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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?
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
_________________
Re: If 0.01 < a < 0.02, and 100 < b < 200, which of the following can be
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04 Jul 2017, 18:19
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.
If 0.01 < a < 0.02, and 100 < b < 200, which of the following can be
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05 Jul 2017, 13:14
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 \(y * 10^2\)
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.
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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?
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52% (02:19) correct 
