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Bunuel
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I believe the answer is D.

1. (k-1)/k = 1
Rearrange to have k-1 = k

Factor original equation to k(k-1)
Substitute k with k -1 to get (k-1)(k-1) = 0
k =1, sufficient

2. Solve for k
k = (sqrt(5) + 1)/2
No need to solve as it will be a value that you plug into original equation. Sufficient
sqrt(5) will be a positive number as it's an even root, so no need to worry about negative value.
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Solution: Determine K^2 - k ??

Statement 1: The value of k - 1/k is 1
k^2 - 1 = k
k^2 - k = 1 --> Sufficient

Statement 2: The value of 2k –1 is \sqrt{5}
Substituting this value of k, we can determine k^2 - k --> Sufficient

Each statement alone is sufficient, Answer D
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Original Statement K^2 - K = ?
K(K-1) = ?

Statement 1:
k- (1/K) = 1
(K^2 -1) / K = 1
K^2 = K +1
Putting Value of K^2 in original, we will get answer ---- Statement 1 Sufficient

Statement 2
2K - 1 = sqrt(5)
Statement 2 sufficient to find unique value of K and hence can answer the original Statement.

Answer : Each one is sufficient - Option D
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its not a very tough question. The only place where i got tricked was i read choice St1 as

\(\frac{k-1}{k}=1\) instead of \(k-\frac{1}{k}=1\)

Hope this won't be the case in real GMAT.
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same with me. I tried very hard to understand how others reached the solution. But in the end, I saw.....

I think in Real Gmat, Test people will ensure there are no ambiguities.

But in the end a good question By Bunuel.

akhilbajaj
its not a very tough question. The only place where i got tricked was i read choice St1 as

\(\frac{k-1}{k}=1\) instead of \(k-\frac{1}{k}=1\)

Hope this won't be the case in real GMAT.
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here we need the value of k^2-k
Statement 1
k-1/k=1
k^2-1=k
k^-k=1
hence sufficient
Statement 2
here we have k
so we can get k^2
so we can get k^2-k
hence sufficient

Hence D
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Oh! Can the question be re-written?
If I think of St 1: (k-1)/ (whole) over K the answer is insuff
If it is k - (1/k) it is suff

Thank you!
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Madhavi1990
Oh! Can the question be re-written?
If I think of St 1: (k-1)/ (whole) over K the answer is insuff
If it is k - (1/k) it is suff

Thank you!

You should be familiar with basic math notations.

k - 1/k can ONLY mean \(k - \frac{1}{k}\). If it were \(\frac{k - 1}{k}\), it would have been written as (k - 1)/k.
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Bunuel
Madhavi1990
Oh! Can the question be re-written?
If I think of St 1: (k-1)/ (whole) over K the answer is insuff
If it is k - (1/k) it is suff

Thank you!

You should be familiar with basic math notations.

k - 1/k can ONLY mean \(k - \frac{1}{k}\). If it were \(\frac{k - 1}{k}\), it would have been written as (k - 1)/k.

Hi Bunuel,

Thanks for a great question. I thought I shuld not get take the K to other side unless I know the sign. What if it K was negative. Is the rule of not taking variable applicable only in case of equality signs (<,>). Can I take the K to other side when it is equal to(=).
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kamalakarthi
Bunuel
Madhavi1990
Oh! Can the question be re-written?
If I think of St 1: (k-1)/ (whole) over K the answer is insuff
If it is k - (1/k) it is suff

Thank you!

You should be familiar with basic math notations.

k - 1/k can ONLY mean \(k - \frac{1}{k}\). If it were \(\frac{k - 1}{k}\), it would have been written as (k - 1)/k.

Hi Bunuel,

Thanks for a great question. I thought I shuld not get take the K to other side unless I know the sign. What if it K was negative. Is the rule of not taking variable applicable only in case of equality signs (<,>). Can I take the K to other side when it is equal to(=).

We cannot multiply/divide an inequality by the variable if we don't know its sign but we can add/subtract whatever we want to/from both sides of an inequality.

For example, we cannot divide x > y by x unless we know the sign of x. If x is positive, then we'll get 1 > y/x but if x is negative, then we'll get 1 < y/x (flip the sign when multiplying/dividing by negative number). On the other hand we can subtract x from both sides of x - x > y - x to get 0 > y - x.

9. Inequalities



For more check Ultimate GMAT Quantitative Megathread



Hope it helps.
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