AMC8 ARGHHHHH HARD QUESTIONS

I just went backwards...

:

T=32-3=29 so volume at that time is=24-4=20

T=29-3=26 , v=16

T=23, v=12

T=20, v =8 (answer)

It didn't take time so it could be a good strategy for this specific question.

Using arthimetic sequence(Really isn't needed and takes a lot of time for this specific question but good if the numbers were far from each other)

We don't know the current temperature (Start) 

So let Temperature= T 

To find T , we know that at somepoint the temperature was 32 and volume was 24 

So T=32-3L 

Where did the 3 come from? "For every 3 deg rise in Temp" So going backward from 32 degrees We wouldn't add 3. We subtract  then

so 32-3L

L is a specific number of terms.. For example, If 32 was for example, the 5th term and i want the 2nd term 

therefore 32-3 -3 -3 etc..

Also could be said as: 32-3L

So Now we know that temperature is equal to , T=32-3L

Same thing goes for the volume.

V=The point which was given with T (t=32 degrees, v=24 cubic cen v) -4L

So 

V=24-4L

where did the 4 come from?

for every 3 degrees, the volume expands by 4.

Going backwards, for decreasing 3 degrees, the volume shrinks(Decreases) by 4

So 24-4L 

L no. of terms backwards as well. 

We now know that:

T=32-3L

V=24-4L

Question:

"What was the volume of the gas in cubic cent. when the temperature was 20 degrees?"

So T=32-3L=20 (deg)

Solve the equation for L

-3L=20-32

-3L=-12

L=-12/-3=4

Subsituite L=4 in Volume equation

V=24-4(4) = 8 

Again, the strategy I used earlier above would save a lot of time but who knows what they would put in the exam.. It wouldn't work for far numbers apart. So Using arithmetic sequence is better.  

There might be a better answer. That's just my answer. 

Hope that helped :).

By the way that site you linked is really great :)