Yearly Archives: 2022

Ex 10.1 Class 6 Maths Question 1.
Find the perimeter of each of the following figures:

Solution:
(a) Perimeter = the sum of the lengths of sides
= 5 cm +1 cm +2 cm + 4 cm = 12cm
(b) Perimeter = the sum of the lengths of sides
= 40 cm +35 cm + 23 cm +35 cm = 133 cm
(c) Perimeter = 4x the length of one side
= 4 x 15 cm = 60 cm
(d) Perimeter = 5 x the length of one side
4 = 5 x 4 cm = 20 cm
(e) Perimeter = the sum of the lengths of sides
= 4 cm + 0.5 cm + 2.5 cm + 2.5 cm + 0.5 cm + 4 cm + 1 cm = 15 cm
(f) Perimeter = the sum of the lengths of sides
= 4 cm + 3 cm + 2 cm + 3 cm + 1 cm + 4 cm + 3 cm + 2 cm + 3 cm + 1 cm + 4 cm + 3 cm + 2 cm +3 cm +1 cm + 4 cm + 3 cm + 2 cm + 3 cm + 1 cm = 52 cm

Ex 10.1 Class 6 Maths Question 2.
The lid of a rectangular box of sides 40 cm by 10 cm is sealed all round with tape. What is the length of the tape required?
Solution:
Length of the tape required
= Perimeter of the lid of a rectangular box = 2 x ( length + breadth)
= 2 x (40 cm +10 cm)
= 2 x 50 cm = 100 cm or 1 m

Ex 10.1 Class 6 Maths Question 3.
A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table-top?
Solution:
Perimeter of the table-top
= 2 x (length + breadth)
= 2 x (2 m 25 cm +1 m 50 cm)
= 2 x (3 m 75 cm)
= 2 x 3.75 m = 7.50 m

Ex 10.1 Class 6 Maths 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:
The length of the wooden strip required to frame a photograph is the perimeter of the photograph.
Perimeter of the photograph = 2 x (length + breadth)
= 2 x (32 cm +21 cm)
= 2 x 53 cm = 106 cm
∴ The length of the wooden strip required is 106 cm.

Ex 10.1 Class 6 Maths 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:
We have to cover 4 times the perimeter of the land measuring 0.7 km by 0.5 km.
∴ Total length of wire required is 4 times its perimeter.
Perimeter of the land = 2 x (length + breadth)
= 2x (0.7 km+0.5 km)
= 2 x 1.2 km = 2.4 km
∴ Total length of wire required = 4 x 2.4 km = 9.6 km

Ex 10.1 Class 6 Maths 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 and third side 6 cm.
Solution:
(a) Perimeter = the sum of the sides
= 3 cm + 4 cm + 5cm = 12 cm
(b) Perimeter = 3 x the length of one side
= 3 x 9 cm = 27 cm
(c) Perimeter = the sum of the lengths of the sides
= 8 cm + 8 cm + 6 cm = 22 cm

Ex 10.1 Class 6 Maths Question 7.
Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.
Solution:
Perimeter of the triangle = the sum of the lengths of its sides
= 10 cm + 14 cm + 15 cm = 39 cm

Ex 10.1 Class 6 Maths Question 8.
Find the perimeter of a regular hexagon with each side measuring 8 m.
Solution:
A regular hexagon has 6 sides, so its perimeter
= 6 x length of its one side = 6 x 8 m
= 48 m

Ex 10.1 Class 6 Maths Question 9.
Find the side of the square whose perimeter is 20 m.
Solution:
Perimeter = 20 m
A square has 4 equal sides, so we can divide the perimeter by 4 to get the length of one side.
One side of the square = 20 m + 4 = 5m

Ex 10.1 Class 6 Maths Question 10.
The perimeter of a regular pentagon is 100 cm. How long is its each side?
Solution:
Perimeter = 100 cm
A regular pentagon has 5 equal sides, so we can divide the perimeter by
5 to get the length of one side
∴ length of one side = 100 cm + 5 = 20 cm

Ex 10.1 Class 6 Maths Question 11.
A piece of string is 30 cm long. What will be the length of each side if the string is used to forms
(a) a square?
(b) an equilateral triangle?
(c) a regular hexagon?
Solution:
(a) Perimeter = Length of the string = 30 cm
A square has 4 equal sides, so we can divide the perimeter by 4 to get the length of one side.
One side of the square = 30 cm + 4 = 7.5 cm

(b) Perimeter = Length of the string = 30 cm
An equilateral triangle has 3 equal sides, so we can divide the perimeter by 3 to get the length of one side.
∴ One side of an equilateral triangle = 30 cm ÷ 3 = 10 cm
(c) Perimeter = Length of the string = 30 cm
A regular hexagon has 6 equal sides, so we can divide the perimeter by
6 to get the length of one side.
One side of a regular hexagon = 30 cm + 6 = 5 cm

Ex 10.1 Class 6 Maths Question 12.
Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is its third side?
Solution:
Let ABC be the given triangle such that AB =12 cm, BC = 14 cm and its perimeter = 36 cm.
i.e., AB + BC + CA = 36 cm
or 12 cm + 14 cm + CA = 36 cm
or 26 cm + CA = 36 cm
or CA =36 cm – 26 cm = 10 cm
∴ The third side of triangle is 10 cm.

Ex 10.1 Class 6 Maths Question 13.
Find the cost offencing a square park ofside 250 mat the rate of ? 20 per metre.
Solution:
Side of the square park = 250 m
∴ Perimeter of the square park
= 4 x side
= 4 x 250 m = 1000 m
∴ Cost of fencing = ₹ (1000 x 20)
= ₹ 20000

Ex 10.1 Class 6 Maths Question 14.
Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of? 12 per metre.
Solution:
Length of the rectangular park = 175 m Breadth of the rectangular park 125 m .-. Perimeter of the park = 2 x (length + breadth)
= 2 x (175m + 125m)
= 2 x 300 m = 600 m
∴ Cost of fencing = ₹ (600 x 12) = ₹ 7200

Ex 10.1 Class 6 Maths 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:
The distance each girl covers in one round is the same as the perimeter of the respective field. Therefore, the distance that Sweety covers in one round
= 4 x side
= 4 x 75 m = 300 m
Also, the distance that Bulbul covers in one round
= 2 x (length + breadth)
= 2 x (60 m + 45 m)
= 2 x 105 m = 210 m
This shows that Bulbul covers less distance than Sweety.

Ex 10.1 Class 6 Maths Question 16.
What is the perimeter of each of the following figures? What do you infer from the answer?


Solution:
(a) Perimeter = 4 x side
= 4 x 25 cm
= 100 cm
(b) Perimeter = 2 x (length + breadth)
= 2 x (40 cm +10 cm)
= 2 x 50 cm = 100 cm
(c) Perimeter = 2 x (length + breadth)
2 x (30 cm +20 cm)
= 2 x 50 cm = 100 cm
(d) Perimeter = 30 cm +30 cm +40 cm = 100 cm
Thus, we observe that the perimeter of each figure is 100 cm i.e., they have equal perimeters.

Ex 10.1 Class 6 Maths Question 17.
Avneet buys 9 square paving slabs, each with a side of 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 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) In case of Avneet’s arrangement:

(b) In case of Shari’s arrangement:

(c) Clearly, perimeter in case of Shari is greater.
(d) Yes, there is a way shown in the figure in which we get a greater perimeter

Perimeter = 2 x (9 + 1) units = 2 x 10 units = 20 units

Ex 10.2 Class 6 Maths Question 1.
Find the areas of the following figures by counting square:

Solution:
Placing these figures on the centimetre square, we have
(a) Full squares = 9
∴ Area covered by the figure = 9 x 1 sq. cm = 9 sq. cm

(b) Full squares = 5
∴ Area covered by the figure = 5 x 1 sq. cm = 5 sq. cm

(c) Full squares = 2
Half squares = 4
∴ Area covered by the figure (2×1+4×12) sq. cm
= (2 + 2) sq cm = 4 sq. cm

(d) Full squares = 10
∴ Area covered by the figure = 10 x 1 sq. cm = 10 sq. cm

(e) Full squares = 10
∴ Area covered by the figure = 10 x 1 sq. cm = 10 sq. cm

(f) Full squares = 2
Half squares = 4
∴ Area covered by the figure = (2×1+4×12) sq. cm
= (2 + 2)sq. cm = 4 sq. cm

(g) Full squares = 4
Half squares = 4
∴ Area covered by the figure = (2×1+4×12) sq. cm
= (4 + 2)sq. cm = 6 sq. cm

(h) Full squares = 5
∴ Area covered by the figure = 5 x 1 sq. cm
= 5 sq. cm

(i) Full squares = 9
∴ Area covered by the figure = 9 x 1 sq. cm
= 9 sq. cm

(j) Full squares = 2
Half squares = 4
∴ Area covered by the figure = (2×1+4×12) sq. cm
= (2 + 2)sq. cm = 4 sq. cm

(k) Full squares = 4
Half squares = 2
∴ Area covered by the figure = (4×1+2×12) sq. cm
= (4 + 1)sq. cm = 5 sq. cm

(l) Full squares = 4
More than half squares = 3
Half square = 2
∴ Area covered by the figure = (4×1+3×1+2×12) sq. cm
= (4 + 3 + 1)sq. cm
= 8 sq. cm

(m) Full squares = 7
More than half squares = 7
Half square = 0
∴ Area covered by the figure = (7×1+7×1+0×12) sq. cm
= (7 + 7 + 0)sq. cm
= 14 sq. cm

(n) Full squares = 10
More than half squares = 8
Half square = 0
∴ Area covered by the figure = (10×1+8×1+0×12) sq. cm
= (10 + 8 + 0)sq. cm
= 18 sq. cm

Ex 10.3 Class 6 Maths Question 1.
Find the areas of the rectangles whose sides are:
(a) 3 cm and 4 cm
(b) 12 m and 21 m
(c) 2 km and 3 km
(d) 2 m and 70 cm
Solution:
(a) Length of the rectangle = 4 cm
Breadth of the rectangle = 3 cm
Area = Length x Breadth
= 4 cm x 3 cm
= 12 sq. cm

(b) Length of the rectangle = 21 m
Breadth of the rectangle = 12 m
Area = Length x Breadth
= 21 m xl2 m
= 252 sq. m.

(c) Length of the rectangle = 3 km
Breadth of the rectangle =2 km
Area = Length x Breadth
= 3 km x 2km
= 6 sq. km

(d) Length of the rectangle =2m = 2 x l00= 200 cm
Breadth of the rectangle = 70 cm
Area = Length x Breadth
= 200 cm x 70 cm
= 14000 sq. cm

Ex 10.3 Class 6 Maths Question 2.
Find the areas of the squares whose sides are:
(a) 10 cm
(b) 14 cm
(c) 5 m
Solution:
(a) Side of the square = 10 cm
Area = (side)²
= (10)² sq. cm
= 100 sq. cm

(b) Side of the square = 14 cm
Area = (side)²
= (14)² sq. cm
= 196 sq. cm

(c) Side of the square = 5 m
Area = (side)²
= (5)² sq. m = 25sq. m

Ex 10.3 Class 6 Maths Question 3.
The length and breadth of three rectangles are as given below:
(a) 9 m and 6 m
(b) 17 m and 3 m
(c) 4 m and 14 m
Which one has the largest area and which one has the smallest?
Solution:
In rectangle (a), Area = (9 x 6) sq. m
= 54 sq. m
In rectangle (b), Area =(3 x 17) sq. m = 51 sq. m
In rectangle (c), Area = (4 x 14) sq. m = 56 sq. m
Clearly, 56 > 54 > 51 i.e., 56 is the largest number and 51 is the smallest number.
∴ The rectangle (c) has the largest area and the rectangle (a) has the smallest area.

Ex 10.3 Class 6 Maths Question 4.
The area of a rectangular garden 50 m long is 300 sq m. Find the width of the garden.
Solution:
We use the formula A = l x b, where l is the length and b is the breadth of the rectangle in metre and its area in m2.
Here, A = 300 m², and l = 50 m
Therefore, 300 =50 x b [ ∵ A = l x b]
or b=30050=6
Hence the breadth (width) of the rectangle is 6 m.

Ex 10.3 Class 6 Maths Question 5.
What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of ₹ 8 per hundred sq. m?
Solution:
Area of rectangular piece of land = 500 x 200 sq. m
= 100000 sq. m
Cost of tiling the land at the rate of ₹ 8 per hundred sq. m
= ₹ (100000×1100) = ₹ 8000

Ex 10.3 Class 6 Maths Question 6.
A table-top measures 2 m by 1 m 50 cm. What is its area in square metres?
Solution:
Length of table’s top = 2 m
Breadth of table’s top = 1 m 50 cm = 1.50 m
∴ Area = Length x Breadth
= (2 x 1.50) sq m = 3 sq m

Ex 10.3 Class 6 Maths Question 7.
A room is 4 m long and 3 m 50 cm wide. How many square metres 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
Carpet needed to cover the floor of the room
= Area of the floor of the room
= Length x Breadth
= (4 x 3.50) sq. m = 14 sq. m

Ex 10.3 Class 6 Maths 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:

Area of the floor of the room = (5 x 4) sq. m
= 20 sq. m
Area of the carpet = (3 x 3) sq. m = 9 sq. m
Area of the floor not carpeted = (20 – 9) sq. m
= 11 sq. m

Ex 10.3 Class 6 Maths 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 the land?
Solution:
Area of the piece of land =(5 x 4) sq. m
= 20 sq. m
Area of square flower bed =(1 x 1) sq. m
= 1 sq. m
Area of 5 such flower beds =(5 x 1) sq. m
= 5 sq. m
∴ Area of the remaining part of land
= (20 – 5) sq. m = 15 sq. m

Ex 10.3 Class 6 Maths Question 10.
By splitting the following figures into rectangles, find their areas (The measures are given in centimetres).

Solution:
(a) Let the figure may be divided into rectangles marked as A, B, C etc. Area of rectangle (A) = (3 x 3) sq. cm = 9 sq. cm

Area of rectangle (B) = (1 x 2) sq. cm = 2 sq. cm
Area of rectangle (C) = (3 x 3) sq. cm = 9 sq. cm
Area of rectangle (D) = (4 x 2) sq. cm = 8 sq. cm
∴ The total area of the figure = (9 + 2 + 9 + 8) sq. cm = 28 sq. cm
(b) Let the figure may be divided into rectangle marked as shown.

Area of rectangle (A) = (2 x 1) sq cm = 2 sq cm
Area of rectangle (B) = (5 x 1) sq cm = 5 sq cm
Area of rectangle (C) = (2 x 1) sq cm = 2 sq cm
∴ Total area of the figure = (2 + 5 + 2) sq cm = 9 sq cm

Ex 10.3 Class 6 Maths Question 11.
Split the following shapes into rectangles and find their areas. (The measures are given in centimetres)

Solution:
(a) Splitting the given shape into rectangles as marked A and B.

Area of the rectangle (A) = (10 x 2) sq cm
= 20 sq cm
Area of the rectangle (B) = (10 x 2) sq cm
= 20 sq cm
∴ Total area of given shape = (20 +20) sq cm
= 40 sq cm
(b) Splitting the given shape into five squares each of side 7 cm.

Area of the given shape
= 5 x Area of one square of side 7 cm = 5 x (7 x 7) sq. cm
= (5 x 49)sq cm 7
= 245 sq cm
(c) Splitting the given shape into two rectangles named A and B. Area of rectangle (A) = (5 x 1) sq cm = 5 sq cm
Area of rectangle (B) = (4 x 1) sq cm = 4 sq cm
∴ Total area of given shape = (5 + 4) sq cm = 9 sq cm

Ex 10.3 Class 6 Maths 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) Area of the rectangular region
= (100 x 144) sq cm
= 14400 sq cm
Area of one tile = (5 x 12) sq cm
= 60 sq cm

(b) Area of the rectangular region
= (70 x 36) sq cm
= 2520 sq cm

Ex 9.1 Class 6 Maths 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:
In the first column of the table, we write all the values of marks scored by the students starting from the lowest to the highest. In the second column, a vertical bar (|) called the tally mark is put against the number, whenever it occurs. For our convenience, we shall keep the tally marks in bunches of five, the fifth mark being drawn diagonally across the first four. We continue this process for all the values of the first column. Finally, we count the number of tally marks corresponding to each observation and write the number in the third column to represent the number of students.
Thus, we have the table as under:


From the above, clearly the number of students who obtained marks equal to or more than 7 marks are
5 + 4+3=12
Clearly, from the above table the number of students scoring marks below 4 are
2 + 3 + 3 = 8

Ex 9.1 Class 6 Maths 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) The required table is as under:

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

Ex 9.2 Class 6 Maths 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:
(a) How many symbols represent animals of village E?
(b) Which village has the maximum number of animals?
(c) Which village has more animals: village A or village C?
Solution:
The given data can be represented by a pictograph as given below:

(i) 6 symbols represent the animals of village E.
(ii) Village B has the maximum number of animals.
(iii) Village C has more animals than that of village A.

Ex 9.2 Class 6 Maths Question 2.
Total number of students of a school in different years is shown in the following table:

A. Prepare a pictograph of students using one symbol A to represent 100 students and answer the following questions:
(a) How many symbols represent total number of students in the year 2002?
(b) 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:
The given data can be represented by a pictograph as given below:

(a) 6 symbols represent total number of students in the year 2002.
(b) 5 complete symbols and one incomplete symbol of 35 students
B. Let us prepare another pictograph of students using symbol  each representing 50 students.

Second pictograph is more informative as it is accurate.

Ex 9.3 Class 6 Maths Question 1.
The bar graph given alongside shows the amount of wheat purchased by government during the year 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?

Solution:
The given bar graph represents the amount of wheat (in thousand tonnes) purchased by government during the year 1998-2002. The amount of wheat purchased during 1998 – 2002 = 15 + 25 + 20 +20 +30 = 110 (in thousand tonnes).
(a) The maximum wheat production was in the year 2002.
(b) The minimum wheat production was in the year. 1998.

Ex 9.3 Class 6 Maths Question 2.
Observe this bar graph which is showing the sale of shirts in a ready made shop from Monday to Saturday.

Now answer the following questions:
(a) What information does the above bar graph given?
(b) What is the scale chosen on the horizontal line representing number of shirts?
(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) The given bar graph represents the number of shirts sold from Monday to Saturday.
(b) Scale: 1 unit length = 5 shirts.
(c) Saturday and 60 in number.
(d) Tuesday.
(e) 35

Ex 9.3 Class 6 Maths Question 3.
Observe this bar graph which shows the marks obtained by Aziz in half yearly examination in different subjects.

Answer the given questionsn.
(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 names of the subjects and marks obtained in each of them.
Solution:
(a) The given bar graph represents the marks obtained by Aziz in half yearly examination in different subjects.
(b) Hindi
(c) Social Studies.
(d) Hindi-80, English-60, Mathematics-70, Science-50 and Social Studies-40.

Ex 9.4 Class 6 Maths Question 1.
A survey of 120 school students was done to find which activity they prefer to do in their free time:

Draw a bar graph to illustrate the above data taking scale of 1 unit length = 5 students.
Which activity is preferred by most of the students other than playing?
Solution:
Here, 5 values of the data are given. So, mark 5 points on the horizontal axis at equal distances and erect rectangles of the same width whose heights are proportional to the values of the numerical data.

The activity of ‘reading story books’ is preferred by most of the students other than playing.

Ex 9.4 Class 6 Maths Question 2.
The number of Mathematics books sold by a shopkeeper on six consecutive days is shown below:

Draw a bar graph to represent the above information choosing the scale of your choicer
Solution:
Here 6 values of the data are given. So, mark 6 points on the horizontal axis at equal distances and erect rectangles of the same width whose heights are proportional to the values of the numerical data.

Ex 9.4 Class 6 Maths 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.

(a) In which year were the maximum number of bicycles manufactured?
(b) In which year were the minimum number of bicycles manufactured.
Solution:
Here, 5 values of the data are given. So, mark 5 points on the horizontal axis at equal distances and erect rectangles of the same width whose heights are proportional to the values of the numerical data.

(a) The maximum number of bicycles were manufactured in the year 2002.
(b) The minimum number of bicycles were manufactured in the year 1999.

Ex 9.4 Class 6 Maths Question 4.
Number of persons in various age groups in a town is given in the following table.

Draw a bar graph to represent , the above information and answer the following questions.
(take 1 unit length = 20 thousands)
(a) Which two age groups have same population?
(b) All persons in the’ age group of 60 and above are called senior citizens. How many senior citizens are there in the town?
Solution:
Here, 6 values of the data are given. So, mark 6 points on the horizontal axis at equal distances and erect rectangles of the same width whose heights are as under:
Age Group

(a) Age groups 30 – 44 and 45 – 59 have the same population.
(b) Number of senior citizens are 80000 + 40000 = 120000.

Ex 8.1 Class 6 Maths Question 1.
Write the following as numbers in the given table:

Solution:
On filling the given table, we have

Ex 8.1 Class 6 Maths Question 2.
Write the following decimals in the place value table:
(a) 19.4
(b) 0.3
(c) 10.6
(d) 205
Solution:
Let us make a common place value table, assigning appropriate place value to the digits in the given numbers. We have,

Ex 8.1 Class 6 Maths 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 htihdred point eight
Solution:
(a) Seven-tenths = 710 = 0.7
(b) Two tens, 9 – tenths = 20 + 910 = 20.9
(c) Fourteen point six = 14.6
(d) One Hundred and 2-ones = 1 x 100 + 0 x 10 + 2 x 1 + 0 x 110
= 100 + 0 + 2 = 102.0
(e) Six hundred point eight = 600.8

Ex 8.1 Class 6 Maths Question 4.
Write each of the following as decimals:

Solution:

Ex 8.1 Class 6 Maths Question 5.
Write the following decimals as fractions. Reduce the fractions to lowest form:
(a) 0.6
(b) 2.5
(c) 1.0
(d) 3.8
(e) 13.7
(f) 21.2
(g) 6.4
Solution:

Ex 8.1 Class 6 Maths Question 6.
Express the following as cm vising decimals:
(a) 2 mm
(b) 30 mm
(c) 116 mm
(d) 4 cm 2mm
(e) 162 mm
(f) 83 mm
Solution:

Ex 8.1 Class 6 Maths Question 7.
Between which two whole numbers on the number line are the given numbers lie?
Which of these whole numberes is nearer the number?

(a) 0.8
(b) 5.1
(c) 2.6
(d) 6.4
(e) 9.1
(f) 4.9
Solution:

Ex 8.1 Class 6 Maths Question 8.
Show the following numbers on the number line.
(a) 0.2
(b) 1.9
(c) 1.1
(d) 2.5
Solution:
(a) Since 0.2 > 0 but <1. There are 2 tenth in it. Divide the unit length between 0 and 1 into 10 equal parts and take 2 parts as shown below. Thus, A represents 0.2.

(b) Since 1 < 1.9 < 2. Divide the unit length between 1 and 2 into 10 equal parts and take 9 parts as shown below. Thus, A represents 1.9.

(c) Since 1 < 1.1 < 2. Divide the unit length between 1 and 2 into 10 equal parts and take 1 part as shown below. Thus, A represents 1.1.

(d) Since 2 < 2.5 < 3. Divide the unit length between 2 and 3 into 10 equal parts and take 5 parts as shown below. Thus, A represents 2.5.

Ex 8.1 Class 6 Maths Question 9.
Write the decimal number represented by the points A, B, C, D on the given number line

Solution:
Since A lies between 0 and 1 and the unit length between 0 and 1 stands divided in 10 equal parts. Also A is at the 8th point.
Thus, A represents 0.8.
Since, B lies between 1 and 2 and the unit length between 1 and 2 stands divided in 10 equal parts. Also, B lies 3 points ahead of 1. Therefore, B represents 1.3.
Since C and D lies between 2 and 3 and the unit length between 2 and 3 stands divided between 2 and 3 in 10 equal parts. Also C and D respectively lies 2 and 9 points ahead of 2.
∴ C represents 2.2 and D represents 2.9.

Ex 8.1 Class 6 Maths 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

(b) Length of gram plant

Ex 8.2 Class 6 Maths Question 1.
Complete the table with the help of these boxes and use decimals to write the number

Solution:
Completing the table, we have

Ex 8.2 Class 6 Maths Question 2.
Write the numbers given in the following place value table in decimal form:

Solution:

Ex 8.2 Class 6 Maths Question 3.
Write the following decimals in the place value table:
(a) 0.29
(b) 2.08
(c) 19.60
(d) 148.32
(e) 200.812
Solution:
Let us make a common place value table, assigning appropriate place value to the digits in the given numbers. We have,

Ex 8.2 Class 6 Maths Question 4.
Write each of the following as decimals:

Solution:

Ex 8.2 Class 6 Maths Question 5.
Write each of the following decimals in words:
(a) 0. 03
(b) 1.20
(c) 108.56
(d) 10.07
(e) 0. 032
(f) 5.008
Solution:
(a) 0.03 = Zero point zero three or three hundredths.
(b) 1.20 = One point two zero.
(c) 108.56 = One hundred eight point five six.
(d) 10.7 = Ten point zero seven.
(e) 0. 032 = Zero point zero three two.
(f) 5.008 = Five point zero zero eight.

Ex 8.2 Class 6 Maths Question 6.
Between which two numbers in tenths place on the number line does each of the given number lie?
(a) 0. 06
(b) 0.45
(c) 0.19
(d) 0.66
(e) 0.92
(f) 0.57
Solution:
All these points lies between 0 and 1.

Ex 8.2 Class 6 Maths Question 7.
Write as fractions in lowest terms:
(a) 0.60
(b) 0.05
(c) 0.75
(d) 0.25
(f) 0.125
(g) 0.066
Solution:

Ex 8.3 Class 6 Maths 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:



In this case, the two numbers have the same parts upto tenths. The hundredth part of 5.64 is greater than that of 5.603.
∴ 5.64 > 5.603

Ex 8.3 Class 6 Maths Question 2.
Make five more examples and find the greater number from them.
(i) 0.5 and 0.8
(ii) 2.0 or 0.9
(iii) 0.042 or 0.22
(iv) 3.012 or 2.99
(v) 0.055 or 0.15
Solution:
Their solutions are:

Ex 8.4 Class 6 Maths Question 1.
Express as rupees using decimals:
(a) 5 paise
(b) 75 paise
(c) 20 paise
(d) 50 rupees 90 paise
(e) 725 paise
Solution:

Ex 8.4 Class 6 Maths Question 2.
Express as metres using decimals:
(a) 15 cm
(b) 6 cm
(c) 2 m 45 m
(d) 9 m 7 cm
(e) 419 cm
Solution:

Ex 8.4 Class 6 Maths Question 3.
Express as cm using decimals:
(a) 5 nun
(b) 60 mm
(c) 164 mm
(d) 9 cm 8 mm
(e) 93 mm
Solution:

Ex 8.4 Class 6 Maths Question 4.
Express as km using decimals:
(a) 8 m
(b) 88 m
(c) 8888 m
(d) 70 km 5 m
Solution:

Ex 8.4 Class 6 Maths Question 5.
Express as kg using decimals:
(a) 2 g
(b) 100 g
(c) 3750 g
(d) 5 kg 8 g
(e) 26 kg 50 g
Solution:

Ex 8.5 Class 6 Maths 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:

Ex 8.5 Class 6 Maths Question 2.
Rashid spent ₹ 35.75 for Maths book and ₹ 32.60 for Science book. Find the total amount spent by Rashid.
Solution:
Money spent on Maths book = ₹ 35.75
Money spent on Science book=₹ 32.60
∴ Total money spent is =₹ 35.75 +₹ 32.60 = ₹ 68.35

Ex 8.5 Class 6 Maths 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 Radhika’s mother = ₹ 10.50
Money given by Radhika’s father = ₹ 15.80
∴ Total amount given to Radhika = ₹ 10.50 + ₹ 15.80 = ₹ 26.30

Ex 8.5 Class 6 Maths 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 by Nasreen for her shirt = 3 m 20 cm
Cloth bought by Nasreen for her trouser = 2 m 5 cm
∴ Total cloth bought by Nasreen = 3 m 20 cm + 2 m 5 cm
= 5 m 25 cm

Ex 8.5 Class 6 Maths 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 in the morning = 2 km 35 m = 2.035 km
Distance walked in the evening = 1 km 7 m = 1.007 km
∴ Total distance walked by Naresh = 2.035 km +1.007 km = 3.042 km

Ex 8.5 Class 6 Maths Question 6.
Sunita travelled 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:
Distance travelled by bus = 15 km 268 m = 15.268 km
Distance travelled by car = 7 km 7 m = 7.007 km
Distance travelled on foot = 500 m = 0.500 km
Total distance of school from her residence
= 15.268 km + 7.007 km + 0.500 km = 22.775km

Ex 8.5 Class 6 Maths Question 7.
Ravi purchased 5 kg 400 g rice, 2 kg 20 g sugar and 10 kg 850 g flour. Find the total weight of his purchases.
Solution:
Weight of rice purchased = 5 kg 400 g = 5.400 kg
Weight of sugar purchased = 2 kg 20 gm = 2.020 kg
Weight of flour purchased = 10 kg 850 g = 10.850 kg
Total weight of purchases = 5.400 kg +2.020 kg +10.850 kg = 18.270 kg

Ex 8.6 Class 6 Maths Question 1.
Subtract:
(a) ₹ 18.25 from ₹ 20.75
(b) ₹ 202.54 m from 250 m
(c) ₹ 5.36 from ₹ 8.40
(d) 2.051 km from 5.206 km
(e) 0.314 kg from 2.107 kg
Solution:

Ex 8.6 Class 6 Maths 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.
Solution:

Ex 8.6 Class 6 Maths Question 3.
Raju bought a book for ₹ 35.65. He gave ₹ 50 to the shopkeeper. How much money did he get back from the shopkeeper?
Solution:
Cost of the book = ₹ 35.65
Money given to the shopkeeper by Raju = ₹ 50
Money he got back = ₹ 50 – ₹ 35.65 = ₹ 14.35

Ex 8.6 Class 6 Maths Question 4.
Rani had ₹ 18.50. She bought one ice-cream for ₹ 11.75. How much money does she have now?
Solution:
Money with Rani = ₹ 18.50
Money spent on ice-cream=₹ 11.75
Money left with Rani = ₹ 18.50 – ₹ 11.75 = ₹ 6.75

Ex 8.6 Class 6 Maths 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:
Total length of cloth = 20 m 5 cm = 20.05 m
Length cut out for curtain = 4 m 50 cm = 4.50 m
Cloth left over = 20.05 m – 4.50 m = 15.55 m

Ex 8.6 Class 6 Maths Question 6.
Namita travels 20 km 50 m every day. 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:
Total distance travelled by Namita
20 km 50 m = 20.050 km
Distance travelled by bus = 10 km 200 m = 10.200 km
∴ Distance travelled by auto = 20.050 km – 10.200 km
= 9.850 km

Ex 8.6 Class 6 Maths 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:
Total weight of vegetables bought = 10 kg
Weight of onions = 3 kg 500 g = 3.500 kg
Weight of tomato= 2 kg 75 g = 2.075 kg
∴ Total weight of these vegetables =3.500 kg + 2.075 kg = 5.575 kg
As per question
Weight of potatoes = 10 kg – 5.575 kg = 4.425 kg

Ex 7.1 Class 6 Maths Question 1.
Write the fraction representing the shaded portion.

Solution:
The fraction representing the shaded portion in the given diagram are as under:

Ex 7.1 Class 6 Maths Question 2.
Colour the part according to the given fraction


Solution:
The figures are shaded as per indicated fraction shown against them.

Ex 7.1 Class 6 Maths Question 3.
Identify the error, if any?

Solution:
In all the figures, the shaded portions do not represent the given fractions.

Ex 7.1 Class 6 Maths Question 4.
What fraction of a day is 8 hours?
Solution:
Since, there are 24 hours in a day, therefore, the required fraction
=824=13.

Ex 7.1 Class 6 Maths Question 5.
What fraction of an hour is 40 minutes?
Solution:
Since, there are 60 minutes in an hour, therefore, the required fraction =4060=23.

Ex 7.1 Class 6 Maths Question 6.
Arya, Abhimanyu and Vivek shared lunch. Arya has brought 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) To divide 2 sandwiches brought by Arya equally among Arya, Abhimanyu and Vivek (i.e., 3 persons), we have to divide each sandwich into 3 equal parts. Then two parts thus obtained be given to each of them.
(b) 13 of a sandwich will be received by each boy but in total each will get 23.

Ex 7.1 Class 6 Maths 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:
Total number of dresses = 30
Work finished = 20
∴ The fraction of dresses she has finished ==2030=23.

Ex 7.1 Class 6 Maths Question 8.
Write the natural numbers from 2 to 12. What fraction of them are prime numbers?
Solution:
Natural numbers from 2 to 12 are:
2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 i.e., 11 in number.
Prime numbers out of these numbers are:
2, 3, 5, 7, 11 i.e., 5 in number.
∴ Required fraction = =412=13

Ex 7.1 Class 6 Maths Question 9.
Write the natural numbers from 102 to 113. What fraction of them are prime numbers?
Solution:
Natural numbers from 102 to 113 are:
102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113 i.e., 12 in number.
Prime numbers out of the above numbers are:
103, 107, 109, 113 i.e., 4 in number.
∴ Required fraction ==412=13

Ex 7.1 Class 6 Maths Question 10.
What fraction of these circles have X’s in them?

Solution:
∴Required fraction = =48=12

Ex 7.1 Class 6 Maths 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:
Total number of CDs = 3 + 5 = 8
Number of CDs purchased = 3
Fraction of CDs purchased = 38
Fraction of CDs received as gift = 58 .

Ex 7.2 Class 6 Maths Question 1.
Draw number lines and locate the points on them:
(a) 12,14,34,44
(b) 18,28,38,78
(c) 25,35,85,45
Solution:
(a) We know that 12 is greater than zero and less than 1, so it should lie 2 between 0 and 1.
To show 12 on the number line, we divide the gap between 0 and 1 into two equal parts and the point P represents the fraction 12 .

To represent 14,34,44 on the number line, we divide the gap between 0 and 1 into 4 equal parts, then the points P, R and A respectively represent the fraction 14,34 and 44 .

(b) To rfepresent 18,28,38 and 78 on the number line, we divide the gap between 0 and 1 into 8 equal parts, then the points P, Q, R and S respectively represent the fraction 18,28,38 and 78 .

(c) To represent 25,35 and 45 on the number line, we divide the gap between 0 and 1 into 5 equal parts, then the points P, Q and R respectively represent the fraction 25,35 and 45 .

For representing 85
Draw a line. Take a point 0 on it. Let it represent 0.
Now, 85=135=1+35

From O, set off unit distances to the right. Let these segments be OA, AB. Then, clearly the points A and B represent 1 and 2 respectively. Take 1 full unit length to the right of O. Divide the 2nd unit AB into 5 equal parts. Take 3 parts out of these 5 parts to reach P. Then, P represents 85.

Ex 7.2 Class 6 Maths Question 2.
Express the following as mixed fractions:

Solution:

Ex 7.2 Class 6 Maths Question 3.
Express the following as improper fractions:

Solution:

Ex 7.3 Class 6 Maths Question 1.

Write the fractions. Are all these fractions equivalent?

Solution:

Ex 7.3 Class 6 Maths Question 2.
Write the fractions and pair up the equivalent fractions from each row.

Solution:

Ex 7.3 Class 6 Maths Question 3.
Replace □ in each of the following by the correct number:

Solution:

Ex 7.3 Class 6 Maths Question 4.
Find the equivalent fraction of 35 having
(a) denominator 20
(b) numerator 9
(c) denominator 30
(d) numerator 27
Solution:

Ex 7.3 Class 6 Maths Question 5.
Find the equivalent fraction of 3648 with
(a) numerator 9
(b) denominator 4
Solution:
(a) ∵ 36 ÷ 9 = 4
∴ We divide the numerator and denominator by 4
3648=36÷448÷4=912
(b) ∵ 48 ÷ 4 = 12
∴ We divide the numerator and denominator by 12
3648=36÷1248÷12=34

Ex 7.3 Class 6 Maths Question 6.
Check whether the given fractions are equivalent:

Solution:

Ex 7.3 Class 6 Maths Question 7.
Reduce the following fractions to simplest form:

Solution:

Ex 7.3 Class 6 Maths Question 8.
Ramesh had 20 pencils, Sheelu had 50 pencils and Jamaal had 80 pencils. After 4 months, Ramesh used up 10 pencils, Sheelu used up 25 pencils and Jamaal used up 40 pencils. What fraction did each use up? Check if each has used up an equal fraction of their of her/his pencils?
Solution:
Fractions of pencils used by Ramesh, Sheelu and Jamaal are 1020,2550 and 4080 respectively.

Thus, these fractions are equal.

Ex 7.3 Class 6 Maths Question 9.
Match the equivalent fractions and write two more for each:

Solution:

Ex 7.4 Class 6 Maths Question 1.
Write shaded portion as fraction. Arrange them in ascending and descending order using correct sign ‘>’, ‘=’, ‘>’ between the fractions:


Solution:

Ex 7.4 Class 6 Maths Question 2.
Compare the fractions and put an appropriate sign.

Solution:

Ex 7.4 Class 6 Maths Question 3.
Make five more such pairs and put appropriate signs.
Solution:
Five more such pairs may be taken as under:
Compare the fractions and put an appropriate sign.

Ex 7.4 Class 6 Maths Question 4.
Look at the figures and write ‘<’ or ‘>’, ‘=’ between the given pairs of fractions.

Make five more such problems and solve them with your friends.
Solution:

Ex 7.4 Class 6 Maths Question 5.
Five more such problems are as under:
Look at the figures given in question 5 and write ‘<’ or ‘>’ or ‘=’ between the pairs of fractions.

Solution:

Ex 7.4 Class 6 Maths Question 6.
How quickly can you do this? Fill appropriate sign. (<, =, >)

Solution:

Ex 7.4 Class 6 Maths Question 7.
The following fractions represent just three different numbers. Separate them into three groups of equivalent fractions, by changing each one to its simplest form.

Solution:

On separating them into three groups of equivalent fractions, we have

Ex 7.4 Class 6 Maths Question 8.
Find answers to the following. Write and indicate how you solved them.

Solution:

Ex 7.4 Class 6 Maths Question 9.
Ila read 25 pages of a book containing 100 pages. Lalita read 25 of the same book. Who read less?
Solution:

Ex 7.4 Class 6 Maths Question 10.
Rafiq exercised for 36 of an hour, while Rohit exercised for 34 of
an hour. Who exercised for a longer time?
Solution:

Ex 7.4 Class 6 Maths Question 11.
In a class A of 25 students, 20 passed in first class; in another class B of 30 students, 24 passed in first class. In which class was a greater fraction of students getting first class?
Solution:
Fraction of students getting first class in class A
=2025=20÷525÷5=45
Fraction of students getting first class in class B
=2430=24÷630÷6=45
Clearly, an equal number of students got 1st class in both the classes.

Ex 7.5 Class 6 Maths Question 1.
Write these fractions appropriately as additions or subtractions:

Solution:

Ex 7.5 Class 6 Maths Question 2.
Solve:

Solution:

Ex 7.5 Class 6 Maths Question 3.
Shubham painted 23 of the wall space in his room. His sister Madhavi helped and painted 13 of the wall space. How much did they paint together?
Solution:
Portion of wall painted by Shubham = 23
Portion of wall painted by Madhavi = 13
Wall painted by both of them = 23+13=2+13=33=1
i.e., full wall was painted by them.

Ex 7.5 Class 6 Maths Question 4.
Fill in the missing fractions.

Solution:

Ex 7.5 Class 6 Maths Question 5.
Javed was given 57 of a basket of oranges. What fraction of oranges was left in the basket? ‘
Solution:
Portion of oranges received by Javed = 57
Portion of oranges left in the basket

Ex 7.6 Class 6 Maths Question 1.
Solve:

Solution:

Ex 7.6 Class 6 Maths Question 2.
Sarita bought 25 metre of ribbon and Lalita 34 metre of ribbon.
What is the total length of the ribbon they bought?
Solution:
Ribbon bought of Sarita = 25 m
Ribbon bought of Lalita = 34 m
Total length of the ribbon bought = (25+34) m

Ex 7.6 Class 6 Maths Question 3.
Naina was given 112 piece of cake and Najma was given 113piece of cake. Find the total amount of cake was given to both of them.
Solution:
Number of pieces of cake consumed by Naina = 112
Number of pieces of cake consumed by Najma = 113
Total amount of cake consumed = 112+113

Ex 7.6 Class 6 Maths Question 4.
Fill in the boxes:

Solution:

Ex 7.6 Class 6 Maths Question 5.
Complete the addition-subtraction box.

Solution:

Ex 7.6 Class 6 Maths Question 6.
A piece of wire 78 metre long broke into two pieces. One piece was 14 metre long. How long is the other piece?
Solution:
Total length of wire = 78 metre
Length of a piece of wire = 14 metre
∴ Other piece of wire =(78−14) metre

Ex 7.6 Class 6 Maths Question 7.
Nandini’s house is 910 km from her school. She walked some distance and then took a bus for 12 km to reach the school. How
Solution:
Total distance of Nandini’s house from her school = 910 km
Distance for which bus was taken = 12 km
∴ Distance walked by Nandini for home

Ex 7.6 Class 6 Maths Question 8.
Asha and Samuel have bookshelves of the same size partly filled with hooks. Asha’s shelf is 25 th full and Samuel’s shelf is 56 th full. Whose bookshelf is more full? By what fraction?
Solution:

Ex 7.6 Class 6 Maths Question 9.
Jaidev takes 215 minutes to walk across the school ground. Rahul takes 74 minutes to do the same. Who takes less time and by what fraction?
Solution:

Ex 6.1 Class 6 Maths Question 1.
Write opposites of the following:
(a) Increase in weight
(b) 30 km north
(c) 326 BC
(d) Loss of ₹ 700
(e) 100 m above sea level
Solution:
(a) Decrease in weight
(b) 30 km South
(c) 326 AD
(d) Gain of ₹ 700
(e) 100 m below sea level

Ex 6.1 Class 6 Maths Question 2.
Represent the following numbers as integers with appropriate signs.
(a) An aeroplane is flying at a height two thousand metre above the ground.
(b) A submarine is moving at a depth, eight hundred metre below the sea level.
(c) A deposit of rupees two hundred.
(d) Withdrawal of rupees seven hundred.
Solution:
(a) +2000
(b) -800
(c) +200
(d) -700.

Ex 6.1 Class 6 Maths Question 3.
Represent the following numbers on a number line:
(a) +5
(b) -10
(c) +8
(d) -1
(e) -6
Solution:
In order to represent integers on a number line, we draw a line and mark a point O almost in the middle of it as shown. Now, set off equal distances on the right hand side as well as on the left hand side of O. On the right side of O, label the points of subdivisions as 1,2, 3 etc. and the point O as 0. Since negative integers are opposite of positive integers (i.e., natural numbers), so we represent negative integers in the opposite direction Le., on the left side of O on the number line. On the left side of point O, label the points of subdivision by -1, -2, -3 etc. as shown in figure.

Clearly (a) +5 (b) -10 (c) +8 (d) -1 and (e) -6 are represented by the points A, B, C, D and E respectively on the number line.

Ex 6.1 Class 6 Maths Question 4.
Adjacent figure is a vertical number line, representing integers. Observe it and locate the following points:

(a) If point D is+ 8, then which point is -8?
(b) Is point G a negative integer or a positive integer?
(c) Write integers for points B and E.
(d) Which point marked on this number line has the least value?
(e) Arrange all the points in decreasing order of value.
Solution:
From the given vertical number line, clearly
(a) The point F represents -8.
(b) Since G lies below O, so it is a negative integer.
(c) Point B represents +4 and E represents -10.
(d) E has the least value.
(e) Points in decreasing order are as under: D, C, B, A, O, H, G, F, E

Ex 6.1 Class 6 Maths Question 5.
Following is the list of temperatures of five places in India, on a particular day of the year.

(a) Write the temperatures of these places in the form of integers in the blank column.
(b) Following is the number line representing the temperature in degree Celsius.

Plot the name of the city against its temperature.
(c) Which is the coolest place?
(d) Write the names of the places where temperatures are above 10°C.
Solution:
(a) Siachin: -10 °C Shimla: -8°C
Ahmedabad: +30°C Delhi: +35°C
Srinagar: -2°C
(b) Plotting the name of the city against its temperature on the number line representing the temperature in degree Celsius, we have

(c) The coolest place is Siachin.
(d) The places having temperatures, above 10°C are Ahmedabad anti Delhi.

Ex 6.1 Class 6 Maths Question 6.
In each of the following pairs, which number is to the right of the other on the number line?
(a) 2, 9
(b) -3, -8
(c) 0, -1
(d) -11, 10
(e) -6, 6
(f) 1, -100
Solution:
(a) 9 is on the right of 2 on the number line.
(b) -3 is on the right of -8 on the number line.
(c) 0 is on the right of -1 on the number line.
(d) 10 is on the right of -11 on the number line.
(e) 6 is on the right of-6 on the number line.
(f) 1 is on the right of -100 on the number line.

Ex 6.1 Class 6 Maths Question 7.
Write all the integers between the given pairs (write them in the increasing order).
(a) 0 and -7
(b) -4 and 4
(c) -8 and -15
(d) -30 and -23
Solution:
(a) All the integers between 0 and -7 in the increasing order are -6, -5, -4, -3, -2,-1.
(b) All the integers between -4 and 4 in the increasing order are -3, -2, -1, 0, 1, 2, 3.
(c) All the integers between -8 and -15 in the increasing order are -14, -13,-12,-11,-10,-9.
(d) All the integers between -30 and -23 in the increasing order are -29, -28, -27, -26, -25, -24.

Ex 6.1 Class 6 Maths Question 8.
(a) Write four negative integers greater than 20.
(b) Write four negative integers less than 10.
Solution:
(a) Four negative integers greater than -20 are -20 +1, -20 + 2, -20 + 3 and -20 + 4
i. e., -19 , -18, -17 and -16.
(b) Four negative integers less than-10 are -10 -1, -10 -2, -10 -3 and -10 -4
i. e., -11, -12, -13, -14.

Ex 6.1 Class 6 Maths Question 9.
For the following statements, write true (T) or false (F). If the statement is false, correct the statement.
(a) -8 is to the right of -10 on a number line.
(b) -100 is to the right of -50 on a number line.
(c) Smallest negative integer is -1.
(d) -26 is larger than -25.
Solution:
(a) True.
(b) False. The correct statement should be “ -100 is to the left of -50 on a number line”.
(c) False. The correct statement should be “ There is no last or smallest negative number”.
(d) False. The correct statement should be “ -26 is smaller than -25’.

Ex 6.1 Class 6 Maths Question 10.
Draw a number line and answer the following:
(a) Which number will we reach if we move 4 numbers to the right of -2?
(b) Which number will we reach if we move 5 numbers to the left of 1?
(c) If we are at -8 on the number line, in which direction should we move to reach -13?
(d) If we are at -6 on the number line, in which direction should we move to reach -1?
Solution:
(a) We want to obtain integer on moving 4 numbers to the right of -2. So, we start from -2 and proceed 4 units to the right of -2 to obtain 2 as shown in the figure.

Hence, we will reach the number 2.
(b) We want to obtain integer on moving 5 numbers to the left of 1. So, we start from 1 and proceed 5 units to the left of 1 to obtain -4 as shown in the figure.

Hence, we will reach the number -4. ,
(c) Since -13 < -8. So, to reach -13 from -8 on the number line we should move to the left of -8.
(d) Since -1 > -6. So,, to reach -1 from -6 on the number line, we should move to the right of -6.

Ex 6.2 Class 6 Maths Question 1.
Using the number line write the integer which is:
(a) 3 more than 5
(b) 5 more than -5
(c) 6 less than 2
(d) 3 less than -2
Solution:
(a) We want to obtain integer 3 more than 5. So, we start from 5 and proceed 3 units to the right to obtain 8 as shown in the figure.

Hence, 3 more than 5 is 8.
(b) We want to obtain integer 5 more than -5. So, we start from -5 and proceed 5 units to the right to obtain 0 as shown in the figure.

Hence, 5 more than -5 is 0.
(c) We want to obtain an integer 6 less than 2. So, we start from 2 and proceed 6 units to the left of it to obtain -4 as shown in the figure.

Hence, 6 less than 2 is -4.
(d) We want to obtain an integer 3 less than -2. So, we start from -2 and proceed 3 units to the left of it to obtain -5 as shown in the figure.

Hence, 3 less than -2 is -5.

Ex 6.2 Class 6 Maths Question 2.
Use number line and add the following integers:
(a) 9 + (-6)
(b) 5 + (-11)
(c) (-1) + (-7)
(d) (-5) + 10
(e) (-1) + (-2) + (-3)
(f) (-2) + 8 + (-4)
Solution:
(a) First we move to the right of 0 by 9 steps reaching 9. Then we move 6 steps to the left of 9 reaching 3 as shown in the figure.

Thus, 9 + (-6) = 3.
(b) First we move to the right of 0 by 5 steps reaching 5. Then, we move 11 steps to the left of 5 reaching -6 as shown in the figure.

Thus, 5 + (-11) = -6.
(c) First we move to the left of 0 by 1 step reaching -1. Then we move 7 steps to* the left of -1 reaching -8 as shown in the figure.

Thus, (-1) + (-7) = -8.
(d) First we move to the left of 0 by 5 steps reaching -5. Then we move 10 steps to the right of -5 reaching 5 as shown in the figure.

Thus, (-5) + 10 = 5.
(e) First we move to the left of 0 by 1 step reaching -1. Then we move 2 steps to the left of -1 reaching -3. Again, we move 3 steps to the left of -3 reaching -6 as shown in the figure.

Thus, (-1) + (-2) + (-3) = -6.
(f) First we move to the left of 0 by 2 steps reaching -2. Then we move 8 steps to the right of -2 reaching 6. Again, we move 4 steps to the left of 6 reaching -2 as shown in the figure.
Thus, (-2) + 8 + (-4) = 2.

Ex 6.2 Class 6 Maths Question 3.
Add without using number line:
(a) 11 + (-7)
(b) (-13) + (+18)
(c) (-10) + (+19)
(d) (-250) + (+150)
(e) (-380)+ (-270)
(f) (-217) + (-100)
Solution:
(a) 11 + (-7) = 4 + 7 + (-7)
= 4 + 0 = 4
(b) (-13) + (+18) = (-13) + (+13) + (+ 5)
= 0 + (+5) = 5
(c) (-10) + (+19) = (-10) + (+10) + (+9)
= 0 + (+9) = +9
(d) (-250) + (+150) = (-100) + (-150) + (+150)
= (-100) + 0 = -100
(e) (-380) + (-270) = -650
(f) (-217) + (-100) = -317

Ex 6.2 Class 6 Maths Question 4.
Find the sum of:
(a) 137 and -354
(b) -52 and 52
(c) -312, 39 and 192
(d) -50, – 200 and 300
Solution:
(a) (+137) + (-354) = (+137) + (-137) + (-217)
= 0 + (-217) =-217
(b) (-52) + (+52) = 0
(c) (-312) + (+39) + (+192) = (-312) + (+231)
= (-81) + (-231) + (+231)
= (-81) + 0 = -81
(d) (-50) + (-200) + (+300) = (-250) + (+300)
= (-250) + (+250) + (+50)
= 0 + (+50) = +50

Ex 6.2 Class 6 Maths Question 5.
Find the sum:
(a) (-7) + (-9) + 4 + 16
(b) (37) + (-2) + (-65) + (-8)
Solution:
(a) (-7) + (-9) + 4 + 16 = (-16) + 4 + 16
= (-16) + (+16) + 4 = 0 + 4 = 4
(b) (37) + (-2) + (-65) + (-8) = 37 + (-75)
= 37+ (-37)+ (-38)
= 0 + (-38) = -38

Ex 6.3 Class 6 Maths Question 1.
Find:
(a) 35-(20)
(b) 72 (90)
(c) (-15) – (-18)
(d) (-20) – (13)
(e) 23 (-12)
(f) (-32) – (-40)
Solution:
(a) 35 – 20 = (15 + 20) – 20
= 15 + (20 – 20) = 15 + 0 = 15
(b) 72 – 90 = 72 – (72 + 18)
= (72 – 72) + (-18)
= 0 + (-18) = -18 1
(c) (-15) – (-18) = (-15) + (additive inverse -18)
= -15 + 18 = -15 +(15 + 3)
= (-15 + 15) 43 = 0 + 3 = +3
(d) (-20) -13 = -20 -13 = -(20 + 13) = -33
(e) 23 – (-12) = 23 + 12 = 35
(f) (-32) – (-40) = -32 + 40 = (-32 + 32) +8
= 0 + 8 + 8

Ex 6.3 Class 6 Maths Question 2.
Fill in the blanks with >, < or = sign:
(a) (-3) + (-6) …… (-3) -(-6)
(b) (-21) – (-10) …… (-31) + (-11)
(e) 45 – (-11) ……. 57 + (-4)
(d) (-25) – (-42) …….. (-42) – (-25)
Solution:
(a) Here, (-3) + (-6) = -3 – 6 = =9
and (-3)-(-6) = -3 + 6 =3
As -9 < 3
∴ (-3) + (-6) < (-3) – (-6) (b)

(b) Here, (-21) – (-10) = -21 +10 = -11
and (-31)+ (-11) = -31 -11 = -42
As -11 > -42
∴ (-21) – (-10) > (-31) + (-11)

(c) Here, 45 – (-11) = 45 + 11 = 56
and 57 + (-4) = 57-4 = 53
As 56 > 53
∴ 45 – (-11) > 57 + (-4)

(d) Here, -25 – (-42) = -25 + 42
= -25 + (25 + 17)
= (-25 + 25) +17 = 0 + 17 = 17
and, (-42) – (-25) = -42 + 25
= (-17 – 25) + 25
= -17 + (-25 + 25) = -17 + 0 = -17
As 17 > -17
∴ (-25) – (-42) > (-42) – (-25)

Ex 6.3 Class 6 Maths Question 3.
Fil in the blanks:
(a) (-8)+ ….. = 0
(b) 13 + …….. = 0
(c) 12 + (-12) = ……
(d) (-4) + ……. = -12
(e) ………. -15 = -10
Solution:
(a) (-8) + (+8) = 0
(b) 13 + (-13) = 0
(c) 12 + (-12) = 0
(d) (-4) +(-8) = -12 [∵ -12 – (-4) = -12 + 4 = -8]
(e) 5-15 = -10 [∵ -10 + 15 = 5]

Ex 6.3 Class 6 Maths Question 4.
Find:
(a) (-7) – 8- (-25)
(b) (-13) + 32 – 8 – 1
(c) (-7) + (-8) + (-90)
(d) 50 – (-40) – (-2)
Solution:
(a) (-7) – 8 – (-25) = -7 – 8 + 25
= -15 + 25 = 10.
(b) (-13) + 32 – 8-1 = -13 – 8 – 1 + 32
= -22 + 32 = 10
(c) (-7) + (-8) + (-90) = -7 – 8 – 90 = -105.
(d) 50 – (-40) – (-2) = 50+ 40 + 2 = 92.

Ex 5.1 Class 6 Maths Question 1.
What is the disadvantage in comparing line segments by mere observation?
Solution:
The disadvantage in comparing line segments by mere observation is the inaccuracy in judgement.

Ex 5.1 Class 6 Maths Question 2.
Why is it better to use a divider than a ruler, while measuring the length of a line segment?
Solution:
It is better to use a divider with ruler as it given an accurate measurement of the line segment.

Ex 5.1 Class 6 Maths Question 3.
Draw any line segment, say AB. Take any point C lying in between A and B. Measure the lengths of AB, BC and AC. Is AB = AC + CB?
[Note: If A, B, C are any three points on a line, such that AC + CB = AB, then we can be sure that C lies between A and B.]
Solution:
On measuring the lengths of line segments AB, BC and AC using ruler, we find

AB =6.5 cm,
CB = 4.2 cm,
and AC = 2.3 cm
Now, AC +CB = 2.3 cm +4.2 cm = 6.5 cm = AB
AB = AC+CB.

Ex 5.1 Class 6 Maths Question 4.
If A, B, C are three points on a line such that AB = 5 cm, BC = 3 cm and AC = 8 cm, which one of them lies between the other two?
Solution:
Since, AB + BC = 5 cm +3 cm = 8 cm = AC
∴ A, B, C are collinear and B lies between A and C.

Ex 5.1 Class 6 Maths Question 5.
Verify, whether D is the mid point of AG¯¯¯¯¯¯¯¯.

Solution:
Since AD = AB +BC + CD = 3 units
and, DG = DE +EF + FG = 3 units
∵ AD = DG [∵ Each = 3 units]
Thus, D is the mid-point of AG¯¯¯¯¯¯¯¯.

Ex 5.1 Class 6 Maths Question 6.
If B is the mid point of AC¯¯¯¯¯¯¯¯ and C is the mid point of BD¯¯¯¯¯¯¯¯, where A, B, C, D lie on a straight line, say why AB = CD?
Solution:
∵ B is the mid point of AC¯¯¯¯¯¯¯¯. Therefore, AB = BC.
∵ C is the mid-point of BD¯¯¯¯¯¯¯¯. Therefore, BC = CD.
Thus, AB =BC = CD i.e., AB = CD.

Ex 5.1 Class 6 Maths Question 7.
Draw five triangles and measure their sides. Check in each case, if the sum of the lengths of any two sides is always less than the third side.
Solution:
Draw any five triangles, T₁, T₂, T₃, T₄ and T₅. label each one as ∆ ABC. Measure, in each case, the three sides a = BC, b = CA and c = AB.

Let us tabulate the measurements of sides as under:

From the above table, we find that
(i) each value of b + c – a is positive;
(ii) each value of c + a – b is positive;
(iii) each value of a + b – c is positive.
Now, b + c – a is positive ⇒ b + c – a > 0 ⇒ b + c > a
c + a – b is positive ⇒ c + a- b > 0 ⇒ c + a > b
a+ b – c is positive ⇒ a + b – c > 0 ⇒ a + b > c
Thus, it is verified that the sum of any two sides of a triangle is greater than the third side.
∴ The statement ‘the sum of two sides is ever less than the third side’ is never true.

Ex 5.2 Class 6 Maths Question 1.
What fraction of a clockwise revolution does the hour hand of a clock turn through, when it goes from
(a) 3 to 9
(b) 4 to 7
(c) 7 to 10
(d) 12 to 9
(e) 1 to 10
(f) 6 to 3
Solution:
(a) 12 of a revolution
(b) 14 of a revolution
(c) 14 of a revolution
(d) 34 of a revolution
(e) 34 of a revolution
(f) 34 of a revolution

Ex 5.2 Class 6 Maths Question 2.
Where will the hand of a clock stop if it
(a) starts at 12 and makes 12 of a revolution, clockwise?
(b) starts at 2 and makes 12 of a revolution clockwise?
(c) starts at 5 and makes 14 of a revolution, clockwise?
(d) starts at 5 and makes 34 of a revolution, clockwise?
Solution:
(a) If the hand of clock starts at 12 and makes 12 of a revolution clockwise it reaches at 6.
(b) If the hand of clock starts at 2 and makes 12 of a revolution clockwise it reaches at 8.
(c) If the hand of clock starts at 5 and makes 14 of a revolution clockwise it reaches at 8.
(d) If the hand of clock starts at 5 and makes 34 of a revolution clockwise it reaches at 2.

Ex 5.2 Class 6 Maths Question 3.
Which direction will you face if you start facing
(a) east and make 12 of a revolution clockwise?
(b) east and make 112 of a revolution clockwise?
(c) west and make 34 of a revolution anti-clockwise?
(d) south and make one full revolution?
(Should we specify clockwise or anit-clockwise for this last question? Why nor?)
Solution:
(a) Facing east and on making 12 of a revolution clockwise, we will face west.

(b) Facing east and on making 112 of a revolution clockwise, we will face west.

(c) Facing west and on making 34 of a revolution anti-clockwise, we will face north.

(d) Facing south and on making 1 full revolution we will face south again.
There is no need to specify clockwise or anti-clockwise in the last question as turning by one full revolution (i.e., two straight angles) make us to reach the original position.

Ex 5.2 Class 6 Maths Question 4.
What part of a revolution have you turned through if you stand facing
(a) east and turn clockwise to face north?
(b) south and turn clockwise to face east?
(c) west and turn clockwise to face east?
Solution:
Figures drawn will dearly show the amount of revolution:
(a) 34 of a revolution

(b) 34 of a revolution

(c) 12 of a revolution

Ex 5.2 Class 6 Maths Question 5.
Find the number of right angles turned through by the hour hand of clock when it goes from
(a) 3 to 6
(b) 2 to 8
(c) 5 to 11
(d) 10 to 1
(e) 12 to 9
(f) 12 to 6
Solution:
Since turning by two straight angles (or four right angles) in the same direction makes a full turn. Thus for the movement of clock hand from 12 – 1,1 – 2, 2 – 3, … 11 – 12 i.e., 12 in number gives 4 right angles turn. Therefore,
(a) From 3 to 6 : 3 – 4, 4 – 5 and 5 – 6 i.e., 3 in number.
∴ Number of right angles turned through by the hour hand of a clock when it goes from 3 to 6
= (412×3) right angles = 1 right angles

(b) From 2 to 8 :2 – 3,3 – 4, 4 – 5, 5 – 6,6 – 7 and 7 – 8 i.e., 6 in number.
∴ Number of right angles turned through by the hour hand of a clock
when it goes from 2 to 8
= (412×6) right angles = 2 right angles

(c) From 5 to II : 5 – 6, 6 – 7, 7 – 8, 8 – 9, 9 – 10 and 10 – 11 i.e., 6 in number.
∴ Number of right angles turned through by the hour hand of a clock when it goes from 5 to 11
= (412×6) right angles = 2 right angles

(d) From 10 to 1 :10 – 11,11 -12 and 12 – 1 i.e., 3 in number.
∴ Number of right angles turned through by the hour hand of a clock when it goes from 10 to 1
= (412×3) right angles = 1 right angles

(e) From 12 to 9 : 12 – 1, 1 – 2, 2 – 3, 3 – 4, 4 – 5, 5 – 6, 6 – 7, 7 – 8 and 8 – 9 i.e., 9 in number
∴ Number of right angles turned through by the hour hand of a clock when it goes from 12 to 9
= (412×9) right angles = 1 right angles

(f) From 12 to- 6 : 12 – 1, 1 – 2, 2 – 3, 3 – 4, 4 – 5 and 5 – 6 i.e., 6 in number.
∴ Number of right angles turned through by the hour hand of a clock when it goes from 12 to 6
= (412×6) right angles = 2 right angles

Ex 5.2 Class 6 Maths Question 6.
How many right angles do you make if you start facing
(a) south and turn clockwise to west?
(b) north and turn anti-clockwise to east?
(c) west and turn to west?
(d) south and turn to north?
Solution:
Figures drawn will clearly indicate the number of right angles we make if we start facing:
(a) south and turn clockwise to west is 1 right angle.

(b) north and turn anti-clockwise to east are 3 right angles.

(c) west and turn to west are 4 right angles.

(d) south and turn to north are 2 right angles.

Ex 5.2 Class 6 Maths Question 7.
Where will the hour hand qfa clock stop if it starts
(a) from 6 and turns through 1 right angle?
(b) from 8 and turns through 2 right angles?
(c) from 10 and turns through 3 right angles?
(d) from 7 and turns through 2 straight angles?
Solution:
We know that in 12 hours hour hand turns through 4 right angles i.e., in a turn of 1 right angle, the hour hand moves 3 hours.
(a) If the hour hand of a clock starts from 6 and turns through 1 right angle then its hour hand stops at (6 +3) i.e., 9.
(b) If the hour hand of a clock starts from 8 and turns through 2 right
angles then its hour hand stops at (8 + 2 x 3) = 14 i.e., 2. .
(c) If the hour hand of a clock starts from 10 and turns through 3 right angles then its hour hand stops at (10+3×3) = 19 i.e., 7.
(d) If the hour hand of a clock starts from 7 and turns through 2 straight angles i.e., 4 right angles then its hour hand stops at (7 + 4 x 3) = 19 i.e., 7.

Ex 5.3 Class 6 Maths Question 1.
Match the following:
(i) Straight angle                             (a) Less than one-fourth of a revolution
(ii) Right angle                               (b) More than half a revolution
(iii) Acute angle                             (c) Half of a revolution
(iv) Obtuse angle                         (d) One-fourth of a revolution
(v) Reflex angle                         (e) Between 14 and 12 of a revolution
(f) One complete revolution
Solution:
On matching, we find
(i) → (c)
(ii) → (d)
(iii) → (a)
(iv) → (e)
(v) → (b)

Question 2.
Classify each one of the following angles as right, straight, acute, obtuse or reflex:

Solution:
Clearly, we find that
(a) → acute angle
(b) → obtuse angle
(c) → right angle
(d) → reflex angle
(e) → straight angle
(f) → acute angles.

Ex 5.4 Class 6 Maths Question 1.
What is the measure of
(i) a right angle?
(ii) a straight angle?
Solution:
(i) 1 right angle = 90°
(ii) 1 straight angle = 2 right angles = 2 x 90°=180°

Ex 5.4 Class 6 Maths Question 2.
Say True or False:
(a) The measure of an acute angle < 90°.
(b) The measure of an obtuse angle < 90°.
(c) The measure of a reflex angle > 180°.
(d) The measure of one complete revolution = 360°.
(e) If m∠A = 53° and m∠B = 35°, then m∠A > m∠B.
Solution:
(a) True
(b) False
(c) True
(d) True
(e) True

Ex 5.4 Class 6 Maths Question 3.
Write down the measures of
(a) some acute angles.
(b) some obtuse angles.
(give at least two examples of each).
Solution:
(a) Since the measure of an acute angle < 90°. So, two acute angles may be of 60° and 30°.
(b) Since the measure of an obtuse angle > 90°. So, two obtuse angles may be of 110° and 120°.

Ex 5.4 Class 6 Maths Question 4.
Measure the angles given below using the Protractor and write down the measure.

Solution:
On measuring the given angles using protractor, we find that
(a) 45°
(b) 125°
(c) 90°
(d) ∠1 = 40°, ∠2 = 125° and ∠3 = 95°.

Ex 5.4 Class 6 Maths Question 5.
Which angle has a large measure ? First estimate and then measure.
Measure of Angle A =
Measure of Angle B =

Solution:
Observing the given angles, we find that
∠B > ∠A
On measuring, we find that
Measure of Angle A =40°
Measure of Angle B = 65°
∴ ∠B > ∠A.

Ex 5.4 Class 6 Maths Question 6.
From these two angles which has larger measure? Estimate and then confirm by measuring them.

Solution:
On observing, we find that
∠B > ∠A
On measuring, we find that
∠A = 45° and ∠B =60°
Thus, ∠B > ∠A.

Ex 5.4 Class 6 Maths Question 7.
Fill in the blanks with acute, obtuse, right or straight:
(a) An angle whose measure is less than that of a right angle is ……. .
(b) An angle whose measure is greater than that of a right angle is ……… .
(c) An angle whose measure is the sum of the measures of two right angles is ……. .
(d) When the sum of the measures of two angles is that of a right angle, then each one of them is …… .
(e) When the sum of the measures of two angles is that of a straight angle and if one of them is acute then the other should be …… .
Solution:
(a) acute
(b) obtuse
(c) straight
(d) acute
(e) obtuse

Ex 5.4 Class 6 Maths Question 8.
Find the measure of the angle shown in each figure. (First estimate with your eyes and then find the actual measure with a protractor).

Solution:
On estimating without measuring, the measure of the angle shown in each figure is
(a) 45°
(b) 60°
(c) 125°
(d) 135°

On measuring with the protractor, we find their measures as under:
(a) 40°
(b) 65°
(c) 130°
(d) 135°

Ex 5.4 Class 6 Maths Question 9.
Find the angle measure between the hands of the clock in each figure:

Solution:
The angle measure between the hands of the clock in the time shown is as under:
At 9.00 a.m. : 90°
At 1.00 p.m. : 30°
At 6.00 p.m. : 180°

Ex 5.4 Class 6 Maths Question 10.
Investigate
In the given figure, the angle measures 30°. Look at the same figure through a magnifying glass. Does the angle becomes larger? Does the size of the angle change?

Solution:
Looking through a magnifying glass does not make the angle larger and so the size of the angle does not change.

Ex 5.4 Class 6 Maths Question 11.
Measure and classify each angle:

Solution:
The measure of angle and its classification is recorded in the given table as under:

Ex 5.5 Class 6 Maths Question 1.
Which of the following are models for perpendicular lines:
(a) The adjacent edges of a table top.
(b) The lines of a railway track.
(c) The line segments forming the letter ‘L’.
(d) The letter V.
Solution:
(a) Yes
(b) No
(c) Yes
(d) No

Ex 5.5 Class 6 Maths Question 2.
Let PQ¯¯¯¯¯¯¯¯ be the perpendicular to the line segment XY¯¯¯¯¯¯¯¯. Let PQ¯¯¯¯¯¯¯¯ and XY¯¯¯¯¯¯¯¯ intersect in the point A. What is the measure of ∠PAY?
Solution:
The measure of ∠PAY =90° [∵ PQ ⊥ XY, given]

Ex 5.5 Class 6 Maths Question 3.
There are two “set-squares” in your box. What are the measures of the angles that are formed at their corners? Do they have any angle measure that is common?
Solution:
In one, the angles are of 30°, 60°, 90° and in the other the angles are of 90°, 45°, 45°.
Clearly, the common angle = 90°

Ex 5.5 Class 6 Maths Question 4.
Study the diagram. The line l is perpendicular to line m.

(a) Is CE = EG?
(b) Does PE bisect CG?
(c) Identify any two line segments for which PE is the perpendicular bisector. ‘ ’
(d) Are these true?
(i) AC > FG
(ii) CD = GH
(iii) BC< EH
Solution:

Ex 5.6 Class 6 Maths Question 1.
Name the types of following triangles:
(a) Triangle with lengths of sides 7 cm, 8 cm and 9 cm.
(b) ∆AJBC with AB = 8.7 cm, AC = 7 cm and BC = 6 cm.
(c) ∆PQR such that PQ = QR = PR = 5 cm.
(d) ∆DEF with m∠D = 90°
(e) ∆XYZ with m∠Y = 90° and XY = YZ.
(f) ∆LMN with m∠L = 30°, m∠M = 70° and m∠N = 80°
Solution:
(a) Since all the sides are of different lengths
∴ It is scalene triangle.
(b) In ∆ABC, AB ≠ BC ≠ CA.
∴ AABC is scalene triangle.
(c) In ∆PQR, PQ = QR = PR
∴ ∆PQR is equilateral triangle.
(d) In ∆DEF, m∠D =90°
∴ ∆DEF is right angled triangle.
(e) In ∆ XYZ, m∠Y = 90° and XY = YZ
∴ ∆ XYZ is an isosceles right angled.
(f) As each angle of ∆ LMN is < 90°
∴ ∆ LMN is an acute angled triangle.

Ex 5.6 Class 6 Maths Question 2.
Match the following:

Solution:
The matching is as shown:
(i) → (e)
(ii)→ (g)
(iii) → (a)
(iv) → (f)
(v) → (d)
(vi) → (c)
(vii) → (b)

Ex 5.6 Class 6 Maths Question 3.
Name each of the following triangles in two different ways: (you may judge the nature of the angle by observation)

Solution:
(a) Isosceles triangle; Acute angled triangle.
(b) Scalene triangle; Right angled triangle.
(c) Isosceles triangle; Obtuse angled triangle.
(d) Isosceles triangle; Right angled triangle.
(e) Equilateral triangle; Acute angled triangle.
(f) Scalene triangle; Obtuse angled triangle.

Ex 5.6 Class 6 Maths Question 4.
Try to construct triangles match sticks. Some are shown here. Gan you make a triangle with

(a) 3 matchsticks?
(b) 4 matchsticks?
(c) 5 matchsticks?
(d) 6 matchsticks?
(Remember you have to use all the available matchsticks in each case)
Name the type of triangle in each case.
If you cannot make a triangle, think of reasons for it.

Solution:
(a) Yes, an equilateral triangle.
(b) No (as sum of two sides is always > the third side but here 1 + 1 = 2 ≱ ≯ 2).
(c) Yes, an isosceles triangle.
(d) Yes, an equilateral triangle.

Ex 5.7 Class 6 Maths Question 1.
Say True or False:
(a) Each angle of a rectangle is a right angle.
(b) The opposite sides of a rectangle are equal in length.
(c) The diagonals of a square are perpendicular to ont i^Atber.
(d) All the sides of rhombus are of equal length.
(e) All the sides of a parallelogram are of equal length.
(f) The opposite sides of a trapezium are parallel.
Solution:
(a) True
(b) True
(c) True
(d) True
(e) False
(f) False

Ex 5.7 Class 6 Maths Question 2.
Give reasons for the following:
(a) A square can be thought of as a special rectangle.
(b) A rectangle can be thought of as a special parallelogram.
(c) A square can be thought of as a special rhombus.
(d) Squares, rectangles, parallelograms are all quadrilaterals.
(e) Square is also a parallelogram.
Solution:
(a) All the properties of a rectangle are there in a square.
(b) All the properties of a parallelogram are there in a rectangle.
(c) All the properties of a rhombus are there in a square.
(d) All the four sided closed plane figure are known as quadrilateral.
(e) All the properties of a parallelogram are there in a square.

Ex 5.7 Class 6 Maths Question 3.
A figure is said to be regular if its sides are equal in length and angles are equal in measure. Can you identify the regular quadrilateral?
Solution:
A regular quadrilateral is a square, regular pentagon, regular hexagon, regular octagon etc.

Ex 5.8 Class 6 Maths Question 1.
Examine whether the following are polygons. If any one among them is not, say why?

Solution:
(a) No, because it is not closed,
(b) Yes.
(c) No, because it is not drawn by using line segments.
(d) No, because it is not closed by line segments.

Ex 5.8 Class 6 Maths Question 2.
Name each polygon:

make two more examples of each of these.
Solution:
(a) Quadrilateral
(b) Triangle
(c) Pentagon
(d) Octagon

Two more examples of :

Ex 5.8 Class 6 Maths Question 3.
Draw a rough sketch of a regular hexagon. Connecting any three of its vertices, draw a triangle. Identify the type of the triangle you have drawn.
Solution:

Let ABCDEF be a rough sketch of a regular hexagon.
Join AC
A ∆ ABC is obtained in which AB = BC. Thus,
∆ ABC is an isosceles triangle.

Ex 5.8 Class 6 Maths Question 4.
Draw a rough sketch of a regular octagon.
(Use squared paper if you wish). Draw a rectangle by joining exactly four of the vertices of the octagon.
Solution:
Let ABCDEFGH be a rough sketch of a regular octagon. Join AC, CE, EG and GA to obtain a rectangle ACEG.

Ex 5.8 Class 6 Maths Question 5.
A diagonal is a line segment that joins any two vertices of the polygon and is not a side of the polygon. Draw a rough sketch of a pentagon and draw its diagonals.
Solution:
Let ABCDE be a rough sketch of a pentagon. To draw its diagonals, join AC, AD, BD, DE and CE. Then, AC, AD, BD, BE and CE are its diagonals.

Ex 5.9 Class 6 Maths Question 1.
Match the following:


Give two new examples of each shape.
Solution:
Matching is as under:
(a) → (ii)
(b) → (iv)
(c) → (v)
(d) → (iii)
(e) → (i)

Two new examples of

(a) Cone: Conical tent.
(b) Sphere: Tennis ball, a ball of wool.
(c) Cylinder: Measuring jars, gas cylinder.
(d) Cuboid: Brick, Match box.
(e) Pyramid: Pyramids of Egypt.

Ex 5.9 Class 6 Maths Question 2.
What shape is
(a) Your instruments box?
(b) A brick?
(c) A match box?
(d) A road-roller?
(e) A sweet laddu?
Solution:
(a) Cuboid
(b) Cuboid
(c) Cuboid
(d) Cylinder
(e) Sphere

Ex 4.1 Class 6 Maths Question 1.
Use the figure to name:
(a) Five points
(b) A line
(c) Four rays
(d) Five line segments
Solution:
Clearly, from the given figure
(a) 5 points are points, O, C, B, D and E.
(b) A line is [altex]\overleftrightarrow { BD } [/latex]
(c) 4 rays are OC←→,OB←→,OD←→ and ED←→.
(d) 5 line segments are OE¯¯¯¯¯¯¯¯,OD¯¯¯¯¯¯¯¯,DE,¯¯¯¯¯¯¯¯¯OB¯¯¯¯¯¯¯¯ and BD¯¯¯¯¯¯¯¯.

Ex 4.1 Class 6 Maths Question 2.
Name the line given in all possible (twelve) ways, choosing only two letters at a time from the four given.
Solution:
All the possible ways of naming the given by choosing only two letters at a time out of 4 letters are
AB←→,AC←→,AD←→,BC←→,BD←→,CD←→,DC←→,DB←→,CB←→,DA←→,CA←→ and BA←→.

Ex 4.1 Class 6 Maths Question 3.
Use the figure to name :
(a) Line containing point E.
(b) Line passing through A.
(c) Line on which O lies.
(d) Two pairs of intersecting lines.
Solution:
Clearly, from the given figure
(a) Line containing pqint E is AE←→.
(b) Line passing through A is AE←→.
(c) O lies on the line CB←→.
(d) Two pairs of intersecting lines are CO←→,AE←→ and AE←→,EF←→.

Ex 4.1 Class 6 Maths Question 4.
How many lines can pass through
(a) one given point?
(b) two given points?
Solution:
(a) An unlimited number of lines can be drawn passing through a given point.
(b) Exactly one line can be drawn passing through two different given points in a plane.

Ex 4.1 Class 6 Maths Question 5.
Draw a rough figure and label suitably in each of the following cases:
(a) Point P lies on AB¯¯¯¯¯¯¯¯.
(b) XY←→ and PQ←→ intersect at M.
(c) Line l contains E and F but not D.
(d) OP←→ and OQ←→ meet at O.
Solution:
A rough figure and labelled suitably for the given cases are as under:
(a) Point P lies on AB←→ :
(b) XY←→ and PQ←→ intersect at M:
(c) Line l contains E and F but not D:
(d) OP←→ and OQ←→ meet at O.

Ex 4.1 Class 6 Maths Question 6.
Consider the following figure of line MN. Say whether following statements are true or false in context of the given figure.
(a) Q, M, O, N, P are points on the line MN←→− .
(b) M, O, N are points on a line segment MN¯¯¯¯¯¯¯¯¯¯.
(c) M and N are end points of line segment MN¯¯¯¯¯¯¯¯¯¯.
(d) O and N are end points of line segment OP¯¯¯¯¯¯¯¯.
(e) M is one of the end points of line segment QO¯¯¯¯¯¯¯¯.
(f) M is point on ray OP−→−.
(g) Ray OP−→− is different from ray QP−→−.
(h) Ray OP−→− is same as ray OM−→−.
(i) Ray OM−→− is not opposite to ray OP−→−.
(j) O is not an initial point of OP−→−.
(k) N is the initial point of NP−→− and NM−→−.
Solution:
In the context of the given figure, the given statement is:
(a) True
(b) True
(c) True
(d) False
(e) False
(f) False
(g) True
(h) False
(i) True
(j) False
(k) True

Ex 4.2 Class 6 Maths Question 1.
Classify the following curves as
(i) Open or
(ii) Closed.
Solution:
Open curves are (a) and (c) and closed curves are (b), (d) and (e).

Ex 4.2 Class 6 Maths Question 2.
Draw rough diagrams to illustrate the following:
(a) Open curve
(b) Closed curve
Solution:
Rough diagram to illustrate
(a) Open curve is

(b) Closed curve is

Ex 4.2 Class 6 Maths Question 3.
Draw any polygon and shade its interior.
Solution:
We know that in a closed curve, the interior is inside of the curve. Thus, shaded portion of the polygon indicate its interior.

Ex 4.2 Class 6 Maths Question 4.
Consider the given figure and answer the questions:
(a) Is it a curve?
(b) Is it closed?
Solution:
(a) Yes, the given figure represents a curve.
(b) Yes, the curve is closed.

Ex 4.2 Class 6 Maths Question 5.
Illustrate, if possible, each one of the following with a rough diagram:
(a) A closed curve that is not a polygon.
(b) An open curve made up entirely of line segments.
(c) A polygon with two sides.
Solution:
(a) Impossible, as a close plane figure bounded by lines is called a polygon.
(b) Yes, it may be as
(c) No, impossible, as a polygon of two sides cannot be drawn.

Ex 4.3 Class 6 Maths Question 1.
Name the angles in the given figure.
Solution:
The angles in the given figure are named as ∠ABC, ∠BCD, ∠CDA and ∠DAB.

Ex 4.3 Class 6 Maths Question 2.
In the given diagram, name the point(s)
(a) In the interior of ∠DOE
(b) In the exterior of ∠EOF
(c) On ∠EOF
Solution:
Clearly, from the given figure :
(a) The point A lies in the interior of ∠DOE. .
(b) The points A and C lie in the exterior of ∠EOF.
(c) The points E, B, O and F lie on ∠EOF.

Ex 4.3 Class 6 Maths Question 3.
Draw rough diagrams of two angles such that they have
(a) One point in common
(b) Two points in common
(c) Three points in common
(d) Four points in common
(e) One ray in common
Solution:
(a) The diagram is shown as below
Here, ∠ROQ and ∠QOP have one point O in common.
(b) The diagram is shown as below
Here, ∠MON and ∠ONR have two points O and N in common.
(c) Drawing a diagram of two angles, such that they have three points in common, is not possible.
(d) Drawing a diagram of two angles, such that they have four points in common, is not possible.
(e) In the figure given below, ∠SOT and ∠POT have one ray OT−→− in common.

Ex 4.4 Class 6 Maths Question 1.
Draw a rough sketch of a triangle ABC. Mark a point P in its interior and a point Q in its exterior. Is the point A in its exterior or in its interior?
Solution:
Rough sketch of a ∠ABC as shown.
Points P and Q are marked in the interior and exterior of ∆ABC.
Point A lies on the ∆ABC.

Ex 4.4 Class 6 Maths Question 2.
(a) Identify three triangles in the figure.
(b) Write the names of seven angles.
(c) Write the names of six line segments.
(d) Which two triangles have ZB as common?
Solution:
(a) Three triangles are identified in the figure as ∆ ABC, ∆ ABD and ∆ ACD.
(b) The names of seven angles are ∠BAD, ∠BAC, ∠CAD, ∠ABD, ∠ACD, ∠ADC and ∠ADB.
(c) The names of six line segments are AB, BD, DC, CA, BC and AD.
(d) ∆ ABC and ∆ ABD have ∠B as common.

Ex 4.5 Class 6 Maths Question 1.
Draw a rough sketch of a quadrilateral PQRS. Draw its diagonals. Name them. Is the meeting point of the diagonals in the interior or exterior of the quadrilateral?
Solution:
PQRS is a quadrilateral having PR and QS as its diagonals intersecting at the point O, which is in the interior of the quadrilateral PQRS.

Ex 4.5 Class 6 Maths Question 2.
Draw a rough sketch of a quadrilateral KLMN. State,
(a) two pairs of opposite sides,
(b) two pairs of opposite angles,
(c) two pairs of adjacent sides,
(d) two pairs of adjacent angles.
Solution:
KLMN is a quadrilateral.

(a) KL, MN and LM, KN are two pairs of its opposite sides.
(b) ∠K, ∠M and∠L, ∠N are two pairs of opposite angles.
(c) KL, LM and KN, NM are two pairs of adjacent sides.
(d) ∠K, ∠N and ∠L, ∠M are two pairs of adjacent angles.

Ex 4.5 Class 6 Maths Question 3.
Investigate:
Use strips and fasteners to make a triangle and a quadrilateral.
Try to push inward at any one vertex of the triangle. Do the same to the quadrilateral.
Is the triangle distorted? Is the quadrilateral distorted? Is the triangle rigid?
Why is it that structures like electric towers make use of triangular shapes and not quadrilaterals?
Solution:
On pushing inward at any one vertex of the triangle, we find that the triangle is not distorted. Whereas doing so with the quadrilateral we find that it is distorted. Triangle is rigid. Thus, we make use of triangular shapes in structures like electric towers as triangular shapes are more rigid.

Ex 4.6 Class 6 Maths Question 1.
From the figure, identify:
(a) the centre of circle
(b) three radii
(c) a diameter
(d) a chord
(e) two points in the interior
(f) a point in the exterior
(g) a sector
(h) a segment

Solution:
(a) O is the centre of the circle.
(b) OA, OB and OC are three radii of circle.
(c) AC is a diameter of circle.
(d) ED is a chord of circle.
(e) Points O and P are in the interior of circle.
(f) Point Q is in the exterior of circle.
(g) OAB is a sector of circle.
(h) Shaded region in the interior of a circle enclosed by a chord ED.

Ex 4.6 Class 6 Maths Question 2.
(a) Is every diameter of a circle also a chord?
(b) Is every chord of a circle also a diameter?
Solution:
(a) Yes, a diameter is the longest chord.
(b) Not always.

Ex 4.6 Class 6 Maths Question 3.
Draw any circle and mark
(a) its centre
(b) a radius
(c) a diameter
(d) a Sector
(e) a segment
(f) a point in its interior
(g) a point in its exterior
(h) an arc
Solution:
Draw any circle and its various parts are as under :

(a) Its centre is O.
(b) Its radii are OA, OB and OC.
(c) Its diameter is AOC.
(d) Its sector is OAB.
(e) Shaded region in the interior of a circle enclosed by a chord PQ.
(f) M is a point in its interior.
(g) N is a point in its exterior.
(h) BC is an arc.

Ex 4.6 Class 6 Maths Question 4.
Say true or false:
(a) Two diameters of a circle will necessarily intersect.
(b) The centre of a circle is always in its interior.
Solution:
(a) True, because each diameter passes through the centre of a circle.
(b) True.

Ex 3.1 Class 6 Maths 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
Solution:
(a) We have,
24 = 1 x 24
24 = 2 x 12
24 = 3 x 8
24 = 4 x 6
24 = 6 x 4
Stop here, because 4 and 6 have occurred earlier.
Thus, all the factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24.

(b) We have,
15 = 1 x 15
15 = 3 x 5
15 = 5 x 3
Stop here, because 3 and 5 have occurred earlier.
Thus, all the factors of 15 are 1, 3, 5 and 15.

(c) We have,
21 = 1 x 21
21 = 3 x 7
21 = 7 x 3
Stop here, because 3 and 7 have occurred earlier.
Thus, all the factors of 21 are 1, 3, 7 and 21.

(d) We have,
27 = 1 x 27
27 = 3 x 9
27 = 9 x 3
Stop here, because 3 and 9 have occurred earlier.
Thus, all the factors of 27 are 1, 3, 9 and 27.

(e) We have,
12 = 1 x 12
12 = 2 x 6
12 = 3 x 4
12 = 4 x 3
Stop here, because 3 and 4 have occurred earlier.
Thus, all the factors of 12 are 1, 2, 3, 4, 6 and 12.

(f) We have,
20 = 1 x 20
20 = 2 x10
20 = 4 x 5
20 = 5 x 4
Stop here, because 4 and 5 have occurred earlier.
Thus, all the factors of 20 are 1, 2, 4, 5, 10 and 20.

(g) We have,
18 = 1 x 18
18 = 2 x 9
18 = 3 x 6
18 = 6 x 3
Stop here, because 3 and 6 have occurred earlier.
Thus, all the factors of 18 are 1, 2, 3, 6, 9 and 18.

(h) We have,
23 = 1 x 23
23 = 23 x 1
Stop here, because 1 and 23 have occurred earlier.
Thus, all the factors of 23 are 1 and 23.

(i) We have,
36 = 1 x 36
36 = 2 x 18
36 = 3 x 12
36 = 4 x 9
36 = 1 x 6
Stop here, because both the factors (6) are same.
Thus, all the factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36.

Ex 3.1 Class 6 Maths Question 2.
Write first five multiples of:
(a) 5
(b) 8
(c) 9
Solution:
(a) In order to obtain first five multiples of 5, we multiply it by 1, 2, 3, 4 and 5 respectively. ,r
We have,
5 x 1 = 5
5 x 2 = 10
5 x 3 = 15
5 x 4=20
5 x 5 = 25
Hence, the first five multiples of 5 are 5, 10, 15, 20 and 25 respectively.

(b) In order to obtain first five multiples of 8, we multiply it by 1, 2, 3, 4 and 5 respectively.
We have,
8 x 1=8
8 x 2=16
8 x 3=24
8 x 4 = 32
8 x 5 = 40
Hence, the first five multiples of 8 are 8, 16, 24, 32 and 40 respectively,

(c) In order to obtain first five multiples of 9, we multiply it by 1, 2, 3, 4 and 5 respectively.
We have,
9 x 1 = 9
9 x 2 = 18
9 x 3 = 27
9 x 4 = 36
9 x 5 = 45
Hence, the first five multiples of 9 are 9, 18, 27, 36 and 45 respectively.

Ex 3.1 Class 6 Maths Question 3.
Match the items in column 1 with the items in column 2.

Solution:
Matching is as under :
(i) → (c) ∵ 35 x 2 = 70
(ii) →(d) ∵ 30 = 15 = 2
(iii) → (a) ∵ 8 x 2 = 16
(iv)→ (f) ∵ 20 = 20 = 1
(v) → (e) ∵ 50 = 25 = 2

Ex 3.1 Class 6 Maths Question 4.
Find all the multiples of 9 upto 100.
Solution:
All the multiples of 9 upto 100 are
9 x 1, 9 x 2, 9 x 3, 9 x 4, 9 x 5, 9 x 6, 9 x 7, 9 x 8, 9 x 9, 9 x 10 and 9 x 11.
i. e., 9, 18, 27, 36, 45, 54, 63, 72, 81, 90 and 99.

Ex 3.2 Class 6 Maths Question 1.
What is the sum of any two
(a) Odd numbers?
(b) Even numbers?
Solution:
(a) Sum of two odd numbers is even.
(b) Sum of two even numbers is even.

Ex 3.2 Class 6 Maths 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.
Solution:
(a) False
(b) True
(c) True
(d) False
(e) False
(f) False
(g) False
(h) True
(i) False
(j) True

Ex 3.2 Class 6 Maths 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 upto 100.
Solution:
By the Sieve of Eratosthenes method find the prime numbers between 1 and 100. We find that these are
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97.
Out of these, a pair of prime numbers having same digits are 13, 31; 17, 71; 37, 73, 79, 97.

Ex 3.2 Class 6 Maths Question 4.
Write down separately the prime and composite numbers less than 20.
Solution:
Prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17 and 19.
Composite numbers less than 20 are 4, 6, 8, 9, 10, 12, 14, 15, 16 and 18.

Ex 3.2 Class 6 Maths Question 5.
What is the greatest prime number between 1 and 10?
Solution:
Prime numbers between 1 and 10 are 2, 5 and 7.
∴ Greatest prime number between 1 and 10 = 7

Ex 3.2 Class 6 Maths Question 6.
Express the following as the sum of two odd primes
(a) 44
(b) 36
(c) 24
(d) 18
Solution:
(a) 44 = 13 + 31
(b) 36 = 5 + 31
(c) 24 = 11 +13
(d) 18 = 7+11
Note : In 1742, mathematician Goldbach had a conjecture (guess) for which he could not provide a proof. It may be stated as “Every even number greater than 4 can be expressed as the sum of two odd prime numbers”.

Ex 3.2 Class 6 Maths Question 7.
Give three pairs of prime numbers whose difference is 2.
Solution:
Three pairs of prime number whose difference is 2 are 3, 5; 5, 7 and 11, 13.
Note : Two prime numbers are known as twin-primes if there is one composite number between them. In other words, two prime numbers whose difference is 2 are called twin-primes.

Ex 3.2 Class 6 Maths Question 8.
Which of the following numbers are prime? ,
(a) 23
(b) 51
(c) 37
(d) 26
Solution:
(a) We find that 23 is not exactly divisible by any of the prime numbers 2, 3, 5, 7 and 11 (i.e., upto half of 23). So, it is a prime number.
(b) We find that 51 is divisible by 3. So, it is not a prime number.
(c) We find that 37 is not exactly divisible by any of the prime numbers 2, 3, 5, 7, 11, 13 and 17 (i.e., upto half of 37). So, it is a prime number.
(d) We find that 26 is exactly divisible by 2 and 13. So, it is not a prime number.

Ex 3.2 Class 6 Maths Question 9.
Write seven consecutive composite numbers less-than 100 so that there is no prime number between them.
Solution:
Seven consecutive composite numbers less than 100 so that there is no prime number between them are 90, 91, 92, 93, 94, 95 and 96.

Ex 3.2 Class 6 Maths Question 10.
Express each of the following numbers as the sum of three odd primes:
(a) 21
(b) 31
(c) 53
(d) 61
Solution:
Expressing the given numbers as the sum of three odd primes, we have
(a) 21 = 3 + 5 + 13
(b) 31 = 3 + 5 + 23
(c) 53 = 3 + 19 + 31
(d) 61 = 3 + 11 + 47

Ex 3.2 Class 6 Maths Question 11.
Write five pairs of prime numbers less than 20 whose sum is divisible by 5.
Solution:
Prime numbers below 20 are 2, 3, 5, 7, 13, 17 and 19.
Possible sum of pairs of these numbers :

Clearly, five pairs of prime numbers whose sum is divisible by 5 are 2, 3; 2, 13; 3, 7; 3,17 and 7, 13.

Ex 3.2 Class 6 Maths 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
Solution:
(a) Prime
(b) composite
(c) prime, composite
(d) 2
(e) 4
(f) 2

Ex 3.3 Class 6 Maths 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 4; by 11 (say yes or no) :

Solution:

Ex 3.3 Class 6 Maths 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:
We know that a number is divisible by 4, if the number formed by its digits in ten’s and unit’s place is divisible by 4.
(a) In 572, 72 is divisible by 4. So, 572 is divisible by 4.
(b) In 726352, 52 is divisible by 4. So, it is divisible by 4.
(c) In 5500, 00 is divisible by 4. So, it is divisible by 4.
(d) In 6000, 00 is divisible by 4. So, it is divisible by 4.
(e) In 12159, 59 is not divisible by 4. So, it is not divisible by 4.
(f) In 14560, 60 is divisible by 4. So, it is divisible by 4.
(g) In 21084, 84 is divisible by 4. So, it is divisible by 4.
(h) In 31795072, 72 is divisible by 4. So, it is divisible by 4.
(i) In 1700,00 is divisible by 4. So, it is divisible by 4.
(j) In 2150, 50 is not divisible by 4. So, it is not divisible by 4.

Also, we know that a number is divisible by 8, if the number formed by its hundred’s, ten’s and unit’s places is divisible by 8.

(a) 572 is not divisible by 8.
(b) In 726352, 352 is divisible by 8. So, it divisible by 8.
(c) In 5500, 500 is not divisible by 8. So, it is not divisible by 8.
(d) In 6000, 000 is divisible by 8. So, it is divisible by 8.
(e) In 12159, 159 is not divisible by 8. So, it is not divisible by 8.
(f) In 14560, 560 is divisible by 8. So, it is divisible by 8.
(g) In 21084, 084 is not divisible by 8. So, it is not divisible by 8.
(h) In 31795072, 072 is divisible by 8. So, it is divisible by 8.
(i) In 1700, 700 is not divisible by 8. So, it is not divisible by 8.
(j) In 2150, 150 is not divisible by 8. So, it is not divisible by 8.

Ex 3.3 Class 6 Maths Question 3.
Using divisibility tests, determine which of the 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 a number is divisible by 6, if it is divisible by 2 and 3 both.
(a) Given number = 297144
Its unit’s digit is 4. So, it is divisible by 2.
Sum of its digits = 2+ 9+ 7 + 1 + 4 + 4 = 27, which is divisible by 3.
∴ 297144 is divisible by 6.

(b) Given number =1258
Its unit’s digit is 8. So, it is divisible by 2.
Sum of its digits = 1+ 2 + 5 + 8 = 16, which is not divisible by 3.
∴ 1258 is not divisible by 6.

(c) Given number = 4335 .
Its unit’s digit is 5. So, it is not divisible by 2.
∴ 4335 is also not divisible by 6.

(d) Given number = 61233
Its unit’s digit is 3. So, it is not divisible by 2.
∴ 61233 is also not divisible by 6.

(e) Given number = 901352
Its unit’s digit is 2. So, it is divisible by 2.
Sum of its digits = 9+ 0 + 1 + 345 + 2 = 20, which is not divisible by 3.
∴ 901352 is not divisible by 6.

(f) Given number = 438750
Its unit’s digit is 0. So, it is divisible by 2.
Sum of its digits = 4+ 3 + 8 + 7 + 5 + 0=27, which is divisible by 3.
∴ 438750 is divisible by 6.

(g) Given number = 1790184
Its unit’s digit is 4. So, it is divisible by 2.
Sum of its digits = 1+ 7 + 9+ 0+ 1+ 8 + 4 = 30, which is divisible by 3.
∴ 1790184 is divisible by 6.

(h) Given number = 12583 .
Its unit’s digit is 3. So, it is not divisible by 2.
∴ 12583 is not divisible by 6.

(i) Given number = 639210
Its unit’s digit is 0. So, it is divisible by 2.
Sum of its digits = 6+ 3+ 9+ 2 + 1 + 0 = 21, which is divisible by 3.
∴ 639210 is divisible by 6.

(j) Given number = 17852
Its unit’s digit is 2. So, it is divisible by 2.
Sum of its digits = 1 + 7 + 8 + 5 + 2 = 23, which is not divisible by 3.
∴ 17852 is not divisible by 6.

Ex 3.3 Class 6 Maths 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:
We know that a number is divisible by 11, if the difference in odd places (from the right) and the sum of its digits in even places (from the right) is either 0 or a multiple of 11.
(a) Given number = 5445
Sum of its digits at odd places = 5 + 4 = 9
Sum of its digit at even places = 4 + 5 = 9
Difference of these two sums = 9 – 9 = 0
∴ 5445 is divisible by 11.

(b) Given number = 10824
Sum of its digits at odd places = 4+ 8 + 1 = 13
Sum of its digits at even places =2 + 0 =2
Difference of these two sums =13 – 2 = 11, which is a multiple of 11.
∴ 10824 is divisible by 11.

(c) Given number = 7138965
Sum of its digits at odd places = 5+ 9+ 3 + 7= 24
Sum of its digits at even places = 6+ 8 + 1 = 15
Difference of these two sums = 24 – 15 = 9, which is not a multiple of 11.
∴ 7138965 is not divisible by 11.

(d) Given number = 70169308
Sum of its digits at odd places = 8 + 3 + 6 + 0=17
Sum of its digits at even places = 0 + 9 + 1 + 7 = 17
Difference of these two sums =17 – 17 = 0
∴ 70169308 is divisible by 11.

(e) Given number = 10000001
Sum of its digits at odd places = 1 + 0 + 0 + 0 = 1
Sum of its digits at even places = 0 + 0 + 0 + 1 = 1
Difference of these two sums = 1 – 1 = 0
∴ 10000001 is divisible by 11.

(f) Given number = 901153
Sum of its digits at odd places = 3 + 1 + 0 = 4
Sum of its digits at even places = 5 + 1 + 9=15
Difference of these two sums =15 – 4 = 11,
which is a multiple of 11.
∴ 901153 is divisible by 11.

Ex 3.3 Class 6 Maths Question 5.
Write the smallest digit and the greatest digit in the blank space of each of the following numbers so that the number formed is divisible by 3 :
(a) … 6724
(b) 4765 … 2
Solution:
We know that a number divisible by 3, if the sum of its digits is divisible by 3.
(a) … 6724
For … 6724, we have 6 + 7 + 2 + 4 =19, we add 2 to 19, the resulting number 21 will be divisible by 3.
∴ The required smallest digit is 2.
Again, if we add 8 to 19, the resulting number 27 will be divisible by 3. .’. The required largest digit is 8.

(b) 4765 … 2
For 4765 … 2, we have 4 + 7 + 6 + 5 + 2 = 24, it is divisible by 3.
Hence the required smallest digit is 0.
Again, if we add 9 to 24, the resulting number 33 will be divisible by 3.
∴ The required largest digit is 9.

Ex 3.3 Class 6 Maths Question 6.
Write a digit in the blank space of each of the following numbers so that the number formed is divisible by 11:
(a) 92 … 389
(b) 8 … 9484
Solution:
We know that a number is divisible by 11, if the difference of the sum of its digits at odd places and the sum of its digits at even places is either 0 or divisible by 11.
(a) 92 … 389
For 92 … 389, sum of the digits at odd places and sum of digits at even places
= 9 + 3 + 2 = 14
= 8 + required digit + 9
= required digit+ 17
Difference between these sums
= required digit + 17 – 14
= required digit + 3
For (required digit + 3) to become 11, we must have the required digit as 8 (∵ 3+ 8 gives 11).
Hence, the required smallest digit = 8 .

(b) 8…9484
For 8 … 9484, sum of the digits at odd places
= 4 + 4 + required digit
= 8 + required digit
and sum of digits at even places
= 8 + 9 + 8 = 25
Difference between these sums
= 25 – (8 + required digit)
= 17- required digit
For (17 — required digit) to become 11 we must have the required digit as 6 (∵ 17-6 =11).
Hence the required smallest digit = 6

Ex 3.4 Class 6 Maths 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) We have, 20 = 1 x 20
= 2 x 10
= 4 x 5
∴ All the factors of 20 are 1, 2, 4, 5, 10 and 20
Again, 28 = 1 x 28
28 = 2 x 14
28 = 4 x 7
∴ All the factors of 28 are 1, 2, 4, 7, 14 and 28.
Out of these 1, 2 and 4 occur in both the lists.
∴ 1, 2 and 4 are common factors of 20 and 28.

(b) We have, 15 = 1 x 15
15 = 3 x 5
∴ All the factors of 15 are 1, 3, 5 and 15.
Again, 25 = 1 x 25
25 = 5 x 5
∴ All the factors of 25 are 1, 5 and 25.
Out of these 1 and 5 occur in both the lists.
∴1 and 5 are common factors of 15 and 25.

(c) We have, 35 = 1 x 35
35 = 5 x 7
∴ All the factors of 35 are 1, 5, 7 and 35.
Again, 50 = 1 x 50
50 = 2 x 25
50 = 5 x 10
∴ All the factors of 50 are 1, 2, 5, 10, 25 and 50. ,
Out of these 1 and 5 occur in both the lists.
∴ 1 and 5 are common factors of 35 and 50.

(d) We have, 56 = 1 x 56
56 = 2 x 28
56 = 4 x 14
56 = 7 x 8
∴ All the factors of 56 are 1, 2, 4, 7, 8, 14, 28 and 56.
Again, 120 = 1 x 120
120 = 2 x 60
120 = 3 x 40
120 = 4 x 30
120 = 5 x 24
120 = 6 x 20
120 = 8 x 15
120 = 10 x 12
∴ All the factors of 120 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60 and 120.
Out of these 1, 2, 4 and 8 occur in both the lists.
∴ 1, 2, 4 and 8 are common factors of 56 and 120.

Ex 3.4 Class 6 Maths Question 2.
Find the common factors of:
(a) 4, 8 and 12
(b) 5, 15 and 25
Solution:
(a) We have, 4=1 x 4
4 = 2 x 2
∴ All the factors of 4 are 1, 2 and 4.
Again, 8 = 1 x 8
8 = 2 x 4
∴All the factors of 8 are 1, 2, 4 and 8.
Again, 12 = 1 x 12
12 = 2 x 6
12 = 3 x 4
∴ All the factors of 12 are 1, 2, 3, 4, 6 and 12.
Out of these 1, 2 and 4 occur in all the three lists.
∴ 1, 2 and 4 are common factors of 4, 8 and 12.

(b) We have, 5 = 1 x 5
∴All the factors of 5 are 1 and 5.
15 = 1 x 15
15 = 3 x 5
∴ All the factors of 15 are 1, 3, 5 and 15.
25 = 1 x 25
25 = 5 x 5
∴ All the factors of 25 are 1, 5 and 25.
Out of these 1 and 5 occur in all the three lists.
∴ 1 and 5 are common factors of 5, 15 and 25.

Ex 3.4 Class 6 Maths Question 3.
Find first three common multiples of:
(a) 6 and 8
(b) 12 and 18
Solution:
(a) Multiples of 6 are 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72,…
Multiples of 8 are 8, 16, 24, 32, 40, 48, 56, 64, 72,…
Out of these 24, 48, 72, … occur in both thfe lists.
∴ The first three common multiples of 6 and’8 are 24, 48 and 72.
(b) Multiples of 12 are 12, 24, 36, 48, 60, 72, 84, 96, 108, …
Multiples of 18 are 18, 36, 54, 72, 90, 108, …
Out of these 36, 72, 108, … occur in both the lists.
∴ The first three common multiples of 12 and 18 are 36, 72 and 108.

Ex 3.4 Class 6 Maths Question 4.
Write all the numbers less than 100 which are corrimon multiples of 3 and 4.
Solution:
Common multiples of 3 and 4 are multiples of 3 x 4 i. e., 12.
∴ Common multiples of 3 and 4 less than 100 are 12, 24, 36,48, 60, 72, 84 and 96.

Ex 3.4 Class 6 Maths 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) Factors of 18 are 1, 2, 3, 6, 9 and 18 and, that of 35 are 1, 5, 7 and 35.
∴ Common factor of 18 and 35 is 1.
Thus, 18 and 35 are co-prime.

(b) Factors of 15 are 1, 3, 5 and 15 and, that of 37 are 1 and 37.
∴ Common factor of 15 and 37 is 1.
Thus, 15 and 37 are co-prime.

(c) Since 5 is a common factor of 30 and 415.
∴ 30 and 415 are not co-prime.

(d) ∴ 68 + 17 = 4 i.e., 17 is a common factor of 17 and 68.
∴ 17 and 68 are not co-prime.

(e) Since 1 is the only common factor of 216 and 215.
∴ 216 and 215 are co-prime.

(f) Since 1 is the only common factor of 81 and 16.
∴ 81 and 16 are co-prime.

Ex 3.4 Class 6 Maths Question 6.
A number is divisible by both 5 and 12. By which other number will that number be always divisible?
Solution:
Since a number is divisible by both 5 and 12.
So, it is also divisible by 5 x 12 i. e., 60.

Ex 3.4 Class 6 Maths 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 and 12.
Since a number is divisible by 12. So, it is also divisible by the factors of 12.
Thus, the number is also divisible by 2, 3 and 4.

Ex 3.5 Class 6 Maths 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 exactly divides the sum of two numbers, it must exactly divide the two numbers separately.
Solution:
Statements (b), (d), (g) and (i) are true.

Ex 3.5 Class 6 Maths Question 2.
Here are two different factor trees for 60. Write the missing numbers.

Solution:

Ex 3.5 Class 6 Maths Question 3.
Which factors are not included in the prime factorisation of a
composite number?
Solution:
1 and composite numbers are not included in the prime factorisation of a composite number.

Ex 3.5 Class 6 Maths Question 4.
Write the greatest 4-digit number and express it in terms of its prime factors.
Solution:
The greatest 4-digit number = 9999
We have,

∴ 9999 = 3 x 3 x 11 x 101

Ex 3.5 Class 6 Maths Question 5.
Write the smallest 5-digit number and express it in the form of its prime factors.
Solution:
Smallest 5-digit number = 10000
We have,

∴ 10000=2 x 2 x 2 x 2 x 5 x 5 x 5 x 5

Ex 3.5 Class 6 Maths 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:
We have,

∴ 1729 = 7 x 13 x 19
Relation between two consecutive prime factors is stated as “difference between two consecutive prime factors is 6”.

Ex 3.5 Class 6 Maths Question 7.
The product of three consecutive numbers is always divisible by 6. Verify this statement with the help of some examples.
Solution:
Let us consider the product of three consecutive numbers as under :

In all the products, we have 6 or 4 or 0 in the unit’s place, so each product is divisible by 2.
Also, sum of digits in these products are divisible by 3. So, each of the product is divisible by 3.
Since 2 and 3 are co-prime, so the product 2 x 3 = 6 divides each of the above products.
Thus, the product of three consecutive numbers is always divisible by 6.

Ex 3.5 Class 6 Maths Question 8.
The sum of two consecutive odd numbers is divisible by 4. Verify this statement with the help of same examples.
Solution:
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.

Ex 3.5 Class 6 Maths Question 9.
In which of the following expressions, prime factorisation has been done ? ‘
(a) 24 = 2 x 3 x 4
(b) 56 = 7 x 2 x 2 x 2
(c) 70 = 2 x 5 x 7
(d) 54 = 2 x 3 x 9
Solution:
In (b) and (c) prime factorisation has been done.

Ex 3.5 Class 6 Maths Question 10.
Determine if 25110 is divisible by 45.
Solution:
Since 45 = 5 x 9, where 5 and 9 are co-primes.
So to check the divisibility of 25110 by 45, test it for 5 and 9.
In 25110, unit’s digit = 0
So, 25110 is divisible by 5.
Sum of its digits = 2 + 5 + 1 + 1+ 0 = 9, which is divisible by 9.
So, 25110 is also divisible by 9.
Hence, 25110 is divisible by 45.

Ex 3.5 Class 6 Maths Question 11.
18 is divisible by both 2 and 3. It is also divisible by 2 x 3 = 6. Similarly, a number is divisible by both 4 and 6. Can we say that the number must also be divisible by 4 x 6 = 24 ? If not, give an example to justify your answer.
Solution:
Not necessarily, as the numbers 12,36,60 etc. are each divisible by both 4 and 6. But these numbers are not divisible by 4 x 6 = 24.

Ex 3.5 Class 6 Maths Question 12.
I am the smallest number, having four different prime factors. Can you find me?
Solution:
Smallest four different prime numbers are 2, 3, 5 and 7.
∴ Required number = 2 x 3 x 5 x 7 = 210.

Ex 3.6 Class 6 Maths Question 1.
Find the HCF of the following numbers :
(a) 18, 48
(b) 30, 42
(c) 18, 60
(d) 27, 63
(e) 36, 84
(f) 34, 102
(g) 70, 105, 175
(h) 91, 112, 49
(i) 18, 54, 81
(j) 12, 45, 75
Solution:
(a) First, we write the prime factorisation of each of the given numbers. We have,

∴ 18 = 2 x 3 x 3
and 48 = 2 x 2 x 2 x 2 x 3
We find that 2 and 3 each occurs as a common factor in the given numbers at least once.
∴ Required HCF = 2 x 3 = 6

(b) First, we write the prime factorisation of each of the given numbers.
We have,

∴ 30 = 2 x 3 x 5
and 42 = 2 x 3 x 7
We find that 2 and 3 each occurs as a common factor in the given numbers at least once.
Required HCF = 2 x 3 = 6

(c) First, we write the prime factorisation of each of the given numbers.
We have,

∴ 18 = 2 x 3 x 3
and 60 = 2 x 2 x 3 x 5
We find that 2 and 3 each occurs as a common factor in

(d) 27,63
Factors of 27 are 1, 3, 9 and 27.
Factors of 63 are 1, 3, 7, 9, 21 and 63.
Common factors of 27 and 63 are 1, 3 and 9.
Highest of these common factors is 9.
∴ H.C.F. of 27 and 63 is 9.

(e) 36,84
Factors of 36 are 1, 2, 3,4, 6, 9,12, 18 and 36.
Factors of 84 are 1,2, 3,4,6,7, 12, 14,21, 28, 42 and 84.
Common factors of 36 and 84 are 1,2, 3,4, 6 and 12.
Highest of these common factors is 12.
∴ H.C.F. of 36 and 84 is 12. if) 34,102 • ‘
Factors of 34 are 1,2, 17 and 34.
Factors of 102 are 1, 2, 3,6,17, 34. 51 and 102.
∴ Common factors of 34 and 102 are 1, 2, 17 and 34.
Highest of these common factors is 34.
∴ H.C.F. of 34 and 102 is 34.

(g) 70,105,175
Factors of 70 are 1. 2, 5. 7, 10. 14, 35 and 70.
Factors of 105 are 1, 3, 5. 7. 15. 21. 35 and 105.
Factors of 175 are 1. 5, 7. 25. 35 and 175. .’. Common factors of 70, 105 and 175 are 1, 5 and 35.
Highest of these common factors is 35.
∴ H.C.F. of 70. 105 and 175 are 35.

(h) 91,112,49
Factors of 91 are 1,7, 13 and 91.
Factors of 112 are 1,2. 4. 7, 8, 14. 16. 28, 56 and 112.
Factors of 49 are 1.7 and 49.
Common factors of 91,112 and 49 are 1 and 7.
Highest of these common factors is 7.
∴ H.C.F. of 91, 112 and 49 is 7.

(i) 18,54,81
Factors of 18 are 1. 2, 3. 6, 9 and 18. Factors of 54 are 1, 2. 3, 6. 9, 18. 27 and 54.
Factors of 81 are 1. 3, 9, 27 and 81.
∴ Common factors of 18,54 and 81 are 1, 3 and 9.
Highest of these common factors is 9.
∴ H.C.F. of 18, 54 and 81 is 9.
(j) 12, 45, 75
Factors of 12 are 1, 2, 3, 4, 6 and 12. Factors of 45 are 1, 3, 5, 9, 15 and 45.
: Factors of 75 are 1, 3, 5, 15, 25 and 75.
∴ Common factors of 12,45 and 75 are 1 and 3.
Highest of these common factors is 3.
H.C.F. of 12. 45 and 75 is 3.

Ex 3.6 Class 6 Maths Question 2.
What is the H.C.F. of two consecutive :
(a) numbers?
(b) even numbers?
(c) odd numbers?
Solution :
(a) The H.C.F. of two consecutive numbers is 1.
(b) The H.C.F. of two consecutive even numbers is 2.
(c) The H.C.F. of two consecutive odd numbers is 1.

Ex 3.6 Class 6 Maths Question 3.
H. C.F. of co-prime numbers 4 and 15 was found as follows factorization: 4 = 2 × 2 and 15 = 3 × 5 since there is no common prime factor, so H.C.F. of 4 and 15 is 0. Is the answer correct? If not, what is the correct H.C.F?
Solution :
No, the answer is not correct. The correct answer is as follows :
H.C.F. of 4 and 15 is 1.

Ex 3.7 Class 6 Maths 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 e×act number of times.
Solution :
Factors of 75 are 1, 3, 5, 15, 25 and 75.
Factors of 69 are 1, 3, 23 and 69.
∴ Common factors of 75 and 69 are 1 and 3.
Highest of these common factors is 3.
∴ H.C.F. of 75 and 69 is 3.
Hence, the maximum value of weight which can measure the weight of the fertilizer e×act number of times is 3 kg.

Ex 3.7 Class 6 Maths 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?
Solution :

∴ L.C.M. of 63, 70 and 77
= 2 × 3 × 3 × 5 × 7 × 11 = 6930.
Hence, the minimum distance each should cover so that all can cover the distance in complete steps is 6930 cm.

Ex 3.7 Class 6 Maths 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.
Solution :
Factors of 825 are 1, 3, 5, 11, 15, 25, 33, 55,75,165, 275 and 825.
Factors of 675 are 1, 3, 5,9,15, 25, 27,45,75, 135, 225 and 675.
Factors of 450 are 1,2,3,5,6,9,10,15,18,25, 30,45, 50, 75, 90, 150, 225 and 450.
∴ Common factors of 825, 675 and 450 are 1,3,5,15, 25 and 75.
Highest of these common factors is 75.
Hence, the length of the longest tape which can measure the three dimensions of the room exactly is 75 cm.

Ex 3.7 Class 6 Maths Question 4.
Determine the smallest 3-digit number which is exactly divisible by 6, 8 and 12.
Solution :

∴ L.C.M. of 6, 8 and 12 = 2 × 2 × 2 × 3 = 24. Multiples of 24 are 24,48,72,96,120,144,
Hence, the smallest 3-digit number which is exactly divisible by 6, 8 and 12 is 120.

Ex 3.7 Class 6 Maths Question 5.
Determine the largest 3-digit number exactly divisible by 8, 10 and 12.
Solution :

∴ L.C.M. of 8,10 and 12 = 2×2×2×3×5
= 120.
Multiples of 120 are :
120 × 1 = 120,120 × 2 = 240,120 × 3 = 360,120 × 4 = 480,120 × 5 = 600,120 × 6 = 720,120 × 7 = 840,
120 × 8 = 960,120 × 9 = 1080,
Hence, the largest 3-digit number exactly divisible by 8, 10 and 12 is 960.

Ex 3.7 Class 6 Maths 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 am at what time will they change simultaneously again?
Solution :

∴ L.C.M. of 48,72 and 108 = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 432.
432 seconds = 7 min 12 seconds.
Hence, they will change simultaneously again after 7 min 12 seconds from 7 a.m.

Ex 3.7 Class 6 Maths Question 7.
Three tankers contain 403 liters, 434 liters and 465 liters of diesel respectively. Find the ma×imum capacity of a container that can measure the diesel of the three containers e×act a number of times.
Solution :
Factors of 403 are 1, 13, 31 and 403. Factors of 434 are 1, 2, 7, 14, 31, 62, 217 and 434.
Factors of 465 are 1, 3, 5, 15, 31, 93, 155 and 465.
Common factors of 403,434 and 465 are 1 and 31.
Highest of these common factors is 31.
∴ H.C.F. of 403. 434 and 465 is 31.
Hence, the maximum capacity of the container that can measure the diesel of the three containers an e×act number of times is 31 litres.

Ex 3.7 Class 6 Maths Question 8.
Find the least number which when divided by 6, 15 and 18 leave remainder 5 in each case.
Solution :

∴ L.C.M. of 6, 15 and 18 = 2 × 3 × 3 × 5 = 90.
Hence, the required number is 90 + 5 i.e., 95.

Ex 3.7 Class 6 Maths Question 9.
Find the smallest four digit number which is divisible by 18, 24 and 32.
Solution :

∴ L.C.M. = 2 × 2 × 2 × 2 × 2 × 3 × 3 = 288.
Multiples of 288 are :
288 × 1 = 288, 288 × 2 = 576, 288 × 3 = 864, 288×4= 1152,
Hence, the smallest four digit number which is divisible by 18, 24 and 32 is 1152.

Ex 3.7 Class 6 Maths 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.M.s. Is L.C.M. the product of two numbers in each case?
Solution :
(a) 9 and 4

∴ L.C.M. of 9 and 4 = 2 × 2 × 3 × 3 = 36 (= 9 × 4).

(b) 12 and 5

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

(c) 6 and 5

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

(d) 15 and 4

∴ L.C.M. of 15 and 4 = 2 × 2 × 3 × 5 = 60 (=15×4).
We observe a common property in the obtained L.C.M.’s that L.C.M. is the product of two numbers in each case.

Ex 3.7 Class 6 Maths Question 11.
Find the L.C.M. of the following numbers in which one number is the factor of the other.
(a) 5, 20
(b) 6, 18
(c) 12, 48
(d) 9, 45.
What do you observe in the results obtained?
Solution :
(a) 5, 20
Prime factorisations of 5 and 20 are as follows: 5 = 5
20 = 2 × 2 × 5 ∴ L.C.M. of 5 and 20
=2×2×5 = 20.

(b) 6, 18
Prime factorisations of 6 and 18 are as follows:
6 = 2×3 18 = 2×3×3
∴ L.C.M. of 6 and 18 = 2×3×3 = 18.

(c) 12, 48
Prime factorisations of 12 and 48 are as follows:
12 = 2 × 2 × 3
48 = 2×2×2×2×3
∴ L.C.M. of 12 and 48 = 2 × 2 × 2 × 2 × 3 = 48.

(d) 9, 45
Prime factorisations of 9 and 45 are as follows:
9 = 3×3
45 = 3 × 3 × 5
∴ L.C.M. of 9 and 45 = 3 × 3 × 5 = 45.
In the results obtained, we observe that L.C.M. of the two numbers in which one number is the factor of the other is the greater number.