Foundation November 2021 Paper 3 Q19

OCRFoundationCurrent spec6 marksCompound Measures Units & Time

(a) Amit says

My normal typing speed is 40 words per minute.
Therefore, I estimate that my normal typing speed is about 210 characters per minute.

Each letter, space and piece of punctuation counts as a character.

How many letters per word is Amit most likely to have used in making the estimate?
Show how you decide. [3]

(b) Amit starts some homework at their normal typing speed.
Amit types 52 words in 1 minute 12 seconds.
What may be true about the length of the words that Amit has just typed?
Show how you decide. [3]

Mark scheme (a)

Answer	Marks	Part marks and guidance
4 with correct working	3	M1 for \(210 - n\) where \(40 \leqslant n \leqslant 50\) soi by 160 to 170 M1 for (their number of characters) \(\div 40\) Alternative M2 for two from [4 letters] \(210 \div 5\), [5 letters] \(210 \div 6\), [6 letters] \(210 \div 7\) or M1 for one from [4 letters] \(210 \div 5\), [5 letters] \(210 \div 6\), [6 letters] \(210 \div 7\) Alternative (trials): M2 for two from \(3 \times 40 +\) [40 to 50], \(4 \times 40 +\) [40 to 50], \(5 \times 40 +\) [40 to 50] or M1 for one from \(3 \times 40 +\) [40 to 50] or \(4 \times 40 +\) [40 to 50] or \(5 \times 40 +\) [40 to 50] OR M1 for \(210 \div 40\) A1 for final answer of 5 If 0 scored and no/insufficient working SC1 for answer 4	Correct working requires at least M1 \(n\) represents an estimate of the number of spaces and/or punctuation, digits, symbols etc Allow \(40 \times 5 = 200\) for M1 Answer “4 to 5” or 5 with no working score 0

Mark scheme (b)

Answer	Marks
72 [seconds]	B1
\(\dfrac{40}{60} \times\) their 72 oe or \(\dfrac{52}{\textit{their } 72} \times 60\) oe	M1
OR
1.2 oe	B1	oe may be \(1\dfrac{12}{60}\) oe
\(40 \times\) their 1.2 oe or \(52 \div\) their 1.2	M1	their 1.2 is not 1.12
AND
48 or 43.[3…] and short/shorter word length oe	A1	Ignore non-contradictory statements but, “He may type faster” is incorrect and, if included, scores A0