# Math Help - algebraic word problems

1. ## algebraic word problems

1)a 60% acid solution is to be mixed with a 80% acid solution to produce 20 liters of a 65% acid solution. how many liters of each solution is needed?

2)the ratio between two positive numbers is 6 to 5. if the difference between the two numbers is 8, what are the 2 numbers?

3)two cars are traveling toward each other, starting at the same time 480km apart. if they meet after 4 hours, and one car is traveling 40kph faster than the other, how fast is each car moving?

4) bill has a collection of 65 nickels and dimes. their total worth is \$5.50. find how many nickels and dimes bill has.

can someone please show me how to do these??????

2. Originally Posted by pspman354

2)the ratio between two positive numbers is 6 to 5. if the difference between the two numbers is 8, what are the 2 numbers?
$\frac{x}{y}= \frac{6}{5}$ ...(1)

$x-y = 8$ ...(2)

Now solve.

3. Originally Posted by pspman354
1)a 60% acid solution is to be mixed with a 80% acid solution to produce 20 liters of a 65% acid solution. how many liters of each solution is needed?

20 Litres of 65% acid solution = 20 x 0.65 = 13L acid

a Litres of 60% acid solution = 0.6a L acid

b Litres of 80% acid solution = 0.8b L acid.

You want
0.6a + 0.8b = 13 Litres of acid
a + b = 20 L solution

Solve for a and b.
Attempt the first one. Then apply the same method to the other questions.

4. $\frac{x}{y}= \frac{6}{5}$ ...(1)

how do i solve for x or y using this?

5. Originally Posted by pspman354
$\frac{x}{y}= \frac{6}{5}$ ...(1)

how do i solve for x or y using this?
you can't you need both equations

$\frac{x}{y}= \frac{6}{5}$ ...(1)

$x-y = 8$ ...(2)

Using (2) $x=8+y$ and back into (1) gives

$\frac{8+y}{y}= \frac{6}{5}$

now finish it.