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Re: M11-09 [#permalink]
What about choice A? {ab, b2, cb} = b{a,b,c)?
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Re: M11-09 [#permalink]
rhio wrote:
What about choice A? {ab, b2, cb} = b{a,b,c)?



Hi Rhio,

Multiplication changes the standard deviation.
Here we are multiplying everything with b which means even the SD is being multiplies with b.

Hope this helps :)

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Re: M11-09 [#permalink]
I think this is a high-quality question and I agree with explanation.
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Re: M11-09 [#permalink]
I think this is a high-quality question and I agree with explanation.
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Re: M11-09 [#permalink]
Bunuel - I have a very basic doubt on solving quant questions generally. How do we know when to pick numbers and when to solve the question directly, just by looking at the question? I tried picking up numbers for this question and wasted a lot of time in the process and didn't even get the answer in the first place.
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Re: M11-09 [#permalink]
I think this is a high-quality question and I agree with the explanation.
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Re: M11-09 [#permalink]
Why is C wrong ? Adding/ subtracting "A" from each should not change the answer?
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Re: M11-09 [#permalink]
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RRJ12 wrote:
Why is C wrong ? Adding/ subtracting "A" from each should not change the answer?


The SD will not change if we ONLY add or ONLY subtract a constant from each term in a set. In C we are subtracting from a from a and c but adding a to b.
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Re: M11-09 [#permalink]
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re M11-09 [#permalink]
I think this is a high-quality question and I agree with explanation. Selected A thought those rules also apply for multiplication nice trap
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Re M11-09 [#permalink]
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