AMC Questions (1 Viewer)

[buchanan]

Spoiler

21. D 22. C 23. E 24. C 25. B 26. D 27. B 28. A 29. B 30. C

 

[kurt.physics]

Thanks SO SO Much

 

[kurt.physics]

Spoiler

21. D 22. C 23. E 24. C 25. B 26. D 27. B 28. A 29. B 30. C

So where did you get these?

 

[Undermyskin]

I can answer!
Q1: this is simple by using similar triangle you can find the length of the hypotenuse is 40 for triangle "24". then let 40*4, you have 160.
Answer: C.

Q2: The reason you have Pi is because when turning around a corner, you still have to remain a distance of 2m from the nearest part of the wall.

From adding the lengths of all the wall, substract 4m due to our turn at the right angle, we can limit the answer to either B or D.

Now we need to find the angle at each vertex P, R and Q.
Pi - ^P
Pi - ^Q
Pi - ^R
3Pi - Pi/2 (We can use sum of interior angel in a quadrilateral is equal 2Pi)

Now we have 5Pi/2. To find the 'circumference', we need to 5Pi/2 * 2=5Pi

Thus the answer is B.

CAn you check?

 

[buchanan]

So where did you get these?

Pederson, D.G. & Sommariva, H.M. (eds), AMC Solutions & Statistics 2002, AMT Publishing, 2002, pp. 42-46.

And just so it's fair for others on this thread, I've also attached questions 21-30 here:

 

[tacogym27101990]

I'm probably wrong, so if i am can someone explain it to me.
but in those official solutions
it says
.5*5*32=160
but it doesnt, it equals 80
is there something els eu have to do?
thanks

 

[buchanan]

It shouldn't have the 0.5

5x32=160

Thanks for the correction. But the correct answer is still C.

 

[buchanan]

Other AMC papers with solutions for 1978-2007 are available from http://www.amtt.com.au

 

[kurt.physics]

Thanks