Rounding with 5. How to round numbers up and down using Excel functions
Having learned to multiply multi-digit numbers “in a column”, we became convinced that this is a very dreary task. Fortunately, we won't be doing this for long. Soon we will do all any complex calculations using a calculator. Now we practice counting solely for educational purposes, in order to better understand and feel the “behavior” of numbers. However, understanding and instinct can be honed with no less success on approximate calculations, which are much simpler. We will now proceed to them.
Let's say we want to buy five chocolates for 19 rubles. We look at our wallet and want to quickly figure out whether we have enough money for this. We reason like this: 19 is approximately 20, and 20 multiplied by 5 is 100. Here we have just over a hundred rubles in our wallet. So there is enough money. A mathematician would say that we rounded nineteen to twenty and did some approximation. But let's start from the beginning.
First of all, let’s make a reservation that at first we will only deal with rounding positive numbers. This can be done in different ways. For example, like this:
The “≈” symbol is read as “approximately equal.” Here, as they say, we rounded the numbers down and, accordingly, received a lower estimate. This is done very simply: we leave the first digit of the number as it is, and replace all subsequent digits with zeros. It is clear that the result of such rounding is always less than or equal to the original number.
On the other hand, numbers can also be rounded up, thus obtaining an upper estimate:
With this rounding, all digits, starting from the second, turn to zero, and the first digit increases by one. A special case arises when the first digit is equal to nine, which is replaced by two digits at once, 1 and 0:
The result of rounding up is always greater than or equal to the original number.
Thus, we have a choice in which direction to round: up or down. Usually they round in the direction that is closest. Obviously, in most cases it is better to round 11 to 10, and 19 to 20. The formal rules are as follows: if the second digit of our number is in the range from zero to 4, then we round down. If this figure is in the range from 5 to 9, then up. Thus:
98 765 ≈ 100 000.
Separately, we should note the situation when the second digit of a number is five, and all subsequent digits are equal to zero, for example 1500. This number is at the same distance from both 2000 and 1000:
2000 − 1500 = 500,
1500 − 1000 = 500.
Therefore, it would seem that it doesn’t matter which way to round it. However, it is customary to round it not anywhere, but only up - so that the rounding rules can be formulated as simply as possible. If we see a five in second place, then this is already enough to make a decision about where to round: we don’t have to be at all interested in subsequent numbers.
Using the rounding of numbers, we can now quickly, albeit approximately, solve multiplication examples of any complexity. Suppose we need to calculate:
We round both factors and in a couple of seconds we get:
6879 ∙ 267 ≈ 7000 ∙ 300 = 2,100,000 ≈ 2,000,000 = 2 million.
For comparison, I will give the exact answer that we calculated when we learned to multiply by column:
6879 ∙ 267 = 1 836 693.
What needs to be done now to understand whether the approximate answer is close or far from the exact one? - Of course, round off the exact answer:
6879 ∙ 267 = 1,836,693 ≈ 2,000,000 = 2 million.
It turned out that after rounding, the exact answer became equal to the approximate one. So our approximate answer is not so bad. However, it should be noted that such accuracy is not always achieved. Let's say we need to calculate 1497∙143. Approximate calculations look like this:
1497 ∙ 143 ≈ 1000 ∙ 100 = 100,000 = 100 thousand.
And here is the exact answer (with subsequent rounding):
1497 ∙ 143 = 214,071 ≈ 200,000 = 200 thousand.
Thus, the exact answer after rounding turned out to be 2 times larger than the approximate one. This, of course, is not very good. But I admit honestly: I deliberately took one of the worst cases. Usually the accuracy of approximate calculations is still better.
However, we have so far rounded numbers and made approximate calculations only in the most, so to speak, rough form. Of all the digits of the number, we left only one unzeroed - the most significant one. They say that we rounded numbers to one significant figure. However, we can round more accurately, for example, to two significant figures:
The rule here is almost the same as before. All digits except the two most senior ones are zeroed. If the first of the zeroed digits contained a number ranging from zero to 4, then we do nothing more. If this figure was in the range from 5 to 9, then add one to the last of the non-zero digits. Note that if there is a nine in the digit to which a unit is added, then this digit is overflowed and reset to zero, and the higher digit “inherits” the one. That is, this is what happens:
195 ≈ 190 + 10 = 200,
or even:
995 ≈ 990 + 10 = 1000.
Rounding to three significant figures, and so on, is defined in the same way.
Let's return to our example. Let's see what happens if we round numbers not to one, but to two significant figures:
1497 ∙ 143 ≈ 1500 ∙ 140 = 210,000 = 210 thousand.
And let’s compare it again with the exact answer:
1497 ∙ 143 = 214,071 ≈ 210,000 ≈ 210 thousand.
Isn't it true that our approximate calculation has become noticeably more accurate?
And here is another familiar example, for which we will write two versions of approximate answers and compare them with the exact answer:
6879 ∙ 267 ≈ 7 000 ∙ 3 00 = 2 100 000 ≈ 2 000 000,
6879 ∙ 267 ≈ 69 00 ∙ 27 0 = 1 863 000 ≈ 1 9 00 000,
6879 ∙ 267 = 1836693 ≈ 1 8 00 000 ≈ 2 000 000.
This is the time to mention this rule: If the factors are rounded to one significant figure, then the approximate answer should be immediately rounded to one significant figure. If the factors are rounded to two significant figures, then the answer must be rounded to two significant figures. In general, as many significant figures as the factors have, the same number of significant figures must remain in the product. Therefore, in the first line, having barely received 2,100,000, we immediately rounded this number to 2,000,000. Likewise in the second line: we did not stop at the intermediate result of 1,863,000, but immediately rounded it to 1,9,00,000 . Why is that? Because in the number 2,100,000, all digits except the very first are still calculated incorrectly. Likewise, in the number 1,863,000, all digits except the first two are incorrectly calculated. Let's take a look at the corresponding calculations done "in a column":
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Here, the exact calculations are reproduced on the left, and the approximate calculations on the right, performed after rounding the factors to two significant figures. Instead of zeros, we wrote circles to emphasize that in fact behind these circles-zeros there are some other numbers that, after rounding, became unknown to us. Without knowing all the numbers in the first two lines, we also cannot calculate all the numbers in the subsequent lines - that’s why there are circles there too. Now let's take a closer look: in the two highest ranks we don't see any circles anywhere. This means that in the response line these bits are calculated more or less accurately. But already in the third highest rank there is one circle, which means a figure unknown to us. Therefore, we actually cannot calculate the third digit in the response line. This is especially true for the fourth and subsequent categories. It is these digits with unknown values that must be set to zero during subsequent rounding.
But what, I wonder, will happen if one of the factors is rounded up to three significant figures, and the other - only up to one? Let's see what the calculation will look like in this case:
We see that only the most significant digit is determined with any certainty, so the answer must be rounded to one significant figure:
6879 ∙ 267 ≈ 6880 ∙ 3 00 = 2 064 000 ≈ 2 000 000
We also see that the significant figure (in this case, 2) may differ from the true figure (in this case, 1), but, as a rule, by no more than one.
IN general case, we must focus on the factor with the smallest number Significant digits: Round your answer to exactly the same number of significant digits.
So far we have only talked about approximate multiplication. What about addition? - Of course, addition can also be approximate. Just rounding the terms, preparing them for approximate addition, is not necessary in exactly the same way as we rounded the factors, preparing them for approximate multiplication. Let's look at an example:
61 238 + 349 = 61 587.
To begin with, let’s round each of the terms to one significant figure:
61 238 + 349 ≈ 60 000 + 300 = 60 300 ≈ 60 000.
Or, if you write it in a column:
61 238 + 349 ≈ 60 000 + 000 = 60 000.
Here we can write 0 instead of the second term, or, as they say, completely neglect it in comparison with the first term. Let's try to increase the accuracy of our calculations. Now round to two significant figures:
61 238 + 349 ≈ 61 000 + 350 = 61 350 ≈ 61 000.
Again, we could immediately neglect the second term and write:
61 238 + 349 ≈ 61 000 + 0 = 61 000.
Only when we increase the rounding precision to three significant figures does the second term begin to play some role:
61 238 + 349 ≈ 61 200 + 349 = 61 549 ≈ 61 500.
However, we again overdid it with the accuracy of the second term: for it, one significant figure would have been enough:
61 238 + 349 ≈ 61 200 + 300 = 61 500.
The following rule applies here: terms, unlike factors, should be rounded not to the same number of significant figures, but to the same digit. To round to the tens place means to round so that the last significant digit of the rounding result is in the tens place. When rounding to the hundreds place, the last significant digit is in the hundreds place, and so on. The approximate answer is immediately rounded to the required accuracy and does not require further rounding. Let's write out our example again, calculating it with varying accuracy:
61,238 + 349 = 61,587 (exact calculation),
61,238 + 349 ≈ 61,240 + 350 = 61,590 (rounded to the nearest ten),
61,238 + 349 ≈ 61,200 + 300 = 61,500 (up to hundreds),
61,238 + 349 ≈ 61,000 + 0 = 61,000 (up to thousands),
61,238 + 349 ≈ 60,000 + 0 = 60,000 (up to tens of thousands),
61,238 + 349 ≈ 100,000 + 0 = 100,000 (up to hundreds of thousands).
It should be noted that when rounding the second term (349) to thousands (and, especially, to higher digits), the result is zero. Here in the last line we also encounter another remarkable case:
61 238 ≈ 100 000,
when a number is rounded to a higher place than those contained in itself - and yet the result of such rounding turns out to be different from zero.
Let us now consider approximate subtraction. We know that subtraction can be thought of simply as a form of addition. Therefore, the rules for approximate subtraction generally coincide with the rules for approximate addition. However, a special situation is possible here, which arises when we calculate the difference between numbers that are close to each other. Let's say you want to roughly estimate what the value of the expression is:
After roughly rounding the difference terms we get:
Let's face it, it didn't turn out very well. The exact value, as can be easily calculated, is:
7654 − 7643 = 11.
Still, there is a considerable difference between zero and eleven! Therefore, even with the roughest estimates, it is customary to round off the difference terms to such a level that the result is still different from zero:
7654 − 7643 ≈ 7650 − 7640 = 10.
Here's another problem that can happen during approximate subtraction:
We got as much as a thousand in the answer, while the exact value of the difference is only one! Here we must look carefully and not allow what is called a formalist approach.
However, situations are possible when the difference value needs to be calculated with an accuracy to some predetermined digit, for example, to the thousand digit. In this case, it is quite acceptable to write exactly like this:
7654 − 7643 ≈ 8000 − 8000 = 0.
2500 − 2499 ≈ 3000 − 2000 = 1000.
Formally, we are absolutely right. We are mistaken in the thousands place by no more than one unit, and this is a completely common thing when we work with such precision that the last significant digit falls exactly in the thousands place. Likewise, to the nearest hundreds:
7654 − 7643 ≈ 7700 − 7600 = 100.
2500 − 2499 ≈ 2500 − 2500 = 0.
Although approximate calculations are a fairly simple thing, you cannot approach it completely thoughtlessly. Each time, the accuracy of the approximation must be chosen based on the task at hand and common sense.
We just have to consider approximate division. Looking ahead, I will say that division can be considered a type of multiplication. Therefore, the rules for approximate division are the same as in the case of multiplication: the dividend and the divisor must be rounded to the same number of significant figures, and the same number of significant figures must remain in the answer.
But we still haven’t really gone through the division. We know how to divide by a whole and divide with a remainder, but we still cannot divide “in an adult way”, without a remainder, one arbitrary number by another. Therefore, for now we will develop, so to speak, temporary rules of approximate division that correspond to our current understanding of the subject. For now we will only divide roughly, with an accuracy of one significant figure.
Suppose we need to calculate approximately:
First of all, round the divisor (324) to one significant figure:
76 464 / 324 ≈ 76 464 / 300.
Now let's compare the only significant digit of the divisor (3) with the first digit of the dividend (7). Here, in principle, two cases are possible. The first case is when the first digit of the dividend is greater than or equal to the only significant digit of the divisor. We will now consider this case, because it is the one that is implemented in this example, since 7 ≥ 3. Now we zero all the digits of the dividend, except for the highest one, and round the value of the highest digit to the nearest number divisible by the significant digit of the divisor:
76 464 / 324 ≈ 76 464 / 300 ≈ 90 000 / 300.
Note that according to standard rounding rules, 76,464 ≈ 80,000, however, since 8 is not evenly divisible by 3, we “went even further up” so that we ended up with 76,464 ≈ 90,000. Next, the dividend and we remove the divider at the same time “from the tail” same number"extra zeros":
76 464 / 324 ≈ 76 464 / 300 ≈ 90 000 / 300 = 900 / 3.
After this, division is not difficult:
76 464 / 324 ≈ 76 464 / 300 ≈ 90 000 / 300 = 900 / 3 = 300.
The approximate answer is ready. Let me give you the exact answer for comparison:
76 464 / 324 = 236 ≈ 200.
As you can see, the discrepancy in the only significant figure of the approximate answer is one unit, which is quite acceptable.
Let us now complete the following approximate calculations:
35 144 / 764 ≈ 35 144 / 800.
This is the second case we have mentioned where the first digit of the dividend is less than the only significant digit of the rounded divisor (3< 8). В этом случае мы зануляем все разряды делимого, кроме двух самых старших, а то число, которое образует эти два старших разряда, «подтягиваем» к ближайшему числу, которое можно поделить нацело на единственную значащую цифру делителя:
35 144 / 764 ≈ 35 144 / 800 ≈ 32 000 / 800.
(If you can “pull up” with equal success in both directions, then “pull up”, for definiteness, upwards.) Now we remove the “extra” zeros and perform division:
35 144 / 764 ≈ 35 144 / 800 ≈ 32 000 / 800 = 320 / 8 = 40.
The exact calculation is:
35 144 / 764 = 46 ≈ 50.
And again, the accuracy of the approximate result is quite acceptable.
It should be noted that even numbers that are not completely divisible by each other can be divided approximately. It is only important (for now) that the dividend be greater than or equal to the divisor.
At the end of this lesson, we just need to figure out how to round negative numbers and how to do approximate calculations with them. In fact, for any negative number we can always write something like this:
−3456 = −(+3456).
Here we have a positive number in brackets. We will round it according to the rules that we have developed for positive numbers. For example, if it needs to be rounded to two significant figures, then we get:
−3456 = −(+3456) ≈ −(+3500) = −3500.
All calculations are just as simple with negative numbers replace with calculations involving only positive numbers. For example,
−234 − 567 = −(234 + 567) ≈ −(200 + 600) = −(800) = −800,
234 − 567 = −(567 − 234) ≈ −(600 − 200) = −(400) = −400,
234 ∙ (−567) = −(234 ∙ 567) ≈ −(200 ∙ 600) = −(120 000) = −120 000.
Let's look at examples of how to round numbers to tenths using rounding rules.
Rule for rounding numbers to tenths.
To round decimal to tenths, you need to leave only one digit after the decimal point, and discard all the other digits following it.
If the first of the discarded digits is 0, 1, 2, 3 or 4, then the previous digit is not changed.
If the first of the discarded digits is 5, 6, 7, 8 or 9, then we increase the previous digit by one.
Examples.
Round to the nearest tenth:
To round a number to tenths, leave the first digit after the decimal point and discard the rest. Since the first digit discarded is 5, we increase the previous digit by one. They read: “Twenty-three point seven five hundredths is approximately equal to twenty three point eight tenths.”
To round this number to tenths, leave only the first digit after the decimal point and discard the rest. The first digit discarded is 1, so we do not change the previous digit. They read: “Three hundred forty-eight point thirty-one hundredths is approximately equal to three hundred forty-one point three tenths.”
When rounding to tenths, we leave one digit after the decimal point and discard the rest. The first of the discarded digits is 6, which means we increase the previous one by one. They read: “Forty-nine point nine, nine hundred sixty-two thousandths is approximately equal to fifty point zero, zero tenths.”
We round to the nearest tenth, so after the decimal point we leave only the first of the digits, and discard the rest. The first of the discarded digits is 4, which means we leave the previous digit unchanged. They read: “Seven point twenty-eight thousandths is approximately equal to seven point zero tenths.”
To round a given number to tenths, leave one digit after the decimal point, and discard all those following it. Since the first digit discarded is 7, therefore, we add one to the previous one. They read: “Fifty-six point eight thousand seven hundred six ten thousandths is approximately equal to fifty six point nine tenths.”
And a couple more examples for rounding to tenths:
You have to round numbers more often in life than many people think. This is especially true for people in professions related to finance. People working in this field are well trained in this procedure. But also in Everyday life process converting values to integer form Not unusual. Many people conveniently forgot how to round numbers immediately after school. Let us recall the main points of this action.
In contact with
Round number
Before moving on to the rules for rounding values, it is worth understanding what is a round number. If we are talking about integers, then it must end with zero.
To the question of where in everyday life such a skill can be useful, you can safely answer - during basic shopping trips.
Using the approximate calculation rule, you can estimate how much your purchases will cost and how much you need to take with you.
It is with round numbers that it is easier to perform calculations without using a calculator.
For example, if in a supermarket or market they buy vegetables weighing 2 kg 750 g, then in a simple conversation with the interlocutor they often do not give the exact weight, but say that they purchased 3 kg of vegetables. When determining the distance between populated areas, the word “about” is also used. This means bringing the result to a convenient form.
It should be noted that some calculations in mathematics and problem solving also do not always use exact values. This is especially true in cases where the response receives infinite periodic fraction. Here are some examples where approximate values are used:
- some values of constant quantities are presented in rounded form (the number “pi”, etc.);
- tabular values of sine, cosine, tangent, cotangent, which are rounded to a certain digit.
Note! As practice shows, approximating values to the whole, of course, gives an error, but only an insignificant one. The higher the rank, the more accurate the result will be.
Getting approximate values
This mathematical operation is carried out according to certain rules.
But for each set of numbers they are different. Note that you can round whole numbers and decimals.
But with ordinary fractions the operation does not work.
First they need convert to decimals, and then proceed with the procedure in the required context.
The rules for approximating values are as follows:
- for integers – replacing the digits following the rounded one with zeros;
- for decimal fractions - discarding all numbers that are beyond the digit being rounded.
For example, rounding 303,434 to thousands, you need to replace hundreds, tens and ones with zeros, that is, 303,000. In decimals, 3.3333 rounding to the nearest ten x, simply discard all subsequent digits and get the result 3.3.
Exact rules for rounding numbers
When rounding decimals it is not enough to simply discard digits after rounded digit. You can verify this with this example. If 2 kg 150 g of sweets are purchased in a store, then they say that about 2 kg of sweets were purchased. If the weight is 2 kg 850 g, then round up, that is, about 3 kg. That is, it is clear that sometimes the rounded digit is changed. When and how this is done, the exact rules will be able to answer:
- If the rounded digit is followed by a digit 0, 1, 2, 3 or 4, then the rounded digit is left unchanged, and all subsequent digits are discarded.
- If the digit being rounded is followed by the number 5, 6, 7, 8 or 9, then the rounded digit is increased by one, and all subsequent digits are also discarded.
For example, how to correct a fraction 7.41 bring closer to units. Determine the number that follows the digit. In this case it is 4. Therefore, according to the rule, the number 7 is left unchanged, and the numbers 4 and 1 are discarded. That is, we get 7.
If the fraction 7.62 is rounded, then the units are followed by the number 6. According to the rule, 7 must be increased by 1, and the numbers 6 and 2 discarded. That is, the result will be 8.
The examples provided show how to round decimals to units.
Approximation to integers
It is noted that you can round to units in the same way as to round to integers. The principle is the same. Let us dwell in more detail on rounding decimal fractions to a certain digit in the whole part of the fraction. Let's imagine an example of approximating 756.247 to tens. In the tenths place there is the number 5. After the rounded place comes the number 6. Therefore, according to the rules, it is necessary to perform next steps:
- rounding up tens per unit;
- in the ones place, the number 6 is replaced;
- digits in the fractional part of the number are discarded;
- the result is 760.
Let us pay attention to some values in which the process of mathematical rounding to integers according to the rules does not reflect an objective picture. If we take the fraction 8.499, then, transforming it according to the rule, we get 8.
But in essence this is not entirely true. If we round up to whole numbers, we first get 8.5, and then we discard 5 after the decimal point and round up.
Rounding numbers is the simplest mathematical operation. To be able to round numbers correctly, you need to know three rules.
Rule 1
When we round a number to a certain place, we must get rid of all the digits to the right of that place.
For example, we need to round the number 7531 to hundreds. This number includes five hundred. To the right of this digit are the numbers 3 and 1. We turn them into zeros and get the number 7500. That is, rounding the number 7531 to hundreds, we got 7500.
When rounding fractional numbers, everything happens the same way, only the extra digits can simply be discarded. Let's say we need to round the number 12.325 to the nearest tenth. To do this, after the decimal point we must leave one digit - 3, and discard all the digits to the right. The result of rounding the number 12.325 to tenths is 12.3.
Rule 2
If to the right of the digit we keep, the digit we discard is 0, 1, 2, 3, or 4, then the digit we keep does not change.
This rule worked in the two previous examples.
So, when rounding the number 7531 to hundreds, the closest digit to the one left was three. Therefore, the number we left - 5 - has not changed. The result of rounding was 7500.
Similarly, when rounding 12.325 to the nearest tenth, the digit we dropped after the three was the two. Therefore, the rightmost digit left (three) did not change during rounding. It turned out to be 12.3.
Rule 3
If the leftmost digit to be discarded is 5, 6, 7, 8, or 9, then the digit to which we round is increased by one.
For example, you need to round the number 156 to tens. There are 5 tens in this number. In the units place, which we are going to get rid of, there is a number 6. This means that we should increase the tens place by one. Therefore, when rounding the number 156 to tens, we get 160.
Let's look at an example with a fractional number. For example, we're going to round 0.238 to the nearest hundredth. According to Rule 1, we must discard the eight, which is to the right of the hundredths place. And according to rule 3, we will have to increase the three in the hundredths place by one. As a result, rounding the number 0.238 to hundredths, we get 0.24.
To round a number to any digit, we underline the digit of this digit, and then we replace all the digits after the underlined one with zeros, and if they are after the decimal point, we discard them. If the first digit replaced by a zero or discarded is 0, 1, 2, 3 or 4, then the underlined number leave unchanged . If the first digit replaced by a zero or discarded is 5, 6, 7, 8 or 9, then the underlined number increase by 1.
Examples.
Round to whole numbers:
1) 12,5; 2) 28,49; 3) 0,672; 4) 547,96; 5) 3,71.
Solution. We underline the number in the units (integer) place and look at the number behind it. If this is the number 0, 1, 2, 3 or 4, then we leave the underlined number unchanged, and discard all the numbers after it. If the underlined number is followed by the number 5 or 6 or 7 or 8 or 9, then we will increase the underlined number by one.
1) 12 ,5≈13;
2) 28 ,49≈28;
3) 0 ,672≈1;
4) 547 ,96≈548;
5) 3 ,71≈4.
Round to the nearest tenth:
6) 0, 246; 7) 41,253; 8) 3,81; 9) 123,4567; 10) 18,962.
Solution. We underline the number in the tenths place, and then proceed according to the rule: we discard everything after the underlined number. If the underlined number was followed by the number 0 or 1 or 2 or 3 or 4, then we do not change the underlined number. If the underlined number was followed by the number 5 or 6 or 7 or 8 or 9, then we will increase the underlined number by 1.
6) 0, 2 46≈0,2;
7) 41,2 53≈41,3;
8) 3,8 1≈3,8;
9) 123,4 567≈123,5;
10) 18.9 62≈19.0. Behind nine there is a six, therefore, we increase nine by 1. (9+1=10) we write zero, 1 goes to the next digit and it will be 19. We just can’t write 19 in the answer, since it should be clear that we rounded to tenths - the number must be in the tenths place. Therefore, the answer is: 19.0.
Round to the nearest hundredth:
11) 2, 045; 12) 32,093; 13) 0, 7689; 14) 543, 008; 15) 67, 382.
Solution. We underline the digit in the hundredths place and, depending on which digit comes after the underlined one, leave the underlined digit unchanged (if it is followed by 0, 1, 2, 3 or 4) or increase the underlined digit by 1 (if it is followed by 5, 6, 7, 8 or 9).
11) 2, 04 5≈2,05;
12) 32,09 3≈32,09;
13) 0, 76 89≈0,77;
14) 543, 00 8≈543,01;
15) 67, 38 2≈67,38.
Important: the last answer should contain a number in the digit to which you rounded.
Mathematics. 6 Class. Test 5 . Option 1 .
1. Infinite decimal non-periodic fractions are called... numbers.
A) positive; IN) irrational; WITH) even; D) odd; E) rational.
2 . When rounding a number to any digit, all digits following this digit are replaced with zeros, and if they are after the decimal point, they are discarded. If the first digit replaced by a zero or discarded is 0, 1, 2, 3 or 4, then the digit preceding it is not changed. If the first digit replaced by a zero or discarded is 5, 6, 7, 8 or 9, then the digit preceding it is increased by one. Round number to tenths 9,974.
A) 10,0;B) 9,9; C) 9,0; D) 10; E) 9,97.
3. Round number to tens 264,85 .
A) 270; B) 260;C) 260,85; D) 300; E) 264,9.
4 . Round to whole number 52,71.
A) 52; B) 52,7; C) 53,7; D) 53; E) 50.
5. Round number to thousandths 3, 2573 .
A) 3,257; B) 3,258; C) 3,28; D) 3,3; E) 3.
6. Round number to hundreds 49,583 .
A) 50;B) 0; C) 100; D) 49,58;E) 49.
7. An infinite periodic decimal fraction is equal to an ordinary fraction whose numerator is the difference between the entire number after the decimal point and the number after the decimal point before the period; and the denominator consists of nines and zeros, and there are as many nines as there are digits in the period, and as many zeros as there are digits after the decimal point before the period. 0,58 (3) to ordinary.
8. Convert an infinite periodic decimal fraction 0,3 (12) to ordinary.
9. Convert an infinite periodic decimal fraction 1,5 (3) into a mixed number.
10. Convert an infinite periodic decimal fraction 5,2 (144) into a mixed number.
11. Any rational number can be written Write down the number 3 as an infinite periodic decimal fraction.
A) 3,0 (0);IN) 3,(0); WITH) 3;D) 2,(9); E) 2,9 (0).
12 . Write down common fraction ½ as an infinite periodic decimal fraction.
A) 0,5; B) 0,4 (9); C) 0,5 (0); D) 0,5 (00); E) 0,(5).
You will find answers to the tests on the “Answers” page.
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