CBSE Class 6

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

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers InText Questions

NCERT In-text Question Page No. 48

Question 1.
Find the possible factors of 45, 30 and 36.
Answer:
The possible factor of
45 = 1,3,5,9,15,45
30 = 1,2,3,5,6,10,15,30
36 = 1,2,3,4,6,9,12,18,36

NCERT In-text Question Page No. 52

Question 1.
Observe that 2 × 3 + 1 = 7 is a prime number. Here, 1 has been added to a multiple of 2 to get a prime number. Can you find some more numbers of this type?
Answer:
We can have:
2 × 2 + 1 = 5, which is a prime number.
2 × 5 + 1 = 11, which is a prime number.
2 × 6 + 1 = 13, which is a prime number.
2 × 8 + 1 = 17, which is a prime number.
2 × 9 + 1 = 19, which is a prime number.
2 × 11 + 1 = 23, which is a prime number.

NCERT In-text Question Page No. 58

Question 1.
Find the common factors of
(a) 8,20
(b) 9,15
Answre:
(a) 8, 20
Since, 8 = 1 × 8
8 = 2 × 4
∴ All the factors of 20 are: 1, 2, 4 and 8 …(i)
Again, 20 = 1 × 20
20 = 2 × 10
20 = 4 × 5
∴ All the factors of 20 are: 1, 2, 4, 5, 10 and 20 …(ii)
From (i) and (ii), common factors of 8 and 20 are: 1, 2 and 4.

(b) 9,15
We have: 9 = 1 × 9
9 = 3 × 3
∴ All the factors of 9 are: 1, 3 and 9 …(i)
Again, 15 = 1 × 15
15 = 3 × 5
∴ All the factors of 15 are: 1, 3, 5 and 15 …(ii)
From (i) and (ii), common factors of 9 and 15 are: 1 and 3.

NCERT In-text Question Page No. 61

Question 1.
Write the prime factorisations of 16, 28, 38.
Answer:
(i) 16
We have:

∴ Prime factorisation of 16 = 2 × 2 × 2 × 2

(ii) 28
We have:

∴ Prime factorisation of 28 = 2 × 2 × 7

(iii) 38
We have:

∴ Prime factorisation of 38 = 2 × 19

NCERT In-text Question Page No. 63

Question 1.
Find the HCF of the following:
(i) 24 and 36
(ii) 15, 25 and 30
(iii) 8 and 12
(iv) 12,16 and 28
Answer:
(i) 24 and 36
∵ Factors of 24 are: 1, 2, 3, 4, 6, 12 and 24
Factors of 36 are: 1, 2, 3, 4, 6, 12, 18 and 36
∵ Common factors are: 1, 2, 3, 4, 6 and 12
Since, the highest of these common factors = 12
∴ HCF of 24 and 36 = 12

(ii) 15, 25 and 30
∵ Factors of 15 are: 1, 3, 5 and 15
Factors of 25 are: 1, 5 and 25
Factors of 30 are: 1, 2, 3, 5, 6, 10, 15 and 30
∴ Common factors are: 1 and 5
Since, the highest common factors is 5,
HCF of 15, 25 and 30 = 5

(iii) 8 and 12
∵ Factors of 8 are: 1, 2, 4 and 8
Factors of 12 are: 1, 2, 3, 4, 6 and 12
∴ Common factors are: 1, 2, 4
Since, the highest common factors is 4,
∴ HCF of 8 and 12 = 4

(iv) 12,16 and 28
∵ Factors of 12 are: 1, 2, 3, 4, 6 and 12
Factors of 16 are: 1, 2, 4, 8 and 16
Factors of 28 are: 1, 2, 4, 7, 14 and 28
∴ Common factors are: 1, 2, 4
Since, the highest common factors is 4,
∴ HCF of 12, 16 and 28 = 4

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

NCERT Solutions for Class 6 Maths Chapter 13 Symmetry

InText Questions

NCERT In-text Question Page No. 262
Question 1.
You have two set-squares in your ‘mathematical instruments box’. Are they symmetric?
Answer:
The two set-squares are given as :

The 30° – 60° – 90° set square is not symmetrical.
The 45° – 45° – 90° set square is not symmetrical.

NCERT In-text Question Page No. 264
Question 1.
Form as many shapes as you can by combining two or more set squares. Draw them on squared paper and note their lines of symmetry?
Answer:
(i) By combining two identical 30° – 60° – 90° set squares, we get the following symmetric figures :

(a) l₂ and l₂ are the lines of symmetry.

(b) Here, there is only line of symmetry which is shown by ‘m’.

(c) Here, the dotted line ‘p’ is the only line of symmetry.

(d) Here, the dotted line ‘q’ is the only line of symmetry.

(ii) By combining two identical 45° – 45° – 90° set squares, we get the following
symmetric figures.
(a) Here l₁, l₂, l₃, l₄, are lines of symmetry.

(b) In the figure formed, there is only one line of symmetry which is shown by ‘r’.

(iii) By combining three 45° – 45° – 90° identical set-squares, we can have an isosceles trapezium as shown here:

Here, the line of symmetry is shown by the dotted line ‘q’.

NCERT In-text Question Page No. 270
Question 1.
If you are 100 cm in front of a mirror, where does your image appear to be? If you move towards the mirror, how does your image move?
Answer:
If we are 100 cm in front of a plane mirror our image will be 200 cm away from it (100 cm behind the mirror +100 cm is the position of the object). If we move towards the mirror our image will also move towards as but no change in the size.

These NCERT Solutions for Class 6 Maths Chapter 13 Symmetry Ex 13.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 13 Symmetry

Exercise 13.3

Question 1.
Find the number of lines of symmetry in each of the following shapes. How will you check your answers?
Answer:

Question 2.
Copy the following drawing on squared paper. Complete each one of them such that the resulting figure has two dotted lines as two lines of symmetry.
How did you go about completing the picture?

Question 3.
How did you go about completing the picture?
Answer:

Question 4.
In each figure below, a letter of alphabet is shown along with a vertical line. Take
the mirror image of the letter in the given line. Find which letters look the same after reflection (i.e., which letters look the same in the image) and which do not. Can you guess why?

Answer:

These NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.7 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Exercise 3.7

Question 1.
Renu purchases two bags of fertilizer of weights 75 kg and 69 kg. Find the maximum value of weight which can measure the weight of the fertilizer exact number of times.
Answer:
For finding maximum weight, we have to find H.C.F. of 75 and 69.
Factors of 75 = 3 × 5 × 5
Factors of 69 = 3 × 23
H.C.F. = 3
Therefore, 3 kg is the maximum value of weight which can measure the weight of the fertiliser exact number of times.

Question 2.
Three boys step off together from the same spot. Their steps measure 63 cm, 70 cm and 77 cm respectively. What is the minimum distance each should cover so that all can cover the distance in complete steps?
Answer:
For finding minimum distance, we have to find L.C.M of 63, 70 and 77.

L.C.M. of 63, 70 and 77
= 7 × 9 × 10 × 11 = 6930 cm.
Therefore, the minimum distance is 6930 cm.

Question 3.
The length, breadth and height of a room are 825 cm, 675 cm and 450 cm respectively. Find the longest tape which can measure the three dimensions of the room exactly.
Answer:
The measurement of longest tape = H.C.F. of 825 cm, 675 cm and 450 cm. Factors of 825 = 3 × 5 × 5 × 11 Factors of 675 = 3 × 5 × 5 × 3 × 3
Factors of 450 = 2 × 3 × 3 × 5 × 5
H.C.F. = 3 × 5 × 5 = 75 cm
Therefore, the longest tape is 75 cm.

Question 4.
Determine the smallest 3-digit number which is exactly divisible by 6, 8 and 12.
Answer:

L.C.M. of 6, 8 and 12
= 2 × 2 × 2 × 3 = 24
The smallest 3-digit number =100
To find the number, we have to divide 100 by 24
100 = 24 × 4 + 4
Therefore, the required number = 100 + (24 – 4) = 120.

Question 5.
Determine the greatest 3-digit number which is exactly divisible by 8,10 and 12.
Answer:

L.C.M. of 8,10, 12 = 2 × 2 × 2 × 3 × 5 = 120
The largest three digit number = 999
To find the number, we have to divide 999 by 120
999 = 120 × 8 + 39
Therefore, the required number = 999 – 39 = 960

Question 6.
The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. If they change simultaneously at 7 a.m. at what time will they change simultaneously again?
Answer:

L.C.M. of 48, 72, 108
= 2 × 2 × 2 × 2 × 3 × 3 × 3 = 432 seconds
After 432 seconds, the lights change simultaneously.
432 seconds = 7 minutes 12 seconds
Therefore, the time = 7 a.m. + 7 minutes 12 seconds = 7:07:12 a.m.

Question 7.
Three tankers contain 403 litres, 434 litres and 465 litres of diesel respectively. Find the maximum capacity of a container that can measure the diesel of three containers exact number of times.
Answer:
The maximum capacity of container = H.C.F. (403, 434, 465)
Factors of 403 = 13 × 31
Factors of 434 = 2 × 7 × 31
Factors of 465 = 3 × 5 × 31
H.C.F. = 31
Therefore, 31 litres of container is required to measure the quantity

Question 8.
Find the least number which when divided by 6, 15 and 18, leave remainder 5 in each case.
Answer:

L.C.M. of 6, 15 and 18
= 2 × 3 × 3 × 5 = 90
Therefore, the required number.
= 90 + 5 = 95.

Question 9.
Find the smallest 4-digit number which is divisible by 18, 24 and 32.
Answer:

L.C.M. of 18, 24 and 32
= 2 × 2 × 2 × 2 × 2 × 3 × 3 = 288
The smallest four-digit number = 1000
To find the number, we have to divide 1000 by 288
1000 = 288 × 3 + 136
Therefore, the required number is 1000+ (288 – 136) = 1152.

Question 10.
Find the L.C.M. of the following numbers:
(a) 9 and 4
(b) 12 and 5
(c) 6 and 5
(d) 15 and 4
Observe a common property in the obtained L.C.Ms. Is L.C.M. the product of two numbers in each case?
Answer:
(a) L.C.M. of 9 and 4
= 2 × 2 × 3 × 3 = 36

(b) L.C.M. of 12 and 5 = 2 × 2 × 3 × 5 = 60

(c) L.C.M. of 6 and 5 = 2 × 3 × 5 = 30 6,5

(d) L.C.M. of 15 and 4 = 2 × 2 × 3 × 5 = 60

Yes, L.C.M. is equal to the product of rbers in each case. And L.C.M. is multiple of 3.

Question 11.
Find the L.C.M. of the following numbers in which one number is the factor of other:
(a) 5,20
(b) 6,18
(c) 12,48
(d) 9,45
What do you observe in the result obtained?
Answer:
(a) L.C.M. of 5 and 20 = 2 × 2 × 5 = 20

(b) L.C.M. of 6 and 18 = 2 × 3 × 3 = 18

(c) L.C.M. of 12 and 48 = 2 × 2 × 2 × 2 × 3 = 48

(d) L.C.M. of 9 and 45 = 3 × 3 × 5 = 45

From these all cases, we can conclude that if the smallest number if the factor of largest number, then the L.C.M. of these two numbers is equal to that of larger number.

These NCERT Solutions for Class 6 Maths Chapter 13 Symmetry Ex 13.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 13 Symmetry

Exercise 13.2

Question 1.
Find the number of lines of symmetry for each of the following shapes:

Answer:
(a) 4
(b) 4
(c) 4
(d) 1
(e) 6
(f) 4
(g) 0
(h) 0
(i) 3

Question 2.
Copy the triangle in each of the following figures on squared paper. In each case, draw the line(s) of symmetry, if any and identify the type of triangle. (Some of you may like to trace the figures and try
paper-folding first!)

Answer:
(a) l₁ is the line of symmetry.
(a) l₁ is the line of symmetry.
(b) l₁ is the line of symmetry.
(c) No line of symmetry.

Question 3.
Complete the following table.
Answer:

Question 4.
Can you draw a triangle which has
(a) exactly one line of symmetry?
(b) exactly two lines of symmetry?
(c) exactly three lines of symmetry?
(d) no lines of symmetry?
Sketch a rough figure in each case.
Answer:
(a) Yes, Isosceles triangle

(b) No, such triangle cannot be formed.

(c) Yes, Equilateral triangle

(d) Yes, Scalene triangle

Question 5.
On a squared paper, sketch the following:
(a) A triangle with a horizontal line of symmetry but no vertical line of symmetry.
(b) A quadrilateral with both horizontal and vertical lines of symmetry.
(c) A quadrilateral with a horizontal line of symmetry but no vertical line of symmetry.
(d) A hexagon with exactly two lines of symmetry.
(e) A hexagon with six lines of symmetry.
(Hint: It will be helpful if you first draw the lines of symmetry and then complete the figures.)
Answer:

Question 6.
Trace each figure and draw the lines of symmetry, if any:


Answer:

Question 7.
Consider the letters of English alphabet, A to Z. List among them the letters which have
(a) vertical lines of symmetry (like A)
(b) horizontal lines of symmetry (like B)
(c) no lines of symmetry (like Q)

Answer:
Vertical lines: A, H, I, M, O, T, U, V, W, X, Y
Horizontal lines: B, C, D, E, H, I, K, O, X
No line of symmetry: F, G, J, N, P, Q, R, S, Z

Question 8.
Given here are figures of a few folded sheets and designs drawn about the fold.
In each case, draw a rough diagram of the complete figure that would be seen when the design is cut off.

Answer:

These NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.5 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Exercise 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 also 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 is exactly divides the sum of two numbers, it must exactly divide the two numbers separately.
Answer:
Statements (b), (d), (g) and (h) are true.

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

Answer:

Since, 6 = 2 × 3
and 10 = 5 × 2

Since, 60 = 30 × 2
30= 10 × 3
10 = 5 × 2

Question 3.
Which factors are not included in the prime factorisation of a composite number?
Answer:
1 and the number it self are not included in the prime factorisation of a composite number.

Question 4.
Write the greatest 4-digit number and express it in terms of its prime factors.
Answer:
The greatest 4-digit number = 9999

The prime factors of 9999 are 3 × 3 × 11 × 101.

Question 5.
Write the smallest 5-digit number and express it in the form of its prime factors.
Answer:
The smallest five digit number is 10000.

The prime factors of 10000 are 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.
Answer:
Prime factors of 1729 are 7 × 13 × 19.

The difference of two consecutive prime factors is 6.

Question 7.
The product of three consecutive numbers is always divisible by 6. Verify this statement with the help of some examples.
Answer:
Among the three consecutive numbers, there must be one even number and one multiple of 3. Thus, the product must be multiple of 6.
Example: (i) 2 × 3 × 4 = 24
(ii) 4 × 5 × 6 = 120

Question 8.
The sum of two consecutive odd numbers is divisible by 4. Verify this statement with the help of some examples.
Answer:
3 + 5 = 8 and 8 is divisible by 4.
5 + 7 = 12 and 12 is divisible by 4.
7 + 9 = 16 and 16 is divisible by 4.
9 + 11 = 20 and 20 is divisible by 4.

Question 9.
In which of the following expressions, prime factorisation has been done?
(a) 24 = 2 × 3 × 4
(b) 56 = 7 × 2 × 2 × 2
(c) 70 = 2 × 5 × 7
(d) 54 = 2 × 3 × 9
Answer:
In expressions (b) and (c), prime factorisation has 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.
Answer:
The prime factorization of 45 = 5 × 9
25110 is divisible by 5 as ‘0’ is at its unit place.
25110 is divisible by 9 as sum of digits is divisible by 9.
Therefore, the number must be 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 4 and 6. Can we say that the number must be divisible by 4 × 6 = 24? If not, give an example to justify your answer.
Answer:
No, Number 12 is divisible by both 6 and 4 but 12 is not divisible by 24.

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