O-level Mathematics nov07 paper 1 questn 13

[dasani]

23 The foot of a mountain is at sea level.
The temperature at the foot of the mountain was 16 °C.
The temperature at a height of 3000 m on the mountain was – 4 °C.
(a) Find the difference between these temperatures.
(b) Given that the temperature fell at a constant rate, find
(i) the temperature at a height of 900 m,
(ii) the height at which the temperature was 0 °C,
(iii) an expression, in terms of x, for the temperature, in °C, at a height of x metres.


PLEASE HELP

 

[XPF MASTER]

(a) 16-(-4) => 16+4 => 20 °C
(b) first find the rate at which it is falling ---> in 3000 m it fell for 20 degrees °C --> so for (1 m = 20/3000 °C)
now u do the rest .........
(i):. for 900m----> 20/3000 x 900---->that will be the no. of °C which have fallen .... so ull have to sbtract the answer from 16 °C!!!
(ii):. in this ull have to reverse the equation i just made b4... ie.. (1 °C = 3000/20 m).....0 °C means that the temp. has decreased from 16 to 0 .. thus 16 °C
so --------> 3000/20 x 16= ____m
(iii):. so now in all these parts the expression which u have been using is the same one asked here -------> 20x/3000.... but remember to subtract it from 16 °C
--------------->[ 16 - ( 20x/3000) ] °C
cleared??