CBSE Class 6

These NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Exercise 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
Answer:
(a) Factors of 20 = 1, 2, 4, 5, 10, 20
Factors of 28 = 1,2, 4, 7, 14, 28
Common factors = 1, 2, 4

(b) Factors of 15 = 1, 3, 5, 15
Factors of 25 = 1,5, 25
Common factors = 1,5

(c) Factors of 35 = 1, 5, 7, 35
Factors of 50 = 1, 2, 5, 10, 25, 50
Common factors = 1, 5

(d) Factors of 56 = 1, 2, 4, 7, 8, 14, 28, 56
Factors of 120 = 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 60, 120
Common factors = 1, 2, 4, 8

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

(b) Factors of 5 = 1, 5
Factors of 15 = 1, 3, 5, 15
Factors of 25 = 1, 5, 25
Common factors of 5, 15 and 25 = 1, 5

Question 3.
Find first three common multiples of:
(a) 6 and 8
(b) 12 and 18
Answer:
(a) Multiple of 6 = 6, 12, 18, 24, 30, 36, 42, 28, 54, 60, 72,…
Multiple of 8 = 8, 16, 24, 32, 40, 48, 56, 64, 72,…
Common multiples of 6 and 8 = 24, 48, 72

(b) Multiple of 12 = 12, 24, 36, 48, 60, 72, 84, 96, 108, 120,…
Multiple of 18 = 18, 36, 54, 72, 90, 108,…
Common multiples of 12 and 18 = 36, 72, 108

Question 4.
Write all the numbers less than 100 which are common multiples of 3 and 4.
Answer:
Multiple of 3 = 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 Multiple of 4 = 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
Common multiples of 3 and 4= 12,24, 36, 48, 60, 72, 84, 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
Answer:
(a) Factors of 18 = 1, 2, 3, 6, 9, 18
Factors of 35 = 1, 5, 7, 35
Common factor = 1
Since, both have only one common factor, i.e., 1, therefore, they are co-prime numbers.

(b) Factors of 15 = 1, 3, 5, 15
Factors of 37 = 1, 37
Common factor = 1
Since, both have only one common factor, i.e., 1, therefore, they are co-prime numbers.

(c) Factors of 30 = 1,2, 3, 5, 6, 15, 30
Factors of 415 = 1, 5,…, 83, 415
Common factors =1,5
Since, both have more than one common factor, therefore, they are not co-prime numbers.

(d) Factors of 17 = 1, 17
Factors of 68 = 1,2, 4, 17, 34, 68
Common factors =1,17
Since, both have more than one common factor, therefore, they are not co-prime numbers.

(e) Factors of 216 = 1, 2, 3, 4, 6, 8, 36, 72, 108,216
Factors of 215 = 1, 5, 43, 215
Common factor = 1
Since, both have only one common factor, i.e., 1, therefore, they are coprime numbers.

(f) Factors of 81 = 1,3, 9, 27, 81
Factors of 16 = 1, 2,4, 8, 16
Common factor = 1
Since, both have only one common factor, i.e., 1, therefore, they are coprime numbers.

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

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

These NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.3 Questions and Answers are prepared by our highly skilled subject experts.

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

Question 1.
Using divisibility tests, determine which of 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):
Answer:

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
Answer:
(a) 572
→ Divisible by 4 as its last two digits are divisible by 4.
→ Not divisible by 8 as its last three digits are not divisible by 8.

(b) 726352
→ Divisible by 4 as its last two digits are divisible by 4.
→ Divisible by 8 as its last three digits are divisible by 8.

(c) 5500
→ Divisible by 4 as its last two digits are divisible by 4.
→ Not divisible by 8 as its last three digits are not divisible by 8.

(d) 6000
→ Divisible by 4 as its last two digits are 0.
→ Divisible by 8 as its last three digits are 0.

(e) 12159
→ Not divisible by 4 and 8 as it is an odd number.

(f) 14560
→ Divisible by 4 as its last two digits are divisible by 4.
→ Divisible by 8 as its last three digits are divisible by 8.

(g) 21084
→ Divisible by 4 as its last two digits are divisible by 4.
→ Not divisible by 8 as its last three digits are not divisible by 8.

(h) 31795072
→ Divisible by 4 as its last two digits are divisible by 4.
→ Divisible by 8 as its last three digits are divisible by 8.

(i) 1700
→ Divisible by 4 as its last two digits are 0.
→ Not divisible by 8 as its last three digits are not divisible by 8.

(j) 2150
→ Not divisible by 4 as its last two digits are not divisible by 4.
→ Not divisible by 8 as its last three digits are 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
Answer:
(a) 297144
→ Divisible by 2 as its unit’s place is an even number.
→ Divisible by 3 as sum of its digits (= 27) is divisible by 3.
Since the number is divisible by both 2 and 3, therefore, it is also divisible

(b) 1258
→ Divisible by 2 as its units place is an even number.
→ Not divisible by 3 as sum of its digits (= 16) is not divisible by 3.
Since the number is not divisible by both 2 and 3, therefore, it is not divisible by 6.

(c) 4335
→ Not divisible by 2 as its unit’s place is not an even number.
→ Divisible by 3 as sum of its digits (= 15) is divisible by 3.
Since the number is not divisible by both 2 and 3, therefore, it is not divisible by 6.

(d) 61233
→ Not divisible by 2 as its unit’s place is not an even number.
→ Divisible by 3 as sum of its digits (= 15) is divisible by 3.
Since the number is not divisible by both 2 and 3, therefore, it is not divisible by 6.

(e) 901352
→ Divisible by 2 as its unit’s place is an even number.
→ Not divisible by 3 as sum of its digits (= 20) is not divisible by 3.
Since the number is not divisible by both 2 and 3, therefore, it is not divisible by 6.

(f) 438750
→ Divisible by 2 as its unit’s place is an even number.
→ Divisible by 3 as sum of its digits (= 27) is divisible by 3.
Since the number is divisible by both 2 and 3, therefore, it is divisible by 6.

(g) 1790184
→ Divisible by 2 as its unit’s place is an even number.
→ Divisible by 3 as sum of its digits (= 30) is divisible by 3.
Since the number is divisible by both 2 and 3, therefore, it is divisible by 6.

(h) 12583
→ Not divisible by 2 as its unit’s place is not an even number.
→ Not divisible by 3 as sum of its digits (= 19) is not divisible by 3.
Since the number is not divisible by both 2 and 3, therefore, it is not divisible by 6.

(i) 639210
→ Divisible by 2 as its unit’s place is an even number.
→ Divisible by 3 as sum of its digits (= 21) is divisible by 3.
Since the number is divisible by both 2 and 3, therefore, it is divisible by 6.

(j) 17852
→ Divisible by 2 as its units place is an even number.
→ Not divisible by 3 as sum of its digits (= 23) is not divisible by 3.
Since the number is not divisible by both 2 and 3, therefore, 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
Answer:
(a) 5445
Sum of the digits at odd places =4+5=9
→ Sum of the digits at even places =4+5=9
→ Difference of both sums = 9-9 = 0 Since the difference is 0, therefore, the number is divisible by 11.

(b) 10824
→ Sum of the digits at odd places = 4 + 8 +1 = 13
→ Sum of the digits at even places =2+0=2
→ Difference of both sums = 13 – 2 = 11 Since the difference is 11, therefore, the number is divisible by 11.

(c) 7138965
→ Sum of the digits at odd places = 5 + 9 + 3 + 7 = 24
→ Sum of the digits at even places = 6 + 8 + 1 = 15
→ Difference of both sums = 24 – 15 = 9
Since the difference is neither 0 nor 11, therefore, the number is not divisible by 11.

(d) 70169308
→ Sum of the digits at odd places = 8 + 3 + 6 + 0 = 17
→ Sum of the digits at even places = 0 + 9 + 1 + 7 = 17
→ Difference of both sums = 17 – 17 = 0
Since the difference is 0, therefore, the number is divisible by 11.

(e) 10000001
→ Sum of the digits at odd places
= 1 + 0 + 0 + 0 = 1
→ Sum of the digits at even places
= 0 + 0 + 0 + 1 = 1
→ Difference of both sums =1-1=0 Since the difference is 0, therefore, the number is divisible by 11.

(f) 901153
Sum of the digits at odd places = 3 + 1 +0 = 4
→ Sum of the digits at even places = 5 + 1 + 9 = 15
→ Difference of both sums =15 – 4 = 11
Since the difference is 11, therefore, the number is divisible by 11.

Question 5.
Write the smallest digit and the greatest digit in the blanks space of each of the following numbers so that the number formed is divisibly by 3:
(a) ………………. 6724
(b) 4765 ………………. 2
Answer:
(a) We know that a number is divisible by 3 if the sum of all digits is divisible by 3.
Therefore, Smallest digit: 2 → 26724 = 2 + 6 + 7 + 2 + 4 = 21
Largest digit: 8 → 86724 = 8 + 6 + 7 + 2 + 4 = 27

(b) We know that a number is divisible by 3 if the sum of all digits is divisible by 3.
Therefore, Smallest digit: 0 → 476502 =4+7+6+5+0+2=24
Largest digit: 9 → 476592 = 9 + 4 + 7 + 6 + 5 + 0 + 2 = 33

Question 6.
Write a digit in the blank space of each of the following numbers so that the number formed is divisibly by 11:
(a) 92 …………… 389
(b) 8 ……………….. 9484
Answer:
(a) We know that a number is divisible by 11 if the difference of the sum of the digits at odd places and that of even places should be either 0 or 11.
Therefore, 928389 →
Odd places = 9 + 8 + 8 = 25
Even places = 2 + 3 + 9 = 14
Difference = 25 – 14 = 11

(b) We know that a number is divisible by 11 if the difference of the sum of the digits at odd places and that of even places should be either 0 or 11.
Therefore, 869484 →
Odd places = 8 + 9 + 8 = 25
Even places = 6 + 4 + 4 = 14
Difference = 25 – 14 = 11

These NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.2 Questions and Answers are prepared by our highly skilled subject experts.

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

Question 1.
What is the sum of any two
(a) Odd numbers?
(b) Even numbers?
Answer:
(a) The sum of any two odd numbers is an even numbers.
Example: 1 + 3 = 4, 3 + 5 = 8
(b) The sum of any two even numbers is also an even numbers.
Example: 2 + 4 = 6, 6 + 8 = 14

Question 2.
State whether the following statements are True or False:
(a) The sum of three odd numbers is even.
(b) The sum of two odd numbers and one even number is even.
(c) The product of three odd numbers is odd.
(d) If an even number is divided by 2, the quotient is always odd.
(e) All prime numbers are odd.
(f) Prime numbers do not have any factors.
(g) Sum of two prime numbers is always even.
(h) 2 is the only even prime number.
(i) All even numbers are composite numbers.
(j) The product of two even numbers is always even.
Answer:
(a) False
(b) True
(c) True
(d) False
(e) False
(f) False
(g) False
(h) True
(i) False
(j) True

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

Question 4.
Write down separately the prime and composite numbers less than 20.
Answer:
Prime numbers : 2, 3, 5, 7, 11, 13, 17, 19 Composite numbers : 4, 6, 8, 9, 10, 12, 14, 15, 16, 18

Question 5.
What is the greatest prime number between 1 and 10?
Answer:
2, 3, 5 and 7 are the prime numbers between 1 and 10.
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
Answer:
(a) 3+ 41 = 44
(b) 5 + 31 = 36
(c) 5 + 19 = 24
(d) 5 + 13 = 18

Question 7.
Give three pairs of prime numbers whose difference is 2.
[Remark: Two prime numbers whose difference is 2 are called twin primes].
Answer:
3 and 5;
5 and 7;
11 and 13

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

Question 9.
Write seven consecutive composite numbers less than 100 so that there is no prime number between them.
Answer:
Seven consecutive composite numbers are: 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
Ans. (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 less than 20 whose sum is divisible by 5. (Hint: 3 + 7=10)
Answer:
2 + 3 = 5;
2 + 13 = 15;
3 + 17 = 20;
7 + 13 = 20;
11 + 19 = 30

Question 12.
Fill in the blanks:
(a) A number which has only two factors is called a ………………
(b) A number which has more than two factors is called a …………………
(c) 1 is neither ………………… nor ……………………..
(d) The smallest prime number is …………………….
(e) The smallest composite number is ……………….
(f) The smallest even number is …………………
Answer:
(a) prime number
(b) composite number
(c) prime number and composite number
(d) 2
(e) 4
(f) 2

These NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Exercise 3.1

Question 1.
Write all the factors of the following numbers:
(a) 24
(b) 15
(c) 21
(d) 27
(e) 12
(f) 20
(g) 18
(h) 23
(i) 36
Answer:
(a) 24 = 1 × 24 = 2 × 12 = 3 × 8
= 4 × 6 = 6 × 4
∴ Factors of 24 = 1, 2, 3, 4, 6, 12, 24

(b) 15 = 1 × 15 = 3 × 5 = 5 × 3
∴ Factors of 15 = 1, 3, 5, 15

(c) 21 = 1 × 21 = 3 × 7 = 7 × 3
∴ Factors of 21 = 1, 3, 7, 21

(d) 27 = 1 × 27 = 3 × 9 = 9 × 3
∴ Factors of 27 = 1, 3, 9, 27

(e) 12 = 1 × 12 = 2 × 6 = 3 × 4 = 4 × 3
∴ Factors of 12 = 1, 2, 3, 4, 6, 12

(f) 20 = 1 × 20 = 2 × 10 = 4 × 5 = 5 × 4
∴ Factors of 20= 1,2,4, 5, 10, 20

(g) 18 = 1 × 18 = 2 × 9 = 3 × 6
∴ Factors of 18 = 1, 2, 3, 6, 9, 18

(h) 23 = 1 × 23
∴ Factors of 23 = 1, 23

(i) 36 = 1 × 36 = 2 × 18 = 3 × 12
= 4 × 9 = 6 × 6
∴ Factors of 36 = 1, 2, 3,4, 6, 9, 12,18, 36

Question 2.
Write first five multiples of:
(a) 5
(b) 8
(c) 9
Answer:
(a) 5 × 1 = 5, 5 × 2 = 10,
5 × 3 = 15, 5 × 4 = 20, 5 × 5 = 25
∴ First five multiples of 5 are 5, 10, 15, 20, 25.

(b) 8 × 1 = 8, 8 × 2 = 16, 8 × 3 = 24, 8 × 4 = 32, 8 × 5 = 40
∴ First five multiples of 8 are 8, 16, 24, 32, 40.

(c) 9 × 1 = 9, 9 × 2 = 18, 9 × 3 = 27, 9 x 4 = 36, 9 × 5 = 45
∴ First five multiples of 9 are 9, 18, 27, 36, 45.

Question 3.
Match the items in column 1 with the items in column 2:
Column 1 — Column 2
(i) 35 — (a) Multiple of 8
(ii) 15 — (b) Multiple of 7
(iii) 16 — (c) Multiple of 70
(iv) 20 — (d) Factor of 30
(v) 25 — (e) Factor of 50
— (f) Factor of 20
Answer:
(i) → (b)
(ii) → (d)
(iii) → (a)
(iv) → (f)
(v) → (e)

Question 4.
Find all the multiples of 9 up to 100.
Answer:
∴ All the multiples of 9 up to 100 are: 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99

These NCERT Solutions for Class 6 Maths Chapter 13 Symmetry Ex 13.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 13 Symmetry

Exercise 13.1

Question 1.
List any four symmetrical objects from your home or school.
Answer:
Notebook, Blackboard, Glass, Inkpot.

Question 2.
For the given figure, which one is the mirror line, l₁ or l₂?

Answer:
l₂ is the mirror line as both sides of the lines are symmetric.

Question 3.
Identify the shapes given below. Check whether they are symmetric or not. Draw the line of symmetry as well.
Answer:

Question 4.
Copy the following on a squared paper. A square paper is what you would have used in your arithmetic notebook in earlier classes. Then complete them such that the dotted line is the line of symmetry.


Answer:

Question 5.
In the figure, l is the line of symmetry. Complete the diagram to make it symmetric

Answer:

Question 6. In the figure, l is the line of symmetry. Draw the image of the triangle and complete the diagram, so that it becomes symmetric.

Answer:

These NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry InText Questions

NCERT In-text Question Page No. 286
Question 1.
In Step 2 of the construction using ruler and compasses, what would happen if we take the length of radius to be smaller than half the length of \(\overline{\mathrm{AB}}\) ?
Answer:
In case we take the radius smaller than half of the length of \(\overline{\mathrm{AB}}\), the arcs will not intersect each other at two points P and Q.

NCERT In-text Question Page No. 289
Question 1.
In Step 2 above, what would happen if we take radius to be smaller than half the length BC?
Answer:
If we take radius to be smaller than half of BC, the arcs drawn with centres B and C will not intersect each other.

NCERT In-text Question Page No. 290
Question 1.
How will your construct at 15° angle?
Answer:
Steps of construction:
Step I: Construct an angle ∠ABC of 60°.
Step II: Bisect ∠ABC to get an angle of 30°
i. e. ∠ABD = 30°.

Step III: Bisect ∠ABD, such that \(\overline{\mathrm{BE}}\) is bisector of ∠ABD.
Thus, ∠ABD = \(\frac { 1 }{ 2 }\)(30°) = 15°.

NCERT In-text Question Page No. 291 
Question 1.
How will you construct a 150° angle?
Answer:
Steps of construction:
Step I: Draw a line I and mark a point O on it.
Step II: With centre O and a convenient radius, draw an arc intersecting I at A.
Step III: With the same radius and centre at A, draw an arc to cut the first arc at B.
Step IV: Again with the same radius and centre at B, draw another arc to intersect the
first arc at C.

Step V : Once again with the same radius and centre at C, draw an arc to cut the first are at D.
Step VI: Now, bisect ∠COD, such that ∠COE = ∠EOD = 30°.
Step VII: Since, 150° = 120° + 30°, therefore ∠AOC + ∠COE = ∠AOE. Thus, ∠AOE is the required angle whose measure is 150°.

Question 2.
How will you construct a 45° angle?
Answer:
Steps of construction :
Step I: Construct an angle of 90° as shown in
the figure. ∠POQ = 90°.

Step II: Draw OR, the angle bisector of ∠POQ such that \(\frac { 1 }{ 2 }\)– [∠POQ] = \(\frac { 1 }{ 2 }\) (90°) = 45° or ∠POR = 45°