5th Grade estimation problem.

Helpmewithmymath

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The're are two estimation problems. I need to know why the 4 digit numbers in each of these problems are rounded differently.
1. A rectangular wall measures 1,620 cm by 68 cm. Estimate the area of the wall. The textbook calculation shows that 1,620 is rounded to 1,600.
2. 3,812 people are seated in a concert hall. There are 48 seats in each row. Estimate the number of rows of seats that are occupied. The textbook calculation shows
that in this problem that 3,812 is rounded to 4,000. What are each not rounded to the nearest thousand?
 
… [Why has the book] not rounded [1620] to the nearest thousand?
Hello HMWMM. Rounding 1620 up to 2000 means we've increased 1620 by 380. But 380 is almost one-fourth of 1620 (one-fourth of 1600 is 400), so adding 380 would increase the number 1620 by too much for a good estimate. In other words, rounding 1620 to 2000 introduces too much error, making 2000×70 a poor estimation for 1620×68.

You can also see what happens, by comparing possibilities.

1620 × 68 = 110160 (exact)

1600 × 70 = 112000 (estimation off by 1840)

2000 × 70 = 140000 (estimation off by 29840)

?
 
The answer to your question is simple. The problem does not state what place to round the numbers to.
 
The're are two estimation problems. I need to know why the 4 digit numbers in each of these problems are rounded differently.
1. A rectangular wall measures 1,620 cm by 68 cm. Estimate the area of the wall. The textbook calculation shows that 1,620 is rounded to 1,600.
2. 3,812 people are seated in a concert hall. There are 48 seats in each row. Estimate the number of rows of seats that are occupied. The textbook calculation shows
that in this problem that 3,812 is rounded to 4,000. What are each not rounded to the nearest thousand?
As I pointed out in the other thread for this same question, the difference is that the first is a multiplication problem, and the second is a division problem. We use slightly different approaches to estimation in the two cases. I'm going to go into excruciating detail this time.

First, you need to be aware that there is no one right way to estimate. Whether you choose to round to two significant digits or something else, is up to you. One method might be more accurate, or easier (or possibly both), but there can be many "correct" methods. So they might just happen to have chosen differently. There are also times when you want to "break the rules" because of your understanding of "number sense". As I see it, estimation means "deliberately getting the wrong answer because reducing the time or effort required is more valuable". But what is better is subjective, as long as you do something reasonable.

The first problem is to estimate 1,620 cm times 68 cm. In multiplication, you don't need to round to the same place as you would in addition; but you tend to want to round to the same number of significant digits (though this is often taught before students know what that even means). It is often better, though, to round one number up and the other down, to avoid compounding the error. In this case, rounding 1620 to two sd means rounding down by a little, and rounding 68 to one sd rounds it up a little, which is a good idea. If we rounded 1620 to one sd, it would have to be rounded too far, to 2000, so that's a bad idea; if we rounded both to two sd, we'd have to multiply 1600*68, which is hard to do in one's head. So rounding to 1600*70 = 112000 seems like a good compromise. And compromise is ultimately what estimation is all about.

The second problem is to estimate 3812 / 48. In division, you want to round one or both numbers to about the same number of sd, but most importantly, to numbers that can be easily divided. I'd definitely want to round 48 to 50, so there's only one sd to divide by. That rounds it up only a little, which is good; but to avoid compounding errors, we'd therefore like to round 3812 also up. (Do you see why rounding both in the same direction for a division is good?) So my first thought would be to round up to 4000, if that feels like not too large a jump. Doing 4000/50 is easy: 80. If I rounded to 3800, I'd have to do 3800/50, which takes a little more thought but can be found by doubling both numbers and dividing 7600 by 100, resulting in 76. Since I've divided a smaller number by a larger number, I know this is an underestimate, which can be a good thing to know. The actual result is 3812/48 = 79.4, so in fact what seems to be a rougher estimate is considerably better, because of rounding in the same direction.

I suspect that the book hasn't explained all these ideas about estimation. I think they should do so; but often it appears that they just give problems where blindly following their examples will not take you far astray; often, instead, they just tell you how to round, and leave the details of how to make that choice for later (if anyone gets around to teaching it at all).

As an aside, the second problem is bad in an unrelated way. Since the audience will normally pick seats anywhere they want, there is no way to know how many rows have anyone sitting in them, or how many are full. The only way the question can make sense is if they are forced to fill up rows starting in the front, so that the division makes sense: N full rows times 48 per row (plus some remainder) has to equal 3812. But I hate problems where you have to make an unnatural assumption, with no justification other than "otherwise you couldn't answer the question". When the main topic is estimation, that is no excuse for the problem itself to be nonsense.
 
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