Video: US-SAT03S4-Q24-137138921290

The sum of three numbers is 747. One of the numbers, π‘₯, is 25% more than the sum of the other two numbers. What is the value of π‘₯?

06:19

Video Transcript

The sum of three numbers is 747. One of the numbers, π‘₯, is 25 percent more than the sum of the other two numbers. What is the value of π‘₯?

This question has told us that the sum of three numbers is 747. One of the numbers is π‘₯. And so we can say that π‘₯ plus the sum of two other numbers equals 747. We could substitute π‘Ž in for the sum of the other two numbers. And then we would have an equation that says π‘₯ plus π‘Ž equals 747. Remember that π‘Ž is the value when we add the other two numbers together. This is our first statement. Our second statement says one of the numbers π‘₯ is 25 percent more than the sum of the other two. π‘₯ is 25 percent more. This means that π‘₯ is the whole value of π‘Ž plus 25 percent of π‘Ž. We write 25 percent in decimal form as 0.25.

In this case, we can combine like terms one π‘Ž plus 0.25π‘Ž equals 1.25π‘Ž. If π‘₯ is 25 percent more than the sum of the other two numbers, we write this as π‘₯ equals 1.25 times the sum of the other two numbers. And we use the variable π‘Ž for the sum of other two numbers. At this point, we now have two equations we can work with. In equation two, we have π‘₯ written in terms of π‘Ž. And we can plug in what we know about π‘₯ from equation two into equation one. We know that π‘₯ equals 1.25π‘Ž. And so we can say that 1.25π‘Ž plus π‘Ž equals 747.

Again, we can combine like terms 1.25π‘Ž plus π‘Ž is the same as 1.25π‘Ž plus one π‘Ž. When we combine those terms, we have 2.25π‘Ž equals 747. To solve for π‘Ž, we divide both sides of the equation by 2.25. 747 divided by 2.25 equals 332. If π‘Ž equals 332, then the sum of the other two numbers equals 332. At this point we can take what we found for π‘Ž and plug it into equation one or into equation two. If we plugged it into equation one, we would get this. π‘₯ plus 332 equals 747. If we plug it into equation two, we will have π‘₯ equals 1.25 times 332. In equation one, we would subtract 332 from both sides of the equation. And 747 minus 332 equals 415. For equation two, 1.25 times 332 also equals 415.

Both ways show us that π‘₯ is 415. I wanna look back at this step from the beginning where we let π‘Ž be equal to the sum of the other two numbers. What would have happened if we didn’t do this? Let’s go back to the beginning when we had π‘₯ plus the sum of the other two numbers equals 747. What if we let π‘Ž be equal to the second number and 𝑏 be equal to the third number. Instead of having one variable for the sum of the two numbers, you could assign each number its own variable. Then we have the three numbers represented by three variables. Our first equation, we’d say π‘₯ plus π‘Ž plus 𝑏 equals 747. And we know that π‘₯ equals 25 percent more than the sum of the other two numbers. We would take π‘Ž plus 𝑏 and multiply that by 1.25.

Remember when we take 25 percent more and write it as a percent, we do 25 percent which is the increase and 100 percent which is what you started with. Written as a decimal, that’s 1.25. Now, we have our second equation, π‘₯ equals 1.25 times π‘Ž plus 𝑏. Just like the first time, we’ll plug in what we found from equation two into equation one. 1.25 times π‘Ž plus 𝑏 plus π‘Ž plus 𝑏 equals 747. To solve this problem, we need to do some regrouping. So we can write π‘Ž plus 𝑏 as one times π‘Ž plus 𝑏. If we factor out π‘Ž plus 𝑏, then we add 1.25 plus one. And we get 2.25 times π‘Ž plus 𝑏 equals 747. Divide both sides by 2.25. And we get that π‘Ž plus 𝑏 equals 332.

Remember, in the first method, we found that π‘Ž equal to 332. But π‘Ž was equal to the sum of the other two numbers. And this time the sum of the other two numbers is represented by π‘Ž plus 𝑏. And so if π‘Ž plus 𝑏 equals 332 and π‘₯ equals 1.25 times π‘Ž plus 𝑏, then π‘₯ equals 1.25 times 332 which is 415. Using a variable to represent each number or using one variable to represent the sum of the other two numbers will both yield the same value for π‘₯.

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