[New] CLASS 6 Maths Chapter 2 Whole Numbers.
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# Class 6 Maths Chapter 2 - Whole Numbers.
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This Chapter contains 3 exercises, and The Content available on this page provides Brief Explanation of this Chapter and solutions for the questions present in the NCERT Exercises. Now, let's look at some of the concepts which we will discussed in this chapter.
Here at MSEducator.in you get Complete FREE Study Material for Class 6 Maths Chapter 2 : Whole Numbers.
Which Includes :-
# Class 6 Maths Chapter 2 NCERT Book. (pdf)
# Class 6 Maths Chapter 2 Video Explanation.
- CHAPTER OVERVIEW -
Class 6 Maths Chapter 2 : Whole Numbers contains 3 exercises, and The Content available on this page provides Brief Explanation of this Chapter and solutions for the questions present in the NCERT Exercises. Now, let's look at some of the concepts which we have discuss in this chapter.
# Introduction to Whole Numbers.
# Predecessors and Successors.
# Whole numbers on the Number Line.
# Properties of Whole Numbers.
# Patterns in Whole Numbers.
↓ SCROLL DOWN to get NCERT Solutions for Class 6 Maths Chapter 2 : Whole Numbers.
# Short Notes :-
➡ The numbers 1, 2, 3,... which we use for counting are known as natural numbers.
➡ If we add the number zero to the collection of natural numbers, we get the collection of whole numbers. Thus, the numbers 0, 1, 2, 3,... form the collection of whole numbers.
➡ If you add 1 to a natural number, we get its successor. If you subtract 1 from a natural number, you get its predecessor.
➡ Every natural number has a successor. Every natural number except 1 has a predecessor.
➡ Every whole number has a successor. Every whole number except zero has a predecessor.
➡ All natural numbers are whole numbers, but all whole numbers are not natural numbers.
➡ We take a line, mark a point on it and label it 0. We then mark out points to the right of 0, at equal intervals. Label them as 1, 2, 3,.... Thus, we have a number line with the whole numbers represented on it. We can easily perform the number operations of addition, subtraction and multiplication on the number line.
➡ Addition corresponds to moving to the right on the number line, whereas subtraction corresponds to moving to the left. Multiplication corresponds to making jumps of equal distance starting from zero.
➡ Adding two whole numbers always gives a whole number. Similarly, multiplying two whole numbers always gives a whole number. We say that whole numbers are closed under addition and also under multiplication. However, whole numbers are not closed under subtraction and under division.
➡ Division by zero is not defined.
➡ Zero is the identity for addition of whole numbers. The whole number 1 is the identity for multiplication of whole numbers.
➡ You can add two whole numbers in any order. You can multiply two whole numbers in any order. We say that addition and multiplication are commutative for whole numbers.
➡ Addition and multiplication, both, are associative for whole numbers.
➡ Multiplication is distributive over addition for whole numbers.
➡ Commutativity, associativity and distributivity properties of whole numbers are useful in simplifying calculations and we use them without being aware of them.
➡ Patterns with numbers are not only interesting, but are useful especially for verbal calculations and help us to understand properties of numbers better.
- NCERT SOLUTIONS -
NCERT Solutions for Class 6 Maths Chapter 2 : Whole Numbers are available here. we have prepared step by step solutions with detailed explanations. We advise students who want's to score good marks in Mathematics, to go through these solutions and strengthen their knowledge.
# NCERT Solutions for Ex - 2.1
Class 6 Maths Chapter 1 Exercise 2.1 Question 1.
Q1. Write the next three natural numbers after 10999.
Solution :
The next three natural numbers after 10999 are :
(i) 10999 + 1 = 11000
(ii) 11000 + 1 = 11001
(iii) 11001 + 1 = 11002
Ans. The next three natural numbers after 10999 are 11000, 11001 and 11002.
⇒ View Explanation
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Class 6 Maths Chapter 1 Exercise 2.1 Question 2.
Q2. Write the three whole numbers occurring just before 10001.
Solution :
The three whole numbers just before 10001 are :
(i) 10001 – 1 = 10000
(ii) 10000 – 1 = 9999
(iii) 9999 – 1 = 9998
Ans. Three whole numbers just before 10001 are 10000, 9999 and 9998.
⇒ View Explanation
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Class 6 Maths Chapter 1 Exercise 2.1 Question 3.
Q3. Which is the smallest whole number?
Ans. The smallest whole number is 0
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Class 6 Maths Chapter 1 Exercise 2.1 Question 4.
Q4. How many whole numbers are there between 32 and 53?
Solution :
The whole numbers between 32 and 53 are :
(32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53)
Ans. There are 20 whole numbers between 32 and 53.
⇒ View Explanation
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Class 6 Maths Chapter 1 Exercise 2.1 Question 5.
Q5. Write the successor of :
(a) 2440701
Solution :
Successor = 244070 + 1 = 244071
Ans. Successor of 244070 is 244071.
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(b) 100199
Solution :
Successor = 100199 + 1 = 100200
Ans. Successor of 100199 is 100200.
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(c) 1099999
Solution :
Successor = 1099999 + 1 = 1100000
Ans. Successor of 1099999 is 1100000.
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(d) 2345670
Solution :
Successor = 2345670 + 1 = 2345671
Ans. Successor of 2345670 is 2345671
⇒ View Explanation
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Class 6 Maths Chapter 1 Exercise 2.1 Question 6.
Q6. Write the predecessor of :
(a) 94
Solution :
Predecessor = 94 – 1 = 93
Ans. Predecessor of 94 is 93.
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(b) 10000
Solution :
Predecessor = 10000 – 1 = 9999
Ans. Predecessor of 1000 is 9999.
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(c) 208090
Solution :
Predecessor = 208090 - 1 = 208089
Ans. Predecessor of 208090 is 208089.
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(d) 7654321
Solution :
Predecessor = 7654321 – 1 = 7654320
Ans. Predecessor of 7654321 is 7654320.
⇒ View Explanation
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Class 6 Maths Chapter 1 Exercise 2.1 Question 7.
Q7. In each of the following pairs of numbers, state which whole number is on the left of the other number on the number line. Also write them with the appropriate sign (>, <) between them.
(a) 530, 503
Solution :
Since 503 is smaller than 530.
Hence, 503 is on the left side of 530 on the number line.
Expression : ( 503 < 530 ) OR ( 530 > 503 )
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(b) 370, 307
Solution :
Since 307 is smaller than 370.
Hence, 307 will be on left side of 370 on number line.
Expression : ( 307 < 370 ) OR ( 370 > 307 )
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(c) 98765, 56789
Solution :
Since 56789 is smaller than 98765.
Hence, 56789 will be on left side of 98765 on number line.
Expression : ( 56789 < 98765 ) OR ( 98765 > 56789 )
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(d) 9830415, 10023001
Solution :
Since 9830415 is smaller than 10023001.
Hence, 9830415 will be on left side of 10023001 on number line.
Expression : ( 9830415 < 10023001 ) OR (10023001 > 9830415 )
⇒ View Explanation
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Class 6 Maths Chapter 1 Exercise 2.1 Question 8.
Q8. Which of the following statements are true (T) and which are false (F)?
(a) Zero is the smallest natural number.
Ans. False (0 is a whole number)
(b) 400 is the predecessor of 399.
Ans. False (The predecessor of 399 is 398 Since, (399 – 1 = 398)
(c) Zero is the smallest whole number.
Ans. True (Zero is the smallest whole number)
(d) 600 is the successor of 599.
Ans. True (Since (599 + 1 = 600)
(e) All natural numbers are whole numbers.
Ans. True (All natural numbers are whole numbers)
(f) All whole numbers are natural numbers.
Ans. False (0 is a whole number but is not a natural number)
(g) The predecessor of a two digit number is never a single digit number.
Ans. False (Example the predecessor of 10 is 9)
(h) 1 is the smallest whole number.
Ans. False (0 is the smallest whole number)
(i) The natural number 1 has no predecessor.
Ans. True (The predecessor of 1 is 0 but is not a natural number)
(j) The whole number 1 has no predecessor.
Ans. False (0 is the predecessor of 1 and is a whole number)
(k) The whole number 13 lies between 11 and 12.
Ans. False (13 does not lie between 11 and 12)
(l) The whole number 0 has no predecessor.
Ans. True (The predecessor of 0 is -1 and is not a whole number)
(m) The successor of a two digit number is always a two digit number.
Ans. False (As the successor of 99 is 100)
⇒ View Explanation
# NCERT Solutions for Ex - 2.2
Class 6 Maths Chapter 1 Exercise 2.2 Question 1.
Q1. Find the sum by suitable rearrangement:
(a) 837 + 208 + 363
Solution :
= (837 + 363) + 208
= 1200 + 208
= 1408
(b) 1962 + 453 + 1538 + 647
Solution :
= (1962 + 1538) + (453 + 647)
= 3500 + 1100
= 4600
⇒ View Explanation
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Class 6 Maths Chapter 1 Exercise 2.2 Question 2.
Q2. Find the product by suitable rearrangement :
(a) 2 × 1768 × 50
Solution :
= 2 × 50 × 1768
= 100 × 1768
= 176800
(b) 4 × 166 × 25
Solution :
= 4 × 25 × 166
= 100 × 166
= 16600
(c) 8 × 291 × 125
Solution :
= 8 × 125 × 291
= 1000 × 291
= 291000
(d) 625 × 279 × 16
Solution :
= 625 × 16 × 279
= 10000 × 279
= 2790000
(e) 285 × 5 × 60
Solution :
= 285 × 300
= 85500
(f) 125 × 40 × 8 × 25
Solution :
= 125 × 8 × 40 × 25
= 1000 × 1000
= 1000000
⇒ View Explanation
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Class 6 Maths Chapter 1 Exercise 2.2 Question 3.
Q3. Find the value of the following :
(a) 297 × 17 + 297 × 3
Solution :
= 297 x (17 + 3)
= 297 x 20
= 297 x 2 x 10
= 594 x 10
= 5940
(b) 54279 × 92 + 8 × 54279
Solution :
= 54279 x (92 + 8)
= 54279 x 100
= 5427900
(c) 81265 × 169 – 81265 × 69
Solution :
= 81265 × (169 – 69)
= 81265 × 100
= 8126500
(d) 3845 × 5 × 782 + 769 × 25 × 218
Solution :
= (3845 × 5 × 782) + (769 × 5 × 5 × 218)
= (3845 × 5 × 782) + (3845 × 5 × 218)
= (3845 × 5) × (782 + 218)
= 19225 × 1000
= 19225000
⇒ View Explanation
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Class 6 Maths Chapter 1 Exercise 2.2 Question 4.
Q4. Find the product using suitable properties.
(a) 738 × 103
Solution :
By using distributive property
= 738 x (100 + 3)
= 738 x 100 + 738 x 3
= 73800 + 2214
= 76014
(b) 854 × 102
Solution :
By using distributive property
= 854 x (100 + 2)
= 854 x 100 + 854 x 2
= 85400 + 1708
= 87108
(c) 258 × 1008
Solution :
By using distributive property
= 258 x (1000 + 8)
= 258 x 1000 + 258 x 8
= 258000 + 2064 = 260064
(d) 1005 × 168
Solution :
By using distributive property
= (1000 + 5) x 168
= 1000 x 168 + 5 x 168
= 168000 + 840 = 168840
⇒ View Explanation
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Class 6 Maths Chapter 1 Exercise 2.2 Question 5.
Q5. A taxi driver filled his car petrol tank with 40 litres of petrol on Monday. The next day, he filled the tank with 50 litres of petrol. If the petrol costs rupees 44 per litre, how much did he spend in all on petrol?
Solution :
Cost of petrol = Rs.44 per litre
Petrol filled on Monday = 40 litre
Petrol filled on Tuesday = 50 litre
Total money spent on petrol -
= (40 + 50) x 44
= 90 x 44
= Rs.3960
Ans. Total money he Spend on petrol is Rs. 3960.
⇒ View Explanation
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Class 6 Maths Chapter 1 Exercise 2.2 Question 6.
Q6. A vendor supplies 32 litres of milk to a hotel in the morning and 68 litres of milk in the evening. If the milk costs rupees 45 per litre, how much money is due to the vendor per day?
Solution :
Cost of milk = Rs.45 per litre
Milk supplied in the morning = 32 litre
Milk supplied in the evening = 68 litre
Money due to the vendor per day -
= (32 + 68) x 45
= 100 x 45
= Rs.4500
Ans. Total money due to the vendor per day is rupees 4500.
⇒ View Explanation
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Class 6 Maths Chapter 1 Exercise 2.2 Question 7.
Q7. Match the following :
Answer.
⇒ View Explanation
# NCERT Solutions for Ex - 2.3
Class 6 Maths Chapter 1 Exercise 2.3 Question 1.
Q1. Which of the following will not represent zero :
(a) 1 + 0
(b) 0 × 0
(c) 0 / 2
(d) (10 – 10) / 2
Solutions :
(a) 1 + 0 = 1
Ans. It does not represent zero
(b) 0 × 0 = 0
Ans. It represents zero
(c) 0 / 2 = 0
Ans. It represents zero
(d) (10 – 10) / 2 = 0 / 2 = 0
Ans. It represents zero
⇒ View Explanation
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Class 6 Maths Chapter 1 Exercise 2.3 Question 2.
Q2. If the product of two whole numbers is zero, can we say that one or both of them will be zero? Justify through examples.
Solution :
Condition I : If the product of two whole numbers is zero, definitely one of them is zero.
Example : 0 × 5 = 0 and 12 × 0 = 0
Ans. Yes, If the product of two whole numbers is zero, definitely one of them is zero
Condition II : If the product of two whole numbers is zero, both of them may be zero.
Example : 0 × 0 = 0
Ans. Yes, if the product of two whole numbers is zero, then both of them may be zero
⇒ View Explanation
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Class 6 Maths Chapter 1 Exercise 2.3 Question 3.
Q3. If the product of two whole numbers is 1, can we say that one or both of them will be 1? Justify through examples.
Solution :
If the product of two whole numbers is 1, both numbers should be equal to 1
Example : 1 × 1 = 1 But 1 × 3 = 3
Ans. It is clear that the product of two whole numbers will be 1, only in situations when both numbers to be multiplied are 1.
⇒ View Explanation
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Class 6 Maths Chapter 1 Exercise 2.3 Question 4.
Q4. Find using distributive property :
(a) 728 × 101
Solution :
= 728 × (100 + 1)
= (728 × 100) + (728 × 1)
= 72800 + 728
= 73528
(b) 5437 × 1001
Solution :
= 5437 × (1000 + 1)
= (5437 × 1000) + (5437 × 1)
= 5437000 + 5437
= 5442437
(c) 824 × 25
Solution :
= 824 x (20 + 5)
= (824 x 20) + (824 x 5)
= 16480 + 4120
= 20600
(d) 4275 × 125
Solution :
= 4275 x (100 + 20 + 5)
= (4275 x 100) + (4275 x 20) + (4275 x 5)
= 427500 + 85500 + 21375
= 534375
(e) 504 × 35
Solution :
= (500 + 4) × 35
= 500 × 35 + 4 × 35
= 17500 + 140
= 17640
⇒ View Explanation
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Class 6 Maths Chapter 1 Exercise 2.3 Question 5.
Q5. Study the pattern:
1 × 8 + 1 = 9
12 × 8 + 2 = 98
123 × 8 + 3 = 987
1234 × 8 + 4 = 9876
12345 × 8 + 5 = 98765
Write the next two steps. Can you say how the pattern works?
(Hint: 12345 = 11111 + 1111 + 111 + 11 + 1)
Solution :
Working pattern :
(1) x 8 + 1 = 9
(12) x 8 + 2 = (11 + 1) x 8 + 2 = 98
(123) x 8 + 3 = (111 + 11 + 1) x 8 + 3 = 987
(1234) x 8 + 4 = (1111 + 111 + 11 + 1) x 8 + 4 = 9876
(12345) x 8 + 5 = (11111 + 1111 + 111 + 11 + 1) x 8 + 5 = 98765
So, the next two steps are :
Step I : 123456 x 8 + 6 = 987654
Step II : 1234567 x 8 + 7 = 9876543
⇒ View Explanation
# Class 6 Maths Chapter 2 : Extra Questions
Q1. Determine the sum of the four numbers as given below :
(a) successor of 32
(b) predecessor of 49
(c) predecessor of the predecessor of 56
(d) successor of the successor of 67
Solution :
(a) successor of 32 = 33
(b) predecessor of 49 = 48
(c) predecessor of the predecessor of 56 = 54
(d) successor of the successor of 67 = 69
Sum = 33 + 48 + 54 + 69 = 204
Ans. The Sum of the Four numbers is 204.
⇒ View Explanation
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Class 6 maths chapter 2 extra questions.
Q2. Determine the sum by suitable rearrangement.
(a) 753 + 807 + 947
(b) 23 + 546 + 377 + 154
Solution :
(a) 753 + 807 + 947
= (753 + 947) + 807
= 1700 + 807
Ans. 2507
(b) 23 + 546 + 377 + 154
= (23 + 377) + ( 546 + 154)
= 400 + 700
Ans. 1100
⇒ View Explanation
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Class 6 maths chapter 2 extra questions.
Q3. Find the product 8739 × 102 using distributive property.
Solution :
8739 × 102
By using Distributive property :
= 8739 × (100 + 2)
= (8739 × 100) + (8739 × 2)
= 873900 + 17478
= 891378
Ans. The product of 8739 × 102 is 891378.
⇒ View Explanation
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Class 6 maths chapter 2 extra questions.
Q4. Add the following in three ways. Indicate the property used.
(a) 25 + 36 + 15
(b) 30 + 18 + 22
Solution :
(a) 25 + 36 + 15
Way I : 25 + (36 + 15) = 25 + 51 = 76
Way II : (25 + 36) + 15 = 61 + 15 = 76
Way III : (25 + 15) + 36 = 40 + 36 = 76
Here, we have used associative property.
(b) 30 + 18 + 22
Way I : 30 + (18 + 22) = 30 + 40 = 70
Way II : (30 + 18) + 22 = 48 + 22 = 70
Way III : (30 + 22) + 18 = 52 + 18 = 70
Here, we have used associative property.
⇒ View Explanation
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Class 6 maths chapter 2 extra questions.
Q5. Find the product of the greatest 3-digit number and the greatest 2-digit number.
Solution :
Greatest 3-digit number = 999
Greatest 2-digit number = 99
Product of 999 and 99 is :
= 999 x 99
= (1000 – 1) x 99
= (1000 x 99 ) – (1 x 99)
= 99000 – 99
= 98901
Ans. The product of the greatest 3-digit and 2 digit number is 98901.
⇒ View Explanation
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Class 6 maths chapter 2 extra questions.
Q6. Which property do the following statements hold?
(a) 6 + 4 = 4 + 6
(b) 3 + 2 = whole number
Solution :
(a) 6 + 4 = 4 + 6
Ans. It holds the commutative property of addition.
(b) 3 + 2 = whole number
Ans. It holds the closure property of whole number.
⇒ View Explanation
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Class 6 maths chapter 2 extra questions.
Q7. Write the Four whole numbers which can be arranged as squares.
Solution :
Ans. The required number are 4, 9, 16 and 25.
⇒ View Explanation
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Class 6 maths chapter 2 extra questions.
Q8. Represent the following on number line :
(a) 3 + 4
Ans. 3 + 4 = 7
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(b) 6 – 2
Ans. 6 - 2 = 4
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(c) 2 × 4
Ans. 2 x 4 = 8
⇒ View Explanation
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Class 6 maths chapter 2 extra questions.
Q9. Solve the following and establish a pattern :
(a) 84 x 9
(b) 84 x 99
(c) 84 x 999
(d) 84 x 9999
Solution :
(a) 84 x 9 = 84 x (10 – 1) = (84 x 10) – (84 x 1) = 840 – 84 = 756.
(b) 84 x 99 = 84 x (100 – 1) = (84 x 100) – (84 x 1) = 8400 – 84 = 8316.
(c) 84 x 999 = 84 x (1000 – 1) = (84 x 1000) – (84 x 1) = 84000 – 84 = 83916.
(d) 84 x 9999 = 84 x (10000 – 1) = (84 x 10000) – (84 x 1) = 840000 – 84 = 839916.
⇒ View Explanation
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Class 6 maths chapter 2 extra questions.
Q10. 50 chairs and 30 blackboard were purchased for school. If each chair cost Rs.165 and black board cost Rs.445. Find the total amount of the bill.
Solution :
Number of chair purchased = 50
Cost of one chair = 165
Cost of 50 chairs = 165 x 50 = 8,250.
Number of blackboard purchased = 30
Cost of one blackboard = 445
Cost of 30 blackboards = 445 x 30 = 13,350.
Total amount of the bill = Total cost of Chair + Blackboard
= 8,250 + 13,350.
= 21,600.
Ans. Total amount of the bill is 21,600.
⇒ View Explanation
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Class 6 maths chapter 2 extra questions.
Q11. A dealer purchased 124 LED sets. If the cost of one set is Rs.38,540, determine their total cost using a suitable property of whole numbers.
Solution :
Cost of one LED set = Rs.38,540
Total cost of 124 LED sets = (38,540 x 124)
= 38,540 x (100 + 20 + 4)
= 38,540 x 100 + 38,540 x 20 + 38,540 x 4
= 38,54,000 + 7,70,800 + 1,54,160
= Rs.47,789,60.
Ans. The Total cost of 124 LED sets is Rs.47,789,60.
⇒ View Explanation
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Class 6 maths chapter 2 extra questions.
Q12. 320 km distance is to be covered partially by bus and partially by train. Bus covers 180 km distance with a speed of 40 km/h and the rest of the distance is covered by the train at a speed of 70 km/h. Find the time taken by a passenger to cover the whole distance.
Solution :
Distance covered by the bus = 180 km
Speed of the bus = 40 km/h
Time taken by the Bus =
= Distance/Speed
= 180/40
= 4.5 hours
= 4 hour 30 minutes
Total distance = 320 km
Distance covered by the Bus = 180km
Distance covered by the train = 320 – 180 = 140 km.
Speed of the train = 70 km/h
Time taken by the train =
= Distance/Speed
= 140/70
= 2 hours
Total time taken by the passenger = Time taken by Bus + Time taken by Train
= 4 hours 30 minutes + 2 hours
= 6 hours 30 min
Ans. Time taken by a passenger to cover the whole distance is 6 hours 30 minutes.
⇒ View Explanation
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Class 6 maths chapter 2 extra questions.
Q13. Ramesh buys 10 containers of juice from one shop and 18 containers of the same juice from another shop. If the capacity of each container is same and the cost of each of the container is Rs.150, find the total money spend by Ramesh.
Solution :
Cost of 1 container of juice = Rs.150
Number of container he bought from 1st shop = 10
Number of container he bought from 2nd shop = 18
Total number of containers he bought = 28
Total money spent by Ramesh = 28 x 150
= (20 + 8) x 150
= (20 x 150) + (8 x 150)
= 3000 x 1200
= Rs.4200.
Ans. Total money spend by Ramesh is Rs.4200.
⇒ View Explanation
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Class 6 maths chapter 2 extra questions.
Q14. A housing complex built by DLF consists of 25 large buildings and 40 small buildings. Each large building has 15 floors with 4 apartments on each floor and each small building has 9 floors with 3 apartments on each floor. How many apartments are there in all?
Solution :
Number of large buildings = 25
Number of floors = 15
Number of apartments on each floor = 4
Total number of apartments in large buildings = 25 x 15 x 4 = 1500.
Number of small building = 40
Number of floors = 9
Number of apartments on each floor = 3
Total number of apartments in small buildings = 40 x 9 x 3 = 1080.
Ans. Total number of apartments = 1500 + 1080 = 2580.
⇒ View Explanation
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Class 6 maths chapter 2 extra questions.
Q15. A school principal places orders for 85 chairs and 25 tables with a dealer. Each chair cost Rs.180 and each table cost Rs.140. If the principal has given Rs.2500 to the dealer as an advance money, then what amount to be paid to the dealer now?
Solution :
Number of chairs = 85
Cost of one chair = ₹ 180
Cost of 85 chairs = (85 x 180) = ₹ 15,300
Number of tables = 25
Cost of one table = ₹140
Cost of 25 tables = (25 x 140) = ₹ 3,500
Total cost of all chairs and tables = 15300 + 3500 = ₹ 18,800.
Money paid in advance = ₹ 2500
Ans. Balance money to be paid to the dealer = 18800 – 2500 = ₹ 16,300.
⇒ View Explanation
Class 6 Maths Chapter 2 Fill - ups :
Q. The smallest whole number is 0 .
Q. Successor of 106159 is 106160 .
Q. Predecessor of 100000 is 99999 .
Q. 400 is the predecessor of 401 .
Q. 1000 is the successor of the largest 3 digit number.
Q. If 0 is subtracted from a whole number, then the result is the number itself .
Q. The smallest 6 digit natural number ending in 5 is 100005 .
Q. Whole numbers are closed under addition and under multiplication .
Q. Natural numbers are closed under addition and under multiplication .
Q. Division of a whole number by 0 is not defined.
Q. Multiplication is distributive over addition for whole numbers.
Q. 2395 × 6195 = 6195 × 2395.
Q. 1001 × 2002 = 1001 × (1001+ 1001 ).
Q. 10001 × 0 = 0 .
Q. 2916 × 0 = 0.
Q. 9128 × 1 = 9128.
Q. 125 + (68+17) = (125 + 68 ) + 17.
Q. 8925 ×1 = 8925 .
Q. 19 × 12 + 19 = 19 × (12 + 1 ).
Q. 24 × 35 = 24 × 18 + 24 × 17 .
Q. 32 × (27 × 19) = (32 × 27 ) × 19.
Q. 786 × 3 + 786 × 7 = 7860 .
Q. 24 × 25 = 24 × 100 .
Class 6 Maths Chapter 2 True - False :
Q. Successor of a one digit number is always a one digit number.
Ans. False
Q. Successor of a 3-digit number is always a 3-digit number.
Ans. False
Q. Predecessor of a two digit number is always a two digit number.
Ans. False
Q. Every whole number has its successor.
Ans. True
Q. Every whole number has its predecessor.
Ans. False
Q. Between any two natural numbers, there is one natural number.
Ans. False
Q. The smallest 4-digit number is the successor of the largest 3-digit number.
Ans. True
Q. Of the given two natural numbers, the one having more digits is greater.
Ans. True
Q. Natural numbers are closed under addition.
Ans. True
Q. Natural numbers are not closed under multiplication.
Ans. False
Q. Natural numbers are closed under subtraction.
Ans. False
Q. Addition is commutative for natural numbers.
Ans. True
Q. 1 is the identity for addition of whole numbers.
Ans. False
Q. 1 is the identity for multiplication of whole numbers.
Ans. True
Q. There is a whole number which when added to a whole number, gives the number itself.
Ans. True
Q. There is a natural number which when added to a natural number, gives the number itself.
Ans. False
Q. If a whole number is divided by another whole number, which is greater than the first one, the quotient is not equal to zero.
Ans. True
Q. Any non-zero whole number divided by itself gives the quotient 1.
Ans. True
Q. The product of two whole numbers need not be a whole number.
Ans. False
Q. A whole number divided by another whole number greater than 1 never gives the quotient equal to the former.
Ans. True
Class 6 Maths Chapter 2 MCQ's :
Q. Which of the following is not defined?
(A) 5 + 0
(B) 5 – 0
(C) 5 × 0
(D) 5 ÷ 0
Q. The product of a non-zero whole number and its successor is always divisible by -
(A) 2
(B) 3
(C) 4
(D) 5
Q. Number of whole numbers between 38 and 68 is
(A) 31
(B) 30
(C) 29
(D) 28
Q. The product of successor and predecessor of 999 is
(A) 999000
(B) 998000
(C) 989000
(D) 1998
Q. The product of a non-zero whole number and its successor is always
(A) an even number
(B) an odd number
(C) a prime number
(D) divisible by 3
Q. A whole number is added to 25 and the same number is subtracted from 25. The sum of the resulting numbers is
(A) 0
(B) 25
(C) 50
(D) 75
Q. Which of the following is not true?
(A) (7 + 8) + 9 = 7 + (8 + 9)
(B) (7 × 8) × 9 = 7 × (8 × 9)
(C) 7 + 8 × 9 = (7 + 8) × (7 + 9)
(D) 7 × (8 + 9) = (7 × 8) + (7 × 9)
Q. By using dot (.) patterns, which of the following numbers can be arranged in all the three ways namely a line, a triangle and a rectangle?
(A) 9
(B) 10
(C) 11
(D) 12
Q. Which of the following statements is not true?
(A) Both addition and multiplication are associative for whole numbers.
(B) Zero is the identity for multiplication of whole numbers.
(C) Addition and multiplication both are commutative for whole numbers.
(D) Multiplication is distributive over addition for whole numbers.
Q. Which of the following statements is not true?
(A) 0 + 0 = 0
(B) 0 – 0 = 0
(C) 0 × 0 = 0
(D) 0 ÷ 0 = 0
Q. The predecessor of 1 lakh is
(A) 99000
(B) 99999
(C) 999999
(D) 100001
Q. The successor of 1 million is
(A) 2 millions
(B) 1000001
(C) 100001
(D) 10001
# Class 6 Maths Chapter 2 : MCQ Test
- SELF EVALUATION TEST -
This is a quiz based test which is consist of 25 Multiple Choice Questions to test your Understanding about Class 6 Maths Chapter 2 : Whole Numbers.
online test for class 6 maths chapter 2
We hope that the given Study Material for Class 6 Maths Chapter 2 whole Numbers will help you to understand the concepts with more clarity. But, If you have any query regarding Class 6 Maths Chapter 2 whole Numbers then just ask your doubts in the section given below.
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CLASS 6 LITERATURE