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How many decorations? You need to buy decorations for a celebration. You are given exactly $100 to spend. You need
to spend the entire $100, and you need to get 100 decorations. There are three types to choose from. One costs
$.50, another costs $9.50, and the third costs $5.50. How many of each decoration must you purchase
(don't add in sales tax)?
The answer is figured out like this:
we're going to do some equation math. Let's call the 50 cent decoration X, the $9.50 decoration Y, and the
$5.50 decoration Z. Now X + Y + Z = 100 decorations, and $.50X + $9.50Y + $5.50Z= $100. To get rid of the X component,
multiply the first equation by -.5, making X + Y + Z = 100 ====> -.5X - .5Y - .5Z = -50. Add it to the second equation:
.50X + 9.50Y + 5.50Z= 100
-.5X - .5Y - .5Z = -50
0X + 9.0Y + 5.0Z = 50 is the result of that, so 9Y + 5Z = 50. Break this down further and you can get
5Z = 50 - 9Y, and then Z = 10 - (9/5)Y when dividing both sides of the equation by 5. Since the decorations must be a
whole number, not a fraction each, Z has to be a whole number. If Y is 0 (zero), then Z = 10, which then would be
make .50 (X) + 9,50 (0) + 5.50 (10) = 100 => .5X = 100 - 55 = 45, so .5X = 45, X = 90. Then you'd buy 10 of the
$5.50 decorations, and 90 of the $.50 decorations. This gives you 100 decorations for $100.
OR if Y is a different number, it can only be 5 so 9/5 will become a whole number. Then Y = 5, and Z = 10 - (9/5)Y
becomes Z = 10 - 9 = 1. If Y = 5 and Z = 1, then .50X + 9.50Y + 5.50Z = 100 becomes .50X + 47.50 + 5.50 = 100, which
breaks down into .5X = 47, making X = 94. Then you'd be buying 94 of the $.5 decoration, 5 of the $9.50
decoration, and 1 of the $5.50 decoration. This gives you 100 decorations for $100.
Any other value of Y will make Z a negative number, and you can't buy less than zero of the Z decorations. These are
the only two solutions to the teaser.
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