Home » Class 6 Maths » NCERT Solution Class 6 Maths Chapter – 2 Whole Numbers

NCERT Solution Class 6 Maths Chapter – 2 Whole Numbers


Exercise 1


Question 1:List out any four natural numbers after 11999.
Answer: Three natural numbers after 11999 are 12000, 12001, and 12002.


Question 2: List out any four whole numbers, which come just before 10001?
Answer: The numbers, which come before 10001, are 10000, 9999, and 9998


Question 3: The smallest whole number is:
Answer: The smallest whole number is zero.


Question 4: What are the number of whole numbers lying between 42 and 65?
Answer: The number of whole numbers lying between 42 and 65 is 22 (65 – 42 – 1 = 22).
They are as follows 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, and 64.


Question 5: What is the predecessor of?
(a) 95 (b) 1000 (c) 302780 (d) 6754300
Answer:
(a) 95 – 1 = 94
(b) 1000 – 1 = 999
(c) 302780 – 1 = 302779
(d) 6754300 – 1 = 6754299


Question 6: What is the successor of?
(a) 88 (b) 9999 (c) 84729 (d) 63513
Answer:
(a) 88 + 1 = 89
(b) 9999 + 1 = 10000
(c) 84729 + 1 = 84730
(d) 63513 + 1 = 63514


Question 7: In the following given pair of numbers, find the whole number which lies on the left side of the other number in on the number line, Also put the appropriate sign (<,>) between them.
(a) 115, 151 (b) 435, 345
(c) 9876567, 9876576 (d) 1234567, 1233567
Answer:
(a) 115 < 151 115 lie on the left side of 151 on the number line.
(b) 435 > 345 345 lie on the left side of 435 on the number line.
(c) 9876567 < 9876576 9876567 lie on to the left of 9876576 on the number line.
(d) 1234567 > 1233567 1233567 lie on the left side of 1234567 on the number line.


Question 8: Mark True (T) or False (F) for the given statements.
(a) Zero is the smallest natural number
(b) 400 is the successor of 399
(c) Smallest whole number is Zero
(d) 500 is the predecessor of 499
(e) All natural numbers are whole numbers
(f) All whole numbers are natural numbers
(g) Predecessor of a two-digit numbers can never be a single digit number
(h) The smallest whole number is 1
(i) The natural 1 number has no predecessor
(j) The whole number 1 has no predecessor
(k) The whole number 21 lies between 25 and 27
(l) The whole number 0 have a predecessor
(m) The successor of a three digit number can never be four digit number
Answer:
(a) False, 0 does not come under natural number
(b) True, as 399 + 1 = 400
(c) True
(d) False, predecessor of 499 is 499 -1 = 488
(e) True
(f) False, 0 is a whole number but it is not a natural number
(g) False, as 10 is a 2 digit number but its predecessor is 9 which is a single digit number
(h) False, the smallest whole number is 0
(i) True, since 0 is the successor of 1 but it’s not a natural number
(j) False, 0 is a whole number
(k) False
(l) False, 0 have a predecessor -1 which is not a whole number
(m) False, 999 is a three digit number and its successor is 10000 and it’s a four digit number


EXERCISE- 2


Question 1: Frame the numbers in proper order and find the sum
a) 820 + 200 +325
b) 1960 + 448 + 1520 + 638
Answer:
a) (800 + 325) + 200 = 1325
b) (1960 +1520) + (448 + 638) = 3480 + 1086 = 4566


Question 2:
Find the product by framing it in suitable order
a) 2× 1750× 45
b) 3× 165×25
c) 6×290×120
d) 600×286×18
e) 286×8×70
f) 130×45×6×30
Answer:
a) 2×1750×45    =  (2×45) ×1750 = 90×1750 = 157,500
b) 3×165×25      =  (3×25) ×165 = 75×165 = 12,375
c) 6×290×120    =  (6×120) ×290 = 720×290 = 208,800
d) 600×286×18  =  (600×18) ×286 = 10800×286 = 3,088,800
e) 286×8×70      =  286× (8×70) = 286×560 = 160,160
f) 130×45×6×30  = (130×6) × (45×30) = 780×1350 = 1,053,000


Question 3: Find the solution of the following
a) 290×15 + 290×5
Answer: Since 290 is repeated, we can take it as a common term
290× (15+5) = 290×20 = 5800


b) 54270×95 + 6×54270
Answer: Since 54270 is repeated, we can take it as a common term

54270× (95+6) = 54270×101 = 5,481,270


c) 8250×62 – 8250×50
Answer: Since 8250 is repeated, we can take it as a common term
8250× (62-50) = 8250×12 = 99,000


d) 3845×5×720 + 769×25×220
Answer: = 3845×5×720 + 769×5×5×220
= 3845×5×720 + (769×5)5×220
= 3845×5×720 + 3845×5×220
Since 3845 and 5 is repeated, we can take it as a common term
= 3845×5× (720 +220)
= 19,225×940
= 18,071,500


Question 4: Use suitable properties to find out the product
a) 642×105
Answer:
= 642× (100+5)
By distributive law
A× (B+C) = A×B + A×C
= 642×100 + 642×5
= 64200 + 3210
= 67,410


b) 850×103
Answer:
= 850× (100+3)
By distributive law,
A× (B+C) = A×B + A×C
= 850×100 + 850×3
= 85000 + 2550
= 87,550


c) 260×1006
Answer:
= 260× (1000+6)
By distributive law
A× (B+C) = A×B + A×C
= 260×1000 + 260×6
= 260000 + 1560
= 261,560


d) 1008×168
Answer:
= (1000×8) ×168
By distributive law
A× (B+C) = A×B + A×C
= 1000×168 + 8×168
= 168000 + 1344
= 169,344


Question 5: A cab driver filled his vehicle tank with 35litres of diesel on Tuesday. The next day, he filled the tank with 45litres of diesel. If the diesel costs Rs.32 per litre. How much did he spend on diesel?
Answer:
The quantity of diesel filled on Tuesday = 35L
The quantity of diesel filled on Wednesday = 45L
The total quantity filled = (35+45)L
Cost of diesel (per L) = Rs.32
The total money spent on diesel = 32× (35+45)
= 32×80
= Rs.2560
The total amount spent on diesel = Rs.2560


Question 6: A milkman supplies 30 liters of milk to a teashop in the morning and 60 liters of milk in the evening. If the cost of milk is 12 per liter, how much does the milkman earn a day?
Answer:
The quantity of milk supplied in the morning = 30L
The quantity of milk supplied in the evening = 60L
The total quantity filled = (30+60)L
Cost of milk (per L) = Rs.12
(L represents litres)
The total money spent on diesel = 12× (30+60)
= 12×90
= Rs.1080
The total amount spent on diesel = Rs.1080


Question 7: Write the answers using correct property
a) 424×128
Answer:
424×128 = 424× (100+20+8)
This is distributive property of multiplication over addition


b) 2×49×50
Answer:
2×48×51 = 2×51×48
]This is commutative property under multiplication


c) 82+2000+20
Answer:
82+2000+20 = 82+20+2000
This is commutative property under addition


EXERCISE- 3


Question 1: Which out of the following questions will represent zero;
a) 2+0
b) 1×0
c) 02
d) 20−202
Answer:
a) 2 + 0 = 2
It does not represent zero, since the answer is 2.
b) 1*0 = 0
It represents zero, since number multiplied with anything is zero.
c) 02 = 0
It represents zero, since number divided with anything is zero.
d) 20-20/2 = 0
It represents zero.


Question 2: The product of two numbers is zero; can we say that one or both of them will be zero or non-zero? State reasons.
Answer: First case
The product of 2 whole numbers will be zero only when one of the number is multiplied with zero
Ex)
5×0 = 0
18×0 = 0
Second case
The product of 2 whole numbers will be zero when both the numbers are multiplied are zero
0×0 = 0
Third case
The product of 2 numbers will not be zero, if both are multiplied with a non-zero number
4×3 = 12
5×4 = 20


Question 3:The product of two whole numbers is 1, can we say that one or both of them will be 1? State reasons.
Answer:
First case
The product of 2 whole numbers will be one, if both the numbers are multiplied with 1.
Ex)
1×1 =1
Second case
The product of 2 whole numbers will not be 1, if it is multiplied by numbers other than 1.
6×5 = 30
So, it is clear that the product of two numbers will be 1 only when the numbers are multiplied with 1.


Question 4: Find the solution using distributive property.
a) 725×102
Answer:
By distributive law
A× (B+C) = A×B + A×C
= 728× (100+2)
= 728×100 + 728×2
= 72800 + 1456
= 74,256


b) 5450×1004
Answer:
By distributive law,
A× (B+C) = A×B + A×C
= 5450× (1000+4)
= (5450×1000) + (5450×4)
= 5450000 + 21800
= 5471800


c) 724×25
Answer:
By distributive law,
A× (B+C) = A×B + A×C
= (700+24) ×25
= (700 + 25 – 1) ×25
= 700×25 + 25×25 – (1×25)
= 17500 + 625 – 25
= 18100


d) 4225×125
Answer:
By distributive law,
A× (B+C) = A×B + A×C
= (4000 + 200 +100 -75) ×125
= (4000×125) + (200×125) + (100×125) – (75×125)

= (500000) + (25000) + (12500) – (9375)
= 528125


e) 508×35
Answer:
By distributive law,
A× (B+C) = A×B + A×C
= (500 + 8) ×35
= (500×35) + (8×35)
= 17500 + 280
= 17780


Question 5: Analyze the pattern
(1×8) + 1 = 91234×8 + 4 = 9876
(12×8) +2 = 981234× 8 + 5 = 98765
(123×8) + 3 = 987
Follow the next two steps and explain the working of the pattern
(Note: 12345 = 11111 + 1111 + 111 + 11 + 1)
Answer:
123456×8 + 6 = 987648 + 6 = 987654
1234567×8 + 7 = 9876536 + 7 =9876543
The answer is yes and the pattern works
W.K.T
123456 = (111111 + 11111 + 1111 + 111 + 11 + 1)
123456×8 = (111111 + 11111 + 1111 + 111 + 11 +1) ×8 = (111111×8) + (11111×8) + (1111×8) + (111×8) + (11×8) + (1×8)
= 888888 + 88888 + 8888 + 888 + 88 + 8
= 987648
So,(123456×8) + 6 = 987648 + 6 = 987654