) is not sufficient.

(2) If Z=1, we can take Y=1 as well, and then either X is a prime or not, so the final product XYZ=X, can be or not a prime. Not sufficient.

(1) and (2) together: If Z=1 and X=-Y, then \(XYZ=-X^2\), which can be either 0 (when X=0) or a negative integer. In either case, we won't get a prime number (by definition, a prime must be a positive integer). So, the answer is a definite NO, therefore sufficient.

Answer C
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Re: The sum of n consecutive positive integers is 45 [#permalink]

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29 Jul 2012, 06:18

Q3. Multiplication of the two digit numbers wx and cx, where w,x and c are unique non-zero digits, the product is a three digit number. What is w+c-x? (1) The three digits of the product are all the same and different from w c and x. (2) x and w+c are odd numbers.

(1) If the product is a three digit number, all digits identical, then the number must be of the form AAA=A*111=A*3*37, where A is a non-zero digit. It means that x = 7, and 3A must be a two digit number, which also ends in 7. It follows that 3A = 27, so the numbers are 37 and 27, 37 * 27 = 999, and w+c-x=5-2, sufficient (it doesn't matter who is w and who is c).

(2) Since w+c is odd, one of them must be even and the other one odd. Fro example, we can take w=2, c=1, x=3 or w=3, c=2, x=1. 23*13=299, 31*21=651, are both three digit numbers, and w+c-x is 0 and 4, respectively. Not sufficient.

Answer A.

Remark: In the body of the question, w,x and c are unique non-zero digits, shouldn't be "distinct non-zero digits"?
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Re: The sum of n consecutive positive integers is 45 [#permalink]

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29 Jul 2012, 06:27

Q4. Is y – x positive? (1) y > 0 (2) x = 1 – y

The question we are asked is in fact is y>x?

(1) Not sufficient, as we don't know anything about x. (2) We have x+y=1, again not sufficient. We can have x=y=0.5, or x=0.25 < y=0.75, or x=0.75 > y=0.25

(1) and (2) not sufficient, we can use the examples from the analysis for (2) above.

Answer E
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