Q2. A juice vendor has 3 cans of juic

Question

Q2. A juice vendor has 3 cans of juice. The first contains 35\% ice and the rest juice. The second contains 40% ice. How much juice should he mix from each of the containers so as to get 22 litres of juice such that the ratio of ice to juice is 3: 5?

Answer 1

So, first thing is that you say nothing about the third can of juice. So the problem can't be solved the way it is intended.

But, maybe we can solve it with just two cans. The cans are in percents but the needed answer is in a ratio. We can convert the ratio to a percent. 3:5*20=60:100 or 60%. So, we want 60% ice in 22 liters of juice mixture.

From here, we look at the two cans. One is 40% ice and the other is 35% ice. Adding one to the other will get us somewhere in between 40% and 35% ice. We can't get to 60% ice unless we have the other can, which is probably going to have an ice percentage higher than 60%. Maybe 80 or 90 percent.

If you did have the last can, let's say at 80% ice, then we can finish the problem. We want the percentage of the result to be at 60%, so we can add 1 part 40% and 1 part 80% to get a mixture of 2 parts 60% ice. So you would do half and half to get the 22 liters of 60%.

You can also try it as a system of equations. The work is shown below.

v1*.8+v2*.4=22*.6

v1+v2=22

v1=22-v2

(22-v2)*.8+v2*.4=22*.6

17.6-.8v2+.4v2=13.2

17.6-13.2=.4v2

4.4/.4=v2

v2=11

22-11=v1

v1=11