Category: Ordering Rounding

Problem 1

Evaluate the expression

21 + 32 ÷ 4²

Solution

Step 1:

We follow the order of operations rule PEMDAS,

Here there are no parentheses.

Step 2:

So, first we evaluate the term with an exponent.

21 + 32 ÷ 4² =

21 + 32 ÷ 16

Step 3:

We do division

21 + 32 ÷ 16 =

21 + 2 =

Step 4:

We then do addition

21 + 2 = 23

Step 5:

So 21 + 32 ÷ 4² = 23

Problem 2

Evaluate the expression

1311⁴ ÷ 4

Solution

Step 1:

We follow the order of operations rule PEMDAS,

Step 2:

So, first we evaluate the expression within parentheses 1311

(13 11)⁴ ÷ 4

2⁴ ÷ 4

Step 3:

Next we evaluate the term with an exponent

(2)⁴ ÷ 4 =

16 ÷ 4 =

Step 4:

We do division

16 ÷ 4 = 4

Step 5:

So 1311⁴ ÷ 4 = 4

Introduction

Math expressions use grouping symbols like brackets [], braces {} and parentheses . We now evaluate expressions involving order of operations with whole numbers using such grouping symbols.

Problem 1

Evaluate the following expression

[3 +15+6 ÷ 7] 4

Solution

Step 1:

We must follow the rule of order of operations PEMDAS.

We start with the innermost grouping, the parentheses 15+6

[3 +(15 + 6) ÷ 7] 4 =

[3 + 21 ÷ 7] 4

Step 2:

We next evaluate the remaining grouping, brackets [3 + 21 ÷ 7]

Step 3:

We perform all multiplication and division before any addition or subtraction

[3 + 21 ÷ 7]

[3 + 3]=

6

Step 4:

We then evaluate the last expression

6 4 = 24

Step 5:

So [3 +15+6 ÷ 7] 4 = 24

Problem 2

Evaluate the following expression

[37 129 3] ÷ 7

Solution

Step 1:

We follow the rule of order of operations PEMDAS.

We start with the innermost grouping, the parentheses 129

[37 (12 9) 3] ÷ 7 =

[37 3 3] ÷ 7

Step 2:

We next evaluate the remaining grouping, brackets [37 3 3]

Step 3:

We perform all multiplication and division before any addition or subtraction

[37 3 3]=

[37 9] =

28

Step 4:

We then evaluate the last expression

28 ÷ 7 = 4

Step 5:

[37 129 3] ÷ 7 = 4

Introduction

The whole numbers are first rounded as specified, i.e., rounded to the nearest ten, hundred and so on. Then the quotient of the rounded whole numbers is found to estimate the quotient of whole numbers.

Problem 1

Estimate the quotient 5873 ÷ 346 by first rounding each number so that it has only one non-zero digit.

Solution

Step 1:

Rounding 5873 such that it has only one non-zero digit means rounding it to nearest thousand. Since the hundreds digit, 8 > 5; 5873 rounds up to 6,000.

Step 2:

Rounding 346 such that it has only one non-zero digit means rounding it to nearest hundred. Since the tens digit, 4 < 5 346 rounds down to 300.

Step 3:

So the estimate of the quotient after rounding

= 6,000 ÷ 300 = 20

Problem 2

Estimate the quotient 2162 × 176 by first rounding each number nearest ten.

Solution

Step 1:

2162: Here the digit in ones place, 2 < 5. So 2162 rounds down to nearest ten as 2160.

Step 2:

176: Here the digit in ones place, 6 > 5. So 176 rounds up to nearest ten as 180.

Step 3:

So the estimate of the quotient after rounding

= 2160 ÷ 180 = 12

Introduction

The whole numbers are first rounded as specified, i.e., rounded to the nearest ten, hundred and so on. Then the product of the rounded whole numbers is found to estimate the product of whole numbers.

Problem 1

Estimate the product 573 94 by first rounding each number so that it has only one non-zero digit.

Solution

Step 1:

We round each number such that it has only one non-zero digit

573 is a three-digit number. So its first digit is going to be the only non-zero digit and the other two digits would be zeros. It means rounding to nearest hundred. Since the tens digit, 7 is greater than 5, we round up 573 to 600.

Step 2:

94 is a two-digit number. Its first digit is going to be the only non-zero digit and the other digit would be zero. It means rounding to nearest ten. Since the ones digit, 4 is less than 5, we round down 94 to 90.

Step 3:

The estimate of the product after rounding

= 600 90 = 54,000

Problem 2

Estimate the product 2092 167 by first rounding each number so that it has only one non-zero digit.

Solution

Step 1:

We round each number such that it has only one non-zero digit

2092 is a four-digit number. So its first digit is going to be the only non-zero digit and the other three digits would be zeros. It means rounding to nearest thousand. Since the hundreds digit, 0 is less than 5, we round down 2092 to 2000.

Step 2:

167 is a three-digit number. Its first digit is going to be the only non-zero digit and the other two digits would be zero. It means rounding to nearest hundred. Since the tens digit, 6 is greater than 5, we round up 167 to 200.

Step 3:

The estimate of the product after rounding

= 2000 200 = 400,000

Introduction

The whole numbers are first rounded as specified, i.e., rounded to the nearest ten, hundred and so on. Then the difference of the rounded whole numbers is found to estimate the difference of whole numbers.

Problem 1

Estimate the difference 6,573 4,536 by first rounding each number to the nearest hundred.

Solution

Step 1:

In 6,573, the tens digit, 7 is greater than 5. So we round up 6,573 to the nearest hundred as 6,600.

Step 2:

In 4,536, the tens digit, 3 is less than 5. So we round down 4,536 to nearest hundred as 4,500.

Step 3:

So the estimated difference is 6,600 4,500 = 2,100.

Problem 2

Estimate the difference 44,904 23,091 by first rounding each number to the nearest thousand.

Solution

Step 1:

In 44,904, the hundreds digit, 9 is greater than 5. So we round up 44,904 to the nearest thousand 45,000.

Step 2:

In 23,091, the hundreds digit, 0 is less than 5. So we round down 23,091 to nearest thousand 23,000.

Step 3:

So the estimated difference is 45,000 − 23,000 = 22,000

Introduction

The whole numbers are first rounded as specified, i.e., rounded to the nearest ten, hundred and so on. Then the sum of the rounded whole numbers is found to estimate the sum of the whole numbers.

Problem 1

Estimate the sum 3,273 &plus; 8,781 &plus; 11,309, by first rounding each number to the nearest thousand.

Solution

Step 1:

In 3,273, the hundreds digit, 2 is less than 5. So we round down 3,273 to the nearest thousand 3,000.

Step 2:

In 8,781, the hundreds digit, 7 is greater than 5. So we round up 8,781 to 9,000.

Step 3:

In 11,309, the hundreds digit, 3 is less than 5. So we round down 11,309 to the nearest thousand 11,000.

Step 4:

So the estimated sum is 3,000 &plus; 9,000 &plus; 11,000 = 23,000.

Problem 2

Estimate the sum 514 &plus; 2,327 &plus; 119, by first rounding each number to the nearest hundred.

Solution

Step 1:

In 514, the tens digit, 1 is less than 5. So we round down 514 to the nearest hundred 500.

Step 2:

In 2,327, the tens digit, 2 is less than 5. So we round down 2,327 to nearest hundred 2,300.

Step 3:

In 119, the tens digit, 1 is less than 5. So we round down 119 to the nearest hundred 100.

Step 4:

So the estimated sum is 500 &plus; 2,300 &plus; 100 = 2,900.

Definition

Rounding of a number means replacing it with another number that is nearly equal to it but easy to represent or write. For example 754 rounded to nearest thousand is 1000.

If the rounding number is 4 or less it is round down. If the rounding number is 5 or more it is rounded up.

For example 417, 421, 430 and 446 round down to 400

456, 465, 472, 481 and 499 round up to 500.

Rules for Rounding to Hundreds or Thousands

• To round a number to the nearest hundred, the digit in tens place is considered. If > = 5 rounded up and if < 5 rounded down.
• To round a number to the nearest thousand, the digit in hundreds place is considered. If > = 5, rounded up and if < 5 rounded down.

Problem 1

Round 3,437 to the nearest hundred.

Solution

Step 1:

First, we look for the rounding place which is the hundreds place.

Step 2:

Rounding given number to the nearest hundred means we either round it down to 3,400 or round it up to 3,500.

Step 3:

Now we look at the digit in tens place, 3.

Step 4:

Since 3 is less than 5, we round down 3,437 to 3,400.

Problem 2

Round 7,842 to the nearest thousand.

Solution

Step 1:

The digit in the thousands place is 7.

Step 2:

Rounding given number to nearest thousand means we can round down either to 7,000 or round up to 8,000.

Step 3:

Now we look at the digit in hundreds place.

Step 4:

Since the digit in hundreds place, 8 is greater than five, we round up

7,842 to 8,000 which is the nearest thousand.

Definition

Rounding of a number means replacing it with another number that is nearly equal to it but easy to represent or write. For example 97 is rounded to 100.

If the rounding number is 4 or less it is round down. If the rounding number is 5 or more it is rounded up.

For example 31, 32, 33 and 34 round down to 30

35, 36, 37, 38 and 39 round up to 40.

Rules for Rounding to Tens or Hundreds

• To round a number to the nearest ten, the digit in ones place is considered. If > = 5, rounded up and if < 5 rounded down.
• To round a number to the nearest hundred, the digit in tens place is considered. If > = 5 rounded up and if < 5 rounded down.

Problem 1

Round 661 to the nearest hundred.

Solution

Step 1:

We know that 661 lies between 600 and 700. To round 661 to the nearest hundred, we have to find if it closer to 600 or to 700.

Step 2:

Using the number line, we see that 661 is closer to 700.

So we round up 661 to 700.

Step 3:

We look at the digit in tens place, 6. We find 6 is more 5. So we round up 661 to 700.

Problem 2

Round 94 to the nearest ten.

Solution

Step 1:

To round 94 to nearest ten, we can use the number line to find if it is closer to 90 or to 100.

From the number line we find that 94 is closer to 90. So, we round 94 to 90.

Step 2:

We can also look at the digit in ones place of 94; that is 4. Since 4 < 5, we round down 94 to 90

Introduction

If a series of large numbers are given, we would be interested in ranking or ordering them. We can either order them from the largest to the smallest or from smallest to the largest.

Rules for ordering large numbers

• The given larger numbers are put in a place value table.
• We know that the smallest place value is ones place and it is on the extreme right. The higher place values will be on left.
• We consider digits in the left most column or highest place value column and after comparing decide which is largest number.
• If there are more than one same digits in highest place value, then we compare the digits in the immediate right column.
• This process is repeated from left to right till we find the numbers from the largest to smallest.

Problem 1

Order the following numbers from the least to the greatest

362,239; 76,231; 4572; 125,987

Solution

Step 1:

We know that a number has more value if has more digits. So the number with the fewest digits will be the least in value while the number with the most number of digits will be the greatest.

Step 2:

Here we have one 4-digit number, one 5-digit number and two 6-digit numbers.

Step 3:

Obviously the 4-digit number 4572 is the least in value.

Step 4:

The 5-digit number 76,231 will be next in the order from least to the greatest.

Step 5:

Of the two 6-digit numbers, since 3 > 1, the number 125,987 will be next in order and lastly the number 362,239 will be the greatest in value.

Step 6:

Ordered from least to greatest the numbers are

4572 < 76,231 < 125,987 < 362,239

Problem 2

Order the following numbers from the least to the greatest

65,147; 7,316; 43,190; 254,873

Solution

Step 1:

We know that a number has more value if has more digits. So the number with the fewest digits will be the least in value while the number with the most number of digits will be the greatest.

Step 2:

Here we have one 4-digit number, two 5-digit numbers and one 6-digit numbers.

Obviously the 4-digit number 7,316 is the least in value.

Step 3:

Of the two 5-digit numbers, since 4 < 6, the number 43,190 will be next in order and followed by 65,147.

Step 4:

Lastly the only 6-digit number 254,873 will be the greatest in value.

Step 5:

Ordered from least to greatest the numbers are

7,316 < 43,190 < 65,147 < 254,873

Definition

A numerical expression is a mathematical statement involving only numbers and one or more arithmetic operation symbols+,−,×,÷.

A numerical expression always represents a particular number.

For example, the numerical expression 5 &plus; 20 − 9 is equal to the number 16 when simplified.

Examples

The following are some examples of numerical expressions.

5 &plus; 20 − 7, 4 &plus; 3 − 7, 6×4 ÷ 20, 5 ÷ 30×3, 9 × 41 &plus; 4

Problem 1

Use <, >, or = to compare the numerical expression and the number.

5

÷

(20

3)

8

Solution

Step 1:

5 ÷ 20×3 = 5 ÷ 60 = 

Step 2:

 < 8

Step 3:

So 5 ÷ 20×3 < 8

Problem 2

Use <, >, or = to compare the numerical expression and the number.

5

x

(42

+

3)

90

Solution

Step 1:

5 × 42 &plus; 3 = 5 × 45 = 225

Step 2:

225 > 90

Step 3:

So 5 × 42 &plus; 3 > 90

Problem 3

Use <, >, or = to compare the numerical expression and the number.

(2

+

3)

–

7

6

Solution

Step 1:

2 &plus; 3 − 7 = 5 − 7 = − 2

Step 2:

− 2 < 6

Step 3:

So 2 &plus; 3 − 7< 6

Problem 4

Use <, >, or = to compare the numerical expression and the number.

4

+

20

–

7

14

Solution

Step 1:

4 &plus; 20 − 7 = 17

Step 2:

17 > 14

Step 3:

So 4 &plus; 20 − 7 > 14