NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.3

NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.3 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.3.

Board CBSE
Textbook NCERT
Class Class 6
Subject Maths
Chapter Chapter 11
Chapter Name Algebra
Exercise  Ex 11.3
Number of Questions Solved 6
Category NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.3

Question 1.
Make up as many expressions with numbers (no variables) as you conform with three numbers 5, 7 and 8. Every number should be used not more than once. Use only addition, subtraction and multiplication.(Hint: Three possible expressions are 5 + (8­7), 5 – (8-7), 5 x 8 + 7; make the other expressions.)
Solution.
The possible expressions are 5 + (8 – 7), 5 – (8 – 7), (5 + 8) + 7, 5 + (8 + 7),
5 x 8 + 7, 5 x 7 + 8, 5 x 8-7, 5 x 7-8,
5 x (8 – 7), 5 x (8 + 7), 8 x (7 – 5), 8 x (7 + 5) etc.

Question 2.
Which out of the following are expressions with numbers only ?
(a) y + 3
(b) 7 x 20- 8z
(c) 5(21 -7)+ 7 x 2
(d) 5
(e) 3x
(f) 5 – 5n
(g) 7 x 20 – 5 x 10 – 45 + p.
Solution.
(c), (d).

Question 3.
Identify the operations (addition, subtraction, division, multiplication) in forming the following expressions and tell how the expressions have been formed:
(a) z + 1, z-1, y + 17, y-17
(b) 17y, \(\frac { y }{ 17 } \), 5z
(c) 2y + 17, 2y – 17
(d) 7m, -7m + 3, -7m- 3.
Solution.
(a) Addition, subtraction, addition, subtraction
(b) Multiplication, division, multiplication.
(c) Multiplication and addition, multiplication and subtraction.
(d) Multiplication, multiplication and addition,multiplication and subtraction.

Question 4.
Give expressions for the following cases :
(a) 7 added to p
(b) 7 subtracted from p
(c) p multiplied by 7
(d) p divided by 7
(e) 7 subtracted from – m
(f) -p multiplied by 5
(g) -p divided by 5.
(h) p multiplied by – 5.
Solution.
(a) p + 7
(b) p-1
(c) 7p
(d) \(\frac { p }{ 7 } \)
(e) -m-1
(f) -5p
(g) – \(\frac { p }{ 5 } \)
(h) – 5p.

Question 5.
Give expressions in the following cases :
(a) 11 added to 2m
(b) 11 subtracted from 2m
(c) 5 times y to which 3 is added
(d) 5 times v from which 3 is subtracted
(e) y is multiplied by – 8
(f) y is multiplied by – 8 and then 5 is added to the result
(i) y is multiplied by 5 and the result is subtracted from 16
(j) y is multiplied by-5 and the result is added to 16.
Solution.
(a) 2m +11
(b) 2m – 11
(c) 5y + 3
(d) 5v – 3
(e) -8v
(f) -8y + 5
(g) 16 – 5y
(h) -5y + 16.

Question 6.
(a) Form expressions using t and 4. Use not more than one number operation. Every expression must have t in it.
(b) Form expressions using y, 2 and 7. Every expression must have y in it. Use only two number operations. These should be different.
Solution.
(a) t + 4, t – 4,4t, \(\frac { t }{ 4 } \) , \(\frac { 4 }{ t } \)  , 4 -1,4 +1 4 t
(b) 2y + 7, 2y – 7, 7y + 2

 

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NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.2

NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.2 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.2.

Board CBSE
Textbook NCERT
Class Class 6
Subject Maths
Chapter Chapter 11
Chapter Name Algebra
Exercise  Ex 11.2
Number of Questions Solved 5
Category NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.2

Question 1.
The side of an equilateral triangle is shown by l. Express the perimeter of the equilateral triangle using l.
Solution.
Perimeter (p) of the equilateral triangle with side l = Sum of the lengths of sides of the equilateral triangle = 1 + 1 + 1 = 31.

Question 2.
The side of a regular hexagon (Fig.) is denoted by l. Express the perimeter of the hexagon using l. (Hint: A regular hexagon has all its six sides equal in length.)
NCERT Solutions for Class 6 Maths Chapter 11 Algebra 14
Solution.
Perimeter (p) of the regular hexagon with side
= Sum of the lengths of all sides of the regular hexagon
=l+l+l+l+l+l
= 6 l.

Question 3.
A cube is a three-dimensional figure as shown in Fig. It has six faces and all of them are identical squares. The length of an edge of the cube is given by l. Find the formula for the total length of the
edges of a cube.
NCERT Solutions for Class 6 Maths Chapter 11 Algebra 15
Solution.
Total length (L) of the edges of a cube
= Sum of the lengths of all (12) edges of the cube
= l+l+l+l+l+l+l+l+l+l+l+l
=12 l

Question 4.
The diameter of a circle is a line which joins two points on the circle and also passes through the center of the circle. (In the adjoining figure (Fig.) AB is a diameter of the circle; C is its center.) Express the diameter of the circle (d) in terms of its radius (r).
NCERT Solutions for Class 6 Maths Chapter 11 Algebra 16
Solution.
AB = AC + CB
⇒ d=r+r ⇒ d=2r

Question 5.
To find the sum of three numbers 14, 27 and 13, we can have two ways.
(a) We may first add 14 and 27 to get 41 and then add 13 to it to get the total sum 54 or
(b) We may add 27 and 13 to get 40 and then add 14 to get the sum 54. Thus,
(14 + 27) + 13 = 14 + (27 + 13)
This can be done for any three numbers. This property is known as the associativity of the addition of
numbers. Express this property which we have already studied in the chapter on Whole Numbers, in a general way, by using variables a, b and c.
Solution.
Let a, b and c be three variables, each of which can take any numerical value. Then.
(a + b) + c = a + (b + c)
This property is known as the associativity of the addition of numbers.

 

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NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.3

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.3 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.3.

Board CBSE
Textbook NCERT
Class Class 6
Subject Maths
Chapter Chapter 10
Chapter Name Mensuration
Exercise  Ex 10.3
Number of Questions Solved 12
Category NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.3

Question 1.
Find the areas of the rectangles whose sides are:
(a) 3 cm and 4 cm
(b) 12 m and 21m
(c) 2 km and 3 km
(d) 2 m and 70 cm.
Solution :
(a) Area of the rectangle
= Length × Breadth = 3 × 4 sq cm = 12 sq cm
(b) Area of the rectangle = Length × Breadth
= 12 m × 21 m = 252 sq m
(c) Area of the rectangle = Length × Breadth
= 2 km × 3 km = 6 sq km
(d) 2 m = 2 × 100 cm = 200 cm
Area of the rectangle = Length × Breadth
= 200 × 70 sq cm
14000 sq cm.

Question 2.
Find the areas of the squares whose sides are:
(a) 10 cm
(b) 14 cm
(c) 5 m.
Solution :
(a) Area of the square = Side × Side
= 10 cm × 10 cm – 100 sq cm
(b) Area of the square = Side × Side
= 14 cm × 14 cm = 196 sq cm
(c) Area of the square = Side × Side
= 5 m × 5 m = 25 sq m.

Question 3.
The length and breadth of three rectangles are as given below:
(a) 9 m and 6 m
(b) 3 m and 17 m
(c) 4 m and 14 m.
Which one has the largest area and which one has the smallest?
Solution :
(a) Area of the rectangle = Length × Breadth = 9 m × 6 m = 54 sq m
(b) Length of the rectangle = Length × Breadth = 3 m × 17 m = 51 sq m
(c) Length of the rectangle = Length × Breadth = 4m × 14m = 56 sq m
The rectangle (c) has the largest area and the rectangle (b) has the smallest area.

Question 4.
The area of a rectangular garden 50 m long is 300 sq. m. Find the width of the garden.
Solution :
Area of the rectangular garden = 300 sq m Length of the rectangular garden = 50 m
∴ Width of the rectangular garden
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration 17

Question 5.
What is the cost of tiling a rectangular piece of land 500 m long and 200 m wide at the rate of ₹ 8 per hundred sq m.
Solution :
Length of the rectangular piece of land
500 m
Breadth of the rectangular piece of land
200 m
∴ Area of the rectangular piece of land
= Length × Breadth
= 500 m × 200 m
= 100000 sq m
Cost of tiling 100 sq m = ₹ \(\frac{ 8 }{ 100 }\)
Cost of tiling 100000 sq m
= ₹ \(\frac{ 8 }{ 100 }\) × 100000
= ₹ 8000.

Question 6.
A table-top measures 2 m by 1 m 50 cm. What is its area in square meters?
Solution :
Length of the table-top = 2 m Breadth of the table-top = 1 m 50 cm = 1.50 m
∴ Area of the table-top
= Length × Breadth = 2 m × 1.50 m
= 3.0 sq m.

Question 7.
A room is 4 m long and 3 m 50 cm wide. How many square meters of carpet is needed to cover the floor of the room?
Solution :
Length of the room = 4 m Breadth of the room = 3 m 50 cm = 3.50 m
∴ Area of the room = Length × Breadth
= 4 × 3.5 sq. m
= 14.0 sq m
Hence, 14.0 square meters of carpet is needed to cover the floor of the room.

Question 8.
A floor is 5 m long and 4 m wide. A square carpet of sides 3 m is laid on the floor. Find the area of the floor that is not carpeted.
Solution :
Length of the floor = 5 m Breadth of the floor = 4 m
∴ Area of the floor = Length × Breadth
= 5m × 4m = 20sqm Area of the square carpet = Side × Side
= 3m × 3m = 9sqm
∴ Area of the floor that is not carpeted = 20 sq m – 9 sq m = 11sq m.

Question 9.
Five square flower beds each of sides 1 m are dug on a piece of land 5 m long and 4 m wide. What is the area of the remaining part of land?
Solution :
Area of square flower bed
= Side × Side = 1 m × 1 m
= 1 sq m
∴ Area of 5 square flower beds
= 5 × 1 sq m = 5 sq m
Length of the piece of land = 5 m
Breadth of the piece of land = 4 m
∴ Area of the piece of land
= Length × Breadth = 5 m × 4 m
= 20 sq m
∴ Area of the remaining part of the land
= 20 sq m – 5 sq m
= 15 sqm.

Question 10.
By splitting the following figures into rectangles, find their areas (The measures are given in centimeters).
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration 18
Solution :
(a) Area of the figure
= (3 × 3 + 1 × 2 + 3 × 3 + 4 × 2) sqcm
= (9 + 2 + 9 + 8) sq cm = 28 sq cm
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration 19
(b) Area of the figure
= (3 × 1 + 3 × 1 + 3 × 1) sqm
= (3 + 3 + 3) sq m
= 9 sqm
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration 20

Question 11.
Split the following shapes into rectangles and find the area of each (The measures are given in centimeters).
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration 21
Solution :
(a) Area of the shape
= (8 × 2 + 12 × 2) sq cm = (16 + 24) sq cm = 40 sq cm
(b) Area of the shape
= (17 × 7 + 7 × 7 + 7 × 7 + 7 × 7 + 7 × 7) sq cm
= (49 + 49 + 49 + 49 + 49) sq cm
= 245 sq cm
(c) Area of the shape = ( 5 × 1 + 4 × 1) sq cm
= (5 + 4) sq cm
= 9 sq cm.

Question 12.
How many tiles whose length and breadth are 12 cm and 5 cm respectively will be needed to fit in a rectangular region whose length and breadth are respectively :
(a) 100 cm and 144 cm?
(b) 70 cm and 36 cm?
Solution :
(a) Length of the region = 100 cm Breadth of the region = 144 cm
∴ Area of the region = Length × Breadth
= 100 cm × 144 cm
= 14400 sq cm
Length of a tile = 12 cm Breadth of a tile = 5 cm
∴ Area of a tile = Length × Breadth = 12 × 5 sq cm
= 60 sq cm
∴ Number of tiles needed to fit the region
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration 22

(a) Length of the region = 70 cm Breadth of the region = 36 cm
∴ Area of the rectangular region = 70 × 36 sq cm
= 2520 sq cm Area of a tile = 60 sq cm
∴ Number of tiles needed to fit the region
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration 23

A challenge!
On a centimeter squared paper, make as many rectangles as you can such that the area of the rectangle is 16 sq cm (consider only whole number lengths).
(a) Which rectangle has the greatest perimeter?
(b) Which rectangle has the least perimeter? If you take a rectangle of area 24 sq. cm., what
will be your answers?
Given any area, is it possible to predict the shape of the rectangle with the greatest perimeter? With the least perimeter? Give example and reason.
Solution :
Three rectangles can be made as follows:
(i) Sides 16 cm and 1 cm. Perimeter
= 2 × (Length + Breadth)
= 2 × (16 + 1) cm = 34 cm
(ii) Sides 8 cm and 2 cm. Perimeter
= 2 × (Length + Breadth)
= 2 × (8 + 2) cm = 20 cm
(iii) Sides 4 cm and 4 cm. Perimeter
= 2 × (Length + Breadth)
= 2 (4 + 4) cm = 16 cm

(a) The rectangle (i) has the greatest perimeter.
(b) The rectangle (iii) has the least perimeter.
For Area 24 sq. cm, four rectangles can be made as follows :
(i) Sides 24 cm and 1 cm. Perimeter
= 2 × (24 + 1)cm = 50 cm [Greatest perimeter]
(ii) Sides 12 cm and 2 cm. Perimeter
= 2 × (12 + 2) cm = 28 cm
(iii) Sides 8 cm and 3 cm. Perimeter
= 2 × (8 + 3) cm = 22 cm
(iv) Sides 6 cm and 4 cm. Perimeter
= 2 × (6 + 4) cm = 20 cm [Least perimeter]
Yes ! it is possible to predict the shape of the rectangle with (i) the greatest perimeter (iv) the least perimeter for any given area.
Reason: The rectangle with the greatest length has the maximum perimeter and the rectangle with the smallest length has the least perimeter, for a given area of the rectangle.

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NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.2

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.2 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.2.

Board CBSE
Textbook NCERT
Class Class 6
Subject Maths
Chapter Chapter 10
Chapter Name Mensuration
Exercise  Ex 10.2
Number of Questions Solved 1
Category NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.2

Question 1.
Find the areas of the following figures:
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration 16
Solution :
(a)

Cover Number estimate Area
(i) Full-filled squares 9 9 squares
(ii) Half-filed squares
(iii) More than half-filed squares
(iv) Less than half-filed squares

∴ Total area of the figure = 9 sq units

(b)

Cover Number estimate Area
(i) Full-filled squares 5 5 squares
(ii) Half-filed squares
(iii) More than half-filed squares
(iv) Less than half-filed squares

∴ Total area of the figure = 9 sq units

(c) Full-filled squares = 2
Half-filled squares = 4
Area covered by full squares
= 2 × 1 sq unit = 2 sq units
Area covered by half squares = 4 × \(\frac{ 1 }{ 2 }\) sq unit = 2 sq units
∴ Total area = 2 sq units + 2 sq units = 4 sq units.

(d) Full-filled squares = 8
Total area = Area covered by full squares = 8 × 1 sq unit = 8 sq units.

(e) Full-filled squares =10
∴ Total area = Area covered by full squares = 10 × 1 sq unit = 10 sq units.

(f) Full-filled squares = 2 Half-filled squares = 4 Area covered by full squares
= 2 × 1 sq unit = 2 sq units Area covered by half squares
= 4 × \(\frac{ 1 }{ 2 }\) sq unit = 2 sq units
Total area = 2 sq units + 2 sq units = 4 sq units.

(g) Full-filled squares = 4 Half-filled squares = 4 Area covered by full squares
= 4 × 1 sq unit = 4 sq units Area covered by half squares
= 4 × \(\frac{ 1 }{ 2 }\) sq unit = 2 sq units
∴ Total area = 4 sq units + 2 sq units.
= 6 sq units

(h) Full-filled squares = 5
∴ Total area = Area covered by full squares = 5 × 1 sq unit = 5 sq units.

(i) Full-filled squares = 9
∴ Total area = Area covered by full squares = 9 × 1 sq unit = 9 sq units.

(j) Full-filled squares = 2 Half-filled squares = 4 Area covered by full squares
= 2 × 1 sq unit = 2 sq units Area covered by half squares
= 4 × \(\frac{ 1 }{ 2 }\) sq unit = 2 sq units
∴ Total area = 2 sq units + 2 sq units = 4 sq units.

(k) Full-filled squares = 4 Half-filled squares = 2 Area covered by full squares
= 4 × 1 sq unit = 4 sq units Area covered by half squares
= 2 × \(\frac{ 1 }{ 2 }\) sq unit = 1 sq unit
∴ Total area = 4 sq units + 1 sq unit = 5 sq units.
(l)

Cover Number estimate Area
(i) Full-filled squares 4 4 × 1 sq unit = 4 sq units
(ii) Half-filed squares 2 2 × \(\frac{ 1 }{ 2 }\) sq unit = 1 sq unit
(iii) More than half-filed squares 3 3 × 1 sq unit = 3 sq units
(iv) Less than half-filed squares 4

(m)

Cover Number estimate Area
(i) Full-filled squares 9 9 × 1 sq unit = 9 sq units
(ii) Half-filed squares 2 2 × \(\frac{ 1 }{ 2 }\) = 1 sq unit
(iii) More than half-filed squares 6 6 × 1 sq unit = 6 sq units
(iv) Less than half-filed squares 4

∴ Total Area = 9 sq units + 1 sq. unit + 6 sq units = 16 sq units
(n)

Cover Number estimate Area
(i) Full-filled squares 11 11 × 1 sq unit = 11 sq units
(ii) Half-filed squares 6 6 × \(\frac{ 1 }{ 2 }\) = 3 sq unit
(iii) More than half-filed squares 9 9 × 1 sq unit = 9 sq units
(iv) Less than half-filed squares

∴ Total area = 11 sq units + 3 sq units + 9 sq units = 23 sq units.

 

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NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Ex 9.4

NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Ex 9.4 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Ex 9.4.

Board CBSE
Textbook NCERT
Class Class 6
Subject Maths
Chapter Chapter 9
Chapter Name Data Handling
Exercise  Ex 9.4
Number of Questions Solved 4
Category NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Ex 9.4

Question 1.
A survey of 120 school students was done to find which activity they prefer to do in their free time:

Preferred activity Number of students
Playing 45
Reading story books 30
Watching T.V. 20
Listening to music 10
Painting 15

Draw a bar graph to illustrate the above data taking the scale of 1 unit length = 5 students.
Which activity is preferred by most of the students other than playing?
Solution.
(i) Draw two perpendicular lines—one vertical and one horizontal.
(ii) Along horizontal line mark the “Preferred activity” and along vertical line mark the “No. of students”
(iii) Take bars of same width keeping the uniform gap between them.
(iv) Take scale of 1 unit length = 5 students along the vertical line and then mark the corresponding values.
(v) Calculate the heights of the bars for various activities preferred as shown below :
Playing                      :  45÷5= 9 units
Reading story books :  30÷5= 6 units
Watching T.V.           :   20÷4 =4 units
Listening to Music    :   10÷5= 2 units
Painting                    :   15÷5=3 units
(vi) Now draw various bars.
NCERT Solutions for Class 6 Maths Chapter 9 Data Handling 16
The activity “Reading story books” is preferred by most of the students other than playing.

Question 2.
The number of Mathematics books sold by a shopkeeper on six consecutive days is shown below:

Days Sun­day Mon­day Tues­day Wednes­day Thurs­day Fri­day
Number of books sold 65 40 30 50 20 70

Draw a bar graph to represent the above information choosing the scale of your choice.
Solution.
(i) Draw two perpendicular lines—one vertical and one horizontal.
(ii) Along horizontal line mark the “days” and along vertical line mark the “number of books sold.”
(iii) Take bars of same width keeping the uniform gap between them.
(iv) Take scale of 1 unit length = 5 books along the vertical line and mark the corresponding values.
(v) Calculate the heights of the bars for various days as shown below:
Sunday   : 65÷5=13 units
Monday  : 40÷5=8 units
Tuesday  : 30÷5=6 units
Wednesday : 50÷5=10 units
Thursday : 20÷5=4 units
Friday : 70÷5=14 units
(vi) Now draw various bars.
NCERT Solutions for Class 6 Maths Chapter 9 Data Handling 17

Question 3.
Following table shows the number of bicycles manufactured in a factory during the years 1998 to 2002. Illustrate this data using a bar graph. Choose a scale of your choice.

 Year Number of bicycles manufactured
1998 800
1999 600
2000 900
2001 1100
2002 1200

(i) In which year was the maximum number of bicycles manufactured?
(ii) In which year was the minimum number of bicycles manufactured?
Solution.
Steps for drawing a bar graph
(i) Draw two perpendicular lines — one vertical and one horizontal.
(ii) Along horizontal line marks the Year and along vertical line mark the “No. of bicycles manufactured”.
(iii) Take bars of same width keeping uniform gaps between them.
(iv) Take scale of 1 unit length = 100 bicycles along the vertical line and then mark the corresponding values.
(v) Calculate the heights of the bars for various years as shown below:
1998: 800÷100=8 units
1999: 600÷100=6 units
2000: 900÷100=9 units
2001: 1100÷100=11 units
2002: 1200÷100=12 units
(vi) Now draw various bars.
NCERT Solutions for Class 6 Maths Chapter 9 Data Handling 18
(i) The maximum number of bicycles were manufactured in the year 2002.
(ii) The minimum number of bicycles were manufactured in the year 1999.

Question 4.
A number of persons in various age groups in a town is given in the following table:

Age group Number of persons
1-14 2 lakhs
15-29 1 lakh 60 thousand
30-44 1 lakh 20 thousand
45-59 1 lakh 20 thousand
60-74 80 thousand
75 and above 40 thousand

Draw a bar graph to represent the above information and answer the following questions. (take 1 unit length = 20 thousands):
(i) Which two age groups have the same population?
(ii) All persons in the age group of 60 and above are called senior citizens. How many senior citizens are there in the town?
Solution.
(i) Draw two perpendicular lines — one vertical and one horizontal.
(ii) Along horizontal line mark the “Age group” and along vertical line mark the “Number of persons”.
(iii) Take bars of same width keeping uniform gap between them.
(iv) Take scale of 1 unit length = 20 thousand along the vertical line and then mark the corresponding values.
(v) Calculate the heights of the bars for various of groups as shown below:
1- 14 : \(\frac { 200000 }{ 20000 } \) =10 units
15- 29 : \(\frac { 160000 }{ 20000 } \) = 8 units
30- 44 : \(\frac {120000 }{ 20000 } \) = 6 units
45- 59 : \(\frac {120000}{ 20000 } \) = 6 units
60- 74 : \(\frac { 80000 }{ 20000 } \) = 4 units
75 and above : \(\frac { 40000 }{ 20000 } \) = 2 units
(vi) Now draw various bars.
NCERT Solutions for Class 6 Maths Chapter 9 Data Handling 19
(i) The two age-groups 30-44 and 45-59 have the same population.
(ii) Number of senior citizens in the town
= 80000 + 40000
= 120000

We hope the NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Ex 9.4 help you. If you have any query regarding. NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Ex 9.4, 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 9.3

NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Ex 9.3 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Ex 9.3.

Board CBSE
Textbook NCERT
Class Class 6
Subject Maths
Chapter Chapter 9
Chapter Name Data Handling
Exercise  Ex 9.3
Number of Questions Solved 3
Category NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Ex 9.3

Question 1.
The bar graph given alongside shows the amount of wheat purchased by the government during the years 1998-2002:
Read the bar graph and write down your observations. In which year was
(a) the wheat production maximum?
(b) the wheat production minimum?
NCERT Solutions for Class 6 Maths Chapter 9 Data Handling 13
Solution.
(a) The wheat production was maximum in the year 2002.
(b) The wheat production was minimum in the year 1998.

Question 2.
Observe this bar graph which is showing the sale of shirts in a ready made shop from Monday to Saturday.
NCERT Solutions for Class 6 Maths Chapter 9 Data Handling 14
Now answer the following questions:
(a)
What information does the above bar graph give?
(b) What is the scale chosen on the horizontal line representing a number of shills?
(c) On which day were the maximum number of shirts sold? How many shirts were sold on that day?
(d) On which day were the minimum number of shirts sold?
(e) How many shirts were sold on Thursday?
Solution.
(a) This bar graph shows the number of shirts sold from Monday to Saturday.
(b) 1 unit = 5 shirts
(c) Saturday, 60
(d) Tuesday
(e) 35 shirts.

Question 3.
Observe this bar graph which shows the marks obtained by Aziz in half-yearly examination in different subjects.
Answer the given questions.
(a) What information does the bar graph give ?
(b) Name the subject in which Aziz scored maximum marks.
(c) Name the subject in which he has scored minimum marks.
(d) State the name of the subjects and marks obtained in each of them
NCERT Solutions for Class 6 Maths Chapter 9 Data Handling 15
Solution.
(a) This bar graph shows the marks obtained by Aziz in different subjects.
(b) Hindi
(c) Social Studies
(d) Hindi-80, English-60, Mathematics-70, Science-50 and Social Studies -40.

 

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NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Ex 9.2

NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Ex 9.2 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Ex 9.2.

Board CBSE
Textbook NCERT
Class Class 6
Subject Maths
Chapter Chapter 9
Chapter Name Data Handling
Exercise  Ex 9.2
Number of Questions Solved 2
Category NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Ex 9.2

Question 1.
Total number of animals in five villages are as follows:
Village A : 80
Village B  : 120
Village C  : 90
Village D  : 40
Village E  : 60
Prepare a pictograph of these animals using one symbol ⊗ to represent 10 animals and answer the following questions:
(i) How many symbols represent animals of village E?
(ii) Which village has the maximum number of animals?
(iii) Which village has more animals: Village A or village C?
Solution.
NCERT Solutions for Class 6 Maths Chapter 9 Data Handling 9
(i) 6 symbols represent animals of village E.
(ii) Village B has the maximum number of animals.
(iii) Village C has more animals than village A.

Question 2.
The total number of students in a school in different years is shown in the following table:

Year Number of students
1996 400
1998 535
2000 472
2002 600
2004 623

A. Prepare a pictograph of students using one symbol to NCERT Solutions for Class 6 Maths Chapter 9 Data Handling 10represent 100 students and answer the following questions:
(i) How many symbols represent total number of students in the year 2002?
(ii) How many symbols represent total number of students for the year 1998?
B. Prepare another pictograph of students using any other symbol each representing 50 students. Which pictograph do you find more informative?
Solution.
(A)
NCERT Solutions for Class 6 Maths Chapter 9 Data Handling 11
(i) 6 symbols represent total number of students in the year 2002.
(ii) 0 5 complete and 1 incomplete symbols represent total number of students for the year 1998.

(B)
NCERT Solutions for Class 6 Maths Chapter 9 Data Handling 12
Pictograph B is more informative because it gives better approximations.

 

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NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.6

NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.6 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.6.

Board CBSE
Textbook NCERT
Class Class 6
Subject Maths
Chapter Chapter 8
Chapter Name  Decimals
Exercise  Ex 8.6
Number of Questions Solved 7
Category NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.6

Question 1.
Subtract

(a) ₹ 18.25 from ₹ 20.75
(b) ₹ 25from ₹ 20.75
(c) 54 mfrom 250 m
(d) ₹ 36 from ₹ 8.40
(e) 2.051 km from 5.206 km
(f) 0.314 kg from 2.107 kg.
Solution.
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 48
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 49

Question 2.
Find the value of:
(a) 9.756 – 6.28
(b) 21.05 – 15.27
(c) 18.5 – 6.79
(d) 11.6- 9.847.
Solution.
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 50

Question 3.
Raju bought a book for ₹ 35.65. Did he give? 50 to the shopkeeper. How much money did he get back from the shopkeeper?
Solution.
Book bought for = ₹ 35.65
Money is given to the shopkeeper = ₹ 50
∴ Money got back from the shopkeeper
= ₹ 50 – ?35.65
= ₹ 14.35
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 51

Question 4.
Rani had ₹ 18.50. She bought one ice-cream for ₹ 11.75. How much money does she have now?
Solution.
Money Rani had = ₹ 18.50
Ice-cream bought for = ₹ 11. 75
∴ Money she has now = ₹ 18.50 – ₹ 11. 75
= ₹ 6.75
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 52

Question 5.
Tina had 20 m 5 cm long cloth. She cuts 4 m 50 cm length of cloth from this for making a curtain. How much cloth is left with her?
Solution.
Length of cloth Tina had
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 53

Question 6.
Namita travels 20 km 50 m everyday. Out of this, she travels 10 km 200 m by bus and the rest by auto. How much distance does she travel by auto?
Solution.
Distance travelled everyday
= 20 km 50 m = 20 km + 50 m
= 20 km +\(\frac { 50 }{ 1000 } \) km
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 54

Question 7.
Aakash bought vegetables weighing 10 kg. Out of this, 3 kg 500 g is onions, 2 kg 75 g is tomatoes and the rest is potatoes. What is the weight of the potatoes?
Solution.
Weight of vegetables bought = 10 kg
Weight of onions = 3 kg 500 g = 3 kg + 500 g
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 55

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NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.5

NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.5 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.5.

Board CBSE
Textbook NCERT
Class Class 6
Subject Maths
Chapter Chapter 8
Chapter Name  Decimals
Exercise  Ex 8.5
Number of Questions Solved 7
Category NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.5

Question 1.
Find the sum in each of the following:
(a) 0.007+8.5+30.08
(b) 15 + 0.632 + 13.8
(c) 27.076 + 0.55 + 0.004
(d) 25.65 + 9.005 + 3.7
(e) 0.75 + 10.425 + 2
(f) 280.69 + 25.2 + 38.
Solution.
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 36
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 37
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 38

Question 2.
Rashid spent Rs.35.75 for Maths book and Rs.32.60 for Science book. Find the total amount spent by Rashid.
Solution.
Money spent for Maths book = Rs. 35.75
Money spent for Science book = Rs. 32.60
Total amount spent by Rashid = Rs. 35.75 +Rs.32.60
= Rs. 68.35.
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 39

Question 3.
Radhika’s mother gave her ? 10.50 and her father gave her ? 15.80, find the total amount given to Radhika by the parents.
Solution.
Money given by mother = Rs. 10.50
Money given by father = Rs. 15.80
Total amount given to Radhika by parents = Rs. 10.50 +Rs.15.80 =Rs. 26.30
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 40

Question 4.
Nasreen bought 3 m 20 cm cloth for her shirt and 2 m 5 cm cloth for her trouser. Find the total length of cloth bought by her.
Solution.
Cloth bought for her shirt
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 41

Question 5.
Naresh walked 2 km 35 m in the morning and 1 km 7 m in the evening. How much distance did he walk in all?
Solution.
Distance walked by Naresh in the morning
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 42
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 43

Question 6.
Sunita travels 15 km 268 m by bus, 7 km 7 m by car and 500 m on foot in order to reach her school. How far is her school from her residence?
Solution.
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 44
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 45

Question 7.
Ravi purchased 5 kg 400 grice, 2 kg 20 g sugar and 10 kg 850 g flour. Find the total weight of his purchases.
Solution.
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 46
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 47

 

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NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.4

NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.4 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.4.

Board CBSE
Textbook NCERT
Class Class 6
Subject Maths
Chapter Chapter 8
Chapter Name  Decimals
Exercise  Ex 8.4
Number of Questions Solved 5
Category NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.4

Question 1.
Express as rupees using decimals.
(a) 5 paise
(b) 75 paise
(c) 20 paise
(d) 50 rupees 90paise
(e) 725 paise.
Solution.
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 23
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 24
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 25

Question 2.
Express as meters using decimals.
(a) 15 cm
(b) 6 cm
(c) 2 m 45 cm
(d) 9 m 7cm
(e) 419 cm.
Solution.
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 26
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 27

Question 3.
Express as cm using decimals.
(a) 5 mm
(b) 60 mm
(c) 164 mm
(d) 9 cm 8 mm
(e) 93mm.
Solution.
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 28
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 29

Question 4.
Expresc as km using decimals.
(a) 8m
(b) 88m
(c) 8888 m
(d) 70 km 5 m.
Solution.
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 30
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 31

Question 5.
Express as kg using decimals.
(a) 2g
(b) 100g
(c) 3750g
(d) 5kg8g
(e) 26 kg 50g
Solution.
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 32
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 33
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 34
NCERT Solutions for Class 6 Maths Chapter 8 Decimals 35

 

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NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.3

NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.3 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.3.

Board CBSE
Textbook NCERT
Class Class 6
Subject Maths
Chapter Chapter 8
Chapter Name  Decimals
Exercise  Ex 8.3
Number of Questions Solved 2
Category NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.3

Question 1.
Which is greater?
(a) 0.3 or 0.4
(b) 0.07 or 0.02
(c) 3 or 0.8
(d) 0.5 or 0.05
(e) 1.23 or 1.2
(f) 0.099 or 0.19
(g) 1.5 or 1.50
(h) 1.431 or 1.490
(i) 3.3 or 3.300
(j) 5.64 or 5.603
Solution.
(a) 0.3 or 0.4
Whole part for both is 0.
Tenth part for 0.3 is 3 and for 0.4 is 4
∵ 4>3
∴ 0.4 > 0.3

(b) 07 or 0.02
Whole part for both is 0.
Tenth part for both is 0.
Hundredth part for 0.07 is 7 and for 0.02 is 2.
∵ 7>2
∴  0.07 > 0.02

(c) 3 or 0.8
Whole part for 3 is 3 and for 0.8 is 0.
∵ 3 >0
∴ 3 >0.8

(d) 5 or 0.05
Whole part for both is 0.
Tenth part for 0.5 is 5 and for 0.05 is 0.
∵ 5 >0
∴ 0.5 > 0.05

(e) 23 or 1.2
Whole part for both is 1.
Tenth part for both is 2.
Hundredth part for 1.23 is 3 and for 1.2 is 0 (as 1.2= 1.20)

(f) 0.099 or 0.19
Whole part for both is 0.
Tenth part for 0.099 is 0 and for 0.19 is 1.
∵ 1>0
∴ 0.19 >0.099

(g)1.5 or 1.50
Whole part for both is 1.
Tenth part for both is 5.
Hundredth part for both is 0 (as 1.5 = 1.50).
∴ 1.5=1.50

(h) 1431 or 490
Whole part for both is 1.
Tenth part for both is 4.
Hundredth part for 1.431 is 3 and for 1.490 is 9
∵ 9>3
∴ 1.490 >1.431

(i) 3 or 3.300
Whole part for both is 3.
Tenth part for both is 3.
Hundredth part for both is 0 (as 3.3 = 3.30).
Thousandth part for both is 0 (as 3.3 = 3.300).
∴ 3.3 = 3.300

(J) 5.64 or 5.603 
Whole part for both is 5.
Tenth part for both is 6.
Hundredth part for 5.64 is 4 and for 5.603 is 0.
∵ 4>0
∴ 5.64 > 5.603

Question 2.
Make five more examples and find the greater number from them.
Solution.
Five more examples are as follows:
Which is greater?
Ex. 1. 0.4 or 0.5
Ex. 2. 0.06 or 0.04
Ex. 3. 4 ox 0.9
Ex. 4. 0.6 or 0.06
Ex. 5. 0.063 or 0.22
Solution.
1.
0.4 or 0.5

Whole part for both = 0
Tenth part for 0.4 = 4
Tenth part for 0.5 = 5
∵ 5 >4
∴ 0.5 > 0.4

2.
0.06 or 0.04

Whole part for both = 0
Tenth part for both = 0
Hundredth part for 0.06 = 6
Hundredth part for 0.04 = 4
∵ 6 >4
∴ 0.06 > 0.04

3.
4 or 0.9
Whole part for 4 = 4
Whole part for 0.9 = 0
∵ 4 > 0
∴ 4 > 0.9

4.
0.6 or 0.06
Whole part for both = 0
Tenth part for 0.6 = 6
Tenth part for 0.06 = 0
∵  6 > 0
∴ 0.6 > 0.06

5.
0.063 or 0.22
Whole part for both = 0
Tenth part for 0.063 = 0
Tenth part for 0.22 = 2
∵  2 > 0
∴ 0.22 > 0.063.

 

We hope the NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.3 help you. If you have any query regarding. NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.3, drop a comment below and we will get back to you at the earliest.