How are you? I'm Luis and today I'm going to talk about math. And if I can understand them, you can too. Today I am going to solve the problem that I proposed in the previous video and we are going to do it in two ways. First, we will use a bar model. And then we'll set up an equation. Ready? Let's start! We pour a quantity of water from a jug into an empty cup so that the water we pour into the cup is one sixth of the volume left in the jug. If we put 50 more milliliters of water in the cup, then the water in the cup is one-fifth of the volume left in the pitcher. How much water was in the jar at the beginning? As always, we start by reading the entire problem, sentence by sentence, and look at the question we have to solve. How much water was in the jar at the beginning? We write the answer leaving the gap.
At first, in the jug bia “I don't know how many” milliliters. Who are the characters in this problem? We can see that it is a problem in which there is an initial situation in which we have a pitcher with a quantity of water and an empty cup. We make a change by pouring a little water into the cup. We make another change by pouring another little water into the cup. It's a before-after problem. We can draw a bar model of the initial situation. The amount of water in the jug is represented by a bar. The amount of water in the cup is zero because it is empty. We now draw the first situation after the first change. The first sentence of the problem tells us that the amount of water in the cup is one sixth of what is left in the pitcher. The following must be taken into account: the bar that represents the jug after the.
First change is shorter than the one we had at the beginning because we have poured water into the cup. In the cup there is one sixth of what is left in the jar, so we divide what is left in the jar into six parts and one of them is what is in the cup. The model looks like this. What relationship do you see? That is. What was initially in the jar are seven parts like the ones I have drawn. The six that are left and the one you put in the cup. We continue reading to see what information we have about the second change. The second sentence tells us that if we put 50 ml more in the cup, then what is in the cup represents one fifth of what is left in the pitcher. The following must be taken into account: the bar that represents the jug after the second change is shorter than the one we had because we have poured more water into the cup.
How much shorter? That is. 50 milliliters. In the cup we have a slightly longer bar than the one we had. How much longer? That is. 50 milliliters. Furthermore, we know that it represents a fifth of what is left in the jar. Therefore, five times what is in the cup is what is left in the pitcher. We already have a model that represents our problem. We have converted the text of the problem into a drawing that allows us to better see the relationships between the information we have. Now we are going to draw the relationships we have to find a way to solve the problem. Let's draw the second change. In the drawing that represents the first time we put water in the cup we can draw each part as another smaller rectangle and a green one that represents 50 mL.
The drawing of the contents of the jar and the cup is like this. Now we apply the second change. We transfer 50 mL to the cup. It would stay like this. This drawing also represents the final situation of the cup and the pitcher. What content do we have in the jar? Five times what's in the cup. We can draw it like this. This drawing also represents the final situation of the jar. We rearrange the rectangles to compare the two drawings and look for a relationship. If you look closely, the yellow rectangle is worth five green rectangles. That is, the yellow rectangle is worth five times 50 mL, which is 250 mL. Can we calculate how much the blue rectangle is worth? Yes. It is a yellow rectangle and a green one. So it is 250 mL + 50 mL = 300 mL. Can we calculate how much was in the jar in the initial situation? Yes.
There were seven blue rectangles, so seven times 300 mL is 2100 mL. Now we can answer the question of the problem. At first, there was 2100 mL in the jar. I am going to set up an equation to solve the problem. I look at the drawing after the first change. Determined that the blue rectangle is the unknown x. So I have 6x in the jar and x in the cup. Now I apply the second change. I have 6x – 50 left in the jar and x + 50 in the cup. We also know that, in the final situation, there are 5 times what is in the cup in the jar. This is 5 times (x + 50). So I have this equation 6x – 50 = 5(x + 50). We solve the equation 6x – 50 = 5x + 250. x = 300. So a blue rectangle is worth 300 mL. I can now find out how much water was in the jug. There were seven blue rectangles. So 7 x 300 mL = 2100 mL.
We can now answer the question in the problem. At first, there was 2100 mL in the jar. As you can see, we have solved the problem using two methods. First, we have used a bar model to solve the problem completely. Next, we have used an equation that we have easily set up from the bar model. The bar model is a very useful tool for problem solving because it helps us see the relationships between the information we have. Always remember that the more methods you know how to use, the better your ability to solve problems. And here is today's content. I hope you found it interesting and see you again. Until next time!