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Sunday, 28 August 2011
Math GRE - #27
Which of the following is the best approximation of $\sqrt{1.5}(266)^{3/2}$?
1000
2700
3200
4100
5300
Solution :
Choice 5 is the answer.
With some simplification, we note that: \[\begin{eqnarray*}
\sqrt{1.5}(266)^{3/2}=\sqrt{\frac{3}{2}}\cdot(266)^{3/2} & = & 266\sqrt{\frac{3\cdot 266}{2}} \\
& = & 266\sqrt{399}\approx266\cdot20\approx 5300.
\end{eqnarray*}\]
Those of us familiar with Taylor series may try approximations with calculus, however this question teaches us that too much knowledge may be a little bit dangerous if misapplied. And that most of the time, there are simpler solutions than you'd think.
2 comments:
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This webpage is LaTeX enabled. To type in-line formulae, type your stuff between two '$'. To type centred formulae, type '\[' at the beginning of your formula and '\]' at the end.
Another technique is to roughly approximate the square of the given formula, which will be slightly greater than
[; 1.5 * 250^3 \sim 1.5 * 16000000 = 24000000.;]
The square root of that is nearly $5000$, so the answer will be $5300$.
@686935176283448507.0
Hmm but you rounded up when you approximated 1.5*16000000 and then you rounded up again when you approximated 24000000 as 25000000 so I'm not entirely convinced.
2 comments:
Post a Comment
This webpage is LaTeX enabled. To type in-line formulae, type your stuff between two '$'. To type centred formulae, type '\[' at the beginning of your formula and '\]' at the end.
[; 1.5 * 250^3 \sim 1.5 * 16000000 = 24000000.;]
The square root of that is nearly $5000$, so the answer will be $5300$.
Hmm but you rounded up when you approximated 1.5*16000000 and then you rounded up again when you approximated 24000000 as 25000000 so I'm not entirely convinced.
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