Foundation November 2017 Paper 2 Q17

AQACurrent spec3 marksSubstitution

17 Here is a formula to convert degrees Celsius (°C) to degrees Fahrenheit (°F).

\[F = 1.8C + 32\]

\(F\) is the number of degrees Fahrenheit
\(C\) is the number of degrees Celsius

(a) Show that \(\qquad\) –40°C = –40°F [2 marks]

(b) The temperature is –15°C

Nick says,

“Because the temperature is negative in Celsius, it must be negative in Fahrenheit.”

Is he correct?

You must show your working. [1 mark]

Mark scheme (a)

Answer	Mark	Comments
Alternative method 1
\(1.8 \times -40 + 32\) or –72	M1	oe eg \(1.8(-40) + 32\)
\(1.8 \times -40 + 32 = -40\) or \(1.8 \times -40 = -72\) and \(-72 + 32 = -40\)	A1	oe eg \(1.8(-40) + 32 = -40\) Full working must be seen oe eg \(1.8 \times -40 = -72\) and \(-40 - 32 = -72\)
Alternative method 2
\(\dfrac{-40 - 32}{1.8}\) or –72	M1
\(\dfrac{-40 - 32}{1.8} = -40\) or \(-40 - 32 = -72\) and \(-72 \div 1.8 = -40\)	A1	Full working must be seen oe eg \(-40 - 32 = -72\) and \(-40 \times 1.8 = -72\)
Alternative method 3
\(F = 1.8F + 32\) and \(F - 1.8F = 32\) or \(0.8F = -32\)	M1	Forms equation in one variable and collects terms correctly using any letter oe eg \(1.8F - F = -32\) or \(-0.8F = 32\)
(\(F =\)) \(-32 \div 0.8\) and \(F = -40\)	A1	Full working must be seen oe eg (\(F =\)) \(32 \div -0.8\) and \(F = -40\)

Mark scheme (b)

Answer	Mark	Comments
No and 5 or No and correctly evaluated counter example	B1

No, anything between –17°C and 0°C is positive in Fahrenheit	B1
No, anything between 0°F and 32°F is negative in Celsius	B1
Unless the range from –17°C to 0°C is given, then the counter example must be evaluated correctly
No because \(1.8 \times -15\) is –27, and 32 – 27 = 4	B0
Any temperature in Celsius between \(-17\dfrac{7}{9}\)°C and 0°C can be used as a counter-example
eg1 \(1.8 \times -10 + 32 = 14\) so No	B1
eg2 \(1.8 \times -1 + 32 = 30.2\) so No	B1
No because 14°F is –10°C	B1
Accept No because –10 = 14	B1
No because –15 is positive in Fahrenheit	B0