Chapter 4 Simple Equations

Solution:

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(i) "The sum of three times x and 11 is 32."

"Three times x" can be written as $3 \times x$ or $3x$.

"The sum of $3x$ and 11" is $3x + 11$.

"is 32" means it is equal to 32.

The equation form is $\mathbf{3x + 11 = 32}$.

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(ii) "If you subtract 5 from 6 times a number, you get 7."

Let the number be represented by a variable, say $y$.

"6 times a number" is $6 \times y$ or $6y$.

"Subtract 5 from 6 times a number" means $6y - 5$.

"you get 7" means it is equal to 7.

The equation form is $\mathbf{6y - 5 = 7}$.

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(iii) "One fourth of m is 3 more than 7."

"One fourth of m" is $\frac{1}{4} \times m$ or $\frac{m}{4}$.

"3 more than 7" is $7 + 3 = 10$.

"is" means it is equal to.

The equation form is $\mathbf{\frac{m}{4} = 7 + 3}$ or $\mathbf{\frac{m}{4} = 10}$.

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(iv) "One third of a number plus 5 is 8."

Let the number be represented by a variable, say $n$.

"One third of a number" is $\frac{1}{3} \times n$ or $\frac{n}{3}$.

"plus 5" means adding 5: $\frac{n}{3} + 5$.

"is 8" means it is equal to 8.

The equation form is $\mathbf{\frac{n}{3} + 5 = 8}$.

Solution:

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(i) $x - 5 = 9$

This equation shows that 5 is subtracted from $x$, and the result is 9.

Statement form: 5 subtracted from x is 9.

(Other possible statements: The difference between x and 5 is 9. Take away 5 from x and you get 9.)

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(ii) $5p = 20$

This equation shows that 5 multiplied by $p$ is equal to 20.

"5 multiplied by p" can be expressed as "5 times p" or "the product of 5 and p".

Statement form: Five times a number p is 20.

(Other possible statements: The product of 5 and p is 20.)

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(iii) $3n + 7 = 1$

This equation shows that 7 is added to 3 times $n$, and the result is 1.

"3 times n" is $3n$.

"7 added to 3 times n" is $3n + 7$.

"is equal to 1" means the result is 1.

Statement form: Seven added to three times a number n is 1.

(Other possible statements: The sum of three times n and 7 is 1.)

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(iv) $\frac{m}{5} - 2 = 6$

This equation shows that 2 is subtracted from $m$ divided by 5, and the result is 6.

"m divided by 5" can be expressed as "m divided by 5" or "one-fifth of m".

"2 subtracted from m divided by 5" is $\frac{m}{5} - 2$.

"is equal to 6" means the result is 6.

Statement form: Two subtracted from m divided by 5 is 6.

(Other possible statements: Two less than one-fifth of m is 6.)

Solution:

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Given information:

Raju's father's age is 5 years more than three times Raju's age.

Raju's father is 44 years old.

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To Find:

Set up an equation to find Raju's age.

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Solution:

Let's represent Raju's age by a variable. Let Raju's age be $r$ years.

"Three times Raju's age" can be written as $3 \times r$ or $3r$.

"5 years more than three times Raju's age" means $3r + 5$.

This expression represents Raju's father's age.

We are also given that Raju's father is 44 years old.

So, we can set the expression for the father's age equal to his actual age:

Raju's father's age = $3r + 5$

Raju's father's age = 44

Therefore, the equation is:

$3r + 5 = 44$

The equation to find Raju's age is $\mathbf{3r + 5 = 44}$.

(To find Raju's age, we would solve this equation for $r$. Subtract 5 from both sides: $3r = 44 - 5 = 39$. Divide both sides by 3: $r = \frac{39}{3} = 13$. So, Raju's age is 13 years.)

Solution:

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Given information:

A large box contains the number of mangoes equal to 8 small boxes plus 4 loose mangoes.

The number of mangoes in a large box is 100.

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To Find:

Set up an equation to find the number of mangoes in each small box.

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Solution:

Let's represent the number of mangoes in one small box by a variable. Let the number of mangoes in each small box be $s$.

The number of mangoes in 8 small boxes is $8 \times s$ or $8s$.

A large box contains 8 small boxes plus 4 loose mangoes. So, the total number of mangoes in a large box can be expressed as $8s + 4$.

We are given that the number of mangoes in a large box is 100.

So, we can set the expression for the number of mangoes in a large box equal to the given number:

Number of mangoes in a large box = $8s + 4$

Number of mangoes in a large box = 100

Therefore, the equation is:

$8s + 4 = 100$

The equation which gives the number of mangoes in each small box is $\mathbf{8s + 4 = 100}$.

(To find the number of mangoes in a small box, we would solve this equation for $s$. Subtract 4 from both sides: $8s = 100 - 4 = 96$. Divide both sides by 8: $s = \frac{96}{8} = 12$. So, each small box contains 12 mangoes.)

Solution:

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To complete the table, we need to substitute the given value of the variable into the equation and check if the left-hand side (LHS) is equal to the right-hand side (RHS). If LHS = RHS, the equation is satisfied (Yes); otherwise, it is not satisfied (No).

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(i) Equation: $x + 3 = 0$, Value: $x = 3$

LHS = $3 + 3 = 6$. RHS = 0.

Since $6 \neq 0$, the equation is not satisfied.

Answer: No

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(ii) Equation: $x + 3 = 0$, Value: $x = 0$

LHS = $0 + 3 = 3$. RHS = 0.

Since $3 \neq 0$, the equation is not satisfied.

Answer: No

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(iii) Equation: $x + 3 = 0$, Value: $x = -3$

LHS = $-3 + 3 = 0$. RHS = 0.

Since $0 = 0$, the equation is satisfied.

Answer: Yes

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(iv) Equation: $x - 7 = 1$, Value: $x = 7$

LHS = $7 - 7 = 0$. RHS = 1.

Since $0 \neq 1$, the equation is not satisfied.

Answer: No

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(v) Equation: $x - 7 = 1$, Value: $x = 8$

LHS = $8 - 7 = 1$. RHS = 1.

Since $1 = 1$, the equation is satisfied.

Answer: Yes

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(vi) Equation: $5x = 25$, Value: $x = 0$

LHS = $5 \times 0 = 0$. RHS = 25.

Since $0 \neq 25$, the equation is not satisfied.

Answer: No

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(vii) Equation: $5x = 25$, Value: $x = 5$

LHS = $5 \times 5 = 25$. RHS = 25.

Since $25 = 25$, the equation is satisfied.

Answer: Yes

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(viii) Equation: $5x = 25$, Value: $x = -5$

LHS = $5 \times (-5) = -25$. RHS = 25.

Since $-25 \neq 25$, the equation is not satisfied.

Answer: No

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(ix) Equation: $\frac{m}{3} = 2$, Value: $m = -6$

LHS = $\frac{-6}{3} = -2$. RHS = 2.

Since $-2 \neq 2$, the equation is not satisfied.

Answer: No

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(x) Equation: $\frac{m}{3} = 2$, Value: $m = 0$

LHS = $\frac{0}{3} = 0$. RHS = 2.

Since $0 \neq 2$, the equation is not satisfied.

Answer: No

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(xi) Equation: $\frac{m}{3} = 2$, Value: $m = 6$

LHS = $\frac{6}{3} = 2$. RHS = 2.

Since $2 = 2$, the equation is satisfied.

Answer: Yes

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Completed Table:

S. No.	Equation	Value	Say, whether the Equation is Satisfied. (Yes/ No)
(i)	$x + 3 = 0$	$x = 3$	No
(ii)	$x + 3 = 0$	$x = 0$	No
(iii)	$x + 3 = 0$	$x = -3$	Yes
(iv)	$x - 7 = 1$	$x = 7$	No
(v)	$x - 7 = 1$	$x = 8$	Yes
(vi)	$5x = 25$	$x = 0$	No
(vii)	$5x = 25$	$x = 5$	Yes
(viii)	$5x = 25$	$x = -5$	No
(ix)	$\frac{m}{3} = 2$	$m = -6$	No
(x)	$\frac{m}{3} = 2$	$m = 0$	No
(xi)	$\frac{m}{3} = 2$	$m = 6$	Yes

Solution:

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To check if a given value is a solution to an equation, we substitute the value into the equation and verify if the left-hand side (LHS) equals the right-hand side (RHS).

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(a) Equation: $n + 5 = 19$, Value: $n = 1$

Substitute $n = 1$ into the LHS:

LHS = $1 + 5 = 6$.

RHS = 19.

Since LHS ($6$) $\neq$ RHS ($19$), the value $n = 1$ is not a solution.

Answer: No

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(b) Equation: $7n + 5 = 19$, Value: $n = – 2$

Substitute $n = -2$ into the LHS:

LHS = $7 \times (-2) + 5 = -14 + 5 = -9$.

RHS = 19.

Since LHS ($-9$) $\neq$ RHS ($19$), the value $n = -2$ is not a solution.

Answer: No

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(c) Equation: $7n + 5 = 19$, Value: $n = 2$

Substitute $n = 2$ into the LHS:

LHS = $7 \times 2 + 5 = 14 + 5 = 19$.

RHS = 19.

Since LHS ($19$) = RHS ($19$), the value $n = 2$ is a solution.

Answer: Yes

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(d) Equation: $4p – 3 = 13$, Value: $p = 1$

Substitute $p = 1$ into the LHS:

LHS = $4 \times 1 - 3 = 4 - 3 = 1$.

RHS = 13.

Since LHS ($1$) $\neq$ RHS ($13$), the value $p = 1$ is not a solution.

Answer: No

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(e) Equation: $4p – 3 = 13$, Value: $p = – 4$

Substitute $p = -4$ into the LHS:

LHS = $4 \times (-4) - 3 = -16 - 3 = -19$.

RHS = 13.

Since LHS ($-19$) $\neq$ RHS ($13$), the value $p = -4$ is not a solution.

Answer: No

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(f) Equation: $4p – 3 = 13$, Value: $p = 0$

Substitute $p = 0$ into the LHS:

LHS = $4 \times 0 - 3 = 0 - 3 = -3$.

RHS = 13.

Since LHS ($-3$) $\neq$ RHS ($13$), the value $p = 0$ is not a solution.

Answer: No

Solution:

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The trial and error method involves substituting different values for the variable until the left-hand side (LHS) of the equation equals the right-hand side (RHS).

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(i) Equation: $5p + 2 = 17$

We need to find a value of $p$ for which $5p + 2 = 17$.

Let's try some values for $p$:

• If $p = 0$, LHS = $5(0) + 2 = 0 + 2 = 2$. $2 \neq 17$. (Too small)
• If $p = 1$, LHS = $5(1) + 2 = 5 + 2 = 7$. $7 \neq 17$. (Too small)
• If $p = 2$, LHS = $5(2) + 2 = 10 + 2 = 12$. $12 \neq 17$. (Still too small)
• If $p = 3$, LHS = $5(3) + 2 = 15 + 2 = 17$. $17 = 17$. (Correct!)

When $p = 3$, the equation $5p + 2 = 17$ is satisfied.

The solution is $\mathbf{p = 3}$.

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(ii) Equation: $3m – 14 = 4$

We need to find a value of $m$ for which $3m - 14 = 4$.

Let's try some values for $m$:

• If $m = 1$, LHS = $3(1) - 14 = 3 - 14 = -11$. $-11 \neq 4$. (Too small)
• If $m = 2$, LHS = $3(2) - 14 = 6 - 14 = -8$. $-8 \neq 4$. (Too small)
• If $m = 3$, LHS = $3(3) - 14 = 9 - 14 = -5$. $-5 \neq 4$. (Too small)
• If $m = 4$, LHS = $3(4) - 14 = 12 - 14 = -2$. $-2 \neq 4$. (Too small)
• If $m = 5$, LHS = $3(5) - 14 = 15 - 14 = 1$. $1 \neq 4$. (Still too small)
• If $m = 6$, LHS = $3(6) - 14 = 18 - 14 = 4$. $4 = 4$. (Correct!)

When $m = 6$, the equation $3m – 14 = 4$ is satisfied.

The solution is $\mathbf{m = 6}$.

Solution:

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(i) "The sum of numbers x and 4 is 9."

The sum of $x$ and 4 is $x + 4$. "is 9" means equal to 9.

Equation: $\mathbf{x + 4 = 9}$.

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(ii) "2 subtracted from y is 8."

Subtracting 2 from $y$ is $y - 2$. "is 8" means equal to 8.

Equation: $\mathbf{y - 2 = 8}$.

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(iii) "Ten times a is 70."

Ten times $a$ is $10 \times a$ or $10a$. "is 70" means equal to 70.

Equation: $\mathbf{10a = 70}$.

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(iv) "The number b divided by 5 gives 6."

The number $b$ divided by 5 is $\frac{b}{5}$. "gives 6" means equal to 6.

Equation: $\mathbf{\frac{b}{5} = 6}$.

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(v) "Three-fourth of t is 15."

Three-fourth of $t$ is $\frac{3}{4} \times t$ or $\frac{3t}{4}$. "is 15" means equal to 15.

Equation: $\mathbf{\frac{3t}{4} = 15}$.

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(vi) "Seven times m plus 7 gets you 77."

Seven times $m$ is $7m$. "plus 7" is $7m + 7$. "gets you 77" means equal to 77.

Equation: $\mathbf{7m + 7 = 77}$.

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(vii) "One-fourth of a number x minus 4 gives 4."

One-fourth of a number $x$ is $\frac{1}{4}x$ or $\frac{x}{4}$. "minus 4" is $\frac{x}{4} - 4$. "gives 4" means equal to 4.

Equation: $\mathbf{\frac{x}{4} - 4 = 4}$.

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(viii) "If you take away 6 from 6 times y, you get 60."

Six times $y$ is $6y$. "Take away 6 from 6 times y" is $6y - 6$. "you get 60" means equal to 60.

Equation: $\mathbf{6y - 6 = 60}$.

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(ix) "If you add 3 to one-third of z, you get 30."

One-third of $z$ is $\frac{1}{3}z$ or $\frac{z}{3}$. "Add 3 to one-third of z" is $\frac{z}{3} + 3$. "you get 30" means equal to 30.

Equation: $\mathbf{\frac{z}{3} + 3 = 30}$.

Solution:

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We convert each equation into a verbal statement.

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(i) $p + 4 = 15$

Statement: The sum of a number p and 4 is 15.

(or: 4 added to a number p is 15.)

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(ii) $m – 7 = 3$

Statement: 7 subtracted from a number m is 3.

(or: The difference between a number m and 7 is 3.)

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(iii) $2m = 7$

Statement: Twice a number m is 7.

(or: 2 times a number m is 7; The product of 2 and a number m is 7.)

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(iv) $\frac{m}{5} = 3$

Statement: A number m divided by 5 is 3.

(or: One-fifth of a number m is 3.)

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(v) $\frac{3m}{5} = 6$

Statement: Three-fifths of a number m is 6.

(or: 3 times a number m divided by 5 is 6.)

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(vi) $3p + 4 = 25$

Statement: 4 added to three times a number p is 25.

(or: The sum of three times a number p and 4 is 25.)

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(vii) $4p – 2 = 18$

Statement: 2 subtracted from four times a number p is 18.

(or: Four times a number p minus 2 is 18.)

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(viii) $\frac{p}{2} + 2 = 8$

Statement: 2 added to half of a number p is 8.

(or: The sum of a number p divided by 2 and 2 is 8.)

Solution:

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(i) Irfan's marbles related to Parmit's marbles.

Let the number of Parmit's marbles be $m$ (as given).

"Five times the marbles Parmit has" is $5 \times m = 5m$.

"7 marbles more than five times the marbles Parmit has" is $5m + 7$. This is the number of marbles Irfan has.

Irfan has 37 marbles (given).

So, we set the expression for Irfan's marbles equal to 37.

Equation: $\mathbf{5m + 7 = 37}$.

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(ii) Laxmi's father's age related to Laxmi's age.

Let Laxmi's age be $y$ years (as given).

"Three times Laxmi's age" is $3 \times y = 3y$.

"4 years older than three times Laxmi's age" is $3y + 4$. This is the father's age.

Laxmi's father is 49 years old (given).

So, we set the expression for the father's age equal to 49.

Equation: $\mathbf{3y + 4 = 49}$.

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(iii) Highest marks related to lowest marks.

Let the lowest score be $l$ (as given).

"twice the lowest marks" is $2 \times l = 2l$.

"twice the lowest marks plus 7" is $2l + 7$. This is the highest score.

The highest score is 87 (given).

So, we set the expression for the highest score equal to 87.

Equation: $\mathbf{2l + 7 = 87}$.

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(iv) Angles in an isosceles triangle.

Let the base angle be $b$ in degrees (as given).

In an isosceles triangle, the two base angles are equal. So, both base angles are $b^\circ$.

The vertex angle is twice either base angle. So, the vertex angle is $2 \times b = 2b^\circ$.

The sum of the angles in a triangle is 180 degrees.

Sum of angles = Base angle 1 + Base angle 2 + Vertex angle = $180^\circ$.

$b^\circ + b^\circ + 2b^\circ = 180^\circ$

$4b^\circ = 180^\circ$

Equation: $\mathbf{4b = 180}$.

Solution:

To solve an equation, we perform operations on both sides of the equation to isolate the variable.

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(a) Equation: $3n + 7 = 25$

Our goal is to isolate $n$.

First, subtract 7 from both sides of the equation to remove the constant term from the LHS:

$3n = 18$

Now, divide both sides by the coefficient of $n$ (which is 3) to isolate $n$:

$n = 6$

To verify the solution, substitute $n = 6$ back into the original equation:

LHS = $3(6) + 7 = 18 + 7 = 25$.

RHS = 25.

Since LHS = RHS, the solution is correct.

The solution is $\mathbf{n = 6}$.

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(b) Equation: $2p – 1 = 23$

Our goal is to isolate $p$.

First, add 1 to both sides of the equation to remove the constant term from the LHS:

$2p = 24$

Now, divide both sides by the coefficient of $p$ (which is 2) to isolate $p$:

$p = 12$

To verify the solution, substitute $p = 12$ back into the original equation:

LHS = $2(12) - 1 = 24 - 1 = 23$.

RHS = 23.

Since LHS = RHS, the solution is correct.

The solution is $\mathbf{p = 12}$.

Solution:

To separate the variable, we perform the inverse operation on both sides of the equation to eliminate the constant term with the variable.

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(a) Equation: $x – 1 = 0$

The constant term with the variable is -1 (subtracted). The inverse operation is adding 1.

First step: Add 1 to both sides of the equation.

$x = 1$

Solution: $\mathbf{x = 1}$.

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(b) Equation: $x + 1 = 0$

The constant term with the variable is +1 (added). The inverse operation is subtracting 1.

First step: Subtract 1 from both sides of the equation.

$x = -1$

Solution: $\mathbf{x = -1}$.

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(c) Equation: $x – 1 = 5$

The constant term with the variable is -1 (subtracted). The inverse operation is adding 1.

First step: Add 1 to both sides of the equation.

$x = 6$

Solution: $\mathbf{x = 6}$.

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(d) Equation: $x + 6 = 2$

The constant term with the variable is +6 (added). The inverse operation is subtracting 6.

First step: Subtract 6 from both sides of the equation.

$x = -4$

Solution: $\mathbf{x = -4}$.

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(e) Equation: $y – 4 = – 7$

The constant term with the variable is -4 (subtracted). The inverse operation is adding 4.

First step: Add 4 to both sides of the equation.

$y = -3$

Solution: $\mathbf{y = -3}$.

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(f) Equation: $y – 4 = 4$

The constant term with the variable is -4 (subtracted). The inverse operation is adding 4.

First step: Add 4 to both sides of the equation.

$y = 8$

Solution: $\mathbf{y = 8}$.

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(g) Equation: $y + 4 = 4$

The constant term with the variable is +4 (added). The inverse operation is subtracting 4.

First step: Subtract 4 from both sides of the equation.

$y = 0$

Solution: $\mathbf{y = 0}$.

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(h) Equation: $y + 4 = – 4$

The constant term with the variable is +4 (added). The inverse operation is subtracting 4.

First step: Subtract 4 from both sides of the equation.

$y = -8$

Solution: $\mathbf{y = -8}$.

Solution:

To separate the variable, we perform the inverse operation on both sides of the equation to eliminate the coefficient or divisor of the variable.

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(a) Equation: $3l = 42$

The variable $l$ is multiplied by 3. The inverse operation is dividing by 3.

First step: Divide both sides of the equation by 3.

$l = 14$

Solution: $\mathbf{l = 14}$.

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(b) Equation: $\frac{b}{2} = 6$

The variable $b$ is divided by 2. The inverse operation is multiplying by 2.

First step: Multiply both sides of the equation by 2.

$b = 12$

Solution: $\mathbf{b = 12}$.

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(c) Equation: $\frac{p}{7} = 4$

The variable $p$ is divided by 7. The inverse operation is multiplying by 7.

First step: Multiply both sides of the equation by 7.

$p = 28$

Solution: $\mathbf{p = 28}$.

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(d) Equation: $4x = 25$

The variable $x$ is multiplied by 4. The inverse operation is dividing by 4.

First step: Divide both sides of the equation by 4.

$x = \frac{25}{4}$

Solution: $\mathbf{x = \frac{25}{4}}$.

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(e) Equation: $8y = 36$

The variable $y$ is multiplied by 8. The inverse operation is dividing by 8.

First step: Divide both sides of the equation by 8.

$y = \frac{36}{8}$

Simplify the fraction: $y = \frac{\cancel{36}^{9}}{\cancel{8}_{2}} = \frac{9}{2}$.

Solution: $\mathbf{y = \frac{9}{2}}$.

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(f) Equation: $\frac{z}{3} = \frac{5}{4}$

The variable $z$ is divided by 3. The inverse operation is multiplying by 3.

First step: Multiply both sides of the equation by 3.

$z = \frac{5 \times 3}{4} = \frac{15}{4}$

Solution: $\mathbf{z = \frac{15}{4}}$.

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(g) Equation: $\frac{a}{5} = \frac{7}{15}$

The variable $a$ is divided by 5. The inverse operation is multiplying by 5.

First step: Multiply both sides of the equation by 5.

$a = \frac{7 \times \cancel{5}^{1}}{\cancel{15}_{3}} = \frac{7 \times 1}{3} = \frac{7}{3}$

Solution: $\mathbf{a = \frac{7}{3}}$.

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(h) Equation: $20t = – 10$

The variable $t$ is multiplied by 20. The inverse operation is dividing by 20.

First step: Divide both sides of the equation by 20.

$t = \frac{-10}{20} = -\frac{10}{20} = -\frac{1}{2}$.

Solution: $\mathbf{t = -\frac{1}{2}}$.

Solution:

To solve these equations, we need to perform operations that isolate the variable on one side of the equation.

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(a) Equation: $3n – 2 = 46$

Step 1: The term '-2' is subtracted from 3n. Add 2 to both sides of the equation to eliminate the constant term from the LHS.

$3n = 48$

Step 2: The variable $n$ is multiplied by 3. Divide both sides by 3 to isolate $n$.

$n = 16$

Solution: $\mathbf{n = 16}$.

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(b) Equation: $5m + 7 = 17$

Step 1: The term '+7' is added to 5m. Subtract 7 from both sides of the equation to eliminate the constant term from the LHS.

$5m = 10$

Step 2: The variable $m$ is multiplied by 5. Divide both sides by 5 to isolate $m$.

$m = 2$

Solution: $\mathbf{m = 2}$.

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(c) Equation: $\frac{20p}{3} = 40$

Step 1: The variable $p$ is multiplied by 20 and divided by 3. First, multiply both sides by 3 to remove the denominator.

$20p = 120$

Step 2: The variable $p$ is multiplied by 20. Divide both sides by 20 to isolate $p$.

$p = 6$

Solution: $\mathbf{p = 6}$.

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(d) Equation: $\frac{3p}{10} = 6$

Step 1: The variable $p$ is multiplied by 3 and divided by 10. First, multiply both sides by 10 to remove the denominator.

$3p = 60$

Step 2: The variable $p$ is multiplied by 3. Divide both sides by 3 to isolate $p$.

$p = 20$

Solution: $\mathbf{p = 20}$.

Solution:

We will solve each equation by isolating the variable using inverse operations.

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(a) $10p = 100$

Divide both sides by 10:

$p = 10$.

Solution: $\mathbf{p = 10}$.

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(b) $10p + 10 = 100$

Subtract 10 from both sides:

$10p = 90$

Divide both sides by 10:

$p = 9$.

Solution: $\mathbf{p = 9}$.

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(c) $\frac{p}{4} = 5$

Multiply both sides by 4:

$p = 20$.

Solution: $\mathbf{p = 20}$.

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(d) $\frac{-p}{3} = 5$}

Multiply both sides by 3:

$-p = 15$

Multiply both sides by -1:

$p = -15$.

Solution: $\mathbf{p = -15}$.

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(e) $\frac{3p}{4} = 6$}

Multiply both sides by 4:

$3p = 24$

Divide both sides by 3:

$p = 8$.

Solution: $\mathbf{p = 8}$.

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(f) $3s = –9$

Divide both sides by 3:

$s = -3$.

Solution: $\mathbf{s = -3}$.

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(g) $3s + 12 = 0$

Subtract 12 from both sides:

$3s = -12$

Divide both sides by 3:

$s = -4$.

Solution: $\mathbf{s = -4}$.

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(h) $3s = 0$

Divide both sides by 3:

$s = 0$.

Solution: $\mathbf{s = 0}$.

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(i) $2q = 6$

Divide both sides by 2:

$q = 3$.

Solution: $\mathbf{q = 3}$.

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(j) $2q – 6 = 0$

Add 6 to both sides:

$2q = 6$

Divide both sides by 2:

$q = 3$.

Solution: $\mathbf{q = 3}$.

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(k) $2q + 6 = 0$

Subtract 6 from both sides:

$2q = -6$

Divide both sides by 2:

$q = -3$.

Solution: $\mathbf{q = -3}$.

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(l) $2q + 6 = 12$

Subtract 6 from both sides:

$2q = 6$

Divide both sides by 2:

$q = 3$.

Solution: $\mathbf{q = 3}$.

Solution:

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Equation: $12p – 5 = 25$

We want to isolate the variable $p$.

Step 1: Eliminate the constant term -5 from the LHS by adding 5 to both sides of the equation.

$12p = 30$

Step 2: Eliminate the coefficient 12 from $p$ by dividing both sides by 12.

$p = \frac{30}{12}$

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 6.

$p = \frac{\cancel{30}^{5}}{\cancel{12}_{2}} = \frac{5}{2}$.

The solution is $\mathbf{p = \frac{5}{2}}$ or $\mathbf{p = 2.5}$.

To verify the solution, substitute $p = \frac{5}{2}$ back into the original equation:

LHS = $12\left(\frac{5}{2}\right) - 5 = \frac{12 \times 5}{2} - 5 = \frac{60}{2} - 5 = 30 - 5 = 25$.

RHS = 25.

Since LHS = RHS, the solution is correct.

Solution:

We will solve each equation by isolating the variable using inverse operations.

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(a) Equation: $4(m + 3) = 18$

Method 1: Divide by the coefficient first.

Step 1: The term $(m+3)$ is multiplied by 4. Divide both sides by 4.

$m + 3 = \frac{18}{4}$

Simplify the fraction: $\frac{18}{4} = \frac{\cancel{18}^{9}}{\cancel{4}_{2}} = \frac{9}{2}$.

$m + 3 = \frac{9}{2}$

Step 2: Subtract 3 from both sides.

$m = \frac{9}{2} - \frac{6}{2} = \frac{9 - 6}{2} = \frac{3}{2}$.

Method 2: Distribute first.

Step 1: Apply the distributive property to the LHS.

$4m + 12 = 18$

Step 2: Subtract 12 from both sides.

$4m = 6$

Step 3: Divide both sides by 4.

$m = \frac{6}{4}$

Simplify the fraction: $m = \frac{\cancel{6}^{3}}{\cancel{4}_{2}} = \frac{3}{2}$.

Both methods give the same solution. The solution is $\mathbf{m = \frac{3}{2}}$.

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(b) Equation: – $2(x + 3) = 8$

Method 1: Divide by the coefficient first.

Step 1: The term $(x+3)$ is multiplied by -2. Divide both sides by -2.

$x + 3 = -4$

Step 2: Subtract 3 from both sides.

$x = -7$.

Method 2: Distribute first.

Step 1: Apply the distributive property to the LHS.

$-2x - 6 = 8$

Step 2: Add 6 to both sides.

$-2x = 14$

Step 3: Divide both sides by -2.

$x = -7$.

Both methods give the same solution. The solution is $\mathbf{x = -7}$.

(a) $2y + \frac{5}{2} = \frac{37}{2}$

To solve the equation, we first transpose $\frac{5}{2}$ from the left-hand side (LHS) to the right-hand side (RHS).

$2y = \frac{37}{2} - \frac{5}{2}$

$2y = \frac{37 - 5}{2}$

$2y = \frac{32}{2}$

$2y = 16$

Now, we divide both sides by 2 to find the value of y.

$y = \frac{16}{2}$

$y = 8$

Therefore, the solution is y = 8.

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(b) 5t + 28 = 10

Transpose 28 from LHS to RHS.

$5t = 10 - 28$

$5t = -18$

Divide both sides by 5.

$t = -\frac{18}{5}$

Therefore, the solution is t = $-\frac{18}{5}$.

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(c) $\frac{a}{5} + 3 = 2$

Transpose 3 from LHS to RHS.

$\frac{a}{5} = 2 - 3$

$\frac{a}{5} = -1$

Multiply both sides by 5.

$a = -1 \times 5$

$a = -5$

Therefore, the solution is a = -5.

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(d) $\frac{q}{4} + 7 = 5$

Transpose 7 from LHS to RHS.

$\frac{q}{4} = 5 - 7$

$\frac{q}{4} = -2$

Multiply both sides by 4.

$q = -2 \times 4$

$q = -8$

Therefore, the solution is q = -8.

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(e) $\frac{5}{2}x = -5$

To find x, we multiply both sides by the reciprocal of $\frac{5}{2}$, which is $\frac{2}{5}$.

$x = -5 \times \frac{2}{5}$

$x = -\frac{\cancel{5} \times 2}{\cancel{5}}$

$x = -2$

Therefore, the solution is x = -2.

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(f) $\frac{5}{2} x = \frac{25}{4}$

Multiply both sides by the reciprocal of $\frac{5}{2}$, which is $\frac{2}{5}$.

$x = \frac{25}{4} \times \frac{2}{5}$

$x = \frac{\cancel{25}^{5}}{\cancel{4}_{2}} \times \frac{\cancel{2}^{1}}{\cancel{5}_{1}}$

$x = \frac{5 \times 1}{2 \times 1}$

$x = \frac{5}{2}$

Therefore, the solution is x = $\frac{5}{2}$.

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(g) $7m + \frac{19}{2} = 13$

Transpose $\frac{19}{2}$ from LHS to RHS.

$7m = 13 - \frac{19}{2}$

To subtract the fractions, we find a common denominator.

$7m = \frac{13 \times 2}{2} - \frac{19}{2}$

$7m = \frac{26 - 19}{2}$

$7m = \frac{7}{2}$

Divide both sides by 7.

$m = \frac{7}{2} \div 7$

$m = \frac{7}{2} \times \frac{1}{7}$

$m = \frac{1}{2}$

Therefore, the solution is m = $\frac{1}{2}$.

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(h) 6z + 10 = -2

Transpose 10 from LHS to RHS.

$6z = -2 - 10$

$6z = -12$

Divide both sides by 6.

$z = \frac{-12}{6}$

$z = -2$

Therefore, the solution is z = -2.

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(i) $\frac{3l}{2} = \frac{2}{3}$

To find l, we multiply both sides by the reciprocal of $\frac{3}{2}$, which is $\frac{2}{3}$.

$l = \frac{2}{3} \times \frac{2}{3}$

$l = \frac{4}{9}$

Therefore, the solution is l = $\frac{4}{9}$.

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(j) $\frac{2b}{3} - 5 = 3$

Transpose -5 from LHS to RHS.

$\frac{2b}{3} = 3 + 5$

$\frac{2b}{3} = 8$

Multiply both sides by the reciprocal of $\frac{2}{3}$, which is $\frac{3}{2}$.

$b = 8 \times \frac{3}{2}$

$b = \frac{8 \times 3}{2}$

$b = \frac{24}{2}$

$b = 12$

Therefore, the solution is b = 12.

Solution:

We will solve each equation by isolating the variable.

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(a) Equation: $2(x + 4) = 12$

Divide both sides by 2:

$x + 4 = 6$

Subtract 4 from both sides:

$x = 2$.

Solution: $\mathbf{x = 2}$.

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(b) Equation: $3(n – 5) = 21$

Divide both sides by 3:

$n - 5 = 7$

Add 5 to both sides:

$n = 12$.

Solution: $\mathbf{n = 12}$.

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(c) Equation: $3(n – 5) = – 21$

Divide both sides by 3:

$n - 5 = -7$

Add 5 to both sides:

$n = -2$.

Solution: $\mathbf{n = -2}$.

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(d) Equation: – $4(2 + x) = 8$

Divide both sides by -4:

$2 + x = -2$

Subtract 2 from both sides:

$x = -4$.

Solution: $\mathbf{x = -4}$.

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(e) Equation: $4(2 – x) = 8$

Divide both sides by 4:

$2 - x = 2$

Subtract 2 from both sides:

$-x = 0$

Multiply both sides by -1:

$x = 0$.

Solution: $\mathbf{x = 0}$.

Solution:

We will solve each equation by isolating the variable.

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(a) Equation: $4 = 5(p – 2)$

Divide both sides by 5:

$\frac{4}{5} = p - 2$

Add 2 to both sides:

$\frac{4}{5} + \frac{10}{5} = p$

$\frac{4 + 10}{5} = p$

$\frac{14}{5} = p$.

Solution: $\mathbf{p = \frac{14}{5}}$.

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(b) Equation: – $4 = 5(p – 2)$

Divide both sides by 5:

$-\frac{4}{5} = p - 2$

Add 2 to both sides:

$-\frac{4}{5} + \frac{10}{5} = p$

$\frac{-4 + 10}{5} = p$

$\frac{6}{5} = p$.

Solution: $\mathbf{p = \frac{6}{5}}$.

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(c) Equation: $16 = 4 + 3(t + 2)$

Subtract 4 from both sides:

$12 = 3(t + 2)$

Divide both sides by 3:

$4 = t + 2$

Subtract 2 from both sides:

$2 = t$.

Solution: $\mathbf{t = 2}$.

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(d) Equation: $4 + 5(p – 1) = 34$

Subtract 4 from both sides:

$5(p - 1) = 30$

Divide both sides by 5:

$p - 1 = 6$

Add 1 to both sides:

$p = 7$.

Solution: $\mathbf{p = 7}$.

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(e) Equation: $0 = 16 + 4(m – 6)$

Subtract 16 from both sides:

$-16 = 4(m - 6)$

Divide both sides by 4:

$-4 = m - 6$

Add 6 to both sides:

$2 = m$.

Solution: $\mathbf{m = 2}$.

Solution:

To construct equations starting with a given value for the variable, we perform the same arithmetic operations on both sides of the initial equality.

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(a) Construct 3 equations starting with $x = 2$.

Start with the equality: $x = 2$.

Equation 1: Add a number to both sides (e.g., add 3).

$x + 3 = 2 + 3$

$\mathbf{x + 3 = 5}$

Equation 2: Multiply both sides by a number (e.g., multiply by 5).

$x \times 5 = 2 \times 5$

$\mathbf{5x = 10}$

Equation 3: Perform multiple operations (e.g., multiply by 2, then subtract 4).

Multiply by 2: $2x = 2 \times 2 = 4$.

Subtract 4: $2x - 4 = 4 - 4$

$\mathbf{2x - 4 = 0}$

Three equations starting with $x = 2$ are $\mathbf{x + 3 = 5}$, $\mathbf{5x = 10}$, and $\mathbf{2x - 4 = 0}$. (Many other correct equations are possible).

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(b) Construct 3 equations starting with $x = – 2$.

Start with the equality: $x = -2$.

Equation 1: Subtract a number from both sides (e.g., subtract 7).

$x - 7 = -2 - 7$

$\mathbf{x - 7 = -9}$

Equation 2: Multiply both sides by a negative number (e.g., multiply by -3).

$x \times (-3) = -2 \times (-3)$

$\mathbf{-3x = 6}$

Equation 3: Perform multiple operations (e.g., divide by 2, then add 5).

Divide by 2: $\frac{x}{2} = \frac{-2}{2} = -1$.

Add 5: $\frac{x}{2} + 5 = -1 + 5$

$\mathbf{\frac{x}{2} + 5 = 4}$

Three equations starting with $x = -2$ are $\mathbf{x - 7 = -9}$, $\mathbf{-3x = 6}$, and $\mathbf{\frac{x}{2} + 5 = 4}$. (Many other correct equations are possible).

Solution:

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Let the unknown number be represented by the variable $n$.

"Three times a number" is $3 \times n = 3n$.

"The sum of three times a number and 11" is $3n + 11$.

"is 32" means the expression equals 32.

The equation representing the given statement is:

$3n + 11 = 32$

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To find the number, we need to solve this equation for $n$.

Step 1: Subtract 11 from both sides of the equation to isolate the term with the variable.

$3n = 21$

Step 2: Divide both sides by the coefficient of $n$, which is 3, to isolate $n$.

$n = 7$

The number is 7.

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To verify, substitute $n=7$ back into the original equation: $3(7) + 11 = 21 + 11 = 32$. This matches the RHS, so the solution is correct.

The number is $\mathbf{7}$.

Solution:

---

Let the unknown number be represented by the variable $x$.

"One-fourth of the number" is $\frac{1}{4} \times x = \frac{x}{4}$.

"3 more than 7" means $7 + 3 = 10$.

The statement "one-fourth of the number is 3 more than 7" can be written as the equation:

$\frac{x}{4} = 10$

---

To find the number, we need to solve this equation for $x$.

The variable $x$ is divided by 4. Multiply both sides by 4 to isolate $x$.

$x = 40$

The number is 40.

---

To verify, one-fourth of 40 is $\frac{1}{4} \times 40 = 10$. And 3 more than 7 is $7 + 3 = 10$. Since $10 = 10$, the solution is correct.

The number is $\mathbf{40}$.

Solution:

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Let Raju's age be $r$ years.

"Three times Raju's age" is $3r$.

"5 years more than three times Raju's age" is $3r + 5$. This represents Raju's father's age.

We are given that Raju's father is 44 years old.

Set up the equation:

$3r + 5 = 44$

---

To find Raju's age, we need to solve this equation for $r$.

Step 1: Subtract 5 from both sides of the equation.

$3r = 39$

Step 2: Divide both sides by 3.

$r = 13$

Raju's age is 13 years.

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To verify, three times Raju's age is $3 \times 13 = 39$. 5 years more than this is $39 + 5 = 44$, which is the father's age. The solution is correct.

Raju's age is $\mathbf{13}$ years.

Solution:

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(a) Let the number be $n$.

"Eight times a number" is $8n$.

"Add 4 to eight times a number" is $8n + 4$.

"you get 60" means $8n + 4 = 60$.

Equation: $\mathbf{8n + 4 = 60}$.

Solving the equation:

$8n + 4 = 60$

Subtract 4 from both sides:

$8n = 56$

Divide both sides by 8:

$n = 7$.

The unknown number is $\mathbf{7}$.

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(b) Let the number be $y$.

"One-fifth of a number" is $\frac{1}{5}y = \frac{y}{5}$.

"minus 4" is $\frac{y}{5} - 4$.

"gives 3" means $\frac{y}{5} - 4 = 3$.

Equation: $\mathbf{\frac{y}{5} - 4 = 3}$.

Solving the equation:

$\frac{y}{5} - 4 = 3$

Add 4 to both sides:

$\frac{y}{5} = 7$

Multiply both sides by 5:

$y = 35$.

The unknown number is $\mathbf{35}$.

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(c) Let the number be $k$.

"three-fourths of a number" is $\frac{3}{4}k$.

"add 3 to it" is $\frac{3}{4}k + 3$.

"I get 21" means $\frac{3}{4}k + 3 = 21$.

Equation: $\mathbf{\frac{3}{4}k + 3 = 21}$.

Solving the equation:

$\frac{3}{4}k + 3 = 21$

Subtract 3 from both sides:

$\frac{3}{4}k = 18$

Multiply both sides by $\frac{4}{3}$ (the reciprocal of $\frac{3}{4}$):

$k = \frac{18 \times 4}{3} = \frac{\cancel{18}^{6} \times 4}{\cancel{3}_{1}} = 6 \times 4 = 24$.

$k = 24$.

The unknown number is $\mathbf{24}$.

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(d) Let the number be $p$.

"twice a number" is $2p$.

"subtracted 11 from twice a number" is $2p - 11$.

"the result was 15" means $2p - 11 = 15$.

Equation: $\mathbf{2p - 11 = 15}$.

Solving the equation:

$2p - 11 = 15$

Add 11 to both sides:

$2p = 26$

Divide both sides by 2:

$p = 13$.

The unknown number is $\mathbf{13}$.

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(e) Let the number of notebooks Munna has be $m$.

"thrice the number of notebooks he has" is $3m$.

"subtracts thrice the number of notebooks he has from 50" is $50 - 3m$.

"he finds the result to be 8" means $50 - 3m = 8$.

Equation: $\mathbf{50 - 3m = 8}$.

Solving the equation:

$50 - 3m = 8$}

Subtract 50 from both sides:

$-3m = -42$

Divide both sides by -3:

$m = 14$.

The number of notebooks Munna has is $\mathbf{14}$.

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(f) Let the number Ibenhal thinks of be $t$.

"she adds 19 to it" is $t + 19$.

"divides the sum by 5" is $\frac{t + 19}{5}$.

"she will get 8" means $\frac{t + 19}{5} = 8$.

Equation: $\mathbf{\frac{t + 19}{5} = 8}$.

Solving the equation:

$\frac{t + 19}{5} = 8$

Multiply both sides by 5:

$t + 19 = 40$

Subtract 19 from both sides:

$t = 21$.

The number Ibenhal thinks of is $\mathbf{21}$.

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(g) Let the number Anwar thinks of be $y$.

"$\frac{5}{2}$ of the number" is $\frac{5}{2}y$.

"takes away 7 from $\frac{5}{2}$ of the number" is $\frac{5}{2}y - 7$.

"the result is 23" means $\frac{5}{2}y - 7 = 23$.

Equation: $\mathbf{\frac{5}{2}y - 7 = 23}$.

Solving the equation:

$\frac{5}{2}y - 7 = 23$

Add 7 to both sides:

$\frac{5}{2}y = 30$

Multiply both sides by the reciprocal of $\frac{5}{2}$, which is $\frac{2}{5}$.

$y = \frac{\cancel{30}^{6} \times 2}{\cancel{5}_{1}} = 6 \times 2 = 12$.

$y = 12$.

The number Anwar thinks of is $\mathbf{12}$.

Solution:

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(a) Let the lowest score be $l$.

"twice the lowest marks" is $2l$.

"twice the lowest marks plus 7" is $2l + 7$. This is the highest score.

The highest score is 87.

Equation: $2l + 7 = 87$.

Solving the equation:

$2l + 7 = 87$

Subtract 7 from both sides:

$2l = 80$

Divide both sides by 2:

$l = 40$.

The lowest score is 40.

The lowest score is $\mathbf{40}$.

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(b) In an isosceles triangle, the base angles are equal.

Let each base angle be $b$ degrees.

The vertex angle is given as 40°.

The sum of the angles in a triangle is 180°.

Sum of angles = Base angle + Base angle + Vertex angle = 180°.

Equation: $b + b + 40 = 180$.

Solving the equation:

$2b + 40 = 180$

Subtract 40 from both sides:

$2b = 140$

Divide both sides by 2:

$b = 70$.

Each base angle is 70°.

The base angles of the triangle are $\mathbf{70^\circ}$ each.

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(c) Let the number of runs scored by Rahul be $r$.

Sachin scored twice as many runs as Rahul, so Sachin's score is $2r$.

"Together, their runs fell two short of a double century."

A double century is 200 runs.

"Two short of a double century" is $200 - 2 = 198$.

"Together, their runs" means the sum of their scores: Rahul's score + Sachin's score = $r + 2r = 3r$.

So, the sum of their runs is equal to 198.

Equation: $3r = 198$.

Solving the equation:

$3r = 198$

Divide both sides by 3:

$r = 66$.

Rahul's score is 66 runs.

Sachin's score is $2r = 2 \times 66 = 132$ runs.

Check: Rahul's score (66) + Sachin's score (132) = $66 + 132 = 198$, which is two short of 200.

Rahul scored $\mathbf{66}$ runs.

Sachin scored $\mathbf{132}$ runs.

Solution:

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(i) Let the number of marbles Parmit has be $m$.

"five times the marbles Parmit has" is $5m$.

"7 marbles more than five times the marbles Parmit has" is $5m + 7$. This is the number of marbles Irfan has.

Irfan has 37 marbles.

Equation: $5m + 7 = 37$.

Solving the equation:

$5m + 7 = 37$

Subtract 7 from both sides:

$5m = 30$

Divide both sides by 5:

$m = 6$.

Parmit has 6 marbles.

Parmit has $\mathbf{6}$ marbles.

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(ii) Let Laxmi's age be $y$ years.

"three times Laxmi’s age" is $3y$.

"4 years older than three times Laxmi’s age" is $3y + 4$. This is the father's age.

Laxmi’s father is 49 years old.

Equation: $3y + 4 = 49$.

Solving the equation:

$3y + 4 = 49$

Subtract 4 from both sides:

$3y = 45$

Divide both sides by 3:

$y = 15$.

Laxmi's age is 15 years.

Laxmi's age is $\mathbf{15}$ years.

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(iii) Let the number of fruit trees planted be $f$.

"three times the number of fruit trees" is $3f$.

"two more than three times the number of fruit trees" is $3f + 2$. This is the number of non-fruit trees.

The number of non-fruit trees planted was 77.

Equation: $3f + 2 = 77$.

Solving the equation:

$3f + 2 = 77$

Subtract 2 from both sides:

$3f = 75$

Divide both sides by 3:

$f = 25$.

The number of fruit trees planted was 25.

The number of fruit trees planted was $\mathbf{25}$.

Solution:

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Let the unknown number be represented by the variable $x$.

Interpret the lines of the riddle as mathematical expressions:

"Take me seven times over" means multiply the number by 7: $7x$.

"And add a fifty" means add 50 to the previous expression: $7x + 50$.

"To reach a triple century" refers to a target value. A triple century is $3 \times 100 = 300$.

"You still need forty!" means the current value ($7x + 50$) is 40 less than 300.

So, $7x + 50 = 300 - 40$.

Simplify the right side:

$300 - 40 = 260$.

The equation representing the riddle is:

$7x + 50 = 260$

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Now, solve the equation to find the identity of the number ($x$).

Step 1: Subtract 50 from both sides of the equation.

$7x = 210$

Step 2: Divide both sides by 7.

$x = 30$.

The number is 30.

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Let's check the riddle with the number 30:

"Take me seven times over": $7 \times 30 = 210$.

"And add a fifty!": $210 + 50 = 260$.

"To reach a triple century (300) You still need forty!": $260 + 40 = 300$. This matches the condition.

The identity of the number is $\mathbf{30}$.