Chapter 2 Linear Equations in One Variable

Given Equation:

$2x - 3 = 7$

---

To Find:

The value of $x$ that satisfies the equation.

---

Solution:

We need to isolate the variable $x$ on one side of the equation.

Step 1: Add 3 to both sides of the equation to eliminate the -3 on the left side.

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

$2x = 10$

Step 2: Divide both sides of the equation by 2 to isolate $x$.

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

$x = 5$

---

Check the solution:

Substitute $x=5$ back into the original equation:

$2(5) - 3 = 10 - 3 = 7$

Since the left side (7) equals the right side (7), the solution is correct.

---

Final Answer:

The solution to the equation $2x - 3 = 7$ is $\mathbf{x = 5}$.

Given Equation:

$2y + 9 = 4$

---

To Find:

The value of $y$ that satisfies the equation.

---

Solution:

We need to isolate the variable $y$ on one side of the equation.

Step 1: Subtract 9 from both sides of the equation to eliminate the +9 on the left side.

$2y + 9 - 9 = 4 - 9$

$2y = -5$

Step 2: Divide both sides of the equation by 2 to isolate $y$.

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

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

---

Check the solution:

Substitute $y = -\frac{5}{2}$ back into the original equation:

$2\left(-\frac{5}{2}\right) + 9 = \frac{-10}{2} + 9 = -5 + 9 = 4$

Since the left side (4) equals the right side (4), the solution is correct.

---

Final Answer:

The solution to the equation $2y + 9 = 4$ is $\mathbf{y = -\frac{5}{2}}$.

Given Equation:

$\frac{x}{3} + \frac{5}{2} = -\frac{3}{2}$

---

To Find:

The value of $x$ that satisfies the equation.

---

Solution:

We need to isolate the variable $x$ on one side of the equation.

Step 1: Subtract $\frac{5}{2}$ from both sides of the equation to isolate the term containing $x$.

$\frac{x}{3} + \frac{5}{2} - \frac{5}{2} = -\frac{3}{2} - \frac{5}{2}$

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

$\frac{x}{3} = \frac{-8}{2}$

Simplify the right side:

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

Step 2: Multiply both sides of the equation by 3 to isolate $x$.

$\frac{x}{3} \times 3 = -4 \times 3$

$x = -12$

---

Check the solution:

Substitute $x = -12$ back into the original equation:

$\frac{-12}{3} + \frac{5}{2} = -4 + \frac{5}{2}$

To add $-4$ and $\frac{5}{2}$, find a common denominator (which is 2):

$-4 = -\frac{4 \times 2}{1 \times 2} = -\frac{8}{2}$

$-\frac{8}{2} + \frac{5}{2} = \frac{-8 + 5}{2} = \frac{-3}{2}$

The left side ($\frac{-3}{2}$) equals the right side ($-\frac{3}{2}$), so the solution is correct.

---

Final Answer:

The solution to the equation $\frac{x}{3} + \frac{5}{2} = -\frac{3}{2}$ is $\mathbf{x = -12}$.

Given Equation:

$\frac{15}{4} - 7x = 9$

---

To Find:

The value of $x$ that satisfies the equation.

---

Solution:

We need to isolate the term containing the variable $x$.

Step 1: Subtract $\frac{15}{4}$ from both sides of the equation.

$\frac{15}{4} - 7x - \frac{15}{4} = 9 - \frac{15}{4}$

$-7x = 9 - \frac{15}{4}$

Step 2: Combine the terms on the right side. Find a common denominator, which is 4.

$9 = \frac{9}{1} = \frac{9 \times 4}{1 \times 4} = \frac{36}{4}$

So, the equation becomes:

$-7x = \frac{36}{4} - \frac{15}{4}$

$-7x = \frac{36 - 15}{4}$

$-7x = \frac{21}{4}$

Step 3: Divide both sides by -7 (or multiply by the reciprocal, $-\frac{1}{7}$) to isolate $x$.

$\frac{-7x}{-7} = \frac{21/4}{-7}$

$x = \frac{21}{4} \times \left(-\frac{1}{7}\right)$

$x = -\frac{21 \times 1}{4 \times 7}$

Simplify the fraction by cancelling the common factor 7:

$x = -\frac{\cancel{21}^3}{4 \times \cancel{7}_1}$

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

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

---

Check the solution:

Substitute $x = -\frac{3}{4}$ back into the original equation:

$\frac{15}{4} - 7\left(-\frac{3}{4}\right) = \frac{15}{4} - \left(\frac{-21}{4}\right)$

$\frac{15}{4} + \frac{21}{4} = \frac{15 + 21}{4} = \frac{36}{4} = 9$

The left side (9) equals the right side (9), so the solution is correct.

---

Final Answer:

The solution to the equation $\frac{15}{4} - 7x = 9$ is $\mathbf{x = -\frac{3}{4}}$.

Given Equation:

$x - 2 = 7$

---

To Find:

The value of $x$.

---

Solution:

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

The equation is $x - 2 = 7$.

Add 2 to both sides of the equation:

$x - 2 + 2 = 7 + 2$

Simplify both sides:

$x = 9$

---

Check the solution:

Substitute $x=9$ into the original equation:

$9 - 2 = 7$

$7 = 7$

The left side equals the right side, so the solution is correct.

---

Final Answer:

The solution is $\mathbf{x = 9}$.

Given Equation:

$y + 3 = 10$

---

To Find:

The value of $y$.

---

Solution:

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

The equation is $y + 3 = 10$.

Subtract 3 from both sides of the equation:

$y + 3 - 3 = 10 - 3$

Simplify both sides:

$y = 7$

---

Check the solution:

Substitute $y=7$ into the original equation:

$7 + 3 = 10$

$10 = 10$

The left side equals the right side, so the solution is correct.

---

Final Answer:

The solution is $\mathbf{y = 7}$.

Given Equation:

$6 = z + 2$

---

To Find:

The value of $z$.

---

Solution:

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

The equation is $6 = z + 2$.

We can rewrite this as $z + 2 = 6$.

Subtract 2 from both sides of the equation:

$z + 2 - 2 = 6 - 2$

Simplify both sides:

$z = 4$

---

Check the solution:

Substitute $z=4$ into the original equation:

$6 = 4 + 2$

$6 = 6$

The left side equals the right side, so the solution is correct.

---

Final Answer:

The solution is $\mathbf{z = 4}$.

Given Equation:

$\frac{3}{7} + x = \frac{17}{7}$

---

To Find:

The value of $x$.

---

Solution:

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

The equation is $\frac{3}{7} + x = \frac{17}{7}$.

Subtract $\frac{3}{7}$ from both sides of the equation:

$\frac{3}{7} + x - \frac{3}{7} = \frac{17}{7} - \frac{3}{7}$

Simplify both sides:

$x = \frac{17 - 3}{7}$

$x = \frac{14}{7}$

Simplify the fraction:

$x = 2$

---

Check the solution:

Substitute $x=2$ into the original equation:

$\frac{3}{7} + 2 = \frac{17}{7}$

Convert 2 to a fraction with denominator 7: $2 = \frac{2 \times 7}{1 \times 7} = \frac{14}{7}$.

$\frac{3}{7} + \frac{14}{7} = \frac{3 + 14}{7} = \frac{17}{7}$

The left side ($\frac{17}{7}$) equals the right side ($\frac{17}{7}$), so the solution is correct.

---

Final Answer:

The solution is $\mathbf{x = 2}$.

Given Equation:

$6x = 12$

---

To Find:

The value of $x$.

---

Solution:

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

The equation is $6x = 12$.

Divide both sides of the equation by 6:

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

Simplify both sides:

$x = 2$

---

Check the solution:

Substitute $x=2$ into the original equation:

$6(2) = 12$

$12 = 12$

The left side equals the right side, so the solution is correct.

---

Final Answer:

The solution is $\mathbf{x = 2}$.

Given Equation:

$\frac{t}{5} = 10$

---

To Find:

The value of $t$.

---

Solution:

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

The equation is $\frac{t}{5} = 10$.

Multiply both sides of the equation by 5:

$\frac{t}{5} \times 5 = 10 \times 5$

Simplify both sides:

$t = 50$

---

Check the solution:

Substitute $t=50$ into the original equation:

$\frac{50}{5} = 10$

$10 = 10$

The left side equals the right side, so the solution is correct.

---

Final Answer:

The solution is $\mathbf{t = 50}$.

Given Equation:

$\frac{2x}{3} = 18$

---

To Find:

The value of $x$.

---

Solution:

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

The equation is $\frac{2x}{3} = 18$.

Step 1: Multiply both sides of the equation by 3 to eliminate the denominator.

$\frac{2x}{3} \times 3 = 18 \times 3$

$2x = 54$

Step 2: Divide both sides of the equation by 2 to isolate $x$.

$\frac{2x}{2} = \frac{54}{2}$

$x = 27$

---

Check the solution:

Substitute $x=27$ back into the original equation:

$\frac{2(27)}{3} = \frac{54}{3}$

Performing the division:

$\frac{54}{3} = 18$

The left side (18) equals the right side (18), so the solution is correct.

---

Final Answer:

The solution is $\mathbf{x = 27}$.

Given Equation:

$1.6 = \frac{y}{1.5}$

---

To Find:

The value of $y$.

---

Solution:

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

The equation is $1.6 = \frac{y}{1.5}$.

We can rewrite this as $\frac{y}{1.5} = 1.6$.

Multiply both sides of the equation by 1.5:

$\frac{y}{1.5} \times 1.5 = 1.6 \times 1.5$

Simplify the left side:

$y = 1.6 \times 1.5$

So, $y = 2.40$ or $y = 2.4$.

---

Check the solution:

Substitute $y=2.4$ back into the original equation:

$1.6 = \frac{2.4}{1.5}$

To divide 2.4 by 1.5, we can multiply the numerator and denominator by 10 to get whole numbers:

$\frac{2.4}{1.5} = \frac{24}{15}$

Divide 24 by 15:

$24 \div 15 = 1.6$

So, $1.6 = 1.6$.

The left side equals the right side, so the solution is correct.

---

Final Answer:

The solution is $\mathbf{y = 2.4}$.

Given Equation:

$7x - 9 = 16$

---

To Find:

The value of $x$.

---

Solution:

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

The equation is $7x - 9 = 16$.

Step 1: Add 9 to both sides of the equation.

$7x - 9 + 9 = 16 + 9$

$7x = 25$

Step 2: Divide both sides of the equation by 7.

$\frac{7x}{7} = \frac{25}{7}$

$x = \frac{25}{7}$

---

Check the solution:

Substitute $x = \frac{25}{7}$ back into the original equation:

$7\left(\frac{25}{7}\right) - 9 = 16$

Cancel the 7s:

$25 - 9 = 16$

$16 = 16$

The left side equals the right side, so the solution is correct.

---

Final Answer:

The solution is $\mathbf{x = \frac{25}{7}}$.

Given Equation:

$14y - 8 = 13$

---

To Find:

The value of $y$.

---

Solution:

We need to isolate the variable $y$ on one side of the equation.

The equation is $14y - 8 = 13$.

Step 1: Add 8 to both sides of the equation.

$14y - 8 + 8 = 13 + 8$

$14y = 21$

Step 2: Divide both sides of the equation by 14.

$\frac{14y}{14} = \frac{21}{14}$

$y = \frac{21}{14}$

Step 3: Simplify the fraction. The greatest common divisor of 21 and 14 is 7.

$y = \frac{21 \div 7}{14 \div 7}$

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

---

Check the solution:

Substitute $y = \frac{3}{2}$ back into the original equation:

$14\left(\frac{3}{2}\right) - 8 = 13$

Multiply 14 by $\frac{3}{2}$:

$\frac{14 \times 3}{2} - 8 = \frac{42}{2} - 8$

$21 - 8 = 13$

$13 = 13$

The left side equals the right side, so the solution is correct.

---

Final Answer:

The solution is $\mathbf{y = \frac{3}{2}}$.

Given Equation:

$17 + 6p = 9$

---

To Find:

The value of $p$.

---

Solution:

We need to isolate the variable $p$ on one side of the equation.

The equation is $17 + 6p = 9$.

Step 1: Subtract 17 from both sides of the equation.

$17 + 6p - 17 = 9 - 17$

$6p = -8$

Step 2: Divide both sides of the equation by 6.

$\frac{6p}{6} = \frac{-8}{6}$

$p = -\frac{8}{6}$

Step 3: Simplify the fraction. The greatest common divisor of 8 and 6 is 2.

$p = -\frac{8 \div 2}{6 \div 2}$

$p = -\frac{4}{3}$

---

Check the solution:

Substitute $p = -\frac{4}{3}$ back into the original equation:

$17 + 6\left(-\frac{4}{3}\right) = 9$

$17 + \frac{6 \times (-4)}{3} = 9$

$17 + \frac{-24}{3} = 9$

$17 + (-8) = 9$

$17 - 8 = 9$

$9 = 9$

The left side equals the right side, so the solution is correct.

---

Final Answer:

The solution is $\mathbf{p = -\frac{4}{3}}$.

Given equation:

---

Solution:

To solve for $x$, we first need to isolate the term containing $x$ ($\frac{x}{3}$) on one side of the equation.

Subtract 1 from both sides of the equation:

To subtract 1 from $\frac{7}{15}$, we express 1 as a fraction with a denominator of 15:

Now, perform the subtraction on the right side:

Next, to solve for $x$, multiply both sides of the equation by 3:

Simplify the equation:

Therefore, the solution to the equation is $x = -\frac{8}{5}$.

Solution:

First, we need to find "twice the rational number $\frac{-7}{3}$".

Twice the number = $2 \times \frac{-7}{3}$

$= \frac{2 \times (-7)}{3}$

$= \frac{-14}{3}$

---

Now, let the number that should be added be 'x'.

According to the question, the sum of this number 'x' and twice the initial number should be $\frac{3}{7}$. We can write this as an equation:

Twice the rational number + $x$ = $\frac{3}{7}$

Substituting the value we found:

$\frac{-14}{3} + x = \frac{3}{7}$

---

To find the value of 'x', we need to isolate it on one side of the equation. We can do this by moving $\frac{-14}{3}$ to the right side, which changes its sign to positive.

$x = \frac{3}{7} - (\frac{-14}{3})$

$x = \frac{3}{7} + \frac{14}{3}$

To add these two fractions, we need to find a common denominator. The least common multiple (LCM) of the denominators 7 and 3 is $7 \times 3 = 21$.

Now, we convert each fraction to an equivalent fraction with a denominator of 21:

$x = \frac{3 \times 3}{7 \times 3} + \frac{14 \times 7}{3 \times 7}$

$x = \frac{9}{21} + \frac{98}{21}$

Now we can add the numerators:

$x = \frac{9 + 98}{21}$

$x = \frac{107}{21}$

---

Therefore, the number that should be added to twice of $\frac{-7}{3}$ to get $\frac{3}{7}$ is $\frac{107}{21}$.

Given:

Perimeter of the rectangle (P) = 13 cm

Width of the rectangle (W) = $2\frac{3}{4}$ cm

---

To Find:

Length of the rectangle (L)

---

Solution:

First, let's convert the width from a mixed number to an improper fraction:

The formula for the perimeter of a rectangle is:

Substitute the given values of P and W into the formula:

To solve for L, first divide both sides of the equation by 2:

Now, subtract $\frac{11}{4}$ from both sides to isolate L:

To perform the subtraction, find a common denominator, which is 4:

Perform the subtraction:

Convert the improper fraction back to a mixed number:

Therefore, the length of the rectangle is $3\frac{3}{4}$ cm.

Given:

1. The present age of Sahil’s mother is three times the present age of Sahil.

2. After 5 years, the sum of their ages will be 66 years.

---

To Find:

The present ages of Sahil and his mother.

---

Solution:

Let Sahil's present age be $x$ years.

According to the first condition, Sahil's mother's present age is three times Sahil's present age.

So, Sahil's mother's present age = $3x$ years.

Now, let's find their ages after 5 years:

Sahil's age after 5 years = (Present age of Sahil) + 5 = $x + 5$ years.

Sahil's mother's age after 5 years = (Present age of mother) + 5 = $3x + 5$ years.

According to the second condition, the sum of their ages after 5 years is 66 years.

So, we can write the equation:

Now, solve this equation for $x$:

Combine like terms on the left side:

Subtract 10 from both sides:

Divide both sides by 4:

So, Sahil's present age ($x$) is 14 years.

Sahil's mother's present age ($3x$) is $3 \times 14 = 42$ years.

Therefore, Sahil's present age is 14 years and his mother's present age is 42 years.

Verification:

Present ages: Sahil = 14, Mother = 42 (42 is 3 times 14).

Ages after 5 years: Sahil = $14+5=19$, Mother = $42+5=47$.

Sum of ages after 5 years = $19 + 47 = 66$. This matches the given condition.

Given:

1. Bansi has two types of coins: $\textsf{₹} 2$ coins and $\textsf{₹} 5$ coins.

2. The number of $\textsf{₹} 2$ coins is 3 times the number of $\textsf{₹} 5$ coins.

3. The total value of all the coins is $\textsf{₹} 77$.

---

To Find:

The number of $\textsf{₹} 2$ coins and the number of $\textsf{₹} 5$ coins Bansi has.

---

Solution:

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

According to the problem, the number of $\textsf{₹} 2$ coins is 3 times the number of $\textsf{₹} 5$ coins.

So, the number of $\textsf{₹} 2$ coins = $3x$.

Now, let's calculate the total value from each type of coin:

Value from $\textsf{₹} 5$ coins = (Number of $\textsf{₹} 5$ coins) $\times$ (Value of one $\textsf{₹} 5$ coin) = $x \times 5 = 5x$ Rupees.

Value from $\textsf{₹} 2$ coins = (Number of $\textsf{₹} 2$ coins) $\times$ (Value of one $\textsf{₹} 2$ coin) = $3x \times 2 = 6x$ Rupees.

The total value of all coins is the sum of the values from each denomination, which is given as $\textsf{₹} 77$.

Therefore, we can write the equation:

Combine the like terms on the left side:

Divide both sides by 11 to solve for $x$:

So, the number of $\textsf{₹} 5$ coins ($x$) is 7.

The number of $\textsf{₹} 2$ coins ($3x$) is $3 \times 7 = 21$.

Thus, Bansi has 7 coins of $\textsf{₹} 5$ and 21 coins of $\textsf{₹} 2$.

Verification:

Value from 7 five-rupee coins = $7 \times \textsf{₹} 5 = \textsf{₹} 35$.

Value from 21 two-rupee coins = $21 \times \textsf{₹} 2 = \textsf{₹} 42$.

Total value = $\textsf{₹} 35 + \textsf{₹} 42 = \textsf{₹} 77$. This matches the given total sum.

Given:

1. We are considering three consecutive multiples of 11.

2. The sum of these three multiples is 363.

---

To Find:

The three consecutive multiples of 11.

---

Solution:

Let the three consecutive multiples of 11 be represented. A common way to represent consecutive multiples is to let the middle multiple be $m$. Since $m$ is a multiple of 11, we can write $m = 11k$ for some integer $k$.

The multiple immediately preceding $11k$ is $11k - 11 = 11(k-1)$.

The multiple immediately succeeding $11k$ is $11k + 11 = 11(k+1)$.

So, the three consecutive multiples are $11(k-1)$, $11k$, and $11(k+1)$.

According to the problem, their sum is 363.

Therefore, we can write the equation:

Expand the terms:

Combine like terms (the constant terms -11 and +11 cancel out):

Divide both sides by 33 to solve for $k$:

Performing the division:

Now we find the three multiples using $k=11$:

First multiple = $11(k-1) = 11(11-1) = 11 \times 10 = 110$.

Second multiple (middle) = $11k = 11 \times 11 = 121$.

Third multiple = $11(k+1) = 11(11+1) = 11 \times 12 = 132$.

Therefore, the three consecutive multiples of 11 are 110, 121, and 132.

---

Alternate Solution:

Let the first multiple of 11 be $11x$.

Then the next consecutive multiple of 11 is $11x + 11 = 11(x+1)$.

The third consecutive multiple of 11 is $(11x + 11) + 11 \ $$ = 11x + 22 \ $$ = 11(x+2)$.

Their sum is 363:

Combine like terms:

Subtract 33 from both sides:

Divide both sides by 33:

Now find the multiples:

First multiple = $11x = 11 \times 10 = 110$.

Second multiple = $11x + 11 = 110 + 11 = 121$.

Third multiple = $11x + 22 = 110 + 22 = 132$.

The multiples are again 110, 121, and 132.

Verification:

The numbers 110, 121, and 132 are consecutive multiples of 11.

Their sum is $110 + 121 + 132 = 363$. This matches the given sum.

Given:

1. Two numbers are whole numbers.

2. The difference between the two numbers is 66.

3. The ratio of the two numbers is 2 : 5.

---

To Find:

The two whole numbers.

---

Solution:

Since the ratio of the two numbers is 2 : 5, let the numbers be $2x$ and $5x$, where $x$ is a common factor.

Since the numbers are whole numbers and their difference is non-zero, $x$ must be a positive number. Also, $5x$ will be the larger number and $2x$ will be the smaller number (assuming $x$ is positive).

The difference between the two numbers is given as 66.

So, we can write the equation:

Combine the like terms on the left side:

Divide both sides by 3 to solve for $x$:

Now, find the two numbers using the value of $x$:

Smaller number = $2x = 2 \times 22 = 44$.

Larger number = $5x = 5 \times 22 = 110$.

Therefore, the two whole numbers are 44 and 110.

Verification:

Difference: $110 - 44 = 66$. (Matches the given difference)

Ratio: $\frac{44}{110}$. To simplify, divide both by their greatest common divisor, which is 22: $\frac{44 \div 22}{110 \div 22} = \frac{2}{5}$. (Matches the given ratio 2 : 5)

The numbers 44 and 110 are whole numbers.

Given:

1. Total amount of money = $\textsf{₹} 590$.

2. Denominations of currency notes = $\textsf{₹} 50$, $\textsf{₹} 20$, $\textsf{₹} 10$.

3. Ratio of the number of $\textsf{₹} 50$ notes to $\textsf{₹} 20$ notes = 3 : 5.

4. Total number of notes = 25.

---

To Find:

The number of notes of each denomination ($\textsf{₹} 50$, $\textsf{₹} 20$, $\textsf{₹} 10$).

---

Solution:

Let the number of $\textsf{₹} 50$ notes and $\textsf{₹} 20$ notes be $3x$ and $5x$ respectively, based on the given ratio 3:5.

Let $N_{50}$ be the number of $\textsf{₹} 50$ notes, $N_{20}$ be the number of $\textsf{₹} 20$ notes, and $N_{10}$ be the number of $\textsf{₹} 10$ notes.

So, $N_{50} = 3x$ and $N_{20} = 5x$.

The total number of notes is 25. Therefore:

Substitute the expressions for $N_{50}$ and $N_{20}$:

From this, we can express the number of $\textsf{₹} 10$ notes in terms of $x$:

Now, consider the total value of the money, which is $\textsf{₹} 590$.

The total value is the sum of the values of each denomination:

Value from $\textsf{₹} 50$ notes = $N_{50} \times 50 = (3x) \times 50 = 150x$

Value from $\textsf{₹} 20$ notes = $N_{20} \times 20 = (5x) \times 20 = 100x$

Value from $\textsf{₹} 10$ notes = $N_{10} \times 10 = (25 - 8x) \times 10 = 250 - 80x$

The sum of these values equals the total amount:

Simplify the equation:

Combine the $x$ terms:

Subtract 250 from both sides:

Divide both sides by 170 to find $x$:

Now, substitute $x = 2$ back into the expressions for the number of notes:

Number of $\textsf{₹} 50$ notes ($N_{50}$) = $3x = 3 \times 2 = 6$.

Number of $\textsf{₹} 20$ notes ($N_{20}$) = $5x = 5 \times 2 = 10$.

Number of $\textsf{₹} 10$ notes ($N_{10}$) = $25 - 8x = 25 - 8(2) = 25 - 16 = 9$.

Therefore, Deveshi has 6 notes of $\textsf{₹} 50$, 10 notes of $\textsf{₹} 20$, and 9 notes of $\textsf{₹} 10$.

Verification:

Total number of notes = $6 + 10 + 9 = 25$. (Matches)

Ratio of $\textsf{₹} 50$ notes to $\textsf{₹} 20$ notes = $6 : 10$, which simplifies to $3 : 5$. (Matches)

Total value = $(6 \times \textsf{₹} 50) + (10 \times \textsf{₹} 20) + (9 \times \textsf{₹} 10) \ $$ = \textsf{₹} 300 + \textsf{₹} 200 + \textsf{₹} 90 \ $$ = \textsf{₹} 590$. (Matches)

Given:

1. A number is considered.

2. $\frac{1}{2}$ is subtracted from this number.

3. The result of the subtraction is multiplied by $\frac{1}{2}$.

4. The final result of this operation is $\frac{1}{8}$.

---

To Find:

The original number.

---

Solution:

Let the unknown number be $x$.

According to the problem statement, first subtract $\frac{1}{2}$ from the number:

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

Next, multiply this result by $\frac{1}{2}$:

$\frac{1}{2} \times \left( x - \frac{1}{2} \right)$

The problem states that this final result is equal to $\frac{1}{8}$. So, we can set up the equation:

To solve for $x$, first multiply both sides of the equation by 2 to remove the $\frac{1}{2}$ multiplier on the left side:

Simplify the fraction on the right side:

Now, add $\frac{1}{2}$ to both sides of the equation to isolate $x$:

To add the fractions, find a common denominator, which is 4:

Perform the addition:

Therefore, the number is $\frac{3}{4}$.

Verification:

Start with the number $\frac{3}{4}$.

Subtract $\frac{1}{2}$: $\frac{3}{4} - \frac{1}{2} = \frac{3}{4} - \frac{2}{4} = \frac{1}{4}$.

Multiply the result by $\frac{1}{2}$: $\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$.

This matches the result given in the problem statement.

Given:

1. Shape: Rectangular swimming pool.

2. Perimeter (P) = 154 m.

3. Relationship between length (L) and breadth (B): Length is 2 m more than twice its breadth, i.e., $L = 2B + 2$.

---

To Find:

The length (L) and the breadth (B) of the pool.

---

Solution:

Let the breadth of the rectangular pool be $B$ meters.

According to the given condition, the length (L) is 2 m more than twice its breadth.

So, $L = 2B + 2$ meters.

The formula for the perimeter of a rectangle is:

We are given that the perimeter P = 154 m. Substitute the values of P and the expression for L into the formula:

Now, solve this equation for B:

First, simplify the expression inside the parenthesis:

Divide both sides of the equation by 2:

Subtract 2 from both sides:

Divide both sides by 3:

So, the breadth of the pool is 25 meters.

Now, find the length using the relationship $L = 2B + 2$:

So, the length of the pool is 52 meters.

Therefore, the breadth of the pool is 25 m and the length of the pool is 52 m.

Verification:

Length = 52 m, Breadth = 25 m.

Is length 2 m more than twice the breadth? $2 \times \text{Breadth} + 2 = 2 \times 25 + 2 = 50 + 2 = 52$ m. Yes, it matches the length.

Perimeter = $2(L + B) = 2(52 + 25) = 2(77) = 154$ m. Yes, it matches the given perimeter.

Given:

1. Triangle type: Isosceles.

2. Base length = $\frac{4}{3}$ cm.

3. Perimeter = $4\frac{2}{15}$ cm.

4. In an isosceles triangle, the two sides other than the base are equal in length.

---

To Find:

The length of each of the two equal sides.

---

Solution:

Let the length of each of the equal sides be $x$ cm.

The perimeter of a triangle is the sum of the lengths of its three sides.

Perimeter = Base + Side 1 + Side 2

In this case, Perimeter = Base + $x + x$.

First, convert the perimeter from a mixed number to an improper fraction:

Now, set up the equation for the perimeter:

To solve for $x$, first isolate the term $2x$ by subtracting $\frac{4}{3}$ from both sides:

To subtract the fractions, find a common denominator, which is 15:

Perform the subtraction:

Now, divide both sides by 2 (or multiply by $\frac{1}{2}$) to find $x$:

Simplify the fraction. Both 42 and 30 are divisible by 6:

Optionally, convert the improper fraction to a mixed number:

Therefore, the length of either of the remaining equal sides is $\frac{7}{5}$ cm or $1\frac{2}{5}$ cm.

Verification:

Check if the perimeter adds up:

Perimeter = Base + Side + Side = $\frac{4}{3} + \frac{7}{5} + \frac{7}{5}$

Common denominator is 15:

Perimeter = $\frac{4 \times 5}{15} + \frac{7 \times 3}{15} + \frac{7 \times 3}{15} = \frac{20}{15} + \frac{21}{15} + \frac{21}{15} = \frac{20 + 21 + 21}{15} = \frac{62}{15}$

Convert back to mixed number: $\frac{62}{15} = 4\frac{2}{15}$ cm.

This matches the given perimeter.

Given:

1. The sum of two numbers is 95.

2. One number is 15 more than the other number.

---

To Find:

The two numbers.

---

Solution:

Let the smaller number be $x$.

Since one number exceeds the other by 15, the larger number is $x + 15$.

The sum of the two numbers is given as 95.

So, we can set up the equation:

Combine the like terms on the left side:

Subtract 15 from both sides of the equation:

Divide both sides by 2 to solve for $x$:

So, the smaller number is $x = 40$.

The larger number is $x + 15 = 40 + 15 = 55$.

Therefore, the two numbers are 40 and 55.

---

Alternate Solution:

Let the larger number be $y$.

Since one number exceeds the other by 15, the smaller number is $y - 15$.

The sum of the two numbers is 95.

Combine like terms:

Add 15 to both sides:

Divide both sides by 2:

So, the larger number is $y = 55$.

The smaller number is $y - 15 = 55 - 15 = 40$.

The two numbers are again 40 and 55.

Verification:

Sum: $40 + 55 = 95$. (Correct)

Difference: $55 - 40 = 15$. (One exceeds the other by 15. Correct)

Given:

1. The ratio of two numbers is 5:3.

2. The difference between the two numbers is 18.

---

To Find:

The two numbers.

---

Solution:

Since the ratio of the two numbers is 5:3, we can represent the numbers as $5x$ and $3x$, where $x$ is a common multiplier.

Assuming $x$ is positive, $5x$ is the larger number and $3x$ is the smaller number.

The difference between the numbers is given as 18.

So, we can set up the equation:

Combine the like terms on the left side:

Divide both sides by 2 to solve for $x$:

Now, find the two numbers using the value of $x$:

First number (larger) = $5x = 5 \times 9 = 45$.

Second number (smaller) = $3x = 3 \times 9 = 27$.

Therefore, the two numbers are 45 and 27.

Verification:

Check the ratio: $\frac{45}{27}$. Divide both numerator and denominator by their greatest common divisor, 9: $\frac{45 \div 9}{27 \div 9} = \frac{5}{3}$. The ratio is 5:3. (Correct)

Check the difference: $45 - 27 = 18$. (Correct)

Given:

1. Three integers are consecutive.

2. The sum of these three integers is 51.

---

To Find:

The three consecutive integers.

---

Solution:

Let the three consecutive integers be $x$, $x+1$, and $x+2$.

According to the problem, their sum is 51.

So, we can set up the equation:

Combine the like terms on the left side:

Subtract 3 from both sides of the equation:

Divide both sides by 3 to solve for $x$:

So, the first integer is $x = 16$.

The second integer is $x+1 = 16 + 1 = 17$.

The third integer is $x+2 = 16 + 2 = 18$.

Therefore, the three consecutive integers are 16, 17, and 18.

---

Alternate Solution:

Let the middle integer be $y$.

Since the integers are consecutive, the integer before $y$ is $y-1$, and the integer after $y$ is $y+1$.

The three consecutive integers are $y-1$, $y$, and $y+1$.

Their sum is 51:

Combine like terms (the -1 and +1 cancel out):

Divide both sides by 3:

So, the middle integer is 17.

The first integer is $y-1 = 17 - 1 = 16$.

The third integer is $y+1 = 17 + 1 = 18$.

The three consecutive integers are again 16, 17, and 18.

Verification:

Check if they are consecutive integers: Yes, 16, 17, 18 are consecutive.

Check if their sum is 51: $16 + 17 + 18 = 33 + 18 = 51$. (Correct)

Given:

1. Three multiples are consecutive multiples of 8.

2. The sum of these three multiples is 888.

---

To Find:

The three consecutive multiples of 8.

---

Solution:

Let the three consecutive multiples of 8 be represented by $8x$, $8(x+1)$, and $8(x+2)$, where $x$ is an integer.

Alternatively, let the first multiple be $m$. Since it's a multiple of 8, let $m=8x$.

The next consecutive multiple of 8 is $m+8 = 8x+8$.

The third consecutive multiple of 8 is $(m+8)+8 = m+16 = 8x+16$.

The sum of these three multiples is given as 888.

So, we can set up the equation:

Combine the like terms on the left side:

Subtract 24 from both sides of the equation:

Divide both sides by 24 to solve for $x$:

Now, find the three multiples:

First multiple = $8x = 8 \times 36 = 288$.

Second multiple = $8x+8 = 288 + 8 = 296$.

Third multiple = $8x+16 = 288 + 16 = 304$.

Therefore, the three consecutive multiples of 8 are 288, 296, and 304.

---

Alternate Solution:

Let the middle multiple of 8 be $y$. Since it's a multiple of 8, $y=8k$ for some integer $k$.

The multiple before $y$ is $y-8$.

The multiple after $y$ is $y+8$.

The three consecutive multiples are $y-8$, $y$, and $y+8$.

Their sum is 888:

Combine like terms (the -8 and +8 cancel out):

Divide both sides by 3:

So, the middle multiple is 296.

The first multiple is $y-8 = 296 - 8 = 288$.

The third multiple is $y+8 = 296 + 8 = 304$.

The three consecutive multiples are again 288, 296, and 304.

Verification:

Check if they are consecutive multiples of 8: Yes, $288=8\times36$, $296=8\times37$, $304=8\times38$.

Check if their sum is 888: $288 + 296 + 304 = 584 + 304 = 888$. (Correct)

Given:

1. Three integers are consecutive.

2. They are taken in increasing order.

3. The first integer is multiplied by 2.

4. The second integer is multiplied by 3.

5. The third integer is multiplied by 4.

6. The sum of these products is 74.

---

To Find:

The three consecutive integers.

---

Solution:

Let the three consecutive integers in increasing order be $x$, $x+1$, and $x+2$.

According to the problem statement:

The first integer multiplied by 2 is $2x$.

The second integer multiplied by 3 is $3(x+1)$.

The third integer multiplied by 4 is $4(x+2)$.

The sum of these products is 74. So, we can set up the equation:

Now, solve this equation for $x$:

Distribute the multipliers:

Combine like terms on the left side:

Subtract 11 from both sides:

Divide both sides by 9:

So, the first integer is $x = 7$.

The second integer is $x+1 = 7 + 1 = 8$.

The third integer is $x+2 = 7 + 2 = 9$.

Therefore, the three consecutive integers are 7, 8, and 9.

Verification:

The integers are 7, 8, 9. They are consecutive and in increasing order.

Multiply them by 2, 3, and 4 respectively:

$2 \times 7 = 14$

$3 \times 8 = 24$

$4 \times 9 = 36$

Add these results: $14 + 24 + 36 = 38 + 36 = 74$.

The sum is 74, which matches the given condition.

Given:

1. The ratio of the present ages of Rahul and Haroon is 5:7.

2. Four years later, the sum of their ages will be 56 years.

---

To Find:

The present ages of Rahul and Haroon.

---

Solution:

Let the present ages of Rahul and Haroon be $5x$ years and $7x$ years respectively, based on the given ratio 5:7.

Now, find their ages after 4 years:

Rahul's age after 4 years = (Present age of Rahul) + 4 = $5x + 4$ years.

Haroon's age after 4 years = (Present age of Haroon) + 4 = $7x + 4$ years.

According to the problem, the sum of their ages after 4 years is 56 years.

So, we can set up the equation:

Combine the like terms on the left side:

Subtract 8 from both sides of the equation:

Divide both sides by 12 to solve for $x$:

Now, find their present ages using the value of $x$:

Rahul's present age = $5x = 5 \times 4 = 20$ years.

Haroon's present age = $7x = 7 \times 4 = 28$ years.

Therefore, Rahul's present age is 20 years and Haroon's present age is 28 years.

Verification:

Present ages are 20 and 28. The ratio is $20:28 = 5:7$. (Correct)

Ages after 4 years: Rahul will be $20+4 = 24$, Haroon will be $28+4 = 32$.

Sum of ages after 4 years: $24 + 32 = 56$. (Correct)

Given:

1. The ratio of the number of boys to the number of girls in a class is 7:5.

2. The number of boys is 8 more than the number of girls.

---

To Find:

The total class strength (total number of students).

---

Solution:

Let the number of boys be $7x$ and the number of girls be $5x$, according to the given ratio 7:5.

It is given that the number of boys is 8 more than the number of girls.

So, we can write the equation:

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

Divide both sides by 2:

Now, find the actual number of boys and girls using $x=4$:

Number of boys = $7x = 7 \times 4 = 28$.

Number of girls = $5x = 5 \times 4 = 20$.

The total class strength is the sum of the number of boys and the number of girls.

Therefore, the total class strength is 48 students.

Verification:

Number of boys = 28, Number of girls = 20.

Ratio = $28:20$. Dividing both by 4 gives $7:5$. (Correct)

Difference = $28 - 20 = 8$. The number of boys is 8 more than the number of girls. (Correct)

Given:

1. Baichung's father is 29 years older than Baichung.

2. Baichung's father is 26 years younger than Baichung's grandfather.

3. The sum of the ages of Baichung, his father, and his grandfather is 135 years.

---

To Find:

The present age of Baichung, his father, and his grandfather.

---

Solution:

Let Baichung's present age be $x$ years.

From condition 1, Baichung's father is 29 years older than Baichung.

So, Father's age = $(x + 29)$ years.

From condition 2, Baichung's father is 26 years younger than Baichung's grandfather. This means the grandfather is 26 years older than the father.

So, Grandfather's age = (Father's age) + 26

Grandfather's age = $(x + 29) + 26 = (x + 55)$ years.

From condition 3, the sum of the ages of all three is 135 years.

We can set up the equation:

Combine the like terms on the left side:

Subtract 84 from both sides of the equation:

Divide both sides by 3 to solve for $x$:

So, Baichung's age ($x$) is 17 years.

Now, find the ages of his father and grandfather:

Father's age = $x + 29 = 17 + 29 = 46$ years.

Grandfather's age = $x + 55 = 17 + 55 = 72$ years.

Therefore, Baichung's age is 17 years, his father's age is 46 years, and his grandfather's age is 72 years.

Verification:

Father's age (46) is 29 years older than Baichung (17)? $17 + 29 = 46$. Yes.

Father's age (46) is 26 years younger than Grandfather (72)? $72 - 26 = 46$. Yes.

Sum of ages = $17 + 46 + 72 = 63 + 72 = 135$. Yes.

Given:

1. Fifteen years from now, Ravi's age will be a certain value.

2. This future age will be four times his present age.

---

To Find:

Ravi's present age.

---

Solution:

Let Ravi's present age be $x$ years.

His age fifteen years from now will be (Present age) + 15 = $x + 15$ years.

The problem states that his age fifteen years from now will be four times his present age.

So, we can set up the equation:

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

Now, divide both sides by 3:

Therefore, Ravi's present age is 5 years.

Verification:

Ravi's present age = 5 years.

Ravi's age after 15 years = $5 + 15 = 20$ years.

Four times Ravi's present age = $4 \times 5 = 20$ years.

Since his age after 15 years (20) is equal to four times his present age (20), the answer is correct.

Given:

1. A rational number is considered.

2. The number is multiplied by $\frac{5}{2}$.

3. $\frac{2}{3}$ is added to the result of the multiplication.

4. The final result is $-\frac{7}{12}$.

---

To Find:

The rational number.

---

Solution:

Let the rational number be $x$.

According to the problem statement, first multiply the number by $\frac{5}{2}$:

Product = $x \times \frac{5}{2} = \frac{5}{2}x$.

Next, add $\frac{2}{3}$ to this product:

Result = $\frac{5}{2}x + \frac{2}{3}$.

The problem states that this result is equal to $-\frac{7}{12}$. So, we can set up the equation:

To solve for $x$, first isolate the term containing $x$ by subtracting $\frac{2}{3}$ from both sides:

To subtract the fractions on the right side, find a common denominator, which is 12:

Combine the fractions on the right side:

Simplify the fraction on the right side by dividing the numerator and denominator by their greatest common divisor, 3:

Now, to isolate $x$, multiply both sides by the reciprocal of $\frac{5}{2}$, which is $\frac{2}{5}$:

Perform the multiplication:

Cancel out the common factor 5:

Simplify the fraction:

Therefore, the rational number is $-\frac{1}{2}$.

Verification:

Multiply the number $-\frac{1}{2}$ by $\frac{5}{2}$: $-\frac{1}{2} \times \frac{5}{2} = -\frac{5}{4}$.

Add $\frac{2}{3}$ to the product: $-\frac{5}{4} + \frac{2}{3}$.

Find a common denominator (12): $-\frac{5 \times 3}{12} + \frac{2 \times 4}{12} = -\frac{15}{12} + \frac{8}{12}$.

Perform the addition: $\frac{-15 + 8}{12} = -\frac{7}{12}$.

This matches the result given in the problem statement.

Given:

1. Denominations of currency notes: $\textsf{₹} 100$, $\textsf{₹} 50$, $\textsf{₹} 10$.

2. Ratio of the number of $\textsf{₹} 100$ notes : $\textsf{₹} 50$ notes : $\textsf{₹} 10$ notes = 2 : 3 : 5.

3. Total cash amount = $\textsf{₹} 4,00,000$.

---

To Find:

The number of notes of each denomination ($\textsf{₹} 100$, $\textsf{₹} 50$, $\textsf{₹} 10$).

---

Solution:

Let the number of $\textsf{₹} 100$, $\textsf{₹} 50$, and $\textsf{₹} 10$ notes be $2x$, $3x$, and $5x$ respectively, based on the given ratio 2:3:5.

Now, calculate the total value contributed by each denomination:

Value from $\textsf{₹} 100$ notes = (Number of $\textsf{₹} 100$ notes) $\times$ 100 = $(2x) \times 100 = 200x$ Rupees.

Value from $\textsf{₹} 50$ notes = (Number of $\textsf{₹} 50$ notes) $\times$ 50 = $(3x) \times 50 = 150x$ Rupees.

Value from $\textsf{₹} 10$ notes = (Number of $\textsf{₹} 10$ notes) $\times$ 10 = $(5x) \times 10 = 50x$ Rupees.

The total cash is the sum of the values from all denominations, which is given as $\textsf{₹} 4,00,000$.

Therefore, we can write the equation:

Combine the like terms on the left side:

Divide both sides by 400 to solve for $x$:

Simplify the division:

Now, find the number of notes of each denomination using $x=1000$:

Number of $\textsf{₹} 100$ notes = $2x = 2 \times 1000 = 2000$.

Number of $\textsf{₹} 50$ notes = $3x = 3 \times 1000 = 3000$.

Number of $\textsf{₹} 10$ notes = $5x = 5 \times 1000 = 5000$.

Thus, Lakshmi has 2000 notes of $\textsf{₹} 100$, 3000 notes of $\textsf{₹} 50$, and 5000 notes of $\textsf{₹} 10$.

Verification:

Check the ratio of notes: $2000 : 3000 : 5000$. Dividing all parts by 1000 gives $2 : 3 : 5$. (Correct)

Check the total value:

Value from $\textsf{₹} 100$ notes: $2000 \times \textsf{₹} 100 = \textsf{₹} 2,00,000$.

Value from $\textsf{₹} 50$ notes: $3000 \times \textsf{₹} 50 = \textsf{₹} 1,50,000$.

Value from $\textsf{₹} 10$ notes: $5000 \times \textsf{₹} 10 = \textsf{₹} 50,000$.

Total value = $\textsf{₹} 2,00,000 + \textsf{₹} 1,50,000 + \textsf{₹} 50,000 \ $$ = \textsf{₹} 4,00,000$. (Correct)

Given:

1. Total amount of money = $\textsf{₹} 300$.

2. Coin denominations = $\textsf{₹} 1$, $\textsf{₹} 2$, $\textsf{₹} 5$.

3. The number of $\textsf{₹} 2$ coins is 3 times the number of $\textsf{₹} 5$ coins.

4. Total number of coins = 160.

---

To Find:

The number of coins of each denomination ($\textsf{₹} 1$, $\textsf{₹} 2$, $\textsf{₹} 5$).

---

Solution:

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

According to condition 3, the number of $\textsf{₹} 2$ coins is 3 times the number of $\textsf{₹} 5$ coins.

So, the number of $\textsf{₹} 2$ coins = $3x$.

Let the number of $\textsf{₹} 1$ coins be $y$.

The total number of coins is 160 (condition 4).

Therefore:

From this, we can express the number of $\textsf{₹} 1$ coins ($y$) in terms of $x$:

Now, consider the total value of the money, which is $\textsf{₹} 300$ (condition 1).

The total value is the sum of the values of each denomination:

Value from $\textsf{₹} 1$ coins = (Number of $\textsf{₹} 1$ coins) $\times$ 1 = $y \times 1 = y$.

Value from $\textsf{₹} 2$ coins = (Number of $\textsf{₹} 2$ coins) $\times$ 2 = $(3x) \times 2 = 6x$.

Value from $\textsf{₹} 5$ coins = (Number of $\textsf{₹} 5$ coins) $\times$ 5 = $x \times 5 = 5x$.

Set up the equation for the total value:

Now, substitute the expression for $y$ from equation (i) into equation (ii):

Simplify and solve for $x$:

Subtract 160 from both sides:

Divide both sides by 7:

So, the number of $\textsf{₹} 5$ coins ($x$) is 20.

Now, find the number of coins of the other denominations:

Number of $\textsf{₹} 2$ coins = $3x = 3 \times 20 = 60$.

Number of $\textsf{₹} 1$ coins = $y = 160 - 4x = 160 - 4(20) = 160 - 80 = 80$.

Therefore, I have 80 coins of $\textsf{₹} 1$, 60 coins of $\textsf{₹} 2$, and 20 coins of $\textsf{₹} 5$.

Verification:

Check total number of coins: $80 (\textsf{₹} 1) + 60 (\textsf{₹} 2) + 20 (\textsf{₹} 5) = 160$. (Matches)

Check relationship between $\textsf{₹} 2$ and $\textsf{₹} 5$ coins: Number of $\textsf{₹} 2$ coins (60) is 3 times the number of $\textsf{₹} 5$ coins (20). $60 = 3 \times 20$. (Matches)

Check total value:

Value from $\textsf{₹} 1$ coins: $80 \times \textsf{₹} 1 = \textsf{₹} 80$.

Value from $\textsf{₹} 2$ coins: $60 \times \textsf{₹} 2 = \textsf{₹} 120$.

Value from $\textsf{₹} 5$ coins: $20 \times \textsf{₹} 5 = \textsf{₹} 100$.

Total value = $\textsf{₹} 80 + \textsf{₹} 120 + \textsf{₹} 100 = \textsf{₹} 200 + \textsf{₹} 100 = \textsf{₹} 300$. (Matches)

Given:

1. Prize money for a winner = $\textsf{₹} 100$.

2. Prize money for a participant who does not win (non-winner) = $\textsf{₹} 25$.

3. Total prize money distributed = $\textsf{₹} 3,000$.

4. Total number of participants = 63.

---

To Find:

The number of winners.

---

Solution:

Let the number of winners be $w$.

The total number of participants is 63.

So, the number of participants who did not win (non-winners) = Total participants - Number of winners

Number of non-winners = $63 - w$.

Now, calculate the total prize money distributed based on the number of winners and non-winners:

Total prize money = (Number of winners $\times$ Prize per winner) + (Number of non-winners $\times$ Prize per non-winner)

We are given that the total prize money distributed is $\textsf{₹} 3,000$.

Set up the equation:

Now, solve this equation for $w$:

Distribute the 25:

Calculate $25 \times 63 = 1575$.

Combine the like terms (w terms):

Subtract 1575 from both sides:

Divide both sides by 75:

Performing the division ($1425 \div 75 = 19$):

Therefore, the number of winners is 19.

Verification:

Number of winners = 19.

Number of non-winners = $63 - 19 = 44$.

Total participants = $19 + 44 = 63$. (Matches)

Prize money for winners = $19 \times \textsf{₹} 100 = \textsf{₹} 1900$.

Prize money for non-winners = $44 \times \textsf{₹} 25 = \textsf{₹} 1100$.

Total prize money = $\textsf{₹} 1900 + \textsf{₹} 1100 = \textsf{₹} 3000$. (Matches)

Given Equation:

---

To Find:

The value of $x$.

---

Solution:

We need to isolate the variable $x$ on one side of the equation.

First, move the variable terms to one side. Subtract $x$ from both sides:

Simplify:

Next, move the constant terms to the other side. Add 3 to both sides:

Simplify:

Therefore, the solution to the equation is $x = 5$.

Verification:

Substitute $x = 5$ into the original equation $2x - 3 = x + 2$.

Left Hand Side (LHS): $2(5) - 3 = 10 - 3 = 7$.

Right Hand Side (RHS): $5 + 2 = 7$.

Since LHS = RHS (7 = 7), the solution is correct.

Given Equation:

---

To Find:

The value of $x$.

---

Solution:

To solve for $x$, we need to gather the terms involving $x$ on one side of the equation and the constant terms on the other side.

Subtract $\frac{3}{2}x$ from both sides:

Combine the $x$ terms on the left side. Find a common denominator (2) for the coefficients:

Now, subtract $\frac{7}{2}$ from both sides to isolate the term with $x$:

Combine the constant terms on the right side. Find a common denominator (2):

To solve for $x$, multiply both sides by the reciprocal of $\frac{7}{2}$, which is $\frac{2}{7}$:

Simplify both sides:

Therefore, the solution to the equation is $x = -5$.

Verification:

Substitute $x = -5$ into the original equation $5x + \frac{7}{2} = \frac{3}{2}x - 14$.

Left Hand Side (LHS): $5(-5) + \frac{7}{2} = -25 + \frac{7}{2} = -\frac{50}{2} + \frac{7}{2} = -\frac{43}{2}$.

Right Hand Side (RHS): $\frac{3}{2}(-5) - 14 = -\frac{15}{2} - 14 = -\frac{15}{2} - \frac{28}{2} = -\frac{43}{2}$.

Since LHS = RHS ($-\frac{43}{2} = -\frac{43}{2}$), the solution is correct.

Given Equation:

---

To Find:

The value of $x$.

---

Solution:

To solve for $x$, we need to collect the terms involving $x$ on one side of the equation and the constant terms on the other side.

Subtract $2x$ from both sides of the equation:

Simplify both sides:

Therefore, the solution to the equation is $x = 18$.

Verification:

Substitute $x = 18$ into the original equation $3x = 2x + 18$.

Left Hand Side (LHS): $3x = 3(18) = 54$.

Right Hand Side (RHS): $2x + 18 = 2(18) + 18 = 36 + 18 = 54$.

Since LHS = RHS (54 = 54), the solution is correct.

Given Equation:

---

To Find:

The value of $t$.

---

Solution:

To solve for $t$, we need to gather the terms involving $t$ on one side of the equation and the constant terms on the other side.

Subtract $3t$ from both sides:

Simplify:

Add 3 to both sides:

Simplify:

Divide both sides by 2:

Therefore, the solution to the equation is $t = -1$.

Verification:

Substitute $t = -1$ into the original equation $5t - 3 = 3t - 5$.

Left Hand Side (LHS): $5t - 3 = 5(-1) - 3 = -5 - 3 = -8$.

Right Hand Side (RHS): $3t - 5 = 3(-1) - 5 = -3 - 5 = -8$.

Since LHS = RHS (-8 = -8), the solution is correct.

Given Equation:

---

To Find:

The value of $x$.

---

Solution:

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

Subtract $3x$ from both sides:

Simplify:

Subtract 9 from both sides:

Simplify:

Divide both sides by 2:

Therefore, the solution to the equation is $x = -2$.

Verification:

Substitute $x = -2$ into the original equation $5x + 9 = 5 + 3x$.

Left Hand Side (LHS): $5x + 9 = 5(-2) + 9 = -10 + 9 = -1$.

Right Hand Side (RHS): $5 + 3x = 5 + 3(-2) = 5 - 6 = -1$.

Since LHS = RHS (-1 = -1), the solution is correct.

Given Equation:

---

To Find:

The value of $z$.

---

Solution:

To solve for $z$, we collect the terms involving $z$ on one side of the equation and the constant terms on the other side.

Subtract $2z$ from both sides:

Simplify:

Subtract 3 from both sides:

Simplify:

Divide both sides by 2:

Therefore, the solution to the equation is $z = \frac{3}{2}$.

Verification:

Substitute $z = \frac{3}{2}$ into the original equation $4z + 3 = 6 + 2z$.

Left Hand Side (LHS): $4z + 3 = 4\left(\frac{3}{2}\right) + 3 = \frac{12}{2} + 3 = 6 + 3 = 9$.

Right Hand Side (RHS): $6 + 2z = 6 + 2\left(\frac{3}{2}\right) = 6 + \frac{6}{2} = 6 + 3 = 9$.

Since LHS = RHS (9 = 9), the solution is correct.

Given Equation:

---

To Find:

The value of $x$.

---

Solution:

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

Add $x$ to both sides:

Simplify:

Add 1 to both sides:

Simplify:

Divide both sides by 3:

Therefore, the solution to the equation is $x = 5$.

Verification:

Substitute $x = 5$ into the original equation $2x - 1 = 14 - x$.

Left Hand Side (LHS): $2x - 1 = 2(5) - 1 = 10 - 1 = 9$.

Right Hand Side (RHS): $14 - x = 14 - 5 = 9$.

Since LHS = RHS (9 = 9), the solution is correct.

Given Equation:

---

To Find:

The value of $x$.

---

Solution:

First, simplify the right-hand side (RHS) of the equation by distributing the 3:

Combine the constant terms on the RHS:

Now, gather the terms involving $x$ on one side and the constant terms on the other side.

Subtract $3x$ from both sides:

Simplify:

Subtract 4 from both sides:

Simplify:

Divide both sides by 5:

Therefore, the solution to the equation is $x = 0$.

Verification:

Substitute $x = 0$ into the original equation $8x + 4 = 3(x - 1) + 7$.

Left Hand Side (LHS): $8x + 4 = 8(0) + 4 = 0 + 4 = 4$.

Right Hand Side (RHS): $3(x - 1) + 7 = 3(0 - 1) + 7 = 3(-1) + 7 \ $$ = -3 + 7 = 4$.

Since LHS = RHS (4 = 4), the solution is correct.

Given Equation:

---

To Find:

The value of $x$.

---

Solution:

To solve for $x$, we first eliminate the fraction by multiplying both sides of the equation by 5:

Simplify:

Distribute the 4 on the right-hand side:

Now, gather the terms involving $x$ on one side. Subtract $4x$ from both sides:

Simplify:

Therefore, the solution to the equation is $x = 40$.

Verification:

Substitute $x = 40$ into the original equation $x = \frac{4}{5}(x + 10)$.

Left Hand Side (LHS): $x = 40$.

Right Hand Side (RHS): $\frac{4}{5}(x + 10) = \frac{4}{5}(40 + 10) = \frac{4}{5}(50)$.

RHS = $4 \times \frac{50}{5} = 4 \times 10 = 40$.

Since LHS = RHS (40 = 40), the solution is correct.

Given Equation:

---

To Find:

The value of $x$.

---

Solution:

To solve for $x$, we first eliminate the fractions. The denominators are 3 and 15. The Least Common Multiple (LCM) of 3 and 15 is 15.

Multiply every term in the equation by the LCM (15):

Simplify each term:

Now, gather the terms involving $x$ on one side and the constant terms on the other side.

Subtract $7x$ from both sides:

Simplify:

Subtract 15 from both sides:

Simplify:

Divide both sides by 3:

Therefore, the solution to the equation is $x = 10$.

Verification:

Substitute $x = 10$ into the original equation $\frac{2x}{3} + 1 = \frac{7x}{15} + 3$.

Left Hand Side (LHS): $\frac{2(10)}{3} + 1 = \frac{20}{3} + 1 = \frac{20}{3} + \frac{3}{3} = \frac{23}{3}$.

Right Hand Side (RHS): $\frac{7(10)}{15} + 3 = \frac{70}{15} + 3 = \frac{14}{3} + 3 = \frac{14}{3} + \frac{9}{3} = \frac{23}{3}$.

Since LHS = RHS ($\frac{23}{3} = \frac{23}{3}$), the solution is correct.

Given Equation:

---

To Find:

The value of $y$.

---

Solution:

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

Add $y$ to both sides:

Simplify:

Subtract $\frac{5}{3}$ from both sides:

Simplify:

Simplify the fraction on the right side:

Divide both sides by 3:

Therefore, the solution to the equation is $y = \frac{7}{3}$.

Verification:

Substitute $y = \frac{7}{3}$ into the original equation $2y + \frac{5}{3} = \frac{26}{3} - y$.

Left Hand Side (LHS): $2y + \frac{5}{3} = 2\left(\frac{7}{3}\right) + \frac{5}{3} = \frac{14}{3} + \frac{5}{3} = \frac{14 + 5}{3} = \frac{19}{3}$.

Right Hand Side (RHS): $\frac{26}{3} - y = \frac{26}{3} - \frac{7}{3} = \frac{26 - 7}{3} = \frac{19}{3}$.

Since LHS = RHS ($\frac{19}{3} = \frac{19}{3}$), the solution is correct.

Given Equation:

---

To Find:

The value of $m$.

---

Solution:

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

Subtract $5m$ from both sides:

Simplify:

Divide both sides by -2:

Simplify the multiplication:

Reduce the fraction to its simplest form by dividing numerator and denominator by 2:

Therefore, the solution to the equation is $m = \frac{4}{5}$.

Verification:

Substitute $m = \frac{4}{5}$ into the original equation $3m = 5m - \frac{8}{5}$.

Left Hand Side (LHS): $3m = 3\left(\frac{4}{5}\right) = \frac{12}{5}$.

Right Hand Side (RHS): $5m - \frac{8}{5} = 5\left(\frac{4}{5}\right) - \frac{8}{5} = \frac{20}{5} - \frac{8}{5} = \frac{20 - 8}{5} = \frac{12}{5}$.

Since LHS = RHS ($\frac{12}{5} = \frac{12}{5}$), the solution is correct.

Given:

1. The number is a two-digit number.

2. The difference between the digits of the number is 3.

3. The sum of the original number and the number formed by interchanging the digits is 143.

---

To Find:

The original two-digit number.

---

Solution:

Let the digit at the units place of the two-digit number be $x$.

Since the difference between the digits is 3, the digit at the tens place can be either $(x+3)$ or $(x-3)$. This gives us two possible cases.

We know that a two-digit number with tens digit 'a' and units digit 'b' can be written as $10a + b$.

---

Case 1: The tens digit is $(x+3)$.

The units digit = $x$

The tens digit = $x+3$

So, the original number = $10 \times (\text{tens digit}) + (\text{units digit})$

$= 10(x+3) + x$

$= 10x + 30 + x$

$= 11x + 30$

When the digits are interchanged:

The new units digit = $x+3$

The new tens digit = $x$

The new number = $10x + (x+3)$

$= 11x + 3$

According to the question, the sum of the original number and the new number is 143.

$(\text{Original Number}) + (\text{New Number}) = 143$

$(11x + 30) + (11x + 3) = 143$

$22x + 33 = 143$

$22x = 143 - 33$

$22x = 110$

$x = \frac{110}{22}$

$x = 5$

So, the units digit is 5.

The tens digit is $x+3 = 5+3 = 8$.

The original number is $10(8) + 5 = 85$.

Check: The digits are 8 and 5, their difference is $8-5=3$. The new number is 58. The sum is $85 + 58 = 143$. This matches the given conditions.

---

Case 2: The tens digit is $(x-3)$.

The units digit = $x$

The tens digit = $x-3$

So, the original number = $10(x-3) + x$

$= 10x - 30 + x$

$= 11x - 30$

When the digits are interchanged:

The new units digit = $x-3$

The new tens digit = $x$

The new number = $10x + (x-3)$

$= 11x - 3$

According to the question:

$(11x - 30) + (11x - 3) = 143$

$22x - 33 = 143$

$22x = 143 + 33$

$22x = 176$

$x = \frac{176}{22}$

$x = 8$

So, the units digit is 8.

The tens digit is $x-3 = 8-3 = 5$.

The original number is $10(5) + 8 = 58$.

Check: The digits are 5 and 8, their difference is $8-5=3$. The new number is 85. The sum is $58 + 85 = 143$. This also matches the given conditions.

---

Therefore, the original number can be either 58 or 85.

Given:

1. Arjun's present age is twice Shriya's present age.

2. Five years ago, Arjun's age was three times Shriya's age at that time.

---

To Find:

The present ages of Arjun and Shriya.

---

Solution:

Let Shriya's present age be $x$ years.

According to condition 1, Arjun's present age is twice Shriya's present age.

So, Arjun's present age = $2x$ years.

Now, consider their ages five years ago:

Shriya's age five years ago = (Shriya's present age) - 5 = $x - 5$ years.

Arjun's age five years ago = (Arjun's present age) - 5 = $2x - 5$ years.

According to condition 2, five years ago, Arjun's age was three times Shriya's age.

We can set up the equation:

Now, solve this equation for $x$:

Distribute the 3 on the right side:

Gather the terms involving $x$ on one side and the constant terms on the other side.

Subtract $2x$ from both sides:

Add 15 to both sides:

So, Shriya's present age ($x$) is 10 years.

Arjun's present age = $2x = 2 \times 10 = 20$ years.

Therefore, Shriya's present age is 10 years and Arjun's present age is 20 years.

Verification:

Present ages: Arjun = 20, Shriya = 10. Is Arjun twice as old as Shriya? $20 = 2 \times 10$. Yes.

Ages 5 years ago: Arjun = $20 - 5 = 15$, Shriya = $10 - 5 = 5$.

Was Arjun's age 5 years ago three times Shriya's age 5 years ago? $15 = 3 \times 5$. Yes.

Both conditions are satisfied.

Given:

1. Amina thinks of a number.

2. She subtracts $\frac{5}{2}$ from the number.

3. She multiplies the result of the subtraction by 8.

4. The final result is equal to 3 times the original number.

---

To Find:

The number Amina thought of.

---

Solution:

Let the number Amina thought of be $x$.

Step 1: Subtract $\frac{5}{2}$ from the number.

Result after subtraction = $x - \frac{5}{2}$.

Step 2: Multiply the result by 8.

Result after multiplication = $8 \times \left( x - \frac{5}{2} \right)$.

Step 3: The final result is 3 times the original number.

Final result = $3x$.

According to the problem, the result after multiplication is equal to 3 times the original number. So, we can set up the equation:

Now, solve this equation for $x$:

Distribute the 8 on the left side:

Simplify the multiplication term:

Gather the terms involving $x$ on one side. Subtract $3x$ from both sides:

Simplify:

Move the constant term to the other side. Add 20 to both sides:

Simplify:

Divide both sides by 5:

Therefore, the number Amina thought of is 4.

Verification:

Number thought of = 4.

Subtract $\frac{5}{2}$: $4 - \frac{5}{2} = \frac{8}{2} - \frac{5}{2} = \frac{3}{2}$.

Multiply the result by 8: $8 \times \frac{3}{2} = \frac{24}{2} = 12$.

Is the result (12) equal to 3 times the original number (4)? $3 \times 4 = 12$. Yes.

The conditions are satisfied.

Given:

1. One positive number is 5 times another number.

2. If 21 is added to both numbers, one of the resulting new numbers is twice the other resulting new number.

---

To Find:

The original two numbers.

---

Solution:

Let the "another number" be $x$.

Since the "positive number" is 5 times another number, the positive number is $5x$.

(The problem states it's a positive number, so $5x > 0$, which implies $x > 0$.)

Now, add 21 to both numbers:

New first number (from $x$) = $x + 21$.

New second number (from $5x$) = $5x + 21$.

According to the problem, after adding 21, one of the new numbers becomes twice the other new number.

Since $x$ is positive, $5x > x$, which means $5x + 21 > x + 21$. Therefore, the larger new number ($5x + 21$) must be twice the smaller new number ($x + 21$).

We can set up the equation:

Now, solve this equation for $x$:

Distribute the 2 on the right side:

Subtract $2x$ from both sides:

Simplify:

Subtract 21 from both sides:

Simplify:

Divide both sides by 3:

So, the "another number" is $x = 7$.

The "positive number" is $5x = 5 \times 7 = 35$.

Therefore, the two numbers are 7 and 35.

Verification:

The numbers are 7 and 35. $35 = 5 \times 7$. (Condition 1 met).

Add 21 to both numbers: $7 + 21 = 28$ and $35 + 21 = 56$.

Check if one new number is twice the other: $56 = 2 \times 28$. (Condition 2 met).

The conditions are satisfied.

Given:

1. A two-digit number.

2. The sum of its digits is 9.

3. When the digits are interchanged, the new number is formed.

4. The new number is greater than the original number by 27.

---

To Find:

The original two-digit number.

---

Solution:

Let the two-digit number be represented by $10a + b$, where $a$ is the digit in the tens place and $b$ is the digit in the units place. ($a \in \{1, ..., 9\}$, $b \in \{0, ..., 9\}$).

From Condition 2: The sum of the digits is 9.

From Condition 3 and 4: When the digits are interchanged, the new number is $10b + a$.

This new number is greater than the original number ($10a + b$) by 27.

So, we can write the equation:

Now, simplify this equation:

Bring variable terms to one side and constants to the other.

Divide the entire equation by 9:

Now we have a system of two linear equations:

1. $a + b = 9$

2. $-a + b = 3$

Add equation (i) and equation (ii):

Divide by 2:

Substitute the value of $b$ into equation (i):

Solve for $a$:

So, the tens digit $a = 3$ and the units digit $b = 6$.

The original two-digit number is $10a + b = 10(3) + 6 = 36$.

Therefore, the two-digit number is 36.

Verification:

The number is 36. Digits are 3 and 6.

Sum of digits: $3 + 6 = 9$. (Matches condition 2)

Interchanged number: 63.

Check if new number is 27 greater than original: $63 - 36 = 27$. (Matches condition 4)

The conditions are satisfied.

Given:

1. A two-digit number.

2. One of its digits is three times the other digit.

3. When the digits are interchanged, a new number is formed.

4. The sum of the original number and the new number is 88.

---

To Find:

The original two-digit number.

---

Solution:

Let the two-digit number be represented as $10a + b$, where $a$ is the digit in the tens place and $b$ is the digit in the units place. ($a \in \{1, ..., 9\}$, $b \in \{0, ..., 9\}$).

From Condition 4: The sum of the original number and the number with interchanged digits is 88.

Original number = $10a + b$.

Number with interchanged digits = $10b + a$.

Their sum is 88:

Simplify this equation:

Divide both sides by 11:

From Condition 2: One digit is three times the other digit.

This gives two possibilities:

Case I: The tens digit is three times the units digit ($a = 3b$).

Case II: The units digit is three times the tens digit ($b = 3a$).

Case I: $a = 3b$

Substitute $a = 3b$ into equation (i):

Now find $a$ using $a = 3b$:

The digits are $a=6$ and $b=2$. The original number is $10a + b = 10(6) + 2 = 62$.

Case II: $b = 3a$

Substitute $b = 3a$ into equation (i):

Now find $b$ using $b = 3a$:

The digits are $a=2$ and $b=6$. The original number is $10a + b = 10(2) + 6 = 26$.

Both numbers 62 and 26 satisfy all the conditions.

Therefore, the original number can be either 62 or 26.

Verification:

For 62: Digits are 6 and 2. $6 = 3 \times 2$. Sum $6+2=8$. Interchanged is 26. Sum = $62 + 26 = 88$. (Correct)

For 26: Digits are 2 and 6. $6 = 3 \times 2$. Sum $2+6=8$. Interchanged is 62. Sum = $26 + 62 = 88$. (Correct)

Given:

1. Shobo's mother's present age is six times Shobo's present age.

2. Shobo's age five years from now will be one-third of his mother's present age.

---

To Find:

The present ages of Shobo and his mother.

---

Solution:

Let Shobo's present age be $x$ years.

From condition 1, Shobo's mother's present age is $6x$ years.

Shobo's age five years from now will be (Present age of Shobo) + 5 = $x + 5$ years.

According to condition 2, Shobo's age five years from now ($x+5$) will be one-third of his mother's present age ($6x$).

We can set up the equation:

Now, solve this equation for $x$:

Simplify the right side:

Subtract $x$ from both sides:

Simplify:

So, Shobo's present age ($x$) is 5 years.

Shobo's mother's present age = $6x = 6 \times 5 = 30$ years.

Therefore, Shobo's present age is 5 years and his mother's present age is 30 years.

Verification:

Mother's present age (30) is six times Shobo's present age (5)? $30 = 6 \times 5$. Yes.

Shobo's age five years from now = $5 + 5 = 10$.

Is Shobo's age five years from now (10) one third of his mother's present age (30)? $\frac{1}{3} \times 30 = 10$. Yes.

Both conditions are satisfied.

Given:

1. Shape of the plot: Rectangular.

2. Ratio of Length (L) to Breadth (B) = 11 : 4.

3. Cost of fencing per metre = $\textsf{₹} 100$.

4. Total cost of fencing the plot = $\textsf{₹} 75,000$.

5. Fencing corresponds to the perimeter of the plot.

---

To Find:

The dimensions (Length and Breadth) of the plot.

---

Solution:

First, let's find the perimeter of the plot using the cost information.

Solve for the Perimeter:

Now, let's use the ratio of the length and breadth.

Since the ratio L : B = 11 : 4, we can represent the length and breadth using a common factor $x$.

Let Length (L) = $11x$ metres.

Let Breadth (B) = $4x$ metres.

The formula for the perimeter of a rectangle is:

Substitute the expressions for L and B, and the calculated perimeter:

Simplify the equation:

Solve for $x$ by dividing both sides by 30:

Now, find the dimensions using $x=25$:

Length (L) = $11x = 11 \times 25 = 275$ metres.

Breadth (B) = $4x = 4 \times 25 = 100$ metres.

Therefore, the dimensions of the plot are Length = 275 m and Breadth = 100 m.

Verification:

Ratio L:B = $275:100$. Divide by 25: $11:4$. (Matches)

Perimeter = $2(L+B) = 2(275+100) = 2(375) = 750$ m.

Total cost = Perimeter $\times$ Rate = $750 \times \textsf{₹} 100 = \textsf{₹} 75,000$. (Matches)

Given:

1. Cost of shirt material = $\textsf{₹} 50$ per metre.

2. Cost of trouser material = $\textsf{₹} 90$ per metre.

3. Ratio of shirt material bought to trouser material bought = 3 : 2.

4. Profit on selling shirt material = 12%.

5. Profit on selling trouser material = 10%.

6. Total sale amount = $\textsf{₹} 36,600$.

---

To Find:

The amount of trouser material Hasan bought (in metres).

---

Solution:

Let the amount of shirt material bought be $3x$ metres and the amount of trouser material bought be $2x$ metres, according to the given ratio 3:2.

Calculate the Cost Price (CP) for each material:

CP of shirt material = (Quantity of shirt material) $\times$ (Cost per metre)

CP of trouser material = (Quantity of trouser material) $\times$ (Cost per metre)

Calculate the Selling Price (SP) for each material:

The formula for Selling Price is SP = CP $\times (1 + \frac{\text{Profit %}}{100})$.

SP of shirt material (12% profit):

SP of trouser material (10% profit):

Use the Total Sale information:

Total Sale = SP of shirt material + SP of trouser material

The total sale is given as $\textsf{₹} 36,600$.

Set up the equation:

Combine the like terms on the left side:

Solve for $x$ by dividing both sides by 366:

The question asks for the amount of trouser material bought.

Amount of trouser material = $2x$.

Substitute $x = 100$:

Therefore, Hasan bought 200 metres of trouser material.

Verification:

If $x=100$, then:

Shirt material bought = $3x = 3 \times 100 = 300$ m.

Trouser material bought = $2x = 2 \times 100 = 200$ m.

SP of shirt material = $168x = 168 \times 100 = \textsf{₹} 16,800$.

SP of trouser material = $198x = 198 \times 100 = \textsf{₹} 19,800$.

Total Sale = $\textsf{₹} 16,800 + \textsf{₹} 19,800 = \textsf{₹} 36,600$. (Matches)

Given:

1. Half ($\frac{1}{2}$) of the herd are grazing.

2. Three-fourths ($\frac{3}{4}$) of the remaining deer are playing.

3. The rest of the deer, which is 9, are drinking water.

---

To Find:

The total number of deer in the herd.

---

Solution:

Let the total number of deer in the herd be $x$.

Number of deer grazing = Half of the total herd = $\frac{1}{2}x$.

Number of deer remaining after grazing = Total deer - Grazing deer

Number of deer playing = Three-fourths of the remaining deer

Number of deer drinking = 9.

The total number of deer is the sum of those grazing, playing, and drinking:

Now, solve this equation for $x$. Move all terms with $x$ to one side:

Find a common denominator (8) for the terms with $x$:

Combine the terms:

Multiply both sides by 8:

Therefore, the total number of deer in the herd is 72.

Verification:

Total deer = 72.

Grazing = $\frac{1}{2} \times 72 = 36$.

Remaining = $72 - 36 = 36$.

Playing = $\frac{3}{4}$ of remaining = $\frac{3}{4} \times 36 = 27$.

Drinking = 9 (given).

Total calculated = Grazing + Playing + Drinking = $36 + 27 + 9 = 72$.

This matches the calculated total number of deer.

Given:

1. Grandfather's present age is ten times his granddaughter's present age.

2. Grandfather is 54 years older than his granddaughter.

---

To Find:

The present ages of the grandfather and the granddaughter.

---

Solution:

Let the granddaughter's present age be $x$ years.

From condition 1, the grandfather's present age is ten times the granddaughter's age.

So, Grandfather's present age = $10x$ years.

From condition 2, the grandfather is 54 years older than the granddaughter.

This means:

Now, substitute the expressions for their ages in terms of $x$ into this relationship:

Solve this equation for $x$. Subtract $x$ from both sides:

Simplify:

Divide both sides by 9:

So, the granddaughter's present age ($x$) is 6 years.

Grandfather's present age = $10x = 10 \times 6 = 60$ years.

Therefore, the granddaughter's present age is 6 years and the grandfather's present age is 60 years.

Verification:

Grandfather's age (60) is ten times granddaughter's age (6)? $60 = 10 \times 6$. Yes.

Grandfather's age (60) is 54 years older than granddaughter's age (6)? $60 = 6 + 54$. Yes.

Both conditions are satisfied.

Given:

1. Aman's present age is three times his son's present age.

2. Ten years ago, Aman's age was five times his son's age at that time.

---

To Find:

The present ages of Aman and his son.

---

Solution:

Let the son's present age be $x$ years.

From condition 1, Aman's present age is $3x$ years.

Now, consider their ages ten years ago:

Son's age ten years ago = (Son's present age) - 10 = $x - 10$ years.

Aman's age ten years ago = (Aman's present age) - 10 = $3x - 10$ years.

According to condition 2, ten years ago, Aman's age was five times his son's age:

Now, solve this equation for $x$:

Distribute the 5 on the right side:

Gather the terms involving $x$ on one side and the constant terms on the other side.

Subtract $3x$ from both sides:

Add 50 to both sides:

Divide both sides by 2:

So, the son's present age ($x$) is 20 years.

Aman's present age = $3x = 3 \times 20 = 60$ years.

Therefore, Aman's present age is 60 years and his son's present age is 20 years.

Verification:

Present ages: Aman = 60, Son = 20. Is Aman's age three times his son's age? $60 = 3 \times 20$. Yes.

Ages 10 years ago: Aman = $60 - 10 = 50$, Son = $20 - 10 = 10$.

Was Aman's age 10 years ago (50) five times his son's age 10 years ago (10)? $50 = 5 \times 10$. Yes.

Both conditions are satisfied.

Given Equation:

---

To Find:

The value of $x$.

---

Solution:

To solve the equation, we first eliminate the fractions. The denominators are 3 and 6. The Least Common Multiple (LCM) of 3 and 6 is 6.

Multiply every term in the equation by the LCM (6):

Simplify each term:

Distribute the 2 on the left side:

Combine the constant terms on the left side:

Gather the terms involving $x$ on one side and the constant terms on the other side.

Subtract $x$ from both sides:

Simplify:

Subtract 8 from both sides:

Simplify:

Divide both sides by 11:

Therefore, the solution to the equation is $x = -1$.

Verification:

Substitute $x = -1$ into the original equation $\frac{6x+1}{3} + 1 = \frac{x - 3}{6}$.

Left Hand Side (LHS): $\frac{6(-1) + 1}{3} + 1 = \frac{-6 + 1}{3} + 1 = \frac{-5}{3} + \frac{3}{3} = -\frac{2}{3}$.

Right Hand Side (RHS): $\frac{-1 - 3}{6} = \frac{-4}{6} = -\frac{2}{3}$.

Since LHS = RHS ($-\frac{2}{3} = -\frac{2}{3}$), the solution is correct.

Given Equation:

---

To Find:

The value of $x$.

---

Solution:

First, simplify both sides of the equation by distributing the terms in the parentheses:

Left Hand Side (LHS):

Right Hand Side (RHS):

Combine the constant terms on the RHS by finding a common denominator:

Now, set the simplified LHS equal to the simplified RHS:

Gather the terms involving $x$ on one side and the constant terms on the other side.

Subtract $x$ from both sides:

Subtract $\frac{3}{2}$ from both sides:

Find a common denominator for the left side:

Divide both sides by 5 (or multiply by $\frac{1}{5}$):

Simplify:

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

Verification:

Substitute $x = \frac{5}{2}$ into the original equation $5x - 2(2x - 7) \ $$ = 2(3x - 1) + \frac{7}{2}$.

LHS: $5\left(\frac{5}{2}\right) - 2\left(2\left(\frac{5}{2}\right) - 7\right) = \frac{25}{2} - 2(5 - 7) = \frac{25}{2} - 2(-2) = \frac{25}{2} + 4 \ $$ = \frac{25}{2} + \frac{8}{2} = \frac{33}{2}$.

RHS: $2\left(3\left(\frac{5}{2}\right) - 1\right) + \frac{7}{2} = 2\left(\frac{15}{2} - \frac{2}{2}\right) + \frac{7}{2} = 2\left(\frac{13}{2}\right) + \frac{7}{2} = 13 + \frac{7}{2} \ $$ = \frac{26}{2} + \frac{7}{2} = \frac{33}{2}$.

Since LHS = RHS ($\frac{33}{2} = \frac{33}{2}$), the solution is correct.

Given Equation:

---

To Find:

The value of $x$.

---

Solution:

To solve the equation, we first eliminate the fractions. The denominators are 2, 5, 3, and 4. The Least Common Multiple (LCM) of 2, 5, 3, and 4 is 60.

Multiply every term in the equation by the LCM (60):

Simplify each term:

Now, gather the terms involving $x$ on one side and the constant terms on the other side.

Subtract $20x$ from both sides:

Simplify:

Add 12 to both sides:

Simplify:

Divide both sides by 10:

Therefore, the solution to the equation is $x = \frac{27}{10}$.

Verification:

Substitute $x = \frac{27}{10}$ into the original equation.

LHS: $\frac{x}{2} - \frac{1}{5} = \frac{27/10}{2} - \frac{1}{5} = \frac{27}{20} - \frac{1}{5} = \frac{27}{20} - \frac{4}{20} = \frac{23}{20}$.

RHS: $\frac{x}{3} + \frac{1}{4} = \frac{27/10}{3} + \frac{1}{4} = \frac{27}{30} + \frac{1}{4} = \frac{9}{10} + \frac{1}{4} = \frac{18}{20} + \frac{5}{20} = \frac{23}{20}$.

Since LHS = RHS ($\frac{23}{20} = \frac{23}{20}$), the solution is correct.

Given Equation:

---

To Find:

The value of $n$.

---

Solution:

To solve the equation, we first eliminate the fractions. The denominators are 2, 4, and 6. The Least Common Multiple (LCM) of 2, 4, and 6 is 12.

Find the LCM:

$\begin{array}{c|cc} 2 & 2 \;, & 4 \;, & 6 \\ \hline 2 & 1 \; , & 2 \; , & 3 \\ \hline 3 & 1 \; , & 1 \; , & 3 \\ \hline & 1 \; , & 1 \; , & 1 \end{array}$

LCM = $2 \times 2 \times 3 = 12$.

Multiply every term in the equation by the LCM (12):

Simplify each term:

Combine the terms involving $n$ on the left side:

Divide both sides by 7:

Therefore, the solution to the equation is $n = 36$.

Verification:

Substitute $n = 36$ into the original equation.

LHS: $\frac{36}{2} - \frac{3(36)}{4} + \frac{5(36)}{6} = 18 - \frac{108}{4} + \frac{180}{6} = 18 - 27 + 30 = -9 + 30 = 21$.

RHS: $21$.

Since LHS = RHS ($21 = 21$), the solution is correct.

Given Equation:

---

To Find:

The value of $x$.

---

Solution:

To solve the equation, we first eliminate the fractions. The denominators are 3, 6, and 2. The Least Common Multiple (LCM) of 3, 6, and 2 is 6.

Multiply every term in the equation by the LCM (6):

Simplify each term:

Combine the terms involving $x$ on the left side:

Gather the terms involving $x$ on one side and the constant terms on the other side.

Add $15x$ to both sides:

Simplify:

Subtract 42 from both sides:

Simplify:

Divide both sides by 5:

Therefore, the solution to the equation is $x = -5$.

Verification:

Substitute $x = -5$ into the original equation.

LHS: $x + 7 - \frac{8x}{3} = (-5) + 7 - \frac{8(-5)}{3} = 2 - \left(\frac{-40}{3}\right) \ $$ = 2 + \frac{40}{3} \ $$ = \frac{6}{3} + \frac{40}{3} \ $$ = \frac{46}{3}$.

RHS: $\frac{17}{6} - \frac{5x}{2} = \frac{17}{6} - \frac{5(-5)}{2} = \frac{17}{6} - \left(\frac{-25}{2}\right) = \frac{17}{6} + \frac{25}{2} = \frac{17}{6} + \frac{25 \times 3}{2 \times 3} \ $$ = \frac{17}{6} + \frac{75}{6} \ $$ = \frac{17 + 75}{6} = \frac{92}{6} = \frac{46}{3}$.

Since LHS = RHS ($\frac{46}{3} = \frac{46}{3}$), the solution is correct.

Given Equation:

---

To Find:

The value of $x$.

---

Solution:

To solve the equation, we can eliminate the fractions by cross-multiplication or by multiplying both sides by the LCM of the denominators (3 and 5), which is 15.

Method 1: Cross-Multiplication

Multiply the numerator of the left side by the denominator of the right side, and the numerator of the right side by the denominator of the left side:

Distribute the numbers on both sides:

Gather the terms involving $x$ on one side and the constant terms on the other side.

Subtract $3x$ from both sides:

Simplify:

Add 25 to both sides:

Simplify:

Divide both sides by 2:

---

Method 2: Multiplying by LCM

The LCM of 3 and 5 is 15. Multiply both sides by 15:

Simplify:

This leads to the same equation as in Method 1, $5x - 25 = 3x - 9$, which solves to $x=8$.

Therefore, the solution to the equation is $x = 8$.

Verification:

Substitute $x = 8$ into the original equation $\frac{x-5}{3} = \frac{x-3}{5}$.

LHS: $\frac{8-5}{3} = \frac{3}{3} = 1$.

RHS: $\frac{8-3}{5} = \frac{5}{5} = 1$.

Since LHS = RHS ($1 = 1$), the solution is correct.

Given Equation:

---

To Find:

The value of $t$.

---

Solution:

To solve the equation, we first eliminate the fractions. The denominators are 4 and 3. The Least Common Multiple (LCM) of 4 and 3 is 12.

Rewrite the equation with $t$ as $\frac{t}{1}$:

Multiply every term in the equation by the LCM (12):

Simplify each term:

Distribute the numbers in parentheses:

Combine like terms on the left side:

Gather the terms involving $t$ on one side and the constant terms on the other side.

Add $12t$ to both sides:

Simplify:

Add 18 to both sides:

Simplify:

Divide both sides by 13:

Therefore, the solution to the equation is $t = 2$.

Verification:

Substitute $t = 2$ into the original equation.

LHS: $\frac{3(2) - 2}{4} - \frac{2(2) + 3}{3} = \frac{6 - 2}{4} - \frac{4 + 3}{3} = \frac{4}{4} - \frac{7}{3} = 1 - \frac{7}{3} = \frac{3}{3} - \frac{7}{3} = -\frac{4}{3}$.

RHS: $\frac{2}{3} - t = \frac{2}{3} - 2 = \frac{2}{3} - \frac{6}{3} = -\frac{4}{3}$.

Since LHS = RHS ($-\frac{4}{3} = -\frac{4}{3}$), the solution is correct.

Given Equation:

---

To Find:

The value of $m$.

---

Solution:

To solve the equation, we first eliminate the fractions. The denominators are 2 and 3. The Least Common Multiple (LCM) of 2 and 3 is 6.

Rewrite the equation with $m$ as $\frac{m}{1}$ and $1$ as $\frac{1}{1}$:

Multiply every term in the equation by the LCM (6):

Simplify each term:

Distribute the numbers in parentheses. Be careful with the negative signs:

Combine like terms on each side:

Gather the terms involving $m$ on one side and the constant terms on the other side.

Add $2m$ to both sides:

Simplify:

Subtract 3 from both sides:

Simplify:

Divide both sides by 5:

Therefore, the solution to the equation is $m = \frac{7}{5}$.

Verification:

Substitute $m = \frac{7}{5}$ into the original equation.

LHS: $m - \frac{m-1}{2} = \frac{7}{5} - \frac{\frac{7}{5} - 1}{2} = \frac{7}{5} - \frac{\frac{7}{5} - \frac{5}{5}}{2} = \frac{7}{5} - \frac{\frac{2}{5}}{2} = \frac{7}{5} - \frac{2}{10} \ $$ = \frac{14}{10} - \frac{2}{10} = \frac{12}{10} = \frac{6}{5}$.

RHS: $1 - \frac{m-2}{3} = 1 - \frac{\frac{7}{5} - 2}{3} = 1 - \frac{\frac{7}{5} - \frac{10}{5}}{3} = 1 - \frac{-\frac{3}{5}}{3} = 1 - \left(-\frac{3}{15}\right) \ $$ = 1 - \left(-\frac{1}{5}\right) = 1 + \frac{1}{5} = \frac{5}{5} + \frac{1}{5} = \frac{6}{5}$.

Since LHS = RHS ($\frac{6}{5} = \frac{6}{5}$), the solution is correct.

Given Equation:

---

To Find:

The value of $t$.

---

Solution:

First, distribute the numbers outside the parentheses on both sides of the equation:

Left Hand Side (LHS):

Right Hand Side (RHS):

Now, set the simplified LHS equal to the simplified RHS:

Gather the terms involving $t$ on one side and the constant terms on the other side.

Subtract $3t$ from both sides:

Simplify:

Subtract 5 from both sides:

Simplify:

Divide both sides by 7:

Therefore, the solution to the equation is $t = -2$.

Verification:

Substitute $t = -2$ into the original equation $3(t - 3) = 5(2t + 1)$.

LHS: $3(t - 3) = 3(-2 - 3) = 3(-5) = -15$.

RHS: $5(2t + 1) = 5(2(-2) + 1) = 5(-4 + 1) = 5(-3) = -15$.

Since LHS = RHS ($-15 = -15$), the solution is correct.

Given Equation:

---

To Find:

The value of $y$.

---

Solution:

First, distribute the constants outside the parentheses:

Remove the parentheses, being careful with the signs:

Combine the like terms (terms with $y$ and constant terms):

Simplify:

Add 12 to both sides:

Divide both sides by 18:

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, 6:

Therefore, the solution to the equation is $y = \frac{2}{3}$.

Verification:

Substitute $y = \frac{2}{3}$ into the original equation.

$15\left(\frac{2}{3} - 4\right) - 2\left(\frac{2}{3} - 9\right) + 5\left(\frac{2}{3} + 6\right)$

$= 15\left(\frac{2}{3} - \frac{12}{3}\right) - 2\left(\frac{2}{3} - \frac{27}{3}\right) + 5\left(\frac{2}{3} + \frac{18}{3}\right)$

$= 15\left(-\frac{10}{3}\right) - 2\left(-\frac{25}{3}\right) + 5\left(\frac{20}{3}\right)$

$= \frac{-150}{3} - \left(-\frac{50}{3}\right) + \frac{100}{3}$

$= -50 + \frac{50}{3} + \frac{100}{3}$

$= -50 + \frac{50 + 100}{3}$

$= -50 + \frac{150}{3}$

$= -50 + 50$

$= 0$

The result is 0, which matches the right side of the original equation. So, the solution is correct.

Given Equation:

---

To Find:

The value of $z$.

---

Solution:

First, simplify both sides of the equation by distributing the constants into the parentheses:

Left Hand Side (LHS):

Remove the parentheses, being careful with the negative sign:

Combine like terms on the LHS:

Right Hand Side (RHS):

Combine like terms on the RHS:

Now, set the simplified LHS equal to the simplified RHS:

Gather the terms involving $z$ on one side and the constant terms on the other side.

Add $3z$ to both sides:

Simplify:

Add 69 to both sides:

Simplify:

Divide both sides by 35:

Therefore, the solution to the equation is $z = 2$.

Verification:

Substitute $z = 2$ into the original equation.

LHS: $3(5(2) - 7) - 2(9(2) - 11) = 3(10 - 7) - 2(18 - 11) \ $$ = 3(3) - 2(7) \ $$ = 9 - 14 = -5$.

RHS: $4(8(2) - 13) - 17 = 4(16 - 13) - 17 = 4(3) - 17 \ $$ = 12 - 17 = -5$.

Since LHS = RHS ($-5 = -5$), the solution is correct.

Given Equation:

---

To Find:

The value of $f$.

---

Solution:

To solve the equation, we can first eliminate the decimals. The decimals involved are 0.25 and 0.05. We can multiply both sides of the equation by 100 to clear the decimals.

Simplify:

Distribute the constants on both sides:

Left Hand Side (LHS):

Right Hand Side (RHS):

Set the simplified LHS equal to the simplified RHS:

Gather the terms involving $f$ on one side and the constant terms on the other side.

Subtract $50f$ from both sides:

Simplify:

Add 75 to both sides:

Simplify:

Divide both sides by 50:

Simplify the fraction:

Alternatively, the answer can be expressed as a decimal:

Therefore, the solution to the equation is $f = \frac{3}{5}$ or $f = 0.6$.

Verification:

Substitute $f = 0.6$ into the original equation $0.25(4f - 3) = 0.05(10f - 9)$.

LHS: $0.25(4(0.6) - 3) = 0.25(2.4 - 3) = 0.25(-0.6) = -0.15$.

RHS: $0.05(10(0.6) - 9) = 0.05(6 - 9) = 0.05(-3) = -0.15$.

Since LHS = RHS ($-0.15 = -0.15$), the solution is correct.

Given Equation:

---

To Find:

The value of $x$.

---

Solution:

This equation involves a proportion (one fraction equals another). We can solve it by cross-multiplication.

Multiply the numerator of the left fraction by the denominator of the right fraction, and set it equal to the product of the numerator of the right fraction and the denominator of the left fraction:

Distribute the constants on both sides:

Gather the terms involving $x$ on one side and the constant terms on the other side.

Subtract $6x$ from both sides:

Simplify:

Subtract 8 from both sides:

Simplify:

Divide both sides by 2:

We should check that this value of $x$ does not make the original denominator zero. $2x + 3 = 2(\frac{1}{2}) + 3 = 1 + 3 = 4 \neq 0$. The solution is valid.

Therefore, the solution to the equation is $x = \frac{1}{2}$.

Verification:

Substitute $x = \frac{1}{2}$ into the original equation $\frac{x+1}{2x+3} = \frac{3}{8}$.

LHS: $\frac{\frac{1}{2} + 1}{2(\frac{1}{2}) + 3} = \frac{\frac{1}{2} + \frac{2}{2}}{1 + 3} = \frac{\frac{3}{2}}{4} = \frac{3}{2} \times \frac{1}{4} = \frac{3}{8}$.

RHS: $\frac{3}{8}$.

Since LHS = RHS ($\frac{3}{8} = \frac{3}{8}$), the solution is correct.

Given:

1. Ratio of present ages of Anu and Raj = 4 : 5.

2. Eight years from now, the ratio of their ages will be 5 : 6.

---

To Find:

The present ages of Anu and Raj.

---

Solution:

Let the present age of Anu be $4x$ years and the present age of Raj be $5x$ years, according to the given ratio 4:5.

Now, consider their ages eight years from now:

Anu's age after 8 years = (Present age of Anu) + 8 = $4x + 8$ years.

Raj's age after 8 years = (Present age of Raj) + 8 = $5x + 8$ years.

According to the problem, the ratio of their ages after 8 years will be 5:6.

So, we can set up the proportion:

Solve this equation for $x$ by cross-multiplication:

Distribute the constants on both sides:

Gather the terms involving $x$ on one side and the constant terms on the other side.

Subtract $24x$ from both sides:

Simplify:

Subtract 40 from both sides:

Simplify:

So, $x = 8$.

Now, find their present ages using $x = 8$:

Anu's present age = $4x = 4 \times 8 = 32$ years.

Raj's present age = $5x = 5 \times 8 = 40$ years.

Therefore, Anu's present age is 32 years and Raj's present age is 40 years.

Verification:

Present ages: Anu = 32, Raj = 40. Ratio $32:40 = 4:5$. (Correct)

Ages after 8 years: Anu = $32+8 = 40$, Raj = $40+8 = 48$.

Ratio after 8 years: $40:48 = 5:6$. (Correct)

Given Equation:

Restriction: The denominator cannot be zero, so $3x \neq 0$, which means $x \neq 0$.

---

To Find:

The value of $x$.

---

Solution:

To solve the equation, we can clear the fraction by multiplying both sides by the denominator $3x$.

Simplify:

Gather the terms involving $x$ on one side and the constant terms on the other side.

Subtract $6x$ from both sides:

Simplify:

Add 3 to both sides:

Divide both sides by 2:

The solution $x = \frac{3}{2}$ does not violate the restriction $x \neq 0$.

Therefore, the solution to the equation is $x = \frac{3}{2}$.

Verification:

Substitute $x = \frac{3}{2}$ into the original equation $\frac{8x - 3}{3x} = 2$.

LHS: $\frac{8\left(\frac{3}{2}\right) - 3}{3\left(\frac{3}{2}\right)} = \frac{\frac{24}{2} - 3}{\frac{9}{2}} = \frac{12 - 3}{\frac{9}{2}} = \frac{9}{\frac{9}{2}} = 9 \times \frac{2}{9} = 2$.

RHS: $2$.

Since LHS = RHS ($2 = 2$), the solution is correct.

Given Equation:

Restriction: The denominator cannot be zero, so $7 - 6x \neq 0$, which implies $6x \neq 7$ or $x \neq \frac{7}{6}$.

---

To Find:

The value of $x$.

---

Solution:

To solve the equation, we can clear the fraction by multiplying both sides by the denominator $(7 - 6x)$.

Simplify:

Distribute the 15 on the right side:

Gather the terms involving $x$ on one side.

Add $90x$ to both sides:

Simplify:

Divide both sides by 99:

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, 3:

The solution $x = \frac{35}{33}$ does not violate the restriction $x \neq \frac{7}{6}$.

Therefore, the solution to the equation is $x = \frac{35}{33}$.

Verification:

Substitute $x = \frac{35}{33}$ into the original equation $\frac{9x}{7 - 6x} = 15$.

LHS: $\frac{9(\frac{35}{33})}{7 - 6(\frac{35}{33})} = \frac{\frac{3 \times 35}{11}}{7 - \frac{2 \times 35}{11}} = \frac{\frac{105}{11}}{7 - \frac{70}{11}} = \frac{\frac{105}{11}}{\frac{77}{11} - \frac{70}{11}} = \frac{\frac{105}{11}}{\frac{7}{11}} \ $$ = \frac{105}{11} \times \frac{11}{7} \ $$ = \frac{105}{7} = 15$.

RHS: $15$.

Since LHS = RHS ($15 = 15$), the solution is correct.

Given Equation:

Restriction: The denominator cannot be zero, so $z + 15 \neq 0$, which means $z \neq -15$.

---

To Find:

The value of $z$.

---

Solution:

This equation involves a proportion. We can solve it by cross-multiplication.

Multiply the numerator of the left fraction by the denominator of the right fraction, and set it equal to the product of the numerator of the right fraction and the denominator of the left fraction:

Simplify and distribute:

Gather the terms involving $z$ on one side.

Subtract $4z$ from both sides:

Simplify:

Divide both sides by 5:

The solution $z = 12$ does not violate the restriction $z \neq -15$.

Therefore, the solution to the equation is $z = 12$.

Verification:

Substitute $z = 12$ into the original equation $\frac{z}{z + 15} = \frac{4}{9}$.

LHS: $\frac{12}{12 + 15} = \frac{12}{27}$. Simplify by dividing numerator and denominator by 3: $\frac{12 \div 3}{27 \div 3} = \frac{4}{9}$.

RHS: $\frac{4}{9}$.

Since LHS = RHS ($\frac{4}{9} = \frac{4}{9}$), the solution is correct.

Given Equation:

Restriction: The denominator cannot be zero, so $2 - 6y \neq 0$, which implies $6y \neq 2$ or $y \neq \frac{1}{3}$.

---

To Find:

The value of $y$.

---

Solution:

This equation involves a proportion. We can solve it by cross-multiplication.

Multiply the numerator of the left fraction by the denominator of the right fraction, and set it equal to the product of the numerator of the right fraction and the denominator of the left fraction:

Distribute the constants on both sides:

Gather the terms involving $y$ on one side and the constant terms on the other side.

Subtract $12y$ from both sides:

Simplify:

Subtract 20 from both sides:

Simplify:

Divide both sides by 3:

The solution $y = -8$ does not violate the restriction $y \neq \frac{1}{3}$.

Therefore, the solution to the equation is $y = -8$.

Verification:

Substitute $y = -8$ into the original equation $\frac{3y + 4}{2 - 6y} = \frac{-2}{5}$.

LHS: $\frac{3(-8) + 4}{2 - 6(-8)} = \frac{-24 + 4}{2 - (-48)} = \frac{-20}{2 + 48} = \frac{-20}{50} = \frac{-2}{5}$.

RHS: $\frac{-2}{5}$.

Since LHS = RHS ($-\frac{2}{5} = -\frac{2}{5}$), the solution is correct.

Given Equation:

Restriction: The denominator cannot be zero, so $y + 2 \neq 0$, which implies $y \neq -2$.

---

To Find:

The value of $y$.

---

Solution:

This equation involves a proportion. We can solve it by cross-multiplication.

Multiply the numerator of the left fraction by the denominator of the right fraction, and set it equal to the product of the numerator of the right fraction and the denominator of the left fraction:

Distribute the constants on both sides:

Gather the terms involving $y$ on one side and the constant terms on the other side.

Add $4y$ to both sides:

Simplify:

Subtract 12 from both sides:

Simplify:

Divide both sides by 25:

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, 5:

The solution $y = -\frac{4}{5}$ does not violate the restriction $y \neq -2$.

Therefore, the solution to the equation is $y = -\frac{4}{5}$.

Verification:

Substitute $y = -\frac{4}{5}$ into the original equation $\frac{7y + 4}{y + 2} = \frac{-4}{3}$.

LHS: $\frac{7(-\frac{4}{5}) + 4}{-\frac{4}{5} + 2} = \frac{-\frac{28}{5} + \frac{20}{5}}{-\frac{4}{5} + \frac{10}{5}} = \frac{-\frac{8}{5}}{\frac{6}{5}} = -\frac{8}{5} \times \frac{5}{6} = -\frac{8}{6} = -\frac{4}{3}$.

RHS: $-\frac{4}{3}$.

Since LHS = RHS ($-\frac{4}{3} = -\frac{4}{3}$), the solution is correct.

Given:

1. Ratio of present ages of Hari and Harry = 5 : 7.

2. Four years from now, the ratio of their ages will be 3 : 4.

---

To Find:

The present ages of Hari and Harry.

---

Solution:

Let the present age of Hari be $5x$ years and the present age of Harry be $7x$ years, according to the given ratio 5:7.

Now, consider their ages four years from now:

Hari's age after 4 years = (Present age of Hari) + 4 = $5x + 4$ years.

Harry's age after 4 years = (Present age of Harry) + 4 = $7x + 4$ years.

According to the problem, the ratio of their ages after 4 years will be 3:4.

So, we can set up the proportion:

Solve this equation for $x$ by cross-multiplication:

Distribute the constants on both sides:

Gather the terms involving $x$ on one side and the constant terms on the other side.

Subtract $20x$ from both sides:

Simplify:

Subtract 12 from both sides:

Simplify:

So, $x = 4$.

Now, find their present ages using $x = 4$:

Hari's present age = $5x = 5 \times 4 = 20$ years.

Harry's present age = $7x = 7 \times 4 = 28$ years.

Therefore, Hari's present age is 20 years and Harry's present age is 28 years.

Verification:

Present ages: Hari = 20, Harry = 28. Ratio $20:28 = 5:7$. (Correct)

Ages after 4 years: Hari = $20+4 = 24$, Harry = $28+4 = 32$.

Ratio after 4 years: $24:32 = 3:4$. (Correct)

Given:

1. A rational number where the denominator is greater than the numerator by 8.

2. If the numerator is increased by 17 and the denominator is decreased by 1, the new number formed is $\frac{3}{2}$.

---

To Find:

The original rational number.

---

Solution:

Let the numerator of the rational number be $x$.

Since the denominator is greater than the numerator by 8, the denominator is $x + 8$.

The original rational number is $\frac{x}{x + 8}$.

Now, modify the numerator and denominator according to the problem statement:

New numerator = Original numerator + 17 = $x + 17$.

New denominator = Original denominator - 1 = $(x + 8) - 1 = x + 7$.

The new rational number formed is $\frac{x + 17}{x + 7}$.

It is given that this new number is equal to $\frac{3}{2}$.

Set up the equation:

Restriction: The denominators cannot be zero. Original denominator $x+8 \neq 0 \implies x \neq -8$. New denominator $x+7 \neq 0 \implies x \neq -7$.

Solve the equation by cross-multiplication:

Distribute the constants on both sides:

Gather the terms involving $x$ on one side and the constant terms on the other side.

Subtract $2x$ from both sides:

Simplify:

Subtract 21 from both sides:

Simplify:

So, the numerator is $x = 13$.

The denominator is $x + 8 = 13 + 8 = 21$.

The original rational number is $\frac{13}{21}$.

(The solution $x=13$ satisfies the restrictions $x \neq -8$ and $x \neq -7$.)

Therefore, the rational number is $\frac{13}{21}$.

Verification:

Original number: $\frac{13}{21}$. Denominator (21) is greater than numerator (13) by 8. Correct.

Increase numerator by 17: $13 + 17 = 30$.

Decrease denominator by 1: $21 - 1 = 20$.

New number: $\frac{30}{20}$.

Is the new number equal to $\frac{3}{2}$? $\frac{30}{20} = \frac{3}{2}$. Correct.