Indices

Topic

Higher June 2025 Paper 2 Q15

EdexcelCurrent spec3 marksIndices

15 Simplify fully \(\left(\dfrac{2a^3}{10a^7c^2}\right)^{-3}\)

(3)

Mark scheme

Scheme	Marks
\(125a^{12}c^6\)	B3
	(3)
	(3 marks)

Notes

B3: for \(125a^{12}c^6\) oe

eg \(\dfrac{125a^{12}}{c^{-6}}\) or \(\dfrac{125c^6}{a^{-12}}\) or \(\dfrac{125}{a^{-12}c^{-6}}\) or \(\dfrac{a^{12}}{125^{-1}c^{-6}}\) or \(\dfrac{1}{125^{-1}a^{-12}c^{-6}}\) or \(\dfrac{a^{12}c^6}{125^{-1}}\)

B2 for 2 correct terms (Allow eg \(125a^{12}\) or \(125c^6\) or \(ka^{12}c^6\) as long as not added to any other terms)

B1 for one correct term (allow eg \(a^{12}\) or \(a^{-12}c^6\) or \(\dfrac{1}{125^{-1}}\) or \(\dfrac{1000a^{21}c^6}{8a^9}\) as long as not added to any other terms)

Higher June 2025 Paper 1R Q15

EdexcelCurrent spec6 marksIndices Solving Simple Equations

(a) Solve \(\dfrac{5a + 8}{3} - \dfrac{2a + 5}{4} = 23\)
Show clear algebraic working. (4)

(b) Express \(\left(\dfrac{\sqrt{y}}{3}\right)^{-1}\) in the form \(cy^n\) where \(c\) and \(n\) are numbers to be found. (2)

Mark scheme (a)

Scheme	Marks
eg \(12 \times \dfrac{5a + 8}{3} - 12 \times \dfrac{2a + 5}{4} = 12 \times 23\) or eg \(4(5a + 8) - 3(2a + 5) = 12 \times 23 (= 276)\) or eg \(\dfrac{4(5a + 8)}{12} - \dfrac{3(2a + 5)}{12}(= 23)\) or eg \(\dfrac{4(5a + 8) - 3(2a + 5)}{12}(= 23)\)	M1
eg \(20a + 32 - 6a - 15 = 12 \times 23 (= 276)\) oe or \(14a + 17 = 276\)	M1
eg \(20a - 6a = 276 - 32 + 15\) oe or \(14a = 259\)	M1
Working required Answer: 18.5	A1
	(4)

Notes

M1: for clear intention to multiply all terms by 12 or a multiple of 12

to express LHS as two fractions over 12 or a multiple of 12

or as a single fraction with a denominator of 12 or a multiple of 12

(If expanded numerator, allow one sign error or one numerical error but not both)

\(\dfrac{20a + 32}{12} - \dfrac{6a + 15}{12}(= 23)\) or \(\dfrac{20a + 32 - 6a + 15}{12}(= 23)\) or

\(\dfrac{20a + 32 - 6a + 15}{12}(= 23)\)

M1: ft for expanding brackets and multiplying both sides by denominator with no more than one error in total leading to a linear equation

Accept a linear equation leading to

\(14a - 17 = 276\) oe or \(14a + 47 = 276\) oe or
\(26a + 17 = 276\)

This mark implies the previous M mark if not already awarded

M1: ft dep on previous M1 for correctly rearranging terms in \(a\) on one side and number terms on the other side

A1: oe dep on M2 eg \(\dfrac{259}{14}\) or \(\dfrac{37}{2}\)

Mark scheme (b)

Scheme	Marks
eg \(\dfrac{3}{\sqrt{y}}\left(= \dfrac{3\sqrt{y}}{y}\right)\) or \(\dfrac{3}{y^{0.5}}\) or \(\dfrac{3}{y^{\frac{1}{2}}}\) or \(\left(\dfrac{y^{0.5}}{3}\right)^{-1}\) or \(\left(\dfrac{y^{\frac{1}{2}}}{3}\right)^{-1}\) oe	M1
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(3y^{-0.5}\)	A1
	(2)
	(6 marks)

Notes

M1: for a correct first step by applying one of the following index rules

\(\sqrt{x} = x^{\frac{1}{2}} = x^{0.5}\) or \(\left(\dfrac{a}{b}\right)^{-1} = \left(\dfrac{b}{a}\right)\)

A1: oe eg \(3y^{-\frac{1}{2}}\), accept \(c = 3\) and \(n = -0.5\) oe

Higher June 2025 Paper 1R Q8

EdexcelCurrent spec5 marksFactorising Indices

(a) Simplify \(a^6 \times a^{10}\) (1)

(b) Simplify \(c^{30} \div c^{12}\) (1)

(c)

(i) Factorise \(y^2 - 10y + 21\) (2)

(ii) Hence, solve \(y^2 - 10y + 21 = 0\) (1)

Scheme	Marks
\(a^{16}\)	B1
	(1)

Scheme	Marks
\(c^{18}\)	B1
	(1)

Mark scheme (c)

Scheme	Marks
(i)	M1
Correct answer scores full marks (unless from obvious incorrect working) Answer: \((y - 3)(y - 7)\)	A1
	(2)
(ii) Answer: 3, 7	B1
	(1)
	(5 marks)

Notes

M1: for \((y \pm 3)(y \pm 7)\)
or for \((y \pm a)(y \pm b)\) with \(ab = 21\) or \(a + b = -10\)

A1: for correct factors

B1: ft dep on factorising in the form \((y \pm p)(y \pm q)\)

Higher June 2025 Paper 2R Q6

EdexcelCurrent spec5 marksIndices

(a) Write down the value of \(5^0\) (1)

\(\dfrac{5^9 \times 5^{-3}}{5^{-2}} = 5^k\)

(b) Find the value of \(k\) (2)

Scheme	Marks
1	B1
	(1)

Mark scheme (b)

Scheme	Marks
eg \((5^9 \times 5^{-3} =)\;5^6\) or \((5^9 \div 5^{-2} =)\;5^{11}\) or \((5^{-3} \div 5^{-2} =)\;5^{-1}\) or \((5^k \times 5^{-2} =)\;5^{k - 2}\) or \(9 - 3 = k - 2\) oe or \(9 - 3 - -2\) or \(9 - 3 + 2\)	M1
Correct answer scores full marks (unless from obvious incorrect working) Answer: 8	A1
	(2)

Notes

M1: for one correct application of an index rule (must be seen in powers of 5)
this could be after an initial mistake – working will need to be clearly seen
or
for forming a correct equation in the indices alone
or
for a complete method for the value of \(k\)

A1: condone \(5^8\)

Scheme	Marks
\(8d^{12}e^{15}\)	B2
	(2)
	(5 marks)

Higher June 2025 Paper 1 Q4

EdexcelCurrent spec4 marksIndices

4 \(x^5 \times x^7 = x^m\)

(a) Find the value of \(m\) (1)

\(y^8 \div y^3 = y^n\)

(b) Find the value of \(n\) (1)

Scheme	Marks
12	B1
	(1)

Scheme	Marks
5	B1
	(1)

Mark scheme (c)

Scheme	Marks
\(125a^{12}r^6\)	B2
	(2)
	(4 marks)

Notes

B2: for \(125a^{12}r^6\)

(B1 for a product in the form \(ka^pr^q\) where 2 from \(k\), \(p\) or \(q\) are correct eg \(5a^{12}r^6\)
Allow \(125a^{12}\) or \(125r^6\) or \(a^{12}r^6\) so as long as not added to any other terms)

Higher November 2024 Paper 2 Q14

EdexcelCurrent spec3 marksIndices

14 Given that \(\dfrac{3^{2n + 3}}{3^4} = 3^3 \times 3^{1 - 2n}\)

find the value of \(n\)
Show your working clearly.

(3)

Mark scheme

Scheme	Marks
eg \(3^{2n + 3 - 4}\left(= 3^3 \times 3^{1 - 2n}\right)\) or \(3^{2n - 1}\left(= 3^3 \times 3^{1 - 2n}\right)\) or \(\left(3^{2n + 3} =\right) 3^7 \times 3^{1 - 2n}\) or \(\left(\dfrac{3^{2n + 3}}{3^4} =\right) 3^{4 - 2n}\) or \(\left(3^{2n + 3} =\right) 3^3 \times 3^{5 - 2n}\) or \(\dfrac{3^{2n}}{3^4} = 3^{1 - 2n}\) (division by 3³) This is not an exhaustive list or \(2n + 3 - 4\) (= ….) or (… =) \(3 + 1 - 2n\) or \(2n - 1\) (=…) or (... =) \(4 - 2n\) (no other options for this)	M1
eg \(3^{2n + 3 - 4} = 3^{3 + 1 - 2n}\) or \(3^{2n + 3} = 3^{8 - 2n}\) or \(3^{2n - 1} = 3^{4 - 2n}\) or \(2n + 3 - 4 = 3 + 1 - 2n\) oe eg \(2n - 1 = 4 - 2n\)	M1
Working required Answer: \(\dfrac{5}{4}\)	A1
	(3)
	(3 marks)

Notes

M1: For one rule of indices used to correctly combine two or more of the given expressions (do not need part in brackets)

(must include algebra)

or

For one of the 4 expressions shown, providing it is clear that they apply to the LHS or to the RHS

M1: A correct single power of 3 on both sides or a correct equation in \(n\) without indices
(some students may go straight to this and gain M2)

A1: oe dep on M1