Algebra

Let n denote the number of letters formed.

(i) The letter T requires 2 matchsticks (one horizontal and one vertical).

Number of matchsticks required to make n T's = 2n.

(ii) The letter V requires 2 matchsticks (two slanting sticks).

Number of matchsticks required to make n V's = 2n.

(iii) The letter Z requires 3 matchsticks (top horizontal, slanting, bottom horizontal).

Number of matchsticks required to make n Z's = 3n.

(iv) The letter U requires 3 matchsticks (left vertical, bottom horizontal, right vertical).

Number of matchsticks required to make n U's = 3n.

Number of mangoes in 1 box = 24

Number of mangoes in b boxes = 24 × b = 24b

Number of dots in 1 row = 8

Number of dots in r rows = 8 × r = 8r

For 12 rows, the number of dots = 8 × 12 = 96.

Meenu's age = x years

Since Anu is 5 years younger than Meenu,

Anu's age = (Meenu's age − 5) years = (x − 5) years.

Number of oranges in 1 smaller box = x

Number of oranges in 3 smaller boxes = 3x

Number of oranges left = 7

Number of oranges in a larger box = 3x + 7

Hence, the number of oranges in the larger box is (3x + 7).

Harsha's Science score = x marks

Three-fourth of her Science score = $\frac{3}{4}x$

Mathematics score = 15 more than three-fourth of Science score

= $\frac{3}{4}x + 15$

Hence, Harsha's score in Mathematics = $\left (\frac{3}{4}x + 15\right)$ marks.

Let x be the number of triangles.

For 1 triangle, matchsticks = 3.

Each additional triangle adds 2 matchsticks (since 1 matchstick is shared with the previous triangle).

For x triangles, matchsticks = 3 + 2(x − 1) = 3 + 2x - 2

= 2x + 1.

Hence, the rule is (2x + 1) matchsticks for x triangles.

Let x be the number of A's formed.

For each letter A, matchsticks = 6.

Remaining A's = x - 1

Each additional A adds 4 matchsticks (since 2 matchsticks are shared with the previous A).

For x number of A's, matchsticks = 6 + 4(x − 1) = 6 + 4x - 4

= 4x + 2.

Hence, the rule is (4x + 2) matchsticks for x number of A's.

An equilateral triangle has 3 equal sides, each of length l.

Perimeter = sum of all sides = l + l + l = 3l.

A regular hexagon has 6 equal sides, each of length l.

Perimeter = sum of all sides = l + l + l + l + l + l = 6l.

A cube has 12 edges, all of equal length l.

Sum of lengths of all edges = 12 × l = 12l.

We know that the diameter of a circle is twice its radius.

Diameter = 2 × radius = 2r units.

(i) The terms of 3 − 7x are 3 and −7x.

(ii) The terms of $2 - 5a + \frac{3}{2}b$ are 2, -5a and $\frac{3}{2}b$.

(iii) The terms of 3x⁵ + 4y³ − 7xy² + 3 are 3x⁵, 4y³, −7xy² and 3.

(iv) The terms of $2x^2 - \frac{3}{x} + \frac{5}{x^2} + 9$ are $2x^2, -\frac{3}{x}, \frac{5}{x^2}$ and 9.

(i) The required expression is 5 − 2x.

(ii) The required expression is −3 + 5x² −7x⁵.

(iii) The required expression is a² − 2b² + 3c² + 7.

(iv) The required expression is $2x - \frac{3}{x} + \frac{7y}{x^2} + 5z - 1$.

(i) The terms are x ÷ 2, y and −3.

Hence, the expression has 3 terms.

(ii) The terms are $\frac{2x}{5}, \frac{y}{5} , \frac{−3}{5}$.

Hence, the expression has 3 terms.

(iii) The terms are 5 x ab and −7.

Hence, the expression has 2 terms.

(iv) The terms are 7 x a, b ÷ 3, −c and 5.

Hence, the expression has 4 terms.

An expression with 1 term is a monomial, 2 terms is a binomial and 3 terms is a trinomial.

(i) x² − 2y² has 2 terms → Binomial.

(ii) xy + yz + zx has 3 terms → Trinomial.

(iii) 5 + 5x has 2 terms → Binomial.

(iv) 5x has 1 term → Monomial.

(v) $\frac{5}{x}$ + 3 has 2 terms → Binomial.

(vi) −7 has 1 term → Monomial.

(vii) a² − b² − c³ + 5abc has 4 terms → Multinomial .

(viii) −5a²b²c² has 1 term → Monomial.

(ix) 1 + x + x² + x³ + x⁴ has 5 terms → Multinomial.

(x) x ÷ 5 + y ÷ 3 has 2 terms → Binomial.

(xi) 5p × ab = 5pab has 1 term → Monomial.

(xii) 3 ÷ a − 2 × b + c has 3 terms → Trinomial.

(i) In −7x:

Numerical coefficient = −7

Literal coefficient = x

(ii) In -2x³y²:

Numerical coefficient = -2

Literal coefficient = x³y²

(iii) In 6abcd²:

Numerical coefficient = 6

Literal coefficient = abcd²

The terms of the given expression are $x^4, -5xy^2, 7x^2, -\frac{2}{3}x, -\frac{1}{2}$

Numerical coefficient of x⁴ is 1 .

Numerical coefficient of - 5xy² is −5.

Numerical coefficient of $7x^2$ is 7.

Numerical coefficient of $-\frac{2}{3}x$ is $-\frac{2}{3}$.

Numerical coefficient of $-\frac{1}{2}$ (constant term) is $-\frac{1}{2}$.

We have −7xy²z³.

(i) Coefficient of x = −7y²z³.

(ii) Coefficient of 7x = −y²z³.

(iii) Coefficient of −xy² = 7z³.

(iv) Coefficient of xy = −7yz³.

(v) Coefficient of z³ = −7xy².

(vi) Coefficient of −7z = xy²z².

(vii) Coefficient of xyz = −7yz².

(viii) Coefficient of 7yz² = −xyz.

(i) True. Both 5y and 5 + y depend on the value of y, so they are variables.

(ii) False. 7x is a single term (a monomial) with 7 as its numerical coefficient and x as its literal coefficient.

(iii) False. 5 + xy has 2 terms, so it is a binomial (not a trinomial).

(iv) False. 7a × bc = 7abc is a single term (a monomial).

(v) True. All the powers of x are non-negative integers.

(vi) True. The expression has 4 terms, so it is a multinomial.

(vii) False. $\frac{3}{x}$ has a negative power of x, so the expression is not a polynomial.

(viii) False. The coefficient of x in −3xy is −3y.

(ix) True. Since 3xy and −2yx have the same literal coefficient, so they are like terms.

Two terms are like terms if they have the same literal (variable) part.

(i) 5x and −2x have the same literal part x → Like terms.

(ii) 5x and 5y have different variables → Unlike terms.

(iii) −2xyz and −3x²y²z² have different powers → Unlike terms.

(iv) 10xyz and −10xyz have the same literal part xyz → Like terms.

(v) 2x²y and 2xy² have different powers → Unlike terms.

(vi) 2xy and −3yx have the same literal part (xy = yx) → Like terms.

(vii) x and x² have different powers → Unlike terms.

(viii) 6 has no variable while 6x has x → Unlike terms.

(ix) 2xy and 3xyz have different variables → Unlike terms.

Hence, the pairs containing like terms are (i), (iv) and (vi).

An algebraic expression is a polynomial only when the powers of the variable are non-negative integers.

(i) $\frac{2}{3}x^2 - \frac{5}{7}x - \frac{1}{4}$

Since the highest power of x in the expression is 2, the expression is a polynomial with degree 2.

(ii) $3x^2 + \frac{2}{x} - 3$

Power of x is negative, so it is not a polynomial.

It is not a polynomial.

(iii) $x^2 - \frac{2}{3}x^7 - 5x^8$

Since the highest power of x is 8, the expression is a polynomial with degree 8.

An expression with numbers only does not contain any variable.

(i) 2y + 3 contains the variable y → Not an expression with numbers only.

(ii) (7 × 20) − 8z contains the variable z → Not an expression with numbers only.

(iii) 5 × (21 − 7) + 9 × 2 contains only numbers → Expression with numbers only.

(iv) 5 − 11n contains the variable n → Not an expression with numbers only.

(v) (5 × 4) − 45 + p contains the variable p → Not an expression with numbers only.

(vi) 3 × (11 + 7) − 24 ÷ 2 contains only numbers → Expression with numbers only.

Hence, (iii) and (vi) are expressions with numbers only.

(i) 11 added to 2m = 2m + 11.

(ii) 11 subtracted from 2m = 2m − 11.

(iii) 3 added to 5 times y = 5y + 3.

(iv) 3 subtracted from 5 times y = 5y − 3.

(v) y multiplied by −8 = −8y. Then 5 added to it = −8y + 5.

(vi) y multiplied by 5 = 5y. Then 5y subtracted from 16 = 16 − 5y.

(i) Thrice the number x = 3x. 6 more than this = 3x + 6.

(ii) 7 taken away from y = y − 7.

(iii) Quotient of x by y = $\frac{x}{y}$, 3 less than this = $\frac{x}{y} - 3$.

Using one operation each between t and 4, six expressions are:

(i) t + 4

(ii) t − 4

(iii) 4 − t

(iv) 4t (= 4 × t)

(v) $\frac{t}{4}$

(vi) $\frac{4}{t}$

Marks scored = x

Marks deducted = 5

Final score = x − 5

Hence, the student's final score in English is (x − 5) marks.

Raju's present age = y years

3 times Raju's age = 3y years

2 more than 3 times Raju's age = 3y + 2

Hence, Raju's father's age is (3y + 2) years.

Mohini's present age = x years.

(i) Mohini's age next year = (x + 1) years.

Three times this = 3(x + 1) years.

(ii) Mohini's age 3 years ago = (x − 3) years.

Four times this = 4(x − 3) = (4x − 12) years.

(iii) Mohini's age 2 years from now = (x + 2) years.

Uncle's age = 5(x + 2) = (5x + 10) years.

Speed of the bus = v km/h

Time travelled = 5 hours

Distance covered in 5 hours = speed × time = 5v km

Remaining distance to Jaipur = 20 km

Total distance from Delhi to Jaipur = Distance covered + Remaining distance = 5v + 20

Hence, the distance from Delhi to Jaipur is (5v + 20) km.

(a) 2x + 5x = (2 + 5)x = 7x.

(b) 7p + (−2p) = (7 − 2)p = 5p.

(c) (2a + 4b) + (9a + 10b)

= (2a + 9a) + (4b + 10b)

= 11a + 14b.

(d) 2xy + 3yx = 2xy + 3xy = (2 + 3)xy = 5xy.

(e) $\frac{2}{3}x + 2x = \frac{2x + 6x}{3} = \frac{8}{3}x$

Hence, the sum is $\frac{8}{3}x$.

(a) 19y − 12y = (19 − 12)y = 7y.

(b) 5z − 6z = (5 − 6)z = −z.

(c) 20p − 10p = (20 − 10)p = 10p.

(d) 23m − 13m = (23 − 13)m = 10m.

(e) 50n − 48n = (50 − 48)n = 2n.

(a) 6x + (9x − 2x) = 6x + 7x = 13x.

(b) 11y + 2y + 3x = 13y + 3x = 13y + 3x.

(c) 11z + 3z + z = (11 + 3 + 1)z = 15z.

(d) 8ab + (−3ab) = (8 − 3)ab = 5ab.

(e) 3x − 4x + 7x = (3 − 4 + 7)x = 6x.

(f) 35b − (16b + 9b) = 35b − 25b = 10b.

(g) (9a − 3a) + 4a = 6a + 4a = 10a.

(a) 5 × b = 5b.

(b) x × y = xy.

(c) 2a × 3b = (2 × 3) × (a × b) = 6ab.

(d) a × a = a².

(e) 6a × 2a = (6 × 2) × (a × a) = 12a².

(a) $12x \div 2x = \frac{12x}{2x} = \frac{12}{2} \times \frac{x}{x}$ = 6.

(b) $24ab \div 8a = \frac{24ab}{8a} = \frac{24}{8} \times \frac{ab}{a}$ = 3b.

(c) $9x \div 3y = \frac{9x}{3y} = \frac{9}{3} \times \frac{x}{y}$ = 3 $\frac{x}{y}$.

(d) $25a^2 \div 5a = \frac{25a^2}{5a} = \frac{25}{5} \times \frac{a^2}{a}$ = 5a.

(e) $49xy \div 7xy = \frac{49xy}{7xy} = \frac{49}{7} \times \frac{xy}{xy}$ = 7.

(i) Substituting x = 3 and y = 2 in the given expression, we get:

3x + 2y

= 3(3) + 2(2)

= 9 + 4

∴ 3x + 2y = 13

(ii) Substituting x = 2 and y = 5 in the given expression, we get:

5x - 3y

= 5(2) - 3(5)

= 10 - 15

∴ 5x - 3y = -5

(iii) Substituting a = 2, b = 3 and c = 1 in the given expression, we get:

a + 2b - 5c

= 2 + 2(3) - 5(1)

= 2 + 6 - 5

∴ a + 2b - 5c = 3

(iv) Substituting p = 1, q = 2 and r = 3 in the given expression, we get:

2p + 3q + 4r + pqr

= 2(1) + 3(2) + 4(3) + (1)(2)(3)

= 2 + 6 + 12 + 6

∴ 2p + 3q + 4r + pqr = 26

(v) Substituting a = 4, b = 5 and c = 2 in the given expression, we get:

3ab + 4bc - 5ca

= 3(4)(5) + 4(5)(2) - 5(2)(4)

= 60 + 40 - 40

∴ 3ab + 4bc - 5ca = 60

(i) Substituting x = 2 in the given expression, we get:

2x² - 3x + 4

= 2(2)² - 3(2) + 4

= 2(4) - 6 + 4

= 8 - 6 + 4

∴ 2x² - 3x + 4 = 6

(ii) Substituting x = 3 in the given expression, we get:

4x³ - 5x² - 6x + 7

= 4(3)³ - 5(3)² - 6(3) + 7

= 4(27) - 5(9) - 18 + 7

= 108 - 45 - 18 + 7

∴ 4x³ - 5x² - 6x + 7 = 52

(iii) Substituting x = 4 in the given expression, we get:

3x³ + 9x² - x + 8

= 3(4)³ + 9(4)² - 4 + 8

= 3(64) + 9(16) - 4 + 8

= 192 + 144 - 4 + 8

∴ 3x³ + 9x² - x + 8 = 340

(iv) Substituting x = 2 in the given expression, we get:

2x⁴ - 5x³ + 7x - 3

= 2(2)⁴ - 5(2)³ + 7(2) - 3

= 2(16) - 5(8) + 14 - 3

= 32 - 40 + 14 - 3

∴ 2x⁴ - 5x³ + 7x - 3 = 3

(i) Substituting x = 5 in the given expression, we get:

6 - 7x²

= 6 - 7(5)²

= 6 - 7(25)

= 6 - 175

∴ 6 - 7x² = -169

(ii) Substituting x = 5 in the given expression, we get:

3x² + 8x - 10

= 3(5)² + 8(5) - 10

= 3(25) + 40 - 10

= 75 + 40 - 10

∴ 3x² + 8x - 10 = 105

(iii) Substituting x = 5 in the given expression, we get:

2x³ - 4x² - 6x + 25

= 2(5)³ - 4(5)² - 6(5) + 25

= 2(125) - 4(25) - 30 + 25

= 250 - 100 - 30 + 25

∴ 2x³ - 4x² - 6x + 25 = 145

(i) Substituting x = 2 and y = 3 in the given expression, we get:

x ÷ y

= 2 ÷ 3

∴ x ÷ y = $\frac{2}{3}$

(ii) Substituting x = 2, y = 3 and z = 1 in the given expression, we get:

$\frac{xy}{z}$

= $\frac{(2)(3)}{1}$

= $\frac{6}{1}$

∴ $\frac{xy}{z}$ = 6

(iii) Substituting x = 2, y = 3 and z = 1 in the given expression, we get:

$$\begin{aligned}\Rightarrow \frac{2x + 3y - 4z}{3x - z} \\[1em] = \frac{2(2) + 3(3) - 4(1)}{3(2) - 1} \\[1em] = \frac{4 + 9 - 4}{6 - 1} \\[1em] = \frac{9}{5} = 1\frac{4}{5}\end{aligned}$$

∴ $\frac{2x + 3y - 4z}{3x - z}$ = $\frac{9}{5} = 1\frac{4}{5}$

Substituting a = 2, b = 3 and c = 1 in the given expression, we get:

a² + b² + c² - 2ab - 2bc - 2ca + 3abc

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

= 4 + 9 + 1 - 12 - 6 - 4 + 18

= 14 - 22 + 18

= -8 + 18

∴ a² + b² + c² - 2ab - 2bc - 2ca + 3abc = 10

Substituting p = 4, q = -3 and r = 2 in the given expression, we get:

p³ + q³ - r³ - 3pqr

= (4)³ + (-3)³ - (2)³ - 3(4)(-3)(2)

= 64 + (-27) - 8 - (-72)

= 64 - 27 - 8 + 72

= 136 - 35

∴ p³ + q³ - r³ - 3pqr = 101.

Substituting m = 1, n = 2 and p = 3 in the given expression, we get:

2mn⁴ - 15m²n - p

= 2(1)(2)⁴ - 15(1)²(2) - 3

= 2(1)(16) - 15(1)(2) - 3

= 32 - 30 - 3

∴ 2mn⁴ - 15m²n - p = -1

(i) Substituting x = 5 and y = 2 in the given result, we get:

LHS = (x + y)²

= (5 + 2)²

= (7)²

= 49

RHS = x² + 2xy + y²

= (5)² + 2(5)(2) + (2)²

= 25 + 20 + 4

= 49

Since, LHS = RHS.

Hence, (x + y)² = x² + 2xy + y² is verified for x = 5 and y = 2.

(ii) Substituting x = 5 and y = 2 in the given result, we get:

LHS = (x - y)²

= (5 - 2)²

= (3)²

= 9

RHS = x² - 2xy + y²

= (5)² - 2(5)(2) + (2)²

= 25 - 20 + 4

= 9

Since, LHS = RHS.

Hence, (x - y)² = x² - 2xy + y² is verified for x = 5 and y = 2.

(iii) Substituting x = 5 and y = 2 in the given result, we get:

LHS = x² - y²

= (5)² - (2)²

= 25 - 4

= 21

RHS = (x + y)(x - y)

= (5 + 2)(5 - 2)

= 7 × 3

= 21

Since, LHS = RHS.

Hence, x² - y² = (x + y)(x - y) is verified for x = 5 and y = 2.

(iv) Substituting x = 5 and y = 2 in the given result, we get:

LHS = (x + y)²

= (5 + 2)²

= (7)²

= 49

RHS = (x - y)² + 4xy

= (5 - 2)² + 4(5)(2)

= (3)² + 40

= 9 + 40

= 49

Since, LHS = RHS.

Hence, (x + y)² = (x - y)² + 4xy is verified for x = 5 and y = 2.

(v) Substituting x = 5 and y = 2 in the given result, we get:

LHS = (x + y)³

= (5 + 2)³

= (7)³

= 343

RHS = x³ + y³ + 3x²y + 3xy²

= (5)³ + (2)³ + 3(5)²(2) + 3(5)(2)²

= 125 + 8 + 3(25)(2) + 3(5)(4)

= 125 + 8 + 150 + 60

= 343

Since, LHS = RHS.

Hence, (x + y)³ = x³ + y³ + 3x²y + 3xy² is verified for x = 5 and y = 2.

(i) 17 + x = 5 is an equation with variable x.

(ii) 2b − 3 = 7 is an equation with variable b.

(iii) (y - 7) > 5 is not an equation (it is an inequality).

(iv) $\frac{9}{3} = 3$ is not an equation with variable.

(v) 7 × 3 − 19 = 2 is not an equation with a variable.

(vi) 5 × 4 − 8 = 3t is an equation with variable t.

(vii) 2p < 15 is not an equation (it is an inequality).

(viii) 7 = 11 × 5 − 12 × 4 is not an equation with a variable.

(ix) $\frac{3}{2}q = 5$ is an equation with variable q.

(i) Given x + 6 = 8

⇒ x = 8 − 6     [Transposing +6 to RHS]

∴ x = 2

(ii) Given x − 2 = −5

⇒ x = −5 + 2     [Transposing −2 to RHS]

∴ x = −3

(iii) Given 4x = 6

⇒ $x = \frac{6}{4}$     [Dividing both sides by 4]

∴ x = $\frac{3}{2}$

(iv) Given $\frac{x}{2} = 5$

⇒ x = 5 × 2

∴ x = 10

(v) Given 2y − 3 = 2

⇒ 2y = 2 + 3     [Transposing −3 to RHS]

⇒ 2y = 5

⇒ $y = \frac{5}{2}$     [Dividing both sides by 2]

∴ y = $\frac{5}{2}$

(vi) Given 5y + 2 = 4

⇒ 5y = 4 − 2     [Transposing +2 to RHS]

⇒ 5y = 2

⇒ $y = \frac{2}{5}$     [Dividing both sides by 5]

∴ y = $\frac{2}{5}$

(i) We have:

5(x + 1) = 25

⇒ x + 1 = 5     [Dividing both sides by 5]

⇒ x = 5 − 1     [Transposing +1 to RHS]

∴ x = 4.

(ii) We have:

2(3x − 1) = 10

⇒ 3x − 1 = 5     [Dividing both sides by 2]

⇒ 3x = 5 + 1     [Transposing −1 to RHS]

⇒ 3x = 6

⇒ $x = \frac{6}{3}$

∴ x = 2.

(iii) We have:

$\frac{3x - 1}{4} = 11$

⇒ 3x − 1 = 11 × 4     [Multiplying both sides by 4]

⇒ 3x − 1 = 44

⇒ 3x = 44 + 1 = 45

⇒ x = $\frac{45}{3}$

⇒ x = 15

∴ x = 15.

(i) We have:

5x − 6 = 12 − x

⇒ 5x + x = 12 + 6     [Transposing −x to LHS and −6 to RHS]

⇒ 6x = 18

⇒ $x = \frac{18}{6}$

∴ x = 3.

(ii) We have:

$$\begin{aligned}\Rightarrow \frac{n}{3} + 1 = 4 - n \\[1em] \Rightarrow \frac{n}{3} + n = 4 - 1 \\[1em] \Rightarrow \frac{n + 3n}{3} = 3 \\[1em] \Rightarrow \frac{4n}{3} = 3 \\[1em] \Rightarrow 4n = 9 \\[1em] \Rightarrow n = \frac{9}{4} = 2\frac{1}{4}\end{aligned}$$

∴ n = $2\frac{1}{4}$

(iii) We have:

5p + 7 = 19 − 2p

⇒ 5p + 2p = 19 − 7     [Transposing −2p to LHS and +7 to RHS]

⇒ 7p = 12

⇒ $p = \frac{12}{7} = 1\frac{5}{7}$

∴ p = $1\frac{5}{7}$.

(iv) We have:

$$\begin{aligned}\Rightarrow 2x + \frac{2}{3} = \frac{5}{2} - x\\[1em] \Rightarrow 2x + x = \frac{5}{2} - \frac{2}{3}\\[1em] \Rightarrow 3x = \frac{15 - 4}{6} \\[1em] \Rightarrow 3x = \frac{11}{6} \\[1em] \Rightarrow x = \frac{11}{18}\end{aligned}$$

∴ x = $\frac{11}{18}$.

(v) We have:

$$\begin{aligned}\Rightarrow \frac{x}{2} - 5 = \frac{x}{3} - 4 \\[1em] \Rightarrow \frac{x}{2} - \frac{x}{3} = -4 + 5 \\[1em] \Rightarrow \frac{3x - 2x}{6} = 1 \\[1em] \Rightarrow \frac{x}{6} = 1 \\[1em] \Rightarrow x = 6\end{aligned}$$

∴ x = 6.

(vi) We have:

$$\begin{aligned}\Rightarrow 18 - \frac{3y}{4} = 11 + y\\[1em] \Rightarrow 18 - 11 = y + \frac{3y}{4} \\[1em] \Rightarrow 7 = \frac{4y + 3y}{4} \\[1em] \Rightarrow 7 = \frac{7y}{4} \\[1em] \Rightarrow 7y = 28 \\[1em] \Rightarrow y = \frac{28}{7} = 4\end{aligned}$$

∴ y = 4.

(i) In algebra, we use letters (literals) to represent variables.

(ii) A symbol or letter which can be given various numerical values is called a variable.

(iii) Jaggu's age 7 years from now = (x + 7) years.

(iv) Cost of 9 pens at ₹x each = ₹9x.

(v) An equation is a statement that the two expressions are equal.

(vi) Numerical coefficient of −5x²y³ is −5.

(vii) Thrice y = 3y. 7 less than this = 3y − 7.

(viii) 3x + 4 = 19 ⇒ 3x = 15 ⇒ x = 5.

(ix) Number of pencils = $\frac{\text{Total money}}{\text{Cost per pencil}}$ = $\frac{x}{2}$.

(x) 3p²q and −5pq² have different powers of p and q, so they are unlike terms.

(xi) Value of 3 − 5x at x = 4 = 3 − 5(4) = 3 − 20 = −17.

(i) True. The value of 5x depends on the value of x, so 5x is a variable.

(ii) True. The value of y − 5 depends on the value of y, so y − 5 is a variable.

(iii) False. The number of angles in any triangle is always 3, which is a constant.

(iv) True. The value of an expression depends on the value of the variable.

(v) False. If breadth is b, then length = 2b and area = 2b², which depends on b. So the area is a variable.

(vi) True. A linear equation in one variable has exactly one solution.

(vii) False. Each child gets $\frac{x}{5}$ toffees, not 5x.

(viii) True. Since 1 minute = 60 seconds, t minutes = 60t seconds.

(ix) True. The additive inverse of a negative integer is positive.

(x) False. At x = 5, 3x + 2 = 3(5) + 2 = 17, which is not equal to 13.

(xi) False. 2y - 7 > 13, it is an inequality, not an equation.

(xii) False. The correct expression is $\frac{x}{3} + x = 8$ (or $x + \frac{x}{3} = 8$).

(xiii) False. The difference between the ages of two persons remains the same throughout their lives, so it is a constant.

Adding 5 to x gives (x + 5). Multiplying the result by 2 gives 2(x + 5). This equals 24.

So, (x + 5) × 2 = 24.

Hence, option 2 is the correct option.

An equation is a statement having an equality sign (=) between two expressions.

1. x + 5 is just an expression (no equality sign).

2. 7x is just an expression.

3. 2y + 3 = 11 has an equality sign — it is an equation.

4. 2p < 7 is an inequality.

Hence, option 3 is the correct option.

Matchsticks in 1 box = 48.

Matchsticks in n boxes = 48 × n.

Hence, option 4 is the correct option.

A regular hexagon has 6 equal sides. If perimeter = x, then:

Length of each side = $\frac{x}{6}$ = (x ÷ 6) metres.

Hence, option 3 is the correct option.

Substitute x = 3 in each equation:

1. 3 + 7 = 10 ≠ 4.

2. 3 + 10 = 13 ≠ 7.

3. 3 + 7 = 10. ✓

4. 3 + 3 = 6 ≠ 7.

Hence, option 3 is the correct option.

We have:

3x − 2 = 10

⇒ 3x = 10 + 2 = 12

⇒ x = $\frac{12}{3}$ = 4

Hence, option 4 is the correct option.

In 5x + $\frac{5}{x}$:

• $\frac{5}{x}$ involves division.
• 5x involves multiplication.
• 5x + $\frac{5}{x}$ involves addition.
• There is no subtraction.

Hence, option 2 is the correct option.

Quotient of x by 3 = $\frac{x}{3}$.

Adding 7 to this = $\frac{x}{3} + 7$.

Hence, option 1 is the correct option.

Persons that can be seated in 1 row = x.

Persons that can be seated in 8 rows = 8 × x = 8x.

Hence, option 3 is the correct option.

Daily savings = earnings − expenditure = ₹(x − y).

Number of days in March = 31.

Total savings in March = 31 × (x − y) = ₹31(x − y).

Hence, option 2 is the correct option.

Breadth = x units.

Length = 3 × breadth = 3x units.

Perimeter = 2(length + breadth) = 2(3x + x) = 2 × 4x = 8x units.

Hence, option 3 is the correct option.

Total amount spent = 800 + 600 + 500 = ₹1900.

Amount left after spending = x − 1900.

Money received as gift = ₹200.

Final amount = (x − 1900) + 200 = x − 1700.

Hence, option 1 is the correct option.

The commutative property of addition states that for any two integers a and b, a + b = b + a.

Hence, option 2 is the correct option.

Let us simplify the given expression:

$$\begin{aligned}\Rightarrow 2x^2y − 3y^2z − 5yx^2 + yz(x + y) + 7 \\[1em] = 2x^2y − 5yx^2 − 3y^2z + yzx + y^2z + 7 \\[1em] = 2x^2y − 5yx^2 − 3y^2z + y^2z + yzx + 7 \\[1em] = -3x^2y - 2y^2z + xyz + 7\end{aligned}$$

The simplified expression has 4 unlike terms: -3x²y, -2y²z, xyz and 7.

Hence, option 2 is the correct option.

Substituting x = 1 in 3x² − 5x + 3 :

= 3(1)² − 5(1) + 3

= 3 − 5 + 3

= 1.

Hence, option 3 is the correct option.

Statement I: The terms of x + 2y + y = x + 3y are x and 3y. So, the expression has 2 terms. Hence, Statement I is false.

Statement II: 3x + 4 = 0 contains the variable x with highest power 1, so it is a linear equation. Hence, Statement II is true.

Hence, option 2 is the correct option.

Statement I: Double of John's age = 2x. Five less than this = 2x − 5. Hence, Statement I is true.

Statement II: 3x × 4y = 12xy, which is a single term (monomial). Hence, Statement II is true.

Hence, option 3 is the correct option.

Statement I: Substituting x = 2 and y = 3 in 4y + 3x:

= 4(3) + 3(2)

= 12 + 6

= 18

The value of the expression is 18, not 22. Hence, Statement I is false.

Statement II: By definition, an algebraic expression is a combination of terms connected by mathematical operations such as addition or subtraction. Hence, Statement II is true.

Hence, option 2 is the correct option.

Each polygon (house-shape) is formed using 5 matchsticks. Two neighbouring polygons have one common matchstick. So, for every new polygon added after the first, 4 more matchsticks are required.

Number of polygons	1	2	3	4	5	...	n
Number of matchsticks	5	9	13	17	21	...	4n + 1

Observe that:

5 = 4 × 1 + 1

9 = 4 × 2 + 1

13 = 4 × 3 + 1

17 = 4 × 4 + 1

21 = 4 × 5 + 1

Hence, the general rule that gives the number of matchsticks is 4n + 1, where n is the number of polygons.

(i) The given expression $\frac{3}{2}x^2y - \frac{1}{2}xy^2 + 6x^2y^2$ has 3 terms.

Since the expression has three terms, it is called a trinomial.

(ii) The terms of the expression are:

$\frac{3}{2}x^2y, -\frac{1}{2}xy^2$ and $6x^2y^2$

(iii) In the term $-\frac{1}{2}xy^2$:

Numerical coefficient = $-\frac{1}{2}$

Literal coefficient = xy²

(iv) In the term $-\frac{1}{2}xy^2$, the coefficient of x is the remaining factor after removing x from the term.

$-\frac{1}{2}xy^2 = -\frac{1}{2} \times x \times y^2$

Hence, the coefficient of x is $-\frac{1}{2}y^2$.

(i) Cost of 1 metre cloth = ₹ x

∴ Cost of 6 metre cloth = ₹ (6 × x)

∴ Cost of 6 metre cloth = ₹ 6x

(ii) Cost of 1 notebook = ₹ x

Cost of 5 notebooks = ₹ 5x

Cost of 1 book = ₹ y

Cost of 2 books = ₹ 2y

∴ Total cost of 5 notebooks and 2 books = ₹ (5x + 2y)

∴ The cost of 5 notebooks and 2 books = ₹ (5x + 2y)

(iii) Ragini's score in English = x marks

Two-third of her score in English = $\frac{2}{3}x$

Score in Mathematics = 23 more than two-third of her score in English

∴ Ragini's score in Mathematics = $\left(\frac{2}{3}x + 23\right)$ marks

(iv) Length of one side of the regular pentagon = x cm

A regular pentagon has 5 equal sides.

Perimeter of the pentagon = 5 × length of one side

= 5 × x

∴ Perimeter of the regular pentagon = 5x cm

Given, x = 4 and y = 2.

(i) Substituting x = 4 and y = 2 in x + y, we get:

x + y = 4 + 2

∴ x + y = 6

(ii) Substituting x = 4 and y = 2 in x - y, we get:

x - y = 4 - 2

∴ x - y = 2

(iii) Substituting x = 4 in x² + 2, we get:

x² + 2 = (4)² + 2

= 4 × 4 + 2

= 16 + 2

∴ x² + 2 = 18

(iv) Substituting x = 4 and y = 2 in x² - 2xy + y², we get:

x² - 2xy + y² = (4)² - 2(4)(2) + (2)²

= 16 - 16 + 4

= 4

∴ x² - 2xy + y² = 4

Given, a = 3, b = 0 and c = 4.

(i) Substituting a = 3, b = 0 and c = 4 in ab + 2bc + 3ca + 4abc, we get:

ab + 2bc + 3ca + 4abc

= (3)(0) + 2(0)(4) + 3(4)(3) + 4(3)(0)(4)

= 0 + 0 + 36 + 0

= 36.

∴ ab + 2bc + 3ca + 4abc = 36.

(ii) Substituting a = 3, b = 0 and c = 4 in a³ + b³ + c³ - 3abc, we get:

a³ + b³ + c³ - 3abc

= (3)³ + (0)³ + (4)³ - 3(3)(0)(4)

= 27 + 0 + 64 - 0

= 91.

∴ a³ + b³ + c³ - 3abc = 91.

(i) We have:

$$\begin{aligned}\phantom{=} 2x - 1\frac{1}{2} = 4\frac{1}{2} \\[1em] \Rightarrow 2x - \frac{3}{2} = \frac{9}{2} \\[1em] \Rightarrow 2x = \frac{9}{2} + \frac{3}{2} \quad\text{[Transposing } -\frac{3}{2} \text{ to RHS]} \\[1em] \Rightarrow 2x = \frac{9 + 3}{2} \\[1em] \Rightarrow 2x = \frac{12}{2} \\[1em] \Rightarrow 2x = 6 \\[1em] \Rightarrow x = \frac{6}{2}\end{aligned}$$

Hence, x = 3.

(ii) We have:

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

⇒ 3y - 3 = 2y + 2 $\quad$[Removing brackets]

⇒ 3y - 2y = 2 + 3 $\quad$[Transposing -3 to RHS and +2y to LHS]

⇒ y = 5

Hence, y = 5.

(iii) We have:

n - 3 = 5n - 5

⇒ n - 5n = -5 + 3 $\quad$[Transposing -3 to RHS and +5n to LHS]

⇒ -4n = -2

⇒ n = $\frac{-2}{-4}$

⇒ n = $\frac{1}{2}$

Hence, n = $\frac{1}{2}$.

(iv) We have:

$$\begin{aligned}\phantom{=} \frac{1}{3}(7x - 1) = \frac{1}{4} \\[1em] \Rightarrow 7x - 1 = \frac{1}{4} \times 3 \quad\text{[Multiplying both sides by 3]} \\[1em] \Rightarrow 7x - 1 = \frac{3}{4} \\[1em] \Rightarrow 7x = \frac{3}{4} + 1 \quad\text{[Transposing -1 to RHS]} \\[1em] \Rightarrow 7x = \frac{3 + 4}{4} \\[1em] \Rightarrow 7x = \frac{7}{4} \\[1em] \Rightarrow x = \frac{7}{4} \times \frac{1}{7}\end{aligned}$$

Hence, x = $\frac{1}{4}$.