Maths  permalink

INT 2

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Messages: 1 - 3 of 3
  • Message 1. 

    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:/

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  • Message 2

    , in reply to message 1.

    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!!

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  • Message 3

    , in reply to message 2.

    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

    Report message3

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