Some examples of writing two ratios and setting them equal to each other to solve proportion word problems. Created by Sal Khan.
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- Would the answers be:
#2: 11.2 apples
#3: 6 eggs
I'm only asking because I'm not sure if I fully understand this unless I get the answer right...(187 votes)
- Lonnie was paid 115.50 for 14 hours of work. work At this rate,how much will Lonnie earn for working 22 hours? How do I set up the problem.(38 votes)
- hey good(7 votes)
- This equations make no sense, I have no idea on how to learn them, for example; here i have one;
5 erasers cost $9.85.
Which equation would help determine the cost of 9 erasers?
Normaly I will divide the price by 5 to determin the cost of one eraser and then multiply by 9 to determin the cost of 9, but here the system gives me this;
5/9=$9.85/x as a final equation that I should have done it ?
Can someone explain to me how this equation makes sense?(6 votes)
- I got 8.96 for question one, 9 divided by 11.50 gives me 1.277 ( repeated sevens) rounded off gives 1.28, times 7 gave me 8.96. Is this incorrect?(3 votes)
- 9 divided by 11.5 does not = 1.2777...
Sounds like you actually did 11.5/9 which does = 1.2777..., which is ok.
Multiply that by 7 and you get the actual answer of 8.94
Your answer is a little off because you rounded too early. Keep 3-4 of the 7's rather than rounding when you did. 1.27777 * 7 = 8.94439, then round to 8.94
Your answer will be more accurate.(7 votes)
- a store manager randomly inspected the sales slip for a sampling for his customers. He found that of the 60 people he inspected, 3 had a mistake on the sales slip. If he expects to have 850 customers a day , how many of those will likely have a mistake on the sales slip?(4 votes)
- You were given one entire ratio: 3 errors to 60 people. And, you were given part of another ratio: 850 people. Make X = the number of errors for the 850 people.
Your proportion becomes: 3/60 = x/850
Cross multiply and solve for X.(2 votes)
I have three word problems here. And what I want to do in this video is not solve the word problems but just set up the equation that we could solve to get the answer to the word problems. And essentially, we're going to be setting up proportions in either case. So in this first problem, we have 9 markers cost $11.50. And then they ask us, how much would 7 markers cost? And let's just set x to be equal to our answer. So x is equal to the cost of 7 markers. So the way to solve a problem like this is to set up two ratios and then set them equal to each other. So you could say that the ratio of 9 markers to the cost of 9 markers, so the ratio of the number of markers, so 9, to the cost of the 9 markers, to 11.50, this should be equal to the ratio of our new number of markers, 7, to whatever the cost of the 7 markers are, to x. Let me do x in green. So this is a completely valid proportion here. The ratio of 9 markers to the cost of 9 markers is equal to 7 markers to the cost of 7 markers. And then you could solve this to figure out how much those 7 markers would cost. And you could flip both sides of this, and it would still be a completely valid ratio. You could have 11.50 to 9. The ratio between the cost of the markers to the number of markers you're buying, 11.50 to 9, is equal to the ratio of the cost of 7 markers to the number of markers, which is obviously 7. So all I would do is flip both sides of this equation right here to get this one over here. You could also think about the ratios in other ways. You could say that the ratio of 9 markers to 7 markers is going to be the same as the ratio of their costs, is going to be equal to the ratio of the cost of 9 markers to the cost of 7 markers. And then, obviously, you could flip both of these sides. Let me do that in the same magenta color. The ratio of 7 markers to 9 markers is the same thing as the ratio of the cost of 7 markers to the cost of 9 markers. So that is 11.50. So all of these would be valid proportions, valid equations that describe what's going on here. And then you would just have to essentially solve for x. So let's do this one right over here. 7 apples cost $5. How many apples can I buy it with $8? So one again, we're going to assume that what they're asking is how many apples-- let's call that x. x is what we want to solve for. So 7 apples costs $5. So we have the ratio between the number of apples, 7, and the cost of the apples, 5, is going to be equal to the ratio between another number of apples, which is now x, and the cost of that other number of apples. And it's going to be $8. And so notice here in this first situation, what was unknown was the cost. So we kind of had the number of apples to cost, the number of apples to cost. Now in this example, the unknown is the number of apples, so number of apples to cost, number of apples to cost. And we could do all of the different scenarios like this. You could also say the ratio between 7 apples and x apples is going to be the same as the ratio between the cost of 7 apples and the cost of 8 apples. Obviously, you can flip both sides of these in either of these equations to get two more equations. And any of these would be valid equations. Now let's do this last one. I'll use new colors here. A cake recipe for 5 people requires 2 eggs. So we want to know how many eggs-- so this we'll call x. And you don't always have to call it x. You could call it e for eggs. Well, e isn't a good idea, because e represents another number once you get to higher mathematics. But you could call them y or z or any variable-- a, b, or c, anything. How many eggs do we need for a 15-person cake? So you could say the ratio of people to eggs is constant. So if we have 5 people for 2 eggs, then for 15 people, we are going to need x eggs. This ratio is going to be constant. 5/2 is equal to 15/x. Or you could flip both sides of this. Or you could say the ratio between 5 and 15 is going to be equal to the ratio between the number of eggs for 5 people-- let me do that in that blue color-- and the number of eggs for 15 people. And obviously, you could flip both sides of this equation. So all of these, we've essentially set up the proportions that describe each of these problems. And then you can go later and solve for x to actually get the answer.