I have 18 different boxes and the boxes all weigh the same. The boxes are eached filled with different numbers of buttons. All the buttons weigh the same but you don't know how much one button weighs. The goal is to figure out how much one button weighs! Someone please help me or give me a hint. Thank you very very much!!!!!!!!
2007-11-03 06:55:06 · 13 answers · asked by SANDala 1 in Education & Reference ➔ Homework Help
I'm going to assume that you mean that all of the boxes weigh the same when empty. The procedure would be to weigh all of the boxes with the buttons in them and find the smallest difference between two button containing boxes. Then check to see if the other differences are multiples of that difference. If they are then you've found two boxes with a button count difference of only one. If not then divide the smallest difference and try again. You would have a difference of two then. Keep trying.
2007-11-03 07:05:33 · answer #1 · answered by Mike B 5 · 0⤊ 0⤋
I don't think you've given, or been given' sufficient information here. However there is an answer.
Each box weighs the same as all the others. Each box is filled with different numbers of buttons and those buttons, the ones in each box, weigh the same. Therefore if you know the weight of the 18 boxes in total divide that total by 18 then divide the result by the number of buttons in any particular box. The only problem there is the weight of the box sop you need to know how much an empty box weighs and how much one filled with buttons weighs. Take the weight of the empty box away from the weight of the full box and divide the result by the number of buttons. That gives you the weight of one button.
2007-11-03 07:09:14 · answer #2 · answered by quatt47 7 · 0⤊ 0⤋
All boxes weigh the same right? Though, each box has a different number of buttons. Then, shouldn't each button be a different kind and have different weights? It doesn't make sense. Give us some numbers please?
2007-11-03 07:02:08 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Weigh a button... there's your answer.
An afterthought....... if all the boxes contain different numbers of buttons, and all the buttons are the same weight, then how can all the boxes weigh the same?
2007-11-03 07:01:34 · answer #4 · answered by LuLu 4 · 0⤊ 0⤋
ok...to respond to all of those questions you should use what i call the airborne dirt and dust formula- Distance = fee * Time (the place Time is in hours). each and every so often that's good to establish a table that might actually assist you spot it less complicated. a million. D = R * T in this difficulty we'd desire to appreciate while Joe's D = Ken's D so all of us understand joe's time is half-hour and ken time is 20 minutes, yet we'd desire to alter that to hours. D = R* T joe's D= 6 ( a million/2 ) ken's D= x ( a million/3 ) now set those = to eachother and remedy for x 6(a million/2) = x (a million/3) 3 = x/3 9 = x 9 mi/h ck you answer to make sure that's smart... 2. as quickly as lower back, those 2 distances are equivalent, yet time there is 40 minutes longer or 2/3 of an hour longer so if T back is x, then T there is x+2/3 D=RT D there 525 ( x + 2/3) D back six hundred ( x ) set them equivalent and remedy for x...yet right here x is a time and additionally you decide on a distance, so which you will possibly plug your answer back in to the back equation (certainly the two, yet back is easier) to locate distance 525(X + 2/3) = six hundred x 525x + 350 = 600x 350 = 75x 14/3 = x now six hundred (14/3) = 2800 miles aside 3. as quickly as lower back distances are equivalent and additionally you decide directly to appreciate while they meet eachother in fact and garcias time is 20 minutes much less considering all of them started 20 minutes later so if smith's T=x, then garcias' T could be x-a million/3 so smith's D = 40 8 ( x ) garcias' D = 60 ( x - a million/3) set them equivalent and remedy for x and then upload that answer to 8am 48x = 60 (x - a million/3) 48x = 60x - 20 20 = 12 x 20/12 = x 5/3 = x that's a million hour and 40 minutes which could be 9:40 wish this facilitates...the base line is that component is in hours so which you should transform minutes to hours and additionally you will possibly be able to desire to appreciate the "airborne dirt and dust" formula- Distance = fee * Time
2016-10-14 21:47:41 · answer #5 · answered by ? 4 · 0⤊ 0⤋
well to work this question out you need to know some sort of weight to begin with, maybe the combined weight of the 18 boxes, and probably how many buttons there are in total, then you can set up some equations.
2007-11-03 06:58:09 · answer #6 · answered by Anonymous · 1⤊ 0⤋
Weigh 100 buttons, then divide by 100.
2007-11-03 06:57:52 · answer #7 · answered by Renaissance Man 5 · 0⤊ 1⤋
YOU CAN'T KNOW THE WEIGHT OF THE BUTTON IF YOU
DON'T KNOW THE WEIGHT OF EACH BOX FIND THAT OUT AND YOU SHOULD BE ABLE TO FIGURE OUT THE REST
2007-11-03 07:01:58 · answer #8 · answered by LaQuana C 1 · 0⤊ 0⤋
See what other info you can add to this,it's hard to answer this with such limited info.
2007-11-03 06:59:13 · answer #9 · answered by TAMMY 3 · 0⤊ 0⤋
umm that is a good question sorry i couldn't help i am only in 8th grade doing 9th grade math!
2007-11-03 06:58:17 · answer #10 · answered by ? 2 · 0⤊ 1⤋