Tag: NCERT Solutions for Class 6 Maths

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.5 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.5.

• Playing With Numbers Class 6 Ex 3.1
• Playing With Numbers Class 6 Ex 3.2
• Playing With Numbers Class 6 Ex 3.3
• Playing With Numbers Class 6 Ex 3.4
• Playing With Numbers Class 6 Ex 3.6
• Playing With Numbers Class 6 Ex 3.7

Board	CBSE
Textbook	NCERT
Class	Class 6
Subject	Maths
Chapter	Chapter 3
Chapter Name	Playing With Numbers
Exercise	Ex 3.6
Number of Questions Solved	12
Category	NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.5

Question 1.
Which of the following statements are true?
(a) If a number is divisible by 3, it must be divisible by 9.
(b) If a number is divisible by 9, it must be divisible by 3.
(c) A number is divisible by 18 if it is divisible by both 3 and 6.
(d) If a number is divisible by 9 and 10 both, then it must be divisible by 90.
(e) If two numbers are co-primes, at least one of them must be prime.
(f) All numbers which are divisible by 4 must be divisible by 8.
(g) All numbers which are divisible by 8 must also be divisible by 4.
(h) If a number exactly divides two numbers separately, it must exactly divide their sum.
(i) If a number exactly divides the sum of two numbers, it must exactly divide the two numbers separately.
Solution :
(a) This statement is false.
(b) This statement is true.
(c) This statement is false.
(d) This statement is true.
(e) This statement is false.
(j) This statement is false.
(g) This statement is true.
(h) This statement is true.
(i) This statement is false.

Question 2.
Here are two different factor trees for 60. Write the missing numbers.
(a)

(b)

Solution :
(a)

(b)

Question 3.
Which factors are not included in the prime factorization of a composite number?
Solution :
1 and the number itself are not included in the prime factorization of a composite number.

Question 4.
Write the greatest 4-digit number and e×press it in terms of its prime factors.
Solution :
The greatest 4 digit number is 9999.

∴ 9999 = 3×3× 11 × 101.

Question 5.
Write the smallest 5-digit number and e×press it into the form of its prime factors.
Solution :
The smallest 5-digit number is 10000.

∴ 10000 = 2×2×2×2×5×5×5×5.

Question 6.
Find all the prime factors of 1729 and arrange them in ascending order. Now state the relation, if any; between two consecutive prime factors.
Solution :

All the prime factors of 1729 are 7, 13 and 19. When arranged in ascending order, these are 7, 13, 19. We observe that 13 – 7 = 6 19 – 13 = 6
Relation: The difference between two consecutive prime factors is 6.

Question 7.
The product of three consecutive numbers is always divisible by 6. E×plain this statement with the help of some examples.
Solution :
Example 1: Take three consecutive numbers 21, 22 and 23.
21 is divisible by 3.
22 is divisible by 2.
∴ 21 × 22 is divisible by 3 × 2 ( = 6)
∴ 21 × 22 × 23 is divisible by 6.

E×ample 2: Take three consecutive numbers 47, 48 and 49.
48 is divisible by 2 and 3 both.
∴ 48 is divisible by 2 × 3 (= 6)
47 × 48 × 49 is divisible by 6.

Question 8.
The sum of two consecutive odd numbers is divisible by 4. Verify this statement with the help of some examples.
Solution :
Example 1: Take two consecutive odd numbers 5 and 7.
Sum of these numbers = 5 + 7=12 12 is divisible by 4.
Example 2: 13 and 15
Sum of 13 and 15 =13+15 = 28
28 is divisible by 4.

Question 9.
In which of the following expressions, prime ffactorizationhas been done: j
(a) 24 = 2 × 3 × 4
(b) 56 = 7 × 2 × 2 × 2
(c) 70 = 2 × 5 × 7
(d) 54 = 2 × 3 × 9
Solution :
(a) Prime factorisation has not been done.
(b) Prime factorisation has been done.
(c) Prime factorisation has been done.
(d) Prime factorisation has not been done.

Question 10.
Determine, if 25110 is divisible by 45. [Hint : 5 and 9 are co-prime numbers. Test the divisibility of the number by 5 and 9]
Solution :
Divisibility of 25110 by 5
Number in the unit’s place of 25110 = 0
∴ 25110 is divisible by 5.
Divisibility of 25110 by 9
Sum of the digits of the number 25110 = 2+ 5 + 1 + 1 + 0 = 9
9 is divisible by 9.
∴ 25110 is divisible by 9
As 25110 is divisible by 5 and 9 both and 5 and 9 are co-prime numbers, so 25110 is divisible by 5 × 9 = 45.

Question 11.
18 is divisible by both 2 and 3. It is also divisible by 2 × 3 = 6. Similarly, a number is divisible by both 4 and 6. Can we say that the number must also be divisible by 4 × 6 = 24 ? If not, give an example to justify your answer.
Solution :
No we cannot say that the number will be divisible by 4 × 6 = 24, if it is divisible by both 4 and 6 because 4 and 6 are not co-prime numbers (they have two common factors 1 and 2).
Example : 36 is divisible by both 4 and 6.
But, 36 is not divisible by 24.

Question 12.
am the smallest number, having four different prime factors. Can you find me ?
Solution :
The smallest four different prime numbers are 2. 3. 5 and 7.
∴ The smallest number, having four different prime factors is 2 × 3 × 5 × 7 = 210.

 

We hope the NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.5 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.5, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.4 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.4.

• Playing With Numbers Class 6 Ex 3.1
• Playing With Numbers Class 6 Ex 3.2
• Playing With Numbers Class 6 Ex 3.3
• Playing With Numbers Class 6 Ex 3.5
• Playing With Numbers Class 6 Ex 3.6
• Playing With Numbers Class 6 Ex 3.7

Board	CBSE
Textbook	NCERT
Class	Class 6
Subject	Maths
Chapter	Chapter 3
Chapter Name	Playing With Numbers
Exercise	Ex 3.4
Number of Questions Solved	7
Category	NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.4

Question 1.
Find the common factors of:
(a) 20 and 28
(b) 15 and 25
(c) 35 and 50
(d) 56 and 120.
Solution :
(a) 20 and 28
Factors of 20 are 1, 2,4, 5, 10 and 20.
Factors of 28 are 1, 2,4, 7, 14 and 28.
Hence, the common factors of 20 and 28 are 1, 2 and 4.

(b) 15 and 25
Factors of 15 are 1, 3, 5 and 15.
Factors of 25 are 1, 5 and 25.
Hence, the common factors of 15 and 25 are 1 and 5.

(c) 35 and 50
Factors of 35 are 1, 5, 7 and 35.
Factors of 50 are 1, 5, 10, 25 and 50.
Hence, the common factors of 35 and 50 are 1 and 5.

(d) 56 and 120
Factors of 56 are 1, 2,4, 7, 8, 14, 28 and 56. Factors of 120 are 1, 2, 3,4, 5, 6, 8,10, 12,15, 20, 24, 30,40, 60 and 120. ,
Hence, the common factors of 56 and 120 are 1, 2,4 and 8.

Question 2.
Find the common factors of:
(a) 4, 8 and 12
(b) 5, 15 and 25.
Solution :
(a) 4,8 and 12 Factors of 4 are 1,2 and 4.
Factors of 8 are 1, 2,4 and 8.
Factors of 12 are 1, 2, 3,4,6 and 12.
Hence, the common factors of 4,8 and 12 are 1, 2 and 4.

(b) 5,15 and 25
Factors of 5 are 1 and 5.
Factors of 15 are 1, 3 and 5.
Factors of 25 are 1,5 and 25.
Hence, the common factors of 5,15 and 25 are 1 and 5.

Question 3.
Find the first three common multiples of:
(a) 6 and 8
(b) 12 and 18.
Solution :
(a) 6 and 8
Multiples of 6 are 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72,78, 84,90,96,
Multiples of 8 are 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96
Common multiples of 6 and 8 are 24, 48, 72, 96,
∴ First three common multiples of 6 and 8 are 24, 48 and 72.

(b) 12 and 18
Multiples of 12 are 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, 132, 144, *
Multiples of 18 are 18,36,54,72,90,108,126, 144,
∴ Common multiples of 12 and 18 are 36, 72,108, 144,
∴ First three common multiples of 12 and 18 are 36, 72 and 108.

Question 4.
Write all the numbers less than 100 which are common multiples of 3 and 4.
Solution :
Multiples of 3 are : 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99, 102, 105, 108,
Multiples of 4 are : 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100, 104, 108,
Common multiples of 3 and 4 are : 12, 24, 36, 48, 60, 72, 84, 96, 108,
∴ All the numbers less than 100 which are common multiples of 3 and 4 are 12, 24, 36, 48, 60, 72, 84 and 96.

Question 5.
Which of the following numbers are co-prime :
(a) 18 and 35
(b) 15 and 37
(c) 30 and 415
(d) 17 and 68
(e) 216 and 215
(f) 81 and 16.
Solution :
(a) 18 and 35
Factors of 18 are 1, 2, 3, 6, 9 and 18.
Factors of 35 are 1, 5, 7 and 35.
∴ A common factor of 18 and 35 is 1.
∴ 18 and 3 5 have only 1 as the common factor
∴ 18 and 35 are co-prime numbers.

(b) 15 and 37
Factors of 15 are 1, 3, 5 and 15.
Factors of 37 are 1 and 37.
∴ A common factor of 15 and 37 is 1. v 15 and 37 have only 1 as the common factor ∴ 15 and 37 are co-prime numbers.

(c) 30 and 415
Factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30.
Factors of 415 are 1, 5, 83 and 415.
∴ Common factors of 30 and 415 are 1 and 5. v
∴ 30 and 415 have two common factors
∴ 30 and 45 are not co-prime numbers.

(d) 17 and 68 Factors of 17 are 1 and 17.
Factors of 68 are 1, 2, 4, 17, 34 and 68.
∴ Common factors of 17 and 68 are 1 and 17.
∴ 17 and 68 have two common factors
∴ 17 and 68 are not co-prime numbers.

(e) 216 and 215
Factors of 216 are 1,2, 3, 4, 6, 8, 9, 12, 18, 24, 27, 36, 54, 72, 108 and 216.
Factors of 215 are 1, 5 and 43
∴ A common factor of 216 and 215 is 1.
∴ 216 and 215 have only 1 as the common factor
∴ 216 and 215 are co-prime numbers.

(f) 81 and 16
Factors of 81 are 1, 3, 9, 27 and 81.
Factors of 16 are 1, 2, 4, 8 and 16.
∴ A common factor of 81 and 16 is 1.
∴ 81 and 16 have only 1 as the common factor
∴ 81 and 16 are co-prime numbers.

Question 6.
A number is divisible by both 5 and 12. By which other numbers will that number be always divisible?
Solution :
The number will be always divisible by 5 × 12 = 60.

Question 7.
A number is divisible by 12. By what other numbers will that number be divisible?
Solution :
Factors of 12 are 1, 2, 3, 4, 6 and 12. So, that number will be divisible by other numbers 1, 2, 3, 4 and 6.

 

We hope the NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.4 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.4, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.3 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.3.

• Playing With Numbers Class 6 Ex 3.1
• Playing With Numbers Class 6 Ex 3.2
• Playing With Numbers Class 6 Ex 3.4
• Playing With Numbers Class 6 Ex 3.5
• Playing With Numbers Class 6 Ex 3.6
• Playing With Numbers Class 6 Ex 3.7

Board	CBSE
Textbook	NCERT
Class	Class 6
Subject	Maths
Chapter	Chapter 3
Chapter Name	Playing With Numbers
Exercise	Ex 3.3
Number of Questions Solved	6
Category	NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.3

Question 1.
Using divisibility tests, determine which the following numbers are divisible by 2; by 3; by 4; by 5; by 6; by 8; by 9; by 10; by 11; (say yes or no):

Solution :

Question 2.
Using divisibility tests, determine which of the following numbers are divisible by 4: by 8:
(a) 572
(b) 726352
(C) 5500
(d) 6000
(e) 12159
(f) 14560
(g) 21084
(h) 31795072
(i) 1700
(j) 2150.
Solution :
(a) 572
(i) Divisibility by 4
The number formed by last two digits = 72

∵ Remainder is 0
∴ 72 is divisible by 4
∴ 572 is divisible by 4.

(ii) Divisibility by 8
The number formed by last three digits = 572

∵ Remainder = 4 ≠ 0
∴ 572 is not divisible by 8.

(b) 726352
(i) Divisibility by 4
The number formed by last two digits = 52

∵ Remainder is 0
∴ 52 is divisible by 4
∴ 726352 is divisible by 4.

(ii) Divisibility by 8
The number formed by last three digits = 352

∵ Remainder is 0
∴ 352 is divisible by 8
∴ 726352 is divisible by 8.

(c) 5500
(i) Divisibility by 4
The number formed by last two digits = 00, which is divisible by 4 5500 is divisible by 4.
∴ 500 is Divisibility by 4

(ii)
The number formed by last three digits = 500

∵ Remainder 4 0
∴ 500 is not divisible by 8
∴ 5500 is not divisible by 8.

(d) 6000
(i) Divisibility by 4
The number formed by last two digits = 00, which is divisible by 4
6000 is divisible by 4.

(ii) Divisibility by 8
The number formed by last three digits = 000, which is divisible by 8 6000 is divisible by 8.
6000 is divisible by 8

(e) 12159
(i) Divisibility by 4
The number formed by last two digits = 59

∵ Remainder = 3 ≠ 0
∴ 59 is not divisible by 4.
∴ 12159 is not divisible by 4.

(ii) Divisibility by 8
The number formed by last three digits =159

∵ Remainder 7 ≠ 0
∴ 159 is not divisible by 8
∴ 12159 is not divisible by 8.

(f) 14560
(i) Divisibility by 4
The number formed by last two digits = 60

∵ Remainder is 0
∴ 60 is divisible by 4
∴ 14560 is divisible by 4.

(ii) Divisibility by 8
The number formed by last three digits = 560

∵ Remainder is 0
∴ 560 is divisible by 8
∴ 14560 is divisible by 8.

(g) 21084
(i) Divisibility by 4
The number formed by last two digits = 84

∵ Remainder is 0
∴ 484 is divisible by 4
∴ 21084 is divisible by 4.

(ii) Divisibility by 8
The number formed by last three digits = 84

∵ The remainder is not 0
∴ 84 is not divisible by 8
∴ 21084 is not divisible by 8.

(h) 31795072
(i) Divisibility by 4
The number formed by last two digits = 72

∵ Remainder is 0
∴ 72 is divisible by 4
∴ 31795072 is divisible by 4.

(ii) Divisibility by 8
The number formed by last three digits = 72

∵ Remainder is 0
∴ 72 is divisible by 8
∴ 31795072 is divisible by 8.

(i) 1700
(i) Divisibility by 4
The number formed by last two digits = 00, which is divisible by 4
∴ 1700 is divisible by

(ii) Divisibility by 8
The number formed by last three digits = 700

∵ Remainder 4 ≠ 0
∴ 700 is not divisible by 8
∴ 1700 is not divisible by 8.

(j) 2150
(i) Divisibility by 4
The number formed by last two digits = 50

∵ Remainder = 2 ≠ 0
∴ 50 is not divisible by 4
∴ 2150 is not divisible by 4.

(ii) Divisibility by 8
The number formed by last three digits =150

∵ Remainder 6 ≠ 0
∴ 150 is not divisible by 8
∴ 2150 is not divisible by 8.

Question 3.
Using divisibility tests, determine which of following numbers are divisible by 6 :
(a)297144
(b) 1258
(c) 4335
(d) 61233
(e) 901352
(f) 438750
(g) 1790184
(h) 12583
(i) 639210
(j) 17852.
Solution :
We know that number is divisible by 6 if it is divisible by 2 and 3 both.
(a) 297144
(i) Divisibility by 2
Unit’s digit = 4
297144 is divisible by 2.

(ii) Divisibility by 3
Sum of the digits = 2 + 9 + 7 + 1+ 4 + 4 = 27, which is divisible by 3 297144 is divisible by 3 Since, 297144 is divisible by 2 and 3 both, so it is divisible by 6.

(b) 1258
(i) Divisibility by 2 Unit’s digit = 8 1258 is divisible by 2.
(ii) Divisibility by 3 Sum of the digits =1+2 + 5 + 8=16 which is not divisible by 3 1258 is not divisible by 3. Since 1258 is divisible by 2 but not by 3. so 1258 is not divisible by 6.

(c) 4335
(i) Divisibility by 2
Unit’s digit = 5, which is not any of the digits 0, 2, 4, 6 or 8 4335 is not divisible by 2 4335 is not divisible by 6.

(d) 61233
(i) Divisibility by 2
Unit’s digit = 3, which is not any of the digits 0, 2, 4, 6 or 8

61233 is not divisible by 2
61233 is not divisible by 6.

(e) 901352
(i) Divisibility by 2 v Unit’s digit = 2
901352 is divisible by 2.
(ii) Divisibility by 3
Sum of the digits = 9 + 0+1+3 + 5 + 2 = 20, which is not divisible by 3
901352 is not divisible by 3 Since, 901352 is divisible by 2 but not by 3, so it is not divisible by 6.

(f) 438750
(i) Divisibility by 2 v Unit’s digit = 0
438750 is divisible by 2.

(ii) Divisibility by 3
Sum of the digits = 4 + 3 + 8 + 7 + 5+ 0 = 27, which is divisible by 3
438750 is divisible by 3 Since, 438750 is divisible by 2 and 3 both, so it is divisible by 6.

(g) 1790184
(i) Divisibility by 2 v Unit’s digit = 4
1790184 is divisible by 2.
(ii) Divisibility by 3
Sum of the digits =1+7 + 9 + 0+1 + 8 + 4=30, which is divisible by 3
1790184 is divisible by 3 Since, 1790184 is divisible by 2 and 3 both, so it is divisible by 6.

(h) 12583,
(i) Divisibility by 2
Unit’s digit = 3, which is not any of the digits 0, 2, 4, 6 or 8
12583 is not divisible by 2
12583 is not divisible by 6.

(i) 639210
(i) Divisibility by 2
Unit’s digit = 0
639210 is divisible by 2.
(ii) Divisibility by 3
Sum of the digits = 6 + 3 + 9 + 2+1+0 = 21, which is divisible by 3
639210 is divisible by 3 Since, 639210 is divisible by 2 and 3 both, so it is divisible by 6.

(j) 17852
(i) Divisibility by 2
Unit’s digit = 2
∴ 17852 is divisible by 2.
(ii) Divisibility by 3 Sum of the digits =1+7 + 8 + 5 + 2 = 23, which is not divisible by 3
17852 is not divisible by 3
Since 17852 is divisible by 2 but not by 3, so it is not divisible by 6.

Question 4.
Using divisibility tests, determine which of the following numbers are divisible by 11:
(a) 5445
(b) 10824
(c) 7138965
(d) 70169308
(e) 10000001
(f) 901153.
Solution :
(a) 5445
Sum of the digits (at odd places) from the right =5+4=9
Sum of the digits (at even places) from the right =4+5=9
Difference of these sums = 9-9 = 0 v 0 is divisible by 11
5445 is divisible by 11.

(b) 10824
Sum of the digits (at odd places) from the right = 4 + 8+1 = 13
Sum of the digits (at even places) from the right = 2 + 0 = 2
Difference of these sums = 13 – 2 = 11
11 is divisible by 11
10824 is divisible by 11.

(c) 7138965
Sum of the digits (at odd places) from the right =5+9+3+7=24
Sum of the digits (at even places) from the right = 6 + 8 + 1 = 15
Difference of these sums = 24 – 15 = 9
Y 9 is not divisible by 11
∴ 7138965 is not divisible by 11.

(d) 70169308
Sum of the digits (at odd places) from the right = 8 + 3+6 + 0=17
Sum of the digits (at even places) from the right = 0 + 9 + 1+7=17 Difference of these sums = 17-17=0
Y 0 is divisible by 11
∴ 70169308 is divisible by 11.

(e) 10000001
Sum of the digits (at odd places) from the right =l+0+0+0=l
Sum of the digits (at even places) from the right =0+0+0+1=1 Difference of these sums =1-1=0 0 is divisible by 11
∴ 10000001 is divisible by 11.

(f) 901153
Sum of the digits (at odd places) from the right = 3+ 1 +0 = 4
Sum of the digits (at even places) from the right = 5+1+9=15
Difference of these sums =15-4=11
Y 11 is divisible by 11
∴ 901153 is divisible by 11.

Question 5.
Write the smallest digit and the largest digit in the blank space of each of the following numbers so that the number is divisible by 3:
(a) __ 6724
(b) 4765 __ 2.
Solution :
(a) __ 6724
(i) Smallest digit
Sum of the given digits = 6 + 7 + 2 + 4=19
19 is not divisible by 3
∴ Smallest digit (non-zero) is 2.

(ii) Largest digit The largest digit is 8.

(b) 4765 _ 2
(i) Smallest digit
Sum of the given digits = 4 + 7 + 6 + 5 + 2 = 24 Y 24 is divisible by 3
∴ Smallest digit is 0.

(ii) Largest digit The largest digit is 9.

Question 6.
Write digit in the blank space of each of the following numbers so that the number is divisible by 11:
(a) 92 __ 389
(b) 8 __ 9484.
Solution :
(a) 92 __ 389
Sum of the given digits (at odd places) from the right = 9 + 3 + 2= 14
Sum of the given digits (at even places) from the right = 8 + required digit + 9 = required digit + 17
Difference of these sums = required digit + 3 For the above difference to be divisible by 11, required digit = 8.
Hence, the required number is 92 8 389.

(b) 8 __ 9484
Sum of the given digits (at odd places) from the right = 4 + 4 + required digit = 8 + required digit
Sum of the given digits (at even places) from the right = 8 + 9 + 8 = 25
Difference of the sums = 17 – required digit For the above difference to be divisible by 11, required digit = 6.
Hence, the required number is 8 6 9484.

 

We hope the NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.3 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.2 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.2.

• Playing With Numbers Class 6 Ex 3.1
• Playing With Numbers Class 6 Ex 3.3
• Playing With Numbers Class 6 Ex 3.4
• Playing With Numbers Class 6 Ex 3.5
• Playing With Numbers Class 6 Ex 3.6
• Playing With Numbers Class 6 Ex 3.7

Board	CBSE
Textbook	NCERT
Class	Class 6
Subject	Maths
Chapter	Chapter 3
Chapter Name	Playing With Numbers
Exercise	Ex 3.2
Number of Questions Solved	12
Category	NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.2

Question 1.
What is the sum of any two :
(a) Odd numbers
(b) Even numbers.
Solution :
(a) The sum of any two odd numbers is an even number.
(b) The sum of any two even numbers is an even number.

Question 2.
State whether the following statements are True or False :

1. The sum of three odd numbers is even.
2. The sum of two odd numbers and one even number is even.
3. The product of three odd numbers is odd.
4. If an even number is divided by 2, the quotient is always odd.
5. All prime numbers are odd.
6. Prime numbers do not have any factors.
7. Sum of two prime numbers is always even.
8. 2 is the only even prime number.
9. All even numbers are composite numbers.
10. The product of two even numbers is always even.

Solution :

1. This statement is false.
2. This statement is true.
3. This statement is true.
4. This statement is false.
5. This statement is false.
6. This statement is false.
7. This statement is false.
8. This statement is true.
9. This statement is false.
10. This statement is true.

Question 3.
The numbers 13 and 31 are prime numbers. Both these numbers have the same digits 1 and 3. Find such pairs of prime numbers up to 100.
Solution :
All other such pairs of prime numbers up to 100 are as follows :
17 and 71; 37 and 73; 79 and 97.

Question 4.
Write down separately the prime and composite numbers less than 20.
Solution :

Question 5.
What is the greatest prime number between 1 and 10?
Solution :
The greatest prime number between 1 and 10 is 7.

Question 6.
Express the following as the sum of two odd primes :
(a) 44
(b) 36
(c) 24
(d) 18.
Solution :
(a) 44 = 3+41
(b) 36 = 5 + 31
(c) 24 = 5+19
(d) 18 = 5+13.

Question 7.
Give three pairs of prime numbers whose difference is 2.
[Remark: Two prime numbers whose difference is 2 are called twin primes],
Solution :
The three pairs of prime numbers, whose difference is 2, are as follows :
3, 5 ; 5, 7 ; 11, 13.

Question 8.
Which of the following numbers are prime?
(a) 23
(b) 51
(c) 37
(d) 26.
Solution :
(a) 23,
(c) 37 are prime numbers.

Question 9.
Write seven consecutive composite numbers less than 100 so that there is no prime number between them.
Solution :
90, 91, 92, 93, 94, 95, 96.

Question 10.
Express each of the following numbers as the sum of three odd primes :
(a) 21
(b) 31
(c) 53
(d) 61.
Solution :
(a) 21 = 3 + 5 + 13
(b) 31 = 3 + 5 + 23
(c) 53 = 13+ 17 + 23
(d) 61 =7;+ 13+41.

Question 11.
Write five pairs of prime numbers below 20 whose sum is divisible by 5. (Hint: 3 + 7 – 10).
Solution :
2, 3 ; 2, 13 ; 3, 17 ; 7, 13 ; 11, 19.

Question 12.
Fill in the blanks in the following :

1. A number which has only two factors is called a
2. A number which has more than two factors is called a
3. 1 is neither nor
4. The smallest prime number is
5. The smallest composite number is
6. The smallest even number is

Solution :

1. A number which has only two factors is called a prime number.
2. A number which has more than two factors is called a composite number.
3. 1 is neither prime number nor composite number.
4. The smallest prime number is 2.
5. The smallest composite number is 4.
6. The smallest even number is 2.

 

We hope the NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.2 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.3 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.3.

• Whole Numbers Class 6 Ex 2.1
• Whole Numbers Class 6 Ex 2.2

Board	CBSE
Textbook	NCERT
Class	Class 6
Subject	Maths
Chapter	Chapter 2
Chapter Name	Whole Numbers
Exercise	Ex 2.3
Number of Questions Solved	5
Category	NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.3

Question 1.
Which of the following will not represent
(a) 1 + 0
(b) 0 × 0
(c) \(\frac { 0 }{ 2 }\)
(d) \(\frac { 10\quad 10 }{ 2 }\)
Solution :
(a) 1 + 0 = 1 ≠ 0
(b) 0 × 0 = 0
(c) \(\frac { 0 }{ 2 }\)
(d) \(\frac { 10-10 }{ 2 } =\frac { 0 }{ 2 } =0\)
Hence, (a) 1 + 0 will not represent zero.

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.
Solution :
Yes! For example :
2 × 0 = 0
0 × 3 = 0
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.
Solution :
Both of them must be ‘ 1’ as 1 × 1 = 1.

Question 4.
Find using distributive property :
(a) 728 × 101
(b) 5437 × 1001
(c) 824 × 25
(d) 4275 × 125
(e) 504 × 35
Solution :

Question 5.
Study the pattern :

Write the next two steps. Can you say how the pattern works ?
(Hint : 12345 =11111 + 1 1 11 + 111 + 11 + 1). Sol. Next two steps are as follows :
Solution :
Next two steps are as follows :
123456 x 8 + 6 = 987654
1234567 x 8 + 7 = 9876543.
Working of the pattern

We hope the NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.3 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.2 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.2.

• Whole Numbers Class 6 Ex 2.1
• Whole Numbers Class 6 Ex 2.3

Board	CBSE
Textbook	NCERT
Class	Class 6
Subject	Maths
Chapter	Chapter 2
Chapter Name	Whole Numbers
Exercise	Ex 2.2
Number of Questions Solved	7
Category	NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.2

Question 1.
Find the sum by suitable rearrangement :
(a) 837 + 208 + 363
(b) 1962 + 453 + 1538 + 647.
Solution :
(a) 837 + 208 + 363
= 837 + 363 + 208 = (837 + 363) + 208 = 1200 + 208 = 1408
(b) 1962+ 453+ 1538 + 647
= 1962 + 1538 + 453 + 647 = (1962 + 1538) + (453 + 647)
= 3500+ 1100 = 4600.

Question 2.
Find the product by a 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
= 2 × 50 × 1768 = (2 × 50) × 1768 = 100 × 1768= 1,76,800
(b) 4 × 166 × 25 = 4 × 25 × 166 = (4 × 25) × 166 = 100 × 166 = 16,600
(c) 8 × 291 × 125 = 8 × 125 × 291
= (8 × 125) × 291 = 1000 × 291 =2,91,000
(d) 625 × 279 × 16 = 625 × 16 × 279
= (625 × 16) × 279 = 10000 ×279 = 27,90,000
(e) 285 × 5 × 60 = 285 × (5 × 60) = 285 × 300
= 85,500
(f) 125 × 40 × 8 × 25 = (125 × 40) × (8 × 25) = 5000 × 200 = 10,00,000.

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.
Solution :
(a) 297 × 17 + 297 × 3 = 297 × (17+ 3)
= 297 × 20 = 5940
(b) 54279 × 92 + 8 × 54279
= 54279 × 92 + 54279 × 8 = 54279 × (92 + 8)
= 54279 × 100 = 54,27,900
(c) 81265 × 169-81265 × 69
= 81265 × (169-69)
= 81265 × 100 = 81,26,500
(d) 3845 × 5 × 782 + 769 × 25 × 218
= 3845 × 5 × 782 + 769 × 5 × 5 × 218 = 3845 × 5 × 782 + (769 × 5) × 5 × 218 = 3845 × 5 × 782 + 3845 × 5 × 218 = 3845 × 5 × (782 + 218)
= 3845 × 5 × 1000 = 19225 × 1000 = 1,92,25,000.

Question 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 = 73,800 + 2,214 = 76,014
(b) 854 × 102
= 854 × (100 + 2)
= 854 × 100 + 854 × 2 = 85,400+ 1,708 = 87,108
(c) 258 × 1008
= 258 × (1000+ 8)
= 258 × 1000 + 258 × 8 = 2,58,000 + 2,064 = 2,60,064
(d) 1005 × 168
= 168 × 1005
= 168 × (1000+ 5)
= 168 × 1000+ 168 × 5
= 1,68,000 + 840 = 1,68,840.

Question 5.
A taxi driver filled his car petrol tank with 40 liters of petrol on Monday. The next day he filled the tank with 50 liters of petrol. If the petrol costs ₹ 44 per liter, how much did he spend all on petrol?
Solution :
Petrol filled on Monday = 40 litres
Petrol filled the next day = 50 litres
∴ Total petrol filled on the two days = 40 litres + 50 litres = 90 litres
∴ Cost of petrol per litre = ₹ 44
∴ Cost of 90 litres of 7 petrol = ₹ 44 × 90 = ? 3960.

Question 6.
A vendor’supplies 32 liters of milk to a hotel in the morning and 68 liters of milk in the evening. If the milk costs ₹ 15 per liter, how much money is due to the vendor per day?
Solution :
Milk supplied in the morning = 32 liters
Milk supplied in the evening = 68 liters
∴ Milk supplied per day = 32 litres + 68 litres = 100 litres
Cost of milk per liter = ₹ 15
Money due to the vendor per day = Cost of
100 litres of milk = ₹ 15 × 100 = ₹ 1500.

Question 7.
Match the following :

Solution :

We hope the NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.2 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.3 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.3.

• Knowing Our Numbers Class 6 Ex 1.1
• Knowing Our Numbers Class 6 Ex 1.2

Board	CBSE
Textbook	NCERT
Class	Class 6
Subject	Maths
Chapter	Chapter 1
Chapter Name	Knowing Our Numbers
Exercise	Ex 1.3
Number of Questions Solved	3
Category	NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.3

Question 1.
Estimate each of the following using general rule:
(a) 730 + 998
(b) 796 – 314
(c) 12,904 + 2,888
(d) 28,292 – 21,496
Make ten more such examples of addition, subtraction and estimation of their outcome.
Solution :
(a) 730 + 998 = 700 + 1000 = 1700
(b) 796 – 314 = 800 – 300 = 500
(c) 12,904 + 2,888 = 13,000 + 3,000 = 16,000
(d) 28,292 – 21,496 = 28,000 – 21,000 = 7,000
Ten more such examples of addition, subtraction and estimation of their outcome are as follows:
Ex. 1.720 + 990
Ex. 2.640 + 880
Ex. 3. 749 + 740
Ex. 4.890 – 420
Ex. 5.680 – 370
Ex. 6.585 – 220
Ex. 7.10803 + 3777
Ex. 8.15663 + 2125
Ex. 9. 30990 – 21660
Ex. 10. 40870-19530
Solution :
1. 720 + 990 = 700 + 1000 = 1700
2. 640 + 880 = 600 + 900 = 1500
3. 749 + 740 = 700 + 700 = 1400
4. 890 – 420 = 900 – 400 = 500
5. 680 – 370 = 700 – 400 = 300
6. 585 – 220 = 600 – 200 = 400
7. 10803 + 3777 = 11000 + 4000 = 15000
8. 15663 + 2125 = 16000 + 2000 = 18000
9. 30990 – 21660 = 31000 – 22000 = 9000
10. 40870 – 19530 = 41000 – 20000 = 21000

Question 2.
Give a rough estimate (by rounding off to nearest hundreds) and also a closer estimate (by rounding off to nearest tens):
(a) 439 + 334 4,317
(b) 1,08,734 – 47,599
(c) 8,325 – 491
(d) 4,89,348-48,365 Make four more such examples.
Solution :
(a)

• Rough estimate (Rounding off to nearest hundreds)
  439 + 334 + 4317
  = 400 + 300 + 4,300 = 5000
• Closer estimate (Rounding off to nearest tens) 439 + 334 + 4317
  = 440 + 330 + 4,320 = 5,090.

(b)

• Rough estimate (Rounding off to nearest hundreds) 1,08,734 – 47,599
  = 1,08,700 – 47,600 = 61,100
• Closer estimate (Rounding off to nearest tens) 1,08,734 – 47,599
  = 1,08,730 – 47,600 = 61,130.

(c)

• Rough estimate (Rounding off to nearest hundreds) 8325 – 491
  = 8300 – 500 = 7800
• Closer estimate (Rounding off to nearest tens)
  8325 – 491
  = 8330 – 490 = 7840.

(d)

• Rough estimate (Rounding off to nearest hundreds)
  4,89,348 – 48,365
  = 4,89,300 – 48,400 = 4,40,900
• Closer estimate (Rounding off to nearest tens)
  4,89,348 – 48,365
  = 4,89,350 – 48,370 = 4,40,980

Four more such examples are as follows:
1. 538 + 432 + 5326
2. 2,09, 849 – 57,698
3. 9426 – 395
4. 5,98,459 – 36,463 Sol.
solution :
1. 538 + 432 + 5326
= 500 + 400 + 5300
(Rough estimate Rounding off to nearest hundreds)
= 6200
538 + 432 + 5326
= 540 + 430 + 5330
(Closer estimate Rounding off to nearest tens) = 6300

2. 2,09,849 – 57.698
= 2.09.800 – 57,700
(Rough estimate Rounding off to nearest hun-dreds)
=152100 2,09, 849 – 57,698
= 2,09, 850 – 57,700
(Closer estimate Rounding off to nearest tens) =152150

3. 9426-395
= 9400 – 400 ..
(Rough estimate Rounding off to nearest hundreds)
= 9000 9426 – 395
= 9430 – 400
(Closer estimate Rounding off to nearest tens) = 9030

4. 5,98,459 – 36,463
= 5,98,500 – 36,500
(Rough estimate Rounding off to nearest hundreds)
= 5,62,000 5,98,459 – 36,463
= 5,98,460 – 36,460
(Closer estimate Rounding off to nearest tens) = 5,62,000

Question 3.
Estimate the following products using the general rule:
(a) 578 ×161
(b) 5281 × 3491
(c) 1291 × 592
(d) 9250 × 29
Make four more such examples.
Solution :
(a) 578 × 161
Estimated product = 600 × 200 = 1,20,000
(b) 5281 × 3491
Estimated product = 5000 × 3500 = 1,75,00,000
(c) 1291 × 592
Estimated product = 1300 × 600 = 7,80,000
(d) 9250 × 29
Estimated product = 10000 × 30 = 3,00,000

Four more such examples are:
1. 678 × 261
2. 4271 × 4391
3. 2391 × 629
4. 8250 × 39

We hope the NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.3 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.2 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.2.

• Knowing Our Numbers Class 6 Ex 1.1
• Knowing Our Numbers Class 6 Ex 1.3

Board	CBSE
Textbook	NCERT
Class	Class 6
Subject	Maths
Chapter	Chapter 1
Chapter Name	Knowing Our Numbers
Exercise	Ex 1.2
Number of Questions Solved	12
Category	NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.2

Question 1.
A book exhibition was held for four days in a school. The number of tickets sold at the counter on the first, second, third and final day was respectively 1094, 1812, 2050 and 2751. Find the total number of tickets sold on all four days.
Solution :
Number of tickets sold on the first day = 1094
Number of tickets sold on the second day =1812
Number of tickets sold on the third day = 2050
Number of tickets sold on the final day = 2751
Total number of tickets sold on all the four days = 1094 + 1812 + 2050 + 2751 = 7707.

Question 2.
Shekhar is a famous cricket player. He has so far scored 6980 runs in test matches. He wishes to complete 10000 runs. How many more runs does he need?
Solution :
Runs scored so far = 6980
Runs wished to be scored = 10000
Runs needed more = 10000 – 6980 = 3020.

Question 3.
In an election, the successful candidate registered 5,77,500 votes and his nearest rival secured 3,48,700 votes. By what margin did the successful candidate win the election?
Solution :
Votes registered by the successful candidate = 5,77,500
Votes secured by the nearest rival = 3,48.700
Margin by which the successful candidate won the election = 5,77,500 – 3,48,700 = 2,28,800.

Question 4.
Kirti bookstore sold books worth? 2,85,891 in the first week of June and books worth ?. 4,00,768 in the second week of the month. How much was the sale for the two weeks together? In which week was the sale greater and by how much?
Solution :
Sale of books in the first week = ₹ 2,85,891
Sale of books in the second week = ₹ 4,00,768
∴ Sale for the two weeks together = ₹ 2,85,891 + ₹ 4,00,768 = ₹ 6,86,659.
The sale was greater in the second week by ₹ 4,00,768 – ₹ 2,85,891 i.e., by ₹ 1,14,877.

Question 5.
Find the difference between the greatest and least numbers that can be written using the digits 6, 2, 7, 4, 3 each only once.
Sol.
Greatest number that can be written using the digits 6, 2, 7, 4, 3 each only once = 76,432
Least number that can be written using the digits 6, 2, 7, 4, 3 each only once = 23,467
∴ Difference between the greatest and least numbers that can be written using the digits 6,2,7,4, 3 each only once = 76,432 – 23,467 = 52,965.

Question 6.
A machine, on an average, manufactures 2,825 screws a day. How many screws did it produce in the month of January 2006?
Solution :
The number of screws manufactured by the machine a day on an average = 2,825.
Number of days in the month of January 2006 = 31
The number of screws produced by the machine in the month of January 2006 = 2,825 × 31 =87,575.

Question 7.
A merchant had ₹ 78,592 with her. She placed an order for purchasing 40 radio sets at ? 1200 each. How much money will remain with her after the purchase?
Sol.
Money which the merchant had = ₹ 78,592
Cost of 40 radio sets at ? 1200 each = ₹ 1200 × 40 = ₹ 48,000
Money that will remain with the merchant after the purchase = ₹ 78,592 – ₹ 48,000 = ₹ 30,592.

Question 8.
A student multiplied 7236 by 65 instead of multiplying by 56. By how much was his answer greater than the. correct answer?
[Hint: Do you need to do both the multiplications?]
Solution :
The wrong answer was greater than the correct answer by
= 7236 × 65 – 7236 × 56
= 7236 × (65 – 56)
= 7236 × 9 = 65.124

Question 9.
To stitch a shirt, 2 m 15 cm cloth is needed. Out of 40 m cloth, how many shirts can be stitched and how much cloth will remain ?
[Hint: convert data in cm.]
Solution :
2 m 15 cm = 2 m + 15 cm = 2 × 100 cm + 15 cm = 200 cm + 15 cm = 215 cm
40 m = 40 × 100 cm = 4000 cm

Hence, 18 shirts can be stitched and 130 cm,
i. e., 1 m 30 cm cloth will remain.

Question 10.
Medicine is packed in boxes, each weighing 4 kg 500 g. Mow many such boxes can be loaded in a van which cannot carry beyond 800 kg?
Solution :
4 kg 500 g = 4 kg + 500 g
= 4 × 1000 g + 500 g
= 4000 g + 500 g
= 4500 g 800 kg
= 800 × 1000 g = 800000 g

Hence, 177 such boxes can be loaded.

Question 11.
The distance between the school and the house of a student’s house is 1 km 875 m. Every day she walks both ways. Find the total distance covered by her in six days.
Solution :
1 km 875 m = 1 km + 875 m
= 1 × 1000 m + 875 m
= 1000 m + 875 m = 1875 m
Distance covered by her in a day in walking both ways between school and home = 1875 × 2 m = 3750 m
∴ Total distance covered by her in six days in walking both ways between school and home = 3750 m × 6 = 22500 m
= 22000 m + 500 m = \(\frac { 22000 }{ 1000 }\) km + 500 m
= 22 km + 500 m = 22 km 500 m.

Question 12.
A vessel has 4 liters and 500 ml of curd. In how many glasses, each of 25 ml capacity, can it be filled?
Solution :
41500 ml = 41 + 500 ml
= 4 × 1000 ml + 500 ml
= 4000 ml + 500 ml = 4500 ml

Hence, it can be filled in 180 glasses.

We hope the NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.2 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.1 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.1.

• Practical Geometry Class 6 Ex 14.2
• Practical Geometry Class 6 Ex 14.3
• Practical Geometry Class 6 Ex 14.4
• Practical Geometry Class 6 Ex 14.5
• Practical Geometry Class 6 Ex 14.6

Board	CBSE
Textbook	NCERT
Class	Class 6
Subject	Maths
Chapter	Chapter 14
Chapter Name	Practical Geometry
Exercise	Ex 14.1
Number of Questions Solved	5
Category	NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.1

Question 1.
Draw a circle of radius 3.2 cm.
Solution :

Steps of Construction

• Open the compasses for the required radius 3.2 cm, by putting the pointer on 0 and opening the pencil upto 3.2 cm.
• Draw a point with a sharp pencil and marks it as O in the centre.
• Place the pointer of the compasses where the centre has been marked.
• Turn the compasses slowly to draw the circle.

Question 2.
With the same centre O, draw two circles of radii 4 cm and 2.5 cm.
Solution :
Steps of Construction
1. For circle of radius 4 cm

• Open the compasses for the required radius 4 cm, by putting the pointer on 0 and opening the pencil upto 4 cm.
• Place the pointer of the compasses at O.
• Turn the compasses slowly to draw the circle.

2. For circle of radius 2.5 cm

• Open the compasses for the required radius 2.5 cm. by putting the pointer on 0 and opening the pencil upto 2.5 cm.
• Place the pointer of the compasses at O.
• Turn the compasses slowly to draw the circle.

Question 3.
Draw a circle and any two of its diameters. If you join the ends of these diameters, what is the figure obtained? What figure is obtained if the diameters are perpendicular to each other? How do you check your answer?
Solution :
(i) On joining the ends of any two diameters of the circle, the figure obtained is a rectangle.

(ii) On joining the ends of any two diameters of the circle, perpendicular to each other, the figure obtained is a square.
To check the answer, we measured the sides and angles of the figure obtained.

Question 4.
Draw any circle and mark points A, B and C such that:
(a) A is on the circle.
(b) B is in the interior of the circle.
(c) C is in the exterior of the circle.
Solution :

Question 5.
Let A, B be the centres of two circles of equal radii; draw them so that each one of them passes We need a ruler here. through the centre of the other. Let them intersect at C and D. Examine whether \(\overline { AB }\) and \(\overline { CD }\) are at right angles.
Solution :

Yes! \(\overline { AB }\) and \(\overline { CD }\) are at right angles.

 

We hope the NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.1 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.1, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.1 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.1.

• Fractions Class 6 Ex 7.2
• Fractions Class 6 Ex 7.3
• Fractions Class 6 Ex 7.4
• Fractions Class 6 Ex 7.5
• Fractions Class 6 Ex 7.6

Board	CBSE
Textbook	NCERT
Class	Class 6
Subject	Maths
Chapter	Chapter 7
Chapter Name	Fractions
Exercise	Ex 7.1
Number of Questions Solved	11
Category	NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.1

Question 1.
Write the fraction representing the shaded portion.



Solution :

Question 2.
Color the part according to the given fraction.

Solution :

Question 3.
Identify the error, if any

Solution :
Neither of the shaded portions represents the corresponding given fractions. 4.

Question 4.
What fraction of a day is 8 hours?
Solution :
1 day = 24 hours
∴ Required fraction = \(\frac { 8 }{ 24 }\)

Question 5.
What fraction of an hour is 40 minutes?
Solution :
1 hour = 60 minutes
∴ Required fraction = \(\frac { 40 }{ 60 }\)

Question 6.
Arya, Abhimanyu, and Vivek shared lunch. Arya brings two sandwiches, one made of vegetable and one of jam. The other two boys forgot to bring their lunch. Arya agreed to share his sandwiches so that each person will have an equal share of each sandwich.
(a) How can Arya divide his sandwiches so that each person has an equal share?
(b) What part of a sandwich will each boy receive?
Solution :
(a) Arya will divide each sandwich into three equal parts, and give one part of each sandwich to each one of them.
(b) Each boy will receive \(\frac { 1 }{ 3 }\) part of a sandwich.

Question 7.
Kanchan dyes dresses. She had to dye 30 dresses. She has so far finished 20 dresses. What fraction of dresses has she finished?
Solution :
She has finished \(\frac { 2 }{ 3 }\) fraction of the dresses.

Question 8.
Write the natural numbers from 2 to 12. What fraction of them are prime numbers?
Solution :
The natural numbers from 2 to 12 are 2, 3, 4,5,6, 7, 8, 9, 10, 11 and 12
Total number of natural numbers = 11
Out of these, the prime numbers are 2, 3, 5, 7, 11

Total number of these prime numbers = 5 5
∴ Required fraction = \(\frac { 5 }{ 11 }\).

Question 9.
Write the natural numbers from 102 to 113. What fraction of them are prime numbers?
Solution :
The natural numbers from 102 to 113 are
102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112 and 113
Total number of natural numbers =12
Out of these, the prime numbers are 103, 107, 109, 113.
Total number of these prime numbers = 4 . 4
∴ Required fraction = \(\frac { 4 }{ 12 }\).

Question 10.
What fraction of these circles have X’s in them?

Solution :
Total number of circles = 8
Number of circles which have X’s in them = 4
∴ Required fraction = \(\frac { 4 }{ 8 }\).

Question 11.
Kristin received a CD player for her birthday. She bought 3 CDs and received 5 others as gifts. What fraction of her total CDs did she buy and what fraction did she receive as gifts?
Solution :
Number of CDs bought = 3
Number of CDs received as gifts = 5
∴ Total number of CDs = 3 + 5 = 8
Fraction of her total CDs that she bought \(\frac { 3 }{ 8 }\) and, fraction of her total CDs that she received as gifts \(\frac { 5 }{ 8 }\).

 

We hope the NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.1 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.1, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.1 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.1.

• Decimals Class 6 Ex 8.2
• Decimals Class 6 Ex 8.3
• Decimals Class 6 Ex 8.4
• Decimals Class 6 Ex 8.5
• Decimals Class 6 Ex 8.6

Board	CBSE
Textbook	NCERT
Class	Class 6
Subject	Maths
Chapter	Chapter 8
Chapter Name	Decimals
Exercise	Ex 8.1
Number of Questions Solved	10
Category	NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.1

Question 1.
Write the following as numbers in the given table:
(a)

(b)

Hundreds (100)	Tens (10)	Ones (1)	Tenths (\(\frac { 1 }{ 10 } \))

Solution.

	Hundreds (100)	Tens (10)	Ones (1)	Tenths (\(\frac { 1 }{ 10 } \))
(a)	0	3	1	2
(b)	1	1	0	4

Question 2.
Write the following decimals in the place value table :
(a) 19.4
(b) 0.3
(c) 10.6
(d) 205.9
Solution.

	Hundreds	Tens	Ones	Tenths
(a)	0	1	9	4
(b)	0	0	0	3
(c)	0	1	0	6
(d)	2	0	5	9

Question 3.
Write each of the following as decimals:
(a) Seven-tenths
(b) Two tens and nine-tenths
(c) Fourteen point six
(d) One hundred and two ones
(e) Six hundred point eight
Solution.
(a) 7 tenths : \(\frac { 7 }{ 10 } \) =0.7.
(b) Two tens and nine-tenths = 20 + \(\frac { 9 }{ 10 } \)  =20.9
(c) Fourteen point six = 14.6
(d) One hundred and two-ones =100 + 2=102
(e) Six hundred point eight = 600.8

Question 4.
Write each of the following as decimals:
(a) \(\frac { 5 }{ 10 } \)
(b) 3 +\(\frac { 7 }{ 10 } \)
(c) 200+60+5+\(\frac { 1 }{ 10 } \)
(d) 70+\(\frac { 8 }{ 10 } \)
(e) 4 \(\frac { 2 }{ 10 } \)
(g) \(\frac { 3 }{ 2 } \)
(h) \(\frac { 2 }{ 5 } \)
(i) \(\frac { 12 }{ 5 } \)
(j) 3 \(\frac { 3 }{ 5 } \)
(k) 4\(\frac { 1 }{ 2 } \)
Solution.

Question 5.
Write the following decimals as fractions. Reduce the fractions to lowest form :
(a) 0.6
(b) 205
(c) 1.0
(d) 3.8
(e) 13.7
(f) 21.2
(g) 6.4
Solution.
(a)
 \(0.6=\frac { 6 }{ 10 } =\frac { 6\div 2 }{ 10\div 5 } \)
∵  H.C.F.(6,10) = 2
= \(\frac { 3 }{ 5 } \)

(b)
\(2.5=\frac { 25 }{ 10 } =\frac { 25\div 5 }{ 10\div 5 } \)
∵  H.C.F.(25,10) = 5
= \(\frac { 5 }{ 2 } \)

(c)
\(1.5=\frac { 10 }{ 10 } =\frac { 10\div 10 }{ 10\div 10 } \)
∵  H.C.F.(10,10) = 10
= \(\frac { 1 }{ 1 } \) =1

(d)
\(3.8 =\frac { 38 }{ 10 } =\frac { 38\div 2 }{ 10\div 2 } \)
∵  H.C.F.(38,10) = 2
= \(\frac { 19 }{ 5 } \)

(e)
\(13.7=\frac { 137 }{ 10 } \)

(f)
\( 21.2=\frac { 212 }{ 10 } =\frac { 212\div 2 }{ 10\div 5 } \)
∵  H.C.F.(212,10) = 5
= \(\frac { 106 }{ 5 } \)

(g)
\(6.4=\frac { 64 }{ 10 } =\frac { 64\div 2 }{ 10\div 5 } \)
∵  H.C.F.(64,10) = 2
= \(\frac { 32 }{ 5 } \)

Question 6.
Express the following as cm using decimals:
(a) 2 mm
(b) 30 mm
(c) 116 mm
(d) 4cm 2mm
(e) 11cm 52mm
(f) 83 mm
Solution.
(a)
\(2mm =\frac { 2 }{ 10 } \)
∵  \( 1mm =\frac { 1 }{ 10 } \)
= 0.2cm

(b)
\(30mm =\frac { 30 }{ 10 } \)
∵  \( 1mm =\frac { 1 }{ 10 } \)
= 3cm = 3.0cm

Question 7.
Between which two whole numbers on the number line are the given numbers lie? Which of these whole numbers is nearer the number?
(a) 0.8
(b) 5.1
(c) 2.6
(d) 6.4
(e) 9.0
(f) 4.9

Solution.
(a) 0.8. The number 0.8 is between the two whole numbers 0 and 1. The whole number 1 is nearer the number 0.8.

(b) 5.1. The number 5.1 is between the two whole numbers 5 and 6. The whole number 5 is nearer the number 5.1.

(c) 2.6. The number 2,6 is between the two whole numbers 2 and 3. The whole number 3 is nearer the number 2.6.

(d) 6.4 The number 6.4 is between the two whole numbers 6 and 7. The whole number 6 is nearer the number 6.4.

(e) 0. 9.0 itself is a whole number.

(f) 4.9. The number 4.9 is between the two whole numbers 4 and 5. The whole number 5 is nearer the number 4.9.

Question 8.
Show the following numbers on the number line:
(a) 2
(b) 1.9
(c) 1
(d) 2.5.
Solution.
(a) 0.2. We know that 0.2 is more than zero but less than one. There are 2-tenths in it. Divide the unit length between 0 and 1 into 10 equal parts and take 2 parts as shown below:

(b) 1.9. We know that 1.9 is more than one but less than two. There are 9-tenths in’it. Divide the unit length between 1 and 2 into 10 equal parts and take 9 parts as shown below:

(c) 1.1. We know that 1.1 is more than one but less than two. There is one-tenth in it. Divide the unit length between 1 and 2 into 10 equal parts and take 1 part as shown below:

(d) 2.5. We know that 2.5 is more than two but less than three. There are 5-tenths in it. Divide the unit length between 2 and 3 into 10 equal parts and take 5 parts as shown below :

Question 9.
Write the decimal number represented by the points A, B, C, D on the given number line :

Solution.
(i) A. The decimal number represented by the point A is 0.8 as the unit length between 0 and 1 has been divided into 10 equal parts and 8 parts have been taken.

(ii) B. The decimal number represented by the point B is 1.3 as the unit length between 1 and 2 has been divided into 10 equal parts and 3 parts have been taken.

(iii) C. The decimal number represented by the point C is 2.2 as the unit length between 2 and 3 has been divided into 10 equal parts and 2 parts have been taken.

(iv) D. The decimal number represented by the point D is 2.9 as the unit length between 2 and 3 has been divided into 10 equal parts and 9 parts have been taken.

Question 10.
(a) The length of Ramesh’s notebook is 9 cm 5 mm. What will be its length in cm?
(b) The length of a young gram plant is 65 mm. Express its length in cm.
Solution.
(a) Length of Ramesh’s notebook = 9 cm 5mm

 

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