Algebra Word Problems

I have a test on these kind of problems. I do not know how to do them. I need and explanation on how to do them with the correct Heading and Equation.


1) A collection on 36 dimes and quarters is worth $4.80. How many dimes are there?

2) Tickets for a play cost $6.00 for adults and $3.00 for students. A total of 1050 tickets worth $4200 were sold. How many adult tickets were sold?

3) There are twice as many nickels as quarters. If the coins are worth $2.80, how many quarters are there?

4) There are six more quarters than nickels. If the coins are worth $5.70, how many nickels are there?

5) Sixty three students bought visors at the baseball game. Plain visors cost $5.00 each and logo visors cost $6.00 each. If the total amount spent on visors was $350.00, how many students bought plain visors?

6) A peach sells for 20 cents and an orange sells for 30 cents. A total of 25 pieces of fruit were sold for $6.00. How many peaches were sold?


Thank You

1) We know that there are 36 dimes and quarters so we can write that as:
36=d+q

We also know that the dimes and quarters together equal $4.80 and we can write that as:

4.8=.1d+.25q
(since the answer is in dollars you want to make sure that the variables also represent their worth)

Re-arrange the first equation so we can eliminate the quarters since they want to know how many dimes there are.

q=36-d

Insert this into the second equation.
4.8=.1d+.25(36-d)
4.8=.1d+9-.25d
4.8=(-.15)d+9
-4.2=(-.15)d
d=28

So there are 28 dimes.

2) We know the total amount of sales was $4200 so we'll write out the equation for that first.
4200=6a+3s

Adults tickets and student tickets equal 1050 so:
a+s=1050

Re-arrange to solve for a.
s=1050-a

And input:
4200=6a+3(1050-a)
4200=6a+3150-3a
4200=3a+3150
1050=3a
a=350

4) There are twice as many nickels as quarters so: