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(.0032)(.028)(100) / (.008)(.07)(50) =

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New post 10 Aug 2019, 14:18
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\(\frac{(.0032)(.028)(100)}{(.008)(.07)(50)}\) =

(A) .16
(B) .32
(C) 3.2
(D) 32
(E) 64

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New post 10 Aug 2019, 16:19
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(.0032)(.028)(100)/(.008)(.07)(50)

=> (32/10^4)(28/10^3)(100)/ (8/10^3)(7/10^2)(50)
=>(32/10^4)(28/10^3)(2) • (10^3/8) (10^2/7)

=> (32•28•2/ 8•7) • (10^3•10^2/10^4•10^3)
(8•4)•(1/10^2) = 32/100 = 0.32

Or

(10^7/10^7)(.0032)(.028)(100)/(.008)(.07)(50)
=> (32)(28)(100)/(80)(70)(50)
=> (4)(4)(2)/(100) =0.32

Answer B

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New post 11 Aug 2019, 10:03
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MikeScarn wrote:
\(\frac{(.0032)(.028)(100)}{(.008)(.07)(50)}\) =

(A) .16
(B) .32
(C) 3.2
(D) 32
(E) 64


I think the easiest way is to look at this as 3 separate fractions.

For each fraction, move the denominator over to right in order to get whole numbers.

\(\frac{.0032}{0.008}\) --> \(\frac{32}{80}\) --> \(\frac{4}{10}\)

\(\frac{0.028}{0.07}\) --> \(\frac{28}{70}\) --> \(\frac{4}{10}\)

\(\frac{100}{50}\) --> \(\frac{2}{1}\)

\(\frac{4}{10}\) * \(\frac{4}{10}\) * \(\frac{2}{1}\) = \(\frac{32}{100}\) = .32
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New post 21 May 2020, 11:31
Another method: use scientific notation to group the decimals together. Rewrite each number as \(x*10^y\), where x is an integer. Group the powers of 10 together to keep track of them, and perform standard division/multiplication of the integers.

\(.0032=32*10^{-4}\)
\(.028=28*10^{-3} \)
\(\frac{100}{50} = 2 \)
\(.008 = 8*10^{-3}\)
\(.07=7*10^{-2}\)

Now rewrite it and group the powers of 10 together.
So now the numerator is: \((32*10^{-4})*(28*10^{-3})*(2)=(10^{-7})*(32)*(28)*(2) \)
and the denominator is : \((8*10^{-3})*(7*10^{-2})=(10^{-5})*(8)*(7)\)

Combine the powers of 10 and simplify by dividing, the entire equation is:
\(\frac{(10^{-7})*(32)*(28)*(2)}{(10^{-5})*(8)*(7)}=(10^{-2})*(4)*(4)*(2)=\frac{32}{10^{2}}=\frac{32}{100}=.32\)
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(.0032)(.028)(100) / (.008)(.07)(50) =   [#permalink] 21 May 2020, 11:31

(.0032)(.028)(100) / (.008)(.07)(50) =

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