500 ds test4 #20

This topic has expert replies
Master | Next Rank: 500 Posts
Posts: 269
Joined: Sun Apr 01, 2007 5:41 am

500 ds test4 #20

by dunkin77 » Thu Apr 05, 2007 4:51 pm
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

User avatar
Community Manager
Posts: 789
Joined: Sun Jan 28, 2007 3:51 pm
Location: Silicon valley, California
Thanked: 30 times
Followed by:1 members

by jayhawk2001 » Thu Apr 05, 2007 9:10 pm
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.

Master | Next Rank: 500 Posts
Posts: 269
Joined: Sun Apr 01, 2007 5:41 am

by dunkin77 » Thu Apr 05, 2007 9:47 pm
:shock: 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)?

Newbie | Next Rank: 10 Posts
Posts: 2
Joined: Sat Mar 10, 2007 9:22 pm

Re: 500 ds test4 #20

by myprepgmat » Sat Apr 07, 2007 4:47 am
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