Q24. Find the square root of 11.0428 correct to four decimal places.
Solution:
Group the digits in pairs:
\[ 11.0428=\overline{11}.\overline{04}\,\overline{28}\,\overline{00}\,\overline{00} \]
\[ \begin{array}{r|l} \sqrt{11.0428} & 3.32307\ldots\\ \hline 3^2=9 & 11-9=2\\ &\downarrow 04\\ &204 \end{array} \]
Double the quotient:
\[ 2\times3=6 \]
Choose the largest digit \(x\) such that
\[ (60+x)\times x\le204. \]
\[ (60+3)\times3=63\times3=189\le204. \]
Hence, \[ \text{Quotient}=3.3 \]
Remainder:
\[ 204-189=15. \]
Bring down the next pair:
\[ 1528. \]
Double the quotient:
\[ 2\times33=66. \]
Choose \(x\) such that
\[ (660+x)\times x\le1528. \]
\[ (660+2)\times2=662\times2=1324\le1528. \]
Hence, \[ \text{Quotient}=3.32 \]
Remainder:
\[ 1528-1324=204. \]
Bring down the next pair:
\[ 20400. \]
Double the quotient:
\[ 2\times332=664. \]
Choose \(x\) such that
\[ (6640+x)\times x\le20400. \]
\[ (6640+3)\times3=6643\times3=19929\le20400. \]
Hence, \[ \text{Quotient}=3.323 \]
Remainder:
\[ 20400-19929=471. \]
Bring down the next pair:
\[ 47100. \]
Double the quotient:
\[ 2\times3323=6646. \]
Since \[ 66460>47100, \] the next digit is \(0\).
Bring down another pair of zeros.
\[ 4710000 \]
\[ 2\times33230=66460. \]
\[ (664600+7)\times7 =664607\times7 =4652249\le4710000. \]
Therefore,
\[ \sqrt{11.0428}=3.32307\ldots \]
Correct to four decimal places:
\[ \mathbf{\sqrt{11.0428}=3.3231} \]
Q25. Find the square root of 11 correct to five decimal places.
Solution:
Group the digits in pairs:
\[ 11=\overline{11}.\overline{00}\,\overline{00}\,\overline{00}\,\overline{00}\,\overline{00} \]
Using the long division method, the successive digits of the square root are:
\[ 3,\;3,\;1,\;6,\;6,\;2,\;4,\ldots \]
Hence,
\[ \sqrt{11}=3.316624\ldots \]
Since the sixth decimal place is \(4<5\), the fifth decimal place remains unchanged.
Correct to five decimal places:
\[ \mathbf{\sqrt{11}=3.31662} \]