If 0.01

Question

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!

(PS, Algebra, Inequality, 49-51) Source: MathRevolution

chetan2u · Accepted Answer

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

Milano2017 · Answer

The way I solved this problem was by picking a smart number for b (100) and a (0.01) and then I used logic. 
Step 1) You know that 0.01x0.01x0.01 will have 6 decimal places so a3∗b= some decimal. 
Step 2) b3∗a = 1,000,000 * 0.01 = 10,000, which is your numerator 
Step 3) a3∗b3 = something with 6 decimal places * 1,000,000 = 1, which will be your denominator
Here I looked at the answer choices and immediately crossed A and B. From here I realized that by increasing b, the value will be only decreasing so I picked the first number closest to 10,000.

Hope this helped.

pushpitkc · Answer

The given expression is as follows :  \frac{a^{3}*b + b^{3}*a}{a^{3}*b^{3}}  
It can be further simplified as  \frac{ab(a^{2} + b^{2})}{ab(a^{2}*b^{2})}  =  \frac{a^{2} + b^{2}}{a^{2}*b^{2}}

Now coming to values that a and b can take : 
a will be of form  x * 10^{-2}  and b will be of form  y * 10^2

Substituting these values,
 \frac{a^{2} + b^{2}}{a^{2}*b^{2}}  =  \frac{(x^2 * 10^{-4}) + (y^2 * 10^{4})}{(x^2 * 10^{-4}) * (y^2 * 10^{4})}  =  \frac{(x^2 * 10^{-4}) + (y^2 * 10^{4})}{(x^2 * 10^{-4}) * (y^2 * 10^{4})}

This equation can be approximated to (since x and y are extremely small)
 \frac{(10^{-4}) + (10^{4})}{(10^{-4}) * (10^{4})}  =  \frac{(10^{-4}) + (10^{4})}{(10^{-4 + 4})}  (because  a^m*a^n = a^{m+n} )
=  \frac{(10^{-4}) + (10^{4})}{(10^{0})}  =  10^4 = 10000  which is the largest value possible for the expression

The value closest to this will be Option E.

sahilvijay · Answer

ekniv 
Equation can be written as 1/a^2 + 1/b^2

I HAVE THE WAY WHICH YOU ARE LOOKING FOR-
 
a belongs (0.01,0.02) 
1/a belongs ( 50,100)
1/a^2 belongs ( 2500, 100 00 )

b belongs (100,200)
1/b^2 belongs (0.25x 10^-4 , 10^-4 )

Lets find min value of given equation.
2500 + 0.25x10^-4 = 2500.0 something
 >2500 which is E

Crytiocanalyst · Answer

equations can be resolved to 
=> 1/a^2 + 1/b^2

1/b^2 is negligible 
1/a^2 can be even >5000 if a=0.15

THerefore IMO E

bumpbot · Answer

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If 0.01 < a < 0.02, and 100 < b < 200, which of the following can be

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