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Math Problems, Help?

E

Ender

Golden Member
Can someone solve these basic math problems? I got them from a friend who couldn't solve any either.

1) A pyramid has a square base which has an area of 1440cm². Each of the pyramid's triangular faces is identical and each has an area of 840cm². The height of the pyram is...

2) Troye and Daniella are running at constant speeds in opposite directions around a circular track. Troye completes one lap every 56 seconds and meets Daniella every 24 seconds. How many seconds does it take Daniella to complete one lap?

3)In how many ways can a, b, c, and d be chosen from the set {0, 1, 2,.... 9} so that a<b<c<d and a+b+c+d is a multiple of three?

Before anyone asks, these aren't homework questions. They're from a math contest here in Canada.

Thanks.
 
I think we need a Homework forum up here. Maybe CallTheFBI/HappyClown could be the tutor for the math class questions.
rolleye.gif
 
Originally posted by: conjur
I think we need a Homework forum up here. Maybe CallTheFBI/HappyClown could be the tutor for the math class questions.
rolleye.gif
Click to expand...

As I said, these aren't homework questions. They're math contests questions. To be more specific, the Cayley math contest which is sponsored by the Canadian Mathematics Competition 🙂
 
For the first one, the pyramid, i got about 40.....not 100% sure if i did it right, if anyone else can check my answer, i'll be moving on the the next one....
 
Originally posted by: Lithium381
For the first one, the pyramid, i got about 40.....not 100% sure if i did it right, if anyone else can check my answer, i'll be moving on the the next one....
Click to expand...

Wow, that's actually one of the answers given... can you elaborate a bit on how you got it? It would be awesome if you did. I've gotten up to the part where I calculate slant height to be 14sqrt10 and side length of base is 12sqrt10... I can't get past that, have't taken math 10 in a year and I forgot all about radicals 😱
 
hm isn't the first one done like bh = 1440cm^2, and 2B+2H = something, and then you make them equal some how to get the sides of the square base. then im sure you do the same for the 1/2bh = 840cm^2
 
1)
Base of pyramid each side is sqrt(1440) cm
the height of each face is. 840*2/base = 1680/sqrt(1440)
Pyth. theorem
height of pyramid is sqrt ( height of face^2 - (base side/2)^2) = sqrt ( 1680^2/1440 - 1440/4)= 40

 
Originally posted by: Merkon
Originally posted by: Lithium381
For the first one, the pyramid, i got about 40.....not 100% sure if i did it right, if anyone else can check my answer, i'll be moving on the the next one....
Click to expand...

Wow, that's actually one of the answers given... can you elaborate a bit on how you got it? It would be awesome if you did. I've gotten up to the part where I calculate slant height to be 14sqrt10 and side length of base is 12sqrt10... I can't get past that, have't taken math 10 in a year and I forgot all about radicals 😱
Click to expand...

Basicly here's what i did:

the base is a square, it's area is 1440cm², so i square rooted that.......got 38cm per side

i carried that over to the area of the side panels, given an area of 840²......the formula for area of a right angle triangle is 1/2(b)H if you split the side down the middle, you get two(2) right angle triangles, so ignore the 1/2 thing.... (b)H = 840cm² and we know B is 19(from the half the side of the base) so solve for H, which gives us 44cm for the total height of the side.

take that and use the pythagorean theorum for the overall triangle, when you know that the base is 19(half) and the hypotonuse is 44....solve and you get about 40(i rounded some numbers)

there you have it, if you want i can make a diagram or something....hope that helped
 
Originally posted by: SuperTool
1)
Base of pyramid each side is sqrt(1440) cm
the height of each face is. 840*2/base = 1680/sqrt(1440)
Pyth. theorem
height of pyramid is sqrt ( height of face^2 - (base side/2)^2) = sqrt ( 1680^2/1440 - 1440/4)= 40
Click to expand...

Thanks! I think that makes reasonable sense.
 
I got the second one. Let us say that Troye's velocity is vt and Daniella's velocity is vd. Let the units be m/s. Let r be the radius of the circle. vt * 56 = 2*pi*r

Also in 24 seconds, the sum of the distance travelled by both will equal 2*pi*r.

So (vt+vd)*24 = 2*pi*r. Substituting vt = (2*pi*r)/56.

We get (2*pi*r*24)/56 + vd*24 = 2*pi*r

Divide throughout by 2*pi*r. We get

24/56 + 24vd/(2*pi*r) = 1

Now remember, 2*pi*r/vd is the time taken by Daniella to complete the lap. So rearrange to get,

(2*pi*r)/vd = 56*24/32

Time = 42 seconds

 
These questions are easy. Just use simple algebra and geometry. Draw a picture and label what you know, and go from there. It really isn't hard at all.

Now if you want to find the first few roots to (1/2 + i*sqrt(3)/2) ^ (1/sqrt(17)), thats a different story.
 
Originally posted by: SuperTool
2)

1/(1/24-1/56) =42
Click to expand...

Can you elaborate a bit?


Originally posted by: Lithium381
Originally posted by: Merkon
Originally posted by: Lithium381
For the first one, the pyramid, i got about 40.....not 100% sure if i did it right, if anyone else can check my answer, i'll be moving on the the next one....
Click to expand...

Wow, that's actually one of the answers given... can you elaborate a bit on how you got it? It would be awesome if you did. I've gotten up to the part where I calculate slant height to be 14sqrt10 and side length of base is 12sqrt10... I can't get past that, have't taken math 10 in a year and I forgot all about radicals 😱
Click to expand...

Basicly here's what i did:

the base is a square, it's area is 1440cm², so i square rooted that.......got 38cm per side

i carried that over to the area of the side panels, given an area of 840²......the formula for area of a right angle triangle is 1/2(b)H if you split the side down the middle, you get two(2) right angle triangles, so ignore the 1/2 thing.... (b)H = 840cm² and we know B is 19(from the half the side of the base) so solve for H, which gives us 44cm for the total height of the side.

take that and use the pythagorean theorum for the overall triangle, when you know that the base is 19(half) and the hypotonuse is 44....solve and you get about 40(i rounded some numbers)

there you have it, if you want i can make a diagram or something....hope that helped
Click to expand...


Thanks for elaborating, makes sense!
 
Originally posted by: Marty
These questions are easy. Just use simple algebra and geometry. Draw a picture and label what you know, and go from there. It really isn't hard at all.

Now if you want to find the first few roots to (1/2 + i*sqrt(3)/2) ^ (1/sqrt(17)), thats a different story.
Click to expand...

easy for some, not for those who havne't taken math in a billion years, though this SHOULD be easy for me, though I'm only in trig now
 
I get an answer of 29.66 cm^2 for the pyramid's height. You need to remember that a pyramid is 3-D, and that the height of the triangle that comprises a face IS NOT the height of a pyramid.

EDIT: Need to do some corrections. Here is how I did it... First, get the length of a single side of the base from the area of the base. Second, get the "height" of the triangle comprising a single side of the pyramid. Do this knowing the area of the triangle and applying 1/2BH = 1/2A where A is the total area of a side, and B is one half the length of a single side of the base. Use H and one half the length of a single side of the base to find the length of the diagonal edge of the pyramid. This is the line originating at a corner of the base and ending at the tip of the pyramid. Use this diagonal as well as the half the length of a diagonal of the base to find the height of the pyramid.

Ryan
 
a+b+c+d =3x and a<b<c<d while a,b,c,d are all between 0-9, then a+b+c+d could only be 6, 9, 12, 15, 18, 21, 24, 27, 30 because when a+b+c+d=6, {a,b,c,d) must be {0,1,2,3}, which makes the smallest 3x. When a+b+c+d=30, {a,b,c,d}must be {6,7,8,9}, which makes the largest 3x possible.
When a+b+c+d=6, there's 1 combination
=9, 3 combinations
=12, 9 combinations
=15, 14 combinations
=18, 18 combinations
=21, 14 combinations
=24, 9 combinations
=27, 3 combinations
=30, 1 combinations
so 1+3+9+14+18+14+9+3+1=72 ways to arrange a,b,c and d.

How is it a "possible answer"? what are other possible answers?
 
Originally posted by: newt
a+b+c+d =3x and a<b<c<d while a,b,c,d are all between 0-9, then a+b+c+d could only be 6, 9, 12, 15, 18, 21, 24, 27, 30 because when a+b+c+d=6, {a,b,c,d) must be {0,1,2,3}, which makes the smallest 3x. When a+b+c+d=30, {a,b,c,d}must be {6,7,8,9}, which makes the largest 3x possible.
When a+b+c+d=6, there's 1 combination
=9, 3 combinations
=12, 9 combinations
=15, 14 combinations
=18, 18 combinations
=21, 14 combinations
=24, 9 combinations
=27, 3 combinations
=30, 1 combinations
so 1+3+9+14+18+14+9+3+1=72 ways to arrange a,b,c and d.

How is it a "possible answer"? what are other possible answers?
Click to expand...

Well they are all multiple choice, but if I told you that I would fear you would just make up some crap that has to do with one of the answers.

 
Originally posted by: Merkon
Originally posted by: newt
a+b+c+d =3x and a<b<c<d while a,b,c,d are all between 0-9, then a+b+c+d could only be 6, 9, 12, 15, 18, 21, 24, 27, 30 because when a+b+c+d=6, {a,b,c,d) must be {0,1,2,3}, which makes the smallest 3x. When a+b+c+d=30, {a,b,c,d}must be {6,7,8,9}, which makes the largest 3x possible.
When a+b+c+d=6, there's 1 combination
=9, 3 combinations
=12, 9 combinations
=15, 14 combinations
=18, 18 combinations
=21, 14 combinations
=24, 9 combinations
=27, 3 combinations
=30, 1 combinations
so 1+3+9+14+18+14+9+3+1=72 ways to arrange a,b,c and d.

How is it a "possible answer"? what are other possible answers?
Click to expand...

Well they are all multiple choice, but if I told you that I would fear you would just make up some crap that has to do with one of the answers.
Click to expand...


Math is fun.....the third question is more of logic though
 
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