Money box problem

• July 14th 2010, 03:15 AM
claffin
Money box problem
Last Christmas i was given a money box. The deal was if i can fill it up by next Christmas with obviously real money I would get a $1,000 bonus so i thought how would I go about using the least amount of money to fill up the money box. The following mesaurments are all in metric The money box dimensions are 340mm high by 200mm wide. To work out the volume of the money box i did the following 200 / 2 = 100mm (Radius) x 3.14 = 3140mm (area) x 34 = 106,760 (volume) I believe this is right however correct me if im wrong please The next stage was how many 5 cent pieces can fit into that amount (i understand it would not be exact but im not looking for exact just a there about number) the 5 cent piece is 19.41mm wide by 1.3mm thick So i worked out the volume as before 19.41 / 2 = 9.705 (radius) x 3.14 = 30.47mm (area) x 1.3 = 39.62 (volume) So i worked out that to fill up the money box with 5 cent pieces i would do the following volume of money box (106,760) / Volume of coin (39.62) = 2,695 coins therefore to put a$1 value to this i would only need

2,695 * 0.05 = $134.75c of 5 cent pieces to fill up the money box. Is this maths correct because it doesn't seem like that amount of money would fill up this money box. Thanks Chris • July 14th 2010, 03:54 AM running-gag Hi 1) Be careful about the units (mm or cm) 2) The volume of a cylinder is $\displaystyle \pi R^2 h$ and not $\displaystyle \pi R h$ 3) Dividing the volume of the box by the volume of one coin will give you the maximum number of coins but since coins are solid and not liquid you cannot fill the box with this maximum number of coins • July 14th 2010, 04:35 AM claffin so by using new formula to find the volume of the money box is v = 3.14 * (100^2) * 340 v = 3.14 * 10,000 * 340 v = 10,676,000 Volume of 5 cent piece v = 3.14 * (9.705^2) * 1.3 v = 3.14 * 94.187025 * 1.3 v = 384 Therefore 27,802 of 5 cent coins can fit into the money box (if liquid) which means that i would need$1,390.10 in order to fill the money box with 5 cent pieces.

It would obvously be a little less than that as there will be gaps between the 5 cent pieces because they are round.

Can you confirm these findings for me so future people when searching the answer to this money box question will know the answer as i couldn't find the answer to this question anywhere online. Thanks
• July 14th 2010, 04:52 AM
claffin
i decided to work using that formula on all coins with the following measurements

0.05 19.41 1.30 9.71 94.19 384.67 27754 $1,387.70 0.10 23.60 2.00 11.80 139.24 874.87 12203$1,220.29
0.20 28.52 2.50 14.26 203.35 1597.09 6685 $1,336.93 0.50 31.51 2.00 15.76 248.22 1559.61 6845$3,422.65
1.00 25.00 3.00 12.50 156.25 1472.62 7250 $7,249.66 2.00 20.50 3.20 10.25 105.06 1056.20 10108$20,215.80
• July 14th 2010, 07:30 AM
running-gag
Using real values (not rounded off) I can find something very close to you

0.05 19.41 1.30 9.71 94.19 384.67 27768 $1,388.40 0.10 23.60 2.00 11.80 139.24 874.87 12209$1,220.90
0.20 28.52 2.50 14.26 203.35 1597.09 6688 $1,337.60 0.50 31.51 2.00 15.76 248.22 1559.61 6848$3,424.00
1.00 25.00 3.00 12.50 156.25 1472.62 7253 $7,253.00 2.00 20.50 3.20 10.25 105.06 1056.20 10113$20,226.00

When you divide the volume of the box by the volume of one coin, you must take the entire part of the result
For instance if you find 12209.1 then the result is 12209
• July 14th 2010, 09:59 AM
SpringFan25
do you have to use american money?
• July 14th 2010, 04:40 PM
claffin
The currency i used was Australian coins. I guess you could use any currency you can find out what the dimensions are on wikipedia.