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Mathematical Problem I - Let's See How You Do With This One! SOLUTION INSIDE 12/25/05
posted Thursday, 3 November 2005
Mathematical Problem I -
Let's See How You Do With This One!
A novelty store has 100 toy animals to sell:
I'm a Toy Cow! I only cost $5 each.
I'm a Toy Horse! I only cost $1 each.
I'm a Toy Sheep! I only cost $.05 each (5-cents).
The total number of toy animals is 100.
The total received in sales for all 100 toy animals is $100.
The question is, "How many of each type of toy animal did the novelty store owner have initially?"
How many cows @$5? How many horses @$1? How many sheep @$.05?
Remember, the total number of toys is 100. Your answer must total 100.
Submit your answer via Comment below. You may identify yourself if you wish or by some pseudonym you select. Please specify how you wish to be identified and whether you want your Comment displayed. No Comment can be displayed until and unless Tabacco approves it.
Either your correct answer Comment, if you wish, or your pseudonym will be published as a Comment by Tabacco.
I will publish the correct answer as your Christmas present for one and all to see.
Good Luck,
T.A.B.A.C.C.O. (Truth About Business And Congressional Crimes Organization)
PS No cheating means no help from the kids. (Smile)
ANSWER TO PROBLEM BELOW THIS LINE:
-------------------------------------------------------------------------------------------------------------------------
SOLUTION TO PROBLEM:
There have been 37 Hits on this problem to date. Audrey answered the problem the day I published it. Bill Amoureux, her ex, must have been pea-green with envy. Nobody else answered the problem.
CONGRATULATIONS TO AUDREY, OUR ONLY WINNER!
So here is the solution:
Sheep are 5 cents each. Therefore only multiples of 20 sheep will give even $-results. The number of sheep can only be 20, 40, 60, or 80. By solving 4 separate conditions we reduce the unknowns to 2 with 2 separate equations. This allows a simple algebraic solution.
1- 20 sheep @5cents = $1
X=Cows @$5
Y=Horses@$1
5X + Y = 99
X + Y = 80 (since there are 20 sheep in this Condition)
Deduct the second equation from the first equation to eliminate the Y-term.
4X = 19 N/A, since X must be an integer! 20 sheep is therefore NOT an answer.
2- 40 sheep @5cents = $2
5X + Y = 98
X + Y = 60 (since there are 40 sheep in this Condition)
4X = 38 N/A, since X must be an integer! 40 sheep is therefore NOT an answer.
3- 60 sheep @5cents = $3
5X + Y = 97
X + Y = 40 (since there are 60 sheep in this Condition)
4X = 57 N/A, since X must be an integer! 60 sheep is therefore NOT an answer.
4- 80 sheep @5cents = $4
5X + Y = 96
X + Y = 20 (since there are 80 sheep in this Condition)
4X = 76
X = 19
If X = 19, Y= 1…QED!
Thus the only Solution to the Math problem is:
19 cows @ $5 each
1 horse @ $1 each
80 sheep @ $.05 each
This yields 100 animals @ $100 total price.
Solution published December 25, 2005 by Tabacco
Let's See How You Do With This One!
A novelty store has 100 toy animals to sell:
I'm a Toy Cow! I only cost $5 each.
I'm a Toy Horse! I only cost $1 each.
I'm a Toy Sheep! I only cost $.05 each (5-cents).
The total number of toy animals is 100.
The total received in sales for all 100 toy animals is $100.
The question is, "How many of each type of toy animal did the novelty store owner have initially?"
How many cows @$5? How many horses @$1? How many sheep @$.05?
Remember, the total number of toys is 100. Your answer must total 100.
Submit your answer via Comment below. You may identify yourself if you wish or by some pseudonym you select. Please specify how you wish to be identified and whether you want your Comment displayed. No Comment can be displayed until and unless Tabacco approves it.
Either your correct answer Comment, if you wish, or your pseudonym will be published as a Comment by Tabacco.
I will publish the correct answer as your Christmas present for one and all to see.
Good Luck,
T.A.B.A.C.C.O. (Truth About Business And Congressional Crimes Organization)
PS No cheating means no help from the kids. (Smile)
ANSWER TO PROBLEM BELOW THIS LINE:
-------------------------------------------------------------------------------------------------------------------------
SOLUTION TO PROBLEM:
There have been 37 Hits on this problem to date. Audrey answered the problem the day I published it. Bill Amoureux, her ex, must have been pea-green with envy. Nobody else answered the problem.
CONGRATULATIONS TO AUDREY, OUR ONLY WINNER!
So here is the solution:
Sheep are 5 cents each. Therefore only multiples of 20 sheep will give even $-results. The number of sheep can only be 20, 40, 60, or 80. By solving 4 separate conditions we reduce the unknowns to 2 with 2 separate equations. This allows a simple algebraic solution.
1- 20 sheep @5cents = $1
X=Cows @$5
Y=Horses@$1
5X + Y = 99
X + Y = 80 (since there are 20 sheep in this Condition)
Deduct the second equation from the first equation to eliminate the Y-term.
4X = 19 N/A, since X must be an integer! 20 sheep is therefore NOT an answer.
2- 40 sheep @5cents = $2
5X + Y = 98
X + Y = 60 (since there are 40 sheep in this Condition)
4X = 38 N/A, since X must be an integer! 40 sheep is therefore NOT an answer.
3- 60 sheep @5cents = $3
5X + Y = 97
X + Y = 40 (since there are 60 sheep in this Condition)
4X = 57 N/A, since X must be an integer! 60 sheep is therefore NOT an answer.
4- 80 sheep @5cents = $4
5X + Y = 96
X + Y = 20 (since there are 80 sheep in this Condition)
4X = 76
X = 19
If X = 19, Y= 1…QED!
Thus the only Solution to the Math problem is:
19 cows @ $5 each
1 horse @ $1 each
80 sheep @ $.05 each
This yields 100 animals @ $100 total price.
Solution published December 25, 2005 by Tabacco
comments (1)
1. Tabacco left...
Friday, 4 November 2005 7:54 pm :: http://tabacco.blog-city.com/
We have a correct answer from Audrey. The 2nd answer is correct. There is
only 1 answer to this problem.