**Re: help on problem solving**
Avatar.Of.Woe said:

i need help about 2 questions so i can figure things out later how to solve the other problems

1st)

A 40% dye solution is to be mixed with a 70% dye solution to get 120 L of a 50% solution.

How many liters of the 40% and 70% solutions will be needed?

2nd)

A boat takes 3 hr to go 24 mi upstream. It can go 36 mi downstream in the same time. Find the speed of the current and the speed of the boat in still water if x = the speed of the boat in still water and y = the speed of the current.

1st) NAME things.

Let x = liters of 40% solution

Now, since you know you will have 120 L in all, then 120 - x will be the number of liters of 70% solution.

Dye in 40% solution + dye in 70% solution = dye in final 50% solution

0.40(x) + 0.70(120 - x) = 0.50(120)

Solve for x, which is the amount of 40% solution needed....and your "final answer" will also require the amount of 70% solution needed, which will be 120 - x.

2nd)

Let r = speed of boat in still water

Let c = speed of current

Going UPSTREAM, the speed of the current will be subtracted from the boat's speed, since the current will slow the boat down. So, going UPSTREAM, the boat's actual speed will be r - c . At this speed, the boat travels 24 miles in 3 hours.

distance = rate * time

24 = (r - c)* 3

24 = 3r - 3c

Going DOWNSTREAM, the speed of the current will be ADDED to the boat's speed in still water. Going downstream, the boat's actual speed will be r + c. We know that in 3 hours, the boat travels 36 miles at this speed. So,

36 = 3(r + c)

36 = 3r + 3c

Ok...you have two equations in two variables:

24 = 3r - 3c

36 = 3r + 3c

That's an easy system to solve....

If you have questions about this, please SHOW us what you have tried.