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The answer is 0.05. I used algebra. But my friends say, why not 0.10, and they also say, it can be that the hammer is 1.04 and the nail 0.06. How do I tell them that 0.05 is the definite answer, nothing else; I need mathematical evidence and proof to convince folks.

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    $\begingroup$ Just hammer it into their head, and if they refuse to understand accuse them of being dumb as a doornail. $\endgroup$ – user139388 May 9 '14 at 19:03
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    $\begingroup$ That's a very affordable hammer, and an expensive nail. $\endgroup$ – Darth Egregious May 9 '14 at 20:15
  • $\begingroup$ Yes,an ideal world that. $\endgroup$ – Ali Gajani May 9 '14 at 20:26
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It seems that your friend is only solving one equation at once.

Let $h$ be the price of a hammer. Let $n$ be the price of a nail. Can you form any equations?

If a hammer and a nail cost $1.10$ then $h+n = 1.1$

If a hammer costs one more dollar than a nail then $h=1+n$

You now have two equations that are both true at the same time.

Solve these two equations simultaneously, and you will find there is a unique value for $h$ and $n$.

It seems that your friend is only solving one equation at once. If $h+n=1.1$ then the hammer could be free and the nail cost $1.10$. The hammer could cost $0.10$ and the nail $1.00$. The hammer could cost $0.21$ and the nail $0.79$.

There are $110$ solutions to $h+n=1.1$ We could have $$h=0, \ 0.01, \ 0.02, \ 0.03, \ \ldots, \ 1.08, \ 1.09, \ 1.10$$

There are infinitely many solutions to $h=1+n$. We could have, for example,

$$h=0,1,2,3,\ldots$$

However, there is only one solution to $h+n=1.1$ and $h=1+n$, namely $$h=1.05 \ \ \text{ and } \ \ n=0.05$$

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  • $\begingroup$ Right on the money, the graph hint was good. :) $\endgroup$ – Ali Gajani May 9 '14 at 18:39
  • $\begingroup$ @AliGajani Thanks for your kind words. I'm pleased I could help :) $\endgroup$ – Fly by Night May 9 '14 at 18:43
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Show your friends that their proposed alternate solutions don't fit the bill.

If the nail is 10 cents, then the hammer is a dollar more: $\$1.10$, and then the total is $\$1.20$.

If the nail is 6 cents and the hammer is $\$1.04$, then that is not a dollar more for the hammer; it's 98 cents more.

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So, (1) Hammer+Nail =$1.10

(2) Hammer-Nail = $1

From equation (1)

Let Hammer = 1.10 - nail

So, Substitute Hammer into equation 2:

1.10 - nail - nail = 1

1.10-1 = 2 nails

0.1 = 2 nails

1 nail = $0.05

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Ali, ask your friends how much more the hammer cost than the nail IF the nail cost $.10?

Or, IF the nail cost $0.06?

They should not get the $1.00 more that the problem requires.

That resolves their alternative answers to your correct answer.

How do you know that the answer is unique? I.e., "0.05 is the definite answer, nothing else; "

Draw the graph (lines) of the two equations you solved with algebra, H + N = 1.10 and H - N = 1.00, and show them the one point where both equations are true.

Definite, nothing else makes those two equations true at the same time.

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The answers above have already clarified your question. However I would like to add that when this question is often asked, most people would reply 10 cents instead of 5 cents (5 cents is correct). Because we instinctively subtract the two numbers without much thought as to what the question was. Tell them to read the question again as it says "The hammer costs 1\$ more than nail". Not "The hammer costs 1\$."

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5 cents If you quickly blurt out “10 cents”, you’re not paying attention. (If the nail cost 10 cents, the hammer would cost a dollar, which would make it 90 cents more expensive than the nail.) Here is how you solve this. From the given information we can write; h + n = 1.1 ….(a) and also h = 1 + n ……(b) 1 + n + n = 1.1 Solving the equation we get; 2n = 0.1 Hence; n = 0.05 Source: http://www.bhavinionline.com/2014/12/morning-trivia-find-price-hammer-nail/

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Given that H+N = 1.1 and H = N+1 we can solve for N for both. H = 1.1-N AND H = 1+N so 1.1-N = 1+N, then 1.1 = 1+2N, so 0.1 = 2N, therefore .1/2 = N, thus .05 = N. Substitute .05 for N either original equation and H = 1.05.

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    $\begingroup$ What does this answer contribute that the other answers do not? $\endgroup$ – HDE 226868 Dec 10 '14 at 21:51

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