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Question 208582: I am very confused with word problems. Can you help me with these?
1. a girl's softball team has won 11 games. they have 11 more games to play. to win at least 50% of all games, how many more must they win?
2.The sides line AB and line AD of a square are extended 10 cm and 6 cm respectively, to form sides line AE and line AF of a rectangle. At most how long is the side of the square if the perimeter of the rectangle is at least twice the perimeter of the square?
3. The three sides of an equilateral triangle are increased by 20 cm, 30 cm, and 40 cm, respectively. The perimeter of the resulting triangle is between twice and three times the perimeter of the original triangle. What can you conclude about the length of a side of the original triangle?
I just need the equation to start out the problem...
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! PROBLEM NUMBER 1
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1. a girl's softball team has won 11 games. they have 11 more games to play. to win at least 50% of all games, how many more must they win?
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the general equation you are looking for is:
ratio of games won that were already played plus ratio of games won to those that will still have to be played is greater than to equal to the ratio of games won to the total games to be played.
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algebraically, this would be:
x * a + y * b >= z * (a + b)
where:
a is number of games already played
x is ratio of wins to games already played
b is number of games still to play
y is ratio of wins to games still to play
(a + b) is total games for the season
z is ratio of wins to total games for the season
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you know a,x,b,z and you are solving for y
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let's look at this equation in relationship to your problem.
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a is the number of games they aleady played. this was not really specified so i assumed the number of games they already played was 11 which was the games they already won.
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x is the ratio of wins to games already played. this would be 11/11 = 1 based on my first assumption.
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b is the number of games still to play which equaled 11.
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y is the ratio of wins to games left to play which is unknown as this time.
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z is the ratio of wins to total games played that they want to equal or exceed. this makes z = .5
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(a+b) is the total games played for the season which is 11 + 11 = 22
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based on this, the formula becomes:
1*11 + y*11 >= .5*22
simplifying, this becomes:
11 + 11y >= 11
subtract 11 from both sides of this equation to get:
11y >= 0
y >= 0
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assuming they only played 11 games already and won all of them, they don't have to win any more games for the season to finish the season with a greater than or equal to 50% win ratio.
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if the number of games they already played is greater than 11, then a would be greater than 11 and x would be smaller than 1.
this would then generate a different value for y.
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PROBLEM NUMBER 2
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2.The sides line AB and line AD of a square are extended 10 cm and 6 cm respectively, to form sides line AE and line AF of a rectangle. At most how long is the side of the square if the perimeter of the rectangle is at least twice the perimeter of the square?
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you start out with a square.
let the side of the square be equal to x
the perimeter of the square is given by the equation P(square) = 4*x
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you extend one side of the square 10 cm to equal the length of the rectangle which we'll call (x+10).
you extend another side of the square 6 cm to equal the width of the rectangle which we'll call (x+6)
the perimeter of the rectangle is given by the equation:
P(rectangle) = 2 * (x+10) + 2 * (x+6)
the equation you are trying to solve is:
P(rectangle) >= 2 * P(square)
this winds up being:
(2*(x+10))+(2*(x+6)) >= 2*(4*x)
you solve for x and that should be your answer.
equation becomes:
2x + 20 + 2x + 12 >= 8x
combining like terms gets:
4x + 32 >= 8x
subtract from both sides gets:
32 >= 4x
divide both sides by 4 to get:
8 >= x
which is the same as:
x <= 8
that would be your answer.
i checked it out and confirmed it's good.
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PROBLEM NUMBER 3
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3. The three sides of an equilateral triangle are increased by 20 cm, 30 cm, and 40 cm, respectively. The perimeter of the resulting triangle is between twice and three times the perimeter of the original triangle. What can you conclude about the length of a side of the original triangle?
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let x = length of the side of the equilateral triangle A.
let:
x+20, x+30, x+40 = length of the sides of triangle B.
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let p = perimeter.
p(A) = 3*x
p(B) = x+20 + x+30 + x+40 = 3x + 90
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if p(B) is between 2 times p(A) and 3 times p(A), your equation would be:
2*p(A) <= p(B) <= 3*p(A)
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p(B) >= 2 times p(A) would be given by the equation:
p(B) >= 2*p(A)
which would be equivalent to:
3x + 90 >= 2*3*x
this would become:
3x + 90 >= 6x
which would become:
90 >= 3x
which would become:
90/3 >= x
which is the same as:
x <= 30
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if p(B) <= 3*p(A), then your equation would become:
3x + 90 <= 3*3*x
this would become:
3x + 90 <= 9x
which would become:
90 <= 6x
which would become:
90/6 <= x
which is the same as:
x >= 15
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your value of x is <= 30 and >= 15
this can be written as:
15 <= x <= 30
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