Challenge Level

$\dfrac{10 000}{2012}\approx\dfrac{10000}{2000}=5$, so $\sqrt{\dfrac{10000}{2012}}\approx\sqrt5$

So one of the options should square to a number close to $5$. Squaring them gives:

$1.9^2=3.61$ which is too small

$2.2^2=4.84$ which is too small

$2.5^2=6.25$ which is too big

so $2.7^2$ will definitely be too big.

$2.2$ is too small and $2.5$ is too big. To find which is closer to $\sqrt5$, test a number in between.

If $2.3$ is too big, then we will know that $\sqrt5$ is between $2.2$ and $2.3$, so it is definitely closer to $2.2$ than to $2.5$.

$2.3^2=2.3\times2+2.3\times0.3=4.6+0.69=5.29$, which is too big.

So $2.2$ is the closest.

$10 000=100^2$

$\dfrac{100}?\times\dfrac{100}?=\dfrac{10000}{?\times?}$ where $?\times?$ is close to $2012$

$40^2=1600$ and $50^2=2500$. Try $45^2=2025$

$\therefore\dfrac{100}{45}\times\dfrac{100}{45}=\dfrac{10000}{2025}\approx\dfrac{10000}{2012}$

So $\sqrt{\dfrac{10000}{2012}}\approx\dfrac{100}{45}=\dfrac{20}{9}=2.\dot2$

So $2.2$ is the closest.

&=10000\left(\frac{2012-2000}{2000\times2012}\right)\\

&=\frac{10000\times12}{2000\times2012}\\

&=\frac{10\times6}{2012}=\frac{30}{1006}<\frac{30}{1000}=0.03\end{align}$$ So really we were looking for a number whose square was somewhere between $5$ and $5.03$. Squaring the options gave results far less precise than this, so this approximation was good enough for this situation.

&=10000\left(\frac{2025-2012}{2012\times2025}\right)\\

&=\frac{10000\times13}{2025\times2012}<\frac{10000\times13}{2000\times2000}=\frac{13}{2\times200}<\frac{16}{400}=\frac4{100}=0.04\end{align}$$

So the square of $2.\dot2$ is too large, by up to $0.04$.

We can write this as $2.\dot2^2-c^2<0.04$, where $c$ is the exact value of $\sqrt{\dfrac{10000}{2012}}$

We can factorise $2.\dot2^2-c^2$ as the difference of two squares:

$2.\dot2^2-c^2=(2.\dot2+c)(2.\dot2-c)$, so $(2.\dot2+c)(2.\dot2-c)<0.04$

$2.\dot2-c$ must be very close to $0$, because the product is close to $0$, and $(2.\dot2+c)$ is not particularly close to $0$. This is good, because if $2.\dot2$ is a good approximation, then $2.\dot2-c$ is close to $0$.

In fact, $(2.\dot2+c)$ must be more than $2$, since $c$ is positive. So $2.\dot2-c$ must less than $0.02$, to give a product of less than $0.04$ (since $2\times0.02=0.04$).

So $2.\dot2$ is an over-estimation by less than $0.02$, which means that the true value of $c$ is somewhere between $2.\dot2-0.02=2.20\dot2$ and $2.\dot2$. So $2.2$ is defiitely the closest of the options. Our approximation was actually far better than the question required.

You can find more short problems, arranged by curriculum topic, in our short problems collection.