Challenge:Balancing problem

I am not sure if this is a popular question but I hope you will respect my wishes and not Google it. If you have already heard it,either move on,or post a hint,or add a followup question.

You have $5$ boxes of coins. $4$ of them contain $10$ gm coins and $1$ of them contains $11$ gm coins.Assuming you can weigh any weight using a balance,can you spot the box with the $11$ gm coins by weighing only once?If yes,how?

Details and Assumptions: You can take coins out of the boxes if you like.

#Algebra #Logic #Balancingproblem #Strategy

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

That's very simple. If suppose the weight of the box weighed is 10 gm, then all the four boxes weighing 10 gm coins have equal number of coins. Just take the coins out of each box and count them.

Deepak Sharma - 7 years, 4 months ago

Can you explain more clearly what you mean?Also,I never said that the number of coins must be equal.Don't make unwarranted assumptions.

Rahul Saha - 7 years, 4 months ago

By saying "You can take coins out of the boxes if you like", do you mean that we can take them out while they're being balanced? Obviously that'd make it damn easy. Explain that statement!

Parth Thakkar - 7 years, 3 months ago

Can I assume the boxes have no mass?

Jihoon Kang - 5 years, 10 months ago

You can, I suppose. Alternatively, you can assume the boxes have equal mass.

Rahul Saha - 5 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$