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We have 2 pieces of metal weighing 1 kg and 2 kg. From these pieces they made 2 other pieces. 0.5 kg which is 40% copper and 2.5 kg which is 88% copper. What percentage of copper is in the original pieces?

user170231
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Gigi Mzhavanadze
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  • We know that total weight of copper im both of them is 2.4 kg so total percent is 80% and x%+2y%=240 but it still doesnt show how can i get percentage in each of them – Gigi Mzhavanadze Oct 08 '23 at 19:49
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    not enough info is provided, you can calculate that you have 2.4 kg of copper, but you cannot know how that is distributed between the original pieces. – WW1 Oct 08 '23 at 19:50
  • There is actually enough information, my math professor did it by testing like 10 possible answers and solved a system for every of them, the answer is 40% and 100%, i have the answer but dont know how to solve it – Gigi Mzhavanadze Oct 08 '23 at 19:54
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    How did they test 10 possible "answers" when you claim there's only one? – user170231 Oct 08 '23 at 20:01
  • Consider the minimum copper content of the 1 kg piece. – Daniel Mathias Oct 08 '23 at 20:01
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    Was the possible answer in which both of the original pieces had 80% copper tested? If so, what is wrong with that? – WW1 Oct 08 '23 at 20:02
  • @WW1 If both of the original pieces have more than 40% copper, any part of either will also have more than 40% copper. – Daniel Mathias Oct 08 '23 at 20:05
  • The 0.5 kg one was half of 1 kg one which had 40% of copper and 2.5 kg one was other 0.5 40% copper one and whole 100% 2kg – Gigi Mzhavanadze Oct 08 '23 at 20:12

2 Answers2

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Apparently, we cannot just melt both original pieces, separate all the copper and recombine them into two new pieces.

We have to accomplish this by cutting both pieces and melting them in pairs together

Let $x$ be the smaller of the pieces into which the 1 kg piece is cut, then we have the restriction $ 0\le x \le 0.5$

The piece of mass $x$ must be combined with a piece of mass $(0.5-x)$ coming from the 2 kg piece.

following @true blue anil, let the original percentages be $a$ and $b$ respectively,

then $a+2b = 2.4 $

We must have $0 \le a,b \le 1$

The copper percentage in the 0.5 kg is 0.4 so.. $$ ax+b(0.5-x)=0.5 (0.4)=0.2 $$

solving for x... $$x = \frac{0.5b-0.2}{b-a}$$

but $a = 2.4-2b$

$$x = 0.5 \bigg( \frac{b-0.4}{3b-2.4} \bigg )$$

so the condition that $x \le 1$ is equivalent to $$b-0.4\le 3b-2.4$$ Which is equivalent to $b\ge 1$ , but since we also have $b\le 1$, the only solution is $b=1$ and $a=0.4$

WW1
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Assumption: $\mathtt{The\; two\; alloy\; pieces\; can\; be\; melted,\; and\; the\; copper\; redistributed\; as\; needed}$

There is not enough information, and the Professor could not have solved it to get an $\mathtt{unconditional unique}$ answer

We shall have to set some conditions, and then a unique answer can be found. Suppose, for example, we set two conditions:

[i] the percentages in the original two pieces are whole numbers

[ii] the percentages are as close to each other as possible

then if the original percentages were $a$ and $b$ respectively, we get $a+2b = 0.5*40 + 2.5*88 =240 \tag1$

and we get the solution $a = b = 80\%$

This didn't need solving $10$ equations, but we don't know what criteria the Professor used !

Added

The criteria chosen has given a very neat solution, but with other figures, the two percentages might not have come as exactly equal.

Actually, the first criterion makes it a Diophantine equation, with many solutions starting with $a=2,b=119$ and going upto $a=100, b=70$, so maybe the Professor had only used the first criterion, and was trying to see which combo he liked !

true blue anil
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  • With $a=b=80%$, any way you divide the original pieces will result in the final pieces each being $80%$ copper. – Daniel Mathias Oct 09 '23 at 01:19
  • @DanielMathias: I have now explicitly stated the assumption I made, as presumably this is what led the Professor to try out many solutions. But I really like WWII's unique solution. Thanks a lot for being my guardian angel, and when you get time would you have a look at my query which may not have reached you at https://math.stackexchange.com/questions/4648839/find-the-number-of-ways-to-arrange-so-that-at-least-two-are-followed/4648874#4648874 – true blue anil Oct 09 '23 at 04:56