GCSE Forming and solving Practice Questions

Free GCSE Forming and solving practice questions with full step-by-step worked solutions. Covers forming an equation, solving one-step equations, inverse operations, money context. Practise exam-style problems and check your method.

forming an equationsolving one-step equationsinverse operationsmoney contextinterpreting the solutionperimeter
GCSE Foundation70 questionsStep-by-step solutions
Question 1
1 markeasy
I think of a number and add 77. The answer is 1212. Using nn for the number, form an equation and solve it to find the number.
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Worked solution

  1. Form the equation

    n+7=12n + 7 = 12

    The number is nn, and adding 77 to it gives 1212.

  2. Solve the equation

    n=12−7=5n = 12 - 7 = 5

    Subtract 77 from both sides to undo the addition.

  3. Interpret the answer

    n=5n = 5

    The number I thought of is 55. Check: 5+7=125 + 7 = 12.

Answer
n=5n = 5
Question 2
2 markseasy
A rectangle has width ww cm. Its length is three times its width. The perimeter is 3232 cm. Form an equation and solve it to find the width.
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Worked solution

  1. Form the equation

    2(w+3w)=322(w + 3w) = 32

    The width is ww and the length is 3w3w; the perimeter is twice their sum.

  2. Solve the equation

    8w=32⇒w=48w = 32 \Rightarrow w = 4

    Collect inside the bracket to get 2×4w=8w2 \times 4w = 8w, then divide by 88.

  3. Interpret the answer in context

    w=4 cmw = 4 \text{ cm}

    The width is 44 cm and the length is 1212 cm. Check: 2(4+12)=322(4 + 12) = 32 cm.

Answer
w=4 cmw = 4 \text{ cm}
Question 3
2 marksintermediate
I think of a number, divide it by 44 and then add 33. The answer is 1010. Form an equation and solve it to find the number.
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Worked solution

  1. Define the letter

    n=the numbern = \text{the number}

    State the unknown clearly.

  2. Translate the instructions

    divide by 4→n4,then add 3→n4+3\text{divide by } 4 \rightarrow \frac{n}{4}, \quad \text{then add } 3 \rightarrow \frac{n}{4} + 3

    The 33 is added after the division.

  3. Form the equation

    n4+3=10\frac{n}{4} + 3 = 10

    The final result is 1010.

  4. Subtract 3 from both sides

    n4=7\frac{n}{4} = 7

    Undo the addition first, working backwards through the operations.

  5. Multiply both sides by 4

    n=28n = 28

    Multiplication undoes division.

  6. Interpret and check

    n=28n = 28

    The number is 2828. Check: 28÷4=728 \div 4 = 7 and 7+3=107 + 3 = 10.

Answer
n=28n = 28
Question 4
4 markshard
A meal deal contains a sandwich, a drink and a packet of crisps. The drink costs half as much as the sandwich. The crisps cost £0.85. Three meal deals cost £13.35 altogether. Work out the cost of a sandwich.
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Worked solution

  1. Define the letter

    s=cost of a sandwich in poundss = \text{cost of a sandwich in pounds}

    Everything else in the deal is described in terms of the sandwich.

  2. Write the cost of the drink

    s2\frac{s}{2}

    The drink costs half as much as the sandwich.

  3. Write the cost of one meal deal

    s+s2+0.85s + \frac{s}{2} + 0.85

    Sandwich ++ drink ++ crisps.

  4. Simplify one meal deal

    1.5s+0.851.5s + 0.85

    s+s2=1.5ss + \frac{s}{2} = 1.5s.

  5. Form the equation

    3(1.5s+0.85)=13.353(1.5s + 0.85) = 13.35

    Three identical meal deals cost £13.35\pounds 13.35.

  6. Expand the bracket

    4.5s+2.55=13.354.5s + 2.55 = 13.35

    Multiply both terms by 33.

  7. Subtract 2.55

    4.5s=10.804.5s = 10.80

    Move the constant across.

  8. Divide by 4.5

    s=10.804.5=2.40s = \frac{10.80}{4.5} = 2.40

    This gives the cost of one sandwich.

  9. Check one meal deal

    2.40+1.20+0.85=£4.452.40 + 1.20 + 0.85 = \pounds 4.45

    The drink costs £1.20\pounds 1.20, so one deal costs £4.45\pounds 4.45.

  10. Check three meal deals

    3×4.45=£13.353 \times 4.45 = \pounds 13.35

    The total matches, so a sandwich costs £2.40\pounds 2.40.

Answer
£2.40\pounds 2.40
Question 5
5 markschallenging
The four angles of a quadrilateral are (x+20)°(x + 20)°, 2x°2x°, (2x−10)°(2x - 10)° and (x+50)°(x + 50)°. Work out the size of the smallest angle.
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Worked solution

  1. State the angle fact

    angles in a quadrilateral=360°\text{angles in a quadrilateral} = 360°

    This fact turns the four expressions into an equation.

  2. List the four angles

    (x+20),  2x,  (2x−10),  (x+50)(x + 20), \; 2x, \; (2x - 10), \; (x + 50)

    Every angle is written in terms of the same unknown xx.

  3. Form the equation

    (x+20)+2x+(2x−10)+(x+50)=360(x + 20) + 2x + (2x - 10) + (x + 50) = 360

    The four angles add up to 360°360°.

  4. Remove the brackets

    x+20+2x+2x−10+x+50=360x + 20 + 2x + 2x - 10 + x + 50 = 360

    The brackets only group terms and can be dropped.

  5. Collect the x terms

    x+2x+2x+x=6xx + 2x + 2x + x = 6x

    There are six xx altogether.

  6. Collect the constants

    20−10+50=6020 - 10 + 50 = 60

    Take care with the −10-10.

  7. Rewrite the equation

    6x+60=3606x + 60 = 360

    The equation is now a simple two-step equation.

  8. Subtract 60

    6x=3006x = 300

    Move the constant across.

  9. Divide by 6

    x=50x = 50

    This is the value of xx, not an angle in the quadrilateral.

  10. Work out the four angles

    70°,  100°,  90°,  100°70°, \; 100°, \; 90°, \; 100°

    Substitute x=50x = 50 into each of the four expressions in turn.

  11. Check the total

    70+100+90+100=36070 + 100 + 90 + 100 = 360

    The four angles do add to 360°360°.

  12. Check the angles are sensible

    each angle>0°\text{each angle} > 0°

    Every angle is positive, so the quadrilateral is possible.

  13. Identify the smallest angle

    x+20=70°x + 20 = 70°

    Comparing 7070, 100100, 9090 and 100100, the smallest is 70°70°.

  14. Beware the tempting answer

    x=50≠an anglex = 50 \ne \text{an angle}

    x=50x = 50 is not one of the angles — the smallest angle is x+20x + 20.

  15. State the answer in context

    70°70°

    The smallest angle of the quadrilateral is 70°70°.

Answer
70°70°

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