NCERT Solutions for Class 6 Maths

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

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 11 Algebra Ex 11.1 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.1.

• Algebra Class 6 Ex 11.2
• Algebra Class 6 Ex 11.3
• Algebra Class 6 Ex 11.4
• Algebra Class 6 Ex 11.5

NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.1

Question 1.
Find the rule which gives the number of matchsticks required to make the following matchstick patterns. Use a variable to write the rule.
(a) A pattern of letter T as
(b) A pattern of letter Z as
(c) A pattern of letter U as
(d) A pattern of letter V as
(e) A pattern of letter E as
(f) A pattern of letter S as
(g) A pattern of letter A as
Solution.
(a) Number of matchsticks required = 2n
(b) Number of matchsticks required = 3n
(c) Number of matchsticks required = 3n
(d) Number of matchsticks required = 2n
(e) Number of matchsticks required = 5n
(f) Number of matchsticks required = 5n
(g) Number of matchsticks required = 6

Question 2.
We already know the rule for the pattern of letters L, C and F. Some of the letters from Q. 1 (given above) give us the same rule as that given by L. Which are these? Why does this happen?
Solution.
These letters are T and V. This happens since the number>of matchsticks require,d in each of them is 2.

Question 3.
Cadets are marching in a parade. There are 5 cadets in a row. What is the rule, which gives the number of cadets, given the number of rows? (Use n for the number of rows.)
Solution.
The number of cadets = 5n.

Question 4.
If there are 50 mangoes in a box, how will you write the total number of mangoes in terms of the number of boxes? (Use b for the number of boxes.)
Solution.
Total number of mangoes = 50b.

Question 5.
The teacher distributes 5 pencils per student. Can you tell how many pencils are needed, given the number of students? (Use for the number of students.)
Solution.
Number of pencils needed = 5s

Question 6.
A bird flies 1 kilometer in one minute. Can you express the distance covered by the bird in tertns of its flying time in minutes? (Use t for flying time in minutes.)
Solution.
Yes! / kilometers
The bird flies in one minute = 1 kilometer
The bird flies in / minutes = 1 x t kilometers
= kilometers

Question 7.
Radha is drawing a dot Rangoli (a beautified pattern of lines joining dots with chalk powder. She has 9 dots in a row. How many dots will her Rangoli have for r rows? How many dots are there if there are 8 rows? If there are 10 rows?

Solution.
∵ Number of dots in 1 row = 9
∴ Number of dots in r rows = 9 x r=9r
Number of dots in 8 rows = 9 x 8 = 72
Number of dots in 10 row = 9 x 10 = 90

Question 8.
Leela is Radha’s younger sister. Leela is 4 years younger than Radha. Can you write Leela’s age in terms of Radha’s age? Take Radha’s age to be x years.
Solution.
Yes! we can write Leela’s age in terms of Radha’s age.
Age of Radha = x years
∵ Leela is 4 years younger than Radha.
∴ Age of Leela = (x – 4) years

Question 9.
Mother has made laddus. She gives some laddus to guests and family members; still, 5 laddus remain. If the number of laddus mother gave away is l, how many laddus did she make?
Solution.
Number of laddus given away to guests and family members = l
Number of laddus remained = 5
∴ Number of laddus she made = 1 + 5

Question 10.
Oranges are to be transferred from larger boxes into smaller boxes. When a large box is emptied, the oranges from it fill two smaller boxes and still, 10 oranges remain, outside. If the number of oranges in a small box is taken to be x, what is the number of oranges in the larger box?
Solution.
Let the number of oranges in a smaller box box
∴ Number of oranges in two smaller boxes = 2x
Number of oranges remained outside = 10
∴ Number of oranges in the larger box = 2x+ 10

Question 11.
(a) Look at the following matchstick pattern of squares (figure). The squares are not separate. Two neighboring squares have a common matchstick. Observe the pattern and find the rule that gives the number of matchsticks in terms of the number of squares.
(Hint: If you remove the vertical stick at the end, you will get a pattern of Cs.)

(b) Figure gives a matchstick pattern of triangles. Av in Exercise 11(a) above, find the general rule that gives the number of matchsticks in terms of the number of triangles.


Solution.
(a)

 Rule: Number of matchsticks required = 3x + I
where x is the number of squares.

(b)

Rule: Number of matchsticks required = 2x + 1,
where x is the number of triangles.

We hope the NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.1 help you. If you have any query regarding. NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.1, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 13 Symmetry Ex 13.1 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 13 Symmetry Ex 13.1.

NCERT Solutions for Class 6 Maths Chapter 13 Symmetry Ex 13.1

Question 1.
List any four symmetrical objects from your home or school.
Solution :
The blackboard, the table top, a pair of scissors, the computer disc.

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

Solution :
I₂ is the mirror line.

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

Solution :
(a) Symmetric

(b) Symmetric

(c) Not symmetric
(d) Symmetric

(e) Symmetric

(f) Symmetric

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.



Solution :

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

Solution :

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.

Solution :

We hope the NCERT Solutions for Class 6 Maths Chapter 13 Symmetry Ex 13.1 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 13 Symmetry Ex 13.1, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.1 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.1.

• Mensuration Class 6 Ex 10.2
• Mensuration Class 6 Ex 10.3

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.1

Question 1.
Find the perimeter of each of the following figures:



Solution :
(a) Perimeter
= 5 cm + I cm + 2 cm + 4 cm
= 12 cm
(b) Perimeter
= 40 cm + 35 cm + 23 cm + 35 cm
= 133 cm
(c) Perimeter
= 15 cm + 15 cm + 15 cm+ 15 cm
= 60 cm
(d) Perimeter
= 4 cm + 4 cm + 4 cm + 4 cm + 4 cm
= 20 cm
(e) Perimeter
= 1 cm + 4 cm + 0.5 cm + 2.5 cm + 2.5 cm + 0.5 cm + 4 cm
= 15 cm
(f) Perimeter = 1 cm + 3 cm + 2 cm + 3 cm + 4 cm + 1 cm + 3 cm + 2 cm + 3 cm + 4 cm + 1 cm + 3 cm + 2 cm + 3 cm + 4 cm + 1 cm + 3 cm + 2 cm + 3 cm + 4 cm
= 52 cm

Question 2.
The lid of a rectangular box of sides 40 cm by 10 cm is sealed all around with tape. What is the length of the tape required?
Solution :
Length of the tape required
= Perimeter of the rectangular box = 2 × (Length + Breadth)
= 2 × (40 cm + 10 cm)
= 2 × (50 cm)
= 100 cm = l m.

Question 3.
A table-top measures 2 m 25 cm by l m 50 cm. What is the perimeter of the table-top?
Solution :
Perimeter of the table-top
= 2 × (Length + Breadth)
= 2 × (2 m 25 cm + 1 m 50 cm)
= 2 × (2.25 m+1.50 m)
= 2 × (3.75 m)
= 7.5 m

Question 4.
What is the length of the wooden strip required to frame a photograph of length and breadth 32 cm and 21 cm respectively?
Solution :
Length of the wooden strip required = 2 × (Length + Breadth)
= 2 × (32 cm + 21 cm)
= 2 × (53 cm)
= 106 cm = 1.06 m.

Question 5.
A rectangular piece of land measures 0.7 km by 0.5 km. Each side is to be fenced with 4 rows of wires. What is the length of the wire needed?
Solution :
Perimeter of the rectangle
= 2 × (Length + Breadth )
= 2 × (0.7 km + 0.5 km)
= 2 × (1.2km)
= 2.4 km
Length of the wire needed
= 4 × Perimeter of the rectangle = 4 × (2.4 km)
= 9.6 km.

Question 6.
Find the perimeter of each of the following shapes:
(a) A triangle of sides 3 cm, 4 cm, and 5 cm
(b) An equilateral triangle of side 9 cm
(c) An isosceles triangle with equal sides 8 cm each a mi third side 6 cm.
Solution :
(a) Perimeter of the triangle = 3 cm + 4 cm + 5 cm = 12 cm
(b) Perimeter of the equilateral triangle = 3 × Length of a side = 3 × (9 cm) = 27 cm
(c) Perimeter of the isosceles triangle = 8 cm + 8 cm + 6 cm = 22 cm.

Question 7.
Find the perimeter of a triangle with sides measuring 10 cm, 14 cm, and 15 cm.
Solution :
Perimeter of the triangle
= Sum of the lengths of its three sides
= 10 cm + 14 cm + 15 cm
= 39 cm.

Question 8.
Find the perimeter of a regular hexagon with each side measuring 8 m.
Solution :
Perimeter of the regular hexagon
= 6 × Length of a side
= 6 × (8m)
= 48 m.

Question 9.
Find the side of the square whose perimeter is 20 m.
Solution :
Perimeter of the square
= 4 × Length of a side
⇒ Length of one side

Question 10.
The perimeter of a regular pentagon is 100 cm. How long is its every side?
Solution :
Perimeter of the regular pentagon = 5 × Length of a side
⇒ Length of one (each) side

Question 11.
A piece of string is 30 cm long. What will be the length of each side if the string is used to form:
(a) a square?
(b) an equilateral triangle?
(c) a regular hexagon?
Solution :
(a) Perimeter of the square = 4 × Length of a side
⇒ Length of a side

(b) Perimeter of the equilateral triangle
= 3 × Length of a side
⇒ Length of a side


(c) Perimeter of the regular hexagon = 6 × Length of a side.
⇒ Length of a side

Question 12.
Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is the third side?
Solution :
Perimeter of a triangle = Sum of the lengths of its three sides
⇒ 36 cm = 12 cm + 14 cm + Length of the third side
⇒ 36 cm = 26 cm + Length of the third side
⇒ Length of the third side = 36 cm – 26 cm = 10 cm.

Question 13.
Find the cost of fencing a square park of side 250 m at the rate of ₹ 20 per meter.
Solution :
Perimeter of the square park = 4 × Length of a side = 4 × (250m)
= 1000 m
∴ Cost of fencing the square park at the rate of?
20 per metre = ₹ 1000 × 20
= ₹ 20,000

Question 14.
Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of? 12 per meter.
Solution :
Perimeter of the rectangular park = 2 × (Length + Breadth)
= 2 × (175m + 125 m)
= 2 × (300 m)
= 600 m
Cost of fencing the rectangular park at the rate of ?
12 per metre = ₹ 600 × 12
= ₹ 7200.

Question 15.
Sweety runs around a square park of side 75 m. Bulbul runs around a rectangular park with length 60 m and breadth 45 m. Who covers less distance?
Solution :
Perimeter of the square park = 4 × Length of a side = 4 × (75 m)
= 300 m
Perimeter of the rectangular park
= 2 × (Length + Breadth)
= 2 × (60 m + 45 m)
= 2 × (105 m)
= 210 m.
Since, the perimeter of the rectangular park is less than the perimeter of the square park, therefore. Bulbul covers less distance.

Question 16.
What is the perimeter of each of the following figures? What do you infer from the answers?


Solution :
(a) Perimeter
= Sum of the lengths of all the sides
= 25 cm + 25 cm + 25 cm + 25 cm
= 100 cm.

(b) Perimeter
= Sum of the lengths of all the sides
= 40 cm + 10 cm + 40 cm + 10 cm
= 100 cm.

(c) Perimeter
= Sum of the lengths of all the sides
= 30 cm + 20 cm + 30 cm + 20 cm
= 100 cm.

(d) Perimeter
= Sum of the lengths of all the sides
= 30 cm + 40 cm + 30 cm
= 100 cm.
The inference from the answers. All the figures have the same perimeter.

Question 17.
Avneet buys 9 square paving slabs, each with a side of \(\frac { 1 }{ 2 }\) m. He lays them in the form of a square.

(a) What is the perimeter of his arrangement (Fig. i)?
(b) Shari does not like his arrangement. She gets him to lay them out like a cross. What is the perimeter of her arrangement (Fig. ii)?
(c) Which has a greater perimeter?
(d) Avneet wonders if there is a way of getting an even greater perimeter. Can you find a way of doing this? (The paving slabs must meet along complete edges i.e. they cannot be broken.)
Solution :
(a) Perimeter of his arrangement = 4 × Length of one side

(b) Perimeter of her arrangement = Sum of the lengths of all the sides


(c) Cross has greater perimeter.
(d) Yes ! there is a way of getting an even greater perimeter. It is shown below:

We hope the NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.1 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.1, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Ex 9.1 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Ex 9.1.

• Data Handling Class 6 Ex 9.2
• Data Handling Class 6 Ex 9.3
• Data Handling Class 6 Ex 9.4

NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Ex 9.1

Question 1.
In a Mathematics test, the following marks were obtained by 40 students. Arrange these marks in a table using, tally marks.

(a) Find how many students obtained marks equal to or more than 7.
(b) How many students obtained marks below 4?
Solution.

(a) 5 + 4 + 3=12 students obtained marks equal to or more than 7.
(b) 3 + 3 + 2 = 8 students obtained marks below 4.

Question 2.
Following is the choice of sweets of 30 students of Class VI
Ladoo, Barfi, Ladoo, Jalebi, Ladoo, Rasgulla, Jalebi, Ladoo, Barfi, Rasgulla, Ladoo, Jalebi Jalebi, Rasgulla, Ladoo, Rasgulla, Jalebi, Ladoo, Rasgulla, Ladoo, Ladoo, Barfi, Rasgulla, Rasgulla, Jalebi, Rasgulla, Ladoo, Rasgulla, Jalebi, Ladoo.
(a) Arrange the names of sweets in a table using tally marks.
(b) Which sweet is preferred by most of the students?
Solution.
(a)

(b) Ladoo is preferred by most of the students.

Question 3.
Make a table and enter the data using tally- marks. Find the number that appeared.
(a) The minimum number of times.
(b) The maximum number of times.
(c) Find those numbers that appear an equal number of limes.
Solution.

(a) The number that appeared the minimum number of times is 4.
(b) The number that appeared the maximum number of times is 5.
(c) The numbers that appeared an equal number of times are 1 and 6.

Question 4.
Following pictograph shows the number of tractors in five villages:

Observe the pictograph and answers the following questions:
(i) Which village has the minimum number of tractors?
(ii) Which village has the maximum number of tractors?
(iii) How many more tractors village C has as compared to village B?
(iv) What is the total number of tractors in all the five villages?
Solution.
(i) Village D has the minimum number of tractors.
(ii) Village C has the maximum number of tractors.
(iii) Village C has 8-5 = 3 more tractors as compared to village B.
(iv) Total number of tractors in all the five villages = 6 +5+ 8 + 3 + 6 = 28.

Question 5.
The number of girl students in each class of a co-educational middle school is depicted by the pictograph:

Observe this pictograph and answer the following questions:
(a) Which class has the minimum number of girl students?
(b) Is the number of girls in class VI less than the number of girls in class V?
(c) How many girls are there in VII class?
Solution.
(a) Class VIII has the minimum no. of girl students.
(b) No! the number of girls in class VI is not less than the number of girls in class V.
(c) Number of girls in class VII – 3 x 4 = 12.

Question 6.
The sale of electric bulbs on different days of a week is shown below:

What can we conclude from the said pictograph?
Observe the pictograph and answer the following questions:
(a) How many bulbs were sold on Friday?
(b) On which day the maximum number of bulbs were sold?
(c) If one bulb was sold at the rate off 10, what was the total sale on Sunday?
(d) Can you find out the total sale of the week?
(e) If one big carton can hold 9 bulbs, how many cartons were needed in the given week?
Solution.
(a) Number of bulbs sold on Friday = 7×2 = 14.
(b) The maximum number of bulbs were sold on Sunday.
(c) Number of bulbs sold on Sunday
= 9 x 2=18.
∴ Total sale on Sunday
= Rs.18 x 10 = Rs. 180.
(d) Total number of bulbs sold in the week
= (6 + 8 + 4 + 5 + 7 + 4 + 9) x 2
= 43 x 2 = 86.

Hence, 10 cartons were needed in the given week.

Question 7.
In a village six fruit merchants sold the following number of fruit baskets in a particular season:

Observe this pictograph and answer the following questions:
(a) Which merchant sold the maximum number of baskets?
(b) How many fruit baskets were sold by Anwar?
(c) The merchants who have sold 600 or number of baskets are planning to buy a godown for the next season. Can you name them?
Solution.
(a) Martin sold the maximum number of baskets.
(b) 7 x 100 = 700 fruit baskets were sold by Anwar.
(c) Yes! Anwar. Martin and Ranjit Singh are planning to buy a godown for the next season.

We hope the NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Ex 9.1 help you. If you have any query regarding. NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Ex 9.1, drop a comment below and we will get back to you at the earliest.