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Posted by Kels (U15215297) on Wednesday, 16th May 2012
Sorry I've gone completely blank, literally. Its 2005 Paper 2.
Q1. In the evening, the temperature in a greenhouse drops by 4% per hour. At 8pm the temperature is 28 degrees celsius.
What will the temperature be at 11pm?
Sorry I've just gone really blank and can't remember:/
Posted by Maths Teacher BBC (U15178780) on Thursday, 17th May 2012
It depends what way you would have been taught it.
This is how I would teach it:
drops by 4% means a subtraction.
Always start with 100% so:
100% - 4% = 96% = 0.96
Between 8pm and 11pm is 3 hours so:
28 x 0.96^3 = 24.772 = 25ᵒ to nearest whole number (or whatever way you have been asked to round it)
Hope this helps!!
Posted by U15280596 (U15280596) on Thursday, 24th May 2012
Hate to be pedantic, but the 'base' for temperature is 0°K not 0°C....
Percentages do not work unless the question defines the base starting point, so this is a fault in the question (which should have said a 4% drop in °C), although one could argue an 'implicit' definition by the sole mention of °C.
If anyone worked it out as a 4% drop in °K, I would regard that as correct too...
(273.15 + 28)x 0.96^3 - 273.15 = - 6.7118 °C
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