Video: AQA GCSE Mathematics Foundation Tier Pack 1 β€’ Paper 2 β€’ Question 30

The value of π‘Ž can be 3 or 7. The value of 𝑏 can be 4 or 16. a) Write down all possible values of π‘Ž/𝑏. b) Calculate the greatest possible value of (π‘Ž + 𝑏)/(π‘Ž βˆ’ 𝑏). You must show how you reached your answer.

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Video Transcript

The value of π‘Ž can be three or seven. The value of 𝑏 can be four or 16. Part a) Write down all possible values of π‘Ž over 𝑏. Part b) Calculate the greatest possible value of π‘Ž plus 𝑏 over π‘Ž minus 𝑏. You must show how you reached your answer.

First, for part a), we can make a two-by-two grid to solve for all of the possible outcomes of π‘Ž over 𝑏. We know that π‘Ž can be three or seven, and variable 𝑏 will be four or 16. We wanna see the outcomes π‘Ž over 𝑏 in all four of these cases.

The first row, the first column will be three over four. The first row and second column will be seven over four. The second row and the first column three over 16. And the second row over the second column seven over 16. And then we list out these four fractions as all possible values of π‘Ž over 𝑏: three-fourths, seven-fourths, three sixteenths, seven sixteenths.

In part b), we’re interested in the greatest possible value of π‘Ž plus 𝑏 over π‘Ž minus 𝑏. For this fraction to be the greatest possible value, we’ll want the fraction as a whole to be positive and not negative. We only have positive values for π‘Ž and 𝑏.

Since variable π‘Ž and variable 𝑏 are both positive, π‘Ž plus 𝑏 is always positive. No matter what we use for π‘Ž or 𝑏, our numerator will be positive. Since we want the fraction as a whole to be positive and we know that the numerator is already positive, we need the value of π‘Ž minus 𝑏 to be positive. What would make π‘Ž minus 𝑏 positive?

If we have a larger number and we subtract a smaller number, we will get a positive result. And if we had a smaller number and subtracted a larger number, the result would be negative. We’re interested in the positive cases. We can say that if π‘Ž is greater than 𝑏, if π‘Ž is the larger number and 𝑏 is the smaller number, then π‘Ž minus 𝑏 will be positive. It will be a positive value. And that would give us a positive numerator and a positive denominator. π‘Ž plus 𝑏 over π‘Ž minus 𝑏 is positive when π‘Ž is greater than 𝑏.

We now need to consider the cases when π‘Ž is larger than 𝑏. Our choices for π‘Ž are three or seven, and our choices for 𝑏 are four or 16. Remember, we’re looking for π‘Ž is greater than 𝑏. We have π‘Ž equals three and 𝑏 equals four. Three is not bigger than four. The next combination would be π‘Ž equals three and 𝑏 equals 16. Three is not larger than 16.

Now we’ll consider the cases where π‘Ž equals seven. If π‘Ž equals seven and 𝑏 equals four, seven is greater than four. Our final case is π‘Ž equals seven and 𝑏 equals 16. Seven is not greater than 16. In our sets of values, the only way for π‘Ž to be greater than 𝑏 is if π‘Ž equals seven and 𝑏 equals four. We say, in order for π‘Ž to be larger than 𝑏, π‘Ž equals seven and 𝑏 equals four.

This means we have seven plus four in the numerator and seven minus four in the denominator. This simplifies to eleven-thirds. This means that 11 over three is the greatest possible value of π‘Ž plus 𝑏 over π‘Ž minus 𝑏.

Remember that part of this question is showing how we reached our answer. In this case, we used logic to eliminate all possibilities except for one. This is how we’ve shown that eleven-thirds is the greatest possible value of π‘Ž plus 𝑏 over π‘Ž minus 𝑏.

Let’s consider a second method for solving part b). We can actually use the grid method that we used for part a) to solve for part b). π‘Ž equals three or seven and 𝑏 equals four or 16. And we’ll solve for π‘Ž plus 𝑏 over π‘Ž minus 𝑏 for each of these four cases.

First, we have three plus four over three minus four, seven over negative one. After that, we’ll have seven plus four over seven minus four, eleven-thirds, three plus 16 over three minus 16, 19 over negative 13, and finally seven plus 16 over seven minus 16, equals 23 over negative nine.

Remember that, in part a), we said the numerator would always be positive. And we see that reflected here, four positive numerators. And we also see that three out of the four are negative fractions. They have a negative in the denominator. The only way for π‘Ž plus 𝑏 over π‘Ž minus 𝑏 to be positive is for π‘Ž to be seven and 𝑏 to be four. Since the other three options produce negative values, the greatest possible value is eleven-thirds.

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