Thread: basic arithmetic

I tried solving this question but I can't seem to get the answer right O.o
I'd appreciate any sort of help , thankyou ^_^

Q. The foot of a mountain is at sealevel.
The temperature at the foot is 16 degree celcius.
The temperature at a height of 3000m on the mountain was -4 degree celcius.
Given that temperature fell at costant rate,find
(i) the temp at a height of 1800m
(ii) the height at which the temperature was 0 degree

Ans: (i)4 degree
(ii) 2400m

You are told that the temperature falls at a constant rate so you can draw a straight line between the two data points you are given (altitude 0, T = 16 C; altitde = 3000, T=-4C).

This gives you the equation:

altitude = 2400 - 150T

So for part one

1800 = 2400 - 150T

T= 4 C

For part two if T= 0

altitude = 2400 - 150*0
=2400.

Thanks man
bless you