Question 11 Mark
Square root of a number $x$ is denoted by $\sqrt{\text{x}}.$
AnswerTrue. Solution: Square root of a number $x$ is denoted by $4x.$
View full question & answer→Question 21 Mark
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 $1, 0$ and $8$ stay the same when viewed upside down, whereas $9$ becomes $6$ and $6$ becomes $9.$
AnswerThree-digit perfect squares are $196$ and $961,$ which looks same when viewed upside down.
View full question & answer→Question 31 Mark
The product of two perfect squares is a perfect square.
AnswerTrue. Solution: e.g. If $4$ and $25$ are the perfect square, then $4 \times 25 = 100$ is also a perfect square. Clearly, the product of two perfect squares is a perfect square.
View full question & answer→Question 41 Mark
The square of $0.7$ is _________.
AnswerThe square of $0.7$ $0.49$ is .
Solution:
Square of $0.7 = (0.7)^2= 07 \times 07 = 0.49$
View full question & answer→Question 51 Mark
The square root of $1521$ is $31.$
AnswerFalse.
Solution:
As, the square of $31 = (31)^2= 31 × 31 = 961$
View full question & answer→Question 61 Mark
If $a^2$ ends in $9,$ then $a^3$ ends in $7.$
AnswerFalse.
Solution:
$\because (7)^2 = 49$ ends in $9$ and $(7)^3$,
i.e. $343$ does not end in $7.$
View full question & answer→Question 71 Mark
The square of every natural number is always greater than the number itself.
Answer$1$ is a natural number and square of $1$ is not greater than $1.$
View full question & answer→Question 81 Mark
The cube of $0.4$ is $0.064.$
AnswerTrue.
Solution:
Cube of $0.4 = (0.4)^2= 0.4 × 0.4 × 0.4 = 0.064$
View full question & answer→Question 91 Mark
The sum of two perfect squares is a perfect square.
Answere.g. $16$ and $25$ are the perfect squares, but $16 + 25 = 41$ is not a perfect square.
View full question & answer→Question 101 Mark
$1000$ is a perfect square.
Answer$1000 = 2 \times 2 \times 2 \times 5 \times 5 \times 5 = 2^2\times 5^2\times 2 \times 5$ Clearly, it is not a perfect square, because it has two unpaired factors $2$ and $5.$
View full question & answer→Question 111 Mark
Using distributive law, find the squares of: $72$
AnswerWe have,
$72^2= 72 × 72 = 72 × (70 + 2) = 5040 + 144 = 5184$
View full question & answer→Question 121 Mark
A number having $7$ at its ones place will have $3$ at the ones place of its cube.
AnswerCube of $7 = 7 \times 7 \times 7 = 343$
Cube of $17 = 17 \times 17 \times 17 = 4913$
Cube of $27 = 27 \times 27 \times 27 = 19683$
and so on.
View full question & answer→Question 131 Mark
The cube root of a number $x$ is denoted by __________.
AnswerThe cube root of a number $x$ is denoted by $\underline{\sqrt[3]{\text{x}}\text{ or}\text{ x}^\frac{1}{3}}.$
Solution: The cube root of a number $x$ is denoted by $\sqrt[3]{\text{x}}\text{ or}\text{ x}^\frac{1}{3}.$
View full question & answer→Question 141 Mark
There are $200$ natural numbers between $100^2$ and $101^2$.
Answernatural numbers between $100^2$ and $101^2$
$= 101^2- 100^2- 1[\because$ Natural numbers between $a$ and $b = b - a - 1]$
$= (101 + 100)(101 - 100) - 1 = 200$
View full question & answer→Question 151 Mark
If one side of a cube is $15m$ in length, find its volume.
AnswerGiven, one side of a cube $= 15\ m$
$\therefore$ Volume of cube $= (Side)^3= (15)^3= 3375\ m^3$
Hence, the volume of cube is $3375\ m^3$.
View full question & answer→Question 161 Mark
The square root of a perfect square of $n$ digits will have $\Big(\frac{\text{n}+1}{2}\Big)$ digits, if $n$ is odd.
AnswerIf the square has $3$ digits, then its square root has $2$ i.e. $\Big(\frac{3+1}{2}\Big)$ digits. Similarly, if the square has $5$ digits, then its square root has $3$ i.e. $\Big(\frac{5+1}{2}\Big)$ digits.
View full question & answer→Question 171 Mark
A number having $7$ at its ones place will have $3$ at the units place of its square.
AnswerSquare of $7 = 7 \times 7 = 49$
Square of $17 = 17 \times 17 = 289$
Square of $27 = 27 \times 27 = 729$ and so on.
View full question & answer→Question 181 Mark
The square root of $0.9$ is $0.3.$
AnswerAs, the square of $0.3 = (0.3)^2 = 0.3 \times 0.3 = 0.09$
View full question & answer→Question 191 Mark
There is no square number between $50$ and $60.$
AnswerNumbers between $50$ and $60$ are $51, 52, 53, 54, 55, 56, 57, 58$ and $59.$ We observed that there is no square number between $50$ and $60.$
View full question & answer→Question 201 Mark
If $x$ and $y$ are integers such that $x^2> y^2$, then $x^3> y^3$.
AnswerSuppose, $1$ and $-2$ are integers
Then, $(-2)^2 > (-2)^2= 4 > 1$
and $(-2)^3 < (-1)^3= - 8 < - 1$
View full question & answer→Question 211 Mark
There are five perfect cubes between $1$ and $100.$
AnswerThere are eight perfect cubes between $1$ and $100,$ i.e. $8, 27, 64, 125, 216, 343, 512$ and $729.$
View full question & answer→Question 221 Mark
The square of $2.8$ is $78.4.$
AnswerFalse.
Solution:
The square of $2.8 = (2.8)^2= 2.8 \times 2.8 = 7.84$
View full question & answer→Question 231 Mark
Using distributive law, find the squares of: $101$
AnswerWe have, $101^2= 101 \times 101$
$= 101 (100 + 1) = 10100 + 101 = 10201$
View full question & answer→Question 241 Mark
Cube roots of $8$ are $+2$ and $-2.$
AnswerCube root of $8$ is $2$ only and cube root of $-8$ is $-2.$
View full question & answer→Question 251 Mark
The square root of $24025$ will have _________ digits.
AnswerThe square root of $24025$ will have $3$ digits.
Solution: Given number is $24025.$
Here, number of digits, $n = 5 ($odd$)$
$\therefore$ Number of digits in the square root of $24025$
$=\frac{\text{n}+1}{2}=\frac{\text{5}+1}{2}=3$
View full question & answer→Question 261 Mark
The cube root of $8000$ is $200.$
AnswerFalse. Solution: We have, $\sqrt[3]{8000}$ $=\sqrt[3]{(2\times2\times2)\times(2\times2\times2)\times(5\times5\times5)}$ $=2\times2\times5=20$
View full question & answer→Question 271 Mark
The sum of first six odd natural numbers is ________.
AnswerThe sum of first six odd natural numbers is $36.$
Solution:
We know that, sum of frist $n$ odd natural numbers $= n^2$
$\therefore$ Sum of frist six odd natural numbers $= (6)^2= 36$
View full question & answer→Question 281 Mark
There are ________ natural numbers between $n^2$ and $(n + 1)^2$
AnswerThere are $2n$ natural numbers between $n^2$ and $(n + 1)^2$
Solution:
Nature numbers between $n^2$ and $(n + 1)^2= [(n + 1)^2- n^2] - 1$
$= [n^2+ 2n + 1 - n^2] - 1$
$= 2n + 1 - 1 = 2n$
View full question & answer→Question 291 Mark
Using prime factorisation, find which of the following are perfect cubes.
$729$
AnswerWe have, $729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3$
Since, the prime factors appear in triplets.
So, $729$ is a perfect cube.
View full question & answer→Question 301 Mark
Let $x$ and $y$ be natural numbers. If $x$ divides $y,$ then $x^3$ divides $y^3$.
AnswerIf $x$ divides $y,$ then $\frac{\text{y}}{\text{x}}$ is a natural number.
$\Rightarrow\Big(\frac{\text{y}}{\text{x}}\Big)^3$ is also natural number.
$\Rightarrow\frac{\text{y}^3}{\text{x}^3}$ is a natural number.
$\Rightarrow\text{x}^3$ divides $\text{y}^3.$
View full question & answer→Question 311 Mark
A perfect square can have $8$ as its units digit.
AnswerA perfect square can never have $8$ as its unit’s digit.
View full question & answer→Question 321 Mark
Can a right triangle with sides $6\ cm, 10\ cm$ and $8\ cm$ be formed$?$ Give reason.
AnswerWe know that, the sum of the square of two smaller sides is equal to the square of longer side.
$\because 10^2=100=64+36=8^2+6^2$
Hence, $6\ cm, 8\ cm$ and $10\ cm$ are the sides of a right angled triangle.
View full question & answer→Question 331 Mark
Write the first five square numbers.
AnswerFirst five square numbers are $1^2, 2^2, 3^2, 4^2$ and $5^2$, i.e. $1,4,9,16$ and $25.$
View full question & answer→Question 341 Mark
The sum of first n odd natural numbers is $n^2$.
AnswerTrue.
Solution:
Sum of odd natural numbers $=\sum(2\text{n}-1)=\frac{2\times\text{n}\times(\text{n}+1)}{2}-2$ $=\text{n}^2+\text{n}-\text{n}=\text{n}^2$
View full question & answer→Question 351 Mark
Cube of an even number is odd.
AnswerFalse.
Solution:
We know that, the cube of an even number is always an even number,
e.g. $2$ is an even number. Then, $2^3= 2 \times 2 \times 2 = 8$
Clearly, 8 is also an even number.
View full question & answer→Question 361 Mark
For every natural number m, $(2m - 1, 2m^2- 2m, 2m^2- 2m + 1)$ is a pythagorean triplet.
AnswerFalse.
Solution:
$ \because(2 m-1)^2 \neq\left(2 m^3-23\right)^2+\left(2 m^2-2 m+1\right)^2 $
$ \left(2 m^3-2 m\right)^2 \neq(2 m-1)^2+\left(2 m^2-2 m+1\right)^2 $
and $\left(2 m^2-2 m+1\right)^2 \neq(2 m-1)^2+\left(2 m^3-2 m\right)^2 $
View full question & answer→Question 371 Mark
$1m^2= \_\_\_\_\_\_\_\_ cm^2$.
Answer$1m^2= 10000\ cm^2$.
Solution:
$1m^2= (100cm)^2$ $[\because1\text{m}=100\text{cm}]$
$ = 10000cm^2$
View full question & answer→Question 381 Mark
The least number by which $72$ be multiplied to make it a perfect cube is __________.
AnswerThe least number by which $72$ be multiplied to make it a perfect cube is $3.$
Solution: Resolving $72$ into prime factors, we get $72 = 2 \times 2 \times 2 \times 3 \times 3$ Grouping the factors in triplets of equal factors, we get $72 = (2 \times 2 \times 2) \times 3 \times 3$ We find that 2 occurs as a prime factor of $72$ thrice, but $3$ occurs as a prime factor only twice. Thus, if we multiply $72$ by $3, 3$ will also occurs as a prime factor thrice and the product will be $2 \times 2 \times 2 \times 3 \times 3 \times 3,$ which is a perfect cube. Hence, the least number, which should be multiplied with $72$ to get perfect cube, is $3.$
View full question & answer→Question 391 Mark
$999$ is a perfect cube.
Answer$\begin{array}{c|c}3 & 999 \\\hline3 & 333\\\hline3&111\\\hline37&37\\\hline&1\end{array}$
Resolving $999$ into prime factors, we get
$999 = 3 \times 3 \times 3 \times 37$
Grouping the factors in triplets of equal factors, we get
$999 = (3 \times 3 \times 3) \times 37$
Clearly, in grouping, the factors in triplets of equal factors, we are left with one factor $37$.
Therefore, $999$ is not a perfect cube.
View full question & answer→Question 401 Mark
A farmer wants to plough his square field of side $150\ m.$ How much area will he have to plough$?$
AnswerGiven, side of square field $= 150m$
$\therefore$ Area of square field $=$ Side $\times $ Side $= 150 \times 150 = 22500\ m^2$
View full question & answer→Question 411 Mark
The cube of a one digit number cannot be a two digit number.
Answere.g. $3$ is a one-digit number and $(3)^3$, i.e. $27$ is two-digit number.
View full question & answer→Question 421 Mark
Write two Pythagorean triplets each having one of the numbers as $5.$
AnswerAs, $5^2=3^2+4^2$
and $13^2=12^2+5^2$
Hence, $3, 4, 5$ and $12, 5, 13$ are the two pythagorean triplets.
View full question & answer→Question 431 Mark
All numbers of a pythagorean triplet are odd.
Answer$3, 4$ and $5$ are the numbers of Pythagorean triplet as $5^2= 4^2+ 3^2$ where, $4$ is not an odd number.
View full question & answer→Question 441 Mark
$\Bigg\{\bigg(6^2+\big(8^2\big)^{\frac{1}{2}}\bigg)\Bigg\}^3$
AnswerWe have, $\bigg\{6^2+\big(8^2\big)^{\frac{1}{2}}\bigg\}^3=\big\{36+8\big\}^3$ $=(44)^3=44\times44\times44=85184$
View full question & answer→Question 451 Mark
The square root of $5.3 \times 5.3$ is _________.
AnswerThe square root of $5.3 \times 5.3$ is $5.3.$
Solution: Square root of $5.3 \times 5.3 =\sqrt{(5.3)^2}$
$=(5.3)^\frac{2}{2}=5.3$
View full question & answer→Question 461 Mark
Using prime factorisation, find which of the following are perfect cubes.$1331$
AnswerWe have, $1331 =11 \times 11 \times 11$
Since, the prime factors appear in triplets. So, $1331$ is a perfect cube.
View full question & answer→Question 471 Mark
$\sqrt{1.96}=$ _________.
Answer$\sqrt{1.96}= \underline{1.4}$
Solution:
We have, $(12)^2$ $=\Big(\frac{12}{10}\Big)^3=\frac{12}{10}\times\frac{12}{10}\times\frac{12}{10}=\frac{1728}{1000}=1.728$
View full question & answer→Question 481 Mark
Cube of an even number is even.
AnswerWe know that, the cube of an even number is always an even number, e.g. $4$ is an even number. Then, $43 = 4 \times 4 \times 4 = 64$ Clearly, $64$ is also an even number.
View full question & answer→Question 491 Mark
$(1.2)^3=$ _________.
Answer$(1.2)^3=$ 1.728.
Solution:
We have, $(1.2)^3$ $=\Big(\frac{12}{10}\Big)^3=\frac{12}{10}\times\frac{12}{10}\times\frac{12}{10}=\frac{1728}{1000}=1.728$
View full question & answer→Question 501 Mark
The square of $86$ will have $6$ at the units place.
AnswerWe know that, the unit’s digit of the square of a number having digit at unit’s place as $4$ or $6$ is $6.$
View full question & answer→