Whole Numbers
Class 6th Mathematics CBSE Solution
- Write the next three natural numbers after 10999.
- Write the three whole numbers occurring just before 10001.
- Which is the smallest whole number?
- How many whole numbers are there between 32 and 53?
- Write successor of: (a) 2440701 (b) 100199 (c) 1099999 (d) 2345670…
- Write the predecessor of: (a) 94 (b) 10000 (c) 208090 (d) 7654321…
- In each of the following pairs of numbers, state which whole number is on the…
- Which of the following statements are true (T) and which are false (F)? (a) Zero…
- Find the sum by suitable rearrangement: (a) 837 + 208 + 363 (b) 1962 + 453 +…
- Find the product by suitable rearrangement: (a) 2 1768 50 (b) 4 166 25 (c) 8 291…
- Find the value of the following: (a) 297 17 + 297 3 (b) 54279 92 + 8 54279 (c)…
- Find the product using suitable properties. (a) 738 103 (b) 854 102 (c) 258 1008…
- A taxidriver filled his car petrol tank with 40 litres of petrol on Monday. The…
- A vendor supplies 32 litres of milk to a hotel in the morning and 68 litres of…
- Match the following:(i) 425 136 = 425 (6+30+100)(a) Commutativity of…
- Which of the following will not represents zero: (a) 1 + 0 (b) 0 0 (c) 0/2 (d)…
- If the product of two whole numbers is zero, can we say that one or both of them…
- If the product of two whole numbers is 1, can we say that one or both of them…
- Find using distributive property: (a) 728 101 (b) 5437 1001 (c) 824 25 (d) 4275…
- Study the pattern: 1 8 + 1 = 9 12 8 + 2 = 98 123 8 + 3 = 987 1234 8 + 4 = 9876…
Exercise 2.1
Question 1.Write the next three natural numbers after 10999.
Answer:
Next three natural numbers after 10999 are;
10999+1, 10999+2, 10999+3
=11000, 11001 and 11002.
Question 2.
Write the three whole numbers occurring just before 10001.
Answer:
Three whole numbers occurring just before 10001 are;
10000, 9999 and 9998.
Question 3.
Which is the smallest whole number?
Answer:
The smallest whole number is 0.
Question 4.
How many whole numbers are there between 32 and 53?
Answer:
We know whole numbers are all counting numbers along with number 0.
So,
Whole numbers between 53 and 32 = (53-32-1) = 20
These are: 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51 and 52.
Question 5.
Write successor of:
(a) 2440701 (b) 100199
(c) 1099999 (d) 2345670
Answer:
The successor of a number is the number obtained by adding one to it, also known as the after number.
For example – successor of 9 will be 9+1 = 10
(a) 2440701+1 = 2440702
(b) 100199+1 = 100200
(c) 1099999+1 = 1100000
(d) 2345670+1 = 2345671
Question 6.
Write the predecessor of:
(a) 94 (b) 10000
(c) 208090 (d) 7654321
Answer:
The predeccessor of a number is the number obtained by substracting one from the given number, also known as the before number.
For example – predecessor of 9 will be 8 (9-1 = 8)
(a) 94 – 1 = 93
(b) 10000 – 1 = 9999
(c) 208090 – 1 = 208089
(d) 7654321 – 1 = 7654320
Question 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 (> or <) between them.
(a) 530, 503
(b) 370, 307
(c) 98765, 56789
(d) 9830415, 10023001
Answer:
(a) 530 ˃ 503
503 is on the left side of the number 530 on the number line.
(b) 370 ˃ 307
307 is on the left side of the number 370 on the number line.
(c) 98765 ˃ 56789
56789 is on the left side of the number 98765 on the number line.
(d) 9830415 ˂ 10023001
9830415 is on the left side of the 10023001 on the number line.
Question 8.
Which of the following statements are true (T) and which are false (F)?
(a) Zero is the smallest natural number.
(b) 400 is the predecessor of 399.
(c) Zero is the smallest whole number.
(d) 600 is the successor of 599.
(e) All natural numbers are whole numbers.
(f) All whole numbers are natural numbers.
(g) The predecessor of a two digit number is never a single digit number.
(h) 1 is the smallest whole number.
(i) The natural number 1 has no predecessor.
(j) The whole number 1 has no predecessor.
(k) The whole number 13 lies between 11 and 12.
(l) The whole number 0 has no predecessor.
(m) The successor of a two digit number is always a two digit number.
Answer:
(a) False.
0 is not a natural number.
The natural numbers are also called as counting numbers which start from 1, 2, 3, 4, 5 and so on….
(b) False.
Beacsuse 400 is the successor of 399.
The predecessor of 399 will be 398 (399-1).
(c) True.
As whole numbers starts from 0,1,2,3,4 and so on.
Yes, 0 is the smallest whole number.
(d) True.
600 is the successor of 599 (599 + 1 = 600).
(e) True.
As we know whole numbers starts from 0, 1, 2, 3, 4, 5… and natural numbers starts from 1, 2, 3, 4, 5..
So, all natural numbers are whole numbers.
(f) False.
As we know, whole numbers starts from 0 and natural numbers starts from 1.
0 is a whole number but not a natural number.
(g) False.
The predecessor of a two-digit number can also be a single digit number.
For example – The predecessor of 10 is 9 which is a single digit number.
(h) False.
As we know whole numbers starts from 0, so, 0 is the smallest whole number.
(i) True.
As 0 is the predecessor of 1 but it’s not the natural number.
(j) False.
0 is the predecessor of 1 and it is a whole number.
(k) False.
The whole number 13 doesn’t lies between 11 and 12. It’s the successor of 12.
(l) True.
The whole number 0 has -1 as its predecessor but it is not a whole number.
(m) False.
It is not necessary that, the successor of a two digit number is always a two digit number as 100 is the successor of 99 which is not a two digit number.
Exercise 2.2
Question 1.Find the sum by suitable rearrangement:
(a) 837 + 208 + 363
(b) 1962 + 453 + 1538 + 647
Answer:
(a) Arrange the numbers in the decreasing order;
837+363+208
Now make pairs,
= (837+363) + 208
= 1200+208
= 1408
(b) Arrange the numbers in decreasing order,
= 1962+1538+647+453
Now make pairs,
= (1962+1538) + (647+453)
= 3500 + 1100
= 4600
Question 2.
Find the product by suitable rearrangement:
(a) 2 × 1768 × 50 (b) 4 × 166 × 25
(c) 8 × 291 × 125 (d) 625 × 279 × 16
(e) 285 × 5 × 60 (f) 125 × 40 × 8 × 25
Answer:
(a) 2×1768×50
Arrange it in increasing order,
= 2×50×1768
= 100×1768
= 176800
(b) 4×166×25
Arrange it in increasing form,
= 4×25×166
= 100×166
= 16600
(c) 8×291×125
= 8×125×291
= 1000×291
= 291000
(d) 625×279×16
= 625×16×279
= 10000×279
= 279000
(e) 285×5×60
= 285×300
= 85500
(f) 125×40×8×25
= 125×8×40×25
= 1000×1000
= 1000000
Question 3.
Find the value of the following:
(a) 297 × 17 + 297 × 3
(b) 54279 × 92 + 8 × 54279
(c) 81265 × 169 – 81265 × 69
(d) 3845 × 5 × 782 + 769 × 25 × 218
Answer:
(a) 297 × 17 + 297 × 3
As we can clearly see it is in the form of;
= ab+ac
= a(b+c)
So,
= 297 × (17+3)
= 297 + 20
= 5940
(b) 54279 × 92 + 8 × 54279
= 54279 × (92+8)
= 54279 × 100
= 5427900
(c) 81265 × 169 – 81265 × 69
= 81265 × (169-69)
= 81265×100
= 8126500
(d) 3845 × 5 × 782 + 769 × 25 × 218
= 3845×5×782 + 769×5×5×218
= 3845×5×782 + 3845×5×218
= 3845 × 5 × (782 + 218)
= 19225×1000
= 19225000
Question 4.
Find the product using suitable properties.
(a) 738 × 103 (b) 854 × 102
(c) 258 × 1008 (d) 1005 × 168
Answer:
(a) 738×103
By using distributive property;
We get,
= 738(100+3)
= 738×100+738×3
= 73800+2214
= 76014
(b) 854 × 102
By using distributive property, we get,
= 854 × (100+2)
= 854×100+854×2
= 85400+1708
= 87108
(c) 258 × 1008
By using distributive property, we get,
= 258 × (1000+8)
= 258×1000+258×8
= 258000+2064
= 260064
(d) 1005×168
By using distributive property, we get,
= (1000+5) ×168
= 1000×168+5×168
= 168000+840
= 168840
Question 5.
A taxidriver 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 Rs. 44 per litre, how much did he spend in all on petrol?
Answer:
Petrol filled on Monday = 40 litres
Petrol filled on Tuesday = 50 litres
Total petrol filled = (40+50) litres
Cost of petrol = 44 Rs per litre
Total money spent on petrol = 44 × (40+50)
= 44 × 40 + 44 × 50
= 1760+2200
= 3960 Rs.
Question 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. 45 per litre, how much money is due to the vendor per day?
Answer:
Milk supplied in the morning = 32 L
Milk supplied in the evening = 68 L
Total milk supplied every day = (32+68) L
Cost of milk = 45 Rs
Total cost per day = 45 × (32+68)
= 45×100
= 4500 Rs
Thus, Rs 4500 is due to the vendor every day
Question 7.
Match the following:
Answer:
Exercise 2.3
Question 1.Which of the following will not represents zero:
(a) 1 + 0 (b) 0 × 0
(c) 
(d) 
Answer:
(a) 1 + 0 = 1
It does not represent 0.
(b) 0×0 = 0
It does represent 0.
(c) 
= 0
It does represent 0.
(d) 
= 0
It does represent 0.
Question 2.
If the product of two whole numbers is zero, can we say that one or both of them will be zero? Justify through examples.
Answer:
Yes, if the product of two whole numbers is zero, then one of them will be 0.
For example-
1×0 = 0
0×7 = 0
Yes, if the product of two whole numbers is zero, then both may also be 0.
For example-
0×0 = 0
Question 3.
If the product of two whole numbers is 1, can we say that one or both of them will be 1? Justify through examples.
Answer:
If the product of two numbers is 1, then both the numbers need to be 1.
Example -
1×1 = 1
However, if only one number is 1 then,
Example –
1×9 = 9
So,
We can say that the product of two whole numbers is 1 only when, both 1.
Question 4.
Find using distributive property:
(a) 728 × 101 (b) 5437 × 1001
(c) 824 × 25 (d) 4275 × 125
(e) 504 × 35.
Answer:
Distributive property is,
a (b+c) = ab+ac
(a) 728×101
= 728×(100+1)
= 728 × 100+728×1
= 72800+728 = 73528
(b) 5437×1001
= 5437×(1000+1)
= 5437×1000+5437×1
= 5437000+5437
= 5442437
(c) 824×25
= (800+24)×25
= (800+25-1)×25
= 800×25+25×25-1×25
= 20000+625-25
= 20000+600
= 20600
(d) 4275×125
= (4000+200+100-25)×125
= 4000×125+200×125+100×125-25×125
= 500000+25000+12500-3125
= 534375
(e) 504×35
= (500+4)×35
= 500×35+4×35
= 17500+140
= 17640
Question 5.
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?
Answer:
From the given pattern, the 1st step is: 1× 8 + 1 = 9
And the 2nd step: 12 × 8 + 2 = 98, which can be written as :
(11 + 1) × 8 + 2
On following distributive property, we get,
= (11 × 8) + (1 × 8) + 2
= 88 + 8 + 2 = 98
Therefore, we can write the 3rd step: 123 × 8 + 3 = 987 as,
= (111 + 11 + 1) × 8 + 3
= 111 × 8 + 11 × 8 + 1 × 8 + 2
= 888 + 88 + 8 + 3 = 987
Similarly, 4th step: 1234 × 8 + 4 = 9876 as,
= (1111 + 111 + 11 + 1) × 8 + 4
= 1111 × 8 + 111 × 8 + 11× 8 + 1 × 8 + 4
= 8888 + 888 + 88 + 8 + 4 = 9876
In the same way, the next steps are:
5th step: 12345 × 8 + 5, can be written as,
= (111111 + 11111 + 1111 + 111 + 11 + 1) × 8 + 5
= 111111 × 8 + 11111 × 8 + 1111 × 8 + 111× 8 + 11 × 8 + 1 × 8 + 5
= 888888 + 88888 + 8888 + 888 + 88 + 8 + 6 = 98765
Thus, the 5th term is : 12345 × 8 + 5 = 98765
Now, the 6th step: 123456 × 8 + 6 can be written as,
= (1111111 + 111111 + 11111 + 1111 + 111 + 11 + 1) × 8 + 7
= 1111111 × 8 + 111111 × 8 + 11111 × 8 + 1111 × 8 + 111× 8 + 11 × 8 + 1 × 8 + 7
= 8888888 + 888888 + 88888 + 8888 + 888 + 88 + 8 + 7
= 9876543
Thus, the 6th step is: 123456 × 8 + 6 = 9876543