Comprehensive mathematics worksheet for 8th-grade students featuring linear equations and various word problems.

Problem Set Analysis and Solutions

The provided worksheet contains a series of problems involving linear equations, number theory, geometry, and word problems. Below are detailed solutions for each problem.

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#### 1. Solve for \( x \):

##### i. \( 4x - 7 - (x + 4) = 3x + 4 - (2x - 1) \)

Step 1: Simplify both sides of the equation.
- Left-hand side: \( 4x - 7 - (x + 4) = 4x - 7 - x - 4 = 3x - 11 \)
- Right-hand side: \( 3x + 4 - (2x - 1) = 3x + 4 - 2x + 1 = x + 5 \)

So the equation becomes:
\[ 3x - 11 = x + 5 \]

Step 2: Solve for \( x \).
- Subtract \( x \) from both sides: \( 3x - x - 11 = 5 \)
- Simplify: \( 2x - 11 = 5 \)
- Add 11 to both sides: \( 2x = 16 \)
- Divide by 2: \( x = 8 \)

Answer: \( x = 8 \)

##### ii. \( \frac{17(2 - x) - 5(x + 12)}{1 - 7x} = 8 \)

Step 1: Simplify the numerator.
- Numerator: \( 17(2 - x) - 5(x + 12) \)
- Expand: \( 17 \cdot 2 - 17x - 5x - 60 = 34 - 17x - 5x - 60 = -22x - 26 \)

So the equation becomes:
\[ \frac{-22x - 26}{1 - 7x} = 8 \]

Step 2: Eliminate the denominator by multiplying both sides by \( 1 - 7x \).
\[ -22x - 26 = 8(1 - 7x) \]

Step 3: Expand the right-hand side.
\[ -22x - 26 = 8 - 56x \]

Step 4: Solve for \( x \).
- Add \( 56x \) to both sides: \( -22x + 56x - 26 = 8 \)
- Simplify: \( 34x - 26 = 8 \)
- Add 26 to both sides: \( 34x = 34 \)
- Divide by 34: \( x = 1 \)

Answer: \( x = 1 \)

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#### 2. The sum of three consecutive even numbers is 30. Find the numbers.

Step 1: Let the three consecutive even numbers be \( x \), \( x + 2 \), and \( x + 4 \).

Step 2: Write the equation for their sum.
\[ x + (x + 2) + (x + 4) = 30 \]

Step 3: Simplify the equation.
\[ 3x + 6 = 30 \]

Step 4: Solve for \( x \).
- Subtract 6 from both sides: \( 3x = 24 \)
- Divide by 3: \( x = 8 \)

Step 5: Find the three numbers.
- First number: \( x = 8 \)
- Second number: \( x + 2 = 10 \)
- Third number: \( x + 4 = 12 \)

Answer: The numbers are \( 8, 10, 12 \).

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#### 3. The sum of three consecutive odd numbers is 63. Find the numbers.

Step 1: Let the three consecutive odd numbers be \( x \), \( x + 2 \), and \( x + 4 \).

Step 2: Write the equation for their sum.
\[ x + (x + 2) + (x + 4) = 63 \]

Step 3: Simplify the equation.
\[ 3x + 6 = 63 \]

Step 4: Solve for \( x \).
- Subtract 6 from both sides: \( 3x = 57 \)
- Divide by 3: \( x = 19 \)

Step 5: Find the three numbers.
- First number: \( x = 19 \)
- Second number: \( x + 2 = 21 \)
- Third number: \( x + 4 = 23 \)

Answer: The numbers are \( 19, 21, 23 \).

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#### 4. The sum of two twin primes is 60. Find the two prime numbers.

Step 1: Twin primes are pairs of primes that differ by 2. Let the two twin primes be \( p \) and \( p + 2 \).

Step 2: Write the equation for their sum.
\[ p + (p + 2) = 60 \]

Step 3: Simplify the equation.
\[ 2p + 2 = 60 \]

Step 4: Solve for \( p \).
- Subtract 2 from both sides: \( 2p = 58 \)
- Divide by 2: \( p = 29 \)

Step 5: Find the twin primes.
- First prime: \( p = 29 \)
- Second prime: \( p + 2 = 31 \)

Answer: The twin primes are \( 29 \) and \( 31 \).

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#### 5. The measures of the angles of a triangle are in the ratio \( 1 : 2 : 3 \). Find the angles.

Step 1: Let the angles be \( x \), \( 2x \), and \( 3x \).

Step 2: Use the fact that the sum of the angles in a triangle is \( 180^\circ \).
\[ x + 2x + 3x = 180^\circ \]

Step 3: Simplify the equation.
\[ 6x = 180^\circ \]

Step 4: Solve for \( x \).
\[ x = 30^\circ \]

Step 5: Find the three angles.
- First angle: \( x = 30^\circ \)
- Second angle: \( 2x = 60^\circ \)
- Third angle: \( 3x = 90^\circ \)

Answer: The angles are \( 30^\circ, 60^\circ, 90^\circ \).

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#### 6. The numerator of a fraction is 3 less than its denominator. If we add 1 to both the numerator and denominator, it becomes equal to \( \frac{1}{2} \). Find the fraction.

Step 1: Let the denominator be \( d \). Then the numerator is \( d - 3 \). The fraction is \( \frac{d - 3}{d} \).

Step 2: After adding 1 to both the numerator and denominator, the new fraction is \( \frac{(d - 3) + 1}{d + 1} = \frac{d - 2}{d + 1} \).

Step 3: According to the problem, this new fraction equals \( \frac{1}{2} \).
\[ \frac{d - 2}{d + 1} = \frac{1}{2} \]

Step 4: Cross-multiply to solve for \( d \).
\[ 2(d - 2) = 1(d + 1) \]
\[ 2d - 4 = d + 1 \]

Step 5: Solve for \( d \).
- Subtract \( d \) from both sides: \( 2d - d - 4 = 1 \)
- Simplify: \( d - 4 = 1 \)
- Add 4 to both sides: \( d = 5 \)

Step 6: Find the original fraction.
- Denominator: \( d = 5 \)
- Numerator: \( d - 3 = 2 \)
- Fraction: \( \frac{2}{5} \)

Answer: The fraction is \( \frac{2}{5} \).

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#### 7. Renu’s mother is four times as old as Renu. After 5 years, her mother will be three times as old as she will then be. Find their present ages.

Step 1: Let Renu’s current age be \( r \). Then her mother’s current age is \( 4r \).

Step 2: After 5 years, Renu’s age will be \( r + 5 \) and her mother’s age will be \( 4r + 5 \).

Step 3: According to the problem, after 5 years, the mother’s age will be three times Renu’s age.
\[ 4r + 5 = 3(r + 5) \]

Step 4: Simplify the equation.
\[ 4r + 5 = 3r + 15 \]

Step 5: Solve for \( r \).
- Subtract \( 3r \) from both sides: \( 4r - 3r + 5 = 15 \)
- Simplify: \( r + 5 = 15 \)
- Subtract 5 from both sides: \( r = 10 \)

Step 6: Find the present ages.
- Renu’s age: \( r = 10 \)
- Mother’s age: \( 4r = 40 \)

Answer: Renu is \( 10 \) years old, and her mother is \( 40 \) years old.

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#### 8. The sum of four consecutive multiples of 7 is 70. Find these multiples.

Step 1: Let the four consecutive multiples of 7 be \( 7x \), \( 7(x + 1) \), \( 7(x + 2) \), and \( 7(x + 3) \).

Step 2: Write the equation for their sum.
\[ 7x + 7(x + 1) + 7(x + 2) + 7(x + 3) = 70 \]

Step 3: Simplify the equation.
\[ 7x + 7x + 7 + 7x + 14 + 7x + 21 = 70 \]
\[ 28x + 42 = 70 \]

Step 4: Solve for \( x \).
- Subtract 42 from both sides: \( 28x = 28 \)
- Divide by 28: \( x = 1 \)

Step 5: Find the four multiples.
- First multiple: \( 7x = 7 \cdot 1 = 7 \)
- Second multiple: \( 7(x + 1) = 7 \cdot 2 = 14 \)
- Third multiple: \( 7(x + 2) = 7 \cdot 3 = 21 \)
- Fourth multiple: \( 7(x + 3) = 7 \cdot 4 = 28 \)

Answer: The multiples are \( 7, 14, 21, 28 \).

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#### 9. The sum of two numbers is 50. If the larger number is divided by the smaller number, we get 3. Find the numbers.

Step 1: Let the two numbers be \( x \) (smaller) and \( y \) (larger). According to the problem:
\[ x + y = 50 \]
\[ \frac{y}{x} = 3 \]

Step 2: From the second equation, express \( y \) in terms of \( x \):
\[ y = 3x \]

Step 3: Substitute \( y = 3x \) into the first equation.
\[ x + 3x = 50 \]
\[ 4x = 50 \]

Step 4: Solve for \( x \).
\[ x = 12.5 \]

Step 5: Find \( y \).
\[ y = 3x = 3 \cdot 12.5 = 37.5 \]

Answer: The numbers are \( 12.5 \) and \( 37.5 \).

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#### 10. The perimeter of a triangle is 49 cm. One side is 7 cm longer than another side and 5 cm shorter than the third side. Find the sides.

Step 1: Let the sides of the triangle be \( a \), \( b \), and \( c \). According to the problem:
- Perimeter: \( a + b + c = 49 \)
- One side is 7 cm longer than another: \( b = a + 7 \)
- One side is 5 cm shorter than the third: \( b = c - 5 \)

Step 2: Express \( c \) in terms of \( a \) using \( b = c - 5 \):
\[ c = b + 5 \]
Since \( b = a + 7 \), substitute \( b \):
\[ c = (a + 7) + 5 = a + 12 \]

Step 3: Substitute \( b \) and \( c \) in the perimeter equation.
\[ a + b + c = 49 \]
\[ a + (a + 7) + (a + 12) = 49 \]

Step 4: Simplify the equation.
\[ a + a + 7 + a + 12 = 49 \]
\[ 3a + 19 = 49 \]

Step 5: Solve for \( a \).
- Subtract 19 from both sides: \( 3a = 30 \)
- Divide by 3: \( a = 10 \)

Step 6: Find \( b \) and \( c \).
- \( b = a + 7 = 10 + 7 = 17 \)
- \( c = a + 12 = 10 + 12 = 22 \)

Answer: The sides of the triangle are \( 10 \) cm, \( 17 \) cm, and \( 22 \) cm.

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Final Answers:

1. i. \( x = 8 \)
ii. \( x = 1 \)
2. \( 8, 10, 12 \)
3. \( 19, 21, 23 \)
4. \( 29, 31 \)
5. \( 30^\circ, 60^\circ, 90^\circ \)
6. \( \frac{2}{5} \)
7. \( 10, 40 \)
8. \( 7, 14, 21, 28 \)
9. \( 12.5, 37.5 \)
10. \( 10, 17, 22 \)

\[
\boxed{8, 10, 12, 19, 21, 23, 29, 31, 30^\circ, 60^\circ, 90^\circ, \frac{2}{5}, 10, 40, 7, 14, 21, 28, 12.5, 37.5, 10, 17, 22}
\]