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.
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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.
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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 Before
10001.
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.
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(k) The Whole Number
13 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.
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
