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Re: Ben has 30 pencils in a box. Each of the pencils is one of 5 [#permalink]
I did not understand any of the solutions here. Hoping someone explain in a more elucidated way.
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Re: Ben has 30 pencils in a box. Each of the pencils is one of 5 [#permalink]
This is a very nice sum. You can solve it without lifting a pen.

So Ben has \(30\) pencils with \(5\) different colors so \(6\) pens of each color.

We need the least no of pens so he has at least \(2\) pencils of each color.

Consider this case:

Ben picks all pencils for \(4\) colors which will be \(6 * 4 = 24\) and then for the last color he didn't lift any.

Now when he picks \(2\) pencils of the last color, he will have \(6\) pencils each of \(4\) colors and \(2\) pencils for the last color. So total \(26\)

\(26\) is the min no of pens he can lift so that he can have \(AT\) \(LEAST\) \(2\) pencils of each color.

Answer C

COolguy101 wrote:
I did not understand any of the solutions here. Hoping someone explain in a more elucidated way.
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Re: Ben has 30 pencils in a box. Each of the pencils is one of 5 [#permalink]
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