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Bunuel
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Bunuel
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I like the solution - it’s helpful.
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I like the solution - it’s helpful.
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Dear sir,

After the first step, where 10^11 becomes 10^4 as the 10s are used in making the decimals into integers, wont 10^11 become 10^5 as there are only 6 10s needed to make the decimals integers?
Bunuel
Official Solution:

\(\sqrt[4]{0.1*0.12*0.123*10^{11}}\) is closest in value to

A. 11
B. 110
C. 121
D. 1100
E. 12,100


\(\sqrt[4]{0.1*0.12*0.123*10^{11}}=\)

\(= \sqrt[4]{1*120*123*10^{4}}=\)

\(= 10\sqrt[4]{120*123} \approx \)

\(\approx 10\sqrt[4]{11^2*11^2}=\)

\(= 10\sqrt[4]{11^4}=\)

\(= 10*11=\)

\(= 110\)


Answer: B
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keshubhturakhia
Dear sir,

After the first step, where 10^11 becomes 10^4 as the 10s are used in making the decimals into integers, wont 10^11 become 10^5 as there are only 6 10s needed to make the decimals integers?


You are right that only six 10s are needed to make the decimals integers, so at that stage \(10^{11}\) becomes \(10^5\).

The solution then rewrites \(12 * 10^5\) as \(120 * 10^4\). So \(10^5\) is reduced to \(10^4\) only because one factor of 10 is moved into 12 to make it 120.
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