Upper & Lower Bounds

Topic

Higher June 2025 Paper 2 Q22

EdexcelCurrent spec4 marksUpper & Lower Bounds

22 The diagram shows a square inside rectangle \(ABCD\)

Rectangle ABCD with a smaller square inside it; the region between the square and the rectangle is shaded

Diagram NOT accurately drawn

The total area of the region shown shaded in the diagram is \(X\) cm²

\(AB = 11.5\) cm correct to the nearest 0.5 cm
\(BC = 9.2\) cm correct to 2 significant figures
side of square = 4.1 cm correct to 2 significant figures

By considering bounds, work out the value of \(X\) to a suitable degree of accuracy.
Show your working clearly.

(4)

Mark scheme

Scheme	Marks
11.25, 11.75, 9.15, 9.25, 4.05, 4.15	B1
\(11.75 \times 9.25 - 4.05^2\) (= 92.285)	M1
\(11.25 \times 9.15 - 4.15^2\) (= 85.715)	M1
Working required Answer: 90	A1
	(4)
	(4 marks)

Notes

B1: for a correct bound

Accept
\(11.74\dot{9}\) for 11.75 or \(11.74\overline{9}\) for 11.75
\(9.24\dot{9}\) for 9.25 or \(9.24\overline{9}\) for 9.25
\(4.14\dot{9}\) for 4.15 or \(4.14\overline{9}\) for 4.15

M1: for a correct method to find the UB of \(X\), allow

\((11.5 \lt AB \leqslant 11.75) \times (9.2 \lt BC \leqslant 9.25) - \left((4.05 \leqslant s \lt 4.1)^2\right)\)

M1: for a correct method to find the LB of \(X\), allow

\((11.25 \leqslant AB \lt 11.5) \times (9.15 \leqslant BC \lt 9.2) - \left((4.1 \lt s \leqslant 4.15)^2\right)\)

A1: dep on M2
90 and both UB and LB correct using correct values 11.25, 11.75, 9.15, 9.25, 4.05 and 4.15

Higher June 2025 Paper 1 Q2

EdexcelCurrent spec6 marksPercentages Upper & Lower Bounds

2 Anna makes cups.
Each cup costs 6 Swiss francs to make.

Anna puts the cups into boxes to sell.
Each box contains 4 cups.

Anna sells 80 boxes of cups for a total of 2160 Swiss francs.

(a) Work out the percentage profit Anna makes.
Show your working clearly. (4)

The height of each cup is 9 cm, correct to the nearest cm

(b) Write down the lower bound of the height. (1)

The weight of each cup is 120 g, correct to the nearest 10 g

Mark scheme (a)

Scheme	Marks
4 × 6 × 80 (= 1920) or 4 × 6 (= 24) or 2160 ÷ 80 (= 27) or 2160 ÷ (80 × 4) (= 6.75)	M1
2160 – “1920” (= 240) or \(\dfrac{2160}{\text{``}{1920}\text{''}}(= 1.125)\) or “27” – “24” (= 3) or \(\dfrac{\text{``}{27}\text{''}}{\text{``}{24}\text{''}}(= 1.125)\) or “6.75” – 6 (= 0.75) or \(\dfrac{\text{``}{6.75}\text{''}}{6}(= 1.125)\)	M1
\(\dfrac{\text{``}{240}\text{''}}{\text{``}{1920}\text{''}}(\times 100)\) or 0.125 (× 100) or \(\left(\dfrac{2160}{\text{``}{1920}\text{''}} - 1\right)(\times 100)\) or (“1.125” – 1) (× 100) or “1.125” × 100 (= 112.5) or \(\dfrac{\text{``}{3}\text{''}}{\text{``}{24}\text{''}}(\times 100)\) or 0.125 (× 100) or \(\left(\dfrac{\text{``}{27}\text{''}}{\text{``}{24}\text{''}} - 1\right)(\times 100)\) or (“1.125” – 1) (× 100) or “1.125” × 100 (= 112.5) or \(\dfrac{\text{``}{0.75}\text{''}}{6}(\times 100)\) or 0.125 (× 100) or \(\left(\dfrac{\text{``}{6.75}\text{''}}{6} - 1\right)(\times 100)\) or (“1.125” – 1) (× 100) or “1.125” × 100 (= 112.5)	M1
Working required Answer: 12.5	A1
	(4)

Notes

M1: for method to work out total income or income from one box or expenditure for one box or income per cup

M1: for working out the profit or income ÷ expenditure

M1: for a method to reach one step from the answer ie getting to \(\dfrac{1}{8}\) oe or 0.125 or 112.5

A1: dep on M1

Scheme	Marks
8.5	B1
	(1)

Scheme	Marks
125	B1
	(1)
	(6 marks)

Higher June 2025 Paper 1R Q2

EdexcelCurrent spec2 marksUpper & Lower Bounds

2 The length of a ship is 142.8 m, correct to 1 decimal place.

(i) Write down the lower bound of the length of the ship. (1)

(ii) Write down the upper bound of the length of the ship. (1)

Scheme	Marks
142.75	B1
	(1)

Scheme	Marks
142.85	B1
	(1)
	(2 marks)

Higher November 2024 Paper 1 Q21

EdexcelCurrent spec3 marksUpper & Lower Bounds

21 \(T = \dfrac{x^2 + y^2}{w}\)

\(x = 28.4\)	correct to 1 decimal place.
\(y = 17\)	correct to 2 significant figures.
\(w = 90\)	correct to the nearest 5

Calculate the upper bound for the value of \(T\)

Give your answer correct to 3 significant figures.
Show your working clearly.

(3)

Mark scheme

Scheme	Marks
28.35 or 28.45 or 16.5 or 17.5 or 87.5 or 92.5	B1
\((T =)\;\dfrac{28.45^2 + 17.5^2}{87.5}\) (= 12.75031429)	M1
Working required Answer: 12.8	A1
	(3)
	(3 marks)

Notes

B1: Accept
\(28.44\dot{9}\) for 28.45
\(17.4\dot{9}\) for 17.5

M1: for substituting the correct bounds into the formula for \(T\)

\((T =)\;\dfrac{UB_x^{\,2} + UB_y^{\,2}}{LB_w}\) where

\(28.4 \lt UB_x \leqslant 28.45\)

\(17 \lt UB_y \leqslant 17.5\)

\(87.5 \leqslant LB_w \lt 90\)

(corrected from the printed mark scheme: the second line is printed as \(17 \lt UB_x \leqslant 17.5\))

A1: awrt 12.8 dep on M1
Answer must come from correct figures (28.45, 17.5 and 87.5)

Higher November 2024 Paper 2 Q2

EdexcelCurrent spec2 marksUpper & Lower Bounds

2 The length of a table is measured as 1.4 metres correct to one decimal place.

(a) Write down the upper bound of the length of the table. (1)

(b) Write down the lower bound of the length of the table. (1)