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## 500 ds test4 #20

This topic has 3 member replies
dunkin77 Master | Next Rank: 500 Posts
Joined
01 Apr 2007
Posted:
269 messages

#### 500 ds test4 #20

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

myprepgmat Newbie | Next Rank: 10 Posts
Joined
10 Mar 2007
Posted:
2 messages
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

dunkin77 Master | Next Rank: 500 Posts
Joined
01 Apr 2007
Posted:
269 messages
Thu Apr 05, 2007 9:47 pm
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)?

jayhawk2001 Community Manager
Joined
28 Jan 2007
Posted:
789 messages
Followed by:
1 members
30
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

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

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