We Distribute, Yet Things Multiply Class 8 Mathematics NCERT Solutions

Q1Figure it Out 1

Observe the multiplication grid below. Each number inside the grid is formed by multiplying two numbers. If the middle number of a $3 \times 3$ frame is given by the expression $pq$ , as shown in the figure, write the expressions for the other numbers in the grid.

Solution

Given: A

3 \times 3

frame from a multiplication grid where the rows and columns are consecutive numbers. The middle cell is the product

p \times q

, which is written as

pq

.

To Find: Expressions for the other eight numbers in the

3 \times 3

frame.

Solution: Since the grid is a multiplication table with consecutive numbers, if the middle cell represents the product of

p

and

q

, then the rows correspond to multipliers

(p-1), p, (p+1)

and the columns correspond to multiplicands

(q-1), q, (q+1)

.

The expressions for each cell in the frame are as follows:

Top-left cell: $(p-1)(q-1) = pq - p - q + 1$
Top-middle cell: $(p-1)q = pq - q$
Top-right cell: $(p-1)(q+1) = pq + p - q - 1$
Middle-left cell: $p(q-1) = pq - p$
Middle cell (given): $pq$
Middle-right cell: $p(q+1) = pq + p$
Bottom-left cell: $(p+1)(q-1) = pq - p + q - 1$
Bottom-middle cell: $(p+1)q = pq + q$
Bottom-right cell: $(p+1)(q+1) = pq + p + q + 1$

Final Answer: The

3 \times 3

grid with expressions is:

$(p-1)(q-1)$	$(p-1)q$	$(p-1)(q+1)$
$p(q-1)$	$pq$	$p(q+1)$
$(p+1)(q-1)$	$(p+1)q$	$(p+1)(q+1)$

Q2Figure it Out 1

Expand the following products.

(i)

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

(ii)

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

(iii)

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

(iv)

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

(v)

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

(vi)

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

Solution

To Find: The expanded form of the given products.

Solution: We use the distributive property

(a+b)(c+d) = ac + ad + bc + bd

.

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

(3+u)(v-3) = 3(v-3) + u(v-3)

= 3v - 9 + uv - 3u

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

\frac{2}{3}(15+6 a) = \left(\frac{2}{3} \times 15\right) + \left(\frac{2}{3} \times 6a\right)

= 2 \times 5 + 2 \times 2a

= 10 + 4a

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

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

= 100ac + 10ad + 10bc + bd

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

(3-x)(x-6) = 3(x-6) - x(x-6)

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

= -x^2 + 9x - 18

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

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

= -5ac - 5ad + bc + bd

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

(5+z)(y+9) = 5(y+9) + z(y+9)

= 5y + 45 + zy + 9z

Final Answer:

(i)

3v - 9 + uv - 3u

(ii)

10 + 4a

(iii)

100ac + 10ad + 10bc + bd

(iv)

-x^2 + 9x - 18

(v)

-5ac - 5ad + bc + bd

(vi)

5y + 45 + zy + 9z

Q3Figure it Out 1

Find 3 examples where the product of two numbers remains unchanged when one of them is increased by 2 and the other is decreased by 4.

Solution

Given: The product of two numbers remains unchanged when one is increased by 2 and the other is decreased by 4.

To Find: Three examples of such pairs of numbers.

Solution: Let the two numbers be

a

and

b

. Their product is

ab

. The new numbers are

(a+2)

and

(b-4)

. Their product is

(a+2)(b-4)

.

According to the condition:

ab = (a+2)(b-4)

ab = ab - 4a + 2b - 8

Subtracting

ab

from both sides:

0 = -4a + 2b - 8

4a - 2b + 8 = 0

2b = 4a + 8

b = 2a + 4

We need to find three pairs of numbers

(a, b)

that satisfy the relation

b = 2a + 4

.

Example 1: Let

a = 1

. Then

b = 2(1) + 4 = 6

. The numbers are 1 and 6. Product =

1 \times 6 = 6

. New numbers are

(1+2)=3

and

(6-4)=2

. New product =

3 \times 2 = 6

. The product is unchanged.

Example 2: Let

a = 5

. Then

b = 2(5) + 4 = 14

. The numbers are 5 and 14. Product =

5 \times 14 = 70

. New numbers are

(5+2)=7

and

(14-4)=10

. New product =

7 \times 10 = 70

. The product is unchanged.

Example 3: Let

a = 10

. Then

b = 2(10) + 4 = 24

. The numbers are 10 and 24. Product =

10 \times 24 = 240

. New numbers are

(10+2)=12

and

(24-4)=20

. New product =

12 \times 20 = 240

. The product is unchanged.

Final Answer: Three examples are the pairs of numbers (1, 6), (5, 14), and (10, 24).

Q4Figure it Out 1

Expand (i) $(a+a b-3 b^{2})(4+b)$ , and (ii) $(4 y+7)(y+11 z-3)$ .

Solution

To Find: The expanded form of the given products.

Solution: We use the distributive property.

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

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

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

= 4a + ab + 4ab + ab^2 - 12b^2 - 3b^3

Combining like terms (

ab

and

4ab

):

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

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

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

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

= 4y^2 + 44yz - 12y + 7y + 77z - 21

Combining like terms (

-12y

and

7y

):

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

Final Answer:

(i)

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

(ii)

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

Q5Figure it Out 1

Expand (i) $(a-b)(a+b)$ , (ii) $(a-b)(a^{2}+a b+b^{2})$ and (iii) $(a-b)(a^{3}+a^{2} b+a b^{2}+b^{3})$ , Do you see a pattern? What would be the next identity in the pattern that you see? Can you check it by expanding?

Solution

To Find: Expand the given products, identify a pattern, state the next identity, and verify it.

Solution:

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

= a(a+b) - b(a+b)

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

= a^2 - b^2

(ii) $(a-b)(a^{2}+a b+b^{2})$

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

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

= a^3 + a^2b + ab^2 - a^2b - ab^2 - b^3

= a^3 - b^3

(iii) $(a-b)(a^{3}+a^{2} b+a b^{2}+b^{3})$

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

= a^4 + a^3b + a^2b^2 + ab^3 - a^3b - a^2b^2 - ab^3 - b^4

= a^4 - b^4

Pattern: The products are of the form

a^n - b^n

. Specifically, we see the pattern:

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

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

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

The general pattern appears to be:

a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + \dots + ab^{n-2} + b^{n-1})

.

Next Identity in the Pattern: The next identity would be for

n=5

:

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

Checking by Expansion: Let's expand the left side (LHS): LHS

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

= a^5 + a^4b + a^3b^2 + a^2b^3 + ab^4 - a^4b - a^3b^2 - a^2b^3 - ab^4 - b^5

= a^5 - b^5

LHS = RHS. The identity is correct.

Final Answer: The expansions are:

(i)

a^2 - b^2

(ii)

a^3 - b^3

(iii)

a^4 - b^4

The pattern is

(a-b)(a^{n-1} + a^{n-2}b + \dots + b^{n-1}) = a^n - b^n

. The next identity is

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

, which is verified by expansion.

Q1Figure it Out 2

Which is greater: $(a-b)^{2}$ or $(b-a)^{2}$ ? Justify your answer.

Solution

To Find: Which of

(a-b)^2

and

(b-a)^2

is greater and provide a justification.

Solution: Let us expand both expressions.

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

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

Since both expressions expand to the same result, they are equal.

Justification: Another way to justify this is to notice that

(b-a)

is the negative of

(a-b)

.

b-a = -1 \times (a-b)

Squaring both sides:

(b-a)^2 = (-1 \times (a-b))^2

(b-a)^2 = (-1)^2 \times (a-b)^2

(b-a)^2 = 1 \times (a-b)^2

(b-a)^2 = (a-b)^2

Final Answer: Neither is greater; they are equal. This is because

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

and

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

, which are identical expressions. Also, the square of a number and the square of its negative are always equal.

Q2Figure it Out 2

Express 100 as the difference of two squares.

Solution

To Find: Two square numbers whose difference is 100.

Solution: We need to find two numbers, say

a

and

b

, such that

a^2 - b^2 = 100

. Using the identity

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

, we have:

(a-b)(a+b) = 100

We need to find two factors of 100, say

x

and

y

, such that

x = a-b

and

y = a+b

. For

a

and

b

to be integers,

x

and

y

must have the same parity (both even or both odd). Let's find pairs of even factors of 100.

Case 1: Factors are 2 and 50.

a-b = 2

a+b = 50

Adding the two equations:

2a = 52 \implies a = 26

. Substituting

a=26

into the second equation:

26 + b = 50 \implies b = 24

. So,

26^2 - 24^2 = 676 - 576 = 100

. This is a valid solution.

Case 2: Factors are 10 and 10.

a-b = 10

a+b = 10

Adding the two equations:

2a = 20 \implies a = 10

. Substituting

a=10

into the second equation:

10 + b = 10 \implies b = 0

. So,

10^2 - 0^2 = 100 - 0 = 100

. This is another valid solution.

Final Answer: 100 can be expressed as the difference of two squares in several ways. One example is

26^2 - 24^2 = 100

.

Q3Figure it Out 2

Find $406^{2}, 72^{2}, 145^{2}, 1097^{2}$ , and $124^{2}$ using the identities you have learnt so far.

Solution

To Find: The values of the given squares using algebraic identities.

Solution: We will use the identities

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

and

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

.

$406^2$ We can write $406 = 400 + 6$ . Using $(a+b)^2$ with $a=400, b=6$ : $406^2 = (400+6)^2 = 400^2 + 2(400)(6) + 6^2 = 160000 + 4800 + 36 = 164836$
$72^2$ We can write $72 = 70 + 2$ . Using $(a+b)^2$ with $a=70, b=2$ : $72^2 = (70+2)^2 = 70^2 + 2(70)(2) + 2^2 = 4900 + 280 + 4 = 5184$
$145^2$ We can write $145 = 150 - 5$ . Using $(a-b)^2$ with $a=150, b=5$ : $145^2 = (150-5)^2 = 150^2 - 2(150)(5) + 5^2 = 22500 - 1500 + 25 = 21025$
$1097^2$ We can write $1097 = 1100 - 3$ . Using $(a-b)^2$ with $a=1100, b=3$ : $1097^2 = (1100-3)^2 = 1100^2 - 2(1100)(3) + 3^2 = 1210000 - 6600 + 9 = 1203409$
$124^2$ We can write $124 = 120 + 4$ . Using $(a+b)^2$ with $a=120, b=4$ : $124^2 = (120+4)^2 = 120^2 + 2(120)(4) + 4^2 = 14400 + 960 + 16 = 15376$

Final Answer:

406^2 = 164836

72^2 = 5184

145^2 = 21025

1097^2 = 1203409

124^2 = 15376

Q4Figure it Out 2

Do Patterns 1 and 2 hold only for counting numbers? Do they hold for negative integers as well? What about fractions? Justify your answer.

Solution

Given: Pattern 1:

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

Pattern 2:

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

To Find: Whether these patterns hold for negative integers and fractions.

Solution: Both patterns are algebraic identities. An algebraic identity is a statement of equality that is true for all possible values of its variables.

Justification: The derivations for these identities rely on the fundamental properties of arithmetic, such as the distributive, commutative, and associative properties of addition and multiplication. These properties are valid not just for counting numbers, but also for all real numbers, which include negative integers and fractions.

Let's verify with an example for each case.

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

Negative Integers: Let $a = -2, b = -3$ . LHS: $2((-2)^2 + (-3)^2) = 2(4+9) = 2(13) = 26$ . RHS: $(-2+(-3))^2 + (-2-(-3))^2 = (-5)^2 + (1)^2 = 25+1 = 26$ . LHS = RHS. It holds.
Fractions: Let $a = 1/2, b = 1/3$ . LHS: $2((\frac{1}{2})^2 + (\frac{1}{3})^2) = 2(\frac{1}{4} + \frac{1}{9}) = 2(\frac{9+4}{36}) = 2(\frac{13}{36}) = \frac{13}{18}$ . RHS: $(\frac{1}{2}+\frac{1}{3})^2 + (\frac{1}{2}-\frac{1}{3})^2 = (\frac{3+2}{6})^2 + (\frac{3-2}{6})^2 = (\frac{5}{6})^2 + (\frac{1}{6})^2 = \frac{25}{36} + \frac{1}{36} = \frac{26}{36} = \frac{13}{18}$ . LHS = RHS. It holds.

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

Negative Integers: Let $a = -5, b = -2$ . LHS: $(-5)^2 - (-2)^2 = 25 - 4 = 21$ . RHS: $(-5+(-2))(-5-(-2)) = (-7)(-3) = 21$ . LHS = RHS. It holds.
Fractions: Let $a = 1/2, b = 1/4$ . LHS: $(\frac{1}{2})^2 - (\frac{1}{4})^2 = \frac{1}{4} - \frac{1}{16} = \frac{4-1}{16} = \frac{3}{16}$ . RHS: $(\frac{1}{2}+\frac{1}{4})(\frac{1}{2}-\frac{1}{4}) = (\frac{2+1}{4})(\frac{2-1}{4}) = (\frac{3}{4})(\frac{1}{4}) = \frac{3}{16}$ . LHS = RHS. It holds.

Final Answer: No, the patterns do not hold only for counting numbers. They are true algebraic identities and therefore hold for all numbers, including negative integers and fractions, because the properties of arithmetic used to prove them apply to these number systems as well.

Q1Figure it Out 3

Compute these products using the suggested identity.

(i)

$46^{2}$ using Identity 1A for $(a+b)^{2}$

(ii)

$397 \times 403$ using Identity 1C for $(a+b)(a-b)$

(iii)

$91^{2}$ using Identity 1B for $(a-b)^{2}$

(iv)

$43 \times 45$ using Identity 1C for $(a+b)(a-b)$

Solution

To Find: The values of the given products using the specified identities.

Solution:

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

a=40

and

b=6

.

46^2 = (40+6)^2 = 40^2 + 2(40)(6) + 6^2

= 1600 + 480 + 36 = 2116

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

397 = 400-3

and

403 = 400+3

. Let

a=400

and

b=3

.

397 \times 403 = (400-3)(400+3) = 400^2 - 3^2

= 160000 - 9 = 159991

(iii) $91^2$ using $(a-b)^2 = a^2-2ab+b^2$ We can write

91 = 100-9

. Let

a=100

and

b=9

.

91^2 = (100-9)^2 = 100^2 - 2(100)(9) + 9^2

= 10000 - 1800 + 81 = 8281

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

43 = 44-1

and

45 = 44+1

. Let

a=44

and

b=1

.

43 \times 45 = (44-1)(44+1) = 44^2 - 1^2

= 1936 - 1 = 1935

(Note:

44^2 = (40+4)^2 = 1600+320+16 = 1936

)

Final Answer:

(i)

46^2 = 2116

(ii)

397 \times 403 = 159991

(iii)

91^2 = 8281

(iv)

43 \times 45 = 1935

Q2Figure it Out 3

Use either a suitable identity or the distributive property to find each of the following products.

(i)

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

(ii)

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

(iii)

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

(iv)

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

(v)

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

(vi)

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

Solution

To Find: The expanded form of the given products.

Solution:

(i) $(p-1)(p+11)$ Using distributive property:

(p-1)(p+11) = p(p+11) - 1(p+11) = p^2 + 11p - p - 11 = p^2 + 10p - 11

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

(x-y)(x+y) = x^2 - y^2

with

x=3a, y=9b

:

(3a)^2 - (9b)^2 = 9a^2 - 81b^2

(iii) $-(2 y+5)(3 y+4)$ First, expand

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

:

2y(3y+4) + 5(3y+4) = 6y^2 + 8y + 15y + 20 = 6y^2 + 23y + 20

Now, apply the negative sign:

-(6y^2 + 23y + 20) = -6y^2 - 23y - 20

(iv) $(6 x+5 y)^{2}$ Using identity

(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

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

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

with

a=2x, b=1/2

:

(2x)^2 - 2(2x)(\frac{1}{2}) + (\frac{1}{2})^2 = 4x^2 - 2x + \frac{1}{4}

(vi) $(7 p) \times(3 r) \times(p+2)$ First, multiply the first two terms:

(7p) \times (3r) = 21pr

Now, distribute this product:

21pr(p+2) = (21pr \times p) + (21pr \times 2) = 21p^2r + 42pr

Final Answer:

(i)

p^2 + 10p - 11

(ii)

9a^2 - 81b^2

(iii)

-6y^2 - 23y - 20

(iv)

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

(v)

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

(vi)

21p^2r + 42pr

Q3Figure it Out 3

For each statement identify the appropriate algebraic expression(s).

(i)

Two more than a square number. $2+s \quad (s+2)^{2} \quad s^{2}+2 \quad s^{2}+4 \quad 2 s^{2} \quad 2^{2} s$

(ii)

The sum of the squares of two consecutive numbers $m^{2}+n^{2} \quad (m+n)^{2} \quad m^{2}+1 \quad m^{2}+(m+1)^{2} \quad m^{2}+(m-1)^{2} \quad (m+(m+1))^{2} \quad (2 m)^{2}+(2 m+1)^{2}$

Solution

To Find: The correct algebraic expression(s) for each statement.

Solution:

(i) Two more than a square number. Let the number be

s

. Its square is

s^2

. Two more than this square is

s^2 + 2

. From the given options,

s^2+2

is the correct expression. Note that

2+s

is two more than a number, not a square number.

(s+2)^2

is the square of a number that is two more than

s

.

Appropriate expression: $s^2+2$

(ii) The sum of the squares of two consecutive numbers. Let the first number be

m

. The next consecutive number is

m+1

. The square of the first number is

m^2

. The square of the second number is

(m+1)^2

. The sum of their squares is

m^2 + (m+1)^2

.

Alternatively, if the second number is

m

, the first consecutive number is

m-1

. The sum of their squares would be

(m-1)^2 + m^2

, which is equivalent to

m^2 + (m-1)^2

.

Let's check the given options:

$m^2 + (m+1)^2$ : Correct.
$m^2 + (m-1)^2$ : Correct.
The other expressions are incorrect. For example, $(2m)^2 + (2m+1)^2$ is the sum of squares of a consecutive even and odd number, not any two consecutive numbers.
Appropriate expressions: $m^{2}+(m+1)^{2}$ and $m^{2}+(m-1)^{2}$

Final Answer:

(i)

s^2+2

(ii)

m^{2}+(m+1)^{2}

and

m^{2}+(m-1)^{2}

Q4Figure it Out 3

Consider any 2 by 2 square of numbers in a calendar, as shown in the figure. Find products of numbers lying along each diagonal. Do this for the other 2 by 2 squares. What do you observe about the diagonal products? Explain why this happens. Hint: Label the numbers in each 2 by 2 square as
$a$ $(a+1)$
$a+7$ $(a+8)$

$a$	$(a+1)$
$a+7$	$(a+8)$

Solution

Given: A 2 by 2 square of numbers from a calendar.

To Find: The relationship between the products of the numbers on the diagonals and explain why it occurs.

Solution: Let's use the hint and represent a generic 2 by 2 square from a calendar algebraically. Let the top-left number be

a

.

The number to its right is $a+1$ .
The number below it is $a+7$ (since there are 7 days in a week).
The number diagonally opposite is $a+8$ .

The 2 by 2 square is:

$a$	$a+1$
$a+7$	$a+8$

Now, let's find the products of the numbers on the two diagonals.

Product of the first diagonal (top-left to bottom-right):

P_1 = a \times (a+8) = a^2 + 8a

Product of the second diagonal (top-right to bottom-left):

P_2 = (a+1) \times (a+7) = a(a+7) + 1(a+7) = a^2 + 7a + a + 7 = a^2 + 8a + 7

Observation: Comparing the two products, we see that

P_2 = P_1 + 7

. The product of the second diagonal (top-right to bottom-left) is always 7 more than the product of the first diagonal (top-left to bottom-right).

Example from the textbook: For the square with top-left number 4:

a=4

. The square is | 4 | 5 | and | 11 | 12 |. First diagonal product:

4 \times 12 = 48

. Second diagonal product:

5 \times 11 = 55

. Difference:

55 - 48 = 7

. The observation holds true.

Explanation: The algebraic expansion shows that

(a+1)(a+7)

simplifies to

a^2+8a+7

, while

a(a+8)

simplifies to

a^2+8a

. The difference between these two expressions is always 7, regardless of the value of

a

.

Final Answer: The product of the numbers on the diagonal from top-right to bottom-left is always 7 greater than the product of the numbers on the diagonal from top-left to bottom-right. This is because if the top-left number is

a

, the products are

a(a+8) = a^2+8a

and

(a+1)(a+7) = a^2+8a+7

, and their difference is always 7.

Q5Figure it Out 3

Verify which of the following statements are true.

(i)

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

(ii)

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

(iii)

Squares of even numbers are multiples of 4, and squares of odd numbers are 1 more than multiples of 8.

(iv)

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

Solution

To Find: Verify if the given statements are true or false.

Solution:

(i) $(k+1)(k+2)-(k+3)$ is always 2. Let's simplify the expression:

(k+1)(k+2)-(k+3) = (k^2 + 2k + k + 2) - (k+3)

= (k^2 + 3k + 2) - k - 3

= k^2 + 2k - 1

Since the result depends on

k

, it is not always 2. For example, if

k=1

, the value is

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

. But if

k=2

, the value is

2^2+2(2)-1 = 7

. The statement is FALSE.

(ii) $(2 q+1)(2 q-3)$ is a multiple of 4. Let's expand the expression:

(2q+1)(2q-3) = 2q(2q-3) + 1(2q-3)

= 4q^2 - 6q + 2q - 3

= 4q^2 - 4q - 3

= 4(q^2 - q) - 3

The term

4(q^2-q)

is a multiple of 4. Subtracting 3 means the result is 3 less than a multiple of 4, so it can never be a multiple of 4. The statement is FALSE.

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

Part 1: Squares of even numbers are multiples of 4. An even number can be written as $2n$ , where $n$ is an integer. The square is $(2n)^2 = 4n^2$ . Since this expression has a factor of 4, it is always a multiple of 4. This part is TRUE.
Part 2: Squares of odd numbers are 1 more than multiples of 8. An odd number can be written as $2n+1$ , where $n$ is an integer. The square is $(2n+1)^2 = 4n^2 + 4n + 1 = 4n(n+1) + 1$ . The product of two consecutive integers, $n(n+1)$ , is always even. So, we can write $n(n+1) = 2k$ for some integer $k$ . Substituting this, we get $4(2k) + 1 = 8k + 1$ . This expression represents a number that is 1 more than a multiple of 8. This part is TRUE. Since both parts are true, the entire statement is TRUE.

(iv) $(6 n+2)^{2}-(4 n+3)^{2}$ is 5 less than a square number. Let's simplify the expression:

(6n+2)^2 - (4n+3)^2 = (36n^2 + 24n + 4) - (16n^2 + 24n + 9)

= 36n^2 + 24n + 4 - 16n^2 - 24n - 9

= 20n^2 - 5

The statement says this is 5 less than a square number. Let the square number be

S

. So,

20n^2 - 5 = S - 5

, which implies

S = 20n^2

. For

S

to be a square number,

20n^2

must be a perfect square. This is not true for all integer values of

n

. For example, if

n=1

,

20(1)^2=20

, which is not a perfect square. The statement is FALSE.

Final Answer:

(i)

False

(ii)

False

(iii)

True

(iv)

False

Q6Figure it Out 3

A number leaves a remainder of 3 when divided by 7, and another number leaves a remainder of 5 when divided by 7. What is the remainder when their sum, difference, and product are divided by 7?

Solution

Given: Let the first number be

N_1

and the second number be

N_2

.

N_1

divided by 7 gives a remainder of 3. So,

N_1 = 7k + 3

for some integer

k

.

N_2

divided by 7 gives a remainder of 5. So,

N_2 = 7m + 5

for some integer

m

.

To Find: The remainder when their sum, difference, and product are divided by 7.

Solution:

1. Sum ( $N_1 + N_2$ ):

N_1 + N_2 = (7k+3) + (7m+5)

= 7k + 7m + 8

= 7(k+m) + 7 + 1

= 7(k+m+1) + 1

This is in the form

7 \times (\text{integer}) + 1

. So, the remainder is 1.

2. Difference ( $N_1 - N_2$ ):

N_1 - N_2 = (7k+3) - (7m+5)

= 7k - 7m + 3 - 5

= 7(k-m) - 2

Since remainders must be non-negative, we rewrite -2 as

7-2-7=5-7

. So:

= 7(k-m) - 7 + 5

= 7(k-m-1) + 5

This is in the form

7 \times (\text{integer}) + 5

. So, the remainder is 5.

3. Product ( $N_1 \times N_2$ ):

N_1 \times N_2 = (7k+3)(7m+5)

= 7k(7m+5) + 3(7m+5)

= 49km + 35k + 21m + 15

= 7(7km + 5k + 3m) + 15

We can write

15 = 14 + 1 = 2 \times 7 + 1

.

= 7(7km + 5k + 3m) + 14 + 1

= 7(7km + 5k + 3m + 2) + 1

This is in the form

7 \times (\text{integer}) + 1

. So, the remainder is 1.

Final Answer:

The remainder of the sum is 1.
The remainder of the difference is 5.
The remainder of the product is 1.

Q7Figure it Out 3

Choose three consecutive numbers, square the middle one, and subtract the product of the other two. Repeat the same with other sets of numbers. What pattern do you notice? How do we write this as an algebraic equation? Expand both sides of the equation to check that it is a true identity.

Solution

To Find: A pattern related to three consecutive numbers, express it as an algebraic identity, and verify it.

Solution: Let's test with a few sets of consecutive numbers.

Set 1: 2, 3, 4 Middle number squared: $3^2 = 9$ . Product of the other two: $2 \times 4 = 8$ . Result: $9 - 8 = 1$ .
Set 2: 9, 10, 11 Middle number squared: $10^2 = 100$ . Product of the other two: $9 \times 11 = 99$ . Result: $100 - 99 = 1$ .
Set 3: 20, 21, 22 Middle number squared: $21^2 = 441$ . Product of the other two: $20 \times 22 = 440$ . Result: $441 - 440 = 1$ .

Pattern: The result is always 1.

Algebraic Equation: Let the three consecutive numbers be represented by

(n-1), n, (n+1)

.

The middle number is $n$ . Its square is $n^2$ .
The other two numbers are $(n-1)$ and $(n+1)$ . Their product is $(n-1)(n+1)$ .

The operation described is: (square of the middle one) - (product of the other two). Algebraically, this is:

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

. The pattern suggests this expression is always equal to 1. So, the algebraic equation (identity) is:

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

Verification: Let's expand the left-hand side (LHS) of the equation. LHS

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

Using the identity

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

, where

a=n

and

b=1

: LHS

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

LHS

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

LHS

= n^2 - n^2 + 1

LHS

= 1

Since LHS = 1 and RHS = 1, the equation is a true identity.

Final Answer: The pattern observed is that the result is always 1. The algebraic identity representing this pattern is

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

. The identity is verified by expanding the left side, which simplifies to 1.

Q8Figure it Out 3

What is the algebraic expression describing the following steps-add any two numbers. Multiply this by half of the sum of the two numbers? Prove that this result will be half of the square of the sum of the two numbers.

Solution

To Find: An algebraic expression for the given steps and prove that it is equal to half of the square of the sum of the two numbers.

Solution:

Step 1: Write the algebraic expression. Let the two numbers be

a

and

b

.

Add the two numbers: $(a+b)$ .
Find half of the sum of the two numbers: $\frac{a+b}{2}$ .
Multiply the sum by half of the sum: $(a+b) \times \frac{a+b}{2}$ .

The resulting algebraic expression is:

\frac{(a+b)^2}{2}

Step 2: Prove the result. We need to prove that the expression from Step 1 is equal to 'half of the square of the sum of the two numbers'.

The sum of the two numbers is $(a+b)$ .
The square of the sum is $(a+b)^2$ .
Half of the square of the sum is $\frac{1}{2}(a+b)^2$ , which can also be written as $\frac{(a+b)^2}{2}$ .

Proof: The expression derived from the steps is

\frac{(a+b)^2}{2}

. The expression for 'half of the square of the sum' is also

\frac{(a+b)^2}{2}

. Since both expressions are identical, the statement is proven.

Final Answer: The algebraic expression is

(a+b) \times \frac{a+b}{2}

or

\frac{(a+b)^2}{2}

. This expression is by definition 'half of the square of the sum of the two numbers', which is

\frac{1}{2}(a+b)^2

. Both expressions are algebraically identical, so the statement is proven.

Q9Figure it Out 3

Which is larger? Find out without fully computing the product.

(i)

$14 \times 26$ or $16 \times 24$

(ii)

$25 \times 75$ or $26 \times 74$

Solution

To Find: Which product is larger in each pair without full computation.

Solution: We can use the identity

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

. This shows that for a fixed sum

2a

, the product

(a-b)(a+b)

is maximized when

b

is minimized, i.e., when the numbers are closest to each other.

(i) $14 \times 26$ or $16 \times 24$ Notice that the sum of the numbers in each pair is the same:

14 + 26 = 40

16 + 24 = 40

Let's express each product using the identity. The average of the numbers is

40/2 = 20

.

$14 \times 26 = (20 - 6)(20 + 6) = 20^2 - 6^2 = 400 - 36$
$16 \times 24 = (20 - 4)(20 + 4) = 20^2 - 4^2 = 400 - 16$

Since we are subtracting a smaller number (16) from 400 in the second case, the result will be larger. Therefore,

16 \times 24

is larger.

(ii) $25 \times 75$ or $26 \times 74$ Again, the sum of the numbers in each pair is the same:

25 + 75 = 100

26 + 74 = 100

The average of the numbers is

100/2 = 50

.

$25 \times 75 = (50 - 25)(50 + 25) = 50^2 - 25^2 = 2500 - 625$
$26 \times 74 = (50 - 24)(50 + 24) = 50^2 - 24^2 = 2500 - 576$

Since we are subtracting a smaller number (576) from 2500 in the second case, the result will be larger. Therefore,

26 \times 74

is larger.

Final Answer:

(i)

16 \times 24

is larger.

(ii)

26 \times 74

is larger.

Q10Figure it Out 3

A tiny park is coming up in Dhauli. The plan is shown in the figure. The two square plots, each of area $g^{2}$ sq. ft ., will have a green cover. All the remaining area is a walking path $w$ ft. wide that needs to be tiled. Write an expression for the area that needs to be tiled.

Solution

Given:

Two square plots for green cover, each with side length $g$ feet.
A walking path of uniform width $w$ feet surrounds the combined green area.

To Find: An expression for the area of the walking path that needs to be tiled.

Solution:

Dimensions of the green area: The two square plots are adjacent, forming a rectangle. The dimensions of this combined rectangular green area are length = $g+g = 2g$ feet and width = $g$ feet. Area of green cover = $2 \times g^2 = 2g^2$ sq. ft.
Dimensions of the total park (green area + path): The path of width $w$ surrounds the green area.
- The total length of the park will be the length of the green area plus the path width on both sides: $2g + w + w = 2g + 2w$ feet.
- The total width of the park will be the width of the green area plus the path width on both sides: $g + w + w = g + 2w$ feet.
Area of the total park: Total Area = (Total Length) $\times$ (Total Width) Total Area = $(2g + 2w)(g + 2w)$
Area to be tiled (walking path): Area of path = Total Area - Area of green cover Area of path = $(2g + 2w)(g + 2w) - 2g^2$ Let's expand the first term using the distributive property: $= 2g(g+2w) + 2w(g+2w) - 2g^2$ $= (2g^2 + 4gw) + (2gw + 4w^2) - 2g^2$ $= 2g^2 + 4gw + 2gw + 4w^2 - 2g^2$ Combine like terms: $= (2g^2 - 2g^2) + (4gw + 2gw) + 4w^2$ $= 6gw + 4w^2$

Final Answer: The expression for the area that needs to be tiled is

6gw + 4w^2

sq. ft.

Q11Figure it Out 3

For each pattern shown below,

(i)

Draw the next figure in the sequence.

(ii)

How many basic units are there in Step 10?

(iii)

Write an expression to describe the number of basic units in Step $y$ .

Solution

To Find: For each pattern, draw the next figure, find the number of units in Step 10, and write a general expression for Step

y

.

Solution:

Pattern (a): L-shape

Description: The figure at Step $y$ consists of a central square unit with two arms of length $y$ extending from it, forming an 'L' shape. The total number of units is $y$ units in the vertical arm, $y$ units in the horizontal arm, plus the corner unit. Total units = $y+y+1 = 2y+1$ . Let's check: Step 1: $2(1)+1=3$ . Step 2: $2(2)+1=5$ . This is incorrect. A better model is: a column of $y+1$ units and a row of $y$ units attached to its base. Total units: $(y+1) + y = 2y+1$ . Still incorrect. Let's look at the images from the book. Step 1 has 3 units (a 2x2 square with one missing). Step 2 has 5 units (a 3x2 rectangle with one missing). Step 3 has 7 units (a 4x2 rectangle with one missing). No, the image is an L-shape. Step 1 is a 2x2 L-shape (3 blocks). Step 2 is a 3x3 L-shape (5 blocks). Step 3 is a 4x4 L-shape (7 blocks). The number of units in Step y is $(y+1) + y = 2y+1$ . Let's recheck. Step 1: 2(1)+1=3. Correct. Step 2: 2(2)+1=5. Correct. Step 3: 2(3)+1=7. Correct. *

(i)

The next figure (Step 4) will be a 5x5 L-shape, with 5 units down and 5 units across, sharing one corner. It will have

2(4)+1 = 9

units. *

(ii)

In Step 10, there will be

2(10)+1 = 21

basic units. *

(iii)

The expression for the number of units in Step

y

is

2y+1

.

Pattern (b): Plus-shape

Description: The figure at Step $y$ consists of a central square unit with four arms, each of length $y$ , extending up, down, left, and right. Total units = $4 \times y + 1 = 4y+1$ . Let's check: Step 1: $4(1)+1=5$ . Correct. Step 2: $4(2)+1=9$ . Correct. Step 3: $4(3)+1=13$ . Correct. *

(i)

The next figure (Step 4) will have a central unit and four arms of length 4. It will have

4(4)+1 = 17

units. *

(ii)

In Step 10, there will be

4(10)+1 = 41

basic units. *

(iii)

The expression for the number of units in Step

y

is

4y+1

.

Pattern (c): Growing rectangle

Description: The figure at Step $y$ is a rectangle with dimensions $y$ by $(y+1)$ . The total number of units is $y(y+1)$ . Let's check: Step 1: $1(1+1)=2$ . Correct. Step 2: $2(2+1)=6$ . Correct. Step 3: $3(3+1)=12$ . Correct. *

(i)

The next figure (Step 4) will be a rectangle of size

4 \times 5

. It will have

4 \times 5 = 20

units. *

(ii)

In Step 10, there will be

10 \times (10+1) = 10 \times 11 = 110

basic units. *

(iii)

The expression for the number of units in Step

y

is

y(y+1)

or

y^2+y

.

Final Answer:

Pattern (a):

(i)

The next figure has 9 units.

(ii)

Step 10 has 21 units.

(iii)

The expression is

2y+1

.

Pattern (b):

(i)

The next figure has 17 units.

(ii)

Step 10 has 41 units.

(iii)

The expression is

4y+1

.

Pattern (c):

(i)

The next figure has 20 units.

(ii)

Step 10 has 110 units.

(iii)

The expression is

y(y+1)

or

y^2+y

.

We Distribute, Yet Things MultiplyClass 8 Mathematics NCERT Solutions

Observe the multiplication grid below. Each number inside the grid is formed by multiplying two numbers. If the middle number of a 3×33 \times 33×3 frame is given by the expression pqpqpq, as shown in the figure, write the expressions for the other numbers in the grid.

Solution

Expand the following products.

(3+u)(v−3)(3+u)(v-3)(3+u)(v−3)

23(15+6a)\frac{2}{3}(15+6 a)32​(15+6a)

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

(3−x)(x−6)(3-x)(x-6)(3−x)(x−6)

(−5a+b)(c+d)(-5 a+b)(c+d)(−5a+b)(c+d)

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

Solution

Find 3 examples where the product of two numbers remains unchanged when one of them is increased by 2 and the other is decreased by 4.

Solution

Expand (i) (a+ab−3b2)(4+b)(a+a b-3 b^{2})(4+b)(a+ab−3b2)(4+b), and (ii) (4y+7)(y+11z−3)(4 y+7)(y+11 z-3)(4y+7)(y+11z−3).

Solution

Solution

Which is greater: (a−b)2(a-b)^{2}(a−b)2 or (b−a)2(b-a)^{2}(b−a)2 ? Justify your answer.

Solution

Express 100 as the difference of two squares.

Solution

Find 4062,722,1452,10972406^{2}, 72^{2}, 145^{2}, 1097^{2}4062,722,1452,10972, and 1242124^{2}1242 using the identities you have learnt so far.

Solution

Do Patterns 1 and 2 hold only for counting numbers? Do they hold for negative integers as well? What about fractions? Justify your answer.

Solution

Compute these products using the suggested identity.

46246^{2}462 using Identity 1A for (a+b)2(a+b)^{2}(a+b)2

397×403397 \times 403397×403 using Identity 1C for (a+b)(a−b)(a+b)(a-b)(a+b)(a−b)

91291^{2}912 using Identity 1B for (a−b)2(a-b)^{2}(a−b)2

43×4543 \times 4543×45 using Identity 1C for (a+b)(a−b)(a+b)(a-b)(a+b)(a−b)

Solution

Use either a suitable identity or the distributive property to find each of the following products.

(p−1)(p+11)(p-1)(p+11)(p−1)(p+11)

(3a−9b)(3a+9b)(3 a-9 b)(3 a+9 b)(3a−9b)(3a+9b)

−(2y+5)(3y+4)-(2 y+5)(3 y+4)−(2y+5)(3y+4)

(6x+5y)2(6 x+5 y)^{2}(6x+5y)2

(2x−12)2(2 x-\frac{1}{2})^{2}(2x−21​)2

(7p)×(3r)×(p+2)(7 p) \times(3 r) \times(p+2)(7p)×(3r)×(p+2)

Solution

For each statement identify the appropriate algebraic expression(s).

Two more than a square number. 2+s(s+2)2s2+2s2+42s222s2+s \quad (s+2)^{2} \quad s^{2}+2 \quad s^{2}+4 \quad 2 s^{2} \quad 2^{2} s2+s(s+2)2s2+2s2+42s222s

The sum of the squares of two consecutive numbers m2+n2(m+n)2m2+1m2+(m+1)2m2+(m−1)2(m+(m+1))2(2m)2+(2m+1)2m^{2}+n^{2} \quad (m+n)^{2} \quad m^{2}+1 \quad m^{2}+(m+1)^{2} \quad m^{2}+(m-1)^{2} \quad (m+(m+1))^{2} \quad (2 m)^{2}+(2 m+1)^{2}m2+n2(m+n)2m2+1m2+(m+1)2m2+(m−1)2(m+(m+1))2(2m)2+(2m+1)2

Solution

Solution

Verify which of the following statements are true.

(k+1)(k+2)−(k+3)(k+1)(k+2)-(k+3)(k+1)(k+2)−(k+3) is always 2.

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

Squares of even numbers are multiples of 4, and squares of odd numbers are 1 more than multiples of 8.

(6n+2)2−(4n+3)2(6 n+2)^{2}-(4 n+3)^{2}(6n+2)2−(4n+3)2 is 5 less than a square number.

Solution

A number leaves a remainder of 3 when divided by 7, and another number leaves a remainder of 5 when divided by 7. What is the remainder when their sum, difference, and product are divided by 7?

Solution

Choose three consecutive numbers, square the middle one, and subtract the product of the other two. Repeat the same with other sets of numbers. What pattern do you notice? How do we write this as an algebraic equation? Expand both sides of the equation to check that it is a true identity.

Solution

What is the algebraic expression describing the following steps-add any two numbers. Multiply this by half of the sum of the two numbers? Prove that this result will be half of the square of the sum of the two numbers.

Solution

Which is larger? Find out without fully computing the product.

14×2614 \times 2614×26 or 16×2416 \times 2416×24

25×7525 \times 7525×75 or 26×7426 \times 7426×74

Solution

A tiny park is coming up in Dhauli. The plan is shown in the figure. The two square plots, each of area g2g^{2}g2 sq. ft ., will have a green cover. All the remaining area is a walking path www ft. wide that needs to be tiled. Write an expression for the area that needs to be tiled.

Solution

For each pattern shown below,

Draw the next figure in the sequence.

How many basic units are there in Step 10?

Write an expression to describe the number of basic units in Step yyy.

Solution

Observe the multiplication grid below. Each number inside the grid is formed by multiplying two numbers. If the middle number of a $3 \times 3$ frame is given by the expression $pq$ , as shown in the figure, write the expressions for the other numbers in the grid.

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

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

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

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

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

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

Expand (i) $(a+a b-3 b^{2})(4+b)$ , and (ii) $(4 y+7)(y+11 z-3)$ .

Which is greater: $(a-b)^{2}$ or $(b-a)^{2}$ ? Justify your answer.

Find $406^{2}, 72^{2}, 145^{2}, 1097^{2}$ , and $124^{2}$ using the identities you have learnt so far.

$46^{2}$ using Identity 1A for $(a+b)^{2}$

$397 \times 403$ using Identity 1C for $(a+b)(a-b)$

$91^{2}$ using Identity 1B for $(a-b)^{2}$

$43 \times 45$ using Identity 1C for $(a+b)(a-b)$

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

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

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

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

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

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

Two more than a square number. $2+s \quad (s+2)^{2} \quad s^{2}+2 \quad s^{2}+4 \quad 2 s^{2} \quad 2^{2} s$

The sum of the squares of two consecutive numbers $m^{2}+n^{2} \quad (m+n)^{2} \quad m^{2}+1 \quad m^{2}+(m+1)^{2} \quad m^{2}+(m-1)^{2} \quad (m+(m+1))^{2} \quad (2 m)^{2}+(2 m+1)^{2}$

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

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

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

$14 \times 26$ or $16 \times 24$

$25 \times 75$ or $26 \times 74$

A tiny park is coming up in Dhauli. The plan is shown in the figure. The two square plots, each of area $g^{2}$ sq. ft ., will have a green cover. All the remaining area is a walking path $w$ ft. wide that needs to be tiled. Write an expression for the area that needs to be tiled.

Write an expression to describe the number of basic units in Step $y$ .