Question Detail The largest four digit number which is a perfect cube, is: 7000800092619999 Answer: Option CExplanation:21*21*21 = 9261 Similar Questions : 1. The least perfect square, which is divisible by each of 21, 36 and 66 is 213414213424213434213444 Answer: Option DExplanation: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 2. The largest four digit number which is a perfect cube, is: 7000800092619999 Answer: Option CExplanation:21*21*21 = 9261 3. if a = 0.1039, then the value of \begin{aligned} \sqrt{4a^2 - 4a + 1} + 3a \end{aligned} 12.0391.203911.0391.1039 Answer: Option DExplanation:Tip: Please check the question carefully before answering. As 3a is not under the root we can convert it into a formula , lets evaluate now : \begin{aligned} = \sqrt{4a^2 - 4a + 1} + 3a \end{aligned} \begin{aligned} = \sqrt{(1)^2 + (2a)^2 - 2x1x2a} + 3a \end{aligned} \begin{aligned} = \sqrt{(1-2a)^2} + 3a \end{aligned} \begin{aligned} = (1-2a) + 3a \end{aligned} \begin{aligned} = (1-2a) + 3a \end{aligned} \begin{aligned} = 1 + a = 1 + 0.1039 = 1.1039 \end{aligned} 4. \begin{aligned} \sqrt{\frac{0.081 * 0.484 }{0.0064 * 6.25}} \end{aligned} 0.660.770.880.99 Answer: Option D 5. Evaluate \begin{aligned} \sqrt{1\frac{9}{16}} \end{aligned} \begin{aligned} 1\frac{1}{6} \end{aligned}\begin{aligned} 1\frac{1}{5} \end{aligned}\begin{aligned} 1\frac{1}{4} \end{aligned}\begin{aligned} 1\frac{1}{3} \end{aligned} Answer: Option CExplanation:\begin{aligned} = \sqrt{\frac{25}{16}} \end{aligned} \begin{aligned} = \frac{\sqrt{25}}{\sqrt{16}} \end{aligned} \begin{aligned} = \frac{5}{4} \end{aligned} \begin{aligned} = 1\frac{1}{4} \end{aligned} Read more from - Square Root and Cube Root Questions Answers