Solutions to Algebra Questions and Problems for Grade 7

Solutions to these Algebra Questions and Problems are presented.

  1. 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. \(2 + 2|(-4) - 4| = 2 + 2| -8 | = 2 + 2 \times 8 = 2 + 16 = 18\)

  2. 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. \(4(-x + 2 - 3(x - 2)) = -4x + 8 - 12x + 24 = (-4x - 12x) + (8 + 24) = -16x + 32\)

  3. Solution

    Simplify using rules for fractions, multiplication and division.
    1. \(\dfrac{x}{y} + \dfrac{4}{y} = \dfrac{x + 4}{y}\)
    2. \(\left(\dfrac{2x}{4}\right) \times \dfrac{1}{2} = \dfrac{2x \times 1}{4 \times 2} = \dfrac{2x}{8} = \dfrac{x}{4}\)
    3. \(\dfrac{3x}{5} \div \dfrac{x}{5} = \dfrac{3x}{5} \times \dfrac{5}{x} = \dfrac{15x}{5x} = 3\)

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

  5. Solution

    Find common factors and factor using distribution backwards.
    1. \(9x - 3 = 3(3x - 1)\)
    2. \(24x + 18y = 6(4x + 3y)\)
    3. \(bx + dx = x(b + d)\)

  6. Solution

    Solve equations step-by-step by performing inverse operations.
    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. \[ \begin{aligned} 3(2x + 2) + 2 &= 20 \\ 6x + 6 + 2 &= 20 \\ 6x + 8 &= 20 \\ 6x &= 12 \\ x &= 2 \end{aligned} \]

  7. Solution

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

  8. Solution

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

  9. Solution

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

  10. Solution

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

  11. Solution

    Distance = time × speed; solve for \(x\). \[ \begin{aligned} 300 &= 3(x + 30) \\ 300 &= 3x + 90 \\ 3x &= 210 \\ x &= 70 \end{aligned} \]

  12. Solution

    Solve proportion by cross-multiplication. \[ \begin{aligned} \dfrac{4}{5} &= \dfrac{a}{16} \\ 4 \times 16 &= 5 \times a \\ 64 &= 5a \\ a &= \dfrac{64}{5} = 12.8 \end{aligned} \]

  13. Solution

    Substitute ordered pair into equation and solve for \(a\). \[ \begin{aligned} 2(2) + 2(a + 2) &= 10 \\ 4 + 2a + 4 &= 10 \\ 2a + 8 &= 10 \\ 2a &= 2 \\ a &= 1 \end{aligned} \]

  14. Solution

    List factors and find greatest common factor. 25: 1, 5, 25 45: 1, 3, 5, 9, 15, 45 Greatest common factor is \(5\).

  15. Solution

    Write the number in digits: \(1234750002\)

  16. Solution

    Write the number in words: three hundred ninety-three billion, two hundred thirty-four million, thirty-four

  17. Solution

    Find multiples and lowest common multiple (LCM):

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

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

    LCM is \(105\).

  18. Solution

    Solve for \(x\): \[ \begin{aligned} \dfrac{2}{3} \times x &= 30 \\ 2x &= 90 \quad (\times 3) \\ x &= 45 \end{aligned} \]

  19. Solution

    Calculate percentage of a fraction: \[ \begin{aligned} 20\% \times \dfrac{1}{3} &= \dfrac{20}{100} \times \dfrac{1}{3} = \dfrac{20}{300} = \dfrac{1}{15} \end{aligned} \]

  20. Solution

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

  21. Solution

    Set up equation for three consecutive integers and solve: \[ \begin{aligned} x + (x + 1) + (x + 2) &= 96 \\ 3x + 3 &= 96 \\ 3x &= 93 \\ x &= 31 \\ \text{Largest} &= x + 2 = 33 \end{aligned} \]

  22. Solution

    Let \(x\) be geography score. Chemistry is \(2x\). Average of 4 courses is 79: \[ \begin{aligned} \dfrac{93 + 88 + x + 2x}{4} &= 79 \\ 3x + 181 &= 316 \\ 3x &= 135 \\ x &= 45 \\ \text{Geography} &= 45 \\ \text{Chemistry} &= 90 \end{aligned} \]

  23. Solution

    Let \(x\) be English score. Math \(= x + 7\). Physics \(= x + 12\). Total = 265: \[ \begin{aligned} x + (x + 7) + (x + 12) &= 265 \\ 3x + 19 &= 265 \\ 3x &= 246 \\ x &= 82 \\ \text{English} &= 82 \\ \text{Math} &= 89 \\ \text{Physics} &= 94 \end{aligned} \]

  24. Solution

    Calculate number of bicycles and cars from wheels: \[ \begin{aligned} \text{Bicycles} &= \dfrac{100}{2} = 50 \\ \text{Car wheels} &= 300 - 100 = 200 \\ \text{Cars} &= \dfrac{200}{4} = 50 \end{aligned} \]

    Solution

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

    Their sum is given as 69: \[ x + (x + 17) = 69 \]

    Simplify: \[ 2x + 17 = 69 \]

    Subtract 17 from both sides: \[ 2x = 52 \]

    Divide both sides by 2: \[ x = 26 \]

    So the larger number is: \[ x + 17 = 26 + 17 = 43 \]


Links and References