Question 111021: I dont know how to set up these word problems...
[how come they dont have a section for word problems ? O__O]
Find three consecutive odd integers such that the sum of the two least integers is 11 more than the greatest integer..
A rectangular park was 15m(meters) longer than it was wide. When a border of bushes 0.5m(meters) wide was planted around the edges of the park, the area of the park increased by 56m2(meters squared). Find the original dimensions of the park.
Marla has 3 dimes and half as many nickels as pennies. If her coins are worth 50cents in all, how many nickels does she have?
Found 2 solutions by scott8148, oberobic:
Answer by scott8148(6628)   (Show Source):
You can put this solution on YOUR website! let x=smallest, so x+2=middle and x+4=greatest ___ x+x+2-11=x+4
let x=width, so x+15=length ___ (x+1)(x+15+1)=(x)(x+15)+56 ___ x^2+17x+16=x^2+15x+56
let x=nickels, so 2x=pennies ___ 5(x)+1(2x)+3(10)=50
Answer by oberobic(2304)  (Show Source):
You can put this solution on YOUR website! I cannot answer your question about a special word problem section. That is the Webmaster's area.
As a volunteer tutor, I can say that I am not inclined to take on answering a question that has 3 questions embedded in it. I might have time to answer 1 or 2, but perhaps not all 3. Also, by posting 3 questions, you could get lucky and have 3 tutors work in parallel to give you the 3 answers in 1/3 the time.
As to setting up the problems:
1) The consecutive integers would be x, x+1, and x+2. And we are told that
x + (x+1) is 11 more than x+2. So,
2x+1=x+2+11
x=12.
2) A rectangule has area A=w*l, where w=width, l=length.
We are told length is 15m more than w, so we could say l = w+15, or l-15=w.
That means the original area A = w(w+15). We don't know what A is though.
But the 0.5m row of bushes, means the length is increased by 1m and the width is increased by 1m (.5 is added on all sides). When that happens, A is increased by 56.
Letting B = new area, the equation is
B = (w+1)(w+15+1)
A = w(w+15)
B = A +56
So...
(w+1)(w+16) = w(w+15)+56.
3) With money problems, you have to keep track of the count as well as the value of the coins involved. Note that we can call the number of dimes d and the value, in cents, will be 10d. Likewise, nickels can be called n and their value is 5n. Pennies are p and have value p (or 1p, but the 1 is superfluous).
The total number of coins will be d+n+p. Their value will be 10d+5n+p.
We know 10d+5n+p=50 cents. We also know the number of dimes is 3, so we can say
30+5n+p=50. Subtracting 30 from both sides, we have 5n+p=20.
We are told she has half as many nickels as pennies. That means, if we have 2 pennies, we have 1 nickel; 4 pennies, 2 nickels; 6 pennies, 3 nickels; 8 pennies, 4 nickels; etc.
In all cases, the relationship is p=2n.
That means
5n+2n=20
7n = 20, which presents us with a problem. 7 does not go into 20 an even number of times, so there is no solution.
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