Chapter 4 Simple Equations

Given equation is:

$3x + 5 = 0$

To solve for $x$, we need to isolate the term involving $x$.

Subtract 5 from both sides of the equation:

$3x + 5 - 5 = 0 - 5$

$3x = -5$

Divide both sides by 3 to find the value of $x$:

$\frac{3x}{3} = \frac{-5}{3}$

$x = -\frac{5}{3}$

The solution to the equation $3x + 5 = 0$ is $x = -\frac{5}{3}$.

Comparing this solution with the given options:

(a) $\frac{5}{3}$

(b) – 5

(c) $-\frac{5}{3}$

(d) 5

The solution matches option (c).

Thus, the correct option is (c).

We need to check which of the given equations is not satisfied when the variable is replaced by -1.

Let's check each option:

(a) $x + 1 = 0$

Substitute $x = -1$:

$-1 + 1 = 0$

$0 = 0$

This equation is satisfied by $x = -1$. So, -1 is a solution for this equation.

(b) $x – 1 = 2$

Substitute $x = -1$:

$-1 - 1 = 2$

$-2 = 2$

This equation is not satisfied by $x = -1$. So, -1 is not a solution for this equation.

(c) $2y + 3 = 1$

Substitute $y = -1$:

$2(-1) + 3 = 1$

$-2 + 3 = 1$

$1 = 1$

This equation is satisfied by $y = -1$. So, -1 is a solution for this equation.

(d) $2p + 7 = 5$

Substitute $p = -1$:

$2(-1) + 7 = 5$

$-2 + 7 = 5$

$5 = 5$

This equation is satisfied by $p = -1$. So, -1 is a solution for this equation.

Therefore, -1 is not a solution of the equation $x - 1 = 2$.

The correct option is (b).

We are given that $x = 5$ is the solution. We need to find which of the given equations is satisfied when $x = 5$. We will substitute $x = 5$ into each equation and check if the Left Hand Side (LHS) equals the Right Hand Side (RHS).

(a) $2x + 3 = 13$

Substitute $x = 5$ into the LHS:

LHS $= 2(5) + 3 = 10 + 3 = 13$

RHS $= 13$

Since LHS $=$ RHS, the equation $2x + 3 = 13$ is satisfied by $x = 5$.

(b) $3x + 2 = 13$

Substitute $x = 5$ into the LHS:

LHS $= 3(5) + 2 = 15 + 2 = 17$

RHS $= 13$

Since LHS $\neq$ RHS, the equation $3x + 2 = 13$ is not satisfied by $x = 5$.

(c) $x - 5 = 1$

Substitute $x = 5$ into the LHS:

LHS $= 5 - 5 = 0$

RHS $= 1$

Since LHS $\neq$ RHS, the equation $x - 5 = 1$ is not satisfied by $x = 5$.

(d) $4x - 9 = 21$

Substitute $x = 5$ into the LHS:

LHS $= 4(5) - 9 = 20 - 9 = 11$

RHS $= 21$

Since LHS $\neq$ RHS, the equation $4x - 9 = 21$ is not satisfied by $x = 5$.

Only equation (a) is satisfied when $x = 5$. Therefore, the equation that can be formed using the expression $x=5$ is $2x + 3 = 13$.

The correct option is (a).

The question asks for the term that describes a value of the variable which, when substituted into an equation, makes the Left Hand Side (LHS) and the Right Hand Side (RHS) equal.

Consider a simple equation like $x + 2 = 5$.

If we substitute $x = 1$, the LHS is $1 + 2 = 3$. The RHS is $5$. Since $3 \neq 5$, $x = 1$ is not the value that makes both sides equal.

If we substitute $x = 3$, the LHS is $3 + 2 = 5$. The RHS is $5$. Since $5 = 5$, $x = 3$ is the value that makes both sides equal.

This specific value of the variable that satisfies the equation (makes both sides equal) is called the solution of the equation.

Therefore, the correct word to fill in the blank is solution.

Given equation is:

$y - 13 = 9$

We need to find the root of the equation, which means finding the value of the variable $y$ that satisfies the equation.

To isolate $y$, we add 13 to both sides of the equation:

$y - 13 + 13 = 9 + 13$

$y = 22$

Thus, the root of the equation $y - 13 = 9$ is 22.

The completed statement is: The root of the equation y – 13 = 9 is 22.

We are given an equation of the form $2x + \text{______} = 11$ and the information that its solution is $x = -4$. This means that when we substitute $x = -4$ into the equation, the equation must be true.

Let the missing number in the blank be represented by a variable, say $k$. The equation is:

$2x + k = 11$

Substitute the given solution $x = -4$ into this equation:

$2(-4) + k = 11$

$-8 + k = 11$

Now, we solve for $k$ by adding 8 to both sides of the equation:

$-8 + k + 8 = 11 + 8$

$k = 19$

So, the missing number is 19.

The completed statement is: 2x + 19 = 11 has the solution – 4.

We need to check if the value $x = 12$ satisfies the given equation $4x - 5 = 3x + 10$.

Substitute $x = 12$ into the Left Hand Side (LHS) of the equation:

LHS $= 4x - 5$

LHS $= 4(12) - 5$

LHS $= 48 - 5$

LHS $= 43$

Substitute $x = 12$ into the Right Hand Side (RHS) of the equation:

RHS $= 3x + 10$

RHS $= 3(12) + 10$

RHS $= 36 + 10$

RHS $= 46$

Now, compare the values of LHS and RHS:

LHS $= 43$

RHS $= 46$

Since LHS $\neq$ RHS ($43 \neq 46$), the value $x = 12$ does not satisfy the equation.

Therefore, 12 is not a solution of the equation $4x - 5 = 3x + 10$.

The given statement is False.

Let's translate the verbal statement "A number x divided by 7 gives 2" into a mathematical equation.

The phrase "a number x divided by 7" can be written as $\frac{x}{7}$.

The phrase "gives 2" means the result is 2, so it equals 2.

Combining these, the correct equation representing the statement is $\frac{x}{7} = 2$.

Now, let's look at the given equation: $\frac{x \;+\; 1}{7} = 2$.

This equation represents the statement "one added to a number x, and the sum is divided by 7, gives 2".

Comparing the correct equation $\frac{x}{7} = 2$ with the given equation $\frac{x \;+\; 1}{7} = 2$, we see that they are different.

Therefore, the statement that the equation $\frac{x \;+\; 1}{7} = 2$ represents "A number x divided by 7 gives 2" is incorrect.

The given statement is False.

To determine if the two equations have the same solutions, we need to solve each equation independently and compare the values of $x$ obtained.

Equation 1: $x + 2 = 5$

To solve for $x$, subtract 2 from both sides of the equation:

$x + 2 - 2 = 5 - 2$

$x = 3$

The solution for the first equation is $x = 3$.

Equation 2: $3x - 1 = 8$

To solve for $x$, first add 1 to both sides of the equation:

$3x - 1 + 1 = 8 + 1$

$3x = 9$

Now, divide both sides by 3:

$\frac{3x}{3} = \frac{9}{3}$

$x = 3$

The solution for the second equation is $x = 3$.

Comparing the solutions, we found that the solution for the first equation is $x = 3$ and the solution for the second equation is also $x = 3$.

Since both equations yield the same solution ($x=3$), the statement is true.

The given statement is True.

We are asked to verify if $x = 1$ is a solution to the equation $3x + 7 = 10$. To do this, we substitute $x = 1$ into the equation and check if the Left Hand Side (LHS) equals the Right Hand Side (RHS).

Substitute $x = 1$ into the LHS of the equation:

LHS $= 3x + 7$

LHS $= 3(1) + 7$

LHS $= 3 + 7$

LHS $= 10$

The RHS of the equation is:

RHS $= 10$

Compare the values of LHS and RHS:

LHS $= 10$

RHS $= 10$

Since LHS $=$ RHS ($10 = 10$), the value $x = 1$ satisfies the equation $3x + 7 = 10$.

Therefore, 1 is indeed a solution of the equation $3x + 7 = 10$.

The given statement is True.

We are asked to represent the given statement as a mathematical equation.

Let the unknown number be $x$.

The phrase "one fourth of a number" can be written as $\frac{1}{4}$ of $x$, which is $\frac{1}{4}x$ or $\frac{x}{4}$.

The phrase "is 20 less than the number itself" means that the number itself ($x$) minus 20. This can be written as $x - 20$.

The word "is" implies equality ($=$).

Combining these parts, the statement "One fourth of a number is 20 less than the number itself" can be translated into the equation:

$\frac{x}{4} = x - 20$

We can also solve this equation to find the number:

Multiply both sides by 4 to eliminate the denominator:

$4 \times \frac{x}{4} = 4 \times (x - 20)$

$x = 4x - 80$

Subtract $4x$ from both sides:

$x - 4x = 4x - 80 - 4x$

$-3x = -80$

Divide both sides by -3:

$\frac{-3x}{-3} = \frac{-80}{-3}$

$x = \frac{80}{3}$

The mathematical equation representing the statement is $\frac{x}{4} = x - 20$, and the number is $\frac{80}{3}$.

We are asked to represent the given statement as a mathematical equation.

Let the unknown number be $x$.

The phrase "3 times of a number" can be written as $3x$.

The phrase "On subtracting 13 from 3 times of a number" means we take $3x$ and subtract 13 from it. This can be written as $3x - 13$.

The phrase "the result is 8" means that the expression $3x - 13$ is equal to 8.

Combining these parts, the statement "On subtracting 13 from 3 times of a number, the result is 8" can be translated into the equation:

$3x - 13 = 8$

We can also solve this equation to find the number:

Add 13 to both sides of the equation:

$3x - 13 + 13 = 8 + 13$

$3x = 21$

Divide both sides by 3:

$\frac{3x}{3} = \frac{21}{3}$

$x = 7$

The mathematical equation representing the statement is $3x - 13 = 8$, and the number is 7.

We are asked to represent the given statement as a mathematical equation.

Let the unknown number be $x$.

The phrase "Two times a number" can be written as $2x$.

The phrase "increased by 5" means we add 5 to $2x$. This can be written as $2x + 5$.

The word "equals" implies equality ($=$).

Combining these parts, the statement "Two times a number increased by 5 equals 9" can be translated into the equation:

$2x + 5 = 9$

We can also solve this equation to find the number:

Subtract 5 from both sides of the equation:

$2x + 5 - 5 = 9 - 5$

$2x = 4$

Divide both sides by 2:

$\frac{2x}{2} = \frac{4}{2}$

$x = 2$

The mathematical equation representing the statement is $2x + 5 = 9$, and the number is 2.

Let the unknown number be $n$.

The phrase "twice a number" can be written as $2n$.

The phrase "9 added to twice a number" means we add 9 to $2n$. This can be written as $2n + 9$ or $9 + 2n$.

The phrase "gives 13" means the result is 13, so it equals 13.

Combining these parts, the statement "9 added to twice a number gives 13" translates into the equation:

$2n + 9 = 13$

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

Subtract 9 from both sides of the equation:

$2n + 9 - 9 = 13 - 9$

$2n = 4$

Divide both sides by 2:

$\frac{2n}{2} = \frac{4}{2}$

$n = 2$

The number is 2.

We can check this: Twice the number is $2 \times 2 = 4$. 9 added to this is $4 + 9 = 13$, which matches the given information.

Let the unknown number be $x$.

The phrase "one third of a number" can be written as $\frac{1}{3}x$ or $\frac{x}{3}$.

The phrase "1 subtracted from one third of a number" means we take $\frac{x}{3}$ and subtract 1 from it. This can be written as $\frac{x}{3} - 1$.

The phrase "gives 1" means the result is 1, so it equals 1.

Combining these parts, the statement "1 subtracted from one third of a number gives 1" translates into the equation:

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

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

Add 1 to both sides of the equation:

$\frac{x}{3} - 1 + 1 = 1 + 1$

$\frac{x}{3} = 2$

Multiply both sides by 3 to isolate $x$:

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

$x = 6$

The number is 6.

We can check this: One third of the number is $\frac{1}{3} \times 6 = 2$. 1 subtracted from this is $2 - 1 = 1$, which matches the given information.

The given equation in Roman numerals is: $\text{IX} - \text{I} = \text{X}$

In numerical values, this equation is:

$9 - 1 = 10$

$8 = 10$

This is a False statement.

We need to make the equation true by moving only one toothpick.

Consider the first term, $\text{IX}$. This is represented as 'I' followed by 'X'. Let's move the toothpick that forms the 'I'.

Move this single toothpick from the beginning of the number $\text{IX}$ to the end of the number $\text{IX}$. This changes the number from $\text{IX}$ (I followed by X, value 9) to $\text{XI}$ (X followed by I, value 11).

The equation now becomes: $\text{XI} - \text{I} = \text{X}$

In numerical values, this corrected equation is:

$11 - 1 = 10$

$10 = 10$

This is a True statement.

Thus, the correct equation formed by moving one toothpick is $\text{XI} - \text{I} = \text{X}$.

First, we need to solve each of the given equations to find the value of each variable (c, e, r, t, s).

1. Solve for c:

$4c = 16$

Divide both sides by 4:

$\frac{4c}{4} = \frac{16}{4}$

$c = 4$

2. Solve for e:

$4e + 8 = 20$

Subtract 8 from both sides:

$4e + 8 - 8 = 20 - 8$

$4e = 12$

Divide both sides by 4:

$\frac{4e}{4} = \frac{12}{4}$

$e = 3$

3. Solve for r:

$2r – 3 = 7$

Add 3 to both sides:

$2r - 3 + 3 = 7 + 3$

$2r = 10$

Divide both sides by 2:

$\frac{2r}{2} = \frac{10}{2}$

$r = 5$

4. Solve for t:

$3t + 8 = 29$

Subtract 8 from both sides:

$3t + 8 - 8 = 29 - 8$

$3t = 21$

Divide both sides by 3:

$\frac{3t}{3} = \frac{21}{3}$

$t = 7$

5. Solve for s:

$2s + 4 = 4s$

Subtract $2s$ from both sides:

$2s + 4 - 2s = 4s - 2s$

$4 = 2s$

Divide both sides by 2:

$\frac{4}{2} = \frac{2s}{2}$

$s = 2$

The solutions for the variables are:

c = 4

e = 3

r = 5

t = 7

s = 2

Now, let's consider the riddle: "What is too much fun for one, enough for two, and means nothing to three?". This is a classic riddle whose answer is SECRET.

The pattern given is $\frac{ \boxdot }{2}$ , $\frac{ \boxdot }{3}$ , $\frac{ \boxdot }{4}$ , $\frac{ \boxdot }{5}$ , $\frac{e}{ \boxdot }$ , $\frac{ \boxdot }{7}$. This pattern has 6 elements, which matches the number of letters in the word SECRET.

Let's examine how the letters of the word SECRET relate to our solutions and the pattern. It appears the pattern lists the letters of the word in order, and the denominator under each letter (represented by the shape $\boxdot$ or the letter 'e' itself) is the solution value associated with that letter from the equations we solved.

Let's match the letters of SECRET with the solved values and the pattern:

1st letter: S. The corresponding pattern element is $\frac{ \boxdot }{2}$. The denominator is 2. The variable with solution 2 is 's'. This matches.

2nd letter: E. The corresponding pattern element is $\frac{ \boxdot }{3}$. The denominator is 3. The variable with solution 3 is 'e'. This matches.

3rd letter: C. The corresponding pattern element is $\frac{ \boxdot }{4}$. The denominator is 4. The variable with solution 4 is 'c'. This matches.

4th letter: R. The corresponding pattern element is $\frac{ \boxdot }{5}$. The denominator is 5. The variable with solution 5 is 'r'. This matches.

5th letter: E. The corresponding pattern element is $\frac{e}{ \boxdot }$. The numerator is 'e'. The solution for 'e' is 3. The denominator $\boxdot$ must represent this solution value (3). This matches.

6th letter: T. The corresponding pattern element is $\frac{ \boxdot }{7}$. The denominator is 7. The variable with solution 7 is 't'. This matches.

Thus, the letters of the word SECRET correspond to the variables with solutions 2, 3, 4, 5, 3, 7 respectively.

The letter with solution 2 is S.

The letter with solution 3 is E.

The letter with solution 4 is C.

The letter with solution 5 is R.

The letter with solution 7 is T.

Arranging the letters corresponding to the solutions in the order specified by the pattern (denominators 2, 3, 4, 5, $\boxdot$=3, 7) gives the word SECRET.

The answer to the riddle is SECRET.

The given equation is:

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

We need to solve for the variable $t$.

First, apply the distributive property on the right-hand side (multiply 3 by each term inside the parenthesis):

$10 = 4 + (3 \times t) + (3 \times 2)$

$10 = 4 + 3t + 6$

Combine the constant terms on the right-hand side ($4 + 6$):

$10 = (4 + 6) + 3t$

$10 = 10 + 3t$

Now, isolate the term involving $t$ by subtracting 10 from both sides of the equation:

$10 - 10 = 10 + 3t - 10$

$0 = 3t$

Finally, solve for $t$ by dividing both sides by 3:

$\frac{0}{3} = \frac{3t}{3}$

$0 = t$

So, $t = 0$.

The solution to the equation $10 = 4 + 3(t + 2)$ is t = 0.

Verification:

Substitute $t = 0$ back into the original equation:

LHS $= 10$

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

RHS $= 4 + 3(2)$

RHS $= 4 + 6$

RHS $= 10$

Since LHS $=$ RHS ($10 = 10$), the solution is correct.

The given equation is a linear equation in one variable $x$:

$ax + b = 0$

To find the solution, we need to isolate the variable $x$.

First, subtract the constant term $b$ from both sides of the equation:

$ax + b - b = 0 - b$

$ax = -b$

Now, divide both sides of the equation by the coefficient of $x$, which is $a$, assuming $a \neq 0$:

$\frac{ax}{a} = \frac{-b}{a}$

$x = -\frac{b}{a}$

The solution of the equation $ax + b = 0$ is $x = -\frac{b}{a}$.

Comparing this result with the given options:

(a) $\frac{a}{b}$

(b) $-b$

(c) $-\frac{b}{a}$

(d) $\frac{b}{a}$

The solution matches option (c).

The correct option is (c).

The given equation is:

$ax = b$

We are given that $a$ and $b$ are positive integers.

To find the solution, we need to isolate $x$. We divide both sides of the equation by $a$:

$\frac{ax}{a} = \frac{b}{a}$

$x = \frac{b}{a}$

Since $a$ is a positive integer, $a > 0$.

Since $b$ is a positive integer, $b > 0$.

The division of a positive number by another positive number always results in a positive number.

So, $x = \frac{\text{positive}}{\text{positive}} = \text{positive}$.

Therefore, the solution of the equation $ax = b$, where $a$ and $b$ are positive integers, will always be a positive number.

Comparing this result with the given options:

(a) positive number

(b) negative number

(c) 1

(d) 0

The solution matches option (a).

The correct option is (a).

Let's analyze each option regarding the properties of equality:

(a) Adding the same number to both sides: This is a fundamental property of equality. If $a = b$, then $a+c = b+c$ for any number $c$. This operation is always allowed and preserves the equality.

(b) Subtracting the same number from both sides: This is also a fundamental property of equality. If $a = b$, then $a-c = b-c$ for any number $c$. This operation is always allowed and preserves the equality.

(c) Multiplying both sides of the equation by the same non-zero number: This is a fundamental property of equality. If $a = b$, then $a \times c = b \times c$ for any non-zero number $c$. Multiplying by zero can change the equation significantly (e.g., $x=5$ becomes $0=0$), so multiplication must be by a non-zero number to maintain the original solution set.

(d) Dividing both sides of the equation by the same number: Division is defined as multiplication by the reciprocal. Dividing by a number $c$ is equivalent to multiplying by $\frac{1}{c}$. For this operation to be defined and to preserve the equality and the solution set, the number $c$ cannot be zero, because division by zero is undefined. The statement in option (d) does not include the crucial condition that the number must be non-zero.

Therefore, dividing both sides of an equation by 0 is not allowed. Since option (d) does not specify "non-zero number", it includes the case of dividing by zero, which is not allowed.

The correct option is (d).

We need to solve each equation and determine the nature of its solution.

(a) $2x + 6 = 0$

$2x = -6$

$x = \frac{-6}{2}$

$x = -3$

The solution is -3. An integer is a whole number (positive, negative, or zero). -3 is an integer. In standard mathematics, every integer is also a fraction (e.g., $-3 = \frac{-3}{1}$). So, this solution is both an integer and a fraction (rational number).

(b) $3x – 5 = 0$

$3x = 5$

$x = \frac{5}{3}$

The solution is $\frac{5}{3}$. This is a rational number that is not an integer. In standard mathematics, this is called a fraction or a non-integer rational number.

(c) $5x – 8 = x + 4$

Subtract $x$ from both sides:

$5x - x - 8 = x - x + 4$

$4x - 8 = 4$

Add 8 to both sides:

$4x = 4 + 8$

$4x = 12$

Divide by 4:

$x = \frac{12}{4}$

$x = 3$

The solution is 3. This is an integer. It is also a fraction (e.g., $3 = \frac{3}{1}$). So, this solution is both an integer and a fraction (rational number).

(d) $4x + 7 = x + 2$

Subtract $x$ from both sides:

$4x - x + 7 = x - x + 2$

$3x + 7 = 2$

Subtract 7 from both sides:

$3x = 2 - 7$

$3x = -5$

Divide by 3:

$x = -\frac{5}{3}$

The solution is $-\frac{5}{3}$. This is a rational number that is not an integer. In standard mathematics, this is called a fraction or a non-integer rational number.

Summary of solutions:

(a) $x = -3$ (Integer and Fraction)

(b) $x = \frac{5}{3}$ (Fraction, not an integer)

(c) $x = 3$ (Integer and Fraction)

(d) $x = -\frac{5}{3}$ (Fraction, not an integer)

The question asks for a solution that is "neither a fraction nor an integer". In standard mathematics, the set of fractions (rational numbers) includes all integers. Therefore, if a number is not a fraction, it cannot be an integer either. The set of numbers that are "neither a fraction nor an integer" is the set of irrational numbers (like $\sqrt{2}$, $\pi$).

However, linear equations with rational coefficients (which all the given equations are) always have rational solutions. Therefore, none of the solutions (-3, 5/3, 3, -5/3) are irrational. They are all rational numbers (fractions).

Given that this is a multiple-choice question from a textbook, there might be a specific, potentially non-standard, definition of "fraction" implied. It seems the question might be using "fraction" to mean a non-integer rational number.

Let's assume the question intends "integer" to mean numbers like ..., -2, -1, 0, 1, 2, ... and "fraction" to mean rational numbers $\frac{p}{q}$ where $q \neq \pm 1$ when the fraction is in simplest form, AND possibly restricting to positive values for "fraction".

If we assume "fraction" means positive, non-integer rational numbers, then:

(a) $x = -3$: Is an integer. Not a "fraction" (not positive). So it's an integer.

(b) $x = \frac{5}{3}$: Is a positive, non-integer rational number. Is a "fraction" (under this assumption). Not an integer.

(c) $x = 3$: Is an integer. Not a "fraction" (not non-integer). So it's an integer.

(d) $x = -\frac{5}{3}$: Is a negative, non-integer rational number. Not an "integer". Not a "fraction" (under the assumption that "fraction" means positive non-integer rational number). Therefore, under this specific interpretation, $-\frac{5}{3}$ is neither an integer nor a fraction.

Based on this likely intended (though mathematically non-standard) interpretation used in the source material, the solution to equation (d) is neither an integer nor a "fraction" (where "fraction" means positive non-integer rational).

Considering the probable intent of the question based on the provided options and typical exam styles, the correct answer corresponds to the equation whose solution is a negative non-integer rational number, assuming "fraction" refers only to positive non-integer rationals.

The equation $4x + 7 = x + 2$ has the solution $x = -\frac{5}{3}$.

The correct option is (d).

To find the equation that cannot be solved in integers, we need to solve each equation and check if the solution is an integer.

(a) $5y - 3 = -18$

Add 3 to both sides:

$5y - 3 + 3 = -18 + 3$

$5y = -15$

Divide both sides by 5:

$y = \frac{-15}{5}$

$y = -3$

$-3$ is an integer.

(b) $3x - 9 = 0$

Add 9 to both sides:

$3x - 9 + 9 = 0 + 9$

$3x = 9$

Divide both sides by 3:

$x = \frac{9}{3}$

$x = 3$

$3$ is an integer.

(c) $3z + 8 = 3 + z$

Subtract $z$ from both sides:

$3z - z + 8 = 3 + z - z$

$2z + 8 = 3$

Subtract 8 from both sides:

$2z + 8 - 8 = 3 - 8$

$2z = -5$

Divide both sides by 2:

$z = \frac{-5}{2}$

$z = -2.5$

$-2.5$ is not an integer.

(d) $9y + 8 = 4y – 7$

Subtract $4y$ from both sides:

$9y - 4y + 8 = 4y - 4y - 7$

$5y + 8 = -7$

Subtract 8 from both sides:

$5y + 8 - 8 = -7 - 8$

$5y = -15$

Divide both sides by 5:

$y = \frac{-15}{5}$

$y = -3$

$-3$ is an integer.

The solutions are: (a) $y = -3$, (b) $x = 3$, (c) $z = -\frac{5}{2}$, (d) $y = -3$.

The equation whose solution is not an integer is $3z + 8 = 3 + z$, with the solution $z = -\frac{5}{2}$.

The correct option is (c).

The given equation is:

$7x + 4 = 25$

To find the value of $x$, we need to isolate $x$ on one side of the equation.

Subtract 4 from both sides of the equation:

$7x + 4 - 4 = 25 - 4$

$7x = 21$

Now, divide both sides by 7:

$\frac{7x}{7} = \frac{21}{7}$

$x = 3$

The value of $x$ is 3.

Comparing this with the given options:

(a) $\frac{29}{7}$

(b) $\frac{100}{7}$

(c) 2

(d) 3

The solution $x = 3$ matches option (d).

The correct option is (d).

The given equation is:

$3x + 7 = -20$

To find the solution, we need to isolate the variable $x$.

First, subtract 7 from both sides of the equation:

$3x + 7 - 7 = -20 - 7$

$3x = -27$

Now, divide both sides of the equation by 3:

$\frac{3x}{3} = \frac{-27}{3}$

$x = -9$

The solution of the equation $3x + 7 = -20$ is $x = -9$.

Comparing this result with the given options:

(a) $\frac{17}{7}$

(b) – 9

(c) 9

(d) $\frac{13}{3}$

The solution matches option (b).

The correct option is (b).

We are given two expressions, $(y - 15)$ and $(2y + 1)$. We need to find the value of $y$ for which these expressions are equal. This means we set the two expressions equal to each other and solve for $y$.

The equation is:

$y - 15 = 2y + 1$

To solve for $y$, we can gather the terms involving $y$ on one side and the constant terms on the other side.

Subtract $y$ from both sides of the equation:

$y - 15 - y = 2y + 1 - y$

$-15 = y + 1$

Now, subtract 1 from both sides of the equation to isolate $y$:

$-15 - 1 = y + 1 - 1$

$-16 = y$

So, $y = -16$.

The value of $y$ for which the two expressions are equal is -16.

Comparing this with the given options:

(a) 0

(b) 16

(c) 8

(d) – 16

The solution $y = -16$ matches option (d).

The correct option is (d).

We are given the equation:

$k + 7 = 16$

First, we need to find the value of $k$ by solving this equation.

Subtract 7 from both sides of the equation:

$k + 7 - 7 = 16 - 7$

$k = 9$

Now that we have the value of $k$, we need to find the value of the expression $8k - 72$.

Substitute the value of $k = 9$ into the expression:

Value of expression $= 8k - 72$

Value of expression $= 8(9) - 72$

Perform the multiplication:

$8 \times 9 = 72$

Now perform the subtraction:

Value of expression $= 72 - 72$

Value of expression $= 0$

The value of the expression $8k - 72$ is 0.

Comparing this result with the given options:

(a) 0

(b) 1

(c) 112

(d) 56

The calculated value matches option (a).

The correct option is (a).

The given equation is:

$43m = 0.086$

To find the value of $m$, we need to isolate $m$ on one side of the equation.

Divide both sides of the equation by the coefficient of $m$, which is 43:

$\frac{43m}{43} = \frac{0.086}{43}$

$m = \frac{0.086}{43}$

Now, perform the division:

$0.086 \div 43$

We can rewrite $0.086$ as $86 \times 10^{-3}$ or $\frac{86}{1000}$.

So, $m = \frac{86/1000}{43} = \frac{86}{1000 \times 43} = \frac{86}{43000}$

Since $86 = 2 \times 43$, we have:

$m = \frac{2 \times \cancel{43}}{\cancel{43} \times 1000} = \frac{2}{1000}$

$m = 0.002$

The value of $m$ is 0.002.

Comparing this with the given options:

(a) 0.002

(b) 0.02

(c) 0.2

(d) 2

The calculated value matches option (a).

The correct option is (a).

The statement "x exceeds 3 by 7" means that $x$ is greater than 3, and the difference between $x$ and 3 is 7.

Mathematically, this can be written as the difference between $x$ and 3 being equal to 7.

So, we can write:

$x - 3 = 7$

Let's check the meaning of the other options:

(a) $x + 3 = 2$: This means "x increased by 3 is 2" or "the sum of x and 3 is 2".

(b) $x + 7 = 3$: This means "x increased by 7 is 3" or "the sum of x and 7 is 3".

(d) $x - 7 = 3$: This means "x decreased by 7 is 3" or "the difference between x and 7 is 3".

Our derived equation $x - 3 = 7$ directly represents the statement "x exceeds 3 by 7".

Comparing this with the given options, option (c) is $x - 3 = 7$.

The correct option is (c).

We need to check which of the given equations is satisfied when $x = 5$. We will substitute $x = 5$ into each equation and verify if the Left Hand Side (LHS) equals the Right Hand Side (RHS).

(a) $4x + 1 = 2$

Substitute $x = 5$ into the LHS:

LHS $= 4(5) + 1 = 20 + 1 = 21$

RHS $= 2$

Since LHS $\neq$ RHS ($21 \neq 2$), $x = 5$ is not a solution for this equation.

(b) $3 - x = 8$

Substitute $x = 5$ into the LHS:

LHS $= 3 - 5 = -2$

RHS $= 8$

Since LHS $\neq$ RHS ($-2 \neq 8$), $x = 5$ is not a solution for this equation.

(c) $x - 5 = 3$

Substitute $x = 5$ into the LHS:

LHS $= 5 - 5 = 0$

RHS $= 3$

Since LHS $\neq$ RHS ($0 \neq 3$), $x = 5$ is not a solution for this equation.

(d) $3 + x = 8$

Substitute $x = 5$ into the LHS:

LHS $= 3 + 5 = 8$

RHS $= 8$

Since LHS $=$ RHS ($8 = 8$), $x = 5$ is a solution for this equation.

Therefore, the equation having 5 as a solution is $3 + x = 8$.

The correct option is (d).

We need to check which of the given equations is satisfied when the variable (which is $x$ in these options) is replaced by -3. We will substitute $x = -3$ into the Left Hand Side (LHS) of each equation and check if it equals the Right Hand Side (RHS).

(a) $x + 3 = 1$

Substitute $x = -3$ into the LHS:

LHS $= -3 + 3 = 0$

RHS $= 1$

Since LHS $\neq$ RHS ($0 \neq 1$), $x = -3$ is not a solution for this equation.

(b) $8 + 2x = 3$

Substitute $x = -3$ into the LHS:

LHS $= 8 + 2(-3) = 8 - 6 = 2$

RHS $= 3$

Since LHS $\neq$ RHS ($2 \neq 3$), $x = -3$ is not a solution for this equation.

(c) $10 + 3x = 1$

Substitute $x = -3$ into the LHS:

LHS $= 10 + 3(-3) = 10 - 9 = 1$

RHS $= 1$

Since LHS $=$ RHS ($1 = 1$), $x = -3$ is a solution for this equation.

(d) $2x + 1 = 3$

Substitute $x = -3$ into the LHS:

LHS $= 2(-3) + 1 = -6 + 1 = -5$

RHS $= 3$

Since LHS $\neq$ RHS ($-5 \neq 3$), $x = -3$ is not a solution for this equation.

Therefore, the equation having -3 as a solution is $10 + 3x = 1$.

The correct option is (c).

The question asks which of the given equations has $x = 0$ as its solution. To check this, we substitute $x = 0$ into each equation and see if the equality holds true (LHS = RHS).

(a) $2x + 1 = -1$

Substitute $x = 0$ into the LHS:

LHS $= 2(0) + 1 = 0 + 1 = 1$

RHS $= -1$

Since LHS $\neq$ RHS ($1 \neq -1$), $x = 0$ is not a solution for this equation.

(b) $\frac{x}{2} + 5 = 7$

Substitute $x = 0$ into the LHS:

LHS $= \frac{0}{2} + 5 = 0 + 5 = 5$

RHS $= 7$

Since LHS $\neq$ RHS ($5 \neq 7$), $x = 0$ is not a solution for this equation.

(c) $3x - 1 = -1$

Substitute $x = 0$ into the LHS:

LHS $= 3(0) - 1 = 0 - 1 = -1$

RHS $= -1$

Since LHS $=$ RHS ($-1 = -1$), $x = 0$ is a solution for this equation.

(d) $3x - 1 = 1$

Substitute $x = 0$ into the LHS:

LHS $= 3(0) - 1 = 0 - 1 = -1$

RHS $= 1$

Since LHS $\neq$ RHS ($-1 \neq 1$), $x = 0$ is not a solution for this equation.

The equation that has $x = 0$ as a solution is $3x - 1 = -1$.

The correct option is (c).

We need to find the equation for which $x = 7$ is not a solution. We will substitute $x = 7$ into each equation and check if the equality holds (LHS = RHS).

(a) $2x + 1 = 15$

Substitute $x = 7$ into the LHS:

LHS $= 2(7) + 1 = 14 + 1 = 15$

RHS $= 15$

Since LHS $=$ RHS ($15 = 15$), $x = 7$ is a solution for this equation. This equation can be formed using $x=7$ (e.g., start with $x=7$, multiply by 2 to get $2x=14$, add 1 to get $2x+1=15$).

(b) $7x - 1 = 50$

Substitute $x = 7$ into the LHS:

LHS $= 7(7) - 1 = 49 - 1 = 48$

RHS $= 50$

Since LHS $\neq$ RHS ($48 \neq 50$), $x = 7$ is not a solution for this equation. This equation cannot be formed starting with $x=7$ using valid operations.

(c) $x - 3 = 4$

Substitute $x = 7$ into the LHS:

LHS $= 7 - 3 = 4$

RHS $= 4$

Since LHS $=$ RHS ($4 = 4$), $x = 7$ is a solution for this equation. This equation can be formed using $x=7$ (e.g., start with $x=7$, subtract 3 to get $x-3 = 7-3 = 4$).

(d) $\frac{x}{7} - 1 = 0$

Substitute $x = 7$ into the LHS:

LHS $= \frac{7}{7} - 1 = 1 - 1 = 0$

RHS $= 0$

Since LHS $=$ RHS ($0 = 0$), $x = 7$ is a solution for this equation. This equation can be formed using $x=7$ (e.g., start with $x=7$, divide by 7 to get $\frac{x}{7} = 1$, subtract 1 to get $\frac{x}{7}-1 = 1-1 = 0$).

The equation that cannot be formed using the equation $x = 7$ is the one for which $x = 7$ is not a solution, which is $7x - 1 = 50$.

The correct option is (b).

We are given the equation:

$\frac{x}{2} = 3$

To find the value of $x$, we need to isolate $x$. Multiply both sides of the equation by 2:

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

$x = 6$

Now that we have the value of $x$, we need to find the value of the expression $3x + 2$.

Substitute $x = 6$ into the expression:

Value of expression $= 3x + 2$

Value of expression $= 3(6) + 2$

Perform the multiplication and addition:

$3(6) = 18$

$18 + 2 = 20$

The value of the expression is 20.

Comparing this result with the given options:

(a) 20

(b) 11

(c) $\frac{13}{2}$

(d) 8

The calculated value matches option (a).

The correct option is (a).

The given equation is:

$-6 + x = -12$

We need to find the value of $x$ that satisfies this equation. To do this, we solve for $x$.

To isolate $x$, add 6 to both sides of the equation:

$-6 + x + 6 = -12 + 6$

$x = -6$

The value of $x$ that satisfies the equation is -6.

Comparing this result with the given options:

(a) 2

(b) 6

(c) – 6

(d) – 2

The solution $x = -6$ matches option (c).

The correct option is (c).

The question asks for the name of the process where a term is moved from one side of an equation to the other side, and its sign is changed.

Let's consider a simple example:

$x + 5 = 10$

To solve for $x$, we move the term '+5' from the left side to the right side. When we move it, its sign changes from positive to negative:

$x = 10 - 5$

$x = 5$

This process of moving a term from one side of an equation to the other with a change in sign is called transposition.

Let's briefly look at the other options:

(a) Commutativity: This is a property that states the order of operands does not affect the result for certain operations (e.g., $a + b = b + a$, $a \times b = b \times a$). It is not related to moving terms in an equation.

(c) Distributivity: This is a property relating multiplication and addition/subtraction (e.g., $a(b + c) = ab + ac$). It is not related to moving terms in an equation.

(d) Associativity: This is a property that states the grouping of operands does not affect the result for certain operations (e.g., $(a + b) + c = a + (b + c)$, $(a \times b) \times c = a \times (b \times c)$). It is not related to moving terms in an equation.

Therefore, the correct term for shifting a term from one side of an equation to another with a change of sign is transposition.

The correct option is (b).

Part (a):

Given that the sum of the two numbers is 60 and the smaller number is $x$.

Let the other number be $y$. According to the problem, the sum of the two numbers is 60, so we have:

$x + y = 60$

To find the other number ($y$) in terms of $x$, we rearrange the equation:

$y = 60 - x$

The other number is $\underline{60 - x}$.

Part (b):

The difference of the two numbers is the larger number minus the smaller number.

Since the smaller number is $x$ and the other number is $60-x$, and their sum is 60 while their difference is 30, the other number ($60-x$) must be the larger number.

Difference = (Larger Number) - (Smaller Number)

Difference = $(60 - x) - x$

Simplifying the expression:

Difference = $60 - x - x$

Difference = $60 - 2x$.

The difference of numbers in term of x is $\underline{60 - 2x}$.

Part (c):

The difference between the two numbers is given as 30.

From part (b), we found the difference in terms of $x$ is $60 - 2x$.

Equating the expression for the difference to the given difference, we form the equation:

$60 - 2x = 30$.

The equation formed is $\underline{60 - 2x = 30}$.

Part (d):

To find the solution of the equation $60 - 2x = 30$, we need to solve for $x$.

Start with the equation:

$60 - 2x = 30$

Subtract 60 from both sides:

$-2x = 30 - 60$

$-2x = -30$

Divide both sides by -2:

$x = \frac{-30}{-2}$

$x = 15$.

The solution of the equation is $\underline{15}$.

Part (e):

From part (d), we found the value of the smaller number, $x$, to be 15.

The other number is given by $60 - x$ (from part a).

Substitute the value of $x$ into the expression for the other number:

Other number = $60 - 15 = 45$.

The two numbers are 15 and 45.

Let's verify the conditions: Sum = $15 + 45 = 60$. Difference = $45 - 15 = 30$. The conditions are satisfied.

The numbers are $\underline{15}$ and $\underline{45}$.

Part (a):

Let the smaller number be $x$.

Given that one number is twice the other. Since $x$ is the smaller number, the other number must be twice $x$.

Other number = $2 \times x = 2x$.

The other number is $\underline{2x}$.

Part (b):

The sum of the two numbers is given as 81.

The two numbers are $x$ (smaller number) and $2x$ (other number).

Sum of the numbers = $x + 2x$.

Equating the sum to 81, we get the equation:

$x + 2x = 81$

Combining like terms on the left side:

$3x = 81$.

The equation formed is $\underline{3x = 81}$.

Part (c):

To find the solution of the equation $3x = 81$, we need to solve for $x$.

Divide both sides of the equation by 3:

$\frac{3x}{3} = \frac{81}{3}$

$x = 27$.

The solution of the equation is $\underline{27}$.

Part (d):

The smaller number is $x$, which we found to be 27 from part (c).

The other number is $2x$ (from part a).

Substitute the value of $x$ into the expression for the other number:

Other number = $2 \times 27 = 54$.

The two numbers are 27 and 54.

Verification: Sum = $27 + 54 = 81$. One number (54) is twice the other number (27). The conditions are met.

The numbers are $\underline{27}$ and $\underline{54}$.

Part (a):

Let Palak's marks be $x$.

Given that Abha gets twice the marks as that of Palak.

So, Abha's marks are $2 \times x = 2x$.

If Palak gets x marks, Abha gets $\underline{2x}$ marks.

Part (b):

According to the problem, two times Abha's marks and three times Palak's marks make 280.

Two times Abha's marks = $2 \times (2x) = 4x$.

Three times Palak's marks = $3 \times x = 3x$.

The sum of these is 280:

$4x + 3x = 280$.

Combining like terms:

$7x = 280$.

The equation formed is $\underline{7x = 280}$.

Part (c):

To find the solution of the equation $7x = 280$, we solve for $x$.

Divide both sides of the equation by 7:

$\frac{7x}{7} = \frac{280}{7}$

$x = 40$.

The solution of the equation is $\underline{40}$.

Part (d):

Palak's marks are $x$, which is 40.

Abha's marks are $2x$ (from part a).

Substitute the value of $x = 40$ into the expression for Abha's marks:

Abha's marks = $2 \times 40 = 80$.

Verification: Palak's marks = 40, Abha's marks = 80. Abha's marks are twice Palak's. Two times Abha's marks ($2 \times 80 = 160$) plus three times Palak's marks ($3 \times 40 = 120$) equals $160 + 120 = 280$. The conditions are satisfied.

Marks obtained by Abha are $\underline{80}$.

Part (a):

Let the breadth of the rectangle be $x$ cm.

Given that the length of the rectangle is two times its breadth.

Length = $2 \times \text{breadth} = 2 \times x = 2x$ cm.

If the breadth of rectangle is x cm, the length of the rectangle is $\underline{2x}$.

Part (b):

The formula for the perimeter of a rectangle is $2 \times (\text{length} + \text{breadth})$.

Substitute the length ($2x$) and breadth ($x$) in terms of $x$ into the formula:

Perimeter = $2 \times (2x + x)$

Perimeter = $2 \times (3x)$

Perimeter = $6x$ cm.

Perimeter in terms of x is $\underline{6x}$.

Part (c):

The perimeter of the rectangle is given as 60 cm.

From part (b), the perimeter in terms of $x$ is $6x$.

Equating the two expressions for the perimeter, we form the equation:

$6x = 60$.

The equation formed is $\underline{6x = 60}$.

Part (d):

To find the solution of the equation $6x = 60$, we solve for $x$.

Divide both sides of the equation by 6:

$\frac{6x}{6} = \frac{60}{6}$

$x = 10$.

The solution of the equation is $\underline{10}$.

Part (a):

Let $x$ be the number of $\textsf{₹} 5$ coins.

The worth of $x$ coins of $\textsf{₹} 5$ each is the number of coins multiplied by the value of each coin.

Worth = $x \times \textsf{₹} 5 = \textsf{₹} 5x$.

The worth of x coins of $\textsf{₹} 5$ each is $\underline{\textsf{₹} 5x}$.

Part (b):

Given that the number of $\textsf{₹} 2$ coins is equal to the number of $\textsf{₹} 5$ coins. So, there are also $x$ coins of $\textsf{₹} 2$.

The worth of $x$ coins of $\textsf{₹} 2$ each is the number of coins multiplied by the value of each coin.

Worth = $x \times \textsf{₹} 2 = \textsf{₹} 2x$.

The worth of x coins of $\textsf{₹} 2$ each is $\underline{\textsf{₹} 2x}$.

Part (c):

The total worth of the coins in the bag is the sum of the worth of $\textsf{₹} 5$ coins and $\textsf{₹} 2$ coins.

Total worth = (Worth of $\textsf{₹} 5$ coins) + (Worth of $\textsf{₹} 2$ coins)

Total worth = $\textsf{₹} 5x + \textsf{₹} 2x = \textsf{₹} (5x + 2x) = \textsf{₹} 7x$.

The total worth is given as $\textsf{₹} 70$.

Equating the total worth in terms of $x$ to the given total worth, we form the equation:

$7x = 70$.

The equation formed is $\underline{7x = 70}$.

Part (d):

To find the number of $\textsf{₹} 5$ and $\textsf{₹} 2$ coins, we need to solve the equation $7x = 70$ for $x$.

Divide both sides of the equation by 7:

$\frac{7x}{7} = \frac{70}{7}$

$x = 10$.

Since $x$ represents the number of $\textsf{₹} 5$ coins and also the number of $\textsf{₹} 2$ coins, there are 10 $\textsf{₹} 5$ coins and 10 $\textsf{₹} 2$ coins.

Verification: Worth of 10 $\textsf{₹} 5$ coins = $10 \times \textsf{₹} 5 = \textsf{₹} 50$. Worth of 10 $\textsf{₹} 2$ coins = $10 \times \textsf{₹} 2 = \textsf{₹} 20$. Total worth = $\textsf{₹} 50 + \textsf{₹} 20 = \textsf{₹} 70$. The conditions are met.

There are $\underline{10}$ 5 rupee coins and $\underline{10}$ 2 rupee coins.

Part (a):

Given that the total number of prizes is 30.

Let the number of 1st prizes be $x$.

The number of 2nd prizes is the total number of prizes minus the number of 1st prizes.

Number of 2nd prizes = $30 - x$.

If 1st prizes are x in number the number of 2nd prizes are $\underline{30 - x}$.

Part (b):

The value of each 1st prize is $\textsf{₹} 2000$.

The value of $x$ 1st prizes is $x \times \textsf{₹} 2000 = \textsf{₹} 2000x$.

The value of each 2nd prize is $\textsf{₹} 1000$.

The number of 2nd prizes is $30 - x$ (from part a).

The value of $30 - x$ 2nd prizes is $(30 - x) \times \textsf{₹} 1000 = \textsf{₹} (30000 - 1000x)$.

The total value of prizes is the sum of the value of 1st prizes and 2nd prizes.

Total value = $\textsf{₹} 2000x + \textsf{₹} (30000 - 1000x) = \textsf{₹} (2000x + 30000 - 1000x) = \textsf{₹} (1000x + 30000)$.

The total value of prizes in terms of x are $\underline{\textsf{₹} (1000x + 30000)}$.

Part (c):

The total prize money is given as $\textsf{₹} 52,000$.

From part (b), the total value of prizes in terms of $x$ is $\textsf{₹} (1000x + 30000)$.

Equating the two expressions for the total value:

$1000x + 30000 = 52000$.

The equation formed is $\underline{1000x + 30000 = 52000}$.

Part (d):

To find the solution of the equation $1000x + 30000 = 52000$, we solve for $x$.

Subtract 30000 from both sides:

$1000x = 52000 - 30000$

$1000x = 22000$

Divide both sides by 1000:

$x = \frac{22000}{1000}$

$x = 22$.

The solution of the equation is $\underline{22}$.

Part (e):

The number of 1st prizes is $x$, which we found to be 22 from part (d).

The number of 2nd prizes is $30 - x$ (from part a).

Substitute the value of $x = 22$ into the expression for the number of 2nd prizes:

Number of 2nd prizes = $30 - 22 = 8$.

Verification: Number of 1st prizes = 22, Number of 2nd prizes = 8. Total prizes = $22 + 8 = 30$. Value of 1st prizes = $22 \times \textsf{₹} 2000 = \textsf{₹} 44000$. Value of 2nd prizes = $8 \times \textsf{₹} 1000 = \textsf{₹} 8000$. Total prize money = $\textsf{₹} 44000 + \textsf{₹} 8000 = \textsf{₹} 52000$. The conditions are satisfied.

The number of 1st prizes are $\underline{22}$ and the number of 2nd prizes are $\underline{8}$.

Given equation is:

$z + 3 = 5$

To find the value of $z$, we need to isolate $z$ on one side of the equation.

Subtract 3 from both sides of the equation:

$z + 3 - 3 = 5 - 3$

Simplifying both sides:

$z = 2$

Therefore, if $z + 3 = 5$, then $z = \underline{2}$.

We are given the equation:

$3x - 2 = 7$

To solve for $x$, first, add 2 to both sides of the equation:

$3x - 2 + 2 = 7 + 2$

$3x = 9$

Next, divide both sides by 3:

$\frac{3x}{3} = \frac{9}{3}$

$x = 3$

The solution of the equation $3x - 2 = 7$ is $x = 3$.

$\underline{3}$ is the solution of the equation 3x – 2 =7.

We are given the equation:

$3x + 10 = 7$

To solve for $x$, first, subtract 10 from both sides of the equation:

$3x + 10 - 10 = 7 - 10$

$3x = -3$

Next, divide both sides by 3:

$\frac{3x}{3} = \frac{-3}{3}$

$x = -1$

The solution of the equation $3x + 10 = 7$ is $x = -1$.

$\underline{-1}$ is the solution of 3x + 10 = 7.

We are given the equation:

$2x + 3 = 5$

First, we need to find the value of $x$ by solving this equation.

Subtract 3 from both sides:

$2x + 3 - 3 = 5 - 3$

$2x = 2$

Divide both sides by 2:

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

$x = 1$

Now we need to find the value of the expression $3x + 2$ using the value of $x$ we just found.

Substitute $x = 1$ into the expression $3x + 2$:

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

The value of $3x + 2$ is 5.

If 2x + 3 = 5, then value of 3x + 2 is $\underline{5}$.

We are given the equation:

$4x - 1 = 8$

To solve for $x$, add 1 to both sides of the equation:

$4x - 1 + 1 = 8 + 1$

$4x = 9$

Now, divide both sides by 4:

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

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

We are asked about the solution in integers. An integer is a whole number (positive, negative, or zero). The value we found for $x$ is $\frac{9}{4}$, which is a fraction and not an integer.

Therefore, the equation $4x - 1 = 8$ has no solution within the set of integers.

In integers, 4x – 1 = 8 has $\underline{\text{no integer}}$ solution.

We are given the equation:

$4x + 5 = -7$

To solve for $x$, subtract 5 from both sides of the equation:

$4x + 5 - 5 = -7 - 5$

$4x = -12$

Now, divide both sides by 4:

$\frac{4x}{4} = \frac{-12}{4}$

$x = -3$

We are asked about the solution in natural numbers. Natural numbers are positive integers (1, 2, 3, ...). The value we found for $x$ is $-3$, which is a negative integer and not a natural number.

Therefore, the equation $4x + 5 = -7$ has no solution within the set of natural numbers.

In natural numbers, 4x + 5 = – 7 has $\underline{\text{no natural number}}$ solution.

We are given the equation:

$x - 5 = -5$

To solve for $x$, add 5 to both sides of the equation:

$x - 5 + 5 = -5 + 5$

$x = 0$

We are asked about the solution in natural numbers. Natural numbers are typically defined as the positive integers $\{1, 2, 3, ...\}$. Some definitions include 0, but the most common definition in this context excludes 0.

The solution we found is $x = 0$. According to the standard definition, 0 is not a natural number.

Therefore, the equation $x - 5 = -5$ has no solution within the set of natural numbers.

In natural numbers, x – 5 = – 5 has $\underline{\text{no natural number}}$ solution.

We are given the equation:

$x + 8 = 12 - 4$

First, simplify the right side of the equation:

$12 - 4 = 8$

So the equation becomes:

$x + 8 = 8$

To solve for $x$, subtract 8 from both sides of the equation:

$x + 8 - 8 = 8 - 8$

$x = 0$

We are asked about the solution in whole numbers. Whole numbers are the set of non-negative integers $\{0, 1, 2, 3, ...\}$.

The solution we found is $x = 0$, which is indeed a whole number.

Therefore, the equation $x + 8 = 12 - 4$ has a solution in whole numbers.

In whole numbers, x + 8 = 12 – 4 has $\underline{\text{a whole number}}$ solution.

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

According to the problem, "three times a number" is $3x$.

"5 is added to three times a number" can be written as $3x + 5$.

"Four times the same number" is $4x$.

"7 is subtracted from four times the same number" can be written as $4x - 7$.

The problem states that these two expressions "become the same", which means they are equal.

So, the equation that represents this fact is:

$3x + 5 = 4x - 7$

This fact can be represented as $\underline{3x + 5 = 4x - 7}$.

We are given the equation:

$x + 7 = 10$

To find the solution for $x$, subtract 7 from both sides of the equation:

$x + 7 - 7 = 10 - 7$

$x = 3$

The solution of the equation $x + 7 = 10$ is $x = 3$.

x + 7 = 10 has the solution $\underline{3}$.

We are given the equation $3x = 12$.

First, we need to find the value of $x$ by solving this equation.

Divide both sides of the equation by 3:

$\frac{3x}{3} = \frac{12}{3}$

$x = 4$

Now, we need to find the value of the expression $x - 0$ using the value of $x$ we just found.

Substitute $x = 4$ into the expression $x - 0$:

$4 - 0 = 4$

The value of $x - 0$ is 4 when $3x = 12$.

x – 0 = $\underline{4}$; when 3x = 12.

We are given the equation $2x = 2$.

First, we need to find the value of $x$ by solving this equation.

Divide both sides of the equation by 2:

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

$x = 1$

Now, we need to find the value of the expression $x - 1$ using the value of $x$ we just found.

Substitute $x = 1$ into the expression $x - 1$:

$1 - 1 = 0$

The value of $x - 1$ is 0 when $2x = 2$.

x – 1= $\underline{0}$; when 2x = 2.

We are given the equation $\frac{x}{2} = 6$.

First, we need to find the value of $x$ by solving this equation.

Multiply both sides of the equation by 2:

$\frac{x}{2} \times 2 = 6 \times 2$

$x = 12$

Now, we need to find the number that, when subtracted from $x$ (which is 12), results in 15.

Let the blank be represented by the variable $y$. The statement is $x - y = 15$.

Substitute $x = 12$ into the statement:

$12 - y = 15$

To solve for $y$, subtract 12 from both sides:

$-y = 15 - 12$

$-y = 3$

Multiply both sides by -1:

$y = -3$

So the blank should be filled with -3.

x – $\underline{(-3)}$ = 15; when $\frac{x}{2}$ = 6.

Alternatively, if the question means what is the value of $x - 15$, then substitute $x=12$ directly: $12-15 = -3$. The structure of the sentence implies $x - \text{something} = 15$. My interpretation above fits this structure.

Let's re-read the question carefully. "x –_________ = 15". This is an equation structure where the blank is the unknown. Let the blank be $y$. We need to find $y$ such that $x - y = 15$ when $\frac{x}{2} = 6$. We found $x=12$. So, $12 - y = 15$. Solving for $y$, we get $y = 12 - 15 = -3$. The blank is -3.

Final Answer Check:

When $\frac{x}{2} = 6$, $x = 12$.

The statement is $x - \underline{y} = 15$.

$12 - \underline{y} = 15$

$\underline{y} = 12 - 15 = -3$.

x – $\underline{-3}$ = 15; when $\frac{x}{2}$ = 6.

We are given the equation:

$x + 15 = 19$

To find the solution for $x$, subtract 15 from both sides of the equation:

$x + 15 - 15 = 19 - 15$

$x = 4$

The solution of the equation x + 15 = 19 is $\underline{4}$.

In a linear equation, the value of the variable that makes the equation true (i.e., makes the left side equal to the right side) is known as the solution of the equation.

Finding this value is the process of solving the equation.

The value itself is specifically called the solution.

Based on this definition, the blank should be filled with the term that describes this value.

Finding the value of a variable in a linear equation that $\underline{\text{satisfies}}$ the equation is called a $\underline{\text{solution}}$ of the equation.

Considering the structure of the blank in the question, it is asking for what the found value is called.

The blank should be filled with "solution".

Finding the value of a variable in a linear equation that the equation is called a $\underline{\text{solution}}$ of the equation.

When a term is moved from one side of an equation to the other, the operation associated with that term is reversed, which effectively means changing its sign.

For example, if a term is being added on one side, when transposed to the other side, it is subtracted (sign changes from positive to negative).

If a term is being subtracted on one side, when transposed to the other side, it is added (sign changes from negative to positive).

This process is known as transposition, and it relies on changing the sign of the term being moved.

Any term of an equation may be transposed from one side of the equation to the other side of the equation by changing the $\underline{\text{sign}}$ of the term.

We are given the equation:

$\frac{9}{5} x = \frac{18}{5}$

To solve for $x$, we need to isolate $x$ on one side of the equation.

We can multiply both sides of the equation by the reciprocal of $\frac{9}{5}$, which is $\frac{5}{9}$.

$\left(\frac{5}{9}\right) \times \left(\frac{9}{5} x\right) = \left(\frac{5}{9}\right) \times \left(\frac{18}{5}\right)$

On the left side, $\frac{5}{9} \times \frac{9}{5} = 1$, so the left side simplifies to $1x$ or just $x$.

On the right side, we can cancel out common factors:

$\frac{\cancel{5}^{1}}{\cancel{9}_{1}} \times \frac{\cancel{18}^{2}}{\cancel{5}_{1}} = \frac{1 \times 2}{1 \times 1} = 2$

So the equation simplifies to:

$x = 2$

Therefore, if $\frac{9}{5}$ x = $\frac{18}{5}$, then x = $\underline{2}$.

Alternatively, we could multiply both sides by 5 first to eliminate the denominators:

$5 \times \left(\frac{9}{5} x\right) = 5 \times \left(\frac{18}{5}\right)$

$9x = 18$

Then divide both sides by 9:

$\frac{9x}{9} = \frac{18}{9}$

$x = 2$

This gives the same result.

We are given the equation:

$3 - x = -4$

To solve for $x$, we can first isolate the term with $x$. Subtract 3 from both sides of the equation:

$3 - x - 3 = -4 - 3$

$-x = -7$

Now, to find $x$, we need to eliminate the negative sign. Multiply or divide both sides by -1:

$(-1) \times (-x) = (-1) \times (-7)$

$x = 7$

Therefore, if $3 - x = -4$, then $x = \underline{7}$.

We are given the equation:

$x - \frac{1}{2} = -\frac{1}{2}$

To find the value of $x$, we need to isolate $x$ on one side of the equation.

Add $\frac{1}{2}$ to both sides of the equation:

$x - \frac{1}{2} + \frac{1}{2} = -\frac{1}{2} + \frac{1}{2}$

On the left side, $-\frac{1}{2} + \frac{1}{2} = 0$, so the left side simplifies to $x + 0$ or just $x$.

On the right side, $-\frac{1}{2} + \frac{1}{2} = 0$.

So the equation simplifies to:

$x = 0$

Therefore, if x − $\frac{1}{2}$ = $-\frac{1}{2}$, then x = $\underline{0}$.

We are given the equation:

$\frac{1}{6} - x = \frac{1}{6}$

To find the value of $x$, we can isolate the term with $x$. Subtract $\frac{1}{6}$ from both sides of the equation:

$\frac{1}{6} - x - \frac{1}{6} = \frac{1}{6} - \frac{1}{6}$

On the left side, $\frac{1}{6} - \frac{1}{6} = 0$, so the left side becomes $0 - x = -x$.

On the right side, $\frac{1}{6} - \frac{1}{6} = 0$.

So the equation simplifies to:

$-x = 0$

To find $x$, multiply or divide both sides by -1:

$(-1) \times (-x) = (-1) \times 0$

$x = 0$

Therefore, if $\frac{1}{6}$ - x = $\frac{1}{6}$, then x = $\underline{0}$.

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

The phrase "10 less than a number" means we subtract 10 from the number. This can be written as $n - 10$.

The problem states that "10 less than a number is 65". This translates to the equation:

$n - 10 = 65$

To solve for $n$, we need to isolate $n$ on one side of the equation. Add 10 to both sides:

$n - 10 + 10 = 65 + 10$

$n = 75$

The number is 75.

Verification: 10 less than 75 is $75 - 10 = 65$. This matches the given information.

If 10 less than a number is 65, then the number is $\underline{75}$.

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

"A number is increased by 20" means we add 20 to the number. This can be written as $n + 20$.

The problem states that "it becomes 45". This means the result of increasing the number by 20 is equal to 45.

This translates to the equation:

$n + 20 = 45$

To solve for $n$, we need to isolate $n$ on one side of the equation. Subtract 20 from both sides:

$n + 20 - 20 = 45 - 20$

$n = 25$

The number is 25.

Verification: If the number is 25 and it is increased by 20, we get $25 + 20 = 45$. This matches the given information.

If a number is increased by 20, it becomes 45. Then the number is $\underline{25}$.

Let the other number be represented by the variable $m$.

The statement "84 exceeds another number by 12" means that 84 is greater than the other number by an amount of 12.

This can be written as:

$84 = m + 12$

Alternatively, the difference between 84 and the other number is 12:

$84 - m = 12$

We can solve either equation for $m$. Let's use the second equation: $84 - m = 12$.

To isolate $m$, subtract 12 from both sides:

$84 - m - 12 = 12 - 12$

$72 - m = 0$

Add $m$ to both sides:

$72 - m + m = 0 + m$

$72 = m$

Alternatively, using the first equation $84 = m + 12$, subtract 12 from both sides:

$84 - 12 = m + 12 - 12$

$72 = m$

The other number is 72.

Verification: Does 84 exceed 72 by 12? $84 - 72 = 12$. Yes, it does.

If 84 exceeds another number by 12, then the other number is $\underline{72}$.

We are given the equation:

$x - \frac{7}{8} = \frac{7}{8}$

To find the value of $x$, we need to isolate $x$ on one side of the equation.

Add $\frac{7}{8}$ to both sides of the equation:

$x - \frac{7}{8} + \frac{7}{8} = \frac{7}{8} + \frac{7}{8}$

On the left side, $-\frac{7}{8} + \frac{7}{8} = 0$, so the left side simplifies to $x + 0$ or just $x$.

On the right side, we add the fractions:

$\frac{7}{8} + \frac{7}{8} = \frac{7 + 7}{8} = \frac{14}{8}$

Simplify the fraction $\frac{14}{8}$ by dividing both the numerator and the denominator by their greatest common divisor, which is 2:

$\frac{\cancel{14}^{7}}{\cancel{8}_{4}} = \frac{7}{4}$

So the equation simplifies to:

$x = \frac{7}{4}$

Therefore, if x − $\frac{7}{8}$ = $\frac{7}{8}$, then x = $\underline{\frac{7}{4}}$.

To check if 5 is the solution of the equation $3x + 2 = 17$, we substitute $x = 5$ into the equation and see if the left side equals the right side.

Left Hand Side (LHS) = $3x + 2$

Substitute $x = 5$:

LHS = $3(5) + 2$

LHS = $15 + 2$

LHS = $17$

Right Hand Side (RHS) = $17$

Since LHS = RHS ($17 = 17$), the equation is satisfied when $x = 5$.

Therefore, 5 is the solution of the equation $3x + 2 = 17$.

The statement is True.

To check if $\frac{9}{5}$ is the solution of the equation $4x - 1 = 8$, we substitute $x = \frac{9}{5}$ into the equation and see if the left side equals the right side.

Left Hand Side (LHS) = $4x - 1$

Substitute $x = \frac{9}{5}$:

LHS = $4\left(\frac{9}{5}\right) - 1$

LHS = $\frac{4 \times 9}{5} - 1$

LHS = $\frac{36}{5} - 1$

To subtract 1, we write 1 as a fraction with a denominator of 5:

LHS = $\frac{36}{5} - \frac{5}{5}$

LHS = $\frac{36 - 5}{5}$

LHS = $\frac{31}{5}$

Right Hand Side (RHS) = $8$

We can write 8 as a fraction with a denominator of 5 for comparison:

RHS = $\frac{8 \times 5}{5} = \frac{40}{5}$

Since LHS ($\frac{31}{5}$) is not equal to RHS ($\frac{40}{5}$), the equation is not satisfied when $x = \frac{9}{5}$.

Let's find the actual solution of the equation $4x - 1 = 8$ for verification.

$4x - 1 = 8$

Add 1 to both sides:

$4x = 8 + 1$

$4x = 9$

Divide by 4:

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

The actual solution is $\frac{9}{4}$, not $\frac{9}{5}$.

The statement is False.

We need to find the solution of the equation $4x - 5 = 7$ and check if it is an integer.

Given equation:

$4x - 5 = 7$

Add 5 to both sides of the equation:

$4x - 5 + 5 = 7 + 5$

$4x = 12$

Divide both sides by 4:

$\frac{4x}{4} = \frac{12}{4}$

$x = 3$

The solution to the equation is $x = 3$.

An integer is a whole number (positive, negative, or zero). The number 3 is a positive whole number, so it is an integer.

The statement claims that the equation does not have an integer as its solution. However, we found that the solution is 3, which is an integer.

The statement is False.

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

"One third of a number" can be written as $\frac{1}{3} x$ or $\frac{x}{3}$.

"One third of a number added to itself" means adding $\frac{x}{3}$ to the original number $x$. This can be written as $\frac{x}{3} + x$.

"Gives 10" means the result of the operation is equal to 10. So the equation should be:

$\frac{x}{3} + x = 10$

The statement says the fact can be represented as $\frac{x}{3} + 10 = x$.

Comparing our derived equation $\frac{x}{3} + x = 10$ with the given equation $\frac{x}{3} + 10 = x$, we see that they are different.

The given equation $\frac{x}{3} + 10 = x$ means "One third of a number increased by 10 is equal to the number itself". This is not what the problem states.

The statement is False.

To check if $\frac{3}{2}$ is the solution of the equation $8x - 5 = 7$, we substitute $x = \frac{3}{2}$ into the equation and evaluate the left side.

The given equation is $8x - 5 = 7$.

Substitute $x = \frac{3}{2}$ into the Left Hand Side (LHS):

LHS = $8x - 5$

LHS = $8\left(\frac{3}{2}\right) - 5$

LHS = $\frac{8 \times 3}{2} - 5$

LHS = $\frac{24}{2} - 5$

LHS = $12 - 5$

LHS = $7$

The Right Hand Side (RHS) of the equation is $7$.

Since LHS = RHS ($7 = 7$), the equation is satisfied when $x = \frac{3}{2}$.

Therefore, $\frac{3}{2}$ is indeed the solution of the equation $8x - 5 = 7$.

The statement is True.

We need to check if $x = 4$ is the solution to the equation $4x - 7 = 11$. We can do this by substituting $x = 4$ into the equation.

Given equation: $4x - 7 = 11$.

Substitute $x = 4$ into the Left Hand Side (LHS):

LHS = $4x - 7$

LHS = $4(4) - 7$

LHS = $16 - 7$

LHS = $9$

The Right Hand Side (RHS) of the equation is $11$.

Since LHS (9) is not equal to RHS (11), the equation is not satisfied when $x = 4$.

Alternatively, we can solve the equation $4x - 7 = 11$ to find the actual solution.

$4x - 7 = 11$

Add 7 to both sides:

$4x = 11 + 7$

$4x = 18$

Divide both sides by 4:

$x = \frac{18}{4}$

Simplify the fraction:

$x = \frac{\cancel{18}^{9}}{\cancel{4}_{2}} = \frac{9}{2}$

The actual solution is $x = \frac{9}{2}$ or 4.5, not 4.

The statement is False.

We are given the equation:

$\frac{5x \;-\; 7}{2} = y$

We are told that 9 is the solution for the variable $x$. This means we should substitute $x = 9$ into the equation to find the corresponding value of $y$.

Substitute $x = 9$ into the equation:

$y = \frac{5(9) \;-\; 7}{2}$

Perform the multiplication in the numerator:

$y = \frac{45 \;-\; 7}{2}$

Perform the subtraction in the numerator:

$y = \frac{38}{2}$

Perform the division:

$y = 19$

When $x = 9$, the value of $y$ is 19.

The statement claims that the value of $y$ is 28. Since $19 \neq 28$, the statement is incorrect.

The statement is False.

Solving each entry in Column I:

(i) $x + 5 = 9$

Subtract 5 from both sides: $x = 9 - 5 = 4$.

Matches with (C) 4.

(ii) $x – 7 = 4$

Add 7 to both sides: $x = 4 + 7 = 11$.

Matches with (E) 11.

(iii) $\frac{x}{12} = – 5$

Multiply both sides by 12: $x = -5 \times 12 = -60$.

Matches with (F) – 60.

(iv) $5x = 30$

Divide both sides by 5: $x = \frac{30}{5} = 6$.

Matches with (D) 6.

(v) The value of y which satisfies $3y = 5$

Divide both sides by 3: $y = \frac{5}{3}$.

Matches with (B) $\frac{5}{3}$.

(vi) If $p = 2$, then the value of $\frac{1}{3}(1 – 3p)$

Substitute $p = 2$: $\frac{1}{3}(1 – 3(2)) = \frac{1}{3}(1 – 6) = \frac{1}{3}(-5) = -\frac{5}{3}$.

Matches with (A) $-\frac{5}{3}$.

Matching results:

(i) $\to$ (C)

(ii) $\to$ (E)

(iii) $\to$ (F)

(iv) $\to$ (D)

(v) $\to$ (B)

(vi) $\to$ (A)

Let the number be $x$.

Twice of the number is $2x$.

13 subtracted from twice of the number is $2x - 13$.

This "gives 3", which means it is equal to 3.

The equation formed is:

$2x - 13 = 3$

Let the number be $n$.

One-fifth of the number is $\frac{1}{5}n$ or $\frac{n}{5}$.

"5 less than that number" means subtracting 5 from the number. This is $n - 5$.

The statement "One-fifth of a number is 5 less than that number" means that $\frac{n}{5}$ is equal to $n - 5$.

The equation formed is:

$\frac{n}{5} = n - 5$

Let the number be $y$.

One-third of itself is $\frac{1}{3}y$ or $\frac{y}{3}$.

"7 more than one-third of itself" means adding 7 to one-third of the number. This is $\frac{y}{3} + 7$.

The statement "A number is 7 more than one-third of itself" means that the number itself ($y$) is equal to 7 more than one-third of itself ($\frac{y}{3} + 7$).

The equation formed is:

$y = \frac{y}{3} + 7$

Let the number be $a$.

Six times a number is $6a$.

"10 more than the number" means adding 10 to the number. This is $a + 10$.

The statement "Six times a number is 10 more than the number" means that $6a$ is equal to $a + 10$.

The equation formed is:

$6a = a + 10$

Let the number be $k$.

Half of a number is $\frac{1}{2}k$ or $\frac{k}{2}$.

"10 is subtracted from half of a number" means $\frac{k}{2} - 10$.

The statement "the result is 4" means the expression is equal to 4.

The equation formed is:

$\frac{k}{2} - 10 = 4$

The variable is given as $p$.

"Subtracting 5 from p" means $p - 5$.

"The result is 2" means the expression is equal to 2.

The equation formed is:

$p - 5 = 2$

Let the number be $n$.

"Five times a number" is $5n$.

"Increased by 7" means adding 7 to $5n$. This is $5n + 7$.

"Is 27" means the expression is equal to 27.

The equation formed is:

$5n + 7 = 27$

Let Sohan's age be $s$ years.

"Mohan is 3 years older than Sohan" means Mohan's age is Sohan's age plus 3. So, Mohan's age is $s + 3$ years.

"The sum of their ages is 43 years" means Sohan's age plus Mohan's age equals 43.

Sohan's age + Mohan's age = 43

$s + (s + 3) = 43$

Combine like terms on the left side:

$2s + 3 = 43$

The equation formed is:

$2s + 3 = 43$

Let the number be $m$.

"1 is subtracted from a number" means $m - 1$. This is the difference.

"The difference is multiplied by $\frac{1}{2}$" means multiplying $(m - 1)$ by $\frac{1}{2}$. This is $\frac{1}{2}(m - 1)$.

"The result is 7" means the expression is equal to 7.

The equation formed is:

$\frac{1}{2}(m - 1) = 7$

Let the number be $p$.

"A number divided by 2" is $\frac{p}{2}$.

"Then increased by 5" means adding 5 to the previous result. This is $\frac{p}{2} + 5$.

"Is 9" means the expression is equal to 9.

The equation formed is:

$\frac{p}{2} + 5 = 9$

Let the number be $y$.

"Twice a number" is $2y$.

"The sum of twice a number and 4" means adding 4 to $2y$. This is $2y + 4$.

"Is 18" means the expression is equal to 18.

The equation formed is:

$2y + 4 = 18$

Given:

1. The age of Sohan Lal is four times that of his son Amit.

2. The difference of their ages is 27 years.

To Find:

The age of Amit.

Solution:

Let the age of Amit be $x$ years.

According to the first given statement, the age of Sohan Lal is four times that of his son Amit.

Sohan Lal's age = $4 \times (\text{Amit's age}) = 4x$ years.

According to the second given statement, the difference of their ages is 27 years.

The difference between their ages is the age of the older person minus the age of the younger person.

Since Sohan Lal is older (his age is four times Amit's age, assuming ages are positive), the difference is:

Sohan Lal's age - Amit's age = 27

Substitute the expressions for their ages in terms of $x$:

$4x - x = 27$

Now, we solve the equation for $x$:

Combine like terms on the left side:

$3x = 27$

Divide both sides by 3:

$\frac{3x}{3} = \frac{27}{3}$

$x = 9$

So, the age of Amit is 9 years.

We can find Sohan Lal's age to verify: Sohan Lal's age = $4x = 4 \times 9 = 36$ years.

The difference in their ages is $36 - 9 = 27$ years, which matches the given information.

The age of Amit is $\underline{9 \text{ years}}$.

Given:

1. One number exceeds the other by 12.

2. The sum of the two numbers is 72.

To Find:

The two numbers.

Solution:

Let the smaller number be $x$.

According to the first given statement, the other number exceeds the smaller number by 12.

Other number = $x + 12$.

According to the second given statement, the sum of the two numbers is 72.

Smaller number + Other number = 72

$x + (x + 12) = 72$

Now, we solve the equation for $x$:

Combine like terms on the left side:

$x + x + 12 = 72$

$2x + 12 = 72$

Subtract 12 from both sides:

$2x + 12 - 12 = 72 - 12$

$2x = 60$

Divide both sides by 2:

$\frac{2x}{2} = \frac{60}{2}$

$x = 30$

So, the smaller number is $x = 30$.

The other number is $x + 12 = 30 + 12 = 42$.

Verification: The difference between the numbers is $42 - 30 = 12$, which matches the first condition. The sum of the numbers is $30 + 42 = 72$, which matches the second condition.

The two numbers are $\underline{30}$ and $\underline{42}$.

Given:

Seven times a number is 12 less than thirteen times the same number.

To Find:

The number.

Solution:

Let the number be $n$.

"Seven times a number" is $7n$.

"Thirteen times the same number" is $13n$.

"12 less than thirteen times the same number" means subtracting 12 from $13n$. This is $13n - 12$.

The statement "Seven times a number is 12 less than thirteen times the same number" means that $7n$ is equal to $13n - 12$.

$7n = 13n - 12$

Now, we solve the equation for $n$:

Subtract $7n$ from both sides to move the variable terms to one side:

$7n - 7n = 13n - 7n - 12$

$0 = 6n - 12$

Add 12 to both sides to isolate the term with the variable:

$0 + 12 = 6n - 12 + 12$

$12 = 6n$

Divide both sides by 6 to solve for $n$:

$\frac{12}{6} = \frac{6n}{6}$

$2 = n$

So, $n = 2$.

Verification: Seven times the number is $7 \times 2 = 14$. Thirteen times the number is $13 \times 2 = 26$. Is 14 equal to 12 less than 26? $26 - 12 = 14$. Yes, it is.

The number is $\underline{2}$.

Given:

1. The interest received by Karim is $\textsf{₹} 30$ more than that of Ramesh.

2. The total interest received by them is $\textsf{₹} 70$.

To Find:

The interest received by Ramesh.

Solution:

Let the interest received by Ramesh be $\textsf{₹} r$.

According to the first given statement, the interest received by Karim is $\textsf{₹} 30$ more than that of Ramesh.

Karim's interest = Ramesh's interest + $\textsf{₹} 30 = \textsf{₹} (r + 30)$.

According to the second given statement, the total interest received by them is $\textsf{₹} 70$.

Ramesh's interest + Karim's interest = $\textsf{₹} 70$

$r + (r + 30) = 70$

Now, we solve the equation for $r$:

Combine like terms on the left side:

$r + r + 30 = 70$

$2r + 30 = 70$

Subtract 30 from both sides:

$2r + 30 - 30 = 70 - 30$

$2r = 40$

Divide both sides by 2:

$\frac{2r}{2} = \frac{40}{2}$

$r = 20$

So, the interest received by Ramesh is $\textsf{₹} 20$.

We can find the interest received by Karim to verify: Karim's interest = $\textsf{₹} (r + 30) = \textsf{₹} (20 + 30) = \textsf{₹} 50$.

The total interest is $\textsf{₹} 20 + \textsf{₹} 50 = \textsf{₹} 70$, which matches the given information.

The interest received by Ramesh is $\underline{\textsf{₹} 20}$.

Given:

1. The amount paid by Naidu is $\textsf{₹} 125$ more than that of Subramaniam.

2. The total money paid by them is $\textsf{₹} 975$.

To Find:

The amount of money donated by Subramaniam.

Solution:

Let the amount donated by Subramaniam be $\textsf{₹} s$.

According to the first given statement, the amount paid by Naidu is $\textsf{₹} 125$ more than that of Subramaniam.

Naidu's donation = Subramaniam's donation + $\textsf{₹} 125 = \textsf{₹} (s + 125)$.

According to the second given statement, the total money paid by them is $\textsf{₹} 975$.

Subramaniam's donation + Naidu's donation = $\textsf{₹} 975$

$s + (s + 125) = 975$

Now, we solve the equation for $s$:

Combine like terms on the left side:

$s + s + 125 = 975$

$2s + 125 = 975$

Subtract 125 from both sides:

$2s + 125 - 125 = 975 - 125$

$2s = 850$

Divide both sides by 2:

$\frac{2s}{2} = \frac{850}{2}$

$s = 425$

So, the amount of money donated by Subramaniam is $\textsf{₹} 425$.

We can find Naidu's donation to verify: Naidu's donation = $\textsf{₹} (s + 125) = \textsf{₹} (425 + 125) = \textsf{₹} 550$.

The total donation is $\textsf{₹} 425 + \textsf{₹} 550 = \textsf{₹} 975$, which matches the given information.

The amount of money donated by Subramaniam is $\underline{\textsf{₹} 425}$.

Given:

1. The number of girls is 50 more than the number of boys.

2. The total number of students is 1070.

To Find:

The number of girls.

Solution:

Let the number of boys in the school be $b$.

According to the first given statement, the number of girls is 50 more than the number of boys.

Number of girls = Number of boys + 50 = $b + 50$.

According to the second given statement, the total number of students is 1070.

Total students = Number of boys + Number of girls

$1070 = b + (b + 50)$

Now, we solve the equation for $b$:

Combine like terms on the right side:

$1070 = b + b + 50$

$1070 = 2b + 50$

Subtract 50 from both sides:

$1070 - 50 = 2b + 50 - 50$

$1020 = 2b$

Divide both sides by 2:

$\frac{1020}{2} = \frac{2b}{2}$

$510 = b$

So, the number of boys is $b = 510$.

We need to find the number of girls.

Number of girls = $b + 50$

Substitute the value of $b = 510$:

Number of girls = $510 + 50 = 560$.

Verification: Number of boys = 510, Number of girls = 560. Is the number of girls 50 more than boys? $560 - 510 = 50$. Yes. Is the total number of students 1070? $510 + 560 = 1070$. Yes. The conditions are satisfied.

The number of girls is $\underline{560}$.

Given:

Two times a number increased by 5 equals 9.

To Find:

The number.

Solution:

Let the number be $n$.

"Two times a number" is $2n$.

"Increased by 5" means adding 5 to $2n$. This is $2n + 5$.

"Equals 9" means the expression is equal to 9.

The equation formed is:

$2n + 5 = 9$

Now, we solve the equation for $n$:

Subtract 5 from both sides:

$2n + 5 - 5 = 9 - 5$

$2n = 4$

Divide both sides by 2:

$\frac{2n}{2} = \frac{4}{2}$

$n = 2$

The number is 2.

Verification: Two times the number is $2 \times 2 = 4$. Increased by 5 is $4 + 5 = 9$. This matches the given condition.

The number is $\underline{2}$.

Given:

9 added to twice a number gives 13.

To Find:

The number.

Solution:

Let the number be $x$.

"Twice a number" is $2x$.

"9 added to twice a number" means adding 9 to $2x$. This is $2x + 9$.

"Gives 13" means the expression is equal to 13.

The equation formed is:

$2x + 9 = 13$

Now, we solve the equation for $x$:

Subtract 9 from both sides:

$2x + 9 - 9 = 13 - 9$

$2x = 4$

Divide both sides by 2:

$\frac{2x}{2} = \frac{4}{2}$

$x = 2$

The number is 2.

Verification: Twice the number is $2 \times 2 = 4$. 9 added to this is $4 + 9 = 13$. This matches the given condition.

The number is $\underline{2}$.

Given:

1 subtracted from one-third of a number gives 1.

To Find:

The number.

Solution:

Let the number be $n$.

"One-third of a number" is $\frac{n}{3}$.

"1 subtracted from one-third of a number" is $\frac{n}{3} - 1$.

"Gives 1" means the expression is equal to 1.

The equation formed is:

$\frac{n}{3} - 1 = 1$

Now, we solve the equation for $n$:

Add 1 to both sides:

$\frac{n}{3} - 1 + 1 = 1 + 1$

$\frac{n}{3} = 2$

Multiply both sides by 3:

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

$n = 6$

The number is 6.

Verification: One-third of the number is $\frac{1}{3} \times 6 = 2$. 1 subtracted from this is $2 - 1 = 1$. This matches the given condition.

The number is $\underline{6}$.

Given:

1 subtracted from one-third of a number gives 1.

To Find:

The number.

Solution:

Let the number be $n$.

"One-third of a number" is $\frac{n}{3}$.

"1 subtracted from one-third of a number" is $\frac{n}{3} - 1$.

"Gives 1" means the expression is equal to 1.

The equation formed is:

$\frac{n}{3} - 1 = 1$

Now, we solve the equation for $n$:

Add 1 to both sides:

$\frac{n}{3} - 1 + 1 = 1 + 1$

$\frac{n}{3} = 2$

Multiply both sides by 3:

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

$n = 6$

The number is 6.

Verification: One-third of the number is $\frac{1}{3} \times 6 = 2$. 1 subtracted from this is $2 - 1 = 1$. This matches the given condition.

The number is $\underline{6}$.

Given:

1. After 25 years, Rama's age will be 5 times his present age.

To Find:

Rama's present age.

Solution:

Let Rama's present age be $a$ years.

After 25 years, Rama's age will be $a + 25$ years.

According to the given information, after 25 years, his age will be 5 times his present age. So, we can write the equation:

$a + 25 = 5a$

Now, we solve this equation for $a$.

Subtract $a$ from both sides of the equation to collect variable terms on one side:

$a + 25 - a = 5a - a$

$25 = 4a$

Divide both sides by 4 to isolate $a$:

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

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

So, $a = \frac{25}{4}$.

Rama's present age is $\frac{25}{4}$ years, which can also be written as $6.25$ years or $6 \frac{1}{4}$ years.

Verification: Present age = 6.25 years. After 25 years, age = $6.25 + 25 = 31.25$ years. 5 times the present age = $5 \times 6.25 = 31.25$ years. The condition is met.

Rama's present age is $\underline{6.25 \text{ years}}$.

Given:

1. After 20 years, Manoj's age will be 5 times his present age.

To Find:

Manoj's present age.

Solution:

Let Manoj's present age be $m$ years.

After 20 years, Manoj's age will be $m + 20$ years.

According to the given information, after 20 years, his age will be 5 times his present age. So, we can write the equation:

$m + 20 = 5m$

Now, we solve this equation for $m$.

Subtract $m$ from both sides of the equation to collect variable terms on one side:

$m + 20 - m = 5m - m$

$20 = 4m$

Divide both sides by 4 to isolate $m$:

$\frac{20}{4} = \frac{4m}{4}$

$5 = m$

So, $m = 5$.

Manoj's present age is 5 years.

Verification: Present age = 5 years. After 20 years, age = $5 + 20 = 25$ years. 5 times the present age = $5 \times 5 = 25$ years. The condition is met.

Manoj's present age is $\underline{5 \text{ years}}$.

Given:

Present age is 3 times (age 3 years from now) minus 3 times (age 3 years ago).

To Find:

The sister's present age.

Solution:

Let the sister's present age be $a$ years.

Her age 3 years from now will be $a + 3$ years.

Her age 3 years ago was $a - 3$ years.

According to the given statement, her present age ($a$) is equal to 3 times her age 3 years from now ($a+3$) minus 3 times her age 3 years ago ($a-3$).

So, we can write the equation:

$a = 3 \times (a + 3) - 3 \times (a - 3)$

Now, we solve this equation for $a$.

First, distribute the multiplication on the right side:

$a = (3 \times a + 3 \times 3) - (3 \times a - 3 \times 3)$

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

Remove the parentheses. Remember to distribute the negative sign to both terms inside the second parenthesis:

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

$a = 3a + 9 - 3a + 9$

Combine like terms on the right side ($3a$ and $-3a$ cancel each other out, and $9 + 9 = 18$):

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

$a = 0 + 18$

$a = 18$

So, the sister's present age is 18 years.

Verification: Present age = 18 years. Age 3 years from now = $18 + 3 = 21$ years. Age 3 years ago = $18 - 3 = 15$ years. Check the condition: Is 18 equal to $3 \times (21) - 3 \times (15)$?

$3 \times 21 = 63$

$3 \times 15 = 45$

$63 - 45 = 18$. Yes, it is.

Her present age is $\underline{18 \text{ years}}$.

Given:

When 45 is added to half a number, the result is triple the number.

To Find:

The number.

Solution:

Let the number be $n$.

Half a number is $\frac{1}{2}n$ or $\frac{n}{2}$.

"45 is added to half a number" is $\frac{n}{2} + 45$.

"Triple the number" is $3n$.

The statement says "the result is triple the number", meaning $\frac{n}{2} + 45$ is equal to $3n$.

The equation formed is:

$\frac{n}{2} + 45 = 3n$

Now, we solve this equation for $n$.

To eliminate the fraction, multiply the entire equation by 2:

$2 \times \left(\frac{n}{2} + 45\right) = 2 \times (3n)$

$2 \times \frac{n}{2} + 2 \times 45 = 6n$

$n + 90 = 6n$

Subtract $n$ from both sides to collect variable terms on one side:

$n + 90 - n = 6n - n$

$90 = 5n$

Divide both sides by 5 to isolate $n$:

$\frac{90}{5} = \frac{5n}{5}$

$18 = n$

So, $n = 18$.

The number is 18.

Verification: Half the number is $\frac{18}{2} = 9$. Add 45: $9 + 45 = 54$. Triple the number is $3 \times 18 = 54$. The results match, so the condition is met.

The number is $\underline{18}$.

Given:

1. The consumption of wheat is 4 times that of rice.

2. The total consumption of the two cereals is 80 kg.

To Find:

The quantities of rice and wheat consumed.

Solution:

Let the quantity of rice consumed be $r$ kg.

According to the first given statement, the consumption of wheat is 4 times that of rice.

Quantity of wheat consumed = $4 \times (\text{Quantity of rice}) = 4r$ kg.

According to the second given statement, the total consumption of the two cereals is 80 kg.

Quantity of rice + Quantity of wheat = 80

$r + 4r = 80$

Now, we solve the equation for $r$:

Combine like terms on the left side:

$5r = 80$

Divide both sides by 5:

$\frac{5r}{5} = \frac{80}{5}$

$r = 16$

So, the quantity of rice consumed is $r = 16$ kg.

The quantity of wheat consumed is $4r = 4 \times 16 = 64$ kg.

Verification: Quantity of rice = 16 kg, Quantity of wheat = 64 kg. Is the consumption of wheat 4 times that of rice? $64 = 4 \times 16$. Yes. Is the total consumption 80 kg? $16 + 64 = 80$. Yes. The conditions are satisfied.

The quantity of rice consumed is $\underline{16 \text{ kg}}$ and the quantity of wheat consumed is $\underline{64 \text{ kg}}$.

Given:

1. The number of one rupee coins is three times the number of two rupees coins.

2. The total worth of the coins is $\textsf{₹} 120$.

To Find:

The number of $\textsf{₹} 1$ coins.

Solution:

Let the number of $\textsf{₹} 2$ coins be $t$.

According to the first given statement, the number of $\textsf{₹} 1$ coins is three times the number of $\textsf{₹} 2$ coins.

Number of $\textsf{₹} 1$ coins = $3 \times (\text{Number of }\textsf{₹} 2 \text{ coins}) = 3t$.

The value of the $\textsf{₹} 2$ coins is the number of coins multiplied by their value:

Value of $\textsf{₹} 2$ coins = $t \times \textsf{₹} 2 = \textsf{₹} 2t$.

The value of the $\textsf{₹} 1$ coins is the number of coins multiplied by their value:

Value of $\textsf{₹} 1$ coins = $3t \times \textsf{₹} 1 = \textsf{₹} 3t$.

According to the second given statement, the total worth of the coins is $\textsf{₹} 120$.

Value of $\textsf{₹} 2$ coins + Value of $\textsf{₹} 1$ coins = $\textsf{₹} 120$

$2t + 3t = 120$

Now, we solve the equation for $t$:

Combine like terms on the left side:

$5t = 120$

Divide both sides by 5:

$\frac{5t}{5} = \frac{120}{5}$

$t = 24$

So, the number of $\textsf{₹} 2$ coins is $t = 24$.

We need to find the number of $\textsf{₹} 1$ coins.

Number of $\textsf{₹} 1$ coins = $3t = 3 \times 24 = 72$.

Verification: Number of $\textsf{₹} 2$ coins = 24 (worth $\textsf{₹} 24 \times 2 = \textsf{₹} 48$). Number of $\textsf{₹} 1$ coins = 72 (worth $\textsf{₹} 72 \times 1 = \textsf{₹} 72$). Is the number of $\textsf{₹} 1$ coins three times the number of $\textsf{₹} 2$ coins? $72 = 3 \times 24$. Yes. Is the total worth $\textsf{₹} 120$? $\textsf{₹} 48 + \textsf{₹} 72 = \textsf{₹} 120$. Yes. The conditions are satisfied.

The number of 1 rupee coins is $\underline{72}$.

Given:

1. A number is thought of.

2. It is multiplied by 2.

3. 5 is added to the product.

4. The result is 17.

To Find:

The number Anamika thought of.

Solution:

Let the number Anamika thought of be $n$.

She multiplied the number by 2: $2 \times n = 2n$.

She added 5 to the product: $2n + 5$.

She obtained 17 as the result. This means the expression is equal to 17.

The equation formed is:

$2n + 5 = 17$

Now, we solve this equation for $n$:

Subtract 5 from both sides:

$2n + 5 - 5 = 17 - 5$

$2n = 12$

Divide both sides by 2:

$\frac{2n}{2} = \frac{12}{2}$

$n = 6$

The number Anamika thought of is 6.

Verification: Start with the number 6. Multiply by 2: $6 \times 2 = 12$. Add 5: $12 + 5 = 17$. The result matches the given information.

The number she had thought of is $\underline{6}$.

Given:

1. One of the two numbers is twice the other.

2. The sum of the numbers is 12.

To Find:

The two numbers.

Solution:

Let the smaller number be $x$.

According to the first given statement, the other number is twice the smaller number.

Other number = $2 \times x = 2x$.

According to the second given statement, the sum of the two numbers is 12.

Smaller number + Other number = 12

$x + 2x = 12$

Now, we solve the equation for $x$:

Combine like terms on the left side:

$3x = 12$

Divide both sides by 3:

$\frac{3x}{3} = \frac{12}{3}$

$x = 4$

So, the smaller number is $x = 4$.

The other number is $2x = 2 \times 4 = 8$.

Verification: Is one number twice the other? 8 is twice of 4 ($8 = 2 \times 4$). Yes. Is the sum of the numbers 12? $4 + 8 = 12$. Yes. The conditions are satisfied.

The two numbers are $\underline{4}$ and $\underline{8}$.

Given:

The sum of three consecutive integers is 5 more than the smallest of the integers.

To Find:

The three consecutive integers.

Solution:

Let the smallest of the three consecutive integers be $n$.

Since the integers are consecutive, the next integer is $n+1$, and the largest integer is $(n+1)+1 = n+2$.

The three consecutive integers are $n$, $n+1$, and $n+2$.

The sum of the three consecutive integers is $n + (n+1) + (n+2)$.

The smallest of the integers is $n$.

"5 more than the smallest of the integers" is $n + 5$.

According to the given statement, the sum of the integers is equal to 5 more than the smallest integer.

So, the equation is:

$n + (n+1) + (n+2) = n + 5$

Now, we solve this equation for $n$.

Combine like terms on the left side:

$n + n + 1 + n + 2 = n + 5$

$3n + 3 = n + 5$

Subtract $n$ from both sides of the equation:

$3n + 3 - n = n + 5 - n$

$2n + 3 = 5$

Subtract 3 from both sides:

$2n + 3 - 3 = 5 - 3$

$2n = 2$

Divide both sides by 2:

$\frac{2n}{2} = \frac{2}{2}$

$n = 1$

So, the smallest integer is $n = 1$.

The second integer is $n + 1 = 1 + 1 = 2$.

The third integer is $n + 2 = 1 + 2 = 3$.

The three consecutive integers are 1, 2, and 3.

Verification: Sum of the integers = $1 + 2 + 3 = 6$. Smallest integer = 1. Is the sum 5 more than the smallest? $1 + 5 = 6$. Yes. The conditions are satisfied.

The integers are $\underline{1, 2, \text{ and } 3}$.

Given:

A number divided by 6 gives a quotient of 6.

To Find:

The number.

Solution:

Let the unknown number be $N$.

The statement "A number when divided by 6 gives the quotient 6" can be written as a division equation:

$\frac{N}{6} = 6$

Now, we solve this equation for $N$.

To isolate $N$, multiply both sides of the equation by 6:

$\frac{N}{6} \times 6 = 6 \times 6$

$N = 36$

The number is 36.

Verification: If we divide 36 by 6, do we get a quotient of 6? $36 \div 6 = 6$. Yes. The condition is satisfied.

The number is $\underline{36}$.

Note: The relationship between dividend, divisor, quotient, and remainder is Dividend = Divisor $\times$ Quotient + Remainder. In this case, the remainder is not mentioned, implying it is 0. So, Number = $6 \times 6 + 0 = 36$.

Given:

1. The perimeter of a rectangle is 40 m.

2. The length of the rectangle is 4 m less than 5 times its breadth.

To Find:

The length of the rectangle.

Solution:

Let the breadth of the rectangle be $b$ meters.

According to the second given statement, the length is 4 m less than 5 times its breadth.

5 times the breadth is $5b$.

4 m less than 5 times the breadth is $5b - 4$.

So, the length of the rectangle is $l = 5b - 4$ meters.

The formula for the perimeter of a rectangle is $P = 2 \times (\text{length} + \text{breadth})$.

We are given that the perimeter is 40 m.

Substitute the length and breadth in terms of $b$ into the perimeter formula:

$40 = 2 \times ((5b - 4) + b)$

Now, we solve this equation for $b$:

Simplify the expression inside the parentheses:

$40 = 2 \times (5b + b - 4)$

$40 = 2 \times (6b - 4)$

Distribute the 2 on the right side:

$40 = 2 \times 6b - 2 \times 4$

$40 = 12b - 8$

Add 8 to both sides:

$40 + 8 = 12b - 8 + 8$

$48 = 12b$

Divide both sides by 12:

$\frac{48}{12} = \frac{12b}{12}$

$4 = b$

So, the breadth of the rectangle is $b = 4$ meters.

We need to find the length of the rectangle.

Length $l = 5b - 4$.

Substitute the value of $b = 4$:

$l = 5(4) - 4$

$l = 20 - 4$

$l = 16$

The length of the rectangle is 16 meters.

Verification: Breadth = 4 m, Length = 16 m. Is the length 4 m less than 5 times the breadth? 5 times the breadth = $5 \times 4 = 20$. 4 less than this is $20 - 4 = 16$. Yes. Is the perimeter 40 m? Perimeter = $2 \times (16 + 4) = 2 \times 20 = 40$. Yes. The conditions are satisfied.

The length of the rectangle is $\underline{16 \text{ m}}$.

Given:

1. The triangle is isosceles (2 equal sides).

2. Each of the 2 equal sides is twice as large as the third side.

3. The perimeter of the triangle is 30 cm.

To Find:

The length of each side of the triangle.

Solution:

In an isosceles triangle, there are two equal sides and one different side (the base, if the equal sides are the legs).

Let the length of the third side (the unequal side) be $s$ cm.

According to the second given statement, each of the 2 equal sides is twice as large as the third side.

Length of each equal side = $2 \times (\text{length of third side}) = 2s$ cm.

So, the lengths of the three sides of the triangle are $2s$, $2s$, and $s$ cm.

The perimeter of a triangle is the sum of the lengths of its three sides.

We are given that the perimeter is 30 cm.

Perimeter = Side 1 + Side 2 + Side 3

$30 = (2s) + (2s) + (s)$

Now, we solve this equation for $s$:

Combine like terms on the right side:

$30 = 2s + 2s + s$

$30 = 5s$

Divide both sides by 5:

$\frac{30}{5} = \frac{5s}{5}$

$6 = s$

So, the length of the third side is $s = 6$ cm.

The length of each of the equal sides is $2s = 2 \times 6 = 12$ cm.

The lengths of the sides of the triangle are 12 cm, 12 cm, and 6 cm.

Verification: Are the two sides equal and twice the third? $12 = 2 \times 6$. Yes. Is the perimeter 30 cm? Perimeter = $12 + 12 + 6 = 30$. Yes. The conditions are satisfied.

The lengths of the sides of the triangle are $\underline{12 \text{ cm}}, \underline{12 \text{ cm}}$, and $\underline{6 \text{ cm}}$.

Given:

1. Two consecutive multiples of 2.

2. Their sum is 18.

To Find:

The two consecutive multiples of 2.

Solution:

Let the first multiple of 2 be $2n$, where $n$ is an integer.

Since the multiples are consecutive, the next multiple of 2 will be $2n + 2$.

According to the given statement, the sum of these two numbers is 18.

$(2n) + (2n + 2) = 18$

Now, we solve this equation for $n$:

Combine like terms on the left side:

$2n + 2n + 2 = 18$

$4n + 2 = 18$

Subtract 2 from both sides:

$4n + 2 - 2 = 18 - 2$

$4n = 16$

Divide both sides by 4:

$\frac{4n}{4} = \frac{16}{4}$

$n = 4$

Now we find the two numbers using the value of $n=4$:

First number = $2n = 2 \times 4 = 8$.

Second number = $2n + 2 = 2(4) + 2 = 8 + 2 = 10$.

The two consecutive multiples of 2 are 8 and 10.

Verification: Are they consecutive multiples of 2? Yes (8, 10). Is their sum 18? $8 + 10 = 18$. Yes. The conditions are satisfied.

The numbers are $\underline{8}$ and $\underline{10}$.

Given:

1. Two angles are complementary.

2. They differ by $20^\circ$.

To Find:

The measures of the two angles.

Solution:

Two angles are complementary if their sum is $90^\circ$.

Let the smaller angle be $x$ degrees.

Since the two complementary angles differ by $20^\circ$, the larger angle must be $x + 20$ degrees.

The sum of the two angles is $90^\circ$:

Smaller angle + Larger angle = $90^\circ$

$x + (x + 20) = 90$

Now, we solve this equation for $x$:

Combine like terms on the left side:

$x + x + 20 = 90$

$2x + 20 = 90$

Subtract 20 from both sides:

$2x + 20 - 20 = 90 - 20$

$2x = 70$

Divide both sides by 2:

$\frac{2x}{2} = \frac{70}{2}$

$x = 35$

So, the smaller angle is $x = 35^\circ$.

The larger angle is $x + 20 = 35 + 20 = 55^\circ$.

The two angles are $35^\circ$ and $55^\circ$.

Verification: Are they complementary? $35^\circ + 55^\circ = 90^\circ$. Yes. Do they differ by $20^\circ$? $55^\circ - 35^\circ = 20^\circ$. Yes. The conditions are satisfied.

The angles are $\underline{35^\circ}$ and $\underline{55^\circ}$.

Alternate Solution:

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

Since they are complementary, their sum is $90^\circ$:

$a + b = 90$ $\quad$(Assuming $a$ and $b$ are in degrees)

They differ by $20^\circ$. Let's assume $a \geq b$, so their difference is:

$a - b = 20$

We now have a system of two linear equations:

1) $a + b = 90$

2) $a - b = 20$

Add equation (1) and equation (2):

$(a + b) + (a - b) = 90 + 20$

$a + b + a - b = 110$

$2a = 110$

Divide by 2:

$a = \frac{110}{2} = 55$

Substitute the value of $a=55$ into equation (1) to find $b$:

$55 + b = 90$

$b = 90 - 55$

$b = 35$

The angles are $55^\circ$ and $35^\circ$. This matches the previous solution.

Given:

1. The number 150 is divided into two parts.

2. Twice the first part is equal to the second part.

To Find:

The two parts.

Solution:

Let the first part be $p_1$.

Since 150 is divided into two parts, the sum of the two parts is 150.

Let the second part be $p_2$. So, $p_1 + p_2 = 150$.

According to the second given statement, twice the first part is equal to the second part.

$2 \times p_1 = p_2$

$2p_1 = p_2$

Now we have a system of two equations:

1) $p_1 + p_2 = 150$

2) $p_2 = 2p_1$

We can substitute the expression for $p_2$ from equation (2) into equation (1):

$p_1 + (2p_1) = 150$

Combine like terms on the left side:

$3p_1 = 150$

Divide both sides by 3 to find $p_1$:

$\frac{3p_1}{3} = \frac{150}{3}$

$p_1 = 50$

So, the first part is 50.

Now find the second part using $p_2 = 2p_1$:

$p_2 = 2 \times 50 = 100$

The two parts are 50 and 100.

Verification: Do the parts sum to 150? $50 + 100 = 150$. Yes. Is twice the first part equal to the second part? $2 \times 50 = 100$. Yes. The conditions are satisfied.

The two parts are $\underline{50}$ and $\underline{100}$.

Alternate Solution:

Let the first part be $x$.

According to the second condition, the second part is twice the first part, so the second part is $2x$.

According to the first condition, the sum of the two parts is 150.

$x + 2x = 150$

$3x = 150$

$x = \frac{150}{3}$

$x = 50$

The first part is $x = 50$.

The second part is $2x = 2 \times 50 = 100$.

This gives the same result.

Given:

1. Total number of students in the class is 60.

2. The number of girls is one third the number of boys.

To Find:

The number of girls and the number of boys.

Solution:

Let the number of boys in the class be $b$.

According to the second given statement, the number of girls is one third the number of boys.

Number of girls = $\frac{1}{3} \times (\text{Number of boys}) = \frac{1}{3}b$ or $\frac{b}{3}$.

According to the first given statement, the total number of students is 60.

Total students = Number of boys + Number of girls

$60 = b + \frac{b}{3}$

Now, we solve this equation for $b$.

To combine the terms with $b$ on the right side, find a common denominator:

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

$60 = \frac{3b}{3} + \frac{b}{3}$

$60 = \frac{3b + b}{3}$

$60 = \frac{4b}{3}$

Multiply both sides by 3 to eliminate the denominator:

$60 \times 3 = \frac{4b}{3} \times 3$

$180 = 4b$

Divide both sides by 4 to isolate $b$:

$\frac{180}{4} = \frac{4b}{4}$

$45 = b$

So, the number of boys is $b = 45$.

Now find the number of girls.

Number of girls = $\frac{b}{3} = \frac{45}{3} = 15$.

The number of boys is 45 and the number of girls is 15.

Verification: Is the number of girls one third the number of boys? $15 = \frac{1}{3} \times 45 = 15$. Yes. Is the total number of students 60? $45 + 15 = 60$. Yes. The conditions are satisfied.

The number of boys is $\underline{45}$ and the number of girls is $\underline{15}$.

Alternate Solution:

Let the number of girls be $g$.

According to the second condition, the number of girls is one third the number of boys. This means the number of boys is three times the number of girls.

Number of boys = $3 \times (\text{Number of girls}) = 3g$.

According to the first condition, the total number of students is 60.

Number of girls + Number of boys = 60

$g + 3g = 60$

$4g = 60$

$g = \frac{60}{4}$

$g = 15$

The number of girls is $g = 15$.

The number of boys is $3g = 3 \times 15 = 45$.

This gives the same result.

Given:

Two-third of a number is greater than one-third of the number by 3.

To Find:

The number.

Solution:

Let the number be $n$.

Two-third of the number is $\frac{2}{3}n$.

One-third of the number is $\frac{1}{3}n$.

The statement "Two-third of a number is greater than one-third of the number by 3" means that the difference between two-third of the number and one-third of the number is 3.

So, the equation is:

$\frac{2}{3}n - \frac{1}{3}n = 3$

Now, we solve this equation for $n$.

Since the fractions have the same denominator, we can combine the terms on the left side:

$\left(\frac{2}{3} - \frac{1}{3}\right)n = 3$

$\frac{2 - 1}{3}n = 3$

$\frac{1}{3}n = 3$

Multiply both sides by 3 to isolate $n$:

$\frac{1}{3}n \times 3 = 3 \times 3$

$n = 9$

The number is 9.

Verification: Two-third of the number is $\frac{2}{3} \times 9 = 2 \times 3 = 6$. One-third of the number is $\frac{1}{3} \times 9 = 3$. Is 6 greater than 3 by 3? $6 - 3 = 3$. Yes. The condition is satisfied.

The number is $\underline{9}$.

Given:

A number is as much greater than 27 as it is less than 73.

To Find:

The number.

Solution:

Let the unknown number be $n$.

"A number is as much greater than 27" means the difference between the number and 27 is some value. Since the number is greater than 27, this difference is $n - 27$.

"It is less than 73" means the difference between 73 and the number is some value. Since the number is less than 73, this difference is $73 - n$.

The statement says these two differences are equal ("as much greater than... as it is less than...").

So, the equation formed is:

$n - 27 = 73 - n$

Now, we solve this equation for $n$:

Add $n$ to both sides to collect variable terms on one side:

$n - 27 + n = 73 - n + n$

$2n - 27 = 73$

Add 27 to both sides to isolate the term with the variable:

$2n - 27 + 27 = 73 + 27$

$2n = 100$

Divide both sides by 2 to solve for $n$:

$\frac{2n}{2} = \frac{100}{2}$

$n = 50$

The number is 50.

Verification: Is 50 greater than 27 by the same amount as it is less than 73?

Difference from 27 = $50 - 27 = 23$.

Difference from 73 = $73 - 50 = 23$.

The differences are equal (both 23). The condition is satisfied.

The number is $\underline{50}$.

Given:

1. Fraction of journey by train = $\frac{2}{5}$.

2. Fraction of journey by bus = $\frac{1}{3}$.

3. Fraction of journey by car = $\frac{1}{4}$.

4. Remaining distance by foot = 3 km.

To Find:

The total length of the journey.

Solution:

Let the total length of the journey be $J$ km.

Distance travelled by train = $\frac{2}{5}$ of $J = \frac{2}{5}J$ km.

Distance travelled by bus = $\frac{1}{3}$ of $J = \frac{1}{3}J$ km.

Distance travelled by car = $\frac{1}{4}$ of $J = \frac{1}{4}J$ km.

Distance travelled on foot = 3 km.

The sum of the distances travelled by train, bus, car, and foot must equal the total journey length, $J$.

Distance by train + Distance by bus + Distance by car + Distance on foot = Total journey

$\frac{2}{5}J + \frac{1}{3}J + \frac{1}{4}J + 3 = J$

Now, we solve this equation for $J$.

To combine the terms with $J$ on the left side, find a common denominator for the fractions $\frac{2}{5}$, $\frac{1}{3}$, and $\frac{1}{4}$. The least common multiple (LCM) of 5, 3, and 4 is 60.

Rewrite each fraction with a denominator of 60:

$\frac{2}{5}J = \frac{2 \times 12}{5 \times 12}J = \frac{24}{60}J$

$\frac{1}{3}J = \frac{1 \times 20}{3 \times 20}J = \frac{20}{60}J$

$\frac{1}{4}J = \frac{1 \times 15}{4 \times 15}J = \frac{15}{60}J$

Substitute these back into the equation:

$\frac{24}{60}J + \frac{20}{60}J + \frac{15}{60}J + 3 = J$}

Combine the fractions with $J$ on the left side:

$\left(\frac{24}{60} + \frac{20}{60} + \frac{15}{60}\right)J + 3 = J$

$\frac{24 + 20 + 15}{60}J + 3 = J$

$\frac{59}{60}J + 3 = J$

Subtract $\frac{59}{60}J$ from both sides to collect the $J$ terms on one side:

$3 = J - \frac{59}{60}J$

Write $J$ as a fraction with a denominator of 60:

$J = \frac{60}{60}J$

$3 = \frac{60}{60}J - \frac{59}{60}J$}

$3 = \frac{60 - 59}{60}J$

$3 = \frac{1}{60}J$}

Multiply both sides by 60 to solve for $J$:

$3 \times 60 = \frac{1}{60}J \times 60$

$180 = J$

So, the total length of the journey is 180 km.

Verification:

Total journey = 180 km.

Train: $\frac{2}{5} \times 180 = 2 \times 36 = 72$ km.

Bus: $\frac{1}{3} \times 180 = 60$ km.

Car: $\frac{1}{4} \times 180 = 45$ km.

Foot: 3 km.

Total distance travelled = $72 + 60 + 45 + 3 = 177 + 3 = 180$ km. This matches the total journey length. The conditions are satisfied.

The length of his total journey is $\underline{180 \text{ km}}$.

Given:

Twice a number added to half of itself equals 24.

To Find:

The number.

Solution:

Let the number be $n$.

"Twice a number" is $2n$.

"Half of itself" is $\frac{1}{2}n$ or $\frac{n}{2}$.

"Twice a number added to half of itself" is $2n + \frac{n}{2}$.

"Equals 24" means the expression is equal to 24.

The equation formed is:

$2n + \frac{n}{2} = 24$

Now, we solve this equation for $n$.

To combine the terms on the left side, find a common denominator, which is 2.

$2n = \frac{4n}{2}$

$\frac{4n}{2} + \frac{n}{2} = 24$

$\frac{4n + n}{2} = 24$

$\frac{5n}{2} = 24$

Multiply both sides by 2 to eliminate the denominator:

$\frac{5n}{2} \times 2 = 24 \times 2$

$5n = 48$

Divide both sides by 5 to isolate $n$:

$\frac{5n}{5} = \frac{48}{5}$

$n = \frac{48}{5}$

The number is $\frac{48}{5}$ or 9.6.

Verification: Twice the number = $2 \times \frac{48}{5} = \frac{96}{5}$. Half the number = $\frac{1}{2} \times \frac{48}{5} = \frac{24}{5}$.

Sum = $\frac{96}{5} + \frac{24}{5} = \frac{96 + 24}{5} = \frac{120}{5} = 24$. This matches the given condition.

The number is $\underline{\frac{48}{5}}$ or $\underline{9.6}$.