Whole Numbers

The smallest whole number is 0.

No, we cannot write the largest whole number because the process of finding the successor of a whole number does not stop anywhere. For every whole number, there exists a successor which is greater than it. Hence, there is no largest whole number.

The successor of a whole number is 1 more than the given number.

(i) The successor of 3999 = 3999 + 1 = 4000.

(ii) The successor of 378915 = 378915 + 1 = 378916.

(iii) The successor of 5001299 = 5001299 + 1 = 5001300.

The predecessor of a whole number is 1 less than the given number.

(i) The predecessor of 500 = 500 - 1 = 499.

(ii) The predecessor of 38794 = 38794 - 1 = 38793.

(iii) The predecessor of 54789011 = 54789011 - 1 = 54789010.

The required whole number = predecessor of the given number = given number - 1.

(i) The required whole number = 50795 - 1 = 50794.

(ii) The required whole number = 720300 - 1 = 720299.

(iii) The required whole number = 8300000 - 1 = 8299999.

The required whole number = successor of the given number = given number + 1.

(i) The required whole number = 5347 + 1 = 5348.

(ii) The required whole number = 72399 + 1 = 72400.

(iii) The required whole number = 3012999 + 1 = 3013000.

(i) The next three consecutive whole numbers of 79 are:

79 + 1, 79 + 2, 79 + 3 i.e. 80, 81, 82.

(ii) The next three consecutive whole numbers of 598 are:

598 + 1, 598 + 2, 598 + 3 i.e. 599, 600, 601.

(iii) The next three consecutive whole numbers of 35669 are:

35669 + 1, 35669 + 2, 35669 + 3 i.e. 35670, 35671, 35672.

Three consecutive whole numbers occurring just before 320001 are:

320001 - 1, 320001 - 2 and 320001 - 3 i.e., 319998, 319999 and 320000..

(i) The whole numbers between 38 and 68 are:

39, 40, 41, ......, 67

Whole numbers between 68 and 38 = 67 - 39 + 1 = 29.

Hence, there are 29 whole numbers between 38 and 68.

(ii) The whole numbers between 99 and 300 are:

100, 101, 102, ......, 299

Number of these numbers = 299 - 100 + 1 = 200.

Hence, there are 200 whole numbers between 99 and 300.

A whole number which does not change when its digits are written in reverse order is called a palindrome.

A 3-digit palindrome has the form 'aba', where the hundreds digit and the units digit are the same.

For the number to lie between 100 and 200, the hundreds digit must be 1, so the units digit must also be 1.

The tens digit can be any digit from 0 to 9.

Hence, the required whole numbers are:

101, 111, 121, 131, 141, 151, 161, 171, 181, 191.

The two-digit whole numbers between 5 and 92 are:

10, 11, 12, ......, 91

Number of these numbers = 91 - 10 + 1 = 82.

Hence, there are 82 two-digit whole numbers between 5 and 92.

The 3-digit whole numbers between 72 and 407 are:

100, 101, 102, ......, 406

Number of these numbers = 406 - 100 + 1 = 307.

Hence, there are 307 three-digit whole numbers between 72 and 407.

(i) According to the Commutative Property of Addition: a + b = b + a.

378 + 1024 = 1024 + 378.

(ii) According to the Associative Law of Addition: a + (b + c) = (a + b) + c.

337 + (528 + 1164) = (337 + 528) + 1164.

(iii) Using the Commutative and Associative properties of Addition:

(21 + 18) + 13 = (21 + 13) + 18.

(iv) According to the Additive Identity property, adding 0 to any whole number gives the number itself.

3056 + 0 = 3056 = 0 + 3056.

(i) 3189 + 53885

$$\begin{matrix} & & 3 & 1 & 8 & 9 \\ + & 5 & 3 & 8 & 8 & 5 \\ \hline & 5 & 7 & 0 & 7 & 4 \\ \hline \end{matrix}$$

Reversing the order:

$$\begin{matrix} & 5 & 3 & 8 & 8 & 5 \\ + & & 3 & 1 & 8 & 9 \\ \hline & 5 & 7 & 0 & 7 & 4 \\ \hline \end{matrix}$$

In both cases, the sum is the same.

Hence, 3189 + 53885 = 57074 and the result is verified.

(ii) 33789 + 50311

$$\begin{matrix} & 3 & 3 & 7 & 8 & 9 \\ + & 5 & 0 & 3 & 1 & 1 \\ \hline & 8 & 4 & 1 & 0 & 0 \\ \hline \end{matrix}$$

Reversing the order:

$$\begin{matrix} & 5 & 0 & 3 & 1 & 1 \\ + & 3 & 3 & 7 & 8 & 9 \\ \hline & 8 & 4 & 1 & 0 & 0 \\ \hline \end{matrix}$$

In both cases, the sum is the same.

Hence, 33789 + 50311 = 84100 and the result is verified.

(i) 311 + 528 + 289

⇒ (311 + 289) + 528

⇒ 600 + 528

⇒ 1128.

Hence, 311 + 528 + 289 = 1128.

(ii) 723 + 834 + 66 + 277

⇒ (723 + 277) + (834 + 66)

⇒ 1000 + 900

⇒ 1900.

Hence, 723 + 834 + 66 + 277 = 1900.

(iii) 78 + 203 + 435 + 7197 + 422

⇒ (78 + 422) + (203 + 7197) + 435

⇒ 500 + 7400 + 435

⇒ 8335.

Hence, 78 + 203 + 435 + 7197 + 422 = 8335.

(i) According to the Commutative Property of Multiplication: a × b = b × a.

375 × 57 = 57 × 375.

(ii) According to the Associative Law of Multiplication: (a × b) × c = a × (b × c).

(33 × 16) × 25 = 33 × (16 × 25).

(iii) Using the Distributive Law of Multiplication over Addition: a × (b + c) = a × b + a × c.

Since 24 = 18 + 6, we have:

37 × 24 = 37 × 18 + 37 × 6.

(iv) According to the Multiplicative Identity property, any number multiplied by 1 equals itself.

7205 × 1 = 7205 = 1 × 7205.

(v) According to the Multiplication by Zero property, any number multiplied by 0 equals 0.

366 × 0 = 0.

(vi) Any number multiplied by 0 gives 0.

0 × 579 = 0.

(vii) Using the Distributive Law of Multiplication over Addition.

Since 108 = 100 + 8, we have:

473 × 108 = 473 × 100 + 473 × 8.

(viii) Using the Distributive Law of Multiplication over Subtraction.

Since 97 = 100 - 3, we have:

684 × 97 = 684 × (100 - 3) = 684 × 100 - 684 × 3.

(ix) Zero divided by any non-zero whole number is 0.

0 ÷ 5 = 0.

(x) (14 - 14) ÷ 7 = 0 ÷ 7 = 0.

(i) 4 × 528 × 25

⇒ 528 × (4 × 25)

⇒ 528 × 100

⇒ 52800.

Hence, 4 × 528 × 25 = 52800.

(ii) 625 × 239 × 16

⇒ 239 × (625 × 16)

⇒ 239 × 10000

⇒ 2390000.

Hence, 625 × 239 × 16 = 2390000.

(iii) 125 × 40 × 8 × 25

⇒ (125 × 8) × (40 × 25)

⇒ 1000 × 1000

⇒ 1000000.

Hence, 125 × 40 × 8 × 25 = 1000000.

(i) 54279 × 92 + 54279 × 8

Using the Distributive Law of Multiplication over Addition: a × b + a × c = a × (b + c).

⇒ 54279 × (92 + 8)

⇒ 54279 × 100

⇒ 5427900.

Hence, 54279 × 92 + 54279 × 8 = 5427900.

(ii) 60678 × 262 - 60678 × 162

Using the Distributive Law of Multiplication over Subtraction: a × b - a × c = a × (b - c).

⇒ 60678 × (262 - 162)

⇒ 60678 × 100

⇒ 6067800.

Hence, 60678 × 262 - 60678 × 162 = 6067800.

(i) 739 × 102

⇒ 739 × (100 + 2)

Using the Distributive Law of Multiplication over Addition:

⇒ 739 × 100 + 739 × 2

⇒ 73900 + 1478

⇒ 75378.

Hence, 739 × 102 = 75378.

(ii) 1938 × 99

⇒ 1938 × (100 - 1)

Using the Distributive Law of Multiplication over Subtraction:

⇒ 1938 × 100 - 1938 × 1

⇒ 193800 - 1938

⇒ 191862.

Hence, 1938 × 99 = 191862.

(iii) 1005 × 188

⇒ (1000 + 5) × 188

Using the Distributive Law of Multiplication over Addition:

⇒ 1000 × 188 + 5 × 188

⇒ 188000 + 940

⇒ 188940.

Hence, 1005 × 188 = 188940.

$$\begin{array}{l} \phantom{x^2 }{\quad 455} \\ 17\overline{\smash{\big)}7750} \\ \phantom{x}\phantom{}\underline{-68} \\ \phantom{{x^2 }()} 95 \\ \phantom{{x})}\underline{-85} \\ \phantom{{x^2 } ()} 100 \\ \phantom{{x} ()}\underline{-85} \\ \phantom{{x^2 (()}} 15 \\ \end{array}$$

Dividend = 7750

Divisor = 17

Quotient = 455

Remainder = 15

Verification: Dividend = (Divisor × Quotient) + Remainder

Substituting values we get:

(Divisor × Quotient) + Remainder

= (17 × 455) + 15

= 7735 + 15

= 7750

Since L.H.S. = R.H.S.

Hence, the result is verified by the division algorithm.

Given,

Divisor = 38

Quotient = 23

Remainder = 17

Using formula, Dividend = (Divisor × Quotient) + Remainder

= (38 × 23) + 17

= 874 + 17

= 891.

Hence, the required number = 891.

To find the least number to be subtracted from 1000 so that the difference is exactly divisible by 35, we divide 1000 by 35 and find the remainder.

$$\begin{array}{l} \phantom{x^2 }{\quad 28} \\ 35\overline{\smash{\big)}1000} \\ \phantom{x}\phantom{()}\underline{-70} \\ \phantom{{x^2 }+} 300 \\ \phantom{{x}()}\underline{-280} \\ \phantom{{x^2 +(}} 20 \\ \end{array}$$

The remainder when 1000 is divided by 35 is 20.

So, the least number to be subtracted from 1000 = 20.

Therefore, 1000 - 20 = 980, which is exactly divisible by 35.

Hence, the least number to be subtracted from 1000 is 20.

To find the least number to be added to 1000 so that the sum is exactly divisible by 53, we divide 1000 by 53 and find the remainder.

$$\begin{array}{l} \phantom{x^2 }{\quad 18} \\ 53\overline{\smash{\big)}1000} \\ \phantom{x}\phantom{(}\underline{-53} \\ \phantom{{x^2 }()} 470 \\ \phantom{{x}(}\underline{-424} \\ \phantom{{x^2 (()}} 46 \\ \end{array}$$

The remainder when 1000 is divided by 53 is 46.

The least number to be added to 1000 = 53 - 46 = 7.

Therefore, 1000 + 7 = 1007, which is exactly divisible by 53.

Hence, the least number to be added to 1000 is 7.

The largest three-digit number = 999.

To find the largest three-digit number exactly divisible by 47, we divide 999 by 47 and subtract the remainder from 999.

$$\begin{array}{l} \phantom{x^2 }{\quad 21} \\ 47\overline{\smash{\big)}999} \\ \phantom{x}\phantom{}\underline{-94} \\ \phantom{{x^2 }(()} 59 \\ \phantom{{x}()}\underline{-47} \\ \phantom{{x^2 +)}} 12 \\ \end{array}$$

The remainder when 999 is divided by 47 is 12.

Therefore, 999 - 12 = 987.

Hence, the largest three-digit number which is exactly divisible by 47 is 987.

The smallest 5-digit number = 10000.

To find the smallest 5-digit number exactly divisible by 254, we divide 10000 by 254 and add the difference between the divisor and remainder to 10000.

$$\begin{array}{l} \phantom{x^2()}{\quad 39} \\ 254\overline{\smash{\big)}10000} \\ \phantom{x^+(}\phantom{}\underline{-762} \\ \phantom{{x^2 }(())} 2380 \\ \phantom{{x}+}\underline{-2286} \\ \phantom{{x^2 +++}} 94 \\ \end{array}$$

The remainder when 10000 is divided by 254 is 94.

Required number to be added = 254 - 94 = 160.

Number = 10000 + 160 = 10160.

Hence, the smallest five-digit number which is exactly divisible by 254 is 10160.

Quantity of milk supplied in the morning = 72 litres.

Quantity of milk supplied in the evening = 28 litres.

Total quantity of milk supplied per day = 72 + 28 = 100 litres.

Cost of 1 litre of milk = ₹ 60.

Total money due to the vendor per day = 100 × 60 = ₹ 6000.

Hence, the money due to the vendor per day is ₹ 6000.

(i) True.

Reason: If a × b = 0, then either a = 0 or b = 0 (or both). For example, 0 × 5 = 0 and 7 × 0 = 0.

(ii) True.

Reason: The only pair of whole numbers whose product is 1 is (1, 1), since 1 × 1 = 1. No other whole numbers can have a product of 1.

(iii) False.

Reason: If a ≠ 0 and b ≠ 0, then both a and b are non-zero whole numbers. The product of two non-zero whole numbers is always non-zero. Hence, ab cannot be zero.

$$\begin{matrix} & 3 & \text{*} & 7 & \text{*} \\ + & 5 & 2 & \text{*} & 6 \\ \hline & \text{*} & 2 & 5 & 0 \\ \hline \end{matrix}$$

Units column: * + 6 = 0 (with a carry). So, * + 6 = 10, which gives * = 4. (Carry = 1)

Tens column: 7 + * + 1 = 5 (with a carry). So, 7 + * + 1 = 15, which gives * = 7. (Carry = 1)

Hundreds column: * + 2 + 1 = 2 (with a carry). So, * + 3 = 12, which gives * = 9. (Carry = 1)

Thousands column: 3 + 5 + 1 = 9. So, the * in the result = 9.

Hence, the completed sum is:

$$\begin{matrix} & 3 & 9 & 7 & 4 \\ + & 5 & 2 & 7 & 6 \\ \hline & 9 & 2 & 5 & 0 \\ \hline \end{matrix}$$

Verification: 3974 + 5276 = 9250.

$$\begin{matrix} & 6 & 5 & 0 & \text{*} \\ - & \text{*} & 0 & \text{*} & 5 \\ \hline & 4 & \text{*} & 5 & 7 \\ \hline \end{matrix}$$

Units column: * - 5 = 7. Since this requires borrowing, * + 10 - 5 = 7, which gives * = 2. (Borrow 1 from tens)

Tens column: After borrowing, 0 - 1 - * = 5 requires further borrowing. So, 10 + 0 - 1 - * = 5, which gives * = 4. (Borrow 1 from hundreds)

Hundreds column: After borrowing, 5 - 1 - 0 = 4. So, the * in the result = 4.

Thousands column: 6 - * = 4, which gives * = 2.

Hence, the completed subtraction is:

$$\begin{matrix} & 6 & 5 & 0 & 2 \\ - & 2 & 0 & 4 & 5 \\ \hline & 4 & 4 & 5 & 7 \\ \hline \end{matrix}$$

Verification: 6502 - 2045 = 4457.

$$\begin{matrix} & 1 & 7 & 0 & 0 & \text{*} & 4 \\ - & & 8 & \text{*} & \text{*} & 4 & 7 \\ \hline & & \text{*} & 8 & 6 & 6 & \text{*} \\ \hline \end{matrix}$$

Units column: 4 - 7 requires borrowing. So, 10 + 4 - 7 = 7. The * in the result = 7. (Borrow 1)

Tens column: After borrowing, * - 1 - 4 = 6 requires borrowing. So, * + 10 - 1 - 4 = 6, which gives * = 1 (in the minuend). (Borrow 1)

Hundreds column: After borrowing, 0 - 1 - * = 6 requires borrowing. So, 10 + 0 - 1 - * = 6, which gives * = 3 (in the subtrahend). (Borrow 1)

Thousands column: After borrowing, 0 - 1 - * = 8 requires borrowing. So, 10 + 0 - 1 - * = 8, which gives * = 1 (in the subtrahend). (Borrow 1)

Ten thousands column: After borrowing, 7 - 1 - 8 requires borrowing. So, 10 + 7 - 1 - 8 = 8. The * in the result = 8. (Borrow 1)

Hundred thousands column: 1 - 1 = 0, The leftmost digit becomes 0, which is omitted.

Hence, the completed subtraction is:

$$\begin{matrix} & 1 & 7 & 0 & 0 & 1 & 4 \\ - & & 8 & 1 & 3 & 4 & 7 \\ \hline & & 8 & 8 & 6 & 6 & 7 \\ \hline \end{matrix}$$

Verification: 170014 - 81347 = 88667.

(i) 3246 + 9999

⇒ 3246 + (10000 - 1)

⇒ 3246 + 10000 - 1

⇒ 13246 - 1

⇒ 13245.

Hence, 3246 + 9999 = 13245.

(ii) 7501 + 99999

⇒ 7501 + (100000 - 1)

⇒ 7501 + 100000 - 1

⇒ 107501 - 1

⇒ 107500.

Hence, 7501 + 99999 = 107500.

(iii) 5377 - 999

⇒ 5377 - (1000 - 1)

⇒ 5377 - 1000 + 1

⇒ 4377 + 1

⇒ 4378.

Hence, 5377 - 999 = 4378.

(iv) 25718 - 9999

⇒ 25718 - (10000 - 1)

⇒ 25718 - 10000 + 1

⇒ 15718 + 1

⇒ 15719.

Hence, 25718 - 9999 = 15719.

(v) 123 × 999

⇒ 123 × (1000 - 1)

⇒ 123 × 1000 - 123 × 1

⇒ 123000 - 123

⇒ 122877.

Hence, 123 × 999 = 122877.

(vi) 203 × 9999

⇒ 203 × (10000 - 1)

⇒ 203 × 10000 - 203 × 1

⇒ 2030000 - 203

⇒ 2029797.

Hence, 203 × 9999 = 2029797.

(i) The n^th square number is given by n × n = n².

So, the 9^th square number = 9 × 9 = 9² = 81.

(ii) The n^th triangular number is given by $\frac{n(n+1)}{2}$.

So, the 7^th triangular number = $\frac{7 \times (7+1)}{2}$ = $\frac{7 \times 8}{2}$ = $\frac{56}{2}$ = 28.

(i) Yes, a rectangular number can be a square number.

Reason: Yes, a rectangular number can be a square number if it can also be arranged in a square pattern.
For example, 36 can be arranged as 6 × 6, so it is a square number. It can also be arranged as 4 × 9 or 3 × 12, so it is also a rectangular number.

(ii) Yes, a triangular number can be a square number.

Reason: Some numbers are both triangular and square. For example, 1 is both the 1^st triangular number and the 1^st square number. Similarly, 36 is both a triangular number (8^th triangular number = $\frac{8 \times 9}{2}$ = 36) and a square number (6² = 36).

Observing the pattern,

The result consists of as many 1s as the number of digits in the first multiplicand, followed by 0.

1234 × 9 + 4 = 11106 + 4 = 11110.

12345 × 9 + 5 = 111105 + 5 = 111110.

Observing the pattern, each line gives a number consisting of repeated 8.

The number of 8 in the result is equal to the number of digits in the first multiplicand plus 1.

9876 × 9 + 4 = 88888.

98765 × 9 + 3 = 888888.

A rectangular number is a number that can be arranged in a rectangle (with length and breadth being different).

Two more rectangular numbers are:

10 (which can be arranged as 2 × 5 or 5 × 2)

12 (which can be arranged as 3 × 4 or 4 × 3)

(i) A whole number is less than all those whole numbers that lie to its right on the number line.

(ii) One more than a given whole is called its successor.

(iii) There is atleast one whole number between two non-consecutive whole numbers.

(iv) Since 335 = 300 + 30 + 5, we have:

738 × 335 = 738 × (300 + 30 + 5).

(v) Given, a × a = a where a ≠ 0.

Dividing both sides by a (since a ≠ 0):

a = 1.

So, a = 1.

(vi) 0 is the only whole number which is not a natural number.

(vii) The additive identity in whole numbers is 0.

(viii) The largest 3-digit number is 999. Its successor is 999 + 1 = 1000.

So, 1000 is the successor of the largest 3-digit number.

(ix) Division of a whole number by zero (0) is not defined.

(i) False.

Reason: The successor of 9 (a one-digit number) is 10, which is a two-digit number.

(ii) False.

Reason: The predecessor of 99 (a two-digit number) is 98, which is also a two-digit number.

(iii) False.

Reason: The predecessor of 100 (a 3-digit number) is 99, which is a 2-digit number.

(iv) False.

Reason: The successor of 999 (a 3-digit number) is 1000, which is a 4-digit number.

(v) False.

Reason: If a = 0, then a ÷ a = 0 ÷ 0 is not defined. So, the statement is not true for all whole numbers.

(vi) True.

Reason: Zero divided by any non-zero whole number is always 0. For example, 0 ÷ 5 = 0.

(vii) True.

Reason: When two different whole numbers are added, at least one of them is non-zero. So, the sum is at least 1, which is a natural number.

(viii) False.

Reason: Between two consecutive whole numbers (such as 5 and 6), there is no whole number.

(ix) False.

Reason: The number 0, when added to any natural number, gives that number. But 0 is not a natural number. There is no natural number that acts as the additive identity for natural numbers.

(x) True.

Reason: If a × b = 0, then either a = 0 or b = 0 (or both).

(xi) True.

Reason: Any non-zero whole number a satisfies a ÷ a = 1, since a × 1 = a.

The smallest whole number is 0, thus 0 has no predecessor in the whole number system.

Hence, option 1 is the correct option.

The smallest 4-digit number = 1000.

Its predecessor = 1000 - 1 = 999.

Hence, option 2 is the correct option.

1 million = 10,00,000 (1 followed by six zeros).

Its predecessor = 10,00,000 - 1 = 9,99,999.

Hence, option 3 is the correct option.

The greatest 2-digit number = 99.

Predecessor of 99 = 99 - 1 = 98.

Successor of 99 = 99 + 1 = 100.

Product of predecessor and successor = 98 × 100 = 9800.

Hence, option 2 is the correct option.

The greatest 3-digit number = 999.

Successor of 999 = 999 + 1 = 1000.

The smallest 3-digit number = 100.

Predecessor of 100 = 100 - 1 = 99.

Sum = 1000 + 99 = 1099.

Hence, option 4 is the correct option.

The whole numbers between 22 and 54 are:

23, 24, 25, ......, 53.

Number of these numbers = 53 - 23 + 1 = 31.

Hence, option 2 is the correct option.

The smallest whole number = 0.

The greatest 2-digit number = 99.

The whole numbers between 0 and 99 are:

1, 2, 3, ......, 98.

Number of these numbers = 98 - 1 + 1 = 98.

Hence, option 3 is the correct option.

Given, a + a = a

⇒ 2a = a

⇒ 2a - a = 0

⇒ a = 0.

Hence, option 1 is the correct option.

Using the Distributive Law of Multiplication over Addition:

93 × 63 + 93 × 37

⇒ 93 × (63 + 37)

⇒ 93 × 100

⇒ 9300.

Hence, option 2 is the correct option.

1. 0 × 5 = 0.

2. 0 ÷ 5 = 0.

3. (10 - 10) ÷ 5 = 0 ÷ 5 = 0.

4. (5 - 0) ÷ 5 = 5 ÷ 5 = 1.

Option 4 gives 1, which is not equal to zero.

Hence, option 4 is the correct option.

Let us check each option:

1. L.H.S. = 21 - (13 - 5) = 21 - 8 = 13. R.H.S. = (21 - 13) - 5 = 8 - 5 = 3. Since L.H.S. ≠ R.H.S., this statement is false.

2. 21 - 13 = 8, which is a whole number. So, the statement is false.

3. 21 × 1 = 21 and 21 × 0 = 0. Since 21 ≠ 0, the statement is false.

4. 13 - 21 = -8, which is a negative number. Negative numbers are not whole numbers. So, the statement is true.

Hence, option 4 is the correct option.

The sum, sum with a multiple, and product of two whole numbers are always whole numbers (closure property of addition and multiplication).

However, the difference of two whole numbers may not be a whole number. For example, if p = 3 and q = 5, then p - q = 3 - 5 = -2, which is not a whole number.

Hence, option 2 is the correct option.

Given:

Divisor = 9

Quotient = 47

Remainder = 5

By formula, Dividend = (Divisor × Quotient) + Remainder

= (9 × 47) + 5

= 423 + 5

= 428.

Hence, option 2 is the correct option.

Triangular numbers are: 1, 3, 6, 10, 15, ...

A rectangle means at least one arrangement with different rows and columns:

10 = 2 × 5

Hence, option 3 is the correct option.

Statement I: 41 - 1 = 40. This is correct since 41 - 1 = 40.

∴ Statement I is true.

Statement II: Let us take an example. Consider the whole number 41.

Predecessor of 41 = 41 - 1 = 40.

Subtracting 1 from the predecessor: 40 - 1 = 39.

But the original number is 41, not 39. So, the statement is incorrect.

(In fact, when we subtract 1 from the predecessor, we get a number that is 2 less than the original.)

∴ Statement II is false.

Hence, option 1 is the correct option.

Statement I:

L.H.S. = 4 × (5 + 6) = 4 × 11 = 44.

R.H.S. = 4 × 5 + 4 × 6 = 20 + 24 = 44.

Since L.H.S. = R.H.S., Statement I is true.

∴ Statement I is true.

Statement II: This is the Distributive Law of Multiplication over Addition, which is a fundamental property of whole numbers.

∴ Statement II is true.

Both Statement I and Statement II are true.

Hence, option 3 is the correct option.

Statement I: Division of any whole number by 0 is not defined. So, 3 ÷ 0 ≠ 0; in fact, 3 ÷ 0 is undefined.

∴ Statement I is false.

Statement II: Division of a whole number by 0 is not defined; we do not get 0.

∴ Statement II is false.

Hence, option 4 is the correct option.

Statement I: Let us check.

12 × 5 + 6 = 60 + 6 = 66.

So, 66 = 12 × 5 + 6 is correct.

∴ Statement I is true.

Statement II: The correct division algorithm states:

Dividend = Divisor × Quotient + Remainder

The given statement has 'remainder' and 'quotient' interchanged, which is incorrect.

∴ Statement II is false.

Hence, option 1 is the correct option.

The next three consecutive whole numbers of 9998 are:

9998 + 1, 9998 + 2 and 9998 + 3 i.e. 9999, 10000 and 10001.

Three consecutive whole numbers occurring just before 567890 are:

567890 - 1, 567890 - 2 and 567890 - 3 i.e. 567889, 567888 and 567887.

The smallest 3-digit number = 100.

Successor of 100 = 100 + 1 = 101.

Predecessor of 100 = 100 - 1 = 99.

Product = 101 × 99

⇒ 101 × (100 - 1)

⇒ 101 × 100 - 101 × 1

⇒ 10100 - 101

⇒ 9999.

Hence, the product of the successor and the predecessor of the smallest 3-digit number is 9999.

The smallest 2-digit number = 10.

The greatest 2-digit number = 99.

The whole numbers between 10 and 99 are:

11, 12, 13, ...., 98.

Number of these numbers = 98 - 11 + 1 = 88.

Hence, there are 88 whole numbers between the smallest and the greatest 2-digit numbers.

(i) 678 + 1319 + 322 + 5681

⇒ (678 + 322) + (1319 + 5681)

⇒ 1000 + 7000

⇒ 8000.

Hence, 678 + 1319 + 322 + 5681 = 8000.

(ii) 777 + 546 + 1463 + 223 + 537

⇒ (777 + 223) + (1463 + 537) + 546

⇒ 1000 + 2000 + 546

⇒ 3546.

Hence, 777 + 546 + 1463 + 223 + 537 = 3546.

(i) 625 × 437 × 16

⇒ 437 × (625 × 16)

⇒ 437 × 10000

⇒ 4370000.

Hence, 625 × 437 × 16 = 4370000.

(ii) 309 × 25 × 7 × 8

⇒ (309 × 7) × (25 × 8)

⇒ 2163 × 200

⇒ 432600.

Hence, 309 × 25 × 7 × 8 = 432600.

(i) 236 × 414 + 236 × 563 + 236 × 23

Using the Distributive Law of Multiplication over Addition:

⇒ 236 × (414 + 563 + 23)

⇒ 236 × 1000

⇒ 236000.

Hence, 236 × 414 + 236 × 563 + 236 × 23 = 236000.

(ii) 370 × 1587 - 37 × 10 × 587

⇒ 370 × 1587 - 370 × 587 [since 37 × 10 = 370]

Using the Distributive Law of Multiplication over Subtraction:

⇒ 370 × (1587 - 587)

⇒ 370 × 1000

⇒ 370000.

Hence, 370 × 1587 - 37 × 10 × 587 = 370000.

$$\begin{array}{l} \phantom{x^2 }{\quad 225} \\ 29\overline{\smash{\big)}6528} \\ \phantom{x}\phantom{}\underline{-58} \\ \phantom{{x^2 }()} 72 \\ \phantom{{x}(}\underline{-58} \\ \phantom{{x^2 }()} 148 \\ \phantom{{x} (}\underline{-145} \\ \phantom{{x^2 + 3}} 3 \\ \end{array}$$

Dividend = 6528

Divisor = 29

Quotient = 225

Remainder = 3

Verification: Dividend = (Divisor × Quotient) + Remainder

Substituting values we get:

(Divisor × Quotient) + Remainder

= (29 × 225) + 3

= 6525 + 3

= 6528

Since L.H.S. = R.H.S.

Hence, the result is verified by the division algorithm.

The greatest 4-digit number = 9999.

To find the greatest 4-digit number exactly divisible by 357, we divide 9999 by 357 and subtract the remainder from 9999.

$$\begin{array}{l} \phantom{x^2(}{\quad 28} \\ 357\overline{\smash{\big)}9999} \\ \phantom{x^+}\phantom{}\underline{-714} \\ \phantom{{x^2 }(()} 2859 \\ \phantom{{x}()}\underline{-2856} \\ \phantom{{x^2 +(3x}} 3 \\ \end{array}$$

The remainder when 9999 is divided by 357 is 3.

Therefore, 9999 - 3 = 9996.

Hence, the greatest 4-digit number which is exactly divisible by 357 is 9996.

The smallest 5-digit number = 10000.

To find the smallest 5-digit number exactly divisible by 279, we divide 10000 by 279 and add the difference between the divisor and remainder to 10000.

$$\begin{array}{l} \phantom{x^2+}{\quad 35} \\ 279\overline{\smash{\big)}10000} \\ \phantom{x^+(}\phantom{}\underline{-837} \\ \phantom{{x^2+(}} 1630 \\ \phantom{{x}+}\underline{-1395} \\ \phantom{{x^2 ++}} 235 \\ \end{array}$$

The remainder when 10000 is divided by 279 is 235.

Required number to be added = 279 - 235 = 44.

Number = 10000 + 44 = 10044.

Hence, the smallest 5-digit number which is exactly divisible by 279 is 10044.

Height of the pole = 10 m.

In each minute, the insect climbs 5 m and then slips down 4 m.

So, the net gain in height per minute = 5 - 4 = 1 m.

Let us track the insect's progress:

• At the end of 1^st minute: climbs to 5 m, slips down to 1 m.
• At the end of 2^nd minute: climbs to 6 m, slips down to 2 m.
• At the end of 3^rd minute: climbs to 7 m, slips down to 3 m.
• At the end of 4^th minute: climbs to 8 m, slips down to 4 m.
• At the end of 5^th minute: climbs to 9 m, slips down to 5 m.
• During the 6^th minute: starts at 5 m and climbs 5 m to reach 10 m (top of the pole).

Once the insect reaches the top, it does not slip down.

Hence, the insect will reach the top in 6 minutes.

The first twenty whole numbers are: 0, 1, 2, 3, ......, 19.

Sum of first twenty whole numbers:

Sum = 0 + 1 + 2 + 3 + ...... + 19 = 190.

Product of first twenty whole numbers:

Product = 0 × 1 × 2 × 3 × ...... × 19

Since one of the factors is 0, the product is 0.

⇒ Product = 0.

Comparing: 190 > 0.

Hence, the sum of the first twenty whole numbers (190) is greater than their product (0).

A whole number which is divisible by 2 and 4 may not necessarily be divisible by 8.

Reason: Let us consider some examples.

Take the number 12. It is divisible by 2 (12 ÷ 2 = 6) and divisible by 4 (12 ÷ 4 = 3). But 12 is not divisible by 8 (12 ÷ 8 = 1 remainder 4).

Similarly, take 20. It is divisible by 2 (20 ÷ 2 = 10) and divisible by 4 (20 ÷ 4 = 5). But 20 is not divisible by 8 (20 ÷ 8 = 2 remainder 4).

Hence, if a whole number is divisible by 2 and 4 it may or may not be divisible by 8.