Can you find it?

$\dfrac{k}{0.5}+\dfrac{k}{0.2}+\dfrac{k}{0.25} =297 \\ \left ( k \right ) \left ( \dfrac{ 1 }{ \frac{1}{2}}+ \dfrac{1}{\frac{1}{5}}+ \dfrac{1}{\frac{1}{4}} \right ) =297 \\ \left ( k \right ) \left( 2 + 5 + 4 \right )=297 \\ 11k=297 \\ k=27 \\ k^2=27^2=729$

$We\quad have\quad \frac { k }{ 0.5 } +\frac { k }{ 0.2\quad } +\frac { k }{ 0.25 } \quad =\quad 297\\ By\quad taking\quad LCM\quad we\quad get\quad \frac { 2k+5k+4k }{ 1 } =297\\ So\quad after\quad simplifying\quad the\quad equation\quad we\quad get\quad 11k\quad =297\\ So,\quad k=\quad 27\\ Now\quad we\quad have\quad to\quad find\quad { k }^{ 2 }\quad which\quad is\quad equal\quad to\quad { 27 }^{ 2 }\quad =\quad 729.\\ So\quad the\quad answer\quad is\quad 729\\$

just add them by taking lcm and calculate by cross multiplication

As we know,
$0.5=\frac{5}{10}=\frac{1}{2}\Rightarrow \frac{1}{0.5}=2$
$0.2=\frac{2}{10}=\frac{1}{5}\Rightarrow \frac{1}{0.2}=5$
$0.25=\frac{25}{100}=\frac{1}{4}\Rightarrow \frac{1}{0.25}=4$

So, we can write the equation as:
$2k+5k+4k=297$
$\Rightarrow 11k=297$
$\Rightarrow k=\frac{297}{11}=27$
So, we have $\boxed {k=27}$ .
But, we are required to find $k^2$ . Now, instead to calculating $27^2$ , all we have to do is find the square of the last digit of $27$ [ i.e. $7$ ] and then find the option whose last digit matches the last digit of $7^2$ , which would give us the answer $\boxed {729}$ .
P.S.- Matching the last digit worked here only because other options didn't have the same last digit. If we had another option close to $729$ , e.g. $749$ , we'll need to calculate the square of $27$ .

2k/1+5k/1+4k/1=297 11k=297 K=27 Ksq=729

2k + 5k + 4k = 297 k = 27 so, k^2=729

good question!!!! but easy

\frac { k }{ \frac { 1 }{ 2 } } +\frac { k }{ \frac { 1 }{ 5 } } +\frac { k }{ \frac { 1 }{ 4 } } =297\ 2k+5k+4k=297\ 11k=297\ k=27\ { 27 }^{ 2 }=729

Can someone please tell me how to use the formatting thing? I copy and paste it and it has a bunch of brackets and stuff like you see above. Isn't it supposed to convert it to actually look like math, like everyone else's solutions look like? Someone please tell me what I'm doing wrong!

11k=297 so k=27 and k^2=729

2k + 5k + 4k = 297 11k = 297 K = 27 k × k = 729 K

on simplifying...11k=297 k=27 k*k=729

2k+5k+4k=297 11k=297 k=297/11 k=27 k^2 = 729

2k+5k+4k=297 ,k=27 or,k^2 =729 ans.

(k/.5 +k/.2 +k/.25)/100 = 297/100 (2k +5k +4k)/100 =2.97 11k/100 =2.97 11k=297 k=27 k^2 =729

k(2+5+4)=297 11k=297 k=27 k^2=729

2k/1 + 5k/1 = 4k/1= 297

k = 27

27**2 = 729