bonanova Posted July 31, 2008 Report Share Posted July 31, 2008 On a good set of scales, the pans have equal mass, and the arms have equal length. Down at Ralph's Pretty Good Grocery Store, the scales weren't all that good. The pans had equal weight: each weighed m pounds. But the arms had different lengths - the left arm was a cm long and the right arm was b cm long. When Ralph was asked to weigh out 2 pounds of grapes, he had an idea. First, he balanced some grapes in the left pan against a 1-pound weight in the right pan. Then, he balanced some other grapes in the right pan against a 1-pound weight in the left pan. Together, Ralph reasoned, the grapes would weigh 2 pounds. Was Ralph correct? Is there another way? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 31, 2008 Report Share Posted July 31, 2008 On a good set of scales, the pans have equal mass, and the arms have equal length. Down at Ralph's Pretty Good Grocery Store, the scales weren't all that good. The pans had equal weight: each weighed m pounds. But the arms had different lengths - the left arm was a cm long and the right arm was b cm long. When Ralph was asked to weigh out 2 pounds of grapes, he had an idea. First, he balanced some grapes in the left pan against a 1-pound weight in the right pan. Then, he balanced some other grapes in the right pan against a 1-pound weight in the left pan. Together, Ralph reasoned, the grapes would weigh 2 pounds. Was Ralph correct? Is there another way? No, Ralph is not right. A while since I touched algebra but one way is to put a 2-pound weight on, say, the left pan and then add weights to the right pan until the scales are level. Then remove the 2-pound weight from the left pan and replace with grapes until the scales are again level. Ralph may also want to take the opportunity at this point to note the weight ratio between the two pans for future grape weighing. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 31, 2008 Report Share Posted July 31, 2008 this answer assumes that ralph knows the numerical values of a and b (he could just measure them) and that he can use any amount of mass to balance against. to balance, (m+m1)*a=(m+m2)*b where m1 and m2 are the masses in the left and right plates respectively. rearranging gives m2=(m(a+b)+m1*a)/b so if you set m1=2 then m2=(m(a+b)+2a)/b is the amount of mass that you need to put in the right plate to balance two pounds of grapes in the left plate Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 31, 2008 Report Share Posted July 31, 2008 (edited) As others have pointed out, the Ralph's method won't work, but he could simply: Put two pound masses on the shorter arm and add sufficient mass to the longer arm so that it balances (using grapes or scale weights, it doesn't matter which) Leaving the long arm alone, take off the 2 pounds and add enough grapes to balance it. The grapes added must equal the 2 pounds it has replaced. Edit: Oops. just saw leamboy already got there. Oh well! Edited July 31, 2008 by foolonthehill Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 31, 2008 Report Share Posted July 31, 2008 don't we also need to know what m is? my algebra is not good enough to work out if this can be done without resorting to practical application (i.e. start putting weights on the scale) but i'd like to know. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 31, 2008 Report Share Posted July 31, 2008 Griffon: your math is not correct. The basic equation is m1*a=m2*b, m1 is mass one, m2 is mass two, "a" is arm one and "b" is arm two. Basically, m2=m1*a/b. If we have a common mass on both sides "m" then we have a*(m+m1)=b*(m+m2). Solving for m2 would give: a*(m+m1)/b -m. Note the -m is a lone term on the right and can not be "combined" into the first term. (What I wouldn't give for MathML right now.) So, if you knew the ratio of a to b, then you would put a coorespondingly heavier (or lighter) weight on the scale and balance it with grapes on the other side. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted August 1, 2008 Author Report Share Posted August 1, 2008 leamboy and foolonthehill have the correct method. Bonus question: Ralph's method [see OP] was wrong, but was it wrong high? or wrong low? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 1, 2008 Report Share Posted August 1, 2008 I so enjoy the maths puzzles. Have never been a maths person but do enjoy the puzzles. If I can't figure out the answers its nice to be able to watch the forum to be able to figure it out once the puzzle has been answered. You guys are so fast at finding the answer. I look at the question then go away and try to work it out, then when I get back to the forum it has been solved. I don't post as I just feel like a copycat. Maybe one day I will be clever enough to post one of my own. Well done. Maybe it will be my turn next time to solve it first. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 1, 2008 Report Share Posted August 1, 2008 (edited) Naming the grape amounts as g1 and g2: g1 + m = (m+1)b/a g2 + m = (m+1)a/b and a bit of algebra gets us to g1 +g2 = (m+1)(a-b)^2 / ab + 2 clearly, if a=b (so the scales are 'good'), then the first term disappears and g1 + g2 = 2, but otherwise, (m+1)(a-b)^2 is positive, so he clearly it's in his financial interests to buy better scales! (edit: trying to format my formula) Edited August 1, 2008 by foolonthehill Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 1, 2008 Report Share Posted August 1, 2008 Is there another way? Yeah Ralph should buy a new set of scales... cause if he's anything like me, it would take him too long to work out! Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted August 1, 2008 Author Report Share Posted August 1, 2008 I so enjoy the maths puzzles. Have never been a maths person but do enjoy the puzzles. If I can't figure out the answers its nice to be able to watch the forum to be able to figure it out once the puzzle has been answered. You guys are so fast at finding the answer. I look at the question then go away and try to work it out, then when I get back to the forum it has been solved. I don't post as I just feel like a copycat. Maybe one day I will be clever enough to post one of my own. Well done. Maybe it will be my turn next time to solve it first. Hi tearz, Feel free to post your answers, even if you're not first the the finish line. Everyone approaches the solution in his/her own way, so if you outline how you did it, no need to worry about being thought of as a copycat. We're on the honor system here, anyway. No glory in copying... Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted August 1, 2008 Author Report Share Posted August 1, 2008 Yup, altho it might contribute to customer satisfaction. Hint: in edit, you can use the aA button and choose superscript 2 rather than ^2 Naming the grape amounts as g1 and g2: g1 + m = (m+1)b/a g2 + m = (m+1)a/b and a bit of algebra gets us to g1 +g2 = (m+1)(a-b)^2 / ab + 2 clearly, if a=b (so the scales are 'good'), then the first term disappears and g1 + g2 = 2, but otherwise, (m+1)(a-b)^2 is positive, so he clearly it's in his financial interests to buy better scales! (edit: trying to format my formula) Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 1, 2008 Report Share Posted August 1, 2008 Hint: in edit, you can use the aA button and choose superscript 2 rather than ^2 Thanks for the tip, though at work, my IT department strips out most javascript, so the side panel in edit doesn't work for me - any idea of whether there's a tag I can use to get superscripts? Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted August 1, 2008 Author Report Share Posted August 1, 2008 Thanks for the tip, though at work, my IT department strips out most javascript, so the side panel in edit doesn't work for me - any idea of whether there's a tag I can use to get superscripts? a squared can be coded as a(sup)2(/sup). but use [] brackets instead of () to get a2 Others: a(sub)2(/sub) gives a2 (s)delete this(/s) gives delete this Colors: (color=#FF0000")red(/color) comes out red Lists: (list=1)(*)one(*)two(*)three(/list) gives ordered lists:onetwothree(list)(*)bullet(*)bullet(*)bullet(/list) gives unordered lists bulletbulletbulletSpoilers: (spoiler="here's a hint to be hidden")hide this text(/spoiler) does this: hide this textHope that helps. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 2, 2008 Report Share Posted August 2, 2008 a squared can be coded as a(sup)2(/sup). but use [] brackets instead of () to get a2 Others: a(sub)2(/sub) gives a2 (s)delete this(/s) gives delete this Colors: (color=#FF0000")red(/color) comes out red Lists: (list=1)(*)one(*)two(*)three(/list) gives ordered lists:onetwothree(list)(*)bullet(*)bullet(*)bullet(/list) gives unordered lists bulletbulletbulletSpoilers: (spoiler="here's a hint to be hidden")hide this text(/spoiler) does this: hide this textHope that helps. Thanks - these help me too!!! Quote Link to comment Share on other sites More sharing options...
Question
bonanova
On a good set of scales, the pans have equal mass, and the arms have equal length.
Down at Ralph's Pretty Good Grocery Store, the scales weren't all that good.
The pans had equal weight: each weighed m pounds.
But the arms had different lengths - the left arm was a cm long and the right arm was b cm long.
When Ralph was asked to weigh out 2 pounds of grapes, he had an idea.
First, he balanced some grapes in the left pan against a 1-pound weight in the right pan.
Then, he balanced some other grapes in the right pan against a 1-pound weight in the left pan.
Together, Ralph reasoned, the grapes would weigh 2 pounds.
Was Ralph correct?
Is there another way?
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