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

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

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New post 04 Jul 2017, 15:17
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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

[Reveal] Spoiler:
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
[Reveal] Spoiler: OA

Last edited by Bunuel on 04 Jul 2017, 21:35, edited 2 times in total.
Renamed the topic and edited the question.

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Re: If 0.01 < a < 0.02, and 100 < b < 200, which of the following can be [#permalink]

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New post 04 Jul 2017, 18:13
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ekniv wrote:
(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
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Re: If 0.01 < a < 0.02, and 100 < b < 200, which of the following can be [#permalink]

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New post 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.

Hope this helped.

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

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New post 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 \(y * 10^2\)

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.
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Re: If 0.01 < a < 0.02, and 100 < b < 200, which of the following can be [#permalink]

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New post 06 Sep 2017, 01:51
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ekniv wrote:
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

[Reveal] Spoiler:
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





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
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Re: If 0.01 < a < 0.02, and 100 < b < 200, which of the following can be   [#permalink] 06 Sep 2017, 01:51
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