IMAT 2014 Q56 [Evaluation]

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1

Evaluate:

\frac{8\times10^{-5}}{\sqrt{1.6\times10^7}} \times(1.2\times10^3)^2

A. 2.88 × 10^{-­3}
B. 2.88 × 10^­{-2}
C. 2.88 × 10­^{-5}
D. 2.88 × 10^4
E. 2.88 × 10­^{-1}

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\frac{8 \cdot 10^{-5}}{\sqrt{1.6 \cdot 10^7}}\cdot (1.2 \cdot 10^3)^2

The most efficient way to solve this is to divide it into 3 parts.

  1. part:
    8 \cdot 10^{-5}

  2. part:
    \sqrt{1.6 \cdot 10^7} = \sqrt{\frac{16}{10} \cdot 10^7} = \sqrt{16 \cdot 10^{-1} \cdot 10^7} = \sqrt{16}\cdot \sqrt {10^6}
    =4 \cdot (10^6)^\frac{1}{2} =4 \cdot 10^{6\frac{1}{2} }=4 \cdot 10^{\frac{6}{2} }=4 \cdot 10^3

  3. part:
    (1.2 \cdot 10^3)^2 = (\frac{12}{10}\cdot 10^3)^2=(12\cdot 10^{-1}\cdot 10^3)^2 = (12\cdot 10^{3-1})^2
    = (12\cdot 10^2)^2 = 12^2 \cdot (10^2)^2 = 144 \cdot 10^{2\cdot 2} = 144 \cdot 10^4

Putting all the parts together:
\frac{8 \cdot 10^{-5}}{4 \cdot 10^3 }\cdot 144 \cdot 10^4 = 2 \cdot 10^{-5}\cdot ( 10^3)^{-1} \cdot 144 \cdot 10^4 = 288 \cdot 10^{-5 -3 +4}
= 288\cdot 10^{-4} = 2.88 \cdot 10^{2}\cdot 10^{-4} = 2.88 \cdot 10^{2-4} = 2.88 \cdot 10^{-2}.

Which gives us the final answer (B) 2.88 \cdot 10^{-2}

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