We Distribute, Yet Things Multiply — NCERT Solutions

(i) $(p-1)(p+11)$
Using distributive property:
$$= p(p+11) - 1(p+11) = p^2 + 11p - p - 11 = \boxed{p^2 + 10p - 11}$$

(ii) $(3a - 9b)(3a + 9b)$
Using Identity 1C: $(x-y)(x+y) = x^2 - y^2$ with $x = 3a$, $y = 9b$:
$$= (3a)^2 - (9b)^2 = 9a^2 - 81b^2 = \boxed{9a^2 - 81b^2}$$

(iii) $-(2y+5)(3y+4)$
First expand $(2y+5)(3y+4)$:
$$= 2y(3y+4) + 5(3y+4) = 6y^2 + 8y + 15y + 20 = 6y^2 + 23y + 20$$
Now multiply by $-1$:
$$\boxed{= -6y^2 - 23y - 20}$$

(iv) $(6x + 5y)^2$
Using Identity 1A: $(a+b)^2 = a^2 + 2ab + b^2$ with $a = 6x$, $b = 5y$:
$$= (6x)^2 + 2(6x)(5y) + (5y)^2 = 36x^2 + 60xy + 25y^2$$
$$\boxed{= 36x^2 + 60xy + 25y^2}$$

(v) $\left(2x - \dfrac{1}{2}\right)^2$
Using Identity 1B: $(a-b)^2 = a^2 - 2ab + b^2$ with $a = 2x$, $b = \dfrac{1}{2}$:
$$= (2x)^2 - 2(2x)\left(\frac{1}{2}\right) + \left(\frac{1}{2}\right)^2 = 4x^2 - 2x + \frac{1}{4}$$
$$\boxed{= 4x^2 - 2x + \frac{1}{4}}$$

(vi) $(7p) \times (3r) \times (p+2)$
First multiply the constants and variables:
$$7p \times 3r = 21pr$$
Now:
$$21pr \times (p+2) = 21pr \cdot p + 21pr \cdot 2 = 21p^2r + 42pr$$
$$\boxed{= 21p^2r + 42pr}$$

(i) Two more than a square number:

'A square number' means $s^2$ (the square of some number $s$). 'Two more than' means we add 2.

So the expression is $s^2 + 2$.

Also, $2 + s$ is not a square number plus 2 (it's just $s$ plus 2). $(s+2)^2$ means the square of $(s+2)$, not a square plus 2. $s^2 + 4$ would be four more than a square. $2s^2$ is twice a square. $2^2 s = 4s$ is not a square plus 2.

$$\boxed{s^2 + 2}$$

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

Two consecutive numbers can be written as $m$ and $m+1$. Their squares are $m^2$ and $(m+1)^2$. The sum is:
$$m^2 + (m+1)^2$$

Also, $m^2 + (m-1)^2$ represents the sum of squares of $m$ and $m-1$, which are also consecutive numbers. So this is also valid.

And $(2m)^2 + (2m+1)^2$ represents consecutive numbers $2m$ and $2m+1$ (consecutive even and odd), so this is also a valid specific case.

$$\boxed{m^2 + (m+1)^2, \quad m^2 + (m-1)^2, \quad (2m)^2 + (2m+1)^2}$$

Note: $m^2 + n^2$ is the sum of squares of any two numbers, not necessarily consecutive. $(m+n)^2 \neq m^2 + n^2$ in general. $m^2 + 1$ is only valid if the consecutive number is 1, which is too restrictive.

Observation: In every 2×2 square from a calendar, the product of one diagonal is always 7 less than the product of the other diagonal.

Explanation using algebra:

Label the top-left number of any 2×2 square as $a$. Since calendar dates in the same row differ by 1, and dates in the next row differ by 7:

$$\begin{array}{|c|c|}\hline a & a+1 \\ \hline a+7 & a+8 \\ \hline\end{array}$$

Diagonal 1 (top-left × bottom-right):
$$a(a+8) = a^2 + 8a$$

Diagonal 2 (top-right × bottom-left):
$$(a+1)(a+7) = a^2 + 7a + a + 7 = a^2 + 8a + 7$$

Difference:
$$(a+1)(a+7) - a(a+8) = (a^2 + 8a + 7) - (a^2 + 8a) = 7$$

Conclusion: The product of the top-right and bottom-left numbers is always 7 more than the product of the top-left and bottom-right numbers. This is because consecutive dates differ by 1 (horizontally) and 7 (vertically), and the algebraic identity shows the difference is always exactly 7, regardless of which 2×2 square is chosen.

(i) $(k+1)(k+2) - (k+3)$ is always 2.

Expand:
$$(k+1)(k+2) = k^2 + 2k + k + 2 = k^2 + 3k + 2$$
$$k^2 + 3k + 2 - (k+3) = k^2 + 3k + 2 - k - 3 = k^2 + 2k - 1$$

For $k = 1$: $1 + 2 - 1 = 2$. For $k = 2$: $4 + 4 - 1 = 7 \neq 2$.

This statement is FALSE. $(k+1)(k+2) - (k+3) = k^2 + 2k - 1$, which is not always 2.

(ii) $(2q+1)(2q-3)$ is a multiple of 4.

Expand:
$$(2q+1)(2q-3) = 4q^2 - 6q + 2q - 3 = 4q^2 - 4q - 3$$
$$= 4(q^2 - q) - 3$$

$4(q^2 - q)$ is a multiple of 4, but subtracting 3 gives a number that is $\equiv 1 \pmod{4}$ (since $-3 \equiv 1 \pmod 4$).

For $q = 1$: $(3)(-1) = -3$, not a multiple of 4.

This statement is FALSE.

(iii) Squares of even numbers are multiples of 4; squares of odd numbers are 1 more than multiples of 8.

Even numbers: Let $n = 2m$.
$$(2m)^2 = 4m^2$$
This is always a multiple of 4. ✓

Odd numbers: Let $n = 2m+1$.
$$(2m+1)^2 = 4m^2 + 4m + 1 = 4m(m+1) + 1$$
Now, $m(m+1)$ is the product of two consecutive integers, so it is always even. Let $m(m+1) = 2t$.
$$= 4(2t) + 1 = 8t + 1$$
This is always 1 more than a multiple of 8. ✓

This statement is TRUE.

(iv) $(6n+2)^2 - (4n+3)^2$ is 5 less than a square number.

Using identity $(a^2 - b^2) = (a+b)(a-b)$ with $a = 6n+2$, $b = 4n+3$:
$$a + b = (6n+2) + (4n+3) = 10n + 5$$
$$a - b = (6n+2) - (4n+3) = 2n - 1$$
$$(6n+2)^2 - (4n+3)^2 = (10n+5)(2n-1)$$

Expand:
$$= 20n^2 - 10n + 10n - 5 = 20n^2 - 5$$

Now check if this is 5 less than a square:
$$20n^2 - 5 + 5 = 20n^2$$

Is $20n^2$ a perfect square? $20n^2 = 4 \cdot 5 \cdot n^2 = (2n)^2 \cdot 5$. This is a perfect square only when $5$ is a perfect square, which it is not (in general).

For $n = 1$: $20(1) - 5 = 15$. Is $15 + 5 = 20$ a perfect square? No.

This statement is FALSE.

Given:
Let the first number be $a = 7m + 3$ for some integer $m$.
Let the second number be $b = 7n + 5$ for some integer $n$.

Sum:
$$a + b = (7m + 3) + (7n + 5) = 7(m + n) + 8 = 7(m + n + 1) + 1$$
Remainder when sum is divided by 7 = $\boxed{1}$.

Difference:
$$a - b = (7m + 3) - (7n + 5) = 7(m - n) - 2 = 7(m - n - 1) + 5$$
Remainder when difference is divided by 7 = $\boxed{5}$.

(Note: If $b > a$, then $b - a = 7(n-m) + 2$, giving remainder 2.)

Product:
$$a \times b = (7m + 3)(7n + 5)$$
$$= 49mn + 35m + 21n + 15$$
$$= 7(7mn + 5m + 3n + 2) + 1$$
Remainder when product is divided by 7 = $\boxed{1}$.

(Since $3 \times 5 = 15 = 2 \times 7 + 1$, the remainder is 1.)

Exploration:

• Consecutive numbers 4, 5, 6: $5^2 - 4 \times 6 = 25 - 24 = 1$
• Consecutive numbers 7, 8, 9: $8^2 - 7 \times 9 = 64 - 63 = 1$
• Consecutive numbers 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 $(n-1)$, $n$, $(n+1)$.
$$n^2 - (n-1)(n+1) = 1$$

Verification by expansion:

Left side:
$$n^2 - (n-1)(n+1) = n^2 - (n^2 - 1) = n^2 - n^2 + 1 = 1$$

Right side: $1$.

LHS = RHS = 1. ✓

This is a true identity: $n^2 - (n-1)(n+1) = 1$ for all values of $n$.

Let the two numbers be $a$ and $b$.

Step 1: Add the two numbers: $a + b$.

Step 2: Half of the sum of the two numbers: $\dfrac{a+b}{2}$.

Step 3: Multiply: $(a + b) \times \dfrac{a+b}{2}$.

Algebraic expression:
$$\frac{(a+b)(a+b)}{2} = \frac{(a+b)^2}{2}$$

Proof:

We need to show that $(a+b) \times \dfrac{a+b}{2} = \dfrac{(a+b)^2}{2}$.

$$\text{LHS} = (a+b) \times \frac{(a+b)}{2} = \frac{(a+b)^2}{2} = \text{RHS}$$

This is exactly half of the square of the sum of the two numbers $(a+b)^2$.

Hence proved. The result of multiplying the sum of two numbers by half their sum equals half the square of their sum.

We use the identity $(a+b)(a-b) = a^2 - b^2$, which tells us that for a fixed sum, the product is maximised when the two numbers are equal (or as close as possible).

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

Note that $14 + 26 = 40$ and $16 + 24 = 40$. Both pairs have the same sum.

Write each as $(20 - d)(20 + d) = 400 - d^2$:

• $14 \times 26 = (20-6)(20+6) = 400 - 36 = 364$
• $16 \times 24 = (20-4)(20+4) = 400 - 16 = 384$

Since $16 < 36$, we have $400 - 16 > 400 - 36$.

$$\boxed{16 \times 24 > 14 \times 26}$$

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

Note that $25 + 75 = 100$ and $26 + 74 = 100$. Both pairs have the same sum.

Write each as $(50 - d)(50 + d) = 2500 - d^2$:

• $25 \times 75 = (50-25)(50+25) = 2500 - 625 = 1875$
• $26 \times 74 = (50-24)(50+24) = 2500 - 576 = 1924$

Since $576 < 625$, we have $2500 - 576 > 2500 - 625$.

$$\boxed{26 \times 74 > 25 \times 75}$$

Note: The figure is not visible, but based on the description, we assume the park consists of a large square (or rectangle) with two smaller square plots of side $g$ ft. each embedded in it, with a walking path of width $w$ ft. surrounding them.

Assumption (standard version of this problem): The overall park is a rectangle. The two green square plots each have side $g$ ft. (area $g^2$ sq. ft. each). The walking path of width $w$ ft. surrounds the green plots.

If the two green squares are placed side by side, the total green area = $2g^2$ sq. ft.

The overall dimensions of the park (including the path) would be:

• Length: $2g + 3w$ (path on both ends and between the squares)
• Width: $g + 2w$ (path on top and bottom)

Total area of park:
$$(2g + 3w)(g + 2w) = 2g^2 + 4gw + 3gw + 6w^2 = 2g^2 + 7gw + 6w^2$$

Area to be tiled (walking path) = Total area − Green area:
$$= (2g^2 + 7gw + 6w^2) - 2g^2$$
$$\boxed{= 7gw + 6w^2}$$

(The exact expression depends on the figure; the method above illustrates the approach.)

Note: The figure is not visible in the OCR. Based on standard NCERT Class 8 pattern questions of this type, we assume the pattern involves figures where:

• Step 1 has 1 unit
• Step 2 has 3 units (or follows a specific rule)

For a common pattern where each step adds a row/column:

Pattern A (L-shaped or cross pattern): Step $n$ has $n^2$ units or $2n-1$ units.

For the most common version where Step $n$ has $n^2$ basic units:

(i) The next figure (Step 4) would be a $4 \times 4$ arrangement = 16 units.

(ii) Step 10 would have $10^2 = 100$ basic units.

(iii) Expression for Step $y$: $y^2$ basic units.

For a pattern where Step $n$ has $2n + 1$ units (another common type):

(i) Draw the next figure by adding 2 more units.

(ii) Step 10: $2(10) + 1 = 21$ basic units.

(iii) Expression for Step $y$: $2y + 1$ basic units.

Please refer to the actual figure in the textbook to determine the exact pattern and apply the method above accordingly.