Grade 7 Algebra Questions and Problems with Solutions

This page provides a wide range of Grade 7 algebra questions and problems designed to help students, parents, and teachers. You will find step-by-step solutions to exercises covering simplifying expressions, solving equations, operations on fractions, problem-solving strategies, and working with exponents. These practice problems are crafted to strengthen algebra skills and build a solid foundation for higher-level math.

Questions 1

Evaluate each of the expressions for the given value(s) of the variable(s).

  1. \(12 x^{3} + 5 x^{2} + 4 x - 6 \quad \text{for } x = -1\)
  2. \(2 a^{2} + 3 b^{3} - 10 \quad \text{for } a=2 \text{ and } b=-2\)
  3. \(\displaystyle \dfrac{-2x - 1}{x + 3} \quad \text{for } x=2\)
  4. \(2 + 2 |x - 4| \quad \text{for } x = -4\)

Solution:

Substitute each variable by its given numerical value and simplify.

  1. \(12 (-1)^3 + 5 (-1)^2 + 4 (-1) - 6 = 12(-1) + 5(1) - 4 - 6 = -12 + 5 - 4 - 6 = -17\)
  2. \(2 (2)^2 + 3 (-2)^3 - 10 = 2(4) + 3(-8) - 10 = 8 - 24 - 10 = -26\)
  3. \(\dfrac{-2(2) - 1}{2 + 3} = \dfrac{-4 - 1}{5} = \dfrac{-5}{5} = -1\)
  4. \(2 + 2|(-4) - 4| = 2 + 2| -8 | = 2 + 2 \times 8 = 2 + 16 = 18\)

Questions 2

Expand and simplify each of the expressions below.

  1. \(-2(x - 8) + 3(x - 7)\)
  2. \(2 (a + 1) + 5 b + 3 (a + b) + 3\)
  3. \(a (b + 3) + b (a - 2) + 2 a - 5 b + 8\)
  4. \(\dfrac{1}{2} (4x + 4) + \dfrac{1}{3} (6x + 12)\)
  5. \(4 \bigl(-x + 2 - 3(x - 2)\bigr)\)

Solution:

Use the distribution rule \(a(b + c) = ab + ac\) to expand and group like terms.

  1. \(-2(x - 8) + 3(x - 7) = -2x + 16 + 3x - 21 \)
    \( = (-2x + 3x) + (16 - 21) = x - 5\)
  2. \(2(a + 1) + 5b + 3(a + b) + 3 = 2a + 2 + 5b + 3a + 3b + 3 \)
    \( = (2a + 3a) + (5b + 3b) + (2 + 3) = 5a + 8b + 5\)
  3. \(a(b + 3) + b(a - 2) + 2a - 5b + 8 = ab + 3a + ba - 2b + 2a - 5b + 8 \)
    \(= (ab + ba) + (3a + 2a) + (-2b - 5b) + 8 = 2ab + 5a - 7b + 8\)
  4. \(\dfrac{1}{2}(4x + 4) + \dfrac{1}{3}(6x + 12) = 2x + 2 + 2x + 4 \)
    \( = (2x + 2x) + (2 + 4) = 4x + 6\)
  5. \(4(-x + 2 - 3(x - 2)) = -4x + 8 - 12x + 24 \)
    \( = (-4x - 12x) + (8 + 24) = -16x + 32\)

Questions 3

Simplify each of the expressions below.

  1. \(\dfrac{x}{y} + \dfrac{4}{y}\)
  2. \(\left(\dfrac{2x}{4}\right) \times \left(\dfrac{1}{2}\right)\)
  3. \(\left(\dfrac{3x}{5}\right) \div \left(\dfrac{x}{5}\right)\)

Solution:

Simplify using rules for fractions, multiplication and division.

  1. \(\dfrac{x}{y} + \dfrac{4}{y} = \dfrac{x + 4}{y}\)
  2. \(\left(\dfrac{2x}{4}\right) \times \left(\dfrac{1}{2}\right) = \dfrac{2x \times 1}{4 \times 2} = \dfrac{2x}{8} = \dfrac{x}{4}\)
  3. \(\left(\dfrac{3x}{5}\right) \div \left(\dfrac{x}{5}\right) = \dfrac{3x}{5} \times \dfrac{5}{x} = \dfrac{15x}{5x} = 3\)

Questions 4

Simplify each of the expressions below.

  1. \(3 x^{2} \times 5 x^{3}\)
  2. \(\displaystyle \dfrac{(2 y)^{4} \cdot 9 x^{3}}{4 y^{4} (3 x)^{2}}\)

Solution:

Use rules of multiplication and division with exponents.

  1. \(3 x^{2} \times 5 x^{3} = (3 \times 5) x^{2 + 3} = 15 x^{5}\)
  2. \(\dfrac{(2 y)^4 9 x^3}{4 y^4 (3 x)^2} = \dfrac{16 y^4 9 x^3}{4 y^4 9 x^2} = \dfrac{16 \times 9}{4 \times 9} y^{4 - 4} x^{3 - 2} = 4 x\)

Questions 5

Factor fully each of the expressions below.

  1. \(9 x - 3\)
  2. \(24 x + 18 y\)
  3. \(b x + d x\)

Solution:

Find common factors and factor using distribution backwards.

  1. \(9x - 3 = 3 \times 3x + 3 \times 1 = 3(3x - 1)\)
  2. \(24x + 18y = 6 \times 4 x + 6 \times 3 y = 6(4x + 3y) \)
  3. \(bx + dx = x(b + d)\)

Questions 6

Solve each of the equations below.

  1. \(2 x + 5 = 11\)
  2. \(3 x = \dfrac{6}{5}\)
  3. \(3 (2 x + 2) + 2 = 20\)

Solution:

  1. \[ \begin{aligned} 2x + 5 &= 11 \\ 2x &= 6 \\ x &= 3 \end{aligned} \]
  2. \[ \begin{aligned} 3x &= \dfrac{6}{5} \\ 15x &= 6 \quad \text{(multiplying both sides by 5)} \\ x &= \dfrac{6}{15} = \dfrac{2}{5} \end{aligned} \]
  3. \[ \begin{aligned} 3(2x + 2) + 2 &= 20 \\ 6x + 6 + 2 &= 20 \\ 6x + 8 &= 20 \\ 6x &= 12 \\ x &= 2 \end{aligned} \]

Questions 7

Rewrite the expressions \(3 \times a \times a \times a - 5 \times b \times b\) using exponential notation.

Solution:

Rewrite products using exponents. \[ 3 \times a \times a \times a - 5 \times b \times b = 3 a^3 - 5 b^2 \]

Questions 8

A rectangle has a length given by \(2 x + 3\) units, where \(x\) is a variable. The width of the rectangle is given by \(x + 1\) units. Find the value of \(x\) if the perimeter of the rectangle is equal to 32.

Solution:

Use perimeter formula \(P = 2 \times \text{length} + 2 \times \text{width}\), substitute and solve for \(x\). \[ \begin{aligned} 2(2x + 3) + 2(x + 1) &= 32 \\ 4x + 6 + 2x + 2 &= 32 \\ 6x + 8 &= 32 \\ 6x &= 24 \\ x &= 4 \end{aligned} \]

Questions 9

A rectangle has a length given by \(2x - 1\) units, where \(x\) is a variable. The width of the rectangle is equal to 3 units. Find the value of \(x\) if the area of the rectangle is equal to 27.

Solution:

Use area formula \(A = \text{width} \times \text{length}\), substitute and solve. \[ \begin{aligned} 3(2x - 1) &= 27 \\ 6x - 3 &= 27 \\ 6x &= 30 \\ x &= 5 \end{aligned} \]

Questions 10

45% of the students in a school are male. Find the ratio of the number of female to the total number of male students in this school.

Solution:

Since 45% are male, \[ 100\% - 45\% = 55\%\] are female.

Calculate ratio of females to males. \[ \text{Ratio} = \dfrac{55\%}{45\%} = \dfrac{55}{45} = \dfrac{11}{9} \]

Questions 11

A car travels at the speed \(x + 30\) kilometers in one hour, where \(x\) is unknown. Find \(x\) if this car covers 300 kilometers in 3 hours.

Solution:

Distance = time \( \times \) speed. \[300 = 3(x + 30) \] Solve for \( x \). \[ 300 = 3x + 90 \] \[ 3x = 210 \] \[ x = 70 \]

Questions 12

Solve the proportion: \(\dfrac{4}{5} = \dfrac{a}{16}\).

Solution:

Solve proportion by cross-multiplication.

Given \[ \dfrac{4}{5} = \dfrac{a}{16} \] Cross-multiply: \[ 4 \times 16 = 5 \times a \] Simplify \[ 64 = 5a \] Solve for \( a \). \[ a = \dfrac{64}{5} = 12.8 \]

Questions 13

Find \(a\) if the ordered pair \((2, a + 2)\) is a solution to the equation \(2 x + 2 y = 10\).

Solution:

Substitute ordered pair: \( x = 2 \) and \( y = a + 2 \) into the givem equation. \[ 2(2) + 2(a + 2) = 10 \] Expand \[ 4 + 2a + 4 = 10 \] Simplify \[ 2a + 8 = 10 \] \[ 2a = 2 \] Solve for \(a\). \[ a = 1 \]

Questions 14

Find the greatest common factor of the numbers 25 and 45.

Solution:

List all factors of 25 and 45.

Factors of 25 are: 1, 5, 25

Factors of 45 are: 1, 3, 5, 9, 15, 45

The greatest common factor to 25 and 45 is: \(5\).

Questions 15

Write the number "one billion, two hundred thirty four million, seven hundred fifty thousand two" using digits.

Solution:

\[ 1234750002\]

Questions 16

Write the number 393,234,000,034 in words.

Solution:

three hundred ninety-three billion, two hundred thirty-four million, thirty-four

Questions 17

Find the lowest common multiple of the numbers 15 and 35.

Solution:

Find the first few multiples of 15 and 35 till you get one that is common:

Multiples of 15: \(15, 30, 45, 60, 75, 90, 105, 120, 135, \ldots\)

Multiples of 35: \(35, 70, 105, 140, \ldots\)

Select the lowest common multiple (LCM):

LCM is \(105\).

Questions 18

Find \(x\) if \(\dfrac{2}{3}\) of \(x\) is 30.

Solution:

\(\dfrac{2}{3}\) of \(x\) is 30 is mathematically written as: \[ \dfrac{2}{3} \times x = 30 \] Multiply by 3 and simplify \[ 2x = 90 \] Solve for \(x\): \[ x = 90\div 2 = 45 \]

Questions 19

What is 20% of \(\dfrac{1}{3}\)?

Solution:

20% of \(\dfrac{1}{3} \) is written as: \[ 20\% \times \dfrac{1}{3} \] use the fraction \( 20\% = \dfrac{20}{100} \) to simplify: \[ = \dfrac{20}{100} \times \dfrac{1}{3} = \dfrac{20}{300} = \dfrac{1}{15} \]

Questions 20

The difference between two numbers is 17 and their sum is 69. Find the largest of these two numbers.

Solution:

Let the smaller number be \( x \). Then the larger number is \( x + 17 \).

Their sum is 69, hence: \[ x + (x + 17) = 69 \] Simplify: \[ 2x + 17 = 69 \] Subtract 17 from both sides: \[ 2x = 52 \] Divide both sides by 2 to solve for \( x \): \[ x = 26 \] So the larger number is: \[ x + 17 = 26 + 17 = 43 \]

Questions 21

Order \(\dfrac{12}{5} , 250\%, \dfrac{21}{10}\), and \( 2.3 \) from smallest to largest.

Solution:

Convert all to decimals and order: \[ \dfrac{12}{5} = 2.4 \] \[ \quad 250\% = \dfrac{250}{100} = 2.5 \] \[ \dfrac{21}{10} = 2.1 \] \[ 2.3 = 2.3 \] Order from smallest to largest: \(\dfrac{21}{10}, 2.3, \dfrac{12}{5}, 250\%\)

Questions 22

The sum of 3 positive consecutive integers is equal to 96. Find the largest of these numbers.

Solution:

Let \( x \) , \( x + 1 \) and \( x + 2 \) be the 3 consecutive integers.

The sum of the three consecutive integers is 96, hence: \[ x + (x + 1) + (x + 2) = 96 \] Group like terms: \[ 3x + 3 = 96 \] \[ 3x = 93 \] Solve for \( x \) \[ x = 31 \] The largest of these numbers is: \( x + 2 \) and is equal to \[ x + 2 = 33 \]

Questions 23

Dany scored 93 in physics, 88 in mathematics, and a score in chemistry that is double his score in geography. The average score of all 4 courses is 79. What were his scores in chemistry and geography?

Solution:

Let \(x\) be geography score. Chemistry is \(2x\). Average of 4 courses is 79: \[ \dfrac{93 + 88 + x + 2x}{4} = 79 \] Group like termas in numerator \[ \dfrac{3x+181}{4} = 79 \] Multiply left and right sides by 4 and simplify: \[ 3x + 181 = 316 \] \[ 3x = 135 \] Solve for \( x \) \[ x = 45 \] Score in Geography is : \[ x = 45 \] Score in Chemistry is : \[ 2 x = 90 \]

Questions 24

Linda scored a total of 265 points in mathematics, physics and English. She scored 7 more marks in mathematics than in English and she scored 5 more marks in physics than in mathematics. Find her scores in all three subjects.

Solution:

Let \(x\) be the score in English.

The score in Math is: \[ x + 7\] The score in Physics is; \[ x + 12\]. Linda scored a total of 265 points: \[ x + (x + 7) + (x + 12) = 265 \] Group like terms: \[ 3x + 19 = 265 \] Simplify \[ 3x = 246 \] Solve for \( x \): \[ x = 246 \div 3 = 82 \] Score in English is \[ x = 82 \] Score in Math is: \[ x + 7 = 7 + 82 = 89 \] Score in Physics is: \[ x + 12 = 82 + 12 = 94 \]

Questions 25

There are bicycles and cars in a parking lot. There is a total of 300 wheels including 100 small wheels for bicycles. How many cars and how many bicycles are there?

Solution:

Each bicyle has 2 wheels, hence the number of bicycles is: \[\dfrac{100}{2} = 50 \] The number of Car wheels is: \[ 300 - 100 = 200 \] Each car has 4 wheels, hence the number of cars is: \[ \dfrac{200}{4} = 50 \]

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