I am a newbie. Please help me to solve this problem. I would appreciate your help.

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June 22nd 2006, 08:27 PM
maggie

I am a newbie. Please help me to solve this problem. I would appreciate your help.

One person enters the stadium the first mimute, two people the second mimute, three people the third minute, and so on. In how many mimutes will there be 55 people in the stadium?
June 22nd 2006, 08:40 PM
earboth

Quote:

Originally Posted by maggie

One person enters the stadium the first mimute, two people the second mimute, three people the third minute, and so on. In how many mimutes will there be 55 people in the stadium?

Hello,

you want to know the number of summands in a sum:

$1+2+3+...+n=55\Longrightarrow \frac{1}{2}\cdot n\cdot (n+1)=55$

Expand the LHS of this equation and solve the quadratic equation for n:

$\frac{1}{2}n^2+\frac{1}{2}n-55=0$

You'll get x = -11 or x = 10. (-11) persons are a very ugly sight.
So the only possible solution for your problem is: 10 persons.

Greetings

EB
June 22nd 2006, 08:42 PM
CaptainBlack

Other copy of this deleted

Don't make multiple posts of the same question.

RonL
June 22nd 2006, 09:15 PM
maggie

Thank you for your help. However, my son is 11 years old. I don't think he could understand the fomular you just gave. Do you have another way to show my son to understand easier. Thank you so much.
June 22nd 2006, 10:20 PM
earboth

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Quote:

Originally Posted by maggie

Thank you for your help. However, my son is 11 years old. I don't think he could understand the fomular you just gave. Do you have another way to show my son to understand easier. Thank you so much.

Hello, Maggie,

show your son how to calculate in a table.

I've attached a diagram to show you what I mean.

Greetings

EB
June 23rd 2006, 09:13 AM
maggie

Hello Earboth,

Thank you so much for your help. My son is understanding now. It is much easier for him to understand . Thank you again.