Chapter 2 Whole Numbers

Natural numbers are the counting numbers starting from 1: $1, 2, 3, 4, \ldots$.

To find the next natural number after a given natural number, we add 1 to it.

The given number is 10999.

The first natural number after 10999 is obtained by adding 1 to 10999:

$10999 + 1 = 11000$

The second natural number after 10999 is obtained by adding 1 to the previous result (11000):

$11000 + 1 = 11001$

The third natural number after 10999 is obtained by adding 1 to the second result (11001):

$11001 + 1 = 11002$

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Thus, the next three natural numbers after 10999 are 11000, 11001, and 11002.

Whole numbers are the set of non-negative integers starting from 0: $0, 1, 2, 3, \ldots$.

To find the whole number occurring just before a given whole number (greater than 0), we subtract 1 from it.

The given number is 10001.

The first whole number just before 10001 is:

$10001 - 1 = 10000$

The second whole number just before 10001 (which is the number just before 10000) is:

$10000 - 1 = 9999$

The third whole number just before 10001 (which is the number just before 9999) is:

$9999 - 1 = 9998$

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Thus, the three whole numbers occurring just before 10001 are 10000, 9999, and 9998.

Whole numbers are the set of non-negative integers which start from 0.

The set of whole numbers is $\{0, 1, 2, 3, 4, \ldots\}$.

Looking at this set, the first number and the smallest number is 0.

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Therefore, the smallest whole number is 0.

Whole numbers are the set of non-negative integers: $\{0, 1, 2, 3, \ldots\}$.

We need to find the count of whole numbers that lie strictly between 32 and 53.

This means we are looking for whole numbers $n$ such that $32 < n < 53$.

The whole numbers satisfying this condition are $33, 34, 35, \ldots, 51, 52$.

To find the total number of whole numbers in this range, we can subtract the smaller number from the larger number and then subtract 1.

Number of whole numbers = (Last number in the range) - (First number in the range) + 1

In our case, the first number is 33 and the last number is 52.

Number of whole numbers = $52 - 33 + 1$

$= 19 + 1$

$= 20$

Alternatively, using the endpoints:

Number of whole numbers = (Larger endpoint) - (Smaller endpoint) - 1

$= 53 - 32 - 1$

$= 21 - 1$

$= 20$

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Therefore, there are 20 whole numbers between 32 and 53.

The successor of a whole number is the number that comes immediately after it. It is obtained by adding 1 to the number.

Successor = Given Number + 1

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(a) The successor of 24,40,701 is:

$24,40,701 + 1 = 24,40,702$

So, the successor is 24,40,702.

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(b) The successor of 1,00,199 is:

$1,00,199 + 1 = 1,00,200$

So, the successor is 1,00,200.

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(c) The successor of 10,99,999 is:

$10,99,999 + 1 = 11,00,000$

So, the successor is 11,00,000.

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(d) The successor of 23,45,670 is:

$23,45,670 + 1 = 23,45,671$

So, the successor is 23,45,671.

The predecessor of a whole number (other than 0) is the number that comes immediately before it. It is obtained by subtracting 1 from the given number.

Predecessor = Given Number - 1

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(a) The predecessor of 94 is:

$94 - 1 = 93$

So, the predecessor is 93.

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(b) The predecessor of 10,000 is:

$10,000 - 1 = 9,999$

So, the predecessor is 9,999.

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(c) The predecessor of 2,08,090 is:

$2,08,090 - 1 = 2,08,089$

So, the predecessor is 2,08,089.

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(d) The predecessor of 76,54,321 is:

$76,54,321 - 1 = 76,54,320$

So, the predecessor is 76,54,320.

On the number line, the smaller whole number always lies to the left of the larger whole number.

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(a) 530, 503

Comparing the two numbers, we see that $503 < 530$.

Since 503 is smaller than 530, 503 is on the left of 530 on the number line.

The appropriate sign between them is: $530 > 503$.

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(b) 370, 307

Comparing the two numbers, we see that $307 < 370$.

Since 307 is smaller than 370, 307 is on the left of 370 on the number line.

The appropriate sign between them is: $370 > 307$.

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(c) 98765, 56789

Comparing the two numbers, we see that $56789 < 98765$.

Since 56789 is smaller than 98765, 56789 is on the left of 98765 on the number line.

The appropriate sign between them is: $98765 > 56789$.

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(d) 9830415, 10023001

Comparing the two numbers, 9830415 has 7 digits and 10023001 has 8 digits. Thus, $9830415 < 10023001$.

Since 9830415 is smaller than 10023001, 9830415 is on the left of 10023001 on the number line.

The appropriate sign between them is: $9830415 < 10023001$.

(a) False (F)

Natural numbers start from 1 ($1, 2, 3, \ldots$). The smallest natural number is 1, not 0.

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(b) False (F)

The predecessor of 399 is $399 - 1 = 398$. 400 is the successor of 399.

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(c) True (T)

Whole numbers start from 0 ($0, 1, 2, 3, \ldots$). The smallest whole number is 0.

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(d) True (T)

The successor of 599 is $599 + 1 = 600$.

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(e) True (T)

All natural numbers ($1, 2, 3, \ldots$) are included in the set of whole numbers ($0, 1, 2, 3, \ldots$).

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(f) False (F)

The set of whole numbers includes 0, which is not a natural number.

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(g) False (F)

The predecessor of the two-digit number 10 is $10 - 1 = 9$, which is a single-digit number.

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(h) False (F)

The smallest whole number is 0.

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(i) True (T)

The predecessor of 1 would be $1 - 1 = 0$. Since 0 is not a natural number, the natural number 1 has no predecessor within the set of natural numbers.

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(j) False (F)

The predecessor of the whole number 1 is $1 - 1 = 0$. Since 0 is a whole number, the whole number 1 has a predecessor (which is 0) within the set of whole numbers.

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(k) False (F)

A number lying "between 11 and 12" must be greater than 11 and less than 12. There are no whole numbers between 11 and 12. 13 is greater than 12.

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(l) True (T)

The predecessor of 0 would be $0 - 1 = -1$. Since -1 is not a whole number, the whole number 0 has no predecessor within the set of whole numbers.

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(m) False (F)

The successor of the two-digit number 99 is $99 + 1 = 100$, which is a three-digit number.

We need to find the sum: $234 + 197 + 103$.

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Method 1: Direct Addition

We can add the numbers directly in sequence:

$234 + 197 = 431$

$431 + 103 = 534$

Alternatively, using vertical addition:

$\begin{array}{cc} & 2 & 3 & 4 \\ & 1 & 9 & 7 \\ + & 1 & 0 & 3 \\ \hline & 5 & 3 & 4 \\ \hline \end{array}$

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Method 2: Rearranging using Properties

We can use the commutative and associative properties of addition to rearrange the numbers for easier calculation. Notice that adding 197 and 103 gives a round number.

$234 + 197 + 103 = 234 + (197 + 103)$

First, calculate the sum in the bracket:

$197 + 103 = 300$

Now add 234 to this sum:

$234 + 300 = 534$

So, $234 + 197 + 103 = 534$.

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The sum of 234, 197 and 103 is 534.

We need to calculate the sum $14 + 17 + 6$.

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Method 1: Adding from left to right

First, add the first two numbers:

$(14 + 17) + 6$

$31 + 6 = 37$

So, $14 + 17 + 6 = 37$.

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Method 2: Rearranging using properties

We can use the commutative and associative properties to rearrange the numbers. It is easier to add 14 and 6 first as their sum is a multiple of 10.

$14 + 17 + 6 = (14 + 6) + 17$

First, calculate the sum inside the parentheses:

$14 + 6 = 20$

Now, add the remaining number:

$20 + 17 = 37$

So, $(14 + 6) + 17 = 37$.

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Both methods yield the same result. The sum is 37.

We need to calculate the product $12 \times 35$.

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Method 1: Direct Multiplication

We can perform standard multiplication:

$\begin{array}{cc} & & 3 & 5 \\ \times & & 1 & 2 \\ \hline & & 7 & 0 \\ + & 3 & 5 & \times \\ \hline & 4 & 2 & 0 \\ \hline \end{array}$

So, $12 \times 35 = 420$.

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Method 2: Using Distributive Property

We can write 12 as a sum, for example $10 + 2$, and use the distributive property of multiplication over addition: $a \times (b + c) = (a \times b) + (a \times c)$.

$12 \times 35 = (10 + 2) \times 35$

Applying the distributive property:

$= (10 \times 35) + (2 \times 35)$

Calculate the individual products:

$10 \times 35 = 350$

$2 \times 35 = 70$

Now add the results:

$= 350 + 70 = 420$

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Alternatively, we can write 35 as a sum, for example $30 + 5$:

$12 \times 35 = 12 \times (30 + 5)$

$= (12 \times 30) + (12 \times 5)$

$= 360 + 60 = 420$

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The product of 12 and 35 is 420.

To Find:

The value of the product $8 \times 1,769 \times 125$.

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Solution:

We need to calculate $8 \times 1,769 \times 125$.

We can use the commutative property of multiplication ($a \times b = b \times a$) to rearrange the terms and make the calculation easier. It is easier to multiply 8 and 125 first.

$8 \times 1,769 \times 125 = (8 \times 125) \times 1,769$

First, calculate the product inside the parentheses:

$8 \times 125 = 1,000$.

Now, substitute this value back into the expression:

$= 1,000 \times 1,769$

Multiplying by 1,000 is straightforward; we just add three zeros to the end of 1,769:

$= 17,69,000$

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Alternate Method (Direct Calculation):

This method is more computationally intensive.

First, calculate $8 \times 1,769$:

$\begin{array}{cc} & & 1 & 7 & 6 & 9 \\ \times & & & & & 8 \\ \hline & 1 & 4 & 1 & 5 & 2 \\ \hline \end{array}$

So, $8 \times 1,769 = 14,152$.

Now, multiply this result by 125: $14,152 \times 125$.

$\begin{array}{cc} & & & 1 & 4 & 1 & 5 & 2 \\ &\times & & & & 1 & 2 & 5 \\ \hline & & & 7 & 0 & 7 & 6 & 0 \\ & & 2 & 8 & 3 & 0 & 4 & \times \\ + & 1 & 4 & 1 & 5 & 2 & \times & \times \\ \hline & 1 & 7 & 6 & 9 & 0 & 0 & 0 \\ \hline \end{array}$

So, $14,152 \times 125 = 17,69,000$.

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Both methods give the same result. Using the rearrangement property significantly simplifies the calculation.

The value of $8 \times 1,769 \times 125$ is 17,69,000.

Given:

Cost of lunch per day = $\textsf{₹}$ 20

Cost of milk per day = $\textsf{₹}$ 4

Number of days = 5

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To Find:

The total amount of money spent in 5 days on lunch and milk.

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Solution:

Method 1: Calculate the total cost per day first

Find the total cost for one day by adding the cost of lunch and the cost of milk.

Total cost per day = Cost of lunch + Cost of milk

Total cost per day = $\textsf{₹} 20 + \textsf{₹} 4 = \textsf{₹} 24$

Now, find the total cost for 5 days by multiplying the total cost per day by 5.

Total cost for 5 days = Total cost per day $\times$ Number of days

Total cost for 5 days = $\textsf{₹} 24 \times 5 = \textsf{₹} 120$

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Method 2: Calculate the total cost for each item separately

Find the total cost of lunch for 5 days.

Total cost of lunch = Cost of lunch per day $\times$ Number of days

Total cost of lunch = $\textsf{₹} 20 \times 5 = \textsf{₹} 100$

Find the total cost of milk for 5 days.

Total cost of milk = Cost of milk per day $\times$ Number of days

Total cost of milk = $\textsf{₹} 4 \times 5 = \textsf{₹} 20$

Now, find the total money spent by adding the total cost of lunch and the total cost of milk.

Total cost for 5 days = Total cost of lunch + Total cost of milk

Total cost for 5 days = $\textsf{₹} 100 + \textsf{₹} 20 = \textsf{₹} 120$.

This method illustrates the distributive property of multiplication over addition: $5 \times (20 + 4) = (5 \times 20) + (5 \times 4)$.

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Thus, the total money spent in 5 days is $\textsf{₹} 120$.

To Find:

The value of $12 \times 35$ using the distributive property.

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Solution:

The distributive property of multiplication over addition states that for any whole numbers $a$, $b$, and $c$:

$a \times (b + c) = (a \times b) + (a \times c)$

We can use this property to calculate $12 \times 35$.

Method 1: Splitting 12

Write 12 as $10 + 2$.

$12 \times 35 = (10 + 2) \times 35$

Applying the distributive property:

$= (10 \times 35) + (2 \times 35)$

$= 350 + 70$

$= 420$

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Method 2: Splitting 35

Write 35 as $30 + 5$.

$12 \times 35 = 12 \times (30 + 5)$

Applying the distributive property:

$= (12 \times 30) + (12 \times 5)$

$= 360 + 60$

$= 420$

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Method 3: Using Distributivity over Subtraction

We can also express a number as a difference, for example, $12 = 20 - 8$ or $35 = 40 - 5$.

Let's write $35 = 40 - 5$. The property is $a \times (b - c) = (a \times b) - (a \times c)$.

$12 \times 35 = 12 \times (40 - 5)$

$= (12 \times 40) - (12 \times 5)$

$= 480 - 60$

$= 420$

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Using distributivity, we find that $12 \times 35 = \mathbf{420}$.

To Find:

The value of the expression $126 \times 55 + 126 \times 45$.

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Solution:

The given expression is $126 \times 55 + 126 \times 45$.

We observe that this expression is in the form of $(a \times b) + (a \times c)$, where $a = 126$, $b = 55$, and $c = 45$.

We can use the distributive property of multiplication over addition, which states:

$a \times (b + c) = (a \times b) + (a \times c)$

By applying this property in reverse, we can take the common factor (126) out:

$126 \times 55 + 126 \times 45 = 126 \times (55 + 45)$

First, calculate the sum inside the parentheses:

$55 + 45 = 100$

Now substitute this value back into the expression:

$= 126 \times 100$

Multiplying 126 by 100 is simple:

$= 12,600$

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Therefore, the simplified value of $126 \times 55 + 126 \times 45$ is 12,600.

(a) $837 + 208 + 363$

We use the commutative and associative properties of addition to rearrange the numbers for easier calculation. We group 837 and 363 because the sum of their unit digits ($7+3$) is 10, which simplifies addition.

$(837 + 363) + 208$

First, we calculate the sum inside the parentheses:

$\begin{array}{cc} & & 8 & 3 & 7 \\ + & & 3 & 6 & 3 \\ \hline & 1 & 2 & 0 & 0 \\ \hline \end{array}$

Now we add 208 to this sum:

$1200 + 208 = 1408$

Therefore, the sum is 1,408.

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(b) $1962 + 453 + 1538 + 647$

We rearrange and group numbers whose unit digits add up to 10. We pair (1962 and 1538) since $2+8=10$, and (453 and 647) since $3+7=10$.

$(1962 + 1538) + (453 + 647)$

First, calculate the sum of the first pair:

$\begin{array}{cc} & 1 & 9 & 6 & 2 \\ + & 1 & 5 & 3 & 8 \\ \hline & 3 & 5 & 0 & 0 \\ \hline \end{array}$

Next, calculate the sum of the second pair:

$\begin{array}{cc} & & 4 & 5 & 3 \\ + & & 6 & 4 & 7 \\ \hline & 1 & 1 & 0 & 0 \\ \hline \end{array}$

Finally, we add the results of the two sums:

$3500 + 1100 = 4600$

Therefore, the sum is 4,600.

We use the commutative and associative properties of multiplication to rearrange the numbers to form pairs that are easy to multiply (like pairs that result in 100, 1000, etc.).

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(a) $2 \times 1768 \times 50$

We group 2 and 50 as their product is 100.

$(2 \times 50) \times 1768$

$= 100 \times 1768$

$= 1,76,800$

The product is 1,76,800.

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(b) $4 \times 166 \times 25$

We group 4 and 25 as their product is 100.

$(4 \times 25) \times 166$

$= 100 \times 166$

$= 16,600$

The product is 16,600.

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(c) $8 \times 291 \times 125$

We group 8 and 125 as their product is 1000.

$(8 \times 125) \times 291$

$= 1000 \times 291$

$= 2,91,000$

The product is 2,91,000.

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(d) $625 \times 279 \times 16$

We group 625 and 16 as their product is 10,000.

$(625 \times 16) \times 279$

$= 10,000 \times 279$

$= 27,90,000$

The product is 27,90,000.

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(e) $285 \times 5 \times 60$

We group 5 and 60 as their product is 300.

$285 \times (5 \times 60)$

$= 285 \times 300$

$= 85,500$

The product is 85,500.

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(f) $125 \times 40 \times 8 \times 25$

We group (125 and 8) and (40 and 25) as their products are easy to calculate.

$(125 \times 8) \times (40 \times 25)$

$= 1000 \times 1000$

$= 10,00,000$

The product is 10,00,000.

We use the distributive property of multiplication over addition ($a \times b + a \times c = a \times (b + c)$) and subtraction ($a \times b - a \times c = a \times (b - c)$).

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(a) $297 \times 17 + 297 \times 3$

We take 297 as the common factor.

$297 \times (17 + 3)$

$= 297 \times 20$

$= 5,940$

The value is 5,940.

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(b) $54279 \times 92 + 8 \times 54279$

First, we use the commutative property to write the expression as $54279 \times 92 + 54279 \times 8$. Now we can take 54279 as the common factor.

$54279 \times (92 + 8)$

$= 54279 \times 100$

$= 54,27,900$

The value is 54,27,900.

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(c) $81265 \times 169 – 81265 \times 69$

We take 81265 as the common factor.

$81265 \times (169 - 69)$

$= 81265 \times 100$

$= 81,26,500$

The value is 81,26,500.

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(d) $3845 \times 5 \times 782 + 769 \times 25 \times 218$

We look for a common factor. Notice that $3845 = 5 \times 769$.

Let's rewrite the first term:

First term = $(5 \times 769) \times 5 \times 782 \ $$ = 769 \times (5 \times 5) \times 782 \ $$ = 769 \times 25 \ $$ \times 782$

Now the original expression becomes:

$(769 \times 25 \times 782) + (769 \times 25 \times 218)$

The common factor is $(769 \times 25)$. Using the distributive property:

$(769 \times 25) \times (782 + 218)$

Calculate the terms in the parentheses:

$769 \times 25 = 19,225$

$782 + 218 = 1000$

Now multiply the results:

$19,225 \times 1000 = 1,92,25,000$

The value is 1,92,25,000.

We use the distributive property of multiplication over addition: $a \times (b + c) = (a \times b) + (a \times c)$.

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(a) $738 \times 103$

We write 103 as $(100 + 3)$.

$738 \times (100 + 3)$

$= (738 \times 100) + (738 \times 3)$

$= 73,800 + 2,214$

$= 76,014$

The product is 76,014.

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(b) $854 \times 102$

We write 102 as $(100 + 2)$.

$854 \times (100 + 2)$

$= (854 \times 100) + (854 \times 2)$

$= 85,400 + 1,708$

$= 87,108$

The product is 87,108.

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(c) $258 \times 1008$

We write 1008 as $(1000 + 8)$.

$258 \times (1000 + 8)$

$= (258 \times 1000) + (258 \times 8)$

$= 2,58,000 + 2,064$

$= 2,60,064$

The product is 2,60,064.

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(d) $1005 \times 168$

We write 1005 as $(1000 + 5)$.

$(1000 + 5) \times 168$

$= (1000 \times 168) + (5 \times 168)$

$= 1,68,000 + 840$

$= 1,68,840$

The product is 1,68,840.

Given:

Petrol filled on Monday = 40 litres

Petrol filled on the next day = 50 litres

Cost of petrol = $\textsf{₹}$ 44 per litre

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To Find:

Total money spent on petrol.

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Solution:

First, we find the total quantity of petrol filled.

Total petrol filled = Petrol on Monday + Petrol on next day

Total petrol filled = $40 + 50 = 90$ litres

Now, we calculate the total cost.

Total amount spent = Total petrol filled $\times$ Cost per litre

Total amount spent = $90 \times \textsf{₹} 44$

Total amount spent = $\textsf{₹}$ 3,960

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Therefore, the taxidriver spent $\textsf{₹}$ 3,960 in all on petrol.

Given:

Milk supplied in the morning = 32 litres

Milk supplied in the evening = 68 litres

Cost of milk = $\textsf{₹}$ 45 per litre

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To Find:

Total money due to the vendor per day.

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Solution:

First, we find the total quantity of milk supplied per day. The sum of 32 and 68 is easy to calculate, so we use the distributive property.

Total milk supplied per day = Milk in morning + Milk in evening

Total milk supplied per day = $32 + 68 = 100$ litres

Now, we calculate the total money due.

Total money due = Total milk supplied per day $\times$ Cost per litre

Total money due = $100 \times \textsf{₹} 45$

Total money due = $\textsf{₹}$ 4,500

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Therefore, the money due to the vendor per day is $\textsf{₹}$ 4,500.

Let's match each statement with the property it demonstrates.

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(i) 425 × 136 = 425 × (6 + 30 +100)

This statement shows that a number (136) is broken down into a sum (6 + 30 + 100) and then multiplied. This is an application of the Distributivity of multiplication over addition.

So, (i) matches with (c).

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(ii) 2 × 49 × 50 = 2 × 50 × 49

This statement shows that the order of the numbers being multiplied (49 and 50) is changed, but the result remains the same. This property is called Commutativity under multiplication.

So, (ii) matches with (a).

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(iii) 80 + 2005 + 20 = 80 + 20 + 2005

This statement shows that the order of the numbers being added (2005 and 20) is changed, but the result remains the same. This property is called Commutativity under addition.

So, (iii) matches with (b).

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The correct matches are:

(i) $\rightarrow$ (c)

(ii) $\rightarrow$ (a)

(iii) $\rightarrow$ (b)

We need to evaluate each expression to see which one does not result in 0.

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(a) $1 + 0$

Adding 0 to any number results in the number itself (additive identity).

$1 + 0 = 1$

This expression equals 1.

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(b) $0 \times 0$

Multiplying any number by 0 results in 0.

$0 \times 0 = 0$

This expression equals 0.

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(c) $\frac{0}{2}$

Dividing 0 by any non-zero number results in 0.

$\frac{0}{2} = 0$

This expression equals 0.

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(d) $\frac{10 - 10}{2}$

First, evaluate the numerator: $10 - 10 = 0$.

Then, perform the division:

$\frac{0}{2} = 0$

This expression equals 0.

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Comparing the results, only expression (a) evaluates to a number other than 0.

Therefore, (a) $1 + 0$ will not represent zero.

Yes, if the product of two whole numbers is zero, then we can definitely say that at least one of them must be zero. It is also possible that both of them are zero.

This is known as the Zero Product Property for whole numbers.

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Justification:

Whole numbers are $0, 1, 2, 3, \ldots$.

Let the two whole numbers be $a$ and $b$. We are given that their product is zero:

$a \times b = 0$

If $a$ is not zero ($a \ne 0$) and $b$ is not zero ($b \ne 0$), then both $a$ and $b$ must be natural numbers ($1, 2, 3, \ldots$). The product of two natural numbers is always a natural number, which means it cannot be zero. For example, $3 \times 5 = 15 \ne 0$.

Therefore, for the product $a \times b$ to be equal to 0, it is necessary that either $a=0$, or $b=0$, or both $a=0$ and $b=0$.

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Examples:

• One number is zero: Let $a = 5$ and $b = 0$.

  $a \times b = 5 \times 0 = 0$. Here, $b$ is zero.

• The other number is zero: Let $a = 0$ and $b = 9$.

  $a \times b = 0 \times 9 = 0$. Here, $a$ is zero.

• Both numbers are zero: Let $a = 0$ and $b = 0$.

  $a \times b = 0 \times 0 = 0$. Here, both $a$ and $b$ are zero.

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In all cases where the product is zero, at least one of the whole numbers involved in the multiplication is zero.

If the product of two whole numbers is 1, then we must say that both of them are 1.

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Justification:

Let the two whole numbers be $a$ and $b$. We are given that their product is 1:

$a \times b = 1$

The set of whole numbers is $\{0, 1, 2, 3, \ldots\}$.

• If one of the numbers were 0 (e.g., $a=0$), the product would be $0 \times b = 0$, which is not 1. So, neither $a$ nor $b$ can be 0.
• If one of the numbers were greater than 1 (e.g., $a=2$), and the other number $b$ is a whole number, then:
  • If $b=0$, the product is $2 \times 0 = 0 \ne 1$.
  • If $b=1$, the product is $2 \times 1 = 2 \ne 1$.
  • If $b \ge 2$, the product $a \times b$ will be $2 \times b \ge 4$, which is not 1.
  In general, if $a \ge 2$ and $b \ge 1$, then $a \times b \ge 2$. So, if one number is greater than 1, the product cannot be 1.
• The only remaining possibility is that both numbers are 1.

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Example:

• Let $a = 1$ and $b = 1$. Both are whole numbers.

  $a \times b = 1 \times 1 = 1$. This is the only pair of whole numbers whose product is 1.

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Thus, if the product of two whole numbers is 1, it is necessary that both numbers are 1.

The distributive property of multiplication over addition is $a \times (b + c) = (a \times b) + (a \times c)$.

The distributive property of multiplication over subtraction is $a \times (b - c) = (a \times b) - (a \times c)$.

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(a) $728 \times 101$

We write $101$ as $100 + 1$.

$728 \times 101 = 728 \times (100 + 1)$

Applying the distributive property:

$= (728 \times 100) + (728 \times 1)$

$= 72,800 + 728$

$\begin{array}{cc} & 7 & 2 & 8 & 0 & 0 \\ + & & & 7 & 2 & 8 \\ \hline & 7 & 3 & 5 & 2 & 8 \\ \hline \end{array}$

$= 73,528$

Therefore, $728 \times 101 = \mathbf{73,528}$.

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(b) $5,437 \times 1,001$

We write $1,001$ as $1,000 + 1$.

$5,437 \times 1,001 = 5,437 \times (1,000 + 1)$

Applying the distributive property:

$= (5,437 \times 1,000) + (5,437 \times 1)$

$= 54,37,000 + 5,437$

$\begin{array}{cc} & 5 & 4 & 3 & 7 & 0 & 0 & 0 \\ + & & & & 5 & 4 & 3 & 7 \\ \hline & 5 & 4 & 4 & 2 & 4 & 3 & 7 \\ \hline \end{array}$

$= 54,42,437$

Therefore, $5,437 \times 1,001 = \mathbf{54,42,437}$.

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(c) $824 \times 25$

We can write $25$ as $20 + 5$.

$824 \times 25 = 824 \times (20 + 5)$

$= (824 \times 20) + (824 \times 5)$

$= 16,480 + 4,120$

$\begin{array}{cc} & 1 & 6 & 4 & 8 & 0 \\ + & & 4 & 1 & 2 & 0 \\ \hline & 2 & 0 & 6 & 0 & 0 \\ \hline \end{array}$

$= 20,600$

Alternate Method: We can write $824$ as $800 + 24$.

$824 \times 25 = (800 + 24) \times 25$

$= (800 \times 25) + (24 \times 25)$

$= 20,000 + 600 = 20,600$

Therefore, $824 \times 25 = \mathbf{20,600}$.

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(d) $4,275 \times 125$

We can write $125$ as $100 + 20 + 5$.

$4,275 \times 125 = 4,275 \times (100 + 20 + 5)$

Applying the distributive property:

$= (4,275 \times 100) + (4,275 \times 20) + (4,275 \times 5)$

$= 4,27,500 + 85,500 + 21,375$

$= 5,13,000 + 21,375$

$= 5,34,375$

Therefore, $4,275 \times 125 = \mathbf{5,34,375}$.

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(e) $504 \times 35$

We can write $504$ as $500 + 4$.

$504 \times 35 = (500 + 4) \times 35$

Applying the distributive property:

$= (500 \times 35) + (4 \times 35)$

$= 17,500 + 140$

$= 17,640$

Therefore, $504 \times 35 = \mathbf{17,640}$.

Let's look at how the numbers change in each step:

1. The first number (the one multiplied by 8) starts with 1, then adds 2 to become 12, then adds 3 to become 123, and so on (1, 12, 123, 1234, 12345...).

2. The number added at the end is the last digit of the first number in that step (1, 2, 3, 4, 5...).

3. The result starts with 9, then adds 8 to become 98, then adds 7 to become 987, and so on (9, 98, 987, 9876, 98765...). The result numbers are formed by digits $9, 8, 7, \dots$ in decreasing order.

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Next two steps:

Following the pattern for the first number, after 12345 comes 123456.

Following the pattern for the number added, after adding 5 comes adding 6.

Following the pattern for the result, after 98765 comes 987654 (adding the digit 4 to the right).

So, the next step is:

$\mathbf{123456 \times 8 + 6 = 987654}$

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For the step after that:

Following the pattern for the first number, after 123456 comes 1234567.

Following the pattern for the number added, after adding 6 comes adding 7.

Following the pattern for the result, after 987654 comes 9876543 (adding the digit 3 to the right).

So, the step after is:

$\mathbf{1234567 \times 8 + 7 = 9876543}$

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How the pattern works:

The pattern works because each new calculation builds on the one before it.

When we go from $1 \times 8 + 1$ to $12 \times 8 + 2$, the first number changes from $1$ to $12$ (which is $10 \times 1 + 2$). The added number changes from $1$ to $2$.

The calculation for the second step is $12 \times 8 + 2 = (10 \times 1 + 2) \times 8 + 2$.

This is like $(10 \times (1 \times 8)) + (2 \times 8) + 2 \ $$ = 10 \times 8 + 16 + 2 \ $$ = 80 + 18 \ $$ = 98$.

This calculation structure, where the first number grows by adding the next digit and the number added also increases, makes the result neatly add the next decreasing digit (like adding 8 after 9, adding 7 after 98, and so on).

Each time the first number gets one digit longer (1, 12, 123, etc.), the result also gets one digit longer (9, 98, 987, etc.), always adding the next smaller digit at the end.