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. $(100\times 100\times 100)=1000000$

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

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 $90$:

$\underset{―}{2|90}$

$\underset{―}{3|45}$

$\underset{―}{3|15}$

$\underset{―}{5|5}$

$\underset{―}{|1}$

$90=2\times 3\times 3\times 5$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, $432={(40+3)}^2=(40+3)(40+3)=40(40+3)+3(40+3)$

$=40\times 40+40\times 3+3\times 40+3\times 3$

$=1600+240+9$

$=1849$

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 $5832$:

$\underset{―}{2|5832}$

$\underset{―}{2|2916}$

$\underset{―}{2|1458}$

$\underset{―}{3|729}$

$\underset{―}{3|243}$

$\underset{―}{3|81}$

$\underset{―}{3|27}$

$\underset{―}{3|9}$

$\underset{―}{3|3}$

$\underset{―}{|1}$

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

$5832=\underset{―}{2\times 2\times 2}\times \underset{―}{3\times 3\times 3}\times \underset{―}{3\times 3\times 3}$

$=2\times 3\times 3$

$=18$

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 $3,4,5\text{ and }6$ is:

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

$\underset{―}{2|3,4,5,6}$

$\underset{―}{2|3,2,5,3}$

$\underset{―}{3|3,1,5,3}$

$\underset{―}{5|1,1,5,1}$

$\underset{―}{|1,1,1,1}$

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

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

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

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

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

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 $=2450{\text{m}}^2.$

Therefore,

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

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

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

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

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

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

Now the, perimeter of the field $$=2(l+b)$$

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

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

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 $=6250$

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

$\underset{―}{2|1620}$

$\underset{―}{2|810}$

$\underset{―}{5|405}$

$\underset{―}{3|81}$

$\underset{―}{3|27}$

$\underset{―}{3|9}$

$\underset{―}{3|3}$

$\underset{―}{|1}$

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

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

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 $5$

$1620÷5=324$

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

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: 

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

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

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

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

$\Rightarrow {\text{n}}^2$ 

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,

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

$=9x^2+16x^2$ 

$=25x^2$ 

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

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

$\underset{―}{3|567}$

$\underset{―}{3|189}$

$\underset{―}{3|63}$

$\underset{―}{3|21}$

$\underset{―}{7|7}$

$\underset{―}{|1}$ 

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 $\mathbf{17}\mathbf{cm}$ 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

$\underset{―}{5|125}$

$\underset{―}{5|25}$

$\underset{―}{5|5}$

$\underset{―}{|1}$ 

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: $3$

$\underset{―}{2|72}$

$\underset{―}{2|36}$

$\underset{―}{2|18}$

$\underset{―}{3|9}$

$\underset{―}{3|3}$

$\underset{―}{|1}$ 

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: $9$

$\underset{―}{2|72}$

$\underset{―}{2|36}$

$\underset{―}{2|18}$

$\underset{―}{3|9}$

$\underset{―}{3|3}$

$\underset{―}{|1}$ 

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:

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

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

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

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

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

Ans: The given statement is false.

$\underset{―}{2|1000}$

$\underset{―}{2|500}$

$\underset{―}{2|250}$

$\underset{―}{5|125}$

$\underset{―}{5|25}$

$\underset{―}{5|5}$

$\underset{―}{|1}$ 

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 $\mathbf{m},\left(\mathbf{2}\mathbf{m}-\mathbf{1},\mathbf{2}{\mathbf{m}}^{\mathbf{2}}-\mathbf{2}\mathbf{m},\mathbf{2}{\mathbf{m}}^{\mathbf{2}}-\mathbf{2}\mathbf{m}+\mathbf{1}\right)$ 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 ${\text{a}}^3<{\text{a}}^2$.