10 percent sugar

The thing is to translate words in an equation.
In the new composition, you have 3/4 which is 10/100 sugar and 1/4 which is x/100 sugar. You want the total (1) be 16/100 sugar
(3/4)*(10/100) + (1/4)*(x/100)=1*16/100
(30+x)/400=64/400
x=34

pepeprepa,

could you elaborate on how you set the equation up. I am confused on why you did not set it up like:

1/4*10/100 + 3/4*x/100=16/100 Isn't 10% of the solution 1/4 sugar.

Any help would be appreciated, Thanks

Let us imagine that we have 100 liters of a solution 10% sugar.

The new solution is composed whith 75 liters of 10% and 25 liters of x%, obtaining at the end 16% og sugar.
So:
10% of 75 + x% of 25 = 16 % of 100

doing the math: 7.5 + 25x = 16

x=34

Answer: 34%

dbart
that is because "One-forth of a solution that was 10 percent sugar by weight was replaced by a second solution" not three-forth as you wrote in your equation. There is 1/4 of the total new solution for which we do not know the % of sugar. And there is still 3/4 of the solution which has 10% of sugar.
Hope it is ok.

I'm still confused.
"One-forth of a solution that was 10 percent sugar by weight was replaced by a second solution"
I'm confused why 1/4 of the solution is not mult by 10%. Why are we mult 3/4 time the 10%.

Am I not reading the qeustion or is there a hidden trap I'm not seeing

Because this 1/4 disappears and is replaced by another solution.

I knew it had to be easy. This is what I get for trying to look at these problems at work. I can't concentrate.

Thanks a lot

I used to work on maths problems at work but it is totally useless, at this time I made 4 or 5 times more errors than when I am studying alone, you should stop it.

Hey dbart... please look the attachment .. i am sure it will help u out.. this the actual way how u shud work out such problems .. do let me know if u still have doubts..!'

pepeprepa wrote:dbart
And there is still 3/4 of the solution which has 10% of sugar.
Hope it is ok.

I was confused with the question stem as well , especially this wording "One-forth of a solution that was 10 percent sugar by weight". How do we know that the 10% sugar applies to the total solution(say 100lit) and not to just the "One-fourth of a solution"(say 25lit). I did the math based on 10% of sugar by weight in (25lit)and40% of it in 100lit.

Is there any particualr wording in the question that made you guys relaize that the 10% they are talking about is considering the whole solution and not the 1/4th of it. Seems like you have to apply SC rules of the restrictive clause usage of "that" in Math word problems as well

ska7945 wrote:5. One-forth of a solution that was 10 percent sugar by weight was replaced by a second solution, resulting in a solution that was 16% sugar by weight. The second solution was that percent sugar by weight?

1)34% 2)24% 3)22% 4)18% 5)8.5%

Let’s let x = the amount of solution, which initially contains 10% sugar. We remove one-fourth of that solution (.25x), which has 10% sugar. Then we replace that .25x with a solution of unknown percentage (p) of sugar. These actions result in our still having x amount of solution, but now with 16% sugar. We can summarize this in the following equation:

x(0.10) - (0.25x)(0.10) + (0.25x)(p) = x(0.16)

0.10x - 0.025x + 0.25xp = 0.16x

100x - 25x + 250xp = 160x

250xp = 85x

p = 85/250 = .34

Thus, the second solution was 34% sugar by weight.

Answer: A