Telangana SCERT Class 6 Maths Chapter 2 Solution – Whole Numbers. Here in this post we provides Class 6 Maths Whole Numbers Telangana State Board Solution. Telangana State Board English Class VI Medium Students can download this Solution to Solve out Improve Your Learning Questions and Answers.
Telangana State Board Class 6 Maths Chapter 2 Whole Numbers Solution:
EXERCISE – 2.1
1.) Which of the statements are true (T) and which are false (F). Correct the false statements.
i.) There is a natural number that has no predecessor.
ANSWER:
True. There is a natural number that has no predecessor.
ii.) Zero is the smallest whole number.
ANSWER:
True.Zero is the smallest whole number.
iii.) All whole numbers are natural numbers.
ANSWER:
False. The correct statement is
All natural numbers are whole numbers
iv.) A whole number that lies on the number line lies to the right side of another number is the greater number.
ANSWER:
True.A whole number that lies on the number line lies to the right side of another number is the greater number.
v.) A whole number on the left of another number on the number line, is greater.
ANSWER:
False. The correct statement is
The whole number on the left of another number on the number line, is smaller
vi.) We can’t show the smallest whole number on the number line.
ANSWER:
False. The correct statement is
We can show the smallest whole number on the number line.
vii. We can show the greatest whole number on the number line.
ANSWER:
False. The correct statement is
We cannot show the greatest whole number on the number line.
2.) How many whole numbers are there between 27 and 46?
ANSWER:
We have to find whole numbers between 27 and 46.
Whole numbers between 27 and 46 = (46 – 27) – 1
Whole numbers between 27 and 46 = 19 – 1
Whole numbers between 27 and 46 = 18
3.) Find the following using number line.
i.) 6 + 7 + 7
ANSWER:
We know,
6 + 7 + 7 = 20
We have to show this on number line.
ii.) 18 – 9
ANSWER:
We know,
18 – 9 = 9
We have to show this on number line.
iii.) 5 × 3
ANSWER:
We know,
5 × 3 = 15
We have to show this on number line.
4.) In each pair, state which whole number on the number line is on the right of the other number?
i.) 895; 239
ANSWER:
We have to show which whole number on the number line is on the right of the other number.
895 is on the right of 239
ii.) 1001; 10001
ANSWER:
We have to show which whole number on the number line is on the right of the other number.
10001 is on the right of 1001
iii.) 10015678; 284013
ANSWER:
We have to show which whole number on the number line is on the right of the other number.
10015678 is on the right of 284013
5.) Mark the smallest whole number on the number line.
ANSWER:
The smallest whole number on the number line
We know,
The smallest whole number is 0.
6.) Choose the appropriate symbol from < or >
i.) 8 ………. 7
ANSWER:
8 > 7
ii.) 5 ………. 2
ANSWER:
5 > 2
iii.) 0 ………. 1
ANSWER:
0 < 1
iv.) 10 ………. 5
ANSWER:
10 > 5
7.) Place the successor of 11 and predecessor of 5 on the number line.
ANSWER:
We have to show the successor of 11 and predecessor of 5 on the number line.
The successor of 11 = 12
Predecessor of 5 = 4
EXERCISE – 2.2
1.) Give the results without actually performing the operations using the given information.
i.) 28 × 19 = 532 then 19 × 28 =
ANSWER:
Given, 28 × 19 = 532
We know, addition and multiplication are commutative for whole numbers.
19 × 28 =532
ii.) 1 × 47 = 47 then 47 × 1 =
ANSWER:
Given,1 × 47 = 47
We know, addition and multiplication are commutative for whole numbers.
47 × 1 = 47
iii.) a × b = c then b × a =
ANSWER:
Given,a × b = c
We know, addition and multiplication are commutative for whole numbers.
b × a = c
iv.) 58 + 42 = 100 then 42 + 58 =
ANSWER:
Given,58 + 42 = 100
We know, addition and multiplication are commutative for whole numbers.
42 + 58 = 100
v.) 85 + 0 = 85 then 0 + 85 =
ANSWER:
Given,85 + 0 = 85
We know, addition and multiplication are commutative for whole numbers.
0 + 85 =85
vi.) a + b = d then b + a =
ANSWER:
Given,a + b = d
We know, addition and multiplication are commutative for whole numbers.
b + a = d
2.) Find the sum by suitable rearrangement:
i.) 238 + 695 + 162
ANSWER:
Here we have to find sum of 238 + 695 + 162
We know, addition and multiplication are associative over whole numbers.
(238 + 162) + 695
= 400 + 695
238 + 695 + 162 = 1095
ii.) 154 + 197 + 46 + 203
ANSWER:
Here we have to find sum of154 + 197 + 46 + 203
We know, addition and multiplication are associative over whole numbers.
(154 + 46) + (197 +203)
= 200 + 400
154 + 197 + 46 + 203= 600
3.) Find the product by suitable rearrangement.
i.) 25 × 1963 × 4
ANSWER:
25 × 1963 × 4
We know, addition and multiplication are associative over whole numbers.
(25 × 4) × 1963
100 x 1963
25 × 1963 × 4 = 196300
ii.) 20 × 255 × 50 × 6
ANSWER:
20 × 255 × 50 × 6
We know, addition and multiplication are associative over whole numbers.
(20 × 50) × (255 × 6)
1000 x 1530
20 × 255 × 50 × 6 = 1530000
4.) Find the value of the following:
i.) 368 × 12 + 18 × 368
ANSWER:
368 × 12 + 18 × 368
368 is common.
368 x (12 + 18)
368 x 30 = 11040
ii.) 79 × 4319 + 4319 × 11
ANSWER:
79 × 4319 + 4319 × 11
4319is common.
4319 x (79 + 11)
4319 x 90 = 388710
5.) Find the product using suitable properties:
i.) 205 × 1989
ANSWER:
205 × 1989
We use distributive property
(200 + 5) × 1989
200 × 1989 + 5 × 1989
3, 97,800 + 9945
205 × 1989 = 407745
ii.) 1991 × 1005
ANSWER:
1991 × 1005
We use distributive property
1991 × (1000 + 5)
1991x 1000 + 1991 x 5
1991000 + 9955 = 2000955
1991 × 1005 = 2000955
6.) A milk vendor supplies 56 liters of milk in the morning and 44 liters of milk in the evening to a hostel. If the milk costs Rs.30 per liter, how much money he gets per day?
ANSWER:
Given, a milk vendor supplies 56 liters of milk in the morning and 44 liters of milk in the evening to a hostel.
The milk costs Rs.30 per liter
We have to find money he gets per day.
Total milk = 56 liters + 44 liters = 100 liters
The milk costs Rs.30 per liter
Money he gets per day = Rs.30 per liter x 100 liters
Money he gets per day = Rs.3000
7.) Chandana and Venu purchased 12 note books and10 note books respectively. The cost of each note book is Rs.15,then how much amount should they pay to the shop keeper?
ANSWER:
Given,Chandana and Venu purchased 12 note books and10 note books respectively
The cost of each note book is Rs.15
We have to find how much amount they should pay to the shop keeper.
Total number of note books = 12 note books + 10 note books
Total number of note books = 22 note books
The cost of each note book is Rs.15
Amount should they pay to the shop keeper = The cost of each note book x 22 note books
Amount should they pay to the shop keeper = 15 x 22 note books
Amount should they pay to the shop keeper = Rs.330
8.) Match the following
i.) 1991+7 = 7+1991 [ ] a. Additive identity
ii.) 68×50 = 50×68 [ ] b. Multiplicative identity
iii.) 1 [ ] c. Commutative under addition
iv.) 0 [ ] d. Distributive property of multiplication over addition
v.) 879×(100+30) = 879×100+879×30 [ ] e. Commutative under multiplication
ANSWER:
i.) 1991+7 = 7+1991 c. Commutative under addition
ii.) 68×50 = 50×68 e. Commutative under multiplication
iii.) 1 b. Multiplicative identity
iv.) 0 a. Additive identity
v.) 879×(100+30) = 879×100+879×30 d. Distributive property of multiplication over addition
EXERCISE – 2.3
1.) 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 four steps. Can you find out how the pattern works?
ANSWER:
We have to find next four steps of pattern.
123456 × 8 + 6 = 987654
1234567 × 8 + 7 = 9876543
12345678 × 8 + 8 = 98765432
123456789 × 8 + 9 = 987654321
2.) Study the pattern:
91 × 11 × 1 = 1001
91 × 11 × 2 = 2002
91 × 11 × 3 = 3003
Write next seven steps. Check, whether the result is correct.
Try the pattern for 143 × 7 × 1, 143 × 7 × 2…..
ANSWER:
We have to find next seven steps of pattern.
91 × 11 × 4 = 4004
91 × 11 × 5 = 5005
91 × 11 × 6 = 6006
91 × 11 × 7 = 7007
91 × 11 × 8 = 8008
91 × 11 × 9 = 9009
91 × 11 × 10 = 10010
Pattern for
143 × 7 × 1 = 1001
143 × 7 × 2 = 2002
143 × 7 × 3 = 3003
143 × 7 × 4 = 4004
143 × 7 × 5 = 5005
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