# Video: AQA GCSE Mathematics Higher Tier Pack 4 • Paper 1 • Question 8

Three integers are rounded to the nearest 10. The sum of the first two rounded integers is 40, and the difference between this sum and the third rounded number is 20. Work out the minimum possible sum of the three original integers.

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### Video Transcript

Three integers are rounded to the nearest 10. The sum of the first two rounded integers is 40, and the difference between this sum and the third rounded number is 20. Work out the minimum possible sum of the three original integers.

Well, this is quite a tricky question. We’re really gonna have to think about this carefully and write lots of clear definitions and work through our logic. So let’s start off. We’ve got three integers and each of them are rounded to the nearest 10. Let’s call the original integers 𝑎, 𝑏, and 𝑐. And then let’s call 𝑎 10, 𝑏 10, and 𝑐 10 our rounded values. Now, let’s just get ahead around this by going through a simple example.

If 𝑥 is an integer and we’re told that when we round it to the nearest 10 we get 20, then we can say that 𝑥 10 is equal to 20. Now because it was rounded to the nearest 10, we could’ve reported 𝑥 10 as being 10 or 20 or 30. But we didn’t; we said it was 20. So if 𝑥 was as small as 15, we would’ve rounded up to 20. Or if it was up to but not quite including 25 — because if it was 25, we’d have rounded it up to 30 — then we would also rounded it down to 20.

So 𝑥 is somewhere between 15 and 25. In other words, 15 is less than or equal to 𝑥 is less than 25. So we could call the minimum value that 𝑥 could be 𝑥 min and the maximum value that it could be 𝑥 max. And that means that the minimum value that 𝑥 could be is the value rounded to 10 minus five and the maximum value that it could be is the value rounded to 10 plus five.

Back to the question then, the sum of the first two rounded integers is 40. So 𝑎 10 plus 𝑏 10 is equal to 40. And the difference between this sum and the third rounded number is 20. So either 𝑎 10 plus 𝑏 10 take away 𝑐 10 is 20 or 𝑐 10 take away 𝑎 10 plus 𝑏 10 is 20. Now, we know that 𝑎 10 plus 𝑏 10 is equal to 40. And that means that either 40 take away 𝑐 10 is 20, which would make 𝑐 10 equal to 20, or 𝑐 10 take away 40 is equal to 20, which would make 𝑐 10 equal to 60. But we’re trying to work out the minimum possible sum of the three original integers. So let’s discount the fact that 𝑐 10 could be as big as 60. And let’s go with 20 as the minimum possible value.

Now, we saw in our example that the minimum possible value of an integer, that’s been rounded to the nearest 10, is five smaller than that value rounded to 10. So let’s call our minimum values 𝑎 min, 𝑏 min, and 𝑐 min. Now, the minimum possible value for 𝑎 is five lower than 𝑎 10. The minimum value for 𝑏 is five lower than 𝑏 10. And the minimum value for 𝑐 is five lower than 𝑐 10. But we know that 𝑐 10 is equal to 20. So the smallest possible value of that original third integer is 15.

And we know that 𝑎 10 plus 𝑏 10 is equal to 40. So the minimum possible value of 𝑎 is five lower than 𝑎 10. And the minimum possible value of 𝑏 is five lower than 𝑏 [𝑏 10]. So if we took five away from 𝑎 10 and took five away from 𝑏 10, then that sum would be 10 lower than 40. In other words, the smallest possible value for the sum of the first two actual integers is 30. So overall, our minimum possible sum of the original integers is 30 plus 15, which is 45.