Exercise 2.1
Sum No.1:-
Write the next three
natural numbers after 10999.
Solutions:
The next three natural numbers
after 10999 are 11000,11001 and 11002.
Sum No. 2:-
Write three whole numbers
occurring just before 10001.
Solutions:-
10001 - 1= 10000
10000- 1 = 9999
9999 – 1 = 9998
Sum NO 3:-
Which is the smallest whole
number?
Solutions:-
0 is the smallest whole
number.
Sum No 4:-
How many whole numbers are
there between 32 and 53?
Solutions:-
The whole numbers between 32
and 53 are 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48,
49, 50, 51, 52.
Sum No 5:-
Write the successor of:
(a) 2440701 (b)
100199 (c) 1099999 (d) 2345670
Solutions:-
(a) Successor of 244070 is
244070 + 1 = 244071
Hence, successor of 244070 is
244071.
(b) Successor of 100199 is
100199 + 1 = 100200
Hence, successor of 100199 is
100200.
(c) Successor of 1099999 is
1099999 + 1 = 110000,
Hence, successor of 1099999 is
1100000.
(d) Successor of 2345670 is
2345670 + 1 = 2345671
Hence, successor of 2345670 is
2345671
Sum No 6:-
Write the predecessor of:
(a) 94 (b)
10000 (c) 208090 (d) 7654321
Solutions:-
(a) Predecessor of 94 is 94 –
1 = 93.Hence, predecessor of 94 is 93.
(b) Predecessor of 1000 is
10000 – 1 = 9999.Hence, predecessor of 10000 is 9999.
(c) Predecessor of 208090 is
208090 -1 = 208089.Hence, predecessor of 208090 is 208089.
(d) Predecessor of 7654321 is
7654321 – 1 = 7654320. Hence, predecessor of 7654321 is 7654320.
Sum No 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 (b)
370, 307 (c) 98765, 56789 (d)
9830415,10023001
Solutions:-
We know that the smaller
number is always on the left side of the greater number on number line.
(a) 530, 503
Clearly 503 is smaller than
530.
Hence, 503 will be on left
side of 530 on number line.
So, 503 < 530 or 530 >
503
(b) 307 < 370
Clearly 307 is smaller than
370.
Hence, 307 will be on the left
side of 370 on number line.
So, 307 < 370 or 370 >
307.
(c) 98765, 56789
Clearly 56789 is smaller than
98765.
Hence, 56789 will be on left
side of 98765 on number line.
So, 56789 < 98765 or 98765
> 56789.
(d) 9830415, 10023001
Clearly, 9830415 is smaller
than 10023001
Hence, 9830415 will be on the
left side of 10023001 on the number line.
So, 9830415 < 10023001 or
10023001 > 9830415.
Sum No. 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.
Solutions:-
(a) Zero is the smallest
natural number. (F)
(b) 400 is the predecessor of
399. (F)
(c) Zero is the smallest whole
number.(T)
(d) 600 is the successor of
599.(T)
(e) All natural numbers are
whole numbers.(T)
(f) All whole numbers are
natural numbers.(F)
(g) The predecessor of a two-digit
number is never a single-digit number.(F)
(h) 1 is the smallest whole
number.(F)
(i) The natural number 1 has
no predecessor.(T)
(j) The whole number 1 has no
predecessor.(F)
(k) The whole number 13 lies
between 11 and 12.(F)
(l) The whole number 0 has no
predecessor.(T)
(m) The successor of a
two-digit number is always a two-digit number.(F)
Exercise -2.2
Sum No 1:-
Find the sum by suitable rearrangement:
(a) 837 + 208 + 363
(b) 1962 + 453 + 1538 + 647
Solutions:-
(a) Given 837 + 208 + 363
= (837 + 363) + 208
= 1200 + 208
= 1408
(b) Given 1962 + 453 + 1538 + 647
= (1962 + 1538) + (453 + 647)
= 3500 + 1100
= 4600
Sum No.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
Solutions:-
(a) Given 2 × 1768 × 50
= 2 × 50 × 1768
= 100 × 1768
= 176800
(b) Given 4 × 166 × 25
= 4 × 25 × 166
= 100 × 166
= 16600
(c) Given 8 × 291 × 125
= 8 × 125 × 291
= 1000 × 291
= 291000
(d) Given 625 × 279 × 16
= 625 × 16 × 279
= 10000 × 279
= 2790000
(e) Given 285 × 5 × 60
= 285 × 300
= 85500
(f) Given 125 × 40 × 8 × 25
= 125 × 8 × 40 × 25
= 1000 × 1000
= 1000000
Sum No. 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
Solutions:-
(a) Given 297 × 17 + 297 × 3
= 297 × (17 + 3)
= 297 × 20
= 5940
(b) Given 54279 × 92 + 8 × 54279
= 54279 × 92 + 54279 × 8
= 54279 × (92 + 8)
= 54279 × 100
= 5427900
(c) Given 81265 × 169 – 81265 × 69
= 81265 × (169 – 69)
= 81265 × 100
= 8126500
(d) Given 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
Sum .No. 4:-
Find the product using suitable properties.
(a) 738 × 103 (b) 854 × 102 (c) 258 × 1008 (d) 1005 × 168
Solutions:-
(a) Given 738 × 103
= 738 × (100 + 3)
= 738 × 100 + 738 × 3 (using distributive property)
= 73800 + 2214
= 76014
(b) Given 854 × 102
= 854 × (100 + 2)
= 854 × 100 + 854 × 2 (using distributive property)
= 85400 + 1708
= 87108
(c) Given 258 × 1008
= 258 × (1000 + 8)
= 258 × 1000 + 258 × 8 (using distributive
property)
= 258000 + 2064
= 260064
(d) Given 1005 × 168
= (1000 + 5) × 168
= 1000 × 168 + 5 × 168 (using distributive
property)
= 168000 + 840
= 168840
Sum. No.5:-
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 ₹ 44 per litre,
how much did he spend in all on petrol?
Solutions:-
Petrol quantity filled on Monday = 40 litres
Petrol quantity filled on Tuesday = 50 litres
Total petrol quantity filled = (40 + 50) litre
Cost of petrol per litre = ₹ 44
Total money spent = 44 × (40 + 50)
= 44 × 90
= ₹ 3960
Sum No.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 ₹ 45 per
litre, how much money is due to the vendor per day?
Solutions:-
Milk quantity supplied in the morning = 32 litres
Milk quantity supplied in the evening = 68 litres
Cost of milk per litre = ₹ 45
Total cost of milk per day = 45 × (32 + 68)
= 45 × 100
= ₹ 4500
Hence, the money is due to the vendor per day is ₹
4500
Sum No.7:-
|
Match the column |
|
|
(i) 425 × 136 = 425 × (6 + 30 + 100) |
(a) Commutativity under multiplication. |
|
(ii) 2 × 49 × 50 = 2 × 50 × 49 |
(b)
Commutativity under addition.
|
|
(iii) 80 + 2005 + 20 = 80 + 20 + 2005 |
(c)
Distributivity of multiplication over addition.
|
|
Solutions:- Match the column |
|
|
(i) 425 × 136 = 425 × (6 + 30 + 100) |
(c) Distributivity of multiplication over addition |
|
(ii) 2 × 49 × 50 = 2 × 50 × 49 |
(a) Commutativity under multiplication. |
|
(iii) 80 + 2005 + 20 = 80 + 20 + 2005 |
(b)
Commutativity under addition. |
Exercise 2.3
Sum No.1:-
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
Hence, it does not represent zero
(b) 0 × 0 = 0
Hence, it represents zero
(c) 0 / 2 = 0
Hence, it represents zero
(d) (10 – 10) / 2 = 0 / 2 = 0
Hence, it represents zero
Sum No.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.
Solutions:-
If product of two whole numbers is zero, definitely
one of them is zero
Example: 0 × 9 = 0 and 23 × 0 = 0
If product of two whole numbers is zero, both of
them may be zero
Example: 0 × 0 = 0
Yes, if the product of two whole numbers is zero,
then both of them will be zero
Sum No.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.
Solutions:-
If the product of two whole numbers is 1, both the
numbers should be equal to 1
Example: 1 × 1 = 1
But 1 × 8 = 8
Hence, its clear that the product of two whole
numbers will be 1, only in situation when both numbers to be multiplied are 1.
Sum No.4:-
Find using distributive property:
(a) 728 × 101 (b) 5437 × 1001 (c) 824 × 25 (d) 4275 × 125 (e) 504 × 35
Solutions:-
(a) Given 728 × 101
= 728 × (100 + 1)
= 728 × 100 + 728 × 1
= 72800 + 728
= 73528
(b) Given 5437 × 1001
= 5437 × (1000 + 1)
= 5437 × 1000 + 5437 × 1
= 5437000 + 5437
= 5442437
(c) Given 824 × 25
= (800 + 24) × 25
= (800 + 25 – 1) × 25
= 800 × 25 + 25 × 25 – 1 × 25
= 20000 + 625 – 25
= 20000 + 600
= 20600
(d) Given 4275 × 125
= (4000 + 200 + 100 – 25) × 125
= (4000 × 125 + 200 × 125 + 100 × 125 – 25 × 125)
= 500000 + 25000 + 12500 – 3125
= 534375
(e) Given 504 × 35
= (500 + 4) × 35
= 500 × 35 + 4 × 35
= 17500 + 140
= 17640
Sum No.5:-
Study the
pattern:
1 × 8 + 1 = 9 1234 ×
8 + 4 = 9876
12 × 8 + 2 = 98 12345 × 8 +
5 = 98765
123 × 8 + 3 = 987
Write the next two steps. Can you say how the
pattern works?
(Hint: 12345 = 11111 + 1111 + 111 + 11 + 1)
Solutions:-
123456 × 8 + 6 = 987654
1234567 × 8 + 7 = 9876543
Given 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
123456 × 8 + 6 = 987648 + 6
= 987654
Yes, here the pattern works
1234567 × 8 + 7 = 9876543
Given 1234567 = (1111111 + 111111 + 11111 + 1111 +
111 + 11 + 1)
1234567 × 8 = (1111111 + 111111 + 11111 + 1111 +
111 + 11 + 1) × 8
= 1111111 × 8 + 111111 × 8 + 11111 × 8 + 1111 × 8 +
111 × 8 + 11 × 8 + 1 × 8
= 8888888 + 888888 + 88888 + 8888 + 888 + 88 + 8
= 9876536
1234567 × 8 + 7 = 9876536 + 7
= 9876543
Yes, here the pattern works
