Chapter 6 We Distribute, Yet Things Multiply (Class 8 - Latest Maths NCERT (Ganita Prakash I) Solutions)

Given:

A multiplication grid where each entry at the intersection of a row and a column is the product of the row number and the column number. A $3 \times 3$ frame is chosen such that the middle element is $pq$, where $p$ represents the row index and $q$ represents the column index.

To Find:

The algebraic expressions for the remaining $8$ cells in the $3 \times 3$ frame.

Solution:

In a multiplication table, the rows are consecutive integers and the columns are also consecutive integers. If the middle cell corresponds to row $p$ and column $q$, then:

1. The row above the middle row is $(p - 1)$ and the row below it is $(p + 1)$.

2. The column to the left of the middle column is $(q - 1)$ and the column to the right is $(q + 1)$.

We can now calculate the products for all nine cells in the frame as follows:

Therefore, the complete $3 \times 3$ grid around the center $pq$ can be represented as:

To Find:

The expansion of the given algebraic products using the Distributive Property: $a(c + d) = ac + ad$.

Solution:

(i) $(3 + u) (v - 3)$

$3(v - 3) + u(v - 3)$

$(3 \times v) - (3 \times 3) + (u \times v) - (u \times 3)$

$3v - 9 + uv - 3u$

(ii) $\frac{2}{3} (15 + 6a)$

$\frac{2}{3} \times 15 + \frac{2}{3} \times 6a$

$\frac{2 \times \cancel{15}^{5}}{\cancel{3}_{1}} + \frac{2 \times \cancel{6}^{2}a}{\cancel{3}_{1}}$

$(2 \times 5) + (2 \times 2a)$

$10 + 4a$

(iii) $(10a + b) (10c + d)$

$10a(10c + d) + b(10c + d)$

$(10a \times 10c) + (10a \times d) + (b \times 10c) + (b \times d)$

$100ac + 10ad + 10bc + bd$

(iv) $(3 - x) (x - 6)$

$3(x - 6) - x(x - 6)$

$(3x - 18) - (x^2 - 6x)$

$3x - 18 - x^2 + 6x$

$-x^2 + 9x - 18$

(v) $(-5a + b) (c + d)$

$-5a(c + d) + b(c + d)$

$-5ac - 5ad + bc + bd$

(vi) $(5 + z) (y + 9)$

$5(y + 9) + z(y + 9)$

$(5 \times y) + (5 \times 9) + (z \times y) + (z \times 9)$

$5y + 45 + zy + 9z$

Given:

Let the two numbers be $x$ and $y$. The condition is that the product remains unchanged when $x$ is increased by $2$ and $y$ is decreased by $4$.

Solution:

We can write the condition as an algebraic equation:

$xy = (x + 2)(y - 4)$

Expanding the Right Hand Side (RHS):

$xy = x(y - 4) + 2(y - 4)$

$xy = xy - 4x + 2y - 8$

Subtracting $xy$ from both sides:

$0 = -4x + 2y - 8$

Rearranging the equation to solve for $y$:

$2y = 4x + 8$

Dividing by $2$:

$y = 2x + 4$

Now, we can find three examples by choosing any value for $x$ and calculating $y$.

Example 1: Let $x = 1$

$y = 2(1) + 4 = 6$

Checking: Original product $= 1 \times 6 = 6$. New product $= (1+2)(6-4) = 3 \times 2 = 6$. Matches!

Example 2: Let $x = 3$

$y = 2(3) + 4 = 10$

Checking: Original product $= 3 \times 10 = 30$. New product $= (3+2)(10-4) = 5 \times 6 = 30$. Matches!

Example 3: Let $x = 5$

$y = 2(5) + 4 = 14$

Checking: Original product $= 5 \times 14 = 70$. New product $= (5+2)(14-4) = 7 \times 10 = 70$. Matches!

Final Answer: Three pairs of numbers are (1, 6), (3, 10), and (5, 14).

(i) $(a + ab - 3b^2) (4 + b)$

We use the distributive property to multiply each term in the first bracket by each term in the second:

$4(a + ab - 3b^2) + b(a + ab - 3b^2)$

$(4a + 4ab - 12b^2) + (ab + ab^2 - 3b^3)$

Combining the like terms ($4ab$ and $ab$):

$4a + 5ab - 12b^2 + ab^2 - 3b^3$

(ii) $(4y + 7) (y + 11z - 3)$

Expanding the terms:

$4y(y + 11z - 3) + 7(y + 11z - 3)$

$(4y^2 + 44yz - 12y) + (7y + 77z - 21)$

Combining the like terms ($-12y$ and $7y$):

$4y^2 + 44yz - 5y + 77z - 21$

(i) $(a - b) (a + b)$

$a(a + b) - b(a + b)$

$a^2 + ab - ba - b^2$

$a^2 - b^2$

(ii) $(a - b) (a^2 + ab + b^2)$

$a(a^2 + ab + b^2) - b(a^2 + ab + b^2)$

$(a^3 + a^2b + ab^2) - (ba^2 + ab^2 + b^3)$

All middle terms cancel out:

$a^3 - b^3$

(iii) $(a - b)(a^3 + a^2b + ab^2 + b^3)$

$a(a^3 + a^2b + ab^2 + b^3) - b(a^3 + a^2b + ab^2 + b^3)$

$(a^4 + a^3b + a^2b^2 + ab^3) - (ba^3 + a^2b^2 + ab^3 + b^4)$

All middle terms cancel out:

$a^4 - b^4$

Pattern and Generalisation:

We observe that in each case, the expansion results in the difference of $n^{th}$ powers, where $n$ is the total number of terms in the second bracket plus one. The general pattern is:

$(a - b)(a^{n-1} + a^{n-2}b + ... + b^{n-1}) = a^n - b^n$

Next identity in the pattern:

$(a - b)(a^4 + a^3b + a^2b^2 + ab^3 + b^4) = a^5 - b^5$

Verification:

$a(a^4 + a^3b + a^2b^2 + ab^3 + b^4) - b(a^4 + a^3b + a^2b^2 + ab^3 + b^4)$

$(a^5 + a^4b + a^3b^2 + a^2b^3 + ab^4) - (ba^4 + a^3b^2 + a^2b^3 + ab^4 + b^5)$

After cancelling common terms, we are left with:

$a^5 - b^5$

The identity is verified.

Given:

Two algebraic expressions: $(a - b)^2$ and $(b - a)^2$.

To Find:

Which of the two expressions is greater and provide a justification.

Solution:

Let us expand both expressions using the algebraic identity $(x - y)^2 = x^2 - 2xy + y^2$.

Expanding the first expression:

$(a - b)^2 = a^2 - 2ab + b^2$

Expanding the second expression:

$(b - a)^2 = b^2 - 2ba + a^2$

Since multiplication is commutative, we know that $ba = ab$. Thus, the second expression can be rewritten as:

$(b - a)^2 = a^2 - 2ab + b^2$

Comparing the two results, we see that they are exactly the same.

Final Answer: Neither is greater; both expressions are equal.

Given:

The number to be expressed as a difference of two squares is $100$.

To Find:

Two perfect squares $a^2$ and $b^2$ such that their difference is $100$.

Solution:

We use the algebraic identity for the difference of two squares:

We need to find two integers $a$ and $b$. Let $a + b = x$ and $a - b = y$. Then $x \times y = 100$. For $a$ and $b$ to be integers, $x$ and $y$ must both be even or both be odd. Since their product $100$ is even, both factors must be even.

Let us choose the even factors $50$ and $2$ of $100$ ($50 \times 2 = 100$).

By adding equations (ii) and (iii), we get:

Now, we substitute the value of $a$ in equation (ii):

To verify, we calculate the squares of these numbers:

The difference is:

Thus, $100$ can be expressed as $26^2 - 24^2$.

To Find:

The squares of the given numbers using algebraic identities.

Solution:

We will use the identity $(a + b)^2 = a^2 + 2ab + b^2$ or $(a - b)^2 = a^2 - 2ab + b^2$.

1. Solving $406^2$

$406^2 = (400 + 6)^2$

$406^2 = 400^2 + 2(400)(6) + 6^2$

$406^2 = 160000 + 4800 + 36$

$406^2 = 164836$

2. Solving $72^2$

$72^2 = (70 + 2)^2$

$72^2 = 70^2 + 2(70)(2) + 2^2$

$72^2 = 4900 + 280 + 4$

$72^2 = 5184$

3. Solving $145^2$

$145^2 = (140 + 5)^2$

$145^2 = 140^2 + 2(140)(5) + 5^2$

$145^2 = 19600 + 1400 + 25$

$145^2 = 21025$

4. Solving $1097^2$

$1097^2 = (1100 - 3)^2$

$1097^2 = 1100^2 - 2(1100)(3) + 3^2$

$1097^2 = 1210000 - 6600 + 9$

$1097^2 = 1203409$

5. Solving $124^2$

$124^2 = (120 + 4)^2$

$124^2 = 120^2 + 2(120)(4) + 4^2$

$124^2 = 14400 + 960 + 16$

$124^2 = 15376$

To Examine:

Whether Pattern 1 $[2(a^2 + b^2) = (a + b)^2 + (a - b)^2]$ and Pattern 2 $[a^2 - b^2 = (a + b)(a - b)]$ hold for counting numbers, negative integers, and fractions.

Justification for Pattern 1: $2(a^2 + b^2) = (a + b)^2 + (a - b)^2$

Case I: Counting Numbers

Let $a = 4$ and $b = 2$.

$\therefore$ Pattern 1 holds for counting numbers.

Case II: Negative Integers

Let $a = -4$ and $b = -2$.

$\therefore$ Pattern 1 holds for negative integers.

Case III: Fractions

Let $a = \frac{1}{2}$ and $b = \frac{1}{4}$.

$\therefore$ Pattern 1 holds for fractions also.

Justification for Pattern 2: $a^2 - b^2 = (a + b)(a - b)$

Case I: Counting Numbers

Let $a = 5$ and $b = 3$.

$\therefore \text{LHS} = \text{RHS}$. Pattern 2 holds for counting numbers.

Case II: Negative Integers

Let $a = -5$ and $b = -3$.

$\therefore \text{LHS} = \text{RHS}$. Pattern 2 holds for negative integers.

Case III: Fractions

Let $a = \frac{3}{4}$ and $b = \frac{1}{4}$.

$\therefore \text{LHS} = \text{RHS}$. Pattern 2 holds for fractions also.

Conclusion: Both patterns are universal algebraic identities. They hold Always True for all real numbers, including counting numbers, negative integers, and fractions.

(i) $46^2$ using Identity $(a + b)^2 = a^2 + 2ab + b^2$

We can write $46$ as $(40 + 6)$.

Here, $a = 40$ and $b = 6$.

$(40 + 6)^2 = 40^2 + 2(40)(6) + 6^2$

$40^2 = 1600$

$2 \times 40 \times 6 = 480$

$6^2 = 36$

Adding the values: $1600 + 480 + 36$

$46^2 = 2116$

(ii) $397 \times 403$ using Identity $(a - b)(a + b) = a^2 - b^2$

We can write $397$ as $(400 - 3)$ and $403$ as $(400 + 3)$.

Here, $a = 400$ and $b = 3$.

$(400 - 3)(400 + 3) = 400^2 - 3^2$

$400^2 = 160000$

$3^2 = 9$

Subtracting: $160000 - 9$

$397 \times 403 = 159991$

(iii) $91^2$ using Identity $(a - b)^2 = a^2 - 2ab + b^2$

We can write $91$ as $(100 - 9)$.

Here, $a = 100$ and $b = 9$.

$(100 - 9)^2 = 100^2 - 2(100)(9) + 9^2$

$100^2 = 10000$

$2 \times 100 \times 9 = 1800$

$9^2 = 81$

Calculation: $10000 - 1800 + 81$

$91^2 = 8281$

(iv) $43 \times 45$ using Identity $(a - b)(a + b) = a^2 - b^2$

We find the middle number between $43$ and $45$, which is $44$.

We can write $43$ as $(44 - 1)$ and $45$ as $(44 + 1)$.

Here, $a = 44$ and $b = 1$.

$(44 - 1)(44 + 1) = 44^2 - 1^2$

$44^2 = 1936$

$1^2 = 1$

Subtracting: $1936 - 1$

$43 \times 45 = 1935$

(i) $(p - 1)(p + 11)$

Using the distributive property:

$p(p + 11) - 1(p + 11)$

$p^2 + 11p - p - 11$

$p^2 + 10p - 11$

(ii) $(3a - 9b)(3a + 9b)$

Using the identity $(x - y)(x + y) = x^2 - y^2$:

$(3a)^2 - (9b)^2$

$9a^2 - 81b^2$

(iii) $-(2y + 5)(3y + 4)$

First, we multiply the two brackets:

$(2y + 5)(3y + 4) = 2y(3y + 4) + 5(3y + 4)$

$6y^2 + 8y + 15y + 20$

$6y^2 + 23y + 20$

Now, applying the negative sign outside:

$-6y^2 - 23y - 20$

(iv) $(6x + 5y)^2$

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

$(6x)^2 + 2(6x)(5y) + (5y)^2$

$36x^2 + 60xy + 25y^2$

(v) $(2x - \frac{1}{2})^2$

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

$(2x)^2 - 2(2x)(\frac{1}{2}) + ( \frac{1}{2} )^2$

$4x^2 - 2x + \frac{1}{4}$

(vi) $(7p) \times (3r) \times (p + 2)$

First, multiply the monomials:

$7p \times 3r = 21pr$

Now, multiply the result by the binomial $(p + 2)$:

$21pr(p + 2) = (21pr \times p) + (21pr \times 2)$

$21p^2r + 42pr$

(i) Two more than a square number:

Let the number be $s$. Its square is $s^2$. "Two more than" means we add $2$ to this square.

The appropriate expression is: $s^2 + 2$

(ii) The sum of the squares of two consecutive numbers:

Consecutive numbers differ by $1$. If the first number is $m$, the next is $(m + 1)$ and the previous is $(m - 1)$.

1. Using $m$ and $m + 1$: The sum of their squares is $m^2 + (m + 1)^2$.

2. Using $m$ and $m - 1$: The sum of their squares is $m^2 + (m - 1)^2$.

3. Also, $2m$ and $2m + 1$ are always consecutive numbers (an even followed by an odd). The sum of their squares is $(2m)^2 + (2m + 1)^2$.

The appropriate expressions are: $m^2 + (m + 1)^2$, $m^2 + (m - 1)^2$, and $(2m)^2 + (2m + 1)^2$.

Given:

A $2 \times 2$ square of numbers in a calendar. For example, the numbers $4, 5, 11,$ and $12$.

The products of the diagonals are given as:

To Find:

1. Diagonal products for other $2 \times 2$ squares in the calendar.

2. Observation regarding the relationship between these products.

3. Algebraic justification for the observation.

Solution:

Let us choose another $2 \times 2$ square from the calendar, such as the numbers $9, 10, 16,$ and $17$.

Observation:

In both cases, we see that the product of the diagonal starting from the top-right ($5 \times 11 = 55$ and $10 \times 16 = 160$) is exactly $7$ more than the product of the diagonal starting from the top-left ($4 \times 12 = 48$ and $9 \times 17 = 153$).

Algebraic Explanation:

In any calendar, a week consists of $7$ days. If we let the first number in our $2 \times 2$ square be $a$, then the number to its right is $(a + 1)$. The number directly below $a$ represents the same day of the next week, which is $(a + 7)$. The number to the right of $(a + 7)$ is $(a + 8)$.

The general $2 \times 2$ square table is as follows:

Let $P_1$ be the product of the first diagonal (Top-Left to Bottom-Right):

Let $P_2$ be the product of the second diagonal (Top-Right to Bottom-Left):

Expanding the expression for $P_2$:

Now, let us calculate the difference between the two products ($P_2 - P_1$):

Conclusion: Since the algebraic difference is a constant $7$, the product of the second diagonal will always be exactly $7$ more than the product of the first diagonal in any $2 \times 2$ square on a calendar. This happens because the vertical gap between dates in a calendar is always $7$.

(i) Verification of $(k + 1) (k + 2) - (k + 3)$ as a multiple of $2$

First, we simplify the expression:

Let us test for an even value of $k$. Let $k = 2$:

Since $7$ is not a multiple of $2$, the statement is False.

(ii) Verification of $(2q + 1) (2q - 3)$ as a multiple of $4$

Expanding the expression:

The term $4(q^2 - q)$ is always a multiple of $4$, but subtracting $3$ leaves a remainder of $1$ (or $-3$). For example, if $q = 1$, the result is $-3$, which is not a multiple of $4$. Therefore, the statement is False.

(iii) Verification of Squares of Even and Odd numbers

Part A: Squares of even numbers. Let an even number be $2n$.

Part B: Squares of odd numbers. Let an odd number be $2n + 1$.

Since $n(n+1)$ is the product of two consecutive integers, it is always even ($2m$). Substituting this:

Thus, the statement is True.

(iv) Verification of $(6n + 2)^2 - (4n + 3)^2$ as $5$ less than a square number

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

For this to be "$5$ less than a square number" ($x^2 - 5$), the term $20n^2$ must be a perfect square. However, $20$ is not a perfect square ($20 = 2^2 \times 5$). Thus, $20n^2$ is not a square for all $n$. For example, if $n=1$, the value is $15$, and $15+5=20$ (not a square). Therefore, the statement is False.

Conclusion: Only statement (iii) is true. The others are false based on algebraic expansion and testing of values.

Given:

Let the first number be $x$ and the second number be $y$.

$x = 7m + 3$

$y = 7n + 5$

To Find:

Remainders of $(x + y)$, $(y - x)$, and $(xy)$ when divided by $7$.

Solution:

1. For the Sum $(x + y)$:

$x + y = (7m + 3) + (7n + 5)$

$= 7(m + n) + 8$

$= 7(m + n) + 7 + 1$

$= 7(m + n + 1) + 1$

The remainder is $1$.

2. For the Difference $(y - x)$:

$y - x = (7n + 5) - (7m + 3)$

$= 7(n - m) + 2$

The remainder is $2$.

3. For the Product $(xy)$:

$xy = (7m + 3)(7n + 5)$

$= 49mn + 35m + 21n + 15$

$= 7(7mn + 5m + 3n) + 14 + 1$

$= 7(7mn + 5m + 3n + 2) + 1$

The remainder is $1$.

Solution:

Let us test with examples:

Example 1: $4, 5, 6$

$5^2 - (4 \times 6) = 25 - 24 = 1$

Example 2: $10, 11, 12$

$11^2 - (10 \times 12) = 121 - 120 = 1$

Pattern: The result is always $1$.

Algebraic Equation:

Let the three consecutive numbers be $(n - 1), n,$ and $(n + 1)$.

$n^2 - (n - 1)(n + 1) = 1$

Verification:

LHS: $n^2 - (n - 1)(n + 1)$

$= n^2 - (n^2 - 1)$

$= n^2 - n^2 + 1$

$= 1$

RHS: $1$

Since LHS = RHS, the identity is verified.

Step 1: Frame the Algebraic Expression

Let the two numbers be $x$ and $y$.

1. Add the two numbers: $(x + y)$

2. Multiply this by half of the sum: $(x + y) \times \frac{x + y}{2}$

The resulting expression is: $\frac{(x + y)^2}{2}$

Step 2: Proof

To Prove:

The result is half of the square of the sum of the two numbers.

Proof:

Square of the sum $= (x + y)^2$

Half of the square of the sum $= \frac{(x + y)^2}{2}$

From Step 1, our result was:

$(x + y) \times \frac{1}{2}(x + y) = \frac{1}{2} \times (x + y) \times (x + y) = \frac{(x + y)^2}{2}$

Both expressions are identical. Thus, the statement is proved.

To Find:

Which product is larger without performing full multiplication by using algebraic identities.

Solution:

We use the algebraic identity $(a - b)(a + b) = a^2 - b^2$. To compare two products where the sum of the factors is the same, the product where the numbers are closer together will always be larger.

(i) $14 \times 26$ or $16 \times 24$

Both pairs have the same sum: $14 + 26 = 40$ and $16 + 24 = 40$. Their average is $20$.

For $14 \times 26$: $(20 - 6)(20 + 6) = 20^2 - 6^2 = 400 - 36$

For $16 \times 24$: $(20 - 4)(20 + 4) = 20^2 - 4^2 = 400 - 16$

Since $16 < 36$, subtracting $16$ from $400$ leaves a larger result than subtracting $36$.

Therefore, $16 \times 24$ is larger.

(ii) $25 \times 75$ or $26 \times 74$

Both pairs have the same sum: $25 + 75 = 100$ and $26 + 74 = 100$. Their average is $50$.

For $25 \times 75$: $(50 - 25)(50 + 25) = 50^2 - 25^2$

For $26 \times 74$: $(50 - 24)(50 + 24) = 50^2 - 24^2$

Since $24^2$ is smaller than $25^2$, the second expression $(50^2 - 24^2)$ will have a larger value.

Therefore, $26 \times 74$ is larger.

Given:

Area of each square green plot = $g^2$ sq. ft.

Width of the walking path = $w$ ft.

Space between the two square plots = $2w$ ft.

To Find:

The algebraic expression for the total area that needs to be tiled (the walking path).

Solution:

First, we determine the total dimensions of the rectangular park based on the plan.

1. Total Length of the park: The length consists of the path on the left, the first square plot, the space between the plots, the second square plot, and the path on the right.

2. Total Breadth (Height) of the park: The breadth consists of the path at the top, the square plot, and the path at the bottom.

3. Total Area of the park:

Using the distributive property to expand the expression:

4. Area to be tiled: To find the tiled area, we subtract the area of the two green squares from the total area.

Final Answer:

The expression for the area that needs to be tiled is $8gw + 8w^2$ sq. ft.

(a) Yellow Pattern Analysis:

By observing the first pattern, we see that each step forms a square of unit squares where the side of the square increases with each step.

In Step $1$:

In Step $2$:

In Step $3$:

(a) Solutions:

(i) Draw the next figure: Step $4$ will consist of $6$ strips of $6$ units each (a $6 \times 6$ square).

(ii) Number of unit squares in Step $10$:

(iii) Number of unit squares in Step $y$:

(b) Blue Pattern Analysis:

By observing the blue unit squares, we see a different growth pattern.

In Step $1$:

In Step $2$:

In Step $3$:

(b) Solutions:

(i) Draw the next figure: Step $4$ will have $(4 + 1)^2 + 4 = 25 + 4 = 29$ squares.

(ii) Number of unit squares in Step $10$:

(iii) Number of unit squares in Step $y$: