Answer: Option C. -> 1024
Least number of 4 digits is 1000
$$\eqalign{
& \,\,\,\,3|\overline {10} \,\,\overline {00} \,\,(31 \cr
& \,\,\,\,\,\,\,|\,\,\,\,9 \cr
& \,\,\,\,\,\,\,| - - - - - - \cr
& 61|\,\,\,\,\,1\,00 \cr
& \,\,\,\,\,\,\,|\,\,\,\,\,\,61 \cr
& \,\,\,\,\,\,\,| - - - - - - \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,39 \cr
& \therefore {\left( {31} \right)^2} < 1000 < {\left( {32} \right)^2} \cr
& {\text{Hence, required number}} \cr
& = {\left( {32} \right)^2} \cr
& = 1024 \cr} $$

Answer: Option A. -> 4
$$\eqalign{
& \,\,\,\,\,\,\,\,8|\overline {68} \,\,\overline {06} \,\,\overline {21} \,(824 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,|\,\,64 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - - \cr
& \,\,\,162|\,\,\,\,4\,06 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,3\,24 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - - \cr
& 1644|\,\,\,\,\,\,\,\,\,\,82\,21 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,65\,76 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,16\,45 \cr
& \therefore {\text{Number to be added}} \cr
& = {\left( {825} \right)^2} - 680621 \cr
& = 680625 - 680621 \cr
& = 4 \cr} $$

Answer: Option B. -> 9801
$$\eqalign{
& \,\,\,\,\,\,\,9|\overline {99} \,\,\overline {99} \,\,(99 \cr
& \,\,\,\,\,\,\,\,\,\,|\,81 \cr
& \,\,\,\,\,\,\,\,\,\,| - - - - - - \cr
& \,189|\,18\,99 \cr
& \,\,\,\,\,\,\,\,\,\,|\,17\,01 \cr
& \,\,\,\,\,\,\,\,\,\,| - - - - - - \cr
& \,\,\,\,\,\,\,\,\,\,|\,\,\,\,1\,98 \cr
& \therefore {\text{Required number}} \cr
& = \left( {9999 - 198} \right) \cr
& = 9801 \cr} $$

Answer: Option E. -> None of these
$$\eqalign{
& \,\,\,\,\,8|\overline {77} \,\,\overline {00} \,\,(87 \cr
& \,\,\,\,\,\,\,\,\,|\,\,64 \cr
& \,\,\,\,\,\,\,\,\,| - - - - - - \cr
& 167|\,\,\,13\,00 \cr
& \,\,\,\,\,\,\,\,\,|\,\,\,11\,69 \cr
& \,\,\,\,\,\,\,\,\,| - - - - - - \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,31 \cr
& \therefore {\text{Number to be added}} \cr
& = {\left( {88} \right)^2} - 7700 \cr
& = 7744 - 7700 \cr
& = 44 \cr} $$

Answer: Option B. -> 105
$$\eqalign{
& \,\,\,\,\,\,1|\overline 1 \,\,\overline {10} \,\,\overline {25} \,(105 \cr
& \,\,\,\,\,\,\,\,\,|\,\,1 \cr
& \,\,\,\,\,\,\,\,\,| - - - - - - \cr
& \,\,20|\,\,\,\,\,\,10 \cr
& \,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,0 \cr
& \,\,\,\,\,\,\,\,\,| - - - - - - \cr
& 205|\,\,\,\,\,\,\,10\,25 \cr
& \,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,10\,25 \cr
& \,\,\,\,\,\,\,\,\,| - - - - - - - \cr
& \,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{X}} \cr
& {\text{Number of rows}} \cr
& {\text{ = }}\sqrt {10914 + 111} \cr
& = \sqrt {11025} \cr
& = 105 \cr} $$

Answer: Option A. -> 1367631
$$\eqalign{
& {\text{Let,}} \cr
& {\text{ }}99 \times 21 - \root 3 \of x = 1968 \cr
& {\text{Then,}} \cr
& \Leftrightarrow 2079 - \root 3 \of x = 1968 \cr
& \Leftrightarrow \root 3 \of x = 2079 - 1968 \cr
& \Leftrightarrow \root 3 \of x = 111 \cr
& \Leftrightarrow x = {\left( {111} \right)^3} \cr
& \Leftrightarrow x = 1367631 \cr} $$

Answer: Option D. -> 177
$$\eqalign{
& = 262144 \cr
& = 8 \times 8 \times 8 \times 8 \times 8 \times 8 \cr
& = {8^6} \cr
& \therefore \root 3 \of {262144} \cr
& = {8^2} \cr
& = 64 \cr
& {\text{Let, }} \cr
& {\text{1728}} \div \root 3 \of {262144} \times x - 288 = 4491 \cr
& {\text{Then,}} \cr
& \Leftrightarrow 1728 \div 64 \times x - 288 = 4491 \cr
& \Leftrightarrow 27x = 4779 \cr
& \Leftrightarrow x = \frac{{4779}}{{27}} \cr
& \Leftrightarrow x = 177 \cr} $$
$$\eqalign{
& 8|262144 \cr
& - - - - - - - - \cr
& 8|\,\,\,32768 \cr
& - - - - - - - - \cr
& 8|\,\,\,\,4096 \cr
& - - - - - - - - \cr
& 8|\,\,\,\,\,512 \cr
& - - - - - - - - \cr
& 8|\,\,\,\,\,\,64 \cr
& - - - - - - - - \cr
& 8|\,\,\,\,\,\,\,8 \cr
& - - - - - - - - \cr} $$

Answer: Option D. -> 121108
Clearly, the required number must be a perfect square. Since a number having 8 as the unit's digit cannot be a perfect square, so 121108 is not a perfect square.

Answer: Option A. -> 25
Let the number of girls in the group be x
Then, number of oranges given to each girl = 2x
$$\eqalign{
& \therefore x \times 2x = 1250 \cr
& \Leftrightarrow 2{x^2} = 1250 \cr
& \Leftrightarrow {x^2} = 625 \cr
& \Leftrightarrow x = 25 \cr} $$

Answer: Option B. -> 10
$$21600 = {2^5} \times {3^3} \times {5^2}$$
To make it a perfect cube, it must be multiplied by (2 × 5), i.e., 10