# Thread: Real10 - Resolution issues

1. Good post? |
Hi Dkumar,
For the assurance, is it D ?
Tell me dude.

Originally Posted by Dkumar
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.

2. Good post? |
Hi Daniel,
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

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

Originally Posted by uneyoshi
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?

4. Good post? |
Originally Posted by denizen06
Hi Daniel,
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

5. Good post? |

## Daniel take look and confirm me abt the answer

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

6. Good post? |
Hi Denizen06,

Originally Posted by 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

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

7. Good post? |
Thanks Daniel.

8. Good post? |
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?

9. Good post? |

## Real10 Korean Tests

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.

10. Good post? |

## Real10 Korean Tests

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.

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