# Thread: 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 $5.50. find how many nickels and dimes bill has. can someone please show me how to do these?????? 2. Originally Posted by pspman354 2)the ratio between two positive numbers is 6 to 5. if the difference between the two numbers is 8, what are the 2 numbers?$\displaystyle \frac{x}{y}= \frac{6}{5}$...(1)$\displaystyle x-y = 8$...(2) Now solve. 3. Originally Posted by pspman354 1)a 60% acid solution is to be mixed with a 80% acid solution to produce 20 liters of a 65% acid solution. how many liters of each solution is needed? 20 Litres of 65% acid solution = 20 x 0.65 = 13L acid a Litres of 60% acid solution = 0.6a L acid b Litres of 80% acid solution = 0.8b L acid. You want 0.6a + 0.8b = 13 Litres of acid a + b = 20 L solution Solve for a and b. Attempt the first one. Then apply the same method to the other questions. 4.$\displaystyle \frac{x}{y}= \frac{6}{5}$...(1) how do i solve for x or y using this? 5. Originally Posted by pspman354$\displaystyle \frac{x}{y}= \frac{6}{5}$...(1) how do i solve for x or y using this? you can't you need both equations$\displaystyle \frac{x}{y}= \frac{6}{5}$...(1)$\displaystyle x-y = 8$...(2) Using (2)$\displaystyle x=8+y$and back into (1) gives$\displaystyle \frac{8+y}{y}= \frac{6}{5}\$

now finish it.