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Maths 3 Mark Question

Squares and Square Roots · Maharashtra Board STD 8 (MSBSHSE - Semi English Medium). 27 questions.

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Question 13 Marks
Find the smallest number by which 2925 must be divided to obtain a perfect square. Also, find the square root of the perfect square so obtained.
Answer
By prime factorisation, we get
2925 = 3 × 3 × 5 × 5 × 13
So, the given number should be multiplied by 13 to make the product a perfect square.
New number = 2925 ÷ 13 = 225
$\therefore$ 225 = 3 × 3 × 5 × 5
$\sqrt{225}=3\times5=15$
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Question 23 Marks
Evaluate $\sqrt{0.9}$ correct up to two places of decimal.
Answer
$\begin{array}{c|c} &0.948 \\ \hline 9 & 0.\overline{90}\ \overline{00}\ \overline{00}\ \overline{00}\\& -81\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \hline184 &\ \ 900\ \ \ \ \ \ \\ &-736\ \ \ \ \ \ \ \\ \hline1888 &\ 16400 \\ &-15104\ \ \\ \hline &\ \ \ \ 1296 \end{array}$
$\therefore\sqrt{0.9}=0.948=0.95$
(Correct up to two places of decimal)
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Question 33 Marks
Find the least number which must be added to 8400 to obtain a perfect square. Find this perfect square and its square root.
Answer
Finding the square root of 8400 by long division method, we find that 64 is to be added to 8400

We, get 8400 + 64 = 8464

$\begin{array}{c|c} &92 \\ \hline 9 & \overline{84}\ \overline{00}\\& 81 \ \ \ \ \ \\ \hline182 &\ \ 300\\ &-364\\ \hline &\ \ \ \ \ 64 \end{array}$

Least number to be added = 64

Perfect square = 8464

Square root = 92
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Question 43 Marks
Find the greatest number of five digits which is a perfect square. What is the square root of this number?
Answer
The greatest 5 digit number is 99999.
$\begin{array}{c|c}&316\\\hline3&\bar{9}\ \overline{99}\ \overline{99}\\3&9\ \ \ \ \ \ \ \ \ \ \\\hline61&99\\\ \ \ 1&\ \ \ \ 61\ \ \ \ \\\hline626&\ \ \ \ \ 3899\\\ \ \ \ 6&\ \ \ \ \ \ \ \ \ 3756\ \ \ \ \\\hline&\ \ \ \ \ \ \ 143\\\end{array}$
$316<\sqrt{99999}<317$
$316^2=99856$
Thus, this is the greatest 5 digit number.
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Question 53 Marks
Find the least square number which is exactly divisible by each of the numbers 8, 12, 15 and 20.
Answer
The least number divisible by each one of 8, 12, 15 and 20 is their LCM.

Now, LCM of 8, 12, 15 and 20 = (2 × 2 × 3 × 5 × 2) = 120

By prime factorization, we get

120 = 2 × 2 × 2 × 3 × 5

To make it perfect square it must be multiplied by (2 × 3 × 5) i.e., 30

Hence, required number = (120 × 30) = 3600
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Question 63 Marks
Evaluate $\sqrt{3}$ correct up to two places of decimal.
Answer
$\begin{array}{c|c}&1.732\\\hline1&\bar{3}\ \overline{00}\ \overline{00}\ 00\\1&1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\\hline27& 200\ \ \ \ \ \ \ \ \ \ \ \\\ 7&189\ \ \ \ \ \ \ \ \ \ \\\hline343\ \ \ & \ \ \ \ \ \ \ \ 1100\ \ \ \ \ \ \ \ \ \ \ \ \\\ 3&1029\ \ \ \\\hline3462\ \ \ \ \ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 7100\ \ \ \ \ \ \ \ \ \ \ \ \ \\\ 2&\ \ \ \ \ 6924 \\\hline&\ \ \ \ \ \ 176\\\end{array}$
$\sqrt{3}=1.732$
Therefore, the value of $\sqrt{3}$ up to two places of decimal is 1.73.
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Question 73 Marks
Evaluate:
$\sqrt{3\frac{13}{36}}$
Answer
$\sqrt{3\frac{13}{36}}$
$=\sqrt{\frac{3\times36+13}{36}}$
$=\sqrt{\frac{108+13}{36}}$
$=\sqrt{\frac{121}{36}}$
$=\sqrt{\frac{11\times11}{6\times6}}$
$=\frac{11}{6}$
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Question 83 Marks
Evaluate:
$\sqrt{3\frac{33}{289}}$
Answer
$\sqrt{3\frac{33}{289}}$
$=\sqrt{\frac{3\times289+33}{289}}$
$=\sqrt{\frac{867+33}{289}}$
$=\sqrt{\frac{900}{289}}$
$\begin{array}{c|c}2&900\\\hline2&450\\\hline3&225\\\hline3&75\\\hline5&25\\\hline5&5\\\hline&1\end{array}$
$\begin{array}{c|c}17&289\\\hline17&17\\\hline&1\end{array}$
$=\sqrt{\frac{\overline{2\times2}\times\overline{3\times3}\times\overline{5\times5}}{17\times17}}$
$=\frac{2\times3\times5}{17}$
$=\frac{30}{17}$
$=1\frac{13}{17}$
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Question 93 Marks
Evaluate $\sqrt{11236}$
Answer
Using long division method:
$\begin{array}{c|c}&106\\\hline1&\bar{1}\ \overline{12}\ \overline{36}\\1&1\ \ \ \ \ \ \ \ \ \ \\\hline206&\ \ \ 1236\\\ \ \ \ 6&\ \ \ \ 1236\\\hline&\ \ \ \ \ \ \times\\\end{array}$
$\therefore\sqrt{11236}=106$
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Question 103 Marks
Find the least number which must be added to 6203 to obtain a perfect square. Find this perfect square and its square root.
Answer
Finding the square root of 6203 by division method, we find that 38 is to be added to get a perfect square
  1. Least number to be added = 38
  2. Perfect square = 6241
  3. Square root = 79
$\begin{array}{c|c} &79 \\ \hline 7 & \overline{62}\ \overline{03}\\& 49 \ \ \ \ \ \\ \hline149 &\ \ \ \ 1303\\ &-1341\\ \hline &\ \ \ \ -38 \end{array}$
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Question 123 Marks
Evaluate:
$\sqrt{\frac{625}{729}}$
Answer
$\sqrt{\frac{625}{729}}$
$\begin{array}{c|c}5&625\\\hline5&125\\\hline5&25\\\hline5&5\\\hline&1\end{array}$
$\begin{array}{c|c}3&729\\\hline3&243\\\hline3&81\\\hline3&27\\\hline3&9\\\hline3&3\\\hline&1\end{array}$
$=\sqrt{\frac{\overline{5\times5}\times\overline{5\times5}}{\overline{3\times3}\times\overline{3\times3}\times\overline{3\times3}}}$
$=\frac{5\times5}{3\times3\times3}$
$=\frac{25}{27}$
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Question 133 Marks
Find the least number which must be subtracted from 7581 to obtain a perfect square. Find this perfect square and its square root.
Answer
Finding the square root of 7581 by division method, we find that 12 is left as remainder 12 must be subtracted from 7587 to get a perfect square i.e., 7581 - 12 = 7569

$\begin{array}{c|c} &87 \\ \hline 8 & \overline{75}\ \overline{81}\\& 64 \ \ \ \ \ \\ \hline167 &\ \ \ \ 1181\\ &\ \ \ \ 1169\\ \hline &\ \ \ \ \ \ \ 12 \end{array}$
  1. The least number to be subtracted = 12
  2. Perfect square = 7569
  3. square root = 87
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Question 143 Marks
Find the least square number which is exactly divisible by each of the numbers 6, 9, 15 and 20.
Answer
The least number divisible by each one of 6, 9, 15 and 20 is their LCM.

Now, LCM of 6, 9, 15 and 20 = (2 × 3 × 5 × 3 × 2) = 180

By prime factorization, we get

180 = 2 × 2 × 3 × 3 × 5

To make it perfect square it must be multiplied by 5

Hence, required number = (180 × 5) = 900
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Question 153 Marks
Find the least number of four digits which is a perfect square. Also find the square root of the number so obtained.
Answer
Least four digit number = 1000

$\begin{array}{c|c} &32 \\ \hline 3 & \overline{10}\ \overline{00}\\& \ \ 9 \ \ \ \ \ \\ \hline62 &\ \ 100\\ &-124\\ \hline &\ -24 \end{array}$

Finding the square root of 1000 by division method, we find that 24 must be added to get a perfect square of 4 digits.

Perfect square = 1000 + 24 = 1024

Square root of 1024 = 32

$\begin{array}{c|c} &32 \\ \hline 3 & \overline{10}\ \overline{24}\\&\ 9 \ \ \ \ \ \\ \hline62 &\ 124\\ &-124\\ \hline &\ \ \ \times \end{array}$
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Question 163 Marks
Find the smallest number by which 252 must be multiplied to get a perfect square. Also, find the square root of the perfect square so obtained.
Answer
By prime factorisation, we get
252 = 2 × 2 × 3 × 3 × 7
So, the given number should be multiplied by 7 to make the product a perfect square.
New number = 252 × 7 = 1764
$\therefore$ 1764 = 2 × 2 × 3 × 3 × 7 × 7
$\sqrt{1764}=2\times3\times7=42$
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Question 173 Marks
Evaluate $\sqrt{0.2809 }$
Answer
The least number of 4 digit is 1000.
$\begin{array}{c|c}&0.53\\\hline5&\bar0.\overline{28}\ \overline{09}\\5&25\\\hline103&\ \ \ \ \ \ \ \ 3\ 09\\3&\ \ \ \ \ \ \ \ 3\ 09\\\hline&\ \ \ \ \ \ \ 0\end{array}$
$\therefore\sqrt{0.2809 }=0.53$
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Question 183 Marks
Evaluate:
$\sqrt{4\frac{73}{324}}$
Answer
$\sqrt{4\frac{73}{324}}$
$=\sqrt{\frac{4\times324+73}{324}}$
$=\sqrt{\frac{1296+73}{324}}$
$=\sqrt{\frac{1369}{324}}$
$\begin{array}{c|c}37&1369\\\hline37&37\\\hline&1\end{array}$
$\begin{array}{c|c}2&324\\\hline2&162\\\hline3&81\\\hline3&27\\\hline3&9\\\hline3&3\\\hline&1\end{array}$
$=\sqrt{\frac{37\times37}{2\times2\times3\times3\times3\times3}}$
$=\frac{37}{2\times3\times3}$
$=\frac{37}{18}$
$=2\frac{1}{18}$
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Question 193 Marks
Find the least number of four digits which is a perfect square. What is the square root of this number?
Answer
The least number of 4 digit is 1000.
$\begin{array}{c|c}&31\\\hline1&\overline{10}\ \overline{00}\\1&\ \ \ \ \ \ 9\ \ \ \ \ \ \ \ \ \ \\\hline61&\ \ \ 100\ \ \ \ \\\ 1&\ 61\\\hline&\ 39\\\end{array}$
$31<\sqrt{100}<32$
$32^2=1024$
1024 is the least four digit perfect square and its square root is 32.
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Question 213 Marks
Find the greatest number of five digits which is a perfect square. Also find the square root of the number so obtained.
Answer
Greatest number of five digits = 99999

Finding the square root of 99999

We get remainder = 143

$\begin{array}{c|c} &316 \\ \hline 3 & \bar{9}\ \overline{99}\ \overline{99}\\& 9 \ \ \ \ \ \ \ \ \ \ \\ \hline61 &99\ \ \ \\ &61\ \ \ \\ \hline626 &\ 3899\\ &\ 3756\\\hline &\ \ \ \ 143 \end{array}$

Required perfect square = 99999 - 143 = 99856

and square root = 316
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Question 223 Marks
Evaluate $\frac{\sqrt{48}}{\sqrt{243}}$
Answer
$\frac{\sqrt{48}}{\sqrt{243}}$
$=\sqrt{\frac{48}{243}}$
$=\sqrt{\frac{2\times2\times2\times2\times3}{3\times3\times3\times3\times3}}$
$=\frac{\sqrt{2\times2\times2\times2}}{\sqrt{3\times3\times3\times3}}$
$=\frac{2\times2}{3\times3}$
$=\frac{4}{9}$
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Question 243 Marks
Evaluate:
$\sqrt{98}\times\sqrt{162}$
Answer
$\sqrt{98}\times\sqrt{162}$
$=\sqrt{98\times162}$
$=\sqrt{2\times7\times7\times2\times3\times3\times3\times3}$
$=\sqrt{\overline{2\times2}\times\overline{7\times7}\times\overline{3\times3}\times\overline{3\times3}}$
$=2\times7\times3\times3$
$=126$
$\begin{array}{c|c}2&98\\\hline7&29\\\hline7&7\\\hline&1\end{array}$
$\begin{array}{c|c}2&162\\\hline3&81\\\hline3&27\\\hline3&9\\\hline3&3\\\hline&1\end{array}$
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Question 253 Marks
The area of a square field is $60025 m^2$. A man cycles along its boundary at $18 km / h$. In how much time will he return to the starting point?
Answer
Area of a square field $=60025 m^2$

Let its side $= a$

$\therefore\text{a}^2=60025$

$\Rightarrow\text{a}=\sqrt{60025}$

$\Rightarrow\text{a}=245\text{m}$

$\begin{array}{c|c} &245 \\ \hline 2 & \bar{6}\ \overline{00}\ \overline{25}\\& 4 \ \ \ \ \ \ \ \ \ \ \\ \hline44 &200\ \ \ \ \ \\ &176\ \ \ \ \ \\ \hline485 &\ \ 2425\\ &\ \ 2425\\\hline &\ \ \ \times \end{array}$

Perimeter = 4a

= 4 × 245 = 980m

Speed of cycling = 18km/ h

$\therefore$ Time taken to complete its boundary $=\frac{980}{1000}\times\frac{60}{18}=\frac{49}{15}\ \text{minutes}$

$=3\frac{4}{15}\ \text{minutes}$

$=3\ \text{minutes}\ 16\ \text{seconds}$
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Question 263 Marks
Evaluate $\sqrt{2.8}$ correct up to two places of decimal.
Answer
$\begin{array}{c|c} &1.673 \\ \hline 1 &2. \overline{80}\ \overline{00}\ \overline{00}\\& -1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \hline26 &180\ \ \ \ \ \ \ \ \ \ \ \\ &-156\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \hline327 &\ 2400\ \ \ \ \ \ \\ &-2289\ \ \ \ \ \ \ \ \\ \hline \hline3343 &\ \ \ \ \ \ \ 11100\ \ \ \ \ \ \\ &\ \ \ \ -10029\ \ \ \ \ \ \ \ \\ \hline&\ \ \ \ \ \ \ \ \ \times\ \ \ \ \ \ \ \ \end{array}$
$\therefore\sqrt{2.8}=1.673=1.67$
(Correct up to two places of decimal)
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