Hard GCSE Forming and solving Questions

Challenging, exam-style GCSE Forming and solving questions with worked solutions. Stretch yourself on the hardest forming an equation, perimeter, substituting back, area problems.

forming an equationperimetersubstituting backareaconsecutive integersage problems
GCSE Foundation34 questionsStep-by-step solutions
Question 1
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.
Show worked solution

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°
Question 2
6 markschallenging
A charity raises £xx at an event. It gives a third of the money to a food bank and a quarter of the money to a shelter. It keeps the remaining £250. Work out how much money the charity raised.
Show worked solution

Worked solution

  1. Define the letter

    x=amount raised in poundsx = \text{amount raised in pounds}

    The whole amount raised is the unknown.

  2. Write the food bank's share

    x3\frac{x}{3}

    A third of the money goes to the food bank.

  3. Write the shelter's share

    x4\frac{x}{4}

    A quarter of the money goes to the shelter.

  4. Write what is left

    x−x3−x4x - \frac{x}{3} - \frac{x}{4}

    The charity keeps whatever remains after both gifts.

  5. Form the equation

    x−x3−x4=250x - \frac{x}{3} - \frac{x}{4} = 250

    The remainder is £250\pounds 250.

  6. Multiply every term by 12

    12x−4x−3x=300012x - 4x - 3x = 3000

    1212 is the lowest common denominator of 33 and 44; multiply the 250250 as well.

  7. Collect the x terms

    5x=30005x = 3000

    12x−4x−3x=5x12x - 4x - 3x = 5x.

  8. Solve for x

    x=600x = 600

    Divide both sides by 55.

  9. Check the food bank's share

    6003=£200\frac{600}{3} = \pounds 200

    A third of £600\pounds 600 is £200\pounds 200.

  10. Check the shelter's share

    6004=£150\frac{600}{4} = \pounds 150

    A quarter of £600\pounds 600 is £150\pounds 150.

  11. Check the remainder

    600−200−150=£250600 - 200 - 150 = \pounds 250

    The charity keeps £250\pounds 250, exactly as stated.

  12. Check with fractions

    1−13−14=5121 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}

    The charity keeps 512\frac{5}{12} of the money.

  13. Confirm the fraction kept

    512×600=250\frac{5}{12} \times 600 = 250

    Five twelfths of £600\pounds 600 is £250\pounds 250 — the same answer found a second way.

  14. Check the answer is sensible

    £600>£250\pounds 600 > \pounds 250

    The amount raised must be bigger than the amount kept, and it is.

  15. State the answer in context

    £600\pounds 600

    The charity raised £600\pounds 600.

Answer
£600\pounds 600
Question 3
6 markschallenging
A fruit basket holds 33 kg of apples and 22 kg of grapes and costs £14.30. Grapes cost £2.15 more per kilogram than apples. Work out the price of apples per kilogram.
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Worked solution

  1. Define the letter

    a=price of apples in pounds per kga = \text{price of apples in pounds per kg}

    The grapes are described in terms of the apples, so make the apples the unknown.

  2. Write the price of grapes

    a+2.15a + 2.15

    Grapes cost £2.15\pounds 2.15 more per kilogram than apples.

  3. Write the cost of the apples

    3a3a

    33 kg at £a\pounds a per kg.

  4. Write the cost of the grapes

    2(a+2.15)2(a + 2.15)

    22 kg at £(a+2.15)\pounds (a + 2.15) per kg.

  5. Form the equation

    3a+2(a+2.15)=14.303a + 2(a + 2.15) = 14.30

    The two fruits together cost £14.30\pounds 14.30.

  6. Expand the bracket

    3a+2a+4.30=14.303a + 2a + 4.30 = 14.30

    Multiply both terms inside the bracket by 22.

  7. Collect the a terms

    5a+4.30=14.305a + 4.30 = 14.30

    3a+2a=5a3a + 2a = 5a.

  8. Subtract 4.30

    5a=10.005a = 10.00

    Move the constant across.

  9. Divide by 5

    a=2.00a = 2.00

    The apples cost £2.00\pounds 2.00 per kilogram.

  10. Work out the price of grapes

    2.00+2.15=£4.15 per kg2.00 + 2.15 = \pounds 4.15 \text{ per kg}

    Substitute back into a+2.15a + 2.15.

  11. Check the cost of the apples

    3×2.00=£6.003 \times 2.00 = \pounds 6.00

    Three kilograms of apples cost £6.00\pounds 6.00.

  12. Check the cost of the grapes

    2×4.15=£8.302 \times 4.15 = \pounds 8.30

    Two kilograms of grapes cost £8.30\pounds 8.30.

  13. Check the total

    6.00+8.30=£14.306.00 + 8.30 = \pounds 14.30

    The basket total matches the question.

  14. Check the prices are sensible

    £2.00>0,£4.15>0\pounds 2.00 > 0, \quad \pounds 4.15 > 0

    Both prices are positive and given to 22 decimal places.

  15. State the answer in context

    £2.00 per kg\pounds 2.00 \text{ per kg}

    Apples cost £2.00\pounds 2.00 per kilogram.

Answer
£2.00 per kg\pounds 2.00 \text{ per kg}
Question 4
5 markschallenging
The three sides of a triangle are said to be (x+5)(x + 5) cm, (2x−3)(2x - 3) cm and (x−6)(x - 6) cm, with a perimeter of 1616 cm. Which statement correctly explains why no such triangle can exist?
Show worked solution

Worked solution

  1. Write an expression for the perimeter

    (x+5)+(2x−3)+(x−6)(x + 5) + (2x - 3) + (x - 6)

    The perimeter is the sum of the three sides.

  2. Form the equation

    (x+5)+(2x−3)+(x−6)=16(x + 5) + (2x - 3) + (x - 6) = 16

    The stated perimeter is 1616 cm.

  3. Remove the brackets

    x+5+2x−3+x−6=16x + 5 + 2x - 3 + x - 6 = 16

    The brackets only group the terms.

  4. Collect the x terms

    x+2x+x=4xx + 2x + x = 4x

    There are four xx in total.

  5. Collect the constants

    5−3−6=−45 - 3 - 6 = -4

    The constants combine to −4-4.

  6. Rewrite the equation

    4x−4=164x - 4 = 16

    The equation is now in two-step form.

  7. Solve for x

    4x=20⇒x=54x = 20 \Rightarrow x = 5

    Add 44, then divide by 44.

  8. Substitute into the first side

    x+5=10 cmx + 5 = 10 \text{ cm}

    This side is a sensible positive length.

  9. Substitute into the second side

    2x−3=7 cm2x - 3 = 7 \text{ cm}

    This side is also positive.

  10. Substitute into the third side

    x−6=5−6=−1 cmx - 6 = 5 - 6 = -1 \text{ cm}

    The third side comes out as −1-1 cm.

  11. Interpret the negative value

    −1 cm is impossible-1 \text{ cm is impossible}

    A physical length must be positive, so this solution has to be rejected.

  12. Check the algebra was right

    10+7+(−1)=1610 + 7 + (-1) = 16

    The arithmetic does give 1616, which shows that an equation can have a solution the context cannot accept.

  13. State the condition for a valid side

    x−6>0⇒x>6x - 6 > 0 \Rightarrow x > 6

    For all three sides to be positive we would need xx to be greater than 66.

  14. Compare with the solution found

    x=5<6x = 5 < 6

    The only solution of the equation fails this condition.

  15. State the conclusion

    No such triangle exists\text{No such triangle exists}

    The equation solves, but the triangle it describes is impossible.

Answer
x=5 gives a side of −1 cm, so the triangle is impossiblex = 5 \text{ gives a side of } -1 \text{ cm, so the triangle is impossible}
Question 5
5 markschallenging
Two thirds of a number is 1010 more than a quarter of the same number. Form an equation and solve it to find the number.
Show worked solution

Worked solution

  1. Define the letter

    n=the numbern = \text{the number}

    Both fractions refer to the same unknown number.

  2. Translate "two thirds of a number"

    23n\frac{2}{3}n

    "Of" means multiply, so two thirds of nn is 23n\frac{2}{3}n.

  3. Translate "a quarter of the number"

    14n\frac{1}{4}n

    A quarter of the same number nn.

  4. Translate "10 more than"

    14n+10\frac{1}{4}n + 10

    "1010 more than" means add 1010 to the quarter.

  5. Form the equation

    23n=14n+10\frac{2}{3}n = \frac{1}{4}n + 10

    The two thirds equals the quarter plus 1010.

  6. Choose a common denominator

    LCM of 3 and 4=12\text{LCM of } 3 \text{ and } 4 = 12

    Multiplying through by 1212 will clear both fractions at once.

  7. Multiply every term by 12

    12×23n=8n,12×14n=3n,12×10=12012 \times \frac{2}{3}n = 8n, \quad 12 \times \frac{1}{4}n = 3n, \quad 12 \times 10 = 120

    Every term must be multiplied, including the 1010.

  8. Rewrite the equation

    8n=3n+1208n = 3n + 120

    The equation now has no fractions.

  9. Collect the n terms

    5n=1205n = 120

    Subtract 3n3n from both sides.

  10. Solve for n

    n=24n = 24

    Divide both sides by 55.

  11. Check two thirds of the number

    23×24=16\frac{2}{3} \times 24 = 16

    Two thirds of 2424 is 1616.

  12. Check a quarter of the number

    14×24=6\frac{1}{4} \times 24 = 6

    A quarter of 2424 is 66.

  13. Check the relationship

    16=6+1016 = 6 + 10

    Two thirds is indeed 1010 more than a quarter.

  14. Note the common error

    multiply EVERY term by 12\text{multiply EVERY term by } 12

    Forgetting to multiply the constant 1010 by 1212 is the usual slip.

  15. State the answer in context

    n=24n = 24

    The number is 2424.

Answer
n=24n = 24

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