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

Solution:

We know that the square of an even number is always even, and the square of an odd number is always an odd number.

In the given options, we check the units digit of each number:

1. 256: Units digit is 6 (Even number)

2. 361: Units digit is 1 (Odd number)

3. 144: Units digit is 4 (Even number)

4. 400: Units digit is 0 (Even number)

Since 361 is the only odd number among the choices, it must be the square of an odd number ($19^2 = 361$).

The correct option is (b).

Solution:

The units digit of a square number depends only on the units digit of the number being squared.

(a) For $19^2$, the units digit is 9. Since $9 \times 9 = 81$, the units place will be 1.

(b) For $17^2$, the units digit is 7. Since $7 \times 7 = 49$, the units place will be 9.

(c) For $18^2$, the units digit is 8. Since $8 \times 8 = 64$, the units place will be 4.

(d) For $16^2$, the units digit is 6. Since $6 \times 6 = 36$, the units place will be 6.

The correct option is (a).

Solution:

There is a general property that between the squares of two consecutive numbers $n$ and $(n+1)$, there are $2n$ non-perfect square natural numbers.

Here, $n = 18$ and $n + 1 = 19$.

Calculation:

$\text{Number} = 36$

The correct option is (d).

Solution:

A perfect square number can only end with the digits 0, 1, 4, 5, 6, or 9 at its ones place.

A number ending in 2, 3, 7, or 8 is never a perfect square.

Looking at the given numbers:

(a) 361 ends in 1 (Could be a square)

(b) 1156 ends in 6 (Could be a square)

(c) 1128 ends in 8

(d) 1681 ends in 1 (Could be a square)

Since 1128 ends in 8, it is not a perfect square.

The correct option is (c).

Solution:

By observing the squares of digits from 0 to 9:

$0^2=0, 1^2=1, 2^2=4, 3^2=9, 4^2=16, 5^2=25, 6^2=36, $$ 7^2=49, 8^2=64, 9^2=81$.

The units digits of perfect squares are always 0, 1, 4, 5, 6, or 9.

A perfect square never ends in 2, 3, 7, or 8.

From the given options, the digit 3 can never be at the ones place of a perfect square.

The correct option is (d).

To Find:

The value of $\sqrt{176 + \sqrt{2401}}$

Solution:

Step 1: Find the square root of 2401.

So, $\sqrt{2401} = 49$.

Step 2: Substitute this value back into the original expression.

$\sqrt{176 + 49}$ = $\sqrt{225}$

Now, we find $\sqrt{225}$. We know that $15 \times 15 = 225$.

$\sqrt{225} = 15$

The correct option is (b).

Given:

Solution:

We use the given value to find the square roots of the decimals:

1. $\sqrt{0.5625} = \sqrt{\frac{5625}{10000}} = \frac{75}{100} = 0.75$

2. $\sqrt{56.25} = \sqrt{\frac{5625}{100}} = \frac{75}{10} = 7.5$

Addition:

$\begin{array}{cc} & 0 & . & 7 & 5 \\ + & 7 & . & 5 & 0 \\ \hline & 8 & . & 2 & 5 \\ \hline \end{array}$

The sum is 8.25.

The correct option is (c).

Solution:

The perfect squares starting from 1 are:

$1^2 = 1$

$2^2 = 4$

$3^2 = 9$

$4^2 = 16$

$5^2 = 25$

$6^2 = 36$

$7^2 = 49$

$8^2 = 64$

The numbers "between" 1 and 50 are those greater than 1 and less than 50.

These are: 4, 9, 16, 25, 36, and 49.

Counting them, we find there are 6 perfect squares.

Answer: 6

Solution:

The cube of a number is obtained by multiplying the number by itself three times. To find the cube of 100, we perform the following calculation:

$100^3 = 100 \times 100 \times 100$

$10,000 \times 100 = 1,000,000$

The number of zeroes in 1,000,000 is 6.

Generally, if a number has $n$ zeroes, its cube will have $3n$ zeroes. Here $n=2$, so $3 \times 2 = 6$.

Answer: 6

Solution:

To find the square of 6.1, we multiply 6.1 by itself:

$(6.1)^2 = 6.1 \times 6.1$

Since there is one decimal place in each factor, the product will have two decimal places.

$(6.1)^2 = 37.21$

Answer: 37.21

Solution:

To find the cube of 0.3, we multiply 0.3 by itself three times:

$(0.3)^3 = 0.3 \times 0.3 \times 0.3$

Since the total decimal places are three ($1+1+1=3$), we place the decimal accordingly:

$(0.3)^3 = 0.027$

Answer: 0.027

Solution:

The digit at the units place of the square of a number depends only on the units digit of the original number.

The units digit of 68 is 8.

The square of 8 is $8 \times 8 = 64$.

The units digit of 64 is 4.

Therefore, $68^2$ will have 4 at the units place.

Answer: 4

Solution:

In mathematics, the radical sign is used to denote the square root of a value. For any positive number $x$, the positive (principal) square root is represented symbolically.

Answer: $\sqrt{x}$

Solution:

First, find the prime factorisation of 9:

$\begin{array}{c|cc} 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

$9 = 3 \times 3$

To be a perfect cube, every prime factor must appear in a group of three (triplets).

In the factorisation of 9, the factor 3 appears only twice. We need one more 3 to complete the triplet ($3 \times 3 \times 3 = 27$).

Therefore, the least number to be multiplied is 3.

Answer: 3

Solution:

To check the statement, we calculate the square of 0.4:

$(0.4)^2 = 0.4 \times 0.4$

Since there are two decimal places in total ($1 + 1$), the result is $0.16$.

The statement is True (T).

Solution:

We check this by finding the cube of 8:

$8^3 = 8 \times 8 \times 8$

$8 \times 8 = 64$

$64 \times 8 = 512$

Since $8^3 = 512$ and not 729, the statement is incorrect. The cube root of 729 is actually 9 ($9 \times 9 \times 9 = 729$).

The statement is False (F).

Solution:

We know that the number of natural numbers lying between the squares of two consecutive numbers $n$ and $(n + 1)$ is given by the formula $2n$.

Here, $n = 10$ and $n + 1 = 11$.

Number of natural numbers $= 2n$

Number of natural numbers $= 2 \times 10 = 20$

Since there are 20 natural numbers between $10^2$ and $11^2$, the given statement is False (F).

Solution:

The sum of the first $n$ odd natural numbers is always equal to $n^2$.

In this case, $n = 7$.

Sum $= 7^2$

Sum $= 7 \times 7 = 49$

The calculation matches the statement. Therefore, the statement is True (T).

Solution:

There is a mathematical property regarding the number of digits in a square root:

1. If a perfect square has $n$ digits and $n$ is even, its square root contains $\frac{n}{2}$ digits.

2. If $n$ is odd, its square root contains $\frac{n + 1}{2}$ digits.

The statement correctly identifies the condition for an even number of digits. Therefore, the statement is True (T).

Given: The number is 36.

Solution:

We know that $36 = 6^2$. A perfect square $n^2$ can be expressed as the sum of the first $n$ successive odd natural numbers.

Since $n = 6$, we need to add the first 6 odd numbers:

The first six odd numbers are: 1, 3, 5, 7, 9, 11.

Expression:

$36 = 1 + 3 + 5 + 7 + 9 + 11$

Given: The number is 90.

Solution:

First, we find the prime factorisation of 90:

$\begin{array}{c|cc} 2 & 90 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

The prime factorisation is: $90 = 2 \times 3 \times 3 \times 5$

Grouping the identical factors in pairs:

$90 = \underbrace{3 \times 3}_{\text{pair}} \times 2 \times 5$

In a perfect square, all prime factors must exist in pairs. Here, the prime factors 2 and 5 do not have pairs. Therefore, 90 is not a perfect square.

Given: The number is 1728.

Solution:

Performing the prime factorisation of 1728:

$\begin{array}{c|cc} 2 & 1728 \\ \hline 2 & 864 \\ \hline 2 & 432 \\ \hline 2 & 216 \\ \hline 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

Prime factorisation: $1728 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$

Grouping the factors in triplets:

$1728 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (3 \times 3 \times 3)$

Since every prime factor appears in groups of three (triplets), 1728 is a perfect cube.

Specifically, $1728 = (2 \times 2 \times 3)^3 = 12^3$.

To Find: $43^2$ using the distributive law.

Solution:

We can write 43 as $(40 + 3)$.

$43^2 = (40 + 3)^2$

$43^2 = (40 + 3)(40 + 3)$

By applying the distributive law $a(b + c) = ab + ac$:

$43^2 = 40(40 + 3) + 3(40 + 3)$

$43^2 = (40 \times 40) + (40 \times 3) + (3 \times 40) + (3 \times 3)$

$43^2 = 1600 + 120 + 120 + 9$

$43^2 = 1849$

The square of 43 is 1849.

Given: Smallest number $= 6$.

Solution:

For any natural number $m > 1$, the three numbers $2m$, $m^2 - 1$, and $m^2 + 1$ form a Pythagorean triplet.

Let the smallest number be $2m$.

$m = \frac{6}{2} = 3$

Now, we find the other two members of the triplet:

Second member $= m^2 - 1 = 3^2 - 1 = 9 - 1 = 8$

Third member $= m^2 + 1 = 3^2 + 1 = 9 + 1 = 10$

Checking the triplet (6, 8, 10):

$6^2 + 8^2 = 36 + 64 = 100$

$10^2 = 100$

Since $6^2 + 8^2 = 10^2$, the triplet is valid and 6 is indeed the smallest number.

The Pythagorean triplet is (6, 8, 10).

Given: The number is 5832.

To Find: Cube root of 5832 ($\sqrt[3]{5832}$).

Solution:

First, we perform the prime factorisation of 5832:

$\begin{array}{c|cc} 2 & 5832 \\ \hline 2 & 2916 \\ \hline 2 & 1458 \\ \hline 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

The prime factorisation is:

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

Now, we group the identical factors into triplets:

$5832 = (2 \times 2 \times 2) \times (3 \times 3 \times 3) \times (3 \times 3 \times 3)$

Taking one factor from each triplet to find the cube root:

$\sqrt[3]{5832} = 2 \times 3 \times 3$

$\sqrt[3]{5832} = 18$

Therefore, the cube root of 5832 is 18.

Given: The decimal number is 22.09.

To Find: Square root using the long division method.

Solution:

We pair the digits starting from the decimal point. We have 22 in the integer part and 09 in the decimal part.

$\begin{array}{c|cc} & 4 \ . \ 7 & \\ \hline \phantom{()} 4 & \overline{22} \; . \overline{09} \\ + \; 4 & 16\phantom{(........)} \\ \hline \phantom{()} 8 \; 7 & 6 \; 09 \phantom{(.....)} \\ \phantom{()} +7 & 6 \; 09 \phantom{(...)} \\ \hline \phantom{()} & 0 \end{array}$

Steps followed:

1. Find the largest square less than or equal to 22, which is $16 = 4^2$.

2. Subtract 16 from 22 to get 6. Bring down the next pair 09. Put a decimal in the quotient.

3. Double the divisor (4) to get 8. Find a digit $x$ such that $8x \times x$ is less than or equal to 609.

4. $87 \times 7 = 609$.

Therefore, $\sqrt{22.09} = 4.7$.

Given: The numbers are 3, 4, 5, and 6.

To Find: The smallest perfect square divisible by all these numbers.

Solution:

First, we find the LCM (Least Common Multiple) of 3, 4, 5, and 6.

$\begin{array}{c|cc} 2 & 3 \;, & 4 \;, & 5 \;, & 6 \\ \hline 2 & 3 \;, & 2 \;, & 5 \;, & 3 \\ \hline 3 & 3 \;, & 1 \;, & 5 \;, & 3 \\ \hline 5 & 1 \;, & 1 \;, & 5 \;, & 1 \\ \hline & 1 \;, & 1 \;, & 1 \;, & 1 \end{array}$

LCM $= 2 \times 2 \times 3 \times 5 = 60$.

Now, observe the prime factorisation of 60:

$60 = \underbrace{2 \times 2}_{\text{pair}} \times 3 \times 5$

For a number to be a perfect square, all its prime factors must exist in pairs. In the factorisation of 60, the factors 3 and 5 do not have pairs.

To make it a perfect square, we must multiply 60 by 3 and 5.

$\text{Smallest perfect square} = 60 \times 3 \times 5$

$\text{Smallest perfect square} = 60 \times 15 = 900$

The smallest perfect square divisible by 3, 4, 5, and 6 is 900.

Given:

Length of the ladder (Hypotenuse, $AC$) $= 10 \text{ m}$

Distance of the foot from the wall (Base, $BC$) $= 6 \text{ m}$

To Find:

Height of the wall (Perpendicular, $AB$).

Solution:

The wall and the ground form a right angle ($90^\circ$). According to the Pythagoras Theorem:

Substituting the given values:

$10^2 = AB^2 + 6^2$

$100 = AB^2 + 36$

$AB^2 = 100 - 36$

$AB^2 = 64$

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

The height of the wall is 8 m.

Given:

Length of the rectangle ($l$) $= 20 \text{ m}$

Breadth of the rectangle ($b$) $= 15 \text{ m}$

To Find:

Length of the diagonal ($d$).

Solution:

In a rectangle, the length, breadth, and diagonal form a right-angled triangle. We can use the Pythagoras Theorem to find the diagonal.

$d = \sqrt{20^2 + 15^2}$

$d = \sqrt{400 + 225}$

$d = \sqrt{625}$

$d = 25 \text{ m}$

The length of the diagonal of the rectangle is 25 m.

Given:

Area of the rectangular field $= 2450 \text{ m}^2$

Length ($l$) of the field $= 2 \times \text{Breadth } (b)$

To Find:

The perimeter of the field.

Solution:

Let the breadth of the rectangular field be $x$ metres.

Then, the length of the field $= 2x$ metres.

We know that $\text{Area} = \text{Length} \times \text{Breadth}$

$2450 = 2x \times x$

$2450 = 2x^2$

$x^2 = 1225$

$x = \sqrt{1225}$

To find the square root of 1225:

$\begin{array}{c|cc} 5 & 1225 \\ \hline 5 & 245 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

$1225 = 5 \times 5 \times 7 \times 7$

$\sqrt{1225} = 5 \times 7 = 35$

So, Breadth $= 35 \text{ m}$

Length $= 2 \times 35 = 70 \text{ m}$

Now, $\text{Perimeter} = 2(\text{Length} + \text{Breadth})$

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

$\text{Perimeter} = 2(105) = 210 \text{ m}$

The perimeter of the field is $210 \text{ m}$.

Given:

Total number of students $= 6250$

Number of students left out after square arrangement $= 9$

To Find:

The number of children in each row.

Solution:

The students who formed a perfect square are:

$\text{Students in square} = 6250 - 9 = 6241$

Let the number of children in each row be $x$. Since the number of rows is equal to the number of students in each row:

$x \times x = 6241$

$x^2 = 6241$

$x = \sqrt{6241}$

Using long division method to find the square root:

The value of $x$ is 79.

Therefore, there are $79$ children in each row.

Given: The number is 1500.

To Find: The least number to be added to make it a perfect square and its square root.

Solution:

First, we find the square root of 1500 by long division method to find the remainder.

Here, the remainder is 56. This shows that $38^2 < 1500$.

The next perfect square number will be the square of 39.

$39^2 = 1521$

The number to be added $= 39^2 - 1500$

$\text{Number to be added} = 1521 - 1500 = 21$

The least number to be added is $21$.

The perfect square is $1521$ and its square root is $39$.

Given: The number is 1620.

To Find: The smallest number to divide 1620 to make it a perfect square.

Solution:

We start by finding the prime factorisation of 1620:

$\begin{array}{c|cc} 2 & 1620 \\ \hline 2 & 810 \\ \hline 3 & 405 \\ \hline 3 & 135 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 3 & 5 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

The prime factorisation is:

$1620 = 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 5$

Grouping the identical factors in pairs:

$1620 = (2 \times 2) \times (3 \times 3) \times (3 \times 3) \times 5$

We observe that the prime factor 5 does not have a pair.

To make the number a perfect square, we must divide 1620 by 5 so that all remaining factors are in pairs.

$\text{New number} = \frac{1620}{5} = 324$

As $324 = 18^2$, it is a perfect square.

Therefore, the smallest number by which 1620 must be divided is $5$.

Solution:

To find which number's square is 196, we can square each of the given options:

(a) $11^2 = 11 \times 11 = 121$

(b) $12^2 = 12 \times 12 = 144$

(c) $14^2 = 14 \times 14 = 196$

(d) $16^2 = 16 \times 16 = 256$

Comparing the results with 196, we see that $14^2 = 196$.

Therefore, 196 is the square of 14.

The correct option is (c) 14.

Solution:

We know that the square of an even number is always an even number, and the square of an odd number is always an odd number.

Let's examine the given options:

(a) 144 is an even number.

(b) 169 is an odd number.

(c) 441 is an odd number.

(d) 625 is an odd number.

Since the square of an even number must be even, only option (a) 144 can be the square of an even number among the given options.

Let's verify if 144 is a perfect square and if its square root is even.

We find the square root of 144:

$\sqrt{144} = 12$

The number 12 is an even number.

Therefore, 144 is the square of the even number 12.

The correct option is (a) 144.

Solution:

To find the units place of the square of a number, we only need to consider the units place of the number itself.

If a number ends in 9, its units digit is 9.

We find the square of the units digit:

$9^2 = 81$

The units digit of $9^2$ is 1.

Therefore, the units place of the square of any number ending in 9 will be 1.

For example:

$9^2 = 81$ (units digit is 1)

$19^2 = 361$ (units digit is 1)

$29^2 = 841$ (units digit is 1)

The correct option is (c) 1.

Solution:

To find the units place of the square of a number, we only need to consider the units place of the number itself.

Let's check the units digit of the square for each option:

(a) $14^2$: The units digit of 14 is 4. The units digit of $4^2 = 16$ is 6.

(b) $62^2$: The units digit of 62 is 2. The units digit of $2^2 = 4$ is 4.

(c) $27^2$: The units digit of 27 is 7. The units digit of $7^2 = 49$ is 9.

(d) $35^2$: The units digit of 35 is 5. The units digit of $5^2 = 25$ is 5.

We are looking for the number whose square has 4 at the units place. From the above calculations, only $62^2$ has 4 at the units place.

The correct option is (b) 62².

Solution:

We are asked to find the number of natural numbers that lie between $5^2$ and $6^2$.

First, let's calculate the squares of the two numbers:

$5^2 = 5 \times 5 = 25$

$6^2 = 6 \times 6 = 36$

We need to find the number of natural numbers that are greater than 25 and less than 36.

These numbers are 26, 27, 28, 29, 30, 31, 32, 33, 34, and 35.

Counting these numbers, we find there are 10 natural numbers between 25 and 36.

Alternatively:

There is a general formula for the number of non-perfect square natural numbers between the squares of two consecutive natural numbers, $n^2$ and $(n+1)^2$. The number of such numbers is $2n$.

In this question, the consecutive natural numbers are 5 and 6. So, $n=5$.

The number of natural numbers between $5^2$ and $6^2$ is $2 \times 5 = 10$.

The correct option is (b) 10.

Solution:

A perfect square is a number obtained by squaring an integer. The units digit of a perfect square can only be 0, 1, 4, 5, 6, or 9. A number whose units digit is 2, 3, 7, or 8 cannot be a perfect square.

Let's look at the units digit of each option:

(a) 841: The units digit is 1. A number ending in 1 can be a perfect square (e.g., $1^2=1$, $9^2=81$). In fact, $29^2 = 841$, so 841 is a perfect square.

(b) 529: The units digit is 9. A number ending in 9 can be a perfect square (e.g., $3^2=9$, $7^2=49$). In fact, $23^2 = 529$, so 529 is a perfect square.

(c) 198: The units digit is 8. A number ending in 8 cannot be a perfect square.

(d) All of the above: Since (a) and (b) are perfect squares, this option is incorrect.

Based on the units digit rule, 198 cannot be a perfect square because its units digit is 8.

The correct option is (c) 198.

Solution:

To find the one's digit (units digit) of the cube of a number, we only need to find the one's digit of the cube of the one's digit of the original number.

The given number is 23.

The one's digit of 23 is 3.

Now, we cube the one's digit:

$3^3 = 3 \times 3 \times 3 = 9 \times 3 = 27$

The one's digit of 27 is 7.

Therefore, the one's digit of the cube of 23 is 7.

The correct option is (b) 7.

Given:

Area of the square board = 144 square units.

To Find:

The length of each side of the board.

Solution:

The area of a square is calculated by squaring the length of its side.

Area = $(\text{side})^2$

We are given that the area is 144 square units. So,

$(\text{side})^2 = 144$

To find the length of the side, we need to calculate the square root of the area.

$\text{side} = \sqrt{144}$

We know that $12 \times 12 = 144$.

Therefore, $\sqrt{144} = 12$.

The length of each side of the board is 12 units.

The correct option is (b) 12 units.

Given: The value is $\sqrt{25}$.

To Find: The letter representing this value on the number line.

Solution:

First, we calculate the square root of 25:

$\sqrt{25} = \sqrt{5 \times 5}$

$\sqrt{25} = 5$

Now, we observe the given number line to find the point corresponding to the number 5.

1. Point A is located to the left of 0 (representing a negative value).

2. Point B is located at the number 2.

3. Point C is located at the number 5.

4. Point D is located at the number 7.

Since the value of $\sqrt{25}$ is 5, the letter C best represents its location.

The correct option is (c).

Solution:

A Pythagorean triplet consists of three positive integers $a$, $b$, and $c$, such that $a^2 + b^2 = c^2$.

For any natural number $m > 1$, the triplet $(2m, m^2 - 1, m^2 + 1)$ is a Pythagorean triplet because:

$(2m)^2 + (m^2 - 1)^2 = 4m^2 + (m^4 - 2m^2 + 1) = m^4 + 2m^2 + 1 $$ = (m^2 + 1)^2$

The members of this triplet are $2m$, $m^2 - 1$, and $m^2 + 1$.

If one member of a Pythagorean triplet is given as $2m$, then the other two members are $m^2 - 1$ and $m^2 + 1$. Note that the order of the last two members can be switched, as seen in option (b).

Comparing this with the given options, option (b) lists $m^2 + 1$ and $m^2 - 1$ as the other two members.

The correct option is (b) m² + 1 , m² – 1.

Solution:

We need to find the sum of the given successive odd numbers: 1, 3, 5, 7, 9, 11, 13, and 15.

We can find the sum by adding the numbers directly:

$1 + 3 + 5 + 7 + 9 + 11 + 13 + 15$

$= 4 + 5 + 7 + 9 + 11 + 13 + 15$

$= 9 + 7 + 9 + 11 + 13 + 15$

$= 16 + 9 + 11 + 13 + 15$

$= 25 + 11 + 13 + 15$

$= 36 + 13 + 15$

$= 49 + 15$

$= 64$

Alternatively:

The sum of the first $n$ successive odd natural numbers is equal to $n^2$.

Let's count the number of terms in the given sum: 1, 3, 5, 7, 9, 11, 13, 15.

There are 8 terms in the sequence. So, $n=8$.

The sum is equal to $n^2 = 8^2$.

$8^2 = 8 \times 8 = 64$

The sum of the successive odd numbers 1, 3, 5, 7, 9, 11, 13, and 15 is 64.

The correct option is (b) 64.

Solution:

According to the property of square numbers, the sum of the first $n$ odd natural numbers is always equal to the square of $n$.

For example:

Sum of first 1 odd number: $1 = 1^2$

Sum of first 2 odd numbers: $1 + 3 = 4 = 2^2$

Sum of first 3 odd numbers: $1 + 3 + 5 = 9 = 3^2$

In general, for $n$ terms:

$1 + 3 + 5 + ... + (2n - 1) = n^2$

The correct option is (b).

Solution:

A perfect cube is a number that can be obtained by multiplying an integer by itself three times (cubing it). To check if a number is a perfect cube, we can use prime factorization. If the prime factors of a number can be grouped into triplets, then the number is a perfect cube.

Let's find the prime factorization of each option:

(a) 243:

$\begin{array}{c|cc} 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

$243 = 3 \times 3 \times 3 \times 3 \times 3 = 3^5$. The prime factor 3 appears 5 times, which is not a multiple of 3. Thus, 243 is not a perfect cube.

(b) 216:

$\begin{array}{c|cc} 2 & 216 \\ \hline 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

$216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 = 2^3 \times 3^3 = (2 \times 3)^3 = 6^3$. The prime factors 2 and 3 appear in groups of three. Thus, 216 is a perfect cube.

(c) 392:

$\begin{array}{c|cc} 2 & 392 \\ \hline 2 & 196 \\ \hline 2 & 98 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

$392 = 2 \times 2 \times 2 \times 7 \times 7 = 2^3 \times 7^2$. The prime factor 7 appears 2 times, which is not a multiple of 3. Thus, 392 is not a perfect cube.

(d) 8640:

$\begin{array}{c|cc} 2 & 8640 \\ \hline 2 & 4320 \\ \hline 2 & 2160 \\ \hline 2 & 1080 \\ \hline 2 & 540 \\ \hline 2 & 270 \\ \hline 3 & 135 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

$8640 = 2^6 \times 3^3 \times 5^1$. The prime factor 5 appears once, which is not a multiple of 3. Thus, 8640 is not a perfect cube.

From the prime factorizations, only 216 has all prime factors appearing in groups of three.

The correct option is (b) 216.

Given:

Length of one leg of the right triangle = $3x$

Length of the other leg of the right triangle = $4x$

To Find:

The length of the hypotenuse.

Solution:

In a right-angled triangle, the lengths of the sides are related by the Pythagorean theorem, which states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides (the legs).

Let the lengths of the legs be $a$ and $b$, and the length of the hypotenuse be $c$.

According to the Pythagorean theorem:

$a^2 + b^2 = c^2$

Given the lengths of the legs are $3x$ and $4x$, we can substitute these values into the formula:

$(3x)^2 + (4x)^2 = c^2$

Calculate the squares of the terms:

$(3x)^2 = 3^2 \times x^2 = 9x^2$

$(4x)^2 = 4^2 \times x^2 = 16x^2$

Substitute these back into the equation:

$9x^2 + 16x^2 = c^2$

Combine the terms on the left side:

$(9+16)x^2 = c^2$

$25x^2 = c^2$

To find the length of the hypotenuse $c$, take the square root of both sides of the equation:

$c = \sqrt{25x^2}$

$c = \sqrt{25} \times \sqrt{x^2}$

Since the length of a side must be positive, we consider the positive square root. $\sqrt{25} = 5$ and $\sqrt{x^2} = x$ (assuming $x$ represents a positive length).

$c = 5 \times x$

$c = 5x$

Thus, the length of the hypotenuse is $5x$.

The correct option is (a) 5x.

Solution:

Let's examine the given number pattern: 1, 4, 9, 16, 25, ...

We can observe the relationship between each number and its position in the sequence:

The first term is 1, which is $1^2$.

The second term is 4, which is $2^2$.

The third term is 9, which is $3^2$.

The fourth term is 16, which is $4^2$.

The fifth term is 25, which is $5^2$.

The pattern shows that each number is the square of the corresponding natural number representing its position in the sequence.

Therefore, the next two numbers will be the square of the 6th and 7th natural numbers.

The 6th number is $6^2 = 6 \times 6 = 36$.

The 7th number is $7^2 = 7 \times 7 = 49$.

So, the next two numbers in the pattern are 36 and 49.

The correct option is (b) 36, 49.

Solution:

The units digit of the square of a number is determined by the units digit of the number itself.

We need to find which of the given squares ends with the digit 1.

Let's examine the units digit of each number and its square:

(a) For $43^2$, the units digit of 43 is 3. The units digit of $3^2 = 9$ is 9.

(b) For $67^2$, the units digit of 67 is 7. The units digit of $7^2 = 49$ is 9.

(c) For $52^2$, the units digit of 52 is 2. The units digit of $2^2 = 4$ is 4.

(d) For $59^2$, the units digit of 59 is 9. The units digit of $9^2 = 81$ is 1.

The square of a number ends with the digit 1 if the number itself ends with 1 or 9.

Among the given numbers, only 59 ends with 9.

Therefore, $59^2$ will end with the digit 1.

The correct option is (d) 59².

Solution:

The units digit of a perfect square is determined by the units digit of the number being squared.

Let's look at the units digits of the squares of the digits 0 through 9:

$0^2 = 0$ (units digit 0)

$1^2 = 1$ (units digit 1)

$2^2 = 4$ (units digit 4)

$3^2 = 9$ (units digit 9)

$4^2 = 16$ (units digit 6)

$5^2 = 25$ (units digit 5)

$6^2 = 36$ (units digit 6)

$7^2 = 49$ (units digit 9)

$8^2 = 64$ (units digit 4)

$9^2 = 81$ (units digit 1)

The possible units digits of a perfect square are 0, 1, 4, 5, 6, and 9.

This means that a number ending in 2, 3, 7, or 8 can never be a perfect square.

Let's check the given options:

(a) 1 is a possible units digit of a perfect square.

(b) 8 is not a possible units digit of a perfect square.

(c) 0 is a possible units digit of a perfect square.

(d) 6 is a possible units digit of a perfect square.

Therefore, a perfect square can never have the digit 8 in its ones place.

The correct option is (b) 8.

Solution:

A perfect cube is a number that can be obtained by cubing an integer. To determine if a number is a perfect cube, we can find its prime factorization. A number is a perfect cube if and only if the exponents of all prime factors in its prime factorization are multiples of 3.

Let's examine the prime factorization of each option:

(a) 216:

$216 = 6^3 = (2 \times 3)^3 = 2^3 \times 3^3$. The exponents (3 and 3) are multiples of 3. So, 216 is a perfect cube.

(b) 567:

Let's find the prime factorization of 567:

$\begin{array}{c|cc} 3 & 567 \\ \hline 3 & 189 \\ \hline 3 & 63 \\ \hline 3 & 21 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

$567 = 3 \times 3 \times 3 \times 3 \times 7 = 3^4 \times 7^1$. The exponents (4 and 1) are not multiples of 3. So, 567 is not a perfect cube.

(c) 125:

$125 = 5^3$. The exponent (3) is a multiple of 3. So, 125 is a perfect cube.

(d) 343:

$343 = 7^3$. The exponent (3) is a multiple of 3. So, 343 is a perfect cube.

From the analysis, only 567 is not a perfect cube because the exponents in its prime factorization are not multiples of 3.

The correct option is (b) 567.

Solution:

We are asked to find the value of $\sqrt[3]{1000}$.

The cube root of a number is the value that, when multiplied by itself three times, gives the original number.

We need to find a number $x$ such that $x \times x \times x = 1000$, or $x^3 = 1000$.

Let's check the options:

(a) $10^3 = 10 \times 10 \times 10 = 100 \times 10 = 1000$.

(b) $100^3 = 100 \times 100 \times 100 = 10000 \times 100 = 1000000$.

(c) $1^3 = 1 \times 1 \times 1 = 1$.

Since $10^3 = 1000$, the cube root of 1000 is 10.

So, $\sqrt[3]{1000} = 10$.

The correct option is (a) 10.

Solution:

We are given that $m$ is the square of a natural number $n$.

This can be written mathematically as:

$m = n^2$

We need to express $n$ in terms of $m$. To do this, we can take the square root of both sides of the equation:

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

Since $n$ is a natural number, $n$ is positive. The square root of $n^2$ for a positive number $n$ is $n$.

So, $\sqrt{n^2} = n$.

Substituting this back into the equation, we get:

$n = \sqrt{m}$

Thus, if $m$ is the square of a natural number $n$, then $n$ is the square root of $m$.

The correct option is (d) $\sqrt{m}$.

Solution:

There is a mathematical rule to determine the number of digits in the square root of a perfect square:

1. If the number of digits ($n$) in a perfect square is even, the number of digits in its square root is $\frac{n}{2}$.

2. If the number of digits ($n$) in a perfect square is odd, the number of digits in its square root is $\frac{n + 1}{2}$.

Since the question specifies that $n$ is even, the number of digits in the square root will be $\frac{n}{2}$.

The correct option is (b).

Solution:

We are given that $m$ is the cube root of $n$.

This relationship can be written mathematically as:

$m = \sqrt[3]{n}$

To find $n$ in terms of $m$, we need to remove the cube root from $n$. We can do this by cubing both sides of the equation.

$(m)^3 = (\sqrt[3]{n})^3$

When we cube a cube root, the operations cancel each other out:

$m^3 = n$

So, $n$ is equal to $m^3$.

The correct option is (a) m³.

Solution:

We need to find the value of the expression $\sqrt{248 + \sqrt{52+\sqrt{144}}}$.

We evaluate the expression from the innermost square root outwards.

First, evaluate the innermost square root:

$\sqrt{144}$

Since $12^2 = 144$, we have $\sqrt{144} = 12$.

Substitute this value back into the expression:

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

Now, evaluate the expression inside the next square root:

$52 + 12 = 64$

So, the expression becomes:

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

Now, evaluate the square root of 64:

$\sqrt{64}$

Since $8^2 = 64$, we have $\sqrt{64} = 8$.

Substitute this value back into the outermost square root:

$\sqrt{248 + 8}$

Finally, evaluate the expression inside the outermost square root:

$248 + 8 = 256$

The expression simplifies to:

$\sqrt{256}$

We need to find the square root of 256. We know that $16^2 = 16 \times 16 = 256$.

Therefore, $\sqrt{256} = 16$.

The value of the expression is 16.

The correct option is (c) 16.

Given:

$\sqrt{4096} = 64$

To Find:

The value of $\sqrt{4096}$ + $\sqrt{40.96}$

Solution:

We are given the value of $\sqrt{4096} = 64$.

Now, we need to find the value of $\sqrt{40.96}$.

We can write 40.96 as a fraction:

$40.96 = \frac{4096}{100}$

Using the property of square roots, $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$, we have:

$\sqrt{40.96} = \sqrt{\frac{4096}{100}} = \frac{\sqrt{4096}}{\sqrt{100}}$

We know that $\sqrt{4096} = 64$ (given) and $\sqrt{100} = 10$ (since $10 \times 10 = 100$).

Substitute these values:

$\sqrt{40.96} = \frac{64}{10}$

$\sqrt{40.96} = 6.4$

Now, we need to calculate the sum $\sqrt{4096}$ + $\sqrt{40.96}$.

$\sqrt{4096}$ + $\sqrt{40.96}$ = $64 + 6.4$

Perform the addition:

$64.0 + 6.4 = 70.4$

Thus, the value of $\sqrt{4096}$ + $\sqrt{40.96}$ is 70.4.

The correct option is (d) 70.4.

Solution:

The perfect squares from $1$ to $100$ are as follows:

$1^2 = 1$

$2^2 = 4$

$3^2 = 9$

$4^2 = 16$

$5^2 = 25$

$6^2 = 36$

$7^2 = 49$

$8^2 = 64$

$9^2 = 81$

$10^2 = 100$

The question asks for the number of perfect squares between 1 and 100. This means we must exclude the boundary numbers 1 and 100 themselves.

The numbers are: 4, 9, 16, 25, 36, 49, 64, and 81.

Counting these values, we find there are 8 perfect squares.

Answer: 8

Solution:

We need to find the number of perfect cubes that lie between 1 and 1000 (strictly greater than 1 and strictly less than 1000).

A perfect cube is a number that is the cube of an integer.

Let's list the cubes of natural numbers starting from 1:

$1^3 = 1$

$2^3 = 8$

$3^3 = 27$

$4^3 = 64$

$5^3 = 125$

$6^3 = 216$

$7^3 = 343$

$8^3 = 512$

$9^3 = 729$

$10^3 = 1000$

We are looking for perfect cubes greater than 1 and less than 1000.

From the list, the numbers that satisfy this condition are 8, 27, 64, 125, 216, 343, 512, and 729.

Counting these numbers, we find there are 8 perfect cubes between 1 and 1000.

There are 8 perfect cubes between 1 and 1000.

Solution:

To find the units digit of the square of a number, we only need to consider the units digit of the original number.

The units digit of the number 1294 is 4.

Now, we find the square of this units digit:

$4^2 = 4 \times 4 = 16$

The units digit of 16 is 6.

Therefore, the units digit in the square of 1294 is 6.

The units digit in the square of 1294 is 6.

Solution:

We need to find the number of zeroes in the square of 500.

The number 500 can be written as $5 \times 100$, or $5 \times 10^2$.

The square of 500 is $(500)^2$.

$(500)^2 = (5 \times 100)^2 = 5^2 \times (100)^2$

$5^2 = 25$

$(100)^2 = 100 \times 100 = 10000$

So, $(500)^2 = 25 \times 10000 = 250000$

The number 250000 has 4 zeroes at the end.

Alternatively:

A property of squaring numbers ending in zeroes is that the number of zeroes in the square is twice the number of zeroes in the original number.

The number 500 has 2 zeroes.

So, the square of 500 will have $2 \times 2 = 4$ zeroes.

The square of 500 will have 4 zeroes.

Solution:

According to the properties of square numbers, the number of non-perfect square natural numbers lying between the squares of two consecutive natural numbers $n$ and $(n + 1)$ is always equal to $2n$.

For example, if $n = 3$ and $n + 1 = 4$:

$n^2 = 3^2 = 9$

$(n + 1)^2 = 4^2 = 16$

Numbers between 9 and 16 are: 10, 11, 12, 13, 14, 15 (Total 6 numbers).

Using the formula: $2n = 2 \times 3 = 6$.

Answer: $2n$

Given: The number is $24025$.

Solution:

The number of digits in the given perfect square is $n = 5$.

Since $n$ is an odd number, the number of digits in its square root is calculated by the formula:

Substituting $n = 5$:

$\text{Number of digits} = \frac{5 + 1}{2}$

$\text{Number of digits} = \frac{6}{2} = 3$

Alternate Method:

By placing bars over pairs of digits from right to left: $\overline{2} \ \overline{40} \ \overline{25}$. The number of bars is 3, which indicates the number of digits in the square root.

Answer: 3

Solution:

We need to find the square of 5.5.

The square of a number is the result of multiplying the number by itself.

$(5.5)^2 = 5.5 \times 5.5$

We can perform the multiplication:

$5.5 \times 5.5 = 30.25$

Alternatively, we can think of $5.5 = \frac{11}{2}$.

$(5.5)^2 = \left(\frac{11}{2}\right)^2 = \frac{11^2}{2^2} = \frac{121}{4}$

Converting the fraction to a decimal:

$\frac{121}{4} = 30.25$

The square of 5.5 is 30.25.

The square of 5.5 is 30.25.

Solution:

We are asked to find the value of $\sqrt{5.3 \times 5.3}$.

The expression inside the square root is $5.3 \times 5.3$, which is the same as $(5.3)^2$.

So, we need to find the value of $\sqrt{(5.3)^2}$.

The square root operation is the inverse of squaring (for non-negative numbers).

For any non-negative number $a$, $\sqrt{a^2} = a$.

In this case, $a = 5.3$, which is a positive number.

Therefore, $\sqrt{(5.3)^2} = 5.3$.

The value of $\sqrt{5.3 \times 5.3}$ is 5.3.

The square root of 5.3 × 5.3 is 5.3.

Solution:

We need to find the number of zeroes in the cube of 100.

The number 100 can be written as $10^2$.

The cube of 100 is $(100)^3$.

$(100)^3 = (10^2)^3$

Using the property of exponents $(a^m)^n = a^{m \times n}$, we get:

$(10^2)^3 = 10^{2 \times 3} = 10^6$

$10^6$ is equal to 1 followed by 6 zeroes, which is 1,000,000.

The number 1,000,000 has 6 zeroes.

Alternatively:

A property of cubing numbers ending in zeroes is that the number of zeroes in the cube is three times the number of zeroes in the original number.

The number 100 has 2 zeroes.

So, the cube of 100 will have $3 \times 2 = 6$ zeroes.

The cube of 100 will have 6 zeroes.

Solution:

We are asked to convert 1 square meter ($1\text{ m}^2$) to square centimeters ($\text{cm}^2$).

We know the relationship between meters and centimeters for linear measurement:

So, $1\text{ m} = 100\text{ cm}$.

To convert square units, we square the linear conversion factor:

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

Substitute the equivalent value in centimeters for each meter:

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

Multiply the numbers and the units:

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

$1\text{ m}^2 = 10000 \text{ cm}^2$

Thus, 1 square meter is equal to 10000 square centimeters.

1m² = 10000 cm².

Solution:

We are asked to convert 1 cubic meter ($1\text{ m}^3$) to cubic centimeters ($\text{cm}^3$).

We know the basic linear conversion between meters and centimeters:

So, $1\text{ m} = 100\text{ cm}$.

To convert cubic units, we need to cube the linear conversion factor. This means we multiply the conversion factor by itself three times.

$1\text{ m}^3 = (1\text{ m}) \times (1\text{ m}) \times (1\text{ m})$

Substitute the equivalent value in centimeters for each meter:

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

Now, multiply the numbers and the units:

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

$100 \times 100 \times 100 = 10000 \times 100 = 1,000,000$

So, $1\text{ m}^3 = 1,000,000 \text{ cm}^3$

Thus, 1 cubic meter is equal to 1,000,000 cubic centimeters.

1m³ = 1000000 cm³.

Solution:

To find the one's digit (units digit) of the cube of a number, we only need to consider the one's digit of the original number and find its cube's one's digit.

The given number is 38.

The one's digit of 38 is 8.

Now, we find the cube of the one's digit:

$8^3 = 8 \times 8 \times 8 = 64 \times 8 = 512$

The one's digit of 512 is 2.

Therefore, the one's digit in the cube of 38 is 2.

Ones digit in the cube of 38 is 2.

Solution:

We need to find the square of 0.7.

The square of a number is the result of multiplying the number by itself.

$(0.7)^2 = 0.7 \times 0.7$

To multiply decimals, we can multiply them as if they were whole numbers and then place the decimal point in the product.

$7 \times 7 = 49$

The number 0.7 has one decimal place. When we multiply 0.7 by 0.7, the total number of decimal places in the product will be the sum of the decimal places in the numbers being multiplied ($1 + 1 = 2$).

So, we place the decimal point in 49 such that there are two decimal places.

$0.7 \times 0.7 = 0.49$

The square of 0.7 is 0.49.

The square of 0.7 is 0.49.

Solution:

We need to find the sum of the first six odd natural numbers.

The first six odd natural numbers are 1, 3, 5, 7, 9, and 11.

We can add these numbers:

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

$= 4 + 5 + 7 + 9 + 11$

$= 9 + 7 + 9 + 11$

$= 16 + 9 + 11$

$= 25 + 11$

$= 36$

Alternatively:

The sum of the first $n$ odd natural numbers is given by the formula $n^2$.

Here, we are asked for the sum of the first six odd natural numbers, so $n=6$.

Sum $= 6^2 = 6 \times 6 = 36$

The sum of the first six odd natural numbers is 36.

The sum of first six odd natural numbers is 36.

Solution:

To find the digit at the ones place (units digit) of the square of a number, we only need to consider the units digit of the original number.

The units digit of the number 57 is 7.

Now, we find the square of this units digit:

$7^2 = 7 \times 7 = 49$

The units digit of 49 is 9.

Therefore, the digit at the ones place of $57^2$ is 9.

The digit at the ones place of 57² is 9.

Given:

To Find:

The lengths of the other two sides (the legs) of the right-angled triangle.

Solution:

According to the Pythagoras Theorem, in a right-angled triangle, the sum of the squares of the two legs is equal to the square of the hypotenuse.

Substituting the value of the hypotenuse:

$a^2 + b^2 = 17^2$

$a^2 + b^2 = 289$

To find the sides, we use the general formula for a Pythagorean Triplet. For any natural number $m > 1$, the three members of a triplet are given by:

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

Since the hypotenuse is always the longest side, we set the largest member of the triplet equal to 17:

$m^2 = 17 - 1$

$m^2 = 16$

$m = \sqrt{16} = 4$

Now, we find the other two sides using the value of $m = 4$:

First side $(a) = 2m = 2 \times 4 = 8 \text{ cm}$

Second side $(b) = m^2 - 1 = 4^2 - 1 = 16 - 1 = 15 \text{ cm}$

Verification:

We check if $8^2 + 15^2 = 17^2$:

$8^2 = 64$

$15^2 = 225$

$64 + 225 = 289$

$17^2 = 289$

Since the sum of the squares of 8 and 15 is exactly equal to the square of 17, the sides are correct.

Answer: The sides are 8 cm and 15 cm.

Solution:

We need to find the value of $\sqrt{1.96}$.

We can convert the decimal number 1.96 into a fraction:

$1.96 = \frac{196}{100}$

Now, we need to find the square root of this fraction:

$\sqrt{1.96} = \sqrt{\frac{196}{100}}$

Using the property of square roots, $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$, we can write:

$\sqrt{\frac{196}{100}} = \frac{\sqrt{196}}{\sqrt{100}}$

We know that $14^2 = 196$, so $\sqrt{196} = 14$.

We also know that $10^2 = 100$, so $\sqrt{100} = 10$.

Substitute these values into the expression:

$\frac{\sqrt{196}}{\sqrt{100}} = \frac{14}{10}$

Convert the fraction back to a decimal:

$\frac{14}{10} = 1.4$

So, $\sqrt{1.96} = 1.4$.

$\sqrt{1.96}$ = 1.4.

To Find: The value of $(1.2)^3$.

Solution:

The cube of a number is obtained by multiplying the number by itself three times.

$(1.2)^3 = 1.2 \times 1.2 \times 1.2$

We know that $12^3 = 1728$.

Since there is one decimal place in 1.2, its cube will have three decimal places ($1 \times 3 = 3$).

$(1.2)^3 = 1.728$

Answer: 1.728

Solution:

We need to determine whether the cube of an odd number is odd or even.

Recall the properties of multiplying odd and even numbers:

Odd $\times$ Odd = Odd

Odd $\times$ Even = Even

Even $\times$ Even = Even

The cube of a number is the number multiplied by itself three times. If the number is odd, let's call it $O$.

$O^3 = O \times O \times O$

First, $O \times O = \text{Odd} \times \text{Odd} = \text{Odd}$.

Then, the cube is $(\text{Odd} \times \text{Odd}) \times O = \text{Odd} \times \text{Odd}$.

Odd $\times$ Odd = Odd.

Let's check with some examples of odd numbers:

$1^3 = 1$ (Odd)

$3^3 = 27$ (Odd)

$5^3 = 125$ (Odd)

$7^3 = 343$ (Odd)

In all cases, the cube of an odd number is an odd number.

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

Solution:

The cube root of a number $x$ is the value $y$ such that $y^3 = x$.

The standard mathematical notation for the cube root of $x$ is using the radical symbol ($\sqrt{\phantom{x}}$) with a small 3 (called the index) placed above and to the left of the symbol.

So, the cube root of a number $x$ is denoted by $\sqrt[3]{x}$.

The cube root of a number x is denoted by $\sqrt[3]{x}$.

Given: The number is 125.

To Find: The least number to be multiplied to make it a perfect square.

Solution:

First, we find the prime factorisation of 125:

$\begin{array}{c|cc} 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

The prime factorisation is: $125 = \underbrace{5 \times 5}_{\text{pair}} \times 5$

For a number to be a perfect square, every prime factor must exist in a pair. In the factorisation above, one factor 5 is left without a pair.

To make it a perfect square, we must multiply the number by 5.

New number $= 125 \times 5 = 625$ (which is $25^2$).

Answer: 5

Given: The number is 72.

To Find: The least number to be multiplied to make it a perfect cube.

Solution:

First, we find the prime factorisation of 72:

$\begin{array}{c|cc} 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

The prime factorisation is: $72 = \underbrace{2 \times 2 \times 2}_{\text{triplet}} \times \underbrace{3 \times 3}_{\text{incomplete}}$

For a number to be a perfect cube, every prime factor must appear in a group of three (triplets). In the factorisation of 72, the factor 3 appears only twice.

To complete the triplet of 3, we need one more 3.

New number $= 72 \times 3 = 216$ (which is $6^3$).

Answer: 3

Given: The number is 72.

To Find: The least number by which 72 should be divided to make it a perfect cube.

Solution:

Performing the prime factorisation of 72:

$\begin{array}{c|cc} 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

The prime factorisation is: $72 = \underbrace{2 \times 2 \times 2}_{\text{triplet}} \times 3 \times 3$

In the factorisation, 2 forms a complete triplet, but the factors $3 \times 3$ are extra as they do not form a triplet.

To make the number a perfect cube, we must divide 72 by the product of these extra factors.

Divisor $= 3 \times 3 = 9$

New number $= \frac{72}{9} = 8$ (which is $2^3$).

Answer: 9

Solution:

To find the units digit of the cube of a number, we only need to find the units digit of the cube of the units digit of the original number.

The units digit of the original number is 7.

We need to find the units digit of $7^3$.

$7^3 = 7 \times 7 \times 7$

First, calculate $7 \times 7 = 49$. The units digit is 9.

Next, multiply the units digit (9) by 7: $9 \times 7 = 63$. The units digit is 3.

Therefore, the cube of a number ending in 7 will end in the digit 3.

Let's verify with an example:

Consider the number 17, which ends in 7.

$17^3 = 17 \times 17 \times 17 = 289 \times 17$.

The units digit of $289 \times 17$ is the units digit of $9 \times 7 = 63$, which is 3.

Indeed, $17^3 = 4913$, which ends in 3.

Cube of a number ending in 7 will end in the digit 3.

To find the units digit of the square of a number, we only need to consider the units digit of the original number.

The units digit of the number 86 is 6.

We find the square of the units digit: $6^2 = 36$.

The units digit of 36 is 6.

Therefore, the units digit of the square of 86 will be 6.

The statement is True.

Let's consider an example.

Take two perfect squares, $1^2 = 1$ and $2^2 = 4$.

Their sum is $1 + 4 = 5$.

The number 5 is not a perfect square because there is no integer whose square is 5.

Another example: Take $2^2 = 4$ and $3^2 = 9$.

Their sum is $4 + 9 = 13$.

The number 13 is not a perfect square.

While there are instances where the sum of two perfect squares is a perfect square (e.g., $3^2 + 4^2 = 9 + 16 = 25 = 5^2$), the statement claims it is true for *any* two perfect squares, which is not correct.

The statement is False.

Let the two perfect squares be $m^2$ and $n^2$, where $m$ and $n$ are integers.

The product of these two perfect squares is $m^2 \times n^2$.

Using the properties of exponents, we know that $m^2 \times n^2 = (m \times n)^2$.

Since $m$ and $n$ are integers, their product $m \times n$ is also an integer.

Therefore, $(m \times n)^2$ is the square of an integer, which means it is a perfect square.

Let's take an example:

Consider the perfect squares $9$ ($3^2$) and $16$ ($4^2$).

Their product is $9 \times 16 = 144$.

The number 144 is a perfect square because $12^2 = 144$. Note that $12 = 3 \times 4$, which aligns with $(m \times n)^2$.

The statement is True.

We need to check the perfect squares of integers around this range.

Let's list some perfect squares:

$1^2 = 1$

$2^2 = 4$

$3^2 = 9$

$4^2 = 16$

$5^2 = 25$

$6^2 = 36$

$7^2 = 49$

$8^2 = 64$

The perfect square just before 50 is $7^2 = 49$.

The perfect square just after 60 is $8^2 = 64$.

There are no perfect squares between 49 and 64. Since the range is 50 and 60 (exclusive of 50 and 60), there are no perfect squares in this interval.

The statement is True.

To check if the square root of 1521 is 31, we need to calculate the square of 31.

The square of 31 is $31 \times 31$.

$31^2 = 961$.

Since $31^2 = 961$ and not 1521, the statement is incorrect.

(For completeness, the square root of 1521 is 39, as $39^2 = 1521$).

The statement is False.

Statement: Each prime factor appears 3 times in its cube.

Answer: True (T)

Solution:

According to the properties of cubes, when a number is cubed, every prime factor in its prime factorisation is repeated three times (or appears in triplets). For example, if the prime factorisation of a number $x$ contains a prime factor $p$ exactly $n$ times, then in $x^3$, the prime factor $p$ will appear $3n$ times, which is always a multiple of 3.

To check the statement, we need to calculate the square of 2.8.

The square of 2.8 is $(2.8)^2$.

$(2.8)^2 = 2.8 \times 2.8$.

Multiplying 28 by 28 gives 784. Since there is one decimal place in 2.8, there will be $1 + 1 = 2$ decimal places in the product.

So, $2.8 \times 2.8 = 7.84$.

The calculated value of the square of 2.8 is $7.84$.

The statement claims the square of 2.8 is $78.4$.

Since $7.84 \neq 78.4$, the statement is incorrect.

The statement is False.

To verify the statement, we need to calculate the cube of 0.4.

The cube of 0.4 is $(0.4)^3$.

$(0.4)^3 = 0.4 \times 0.4 \times 0.4$.

First, multiply the numbers without considering the decimal point: $4 \times 4 \times 4 = 64$.

Now, count the total number of decimal places in the factors. Each 0.4 has one decimal place. Since we are multiplying three times, the total number of decimal places in the product will be $1 + 1 + 1 = 3$.

Place the decimal point 3 places from the right in the number 64. This gives 0.064.

So, $(0.4)^3 = 0.064$.

The calculated value matches the value given in the statement.

The statement is True.

To check if the square root of 0.9 is 0.3, we need to calculate the square of 0.3.

The square of 0.3 is $(0.3)^2$.

$(0.3)^2 = 0.3 \times 0.3$.

Multiplying the numbers without considering the decimal point: $3 \times 3 = 9$.

Count the total number of decimal places in the factors. Each 0.3 has one decimal place. So, the product will have $1 + 1 = 2$ decimal places.

Place the decimal point 2 places from the right in the number 9. This gives 0.09.

So, $(0.3)^2 = 0.09$.

The statement claims that the square root of 0.9 is 0.3, which means $(0.3)^2$ should be equal to 0.9.

Since $0.09 \neq 0.9$, the statement is incorrect.

The statement is False.

Let $n$ be a natural number. We need to check if $n^2 > n$ for all natural numbers $n$.

Natural numbers are usually considered to be $1, 2, 3, \dots$.

Let's test the statement with the first few natural numbers:

For $n=1$, the square is $1^2 = 1$. Is $1 > 1$? No, $1$ is equal to $1$.

For $n=2$, the square is $2^2 = 4$. Is $4 > 2$? Yes.

For $n=3$, the square is $3^2 = 9$. Is $9 > 3$? Yes.

For any natural number $n > 1$, multiplying $n$ by itself (which is greater than 1) will result in a number larger than $n$. So, $n^2 > n$ for $n > 1$.

However, for $n=1$, we have $1^2 = 1$, which is not strictly greater than 1.

Since the statement claims the square is *always* greater than the number itself for *every* natural number, the case $n=1$ serves as a counterexample.

The statement is False.

Statement: The cube root of 8000 is 200.

Answer: False (F)

Solution:

To verify the statement, we can calculate the cube of 200:

$200^3 = 200 \times 200 \times 200 = 8,000,000$

Now, let's find the actual cube root of 8000:

$\sqrt[3]{8000} = \sqrt[3]{20 \times 20 \times 20}$

$\sqrt[3]{8000} = 20$

Since the cube root is 20 and not 200, the statement is false.

We need to find the perfect cubes of natural numbers and check which ones fall between 1 and 100.

Let's list the first few perfect cubes:

$1^3 = 1$

$2^3 = 8$

$3^3 = 27$

$4^3 = 64$

$5^3 = 125$

We are looking for perfect cubes that are strictly greater than 1 and strictly less than 100.

From the list above:

$1^3 = 1$ is not between 1 and 100.

$2^3 = 8$ is between 1 and 100.

$3^3 = 27$ is between 1 and 100.

$4^3 = 64$ is between 1 and 100.

$5^3 = 125$ is not between 1 and 100.

The perfect cubes between 1 and 100 are 8, 27, and 64.

There are 3 perfect cubes between 1 and 100.

The statement claims there are five perfect cubes between 1 and 100.

The statement is False.

Statement: There are 200 natural numbers between $100^2$ and $101^2$.

Answer: True (T)

Solution:

We know that the number of non-perfect square natural numbers lying between the squares of two consecutive natural numbers $n$ and $(n + 1)$ is given by the formula $2n$.

In this problem, $n = 100$ and $n + 1 = 101$.

$\text{Number of natural numbers} = 2 \times n$

$\text{Number of natural numbers} = 2 \times 100 = 200$

Since the result matches the statement, it is true.

Statement: The sum of first $n$ odd natural numbers is $n^2$.

Answer: True (T)

Solution:

This is a fundamental property of square numbers. The sum of the first $n$ consecutive odd natural numbers starting from 1 is always equal to the square of $n$.

For example:

If $n = 1$, Sum $= 1 = 1^2$

If $n = 2$, Sum $= 1 + 3 = 4 = 2^2$

If $n = 3$, Sum $= 1 + 3 + 5 = 9 = 3^2$

Statement: 1000 is a perfect square.

Answer: False (F)

Solution:

A perfect square ending in zeros must always have an even number of trailing zeros (e.g., 100, 10,000, 1,000,000). The number 1000 has 3 zeros, which is an odd number, so it cannot be a perfect square.

Furthermore, we know that $31^2 = 961$ and $32^2 = 1024$. Since 1000 lies between two consecutive perfect squares, it is not a perfect square itself.

The units digit of a perfect square is determined by the units digit of the number being squared.

Let's examine the units digits of the squares of the digits from 0 to 9:

$0^2 = 0$ (Units digit is 0)

$1^2 = 1$ (Units digit is 1)

$2^2 = 4$ (Units digit is 4)

$3^2 = 9$ (Units digit is 9)

$4^2 = 16$ (Units digit is 6)

$5^2 = 25$ (Units digit is 5)

$6^2 = 36$ (Units digit is 6)

$7^2 = 49$ (Units digit is 9)

$8^2 = 64$ (Units digit is 4)

$9^2 = 81$ (Units digit is 1)

The possible units digits of a perfect square are 0, 1, 4, 5, 6, and 9.

The digit 8 is not among these possible units digits.

Therefore, a number ending in 8 cannot be a perfect square.

The statement is False.

Statement: For every natural number $m$, $(2m – 1, 2m^2 – 2m, 2m^2 – 2m + 1)$ is a pythagorean triplet.

Answer: True (T)

Solution:

Let $a = 2m - 1$, $b = 2m^2 - 2m$, and $c = 2m^2 - 2m + 1$. For these to form a Pythagorean triplet, the condition $a^2 + b^2 = c^2$ must be satisfied.

Observe that $c = b + 1$.

R.H.S: $c^2 = (b + 1)^2 = b^2 + 2b + 1$

L.H.S: $a^2 + b^2 = (2m - 1)^2 + b^2$

Expanding $(2m - 1)^2$: $4m^2 - 4m + 1$

We also know $2b = 2(2m^2 - 2m) = 4m^2 - 4m$.

Therefore, $a^2 = 2b + 1$.

Substituting this in L.H.S: $a^2 + b^2 = (2b + 1) + b^2 = b^2 + 2b + 1$.

Since L.H.S = R.H.S, the numbers form a Pythagorean triplet.

Statement: All numbers of a pythagorean triplet are odd.

Answer: False (F)

Solution:

In a Pythagorean triplet $(a, b, c)$, it is impossible for all three numbers to be odd. This is because the square of an odd number is odd, and the sum of two odd numbers is always even. Thus, if $a$ and $b$ are odd, $a^2 + b^2$ (which equals $c^2$) must be even, meaning $c$ must be even.

Common examples of triplets with even numbers are $(3, 4, 5)$ and $(6, 8, 10)$.

Statement: For an integer $a$, $a^3$ is always greater than $a^2$.

Answer: False (F)

Solution:

We can provide counter-examples where this statement fails:

1. If $a = 1$, then $a^3 = 1^3 = 1$ and $a^2 = 1^2 = 1$. Here, $a^3 = a^2$.

2. If $a = 0$, then $a^3 = 0$ and $a^2 = 0$. Here, $a^3 = a^2$.

3. If $a = -2$ (a negative integer), then $a^3 = (-2)^3 = -8$ and $a^2 = (-2)^2 = 4$. Here, $-8 < 4$, so $a^3 < a^2$.

Statement: If $x$ and $y$ are integers such that $x^2 > y^2$, then $x^3 > y^3$.

Answer: False (F)

Solution:

Let's consider a counter-example using negative integers.

Let $x = -3$ and $y = 2$.

$x^2 = (-3)^2 = 9$

$y^2 = (2)^2 = 4$

Here, $x^2 > y^2$ ($9 > 4$) is True.

Now check the cubes:

$x^3 = (-3)^3 = -27$

$y^3 = (2)^3 = 8$

Here, $x^3 > y^3$ ($-27 > 8$) is False, because $-27$ is less than $8$.

Statement: Let $x$ and $y$ be natural numbers. If $x$ divides $y$, then $x^3$ divides $y^3$.

Answer: True (T)

Solution:

If $x$ divides $y$, it means $y$ can be expressed as a multiple of $x$.

$y = kx$ (where $k$ is some natural number)

Cubing both sides of the equation:

$y^3 = (kx)^3$

$y^3 = k^3 \cdot x^3$

Since $k$ is a natural number, $k^3$ is also a natural number. This shows that $y^3$ is a multiple of $x^3$. Therefore, $x^3$ divides $y^3$.

To Find: Whether the statement is true or false.

Solution:

If $a^2$ ends in $5$, it implies that the units digit of $a$ must be $5$. Let us test this with different numbers ending in $5$.

Case 1: Let $a = 5$

Case 2: Let $a = 15$

In the second case, while $a^2$ ends in $5$, $a^3$ ends in $75$, not $25$. Therefore, the statement is not always true.

Final Answer: False (F)

To Find: Whether the statement is true or false.

Solution:

If $a^2$ ends in $9$, the units digit of $a$ can be either $3$ or $7$ (since $3^2 = 9$ and $7^2 = 49$).

Case 1: If the units digit of $a$ is $3$

Case 2: If the units digit of $a$ is $7$

Since $a^3$ can end in either $7$ or $3$, the statement that it must end in $7$ is false.

Final Answer: False (F)

To Find: Whether the statement regarding the number of digits in a square root is true or false.

Solution:

According to the mathematical rule for perfect squares:

1. If the number of digits $n$ is even, the number of digits in the square root is $\frac{n}{2}$.

2. If the number of digits $n$ is odd, the number of digits in the square root is $\frac{n + 1}{2}$.

Example: Consider the perfect square $625$.

Using the formula:

Since $\sqrt{625} = 25$, which has $2$ digits, the statement is verified.

Final Answer: True (T)

In mathematics, the symbol $\sqrt{\ }$ is used to denote the square root of a number.

For a number $x$, its square root is denoted by $\sqrt{x}$.

Specifically, $\sqrt{x}$ represents the principal (non-negative) square root of $x$ when $x$ is a non-negative real number.

This is the standard and widely accepted notation for the square root.

The statement is True.

To determine the units digit of the square of a number, we only need to consider the units digit of the original number.

The units digit of the number is given as 7.

We need to find the units digit of the square of a number ending in 7. This is the same as finding the units digit of $7^2$.

$7^2 = 49$.

The units digit of 49 is 9.

So, a number having 7 at its ones place will have 9 (not 3) at the units place of its square.

The statement is False.

To determine the units digit of the cube of a number, we only need to consider the units digit of the original number.

The units digit of the number is given as 7.

We need to find the units digit of the cube of a number ending in 7. This is the same as finding the units digit of $7^3$.

$7^3 = 7 \times 7 \times 7 = 49 \times 7$.

To find the units digit of $49 \times 7$, we only need to multiply the units digits: $9 \times 7 = 63$.

The units digit of 63 is 3.

So, the units digit of $7^3$ is 3.

Therefore, a number having 7 at its ones place will have 3 at the ones place of its cube.

The statement is True.

A one-digit number is an integer from 0 to 9. We need to examine the cubes of these numbers and check if any of them result in a two-digit number.

$0^3 = 0$ (One-digit number)

$1^3 = 1$ (One-digit number)

$2^3 = 8$ (One-digit number)

$3^3 = 27$ (Two-digit number)

$4^3 = 64$ (Two-digit number)

$5^3 = 125$ (Three-digit number)

$6^3 = 216$ (Three-digit number)

$7^3 = 343$ (Three-digit number)

$8^3 = 512$ (Three-digit number)

$9^3 = 729$ (Three-digit number)

We can see that the cubes of the one-digit numbers 3 and 4 are 27 and 64, respectively. Both 27 and 64 are two-digit numbers.

Since the cube of a one-digit number can indeed be a two-digit number, the statement is false.

The statement is False.

We need to check the parity (whether a number is odd or even) of the cube of an even number.

An even number can be represented in the form $2k$, where $k$ is an integer.

Let's find the cube of an even number $(2k)$:

$(2k)^3 = (2k) \times (2k) \times (2k)$.

Using the property of multiplication, $(ab)^n = a^n b^n$, we have:

$(2k)^3 = 2^3 \times k^3 = 8 \times k^3$.

Since $8k^3$ is a multiple of 8, it is also a multiple of 2. Any integer that is a multiple of 2 is an even number.

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

Example:

Let the even number be 2. Its cube is $2^3 = 8$. 8 is even.

Let the even number be 4. Its cube is $4^3 = 64$. 64 is even.

Let the even number be 6. Its cube is $6^3 = 216$. 216 is even.

The statement claims that the cube of an even number is odd.

The statement is False.

To Find: Whether the statement is true or false.

Solution:

We know that the product of three odd numbers is always an odd number.

Example: Let the odd number be $5$.

As $125$ is an odd number, the statement that the cube of an odd number is even is incorrect.

Final Answer: False (F)

To Find: Whether the statement is true or false.

Solution:

We know that the product of three even numbers is always an even number.

Example: Let the even number be $4$.

As $64$ is an even number, the statement is correct.

Final Answer: True (T)

To Find: Whether the statement is true or false.

Solution:

According to the properties of numbers, an odd number is of the form $2n + 1$. When we multiply three odd numbers together, the result is always odd.

Example: Let us take the odd number $7$.

Since $343$ is an odd number, the statement is true.

Final Answer: True (T)

A perfect cube is an integer that is the cube of an integer. To determine if 999 is a perfect cube, we can find its prime factorization.

We perform the prime factorization of 999:

Divide 999 by the smallest prime factor, which is 3:

The prime factorization of 999 is $3 \times 3 \times 3 \times 37$.

We can write this using exponents: $999 = 3^3 \times 37^1$.

For a number to be a perfect cube, the exponents of all its prime factors must be multiples of 3.

In the prime factorization of 999, the exponent of the prime factor 3 is 3 (which is a multiple of 3), but the exponent of the prime factor 37 is 1 (which is not a multiple of 3).

Since the exponent of at least one prime factor is not a multiple of 3, 999 is not a perfect cube.

The statement is False.

To Find: Whether the product $363 \times 81$ is a perfect cube.

Solution:

To determine if a number is a perfect cube, we perform Prime Factorisation and check if the prime factors can be grouped into triplets.

First, find the prime factors of $363$:

$\begin{array}{c|cc} 3 & 363 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \end{array}$

Next, find the prime factors of $81$:

$\begin{array}{c|cc} 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

Now, multiply the factors together:

$363 \times 81 = (3 \times 11 \times 11) \times (3 \times 3 \times 3 \times 3)$

Group the same prime factors:

$363 \times 81 = (3 \times 3 \times 3) \times 3 \times 3 \times 11 \times 11$

$363 \times 81 = 3^3 \times 3^2 \times 11^2$

For a number to be a perfect cube, the power of every prime factor must be a multiple of $3$. Here, the powers of $3$ (total power 5) and $11$ are not multiples of $3$.

Final Answer: False (F)

We are asked to determine if the cube roots of 8 are +2 and -2.

A cube root of a number $x$ is a number $y$ such that $y^3 = x$.

Let's check if $(+2)^3 = 8$:

$(+2)^3 = 2 \times 2 \times 2 = 8$.

So, +2 is a cube root of 8.

Now let's check if $(-2)^3 = 8$:

$(-2)^3 = (-2) \times (-2) \times (-2)$.

$(-2) \times (-2) = +4$.

$(+4) \times (-2) = -8$.

So, $(-2)^3 = -8$, which is not equal to 8.

Therefore, -2 is not a cube root of 8.

For real numbers, every positive number has exactly one positive real cube root. The cube root of 8 is the unique real number whose cube is 8.

The notation $\sqrt[3]{8}$ denotes the principal (real) cube root of 8, which is +2.

While there are complex cube roots for any non-zero number, when referring to "the cube root" in this context, it typically means the real cube root.

Since -2 is not a cube root of 8, the statement that the cube roots are +2 and -2 is false.

The statement is False.

We need to evaluate both sides of the given equation and check if they are equal.

Consider the left side of the equation: $\sqrt[3]{8\;+\;27}$.

First, perform the addition inside the cube root: $8 + 27 = 35$.

So, the left side is $\sqrt[3]{35}$.

The number 35 is not a perfect cube (since $3^3 = 27$ and $4^3 = 64$). Thus, $\sqrt[3]{35}$ is an irrational number (or not a simple integer).

Now, consider the right side of the equation: $\sqrt[3]{8}$ + $\sqrt[3]{27}$.

First, find the cube root of 8. Since $2^3 = 8$, the cube root of 8 is 2.

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

Next, find the cube root of 27. Since $3^3 = 27$, the cube root of 27 is 3.

$\sqrt[3]{27} = 3$.

Now, add these cube roots: $\sqrt[3]{8}$ + $\sqrt[3]{27} = 2 + 3 = 5$.

We compare the values of the left side and the right side:

Left side = $\sqrt[3]{35}$.

Right side = 5.

To check if $\sqrt[3]{35} = 5$, we can cube both sides:

$(\sqrt[3]{35})^3 = 35$.

$5^3 = 5 \times 5 \times 5 = 125$.

Since $35 \neq 125$, the statement $\sqrt[3]{35} = 5$ is false.

Therefore, $\sqrt[3]{8\;+\;27} \neq \sqrt[3]{8}$ + $\sqrt[3]{27}$.

In general, for any positive numbers $a$ and $b$, $\sqrt[n]{a+b} \neq \sqrt[n]{a} + \sqrt[n]{b}$ for $n > 1$.

The statement is False.

To Find: Whether the statement is true or false.

Solution:

Unlike square roots of negative numbers (which are not real numbers), the cube root of a negative integer is always defined and is a negative integer.

Example: Let us find the cube root of $-8$.

Since negative integers do have cube roots, the statement is false.

Final Answer: False (F)

To Find: Whether the statement is true or false.

Solution:

The square of any non-zero number (whether positive or negative) is always positive because the product of two negative signs is positive. However, the cube of a number depends on the sign of the original number.

Let the number be $x$.

Case 1: If $x = 2$ (Positive number)

$x^2 = 2^2 = 4$ (Positive)

$x^3 = 2^3 = 8$ (Positive)

Case 2: If $x = -2$ (Negative number)

$x^2 = (-2)^2 = 4$ (Positive)

$x^3 = (-2)^3 = -8$ (Negative)

In Case 2, while the square is positive, the cube is negative. Therefore, the statement is not always true.

Final Answer: False (F)

The first five square numbers are the squares of the first five natural numbers.

The first five natural numbers are 1, 2, 3, 4, and 5.

The square of the first natural number is $1^2 = 1 \times 1 = 1$.

The square of the second natural number is $2^2 = 2 \times 2 = 4$.

The square of the third natural number is $3^2 = 3 \times 3 = 9$.

The square of the fourth natural number is $4^2 = 4 \times 4 = 16$.

The square of the fifth natural number is $5^2 = 5 \times 5 = 25$.

The first five square numbers are 1, 4, 9, 16, and 25.

The first three multiples of 3 are obtained by multiplying 3 by the first three natural numbers (1, 2, and 3).

The first multiple of 3 is $3 \times 1 = 3$.

The second multiple of 3 is $3 \times 2 = 6$.

The third multiple of 3 is $3 \times 3 = 9$.

So, the first three multiples of 3 are 3, 6, and 9.

Now, we need to find the cubes of these numbers.

The cube of the first multiple (3) is $3^3 = 3 \times 3 \times 3 = 27$.

The cube of the second multiple (6) is $6^3 = 6 \times 6 \times 6 = 36 \times 6 = 216$.

The cube of the third multiple (9) is $9^3 = 9 \times 9 \times 9 = 81 \times 9 = 729$.

The cubes of the first three multiples of 3 are 27, 216, and 729.

To Show: 500 is not a perfect square.

Solution:

To determine if a number is a perfect square, we find its prime factors. If any prime factor does not occur in a pair, the number is not a perfect square.

Let us perform the Prime Factorisation of $500$:

$\begin{array}{c|cc} 2 & 500 \\ \hline 2 & 250 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

The prime factors of $500$ are:

Grouping the identical factors into pairs:

We observe that the prime factor $5$ is left without a pair. For a number to be a perfect square, all its prime factors must exist in pairs.

Hence, $500$ is not a perfect square.

We know that the sum of the first $n$ consecutive odd natural numbers is equal to $n^2$.

In this question, the number is 81. We recognize that $81 = 9^2$.

According to the property, $9^2$ is the sum of the first 9 consecutive odd natural numbers.

The first nine consecutive odd natural numbers are:

1, 3, 5, 7, 9, 11, 13, 15, 17.

Let's find the sum of these numbers:

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

We can group them for easier addition:

Sum = $(1 + 17) + (3 + 15) + (5 + 13) + (7 + 11) + 9$

Sum = $18 + 18 + 18 + 18 + 9$

Sum = $4 \times 18 + 9$

Sum = $72 + 9$

Sum = $81$.

So, 81 can be expressed as the sum of the first nine consecutive odd numbers as follows:

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

(a) 484

Prime factorisation of $484$:

$\begin{array}{c|cc} 2 & 484 \\ \hline 2 & 242 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \end{array}$

$484 = 2 \times 2 \times 11 \times 11 = (2 \times 11)^2 = 22^2$

Since all factors are in pairs, 484 is a perfect square.

(b) 11250

Prime factorisation of $11250$:

$\begin{array}{c|cc} 2 & 11250 \\ \hline 3 & 5625 \\ \hline 3 & 1875 \\ \hline 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

$11250 = 2 \times 3 \times 3 \times 5 \times 5 \times 5 \times 5 = 2 \times (3 \times 3) \times (5 \times 5) \times (5 \times 5)$

The factor $2$ does not have a pair. Thus, 11250 is not a perfect square.

(c) 841

Prime factorisation of $841$:

$\begin{array}{c|cc} 29 & 841 \\ \hline 29 & 29 \\ \hline & 1 \end{array}$

$841 = 29 \times 29 = 29^2$

Since the factor $29$ is in a pair, 841 is a perfect square.

(d) 729

Prime factorisation of $729$:

$\begin{array}{c|cc} 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

$729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = (3 \times 3 \times 3)^2 = 27^2$

Since all factors are in pairs, 729 is a perfect square.

A number is a perfect cube if and only if in its prime factorization, all the exponents of the prime factors are multiples of 3.

(a) 128

Find the prime factorization of 128:

The prime factorization of 128 is $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^7$.

The exponent of the prime factor 2 is 7. Since 7 is not a multiple of 3, 128 is not a perfect cube.

(b) 343

Find the prime factorization of 343:

The prime factorization of 343 is $7 \times 7 \times 7 = 7^3$.

The exponent of the prime factor 7 is 3. Since 3 is a multiple of 3, 343 is a perfect cube ($\sqrt[3]{343} = 7$).

(c) 729

Find the prime factorization of 729:

The prime factorization of 729 is $3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^6$.

The exponent of the prime factor 3 is 6. Since 6 is a multiple of 3 ($6 = 3 \times 2$), 729 is a perfect cube ($\sqrt[3]{729} = \sqrt[3]{3^6} = 3^{6/3} = 3^2 = 9$).

(d) 1331

Find the prime factorization of 1331:

The prime factorization of 1331 is $11 \times 11 \times 11 = 11^3$.

The exponent of the prime factor 11 is 3. Since 3 is a multiple of 3, 1331 is a perfect cube ($\sqrt[3]{1331} = 11$).

Based on the prime factorizations, the perfect cubes are 343, 729, and 1331.

The distributive law of multiplication over addition states that $a \times (b + c) = a \times b + a \times c$. We can use this property, often in conjunction with the identity $(x+y)^2 = x^2 + 2xy + y^2$ or $(x-y)^2 = x^2 - 2xy + y^2$, which are derived from the distributive property, to find the squares.

(a) 101

We can write 101 as the sum of two numbers, such as $100 + 1$.

Then, the square of 101 is $101^2 = (100 + 1)^2$.

Using the identity $(x+y)^2 = x^2 + 2xy + y^2$ with $x = 100$ and $y = 1$:

$101^2 = (100)^2 + 2(100)(1) + (1)^2$

$101^2 = 10000 + 200 + 1$

$101^2 = 10201$.

(b) 72

We can write 72 as the sum or difference of two numbers, such as $70 + 2$ or $80 - 8$. Let's use $70 + 2$.

The square of 72 is $72^2 = (70 + 2)^2$.

Using the identity $(x+y)^2 = x^2 + 2xy + y^2$ with $x = 70$ and $y = 2$:

$72^2 = (70)^2 + 2(70)(2) + (2)^2$

$72^2 = 4900 + 280 + 4$

$72^2 = 5180 + 4$

$72^2 = 5184$.

Alternatively, using $80 - 8$:

$72^2 = (80 - 8)^2$.

Using the identity $(x-y)^2 = x^2 - 2xy + y^2$ with $x = 80$ and $y = 8$:

$72^2 = (80)^2 - 2(80)(8) + (8)^2$

$72^2 = 6400 - 1280 + 64$

$72^2 = 5120 + 64$

$72^2 = 5184$.

Given: Sides of the triangle are $6\text{ cm}$, $8\text{ cm}$, and $10\text{ cm}$.

To Find: Whether these sides form a right-angled triangle.

Solution:

A triangle is right-angled if the square of the longest side (hypotenuse) is equal to the sum of the squares of the other two sides. This is known as the Pythagoras Theorem.

Let $a = 6\text{ cm}$, $b = 8\text{ cm}$, and $c = 10\text{ cm}$ (longest side).

Calculating the squares of the sides:

Now, checking the sum of $a^2$ and $b^2$:

Reason: Since the sides satisfy the Pythagoras property, they form a Pythagorean triplet.

Answer: Yes, a right triangle can be formed.

To Find: A Pythagorean triplet with one number as $4$.

Solution:

For any natural number $m > 1$, the general form of a Pythagorean triplet is $2m$, $m^2 - 1$, and $m^2 + 1$.

Let us assume $2m = 4$:

Now, find the other two numbers:

The numbers are $3$, $4$, and $5$. Let us verify:

Final Answer: The Pythagorean triplet is (3, 4, 5).

To find the square root of a number using prime factorization, we first find the prime factorization of the number. Then, we group the identical prime factors into pairs. The square root is obtained by taking one factor from each pair and multiplying them.

(a) 11025

Find the prime factorization of 11025:

The prime factorization of 11025 is $3 \times 3 \times 5 \times 5 \times 7 \times 7$.

Group the prime factors into pairs: $(3 \times 3) \times (5 \times 5) \times (7 \times 7) = 3^2 \times 5^2 \times 7^2$.

To find the square root, take one factor from each pair: $3 \times 5 \times 7$.

Multiply these factors: $3 \times 5 \times 7 = 15 \times 7 = 105$.

So, $\sqrt{11025} = 105$.

(b) 4761

Find the prime factorization of 4761.

The sum of the digits is $4 + 7 + 6 + 1 = 18$, which is divisible by 3, so 4761 is divisible by 3.

The prime factorization of 4761 is $3 \times 3 \times 23 \times 23$.

Group the prime factors into pairs: $(3 \times 3) \times (23 \times 23) = 3^2 \times 23^2$.

To find the square root, take one factor from each pair: $3 \times 23$.

Multiply these factors: $3 \times 23 = 69$.

So, $\sqrt{4761} = 69$.

(a) 512

Prime factorisation of $512$:

$\begin{array}{c|cc} 2 & 512 \\ \hline 2 & 256 \\ \hline 2 & 128 \\ \hline 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$

$512 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2)$

Taking one factor from each triplet for the cube root:

(b) 2197

Prime factorisation of $2197$:

$\begin{array}{c|cc} 13 & 2197 \\ \hline 13 & 169 \\ \hline 13 & 13 \\ \hline & 1 \end{array}$

$2197 = 13 \times 13 \times 13$

Since $13$ forms a complete triplet:

Given: The number is $176$.

To Find: Whether $176$ is a perfect square and the smallest number to multiply it with to make it a perfect square.

Solution:

To check if $176$ is a perfect square, we perform Prime Factorisation:

$\begin{array}{c|cc} 2 & 176 \\ \hline 2 & 88 \\ \hline 2 & 44 \\ \hline 2 & 22 \\ \hline 11 & 11 \\ \hline & 1 \end{array}$

The prime factors of $176$ are:

Grouping the prime factors into pairs:

Since the prime factor $11$ does not have a pair, 176 is not a perfect square.

To make it a perfect square, every factor must be in a pair. Therefore, we must multiply $176$ by $11$.

Verification:

Final Answer: $176$ is not a perfect square. The smallest number by which it should be multiplied is 11.

Given: The number is $9720$.

To Find: Whether $9720$ is a perfect cube and the smallest number to divide it by to make it a perfect cube.

Solution:

To determine if $9720$ is a perfect cube, we perform Prime Factorisation:

$\begin{array}{c|cc} 2 & 9720 \\ \hline 2 & 4860 \\ \hline 2 & 2430 \\ \hline 3 & 1215 \\ \hline 3 & 405 \\ \hline 3 & 135 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

The prime factors of $9720$ are:

Grouping the factors into triplets:

Since the factors $3 \times 3 \times 5$ do not form a complete triplet, 9720 is not a perfect cube.

To make it a perfect cube, we must divide by the extra factors that are not part of a triplet.

If we divide $9720$ by $45$:

Final Answer: $9720$ is not a perfect cube. The smallest number by which it should be divided is 45.

To Find: Two different Pythagorean triplets that contain the number $5$.

Solution:

A Pythagorean triplet consists of three positive integers $a, b,$ and $c$ such that $a^2 + b^2 = c^2$.

Triplet 1: Let us consider the case where $5$ is the largest number (hypotenuse).

We know that $3^2 + 4^2 = 9 + 16 = 25$.

Thus, (3, 4, 5) is the first triplet.

Triplet 2: Let us consider the case where $5$ is the smallest number.

For an odd number $n$, a Pythagorean triplet can be found using the formula: $n, \frac{n^2 - 1}{2}, \frac{n^2 + 1}{2}$.

Substituting $n = 5$:

Verification:

Thus, (5, 12, 13) is the second triplet.

Final Answer: The two Pythagorean triplets are (3, 4, 5) and (5, 12, 13).

Given: The number is $216$.

To Find: The smallest number to divide $216$ to get a perfect square and its square root.

Solution:

First, we find the Prime Factorisation of $216$:

$\begin{array}{c|cc} 2 & 216 \\ \hline 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

The factors are: $216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3$.

Grouping the factors into pairs:

The factors $2$ and $3$ do not have pairs. To make the quotient a perfect square, we must divide $216$ by the product of these unpaired factors.

Now, finding the quotient:

Finding the square root of the quotient:

Final Answer: The smallest number is 6 and the square root of the quotient is 6.

Given: The number is $3600$.

To Find:

1. The smallest number by which $3600$ should be multiplied to make it a perfect cube.

2. The cube root of the resulting product.

Solution:

To find the smallest number required for multiplication, we first determine the Prime Factorisation of $3600$:

$\begin{array}{c|cc} 2 & 3600 \\ \hline 2 & 1800 \\ \hline 2 & 900 \\ \hline 2 & 450 \\ \hline 3 & 225 \\ \hline 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

The prime factors of $3600$ are:

Now, we group these prime factors into triplets:

To make the number a perfect cube, every prime factor must appear in a triplet. Looking at the grouping above, we observe that:

1. The factor $2$ appears once outside the triplet. We need two more $2$s ($2 \times 2$) to complete this triplet.

2. The factor $3$ appears twice. We need one more $3$ to complete this triplet.

3. The factor $5$ appears twice. We need one more $5$ to complete this triplet.

Calculating the Smallest Number:

The smallest number to be multiplied is the product of the missing factors identified above:

Finding the Product and its Cube Root:

Now, multiply $3600$ by $60$ to get the perfect cube:

The cube root of the new product is:

Final Answer: The smallest number by which $3600$ should be multiplied is 60 and the cube root of the resulting product is 60.

(a) 1369

$\begin{array}{c|cc} & 3 \ 7 & \\ \hline \phantom{()} 3 & \overline{13} \ \overline{69} \\ + \; 3 & 9\phantom{(....)} \\ \hline \phantom{()} 6 \ 7 & 4 \ 69 \\ \phantom{()} +7 & 4 \ 69 \\ \hline \phantom{()} & 0 \end{array}$

$\sqrt{1369} = 37$

(b) 5625

$\begin{array}{c|cc} & 7 \ 5 & \\ \hline \phantom{()} 7 & \overline{56} \ \overline{25} \\ + \; 7 & 49\phantom{(....)} \\ \hline \phantom{()} 14 \ 5 & 7 \ 25 \\ \phantom{()} +5 & 7 \ 25 \\ \hline \phantom{()} & 0 \end{array}$

$\sqrt{5625} = 75$

(a) 27.04

$\begin{array}{c|cc} & 5 \ . \ 2 & \\ \hline \phantom{()} 5 & \overline{27} \ . \ \overline{04} \\ + \; 5 & 25\phantom{(....)} \\ \hline \phantom{()} 10 \ 2 & 2 \ 04 \\ \phantom{()} +2 & 2 \ 04 \\ \hline \phantom{()} & 0 \end{array}$

$\sqrt{27.04} = 5.2$

(b) 1.44

$\begin{array}{c|cc} & 1 \ . \ 2 & \\ \hline \phantom{()} 1 & \overline{1} \ . \ \overline{44} \\ + \; 1 & 1\phantom{(....)} \\ \hline \phantom{()} 2 \ 2 & 0 \ 44 \\ \phantom{()} +2 & 44 \\ \hline \phantom{()} & 0 \end{array}$

$\sqrt{1.44} = 1.2$

Solution:

We use the long division method to find the remainder for $1385$.

$\begin{array}{c|cc} & 3 \ 7 & \\ \hline \phantom{()} 3 & \overline{13} \ \overline{85} \\ + \; 3 & 9\phantom{(....)} \\ \hline \phantom{()} 6 \ 7 & 4 \ 85 \\ \phantom{()} +7 & 4 \ 69 \\ \hline \phantom{()} & 16 \end{array}$

The remainder is $16$. This means that $37^2$ is less than $1385$ by $16$.

Now, the perfect square is:

Square root of $1369$:

Final Answer: The least number to be subtracted is 16 and the square root is 37.

Solution:

First, we perform long division on $6200$.

$\begin{array}{c|cc} & 7 \ 8 & \\ \hline \phantom{()} 7 & \overline{62} \ \overline{00} \\ + \; 7 & 49\phantom{(....)} \\ \hline \phantom{()} 14 \ 8 & 13 \ 00 \\ \phantom{()} +8 & 11 \ 84 \\ \hline \phantom{()} & 116 \end{array}$

Here, $78^2 < 6200$. The next perfect square will be $79^2$.

Calculating $79^2$:

The number to be added is the difference between the next perfect square and $6200$:

Final Answer: The least number that should be added is 41.

To Find: The smallest $4$-digit number which is a perfect square.

Solution:

The smallest $4$-digit number is $1000$. To find the least $4$-digit perfect square, we find the square root of $1000$ using the Long Division Method.

$\begin{array}{c|cc} & 3 \ 1 & \\ \hline \phantom{()} 3 & \overline{10} \ \overline{00} \\ + \; 3 & 9\phantom{(....)} \\ \hline \phantom{()} 6 \ 1 & 1 \ 00 \\ \phantom{()} +1 & 61 \\ \hline \phantom{()} & 39 \end{array}$

The remainder is $39$. This shows that $31^2 < 1000$.

The next natural number after $31$ is $32$. Therefore, the smallest $4$-digit perfect square will be the square of $32$.

Final Answer: The least number of four digits that is a perfect square is 1024.

To Find: The largest $3$-digit number which is a perfect square.

Solution:

The greatest $3$-digit number is $999$. We find the square root of $999$ using the Long Division Method to see how much more it is than a perfect square.

$\begin{array}{c|cc} & 3 \ 1 & \\ \hline \phantom{()} 3 & \overline{9} \ \overline{99} \\ + \; 3 & 9\phantom{(....)} \\ \hline \phantom{()} 6 \ 1 & 0 \ 99 \\ \phantom{()} +1 & 61 \\ \hline \phantom{()} & 38 \end{array}$

The remainder is $38$. This means that if we subtract $38$ from $999$, we will get a perfect square.

We can verify this as $31^2 = 961$. The next square is $32^2 = 1024$, which is a $4$-digit number.

Final Answer: The greatest number of three digits that is a perfect square is 961.

To Find: The smallest perfect square number divisible by $3, 4, 5, 6$ and $8$.

Solution:

First, we find the LCM of $3, 4, 5, 6$ and $8$ to find the smallest number divisible by all of them.

$\begin{array}{c|ccccc} 2 & 3 \;, & 4 \;, & 5 \;, & 6 \;, & 8 \\ \hline 2 & 3 \; , & 2 \; , & 5 \; , & 3 \;, & 4 \\ \hline 2 & 3 \; , & 1 \; , & 5 \; , & 3 \;, & 2 \\ \hline 3 & 3 \; , & 1 \; , & 5 \; , & 3 \;, & 1 \\ \hline 5 & 1 \; , & 1 \; , & 5 \; , & 1 \;, & 1 \\ \hline & 1 \; , & 1 \; , & 1 \; , & 1 \;, & 1 \end{array}$

Now, let us look at the prime factorisation of $120$:

To make $120$ a perfect square, all factors must be in pairs. Here, $2, 3,$ and $5$ are not in pairs. Thus, we must multiply $120$ by $(2 \times 3 \times 5) = 30$.

Final Answer: The least square number is 3600.

Given: Diagonal of a square $= 10\text{ cm}$.

To Find: The length of the side of the square.

Solution:

Let the side of the square be $s$. In a square, the diagonal $d$ forms a right-angled triangle with two sides. By Pythagoras Theorem:

Taking square root on both sides:

If we take the value of $\sqrt{2} \approx 1.414$:

Final Answer: The length of the side of the square is $5\sqrt{2}$ cm or approximately 7.07 cm.

Given: Product of a decimal number with itself $= 51.84$.

To Find: The decimal number.

Solution:

Let the decimal number be $x$. According to the question:

Calculating the square root by Long Division Method:

$\begin{array}{c|cc} & 7 \ . \ 2 & \\ \hline \phantom{()} 7 & \overline{51} \ . \ \overline{84} \\ + \; 7 & 49\phantom{(....)} \\ \hline \phantom{()} 14 \ 2 & 2 \ 84 \\ \phantom{()} +2 & 2 \ 84 \\ \hline \phantom{()} & 0 \end{array}$

Final Answer: The decimal number is 7.2.

To Find: The number $x$ such that $x^2 = 84.64$.

Solution:

We need to find the square root of $84.64$.

Using the Long Division Method:

$\begin{array}{c|cc} & 9 \ . \ 2 & \\ \hline \phantom{()} 9 & \overline{84} \ . \ \overline{64} \\ + \; 9 & 81\phantom{(....)} \\ \hline \phantom{()} 18 \ 2 & 3 \ 64 \\ \phantom{()} +2 & 3 \ 64 \\ \hline \phantom{()} & 0 \end{array}$

Final Answer: The decimal fraction is 9.2.

Given that the field is a square.

The length of the side of the square field is 150 m.

To find the area the farmer will have to plough, we need to calculate the area of the square field.

The formula for the area of a square with side length $s$ is:

Substitute the given side length, $s = 150$ m, into the formula:

Calculate the product:

The farmer will have to plough an area of 22500 $\text{m}^2$.

Given:

Total number of unit squares in the square graph paper = $256$

To Find:

The number of unit squares on each side of the graph paper.

Solution:

Since the graph paper is in the shape of a square, let the number of unit squares on each side be $s$.

The total number of unit squares is equal to the area of the square, which is given by:

To find the value of $s$, we need to find the square root of $256$. We can use the Prime Factorisation method:

$\begin{array}{c|cc} 2 & 256 \\ \hline 2 & 128 \\ \hline 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$

Writing $256$ as a product of its prime factors:

$256 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$

Grouping the factors in pairs:

$256 = (2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (2 \times 2)$

Taking one factor from each pair for the square root:

Final Answer:

The number of unit squares on each side of the square graph paper is 16.

Given:

The length of one side of a cube is 15 m.

To Find:

The volume of the cube.

Solution:

The formula for the volume of a cube with side length $s$ is given by:

Substitute the given side length, $s = 15$ m, into the formula:

Calculate the value:

The volume of the cube is 3375 $\text{m}^3$.

Given:

Length of the rectangular field ($l$) = $80 \text{ m}$

Breadth of the rectangular field ($b$) = $18 \text{ m}$

To Find:

The length of its diagonal ($d$).

Solution:

In a rectangle, the length, breadth, and diagonal form a right-angled triangle. By the Pythagoras Theorem, the square of the diagonal is equal to the sum of the squares of its length and breadth.

Substituting the given values in equation (i):

To find the diagonal $d$, we take the square root of $6724$ using the Long Division Method:

$\begin{array}{c|cc} & 8\ . \ 2 & \\ \hline \phantom{()} 8 & \overline{67} \ \overline{24} \\ + \; 8 & 64\phantom{(....)} \\ \hline \phantom{()} 16 \ 2 & 3 \ 24 \\ \phantom{()} +2 & 3 \ 24 \\ \hline \phantom{()} & 0 \end{array}$

Final Answer:

The length of the diagonal of the rectangular field is 82 m.

Given:

The perimeter of the square field is 96 m.

To Find:

The area of the square field.

Solution:

Let the length of each side of the square field be $s$ meters.

The formula for the perimeter of a square is:

We are given that the perimeter is 96 m.

So, we can set up the equation:

To find the side length $s$, divide both sides by 4:

Now that we have the side length, we can find the area of the square field.

The formula for the area of a square is:

Substitute the value of $s = 24$ m into the formula:

Calculate the value:

The area of the square field is 576 $\text{m}^2$.

Given:

The volume of the cube is 512 cm$^3$.

To Find:

The length of each side of the cube.

Solution:

Let the length of each side of the cube be $s$ cm.

The formula for the volume of a cube with side length $s$ is given by:

We are given that the volume $V = 512$ cm$^3$.

Substitute the given volume into the formula:

To find the side length $s$, we need to find the cube root of 512.

We need to find a number which, when multiplied by itself three times, equals 512.

We know that $8 \times 8 \times 8 = 64 \times 8 = 512$.

Therefore, the cube root of 512 is 8.

The length of each side of the cube is 8 cm.

Given:

The ratio of three numbers is 1 : 2 : 3.

The sum of their cubes is 4500.

To Find:

The three numbers.

Solution:

Let the three numbers be $x$, $2x$, and $3x$, where $x$ is a common factor.

According to the problem, the sum of the cubes of these numbers is 4500.

Cube each term:

Combine the terms with $x^3$:

Solve for $x^3$ by dividing both sides by 36:

Simplify the fraction:

Take the cube root of both sides to find $x$:

Now, find the three numbers using the value of $x=5$:

• First number = $x = 5$
• Second number = $2x = 2 \times 5 = 10$
• Third number = $3x = 3 \times 5 = 15$

The three numbers are 5, 10, and 15.

Given:

The room is square-shaped.

The length of the side of the square room is 6.5 m.

To Find:

The area of carpet required to cover the room.

Solution:

The area of carpet required is equal to the area of the square room.

Let the side length of the square room be $s$.

The formula for the area of a square is:

Substitute the given side length, $s = 6.5$ m, into the formula:

Calculate the product:

The number of square metres of carpet required for the room is 42.25 $\text{m}^2$.

Given:

The area of a square is equal to the area of a rectangle.

Dimensions of the rectangle: length ($l$) = 6.4 m, width ($w$) = 2.5 m.

To Find:

The length of the side of the square.

Solution:

First, find the area of the rectangle.

The formula for the area of a rectangle is:

Substitute the given dimensions:

Calculate the product:

Let the side of the square be $s$ meters.

The area of the square is given by the formula:

According to the problem, the area of the square is equal to the area of the rectangle:

To find the side length $s$, take the square root of both sides:

The length of the side of the square is 4 m.

Given:

Difference between two perfect cubes = $189$

Cube root of the smaller number = $3$

To Find:

The cube root of the larger number.

Solution:

Let the larger perfect cube be $x^3$ and the smaller perfect cube be $y^3$.

According to the given condition, we have:

We are given that the cube root of the smaller number is $3$:

Now, find the value of the smaller perfect cube $y^3$:

Substituting the value of $y^3$ from equation (ii) into equation (i):

To find the cube root of the larger number ($x$), we perform the prime factorisation of $216$:

$\begin{array}{c|cc} 2 & 216 \\ \hline 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

The prime factors of $216$ are:

$216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3$

Grouping the factors into triplets:

$216 = (2 \times 2 \times 2) \times (3 \times 3 \times 3)$

Taking the cube root on both sides of equation (iii):

Final Answer:

The cube root of the larger number is 6.

Given:

Total number of plants = $1024$

Condition: Number of plants in each row = Number of rows

To Find:

The number of plants in each row.

Solution:

Let the number of rows be $x$.

Since the number of plants in each row is the same as the number of rows, the number of plants in each row is also $x$.

We find the square root of $1024$ using Long Division Method:

$\begin{array}{c|cc} & 3 \ 2 & \\ \hline \phantom{()} 3 & \overline{10} \ \overline{24} \\ + \; 3 & 9\phantom{(....)} \\ \hline \phantom{()} 6 \ 2 & 1 \ 24 \\ \phantom{()} +2 & 1 \ 24 \\ \hline \phantom{()} & 0 \end{array}$

Final Answer: The number of plants in each row is 32.

Given:

Total capacity (seats) = $2704$

Condition: Number of rows = Number of seats in each row

To Find:

The number of seats in each row.

Solution:

Let the number of seats in each row be $s$.

Since the number of rows is equal to the number of seats in each row, the number of rows is also $s$.

Calculating square root of $2704$ by Long Division Method:

$\begin{array}{c|cc} & 5 \ 2 & \\ \hline \phantom{()} 5 & \overline{27} \ \overline{04} \\ + \; 5 & 25\phantom{(....)} \\ \hline \phantom{()} 10 \ 2 & 2 \ 04 \\ \phantom{()} +2 & 2 \ 04 \\ \hline \phantom{()} & 0 \end{array}$

Final Answer: The number of seats in each row is 52.

Given:

Total number of soldiers = $7500$

To Find:

The number of soldiers left out after forming a square.

Solution:

To find the number of soldiers left out, we need to find the remainder when $7500$ is processed for a square root using the Long Division Method. The remainder will represent the "left out" soldiers.

$\begin{array}{c|cc} & 8 \ 6 & \\ \hline \phantom{()} 8 & \overline{75} \ \overline{00} \\ + \; 8 & 64\phantom{(....)} \\ \hline \phantom{()} 16 \ 6 & 11 \ 00 \\ \phantom{()} +6 & 9 \ 96 \\ \hline \phantom{()} & 104 \end{array}$

From the division, we see that $86^2$ is the largest perfect square less than $7500$.

Final Answer: The number of soldiers left out was 104.

Given:

Total number of students = $8649$

Condition: Number of students in each row = Number of rows

To Find:

The number of students in each row.

Solution:

Let the number of rows be $n$. Then the number of students in each row is also $n$.

Applying the Long Division Method to find the square root:

$\begin{array}{c|cc} & 9 \ 3 & \\ \hline \phantom{()} 9 & \overline{86} \ \overline{49} \\ + \; 9 & 81\phantom{(....)} \\ \hline \phantom{()} 18 \ 3 & 5 \ 49 \\ \phantom{()} +3 & 5 \ 49 \\ \hline \phantom{()} & 0 \end{array}$

Final Answer: There were 93 students in each row.

Given:

Distance walked towards North = $12 \text{ m}$

Distance walked towards West = $35 \text{ m}$

To Find:

The diagonal distance from the friend's house to Rahul's house.

Construction/Diagram:

Solution:

Let Rahul's house be at point $A$. He walks $12 \text{ m}$ North to reach point $B$. Then he turns West and walks $35 \text{ m}$ to reach his friend's house at point $C$.

The path forms a right-angled triangle $ABC$, where $\angle B = 90^\circ$.

According to the Pythagoras Theorem:

Substituting the given values in equation (i):

We find the square root of $1369$ using the Long Division Method:

$\begin{array}{c|cc} & 3 \ 7 & \\ \hline \phantom{()} 3 & \overline{13} \ \overline{69} \\ + \; 3 & 9\phantom{(....)} \\ \hline \phantom{()} 6 \ 7 & 4 \ 69 \\ \phantom{()} +7 & 4 \ 69 \\ \hline \phantom{()} & 0 \end{array}$

Final Answer:

Rahul walked a distance of 37 m while returning.

Given:

Length of the ladder (Hypotenuse, $c$) = $5.5 \text{ m}$

Height reached on the wall (Perpendicular, $a$) = $4.4 \text{ m}$

To Find:

Distance between the wall and the foot of the ladder (Base, $b$).

Construction/Diagram:

Solution:

The ladder, wall, and the ground form a right-angled triangle. According to the Pythagoras Theorem:

Substituting the given values:

We find the square root of $10.89$ using the Long Division Method:

$\begin{array}{c|cc} & 3\ . \ 3 & \\ \hline \phantom{()} 3 & \overline{10} \; . \overline{89} \\ + \; 3 & 9\phantom{(....)} \\ \hline \phantom{()} 6 \; 3 & 1 \; 89 \\ \phantom{()} +3 & 1 \; 89 \\ \hline \phantom{()} & 0 \end{array}$

Final Answer: The distance between the wall and the foot of the ladder is 3.3 m.

Given:

Pattern of coins: $1, 3, 5, 7, \dots$

Duration: $30$ days

To Find:

Total number of coins received in 30 days.

Solution:

The number of coins given each day forms a sequence of successive odd natural numbers starting from 1.

According to the property of square numbers, the sum of the first $n$ odd natural numbers is equal to $n^2$.

Here, $n = 30$.

Final Answer: The advisor will get 900 gold coins in that month.

Given:

Ratio of three numbers = $2 : 3 : 5$

Sum of their squares = $608$

To Find:

The three numbers.

Solution:

Let the common ratio be $x$.

Therefore, the three numbers are $2x, 3x,$ and $5x$.

According to the question:

Now, we find the three numbers:

First number = $2x = 2 \times 4 = 8$

Second number = $3x = 3 \times 4 = 12$

Third number = $5x = 5 \times 4 = 20$

Final Answer: The three numbers are 8, 12, and 20.

To Find: The smallest perfect square divisible by $8, 9,$ and $10$.

Solution:

First, we find the LCM of $8, 9,$ and $10$:

$\begin{array}{c|ccc} 2 & 8 \;, & 9 \;, & 10 \\ \hline 2 & 4 \; , & 9 \; , & 5 \\ \hline 2 & 2 \; , & 9 \; , & 5 \\ \hline 3 & 1 \; , & 9 \; , & 5 \\ \hline 3 & 1 \; , & 3 \; , & 5 \\ \hline 5 & 1 \; , & 1 \; , & 5 \\ \hline & 1 \; , & 1 \; , & 1 \end{array}$

Now, we look at the prime factorisation of $360$:

For $360$ to be a perfect square, all factors must be in pairs. Here, the factors $2$ and $5$ are not in pairs. So, we must multiply $360$ by $2 \times 5 = 10$.

Final Answer: The smallest square number is 3600.

Given:

Area of square plot = $101\frac{1}{400} \text{ m}^2$

To Find:

Length of one side of the plot ($s$).

Solution:

First, convert the mixed fraction into an improper fraction:

We know that $Area = s^2$. Therefore:

Finding $\sqrt{40401}$ using Long Division Method:

$\begin{array}{c|cc} & 2 \ 0 \ 1 & \\ \hline \phantom{()} 2 & \overline{4} \ \overline{04} \ \overline{01} \\ + \; 2 & 4\phantom{(....)} \\ \hline \phantom{()} 40 \ 0 & 0 \ 04 \phantom{(...)} \\ \phantom{()} +0 & 0 \ 00 \phantom{(...)} \\ \hline \phantom{()} 401 \ 1 & 4 \ 01 \\ \phantom{()} +1 & 4 \ 01 \\ \hline \phantom{()} & 0 \end{array}$

And $\sqrt{400} = 20$.

Final Answer: The length of one side of the plot is 10.05 m.

To find the square root of 324 by the method of repeated subtraction, we repeatedly subtract consecutive odd numbers starting from 1 from 324 until the result is 0. The number of steps required to reach 0 is the square root.

Let's perform the repeated subtraction:

Step 1: $324 - 1 = 323$

Step 2: $323 - 3 = 320$

Step 3: $320 - 5 = 315$

Step 4: $315 - 7 = 308$

Step 5: $308 - 9 = 299$

Step 6: $299 - 11 = 288$

Step 7: $288 - 13 = 275$

Step 8: $275 - 15 = 260$

Step 9: $260 - 17 = 243$

Step 10: $243 - 19 = 224$

Step 11: $224 - 21 = 203$

Step 12: $203 - 23 = 180$

Step 13: $180 - 25 = 155$

Step 14: $155 - 27 = 128$

Step 15: $128 - 29 = 99$

Step 16: $99 - 31 = 68$

Step 17: $68 - 33 = 35$

Step 18: $35 - 35 = 0$

We reached 0 in 18 steps.

Therefore, the square root of 324 is the number of steps taken.

The square root of 324 is 18.

Given:

Ratio of three numbers = $2 : 3 : 4$

Sum of their cubes = $0.334125$

To Find:

The three numbers.

Solution:

Let the common ratio be $x$.

The three numbers are $2x, 3x,$ and $4x$.

According to the question, the sum of their cubes is $0.334125$:

To find $x$, we take the cube root of $0.003375$.

Finding the prime factorisation of $3375$:

$\begin{array}{c|cc} 3 & 3375 \\ \hline 3 & 1125 \\ \hline 3 & 375 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

Thus, $\sqrt[3]{3375} = 3 \times 5 = 15$.

Now, we find the three numbers:

First number = $2x = 2 \times 0.15 = 0.30$

Second number = $3x = 3 \times 0.15 = 0.45$

Third number = $4x = 4 \times 0.15 = 0.60$

Final Answer: The numbers are 0.3, 0.45, and 0.6.

To Evaluate: $\sqrt[3]{27} + \sqrt[3]{0.008} + \sqrt[3]{0.064}$

Solution:

Step 1: Find the cube root of $27$.

Step 2: Find the cube root of $0.008$.

Step 3: Find the cube root of $0.064$.

Step 4: Add the results obtained:

Final Answer: The value of the expression is 3.6.

To Evaluate: $\{ ( 5^2 + (12^2)^{\frac{1}{2}} ) \}^{3}$

Solution:

First, simplify the terms inside the parentheses:

Now, substitute these values back into the expression:

Calculating the cube of $37$:

Final Answer: The evaluated result is 50653.

To Evaluate: $\{ ( 6^2 + (8^2)^{\frac{1}{2}} ) \}^{3}$

Solution:

First, simplify the terms inside the parentheses:

Now, substitute these values back into the expression:

Calculating the cube of $44$:

Final Answer: The evaluated result is 85184.

To Find: A 4-digit perfect square $n$ such that its digits follow the pattern: Even, Even, Odd, Even and contains no zeros.

Solution:

Let the 4-digit perfect square be $x^2$. Since it is a 4-digit number, the range of $x$ is from $32$ ($\because 32^2 = 1024$) to $99$ ($\because 99^2 = 9801$).

We are given that the digits are Even, Even, Odd, Even. Let's analyze the properties of squares:

1. If a square ends in an even digit, that digit must be $4$ or $6$ (since $0$ is excluded).

2. Property of squares: If a square ends in $6$, its tens digit must be odd.

3. Property of squares: If a square ends in $4$, its tens digit must be even.

Since our required pattern has an odd digit in the tens place, the number must end in $6$. Thus, the pattern is: $\text{Even, Even, Odd, 6}$.

Now, we check squares of numbers ending in $4$ or $6$ (since only they result in a unit digit of $6$) within the range $80$ to $99$ (as the first two digits must be even and large):

Let us check the digits of $8836$:

First digit: $8$ (Even)

Second digit: $8$ (Even)

Third digit: $3$ (Odd)

Fourth digit: $6$ (Even)

All conditions are satisfied, and there is no zero in the number.

Final Answer: The number is 8836.

To Find: Three different numbers $x, y,$ and $z$ such that $(x+y), (y+z),$ and $(z+x)$ are all perfect squares.

Solution:

Let the three numbers be $x, y,$ and $z$. We want:

Let us try sets of perfect squares. Suppose we want the sums to be $25, 49,$ and $36$.

Adding (i), (ii), and (iii):

Now, subtract each original equation from (iv):

$z = 55 - (x+y) = 55 - 25 = 30$

$x = 55 - (y+z) = 55 - 49 = 6$

$y = 55 - (z+x) = 55 - 36 = 19$

Verification: $6+19=25 (5^2)$, $19+30=49 (7^2)$, $30+6=36 (6^2)$. All three are different numbers.

Final Answer: The three numbers are 6, 19, and 30.

Given:

Perimeter of first square ($P_1$) = $40 \text{ m}$

Perimeter of second square ($P_2$) = $96 \text{ m}$

To Find:

Perimeter of a third square whose area is the sum of the areas of the first two squares.

Solution:

First, find the sides of the two squares:

Now, calculate their areas:

Let the area of the third square be $A_3$:

Find the side of the third square ($Side_3$):

Using the prime factorisation of $676 = 2 \times 2 \times 13 \times 13$, we get $Side_3 = 2 \times 13 = 26 \text{ m}$.

Finally, find the perimeter of the third square ($P_3$):

Final Answer: The perimeter of the new square is 104 m.

To Find: A 3-digit perfect square that remains a perfect square when rotated $180^\circ$ (upside down).

Solution:

According to the hint, we can only use digits $\{0, 1, 6, 8, 9\}$ because other digits like $2, 3, 4, 5, 7$ do not form valid digits when viewed upside down.

List of 3-digit perfect squares using only digits $0, 1, 6, 8, 9$:

1. $100$ ($10^2$)

2. $169$ ($13^2$)

3. $196$ ($14^2$)

4. $676$ (Invalid, contains 7)

5. $961$ ($31^2$)

Now let's view them upside down:

• $100$ becomes $001$ (which is $1^2$, but usually not considered a 3-digit number).

• $169$: Looking upside down, $9$ becomes $6$ (first digit), $6$ becomes $9$ (middle), and $1$ stays $1$. It becomes $691$. (Not a perfect square).

• $196$: Looking upside down, $6$ becomes $9$ (first digit), $9$ becomes $6$ (middle), and $1$ stays $1$. It becomes 961. $961$ is $31^2$.

• $961$: Looking upside down, $1$ stays $1$ (first digit), $6$ becomes $9$ (middle), and $9$ becomes $6$. It becomes 196. $196$ is $14^2$.

Final Answer: The number is 196 (or 961).

To Find: Two pairs of numbers $(a, b)$ such that $b$ is the mirror image of $a$, and $b^2$ is the mirror image of $a^2$.

Solution:

We are looking for numbers where the squaring process does not involve any "carry-over" to other positions, as carry-overs would disrupt the mirror symmetry.

Pair 1: Let's try $12$ and $21$.

Mirror images: $12 \leftrightarrow 21$ and $144 \leftrightarrow 441$. This is a valid pair.

Pair 2: Let's try $112$ and $211$.

Mirror images: $112 \leftrightarrow 211$ and $12544 \leftrightarrow 44521$. This is another valid pair.

Final Answer: Two other such pairs are (12, 21) and (112, 211).