RS Aggarwal Solutions for Class 6 Chapter 4 Integers

 

Exercise 4.1

 

Question 1. Fill in the blanks to make each of the following a true statement:
Answer:
(i) 359 + 476 = 476 + 359 (Commutativity)
(ii) 2008 + 1952 = 1952 + 2008 (Commutativity)
(iii) 90758 + 0 = 90758 (Additive identity)
(iv) 54321 + (489 + 699) = 489 + (54321 + 699) (Associativity)
In simple words: These statements show that addition follows certain rules. You can swap numbers around, add zero to any number and get the same number back, and group numbers in different ways without changing the final sum.

Exam Tip: Remember the three key properties: commutativity (order doesn't matter), additive identity (zero is the identity), and associativity (grouping doesn't matter).

 

Question 2. Add each of the following and check by reversing the order of addends:
Answer:
(i) 5628 + 39784 = 45412
Verification: 39784 + 5628 = 45412
(ii) 923584 + 178 = 923762
Verification: 178 + 923584 = 923762
(iii) 15409 + 112 = 15521
Verification: 112 + 15409 = 15521
(iv) 2359 + 641 = 3000
Verification: 641 + 2359 = 3000
In simple words: When you add two numbers in either sequence, the result stays the same. This shows the commutative property of addition in action.

Exam Tip: Always verify your addition by switching the order of the numbers - if both give the same answer, your work is correct.

 

Question 3. Determine the sum by suitable rearrangements:
Answer:
(i) 953 + 407 + 647
53 + 47 = 100
(953 + 647) + 407 = 1600 + 407 = 2007
(ii) 15409 + 178 + 591 + 322
409 + 91 = 500
78 + 22 = 100
(15409 + 591) + (178 + 322) = (16000) + (500) = 16500
(iii) 2359 + 10001 + 2641 + 9999
59 + 41 = 100
99 + 01 = 100
(2359 + 2641) + (10001 + 9999) = (5000) + (20000) = 25000
(iv) 1 + 2 + 3 + 4 + 1996 + 1997 + 1998 + 1999
99 + 1 = 100
98 + 2 = 100
97 + 3 = 100
96 + 4 = 100
(1 + 1999) + (2 + 1998) + (3 + 1997) + (4 + 1996) = 2000 + 2000 + 2000 + 2000 = 8000
(v) 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20
10 + 20 = 30
1 + 9 = 10
2 + 8 = 10
3 + 7 = 10
4 + 6 = 10
(10 + 20) + (11 + 19) + (12 + 18) + (13 + 17) + (14 + 16) + 15 = 30 + 30 + 30 + 30 + 30 + 15 = 150 + 15 = 165
In simple words: Look for pairs of numbers that add up to round values like 10, 100, or 1000, then combine those pairs first to make the final calculation much easier.

Exam Tip: Strategic regrouping saves time - identify number pairs that form convenient totals before you start adding.

 

Question 4. Which of the following statements are true and which are false?
(i) The sum of two odd numbers is an odd number.
(ii) The sum of two odd numbers is an even number.
(iii) The sum of two even numbers is an even number.
(iv) The sum of two even numbers is an odd number.
(v) The sum of an even number and an odd number is an odd number.
(vi) The sum of an odd number and an even number is an even number.
(vii) Every whole number is a natural number.
(viii) Every natural number is a whole number.
(ix) There is a whole number which when added to a whole number, gives that number
(x) There is a natural number which when added to a natural number, gives that number.
(xi) Commutativity and associativity are properties of whole numbers.
(xii) Commutativity and associativity are properties of addition of whole number.
Answer:
(i) FALSE (3 + 5 = 8; 8 is an even number)
(ii) TRUE (3 + 5 = 8; 8 is an even number)
(iii) TRUE (2 + 4 = 6; 6 is an even number)
(iv) FALSE (2 + 4 = 6; 6 is an even number)
(v) TRUE (2 + 3 = 5; 5 is an odd number)
(vi) FALSE (3 + 2 = 5; 5 is not an even number)
(vii) FALSE - The whole number set is {0, 1, 2, 3, 4 ...}, whereas the natural number set is {1, 2, 3, 4 ...}
(viii) TRUE - The whole number set is {0, 1, 2, 3, 4 ...}, whereas the natural number set is {1, 2, 3, 4 ...}
(ix) TRUE - That number is zero.
(x) FALSE
(xi) FALSE
(xii) TRUE
In simple words: Test each claim with examples. Whole numbers include zero but natural numbers don't. Adding two numbers of the same type (both odd or both even) follows predictable patterns. These properties describe how addition behaves, not the numbers themselves.

Exam Tip: Always verify true-false claims with concrete examples first - one counterexample proves a statement false, but testing a few cases suggests it may be true.

 

Exercise 4.2

 

Question 1. A magic square is an array of numbers having the same number of rows and columns and the sum of numbers in each row, column or diagonal being the same. Fill in the blank cells of the following magic squares:
(i)

(ii)

Answer:
(i) Looking at the diagonal: 13 + 12 + 11 = 36
First cell of first row = 36 - (8 + 13) = 15
First cell of second row = 36 - (15 + 11) = 10
Third cell of second row = 36 - (10 + 12) = 14
Second cell of third row = 36 - (8 + 12) = 16
Third cell of third row = 36 - (11 + 16) = 9

(ii) Looking at the diagonal: 20 + 19 + 18 + 17 + 16 = 90
Second cell of first row = 90 - (22 + 6 + 13 + 20) = 29
First cell of second row = 90 - (22 + 9 + 15 + 16) = 28
Fifth cell of second row = 90 - (28 + 10 + 12 + 19) = 21
Fifth cell of third row = 90 - (9 + 11 + 18 + 25) = 27
Fifth cell of fourth row = 90 - (15 + 17 + 24 + 26) = 8
Second cell of fifth row = 90 - (29 + 10 + 11 + 17) = 23
Third cell of fifth row = 90 - (6 + 12 + 18 + 24) = 30

In simple words: In a magic square, every row, column, and diagonal adds to the same total. Find one complete diagonal or row, add up those numbers to get the target sum, then use that target to find missing numbers one by one.

Exam Tip: Start by identifying a row or diagonal that has the fewest missing numbers - it gives you the target sum fastest.

 

Question 2. Perform the following subtractions and check your results by performing corresponding additions:
Answer:
(i) 57839 - 2983 = 54856
Verification: 54856 + 2983 = 57839
(ii) 92507 - 10879 = 81628
Verification: 81628 + 10879 = 92507
(iii) 400000 - 98798 = 301202
Verification: 301202 + 98798 = 400000
(iv) 5050501 - 969696 = 4080805
Verification: 4080805 + 969696 = 5050501
(v) 200000 - 97531 = 102469
Verification: 102469 + 97531 = 200000
(vi) 3030301 - 868686 = 2161615
Verification: 2161615 + 868686 = 3030301
In simple words: After subtracting, add your answer to the number you subtracted. If you get back the original number, your subtraction was correct.

Exam Tip: Always verify subtraction using addition - if (difference + subtracted number = original number), your answer is right.

 

Question 3. Replace each * by the correct digit in each of the following:
Answer:
(i) In the units digit, 6 - * = 7, which means 1 is borrowed from 7 at tens place, and 6 at unit digit becomes 16, then 16 - 9 = 7.
Now, when 7 gives 1 to 6, it becomes 6, so 6 - 3 = 3.
Also, it is readily deduced that in (8 - * = 6), the value of * is 2.

(ii) In the units place, 9 - 4 = 5; and in the tens place, 8 - 3 = 5.
We can now readily locate the other missing blanks by subtracting 3455 from 8989. The result (difference) = 3455
Therefore, the correct answer is:

(iii) In the units digit, 17 - 8 = 9; in the tens digit, 9 - 7 = 2; in the hundreds place, 10 - 9 = 1; and in the thousands place, 9 - 8 = 1.
Result difference = 5061129.
So, in order to get the minuend, we will subtract 5061129 from 6000107.

Therefore, the correct answer is:

(iv) In the units place, 10 - 1 = 9; also, in the lakhs place, 9 - 0 = 9.
Result difference = 970429.
So, in order to get the minuend, we will subtract 970429 from 1000000.

Therefore, the correct answer is:

(v) In the units digit, 13 - 7 = 6; in the tens digit, 9 - 8 = 1; in the hundreds place, 9 - 9 = 0; and in the thousands place, 10 - 6 = 4.
Result difference = 4844016.
So, in order to get the minuend, we will subtract 4844016 from 5001003.

(vi) It is clear from the units place that 11 - 9 = 2.
Result difference = 54322.
To get the other minuend, we will subtract 54322 from 111111.

Therefore, the correct answer is:

In simple words: Figure out the missing digits by working with one position at a time. Use place value logic - if borrowing occurs in subtraction, the digit above decreases by one. Then check your answer by adding the difference and subtrahend to recover the minuend.

Exam Tip: When digits are hidden, solve one column at a time using place value knowledge. Pay close attention to borrowing - it changes the digit you borrow from.

 

Question 4. What is the difference between the largest number of five digits and smallest number of six digits?
Answer: The largest five-digit number is 99999. The smallest six-digit number is 100000. Therefore, difference between them = 100000 - 99999 = 1
In simple words: The biggest five-digit number and the smallest six-digit number are consecutive whole numbers, so they differ by just 1.

Exam Tip: Remember that the largest n-digit number is always one less than the smallest (n+1)-digit number - their difference is always 1.

 

Question 5. Find the difference between the largest number of 4 digits and the smallest number of 6 digits.
Answer: The largest four-digit number is 9999. The smallest seven-digit number is 1000000. Therefore, difference between them = 1000000 - 9999 = 990001
In simple words: Identify the bounds: the biggest four-digit value is 9999, and the tiniest seven-digit value is 1000000. When you subtract the smaller from the larger, you get the gap between these two numbers.

Exam Tip: Be careful with the number of digits specified - verify you have the right "largest" and "smallest" before subtracting.

 

Question 6. Rohit deposited Rs 125000 in his savings bank account. Later he withdrew Rs 35425 from it. How much money was left in his account?
Answer: Money deposited by Rohit = Rs 125000
Money withdrawn by Rohit = Rs 35425
Therefore, money remaining in the account = Rs (125000 - 35425) = Rs 89575
In simple words: Start with what he put in the bank. Subtract what he took out. What's left is your answer.

Exam Tip: For word problems involving deposits and withdrawals, write down what goes in and what comes out, then subtract to find the balance.

 

Question 7.
Answer: [No content provided in source]
In simple words: [Awaiting full question text]

Exam Tip: Check that all problem information is complete before solving.

 

Question 1. The population of a town is 96209. If the number of men is 29642 and that of women is 29167, determine the number of children.
Answer: The total population of the town is 96209. The number of men is 29642 and the number of women is 29167. Adding these together: 29642 + 29167 = 58809. To find the number of children, subtract the combined number of men and women from the total population: 96209 - 58809 = 37400. Therefore, there are 37400 children in the town.
In simple words: Add up all the men and women, then take that away from the total population to get how many children there are.

Exam Tip: Always check that the sum of men, women, and children equals the total population - it's the quickest way to verify your answer is correct.

 

Question 2. The digits of 6 and 9 of the number 36490 are interchanged. Find the difference between the original number and the new number.
Answer: The original number is 36490. When the digits 6 and 9 are swapped, the new number becomes 39460. The difference between these two numbers is: 39460 - 36490 = 2970. Therefore, the difference between the original number and the new number is 2970.
In simple words: Switch the 6 and the 9 in the original number to get a new number. Then subtract one from the other.

Exam Tip: Be careful to identify which digits are being interchanged and ensure you write down the correct new number before calculating the difference.

 

Question 3. The population of a town was 59000. In one year it was increased by 4563 due to new births. However, 9218 persons died or left the town during the year. What was the population at the end of the year?
Answer: Starting with an initial population of 59000, the town gained 4563 new people through births. This brings the population to 59000 + 4563 = 63563. However, 9218 people left or died during the year. Subtracting this from the increased population: 63563 - 9218 = 54345. Therefore, the population at the end of the year is 54345.
In simple words: Start with the original population, add the new births, then subtract the people who left or died.

Exam Tip: When dealing with population changes, remember to add increases (births/migration in) and subtract decreases (deaths/migration out) in the correct order.

 

Exercise 4.3

 

Question 4. Fill in the blanks to make each of the following a true statement:
(i) 785 × 0 = 0
(ii) 4567 × 1 = 4567 (Multiplicative Identity)
(iii) 475 × 129 = 129 × 475 (Commutativity)
(iv) 1243 × 8975 = 8975 × 1243 (Commutativity)
(v) 10 × 100 × 10 = 10000
(vi) 27 × 18 = 27 × 9 + 27 × 4 + 27 × 5
(vii) 12 × 45 = 12 × 50 - 12 × 5
(viii) 78 × 89 = 78 × 100 - 78 × 16 + 78 × 5
(ix) 66 × 85 = 66 × 90 - 66 × 4 - 66
(x) 49 × 66 + 49 × 34 = 49 × (66 + 34)
Answer: The statements above demonstrate key properties of whole numbers. Each blank is filled using properties like the additive identity (multiplying by 0), multiplicative identity (multiplying by 1), commutativity (changing order doesn't change the result), and distributivity (breaking numbers into parts for easier calculation).
In simple words: These statements show rules about how numbers behave when you multiply them. The answers use basic number properties to make the equations true.

Exam Tip: Recognize which property applies to each statement - identity, commutativity, or distributivity - as this helps verify the correct filling of blanks.

 

Question 5. Determine each of the following products by suitable rearrangements:
(i) 2 × 1497 × 50
(ii) 4 × 358 × 25
(iii) 495 × 625 × 16
(iv) 625 × 20 × 8 × 50
Answer:
(i) Rearranging: (2 × 50) × 1497 = 100 × 1497 = 149700
(ii) Rearranging: (4 × 25) × 358 = 100 × 358 = 35800
(iii) Rearranging: (625 × 16) × 495 = 10000 × 495 = 4950000
(iv) Rearranging: (625 × 8) × (20 × 50) = 5000 × 1000 = 5000000
In simple words: Group the numbers strategically so that some pairs multiply to give easy numbers like 100 or 1000, then finish the multiplication.

Exam Tip: Look for factors that combine to form powers of 10 (like 2 × 50 = 100) - this makes the final multiplication much quicker and reduces errors.

 

Question 6. Using distributivity of multiplication over addition of whole numbers, find each of the following products:
(i) 736 × 103
(ii) 258 × 1008
(iii) 258 × 1008
Answer:
(i) 736 × 103 = 736 × (100 + 3) = (736 × 100) + (736 × 3) = 73600 + 2208 = 75808
(ii) 258 × 1008 = 258 × (1000 + 8) = (258 × 1000) + (258 × 8) = 258000 + 2064 = 260064
(iii) 258 × 1008 = 258 × (1000 + 8) = (258 × 1000) + (258 × 8) = 258000 + 2064 = 260064
In simple words: Break the larger number into a sum of simpler parts, multiply each part separately, and then add the results together.

Exam Tip: Choose how to break apart the number so that at least one part is a simple multiple of 10, 100, or 1000 - this keeps calculations manageable.

 

Question 7. Find each of the following products:
(i) 736 × 93
(ii) 816 × 745
(iii) 2032 × 613
Answer:
(i) Since 93 = (100 - 7), we have 736 × (100 - 7) = (736 × 100) - (736 × 7) = 73600 - 5152 = 68448
(ii) Since 745 = (750 - 5), we have 816 × (750 - 5) = (816 × 750) - (816 × 5) = 612000 - 4080 = 607920
(iii) Since 613 = (600 + 13), we have 2032 × (600 + 13) = (2032 × 600) + (2032 × 13) = 1219200 + 26416 = 1245616
In simple words: Express the number as a sum or difference from a simpler number (like 100 or 600), apply distributivity, and calculate each part separately before combining.

Exam Tip: Check if the number is close to a multiple of 10 or 100 - if so, expressing it as that multiple plus or minus a small amount makes the calculation faster.

 

Question 8. Find the values of each of the following using properties:
(i) 493 × 8 + 493 × 2
(ii) 24579 × 93 + 7 × 24579
(iii) 1568 × 184 - 1568 × 84
(iv) 15625 × 15625 - 15625 × 5625
Answer:
(i) Using distributivity: 493 × (8 + 2) = 493 × 10 = 4930
(ii) Using distributivity: 24579 × (93 + 7) = 24579 × 100 = 2457900
(iii) Using distributivity: 1568 × (184 - 84) = 1568 × 100 = 156800
(iv) Using distributivity: 15625 × (15625 - 5625) = 15625 × 10000 = 156250000
In simple words: When a common factor appears in multiple terms, factor it out and work with the remaining numbers first before multiplying.

Exam Tip: Always scan for a repeated factor in addition or subtraction expressions - factoring it out often leads to much simpler arithmetic.

 

Question 9. Determine the product of:
(i) the greatest number of four digits and the smallest number of three digits.
(ii) the greatest number of five digits and the greatest number of three digits.
Answer:
(i) The greatest four-digit number is 9999 and the smallest three-digit number is 100. Therefore, their product = 9999 × 100 = 999900
(ii) The greatest five-digit number is 99999 and the greatest three-digit number is 999. Their product = 9999 × 999 = 9999 × (1000 - 1) = (9999 × 1000) - (9999 × 1) = 9999000 - 9999 = 9989001
In simple words: First identify the largest or smallest numbers with the specified number of digits, then multiply them together using methods that simplify the work.

Exam Tip: Remember that the largest n-digit number uses all 9s, while the smallest n-digit number is 1 followed by (n-1) zeros.

 

Question 10. In each of the following, fill in the blanks, so that the statement is true:
(i) (500 + 7) (300 - 1)
(ii) 888 + 777 + 555
(iii) 75 × 425
(iv) 89 × (100 - 2)
(v) (15 + 5) (15 - 5)
(vi) 9 × (10000 + 974)
Answer:
(i) (500 + 7) (300 - 1) = 507 × 299 = 299 × 507 (Commutativity)
(ii) 888 + 777 + 555 = 111 (8 + 7 + 5) = 111 × 20 (Distributivity)
(iii) 75 × 425 = (70 + 5) (340 + 85) or other valid decompositions
(iv) 89 × (100 - 2) = 89 × 98 = 98 × 89 (Commutativity)
(v) (15 + 5) (15 - 5) = 20 × 10 = 200 or 225 - 25 (Difference of squares)
(vi) 9 × (10000 + 974) = 98766
In simple words: Use properties like commutativity, distributivity, and difference of squares to complete the statements correctly.

Exam Tip: Identify which property is being illustrated in each blank - this guides you toward the correct simplification or rearrangement.

 

Question 11. A dealer purchased 125 color television sets. If the cost of each set is Rs 19820, determine the cost of all sets together.
Answer: The cost of one television set is Rs 19820. To find the total cost of 125 sets, multiply: Rs 19820 × 125. Breaking this down using distributivity: Rs 19820 × (100 + 25) = (Rs 19820 × 100) + (Rs 19820 × 25) = Rs 1982000 + Rs 495500 = Rs 2477500. Therefore, the total cost of all 125 television sets is Rs 2477500.
In simple words: Multiply the cost of one set by the total number of sets, breaking the multiplication into easier parts.

Exam Tip: When multiplying large numbers, always look for ways to decompose one factor into parts that make computation easier, like breaking 125 into 100 + 25.

 

Question 12. The annual fee charged from a student of class 6th in a school is Rs 8880. If there are, in all, 235 students in class 6th, find the total collection.
Answer: The annual fee per student is Rs 8880 and the total number of students is 235. The total collection equals: Rs 8880 × 235 = Rs 2086800. Therefore, the total fee collection from all 235 class 6 students is Rs 2086800.
In simple words: Multiply the fee per student by the total number of students to get the overall amount collected.

Exam Tip: Double-check your multiplication, especially with money amounts - a single error in computation can give a significantly incorrect total.

 

Question 13. A group housing society constructed 350 flats. If the cost of construction for each flat is Rs 993570, what is the total cost of construction of all the flats.
Answer: The construction cost per flat is Rs 993570 and the total number of flats is 350. The total construction cost = Rs 993570 × 350 = Rs 347749500. Therefore, the total cost of constructing all 350 flats is Rs 347749500.
In simple words: Multiply the cost per flat by the total number of flats to find the complete construction expense.

Exam Tip: When dealing with large sums of money, organize your multiplication carefully and verify the final result by checking the number of digits and reasonableness.

 

Question 14. The product of two whole numbers is zero. What do you conclude?
Answer: If the product of two whole numbers equals zero, then one of the numbers must be zero, or both of them are zero. This is because zero is the only whole number that, when multiplied by any other whole number, produces a result of zero. This property is known as the zero property of multiplication.
In simple words: Whenever you multiply two numbers and get zero as the answer, at least one of those numbers has to be zero.

Exam Tip: Remember that in whole numbers, the only way to get a product of zero is if one (or both) of the factors is zero - this is a fundamental property useful in solving equations.

 

Question 15. What are the whole numbers which when multiplied with itself gives the same number?
Answer: There are two whole numbers that, when multiplied by themselves, give the same number. These are 0 and 1. When we multiply 0 by itself (0 × 0), we get 0. When we multiply 1 by itself (1 × 1), we get 1. For all other whole numbers, multiplying them by themselves produces a larger number. For example, 2 × 2 = 4, which is not equal to 2.
In simple words: Only 0 and 1 stay the same when you multiply them by themselves. Every other number gets bigger.

Exam Tip: These special numbers (0 and 1) have unique multiplicative properties - 0 is the absorbing element and 1 is the multiplicative identity, concepts that appear frequently in algebra.

 

Question 16. In a large housing complex, there are 15 small buildings and 22 large building. Each of the large buildings has 10 floors with 2 apartments on each floor. Each of the small buildings has 12 floors with 3 apartments on each floor. How many apartments are there in all.
Answer: For large buildings: Each has 10 floors with 2 apartments per floor, giving 10 × 2 = 20 apartments per large building. With 22 large buildings, the total is 22 × 20 = 440 apartments. For small buildings: Each has 12 floors with 3 apartments per floor, giving 12 × 3 = 36 apartments per small building. With 15 small buildings, the total is 15 × 36 = 540 apartments. The total number of apartments in the entire complex is 440 + 540 = 980 apartments.
In simple words: Find apartments in one building of each type, multiply by the number of buildings, then add them all together.

Exam Tip: Break complex problems into stages - calculate apartments per building first, then multiply by the number of buildings, and finally add the two types of buildings together.

 

Exercise 4.4

 

Question 17. Does there exists a whole number 'a' such that a/a = a?
Answer: Yes, there exists a whole number 'a' such that a/a = a. The whole number is 1. When we divide 1 by 1, we get: 1/1 = 1. This works because any non-zero number divided by itself always gives 1. However, we must note that a cannot be 0, since division by zero is undefined. Therefore, 1 is the only whole number for which a/a = a holds true.
In simple words: Yes, the number 1 works because 1 divided by 1 gives 1. Any other non-zero number divided by itself gives 1, not itself.

Exam Tip: Remember that division by zero is always undefined in mathematics, so conditions like a/a always exclude the case where a = 0.

 

Question 18. Find the value of:
(i) 23457 / 1
(ii) 0 / 97
(iii) 476 + (840 / 84)
(iv) 964 - (425 / 425)
(v) (2758 / 2758) - (2758 + 2758)
(vi) 72450 / (583 - 58)
Answer:
(i) 23457 / 1 = 23457
(ii) 0 / 97 = 0
(iii) 476 + (840 / 84) = 476 + 10 = 486
(iv) 964 - (425 / 425) = 964 - 1 = 963
(v) (2758 / 2758) - (2758 + 2758) = 1 - 1 = 0. Note: The expression should be (2758 / 2758) - (2758 / 2758) = 1 - 1 = 0
(vi) 72450 / (583 - 58) = 72450 / 525 = 138
In simple words: Follow the order of operations - do divisions and operations in parentheses first, then additions and subtractions from left to right.

Exam Tip: Always perform operations inside parentheses before working with the rest of the expression, and remember key division facts like a/a = 1 and 0/a = 0.

 

Question 19. Which of the following statements are true:
(i) 10 / (5 × 2) = (10 / 5) × (10 / 2)
(ii) (35 - 14) / 7 = 35 / 7 - 14 / 7
(iii) 35 - 14 / 7 = 35 / 7 - 14 / 7
(iv) (20 - 5) / 5 = 20 / 5 - 5
(v) 12 × (14 / 7) = (12 × 14) / (12 × 7)
(vi) (20 / 5) / 2 = 20 / (5 / 2)
Answer:
(i) False - LHS: 10 / (5 × 2) = 10 / 10 = 1. RHS: (10 / 5) × (10 / 2) = 2 × 5 = 10. Since 1 ≠ 10, the statement is false.
(ii) True - LHS: (35 - 14) / 7 = 21 / 7 = 3. RHS: 35 / 7 - 14 / 7 = 5 - 2 = 3. Since both sides equal 3, the statement is true.
(iii) False - LHS: 35 - 14 / 7 = 35 - 2 = 33. RHS: 35 / 7 - 14 / 7 = 5 - 2 = 3. Since 33 ≠ 3, the statement is false. (Note: division before subtraction per order of operations)
(iv) False - LHS: (20 - 5) / 5 = 15 / 5 = 3. RHS: 20 / 5 - 5 = 4 - 5 = -1. Since 3 ≠ -1, the statement is false.
(v) False - LHS: 12 × (14 / 7) = 12 × 2 = 24. RHS: (12 × 14) / (12 × 7) = 168 / 84 = 2. Since 24 ≠ 2, the statement is false.
(vi) True - LHS: (20 / 5) / 2 = 4 / 2 = 2. RHS: 20 / (5 / 2) = 20 × (2 / 5) = 40 / 5 = 8. Actually, LHS = 2 and RHS = 8, so this is False. Correction: This statement is false.
In simple words: Test each statement by computing both sides carefully, watching the order of operations. Division does not follow all the same distributive rules as multiplication does.

Exam Tip: Division has different properties than multiplication - it does not distribute over multiplication or follow the same associative rules, so verify each side separately before concluding.

 

Question 4. Divide and check the quotient and remainder:
(i) 7777 / 58 = 134
Answer: The division calculation shows that 7777 split by 58 produces a quotient of 134 with a remainder of 0. To verify this result, we apply the formula: Dividend = Divisor × Quotient + Remainder, which gives us 7772 = 58 × 134 + 0 = 7772, confirming that LHS equals RHS.
In simple words: When you divide 7777 by 58, you get 134 as your answer with nothing left over. You can check it by multiplying 58 × 134, and you should get 7772 back.

Exam Tip: Always verify your division answer by plugging values into the formula Dividend = Divisor × Quotient + Remainder—if both sides match, your answer is correct.

 

Question 4. (ii) 6906/35 gives quotient = 197 and remainder = 11
Answer: When dividing 6906 by 35, the quotient obtained is 197 and the remainder is 11. Using the verification formula Dividend = Divisor × Quotient + Remainder: 6906 = 35 × 197 + 11 = 6895 + 11 = 6906, we confirm that LHS equals RHS.
In simple words: Dividing 6906 by 35 gives you 197 times with 11 left over. You can double-check by multiplying 35 × 197 and adding 11, and you'll get back to 6906.

Exam Tip: Ensure the remainder is always smaller than the divisor—if not, you need to recalculate your quotient.

 

Question 4. (iii) 16135 / 875 gives quotient = 18 and remainder = 385
Answer: Dividing 16135 by 875 yields a quotient of 18 with a remainder of 385. The verification uses Dividend = Divisor × Quotient + Remainder: 16135 = 875 × 18 + 385 = 15750 + 385 = 16135, proving LHS matches RHS.
In simple words: Splitting 16135 by 875 gives you 18 groups with 385 remaining. Multiply 875 × 18 and add 385 to get back to 16135.

Exam Tip: Write out the division steps clearly and perform the verification check every time to avoid calculation errors.

 

Question 4. (iv) 16025/1000 gives quotient and remainder = 25
Answer: When 16025 is divided by 1000, the quotient is 16 and the remainder is 25. The verification formula Dividend = Divisor × Quotient + Remainder shows: 16025 = 1000 × 16 + 25 = 16000 + 25 = 16025, so LHS equals RHS.
In simple words: Dividing 16025 by 1000 results in 16 with 25 left over. Verify by computing 1000 × 16 plus 25 to return to 16025.

Exam Tip: Pay careful attention to place values when dividing by powers of 10, as these make mental calculation faster and errors easier to spot.

 

Question 5. Find a number which when divided by 35 gives the quotient 20 and remainder 18.
Answer: Using the formula Dividend = Divisor × Quotient + Remainder, we substitute the provided values: Dividend = 35 × 20 + 18 = 700 + 18 = 718. Therefore, the required number is 718.
In simple words: If you divide 718 by 35, you will get exactly 20 as the quotient with 18 remaining. The divisor, quotient, and remainder always combine this way.

Exam Tip: Memorize the division formula and apply it directly—this is the fastest path to finding an unknown dividend.

 

Question 6. Find the number which when divided by 58 gives a quotient 40 and remainder 31.
Answer: Applying the formula Dividend = Divisor × Quotient + Remainder with the given values: Dividend = 58 × 40 + 31 = 2320 + 31 = 2351. The required number is therefore 2351.
In simple words: The number you're looking for is 2351. If you divide it by 58, you get 40 times with 31 left as the remainder.

Exam Tip: Always substitute values into the formula systematically to avoid arithmetic mistakes in multi-step problems.

 

Question 7. The product of two numbers is 504347. If one of the numbers is 1591, find the other.
Answer: To locate the second number, we divide the product by the known number. We perform the division 504347 ÷ 1591. The long division shows that 504347 split by 1591 produces a quotient of 317 with no remainder. We can verify: 1591 × 317 = 504347, confirming our result. Therefore, the other number is 317.
In simple words: Divide the total product (504347) by the number you already know (1591), and you get the missing number, which is 317.

Exam Tip: When finding an unknown factor given the product and one factor, always set up division carefully and verify by multiplying back.

 

Question 8. On dividing 59761 by a certain number, the quotient is 189 and the remainder is 37. Find the divisor.
Answer: We use the rearranged division formula: Dividend = Divisor × Quotient + Remainder. Substituting known values: 59761 = A × 189 + 37, where A is the divisor we seek. Rearranging: 59761 - 37 = A × 189, which gives 59724 = A × 189. Dividing both sides by 189: A = 59724 ÷ 189 = 316. Therefore, the divisor is 316.
In simple words: Subtract the remainder from the dividend, then divide the result by the quotient. This gives you the divisor you were hunting for.

Exam Tip: Rearrange the division formula before dividing to isolate the unknown divisor and reduce calculation errors.

 

Question 9. On dividing 55390 by 299, the remainder is 75. Find the quotient.
Answer: Using the formula Dividend = Divisor × Quotient + Remainder, we substitute: 55390 = 299 × A + 75, where A is the quotient. Rearranging: 55390 - 75 = 299 × A, giving us 55315 = 299 × A. Solving for A: A = 55315 ÷ 299 = 185. Therefore, the quotient is 185.
In simple words: Remove the remainder from the dividend, then split the remaining value by the divisor to uncover the quotient.

Exam Tip: Always isolate the unknown term algebraically before performing division to keep the calculation systematic and error-free.

 

Exercise 4.5

 

Question 1. Without drawing a diagram, find:
Answer:
(i) 10th square number: A square number follows the rule where the nth square number = n × n. So the 10th square number = 10 × 10 = 100.
(ii) 6th triangular number: A triangular number follows the rule where the nth triangular number = n × (n + 1)/2. Therefore, the 6th triangular number = 6 × (6 + 1)/2 = 6 × 7/2 = 21.
In simple words: To find any square number, multiply the position by itself. For triangular numbers, multiply the position by the next whole number and cut it in half.

Exam Tip: Memorize both formulas (n² for square numbers and n(n+1)/2 for triangular numbers) as these are tested frequently.

 

Question 2. (i) Can a rectangle number also be a square number?
Answer: Yes, a rectangular number can also function as a square number. For instance, 16 is considered a square number (since 4 × 4 = 16) and also a rectangular number (as it can be arranged in a 2 × 8 or 1 × 16 rectangle pattern). Square numbers form a special subset of rectangular numbers where length and width happen to be equal.
In simple words: Some numbers can be both square and rectangular. A 4 × 4 arrangement is square-shaped, but it's also technically a rectangle because rectangles include squares.

Exam Tip: Recall that squares are special cases of rectangles, so every square number is automatically a rectangular number too.

 

Question 2. (ii) Can a triangular number also be a square number?
Answer: Yes, there is exactly one triangular number that is also a square number, and that number is 1. The number 1 satisfies both conditions: it equals 1 × 1 (making it the 1st square number) and it equals 1 × (1 + 1)/2 = 1 (making it the 1st triangular number). No other number holds this dual property.
In simple words: Only the number 1 works as both a triangular and a square number. All other numbers fall into one category or the other, but not both.

Exam Tip: Remember this unique property of the number 1—it is a standard exam question to test understanding of pattern definitions.

 

Question 3. Write the first four products of two numbers with difference 4 starting from in the following order: 1, 2, 3, 4, 5, 6, ........... Identify the pattern in the products and write the next three products.
Answer:
1 × 5 = 5 (difference: 5 - 1 = 4)
2 × 6 = 12 (difference: 6 - 2 = 4)
3 × 7 = 21 (difference: 7 - 3 = 4)
4 × 8 = 32 (difference: 8 - 4 = 4)

The pattern shows that each product increases by consecutive odd numbers (5, 12, 21, 32...). The differences between consecutive products are 7, 9, 11, following a pattern of odd numbers increasing by 2 each time. The next three products are:
5 × 9 = 45
6 × 10 = 60
7 × 11 = 77
In simple words: Two numbers that differ by 4, when multiplied together, create a sequence. Each result jumps up by an odd number that grows bigger each step.

Exam Tip: Always compute the differences between consecutive answers to spot the underlying pattern more easily.

 

Question 4. Observe the pattern in the following and fill in the blanks:
Answer:
9 × 9 + 7 = 88
98 × 9 + 6 = 888
987 × 9 + 5 = 8888
9876 × 9 + 4 = 88888
98765 × 9 + 3 = 888888
987654 × 9 + 2 = 8888888
9876543 × 9 + 1 = 88888888

The pattern reveals that as the first number gains more descending digits (9, 98, 987, ...), we multiply by 9 and then add a number that decreases by 1 each step (7, 6, 5, ...). The result is always a string of repeated 8s, with the count matching the number of digits in the first factor.
In simple words: Each row adds one more digit to the starting number (going down from 9), multiplies by 9, and adds a smaller number. You always end up with a long chain of 8s.

Exam Tip: Look for what changes and what stays the same in each step—incremental growth in one part paired with decrease in another often signals a structured pattern.

 

Question 5. Observe the following pattern and extend it to three more steps:
Answer:
6 × 2 - 5 = 7
7 × 3 - 12 = 9
8 × 4 - 21 = 11
9 × 5 - 32 = 13
10 × 6 - 45 = 15
11 × 7 - 60 = 17
12 × 8 - 77 = 19

The sequence starts with 6 × 2 and each subsequent step raises the first number by 1 and the second by 1. The amount subtracted follows the pattern of products with difference 4 (from Question 3). The results form consecutive odd numbers beginning at 7.
In simple words: The first numbers climb by 1 each line, the second numbers climb by 1 each line, and what you subtract gets bigger in a special way. Your answers are all odd numbers in order.

Exam Tip: Recognize how multiple sub-patterns interact within a single sequence—this layered approach is common in advanced pattern questions.

 

Question 6. Study the following pattern:
1 + 3 = 2 × 2
1 + 3 + 5 = 3 × 3
1 + 3 + 5 + 7 = 4 × 4
1 + 3 + 5 + 7 + 9 = 5 × 5

By observing the above pattern, find:
Answer:
(i) 1 + 3 + 5 + 7 + 9 + 11 = 6 × 6 = 36
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 8 × 8 = 64
(iii) 21 + 23 + 25 + ... + 51 = (21 + 23 + 25 +...+ 51) can be rewritten as (1 + 3 + 5 + 7 +...+ 49 + 51) - (1 + 3 + 5 +...+ 17 + 19). The sum (1 + 3 + 5 +...+ 49 + 51) = 26 × 26 = 676, and (1 + 3 + 5 +...+ 17 + 19) = 10 × 10 = 100. Therefore, (21 + 23 + 25 +...+ 51) = 676 - 100 = 576.
In simple words: The sum of the first n odd numbers always equals n squared. To find a sum of odd numbers that doesn't start at 1, find the full sum up to your last number, then subtract the full sum up to the number before where you start.

Exam Tip: This is a key pattern in number theory—consecutive odd numbers always sum to a perfect square, a fact that saves time on many exam problems.

 

Question 7. Study the following pattern:
1 × 1 + 2 × 2 = \( \frac{2 \times 3 \times 5}{6} \)

1 × 1 + 2 × 2 + 3 × 3 = \( \frac{3 \times 4 \times 7}{6} \)

1 × 1 + 2 × 2 + 3 × 3 + 4 × 4 = \( \frac{4 \times 5 \times 9}{6} \)

By observing the above pattern, write next two steps.
Answer: Following the established pattern, the subsequent two steps are:

1 × 1 + 2 × 2 + 3 × 3 + 4 × 4 + 5 × 5 = \( \frac{5 \times 6 \times 11}{6} \) = 55

1 × 1 + 2 × 2 + 3 × 3 + 4 × 4 + 5 × 5 + 6 × 6 = \( \frac{6 \times 7 \times 13}{6} \) = 91

The pattern shows that the sum of squares equals a fraction where the numerator has the form n × (n + 1) × (2n + 1) and the denominator is always 6.
In simple words: To find the sum of the first n perfect squares (1² + 2² + 3² + ...), use the formula n(n+1)(2n+1)/6. Plug in your n value and simplify.

Exam Tip: This is the standard formula for the sum of squares—memorize it as it appears repeatedly in algebra and number theory exams.

 

Question 8. Study the following pattern:
\( 1 = \frac{1 \times 2}{2} \)

\( 1 + 2 = \frac{2 \times 3}{2} \)

\( 1 + 2 + 3 = \frac{3 \times 4}{2} \)

\( 1 + 2 + 3 + 4 = \frac{4 \times 5}{2} \)

By observing the above pattern, find:
Answer:
(i) 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = \( \frac{10 \times 11}{2} \) = 55

(ii) 50 + 51 + 52 + ... + 100. This can be rewritten as (1 + 2 + 3 + ... + 99 + 100) - (1 + 2 + 3 + 4 + ... + 47 + 49). Now, (1 + 2 + 3 + ... + 99 + 100) = \( \frac{100 \times 101}{2} \) = 5050, and (1 + 2 + 3 + 4 + ... + 47 + 49) = \( \frac{49 \times 50}{2} \) = 1225. Therefore, (50 + 51 + 52 + ... + 100) = 5050 - 1225 = 3825.

(iii) 2 + 4 + 6 + 8 + 10 + ... + 100. This can be rewritten as 2 × (1 + 2 + 3 + 4 + ... + 49 + 50). Now, (1 + 2 + 3 + 4 + ... + 49 + 50) = \( \frac{50 \times 51}{2} \) = 1275. Therefore, (2 + 4 + 6 + 8 + 10 + ... + 100) = 2 × 1275 = 2550.
In simple words: The sum of the first n counting numbers equals n(n+1)/2. To find a partial sum, compute the full sum up to your last number, then subtract the full sum up to the number before your start. For even numbers, factor out 2 and apply the formula to whole numbers.

Exam Tip: The formula n(n+1)/2 is fundamental—use it strategically to convert complex-looking sums into quick calculations.

Exercise 4.6

 

Question 1. Which one of the following is the smallest whole number?
(a) 1 (b) 2 (c) 0 (d) None of these
Answer: (c) 0
In simple words: Whole numbers begin from 0 and continue as 0, 1, 2, 3, and so on. Among all these, 0 is the smallest.

Exam Tip: Remember that whole numbers always start from 0, unlike natural numbers which start from 1.

 

Question 2. Which one of the following is the smallest even whole number?
(a) 0 (b) 1 (c) 2 (d) None of these
Answer: (c) 2
In simple words: Even numbers are those that can be divided by 2. Among whole numbers 0, 1, 2, 3, 4, ..., the number 2 is the smallest even one.

Exam Tip: An even whole number must be divisible by 2; don't confuse this with the smallest whole number overall.

 

Question 3. Which one of the following is the smallest odd whole number?
(a) 0 (b) 1 (c) 3 (d) 5
Answer: (b) 1
In simple words: Odd numbers are those that cannot be divided evenly by 2. Looking at the whole numbers 0, 1, 2, 3, 4, ..., the number 1 is the smallest odd one.

Exam Tip: Odd numbers are not divisible by 2; the first odd whole number is 1, not 3 or any higher number.

 

Question 4. How many whole numbers are between 437 and 487?
(a) 50 (b) 49 (c) 51 (d) None of these
Answer: (b) 49
In simple words: The whole numbers between 437 and 487 are 438, 439, 440, ..., 485, 486. To count them, subtract the smaller from the larger and then subtract 1 more: (487 - 437) - 1 = 49.

Exam Tip: Always remember to subtract 1 after finding the difference when counting numbers between two endpoints (not including the endpoints themselves).

 

Question 5. The product of the successor 999 and predecessor of 1001 is:
(a) one lakh (b) one billion (c) one million (d) one crore
Answer: (c) one million
In simple words: The successor of 999 is 1000, and the predecessor of 1001 is also 1000. Multiplying them: 1000 × 1000 = 1,000,000, which equals one million.

Exam Tip: Know the difference between successor (next number) and predecessor (previous number); also memorize that 1 million = 1,000,000.

 

Question 6. Which one of the following whole numbers does not have a predecessor?
(a) 1 (b) 0 (c) 2 (d) None of these
Answer: (b) 0
In simple words: Whole numbers form the set 0, 1, 2, 3, 4, and so on. Since 0 is the first whole number, there is no whole number before it, so it has no predecessor.

Exam Tip: Every whole number except 0 has a predecessor within the whole numbers system.

 

Question 7. The number of whole numbers between the smallest whole number and the greatest 2 digit number is:
(a) 101 (b) 100 (c) 99 (d) 98
Answer: (d) 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. To find how many: (99 - 0) - 1 = 98.

Exam Tip: When counting numbers between two values, always subtract 1 from the difference to exclude the endpoints.

 

Question 8. If n is a whole number such that n + n = n, then n = ?
(a) 1 (b) 2 (c) 3 (d) None of these
Answer: (d) None of these
In simple words: The equation n + n = n simplifies to 2n = n. This is only true when n = 0, since 0 + 0 = 0. The answer 0 is not listed in options (a), (b), or (c).

Exam Tip: When an equation seems to have no solution among the given choices, always check option (d) None of these.

 

Question 9. The predecessor of the smallest 3 digit number is:
(a) 999 (b) 99 (c) 100 (d) 101
Answer: (b) 99
In simple words: The smallest 3-digit number is 100. The predecessor (the number just before it) is 100 - 1 = 99.

Exam Tip: Predecessor means the number immediately before a given number; for 100, that is always 99.

 

Question 10. The least number of 4 digits which is exactly divisible by 9 is:
(a) 1008 (b) 1009 (c) 1026 (d) 1018
Answer: (a) 1008
In simple words: The smallest 4-digit number is 1000. To find the smallest 4-digit number divisible by 9, we add the amount needed: 1000 + (9 - 1) = 1008. This ensures 1008 divides evenly by 9.

Exam Tip: For finding the smallest number after a given value that is divisible by a divisor, add the remainder that makes it fit.

 

Question 11. The number which when divided by 53 gives 8 as quotient and 5 as remainder is:
(a) 424 (b) 419 (c) 429 (d) None of these
Answer: (c) 429
In simple words: We use the division formula: Dividend = Divisor × Quotient + Remainder. Here: Dividend = 53 × 8 + 5 = 424 + 5 = 429.

Exam Tip: Always use the division algorithm formula correctly; make sure the remainder is less than the divisor for validity.

 

Question 12. The whole number n satisfying n + 35 = 101 is:
(a) 65 (b) 67 (c) 64 (d) 66
Answer: (d) 66
In simple words: Solve the equation by subtracting 35 from both sides: n + 35 - 35 = 101 - 35, which gives n = 66.

Exam Tip: When solving equations, perform the same operation on both sides to isolate the variable.

 

Question 13. The value of 4 x 378 x 25 is:
(a) 37800 (b) 3780 (c) 9450 (d) 30078
Answer: (a) 37800
In simple words: Rearrange the multiplication by grouping: 4 × 25 × 378. First multiply 4 and 25 to get 100, then multiply by 378: 100 × 378 = 37,800.

Exam Tip: Look for factors that multiply to simple numbers like 10, 100, or 1000 to make calculation faster and easier.

 

Question 14. The value of 1735 x 1232 - 1735 x 232 is:
(a) 17350 (b) 173500 (c) 1735000 (d) 173505
Answer: (c) 1735000
In simple words: Factor out 1735 from both terms: 1735 × (1232 - 232) = 1735 × 1000 = 1,735,000.

Exam Tip: Use the distributive property to factor out common terms; this dramatically simplifies multiplication and subtraction problems.

 

Question 15. The value of 47 x 99 is:
(a) 4635 (b) 4653 (c) 4563 (d) 6453
Answer: (b) 4653
In simple words: Rewrite 99 as 100 - 1, so 47 × 99 = 47 × (100 - 1) = 4700 - 47 = 4653.

Exam Tip: When multiplying by numbers close to 100, rewrite them as 100 plus or minus a small number to simplify the work.