500 ds test4 #20

dunkin77 · Posted: Thu Apr 05, 2007 5:51 pm Post subject: 500 ds test4 #20

Hi,

The answer is D) but I got different answers for 1) and 2)... but I think if the answer is D), both 1) and 2) should have the same number for k...

1) k-1/k=1
k=1/k
therefore k=1??

2) 2k-1=root5
2k=root+1
k=root5/2+1/2

What is the value of k2-k?
(1) The value of k-1/k is 1.
(2) The value of 2k –1 is root5

jayhawk2001 · Posted: Thu Apr 05, 2007 10:10 pm Post subject:

You don't need the _same_ answer for 1 and 2. It is sufficient if 1 and 2
each yield a unique answer.

Bottom-line you should be able to arrive at a unique value for k^2 - k
using 1 or 2 or both.

dunkin77 · Posted: Thu Apr 05, 2007 10:47 pm Post subject:

Okay, so 1) and 2) has different numbers for K, (if my calcuation was correct), K^2-k would be also different numbers - but it is still sufficient to answer D)?

myprepgmat · Posted: Sat Apr 07, 2007 5:47 am Post subject: Re: 500 ds test4 #20

No need to calculate the value of k. we can approach the question in a different way

1) k - 1/k =1

multiply with k gives k^2 - 1 =k

now rearraange

K^2-k =1 >>Ans

Now take case 2)

2k-1 = root5

Square the equation ie (2k-1)^2 =5

i.e.

4k^2-4k+1=5

4k^2-4k-4=0 (after re-arranging)

Divide by 4 >> k^2-k-1=0

Re-arranging gives k^2-k=1 >> Ans.

Both cases give the same result ... so Ans is D