Decimals: WJEC GCSE Intermediate June 2024 Unit 1 Q10

Intermediate June 2024 Unit 1 Q10

WJECIntermediateCurrent spec6 marksDecimals Fractions Rounding

10.

(a) Write the reciprocal of 4 as a decimal. [1]

(b) Estimate the value of \(\dfrac{79{\cdot}34}{40{\cdot}1 \times 0{\cdot}48}\).
You must show all your approximations in your working. [2]

Mark scheme (a)

Scheme	Marks	Comments
0·25	B1

Mark scheme (b)

Scheme

Marks

Comments

\(\dfrac{80}{40 \times 0.5}\)	4
\(\dfrac{2}{0.5}\)	4
\(\dfrac{80}{20}\)	4
\(\dfrac{79}{40 \times 0.5}\)	3·95 or 4 or \(3\frac{19}{20}\)
\(\dfrac{79}{20}\)	3·95 or 4 or \(3\frac{19}{20}\)
\(\dfrac{79.3}{20}\)	3·965 or 3·97 or 4
\(\dfrac{79.34}{20}\)	3·967 or 3·97 or 4

M1

A1

May be seen in stages.

Award M1 for appropriate calculation seen.

Award A1 for the correct estimate for the calculation seen.

An unsupported answer is M0A0.

Mark scheme (c)

Scheme	Marks	Comments
\(4\frac{1}{2}\) or 4·5 or \(\dfrac{9}{2}\)	B3	Mark final answer. Award B2 for an unsimplified evaluation as a single fraction or mixed number e.g. \(4\frac{7}{14}\) \(3\frac{21}{14}\) \(\dfrac{63}{14}\) \(\dfrac{441}{98}\) or equivalent 4 + ½. FT for one of the following: adding ‘their improper fractions’ (which incorporate the entire numbers) provided fractions have a common denominator (one numerator must be correct) and answer given in a (proper or improper) simplified form 3 + ‘their fractions’ evaluated correctly and in a simplified form, provided fractions have a common denominator (one numerator must be correct) ‘their 3’ + 1·5 (or equivalent) in a simplified form \(a\) + ½ (or equivalent in its simplified form) evaluated provided ½ has come from two fractions with a common denominator (e.g. \(\frac{10}{14} + \frac{11}{14} = \frac{21}{14} = 1\frac{7}{14} + 3 = 3\frac{1}{2}\)). Award B1 for sight of two fractions with a common denominator (allow an error in one numerator) e.g. (1) \(\dfrac{10}{14}\) + (2) \(\dfrac{11}{14}\) \(\dfrac{24}{14} + \dfrac{39}{14}\) \(\dfrac{168}{98} + \dfrac{273}{98}\) or equivalent. An unsupported answer of \(4\frac{1}{2}\) or 4·5 or \(\dfrac{9}{2}\) is awarded B3.

Scheme

Marks

Comments

\(4\frac{1}{2}\) or 4·5 or \(\dfrac{9}{2}\)

Mark final answer.
Award B2 for an unsimplified evaluation as a single fraction or mixed number
e.g.

\(4\frac{7}{14}\)
\(3\frac{21}{14}\)
\(\dfrac{63}{14}\)
\(\dfrac{441}{98}\) or equivalent
4 + ½.

FT for one of the following:

adding ‘their improper fractions’ (which incorporate the entire numbers) provided fractions have a common denominator (one numerator must be correct) and answer given in a (proper or improper) simplified form
3 + ‘their fractions’ evaluated correctly and in a simplified form, provided fractions have a common denominator (one numerator must be correct)
‘their 3’ + 1·5 (or equivalent) in a simplified form
\(a\) + ½ (or equivalent in its simplified form) evaluated provided ½ has come from two fractions with a common denominator
(e.g. \(\frac{10}{14} + \frac{11}{14} = \frac{21}{14} = 1\frac{7}{14} + 3 = 3\frac{1}{2}\)).

Award B1 for sight of two fractions with a common denominator (allow an error in one numerator)
e.g.

(1) \(\dfrac{10}{14}\) + (2) \(\dfrac{11}{14}\)
\(\dfrac{24}{14} + \dfrac{39}{14}\)
\(\dfrac{168}{98} + \dfrac{273}{98}\) or equivalent.

An unsupported answer of \(4\frac{1}{2}\) or 4·5 or \(\dfrac{9}{2}\) is awarded B3.