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**Pete1967****Member**- Registered: 2014-04-08
- Posts: 2

I came to an easier solution to the following problem (Sam's Loyd):When they started off on the great annual picnic every wagon in town was pressed into service.

Half way to the picnic ground ten wagons broke down, so it was necessary for each of the remaining wagons to carry one more person.

When they started for home it was discovered that fifteen more wagons were out of commission, so on the return trip there were three persons more in each wagon than when they started out in the morning.

Now who can tell how many people attended the great annual picnic?

To me: 50 wagons with 4 people in each wagon. 10 break down so the remaining 40 sees 5 people in each (4+1), then another 15 break down so 15 x 5=75 people, divided amongst the twenty five wagons left, (75/25)=an extra three in each wagon.

Or is my logic wrong somewhere?

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,400

Hi, Pete1967, and welcome to the forum!

Yes, there is an error in your logic.

After the second breakdown you added 3 more persons to number of persons in each wagon that remained after the first breakdown, in contradiction of "than when they started out in the morning".

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**Pete1967****Member**- Registered: 2014-04-08
- Posts: 2

thank you phrontister. I'm new at this. I guess I'm going to have to pay attention to the small print in the future.

Thank you gain

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