4MA0 Higher January 2018 Paper 4 Q3

EdexcelHigherOld spec5 marksFractions

(a) Show that \(\dfrac{2}{7} \div \dfrac{4}{5} = \dfrac{5}{14}\) (2)

(b) Show that \(3\dfrac{1}{6} - 1\dfrac{2}{3} = 1\dfrac{1}{2}\) (3)

Mark scheme (a)

Scheme	Marks
\(\dfrac{2}{7} \times \dfrac{5}{4}\)	M1
\(\dfrac{2}{7} \times \dfrac{5}{4} = \dfrac{10}{28} = \dfrac{5}{14}\) or show cancelling giving \(\dfrac{1}{7} \times \dfrac{5}{2} = \dfrac{5}{14}\) Answer: \(\dfrac{5}{14}\)	A1
	(2)

A1: answer from correct working with \(\dfrac{10}{28}\) oe seen

or \(\dfrac{5}{14}\) from correct cancelling

Scheme	Marks
\(\dfrac{10}{35} \div \dfrac{28}{35}\) oe or \(\dfrac{2}{4} = 0.5\;\; \dfrac{7}{5} = 1.4\)	M1
\(\dfrac{10}{28} = \dfrac{5}{14}\) oe or \(\dfrac{0.5}{1.4} = \dfrac{5}{14}\) Answer: \(\dfrac{5}{14}\)	A1

A1: answer from correct working with \(\dfrac{10}{28}\) oe seen or from use of decimals with \(\dfrac{0.5}{1.4}\) seen.

Mark scheme (b)

Scheme	Marks
eg \(\dfrac{19}{6} - \dfrac{5}{3}\) eg \(\dfrac{19}{6} - \dfrac{10}{6}\) or \((2)\dfrac{1}{6} - \dfrac{4}{6}\) \(2 - \dfrac{3}{6}\) or \(3\dfrac{1}{6} - 1\dfrac{4}{6}\) \(2 - \dfrac{3}{6}\) or \(\dfrac{7}{6} - \dfrac{4}{6}\) \(1\dfrac{7}{6} - \dfrac{4}{6}\)	M1
	M1
eg \(\dfrac{9}{6} = 1\dfrac{3}{6} = 1\dfrac{1}{2}\) or \(1\dfrac{3}{6} = 1\dfrac{1}{2}\) or \(1\dfrac{3}{6} = 1\dfrac{1}{2}\) or \(1\dfrac{3}{6} = 1\dfrac{1}{2}\) Answer: \(1\dfrac{1}{2}\)	A1
	(3)

M1: common denominator used for subtraction or improper fractions

M1: Method which would lead to \(\dfrac{9}{6}\) oe or \(1\dfrac{3}{6}\) oe

dep on first M1

A1: answer from correct working with all steps seen