Bunuel
What's the nearest value of \(\frac{0.888888^{27}*0.333333^6}{0.592592^{20}*0.444444}\) ?
A. 4.5
B. 5.0
C. 6.3
D. 8.4
E. 10.2
Let us write numbers in factors that help us.
\(0.888888^{27}=8^{27}*0.111111^{27}=2^{81}*111^{20}*0.001001^{20}*0.111111^7= 2^{81}*37^{20}*3^{20}*0.001001^{20}*0.111111^7\)
\(0.592592^{20}=592^{20}*0.001001^{20}= 16^{20}*37^{20}*0.001001^{20}\)
\(\frac{0.888888^{27}*0.333333^6}{0.592592^{20}*0.444444}\)
\(\frac{ 2^{81}*37^{20}*3^{20}*0.001001^{20}*0.111111^7*0.333333^6}{16^{20}*37^{20}*0.001001^{20}*0.444444}\)
\(\frac{ 2^{81}*37^{20}*3^{20}*0.001001^{20}*0.111111^7*0.333333^6}{2^{80}*37^{20}*0.001001^{20}*4*0.111111}\)
\(\frac{2*3^{20}*0.111111^6*0.333333^6}{4}\)
Now, 0.333333*0.333333 is nearly equal to 0.111~0.111111
So numerator = \( 2*3^{20}*0.111111^6*0.333333^6= 2*3^{14}*0.333333^6*0.111111^3= 2*3^{8}*0.333333^3*0.333333^3=2*3^8*0.111111^3\)
\(2*3^5*0.333333^3=2*3^5*0.111111*0.333333=2*3^4*0.111=2*81*0.111\)
Thus fraction becomes \(\frac{2*81*0.111}{4}=40.5*0.111=4.49\) or 4.5
As we have taken 0.111 as 0.111111, that is increased the value slightly, the answer will be slightly less than 44.9.
Closest is 4.5
A
Pretty calculation intensive. Not likely to be seen on GMAT.