MHB Please, I fast. I'm totally lost....

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The problem involves calculating the number of green candies in a basket containing yellow, green, and red candies, totaling 40. By setting up equations based on the conditions given, it is determined that if yellow candies are replaced with green, there would be seven times more green than red. Additionally, if red candies are replaced with green, there would be four times more green than yellow. Solving these equations reveals that there are 27 green candies in the basket. The calculations confirm that all conditions are satisfied with this solution.
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Problem: "There are 40 yellow, green and red candies in the basket. If all the yellow candies were replaced with green candies, there would be 7 times more green than red candies in the basket. If we replaced red candies with green candies, there would be 4 times more green candies in the basket than yellow candies. How many green candies are in the basket?"

Possible answers given: a)22 b)25 c)27 d)29
 
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Let y be the number of yellow candies, g the number of green candies, r the number of red candies.

"There are 40 yellow, green and red candies in the basket."
So y+ g+ r= 40

"If all the yellow candies were replaced with green candies, there would be 7 times more green than red candies in the basket."
If all the yellow candies were replaced with green candies there would be y+ g green candies.
y+ g= 7r.

"If we replaced red candies with green candies, there would be 4 times more green candies in the basket than yellow candies."
If all the red candies were replaced with green candies there would be g+ r green candies.
g+ r= 4y.

From g+ r= 4y, g= 4y- r so y+ g+ r= y+ 4y- r+ r= 5y= 40. y= 8. y+ g= 8+ g= 7r, g= 7r- 8. r+ g= r+ 7r- 8= 8r- 8= 4y= 32. 8r=40. r= 40/8= 5. g= 7r- 8= 35- 8= 27.

There are 27 green candies.Check: If y= 8, g= 27, and r= 5 then
y+ g+ r= 8+ 27+ 5= 35+ 5= 40.
y+ g= 8+ 27= 35= 7(5)= 7r.
r+ g= 5+ 27= 32= 4(8)= 4y.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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