ekniv
(PS, Algebra, Inequality, 49-51) Source: MathRevolution
If 0.01 < a < 0.02, and 100 < b < 200, which of the following can be the value of:
\(\frac{a^{3}*b + b^{3}*a}{a^{3}*b^{3}}\)
A. 0.02
B. 0.05
C. 775
D. 1,525
E. 5,725
After ~10+ minutes I got to the answer, but I can clearly not afford to spend that much in a problem. How do you solve this problem under 2.5 minutes? What is the trick?
Thanks!
Hi,
Firstly please give the topic name as first few words of the Q..
Now for the Q...
Simplify and find RANGE..
\(\frac{a^{3}*b + b^{3}*a}{a^{3}*b^{3}}\)=\(\frac{ab(a^{2} + b^{2}}{ab(a^{2}*b^{2}}\)=\(\frac{a^{2} + b^{3}}{a^{2}*b^{2}}\)..
1) You can work from here
Now the NUMERATOR will be just b^2 but denominator would reduce drastically..
2) further simplify..
\(\frac{a^2+b^2}{a^2*b^2}\)..
a^2 will be very small as compared to b^2 so neglect in the numerator..
\(\frac{b^2}{a^2b^2}=\frac{1}{a^2}\)..
Now let's find the range of this..
\(\frac{1}{(0.02)^2}=\frac{1}{0.0004}=\frac{10000}{4}=2500\)..
\(\frac{1}{(0.01)^2}=\frac{1}{0.0001}=\frac{10000}{1}=10000\)..
Range is 2500 to 10000..
Only E remains.
E