Chapter 2: Whole Numbers
Ex-2.1
1.Write the next three natural numbers after 10999.
Solution:
Three natural number = 1,2,3.
The next three natural number after 10999
10999 + 1 = 11000
10999 + 2 = 11001
10999+ 3 = 11002
2. Write the three whole numbers occurring just before 10001.
Solution:
There whole number occurring before 10,001 are
= 10,001 – 1 = 10,000
=10,0001 – 2 = 9,999
= 10,001 – 3 = 9,998
3. Which is the smallest whole number?
Solution:
Zero (0) is the smallest whole number.
4. How many whole numbers are there between 32 and 53?
Solution:
33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52.
There are 20 number between 32 and 53.
5. Write the successor of:
(a) 2440701 (b) 100199 (c) 1099999 (d) 2345670
Solution:
a. 24,40,702
b. 1,00,200
c. 11,00,000
d. 23,45,671
6. Write the predecessor of:
(a) 94 (b) 10000 (c) 208090 (d) 7654321
Solution:
a. 93
b. 9,999
c. 2,08,089
d. 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 (b) 370, 307 (c) 98765, 56789 (d) 9830415, 10023001
Solution:
a. 503 is on the left of 530; 503 < 530
b. 307 is on the left of 370; 307 < 370
c. 56,789 is on the left of 98,765; 56,789 < 98,765
d. 98,30,415 is on the left of 1,00,23,001; 98,30,415 < 1,00,23,001
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.
Solution:
(a) F
(b) F
(c) T
(d) T
(e) T
(f) F
(g) F
(h) F
(i) T
(j) F
(k) F
(l) T
(m) F
EXERCISE 2.2
1. Find the sum by suitable rearrangement:
(a) 837 + 208 + 363 (b) 1962 + 453 + 1538 + 647
Solution:
a. 837 + 208+ 363
= 1408
b. 1962 + 453 + 1538 + 647
= 4600
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
Solution:
(a) 2 × 1768 × 50
= 1,76,800
(b) 4 × 166 × 25
= 16,600
(c) 8 × 291 × 125
= 2,91,000
(d) 625 × 279 × 16
= 1,92,25000
(e) 285 × 5 × 60
= 85,500
(f) 125 × 40 × 8 × 25
= 10,00,000
4. Find the product using suitable properties.
(a) 738 × 103
(b) 854 × 102
(c) 258 × 1008
(d) 1005 × 168
Solution:
(a) 738 × 103
=738 * (100 + 3)
= (738 * 100) + (738 * 3)
= 73800 + 2214
= 76014
(b) 854 × 102
= 854 * (100 + 2)
= (854*100) + (854*2)
= 85400 + 1708
= 87108
(c) 258 × 1008
= 258 * (1000 + 8)
= (258*1000) + (258*8)
=258000 + 2064
= 260064
(d) 1005 × 168
= (1000 + 5) * 168
= (1000 *168) + (168*5)
= 168000 + 840
= 168840
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 RS. 44 per litre, how much did he spend in all on petrol?
Solution:
Petrol filled on Monday = 40 lit
Petrol filled next day = 50 lit
Rate of 1 lit of petrol = RS.44
Total rate of petrol
1 lit = RS.44
40 lit = RS.44 * 40
50 lit = RS.44 * 50
Now,
= 40 * 44 + 50 * 44
= 44 (40 + 50)
= 44(90)
= 3960 RS.
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?
Solution:
Milk supplied in morning = 32 lit
Milk supplied in evening = 68 lit
Cost of 1 lit of milk = 45 RS.
Money due to the vendor per day = X
Cost of milk in morning = 32 * 45 RS.
Cost of milk in evening = 68 * 45 RS.
Now,
X = 32 * 45 + 68 * 45
X = 45 (32 + 68)
X = 45 (100)
X = 4500
Money due to the vendor per day = 4500 RS.
7. Match the following:
(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) Distributivism of multiplication over addition.
Solution:
(i) →(c)
(ii) → (a)
(iii) → (b)
EXERCISE 2.3
1. Which of the following will not represent zero:
(a) 1 + 0 (b) 0 × 0 (c) 0/ 2 (d) 10 – 10/ 2
Solution:
(a) 1 + 0
= 1 + 0
= 1 (any no. + 0 = that no. itself)
(b) 0 × 0
= 0*0
= 0 (any no. *0 = 0)
(c) 0/ 2
= 0/2
= 0 ( 0 / any no. = 0)
(d) 10 – 10/ 2
= 10 – 10 /2
= 0/ 2
= 0 ( 0 / any no. = 0)
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.
Solution:
Yes
Example:
Whole no. 1 * Whole no. = 0
0 * 0 = 0
1 * 0 = 0
2 * 0 = 0
3 * 0 = 0
4 * 5 = 20
5 * 0 = 0
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:
Solution:
Both number should be 1
For example:
Whole no. 1 * Whole no.2
1*1 = 1
1*2 = 2
1*0 =0
1*3 =3
4. Find using distributive property:
(a) 728 × 101 (b) 5437 × 1001 (c) 824 × 25 (d) 4275 × 125 (e) 504 × 35
Solution:
(a) 728 × 101
= 728 * (100+1)
= a*(b + c) =(a*b) +(a*c)
= (a = 728, b = 100, c = 1)
= (728 * 100) + (728 * 1)
= 72800 + 728
= 73528
(b) 5437 × 1001
= 5437 * (1000 + 1)
= a*(b + c) = (a*b) + (a*c)
= (a = 5437, b = 1000, c = 1)
= (5437*1000) + (5437*1)
= 5437000 + 5437
= 5442437
(c) 824 × 25
= (800 + 24) * 25
= (800 + 25 – 1) *25
= (800*25) + (25*25) – (25*1)
= 20000 + 625 – 25
= 20000 + 600
= 20600
(d) 4275 × 125
= (4000 + 200 + 75) * 125
= (4000 + 200 + 100 – 25) *125
= (4000*125) + (200*125) + (100*125) – (125*25)
= 500000 + 25000 + 12500 – 3125
= 534,375
(e) 504 × 35
= (500 + 4) * 35
= (500*35) + (35*4)
= 17500 + 140
= 17640
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).
Solution:
123456 × 8 + 6 = 987654
1234567 × 8 + 7 = 9876543
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