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please help me with these tricky DS

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please help me with these tricky DS

by saeed » Mon Aug 25, 2008 5:20 am
Q1. 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

Q2. K is a set of integers such that if the integer r is in the K, then r+1 is also in K. Is 100 in K?
1. 50 is in K
2. 150 is in K
Q3. Is x>k?
1. (2x)(2k)=4
2. (9x)(3k)=81.

Please help me with most efficient approach.
Thanks in advance.
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Re: please help me with these tricky DS

by vish150783 » Mon Aug 25, 2008 7:49 am
Q1. 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

ANSWER: C because we have two variables and one equation.
When we use both we can summarize the above as:

d^2 + x(2d-b) = c


Q2. K is a set of integers such that if the integer r is in the K, then r+1 is also in K. Is 100 in K?
1. 50 is in K
2. 150 is in K

ANSWER: E because if 50 is K that means either 51 or 49 needs to be in K. If 150 is in K that means either 149 or 151 needs to be in K. Both of these dont mean that 100 does not need to be in K because we dont know how many numbers need to be in K.


Q3. Is x>k?
1. (2x)(2k)=4
2. (9x)(3k)=81.

ANSWER: E because

1. xk = 1;
if k>0; x<0 -> implies NO
if k<0; x>0 -> implies YES.

so, A is not sufficient.

2. xk = 3;
if k>3; x<k implies NO
if k<3; x>k implies YES

I am not sure if its the most efficient, I think its logical..
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by pepeprepa » Mon Aug 25, 2008 8:25 am
I think you mean E for Q1

I am surprised by Q3, in my view it can't be a DS given the two sentences contradict.
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by Ian Stewart » Mon Aug 25, 2008 11:56 am
I'm pretty sure Q3 is supposed to read:

Q3. Is x>k?
1. (2^x)(2^k)=4
2. (9^x)(3^k)=81.

Then the statements are not contradictory.
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by Suyog » Mon Aug 25, 2008 12:58 pm
@Ian

Can you comment on Q2. Thanks!
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by bha » Mon Aug 25, 2008 1:08 pm
@Q1
from Q:
x^2+bx+c=(x+d)^2
x^2+bx+c=x^2+2dx+d^2
b=2d
c=d^2

from 1,
d=3..c=d^2 hence c=9..sufficient..

from 2
b=6
b=2d...d=b/2= 6/2=3...sufficient..

Hence 'D'..what do u guys think?? Please correct me if I am making an mistake...
what is the OA?
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by pepeprepa » Mon Aug 25, 2008 1:18 pm
About Q2, here is a similar one
https://www.beatthegmat.com/set-of-numbers-t14261.html
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by Ian Stewart » Mon Aug 25, 2008 1:58 pm
Suyog wrote:@Ian

Can you comment on Q2. Thanks!
Sure.


Q2. K is a set of integers such that if the integer r is in the K, then r+1 is also in K. Is 100 in K?
1. 50 is in K
2. 150 is in K

If r is in K, r+1 is in K. So if 50 is in K, 51 is in K. But if 51 is in K, so is 52, and then so is 53, and so on. Indeed, if you know 50 is in K, you know that every positive integer larger than 50 must also be in K. So 1) is sufficient.

If 150 is in K, all we know for certain is that every positive integer greater than or equal to 150 must be in K. We don't know if 100 is in K. So 2) is insufficient.

Thus, A.
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by Suyog » Mon Aug 25, 2008 2:26 pm
yes..
for a moment I thot I'm crazy to think like that... :P
thank you!
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by bha » Mon Aug 25, 2008 3:26 pm
Ian Stewart wrote:
Suyog wrote:@Ian

Can you comment on Q2. Thanks!
Sure.


Q2. K is a set of integers such that if the integer r is in the K, then r+1 is also in K. Is 100 in K?
1. 50 is in K
2. 150 is in K

If r is in K, r+1 is in K. So if 50 is in K, 51 is in K. But if 51 is in K, so is 52, and then so is 53, and so on. Indeed, if you know 50 is in K, you know that every positive integer larger than 50 must also be in K. So 1) is sufficient.

If 150 is in K, all we know for certain is that every positive integer greater than or equal to 150 must be in K. We don't know if 100 is in K. So 2) is insufficient.

Thus, A.
Ian,
Why can't we trace back in reverse direction for stmt2 just like u did for stmt1...say 150 is in K..149 should also be in K as 150 = 149+1..hence B is also sufficient?
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by bha » Mon Aug 25, 2008 3:27 pm
bha wrote:@Q1
from Q:
x^2+bx+c=(x+d)^2
x^2+bx+c=x^2+2dx+d^2
b=2d
c=d^2

from 1,
d=3..c=d^2 hence c=9..sufficient..

from 2
b=6
b=2d...d=b/2= 6/2=3...sufficient..

Hence 'D'..what do u guys think?? Please correct me if I am making an mistake...
what is the OA?
guye...any comment on my approach?? I am not sure whether it is a correct approach??
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by Ian Stewart » Mon Aug 25, 2008 4:08 pm
bha wrote: Ian,
Why can't we trace back in reverse direction for stmt2 just like u did for stmt1...say 150 is in K..149 should also be in K as 150 = 149+1..hence B is also sufficient?
That would be the trap in the question. Note the question only says that "if the integer r is in K, then r+1 is also in K". It does not say that "if the integer r is in K, then r-1 is also in K". So if you know that some integer x is in K, you can be sure any integer larger than x is also in K, but you can't be sure about anything less than x. That is, if we know 150 is in K, the set could be:

{150, 151, 152, 153, 154, 155, ...}

and you'll see that it's true: for any x in the set, x+1 is also in the set.
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by Ian Stewart » Mon Aug 25, 2008 4:14 pm
bha wrote:
bha wrote:@Q1
from Q:
x^2+bx+c=(x+d)^2
x^2+bx+c=x^2+2dx+d^2
b=2d
c=d^2

from 1,
d=3..c=d^2 hence c=9..sufficient..

from 2
b=6
b=2d...d=b/2= 6/2=3...sufficient..

Hence 'D'..what do u guys think?? Please correct me if I am making an mistake...
what is the OA?
guye...any comment on my approach?? I am not sure whether it is a correct approach??
Looks great to me. If anyone wants a bit more detail:

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

We know x^2+bx+c=(x+d)^2 for all values of x. It is useful to rewrite this as bha did:

x^2+bx+c=x^2+2dx+d^2

You can subtract x^2 from both sides:

bx + c = 2dx + d^2

If these are equal for all values of x, then they're certainly equal for x = 0, so c = d^2

And if c = d^2, we know that

bx + c = 2dx + c, or
bx = 2dx

for all x, including x = 1, so b = 2d

As bha points out, each statement is sufficient to find c.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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by bha » Mon Aug 25, 2008 6:09 pm
Ian Stewart wrote:
bha wrote: Ian,
Why can't we trace back in reverse direction for stmt2 just like u did for stmt1...say 150 is in K..149 should also be in K as 150 = 149+1..hence B is also sufficient?
That would be the trap in the question. Note the question only says that "if the integer r is in K, then r+1 is also in K". It does not say that "if the integer r is in K, then r-1 is also in K". So if you know that some integer x is in K, you can be sure any integer larger than x is also in K, but you can't be sure about anything less than x. That is, if we know 150 is in K, the set could be:

{150, 151, 152, 153, 154, 155, ...}

and you'll see that it's true: for any x in the set, x+1 is also in the set.
Thank You Ian.
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by jeffxujian » Tue Aug 26, 2008 1:21 am
Ian Stewart wrote:I'm pretty sure Q3 is supposed to read:

Q3. Is x>k?
1. (2^x)(2^k)=4
2. (9^x)(3^k)=81.

Then the statements are not contradictory.
Ian:

Is the correct answer of the above question C?

1. x+k=4 insufficient, because x, k can be any numbers
2.2x+k=4 insufficient, same drill as above

combine 1 with 2, x=0 and K=4, thus the result is sufficient to answer question whether x is greater than k.

Please correct me if i made any mistakes. Thank!
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