Chapter 1 A Square and A Cube (Class 8 - Latest Maths NCERT (Ganita Prakash I) Solutions)

Property: A perfect square number never ends with the digits $2$, $3$, $7$, or $8$ at the unit's place. It always ends with $0$, $1$, $4$, $5$, $6$, or $9$.

Analysis:

(i) $2032$: This number ends with the digit $2$. According to the property, numbers ending in $2$ are not perfect squares.

(ii) $2048$: This number ends with the digit $8$. Therefore, it is not a perfect square.

(iii) $1027$: This number ends with the digit $7$. Therefore, it is not a perfect square.

(iv) $1089$: This number ends with $9$, so it could be a perfect square. Let us check using prime factorisation:

$$\begin{array}{c|cc} 3 & 1089 \\ \hline 3 & 363 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \end{array}$$

Since $1089 = 3 \times 3 \times 11 \times 11 = (3 \times 11)^2 = 33^2$, it is a perfect square.

Conclusion:

The numbers $2032$, $2048$, and $1027$ are not perfect squares.

To find the last digit of the square of a number, we only need to square the unit digit of that number.

Calculation:

Conclusion:

From the table above, the squares that have the last digit $4$ are $108^2$ and $292^2$.

Given:

$125^2 = 15625$

To Find:

The value of $126^2$ in the form of the given options.

Solution:

We can write $126$ as the sum of $125$ and $1$.

$126^2 = (125 + 1)^2$

Now, we apply the algebraic identity $(a + b)^2 = a^2 + 2ab + b^2$ to solve this.

By substituting the given value of $125^2$:

$126^2 = 15625 + 250 + 1$

$126^2 = 15625 + 251$

Conclusion:

Comparing this with the given options, the correct value is (iv) $15625 + 251$.

Given:

Area of the square = $441 \text{ m}^2$

To Find:

Length of the side of the square.

Formula:

Therefore, $\text{side} = \sqrt{\text{Area}}$

Solution:

We need to find $\sqrt{441}$ using the long division method:

$$\begin{array}{c|cc} & 2 \ 1 & \\ \hline \phantom{()} 2 & \overline{4} \ \overline{41} \\ + \ 2 & 4 \phantom{(..)} \\ \hline \phantom{()} 4 \ 1 & \phantom{.} 0 \ 41 \\ \phantom{()} + 1 & 41 \\ \hline & 0 \end{array}$$

Thus, $\sqrt{441} = 21$.

Conclusion:

The length of the side of the square is $21 \text{ m}$.

Alternate Solution:

Using Prime Factorisation:

$$\begin{array}{c|cc} 3 & 441 \\ \hline 3 & 147 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$$

$441 = 3 \times 3 \times 7 \times 7$

$\sqrt{441} = \sqrt{(3 \times 7)^2} = 3 \times 7 = 21$

Given:

Numbers: $4, 9,$ and $10$.

To Find:

Smallest square number divisible by these numbers.

Solution:

First, we find the Least Common Multiple (LCM) of $4, 9,$ and $10$.

$$\begin{array}{c|cc} 2 & 4 \; , & 9 \; , & 10 \\ \hline 2 & 2 \; , & 9 \; , & 5 \\ \hline 3 & 1 \; , & 9 \; , & 5 \\ \hline 3 & 1 \; , & 3 \; , & 5 \\ \hline 5 & 1 \; , & 1 \; , & 5 \\ \hline & 1 \; , & 1 \; , & 1 \end{array}$$

LCM of $4, 9,$ and $10 = 2 \times 2 \times 3 \times 3 \times 5 = 180$.

Now, we check the prime factors of $180$ to see if they are in pairs:

$180 = \underline{2 \times 2} \times \underline{3 \times 3} \times 5$

Here, the prime factor $5$ does not have a pair. Therefore, $180$ is not a perfect square.

To make $180$ a perfect square, we must multiply it by $5$.

The smallest square number $= 180 \times 5$

$= 900$

Conclusion:

The smallest square number divisible by $4, 9,$ and $10$ is $900$.

Given:

Number $= 9408$

To Find:

1. Smallest multiplier to make it a perfect square.

2. Square root of the resulting product.

Solution:

We start by finding the prime factorisation of $9408$.

$$\begin{array}{c|cc} 2 & 9408 \\ \hline 2 & 4704 \\ \hline 2 & 2352 \\ \hline 2 & 1176 \\ \hline 2 & 588 \\ \hline 2 & 294 \\ \hline 3 & 147 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$$

Prime factorisation of $9408 = \underline{2 \times 2} \times \underline{2 \times 2} \times \underline{2 \times 2} \times 3 \times \underline{7 \times 7}$.

The prime factor $3$ is left unpaired.

Thus, $9408$ must be multiplied by $3$ to become a perfect square.

New product $= 9408 \times 3 = 28224$.

Now, we find the square root of $28224$.

$\sqrt{28224} = 2 \times 2 \times 2 \times 7 \times \sqrt{3 \times 3}$

$\sqrt{28224} = 2 \times 2 \times 2 \times 7 \times 3$

$\sqrt{28224} = 168$

Conclusion:

The smallest number to be multiplied is $3$ and the square root of the product is $168$.

Property: Between the squares of two consecutive numbers $n$ and $(n+1)$, there are $2n$ non-perfect square numbers.

(i) $16$ and $17$:

Here, $n = 16$.

Number of non-perfect square numbers between $16^2$ and $17^2$ $= 2n$

$= 2 \times 16 = 32$

(ii) $99$ and $100$:

Here, $n = 99$.

Number of non-perfect square numbers between $99^2$ and $100^2$ $= 2n$

$= 2 \times 99 = 198$

Conclusion:

(i) There are $32$ numbers between $16^2$ and $17^2$.

(ii) There are $198$ numbers between $99^2$ and $100^2$.

Pattern Analysis:

Let the numbers be $a^2 + b^2 + c^2 = d^2$.

Observing the given rows:

1. The third number '$c$' is the product of the first two numbers: $c = a \times b$.

2. The fourth number '$d$' is one more than the third number: $d = c + 1$.

Applying the pattern:

For the fourth row:

$a = 4, b = 5$

$c = 4 \times 5 = 20$

$d = 20 + 1 = 21$

So, $4^2 + 5^2 + 20^2 = \mathbf{21^2}$.

For the fifth row:

$a = 9, b = 10$

$c = 9 \times 10 = 90$

$d = 90 + 1 = 91$

So, $9^2 + 10^2 + \mathbf{90^2} = \mathbf{91^2}$.

Final Filled Pattern:

$1^2 + 2^2 + 2^2 = 3^2$

$2^2 + 3^2 + 6^2 = 7^2$

$3^2 + 4^2 + 12^2 = 13^2$

$4^2 + 5^2 + 20^2 = (\mathbf{21})^2$

$9^2 + 10^2 + (\mathbf{90})^2 = (\mathbf{91})^2$

Given:

1. A large image containing a grid of pattern blocks.

2. Number of pattern blocks in each row = $9$.

3. Number of pattern blocks in each column = $9$.

4. Each individual pattern block is composed of a $5 \times 5$ grid of tiny squares.

To Find:

1. Total number of tiny squares in the picture.

2. The prime factorisation of the total number of tiny squares.

Solution:

First, we calculate the total number of pattern blocks present in the image:

Next, we determine the number of tiny squares within each single pattern block. Each block is a square arrangement of $5$ tiny squares by $5$ tiny squares:

Now, to find the total number of tiny squares in the entire picture, we multiply the total number of blocks by the number of squares per block:

$\text{Total Tiny Squares} = 81 \times 25$

Performing the multiplication:

Prime Factorisation:

We will now find the prime factorisation of $2025$ using the division method:

$$\begin{array}{c|cc} 3 & 2025 \\ \hline 3 & 675 \\ \hline 3 & 225 \\ \hline 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$$

The prime factorisation of $2025$ is:

$2025 = 3 \times 3 \times 3 \times 3 \times 5 \times 5$

In index form, it can be written as $3^4 \times 5^2$.

Conclusion:

There are $2025$ tiny squares in the picture, and its prime factorisation is $3 \times 3 \times 3 \times 3 \times 5 \times 5$.

To Find:

The cube root ($\sqrt[3]{\quad}$) of the given numbers $27000$ and $10648$.

Solution for (i) $27000$:

We use the prime factorisation method:

$$\begin{array}{c|cc} 2 & 27000 \\ \hline 2 & 13500 \\ \hline 2 & 6750 \\ \hline 3 & 3375 \\ \hline 3 & 1125 \\ \hline 3 & 375 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$$

$27000 = \underline{2 \times 2 \times 2} \times \underline{3 \times 3 \times 3} \times \underline{5 \times 5 \times 5}$

$\sqrt[3]{27000} = 2 \times 3 \times 5$

$\sqrt[3]{27000} = 30$

Solution for (ii) $10648$:

We use the prime factorisation method:

$$\begin{array}{c|cc} 2 & 10648 \\ \hline 2 & 5324 \\ \hline 2 & 2662 \\ \hline 11 & 1331 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \end{array}$$

$10648 = \underline{2 \times 2 \times 2} \times \underline{11 \times 11 \times 11}$

$\sqrt[3]{10648} = 2 \times 11$

$\sqrt[3]{10648} = 22$

Conclusion:

The cube root of $27000$ is $30$ and the cube root of $10648$ is $22$.

Given:

Number = $1323$

To Find:

The smallest number to be multiplied by $1323$ to make it a perfect cube.

Solution:

First, we find the prime factorisation of $1323$.

$$\begin{array}{c|cc} 3 & 1323 \\ \hline 3 & 441 \\ \hline 3 & 147 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$$

The prime factorisation of $1323$ is:

$1323 = \underline{3 \times 3 \times 3} \times 7 \times 7$

In a perfect cube, every prime factor must appear in triples (groups of three). In the factorisation above, the factor $3$ is in a triple, but the factor $7$ appears only twice.

To make it a triple, we need one more $7$.

Conclusion:

The smallest number by which $1323$ must be multiplied is $7$.

The resulting perfect cube would be $1323 \times 7 = 9261$, which is $21^3$.

(i) The cube of any odd number is even.

False. The cube of an odd number is always odd. For example, $3$ is an odd number and its cube $3^3 = 27$ is also an odd number.

(ii) There is no perfect cube that ends with $8$.

False. A perfect cube can end with any digit. For example, the cube of $2$ is $2^3 = 8$, which ends with the digit $8$. Also, the cube of $12$ is $1728$, which ends with $8$.

(iii) The cube of a $2$-digit number may be a $3$-digit number.

False. The smallest $2$-digit number is $10$. Its cube is $10^3 = 1000$. Since $1000$ is a $4$-digit number, the cube of any $2$-digit number must have at least $4$ digits.

(iv) The cube of a $2$-digit number may have seven or more digits.

False. The largest $2$-digit number is $99$. Its cube is $99^3 = 970299$. This is a $6$-digit number. Since the smallest $3$-digit number is $100$ and its cube is $100^3 = 1,000,000$ (a $7$-digit number), no $2$-digit number's cube can have $7$ or more digits.

(v) Cube numbers have an odd number of factors.

False. Only perfect square numbers have an odd number of factors. A perfect cube that is not a perfect square will have an even number of factors. For example, $8$ is a perfect cube and its factors are $1, 2, 4,$ and $8$. The number of factors is $4$, which is even.

Concept: To find the cube root of a perfect cube by estimation, we group the digits into threes starting from the unit's place and use the property of cubes of numbers from $1$ to $10$.

Solution for $1331$:

Step 1: Group the digits from right to left: $\overline{1}, \overline{331}$.

Step 2: Take the first group $331$. It ends in $1$. We know that the cube of a number ending in $1$ also ends in $1$ ($1^3 = 1$). So, the unit digit of the cube root is $1$.

Step 3: Take the second group $1$. Since $1^3 = 1$, the tens digit of the cube root is $1$.

Thus, $\sqrt[3]{1331} = 11$.

Solution for $4913$:

Step 1: Grouping: $\overline{4}, \overline{913}$.

Step 2: Group $913$ ends in $3$. We know that a number ending in $7$ has its cube ending in $3$ ($7^3 = 343$). So, the unit digit is $7$.

Step 3: Group $4$ lies between $1^3 = 1$ and $2^3 = 8$. We take the smaller number. So, the tens digit is $1$.

Thus, $\sqrt[3]{4913} = 17$.

Solution for $12167$:

Step 1: Grouping: $\overline{12}, \overline{167}$.

Step 2: Group $167$ ends in $7$. A number ending in $3$ has its cube ending in $7$ ($3^3 = 27$). So, the unit digit is $3$.

Step 3: Group $12$ lies between $2^3 = 8$ and $3^3 = 27$. We take the smaller number. So, the tens digit is $2$.

Thus, $\sqrt[3]{12167} = 23$.

Solution for $32768$:

Step 1: Grouping: $\overline{32}, \overline{768}$.

Step 2: Group $768$ ends in $8$. A number ending in $2$ has its cube ending in $8$ ($2^3 = 8$). So, the unit digit is $2$.

Step 3: Group $32$ lies between $3^3 = 27$ and $4^3 = 64$. We take the smaller number. So, the tens digit is $3$.

Thus, $\sqrt[3]{32768} = 32$.

Conclusion:

The estimated cube roots are $11$, $17$, $23$, and $32$ respectively.

Analysis of the expressions:

We will calculate the value of each expression using the identities for the difference of consecutive squares and cubes.

(iv) Solution for $43^2 - 42^2$:

Using the identity $a^2 - b^2 = (a - b)(a + b)$:

$43^2 - 42^2 = (43 - 42)(43 + 42)$

$= 1 \times (43 + 42)$

Value of (iv) = $85$.

(iii) Solution for $67^2 - 66^2$:

Using the identity $a^2 - b^2 = (a - b)(a + b)$:

$67^2 - 66^2 = (67 - 66)(67 + 66)$

$= 1 \times (67 + 66)$

Value of (iii) = $133$.

(ii) Solution for $43^3 - 42^3$:

Using the property for the difference of consecutive cubes: $n^3 - (n - 1)^3 = 1 + 3n(n - 1)$.

$43^3 - 42^3 = 1 + 3 \times 43 \times 42 = 5419$

Value of (ii) = $5419$.

(i) Solution for $67^3 - 66^3$:

Using the property for the difference of consecutive cubes: $n^3 - (n - 1)^3 = 1 + 3n(n - 1)$.

$67^3 - 66^3 = 1 + 3 \times 67 \times 66 = 13267$

Value of (i) = $13267$.

Conclusion:

Comparing all the calculated values:

(i) $13267$

(ii) $5419$

(iii) $133$

(iv) $85$

Since $13267 > 5419 > 133 > 85$, the expression (i) $67^3 - 66^3$ is the greatest.