Real10 - Resolution issues

[denizen06]

Hi Dkumar,
For the assurance, is it D ?
Tell me dude.

1.
n, r, s and t are positive integers.

(n/4)+(r/8) = (s/8) + (t/6)

[A] 2n+r
[B] 2s+t

I guess you are right in this question.

[denizen06]

Hi Daniel,
What's it's answer?
Is it D?
Look at my reasonings....
We got,
(2n+r)/8=(2s+(8/3)t)/16
2n + r = s+ 4/3 t
2n+r - (2s + t ) = s + 4/3 t - 2s - t
= t/3 - s
= (t - 3 s)/3
t - 3s is multiple of 3
could be -ve , + ve or 0
So I think D
por favor, el señor comment sharp

(n/4)+(r/8) = (s/8) + (t/6)

n=4
r=8
s=12
t=3

[A] 2n+r = 16
[b] 2s+t = 27


Suppose that n, r, s and t are necessarily different.

(2n+r)/8=(6s+8t)/48

therefore

(2n+r)/8=(2s+(8/3)t)/16

Since t > 0,

(2n+r)/8 > (2s+t)/16

then

(2n+r) > (2s+t)/2

How do you obtain the relation

2n + r > 2s + t

?

Best regards,
Daniel

[denizen06]

Hi Somebody assure me,
I am so much confused. Is it really D ?

GRE10-6, Section3, Q12.
QC

Figure attached below (GRE10-6S3Q12.jpg)

A. The area of the circular region with radius OP
B. The area of the shaded sector ROQ

Official Answer. C. I thought that as no measures were provided (radius, proportions among areas or distances), there should be no way to compute both areas. Am I wrong?

[Daniel Sadoc]

Hi Daniel,
What's it's answer?
Is it D?
Look at my reasonings....
We got,
(2n+r)/8=(2s+(8/3)t)/16
2n + r = s+ 4/3 t
2n+r - (2s + t ) = s + 4/3 t - 2s - t
= t/3 - s
= (t - 3 s)/3
t - 3s is multiple of 3
could be -ve , + ve or 0
So I think D
por favor, el señor comment sharp

Hi,

What do you mean by -ve, +ve or 0?

Plug the values below, and see what happens:

n=4
r=8
s=12
t=3

Best regards,
Daniel

[denizen06]

We got,
(2n+r)/8=(2s+(8/3)t)/16
2n + r = s+ 4/3 t
2n+r - (2s + t ) = s + 4/3 t - 2s - t
= t/3 - s
= (t - 3 s)/3
t - 3s is multiple of 3
could be -ve , + ve or 0
2n+r - (2s + t ) = (t - 3 s)/3 -----------[1]
I want to say here
we must make t - 3s as a multiple of 3
hence 3 is also a multiple of 3

If we let t = 3s
we get C

I we let t > 3s
then we can choose n and r accordingly to keep the eqn [1] satisfied
we get A

Now if we let t<3s
of course we can choose n and r accordingly to keep the condition [1] straight
we get B

That's why I said D

Make me known about faults in my reasonings. por favor, el señor

[Daniel Sadoc]

Hi Denizen06,

We got,
(2n+r)/8=(2s+(8/3)t)/16
2n + r = s+ 4/3 t
2n+r - (2s + t ) = s + 4/3 t - 2s - t
= t/3 - s
= (t - 3 s)/3
t - 3s is multiple of 3
could be -ve , + ve or 0
2n+r - (2s + t ) = (t - 3 s)/3 -----------[1]

I agree up to this point.

I want to say here
we must make t - 3s as a multiple of 3

If we let t = 3s
we get C

2n + r = 2s + 3s = 5s

Ok.

I we let t > 3s
then we can choose n and r accordingly to keep the eqn [1] satisfied
we get A

For instance, if we choose t = 10, s = 3, then

2n+r - (2s + t ) = (t - 3 s)/3

2n+r - (6 + 10 ) = (10 - 9)/3

2n+r - 16 = 1/3

6n + 3r - 48 = 1

6n + 3r = 49

3(2n+r) = 49

But n and r are integers! It is impossible to solve this equation, because 49 is not divisible by 3.

2n+r = 16 + 1/3

I agree with you that if t - 3s is multiple of 3, for instance,t = 12, s = 3, then indeed we have

2n+r - (2s + t ) = (12 - 9)/3 = 1

2n + r = 2s + t + 1

hence

2n + r > 2s + t

Now if we let t<3s
of course we can choose n and r accordingly to keep the condition [1] straight
we get B

That's why I said D

Make me known about faults in my reasonings. por favor, el señor

The answer is (D).

Notice that if n = r = s = t = 0 then 2n+r = (2s + t )

[denizen06]

Thanks Daniel.

[economicus]

Hey, mates, how valuable (pertinent) do you find these 10 tests (especially for quant)? My gre is on tue. and so I am a bit economical with my time. Are they worth the effort = what is the degree of similarity with the real gre?

[uneyoshi]

Dear fellows,

I have been receiving some e-mails of people complaining they did not manage to download the Real10 Korean Tests from http://kesaa.glwb.info/ezupload/files/.

I myself ended up experiencing the same. In order to be more democratic, instead of sending individually the tests, I will post them here, since the uploading devices of our forum have not been that helpful for some. Below, comes the first batch!

Hope it helps! Good luck to all!

Best regards,
uneyoshi.

[uneyoshi]

Now, comes the second and final batch.

The Official Answer's can be found in the first posts of this thread: http://www.urch.com/forums/showthread.php?t=35429 (Real10 - Resolution issues) .

Cheers,
uneyoshi.