NCERT Exemplar for Class 8 Maths Solutions Chapter 3 Square-Square Root & Cube-Cube Root

In Examples 1 to 7, out of given four Choices Multiple only one is Correct. Write the Correct Answer

Example 1: Which of the following is the square of an odd number?

(a) 256

(b) 361

(c) 144

(d) 400

Ans: (b) because $19\times 19=361$

Example 2: Which of the following will have $\mathbf{1}$ at its unit’s place?

(a) $\mathbf{1}{{\mathbf{9}}^{\mathbf{2}}}$

(b) $\mathbf{1}{{\mathbf{7}}^{\mathbf{2}}}$

(c) $\mathbf{1}{{\mathbf{8}}^{\mathbf{2}}}$

(d) $\mathbf{1}{{\mathbf{6}}^{\mathbf{2}}}$

Ans: (a) because $19\times 19=361$

Example 3: How many natural numbers lie between $\mathbf{1}{{\mathbf{8}}^{\mathbf{2}}}$ and $\mathbf{1}{{\mathbf{9}}^{\mathbf{2}}}$?

(a) $\mathbf{30}$

(b) $\mathbf{37}$

(c) $\mathbf{35}$

(d) $\mathbf{36}$

Ans: (d) As ${{\left( 18 \right)}^{2}}=324\text{and}{{\left( 19 \right)}^{2}}=361$

Therefore, $361324=37$

Example 4: Which of the following is not a perfect square?

(a) $\mathbf{361}$

(b) $\mathbf{1156}$

(c) $\mathbf{1128}$

(d) $\mathbf{1681}$

Ans: (c) As in $1128=2\times 2\times 2\times 141$ here 2 and 141 have incomplete pair.

Example 5: A perfect square can never have the following digit at one’s place.

(a) $\mathbf{1}$

(b) $\mathbf{6}$

(c) $\mathbf{5}$

(d) $\mathbf{3}$

Ans: (d) digits ending with $1,6,5\text{ and }3$ have unit digit $1,6,5\text{ and }9$ squares respectively

Example 6: The value of $\sqrt{\mathbf{176}+\sqrt{\mathbf{2401}}}$ is

(a) $\mathbf{14}$

(b) $\mathbf{15}$

(c) $\mathbf{16}$

(d) $\mathbf{17}$

Ans: (b) $\because 49\times 49=2401$

$\therefore \sqrt{176+\sqrt{2401}}=\sqrt{176+49}=\sqrt{225}=15$

Example 7: Given that $\sqrt{\mathbf{5625}}=\mathbf{75}$, the value of $\sqrt{\mathbf{0}.\mathbf{5625}}+\sqrt{\mathbf{56}.\mathbf{25}}$ is:

(a) $\mathbf{82}.\mathbf{5}$

(b) $\mathbf{0}.\mathbf{75}$

(c) $\mathbf{8}.\mathbf{25}$

(d) $\mathbf{75}.\mathbf{05}$

Ans: (c) $\because 75\times 75=5625$and $0.75\times 0.75=0.5625$

$7.5\times 7.5=56.25$

Therefore, $\sqrt{0.5625}+\sqrt{56.25}=0.75+7.5=8.25$

In examples 8 to 14, fill in the blanks to make the statements true.

Example 8: There are __________ perfect squares between $\mathbf{1}$ and $\mathbf{50}$.

Ans: There are $6$ perfect squares $4,9,16,25,36,49$.

Example 9: The cube of $\mathbf{100}$ will have __________ zeroes.

Ans: Cube of $100$ will have $6$zeroes. $\left( 100\times 100\times 100 \right)=1000000$

Example 10: The square of $\mathbf{6}.\mathbf{1}$ is ____________.

Ans: $6.1\times 6.1=37.21$

Example 11: The cube of $\mathbf{0}.\mathbf{3}$is ____________.

Ans: $0.3\times 0.3\times 0.3=0.027$

Example 12: $\mathbf{6}{{\mathbf{8}}^{\mathbf{2}}}$ will have __________ at the unit’s place.

Ans: Because $68$ has $8$ at its unit place and ${{\left( 8 \right)}^{2}}=64$. So, ${{\left( 68 \right)}^{2}}$ will have $4$ at unit’s place.

Example 13: The positive square root of a number $\mathbf{x}$ is denoted by__________.

Ans: $\sqrt{\text{x}}$

Example 14: The least number to be multiplied with $\mathbf{9}$ to make it a perfect cube is _______________.

Ans: $3\because \left( 9\times 3=27 \right)$and $\left( 3\times 3\times 3=27 \right)$.

In examples 15 to 19, state whether the statements are true (T) or false (F)

Example 15: The square of $\mathbf{0}.\mathbf{4}$ is $\mathbf{0}.\mathbf{16}$.

Ans: True, because $0.4\times 0.4=0.16$

Example 16: The cube root of $\mathbf{729}$is $\mathbf{8}$.

Ans: False because $9\times 9\times 9=729$

Example 17: There are $\mathbf{21}$ natural numbers between $\mathbf{102}$ and $\mathbf{112}$.

Ans: False because $112-102=10$not $21$.

Example 18: The sum of first $\mathbf{7}$ odd natural numbers is $\mathbf{49}$.

Ans: True, because first seven odd natural numbers are $=1,3,5,7,9,11\text{and 13}$

And their sum:

$=1+3+5+7+9+11+13$

$=49$

Example 19: The square root of a perfect square of $\mathbf{n}$ digits will have $\frac{\mathbf{n}}{\mathbf{2}}$ digits if $\mathbf{n}$ is even.

Ans: True, as a square root of a perfect square of $\text{n}$digit will have $\frac{\text{n}}{2}$digits.

Example 20: Express \[\mathbf{36}\] as a sum of successive odd natural numbers.

Ans: successive odd natural numbers are $=1,3,5,7,9\text{ and 11}$ and $1+3+5+7+9+11=36$.

Example 21: Check whether $\mathbf{90}$ is a perfect square or not by using prime factorisation.

Ans: Taking factors of $90$:

$\underline{2|90}$

$\underline{3|45}$

$\underline{3|15}$

$\underline{5|5}$

$\underline{|1}$

$90=2\times 3\times 3\times 5$as $2$ and $5$ have incomplete pair. Therefore, $90$ is not a perfect square.

Example 22: Check whether $\mathbf{1728}$ is a perfect cube by using prime factorisation.

Ans: Taking factors of $1728$:

So, $1728=2\times 2\times 2\times 2\times 2\times 2\times 3\times 3\times 3$

As all prime factors forms triplets. So, $1728$ is a perfect cube

Example 23: Using distributive law, find the square of$\mathbf{43}$.

Ans: Finding square using distributive law:

As $43$ can be written as

$43=40+3$

So, $432={{\left( 40+3 \right)}^{2}}=\left( 40+3 \right)\left( 40+3 \right)=40\left( 40+3 \right)+3\left( 40+3 \right)$

$=\text{ }40\times 40+40\times 3+3\times 40+3\times 3$

$=\text{ }1600+240+9$

$=\text{ }1849$

Therefore, \[{{\left( 43 \right)}^{2}}=1849\].

Example 24: Write a Pythagorean triplet whose smallest number is $\mathbf{6}$.

Ans: Given that smallest number is $6$

Pythagorean triplet is given by: $2\text{m,}{{\text{m}}^{2}}-1\text{and}{{\text{m}}^{2}}+1$

Now, $2\text{m}=6\Rightarrow \text{m}=3$

${{\text{m}}^{2}}+1=32+1=9+1=10$

${{\text{m}}^{2}}1=321=91=8$

Therefore, the Pythagorean triplet is$6,8,10$.

Example 25: Using prime factorisation, find the cube root of$\mathbf{5832}$.

Ans: Prime factors of $5832$:

$\underline{2|5832}$

$\underline{2|2916}$

$\underline{2|1458}$

$\underline{3|729}$

$\underline{3|243}$

$\underline{3|81}$

$\underline{3|27}$

$\underline{3|9}$

$\underline{3|3}$

$\underline{|1}$

$5832=2\times 2\times 2\times 3\times 3\times 3\times 3\times 3\times 3$

$5832=\underline{2\times 2\times 2}\times \underline{3\times 3\times 3}\times \underline{3\times 3\times 3}$

$=2\times 3\times 3$

$=18$

Example 26: Evaluate the square root of $\mathbf{22}.\mathbf{09}$by long division method.

Ans: Square root of $22.09$by long division method is$4.7$.

Example 27: Find the smallest perfect square divisible by 3, 4, 5 and 6.

Ans: The least number divisible by $3,\text{ }4,\text{ }5\text{ and }6$ is:

The LCM of $3,\text{ }4,\text{ }5\text{ and }6$:

$\underline{2|3,4,5,6}$

$\underline{2|3,2,5,3}$

$\underline{3|3,1,5,3}$

$\underline{5|1,1,5,1}$

$\underline{|1,1,1,1}$

Now, $60=2\times 2\times 3\times 5.$

As $5\text{ and }3$does not form pairs. So, $60$ is not a perfect square. Hence, $60$ should bemultiplied by $5\times 3=15$to get a perfect square.

So, the required least square number is $=60\times 15=900.$

Example 28: A ladder $\mathbf{10}\mathbf{m}$ long rests against a vertical wall. If the foot of the ladder is 6m away from the wall and the ladder just reaches the top of the wall, how high is the wall?

Ans: Let $\text{BC}$ be the distance between the foot of the ladder and the wall. And $\text{AC}$ be the ladder.

Given: $\text{AC}$ \[=10\text{m}\]and $\text{BC}$ \[=6\text{m}\]

Because $\Delta \text{ABC}$ forms a right-angled triangle, Therefore, using Pythagoras theorem:

$\text{A}{{\text{C}}^{2}}=\text{A}{{\text{B}}^{2}}+\text{B}{{\text{C}}^{2}}$

${{10}^{2}}=\text{A}{{\text{B}}^{2}}+\text{ }{{6}^{2}}$

$\Rightarrow \text{A}{{\text{B}}^{2}}={{10}^{2}}{{6}^{2}}=10036\text{ }=64$

$\Rightarrow \text{AB}=64\text{ }=8\text{m}$

Hence, the wall is $8\text{m}$high.

Example 29: Find the length of a diagonal of a rectangle with dimensions $\mathbf{20}\mathbf{m}$ by $\mathbf{15}\mathbf{m}$.

Ans:

By using Pythagoras theorem, we know

Length of diagonal of the rectangle is given by $=\sqrt{{{l}^{2}}+{{b}^{2}}}$units

$\Rightarrow \sqrt{{{\left( 20 \right)}^{2}}+{{\left( 15 \right)}^{2}}}\text{m}$

$\Rightarrow \sqrt{400+225}\text{m}$

$\Rightarrow \sqrt{625}\text{m}$

$\Rightarrow \text{25}\text{m}$

Thus, the length of diagonal is $25\text{m}$.

Example 30: The area of a rectangular field whose length is twice its breadth is $\mathbf{2450}{{\mathbf{m}}^{\mathbf{2}}}$. Find the perimeter of the field.

Ans: If the breadth of the field is $\text{x}$ metres then the length of the field is $2\text{x}$ metres.

We know, area of the rectangular field $=\text{ length}\times \text{breadth}$

Also given, area is $=\text{ }2450\text{ }{{\text{m}}^{2}}.$

Therefore,

$\left( 2\text{x} \right)\left( \text{x} \right)=2{{\text{x}}^{2}}=\text{ }2450\text{ }{{\text{m}}^{2}}$

$\Rightarrow 2{{\text{x}}^{2}}=2450$

$\Rightarrow {{\text{x}}^{2}}=\frac{2450}{2}=1225\text{m}$

$\Rightarrow \text{ x}\text{=}\sqrt{1225}\text{m}$

$\Rightarrow \text{x}=35\text{m}$

Therefore, breadth $=35\text{m}$ and length \[=35\times 2=70\text{m}\]

Now the, perimeter of the field \[=2\left( l+b \right)\]

$=\text{ }2\left( 70+35 \right)\text{m}$

$=\text{ }2\times 105\text{m }=210\text{m}$

Example 31: During a mass drill exercise, $\mathbf{6250}$ students of different schools are arranged in rows such that the number of students in each row is equal to the number of rows. In doing so, the instructor finds out that $\mathbf{9}$ children are left out. Find the number of children in each row of the square.

Ans: Given total number of students $=\text{ }6250$

students forming a square $=62509=6241$

Thus, $6241$ students form a big square that has the number of rows equal to the number of students in each row.

Now let the number of students in each row be $x$, then the number of rows $=x$

Therefore, $x\times x=6241$

or

${{x}^{2}}=6241$

$x=\sqrt{6241}\text{= }79$

So, there are $79$ students in each row of the square formed.

Example 32: Find the least number that must be added to $\mathbf{1500}$ ;.so as to get a perfect square. Also find the square root of the perfect square.

Ans: 

We can see $1500>{{38}^{2}}$ and  ${{39}^{2}}>1500$

Hence, the number to be added$={{39}^{2}}1500$

   =15211500

   =21 

Therefore, the perfect square $=1500+21=1521$

$\therefore \sqrt{1521}=39$

Thus, the required number is $21$ and the square root is $39$.

Example 33: Application of problem solving strategies

Finding the smallest number by which $1620$ must be divided to get a perfect square.

Ans: Understand and Explore:

• What information is given in the question? – A number which is not a perfect square.

• What are we trying to find? – The smallest number by which $1620$must be divided to get a perfect square. Plan a strategy

• Using prime factorisation to find the product of prime factors of $1620$.

• Pairing the prime factors to see if any factor is left unpaired.

• Then the unpaired factor will be the smallest number that must be divided to get a perfect square.

Solve: Prime factorisation of $1620$ is

$\underline{2|1620}$

$\underline{2|810}$

$\underline{5|405}$

$\underline{3|81}$

$\underline{3|27}$

$\underline{3|9}$

$\underline{3|3}$

$\underline{|1}$

The product of prime factors $=\text{2}\times 2\times 5\times 3\times 3\times 3\times 3$

After Pairing these prime factors $=\underline{\text{2}\times 2}\times 5\times \underline{3\times 3}\times \underline{3\times 3}$

We can see that the factor $5$ is left unpaired. So, the required smallest number is $5$.

To check if it is a perfect square. Divide $1620$ by $5$

$1620\div 5=324$

We can see that on dividing $1620$ by $5$ no remainder is left, therefore, $324$ is a perfect square, hence our answer is verified.

Multiple question Answer

In each of the questions, 1 to 24, write the correct Answer from the given four options:

1. ${\mathbf{196}}$ is the square of 

a) ${\mathbf{11}}$

b) ${\mathbf{12}}$ 

c) ${\mathbf{14}}$

d) ${\mathbf{16}}$

Ans: (c), $14$

Breaking $196$ into factors = $\sqrt {196}  = \sqrt {2 \times 2 \times 7 \times 7}$

Taking pairs out $2 \times 7 = 14$

Therefore, square root of 196 is 14.

2. Which of the following is a square of an even number? 

a) $\mathbf{144}$

b) $\mathbf{169}$

c) $\mathbf{441}$

d) $\mathbf{625}$

Ans: (a), $144$ 

Therefore, square of an even number is:

$144 = 12 \times 12 = {\left( {12} \right)^2}$

3. A number ending in ${\mathbf{9}}$ will have the units place of its square as 

a) ${\mathbf{3}}$

b) ${\mathbf{9}}$

c) ${\mathbf{1}}$

d) ${\mathbf{6}}$

Ans: (c), $1$. 

We know $9 \times 9 = 81$

Therefore, unit place of a number ending with $9$ is $1$.

4. Which of the following will have 4 at the unit place? 

a) ${\mathbf{1}}{{\mathbf{4}}^{\mathbf{2}}}$

b)${\mathbf{6}}{{\mathbf{2}}^{\mathbf{2}}}$

c) ${\mathbf{2}}{{\mathbf{7}}^2}$

d) ${\mathbf{3}}{{\mathbf{5}}^2}$

Ans: (b), ${62^2}$

Because, unit digit of ${62^2} = 3844$

Square of $2 = 4$.

5. How many natural numbers lie between ${{\mathbf{5}}^{\mathbf{2}}}$ and ${{\mathbf{6}}^{\mathbf{2}}}$? 

a) 9 

b) 10 

c) 11 

d) 12 

Ans: (b), 10.

We know, natural numbers lie between ${5^2} = 25$ and ${6^2} = 36$ are 26, 27, 28, 29, 30, 

31, 32, 33, 34, 35. So there are $10$ natural numbers.

6. Which of the following cannot be a perfect square? 

a) ${\mathbf{841}}$

b) ${\mathbf{529}}$

c) ${\mathbf{198}}$

d) All of the above 

Ans: (c), 198.

$\sqrt {198}  = \sqrt {2 \times 3 \times 3 \times 11}  = 3\sqrt {22}$ because no perfect pairs can be taken out therefore, $198$ can’t be a perfect square.

7. The one’s digit of the cube of 23 is 

a) ${\mathbf{6}}$ 

b) ${\mathbf{7}}$

c) ${\mathbf{3}}$ 

d) ${\mathbf{9}}$ 

Ans: (b), $7$

We know, unit digit of $23 = 3$

And cube of $3 = 27$

8. A square board has an area of ${\mathbf{144}}$ square units. How long is each side of the board? 

a) ${\mathbf{11}}$ units 

b) ${\mathbf{12}}$ units 

c) ${\mathbf{13}}$ units 

d) ${\mathbf{14}}$ units 

Ans: (b), $12$ units

We know, $144$ is square of $12$${\text{i}}{\text{.e}}$$12 \times 12$ unit.

9. Which letter best represents the location of $25$ on a number line? 

a) A 

b) B 

c) C 

d) D

Ans: (c), C

We know, square of $5 = 5 \times 5 = 25$

10. If one member of a Pythagorean triplet is ${\mathbf{2}}$m, then the other two members are 

a) ${{\mathbf{m}}^{\mathbf{2}}},{{\mathbf{m}}^{\mathbf{2}}} + {\mathbf{1}}$

b) ${{\mathbf{m}}^{\mathbf{2}}} + {\mathbf{1}},{{\mathbf{m}}^{\mathbf{2}}} - {\mathbf{1}}$

c) ${{\mathbf{m}}^{\mathbf{2}}},{{\mathbf{m}}^{\mathbf{2}}} - {\mathbf{1}}$

d) ${{\mathbf{m}}^2},{\mathbf{m}} + {\mathbf{1}}$

Ans: (b), ${{\text{m}}^2} + 1,{{\text{m}}^2} - 1$

Therefore, the formula of Pythagorean triplet is:

${{\text{m}}^2} + 1,{{\text{m}}^2} - 1$.

11. The sum of successive odd numbers ${\mathbf{1}}$,${\mathbf{3}}$,${\mathbf{5}}$,${\mathbf{7}}$,${\mathbf{9}}$,${\mathbf{11}}$,${\mathbf{13}}$ and ${\mathbf{15}}$ is 

a) ${\mathbf{81}}$

b) ${\mathbf{64}}$

c) ${\mathbf{49}}$

d) ${\mathbf{36}}$

Ans: (b), $64$

Total numbers are $ = 8$

successive odd numbers $ = 1,3,5,7,9,11,13,15$

sum of successive odd numbers is $= $ square of $8 = 8 \times 8 = 64$

12. The sum of first n odd natural numbers is 

a) ${\mathbf{2n}} + {\mathbf{1}}$

b) ${{\mathbf{n}}^{\mathbf{2}}}$

c) ${{\mathbf{n}}^{\mathbf{2}}} - {\mathbf{1}}$

d) ${{\mathbf{n}}^{\mathbf{2}}} + {\mathbf{1}}$

Ans: (b), ${{\text{n}}^2}$

We know, sum of first n odd natural numbers is given by: 

$\Sigma (2{\text{n}} - {\text{1)}}{\text{ = }}{\text{2}}\Sigma {\text{n}} - {\text{n}}$

$\Rightarrow \frac{{2 \times {\text{n(n + 1)}}}}{2} - {\text{n}}$

$\Rightarrow {\text{n(n + 1)}} - {\text{n}}$

$\Rightarrow {{\text{n}}^2} + {\text{n}} - {\text{n}}$

$\Rightarrow {{\text{n}}^2}$ 

13. Which of the following numbers is a perfect cube? 

a) ${\mathbf{243}}$

b)${\mathbf{216}}$

c) ${\mathbf{392}}$

d) ${\mathbf{8640}}$

Ans:(b), $216$

As square of 6 is $ = 6 \times 6 = 36$

14. The hypotenuse of a right triangle with its legs of lengths ${\mathbf{3}}x \times {\mathbf{4}}x$ is 

a) ${\mathbf{5}}x$

b) ${\mathbf{7}}x$

c) ${\mathbf{16}}x$

d) ${\mathbf{25}}x$

Ans: (a), $5x$

We know,

${{{({\text{Hypotenuse)}}}^2} = {{\left( {3x} \right)}^2} + {{\left( {4x} \right)}^2}}$ 

${ = {\text{ }}9{x^2} + 16{x^2}}$ 

${ = {\text{ }}25{x^2}}$ 

${\text{Hypotenuse}}{\text{ = }}\sqrt {25{x^2}}  = 5x$

15. The next two numbers in the number pattern ${\mathbf{1}},{\text{ }}{\mathbf{4}},{\text{ }}{\mathbf{9}},{\text{ }}{\mathbf{16}},{\text{ }}{\mathbf{25}}...$are 

a) ${\mathbf{35}},{\mathbf{48}}$

b) ${\mathbf{36}},{\mathbf{49}}$

c) ${\mathbf{36}},{\mathbf{38}}$

d) ${\mathbf{35}},{\mathbf{49}}$

Ans: (b), $36,49$

As the given pattern is a series of perfect squares so the next term will be $36,49$.

16. Which among ${\mathbf{4}}{{\mathbf{3}}^2}$,${\mathbf{6}}{{\mathbf{7}}^2}$,${\mathbf{5}}{{\mathbf{2}}^2}$,${\mathbf{5}}{{\mathbf{9}}^2}$ would end with digit ${\mathbf{1}}$? 

a) ${\mathbf{4}}{{\mathbf{3}}^2}$

b) ${\mathbf{6}}{{\mathbf{7}}^2}$

c) ${\mathbf{5}}{{\mathbf{2}}^2}$

d) ${\mathbf{5}}{{\mathbf{9}}^2}$

Ans: (d), ${59^2}$

Because unit digit of $59$ is $9$

And square of $9 \times 9 = 81$

17. A perfect square can never have the following digit in its ones place. 

a) ${\mathbf{1}}$

b) ${\mathbf{8}}$

c) ${\mathbf{0}}$

d) ${\mathbf{6}}$

Ans: (b), $8$.

$2,3,7,8$ are the numbers perfect square ends.

18. Which of the following numbers is not a perfect cube? 

a) ${\mathbf{216}}$

b) ${\mathbf{567}}$

c) ${\mathbf{125}}$

d) ${\mathbf{343}}$

Ans: (b), $567$.

Factors of $567$are $3 \times 3 \times 3 \times 3 \times 7$which do not form perfect pairs for cube. So $567$is not a perfect cube.

$\underline {3|567}$

$\underline {3|189}$

$\underline {3|63}$

$\underline {3|21}$

$\underline {7|7}$

$\underline {|1}$ 

19. $\sqrt[{\mathbf{3}}]{{{\mathbf{1000}}}}$ is equal to 

a) ${\mathbf{10}}$

b) ${\mathbf{100}}$

c) ${\mathbf{1}}$

d) None of these

Ans: (a), $10$

$\sqrt[3]{{1000}} = \sqrt[3]{{10 \times 10 \times 10}} = 10$

20. If m is the square of a natural number n, then n is 

a) The square of m 

b) Greater than m 

c) Equal to m 

d) $\sqrt {\mathbf{m}}$

Ans: (d), $\sqrt {\text{m}}$

Square of ${\text{n}} = {\left( {\text{n}} \right)^2}$

According to the question:

${{\text{n}}^2} = {\text{m}}$

$\Rightarrow {\text{n}}{\text{ = }}\sqrt {\text{m}}$

21. A perfect square number having n digits where n is even will have square root with

a) ${\mathbf{n}} + {\mathbf{1}}$ digit 

b) $\frac{{\mathbf{n}}}{{\mathbf{2}}}$ digit

c) $\frac{{\mathbf{n}}}{{\mathbf{3}}}$ digit

d) $\frac{{{\mathbf{n}} + {\mathbf{1}}}}{{\mathbf{2}}}$digit

Ans: (b), $\frac{{\text{n}}}{2}$ digit

22. If m is the cube root of n, then n is

a) ${{\mathbf{m}}^3}$

b) $\sqrt {\mathbf{m}}$

c) $\frac{{\mathbf{m}}}{{\mathbf{3}}}$

d) $\sqrt[3]{{\mathbf{m}}}$

Ans: (a), ${{\text{m}}^3}$

${\text{m}} \times {\text{m}} \times {\text{m}}{\text{ =  }}{{\text{m}}^3}$

$\Rightarrow {{\text{m}}^3} = {\text{n}}$ 

23. The value of $\sqrt {248 + \sqrt {52 + \sqrt {144} } } $ is

a) ${\mathbf{14}}$

b) ${\mathbf{12}}$

c) ${\mathbf{16}}$

d) ${\mathbf{13}}$

Ans: (c), $16$

$\sqrt {248 + \sqrt {52 + \sqrt {144} } }$

$\Rightarrow \sqrt {248 + \sqrt {52 + 12} }$

$\Rightarrow \sqrt {248 + \sqrt {64} }$

$\Rightarrow \sqrt {248 + 8}$

$\Rightarrow \sqrt {256}$

$\Rightarrow \sqrt {16 \times 16}  = 16$ 

$\because \sqrt {144}  = 12$

$\sqrt {64}  = 8$ 

24. Given that $\sqrt {{\mathbf{4096}}}  = {\mathbf{64}}$, the value of $\sqrt {{\mathbf{4096}}}  + \sqrt {{\mathbf{40}}.{\mathbf{96}}}$ is 

a) ${\mathbf{74}}$

b) ${\mathbf{60}}.{\mathbf{4}}$

c) ${\mathbf{64}}.{\mathbf{4}}$

d)${\mathbf{70}}.{\mathbf{4}}$

Ans: (d), $70.4$

As $\sqrt {4096}  = 64$

Also,$40.96$ can be written as $\frac{{1024}}{{25}}$

And, $\sqrt {\frac{{1024}}{{25}}}  = \sqrt {\frac{{32 \times 32}}{{5 \times 5}}}  = \frac{{32}}{5}$

Then,

$\sqrt {4096}  + \sqrt {40.64}$

$\Rightarrow 64 + \frac{{32}}{5}$

$\Rightarrow 64 + 6.4$

$\Rightarrow 70.4$ 

In questions 25 to 48, fill in the blanks to make the statements true. 

25. There are _________ perfect squares between ${\mathbf{1}}$ and ${\mathbf{100}}$. 

Ans: $8$

Perfect square between $1$ and $100$are:

$2 \times 2 = 45 \times 5 = 258 \times 8 = 64$

$3 \times 3 = 96 \times 6 = 369 \times 9 = 81$

$4 \times 4 = 167 \times 7 = 49$

26. There are _________ perfect cubes between ${\mathbf{1}}$ and ${\mathbf{1000}}$. 

Ans: 8

There are $8$ perfect cubes between $1$and $1000$

$2 \times 2 \times 2 = 85 \times 5 \times 5 = 1258 \times 8 \times 8 = 512$

$3 \times 3 \times 3 = 276 \times 6 \times 6 = 369 \times 9 \times 9 = 729$

$4 \times 4 \times 4 = 647 \times 7 \times 7 = 343$ 

27. The unit digit in the square of ${\mathbf{1294}}$ is _________.

Ans: $6$

Unit digit in $1294$ = 4

$4 \times 4 = 16$

Therefore, the square of $1294$ will have $6$ at its unit place.

28. The square of ${\mathbf{500}}$ will have _________ zeroes. 

Ans: $4$

$500 \times 500 = 250000$

29. There are _________ natural numbers between ${{\mathbf{n}}^2}$ and ${\left( {{\mathbf{n}} + {\mathbf{1}}} \right)^2}$

Ans: $2{\text{n}}$

$[{({\text{n + 1)}}^2} - {{\text{n}}^2}] - 1 = [{{\text{n}}^2} + 2{\text{n}}{\text{ + 1}} - {{\text{n}}^2}] - 1$

$\Rightarrow {{\text{n}}^2}{\text{ + }}{\text{2n  + 1}} - {{\text{n}}^2} - 1$

$\Rightarrow {\text{2n + 1}} - 1$

$\Rightarrow {\text{2n }}$ 

30. The square root of ${\mathbf{24025}}$ will have _________ digits. 

Ans: $3$

Number of digits in a square root for odd digit is given by $= \frac{{{\text{n}} + 1}}{2}$

Number of digits in a given number is $5({\text{odd)}}$

Therefore

${\text{n  =  5}}$

$\therefore \frac{{{\text{n + 1}}}}{2} = \frac{{5 + 1}}{2} = \frac{6}{2} = 3$ 

31. The square of ${\mathbf{5}}.{\mathbf{5}}$ is _________. 

Ans: $30.25$

$5.5 \times 5.5 = 30.25$

32. The square root of ${\mathbf{5}}.{\mathbf{3}} \times {\mathbf{5}}.{\mathbf{3}}$ is _________. 

Ans: 28.09

$5.3 \times 5.3 = 28.09$

33. The cube of ${\mathbf{100}}$ will have _________ zeroes. 

Ans: $6$

${100^3} = 100 \times 100 \times 100 = 1000000$

34. ${\mathbf{1}}{{\mathbf{m}}^{\mathbf{2}}} =$ _________ ${\mathbf{c}}{{\mathbf{m}}^{\mathbf{2}}}$.

Ans: $10000$

$1{\text{m}}{\text{ = 100 cm}}$

$\therefore {\text{1}}{{\text{m}}^2} = 100{\text{cm}} \times {\text{100cm}}$

${\text{ = }}{\text{10000c}}{{\text{m}}^2}$ 

35. ${\mathbf{1}}{{\mathbf{m}}^{\mathbf{3}}} =$ _________ ${\mathbf{c}}{{\mathbf{m}}^{\mathbf{3}}}$. 

Ans: $1000000$

$1{\text{m}}{\text{ = 100 cm}}$

$\therefore {\text{1}}{{\text{m}}^3} = 100{\text{cm}} \times {\text{100cm}} \times 100{\text{cm}}$

${\text{ = }}{\text{1000000c}}{{\text{m}}^3}$ 

36. Ones digit in the cube of ${\mathbf{38}}$ is _________. 

Ans: $2$

Unit digit is $8$ in $38$

$\therefore 8 \times 8 \times 8 = 512$

Therefore, unit digit in the cube of $38$ is $2$.

37. The square of ${\mathbf{0}}.{\mathbf{7}}$ is _________. 

Ans: $0.49$

Therefore, square of $0.7$ is

$0.7 \times 0.7 = 0.49$

38. The sum of first six odd natural numbers is _________. 

Ans: $36$

First six odd natural numbers are $1,3,5,7,9,11$

Sum of First six odd natural numbers are $= 1 + 3 + 5 + 7 + 9 + 11$

$= 36$

39. The digit at the ones place of ${\mathbf{5}}{{\mathbf{7}}^{\mathbf{2}}}$ is _________. 

Ans: $9$

As unit digit of $57$ is $7$

$\therefore 7 \times 7 = 49$

40. The sides of a right triangle whose hypotenuse is ${\mathbf{17cm}}$ are _________ and _________.

Ans: $8$ and v $15$

For Pythagorean triplet sides are given by $2{\text{m,}}{{\text{m}}^2} - 1,{{\text{m}}^2} + 1$

${{\text{m}}^2} + 1 = {(2{\text{m)}}^2} + {({{\text{m}}^2} - 1)^2}$

$\because {{\text{m}}^2} + 1 = 17$

$\Rightarrow {{\text{m}}^2} = 17 - 1$

$\Rightarrow {{\text{m}}^2} = 16$

$\Rightarrow {\text{m}}{\text{ =  4}}$ 

Then,

$2{\text{m}}{\text{ = }}{\text{2}} \times {\text{4 = 8}}$

${{\text{m}}^2} - 1 = {\left( 4 \right)^2} - 1$

$\Rightarrow 16 - 1 = 15$ 

41. $\sqrt {{\mathbf{1}}.{\mathbf{96}}}$______. 

Ans: $1.4$

$1.96$ can be written as $\frac{{196}}{{100}}$

$\therefore \sqrt {\frac{{196}}{{100}}}  = \sqrt {\frac{{14 \times 14}}{{10 \times 10}}}  = \frac{{14}}{{10}} = 1.4$

42. ${\left( {{\mathbf{1}}.{\mathbf{2}}} \right)^{\mathbf{3}}} =$_____.

Ans: $1.728$

$1.2$ can be written as $\frac{{12}}{{10}}$

$\therefore {\left( {1.2} \right)^3} = \frac{{12}}{{10}} \times \frac{{12}}{{10}} \times \frac{{12}}{{10}} = \frac{{1728}}{{1000}} = 1.728$

43. The cube of an odd number is always an _________ number. 

Ans: always an odd number

44. The cube root of a number x is denoted by _________. 

Ans:  $\sqrt[3]{x}$

45. The least number by which ${\mathbf{125}}$ be multiplied to make it a perfect square is _____________. 

Ans: 5

$\underline {5|125}$

$\underline {5|25}$

$\underline {5|5}$

$\underline {|1}$ 

On taking L.C.M we have factors of $125 = 5 \times 5 \times 5$

On grouping these factors in double of equal factors,

$125 = 5 \times 5 \times 5$

We have only one $5$ without a pair. 

So, $125$ to be multiplied by $5$ to make it a perfect square. 

46. The least number by which ${\mathbf{72}}$ be multiplied to make it a perfect cube is _____________. 

Ans: $3$

$\underline {2|72}$

$\underline {2|36}$

$\underline {2|18}$

$\underline {3|9}$

$\underline {3|3}$

$\underline {|1}$ 

On taking L.C.M we have factors of $72 = 2 \times 2 \times 2 \times 3 \times 3$

On grouping these factors in triple of equal factors,

$72 = 2 \times 2 \times 2 \times 3 \times 3$

We have only one $3$ without a pair. 

So, $72$ to be multiplied by $3$ to make it a perfect cube.

47. The least number by which 72 be divided to make it a perfect cube is _____________. 

Ans: $9$

$\underline {2|72}$

$\underline {2|36}$

$\underline {2|18}$

$\underline {3|9}$

$\underline {3|3}$

$\underline {|1}$ 

On taking L.C.M we have factors of $72 = 2 \times 2 \times 2 \times 3 \times 3$

On grouping these factors in triple of equal factors,

$72 = 2 \times 2 \times 2 \times 3 \times 3$

We have only $3$ which is not a triplet. 

So, $72$ to be divided by $9$ to make it a perfect cube ${\text{i}}{\text{.e}}$$8$

48. Cube of a number ending in ${\mathbf{7}}$ will end in the digit _______________.

Ans: $3$

On taking the cube of number $7$ itself we have,

$7 \times 7 \times 7 = 343$

Therefore, we have $3$ at the end.

In questions 49 to 86, state whether the statements are true (T) or false (F). 

49. The square of ${\mathbf{86}}$ will have ${\mathbf{6}}$ at the units place. 

Ans: The given statement is true.

Because square of the number ending with $4{\text{or 6}}$ have $6$ at unit place.

50. The sum of two perfect squares is a perfect square. 

Ans: The given statement is false.

For example: taking two perfect squares, 

Here, 

$4 + 9 = 13$

So, $13$ is not a perfect square.

51. The product of two perfect squares is a perfect square.

Ans: The given statement is true.

For example, taking two perfect squares, 

Here, 

$4 \times 9 = 36$

So, $36$ is a perfect square.

52. There is no square number between ${\mathbf{50}}$ and ${\mathbf{60}}.$

Ans: The given statement is true.

53. The square root of ${\mathbf{1521}}$ is ${\mathbf{31}}$. 

Ans: The given statement is false.

because square of $31 = {\left( {31} \right)^2} = 31 \times 31 = 961$

54. Each prime factor appears ${\mathbf{3}}$ times in its cube. 

Ans: The given statement is true.

As cubes are represented as the product of triplets of prime factors.

55. The square of ${\mathbf{2}}.{\mathbf{8}}$ is ${\mathbf{78}}.{\mathbf{4}}$. 

Ans: The given statement is false.

because square of $2.8 = {\left( {2.8} \right)^2}$

$= 2.8 \times 2.8 = 7.84$

56. The cube of ${\mathbf{0}}.{\mathbf{4}}$ is ${\mathbf{0}}.{\mathbf{064}}$. 

Ans: The given statement is true.

${\left( {0.4} \right)^3} = \frac{{4 \times 4 \times 4}}{{10 \times 10 \times 10}} = \frac{{64}}{{1000}} = 0.064$

57. The square root of ${\mathbf{0}}.{\mathbf{9}}$ is ${\mathbf{0}}.{\mathbf{3}}$. 

Ans: The given statement is false.

because square of $0.3 = {\left( {0.3} \right)^2}$

= $0.3 \times 0.3 = 0.09$

58. The square of every natural number is always greater than the number itself. 

Ans: The given statement is false.

Because the square of ${\left( 1 \right)^2} = 1$which is not greater than $1$

59. The cube root of ${\mathbf{8000}}$ is ${\mathbf{200}}$. 

Ans: The given statement is false.

 As, $\sqrt[3]{{8000}} = \sqrt {2 \times 2 \times 2 \times 10 \times 10 \times 10}  = 2 \times 10 = 20$

60. There are five perfect cubes between ${\mathbf{1}}$ and ${\mathbf{100}}$. 

Ans: The given statement is false.

There are 8 perfect cubes between $1$ and $100$

$2 \times 2 = 45 \times 5 = 258 \times 8 = 64$

$3 \times 3 = 96 \times 6 = 369 \times 9 = 81$

$4 \times 4 = 167 \times 7 = 49$

61. There are ${\mathbf{200}}$ natural numbers between ${\mathbf{10}}{{\mathbf{0}}^{\mathbf{2}}}$ and${\mathbf{10}}{{\mathbf{1}}^{\mathbf{2}}}$.

Ans: The given statement is true.

${\left( {101} \right)^2} = 10201$

${\left( {100} \right)^2} = 10000$

$\therefore {\left( {101} \right)^2} - {\left( {100} \right)^2} = 10201 - 10000 = 201$ 

62. The sum of first ${\text{n}}$odd natural numbers is ${{\text{n}}^{\mathbf{2}}}$. 

Ans: The given statement is true.

Sum of first ${\text{n}}$ odd natural number is given by:

$\Rightarrow \Sigma \left( {2{\text{n - 1}}} \right)$

$\Rightarrow \left( {2 \times {\text{n}} \times \left( {{\text{n + 1}}} \right)} \right)/2 - {\text{n}}$

$\Rightarrow {{\text{n}}^2} + {\text{n}} - {\text{n}}$

$\Rightarrow {{\text{n}}^2}$

63. ${\mathbf{1000}}$ is a perfect square. 

Ans: The given statement is false.

$\underline {2|1000}$

$\underline {2|500}$

$\underline {2|250}$

$\underline {5|125}$

$\underline {5|25}$

$\underline {5|5}$

$\underline {|1}$ 

Factors of $1000$

$2 \times 2 \times 2 \times 5 \times 5 \times 5 = 1000$, So clearly $1000$ is a perfect cube but not square.

So, 1000 is not a perfect square. 

64. A perfect square can have ${\mathbf{8}}$ as its unit’s digit. 

Ans: The given statement is false.

Because for every digit that ends with an even number either has $4$ or $6$ at its unit place.

65. For every natural number ${\mathbf{m}},\left( {{\mathbf{2m}}-{\mathbf{1}},{\mathbf{2}}{{\mathbf{m}}^{\mathbf{2}}}-{\mathbf{2m}},{\mathbf{2}}{{\mathbf{m}}^{\mathbf{2}}}-{\mathbf{2m}} + {\mathbf{1}}} \right)$ is a Pythagorean triplet. 

Ans: The given statement is false.

$\left( {2{\text{m}},{{\text{m}}^2} - 1{\text{, }}{{\text{m}}^2}{\text{ + 1}}} \right)$ is a Pythagorean triplet for every natural number ${\text{m}}{\text{ > 1}}$.

66. All numbers of a Pythagorean triplet are odd. 

Ans: The given statement is false.

Condition for Pythagorean triplet is = square of one should be equal to sum of square of other two.

For example, ${3^2} = {5^2} + {4^2}$

Hence, $4$ is not an odd number.

67. For an integer ${\mathbf{a}}$, ${{\mathbf{a}}^{\mathbf{3}}}$ is always greater than ${{\text{a}}^{\mathbf{2}}}$. 

Ans: The given statement is false. 

If ${\text{a}}$ is a negative integer then ${{\text{a}}^{3}} < {{\text{a}}^2}$.

68. If $X$ and $Y$ are integers such that ${X^2} > {Y^2}$, then${X^3} > {Y^3}$. 

Ans: The given statement is false.

If $X$ is a negative integer then ${X^{3}} < {Y^3}$

Suppose, $X =  - 1$ then,

${X^2} = {\left( { - 1} \right)^2} = 1$

${X^3} = {\left( { - 1} \right)^3} =  - 1$ 

And $Y = 2$ then,

${\left( Y \right)^2} = {\left( 2 \right)^2} = 4$

${\left( Y \right)^3} = {\left( 2 \right)^3} = 8$ 

Therefore, clearly ${X^{3}} < {Y^3}$.

69. Let $x$ and $y$ be natural numbers. If  $x$ divides $y$ , then ${{\mathbf{x}}^3}$ divides ${{\mathbf{y}}^3}$. 

Ans: The given statement is true.

If $x$ and $y$ are natural numbers then ${x^3}$ and ${y^3}$ will also be natural numbers as they are nothing but multiple of $x$ and $y$ respectively.

70. If ${{\mathbf{a}}^{\mathbf{3}}}$ends in ${\mathbf{5}}$, then  ${{\mathbf{a}}^{\mathbf{3}}}$ends in ${\mathbf{25}}$.

Ans: The given statement is false. 

If ${{\text{a}}^2} = {35^2} = 1225$

Then,

${{\text{a}}^3} = {\left( {35} \right)^3} = 42875$

71. If ${{\mathbf{a}}^{\mathbf{2}}}$ ends in ${\mathbf{9}}$, then ${{\mathbf{a}}^{\mathbf{3}}}$ ends in ${\mathbf{7}}$.

Ans: The given statement is false.

If ${{\text{a}}^2} = {7^2} = 49$

Then,

${{\text{a}}^3} = {7^3} = 343$

72. The square root of a perfect square of ${\mathbf{n}}$ digits will have $\frac{{{\mathbf{n}} + {\mathbf{1}}}}{{\mathbf{2}}}$ digits, if ${\mathbf{n}}$is odd. 

Ans: The given statement is true.

For example, $10000$ is a perfect square with $5$ digit then its square root will have 

$\Rightarrow \frac{{{\text{n + 1}}}}{2} = \frac{{5 + 1}}{2} = \frac{6}{2} = 3$ which is true as we know $10000$ is a perfect square of $100$ which is a $3$ digit number.

73. Square root of a number ${\mathbf{x}}$ is denoted by $\sqrt {\mathbf{x}}$.

Ans: The given statement is true.

74. A number having ${\mathbf{7}}$ at its one’s place will have ${\mathbf{3}}$ at the unit place of its square.

Ans: The given statement is false.

As square of $7 = 7 \times 7 = 49$

Square of $27 = 27 \times 27 = 729$

Square of $47 = 47 \times 47 = 2209$

And so on.

75. A number having ${\mathbf{7}}$ at its one’s place will have ${\mathbf{3}}$ at the unit place of its cube.

Ans: The given statement is true.

As cube of $17 = 17 \times 17 \times 17 = 4913$

cube of $27 = 27 \times 27 \times 27 = 19683$

cube of $37 = 37 \times 37 \times 37 = 50,653$

And so on. 

76. The cube of a one-digit number cannot be a two-digit number.

Ans: The given statement is false.

e.g. $5$ is a one-digit number, and ${\left( 5 \right)^3} = 125$ (Two-digit number)

77. Cube of an even number is odd.

Ans: The given statement is false.

For example: 

${\left( 2 \right)^3} = 8$

${\left( 6 \right)^3} = 216$ 

Therefore, cube of an even number is always an even number.

78. Cube of an odd number is even.

Ans: The given statement is false

For example:

${\left( 3 \right)^3} = 27$

${\left( 5 \right)^3} = 125$ 

Therefore, cube of an odd number is always an odd number.

79. Cube of an even number is even.

Ans: The given statement is true.

For example: 

${\left( 2 \right)^3} = 8$

${\left( 6 \right)^3} = 216$ 

80. Cube of an odd number is odd.

Ans: The given statement is true.

For example:

${\left( 3 \right)^3} = 27$

${\left( 5 \right)^3} = 125$ 

81. ${\mathbf{999}}$ is a perfect cube.

Ans: The given statement is false.

$\underline {3|999}$

$\underline {3|333}$

$\underline {3|111}$

$\underline {37|37}$

$\underline {|1}$ 

Factors of 999 = $3 \times 3 \times 3 \times 37$

Therefore, $999$ is not a perfect cube.

82. ${\mathbf{363}} \times {\mathbf{81}}$ is a perfect cube.

Ans: The given statement is false.

Breaking $363 \times 81$into factors we have:

$= 363 \times 81$

$= 3 \times 11 \times 11 \times 3 \times 3 \times 3 \times 3$ 

There is only one triplet pair therefore, $363 \times 81$ is not a perfect cube.

83. Cube roots of ${\mathbf{8}}$ are $ + {\mathbf{2}}$ and $-{\mathbf{2}}$.

Ans: The given statement is false.

Because $\sqrt[3]{8} = 2$

84.$\sqrt[3]{{{\mathbf{8}} + {\mathbf{27}}}}{\text{  = }}\sqrt[{\mathbf{3}}]{{\mathbf{8}}} + \sqrt[{\mathbf{3}}]{{{\mathbf{27}}}}$

Ans: The given statement is false.

As $\sqrt[3]{{8 + 27}} = \sqrt[3]{8} + \sqrt[3]{{27}}$

$\Rightarrow \sqrt[3]{{35}} = 2 + 3$

$\Rightarrow \sqrt[3]{{35}} = 5$ 

Which is not equal.

85. There is no cube root of a negative integer.

Ans: The given statement is false.

Let, $-8$ is a negative number then,

$\sqrt[3]{{ - 8}} =  - 2$.

86. Square of a number is positive, so the cube of that number will also be positive.

Ans: The given statement is false.

Let, $-2$ is negative number.

Square of $-2 = 4$

Cube of $- 2 =  - 8$

Solve the Following Questions

87. Write the first five square numbers.

Ans: First five square numbers are:

${\left( 1 \right)^2} = 1,{\left( 2 \right)^2} = 4,{\left( 3 \right)^2} = 9,{\left( 4 \right)^2} = 16,{\left( 5 \right)^2} = 25$

88. Write cubes of first three multiples of ${\mathbf{3}}$.

Ans: $3,6$ and$9$ are the first three multiple of $3$ then,

Cube of $3 = 3 \times 3 \times 3 = 27$

Cube of $6 = 6 \times 6 \times 6 = 216$

Cube of $9 = 9 \times 9 \times 9 = 729$

89. Show that ${\mathbf{500}}$ is not a perfect square.

Ans: $500$ is not a perfect square because

$\sqrt {500}  = \sqrt {\underline {10 \times 10}  \times 5}$

There is no pair for $5$.

90. Express ${\mathbf{81}}$ as the sum of first nine consecutive odd numbers.

Ans: $1,3,5,7,9,11,13,15$ and $17$ are the first nine consecutive odd numbers

Now, Sum of first nine consecutive odd numbers

$= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 = 81$

91. Using prime factorisation, find which of the following perfect squares are.

a) ${\mathbf{484}}$

b) ${\mathbf{11250}}$

c) ${\mathbf{841}}$

d) ${\mathbf{729}}$

Ans: 

1. $484$

$\underline {2|484}$

$\underline {2|242}$

$\underline {11|121}$

$\underline {11|11}$

$\underline {|1}$ 

On grouping in pairs, there is no unpaired factor left 

$484 = 2 \times 2 \times 11 \times 11$

So, $484$ is a perfect square.

2. $11250$

$\underline {2|11250}$

  $\underline {3|5625}$

$\underline {3|1875}$

$\underline {5|625}$

$\underline {5|125}$

  $\underline {5|25}$

$\underline {5|5}$

$\underline {|1}$ 

$11250{\text{ }} = {\text{ }}2 \times \underline {3 \times 3}  \times \underline {5 \times 5}  \times \underline {5 \times 5}$ on grouping in pair, $2$ have zero pair.

Therefore, $11250$ is not a perfect square.

3. $841$

$\underline {29|841}$

   $\underline {29|29}$

   $\underline {|1}$ 

\[841 = 29 \times 29\]

On grouping in pair, there is no unpaired factor left over. Therefore, $841$ is a perfect square.

4. $729$

$\underline {3|729}$

$\underline {3|243}$

   $\underline {3|81}$

   $\underline {3|27}$

   $\underline {3|9}$

   $\underline {3|3}$

   $\underline {|1}$ 

$729 = \underline {3 \times 3}  \times \underline {3 \times 3}  \times \underline {3 \times 3}$

On grouping in pair, there is no unpaired factor left over. Therefore, $729$ is a perfect square.

92. Using prime factorisation, find which of the following are perfect cubes.

a) ${\mathbf{128}}$

b) ${\mathbf{343}}$

c) ${\mathbf{729}}$

d) ${\mathbf{1331}}$

Ans: 

1. $128$

   $\underline {2|128}$

   $\underline {2|64}$

   $\underline {2|32}$

   $\underline {2|16}$

   $\underline {2|8}$

   $\underline {2|4}$

   $\underline {2|2}$

   $\underline {|1}$ 

Hence, factors of $128{\text{ }} = {\text{ }}\underline {2 \times 2 \times 2}  \times \underline {2 \times 2 \times 2}  \times 2$ on grouping in triplets for perfect cube, $2$ does not form triplet. Therefore, $128$is not a perfect cube.

2. $343$

   $\underline {7|343}$

   $\underline {7|49}$

   $\underline {7|7}$

   $\underline {|1}$ 

Hence, factors of $343 = \underline {7 \times 7 \times 7}$ on grouping in triplets for perfect cube, $7$ forms triplet. Therefore, $343$is a perfect cube.

3. $729$

    $\underline {3|729}$

   $\underline {3|243}$

   $\underline {3|81}$

   $\underline {3|27}$

   $\underline {3|9}$

   $\underline {3|3}$

   $\underline {|1}$ 

Hence, factors of $729 = \underline {3 \times 3 \times 3}  \times \underline {3 \times 3 \times 3}$ on grouping in triplets for perfect cube, $3$ forms triplet. Therefore, $729$ is a perfect cube

4. $1331$

        $\underline {11|1331}$

   $\underline {11|121}$

   $\underline {11|11}$

   $\underline {|1}$ 

Hence, factors of $1331 = \underline {11 \times 11 \times 11}$ on grouping in triplets for perfect cube, $11$ forms triplet. Therefore, $1331$ is a perfect cube.

93. Using distributive law, find the squares of

a) ${\mathbf{101}}$

b) ${\mathbf{72}}$

Ans: 

1. $101$

Finding square using distributive law:

$101 = 100 + 1$

Therefore, ${\left( {101} \right)^2} = {\left( {100 + 1} \right)^2}$

$\Rightarrow \left( {100 + 1} \right)\left( {100 + 1} \right)$

$\Rightarrow \left( {100 \times 100} \right) + 100 + 100 + \left( {1 \times 1} \right)$

    $\Rightarrow 10000 + 200 + 1$

    $\Rightarrow 10201$ 

$\therefore {\left( {101} \right)^2} = 10201$

2. $72$

Finding square using distributive law:

$72 = 70 + 2$

Therefore, ${\left( {72} \right)^2} = {\left( {70 + 2} \right)^2}$

$\Rightarrow \left( {70 + 2} \right)\left( {70 + 2} \right)$

$\Rightarrow \left( {70 \times 70} \right) + \left( {70 \times 2} \right) + \left( {70 \times 2} \right) + \left( {2 \times 2} \right)$

$\Rightarrow 4900 + 140 + 140 + 4$

  $\Rightarrow 5184$ 

$\therefore {\left( {72} \right)^2} = 5184$

94. Can a right triangle with sides ${\mathbf{6cm}}$, ${\mathbf{10cm}}$ and ${\mathbf{8cm}}$ be formed? Give reason.

Ans: We know the sum of two smaller sides of a triangle is always equal to the square of longer side of a triangle. 

Given, Smaller sides $= 6{\text{cm}}$ and $8{\text{cm}}$

Longer side $ = 10{\text{cm}}$

As per the rule,

${\left( {10} \right)^2} = {\left( 6 \right)^2} + {\left( 8 \right)^2}$

$100 = 36 + 64$

$100 = 100$ 

Hence, ${\text{L}}{\text{.H}}{\text{.S}}{\text{ = }}{\text{R}}{\text{.H}}{\text{.S}}$

95. Write the Pythagorean triplet whose one of the numbers is ${\mathbf{4}}$.

Ans: Pythagorean triplets for any natural number greater than $1$ are

$2{\text{m,}}{{\text{m}}^2} - 1{\text{and}}{{\text{m}}^2} + 1$

So, if one number is $2{\text{m}}$, then another two numbers will be ${{\text{m}}^2} - 1{\text{and}}{{\text{m}}^2} + 1$

Given, one number = 4

Then Pythagorean triplets are:

$2{\text{m}}{\text{ = }}{\text{4}}$

$\therefore {\text{m}}{\text{ = }}{\text{2}}$ 

Then, ${{\text{m}}^2} - 1 = {\left( 2 \right)^2} - 1 = 4 - 1 = 3$

Also, ${{\text{m}}^2} + 1 = {\left( 2 \right)^2} + 1 = 4 + 1 = 5$

Therefore, $3$, $4$ and \[5\] are Pythagorean triplets.

96. Using prime factorisation, find the square roots of

a) ${\mathbf{11025}}$

b) ${\mathbf{4761}}$

Ans: 

1. $11025$

$\underline {3|11025}$

   $\underline {3|3675}$

   $\underline {5|1225}$

   $\underline {5|245}$

   $\underline {7|49}$

   $\underline {7|7}$

   $\underline {|1}$ 

Hence, factors of $11025 = \underline {3 \times 3} \times \underline {5 \times 5}  \times \underline {7 \times 7}$ on grouping in pair, therefore, $11025$ is a perfect square.

Hence, $\sqrt {11025}  = \sqrt {\underline {3 \times 3} \times \underline {5 \times 5}  \times \underline {7 \times 7} }  = 3 \times 5 \times 7 = 105$

2. $4761$

   $\underline {3|4761}$

   $\underline {3|1587}$

   $\underline {23|529}$

   $\underline {23|23}$

   $\underline {|1}$ 

Hence, factors of $4761 = \underline {3 \times 3} \times \underline {23 \times 23}$ on grouping in pair, therefore, $4761$ is a perfect square.

Hence, $\sqrt {4761}  = \sqrt {\underline {3 \times 3} \times \underline {23 \times 23} }  = 3 \times 23 = 69$

97. Using prime factorisation, find the cube roots of

a) ${\mathbf{512}}$

b) ${\mathbf{2197}}$

Ans: 

1. $512$

   $\underline {2|512}$

   $\underline {2|256}$

   $\underline {2|128}$

   $\underline {2|64}$

   $\underline {2|32}$

   $\underline {2|16}$

   $\underline {2|8}$

   $\underline {2|4}$

   $\underline {2|2}$

   $\underline {|1}$ 

Hence, factors of $512 = \underline {2 \times 2 \times 2} \times \underline {2 \times 2 \times 2}  \times \underline {2 \times 2 \times 2}$ on grouping in triplets for perfect cube, all number forms triplet. Therefore, $512$ is a perfect cube

Therefore, cube root of is $\sqrt[3]{{512}} = \sqrt {\underline {2 \times 2 \times 2}  \times \underline {2 \times 2 \times 2}  \times \underline {2 \times 2 \times 2} }  = 2 \times 2 \times 2 = 8$

2. $2197$

   $\underline {13|2197}$

   $\underline {13|169}$

   $\underline {13|13}$

   $\underline {|1}$ 

Hence, factors of $2197 = \underline {13 \times 13 \times 13}$ on grouping in triplets for perfect cube, all number forms triplet. Therefore, \[2197\] is a perfect cube.

Therefore, cube root of is $\sqrt[3]{{2197}} = \sqrt {\underline {13 \times 13 \times 13} }  = 13$

98. Are ${\mathbf{176}}$ a perfect square? If not, find the smallest number by which it should be multiplied to get a perfect square.

Ans: Taking factors of $176$

$\underline {2|176}$

$\underline {2|88}$

$\underline {2|44}$

$\underline {2|22}$

$\underline {11|11}$

$\underline {|1}$ 

\[176 = \underline {2 \times 2}  \times \underline {2 \times 2}  \times 11\] on grouping in pair, $11$ have a incomplete pair.

Therefore, $176$ is not a perfect square.

Therefore, \[11\] is the smallest number by which $176$ should be multiplied to get a perfect square. Now,

$176 \times 11 = 2 \times 2 \times 2 \times 2 \times 11 \times 11$

$= 4 \times 4 \times 121$

$= 1936 = {\left( {44} \right)^2}$ 

Hence, $1936$ is a perfect square of \[44\].

99. Is ${\mathbf{9720}}$ a perfect cube? If not, find the smallest number by which it should be divided to get a perfect cube.

Ans: Taking factors of $9720$

$\underline {2|9720}$

$\underline {2|4860}$

$\underline {2|2430}$

$\underline {3|1215}$

$\underline {3|405}$

$\underline {3|135}$

$\underline {3|45}$

$\underline {3|15}$

$\underline {5|5}$

$\underline {|1}$ 

$9720 = \underline {2 \times 2 \times 2}  \times \underline {3 \times 3 \times 3}  \times 3 \times 3 \times 5$ on grouping in triplet, $3$ and $5$ have incomplete triplet.

Therefore, $9726$ is not a perfect cube.

Therefore, on dividing the number by incomplete group of triplet. So, to make $9726$ a perfect cube:

$9726 \div \left( {3 \times 3 \times 5} \right) = 9726 \div 45$

$= 216 = {\left( 6 \right)^3}$ 

Now, 216 is a perfect cube. Hence, 45 is the smallest number by which it should be 

divided to get a perfect cube.

100. Write two Pythagorean triplets each having one of the numbers as ${\mathbf{5}}$.

Ans: Sum of two smaller sides in Pythagorean triplet is always equal to the square of longer side. If one of the numbers as $5$ then:

${\left( 5 \right)^2} = {\left( 3 \right)^2} + {\left( 4 \right)^2}$ and the other Pythagorean triplet is ${\left( {13} \right)^2} = {\left( {12} \right)^2} + {\left( 5 \right)^2}$

101. By what smallest number should ${\mathbf{216}}$ be divided so that the quotient is a perfect square? Also find the square root of the quotient.

Ans: Taking factors of $216$:

$\underline {2|216}$

$\underline {2|108}$

$\underline {2|54}$

$\underline {3|27}$

$\underline {3|9}$

$\underline {3|3}$

$\underline {|1}$ 

$216 = \underline {2 \times 2}  \times 2 \times \underline {3 \times 3}  \times 3$ Here, $2$ and $3$ have incomplete pair. So, $216$ is not a Perfect square. To make it a perfect square the number should be divided by the product of numbers which have incomplete pair ${\text{i}}{\text{.e}}{\text{.}} 2$ and $3$ therefore,

$\Rightarrow 216 \div 6 = 36$

And factors of $36 = \underline {2 \times 2} \times \underline {3 \times 3}$. Now, 36 is a perfect square. 

Therefore, the smallest number is 36 that should divide $216$ so that the quotient is a perfect square.

102. By what smallest number should ${\mathbf{3600}}$ be multiplied so that the quotient is a perfect cube? Also find the cube root of the quotient.

Ans: Taking factors of $3600$

$\underline {2|3600}$

$\underline {2|1800}$

$\underline {2|900}$

$\underline {2|450}$

$\underline {3|225}$

$\underline {3|75}$

$\underline {5|25}$

$\underline {5|5}$

$\underline {|1}$ 

$3600 = \underline {2 \times 2 \times 2}  \times 2 \times 3 \times 3 \times 5 \times 5$ Here, $2$, $3$ and 5 have incomplete group of triplets. So, $3600$ is not a Perfect cube. To make it a perfect cube the number should be multiplied by the product of numbers which have incomplete group of triplets ${\text{i}}{\text{.e}}{\text{.}}$ $2$, $3$ and 5 therefore,

$\Rightarrow 3600 \times \left( {2 \times 2 \times 3 \times 5} \right)$

$\Rightarrow 3600 \times 60 = 216000 = {\left( {60} \right)^3}$ 

And factors of $216000 = 60 \times 60 \times 60$. Now, $216000$ is a perfect cube. 

Therefore, the smallest number is 60 that should be multiplied so that the quotient is a perfect square.

103. Find the square root of the following by long division method.

a) ${\mathbf{1369}}$

b) ${\mathbf{5625}}$

Ans: finding square root using long division method:

1. $1369$

Therefore, square root of $1369$ $\Rightarrow \sqrt {1369}  = 37$

2.  $5625$

Therefore, square root of $5625$ $\Rightarrow \sqrt {5625}  = 75$

104. Find the square root of the following by long division method.

a) ${\mathbf{27}}.{\mathbf{04}}$

b) ${\mathbf{1}}.{\mathbf{44}}$

Ans: 

1. $27.04$

Finding square root using long division method:

Therefore, square root of $27.04$$ \Rightarrow $$\sqrt {27.04}  = 5.2$

2. $1.44$

Finding square root using long division method:

Therefore, square root of $1.44 \Rightarrow \sqrt {1.44}  = 1.2$

105. What is the least number that should be subtracted from ${\mathbf{1385}}$ to get a perfect square? Also find the square root of the perfect square.

Ans: finding square root of $1385$:

We can see that the remainder left is $16$ therefore, the least number that should be subtracted from $1385$ to get a perfect square is $16$. 

Hence, the required perfect square number $\Rightarrow 1385-16 = 1369$

As $1369 = 37 \times 37$.

106. What is the least number that should be added to ${\mathbf{6200}}$ to make it a perfect square?

Ans: finding square root of $6200$:

We can see that ${\left( {78} \right)^2} = 6084$ which is less than $6200$ therefore, the next perfect square is ${\left( {79} \right)^2} = 6241$ which is greater than 6000. So, the least number is the number that

should be added to $6200$ to get a perfect square.

$\Rightarrow \left( {6241-6200 = 41} \right)$

Therefore, 

$6200 + 41 = 6241$

$\Rightarrow 6241 = {\left( {79} \right)^2}$ 

Which is a perfect square.

107. Find the least number of four digits that is a perfect square.

Ans: Least of four-digit number $= 1000$

finding square root of $1000$ by long division method:

So, $24$ to be added in $1000$ to make it a perfect square $=$ 1000 + 24 = 1024.

108. Find the greatest number of three digits that is a perfect square.

Ans: Greatest three-digit number $= 999$

Finding square root of $999$

So, $38$ to be subtracted from 999 to get a perfect square. Therefore,

$\Rightarrow 999-38{\text{ }} = 961$ 

And $\sqrt {961}  = 31$

109. Find the least square number which is exactly divisible by ${\mathbf{3}}$,${\mathbf{4}}$,${\mathbf{5}}$,${\mathbf{6}}{\mathbf{and}}{\mathbf{8}}.$

Ans: Taking L.C.M of $3,4,5,6{\text{ and}}8$:

$\underline {2|3,4,5,6,8}$

$\underline {2|3,2,5,3,4}$

$\underline {2|3,1,5,3,2}$

$\underline {3|3,1,5,3,1}$

$\underline {5|1,1,5,1,1}$

$\underline {|1,1,1,1,1}$ 

L.C.M of $3,4,5,6{\text{ and}}8 = 2 \times 2 \times 2 \times 3 \times 5= 120$

As pair of $2,{\text{ }}3{\text{ and }}5$ is not completed. To make it a perfect square, the number should be multiplied with $2 \times 3 \times 5 = 30$

$\Rightarrow 120 \times 30 = 3600$

Hence, the least square number which is exactly divisible by $3,4,5,6{\text{ and}}8$ is $3600$. 

110. Find the length of the side of a square if the length of its diagonal is${\mathbf{10cm}}$.

Ans: Length of diagonal $= 10{\text{cm}}$

Now, let the length of side of a square is $= x{\text{cm}}$

Then by Pythagoras theorem:

$\Rightarrow {\left( {10} \right)^2} = {\left( x \right)^2} + {\left( x \right)^2}$

$\Rightarrow 100 = 2{x^2}$

$\Rightarrow {x^2} = 50$

$\Rightarrow x = 5\sqrt 2$ 

Therefore, the length of the side of square is $\sqrt {50} {\text{or}}{\text{5}}\sqrt 2$.

111. A decimal number is multiplied by itself. If the product is${\mathbf{51}}.{\mathbf{84}}$, find the number.

Ans: Given, the product of the number is $= 51.84$

Now, let the number be $x$ and the product of number $= x \times x = {x^2}$

According to the question:

${x^2} = 51.84$

Now, finding the square root of $51.84$

$\therefore \sqrt {51.84}  = 7.2$

112. Find the decimal fraction which when multiplied by itself gives 84.64.

Ans: Given, the product of the number is $= 84.64$

Now, let the number be $x$ and the product of number $= x \times x = {x^2}$

According to the question:

${x^2} = 84.64$

Now, finding the square root of $84.64$

$\sqrt {84.64}  = 9.2$ therefore, the required decimal fraction is $9.2$.

113. A farmer wants to plough his square field of side 150m. How much area will he have to plough?

Ans: Given, side of square field \[ = 150{\text{m}}\]

We know, area of square =  side x side

$\therefore 150{\text{m}} \times {\text{150}}{\text{m}} = 2250{{\text{m}}^2}$

114. What will be the number of unit squares on each side of a square graph paper if the total number of unit squares is ${\mathbf{256}}$?

Ans: Given, 

Total number of unit squares $= 256$

If the number is $x$ then:

The number of unit square $\Rightarrow $ $x \times x = {x^2}$

$\therefore {x^2} = 256$

$\Rightarrow x = \sqrt {256}$

$\Rightarrow x = 16$ 

So, the number of unit square $= 16$

115. If one side of a cube is ${\mathbf{15}}{\mathbf{m}}$in length, find its volume.

Ans: Given, side of a cube $= 15{\text{m}}$

We know, Volume of cube = ${\text{sid}}{{\text{e}}^3}$

$\therefore $ Volume of cube $= {\left( {15{\text{m}}} \right)^3} = 3375{{\text{m}}^3}$.

116. The dimensions of a rectangular field are ${\mathbf{80}}{\mathbf{m}}$ and ${\mathbf{18}}{\mathbf{m}}$. Find the length of its diagonal.

Ans: Given, 

length of a rectangular field (l) $= 80{\text{m}}$

breath of a rectangular field (b) $= 18{\text{m}}$

We know, length of diagonal $= \sqrt {{l^2} + {b^2}}$

$\Rightarrow \sqrt {{{\left( {80} \right)}^2} + {{\left( {18} \right)}^2}}$

$\Rightarrow \sqrt {6400 + 324}$

$\Rightarrow \sqrt {6724}  = 82{\text{m}}$ 

117. Find the area of a square field if its perimeter is ${\mathbf{96}}{\mathbf{m}}.$

Ans: Given, Perimeter of square $= 96{\text{m}}$

We know, Perimeter of a square $= 4 \times {\text{side}}$

$\Rightarrow 4 \times {\text{side}} = 96{\text{m}}$

$\Rightarrow {\text{side  = }}\frac{{96}}{4} = 24{\text{m}}$ 

Now, the area of square $= {\text{side}} \times {\text{side}}$

$\Rightarrow 24{\text{m}} \times {\text{24}}{\text{m}}$

$\Rightarrow {\text{576 }}{{\text{m}}^2}$ 

118. Find the length of each side of a cube if its volume is 512 cm3.

Ans: Given, Side of a cube $= 512{\text{c}}{{\text{m}}^3}$

We know, Volume of cube = ${\text{sid}}{{\text{e}}^3}$

Therefore, 

${\text{sid}}{{\text{e}}^3} = 512$

${\text{side  = }}\sqrt[3]{{512}} = 8{\text{cm}}$ 

119. Three numbers are in the ratio ${\mathbf{1}}:{\mathbf{2}}:{\mathbf{3}}$ and the sum of their cubes are ${\mathbf{4500}}.$ Find the numbers.

Ans: Let the numbers be ${\text{a}},2{\text{a}}{\text{and 3a}}$

According to the question:

$\Rightarrow {\left( {\text{a}} \right)^3} + {\left( {2{\text{a}}} \right)^3} + {\left( {3{\text{a}}} \right)^3} = 4500$

$\Rightarrow {{\text{a}}^3} + 8{{\text{a}}^3} + 27{{\text{a}}^3} = 4500$

$\Rightarrow 36{{\text{a}}^3} = 4500$

$\Rightarrow {{\text{a}}^3} = \frac{{4500}}{{36}} = 125$

$\Rightarrow {\text{a}} = \sqrt[3]{{125}} = 5$

$\therefore {\text{a}}{\text{ = }}5$ 

Then the numbers are:

${\text{a}} = 5$

$2{\text{a}}{\text{ =  5}} \times {\text{2}}{\text{ = 10}}$

${\text{and 3a}} = 3 \times 5 = 15$ 

120. How many square metres of carpet will be required for a square room of side ${\mathbf{6}}.{\mathbf{5}}{\mathbf{m}}$ to be carpeted?

Ans: side of square $= 6.5{\text{m}}$

Now, 

Area of square \[ = {\left( {{\text{side}}} \right)^2} = {\left( {6.5{\text{m}}} \right)^2} = 42.25{\text{ }}{{\text{m}}^2}\]

121. Find the side of a square whose area is equal to the area of a rectangle with sides ${\mathbf{6}}.{\mathbf{4}}{\mathbf{m}}{\text{ }}{\mathbf{and}}{\text{ }}{\mathbf{2}}.{\mathbf{5}}{\mathbf{m}}.$

Ans: Given, length of a rectangle $= {\text{ }}6.4{\text{ m}}$

 Breadth of a rectangle $= {\text{ }}2.5{\text{ m}}$

We know,

Area of a rectangle $= {\text{ length}} \times {\text{breadth}} = 6.4 \times 2.5 = 16{{\text{m}}^2}$

Also given, area of square = area of rectangle 

Therefore,

$\Rightarrow {\left( {{\text{Side}}} \right)^2} = 16{{\text{m}}^2}[\because {\text{area}}{\text{of square  =  side}} \times {\text{side]}}$

$\Rightarrow {\text{side}} = \sqrt {16{{\text{m}}^2}}$

$\Rightarrow {\text{side}} = 4{\text{m}}$ 

Hence, side $= 4{\text{m}}$ 

122. Difference of two perfect cubes is ${\mathbf{189}}$. If the cube root of the smaller of the two numbers is ${\mathbf{3}}$, find the cube root of the larger number.

Ans: Given, that the difference of two perfect cubes is $189$.

Cube root of the smaller of the two numbers $= 3$

$\therefore $ Cube of ${\left( 3 \right)^3} = 27$

If cube root of larger number is $x$ then cube will be $ = {x^3}$

Therefore, 

$\Rightarrow {\left( x \right)^3} - {\left( 3 \right)^3} = 189$

$\Rightarrow {x^3} - 27 = 189$

$\Rightarrow {x^3} = 189 + 27$

$\Rightarrow {x^3} = 216$

$\Rightarrow x = \sqrt[3]{{216}} = 6$

$\therefore x = 6$ 

Hence, the cube root of the larger number $= 6$

123. Find the number of plants in each row if ${\mathbf{1024}}$ plants are arranged so that number of plants in a row is the same as the number of rows.

Ans: Total numbers of plants arranged $= 1024$

Let the number of plants is each row $= {\text{a}}$

Given, that the number of rows = number of plants in each row $= {\text{a}}$

Therefore, the total plants $ = {\text{a}} \times {\text{a}} = {{\text{a}}^2}$

$\Rightarrow 1024 = {{\text{a}}^2}$

$\Rightarrow {\text{a}} = \sqrt {1024}$

$\therefore {\text{a}} = 32$ 

So, the number of plants in each row is $32$.

124. A hall has a capacity of ${\mathbf{2704}}$ seats. If the number of rows is equal to the number of seats in each row, then find the number of seats in each row.

Ans: Total capacity of hall $= 2704$

Let the number of seats in each row $= {\text{a}}$

Given, that the number of rows = number of seats in each row $= {\text{a}}$

Therefore, the total seats $= {\text{a}} \times {\text{a}} = {{\text{a}}^2}$

$\Rightarrow 2704 = {{\text{a}}^2}$

$\Rightarrow {\text{a}} = \sqrt {2704}  = \sqrt {52 \times 52}$

$\therefore {\text{a}} = 52$ 

Hence, seats in each row $= 52$

125. A General, wishes to draw up his ${\mathbf{7500}}$ soldiers in the form of a square. After arranging, he found out that some of them are left out. How many soldiers were left out?

Ans:  Total number of soldiers $= 7500$

number of soldiers left:

Therefore, number of soldiers left $ = 104$

126. ${\mathbf{8649}}$ students were sitting in a lecture room in such a manner that there were as many students in the row as there were rows in the lecture room. How many students were there in each row of the lecture room?

Ans: Total number of students in each row of the lecture room $= {\text{x}}$

Also, the number of rows $= {\text{x}}$

Therefore, total students $= {\text{x}} \times {\text{x}} = {{\text{x}}^2}$

As per question, 

$\Rightarrow {{\text{x}}^2} = 8649$

$\Rightarrow {\text{x}} = \sqrt {8649}  = \sqrt {93 \times 93}$

$\therefore {\text{x}} = 93$ 

So, the students in each row $= 93$

127. Rahul walks ${\mathbf{12}}{\mathbf{m}}$ north from his house and turns west to walk ${\mathbf{35}}{\mathbf{m}}$ to reach his friend’s house. While returning, he walks diagonally from his friend’s house to reach back to his house. What distance did he walk while returning?

Ans: If Rahul walked ${\text{x}}{\text{m}}$.

By using Pythagoras theorem, in triangle ${\text{PQR}}$,

$\Rightarrow {\text{P}}{{\text{R}}^2} = {\text{P}}{{\text{Q}}^2} + {\text{Q}}{{\text{R}}^2}$

$\Rightarrow {\text{P}}{{\text{R}}^2} = {\left( {12} \right)^2} + {\left( {35} \right)^2}$

$\Rightarrow {\text{P}}{{\text{R}}^2} = 144 + 1225$

$\Rightarrow {\text{P}}{{\text{R}}^2} = 1369$

$\Rightarrow {\text{PR = }}\sqrt {1369}$

$\therefore {\text{PR}}{\text{ =  37}}{\text{m}}$ 

So, the distance is $37{\text{m}}$. 

128. A ${\mathbf{5}}.{\mathbf{5}}{\mathbf{m}}$ long ladder is leaned against a wall. The ladder reaches the wall to a height of ${\mathbf{4}}.{\mathbf{4}}{\mathbf{m}}$. Find the distance between the wall and the foot of the ladder.

Ans: Let the distance between wall and foot of the ladder $= {\text{AB}}$= ${\text{x}}{\text{m}}$ 

By using Pythagoras theorem,

$\Rightarrow {\text{B}}{{\text{C}}^2} = {\text{A}}{{\text{B}}^2} + {\text{C}}{{\text{A}}^2}$

$\Rightarrow {\left( {5.5} \right)^2} = {\text{A}}{{\text{B}}^2} + {\left( {4.4} \right)^2}$

$\Rightarrow 30.25 = {\text{A}}{{\text{B}}^2} + 19.36$

$\Rightarrow {\text{A}}{{\text{B}}^2} = 30.25 - 19.36 = 10.89$

$\Rightarrow {\text{AB = }}\sqrt {10.89}$ 

So, the distance between the wall and the foot of the ladder is 3.3.

129. A king wanted to reward his advisor, a wise man of the kingdom. So he asked the wiseman to name his own reward. The wiseman thanked the king but said that he would ask only for some gold coins each day for a month. The coins were to be counted out in a pattern of one coin for the first day, $3$ coins for the second day, 5 coins for the third day and so on for $30$ days. Without making calculations, find how many coins will the advisor get in that month?

Ans: As per question,

Total question at the end $= 1 + 3 + 5 + ...$

As this is an odd number series. 

The number of term ,$\left( {\text{n}} \right) = 30$

We know, Sum of an odd natural numbers $= {{\text{n}}^2} = {\left( {30} \right)^2} = 900$

So, the advisor will get $900$ coins at the end. 

130. Find three numbers in the ratio ${\mathbf{2}}:{\mathbf{3}}:{\mathbf{5}}$, the sum of whose squares is ${\mathbf{608}}.$

Ans: let the three numbers be $= 2{\text{a, }}3{\text{a and }}5{\text{a}}$

According to the question, 

$\Rightarrow {\left( {2{\text{a}}} \right)^2} + {\left( {{\text{3a}}} \right)^2} + {\left( {{\text{5a}}} \right)^2} = 608$

$\Rightarrow 4{{\text{a}}^2} + 9{{\text{a}}^2} + 25{{\text{a}}^2} = 608$

$\Rightarrow 38{{\text{a}}^2} = 608$

$\Rightarrow {{\text{a}}^2} = \frac{{608}}{{38}} = 16$

$\Rightarrow {{\text{a}}^2} = 16$

$\therefore {\text{a}} = \sqrt {16}  = 4$ 

Therefore, the three numbers are:

${2{\text{a}} = 2 \times 4 = 8}$ 

${3{\text{a}} = 3 \times 4 = 12}$ 

${5{\text{a }} = 5 \times 4 = 20}$ 

131. Find the smallest square number divisible by each one of the number ${\mathbf{8}},{\text{ }}{\mathbf{9}}{\text{ }}{\mathbf{and}}{\text{ }}{\mathbf{10}}.$

Ans: Taking L.C.M. of $8,9{\text{and 10}}$

$\underline {2|8,9,10}$

$\underline {2|4,9,5}$

$\underline {2|2,9,5}$

$\underline {3|1,9,5}$

$\underline {3|1,3,5}$

$\underline {5|1,1,5}$

$\underline {|1,1,1}$ 

$\therefore$ L.C.M $= 2 \times 2 \times 2 \times 3 \times 3 \times 5 = 360$

Taking pairs $ = \underline {2 \times 2}  \times 2 \times \underline {3 \times 3}  \times 5 = 360$ as $2{\text{and 5}}$ have incomplete pair.

For making it a perfect square, we have to multiple it by the product of incomplete pair ${\text{i}}{\text{.e}}2 \times 5 = 10$

Now, $360 \times 10 = 3600$

Hence, the smallest square number is $3600$.

132. The area of a square plot is ${\mathbf{101}}\dfrac{{\mathbf{1}}}{{{\mathbf{400}}}}{{\mathbf{m}}^{\mathbf{2}}}$. Find the length of one side of the plot.

Ans: Given, Area of a square plot $= 101\frac{1}{{400}}{{\text{m}}^2}$

Let the length of square plot $= {\text{a}}$

And the area of square $= {{\text{a}}^2}$

According to the question:

${{\text{a}}^2} = 101\frac{1}{{400}}{{\text{m}}^2}$

$\Rightarrow {{\text{a}}^2} = \frac{{40401}}{{400}}{{\text{m}}^2}$

$\Rightarrow {\text{a}} = \sqrt {\frac{{40401}}{{400}}}  = \sqrt {\frac{{201 \times 201}}{{20 \times 20}}}  = \frac{{201}}{{20}}$

$\Rightarrow {\text{a}}{\text{ = }}{\text{10}}\frac{1}{{20}}{\text{m}}$ 

Hence, ${\text{a}}{\text{ = }}{\text{10}}\frac{1}{{20}}{\text{m}}$

133. Find the square root of ${\mathbf{324}}$ by the method of repeated subtraction.

Ans: square root of $324$ by the method of repeated subtraction

Now, subtract successive odd numbers starting from 1 as:

${324-1 = 323,{\text{ }}323{\text{ }}-3{\text{ }} = 320,}$ 

${320-5 = 315,315-7 = 308,}$ 

${308-9{\text{ }} = 299,{\text{ }}299-11 = 288,}$ 

${288-13{\text{ }} = 275,{\text{ }}275-15 = 260}$ 

${260-17{\text{ }} = 243,{\text{ }}243-19 = 224}$ 

${224-21 = 203,{\text{ }}203-23 = 180}$ 

${180-25 = 155,155-27 = 128}$ 

${128-29{\text{ }} = 99,{\text{ }}99-31{\text{ }} = 68}$ 

$68-33 = 35,{\text{ }}35-35 = 0$

The number $324$ was reduced to zero after subtracting $18$ odd numbers. 

So, the $324$ is a perfect square of $18$.

134. Three numbers are in the ratio 2:3:4. The sum of their cubes is 0.334125. Find the numbers.

Ans: Suppose the numbers $= {\text{ }}2{\text{a}},3{\text{a}},{\text{4a}}$

Given, the sum of their cubes $= 0.334125$

According to the question,

$\Rightarrow {\left( {2{\text{a}}} \right)^3} + {\left( {{\text{3a}}} \right)^3} + {\left( {{\text{4a}}} \right)^3} = 0.334125$

$\Rightarrow 8{{\text{a}}^3} + 27{{\text{a}}^3} + 64{{\text{a}}^3} = 0.334125$

$\Rightarrow 99{{\text{a}}^3} = 0.334125$

$\Rightarrow {{\text{a}}^3} = \frac{{0.334125}}{{99}} = 0.003375$

$\Rightarrow {\text{a}} = \sqrt[3]{{0.003375}}$

$\therefore {\text{a}} = \sqrt {\frac{{3375}}{{1000000}}}  = \sqrt {\frac{{15 \times 15 \times 15}}{{10 \times 10 \times 10 \times 10 \times 10 \times 10}}}  = \frac{{15}}{{10 \times 10 \times 10}} = 0.015$ 

Hence, the required numbers are:

${2 \times 0.015{\text{ }} = 0.03}$ 

${3 \times 0.015{\text{ }} = 0.045}$ 

${4 \times 0.015{\text{ }} = 0.06}$ 

135. Evaluate: $\sqrt[{\mathbf{3}}]{{{\mathbf{27}}}} + \sqrt[{\mathbf{3}}]{{{\mathbf{0}}.{\mathbf{008}}}} + \sqrt[{\mathbf{3}}]{{{\mathbf{0}}.{\mathbf{064}}}}$

Ans: $\sqrt[3]{{27}} = \sqrt[3]{{3 \times 3 \times 3}} = 3$

$\sqrt[3]{{0.008}} = \sqrt[3]{{0.2 \times 0.2 \times 0.2}} = 0.2$

$\sqrt[3]{{0.064}} = \sqrt[3]{{0.4 \times 0.4 \times 0.4}} = 0.4$

Now, $3 + 0.2 + 0.4 = 3.6$

Hence, Answer is $3.6$

136. ${\{ ({{\mathbf{5}}^2} + {\left( {{\mathbf{1}}{{\mathbf{2}}^2}} \right)^{\frac{1}{2}}})\} ^3}$

Ans: According to the question:

$\Rightarrow {\{ ({5^2} + {\left( {{{12}^2}} \right)^{\frac{1}{2}}})\} ^3}$

$\Rightarrow {\{ ({5^2} + {\left( {144} \right)^{\frac{1}{2}}})\} ^3}$

$\Rightarrow {\{ {5^2} + 12\} ^3} = {\left( {25 + 12} \right)^3}$

$\Rightarrow {\left( {37} \right)^3} = 37 \times 37 \times 37 = 50653$ 

137. ${\left\{ {\left( {{{\mathbf{6}}^{\mathbf{2}}} + {{\left( {{{\mathbf{8}}^{\mathbf{2}}}} \right)}^{\frac{{\mathbf{1}}}{{\mathbf{2}}}}}} \right)} \right\}^{\mathbf{3}}}$

Ans: According to the question:

$\Rightarrow {\left\{ {\left( {{6^2} + {{\left( {{8^2}} \right)}^{\frac{1}{2}}}} \right)} \right\}^3}$

$\Rightarrow {\{ (36 + {\left( {64} \right)^{1/2}})\} ^3}$

$\Rightarrow {\{ 36 + 8\} ^3}$

$\Rightarrow {\{ 44\} ^3} = 44 \times 44 \times 44 = 85184$ 

138. A perfect square number has four digits, none of which is zero. The digits from left to right have values that are: even, even, odd, even. Find the number.

Ans: ABCD is a perfect square. 

Here, A = an even number

B = an even number, 

C = an odd number,

D = an even number

So, $8836$ is a perfect square.

139. Put three different numbers in the circles so that when you add the numbers at the end of each line you always get a perfect square.

Ans: Three numbers which we add at the end of the line we always give perfect square are $6, 19$ and $30$

As, $6 + 19 = 25$ (It is a perfect square)

$19 + 30 = {\text{4}}0$ (It is a perfect square)

and $30 + 6 = 36$ (It is a perfect square)

Therefore, the complete triangle is

140. The perimeters of two squares are ${\mathbf{40}}{\text{ }}{\mathbf{and}}{\mathbf{96}}$ metres respectively. Find the perimeter of another square equal in area to the sum of the first two squares.

Ans: Given, that the perimeter of one square $= {\text{ }}40{\text{ m}}$

If the length of square $= {\text{ }}{{\text{a}}_1}$

Then, $4{{\text{a}}_1} = 40{\text{m}}$

So, side  $\left( {{{\text{a}}_1}} \right) = \frac{{40}}{4} = 10{\text{m}}$

Now, area of a square $= {\text{ }}{\left( {{{\text{a}}_1}} \right)^2} = 10 \times 10 = 100{\text{ }}{{\text{m}}^2}$

Given, perimeter of second square $= {\text{ }}96{\text{ m}}$

And if the length of second square $= {\text{ }}{{\text{a}}_2}$

Then, $4{{\text{a}}_2} = 96{\text{m}}$

So, side  $\left( {{{\text{a}}_2}} \right) = \frac{{96}}{4} = 24{\text{m}}$

Now, area of a square $= {\text{ }}{\left( {{{\text{a}}_2}} \right)^2} = 24 \times 24 = 576{\text{ }}{{\text{m}}^2}$

Now the perimeter of another square is equal in area to the sum of the first two square. 

So, area of another square $= 100 + 576 = 676{\text{ m}}$

Then, area of a square $= {({{\text{a}}_3})^2}$

$\Rightarrow {\left( {{{\text{a}}_3}} \right)^2} = 676{\text{m}}$

$\Rightarrow {{\text{a}}_3} = \sqrt {676} {\text{m}}$

$\Rightarrow {{\text{a}}_3} = 26$ 

Hence, perimeter $= 4{{\text{a}}_3} \Rightarrow 4\left( {26} \right) = 104{\text{m}}$

141. A three-digit perfect square is such that if it is viewed upside down, the number seen is also a perfect square. What is the number? (Hint: The digits ${\mathbf{1}}$, ${\mathbf{0}}{\mathbf{and}}{\mathbf{8}}$ stay the same when viewed upside down, whereas ${\mathbf{9}}$ becomes ${\mathbf{6}}$ and ${\mathbf{6}}$ becomes ${\mathbf{9}}$.)

Ans: $196$ and $961$ is a three-digit perfect squares that looks similar when viewed upside to down. 

142. ${\mathbf{13}}$ and ${\mathbf{31}}$ is a strange pair of numbers such that their squares ${\mathbf{169}}$and ${\mathbf{961}}$are also mirror images of each other. Can you find two other such pairs?

Ans:

i) First pair of number $= 12$ and $21$

Also, there square $= {\left( {12} \right)^2} = 144$ mirror image of which is $441$.

ii) Second pair of number $= 102$ and $201$

Also, there square $= {\left( {102} \right)^2} = 10404$ mirror image of which is $40401$.

NCERT Exemplar for Class 8 Maths Solutions Chapter - 3

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In Examples 1 to 7, out of given four Choices Multiple only one is Correct. Write the Correct Answer

Example 1: Which of the following is the square of an odd number?

(a) 256

(b) 361

(c) 144

(d) 400

Ans: (b) because $19\times 19=361$

Example 2: Which of the following will have $\mathbf{1}$ at its unit’s place?

(a) $\mathbf{1}{{\mathbf{9}}^{\mathbf{2}}}$

(b) $\mathbf{1}{{\mathbf{7}}^{\mathbf{2}}}$

(c) $\mathbf{1}{{\mathbf{8}}^{\mathbf{2}}}$

(d) $\mathbf{1}{{\mathbf{6}}^{\mathbf{2}}}$

Ans: (a) because $19\times 19=361$

Example 3: How many natural numbers lie between $\mathbf{1}{{\mathbf{8}}^{\mathbf{2}}}$ and $\mathbf{1}{{\mathbf{9}}^{\mathbf{2}}}$?

(a) $\mathbf{30}$

(b) $\mathbf{37}$

(c) $\mathbf{35}$

(d) $\mathbf{36}$

Ans: (d) As ${{\left( 18 \right)}^{2}}=324\text{and}{{\left( 19 \right)}^{2}}=361$

Therefore, $361324=37$

Example 4: Which of the following is not a perfect square?

(a) $\mathbf{361}$

(b) $\mathbf{1156}$

(c) $\mathbf{1128}$

(d) $\mathbf{1681}$

Ans: (c) As in $1128=2\times 2\times 2\times 141$ here 2 and 141 have incomplete pair.

Example 5: A perfect square can never have the following digit at one’s place.

(a) $\mathbf{1}$

(b) $\mathbf{6}$

(c) $\mathbf{5}$

(d) $\mathbf{3}$

Ans: (d) digits ending with $1,6,5\text{ and }3$ have unit digit $1,6,5\text{ and }9$ squares respectively

Example 6: The value of $\sqrt{\mathbf{176}+\sqrt{\mathbf{2401}}}$ is

(a) $\mathbf{14}$

(b) $\mathbf{15}$

(c) $\mathbf{16}$

(d) $\mathbf{17}$

Ans: (b) $\because 49\times 49=2401$

$\therefore \sqrt{176+\sqrt{2401}}=\sqrt{176+49}=\sqrt{225}=15$

Example 7: Given that $\sqrt{\mathbf{5625}}=\mathbf{75}$, the value of $\sqrt{\mathbf{0}.\mathbf{5625}}+\sqrt{\mathbf{56}.\mathbf{25}}$ is:

(a) $\mathbf{82}.\mathbf{5}$

(b) $\mathbf{0}.\mathbf{75}$

(c) $\mathbf{8}.\mathbf{25}$

(d) $\mathbf{75}.\mathbf{05}$

Ans: (c) $\because 75\times 75=5625$and $0.75\times 0.75=0.5625$

$7.5\times 7.5=56.25$

Therefore, $\sqrt{0.5625}+\sqrt{56.25}=0.75+7.5=8.25$

In examples 8 to 14, fill in the blanks to make the statements true.

Example 8: There are __________ perfect squares between $\mathbf{1}$ and $\mathbf{50}$.

Ans: There are $6$ perfect squares $4,9,16,25,36,49$.

Example 9: The cube of $\mathbf{100}$ will have __________ zeroes.

Ans: Cube of $100$ will have $6$zeroes. $\left( 100\times 100\times 100 \right)=1000000$

Example 10: The square of $\mathbf{6}.\mathbf{1}$ is ____________.

Ans: $6.1\times 6.1=37.21$

Example 11: The cube of $\mathbf{0}.\mathbf{3}$is ____________.

Ans: $0.3\times 0.3\times 0.3=0.027$

Example 12: $\mathbf{6}{{\mathbf{8}}^{\mathbf{2}}}$ will have __________ at the unit’s place.

Ans: Because $68$ has $8$ at its unit place and ${{\left( 8 \right)}^{2}}=64$. So, ${{\left( 68 \right)}^{2}}$ will have $4$ at unit’s place.

Example 13: The positive square root of a number $\mathbf{x}$ is denoted by__________.

Ans: $\sqrt{\text{x}}$

Example 14: The least number to be multiplied with $\mathbf{9}$ to make it a perfect cube is _______________.

Ans: $3\because \left( 9\times 3=27 \right)$and $\left( 3\times 3\times 3=27 \right)$.

In examples 15 to 19, state whether the statements are true (T) or false (F)

Example 15: The square of $\mathbf{0}.\mathbf{4}$ is $\mathbf{0}.\mathbf{16}$.

Ans: True, because $0.4\times 0.4=0.16$

Example 16: The cube root of $\mathbf{729}$is $\mathbf{8}$.

Ans: False because $9\times 9\times 9=729$

Example 17: There are $\mathbf{21}$ natural numbers between $\mathbf{102}$ and $\mathbf{112}$.

Ans: False because $112-102=10$not $21$.

Example 18: The sum of first $\mathbf{7}$ odd natural numbers is $\mathbf{49}$.

Ans: True, because first seven odd natural numbers are $=1,3,5,7,9,11\text{and 13}$

And their sum:

$=1+3+5+7+9+11+13$

$=49$

Example 19: The square root of a perfect square of $\mathbf{n}$ digits will have $\frac{\mathbf{n}}{\mathbf{2}}$ digits if $\mathbf{n}$ is even.

Ans: True, as a square root of a perfect square of $\text{n}$digit will have $\frac{\text{n}}{2}$digits.

Example 20: Express \[\mathbf{36}\] as a sum of successive odd natural numbers.

Ans: successive odd natural numbers are $=1,3,5,7,9\text{ and 11}$ and $1+3+5+7+9+11=36$.

Example 21: Check whether $\mathbf{90}$ is a perfect square or not by using prime factorisation.

Ans: Taking factors of $90$:

$\underline{2|90}$

$\underline{3|45}$

$\underline{3|15}$

$\underline{5|5}$

$\underline{|1}$

$90=2\times 3\times 3\times 5$as $2$ and $5$ have incomplete pair. Therefore, $90$ is not a perfect square.

Example 22: Check whether $\mathbf{1728}$ is a perfect cube by using prime factorisation.

Ans: Taking factors of $1728$:

So, $1728=2\times 2\times 2\times 2\times 2\times 2\times 3\times 3\times 3$

As all prime factors forms triplets. So, $1728$ is a perfect cube

Example 23: Using distributive law, find the square of$\mathbf{43}$.

Ans: Finding square using distributive law:

As $43$ can be written as

$43=40+3$

So, $432={{\left( 40+3 \right)}^{2}}=\left( 40+3 \right)\left( 40+3 \right)=40\left( 40+3 \right)+3\left( 40+3 \right)$

$=\text{ }40\times 40+40\times 3+3\times 40+3\times 3$

$=\text{ }1600+240+9$

$=\text{ }1849$

Therefore, \[{{\left( 43 \right)}^{2}}=1849\].

Example 24: Write a Pythagorean triplet whose smallest number is $\mathbf{6}$.

Ans: Given that smallest number is $6$

Pythagorean triplet is given by: $2\text{m,}{{\text{m}}^{2}}-1\text{and}{{\text{m}}^{2}}+1$

Now, $2\text{m}=6\Rightarrow \text{m}=3$

${{\text{m}}^{2}}+1=32+1=9+1=10$

${{\text{m}}^{2}}1=321=91=8$

Therefore, the Pythagorean triplet is$6,8,10$.

Example 25: Using prime factorisation, find the cube root of$\mathbf{5832}$.

Ans: Prime factors of $5832$:

$\underline{2|5832}$

$\underline{2|2916}$

$\underline{2|1458}$

$\underline{3|729}$

$\underline{3|243}$

$\underline{3|81}$

$\underline{3|27}$

$\underline{3|9}$

$\underline{3|3}$

$\underline{|1}$

$5832=2\times 2\times 2\times 3\times 3\times 3\times 3\times 3\times 3$

$5832=\underline{2\times 2\times 2}\times \underline{3\times 3\times 3}\times \underline{3\times 3\times 3}$

$=2\times 3\times 3$

$=18$

Example 26: Evaluate the square root of $\mathbf{22}.\mathbf{09}$by long division method.

Ans: Square root of $22.09$by long division method is$4.7$.

Example 27: Find the smallest perfect square divisible by 3, 4, 5 and 6.

Ans: The least number divisible by $3,\text{ }4,\text{ }5\text{ and }6$ is:

The LCM of $3,\text{ }4,\text{ }5\text{ and }6$:

$\underline{2|3,4,5,6}$

$\underline{2|3,2,5,3}$

$\underline{3|3,1,5,3}$

$\underline{5|1,1,5,1}$

$\underline{|1,1,1,1}$

Now, $60=2\times 2\times 3\times 5.$

As $5\text{ and }3$does not form pairs. So, $60$ is not a perfect square. Hence, $60$ should bemultiplied by $5\times 3=15$to get a perfect square.

So, the required least square number is $=60\times 15=900.$

Example 28: A ladder $\mathbf{10}\mathbf{m}$ long rests against a vertical wall. If the foot of the ladder is 6m away from the wall and the ladder just reaches the top of the wall, how high is the wall?

Ans: Let $\text{BC}$ be the distance between the foot of the ladder and the wall. And $\text{AC}$ be the ladder.

Given: $\text{AC}$ \[=10\text{m}\]and $\text{BC}$ \[=6\text{m}\]

Because $\Delta \text{ABC}$ forms a right-angled triangle, Therefore, using Pythagoras theorem:

$\text{A}{{\text{C}}^{2}}=\text{A}{{\text{B}}^{2}}+\text{B}{{\text{C}}^{2}}$

${{10}^{2}}=\text{A}{{\text{B}}^{2}}+\text{ }{{6}^{2}}$

$\Rightarrow \text{A}{{\text{B}}^{2}}={{10}^{2}}{{6}^{2}}=10036\text{ }=64$

$\Rightarrow \text{AB}=64\text{ }=8\text{m}$

Hence, the wall is $8\text{m}$high.

Example 29: Find the length of a diagonal of a rectangle with dimensions $\mathbf{20}\mathbf{m}$ by $\mathbf{15}\mathbf{m}$.

Ans:

By using Pythagoras theorem, we know

Length of diagonal of the rectangle is given by $=\sqrt{{{l}^{2}}+{{b}^{2}}}$units

$\Rightarrow \sqrt{{{\left( 20 \right)}^{2}}+{{\left( 15 \right)}^{2}}}\text{m}$

$\Rightarrow \sqrt{400+225}\text{m}$

$\Rightarrow \sqrt{625}\text{m}$

$\Rightarrow \text{25}\text{m}$

Thus, the length of diagonal is $25\text{m}$.

Example 30: The area of a rectangular field whose length is twice its breadth is $\mathbf{2450}{{\mathbf{m}}^{\mathbf{2}}}$. Find the perimeter of the field.

Ans: If the breadth of the field is $\text{x}$ metres then the length of the field is $2\text{x}$ metres.

We know, area of the rectangular field $=\text{ length}\times \text{breadth}$

Also given, area is $=\text{ }2450\text{ }{{\text{m}}^{2}}.$

Therefore,

$\left( 2\text{x} \right)\left( \text{x} \right)=2{{\text{x}}^{2}}=\text{ }2450\text{ }{{\text{m}}^{2}}$

$\Rightarrow 2{{\text{x}}^{2}}=2450$

$\Rightarrow {{\text{x}}^{2}}=\frac{2450}{2}=1225\text{m}$

$\Rightarrow \text{ x}\text{=}\sqrt{1225}\text{m}$

$\Rightarrow \text{x}=35\text{m}$

Therefore, breadth $=35\text{m}$ and length \[=35\times 2=70\text{m}\]

Now the, perimeter of the field \[=2\left( l+b \right)\]

$=\text{ }2\left( 70+35 \right)\text{m}$

$=\text{ }2\times 105\text{m }=210\text{m}$

Example 31: During a mass drill exercise, $\mathbf{6250}$ students of different schools are arranged in rows such that the number of students in each row is equal to the number of rows. In doing so, the instructor finds out that $\mathbf{9}$ children are left out. Find the number of children in each row of the square.

Ans: Given total number of students $=\text{ }6250$

students forming a square $=62509=6241$

Thus, $6241$ students form a big square that has the number of rows equal to the number of students in each row.

Now let the number of students in each row be $x$, then the number of rows $=x$

Therefore, $x\times x=6241$

or

${{x}^{2}}=6241$

$x=\sqrt{6241}\text{= }79$

So, there are $79$ students in each row of the square formed.

Example 32: Find the least number that must be added to $\mathbf{1500}$ ;.so as to get a perfect square. Also find the square root of the perfect square.

Ans: 

We can see $1500>{{38}^{2}}$ and  ${{39}^{2}}>1500$

Hence, the number to be added$={{39}^{2}}1500$

   =15211500

   =21 

Therefore, the perfect square $=1500+21=1521$

$\therefore \sqrt{1521}=39$

Thus, the required number is $21$ and the square root is $39$.

Example 33: Application of problem solving strategies

Finding the smallest number by which $1620$ must be divided to get a perfect square.

Ans: Understand and Explore:

• What information is given in the question? – A number which is not a perfect square.

• What are we trying to find? – The smallest number by which $1620$must be divided to get a perfect square. Plan a strategy

• Using prime factorisation to find the product of prime factors of $1620$.

• Pairing the prime factors to see if any factor is left unpaired.

• Then the unpaired factor will be the smallest number that must be divided to get a perfect square.

Solve: Prime factorisation of $1620$ is

$\underline{2|1620}$

$\underline{2|810}$

$\underline{5|405}$

$\underline{3|81}$

$\underline{3|27}$

$\underline{3|9}$

$\underline{3|3}$

$\underline{|1}$

The product of prime factors $=\text{2}\times 2\times 5\times 3\times 3\times 3\times 3$

After Pairing these prime factors $=\underline{\text{2}\times 2}\times 5\times \underline{3\times 3}\times \underline{3\times 3}$

We can see that the factor $5$ is left unpaired. So, the required smallest number is $5$.

To check if it is a perfect square. Divide $1620$ by $5$

$1620\div 5=324$

We can see that on dividing $1620$ by $5$ no remainder is left, therefore, $324$ is a perfect square, hence our answer is verified.

Multiple question Answer

In each of the questions, 1 to 24, write the correct Answer from the given four options:

1. ${\mathbf{196}}$ is the square of 

a) ${\mathbf{11}}$

b) ${\mathbf{12}}$ 

c) ${\mathbf{14}}$

d) ${\mathbf{16}}$

Ans: (c), $14$

Breaking $196$ into factors = $\sqrt {196}  = \sqrt {2 \times 2 \times 7 \times 7}$

Taking pairs out $2 \times 7 = 14$

Therefore, square root of 196 is 14.

2. Which of the following is a square of an even number? 

a) $\mathbf{144}$

b) $\mathbf{169}$

c) $\mathbf{441}$

d) $\mathbf{625}$

Ans: (a), $144$ 

Therefore, square of an even number is:

$144 = 12 \times 12 = {\left( {12} \right)^2}$

3. A number ending in ${\mathbf{9}}$ will have the units place of its square as 

a) ${\mathbf{3}}$

b) ${\mathbf{9}}$

c) ${\mathbf{1}}$

d) ${\mathbf{6}}$

Ans: (c), $1$. 

We know $9 \times 9 = 81$

Therefore, unit place of a number ending with $9$ is $1$.

4. Which of the following will have 4 at the unit place? 

a) ${\mathbf{1}}{{\mathbf{4}}^{\mathbf{2}}}$

b)${\mathbf{6}}{{\mathbf{2}}^{\mathbf{2}}}$

c) ${\mathbf{2}}{{\mathbf{7}}^2}$

d) ${\mathbf{3}}{{\mathbf{5}}^2}$

Ans: (b), ${62^2}$

Because, unit digit of ${62^2} = 3844$

Square of $2 = 4$.

5. How many natural numbers lie between ${{\mathbf{5}}^{\mathbf{2}}}$ and ${{\mathbf{6}}^{\mathbf{2}}}$? 

a) 9 

b) 10 

c) 11 

d) 12 

Ans: (b), 10.

We know, natural numbers lie between ${5^2} = 25$ and ${6^2} = 36$ are 26, 27, 28, 29, 30, 

31, 32, 33, 34, 35. So there are $10$ natural numbers.

6. Which of the following cannot be a perfect square? 

a) ${\mathbf{841}}$

b) ${\mathbf{529}}$

c) ${\mathbf{198}}$

d) All of the above 

Ans: (c), 198.

$\sqrt {198}  = \sqrt {2 \times 3 \times 3 \times 11}  = 3\sqrt {22}$ because no perfect pairs can be taken out therefore, $198$ can’t be a perfect square.

7. The one’s digit of the cube of 23 is 

a) ${\mathbf{6}}$ 

b) ${\mathbf{7}}$

c) ${\mathbf{3}}$ 

d) ${\mathbf{9}}$ 

Ans: (b), $7$

We know, unit digit of $23 = 3$

And cube of $3 = 27$

8. A square board has an area of ${\mathbf{144}}$ square units. How long is each side of the board? 

a) ${\mathbf{11}}$ units 

b) ${\mathbf{12}}$ units 

c) ${\mathbf{13}}$ units 

d) ${\mathbf{14}}$ units 

Ans: (b), $12$ units

We know, $144$ is square of $12$${\text{i}}{\text{.e}}$$12 \times 12$ unit.

9. Which letter best represents the location of $25$ on a number line? 

a) A 

b) B 

c) C 

d) D

Ans: (c), C

We know, square of $5 = 5 \times 5 = 25$

10. If one member of a Pythagorean triplet is ${\mathbf{2}}$m, then the other two members are 

a) ${{\mathbf{m}}^{\mathbf{2}}},{{\mathbf{m}}^{\mathbf{2}}} + {\mathbf{1}}$

b) ${{\mathbf{m}}^{\mathbf{2}}} + {\mathbf{1}},{{\mathbf{m}}^{\mathbf{2}}} - {\mathbf{1}}$

c) ${{\mathbf{m}}^{\mathbf{2}}},{{\mathbf{m}}^{\mathbf{2}}} - {\mathbf{1}}$

d) ${{\mathbf{m}}^2},{\mathbf{m}} + {\mathbf{1}}$

Ans: (b), ${{\text{m}}^2} + 1,{{\text{m}}^2} - 1$

Therefore, the formula of Pythagorean triplet is:

${{\text{m}}^2} + 1,{{\text{m}}^2} - 1$.

11. The sum of successive odd numbers ${\mathbf{1}}$,${\mathbf{3}}$,${\mathbf{5}}$,${\mathbf{7}}$,${\mathbf{9}}$,${\mathbf{11}}$,${\mathbf{13}}$ and ${\mathbf{15}}$ is 

a) ${\mathbf{81}}$

b) ${\mathbf{64}}$

c) ${\mathbf{49}}$

d) ${\mathbf{36}}$

Ans: (b), $64$

Total numbers are $ = 8$

successive odd numbers $ = 1,3,5,7,9,11,13,15$

sum of successive odd numbers is $= $ square of $8 = 8 \times 8 = 64$

12. The sum of first n odd natural numbers is 

a) ${\mathbf{2n}} + {\mathbf{1}}$

b) ${{\mathbf{n}}^{\mathbf{2}}}$

c) ${{\mathbf{n}}^{\mathbf{2}}} - {\mathbf{1}}$

d) ${{\mathbf{n}}^{\mathbf{2}}} + {\mathbf{1}}$

Ans: (b), ${{\text{n}}^2}$

We know, sum of first n odd natural numbers is given by: 

$\Sigma (2{\text{n}} - {\text{1)}}{\text{ = }}{\text{2}}\Sigma {\text{n}} - {\text{n}}$

$\Rightarrow \frac{{2 \times {\text{n(n + 1)}}}}{2} - {\text{n}}$

$\Rightarrow {\text{n(n + 1)}} - {\text{n}}$

$\Rightarrow {{\text{n}}^2} + {\text{n}} - {\text{n}}$

$\Rightarrow {{\text{n}}^2}$ 

13. Which of the following numbers is a perfect cube? 

a) ${\mathbf{243}}$

b)${\mathbf{216}}$

c) ${\mathbf{392}}$

d) ${\mathbf{8640}}$

Ans:(b), $216$

As square of 6 is $ = 6 \times 6 = 36$

14. The hypotenuse of a right triangle with its legs of lengths ${\mathbf{3}}x \times {\mathbf{4}}x$ is 

a) ${\mathbf{5}}x$

b) ${\mathbf{7}}x$

c) ${\mathbf{16}}x$

d) ${\mathbf{25}}x$

Ans: (a), $5x$

We know,

${{{({\text{Hypotenuse)}}}^2} = {{\left( {3x} \right)}^2} + {{\left( {4x} \right)}^2}}$ 

${ = {\text{ }}9{x^2} + 16{x^2}}$ 

${ = {\text{ }}25{x^2}}$ 

${\text{Hypotenuse}}{\text{ = }}\sqrt {25{x^2}}  = 5x$

15. The next two numbers in the number pattern ${\mathbf{1}},{\text{ }}{\mathbf{4}},{\text{ }}{\mathbf{9}},{\text{ }}{\mathbf{16}},{\text{ }}{\mathbf{25}}...$are 

a) ${\mathbf{35}},{\mathbf{48}}$

b) ${\mathbf{36}},{\mathbf{49}}$

c) ${\mathbf{36}},{\mathbf{38}}$

d) ${\mathbf{35}},{\mathbf{49}}$

Ans: (b), $36,49$

As the given pattern is a series of perfect squares so the next term will be $36,49$.

16. Which among ${\mathbf{4}}{{\mathbf{3}}^2}$,${\mathbf{6}}{{\mathbf{7}}^2}$,${\mathbf{5}}{{\mathbf{2}}^2}$,${\mathbf{5}}{{\mathbf{9}}^2}$ would end with digit ${\mathbf{1}}$? 

a) ${\mathbf{4}}{{\mathbf{3}}^2}$

b) ${\mathbf{6}}{{\mathbf{7}}^2}$

c) ${\mathbf{5}}{{\mathbf{2}}^2}$

d) ${\mathbf{5}}{{\mathbf{9}}^2}$

Ans: (d), ${59^2}$

Because unit digit of $59$ is $9$

And square of $9 \times 9 = 81$

17. A perfect square can never have the following digit in its ones place. 

a) ${\mathbf{1}}$

b) ${\mathbf{8}}$

c) ${\mathbf{0}}$

d) ${\mathbf{6}}$

Ans: (b), $8$.

$2,3,7,8$ are the numbers perfect square ends.

18. Which of the following numbers is not a perfect cube? 

a) ${\mathbf{216}}$

b) ${\mathbf{567}}$

c) ${\mathbf{125}}$

d) ${\mathbf{343}}$

Ans: (b), $567$.

Factors of $567$are $3 \times 3 \times 3 \times 3 \times 7$which do not form perfect pairs for cube. So $567$is not a perfect cube.

$\underline {3|567}$

$\underline {3|189}$

$\underline {3|63}$

$\underline {3|21}$

$\underline {7|7}$

$\underline {|1}$ 

19. $\sqrt[{\mathbf{3}}]{{{\mathbf{1000}}}}$ is equal to 

a) ${\mathbf{10}}$

b) ${\mathbf{100}}$

c) ${\mathbf{1}}$

d) None of these

Ans: (a), $10$

$\sqrt[3]{{1000}} = \sqrt[3]{{10 \times 10 \times 10}} = 10$

20. If m is the square of a natural number n, then n is 

a) The square of m 

b) Greater than m 

c) Equal to m 

d) $\sqrt {\mathbf{m}}$

Ans: (d), $\sqrt {\text{m}}$

Square of ${\text{n}} = {\left( {\text{n}} \right)^2}$

According to the question:

${{\text{n}}^2} = {\text{m}}$

$\Rightarrow {\text{n}}{\text{ = }}\sqrt {\text{m}}$

21. A perfect square number having n digits where n is even will have square root with

a) ${\mathbf{n}} + {\mathbf{1}}$ digit 

b) $\frac{{\mathbf{n}}}{{\mathbf{2}}}$ digit

c) $\frac{{\mathbf{n}}}{{\mathbf{3}}}$ digit

d) $\frac{{{\mathbf{n}} + {\mathbf{1}}}}{{\mathbf{2}}}$digit

Ans: (b), $\frac{{\text{n}}}{2}$ digit

22. If m is the cube root of n, then n is

a) ${{\mathbf{m}}^3}$

b) $\sqrt {\mathbf{m}}$

c) $\frac{{\mathbf{m}}}{{\mathbf{3}}}$

d) $\sqrt[3]{{\mathbf{m}}}$

Ans: (a), ${{\text{m}}^3}$

${\text{m}} \times {\text{m}} \times {\text{m}}{\text{ =  }}{{\text{m}}^3}$

$\Rightarrow {{\text{m}}^3} = {\text{n}}$ 

23. The value of $\sqrt {248 + \sqrt {52 + \sqrt {144} } } $ is

a) ${\mathbf{14}}$

b) ${\mathbf{12}}$

c) ${\mathbf{16}}$

d) ${\mathbf{13}}$

Ans: (c), $16$

$\sqrt {248 + \sqrt {52 + \sqrt {144} } }$

$\Rightarrow \sqrt {248 + \sqrt {52 + 12} }$

$\Rightarrow \sqrt {248 + \sqrt {64} }$

$\Rightarrow \sqrt {248 + 8}$

$\Rightarrow \sqrt {256}$

$\Rightarrow \sqrt {16 \times 16}  = 16$ 

$\because \sqrt {144}  = 12$

$\sqrt {64}  = 8$ 

24. Given that $\sqrt {{\mathbf{4096}}}  = {\mathbf{64}}$, the value of $\sqrt {{\mathbf{4096}}}  + \sqrt {{\mathbf{40}}.{\mathbf{96}}}$ is 

a) ${\mathbf{74}}$

b) ${\mathbf{60}}.{\mathbf{4}}$

c) ${\mathbf{64}}.{\mathbf{4}}$

d)${\mathbf{70}}.{\mathbf{4}}$

Ans: (d), $70.4$

As $\sqrt {4096}  = 64$

Also,$40.96$ can be written as $\frac{{1024}}{{25}}$

And, $\sqrt {\frac{{1024}}{{25}}}  = \sqrt {\frac{{32 \times 32}}{{5 \times 5}}}  = \frac{{32}}{5}$

Then,

$\sqrt {4096}  + \sqrt {40.64}$

$\Rightarrow 64 + \frac{{32}}{5}$

$\Rightarrow 64 + 6.4$

$\Rightarrow 70.4$ 

In questions 25 to 48, fill in the blanks to make the statements true. 

25. There are _________ perfect squares between ${\mathbf{1}}$ and ${\mathbf{100}}$. 

Ans: $8$

Perfect square between $1$ and $100$are:

$2 \times 2 = 45 \times 5 = 258 \times 8 = 64$

$3 \times 3 = 96 \times 6 = 369 \times 9 = 81$

$4 \times 4 = 167 \times 7 = 49$

26. There are _________ perfect cubes between ${\mathbf{1}}$ and ${\mathbf{1000}}$. 

Ans: 8

There are $8$ perfect cubes between $1$and $1000$

$2 \times 2 \times 2 = 85 \times 5 \times 5 = 1258 \times 8 \times 8 = 512$

$3 \times 3 \times 3 = 276 \times 6 \times 6 = 369 \times 9 \times 9 = 729$

$4 \times 4 \times 4 = 647 \times 7 \times 7 = 343$ 

27. The unit digit in the square of ${\mathbf{1294}}$ is _________.

Ans: $6$

Unit digit in $1294$ = 4

$4 \times 4 = 16$

Therefore, the square of $1294$ will have $6$ at its unit place.

28. The square of ${\mathbf{500}}$ will have _________ zeroes. 

Ans: $4$

$500 \times 500 = 250000$

29. There are _________ natural numbers between ${{\mathbf{n}}^2}$ and ${\left( {{\mathbf{n}} + {\mathbf{1}}} \right)^2}$

Ans: $2{\text{n}}$

$[{({\text{n + 1)}}^2} - {{\text{n}}^2}] - 1 = [{{\text{n}}^2} + 2{\text{n}}{\text{ + 1}} - {{\text{n}}^2}] - 1$

$\Rightarrow {{\text{n}}^2}{\text{ + }}{\text{2n  + 1}} - {{\text{n}}^2} - 1$

$\Rightarrow {\text{2n + 1}} - 1$

$\Rightarrow {\text{2n }}$ 

30. The square root of ${\mathbf{24025}}$ will have _________ digits. 

Ans: $3$

Number of digits in a square root for odd digit is given by $= \frac{{{\text{n}} + 1}}{2}$

Number of digits in a given number is $5({\text{odd)}}$

Therefore

${\text{n  =  5}}$

$\therefore \frac{{{\text{n + 1}}}}{2} = \frac{{5 + 1}}{2} = \frac{6}{2} = 3$ 

31. The square of ${\mathbf{5}}.{\mathbf{5}}$ is _________. 

Ans: $30.25$

$5.5 \times 5.5 = 30.25$

32. The square root of ${\mathbf{5}}.{\mathbf{3}} \times {\mathbf{5}}.{\mathbf{3}}$ is _________. 

Ans: 28.09

$5.3 \times 5.3 = 28.09$

33. The cube of ${\mathbf{100}}$ will have _________ zeroes. 

Ans: $6$

${100^3} = 100 \times 100 \times 100 = 1000000$

34. ${\mathbf{1}}{{\mathbf{m}}^{\mathbf{2}}} =$ _________ ${\mathbf{c}}{{\mathbf{m}}^{\mathbf{2}}}$.

Ans: $10000$

$1{\text{m}}{\text{ = 100 cm}}$

$\therefore {\text{1}}{{\text{m}}^2} = 100{\text{cm}} \times {\text{100cm}}$

${\text{ = }}{\text{10000c}}{{\text{m}}^2}$ 

35. ${\mathbf{1}}{{\mathbf{m}}^{\mathbf{3}}} =$ _________ ${\mathbf{c}}{{\mathbf{m}}^{\mathbf{3}}}$. 

Ans: $1000000$

$1{\text{m}}{\text{ = 100 cm}}$

$\therefore {\text{1}}{{\text{m}}^3} = 100{\text{cm}} \times {\text{100cm}} \times 100{\text{cm}}$

${\text{ = }}{\text{1000000c}}{{\text{m}}^3}$ 

36. Ones digit in the cube of ${\mathbf{38}}$ is _________. 

Ans: $2$

Unit digit is $8$ in $38$

$\therefore 8 \times 8 \times 8 = 512$

Therefore, unit digit in the cube of $38$ is $2$.

37. The square of ${\mathbf{0}}.{\mathbf{7}}$ is _________. 

Ans: $0.49$

Therefore, square of $0.7$ is

$0.7 \times 0.7 = 0.49$

38. The sum of first six odd natural numbers is _________. 

Ans: $36$

First six odd natural numbers are $1,3,5,7,9,11$

Sum of First six odd natural numbers are $= 1 + 3 + 5 + 7 + 9 + 11$

$= 36$

39. The digit at the ones place of ${\mathbf{5}}{{\mathbf{7}}^{\mathbf{2}}}$ is _________. 

Ans: $9$

As unit digit of $57$ is $7$

$\therefore 7 \times 7 = 49$

40. The sides of a right triangle whose hypotenuse is ${\mathbf{17cm}}$ are _________ and _________.

Ans: $8$ and v $15$

For Pythagorean triplet sides are given by $2{\text{m,}}{{\text{m}}^2} - 1,{{\text{m}}^2} + 1$

${{\text{m}}^2} + 1 = {(2{\text{m)}}^2} + {({{\text{m}}^2} - 1)^2}$

$\because {{\text{m}}^2} + 1 = 17$

$\Rightarrow {{\text{m}}^2} = 17 - 1$

$\Rightarrow {{\text{m}}^2} = 16$

$\Rightarrow {\text{m}}{\text{ =  4}}$ 

Then,

$2{\text{m}}{\text{ = }}{\text{2}} \times {\text{4 = 8}}$

${{\text{m}}^2} - 1 = {\left( 4 \right)^2} - 1$

$\Rightarrow 16 - 1 = 15$ 

41. $\sqrt {{\mathbf{1}}.{\mathbf{96}}}$______. 

Ans: $1.4$

$1.96$ can be written as $\frac{{196}}{{100}}$

$\therefore \sqrt {\frac{{196}}{{100}}}  = \sqrt {\frac{{14 \times 14}}{{10 \times 10}}}  = \frac{{14}}{{10}} = 1.4$

42. ${\left( {{\mathbf{1}}.{\mathbf{2}}} \right)^{\mathbf{3}}} =$_____.

Ans: $1.728$

$1.2$ can be written as $\frac{{12}}{{10}}$

$\therefore {\left( {1.2} \right)^3} = \frac{{12}}{{10}} \times \frac{{12}}{{10}} \times \frac{{12}}{{10}} = \frac{{1728}}{{1000}} = 1.728$

43. The cube of an odd number is always an _________ number. 

Ans: always an odd number

44. The cube root of a number x is denoted by _________. 

Ans:  $\sqrt[3]{x}$

45. The least number by which ${\mathbf{125}}$ be multiplied to make it a perfect square is _____________. 

Ans: 5

$\underline {5|125}$

$\underline {5|25}$

$\underline {5|5}$

$\underline {|1}$ 

On taking L.C.M we have factors of $125 = 5 \times 5 \times 5$

On grouping these factors in double of equal factors,

$125 = 5 \times 5 \times 5$

We have only one $5$ without a pair. 

So, $125$ to be multiplied by $5$ to make it a perfect square. 

46. The least number by which ${\mathbf{72}}$ be multiplied to make it a perfect cube is _____________. 

Ans: $3$

$\underline {2|72}$

$\underline {2|36}$

$\underline {2|18}$

$\underline {3|9}$

$\underline {3|3}$

$\underline {|1}$ 

On taking L.C.M we have factors of $72 = 2 \times 2 \times 2 \times 3 \times 3$

On grouping these factors in triple of equal factors,

$72 = 2 \times 2 \times 2 \times 3 \times 3$

We have only one $3$ without a pair. 

So, $72$ to be multiplied by $3$ to make it a perfect cube.

47. The least number by which 72 be divided to make it a perfect cube is _____________. 

Ans: $9$

$\underline {2|72}$

$\underline {2|36}$

$\underline {2|18}$

$\underline {3|9}$

$\underline {3|3}$

$\underline {|1}$ 

On taking L.C.M we have factors of $72 = 2 \times 2 \times 2 \times 3 \times 3$

On grouping these factors in triple of equal factors,

$72 = 2 \times 2 \times 2 \times 3 \times 3$

We have only $3$ which is not a triplet. 

So, $72$ to be divided by $9$ to make it a perfect cube ${\text{i}}{\text{.e}}$$8$

48. Cube of a number ending in ${\mathbf{7}}$ will end in the digit _______________.

Ans: $3$

On taking the cube of number $7$ itself we have,

$7 \times 7 \times 7 = 343$

Therefore, we have $3$ at the end.

In questions 49 to 86, state whether the statements are true (T) or false (F). 

49. The square of ${\mathbf{86}}$ will have ${\mathbf{6}}$ at the units place. 

Ans: The given statement is true.

Because square of the number ending with $4{\text{or 6}}$ have $6$ at unit place.

50. The sum of two perfect squares is a perfect square. 

Ans: The given statement is false.

For example: taking two perfect squares, 

Here, 

$4 + 9 = 13$

So, $13$ is not a perfect square.

51. The product of two perfect squares is a perfect square.

Ans: The given statement is true.

For example, taking two perfect squares, 

Here, 

$4 \times 9 = 36$

So, $36$ is a perfect square.

52. There is no square number between ${\mathbf{50}}$ and ${\mathbf{60}}.$

Ans: The given statement is true.

53. The square root of ${\mathbf{1521}}$ is ${\mathbf{31}}$. 

Ans: The given statement is false.

because square of $31 = {\left( {31} \right)^2} = 31 \times 31 = 961$

54. Each prime factor appears ${\mathbf{3}}$ times in its cube. 

Ans: The given statement is true.

As cubes are represented as the product of triplets of prime factors.

55. The square of ${\mathbf{2}}.{\mathbf{8}}$ is ${\mathbf{78}}.{\mathbf{4}}$. 

Ans: The given statement is false.

because square of $2.8 = {\left( {2.8} \right)^2}$

$= 2.8 \times 2.8 = 7.84$

56. The cube of ${\mathbf{0}}.{\mathbf{4}}$ is ${\mathbf{0}}.{\mathbf{064}}$. 

Ans: The given statement is true.

${\left( {0.4} \right)^3} = \frac{{4 \times 4 \times 4}}{{10 \times 10 \times 10}} = \frac{{64}}{{1000}} = 0.064$

57. The square root of ${\mathbf{0}}.{\mathbf{9}}$ is ${\mathbf{0}}.{\mathbf{3}}$. 

Ans: The given statement is false.

because square of $0.3 = {\left( {0.3} \right)^2}$

= $0.3 \times 0.3 = 0.09$

58. The square of every natural number is always greater than the number itself. 

Ans: The given statement is false.

Because the square of ${\left( 1 \right)^2} = 1$which is not greater than $1$

59. The cube root of ${\mathbf{8000}}$ is ${\mathbf{200}}$. 

Ans: The given statement is false.

 As, $\sqrt[3]{{8000}} = \sqrt {2 \times 2 \times 2 \times 10 \times 10 \times 10}  = 2 \times 10 = 20$

60. There are five perfect cubes between ${\mathbf{1}}$ and ${\mathbf{100}}$. 

Ans: The given statement is false.

There are 8 perfect cubes between $1$ and $100$

$2 \times 2 = 45 \times 5 = 258 \times 8 = 64$

$3 \times 3 = 96 \times 6 = 369 \times 9 = 81$

$4 \times 4 = 167 \times 7 = 49$

61. There are ${\mathbf{200}}$ natural numbers between ${\mathbf{10}}{{\mathbf{0}}^{\mathbf{2}}}$ and${\mathbf{10}}{{\mathbf{1}}^{\mathbf{2}}}$.

Ans: The given statement is true.

${\left( {101} \right)^2} = 10201$

${\left( {100} \right)^2} = 10000$

$\therefore {\left( {101} \right)^2} - {\left( {100} \right)^2} = 10201 - 10000 = 201$ 

62. The sum of first ${\text{n}}$odd natural numbers is ${{\text{n}}^{\mathbf{2}}}$. 

Ans: The given statement is true.

Sum of first ${\text{n}}$ odd natural number is given by:

$\Rightarrow \Sigma \left( {2{\text{n - 1}}} \right)$

$\Rightarrow \left( {2 \times {\text{n}} \times \left( {{\text{n + 1}}} \right)} \right)/2 - {\text{n}}$

$\Rightarrow {{\text{n}}^2} + {\text{n}} - {\text{n}}$

$\Rightarrow {{\text{n}}^2}$

63. ${\mathbf{1000}}$ is a perfect square. 

Ans: The given statement is false.

$\underline {2|1000}$

$\underline {2|500}$

$\underline {2|250}$

$\underline {5|125}$

$\underline {5|25}$

$\underline {5|5}$

$\underline {|1}$ 

Factors of $1000$

$2 \times 2 \times 2 \times 5 \times 5 \times 5 = 1000$, So clearly $1000$ is a perfect cube but not square.

So, 1000 is not a perfect square. 

64. A perfect square can have ${\mathbf{8}}$ as its unit’s digit. 

Ans: The given statement is false.

Because for every digit that ends with an even number either has $4$ or $6$ at its unit place.

65. For every natural number ${\mathbf{m}},\left( {{\mathbf{2m}}-{\mathbf{1}},{\mathbf{2}}{{\mathbf{m}}^{\mathbf{2}}}-{\mathbf{2m}},{\mathbf{2}}{{\mathbf{m}}^{\mathbf{2}}}-{\mathbf{2m}} + {\mathbf{1}}} \right)$ is a Pythagorean triplet. 

Ans: The given statement is false.

$\left( {2{\text{m}},{{\text{m}}^2} - 1{\text{, }}{{\text{m}}^2}{\text{ + 1}}} \right)$ is a Pythagorean triplet for every natural number ${\text{m}}{\text{ > 1}}$.

66. All numbers of a Pythagorean triplet are odd. 

Ans: The given statement is false.

Condition for Pythagorean triplet is = square of one should be equal to sum of square of other two.

For example, ${3^2} = {5^2} + {4^2}$

Hence, $4$ is not an odd number.

67. For an integer ${\mathbf{a}}$, ${{\mathbf{a}}^{\mathbf{3}}}$ is always greater than ${{\text{a}}^{\mathbf{2}}}$. 

Ans: The given statement is false. 

If ${\text{a}}$ is a negative integer then ${{\text{a}}^{3}} < {{\text{a}}^2}$.

68. If $X$ and $Y$ are integers such that ${X^2} > {Y^2}$, then${X^3} > {Y^3}$. 

Ans: The given statement is false.

If $X$ is a negative integer then ${X^{3}} < {Y^3}$

Suppose, $X =  - 1$ then,

${X^2} = {\left( { - 1} \right)^2} = 1$

${X^3} = {\left( { - 1} \right)^3} =  - 1$ 

And $Y = 2$ then,

${\left( Y \right)^2} = {\left( 2 \right)^2} = 4$

${\left( Y \right)^3} = {\left( 2 \right)^3} = 8$ 

Therefore, clearly ${X^{3}} < {Y^3}$.

69. Let $x$ and $y$ be natural numbers. If  $x$ divides $y$ , then ${{\mathbf{x}}^3}$ divides ${{\mathbf{y}}^3}$. 

Ans: The given statement is true.

If $x$ and $y$ are natural numbers then ${x^3}$ and ${y^3}$ will also be natural numbers as they are nothing but multiple of $x$ and $y$ respectively.

70. If ${{\mathbf{a}}^{\mathbf{3}}}$ends in ${\mathbf{5}}$, then  ${{\mathbf{a}}^{\mathbf{3}}}$ends in ${\mathbf{25}}$.

Ans: The given statement is false. 

If ${{\text{a}}^2} = {35^2} = 1225$

Then,

${{\text{a}}^3} = {\left( {35} \right)^3} = 42875$

71. If ${{\mathbf{a}}^{\mathbf{2}}}$ ends in ${\mathbf{9}}$, then ${{\mathbf{a}}^{\mathbf{3}}}$ ends in ${\mathbf{7}}$.

Ans: The given statement is false.

If ${{\text{a}}^2} = {7^2} = 49$

Then,

${{\text{a}}^3} = {7^3} = 343$

72. The square root of a perfect square of ${\mathbf{n}}$ digits will have $\frac{{{\mathbf{n}} + {\mathbf{1}}}}{{\mathbf{2}}}$ digits, if ${\mathbf{n}}$is odd. 

Ans: The given statement is true.

For example, $10000$ is a perfect square with $5$ digit then its square root will have 

$\Rightarrow \frac{{{\text{n + 1}}}}{2} = \frac{{5 + 1}}{2} = \frac{6}{2} = 3$ which is true as we know $10000$ is a perfect square of $100$ which is a $3$ digit number.

73. Square root of a number ${\mathbf{x}}$ is denoted by $\sqrt {\mathbf{x}}$.

Ans: The given statement is true.

74. A number having ${\mathbf{7}}$ at its one’s place will have ${\mathbf{3}}$ at the unit place of its square.

Ans: The given statement is false.

As square of $7 = 7 \times 7 = 49$

Square of $27 = 27 \times 27 = 729$

Square of $47 = 47 \times 47 = 2209$

And so on.

75. A number having ${\mathbf{7}}$ at its one’s place will have ${\mathbf{3}}$ at the unit place of its cube.

Ans: The given statement is true.

As cube of $17 = 17 \times 17 \times 17 = 4913$

cube of $27 = 27 \times 27 \times 27 = 19683$

cube of $37 = 37 \times 37 \times 37 = 50,653$

And so on. 

76. The cube of a one-digit number cannot be a two-digit number.

Ans: The given statement is false.

e.g. $5$ is a one-digit number, and ${\left( 5 \right)^3} = 125$ (Two-digit number)

77. Cube of an even number is odd.

Ans: The given statement is false.

For example: 

${\left( 2 \right)^3} = 8$

${\left( 6 \right)^3} = 216$ 

Therefore, cube of an even number is always an even number.

78. Cube of an odd number is even.

Ans: The given statement is false

For example:

${\left( 3 \right)^3} = 27$

${\left( 5 \right)^3} = 125$ 

Therefore, cube of an odd number is always an odd number.

79. Cube of an even number is even.

Ans: The given statement is true.

For example: 

${\left( 2 \right)^3} = 8$

${\left( 6 \right)^3} = 216$ 

80. Cube of an odd number is odd.

Ans: The given statement is true.

For example:

${\left( 3 \right)^3} = 27$

${\left( 5 \right)^3} = 125$ 

81. ${\mathbf{999}}$ is a perfect cube.

Ans: The given statement is false.

$\underline {3|999}$

$\underline {3|333}$

$\underline {3|111}$

$\underline {37|37}$

$\underline {|1}$ 

Factors of 999 = $3 \times 3 \times 3 \times 37$

Therefore, $999$ is not a perfect cube.

82. ${\mathbf{363}} \times {\mathbf{81}}$ is a perfect cube.

Ans: The given statement is false.

Breaking $363 \times 81$into factors we have:

$= 363 \times 81$

$= 3 \times 11 \times 11 \times 3 \times 3 \times 3 \times 3$ 

There is only one triplet pair therefore, $363 \times 81$ is not a perfect cube.

83. Cube roots of ${\mathbf{8}}$ are $ + {\mathbf{2}}$ and $-{\mathbf{2}}$.

Ans: The given statement is false.

Because $\sqrt[3]{8} = 2$

84.$\sqrt[3]{{{\mathbf{8}} + {\mathbf{27}}}}{\text{  = }}\sqrt[{\mathbf{3}}]{{\mathbf{8}}} + \sqrt[{\mathbf{3}}]{{{\mathbf{27}}}}$

Ans: The given statement is false.

As $\sqrt[3]{{8 + 27}} = \sqrt[3]{8} + \sqrt[3]{{27}}$

$\Rightarrow \sqrt[3]{{35}} = 2 + 3$

$\Rightarrow \sqrt[3]{{35}} = 5$ 

Which is not equal.

85. There is no cube root of a negative integer.

Ans: The given statement is false.

Let, $-8$ is a negative number then,

$\sqrt[3]{{ - 8}} =  - 2$.

86. Square of a number is positive, so the cube of that number will also be positive.

Ans: The given statement is false.

Let, $-2$ is negative number.

Square of $-2 = 4$

Cube of $- 2 =  - 8$

Solve the Following Questions

87. Write the first five square numbers.

Ans: First five square numbers are:

${\left( 1 \right)^2} = 1,{\left( 2 \right)^2} = 4,{\left( 3 \right)^2} = 9,{\left( 4 \right)^2} = 16,{\left( 5 \right)^2} = 25$

88. Write cubes of first three multiples of ${\mathbf{3}}$.

Ans: $3,6$ and$9$ are the first three multiple of $3$ then,

Cube of $3 = 3 \times 3 \times 3 = 27$

Cube of $6 = 6 \times 6 \times 6 = 216$

Cube of $9 = 9 \times 9 \times 9 = 729$

89. Show that ${\mathbf{500}}$ is not a perfect square.

Ans: $500$ is not a perfect square because

$\sqrt {500}  = \sqrt {\underline {10 \times 10}  \times 5}$

There is no pair for $5$.

90. Express ${\mathbf{81}}$ as the sum of first nine consecutive odd numbers.

Ans: $1,3,5,7,9,11,13,15$ and $17$ are the first nine consecutive odd numbers

Now, Sum of first nine consecutive odd numbers

$= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 = 81$

91. Using prime factorisation, find which of the following perfect squares are.

a) ${\mathbf{484}}$

b) ${\mathbf{11250}}$

c) ${\mathbf{841}}$

d) ${\mathbf{729}}$

Ans: 

1. $484$

$\underline {2|484}$

$\underline {2|242}$

$\underline {11|121}$

$\underline {11|11}$

$\underline {|1}$ 

On grouping in pairs, there is no unpaired factor left 

$484 = 2 \times 2 \times 11 \times 11$

So, $484$ is a perfect square.

2. $11250$

$\underline {2|11250}$

  $\underline {3|5625}$

$\underline {3|1875}$

$\underline {5|625}$

$\underline {5|125}$

  $\underline {5|25}$

$\underline {5|5}$

$\underline {|1}$ 

$11250{\text{ }} = {\text{ }}2 \times \underline {3 \times 3}  \times \underline {5 \times 5}  \times \underline {5 \times 5}$ on grouping in pair, $2$ have zero pair.

Therefore, $11250$ is not a perfect square.

3. $841$

$\underline {29|841}$

   $\underline {29|29}$

   $\underline {|1}$ 

\[841 = 29 \times 29\]

On grouping in pair, there is no unpaired factor left over. Therefore, $841$ is a perfect square.

4. $729$

$\underline {3|729}$

$\underline {3|243}$

   $\underline {3|81}$

   $\underline {3|27}$

   $\underline {3|9}$

   $\underline {3|3}$

   $\underline {|1}$ 

$729 = \underline {3 \times 3}  \times \underline {3 \times 3}  \times \underline {3 \times 3}$

On grouping in pair, there is no unpaired factor left over. Therefore, $729$ is a perfect square.

92. Using prime factorisation, find which of the following are perfect cubes.

a) ${\mathbf{128}}$

b) ${\mathbf{343}}$

c) ${\mathbf{729}}$

d) ${\mathbf{1331}}$

Ans: 

1. $128$

   $\underline {2|128}$

   $\underline {2|64}$

   $\underline {2|32}$

   $\underline {2|16}$

   $\underline {2|8}$

   $\underline {2|4}$

   $\underline {2|2}$

   $\underline {|1}$ 

Hence, factors of $128{\text{ }} = {\text{ }}\underline {2 \times 2 \times 2}  \times \underline {2 \times 2 \times 2}  \times 2$ on grouping in triplets for perfect cube, $2$ does not form triplet. Therefore, $128$is not a perfect cube.

2. $343$

   $\underline {7|343}$

   $\underline {7|49}$

   $\underline {7|7}$

   $\underline {|1}$ 

Hence, factors of $343 = \underline {7 \times 7 \times 7}$ on grouping in triplets for perfect cube, $7$ forms triplet. Therefore, $343$is a perfect cube.

3. $729$

    $\underline {3|729}$

   $\underline {3|243}$

   $\underline {3|81}$

   $\underline {3|27}$

   $\underline {3|9}$

   $\underline {3|3}$

   $\underline {|1}$ 

Hence, factors of $729 = \underline {3 \times 3 \times 3}  \times \underline {3 \times 3 \times 3}$ on grouping in triplets for perfect cube, $3$ forms triplet. Therefore, $729$ is a perfect cube

4. $1331$

        $\underline {11|1331}$

   $\underline {11|121}$

   $\underline {11|11}$

   $\underline {|1}$ 

Hence, factors of $1331 = \underline {11 \times 11 \times 11}$ on grouping in triplets for perfect cube, $11$ forms triplet. Therefore, $1331$ is a perfect cube.

93. Using distributive law, find the squares of

a) ${\mathbf{101}}$

b) ${\mathbf{72}}$

Ans: 

1. $101$

Finding square using distributive law:

$101 = 100 + 1$

Therefore, ${\left( {101} \right)^2} = {\left( {100 + 1} \right)^2}$

$\Rightarrow \left( {100 + 1} \right)\left( {100 + 1} \right)$

$\Rightarrow \left( {100 \times 100} \right) + 100 + 100 + \left( {1 \times 1} \right)$

    $\Rightarrow 10000 + 200 + 1$

    $\Rightarrow 10201$ 

$\therefore {\left( {101} \right)^2} = 10201$

2. $72$

Finding square using distributive law:

$72 = 70 + 2$

Therefore, ${\left( {72} \right)^2} = {\left( {70 + 2} \right)^2}$

$\Rightarrow \left( {70 + 2} \right)\left( {70 + 2} \right)$

$\Rightarrow \left( {70 \times 70} \right) + \left( {70 \times 2} \right) + \left( {70 \times 2} \right) + \left( {2 \times 2} \right)$

$\Rightarrow 4900 + 140 + 140 + 4$

  $\Rightarrow 5184$ 

$\therefore {\left( {72} \right)^2} = 5184$

94. Can a right triangle with sides ${\mathbf{6cm}}$, ${\mathbf{10cm}}$ and ${\mathbf{8cm}}$ be formed? Give reason.

Ans: We know the sum of two smaller sides of a triangle is always equal to the square of longer side of a triangle. 

Given, Smaller sides $= 6{\text{cm}}$ and $8{\text{cm}}$

Longer side $ = 10{\text{cm}}$

As per the rule,

${\left( {10} \right)^2} = {\left( 6 \right)^2} + {\left( 8 \right)^2}$

$100 = 36 + 64$

$100 = 100$ 

Hence, ${\text{L}}{\text{.H}}{\text{.S}}{\text{ = }}{\text{R}}{\text{.H}}{\text{.S}}$

95. Write the Pythagorean triplet whose one of the numbers is ${\mathbf{4}}$.

Ans: Pythagorean triplets for any natural number greater than $1$ are

$2{\text{m,}}{{\text{m}}^2} - 1{\text{and}}{{\text{m}}^2} + 1$

So, if one number is $2{\text{m}}$, then another two numbers will be ${{\text{m}}^2} - 1{\text{and}}{{\text{m}}^2} + 1$

Given, one number = 4

Then Pythagorean triplets are:

$2{\text{m}}{\text{ = }}{\text{4}}$

$\therefore {\text{m}}{\text{ = }}{\text{2}}$ 

Then, ${{\text{m}}^2} - 1 = {\left( 2 \right)^2} - 1 = 4 - 1 = 3$

Also, ${{\text{m}}^2} + 1 = {\left( 2 \right)^2} + 1 = 4 + 1 = 5$

Therefore, $3$, $4$ and \[5\] are Pythagorean triplets.

96. Using prime factorisation, find the square roots of

a) ${\mathbf{11025}}$

b) ${\mathbf{4761}}$

Ans: 

1. $11025$

$\underline {3|11025}$

   $\underline {3|3675}$

   $\underline {5|1225}$

   $\underline {5|245}$

   $\underline {7|49}$

   $\underline {7|7}$

   $\underline {|1}$ 

Hence, factors of $11025 = \underline {3 \times 3} \times \underline {5 \times 5}  \times \underline {7 \times 7}$ on grouping in pair, therefore, $11025$ is a perfect square.

Hence, $\sqrt {11025}  = \sqrt {\underline {3 \times 3} \times \underline {5 \times 5}  \times \underline {7 \times 7} }  = 3 \times 5 \times 7 = 105$

2. $4761$

   $\underline {3|4761}$

   $\underline {3|1587}$

   $\underline {23|529}$

   $\underline {23|23}$

   $\underline {|1}$ 

Hence, factors of $4761 = \underline {3 \times 3} \times \underline {23 \times 23}$ on grouping in pair, therefore, $4761$ is a perfect square.

Hence, $\sqrt {4761}  = \sqrt {\underline {3 \times 3} \times \underline {23 \times 23} }  = 3 \times 23 = 69$

97. Using prime factorisation, find the cube roots of

a) ${\mathbf{512}}$

b) ${\mathbf{2197}}$

Ans: 

1. $512$

   $\underline {2|512}$

   $\underline {2|256}$

   $\underline {2|128}$

   $\underline {2|64}$

   $\underline {2|32}$

   $\underline {2|16}$

   $\underline {2|8}$

   $\underline {2|4}$

   $\underline {2|2}$

   $\underline {|1}$ 

Hence, factors of $512 = \underline {2 \times 2 \times 2} \times \underline {2 \times 2 \times 2}  \times \underline {2 \times 2 \times 2}$ on grouping in triplets for perfect cube, all number forms triplet. Therefore, $512$ is a perfect cube

Therefore, cube root of is $\sqrt[3]{{512}} = \sqrt {\underline {2 \times 2 \times 2}  \times \underline {2 \times 2 \times 2}  \times \underline {2 \times 2 \times 2} }  = 2 \times 2 \times 2 = 8$

2. $2197$

   $\underline {13|2197}$

   $\underline {13|169}$

   $\underline {13|13}$

   $\underline {|1}$ 

Hence, factors of $2197 = \underline {13 \times 13 \times 13}$ on grouping in triplets for perfect cube, all number forms triplet. Therefore, \[2197\] is a perfect cube.

Therefore, cube root of is $\sqrt[3]{{2197}} = \sqrt {\underline {13 \times 13 \times 13} }  = 13$

98. Are ${\mathbf{176}}$ a perfect square? If not, find the smallest number by which it should be multiplied to get a perfect square.

Ans: Taking factors of $176$

$\underline {2|176}$

$\underline {2|88}$

$\underline {2|44}$

$\underline {2|22}$

$\underline {11|11}$

$\underline {|1}$ 

\[176 = \underline {2 \times 2}  \times \underline {2 \times 2}  \times 11\] on grouping in pair, $11$ have a incomplete pair.

Therefore, $176$ is not a perfect square.

Therefore, \[11\] is the smallest number by which $176$ should be multiplied to get a perfect square. Now,

$176 \times 11 = 2 \times 2 \times 2 \times 2 \times 11 \times 11$

$= 4 \times 4 \times 121$

$= 1936 = {\left( {44} \right)^2}$ 

Hence, $1936$ is a perfect square of \[44\].

99. Is ${\mathbf{9720}}$ a perfect cube? If not, find the smallest number by which it should be divided to get a perfect cube.

Ans: Taking factors of $9720$

$\underline {2|9720}$

$\underline {2|4860}$

$\underline {2|2430}$

$\underline {3|1215}$

$\underline {3|405}$

$\underline {3|135}$

$\underline {3|45}$

$\underline {3|15}$

$\underline {5|5}$

$\underline {|1}$ 

$9720 = \underline {2 \times 2 \times 2}  \times \underline {3 \times 3 \times 3}  \times 3 \times 3 \times 5$ on grouping in triplet, $3$ and $5$ have incomplete triplet.

Therefore, $9726$ is not a perfect cube.

Therefore, on dividing the number by incomplete group of triplet. So, to make $9726$ a perfect cube:

$9726 \div \left( {3 \times 3 \times 5} \right) = 9726 \div 45$

$= 216 = {\left( 6 \right)^3}$ 

Now, 216 is a perfect cube. Hence, 45 is the smallest number by which it should be 

divided to get a perfect cube.

100. Write two Pythagorean triplets each having one of the numbers as ${\mathbf{5}}$.

Ans: Sum of two smaller sides in Pythagorean triplet is always equal to the square of longer side. If one of the numbers as $5$ then:

${\left( 5 \right)^2} = {\left( 3 \right)^2} + {\left( 4 \right)^2}$ and the other Pythagorean triplet is ${\left( {13} \right)^2} = {\left( {12} \right)^2} + {\left( 5 \right)^2}$

101. By what smallest number should ${\mathbf{216}}$ be divided so that the quotient is a perfect square? Also find the square root of the quotient.

Ans: Taking factors of $216$:

$\underline {2|216}$

$\underline {2|108}$

$\underline {2|54}$

$\underline {3|27}$

$\underline {3|9}$

$\underline {3|3}$

$\underline {|1}$ 

$216 = \underline {2 \times 2}  \times 2 \times \underline {3 \times 3}  \times 3$ Here, $2$ and $3$ have incomplete pair. So, $216$ is not a Perfect square. To make it a perfect square the number should be divided by the product of numbers which have incomplete pair ${\text{i}}{\text{.e}}{\text{.}} 2$ and $3$ therefore,

$\Rightarrow 216 \div 6 = 36$

And factors of $36 = \underline {2 \times 2} \times \underline {3 \times 3}$. Now, 36 is a perfect square. 

Therefore, the smallest number is 36 that should divide $216$ so that the quotient is a perfect square.

102. By what smallest number should ${\mathbf{3600}}$ be multiplied so that the quotient is a perfect cube? Also find the cube root of the quotient.

Ans: Taking factors of $3600$

$\underline {2|3600}$

$\underline {2|1800}$

$\underline {2|900}$

$\underline {2|450}$

$\underline {3|225}$

$\underline {3|75}$

$\underline {5|25}$

$\underline {5|5}$

$\underline {|1}$ 

$3600 = \underline {2 \times 2 \times 2}  \times 2 \times 3 \times 3 \times 5 \times 5$ Here, $2$, $3$ and 5 have incomplete group of triplets. So, $3600$ is not a Perfect cube. To make it a perfect cube the number should be multiplied by the product of numbers which have incomplete group of triplets ${\text{i}}{\text{.e}}{\text{.}}$ $2$, $3$ and 5 therefore,

$\Rightarrow 3600 \times \left( {2 \times 2 \times 3 \times 5} \right)$

$\Rightarrow 3600 \times 60 = 216000 = {\left( {60} \right)^3}$ 

And factors of $216000 = 60 \times 60 \times 60$. Now, $216000$ is a perfect cube. 

Therefore, the smallest number is 60 that should be multiplied so that the quotient is a perfect square.

103. Find the square root of the following by long division method.

a) ${\mathbf{1369}}$

b) ${\mathbf{5625}}$

Ans: finding square root using long division method:

1. $1369$

Therefore, square root of $1369$ $\Rightarrow \sqrt {1369}  = 37$

2.  $5625$

Therefore, square root of $5625$ $\Rightarrow \sqrt {5625}  = 75$

104. Find the square root of the following by long division method.

a) ${\mathbf{27}}.{\mathbf{04}}$

b) ${\mathbf{1}}.{\mathbf{44}}$

Ans: 

1. $27.04$

Finding square root using long division method:

Therefore, square root of $27.04$$ \Rightarrow $$\sqrt {27.04}  = 5.2$

2. $1.44$

Finding square root using long division method:

Therefore, square root of $1.44 \Rightarrow \sqrt {1.44}  = 1.2$

105. What is the least number that should be subtracted from ${\mathbf{1385}}$ to get a perfect square? Also find the square root of the perfect square.

Ans: finding square root of $1385$:

We can see that the remainder left is $16$ therefore, the least number that should be subtracted from $1385$ to get a perfect square is $16$. 

Hence, the required perfect square number $\Rightarrow 1385-16 = 1369$

As $1369 = 37 \times 37$.

106. What is the least number that should be added to ${\mathbf{6200}}$ to make it a perfect square?

Ans: finding square root of $6200$:

We can see that ${\left( {78} \right)^2} = 6084$ which is less than $6200$ therefore, the next perfect square is ${\left( {79} \right)^2} = 6241$ which is greater than 6000. So, the least number is the number that

should be added to $6200$ to get a perfect square.

$\Rightarrow \left( {6241-6200 = 41} \right)$

Therefore, 

$6200 + 41 = 6241$

$\Rightarrow 6241 = {\left( {79} \right)^2}$ 

Which is a perfect square.

107. Find the least number of four digits that is a perfect square.

Ans: Least of four-digit number $= 1000$

finding square root of $1000$ by long division method:

So, $24$ to be added in $1000$ to make it a perfect square $=$ 1000 + 24 = 1024.

108. Find the greatest number of three digits that is a perfect square.

Ans: Greatest three-digit number $= 999$

Finding square root of $999$

So, $38$ to be subtracted from 999 to get a perfect square. Therefore,

$\Rightarrow 999-38{\text{ }} = 961$ 

And $\sqrt {961}  = 31$

109. Find the least square number which is exactly divisible by ${\mathbf{3}}$,${\mathbf{4}}$,${\mathbf{5}}$,${\mathbf{6}}{\mathbf{and}}{\mathbf{8}}.$

Ans: Taking L.C.M of $3,4,5,6{\text{ and}}8$:

$\underline {2|3,4,5,6,8}$

$\underline {2|3,2,5,3,4}$

$\underline {2|3,1,5,3,2}$

$\underline {3|3,1,5,3,1}$

$\underline {5|1,1,5,1,1}$

$\underline {|1,1,1,1,1}$ 

L.C.M of $3,4,5,6{\text{ and}}8 = 2 \times 2 \times 2 \times 3 \times 5= 120$

As pair of $2,{\text{ }}3{\text{ and }}5$ is not completed. To make it a perfect square, the number should be multiplied with $2 \times 3 \times 5 = 30$

$\Rightarrow 120 \times 30 = 3600$

Hence, the least square number which is exactly divisible by $3,4,5,6{\text{ and}}8$ is $3600$. 

110. Find the length of the side of a square if the length of its diagonal is${\mathbf{10cm}}$.

Ans: Length of diagonal $= 10{\text{cm}}$

Now, let the length of side of a square is $= x{\text{cm}}$

Then by Pythagoras theorem:

$\Rightarrow {\left( {10} \right)^2} = {\left( x \right)^2} + {\left( x \right)^2}$

$\Rightarrow 100 = 2{x^2}$

$\Rightarrow {x^2} = 50$

$\Rightarrow x = 5\sqrt 2$ 

Therefore, the length of the side of square is $\sqrt {50} {\text{or}}{\text{5}}\sqrt 2$.

111. A decimal number is multiplied by itself. If the product is${\mathbf{51}}.{\mathbf{84}}$, find the number.

Ans: Given, the product of the number is $= 51.84$

Now, let the number be $x$ and the product of number $= x \times x = {x^2}$

According to the question:

${x^2} = 51.84$

Now, finding the square root of $51.84$

$\therefore \sqrt {51.84}  = 7.2$

112. Find the decimal fraction which when multiplied by itself gives 84.64.

Ans: Given, the product of the number is $= 84.64$

Now, let the number be $x$ and the product of number $= x \times x = {x^2}$

According to the question:

${x^2} = 84.64$

Now, finding the square root of $84.64$

$\sqrt {84.64}  = 9.2$ therefore, the required decimal fraction is $9.2$.

113. A farmer wants to plough his square field of side 150m. How much area will he have to plough?

Ans: Given, side of square field \[ = 150{\text{m}}\]

We know, area of square $=  side \times side$ 

$\therefore 150{\text{m}} \times {\text{150}}{\text{m}} = 2250{{\text{m}}^2}$

114. What will be the number of unit squares on each side of a square graph paper if the total number of unit squares is ${\mathbf{256}}$?

Ans: Given, 

Total number of unit squares $= 256$

If the number is $x$ then:

The number of unit square $\Rightarrow $ $x \times x = {x^2}$

$\therefore {x^2} = 256$

$\Rightarrow x = \sqrt {256}$

$\Rightarrow x = 16$ 

So, the number of unit square $= 16$

115. If one side of a cube is ${\mathbf{15}}{\mathbf{m}}$in length, find its volume.

Ans: Given, side of a cube $= 15{\text{m}}$

We know, Volume of cube = ${\text{sid}}{{\text{e}}^3}$

$\therefore $ Volume of cube $= {\left( {15{\text{m}}} \right)^3} = 3375{{\text{m}}^3}$.

116. The dimensions of a rectangular field are ${\mathbf{80}}{\mathbf{m}}$ and ${\mathbf{18}}{\mathbf{m}}$. Find the length of its diagonal.

Ans: Given, 

length of a rectangular field (l) $= 80{\text{m}}$

breath of a rectangular field (b) $= 18{\text{m}}$

We know, length of diagonal $= \sqrt {{l^2} + {b^2}}$

$\Rightarrow \sqrt {{{\left( {80} \right)}^2} + {{\left( {18} \right)}^2}}$

$\Rightarrow \sqrt {6400 + 324}$

$\Rightarrow \sqrt {6724}  = 82{\text{m}}$ 

117. Find the area of a square field if its perimeter is ${\mathbf{96}}{\mathbf{m}}.$

Ans: Given, Perimeter of square $= 96{\text{m}}$

We know, Perimeter of a square $= 4 \times {\text{side}}$

$\Rightarrow 4 \times {\text{side}} = 96{\text{m}}$

$\Rightarrow {\text{side  = }}\frac{{96}}{4} = 24{\text{m}}$ 

Now, the area of square $= {\text{side}} \times {\text{side}}$

$\Rightarrow 24{\text{m}} \times {\text{24}}{\text{m}}$