Pans Problem
I have two pans similar in shape and thickness. However one of them is eight times more capacious then the other one.
Can you tell me how heavier is the bigger one than the other ?
Can you tell me how heavier is the bigger one than the other ?
Labels: mathemagic, thinktank
posted by Rajesh Lal at
8:29 AM
23 Comments:
I'm not very good at mathematics so to make the problem easier for me I will assume square pans. I assume the first pan is a cube with the top open that is one foot long on each side (one cubic foot). Each side is one foot square and there are five sides so the total area of the pan material is five square feet.
The larger pan is two feet long on each side (eight cubic feet) so each side is four square feet and there are five sides so the total area of the pan material is twenty square feet.
The material is of equal thickness, so the larger pan is four times heavier than the smaller pan.
I think you have the right thinking but i think there might be a mistake in your math
the only thing i find wrong with Black Knights, is that i've never seen a cube-shaped pan in my life
The first pan/pot is a 1qt sauce pan, the other is an 8qt soup pot.
Both are round and have the same thickness of walls.
Their weight is equal to their size.
The pot that has eight time more capacity is also eight times heavier than it smaller counter-part.
That ius, as long as they are empty.
If the pans have identical wall thicknesses, the pan which is 8 times more capacious will be 8 times heavier.
I assumed flat type pans, circular in shape, similar to a cookie sheet, with identical wall thickness. For me, capacity is simply the surface area of the pan. So, I assumed a radius and thickness for the small pan and a density of the material. I then calculated the surface area, then the volume (by multiplying by the thickness for this type of pan), and finally the weight.
For the large pan, I did the same thing, but varied the radius to get a surface area 8 times larger. Then, by taking the ratio of the two weights, I have the "number of times heavier the large pan is than the smaller pan".
This number turned out to be 8. Thus, with a little bit of unnecessary work, one can actually prove that this is the case, given some reasonable assumptions.
Try it for yourself for any shape of pan!
what if the pans are people named pan??? i dont no im just guessing!
- Capacious is related to volume = cubic / power of 3.
- Since both pans have the same wall thickness, we are dealing with wall area = square.
- If vol 1= 8xVol 2 =2^3.Vol 2,
Area 1 = 2^2 x Area 2.
So, I agree with Black Knight.
if you want a pan 8x capacity you need a pan that's 2x higher, 2x wider, and 2x longer (if they are the same shape). because of this, any dimension doubles so the surface area quadruples. the 8x pan will be 4x heavier with the same thickness.
To hold 8x more one would only have to make the bottom of the pan 8x larger. The sides would stay the same height.
If the bottom weighs b and the sides weighs s then the weight of the larger pan would be 8b+s.
jp, if he was talking about people named pan, he probably would have capatalized it, don't you think?
Paricks answer sounds the most logical!!! Volume is not directly proportional to weight!!!
The big pan is twice as big as the small pan
The pans are the same shape. So you can't have the base increase 8x and the side not increase, it has to be 2x width, 2x lenght, 2x height
Patrick, aren't you just too sexy with that Algebra!! Love it. Reminds me of school...sigh.....
jasper
Is the answer simple? Is it 7 times heavier?
right we want 2 pans. one holds 8 times more, this means it is 8 times the size, so u'd think it would need to be eight times the weight, but that is where your wrong. if i have 2 cubes one 1cmx1cmx1cm it holds 1cm^3 right so i want it to hold 8cm^3
well 2x1x1 would hold 2cm^3
2x2x1 would hold 4cm^3
so i want it to be 2cmx2cmx2cm which holds 8cm^3
so the weight would have to be four times the weight as it uses four times as much metal on each side.
ha i didnt think i would get that :D
sorry didnt show any logic there at all, i need paper and a pen.
draw two cubes one 1cmx1cmx1cm count the cubes it makes
now draw another one 2cmx2cmx2cm, now seperate it into smaller cubes of 1cmx1cmx1cm, how many have you got?
now how much bigger is each side
and if each side is the same thickness, what should happen to the weight???
sanket i no but... i was to stunned to say anything smart!
sanket i no but... i was to stunned to say anything smart!
sanket i no but... i was to stunned to say anything smart!
Black Knight is right, even if we persume a circular bottom, or any other shaped bottom for that matter. Here's the maths for a circular bottom:
If the small pan is cylinder shaped with a radius r and a height h, then it's volumne V1 = (pi)r²*h and it's surfice area A1 = (pi)r²+2(pi)r*h. By doubling the dimensions (r -> 2r; h -> 2h), we get a pan with volume and surfice of
V2 = (pi)(2r)²*(2h) = (pi)4r²*2h = 8(pi)r²*h = 8*V1
A2 = (pi)(2r)²+2(pi)(2r)*(2h) = 4(pi)r²+4*2(pi)r*h = 4[(pi)r²+2(pi)r*h] = 4*A1
As the thickness stays the same, surfice area and weigth are equivalent, so the pan with 8 times the capacity is 4 times as heavy. If the thickness were to be doubled also, then you get the 8 times heavier pan.
ANSWER : 4 times heavier
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GOT RIGHT
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Black Knight, Ydrameos , mo
Pans?
Let's suppose they are cylinders, since they are similar, we can assume the depth and radius vary by the same ratio.
If pan 1 has as bottom radius of 6 inches and a depth of 2 inchs, the volume would be 226.18 cu inches.
The bottom and sides would have a surface area of 188.49 sq inches.
For pan2 to have 8 times the volume, and the same radius to depth ratio, it would have a radius of 12 inches and depth of 4 inches. It's surface area would be 753.98 sq inches. This is exactly 4 times the surface area, so the weight would be 4 times the weight of pan 1 since the weight is related to the surface area.
volume = v ; radius = r ; sa=surface area
If the ratio of vol2 / vol1 = 8
then the dimensions of pan2 to pan1 will always be 2 ie. d2/d1 = 2 and r2/r1 = 2 and sa2/sa1 will always be 4.
hjg
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