Algebra (all content)
Sal solves the following problem: Drink A with 40% sugar is mixed with Drink B to obtain Drink C which has 25% sugar. What is the percentage of sugar of Drink B? Created by Sal Khan and Monterey Institute for Technology and Education.
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- How does Sal come up with what should be in the third column for each of these problems? I understand the math well enough but don't think I would be able to come up with what should be in that column from the question.(6 votes)
- I don't know if you still need an answer, but here one is:
Sal is changing the 'percent of sugar' into a decimal number, and multiplying.
So for the first one, he takes 2 L(the first amount of drink), and multiplies it by .4 (for 40% sugar content) to get 0.8 L in the third column, for total sugar amount.(6 votes)
- I understood how Sal did it, but i still have lots of trouble making my own tables. Where can I get more table problems?(7 votes)
- Why is it that (at1:59) all of a sudden we add Drink A and B together? When did the problem say that we added the two drinks together and end up with C?(2 votes)
- Reread the problem statement. In the 2nd sentence it says (somewhat condensed): "he takes 2 liters of Drink A and ADDS 1.2 liters of Drink B". Then, in the 3rd sentence, the problem tells you that this new drink is Drink C.
Hope this helps.(3 votes)
- When Sal is computing the 'amount' of sugar in the drinks, it isn't strictly necessary for figuring out this problem (or problems of this type) is it? The only columns he needed was drink volume, and sugar percentage?(2 votes)
- How can amount of sugar be in litres?(1 vote)
- I don't get these problems. I don't understand how Sal intuitively knows to make a column for "amount of sugar." Reading the problem, it seems only clear that there's volume (liters) and % of sugar. How does he know to also use "amount of sugar" when the question just asks you to find the % of sugar in drink B?(1 vote)
- I have an equation; a cook has two vinegar solutions, a 10% and a 5%. He needs 50 mL of a 9% solution for his recipe. How many mL of each solution does he need. Solve by elimination(1 vote)
Make a table and solve. A biologist is researching the impact of three different water-based sugar drinks on bees ability to make honey. He takes 2 liters of Drink A, which contains 40% sugar. So let me write this down. Let me make our table and then we can solve it. So let's take amount of drink. And then we'll say percent sugar. And then we can say sugar quantity, so the actual physical quantity of sugar. Maybe I should say sugar amount, or amount of sugar. Now this first drink, Drink A, it says he takes 2 liters of Drink A, which contains 40% sugar. The first column will be which drink we're talking about, so Drink A, he takes 2 liters of it. It's 40% sugar. So if we want the actual amount of sugar in liters, we just multiply 2 liters times 40%, or times 0.4. Let me write times with a dot so you don't think it's an x. 2 times 0.4, which is equal to 0.8 liters of sugar. So you have 0.8 liters of sugar. 1.2 liters of I guess the other stuff in there is water. But it's 0.8 of the 2 liters is sugar, which is 40%. Now,, he adds 1.2 liters of Drink B. He finds that bees prefer this new solution, Drink C. So when you add these two together, you end up with Drink C. And we end up with how much of Drink C? 2 plus 1.2 is 3.2 liters of Drink C, which has 25% sugar content. So this is 25% sugar, which also says we know the amount of sugar in it. Because if we have 3.2 liters of it and it's 25% sugar, or it's 1/4 sugar, that means that we have 0.8 liters of sugar here. So this is 0.8 liters of sugar. Well, that I already wrote in the column name. That's the amount of sugar. It's 25% sugar. We have 3.2 liters of it. Now, they want to know what is the percentage of sugar in Drink B? So let's just call that x. So that's right over here. Now, if it's x percent sugar here, or this is the decimal equivalent, that's x, how much sugar do we have? We have 1.2 liters times the decimal equivalent of sugar, so this is going to be 1.2 times x. Now let's think about it. We have 0.8 liters of sugar in Drink A, and when you add this amount to it, you still have 0.8 liters of actual sugar in Drink C. So this thing has to be equal to zero. We could set up an equation here. We could write 0.8 plus 1.2x is equal to 0.8. You subtract 0.8 from both sides. You get 1.2x is equal to 0. x has got to be equal to 0. So this thing right here has got to be zero. There's no sugar in Drink B. It's just got to be like 1.2 liters. I guess the solution is water. So it's 1.2 liters of water. There's no sugar in Drink B. It is 0% sugar.