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# What is the value of ((1.06 - 1/3)+1)^2

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Re: What is the value of ((1.06 - 1/3)+1)^2 [#permalink]
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DrAnkita91 wrote:

Hi, let me help you. Firstly, 1.06=106/100. √106 can be assumed to be in between 10^2(=100) and 11^2(=121). Therefore, 100<106<121 or, √100<√106<√121 or,
10<√106<11. So, √106 will be of the form 10.x. Let's assume x=3, so, 10.3^2=106.09~106. So, √106=10.3. Therefore, √1.06=1.03.
Now, ((1.03-1)/3)+1=(0.03+3)/3=1.01. Finally, 1.01^2=1.0201~1.02(A).
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What is the value of ((1.06 - 1/3)+1)^2 [#permalink]
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$$(1+x)^n=1+nx$$  if $$x<<1$$,

$$(1+0.06)^{0.5}-1=(1+0.03)-1=0.03$$

$$(0.03/3+1)^2=(1+0.01)^2=1+0.02=1.02$$

A)­
What is the value of ((1.06 - 1/3)+1)^2 [#permalink]
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