Hard GCSE Algebraic notation Questions

Challenging, exam-style GCSE Algebraic notation questions with worked solutions. Stretch yourself on the hardest translating words, brackets, powers, collecting like terms problems.

translating wordsbracketspowerscollecting like termscoefficientsdivision as a fraction
GCSE Foundation34 questionsStep-by-step solutions
Question 1
6 markschallenging
A trapezium has parallel sides aa and bb and height hh. Write an expression for its area.
Show worked solution

Worked solution

  1. Recall the trapezium area rule

    average the parallel sides, times the height\text{average the parallel sides, times the height}

    Add the parallel sides, halve, then multiply by the height.

  2. Add the parallel sides

    a+ba + b

    Sum of the two parallel sides.

  3. Halve the sum

    a+b2\frac{a + b}{2}

    This is the average width.

  4. Multiply by the height

    a+b2×h\frac{a + b}{2} \times h

    Average width times height.

  5. Write neatly as one fraction

    (a+b)h2\frac{(a + b)h}{2}

    The height multiplies the whole numerator.

  6. Keep the sum bracketed

    (a+b)h(a + b)h

    The brackets show both sides are averaged.

  7. Warn about a common error

    (a+b)h2≠ah+b2\frac{(a+b)h}{2} \ne \frac{ah + b}{2}

    The height multiplies both a and b.

  8. Interpret

    a=3,b=5,h=4: 8×42=16a=3,b=5,h=4:\ \frac{8 \times 4}{2} = 16

    Area is 16 square units. ✓

  9. Check the average width

    3+52=4\frac{3+5}{2} = 4

    The average of the parallel sides is 4.

  10. Multiply out the check

    4×4=164 \times 4 = 16

    Average width 4 times height 4 is 16. ✓

  11. Note an equivalent form

    12(a+b)h\frac{1}{2}(a+b)h

    A common alternative way to write it.

  12. Check units idea

    (a+b)h2 is an area\frac{(a+b)h}{2} \text{ is an area}

    Measured in square units.

  13. Sense-check against a rectangle

    a=b: 2a⋅h2=aha=b:\ \frac{2a \cdot h}{2} = ah

    If both sides are equal it becomes a rectangle, area ah. ✓

  14. Reflect

    average width×height\text{average width} \times \text{height}

    The formula is a mean width times the height.

  15. State the answer

    (a+b)h2\frac{(a + b)h}{2}

    So the area is (a + b)h over 2.

Answer
(a+b)h2\frac{(a + b)h}{2}
Question 2
5 markschallenging
nn items each cost pp pounds. There is £4 off the whole order. Write an expression, in pounds, for the total cost.
Show worked solution

Worked solution

  1. Cost before the discount

    n×p=npn \times p = np

    n items at p pounds each cost np pounds.

  2. Read '£4 off the whole order'

    subtract 4\text{subtract } 4

    £4 is taken off the total.

  3. Combine

    np−4np - 4

    Subtract the discount from the total cost.

  4. Check the order of letters

    np

    n before p, written in alphabetical order.

  5. Warn about a common error

    np−4≠p(n−4)np - 4 \ne p(n - 4)

    The £4 is off the total, not off each item.

  6. Interpret

    p=5,n=3: 15−4=11p=5,n=3:\ 15 - 4 = 11

    Three items at £5 with £4 off is £11. ✓

  7. Check the wrong version

    5(3−4)=−55(3-4) = -5

    p(n − 4) gives −5 — clearly wrong.

  8. Note the discount is constant

    −4-4

    The £4 does not depend on the number of items.

  9. Note the units

    np−4 poundsnp - 4 \text{ pounds}

    The total is in pounds.

  10. Sense-check

    np−4>0 when np>4np - 4 > 0 \text{ when } np > 4

    The order must cost more than £4.

  11. Re-check

    p=10,n=1: 10−4=6p=10,n=1:\ 10 - 4 = 6

    One £10 item with £4 off is £6. ✓

  12. Contrast per-item discount

    p(n−4) would be per-itemp(n-4) \text{ would be per-item}

    Only p(n − 4) reduces the count, which is not asked.

  13. Reflect

    cost=np−4\text{cost} = np - 4

    Multiply first, then subtract the fixed discount.

  14. Confirm

    np−4np - 4

    Total minus the discount.

  15. State the answer

    np−4np - 4

    So the total cost is np − 4 pounds.

Answer
np−4np - 4
Question 3
5 markschallenging
Write '7 divided by the sum of xx and 2' as an expression.
Show worked solution

Worked solution

  1. Read 'the sum of x and 2'

    x+2x + 2

    Form the sum first — this is the denominator.

  2. Read '7 divided by'

    7÷(x+2)7 \div (x + 2)

    7 is divided by the whole sum.

  3. Write as a fraction

    7x+2\frac{7}{x + 2}

    7 on top, the sum on the bottom.

  4. Keep the sum together

    7x+2\frac{7}{x+2}

    The fraction bar groups the denominator.

  5. Warn about a common error

    7x+2≠7x+2\frac{7}{x+2} \ne \frac{7}{x} + 2

    The +2 is part of the denominator.

  6. Warn about order too

    7x+2≠7x+72\frac{7}{x+2} \ne \frac{7}{x} + \frac{7}{2}

    You cannot split a sum in the denominator.

  7. Interpret

    x=5: 77=1x=5:\ \frac{7}{7} = 1

    7 divided by (5 + 2) is 1. ✓

  8. Check the wrong version

    75+2=3.4\frac{7}{5} + 2 = 3.4

    The other reading gives 3.4 — different.

  9. Note the bar groups the bottom

    7x+2\frac{7}{x + 2}

    Everything below the bar is the divisor.

  10. Consider what values are excluded

    x≠−2x \ne -2

    The denominator must not be zero.

  11. Sense-check

    x=12: 714=0.5x=12:\ \frac{7}{14} = 0.5

    A sensible value. ✓

  12. Reflect

    fraction bar = bracket\text{fraction bar = bracket}

    The bar keeps the sum in the denominator together.

  13. Confirm

    7x+2\frac{7}{x + 2}

    7 over the sum x + 2.

  14. Re-check

    x=−1: 71=7x=-1:\ \frac{7}{1} = 7

    Still defined here. ✓

  15. State the answer

    7x+2\frac{7}{x + 2}

    So the expression is 7 over (x + 2).

Answer
7x+2\frac{7}{x + 2}
Question 4
6 markschallenging
A cuboid has dimensions xx, 2x2x and 3x3x. Write an expression for its volume.
Show worked solution

Worked solution

  1. Recall the volume rule

    V=length×width×heightV = \text{length} \times \text{width} \times \text{height}

    Volume of a cuboid is the product of its three dimensions.

  2. Substitute the dimensions

    x×2x×3xx \times 2x \times 3x

    The three edges.

  3. Collect the numbers

    1×2×3=61 \times 2 \times 3 = 6

    Number parts: x has coefficient 1.

  4. Collect the letters

    x×x×x=x3x \times x \times x = x^3

    Three x's multiplied is x³.

  5. Combine

    6×x3=6x36 \times x^3 = 6x^3

    6 is the coefficient of x³.

  6. Warn about a common error

    6x3≠(6x)36x^3 \ne (6x)^3

    Only the x's are cubed, not the 6.

  7. Interpret

    x=2: 6×8=48x=2:\ 6 \times 8 = 48

    Volume is 48 cubic units when x=2x = 2.

  8. Check directly

    2×4×6=482 \times 4 \times 6 = 48

    Dimensions 2, 4, 6 multiply to 48. ✓

  9. Contrast with surface area

    volume uses x3\text{volume uses } x^3

    Volume gives a cube; area would give a square.

  10. Check units idea

    6x3 is cubic units6x^3 \text{ is cubic units}

    Volume is measured in cubic units.

  11. Track the coefficient

    1×2×3=61 \times 2 \times 3 = 6

    The 6 comes from multiplying the number parts.

  12. Track the power

    x1+1+1=x3x^{1+1+1} = x^3

    The powers of x add when multiplied.

  13. Reflect

    multiply all three dimensions\text{multiply all three dimensions}

    Numbers multiply and powers of x add.

  14. Re-check

    x=1: 6×1=6x=1:\ 6 \times 1 = 6

    Volume 6 when x=1x = 1. ✓

  15. State the answer

    6x36x^3

    So the volume is 6x³.

Answer
6x36x^3
Question 5
5 markschallenging
Adult tickets cost £12 and child tickets cost £8. Write an expression for the total cost, in pounds, of xx adult tickets and yy child tickets.
Show worked solution

Worked solution

  1. Cost of the adult tickets

    12×x=12x12 \times x = 12x

    x adults at £12 each.

  2. Cost of the child tickets

    8×y=8y8 \times y = 8y

    y children at £8 each.

  3. Add the two amounts

    12x+8y12x + 8y

    Total cost is the sum.

  4. Check the terms are unlike

    12x+8y12x + 8y

    x and y are different, so they stay separate.

  5. Warn about a common error

    12x+8y≠20xy12x + 8y \ne 20xy

    Do not combine the ticket counts.

  6. Interpret

    x=2,y=3: 24+24=48x=2,y=3:\ 24 + 24 = 48

    Cost is £48. ✓

  7. Note the coefficients' meaning

    12 per adult, 8 per child12 \text{ per adult},\ 8 \text{ per child}

    The coefficients are the prices.

  8. Note the units

    12x+8y pounds12x + 8y \text{ pounds}

    The total is in pounds.

  9. Consider a factor

    12x+8y=4(3x+2y)12x + 8y = 4(3x + 2y)

    It can be factorised by 4, but need not be.

  10. Re-check

    x=1,y=1: 12+8=20x=1,y=1:\ 12 + 8 = 20

    One of each costs £20. ✓

  11. Sense-check the size

    12x+8y>012x + 8y > 0

    A positive total for positive ticket counts.

  12. Reflect

    price×count, then add\text{price} \times \text{count, then add}

    Each ticket type is priced separately.

  13. Confirm

    12x+8y12x + 8y

    Adults plus children.

  14. Re-state

    12x+8y12x + 8y

    The total in pounds.

  15. State the answer

    12x+8y12x + 8y

    So the total cost is 12x + 8y pounds.

Answer
12x+8y12x + 8y

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