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705-805 (Hard)|   Arithmetic|      
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Ravixxx
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What is the value of ((1.06 - 1/3)+1)^2 16 Apr 2020, 09:50
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85% (hard)

Question Stats:

48% (02:23) correct 52% (02:26) wrong based on 197 sessions

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What is the approximate value of \((\frac{\sqrt{1.06} - 1}{3}+1)^{2}\)

(A) 1.02
(B) 1.03
(C) 1.04
(D) 1.05
(E) 1.06­
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A
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sidahuja
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What is the value of ((1.06 - 1/3)+1)^2 16 Apr 2020, 09:55
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Ravixxx
What is the value of (\(\frac{√1.06 - 1}{3}+1)^{2}\)

(A) 1.02
(B) 1.03
(C) 1.04
(D) 1.05
(E) 1.06
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[(1.03-1)/3 +1]^2 = 1.01^2 = 1.02

IMO A
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What is the value of ((1.06 - 1/3)+1)^2 14 Oct 2023, 02:41
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PLEASE CAN SOMEONE EXPLAIN THE ANSWER?
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samarpan.g28
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What is the value of ((1.06 - 1/3)+1)^2 16 Jan 2024, 08:16
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DrAnkita91
PLEASE CAN SOMEONE EXPLAIN THE ANSWER?
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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 02 May 2024, 11:23
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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)­
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