Bunuel
\(5√2\) percent of \(\frac{1}{√200} =\)
A. 0.005
B. 0.02
C. 0.05
D. 0.2
E. 0.5
I. ArithmeticChange the denominator either initially* or while solving, as here
Setup: 10 percent of \(\frac{1}{2}=\)
LHS: \((\frac{10}{100}*\frac{1}{2})\)
\(5√2\) percent of \(\frac{1}{√200}\):
\((\frac{5\sqrt{2}}{100}*\frac{1}{\sqrt{200}})=\frac{5\sqrt{2}}{100*\sqrt{200}}=\)
\(\frac{5\sqrt{2}}{100*\sqrt{100*2}}=\)
\(\frac{5\sqrt{2}}{100*10\sqrt{2}}=\)
\(\frac{5}{100*10}=\frac{1}{100*2}=\frac{1}{200}=\)
\(0.005\)
Answer A
II. Approximate in pieces \(5√2\) percent of \(\frac{1}{√200}\)?
(1) \(5\sqrt{2}\approx{(5*1.4)}\approx{7}\)
\(7\) percent =\(0.07\)
(2) \(\frac{1}{\sqrt{200}}\):
the denominator is between 14 and 15. \(14^2=196\), and \(15^2=225\)
\(\sqrt{196}<\sqrt{200}<\sqrt{225}\)
\(\sqrt{200}\) is closer to 14:
\(\frac{1}{\sqrt{200}}\approx{\frac{1}{14}}\approx{0.07}\)
(3) Together: 7 percent of 0.07 =
\((0.07*0.07)=0.0049\)
\(\approx{0.005}\)
Answer A
*
\(\frac{1}{\sqrt{200}}\)
Change the denominator
\(\sqrt{200}=\sqrt{100*2}=10\sqrt{2}=>\)
\(\frac{1}{\sqrt{200}}=\\
\frac{1}{10\sqrt{2}}\)