PSEB Class 6 Maths Chapter 2 Whole Numbers Questions and Answers Solution
PSEB Punjab Board Class 6 Maths Textbook Solution Chapter 2 Whole Numbers Exercise Questions and Answers. We have also included some Additional questions, so that students can prepare more strongly.
Exercise 2.1
1.) Answer the following questions.
(a) Write the smallest whole number.
ANSWER:
0 is the smallest whole number.
(b) Write the smallest natural number.
ANSWER:
1 is the smallest natural number.
(c) Write the successor of 0 in whole numbers.
ANSWER:
Successor is next number of given number.
1 is the successor of 0 in whole numbers.
(d) Write the predecessor of 0 in whole numbers.
ANSWER:
Predecessor is previous number of given number.
The predecessor of 0 in whole numbers not possible.
(e) Largest whole number.
ANSWER:
Largest whole number can’t determine.
2.) Which of the following statements are True (T) and which are False (F)?
(a) Zero is the smallest natural number.
ANSWER:
False.
1 is the smallest natural number.
(b) Zero is the smallest whole number.
ANSWER:
True. Zero is the smallest whole number.
(c) Every whole number is a natural number.
ANSWER:
False.
Every whole number is not a natural number.
(d) Every natural number is a whole number.
ANSWER:
True. Every natural number is a whole number.
(e) 1 is the smallest whole number.
ANSWER:
False.
0 is the smallest whole number.
(f) The natural number 1 has no predecessor in natural numbers.
ANSWER:
True. The natural number 1 has no predecessor in natural numbers.
(g) The whole number 1 has no predecessor in whole numbers.
ANSWER:
False.
The whole number 0 has no predecessor in whole numbers.
(h) Successor of the largest two digit number is smallest three digit number.
ANSWER:
True. Successor of the largest two digit number is smallest three digit number.
(i) The successor of a two digit number is always a two digit number.
ANSWER:
False.
The successor of a two digit number is not always a two digit number.
(j) 300 is the predecessor of 299.
ANSWER:
False.
300 is the successor of 299.
(k) 500 is the successor of 499.
ANSWER:
True. 500 is the successor of 499.
(l) The predecessor of a two digit number is never a single digit number.
ANSWER:
False.
The predecessor of a two digit number will be a single digit number.
3.) Write the successor of each of following:
(a) 100909
ANSWER:
We have to write successor of 100909
Successor of 100909 = 100910
(b) 4630999
ANSWER:
We have to write successor of 4630999
Successor of 4630999 = 4631000
(c) 830001
ANSWER:
We have to write successor of 830001
Successor of 830001 = 830002
(d) 99999
ANSWER:
We have to write successor of 99999
Successor of 99999 = 100000
4.) Write the predecessor of each of following:
(a) 1000
ANSWER:
We have to write predecessor of 1000
Predecessor of 1000 = 999
(b) 208090
ANSWER:
We have to write predecessor of 208090
Predecessor of 208090 = 208089
(c) 7654321
ANSWER:
We have to write predecessor of 7654321
Predecessor of 7654321 = 7654320
(d) 12576
ANSWER:
We have to write predecessor of 12576
Predecessor of 12576 = 12575
5.) Represent the following numbers on the number line. 2, 0, 3, 5, 7, 11, 15
ANSWER:
We have to represent 2, 0, 3, 5, 7, 11, 15 on number line.
6.) How many whole numbers are there between 22 and 43?
ANSWER:
We have to find whole numbers between 22 and 43.
Whole numbers between 22 and 43 = (43 – 22) – 1
Whole numbers between 22 and 43 = 21 – 1
Whole numbers between 22 and 43 = 20
7.) Draw a number line to represent each of following on it.
(a) 3 + 2
ANSWER:
We have to represent given on number line.
3 + 2 = 5
(b) 4 + 5
ANSWER:
We have to represent given on number line.
4 + 5 = 9
(c) 6 + 2
ANSWER:
We have to represent given on number line.
6 + 2 = 8
(d) 8 – 3
ANSWER:
We have to represent given on number line.
8 – 3 = 5
(e) 7 – 4
ANSWER:
We have to represent given on number line.
7 – 4 = 3
(f) 7 – 2
ANSWER:
We have to represent given on number line.
7 – 2 = 5
(g) 3 x 3
ANSWER:
We have to represent given on number line.
3 x 3 = 9
(h) 2 x 5
ANSWER:
We have to represent given on number line.
2 x 5 = 10
(i) 3 x 5
ANSWER:
We have to represent given on number line.
3 x 5 = 15
(j) 9/3
ANSWER:
We have to represent given on number line.
9/3 = 3
(k) 12/4
ANSWER:
We have to represent given on number line.
12/4 = 3
(l) 10/2
ANSWER:
We have to represent given on number line.
10/2 = 5
8.) Fill in the blanks with the appropriate symbol < or>:
(a) 25 —– 205
ANSWER:
We have to fill appropriate symbol in blank.
25 < 205
(b) 170 ——- 107
ANSWER:
We have to fill appropriate symbol in blank.
170 > 107
(c) 415———514
ANSWER:
We have to fill appropriate symbol in blank.
415 < 514
(d) 10001 ———- 9999
ANSWER:
We have to fill appropriate symbol in blank.
10001 > 9999
(e) 2300014 ———– 2300041
ANSWER:
We have to fill appropriate symbol in blank.
2300014 < 2300041
(f) 99999 ————– 888888
ANSWER:
We have to fill appropriate symbol in blank.
99999 > 888888
Exercise 2.2
1.) Find the sum by suitable arrangement of terms:
(a) 837 + 208 + 363
ANSWER:
We have to find the sum by suitable arrangement of terms.
(837 + 363)+ 208
1200 + 208
837 + 208 + 363 = 1408
(b) 1962 + 453 + 1538 + 647
ANSWER:
We have to find the sum by suitable arrangement of terms.
1962 + 453 + 1538 + 647
(1962 + 1538) + (647 + 453)
3500 + 1100
1962 + 453 + 1538 + 647 = 4600
2.) Find the product by suitable arrangement of terms:
(a) 2 x 1497 x 50
ANSWER:
We have to find the product by suitable arrangement of terms
2 x 1497 x 50 = 1497 x (2 x 50)
1497 x (2 x 50) = 1497 x 100
2 x 1497 x 50 = 149700
(b) 4 x 263 x 25
ANSWER:
We have to find the product by suitable arrangement of terms
4 x 263 x 25 = 263 x (4 x 25)
263 x (4 x 25) = 263 x 100
4 x 263 x 25 = 26300
(c) 8 x 163 x 125
ANSWER:
We have to find the product by suitable arrangement of terms
8 x 163 x 125 = 163 x (8 x 125)
163 x (8 x 125) = 163 x 1000
8 x 163 x 125 = 163000
(d) 963 x 16 x 25
ANSWER:
We have to find the product by suitable arrangement of terms
963 x 16 x 25 = 963 x 4 x 4 x 25
(963 x 4) x 100 = 3852 x 100
963 x 16 x 25 = 385200
(e) 5 x 171 x 60
ANSWER:
We have to find the product by suitable arrangement of terms.
5 x 171 x 60 = 171 x (5 x 60)
171 x (5 x 60) = 171 x 300
= 513 x 100
5 x 171 x 60 = 51300
(f) 125 x 40 x 8 x 25
ANSWER:
We have to find the product by suitable arrangement of terms
125 x 40 x 8 x 25 = (125 x 8) x (25 x 40)
= 1000 x 1000
125 x 40 x 8 x 25 = 1000000
(g) 30921 x 25 x 40 x 2
ANSWER:
We have to find the product by suitable arrangement of terms
30921 x 25 x 40 x 2 = (30921 x 2) x (25 x 40)
(30921 x 2) x (25 x 40) = 61842 x 1000
30921 x 25 x 40 x 2 = 61842000
(h) 4 x 2 x 1932 x 125
ANSWER:
We have to find the product by suitable arrangement of terms
4 x 2 x 1932 x 125 = (4 x 2 x 125) x 1932
(4 x 2 x 125) x 1932 = 1000 x 1932
4 x 2 x 1932 x 125 = 1932000
(i) 5462 x 25 x 4 x 2
ANSWER:
We have to find the product by suitable arrangement of terms
5462 x 25 x 4 x 2 = (5462 x 2) x (25 x 4)
(5462 x 2) x (25 x 4) = 10924 x 100
5462 x 25 x 4 x 2 = 1092400
3.) Find the value of each of the following using distributive property:
(a) (649 x 8) + (649 x 2)
ANSWER:
We have to find value of given using distributive property.
(649 x 8) + (649 x 2), here, 649 is common.
We take outside.
649 x (8 + 2)
649 x 10
(649 x 8) + (649 x 2) = 6490
(b) (6524 x 69) + (6524 x 31)
ANSWER:
We have to find value of given using distributive property.
(6524 x 69) + (6524 x 31) here, 6524 is common.
We take outside.
6524 x (69 + 31)
6524 x 100
(6524 x 69) + (6524 x 31) = 652400
(c) (2986 x 35) + (2986 x 65)
ANSWER:
We have to find value of given using distributive property.
(2986 x 35) + (2986 x 65) here, 2986 is common.
We take outside.
2986 x (35 + 65)
2986 x 100
(2986 x 35) + (2986 x 65) = 298600
(d) (6001 x 172) – (6001 x 72)
ANSWER:
We have to find value of given using distributive property.
(6001 x 172) – (6001 x 72) here, 6001 is common.
We take outside.
6001 x (172 – 72)
6001 x 100
(6001 x 172) – (6001 x 72) = 600100
4.) Find the value of the following:
(a) 493 x 8 + 493 x 2
ANSWER:
We have to find value of 493 x 8 + 493 x 2
Here, 493 is common.
We take outside.
493 x (8 + 2)
493 x 10
493 x 8 + 493 x 2 = 4930
(b) 24579 x 93 + 7 x 24579
ANSWER:
We have to find value of 24579 x 93 + 7 x 24579
Here, 24579 is common.
We take outside.
24579 x (93 + 7)
24579 x 100
24579 x 93 + 7 x 24579 = 2457900
(c) 3845 x 5 x 782 + 769 x 25 x 218
ANSWER:
We have to find value of 3845 x 5 x 782 + 769 x 25 x 218
We done arrangements.
3845 x 5 x 782 + 769 x 25 x 218 = [(3845 x 5) + (769 x 25)] x (782 + 218)
[(3845 x 5) + (769 x 25)] x (782 + 218) = [(3845 x 5) + (769 x 25)] x 1000
But, (3845 x 5) and (769 x 25) = 19225 is common.
(3845 x 5) x 1000 = 19225 x 1000
3845 x 5 x 782 + 769 x 25 x 218 = 19225000
(d) 3297 x 999 + 3297
ANSWER:
We have to find value of 3297 x 999 + 3297
Here, 3297 is common.
We take outside.
3297 x (999 + 1)
3297 x 1000
3297 x 999 + 3297 = 3297000
5.) Find the product using suitable properties:
(a) 738 x 103
ANSWER:
We done suitable arrangements.
738 x 103 = 738 x (100 + 3)
738 x 100 + 738 x 3
73800 + 2214
738 x 103 = 76014
(b) 854 x 102
ANSWER:
We done suitable arrangements.
854 x 102 = 854 x (100 + 2)
854 x 100 + 854 x 2
85400 + 1708
854 x 102 = 87108
(c) 258 x 1008
ANSWER:
We done suitable arrangements.
258 x 1008 = 258 x 1000 + 258 x 8
258 x 1000 + 258 x 8 =
258000 + 2064
258 x 1008 = 260064
(d) 736 x 93
ANSWER:
We done suitable arrangements.
736 x 93 =736 x (100 – 7)
736 x 100 – 736 x 7
= 73600 – 5152
736 x 93 = 68448
(e) 816 x 745
ANSWER:
We done suitable arrangements.
816 x 745 = 816 x (800 – 55)
816 x 800 – 816 x 55
= 652800 – 44880
816 x 745 = 607920
(f) 2032 x 613
ANSWER:
We done suitable arrangements.
2032 x 613 = 2032 x (600 + 13)
2032 x 600 + 2032 x 13
1219200 + 26416
2032 x 613 = 1245616
6.) 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 78 per litre, how much he spend in all on petrol?
ANSWER:
Given that,
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.
The petrol costs 78 per litre
We have to find how much he spend in all on petrol.
Total petrol filled in his car = 40 litres + 50 litres
Total petrol filled in his car = 90 litres
Money he spend in all on petrol = 78 per litre x 90
Money he spend in all on petrol = Rs.7020
7.) A vendor supplies 32 litres of milk to a hotel in morning and 68 litres of milk in the evening. If the milk costs ₹35 per litre, how much money is due to the vendor per day?
ANSWER:
Given that,
A vendor supplies 32 litres of milk to a hotel in morning and 68 litres of milk in the evening.
If the milk costs ₹35 per litre.
We have to find money is due to the vendor per day.
1st we find total amount of milk supplies to hotel = 32 litres + 68 litres
Total amount of milk supplies to hotel = 100 litres
Money is due to the vendor per day = 100 litres x ₹35 per litre
Money is due to the vendor per day =₹3500
8.) We know that 0 x 0 = 0 Is there any other whole number which when multiplied by itself gives the product equal to the number itself? Find out the number.
ANSWER:
We know,
0 x 0 = 0
Whole number which when multiplied by itself gives the product equal to the number itself is 1.
1 x 1 = 1
9.) Fill in the blanks:
(a) 15 x 0 = —
ANSWER:
We have to fill blank spaces.
15 x 0 = 0
(b) 15 + 0 = —
ANSWER:
We have to fill blank spaces.
15 + 0 = 15
(c) 15 – 0 = —
ANSWER:
We have to fill blank spaces.
15 – 0 = 15
(d) 15 / 0 = —
ANSWER:
We have to fill blank spaces.
15 / 0 = Not defined
(e) 0 x 15 = —
ANSWER:
We have to fill blank spaces.
0 x 15 = 0
f) 0 + 15 = —
ANSWER:
We have to fill blank spaces.
0 + 15 = 15
(g) 0/15 = —
ANSWER:
We have to fill blank spaces.
0/15 = 0
(h)15 x 1 = —
ANSWER:
We have to fill blank spaces.
15 x 1 = 15
(i) 15 / 1 = —
ANSWER:
We have to fill blank spaces.
15 / 1 = 15
(j) 1 / 1 = —
ANSWER:
We have to fill blank spaces.
1 / 1 = 1
10.) The product of two Whole numbers is zero. What do you conclude? Explain with example.
ANSWER:
Suppose 1 whole number is 12 and 2nd whole number is 0.
1 whole number x 2nd whole number = 12 x 0
1 whole number x 2nd whole number = 0
11.) Match the following:
(i) 537 x 106 = 537 x 100 + 537 x 6 (a) Commutativity under multiplication
(ii) 4 x 47 x 25 = 4 x 25 x 47 (b) Commutativity under addition
(iii) 70 + 1923 + 30 = 70 + 30 + 1923 (c) Distributivity of multiplication over addition.
ANSWER:
(i) 537 x 106 = 537 x 100 + 537 x 6 ==(c) Distributivity of multiplication over addition.
(ii) 4 x 47 x 25 = 4 x 25 x 47 ==(a) Commutativity under multiplication
(iii) 70 + 1923 + 30 = 70 + 30 + 1923 ==(b) Commutativity under addition
Exercise 2.3
1.) 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. We can say that one or both of them will be zero, when the product of two whole numbers is zero.
For example,
Suppose 1 whole number is 54 and 2nd whole number is 0.
1 whole number x 2nd whole number = 54 x 0
1 whole number x 2nd whole number = 0
2.) 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:
Yes, we can say that both of them will be 1, when the product of two whole numbers is 1.
For example,
Suppose 1 whole number is 1 and 2nd whole number is 1.
1 whole number x 2nd whole number = 1 x 1
1 whole number x 2nd whole number = 1
- Observe the pattern in the following and fill in the blanks:
1 x 1 = 1
11 x 11 = 121
111 x 111 = 12321
1111 x 1111 = ———
11111 x 1111 = —————–
ANSWER:
By observing pattern, we fill blank.
1 x 1 = 1
11 x 11 = 121
111 x 111 = 12321
1111 x 1111 =1234321
11111 x 11111 = 123454321
4.) Observe the pattern and fill in the blanks:
1 x 9 + 1 = 10
12 x 9 + 2 = 110
123 x 9 + 3 = 1110
1234 x 9 + 4 = 11110
12345 x 9 + 5 = ———-
123456 x 9 + 6 = ———————–
ANSWER:
By observing pattern, we fill blank.
1 x 9 + 1 = 10
12 x 9 + 2 = 110
123 x 9 + 3 = 1110
1234 x 9 + 4 = 11110
12345 x 9 + 5 = 111110
123456 x 9 + 6 = 1111110
5.) Represent numbers from 24 to 30 according to rectangular, square or triangular pattern.
ANSWER:
25 as square pattern.
6.) Study the following pattern:
1= 1 x 1 = 1
1 + 3 = 2 x 2 = 4
1 + 3 + 5 = 3 x 3 = 9
1 + 3 + 5 + 7 = 4 x 4 = 16
Hence find the sum of
(a) First 12 odd numbers
ANSWER:
By observing pattern, we have to find the sum of First 12 odd numbers.
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = 12 x 12 = 144
(b) First 50 odd numbers.
ANSWER:
By observing pattern, we have to find the sum of First 50 odd numbers.
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + ——————— + 99 = 50 x 50 = 2500
Multiple Choice Questions
1.) The smallest whole number is
(a) 0
(b) 1
(c) 2
(d) 3
ANSWER: (a) 0
2.) The smallest natural number is
(a) 0
(b) 1
(c) 2
(d) 3
ANSWER: (b) 1
3.) The successor of 38899 is
(a) 39000
(b) 38900
(c) 39900
(d) 38800
ANSWER: (b) 38900
4.) The predecessor of 24100 is
(a) 24999
(b) 24009
(c) 24199
(d) 24099
ANSWER: (d) 24099
5.) The statement 4 + 3 = 3 + 4 represents
(a) Closure
(b) Associative
(c) Commutative property
(d) Identity
ANSWER: (c) Commutative property
6.) Which of the following is the additive identity?
(a) 0
(b) 1
(c) 2
(d) 3
ANSWER: (a) 0
7.) The multiplicative identity is
(a) 0
(b) 1
(c) 2
(d) 3
ANSWER: (b) 1
8.) 15 x 32 + 15 x 68 =
(a) 1400
(b) 1600
(c) 1700
(d) 1500
ANSWER: (d) 1500
9.) The largest 4 digit number divisible by 13 is
(a) 9997
(b) 9999
(c) 9995
(d) 9991
ANSWER: (a) 9997
10.) The successor of 3 digit largest number is
(a) 100
(b) 998
(c) 1001
(d) 1000
ANSWER: (d) 1000
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