NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers PDF
NCERT Solutions for Class 6 Maths Chapter 2 have been provided in an easy-to-understand way for CBSE class 6 students. Students can check the complete NCERT solutions for Maths Chapter 2 on this page
Jasdeep Bhatia3 Jan, 2024
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NCERT Solutions for Class 6 Maths Chapter 2
NCERT Solutions for Class 6 Maths Chapter 2: NCERT Solutions for Class 6 Maths Chapter 2 aims to share more knowledge about numbers, building upon the concepts introduced in the first chapter for 6th-grade students. PhysicsWallah provides well-organised, verified, and structured materials in its NCERT Solutions for Class 6 Maths Chapter 2. These resources are helpful for students like Victor to solve various problems, and they can also be used for competitive exams. PhysicsWallah makes studying simple and interesting by offering NCERT Solutions for subjects such as Science, Maths, English, and Hindi. Accessing these solutions makes it easier to study and understand the topics.NCERT Solutions for Class 6 Maths Chapter 2 Overview
NCERT Solutions for Class 6 Maths Chapter 2: NCERT Solutions for Class 6 Maths Chapter 2 Exercise 2.1 assist students in understanding fundamental concepts such as Whole Numbers, Natural Numbers, and Number Line. Numbers play a key role in counting the things we encounter in our daily lives. The problems from the first exercise of this chapter have been expertly solved by the team at PhysicsWallah. These NCERT Solutions are explained in a simple manner to help students grasp problem-solving techniques more effectively.NCERT Solutions For Class 6 Maths Chapter 2 Whole Numbers
For Class 6 Maths Chapter 2 Whole Numbers, you can find clear step-by-step explanations here. These solutions are widely favoured by Class 6 students, serving as useful tools for completing homework and getting ready for exams in a timely manner. All the questions and answers from the NCERT Book of Class 6 Maths Chapter 2 are provided here for free. Check the complete NCERT Solutions for Class 6 Maths Chapter 2 below:Exercise (2.1)
1. Write the Next Three Natural Numbers After
10999.
Ans: Natural numbers are those numbers which start from positive integers and go on till infinity. To find the next three natural numbers, just add 1 to every preceding integer. So, 10,999+1=11,000 11,000+1=11,001 11,001+1=11,002 Thus the three natural numbers after 10,999 are 11000,11001,11002 . 2. Write the Three Whole Numbers Occurring Just Before10001.
Ans: Whole numbers are those numbers which start from zero and go on till infinity. To find the three whole numbers occurring before the number, just subtract 1 from every preceding integer. So, 10,001−1=10,000 10,000−1=9,999 9,999−1=9,998 Thus the three whole numbers occurring before 10,001 are 10000,9999,9998 . 3. Which is the Smallest Whole Number? Ans: Whole numbers are those numbers which start from zero and go on till infinity. Since the whole numbers start with zero, the smallest whole number is zero. So 0 is the smallest number. 4. How Many Whole Numbers are There Between 32 and 53 ? Ans: To find the number of whole numbers between two numbers, we have to list out the numbers between 32 and 53. The numbers are: 33,34,35,...,52 . Numbers between 53 and 32=(53−32)−1 =20 So there are 20 whole numbers between 32 and 53 . 5. Write the Successor of (a) 2440701 Ans: The successor is the number that comes after the given number. It can be found by adding 1 to the given number. So, 2440701+1=2440702 So the successor of 2440701 is 2440702. (b) 100199 . Ans: The successor is the number that comes after the given number. It can be found by adding 1 to the given number. So, 100199+1=100200 So the successor of 100199 is 100200. (c) 1099999 . Ans: The successor is the number that comes after the given number. It can be found by adding 1 to the given number. So, 1099999+1=1100000 So the successor of 1099999 is 1100000. (d) 2345670 . Ans: The successor is the number that comes after the given number. It can be found by adding 1 to the given number. So, 2345670+1=2345671 So the successor of 2345670 is 2345671. 6. Write the Predecessor of (a) 94 Ans: The predecessor is the number that comes before the number. It can be found by subtracting 1 from the given number. So, 94−1=93 So the predecessor of 94 is 93. (b) 10000 Ans: The predecessor is the number that comes before the number. It can be found by subtracting 1 from the given number. So, 10,000−1=9,999 So the predecessor of 10,000 is 9,999. (c) 208090 Ans: The predecessor is the number that comes before the number. It can be found by subtracting 1 from the given number. So, 2,08,090−1=2,08,089 So the predecessor of 2,08,090 is 2,08,089. (d) 7654321 Ans: The predecessor is the number that comes before the number. It can be found by subtracting 1 from the given number. So, 76,54,321−1=76,54,320 So the predecessor of 76,54,321 is 76,54,320.7. 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 Ans: Numbers in the number line always increase from left to right. Here the smaller number is 503. So 503 lies on the left of 530. And 530>503 Hence, 503 lies on the left of 530 on the number line and 530>503. (b) 370,307 Ans: Numbers in the number line always increase from left to right. Here the smaller number is 307. So 307 lies on the left of 370. And 370>307 Hence, 307 lies on the left of 370 on the number line and 370>307. (c) 98765,56789 Ans: Numbers in the number line always increase from left to right. Here the smaller number is 56789. So 56789 lies on the left of 98765. And 98765>56789 Hence, 56789 lies on the left of 98765 on the number line and 98765>56789. (d) 9830415,10023001 Ans: Numbers in the number line always increase from left to right. Here the smaller number is 9830415. So 9830415 lies on the left of 10023001. And 9830415<10023001 Hence, 9830415 lies on the left of 10023001 on the number line and 9830415<10023001. 8. Which of the Following Statements are True (T) and Which are False (F): (a) Zero is the Smallest Natural Number. Ans: Natural numbers are those numbers which start from positive integers and go on till infinity. So the smallest natural number is 1 and not zero. So the given statement “Zero is the smallest natural number” is false.(b) 400 is the Predecessor of
399 . Ans: The predecessor is the number that comes before the number. It can be found by subtracting 1 from the given number. So, 399−1=398 So the predecessor of 399 is 398. So the given statement “ 400 is the predecessor of 399” is false. (c) Zero is the Smallest Whole Number. Ans: Whole numbers are those numbers which start from zero and go on till infinity. So the smallest whole number is zero. So the given statement “Zero is the smallest whole number” is true.(d) 600 is the successor of 599.
Ans: The successor is the number that comes after the given number. It can be found by adding 1 to the given number. So, 599+1=600 So the successor of 599 is 600. So the given statement “ 600 is the successor of 599:” is true. (e) All Natural Numbers are Whole Numbers. Ans: Natural numbers are those which start from positive integers to infinity. Whole numbers are those numbers which start from zero and go on till infinity. So natural numbers will also come under whole numbers. So all natural numbers are whole numbers. So the given statement “All natural numbers are whole numbers” is true. (f) All Whole Numbers are Natural Numbers. Ans: Natural numbers are those which start from positive integers to infinity. Whole numbers are those numbers which start from zero and go on till infinity. So all natural numbers are whole numbers but not all whole numbers are natural numbers since natural will miss zero from whole numbers. So the given statement “All whole numbers are natural numbers” is false. (g) The Predecessor of a Two Digit Number is Never a Single Digit Number. Ans: Let us consider a two digit number. Let it be 10. The predecessor is the number that comes before the number. It can be found by subtracting 1 from the given number. So, 10−1=9 So the predecessor of 10 is 9. Thus the predecessor of a two digit number is a single digit number in this case. So the given statement “The predecessor of a two digit number is never a single digit number’ is false. (h ) 1 is the Smallest Whole Number. Ans: Whole numbers are those numbers which start from zero and go on till infinity. So the smallest whole number is 0. So the given statement “ 1 is the smallest whole number” is false. (i) The Natural number 1 has no predecessor. Ans: Natural numbers are those which start from positive integers and go on till infinity. So the smallest natural number is 1 and it has no predecessor. So the given statement “The natural number 1 has no predecessor” is true. (j) The Whole Number 1 has no Predecessor. Ans: Whole numbers are those numbers which start from zero and go on till infinity. The predecessor is the number that comes before the number. It can be found by subtracting 1 from the given number. So, 1−1=0 So the predecessor of 1 is 0 which is a whole number. Thus the given statement “The whole number 1 has no predecessor” is false.CBSE Board Exam Centre List 2024
(k) The Whole Number13 Lies Between 11 and 12.
Ans: Whole numbers are those numbers which start from zero and go on till infinity. 0,1,2,3,...,11,12,13,... The whole number 13 lies after 11, 12 and not between them. So the given statement “The whole number 13 lies between 11 and 12 “ is false. (l) The whole Numbe r 0 has no Predecessor. Ans: Whole numbers are those numbers which start from zero and go on till infinity. So the smallest number in the whole number is zero and it has no predecessor. So the given statement “The whole number 0 has no predecessor” is true. (m) The Successor of Two Digit Number is Always a Two Digit Number” Ans: Let us consider a two digit number. Let it be 99. The successor is the number that comes after the given number. It can be found by adding 1 to the given number. So, 99+1=100 So the successor of 99 is 100. Thus the successor of a two digit number is a three digit number in this case. So the given statement “The successor of a two digit number is always a two digit number’ is false.Exercise (2.2)
1. Find the Sum by Suitable Rearrangement:
(a) 837+208+363 Ans: Add unit places of a two digit number and check if we get an easier number. And then add those numbers and simplify them like this to get the final answer. 837+208+363=(837+363)+208 =1200+208 =1408 So the sum of the numbers 837+208+363 by suitable arrangement is 1408. (b) 1962+453+1538+647 Ans: Add unit places of a two digit number and check if we get an easier number. And then add those numbers and simplify them like this to get the final answer. 1962+453+1538+647=(1962+1538)+(453+647) =3500+1100 =4600 So the sum of the numbers 1962+453+1538+647 by suitable arrangement is 4600. 2. Find the Product by Suitable Arrangement: (a) 2×1768×50 Ans: Arrange the numbers in such a way that we get easier numbers while multiplying them. And multiply them again to get the final answer. 2×1768×50=(2×50)×1768 =(100×1768) =176800 So the product of 2×1768×50 by suitable arrangement is 176800. (b) 4×166×25 Ans: Arrange the numbers in such a way that we get easier numbers while multiplying them. And multiply them again to get the final answer. 4×166×25=(4×25)×166 =(100×166) =16600 So the product of 4×166×25 by suitable arrangement is 16600. (c) 8×291×125 Ans: Arrange the numbers in such a way that we get easier numbers while multiplying them. And multiply them again to get the final answer. 8×291×125=(125×8)×291 =(1000×291) =291000 So the product of 8×291×125 by suitable arrangement is 291000. (d) 625×279×16 Ans: Arrange the numbers in such a way that we get easier numbers while multiplying them. And multiply them again to get the final answer. 625×279×16=(625×16)×279 =(10000×279) =2790000 So the product of 625×279×16 by suitable arrangement is 2790000. (e) 285×5×60 Ans: Arrange the numbers in such a way that we get easier numbers while multiplying them. And multiply them again to get the final answer. 285×5×60=285×(5×60) =(285×300) =85500 So the product of 285×5×60 by suitable arrangement is 85500. (f) 125×40×8×25 Ans: Arrange the numbers in such a way that we get easier numbers while multiplying them. And multiply them again to get the final answer. 125×40×8×25=(125×8)×(40×25) =(1000×1000) =1000000 So the product of 125×40×8×25 by suitable arrangement is 1000000. 3. Find the Value of the Following: (a) 297×17+297×3 Ans: It is in the form of ab+ac ab+ac. So we can use distributive property over addition, that is, a(b+c)=ab+ac 297×17+297×3=297(17+3) =297×20 =5940 So the value of 297×17+297×3 is 5940. (b) 54279×92+8×54279 Ans: It is in the form of ab+ac. So we can use distributive property over addition, that is, a(b+c)=ab+ac 54279×92+8×54279=54279(92+8) =54279×100 =5427900 So the value of 54279×92+8×54279 is 5427900. (c) 81265×169−81265×69 Ans: It is in the form of ab−ac. So we can use distributive property over subtraction, that is, a(b−c)=ab−ac 81265×169−81265×69=81265(169−69) =81265×100 =8126500 So the value of 81265×169−81265×69 is 8126500. (d) 3845×5×782+769×25×218 Ans: The equation can be reduced as, 3845×5×782+769×25×218=3845×5×782+769×5×5×218 =3845×5×782+3845×5×218 It is in the form of ab+ac. So we can use distributive property over addition, that is, a(b+c)=ab+ac 3845×5×782+769×25×218=3845×5(782+218) =3845×5×1000 =19225000 So the value of 3845×5×782+769×25×218 is 19225000. 4. Find the Product Using Suitable Properties: (a) 738×103 Ans: Write the given number 103 as (100+3) 738×103=738×(100+3) Use the distributive law over addition, that is, a(b+c)=ab+ac 738×103=(738×100)+(738×3) =73800+2214 =76014 The product of 738×103 by using proper identity is 76014. (b) 854×102 Ans: Write the given number 102 as (100+2) 854×102=854×(100+2) Use the distributive law over addition, that is, a(b+c)=ab+ac 854×102=(854×100)+(854×2) =85400+1708 =87108 The product of 854×102 by using proper identity is 87108. (c) 258×1008 Ans: Write the given number 1008 as (1000+8) 258×1008=258×(1000+8) Use the distributive law over addition, that is, a(b+c)=ab+ac 258×1008=(258×1000)+(258×8) =258000+2064 =260064 The product of 258×1008 by using proper identity is 260064. (d) 1005×168 Ans: Write the given number 1005 as (1000+5) 1005×168=(1000+5)×168 Use the distributive law over addition, that is, a(b+c)=ab+ac 1005×168=(1000×168)+(5×168) =168000+840 =168840 The product of 1005×168 by using proper identity is 168840.| CBSE Syllabus Class 6 | |
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5. A taxi Driver Filled his Car's 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 Rs.44 Per Litre, How Much did he Spend on Petrol?
Ans: Amount of Petrol filled on Monday =40 litres Amount of petrol filled on next day =50 litres Total amount of petrol filled =40+50 =90 litres Cost of one litre petrol =Rs.44 Cost of 90 litre petrol =44×90 =Rs.3960 Thus the taxi driver spent Rs.3960 on petrol.6. 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
Rs.15 per litre, how much money is due to the vendor per day?
Ans: Amount of milk supply in the morning =32 litres Amount of milk supply in the evening =68 litres Total amount of milk supply =32+68 =100 litres Cost of one litre milk =Rs.15 Cost of 100 litre milk =15×100 =Rs.1500 Amount of money due to the vendor per day is Rs.1500 7. Match the Following: (i) 425×136=425×(6+30+100) Ans: 425×136=425×(6+30+100) This is in the form of a(b+c)=ab+ac which is a distributivity multiplication under addition which is in the option (c). (ii) 2×48×50=2×50×48 Ans: 2×48×50=2×50×48 This is in the form of a×b×c=a×c×b which is commutative under multiplication which is in the option (a). (iii) 80+2005+20=(80+20+2005) Ans: 425×136=425×(6+30+100) This is in the form of a+b+c=a+c+b which is commutative under addition which is in the option (b).Exercise (2.3)
1. Which of the Following will not Represent Zero: (a) 1+0 Ans: The given numbers represent the sum the numbers, So, 1+0=1 So it gives an answer which is a non-zero number. So 1+0 does not represent zero. (b) 0×0 Ans: The given numbers represent the multiplication of the numbers, So, 0×0=0 So 0×0 represents zero. (c) 0 2 02 Ans: The given numbers represent the division of the numbers, So, 0 2 =0 02=0 So 0 2 02 represents zero. (d) 10−10 2 10−102 Ans: The given numbers represent the difference of the numbers followed by division. So, 10−10 2 =0 10−102=0 So 10−10 2 10−102 represents zero. 2. If the Product of Two Whole Numbers is Zero, Can We Say That One or Both of Them Will Be Zero? Justify Your Examples. Ans: Let us consider two numbers as 4,0in which one of them is zero. Product of these numbers is 4×0=0 If we consider both the numbers to be zero, then 0×0=0. Yes, we can say that one or both of the whole numbers will be zero.3. If the Product of Two Whole Numbers is 1, Can We Say That One or Both of the Numbers will be1? Justify Through Examples.
Ans: Let us consider both the whole number to be 1. 1×1=1 Consider either of the numbers to be any other whole number, let it be 3. 3×1=3 So the product of both the whole number will not always be equal to 1. No, we cannot say that the product of one or both the numbers will be 1. 4. Find Using Distributive Property: (a) 728×101 Ans: We can 101 as 101=(100+1) 728×(100+1) So we can use distributive property over addition, that is, a(b+c)=ab+ac 728×(100+1)=728×100+728×1 =72800+728 =73528 So the value of 728×101 is 73528. (b) 5437×1001 Ans: We can 1001 as 1001=(1000+1) 5437×(1000+1) So we can use distributive property over addition, that is, a(b+c)=ab+ac 5437×(1000+1)=5437×1000+5437×1 =5437000+5437 =5442437 So the value of 5437×1001 is 5442437. (c) 824×25 Ans: We can 25 25 as 25=(20+5) 824×(20+5) So we can use distributive property over addition, that is, a(b+c)=ab+ac 824×(20+5)=(824×20)+(824×5) =16480+4120 =20600 So the value of 824×25 is 20600. (d) 4275×125 Ans: We can 125 as 125=(100+20+5) 4275×(100+20+5) So we can use distributive property over addition, that is, a(b+c)=ab+ac 4275×(100+20+5)=(4275×100)+(4275×20)+(4275×5) =427500+85500+21375 =534375 So the value of 4275×125 is 534375. (e) 504×35 Ans: We can 504 as 504=(500+4) (500+4)×35 So we can use distributive property over addition, that is, a(b+c)=ab+ac (500+4)×35=500×35+4×35 =17500+140 =17640 So the value of 504×35 is 17640. 5. Study the pattern: 1×8+1=9 12×8+2=98 123×8+3=987 1234×8+4=9876 12345×8+5=98765 Ans: Since it was like 12...n×8+n=n...1 The next two terms are 123456×8+6=987654 1234567×8+7=9876543 And the pattern will be like: 1×8+1=9 12×8+2=98 123×8+3=987 1234×8+4=9876 12345×8+5=98765 The next two terms are 987654,9876543.NCERT Solutions for Class 6 Maths Chapter-wise Solutions
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