ML Aggarwal Class 6 Maths Solutions Chapter 02 Whole Numbers

 

Exercise 2.1

 

Question 1. Write the smallest whole number. Can you write the largest whole number?
Answer: The smallest whole number is 0. No, you cannot write the largest whole number because you can always find a bigger whole number by adding 1. For every whole number, there is always a larger one that follows it. So there is no largest whole number.
In simple words: Zero is the smallest whole number. There is no biggest whole number — you can always make one bigger.

Exam Tip: Remember that 0 is included as a whole number, and there is no upper limit — this concept is tested frequently.

 

Question 2. Write the successor of each of the following numbers:
(i) 3999
(ii) 378915
(iii) 5001299
Answer: To find the successor, add 1 to 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
In simple words: The successor is the number that comes right after. Just add 1 to any number to get its successor.

Exam Tip: Successor means "one more" — always add 1 to the given number. This is a quick, straightforward operation.

 

Question 3. Write the predecessor of each of the following numbers:
(i) 500
(ii) 38794
(iii) 54789011
Answer: To find the predecessor, subtract 1 from 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
In simple words: The predecessor is the number that comes right before. Just subtract 1 from any number to get its predecessor.

Exam Tip: Predecessor means "one less" — always subtract 1 from the given number.

 

Question 4. Write the whole number (in each of the following) whose successor is:
(i) 50795
(ii) 720300
(iii) 8300000
Answer: If a number's successor is given, find the original number by subtracting 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
In simple words: If you know the successor, subtract 1 to find the original number.

Exam Tip: This is the reverse of finding a successor — subtract 1 from the given number to recover the original.

 

Question 5. Write the whole number (in each of the following) whose predecessor is:
(i) 5347
(ii) 72399
(iii) 3012999
Answer: If a number's predecessor is given, find the original number by adding 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
In simple words: If you know the predecessor, add 1 to find the original number.

Exam Tip: This is the reverse of finding a predecessor — add 1 to the given number to recover the original.

 

Question 6. Write next three consecutive whole numbers of the following numbers:
(i) 79
(ii) 598
(iii) 35669
Answer: To find three consecutive whole numbers, add 1, 2, and 3 to the given number.
(i) The next three consecutive whole numbers of 79 are:
79 + 1, 79 + 2, 79 + 3, which gives 80, 81, 82
(ii) The next three consecutive whole numbers of 598 are:
598 + 1, 598 + 2, 598 + 3, which gives 599, 600, 601
(iii) The next three consecutive whole numbers of 35669 are:
35669 + 1, 35669 + 2, 35669 + 3, which gives 35670, 35671, 35672
In simple words: Consecutive numbers follow one after another. Add 1, 2, and 3 to the starting number to get the next three.

Exam Tip: Consecutive means one right after another — there are no gaps between them.

 

Question 7. Write three consecutive whole numbers occurring just before 320001
Answer: To find three consecutive numbers before a given number, subtract 3, 2, and 1 from it.
Three consecutive whole numbers occurring just before 320001 are:
320001 - 1, 320001 - 2, 320001 - 3, which gives 320000, 319999, 319998
In simple words: To find the three numbers right before a number, subtract 1, 2, and 3 from it.

Exam Tip: When finding numbers "before" a given number, work backwards by subtracting from the given number.

 

Question 8. (i) How many whole numbers are there between 38 and 68?
(ii) How many whole numbers are there between 99 and 300?
Answer:
(i) The whole numbers between 38 and 68 are: 39, 40, 41, ..., 67
To count them, use the formula: Last number - First number + 1 = 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
To count them, use the formula: Last number - First number + 1 = 299 - 100 + 1 = 200
Hence, there are 200 whole numbers between 99 and 300.
In simple words: To count numbers between two numbers, find the last one, find the first one, subtract and add 1.

Exam Tip: The formula (Last - First + 1) always gives the count. Remember to add 1 to include both boundary points in the count.

 

Question 9. Write all whole numbers between 100 and 200 which do not change if the digits are written in reverse order.
Answer: A whole number that remains the same when its digits are written backwards is called a palindrome. A 3-digit palindrome has the form 'aba', where the first and last digits are the same. For a number to fall between 100 and 200, the first digit must be 1, so the last digit must also be 1. The middle 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
In simple words: These are numbers that read the same forwards and backwards. Between 100 and 200, they all have 1 at both ends and any digit in the middle.

Exam Tip: Know the term "palindrome" — it appears in many questions. For a 3-digit palindrome aba, the outer digits must match.

 

Question 10. How many 2-digit whole numbers are there between 5 and 92?
Answer: The two-digit whole numbers between 5 and 92 are: 10, 11, 12, ..., 91
To count them, use the formula: Last number - First number + 1 = 91 - 10 + 1 = 82
Hence, there are 82 two-digit whole numbers between 5 and 92.
In simple words: Two-digit numbers start from 10 and go up to 91 in this range. Count them using the formula.

Exam Tip: When asked for numbers with a certain number of digits in a range, identify the first and last such numbers, then use the counting formula.

 

Question 11. How many 3-digit whole numbers are there between 72 and 407?
Answer: The 3-digit whole numbers between 72 and 407 are: 100, 101, 102, ..., 406
To count them, use the formula: Last number - First number + 1 = 406 - 100 + 1 = 307
Hence, there are 307 three-digit whole numbers between 72 and 407.
In simple words: Three-digit numbers in this range go from 100 to 406. Use the counting formula to find how many there are.

Exam Tip: Always identify the smallest and largest numbers of the required type (e.g., 3-digit) within the given range, then apply the counting formula.

 

Exercise 2.2

 

Question 1. Fill in the blanks to make each of the following a true statement:
(i) 378 + 1024 = 1024 + ....
(ii) 337 + (528 + 1164) = (337 + ....) + 1164
(iii) (21 + 18) + .... = (21 + 13) + 18
(iv) 3056 + 0 = .... = 0 + 3056
Answer:
(i) By the Commutative Property of Addition (a + b = b + a):
378 + 1024 = 1024 + 378

(ii) By 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) By the Additive Identity property (a + 0 = a = 0 + a):
3056 + 0 = 3056 = 0 + 3056
In simple words: These properties show different ways to add numbers. You can swap the order, regroup them, or add zero without changing the result.

Exam Tip: Know the names of these properties — Commutative, Associative, and Identity — as they are tested by name in many questions.

 

Question 2. Add the following numbers and check by reversing the order of addends:
(i) 3189 + 53885
(ii) 33789 + 50311
Answer:
(i) 3189 + 53885

First addition:
3189
+ 53885
________
57074

Reversing the order (53885 + 3189):
53885
+ 3189
________
57074

In both cases, the sum is the same. Hence, 3189 + 53885 = 57074 and the result is verified.

(ii) 33789 + 50311

First addition:
33789
+ 50311
________
84100

Reversing the order (50311 + 33789):
50311
+ 33789
________
84100

In both cases, the sum is the same. Hence, 33789 + 50311 = 84100 and the result is verified.
In simple words: When you add two numbers in either order, you get the same answer. This is the Commutative Property in action.

Exam Tip: Reversing the order is a simple way to check addition. If both orders give the same sum, your answer is correct.

 

Question 3. By suitable arrangements, find the sum of:
(i) 311, 528, 289
(ii) 723, 834, 66, 277
(iii) 78, 203, 435, 7197, 422
Answer:
(i) 311 + 528 + 289

Rearranging to group numbers that add nicely:
(311 + 289) + 528 = 600 + 528 = 1128

(ii) 723 + 834 + 66 + 277

Rearranging:
(723 + 277) + (834 + 66) = 1000 + 900 = 1900

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

Rearranging:
(78 + 422) + (203 + 7197) + 435 = 500 + 7400 + 435 = 8335
In simple words: Rearrange the numbers so you add ones that make round numbers or easy sums. This makes the calculation faster and easier.

Exam Tip: Look for pairs of numbers that add up to 10, 100, 1000, etc. — these make mental math much faster and reduce mistakes.

 

Question 4. Fill in the blanks to make each of the following a true statement:
(i) 375 × 57 = 57 × ....
(ii) (33 × 16) × 25 = 33 × (.... × 25)
(iii) 37 × 24 = 37 × 18 + 37 × ....
(iv) 7205 × 1 = .... = 1 × 7205
(v) 366 × 0 = ....
(vi) .... × 579 = 0
(vii) 473 × 108 = 473 × 100 + 473 × ....
(viii) 684 × 97 = 684 × 100 - .... × 3
(ix) 0 ÷ 5 = ....
(x) (14 - 14) ÷ 7 = ....
Answer:
(i) By the Commutative Property of Multiplication (a × b = b × a):
375 × 57 = 57 × 375

(ii) By the Associative Law of Multiplication ((a × b) × c = a × (b × c)):
(33 × 16) × 25 = 33 × (16 × 25)

(iii) Using the Distributive Law (a × (b + c) = a × b + a × c). Since 24 = 18 + 6:
37 × 24 = 37 × 18 + 37 × 6

(iv) By the Multiplicative Identity property (a × 1 = a = 1 × a):
7205 × 1 = 7205 = 1 × 7205

(v) By the Multiplication by Zero property (a × 0 = 0):
366 × 0 = 0

(vi) Any number times 0 equals 0:
0 × 579 = 0

(vii) Using the Distributive Law. Since 108 = 100 + 8:
473 × 108 = 473 × 100 + 473 × 8

(viii) Using the Distributive Law of Multiplication over Subtraction. Since 97 = 100 - 3:
684 × 97 = 684 × (100 - 3) = 684 × 100 - 684 × 3

(ix) Zero divided by any non-zero whole number equals 0:
0 ÷ 5 = 0

(x) Since 14 - 14 = 0:
(14 - 14) ÷ 7 = 0 ÷ 7 = 0
In simple words: These properties tell us how multiplication and division behave. Using them makes calculations easier and helps check our work.

Exam Tip: Learn these property names and what they mean. Questions often test whether you can identify and apply the right property to simplify a calculation.

 

Question 5. Determine the following products by suitable arrangement:
(i) 4 × 528 × 25
(ii) 625 × 239 × 16
(iii) 125 × 40 × 8 × 25
Answer:
(i) 4 × 528 × 25

Rearrange to group numbers that multiply to make round numbers:
528 × (4 × 25) = 528 × 100 = 52800

(ii) 625 × 239 × 16

Rearrange:
239 × (625 × 16) = 239 × 10000 = 2390000

(iii) 125 × 40 × 8 × 25

Rearrange:
(125 × 8) × (40 × 25) = 1000 × 1000 = 1000000
In simple words: Group the numbers so you multiply ones that make 10, 100, 1000, or other round numbers. This makes the problem much easier.

Exam Tip: Look for pairs like (4 × 25 = 100), (125 × 8 = 1000), or (625 × 16 = 10000) — recognizing these shortcuts saves time on exams.

 

Question 6. Find the value of the following:
(i) 54279 × 92 + 54279 × 8
(ii) 60678 × 262 - 60678 × 162
Answer:
(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

(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
In simple words: When the same number multiplies multiple terms being added or subtracted, factor it out. This simplifies the calculation greatly.

Exam Tip: The Distributive Law is powerful for mental math — always look for a common factor you can pull out to simplify the problem.

 

Question 7. Find the following products by using suitable properties:
(i) 739 × 102
(ii) 1938 × 99
(iii) 1005 × 188
Answer:
(i) 739 × 102

⇒ 739 × (100 + 2)

Using the Distributive Law of Multiplication over Addition:

⇒ 739 × 100 + 739 × 2

⇒ 73900 + 1478

⇒ 75378

Therefore, 739 × 102 = 75378.

(ii) 1938 × 99

⇒ 1938 × (100 - 1)

Using the Distributive Law of Multiplication over Subtraction:

⇒ 1938 × 100 - 1938 × 1

⇒ 193800 - 1938

⇒ 191862

Therefore, 1938 × 99 = 191862.

(iii) 1005 × 188

⇒ (1000 + 5) × 188

Using the Distributive Law of Multiplication over Addition:

⇒ 1000 × 188 + 5 × 188

⇒ 188000 + 940

⇒ 188940

Therefore, 1005 × 188 = 188940.
In simple words: Break the number into easier pieces (like 102 = 100 + 2), multiply each piece separately, then add or subtract the results together.

Exam Tip: The Distributive Property is key - always rewrite numbers close to powers of 10 (100, 1000, 99, 999) as addition or subtraction to make mental calculation faster.

 

Question 8. Divide 7750 by 17 and check the result by division algorithm.
Answer:
Dividend = 7750

Divisor = 17

Quotient = 455

Remainder = 15

Verification: Dividend = (Divisor × Quotient) + Remainder

Substituting values:

(Divisor × Quotient) + Remainder = (17 × 455) + 15

= 7735 + 15

= 7750

Since L.H.S. = R.H.S., the result is verified by the division algorithm.
In simple words: After you divide and get a quotient and remainder, multiply the quotient by the divisor and add the remainder back - you should get the original number again.

Exam Tip: Always verify division using the formula: Dividend = (Divisor × Quotient) + Remainder. This checks that your division work is correct.

 

Question 9. Find the number which when divided by 38 gives the quotient 23 and remainder 17
Answer:
Given:

Divisor = 38

Quotient = 23

Remainder = 17

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

= (38 × 23) + 17

= 874 + 17

= 891

Therefore, the required number = 891.
In simple words: Multiply the divisor by the quotient, then add the remainder to find the original number that was divided.

Exam Tip: Memorize and use the division algorithm formula correctly - this is a direct application question that tests understanding of division structure.

 

Question 10. Which least number should be subtracted from 1000 so that the difference is exactly divisible by 35?
Answer:
To find the least number to be subtracted from 1000 so that the difference becomes exactly divisible by 35, we need to divide 1000 by 35 and find the remainder.

When 1000 is divided by 35, the remainder is 20.

Therefore, the least number to subtract from 1000 = 20.

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

Hence, the least number to be subtracted from 1000 is 20.
In simple words: Divide 1000 by 35 and look at the remainder - that remainder is the number you need to take away to get a result divisible by 35.

Exam Tip: The remainder when a number is divided is always the amount you need to subtract (or add) to make it divisible by that divisor.

 

Question 11. Which least number should be added to 1000 so that 53 divides the sum exactly?
Answer:
To find the least number to be added to 1000 so that the sum becomes exactly divisible by 53, we divide 1000 by 53 and find the remainder.

When 1000 is divided by 53, the remainder 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.
In simple words: Divide 1000 by 53 to find the remainder. Subtract that remainder from the divisor (53) to get the number you must add.

Exam Tip: When adding: use (Divisor - Remainder). When subtracting: use the Remainder itself. This distinction is critical for these divisibility problems.

 

Question 12. Find the largest three-digit number which is exactly divisible by 47
Answer:
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.

When 999 is divided by 47, the remainder is 12.

Therefore, 999 - 12 = 987.

Hence, the largest three-digit number which is exactly divisible by 47 is 987.
In simple words: Start with the biggest three-digit number (999), divide it by 47, and subtract the remainder to get an answer that 47 divides evenly.

Exam Tip: For the "largest number divisible by X," always start with the largest value in that range and subtract the remainder from division.

 

Question 13. Find the smallest five-digit number which is exactly divisible by 254
Answer:
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 the remainder to 10000.

When 10000 is divided by 254, the remainder is 94.

The required number to be added = 254 - 94 = 160.

Therefore, the number = 10000 + 160 = 10160.

Hence, the smallest five-digit number which is exactly divisible by 254 is 10160.
In simple words: Start with the smallest five-digit number (10000), divide it by 254, then add (Divisor - Remainder) to reach the next number divisible by 254.

Exam Tip: For the "smallest number divisible by X," begin with the smallest value in that range and add (Divisor - Remainder) to find the answer.

 

Question 14. A vendor supplies 72 litres of milk to a student's hostel in the morning and 28 litres of milk in the evening every day. If the milk costs Rs. 60 per litre, how much money is due to the vendor per day?
Answer:
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 = Rs. 60.

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

Hence, the money due to the vendor per day is Rs. 6000.
In simple words: Add the morning and evening milk amounts, then multiply the total by the cost per litre to find the daily payment.

Exam Tip: Break word problems into clear steps - find the total quantity first, then multiply by the unit cost. Always include units (litres, rupees) in your working.

 

Question 15. State whether the following statements are true (T) or false (F):
(i) If the product of two whole numbers is zero, then at least one of them will be zero.
(ii) If the product of two whole numbers is 1, then each of them must be equal to 1.
(iii) If a and b are whole numbers such that a ≠ 0 and b ≠ 0, then ab may be zero.
Answer:
(i) True.

Reason: When multiplying whole numbers, if the product equals zero, at least one factor must be zero. For example, 0 × 5 = 0 and 7 × 0 = 0.

(ii) True.

Reason: The only pair of whole numbers whose product is 1 consists of both numbers being 1, because 1 × 1 = 1. No other whole numbers produce a product of 1.

(iii) False.

Reason: When a ≠ 0 and b ≠ 0, both a and b are non-zero whole numbers. Multiplying two non-zero whole numbers always gives a non-zero result. Therefore, ab cannot equal zero.
In simple words: (i) A product is zero only if at least one number is zero. (ii) Only 1 × 1 = 1 among whole numbers. (iii) Two non-zero numbers never multiply to give zero.

Exam Tip: These statements test understanding of multiplicative properties - particularly the zero product property and the identity for multiplication. State true/false clearly and always give a mathematical reason.

 

Question 16(i). Replace each * by the correct digit in each of the following:

Answer:
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:


Verification: 3974 + 5276 = 9250.
In simple words: Work through each column from right to left. When a sum is 10 or more, write the ones digit and carry 1 to the next column.

Exam Tip: Always solve these column by column, starting from the units place. Check carries carefully - they are the most common source of errors in these problems.

 

Question 16(ii). Replace each * by the correct digit in each of the following:

Answer:
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:


Verification: 6502 - 2045 = 4457.
In simple words: Work right to left. When the top digit is smaller than the bottom digit, borrow 10 from the next column to the left.

Exam Tip: In subtraction problems with blanks, work carefully through borrowing - track which columns borrowed and which columns lent. Verify by adding the answer and the subtrahend to get the original number.

 

Question 16(iii). Replace each * by the correct digit in each of the following:

Answer:
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, which is omitted as the leftmost digit.

Hence, the completed subtraction is:


Verification: 170014 - 81347 = 88667.
In simple words: This is a multi-step borrowing problem. At each column where the top digit is smaller, borrow from the next column left, then continue solving.

Exam Tip: For long subtraction with multiple borrows, mark each borrow clearly as you move left. Double-check by adding: Difference + Subtrahend should equal Minuend.

 

Exercise 2.3

 

Question 1. Using shorter method, find:
(i) 3246 + 9999
(ii) 7501 + 99999
(iii) 5377 - 999
(iv) 25718 - 9999
(v) 123 × 999
(vi) 203 × 9999
Answer:
(i) 3246 + 9999

⇒ 3246 + (10000 - 1)

⇒ 3246 + 10000 - 1

⇒ 13246 - 1

⇒ 13245

Therefore, 3246 + 9999 = 13245.

(ii) 7501 + 99999

⇒ 7501 + (100000 - 1)

⇒ 7501 + 100000 - 1

⇒ 107501 - 1

⇒ 107500

Therefore, 7501 + 99999 = 107500.

(iii) 5377 - 999

⇒ 5377 - (1000 - 1)

⇒ 5377 - 1000 + 1

⇒ 4377 + 1

⇒ 4378

Therefore, 5377 - 999 = 4378.

(iv) 25718 - 9999

⇒ 25718 - (10000 - 1)

⇒ 25718 - 10000 + 1

⇒ 15718 + 1

⇒ 15719

Therefore, 25718 - 9999 = 15719.

(v) 123 × 999

⇒ 123 × (1000 - 1)

⇒ 123 × 1000 - 123 × 1

⇒ 123000 - 123

⇒ 122877

Therefore, 123 × 999 = 122877.

(vi) 203 × 9999

⇒ 203 × (10000 - 1)

⇒ 203 × 10000 - 203 × 1

⇒ 2030000 - 203

⇒ 2029797

Therefore, 203 × 9999 = 2029797.
In simple words: Rewrite 9999 as 10000 - 1, 999 as 1000 - 1, and so on. Then use the Distributive Property to break the calculation into simpler steps.

Exam Tip: The "shorter method" is about recognizing patterns with 9s - always express 999, 9999 as (power of 10 - 1) to simplify operations.

 

Question 2. Without using a diagram, find:
(i) 9th square number
(ii) 7th triangular number
Answer: (i) The formula for the nth square number is n × n = n². So, the 9th square number = 9 × 9 = 9² = 81.
(ii) The nth triangular number is given by \( \frac{n(n + 1)}{2} \). So, the 7th triangular number = \( \frac{7 \times (7 + 1)}{2} = \frac{7 \times 8}{2} = \frac{56}{2} = 28 \).
In simple words: To find the 9th square number, multiply 9 by itself to get 81. To find the 7th triangular number, use the formula and get 28.

Exam Tip: Remember that square numbers follow n², while triangular numbers follow \( \frac{n(n+1)}{2} \). Always apply the correct formula for the given position.

 

Question 3. (i) Can a rectangular number be a square number?
(ii) Can a triangular number be a square number?
Answer: (i) Yes, a rectangular number may also be a square number. The reason is that a number can be arranged in both a rectangular and a square pattern. For instance, 36 fits both patterns: it can be arranged as 6 × 6 (square), and also as 4 × 9 or 3 × 12 (rectangular).
(ii) Yes, a triangular number may also be a square number. Some numbers belong to both categories. For example, 1 serves as both the 1st triangular number and the 1st square number. Similarly, 36 functions as both a triangular number (the 8th triangular number = \( \frac{8 \times 9}{2} = 36 \)) and a square number (6² = 36).
In simple words: Yes to both questions. Some numbers can be arranged in more than one way - both as rectangles and squares, or both as triangles and squares. The number 36 is one example that works both ways.

Exam Tip: When answering questions about overlapping number categories, always provide a concrete example (like 36) to demonstrate the concept clearly.

 

Question 4. Observe the following pattern and fill in the blanks:
1 × 9 + 1 = 10
12 × 9 + 2 = 110
123 × 9 + 3 = 1110
1234 × 9 + 4 = .......
12345 × 9 + 5 = .......
Answer: By observing the pattern, the result contains 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.
In simple words: Look at how many digits are in the first number you multiply by 9. Your answer will have that many 1s, then a 0. When you add the final number, it becomes all 1s.

Exam Tip: Pattern questions reward careful observation. Always list at least 2-3 examples to identify the rule before writing your answer.

 

Question 5. Observe the following pattern and fill in the blanks:
9 × 9 + 7 = 88
98 × 9 + 6 = 888
987 × 9 + 5 = 8888
9876 × 9 + 4 = .......
98765 × 9 + 3 = .......
Answer: By observing the pattern, each line gives a result that is made of repeated 8s. The count of 8s equals the number of digits in the first multiplicand plus 1.

9876 × 9 + 4 = 88888.

98765 × 9 + 3 = 888888.
In simple words: Count the digits in the first number. Your answer will have that many 8s, plus one more 8. So five digits gives you six 8s.

Exam Tip: In pattern completion, verify your answer by working backwards - multiply the given 8s by 9 and subtract the final digit to check if you get the starting number.

 

Question 6. Look at the following figures made up of dots: These figures show the arrangement of the numbers 2 × 3 or 3 × 2; 2 × 4 or 4 × 2 i.e. the numbers 6 and 8. The numbers 6 and 8 are called rectangular numbers. Can you write two more rectangular numbers?
Answer: A rectangular number is a number that can be arranged in a rectangle where the length and breadth are different from each other.

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)
In simple words: A rectangular number is one you can arrange into a rectangle shape by putting dots in rows and columns. The number 10 and the number 12 are two examples of rectangular numbers.

Exam Tip: When identifying rectangular numbers, confirm that both arrangements (a × b and b × a) produce the same result, and that the length and breadth are unequal.

 

Mental Maths

 

Question 1. Fill in the blanks:
(i) A whole number is less than all those whole numbers that lie to its .... on the number line.
(ii) One more than a given whole is called its ....
(iii) There is at least one whole number between two .... whole numbers.
(iv) 738 × 335 = 738 × (300 + 30 + ....)
(v) If a is a non-zero whole number and a × a = a, then a = ....
(vi) .... is the only whole number which is not a natural number.
(vii) The additive identity in whole numbers is ....
(viii) .... is the successor of the largest 3-digit number.
(ix) Division of a whole number by .... is not defined.
Answer: (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 at least 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.
In simple words: Whole numbers increase as you move right on the number line. Zero is not a natural number. You cannot divide by zero. The successor of 999 is 1000.

Exam Tip: These fundamental definitions appear frequently on exams. Memorise the terms "successor", "predecessor", "additive identity", and remember that division by zero is undefined.

 

Question 2. State whether the following statements are true (T) or false (F):
(i) The successor of a one-digit number is always a one-digit number.
(ii) The predecessor of every two-digit number is a one-digit number.
(iii) The predecessor of a 3-digit number is always a 3-digit number.
(iv) The successor of a 3-digit number is always a 3-digit number.
(v) If a is any whole number, then a ÷ a = 1
(vi) If a is any non-zero whole number, then 0 ÷ a = 0
(vii) On adding two different whole numbers, we always get a natural number.
(viii) Between two whole numbers there is a whole number.
(ix) There is a natural number which when added to a natural number, gives that number.
(x) If the product of two whole numbers is zero, then at least one of them is zero.
(xi) Any non-zero whole number divided by itself gives the quotient 1.
Answer: (i) False. The successor of 9 (a one-digit number) is 10, which is a two-digit number.

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

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

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

(v) False. If a = 0, then a ÷ a = 0 ÷ 0 is not defined. So, the statement does not hold for all whole numbers.

(vi) True. Zero divided by any non-zero whole number is always 0. For example, 0 ÷ 5 = 0.

(vii) True. When two different whole numbers are added, at least one of them must be non-zero. So, the sum is at least 1, which makes it a natural number.

(viii) False. Between two consecutive whole numbers (such as 5 and 6), there is no whole number at all.

(ix) False. The number 0, when added to any natural number, gives that number. However, 0 itself is not a natural number. No natural number acts as the additive identity for natural numbers.

(x) True. If a × b = 0, then either a = 0 or b = 0 (or both).

(xi) True. Any non-zero whole number a satisfies a ÷ a = 1, since a × 1 = a.
In simple words: When checking these statements, look for counterexamples (cases where the rule breaks). For instance, 9 + 1 = 10 (two digits), so statement (i) is false. Between 5 and 6, there is no whole number, so statement (viii) is false.

Exam Tip: For true/false questions with reasons, always provide a concrete example or calculation as proof. This makes your answer clear and earns full marks.

 

Question 3. The whole number which does not have a predecessor in whole number system is
(a) 0
(b) 1
(c) 2
(d) none of the options
Answer: (a) 0
In simple words: The smallest whole number is 0, so it has no whole number before it. Therefore, 0 has no predecessor in the whole number system.

Exam Tip: Remember that 0 is the starting point of whole numbers. It has no predecessor, but it has a successor (which is 1).

 

Question 4. The predecessor of the smallest 4-digit number is
(a) 99
(b) 999
(c) 1000
(d) 1001
Answer: (b) 999
In simple words: The smallest 4-digit number is 1000. One less than 1000 is 999, which is the predecessor.

Exam Tip: For predecessor and successor questions, identify the target number first, then subtract 1 (for predecessor) or add 1 (for successor).

 

Question 5. The predecessor of 1 million is
(a) 9999
(b) 99999
(c) 999999
(d) 1000001
Answer: (c) 999999
In simple words: One million is 10,00,000 (1 followed by six zeros). One less than this is 9,99,999 (six 9s), which is the predecessor.

Exam Tip: When working with large numbers in Indian notation (like 10,00,000), count the digits carefully to avoid mistakes.

 

Question 6. The product of the predecessor and the successor of the greatest 2-digit number is
(a) 9900
(b) 9800
(c) 9700
(d) none of the options
Answer: (b) 9800
In simple words: The greatest 2-digit number is 99. Its predecessor is 98, and its successor is 100. Multiplying these: 98 × 100 = 9800.

Exam Tip: Break multi-step problems into stages: first find the target number, then find its predecessor and successor, then perform the required operation.

 

Question 7. The sum of the successor of the greatest 3-digit number and the predecessor of the smallest 3-digit number is
(a) 1000
(b) 1100
(c) 1101
(d) 1099
Answer: (d) 1099
In simple words: The greatest 3-digit number is 999, and its successor is 1000. The smallest 3-digit number is 100, and its predecessor is 99. Adding: 1000 + 99 = 1099.

Exam Tip: Write down each step clearly: identify the two numbers, find their predecessor/successor, then perform the final addition or multiplication.

 

Question 8. The number of whole numbers between 22 and 54 is
(a) 30
(b) 31
(c) 32
(d) 42
Answer: (b) 31
In simple words: The whole numbers between 22 and 54 are 23, 24, 25, ..., 53. To count them, subtract: 53 - 23 + 1 = 31.

Exam Tip: When counting numbers in a range, use the formula: (last number - first number + 1). The +1 accounts for including both the first and last numbers in the count.

 

Question 9. The number of whole numbers between the smallest whole number and the greatest 2-digit number is
(a) 100
(b) 99
(c) 98
(d) 88
Answer: (c) 98
In simple words: The smallest whole number is 0, and the greatest 2-digit number is 99. The whole numbers between them are 1, 2, 3, ..., 98. Counting: 98 - 1 + 1 = 98.

Exam Tip: Pay attention to the phrase "between" - it typically excludes the boundary numbers (0 and 99 in this case), so count only from 1 to 98.

 

Question 10. If a is whole number such that a + a = a, then a is equal to
(1) 0
(2) 1
(3) 2
(4) none of these
Answer: (1) 0
In simple words: When you add a number to itself and get the same number back, that number must be zero. This is the only whole number that works this way.

Exam Tip: Test the statement a + a = a by rearranging to 2a - a = 0, which directly gives a = 0. This algebraic approach is faster than checking each option.

 

Question 11. The value of (93 × 63 + 93 × 37) is
(1) 930
(2) 9300
(3) 93000
(4) none of these
Answer: (2) 9300
In simple words: Both terms have 93 in them. Pull out the 93, add what is left (63 + 37 = 100), then multiply: 93 × 100 = 9300.

Exam Tip: Look for a common factor in all terms — here, 93 appears in both products. Use the distributive law to factor it out, then simplify the remaining numbers inside the brackets.

 

Question 12. Which of the following is not equal to zero?
(1) 0 × 5
(2) 0 ÷ 5
(3) (10 - 10) ÷ 5
(4) (5 - 0) ÷ 5
Answer: (4) (5 - 0) ÷ 5
In simple words: Evaluate each choice step by step. The first three all give zero. The last one gives 5 ÷ 5 = 1, which is not zero.

Exam Tip: When checking each option, simplify the brackets first, then carry out the division or multiplication. Always watch the order of operations.

 

Question 13. Which of the following statement is true?
(1) 21 - (13 - 5) = (21 - 13) - 5
(2) 21 - 13 is not a whole number
(3) 21 × 1 = 21 × 0
(4) 13 - 21 is not a whole number
Answer: (4) 13 - 21 is not a whole number
In simple words: Work through each statement. Option (4) is correct because 13 - 21 = -8, which is negative and not a whole number. Whole numbers start from zero and go up.

Exam Tip: Remember that the set of whole numbers {0, 1, 2, 3, ...} does not include negative numbers. Subtracting a larger number from a smaller one takes you outside the whole numbers.

 

Question 14. If p and q are two whole numbers, then which of the following may not be a whole number?
(1) p + q
(2) p - q
(3) p + 2q
(4) p × q
Answer: (2) p - q
In simple words: Addition and multiplication of whole numbers always give whole numbers. Adding a multiple also keeps you in whole numbers. But subtracting might not — for example, 3 - 5 = -2, which is not a whole number.

Exam Tip: Look for closure properties. Addition, multiplication, and addition with multiples are closed under whole numbers, but subtraction is not — whenever you subtract a larger number from a smaller one, you step outside the whole number set.

 

Question 15. On dividing a number by 9 we get 47 as quotient and 5 as remainder. The number is
(1) 418
(2) 428
(3) 429
(4) none of these
Answer: (2) 428
In simple words: Use the division formula: the original number equals divisor times quotient, plus the remainder. So: (9 × 47) + 5 = 423 + 5 = 428.

Exam Tip: Always apply the division algorithm formula correctly: Dividend = (Divisor × Quotient) + Remainder. Verify your answer by dividing it back by the divisor to check you get the same quotient and remainder.

 

Question 16. By using dot (•) pattern, which of the following numbers can be arranged in two ways namely a triangle and a rectangle?
(1) 12
(2) 11
(3) 10
(4) 9
Answer: (3) 10
In simple words: Triangular numbers (1, 3, 6, 10, 15, ...) can be arranged as a triangle shape with dots. Ten is also a triangular number and can be arranged in a rectangle as 2 rows of 5 dots (10 = 2 × 5). So 10 works both ways.

Exam Tip: Triangular numbers follow the pattern 1 + 2 + 3 + ... + n. For a number to be both triangular and rectangular, it must be a triangular number AND have at least two different factor pairs (rows and columns). Check 10: it is triangular, and 10 = 2 × 5 is a valid rectangle.

 

Question 17. Statement I: 41 - 1 = 40
Statement II: When we subtract 1 from the predecessor of any whole number, we get the original number.
(1) Statement I is true but statement II is false.
(2) Statement I is false but statement II is true.
(3) Both Statement I and statement II are true.
(4) Both Statement I and statement II are false.
Answer: (1) Statement I is true but statement II is false
In simple words: Statement I is clearly correct: 41 - 1 = 40. For Statement II, take the number 41. Its predecessor is 40. When you subtract 1 from 40, you get 39, not the original 41. So Statement II is wrong.

Exam Tip: Understand the terms: predecessor of n is (n - 1). When you subtract 1 from the predecessor, you get (n - 1) - 1 = n - 2, which is 2 less than the original, not equal to it. Test with a concrete example to verify before choosing your answer.

 

Question 18. Statement I: 4 × (5 + 6) = 4 × 5 + 4 × 6
Statement II: If a, b and c are three whole numbers then a × (b + c) = a × b + a × c
(1) Statement I is true but statement II is false.
(2) Statement I is false but statement II is true.
(3) Both Statement I and statement II are true.
(4) Both Statement I and statement II are false.
Answer: (3) Both Statement I and statement II are true
In simple words: Statement I checks out: 4 × 11 = 44, and 20 + 24 = 44. Statement II is the distributive law, a core rule for whole numbers. Both statements are correct.

Exam Tip: The distributive law (Statement II) is one of the fundamental properties of whole numbers. Verify Statement I by computing both sides separately to confirm they match, then recognize that Statement II is the general form of this same property.

 

Question 19. Statement I: 3 ÷ 0 = 0
Statement II: When we divide a whole number by 0, we get 0.
(1) Statement I is true but statement II is false.
(2) Statement I is false but statement II is true.
(3) Both Statement I and statement II are true.
(4) Both Statement I and statement II are false.
Answer: (4) Both Statement I and statement II are false
In simple words: Division by zero is not allowed in math. You cannot divide any number by zero — it is undefined, not zero. Both statements incorrectly claim you can divide by zero and get a result.

Exam Tip: Never forget: division by zero is undefined. It is not zero, it is not infinity, it is simply not defined. This is a fundamental rule that appears in every algebra course and on almost every exam.

 

Question 20. Statement I: 66 = 12 × 5 + 6
Statement II: Dividend = divisor × remainder + quotient, for all whole numbers where divisor ≠ 0
(1) Statement I is true but statement II is false.
(2) Statement I is false but statement II is true.
(3) Both Statement I and statement II are true.
(4) Both Statement I and statement II are false.
Answer: (1) Statement I is true but statement II is false
In simple words: Statement I is correct: 12 × 5 = 60, and 60 + 6 = 66. Statement II has the roles of quotient and remainder swapped. The correct formula is: Dividend = Divisor × Quotient + Remainder, not divisor times remainder plus quotient.

Exam Tip: The division algorithm is critical: Dividend = Divisor × Quotient + Remainder (in that order). If quotient and remainder are reversed, the formula is wrong. Always write it out carefully to avoid swapping these terms.

 

Check Your Progress

 

Question 1. Write next three consecutive whole numbers of the number 9998
Answer: The next three consecutive whole numbers after 9998 are obtained by adding 1, 2, and 3 to 9998. These are 9999, 10000, and 10001.
In simple words: Start at 9998 and count up three more numbers: 9999, 10000, 10001.

Exam Tip: To find consecutive whole numbers after a given number, simply add 1, 2, 3, ... successively. Make sure you list them in order and do not skip any.

 

Question 2. Write three consecutive whole numbers occurring just before 567890
Answer: The three consecutive whole numbers that come just before 567890 are found by subtracting 1, 2, and 3 from 567890. These are 567889, 567888, and 567887 (in descending order).
In simple words: Count backwards from 567890: the three numbers before it are 567889, 567888, 567887.

Exam Tip: To find consecutive whole numbers before a given number, subtract 1, 2, 3, ... in order. List them either in descending order (as they appear before the target) or in ascending order (if the question asks for them in sequence).

 

Question 3. Find the product of the successor and the predecessor of the smallest number of 3-digits.
Answer: The smallest 3-digit number is 100. Its successor is 101, and its predecessor is 99. The product is 101 × 99. Using the distributive property: 101 × 99 = 101 × (100 - 1) = 10100 - 101 = 9999.
In simple words: Find the number one more than 100 (which is 101) and one less than 100 (which is 99). Multiply them together to get 9999.

Exam Tip: Recognize that products of the form (n + 1) × (n - 1) = n² - 1. Here, 101 × 99 = 100² - 1 = 10000 - 1 = 9999. This shortcut saves time over direct multiplication.

 

Question 4. Find the number of whole numbers between the smallest and the greatest numbers of 2-digits.
Answer: The smallest 2-digit number is 10 and the greatest is 99. The whole numbers between them (not including 10 and 99 themselves) are 11, 12, 13, ..., 98. The count is 98 - 11 + 1 = 88.
In simple words: Count all numbers from 11 to 98: that is 88 numbers in total.

Exam Tip: When counting from a to b (inclusive), the formula is b - a + 1. Here, from 11 to 98: 98 - 11 + 1 = 88. This avoids manual counting and is the fastest method.

 

Question 5. Find the following sum by suitable arrangements:
(i) 678 + 1319 + 322 + 5681
(ii) 777 + 546 + 1463 + 223 + 537
Answer:
(i) Rearrange as (678 + 322) + (1319 + 5681) = 1000 + 7000 = 8000.
(ii) Rearrange as (777 + 223) + (1463 + 537) + 546 = 1000 + 2000 + 546 = 3546.
In simple words: Look for pairs of numbers that add up to 1000 or other round numbers. Group them first, then add the remaining terms to make the calculation easier.

Exam Tip: Always scan for "complements" — pairs that sum to powers of 10 (like 678 + 322 = 1000). Rearrange and group these first using the commutative and associative properties of addition. This strategy reduces arithmetic errors.

 

Question 6. Determine the following products by suitable arrangements:
(i) 625 × 437 × 16
(ii) 309 × 25 × 7 × 8
Answer:
(i) Rearrange as 437 × (625 × 16) = 437 × 10000 = 4370000.
(ii) Rearrange as (309 × 7) × (25 × 8) = 2163 × 200 = 432600.
In simple words: Look for pairs of numbers whose product is a power of 10 or a round number. Group those pairs first, multiply them, then multiply by the remaining term.

Exam Tip: Identify "factor pairs" that make round numbers: 625 × 16 = 10000, and 25 × 8 = 200. Rearrange using the commutative and associative properties to bring these pairs together. This reduces complex multiplication to simpler steps.

 

Question 7. Find the value of the following by using suitable properties:
(i) 236 × 414 + 236 × 563 + 236 × 23
(ii) 370 × 1587 - 37 × 10 × 587
Answer:
(i) Factor out 236: 236 × (414 + 563 + 23) = 236 × 1000 = 236000.
(ii) Rewrite 370 as 37 × 10 and factor: 37 × 10 × 1587 - 37 × 10 × 587 = 37 × 10 × (1587 - 587) = 370 × 1000 = 370000.
In simple words: In (i), the number 236 appears in every term, so pull it out as a common factor. In (ii), both terms share 37 × 10, so factor that out first, then subtract the remaining numbers inside the brackets.

Exam Tip: Always look for a common factor in all terms. Use the distributive law in reverse: a × b + a × c = a × (b + c). This transforms lengthy multiplications into simpler additions or subtractions inside brackets, followed by one final multiplication by the common factor.

 

Question 8. Divide 6528 by 29 and check the result by division algorithm.
Answer: When we divide 6528 by 29, we get a quotient of 225 with a remainder of 3. To verify using the division algorithm, we apply the formula: Dividend = (Divisor × Quotient) + Remainder. Substituting our values: (29 × 225) + 3 = 6525 + 3 = 6528. Since the left side equals the right side, our division result is confirmed.
In simple words: Divide 6528 by 29 to get 225 with 3 left over. Check it by multiplying 29 × 225 and adding 3 - you should get 6528 back.

Exam Tip: Always show the long division working clearly and then verify using the division algorithm formula - examiners award marks for showing both the calculation and the verification step.

 

Question 9. Find the greatest 4-digit number which is exactly divisible by 357.
Answer: The largest 4-digit number available is 9999. To find the greatest 4-digit number that divides evenly by 357, we divide 9999 by 357. This gives us a quotient of 28 with a remainder of 3. We then subtract this remainder from 9999: 9999 - 3 = 9996. Therefore, 9996 is the greatest 4-digit number exactly divisible by 357.
In simple words: Start with 9999. Divide it by 357 and see what's left over. Take away that leftover amount from 9999 to get your answer.

Exam Tip: When finding the greatest number divisible by a given divisor, always subtract the remainder from the largest possible number - never add anything.

 

Question 10. Find the smallest 5-digit number which is exactly divisible by 279.
Answer: The smallest 5-digit number is 10000. To find the smallest 5-digit number that divides evenly by 279, we divide 10000 by 279. This produces a quotient of 35 with a remainder of 235. We then find the difference: 279 - 235 = 44. Adding this to 10000 gives us: 10000 + 44 = 10044. Therefore, 10044 is the smallest 5-digit number exactly divisible by 279.
In simple words: Start with 10000. Divide it by 279 and find what's left over. Subtract that leftover from 279, then add the result to 10000.

Exam Tip: When finding the smallest number divisible by a given divisor, you must add the adjustment amount to the starting number - the opposite of the greatest number question.

 

Question 11. The height of a slippery pole is 10 m and an insect is trying to climb the pole. The insect climbs 5 m in one minute and then slips down by 4 m. In how much time will insect reach the top?
Answer: The pole measures 10 m in height. Each minute, the insect goes up 5 m but then slides back 4 m. The net progress each minute is 5 - 4 = 1 m. Tracking the insect's movement minute by minute: after 1 minute it reaches 5 m then falls to 1 m; after 2 minutes it climbs to 6 m then drops to 2 m; after 3 minutes it goes to 7 m then slips to 3 m; after 4 minutes it reaches 8 m then slides to 4 m; after 5 minutes it climbs to 9 m then falls to 5 m. During the 6th minute, starting from 5 m, the insect climbs 5 m upward to reach 10 m - the top of the pole. Once at the top, the insect stays there and does not slip back down. Therefore, the insect takes 6 minutes to reach the top.
In simple words: The insect gains 1 m of height each minute after slipping back. After 5 minutes it is at 5 m high. In the 6th minute it climbs 5 m more and reaches 10 m - done.

Exam Tip: The key insight is recognizing that once the insect reaches the top, it no longer slips back - so on the final minute, only the upward climb counts, not the slip.

 

Question 12. Which is greater, the sum of first twenty whole numbers or the product of first twenty whole numbers?
Answer: The first twenty whole numbers are 0, 1, 2, 3, ..., 19. When we add these together: 0 + 1 + 2 + 3 + ... + 19 = 190. When we multiply these together: 0 × 1 × 2 × 3 × ... × 19 = 0 (because one of the factors is 0, the entire product becomes 0). Comparing the two results: 190 is much greater than 0. Therefore, the sum of the first twenty whole numbers equals 190, which is greater than their product which equals 0.
In simple words: Adding the numbers 0 through 19 gives 190. Multiplying them gives 0 because zero is one of the numbers. 190 wins easily.

Exam Tip: Remember that whenever 0 appears as a factor in multiplication, the entire product becomes 0 - this is a trick many questions use to test understanding.

 

Question 13. If a whole number is divisible by 2 and 4, is it divisible by 8 also?
Answer: A number that divides evenly by 2 and 4 does not necessarily divide evenly by 8. Consider the number 12: it divides evenly by 2 (12 ÷ 2 = 6) and by 4 (12 ÷ 4 = 3), but when we divide 12 by 8 we get 1 with a remainder of 4, so it is not divisible by 8. Similarly, take 20: it divides evenly by 2 (20 ÷ 2 = 10) and by 4 (20 ÷ 4 = 5), but when we divide 20 by 8 we get 2 with a remainder of 4, so it does not divide evenly by 8. These examples show that divisibility by 2 and 4 does not guarantee divisibility by 8.
In simple words: Some numbers can be divided by 2 and 4 evenly but not by 8. The numbers 12 and 20 are both examples of this.

Exam Tip: Use concrete counterexamples (like 12 and 20) when answering "is it always true" questions - one example that breaks the rule is enough to prove the answer is no.