Question Detail

The largest four digit number which is a perfect cube, is:

7000
8000
9261
9999

Answer: Option C

Explanation:

21*21*21 = 9261

1. The largest four digit number which is a perfect cube, is:

7000
8000
9261
9999

Answer: Option C

Explanation:

21*21*21 = 9261

2. Evaluate
\begin{aligned} \sqrt[3]{4\frac{12}{125}} \end{aligned}

\begin{aligned} 1\frac{2}{5} \end{aligned}
\begin{aligned} 1\frac{3}{5} \end{aligned}
\begin{aligned} 1\frac{4}{5} \end{aligned}
1

Answer: Option B

Explanation:

\begin{aligned}
= \sqrt[3]{\frac{512}{125}} \end{aligned}
\begin{aligned}
= (\frac{8*8*8}{5*5*5})^{\frac{1}{3}} \end{aligned}

\begin{aligned} = \frac{8}{5} = 1\frac{3}{5} \end{aligned}

3. The least perfect square, which is divisible by each of 21, 36 and 66 is

213414
213424
213434
213444

Answer: Option D

Explanation:

L.C.M. of 21, 36, 66 = 2772

Now, 2772 = 2 x 2 x 3 x 3 x 7 x 11

To make it a perfect square, it must be multiplied by 7 x 11.

So, required number = 2 x 2 x 3 x 3 x 7 x 7 x 11 x 11 = 213444

4. Evaluate
\begin{aligned}
\sqrt{53824}
\end{aligned}

Answer: Option B

5. \begin{aligned}
\sqrt{41 - \sqrt{21 + \sqrt{19 - \sqrt{9}}}}
\end{aligned}

Answer: Option D

Explanation:

\begin{aligned}
= \sqrt{41 - \sqrt{21 + \sqrt{19 - 3}}}
\end{aligned}

\begin{aligned}
= \sqrt{41 - \sqrt{21 + \sqrt{16}}}
\end{aligned}

\begin{aligned}
= \sqrt{41 - \sqrt{21 + 4}}
\end{aligned}

\begin{aligned}
= \sqrt{41 - \sqrt{25}}
\end{aligned}

\begin{aligned}
= \sqrt{41 - \sqrt{25}}
\end{aligned}

\begin{aligned}
= \sqrt{41 - 5}
\end{aligned}

\begin{aligned}
= \sqrt{36} = 6
\end{aligned}