Math Help?

[Chanman]

I am preparing for a math contest...and on last year's contest there were these questions. I am wondering, is there an easy way to solve these questions?


21. The positive integers are arranged in increasing order in a triangle, as shown. Each row contains one more number than the previous row. The sum of the numbers in the
row that contains the number 400 is

(A) 10 990 (B) 12 209 (C) 9855
(D) 10 976 (E) 11 368

1
2 3
4 5 6
7 8 9 10
11 12 etc


22. The number of pairs of positive integers (p; q), with p + q 100, that satisfy the equation
p + q^-1/ p^1 + q = 17 is

(A) 0 (B) 1 (C) 2 (D) 4 (E) 5

23. Dolly, Molly and Polly each can walk at 6 km/h. Their one motorcycle, which travels at 90 km/h, can accommodate at most two of them at once (and cannot drive by itself !). Let t hours be the time taken for all three of them to reach a point 135 km away. Ignoring the time required to start, stop or change directions, what is true about the smallest possible value of t?

(A) t < 3.9 (B) 3.9 t < 4.1 (C) 4.1 t < 4.3
(D) 4:3 t < 4:5 (E) 4:5 t

24. Four numbers w; x; y; z satisfy w < x < y < z. Each of the six possible pairs of distinct numbers has a dierent sum. The four smallest sums are 1; 2; 3, and 4. What is the sum of all possible values of z?

(A) 4 (B)13/2 (C)17/2(D)15/2(E) 7

25. A pyramid has a square base with side length 20. A right circular cylinder has a diameter of 10 and a length of 10. The cylinder is lying on its side, completely inside the pyramid. The central axis of the cylinder lies parallel to and directly above a diagonal of the pyramid's base. The midpoint of the central axis lies directly above the centre of the square base of the pyramid.
The smallest possible height of the pyramid is closest to

(A) 15.3 (B) 22.1 (C) 21.9 (D) 21.7 (E) 15.5

edit: found the solutions, but maybe you guys can try them out lol

link to questions, part c

http://www.cemc.uwaterloo.ca/contests/past_contests/2011/2011FermatContest.pdf

 

[luxraygarchomp]

this hurts my brain looking at ahhhh wrong forum i thought i was on pokegym not at school

 

[sdrawkcab]

I used to take these math tests in jr.high.....I'm sure there is a faster way for 21 but this is how i did it.

As a proof of concept lets try looking for the sum of the row which contains 9. (Since it's easy to verify)

When you add the number of integers in each row together, the total gives you the last number in that row (and the last number you added is the number of numbers in that row......

So the first row has 1 number, the second row has 2 numbers (2 & 3). 1 + 2 = 3 so we just keep going until we pass 9.

1+2+3+4 = 10 meaning the last number in row 4 is 10, and there are 4 numbers. 10 9 8 7. The sum = 34. Which if we look above is correct.

So expanding this to 400.....you have 1+2+3+4+5..........+26+27+28 = 406 the last number in the row is 406 with 28 numbers...

406 + 405 + 404 + 403 + 402.......+381+380+379 = 10990 = A




I`m pretty sure that 22) is 0. Because if p+q = 100, then p^(-1)/q^1 would have to be a negative number which doesn't work.

---------- Post added 02/11/2012 at 02:45 PM ----------

#23)

2 people take the bike while one starts walking.
One person gets off at x-km from the finish line and continues walking towards the finish line, while the other goes back to pick up the person who started walking. By which time he already walked y-km. Ideally you spread the distances so the bike and the person who was first dropped off reaches the finish line at the same time.

Thankfully you don't need an exact answer for this question and can get away with some estimations. And after you adjust the drop off points a couple of times you'll have a better idea of where the drop of points should be.

This equation should help t = 3((135-x)/90) + x/90 where t = total time and x is the total distance walked by either person.

This one is either B or C my first estimation was C, but I feel that with better placement I could get it down to under 4.1 hours

 

[bullados]

22. That middle bit is 1/(pq). And somehow that bit needs to be -83. Enjoy solving for that ^_^

25. Geometry!!! OK, you know what the diagonal of the base of the pyramid is for the purposes of this, right (call this A))? You then subtract the length of the cylinder from the diagonal, then divide that number by 2 (call this B) (cuz there are 2 empty spots that the cylinder doesn't cover). Now, you're basically dealing with two triangles that have the same vertex and angle. So it's a simple relationship. Basically, the relationship is A / B = X / 10. I think this simple version is enough to find the correct answer. If it isn't, you'll have to calculate the pyramid diagonal using B and 10 (call this C), and then use a similar relationship of C / X = A / B, and then pyth out the height using X and A as your hyp and one leg.

 

[desufnoc]

OMG you guys should be using this math for pokemon lol. Like whats the chance of a first turn collector with the following cards in your deck: 4 collector or 3 collector and 2 pokegear or 2 collector and 3 pokegear. Thats a math answer that I would like to know.

 

[Alazor]

21) sdrawkcab has it right.

22) There is at least 1 pair. (p=17, q=1) is one of them. So not (A). (It's P+Q < or equal to 100). Not P +100Q.

23) I don't want to go over all of this, but I used diagrams to keep track of the work:

P = Polly
D = Dolly
M = Molly

Initial Diagram

Car [D,M]

P,D,M

|------------------------------|

135 km

1st iteration

...............................................Car [D]

.........P ...................................D,M time = 1.5 hr

|----|--------------------------|

9 km

2nd iteration


.............................Car [D]

.........P....................D.................M time = 2.5 hr

|----|------------|-------------|

15 km 30 km 90 km


3rd iteration


..............Car [P,D]

..............P,D................................M time = 2.81 hr

|--------|-----------------------|

17 km 118 km


Final iteration

...................................................[Car]

...................................................P,D,M time = 4.12 hr

|--------------------------------|

17 km 118 km

The answer is (C).

24)

Set-up:

w+x = 1
w + y = ?
w + z = ?
x +y = ?
y+z = ?
x + z = ?

w+x+y+z=7 ???

If it were me, I would guess D. Because the other sums could be 5 and 6, so z-w = 5 and somehow
all values of z is 15/2. This could be wrong though.

Good luck.

 

[sdrawkcab]

Alazor, I assumed he forgot the = sign, I didn't verify the question =p

For number 23, the bike doesn't have to go all the way to the finish line.....it could stay 9-12 km away and drop off someone there who could walk the rest of the way while you pick up the person who started walking in the first place.....this should save you enough time to change the answer to B ( 3.9 < t < 4.1)

unless I misinterpreted your diagram

 

[Alazor]

Alazor, I assumed he forgot the = sign, I didn't verify the question =p

For number 23, the bike doesn't have to go all the way to the finish line.....it could stay 9-12 km away and drop off someone there who could walk the rest of the way while you pick up the person who started walking in the first place.....this should save you enough time to change the answer to B ( 3.9 < t < 4.1)

unless I misinterpreted your diagram

Yeah, the spacing didn't show in the diagram. The last car and variables are stretched too far to the right. It makes sense to keep one person walking all the way. The rate of change is greater with the car vs. walking, so you want to use the car.

To do number 23 quicker, you can add the lowest number 379 with the highest number 406 and add the them up. (379+406, 380,+405, 381+404, etc.) Do that with the next highest and the next lowest and you should get the same number. Count the total amount of those numbers (how many numbers from 379-406?)
If it's odd # of numbers, you need to find the median number. Divide the number of #s by 2 and multiply by (379+406). And if it's an odd amount of #s, then add the median #. That's a similar type of problem that Gauss did in kindergarten.

 

[sdrawkcab]

Yeah, the spacing didn't show in the diagram. The last car and variables are stretched too far to the right. It makes sense to keep one person walking all the way. The rate of change is greater with the car vs. walking, so you want to use the car.

I don't see how it would be faster to take one person all the way one the bike rather then dropping him off early?