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I don't need help with this math question but i thought that i should post it as the concept is very important and the question was good. It's a quantitative comparison question (QC), so tell if column a is always larger than b, or b is always larger than a, or are they always equal, or cannot be determined from the given information.

Given: f > g > 0

Column A: (f+1) / f
Column B: (g+1)/ g

just thought this question would be interesting to those of you are still prepping for your SATs - if you do it by substituting numbers u might get the right answer but its cool when you see the real way done, or atleast the way i see it

By Serene (Serene) on Wednesday, August 13, 2003 - 10:41 pm: Edit

=)

By Serene (Serene) on Wednesday, August 13, 2003 - 10:42 pm: Edit

well i definitely agree that it's important to think conceptually on QC questions. Picking numbers is dangerous because the answer may well be D, cannot be determined.

By Wjk323 (Wjk323) on Wednesday, August 13, 2003 - 11:09 pm: Edit

is the answer B?

By Mattman (Mattman) on Wednesday, August 13, 2003 - 11:14 pm: Edit

it's D.

f=10, g=5
A: 11/10, B: 6/5 (B is more)

f=1, g=-1
A: 2/1 ,B: 0/-1 (A is more)

By Serene (Serene) on Wednesday, August 13, 2003 - 11:19 pm: Edit

Given: f > g > 0

both are positive.

By Nutmag345 (Nutmag345) on Wednesday, August 13, 2003 - 11:20 pm: Edit

I think the answer is B too. Am i right? The way I see this question is, both columns are just a numerator divided by a denominator that is one less than the numerator. As the numbers get larger and larger, the difference of one becomes less significant and so the value of the aforementioned faction becomes smaller. Since f is always larger than g, the value of (g+1)/g is larger than (f+1)/f.

By Serene (Serene) on Wednesday, August 13, 2003 - 11:29 pm: Edit

1+1/g v. 1+1/f ==> 1/g v. 1/f
f>g thus 1/f<1/g

though nutmaq's way of thinking is really cool ^^

By Emmittsmith (Emmittsmith) on Wednesday, August 13, 2003 - 11:39 pm: Edit

ahh limits! :P

By Nutmag345 (Nutmag345) on Wednesday, August 13, 2003 - 11:54 pm: Edit

A compliment from the queen of mathematics really means something Serene.

By Tootall (Tootall) on Wednesday, August 13, 2003 - 11:58 pm: Edit

Lol Nutmaq, yeah she's awesome at math. You're good too, i saw your logic on the clock one, nice work.

By Tootall (Tootall) on Thursday, August 14, 2003 - 12:00 am: Edit

What number was this? yeah the key is that f/f and g/g equal one. This leaves 1/g and 1/f , since f is bigger 1/g > 1/f. My sister will be a sophomore and I like to give her problems, but i like to tell her the difficulty before.

By Wjk323 (Wjk323) on Thursday, August 14, 2003 - 12:30 am: Edit

SO WHATS THE ANSWER???

By Serene (Serene) on Thursday, August 14, 2003 - 12:36 am: Edit

wjk: I think we've already agreed on B.

By Bigman82085 (Bigman82085) on Thursday, August 14, 2003 - 12:39 am: Edit

Definitely B. Each column simplifies to 1+1/f and 1+1/g, respectively. Since f is greater than g, the reciprocal of f is less than the reciprocal of g, therefore making column B bigger.

By Wjk323 (Wjk323) on Thursday, August 14, 2003 - 01:36 am: Edit

hehe...i got it right!! =)

By Aoe2guy (Aoe2guy) on Thursday, August 14, 2003 - 12:23 pm: Edit

B is correct - good job guys, nutmag and serene's way of thinkingt was really the way i was trying to get you guys to do the problem. You can also do:
given: f > g
so: 1/f < 1/g

i liked that question...let's try another:

Given: "a" is a positive integer.
Column A: The remainder when a is divided by 7.
Column B: The remainder when a^2 is divided by 7.

this problem can work well with plugging in numbers, but i started working on an "algebraic" approach and hopefully someone else can post how they worked on their approach as well.

By Serene (Serene) on Thursday, August 14, 2003 - 12:28 pm: Edit

wwwwhat????????????!!!!!!!!!!!!


1/f<1/g<0 ???!!!

hold there, aoe, check your own way of thinking!

By Serene (Serene) on Thursday, August 14, 2003 - 12:29 pm: Edit

Aoe2: cannot be determined.

if a=1(7), then a^2=1(7)
if a=2(7), then a^2=4(7)

from there we can decide the answer is D.

By Aoe2guy (Aoe2guy) on Thursday, August 14, 2003 - 12:29 pm: Edit

ooopss haha sorry serene.....i meant this::

f > g so 1/f < 1/g

By Fairyofwind (Fairyofwind) on Thursday, August 14, 2003 - 12:30 pm: Edit

We need only consider the set {0,1,2,3,4,5,6}, a complete residue system (mod 7). 0 = 0^2 (mod 7). 3 greater than 3^2 = 2 (mod 7). So, D.

By Aoe2guy (Aoe2guy) on Thursday, August 14, 2003 - 12:33 pm: Edit

yea serene your answer is correct but i wanted to know if anyone can come up with an algebraic backup for the answer..so far i've got:

note: </= means less than or equal to

0</= r </= 6
where r is the remainder...it's basically nothing but i need to add more to it.

By Serene (Serene) on Thursday, August 14, 2003 - 12:43 pm: Edit

Aoe2guy: Since the answer is D, we have to show that different possibilities. So usually that'd mean we just have to use two different cases. =) Sometimes there isn't an algebraic solution... hmm.... fairy, am i right? that there is no intrinsic relationship (other than the squaring part) between residue and its quadratic residue...?

By Fairyofwind (Fairyofwind) on Thursday, August 14, 2003 - 12:46 pm: Edit

For every number x, x-1 is relatively prime to x, and thus = -1 (mod x). However, (x-1)^2 = 1 (mod x). Clearly (x-1)^2 < (x-1) (mod x), for x>2. Consider kx, a multiple of x. kx = 0 (mod x). Yet (kx)^2=0 (mod x). Therefore, for any number x>2, we can find a number y such that y^2 < y (mod x) and for any number x, a number z such that z^2 = z (mod x). And for all x>4, we can find a number w such that w^2 > w (mod x) (let w=2, w^2=4>2 (mod x)).

By Fairyofwind (Fairyofwind) on Thursday, August 14, 2003 - 01:03 pm: Edit

But then again, this is still using specific cases. No magical formula for determining the quadratic residue of a number from the residue, though from Euler's theorem, we do know a^(totient(b))=1 (mod b), if a is relatively prime to b, and of course, all residues in a specific congruence class (mod k) have identical quadratic residues (mod k).

By Sup4 (Sup4) on Thursday, August 14, 2003 - 01:15 pm: Edit

let a = 7x + r
where r is the remainder

now consider a^2

therefore
a^2 = 49.x^2 + 14xr + r^2

here r^2 is the remainder and it also may be possible that r^2 is also div by 7

or be greater than 7 by some no.

therefore the ans is d

By Serene (Serene) on Thursday, August 14, 2003 - 01:30 pm: Edit

And as fairy said, (x-1)^2=1=1^2(mod x), so at least two different numbers will have the same quadratic residue for x>2.