Class 6

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 12 Data Handling Unit Exercise Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 12th Lesson Handling Unit Exercise

Question 1.
Given below are the ages of 20 Students of Class VI in a School.
i) Organise the data and represent in the form of a frequency distribution table using tally marks.
ii) Find out the age having more number of students.
iii) How many students are there in 10 Years age?
iv) Find out No. of Students who are having more age.
13, 10, 11, 12, 10, 11, 11, 13, 12, 11
10, 11, 12,11, 13, 11, 10, 13, 10, 12
Solution:
i)

ii) More number of students have 11 years age.
iii) 5 students are there in 10 years age.
iv) 7 students having more age.

Question 2.
A dice was thrown 30 times and following scores were obtained

i) Prepare a frequency table of the scores.
ii) Which number obtained more times?
iii) How many times was a score greater than 4 obtained,
iv) Find the total number of times an odd number obtained.
Solution:

ii) 2 and 3 numbers obtained more times.
iii) Total numbers of times obtained greater than 4 are 22,
iv) Two times an odd number is obtained.

Question 3.
Following is the data regarding pass percentage of students in different classes.

Draw a vertical bar graph to represent the above data.
Solution:

Bar graph showing pass percentage of different classes from VI to X.

Steps of construction:

1. Draw two mutually perpendicular lines on a graph sheet – one horizontal (X-axis) and one vertical (Y-axis) mark. Number of class on the X-axis and their pass percentage on the Y-axis.
2. Take the 10 as scale on the Y-axis i.e., 1 cm = 10%
3. Calculate the lengths or heights of the bars by dividing the pass percentage with the scale.
   Class VI = 65 ÷ 10 = 6.5 cm
   Class VII = 75 ÷ 10 = 7.5 cm
   Class VIII = 85 ÷ 10 = 8.5 cm
   Class IX = 60 ÷ 10 = 6 cm
   Class X = 80 ÷ 10 = 8 cm
4. Draw rectangular vertical bars of same width (I cm) on the X-axis with their calculated heights.
   Hence, required vertical bar graph was constructed.

Question 4.
The number of Mathematics books sold by a shopkeeper on Six consecutive days is shown below.

Draw a Horizontal Bar graph to represent the above data.
Solution:

Bar graph showing the percentage of books sold In a week days.

Steps of construction:

1. Draw two mutually perpendicular lines on a graph sheet – one horizontal (X-axis) and one vertical (Y-axis).
2. Mark name of the day on the Y-axis and the number of books sold on the X-axis.
3. Take the 10 as scale on the X-axis i.e., 1cm = 5 books.
   On Monday = 65 ÷ 5 = 13 cm
   On Wednesday = 30 ÷ 5 = 6 cm
   On Friday = 70 ÷ 5 = 14 cm
   On Tuesday = 40 ÷ 5 = 8 cm
   On Thursday = 50 ÷ 5 = 10 cm
   On Saturday = 20 ÷ 5 = 4 cm
4. Calculate the lengths of the bars by dividing the number of books with the scale.
5. Draw rectangular horizontal bars of same width (1cm) on the Y-axis with their calculated lengths.
   Hence, required horizontal bar graph was constructed.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 12 Data Handling Ex 12.3 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 12th Lesson Data Handling Ex 12.3

Question 1.
The life spans of some animals are given below:
Bear – 40 years, Camel – 50 years, Cat – 25 years, Donkey – 45 years, Goat -15 years, Horse -10 years, Elephant – 70 years.
Draw a horizontal bar graph to represent the data.
Solution:

Steps of construction:

1. Draw two mutually perpendicular lines on a graph sheet – one horizontal (X-axis) and one vertical (Y – axis).
2. Mark name of the animal on the Y-axis and their life span (Number of years) on the X-axis.
3. Take the GCD or HCF of given numbers as scale on the X-axis i.e., 1cm = 5 years.
4. Calculate lengths or heights of the bars by dividing the life spans with the scale :
   Bear = 40 ÷ 5 = 8 cm; Camel = 50 ÷ 5 = 10 cm
   Cat = 25 ÷ 5 = 5 cm; Donkey = 45 ÷ 5 = 9 cm
   Goat = 15 ÷ 5 = 3 cm; Horse = 10 ÷ 5 = 2 cm
   Elephant = 70 ÷ 5 = 14 cm
5. Draw rectangular horizontal bars of same width (1cm) on the Y-axis with their calculated heights.
   Hence, required horizontal bar graph was constructed.

Question 2.
Travelling time from Hyderabad to Thirupathi by different means of transport are- Car – 8 hours, Bus – 15 hours, Train – 12 hours, Aeroplane – 1 hour. Represent the information using a bar diagram.
Solution:

Bar graph showing travel time by different vehicles from Hyderabad to Tirupathi.

Steps of construction :

1. Draw two mutually perpendicular lines on a graph sheet – one horizontal (X-axis) and one vertical (Y-axis).
2. Mark name of the vehicle on the X-axis and their travelling time (Number of hours) on the Y-axis.
3. Take the 2 as scale on the Y-axis i.e., 1 cm = 2 hours.
4. Calculate lengths or heights of the bars by dividing the travelling time with the scale.
   Car = 8 ÷ 2 = 4 cm
   Bus = 15 ÷ 2 = 7.5 cm
   Train = 12 ÷ 2 = 6 cm
   Aeroplane = 1 ÷ 2 = 0.5 cm
5. Draw rectangular vertical bars of same width (1cm) on the X-axis, with their calculated heights.
   Hence, required vertical bar graph was constructed.

Question 3.
A survey of 120 school students was conducted to find which activity they prefer to do in their free time.

Draw a bar graph to illustrate the above data.
Solution:

Bar graph showing the students can prefer to do Activities in free time.

Steps of construction :

1. Draw two mutually perpendicular lines on a graph sheet – one horizontal (X-axis) and one vertical (Y-axis).
2. Mark name of the activity on the X-axis and the number of students on the Y-axis.
3. Take the 5 as scale on the Y-axis i.e., 1cm = 5 students.
4. Calculate length or heights of the bars by dividing the number of students-with the scale.
   Playing = 25 ÷ 5 = 5 cm
   Reading story books = 10 ÷ 5 = 2 cm
   Watching TV = 40 ÷ 5 = 8 cm
   Listening to music = 10 ÷ 5 = 2 cm
   Painting = 15 ÷ 5 = 3 cm
5. Draw rectangular vertical bars of same width (1cm) on the X-axis with their calculated heights.
   Hence, required vertical bar graph was constructed.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 12 Data Handling Ex 12.2 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 12th Lesson Data Handling Ex 12.2

Question 1.
The number of wrist watches as manufactured by a factory in a week are as fallows.

Represent the data using a pictograph. Choose a suitable scale.
Solution:
GCD of the given numbers is 25.
So, Scale : 1 watch symbol = 25 watches.

Note:
To represent 300 watches. \(\frac{300}{25}\) = 12 symbols.
To represent 350 watches, \(\frac{350}{25}\) = 14 symbols.
To represent 250 watches, \(\frac{250}{25}\) = 10 symbols.
To represent 400 watches, \(\frac{400}{25}\) = 16 symbols.
To represent 400 watches, \(\frac{300}{25}\) = 12 symbols.
To represent 400 watches, \(\frac{275}{25}\) = 11 symbols.

Question 2.
Votes polled for various candidates in a Sarpanch election are shown below, against their symbols in the following table.

Represent the data using a pictograph. Choose a suitable scale. Answer the following questions:
(i) Which symbol got least votes?
(ii) Which symbol candidate won in the election?
Solution:
GCD or HCF of the given numbers is 50.
So, scale : 1 symbol : 50 votes
To represent 400 votes, \(\frac{400}{50}\) = 8 symbols of Sun.
To represent 550 votes, \(\frac{550}{50}\)= 11 symbols of Pot.
To represent 350 votes, \(\frac{350}{50}\)= 7 symbols of Tree.
To represent 200 votes, \(\frac{200}{50}\) = 4 symbols of Watch

i) Watch symbol got least votes than the other.
ii) Among all symbols pot symbol got more votes that is 550. So, Pot symbol candidate won in the election.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 12 Data Handling Ex 12.1 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 12th Lesson Data Handling Ex 12.1

Question 1.
The favourite colours of 25 students in a class are given below:
Blue, Red, Green, White, Blue, Green, White, Red, Orange, Green, Blue, White, Blue, Orange, Blue, Blue, White, Red, White, White, Red, Green, Blue, Blue, White. Write a frequency distribution table using tally marks for the data. Which is the east favourite colour for the students?
Solution:

From the above distribution table least favourite colour is Orange. Because, less number of students liked Orange colour.

Question 2.
A TV channel invited a SMS poll on ‘Ban of Liquor’ giving options :
A – Complete ban
B – Partial ban
C – Continue sales
They received the following SMS, in the first hour.

Represent the data in a frequency distribution table using tally marks.
Solution:

Question 3.
Vehicles that crossed a checkpost between 10 AM and 11 AM are as follows:
car, lorry, bus, lorry, auto, lorry, lorry, bus, auto, bike, bus, lorry, lorry, jeep, lorry, bus, jeep, car, bike, bus, car, lorry, bus, lorry, bus, bike, car, jeep, bus, lorry, lorry, bus, car, car, bike, auto.
Represent the data in a frequency distribution table using tally marks.
Solution:

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 11 Perimeter and Area Unit Exercise Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 11th Lesson Perimeter and Area Unit Exercise

Question 1
Find the area of the square whose perimeter is 48 cm.
Solution:
Given perimeter of the square = 4 x s = 48 cm
Divide with 4 on both sides, \(\frac{4 \times \mathrm{s}}{4}=\frac{48}{4}\)
side (s) = 12 cm
Area of the square A = s x s = 12 x 12 = 144 sq. cm

Question 2.
If the length of a rectangle is 14cm and its perimeter is 3 times of its length. Find its area.
Solution:
Given the length of a rectangle l = 14 cm
Breadth of a rectangle b = ? cm
Perimeter of the rectangle = 2(l + b) = 3 times of length.
2(14 + b) = 3 x 14
Divide with 2 on both sides,
\(\frac{2(14+b)}{2}=\frac{3 \times 14}{2}\)
14 + b = 21
Subtract with 14 on both sides.
14 + b – 14 = 21 – 14
Breadth b = 7cm
Area of the rectangle A = l x b = 14 x 7 = 98 sq.cm

Question 3.
Find circumference of the circle whose diameter is 14 cm.
Solution:
Given the diameter of circle d = 14 cm
Circumference of the circle C = π.d
\(\frac{22}{7}\) x 14 =44 cm

Question 4.
14cm and 12cm are the length and breadth of a rectangle. If the breadth is increased by 6cm and length is decreased by 6cm, find the difference in areas.
Sol. Given length of the rectanglel = 14 cm
Breadth of the rectangle b = 12 cm
Area of the rectangle A₁ = l x b
= 14 x 12 = 168 sq. cm
If length decreased by 6 cm, then length l = 14-6 = 8 cm
If breadth increased by 6cm, then breadth b = 12 + 6 = 18 cm
Then, the area of the rectangle A2 = l x b = 8 x 18 = 144 sq. cm
Difference of the areas = A₁ – A₂ = 168 – 144 = 24 sq. cm

Question 5.
Find the perimeter of the following figures. What did you observe?

Solution:
i) Perimeter =12 cm + 8cm + 12cm + 8 cm = 40 cm
ii) Perimeter = 3cm + 2cm + 3cm + 2cm + 3cm + 2cm + 3cm + 2cm + 12cm + 8 cm = 40 cm
iii) Perimeter = 5cm + 3cm + 2cm + 3cm + 5cm + 2cm + 5cm + 3cm + 2cm + 3cm + 5cm + 2cm = 40 cm
By observing the perimeters of the above figures perimeters are same for the different shaped figures.

Question 6.
A square sheet of 8cm side was taken and made into 64 equal small squares. Find the perimeter of square sheet and also find the sum of the perimeters of all 64 small squares. What did you observe?
Solution:
Side of the square sheet s = 8cm
Perimeter of the square sheet = 4 x s = 4 x 8 = 32 cm
Side of the small square = 1cm
Perimeter of each small square = 4 x side = 4 x 1 = 4 cm
Perimeter of 64 small squares = 4 x 64 = 256 cm.
By observing sum of perimeters of all 64 small squares = 8 x perimeter of all squares.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.3 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 11th Lesson Perimeter and Area Ex 11.3

Question 1.
Find the area of the rectangle of measurements 15 cm and 8 cm as length and breadth respectively.
Solution:
Given length of the rectangle l = 15 cm
breadth of the rectangle b = 8 cm
Area of the rectangle A = l x b
= 15 x 8 = 120 sq.cm

Question 2.
Find the area of a square whose perimeter is 64 m.
Solution:
Given perimeter of a square = 4 x side = 64 m
Divide with 4 on both sides,
\(\frac{4 \times \operatorname{side}}{4}=\frac{64}{4}\)
side (s) = 16 m
Area of the square = s.s = 16xl6 = 256 sq.m

Question 3.
Perimeters of a rectangle and square are equal. If the length of the rectangle is 14 cm and the perimeter of the square is 44 cm, find the area of the rectangle.
Solution:
Given the length of the rectangle l = 14 cm
breadth of the rectangle b = ?
Perimeter of the square = 44 cm
Given perimeter of the rectangle = Perimeter of the square ‘
2(l + b) = 44
2(14 + b) = 44
Divide with 2 on both sides,
\(\frac{2(14+b)}{2}=\frac{44}{2}\)
14 + b = 22
Subtract 14 on both sides,
14 + b – 14 = 22 – 14
Breadth of the rectangle b = 8 cm
Area of the rectangle = l x b = 14 x 8 = 112 sq. cm

Question 4.
Find the perimeters and areas of the following and answer the questions.
A) A rectangle with length and breadth as 16 cm and 8 cm respectively.
B) A rectangle with length and breadth as 14 cm and 10 cm respectively.
C) A square with side 12 cm.
(i) Which of the above perimeters are equal ?
(ii) Are all these areas equal? If not, which one has the bigger area?
Solution:
A) Given length of the rectangle l =16 cm
– Breadth of the rectangle b = 8 cm
Perimeter of the rectangle P = 2(1 + b)
= 2(16 + 8) = 2×24 = 48 cm
Area of the rectangle A = l x b
= 16 x 8 = 128 sq. cm

B) Given length of the rectangle l = 14 cm
Breadth of the rectangle b = 10 cm
Perimeter of the rectangle P = 2(l + b)
= 2(14 + 10) = 2 x 24 = 48 cm
Area of the rectangle A = l x b
= 14 x 10 = 140 sq.cm

C) Given side of the square s = 12 cm
Perimeter of the square P = 4 x s = 4 x 12 = 48 cm
Area of the square A = s x s = 12 x 12 = 144 sq. cm
i) Perimeters of rectangle A and rectangle B and square are equal.
ii) No. Areas are not equal and area of the square is greater than the areas of rectangles.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 7 Introduction to Algebra InText Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 7th Lesson Introduction to Algebra InText Questions

Let’s Explore (Page No. 102)

Question 1.
Arrange 2 matchsticks to form the shape Continue the same shape for 2 times, 3 times and 4 times. Frame the rule for repeating the pattern.
Solution:
To make the given shape 2 matchsticks are needed.
To make the given 2 shapes 4 matchsticks are needed.
To make the given 3 shapes 6 matchsticks are needed.
To make the given 4 shapes 8 matchsticks are needed.
Continue and arrange the information in the following table.

Number of matchsticks required = 2 × Number of shapes to be formed
= 2 × x = 2x

Question 2.
Rita took matchsticks to form the shape
She repeated the pattern and gave a rule.
Number of matchsticks needed = 6.y, where y is the number of shapes to be formed. Is it correct ? Explain.
What is the number of sticks needed to form 5 such shapes ?
Solution:

To make the given shape 6 matchsticks are needed.
To make the given 2 shapes 12 matchsticks are needed.
To make the given 3 shapes 18 matchsticks are needed.
Continue and arrange the information in the following table.

Yes, it is correct.
Number of matchsticks required = 2 × Number of shapes to be formed
= 2 × y = 2y
Number of matchsticks needed to form 5 such shapes = 6 × 5 = 30

Let’s Explore (Page No. 103)

Question 1.
A line of shapes is constructed using matchsticks.

Shape-1 Shape-2 Shape-3 Shape-4
i) Find the rule that shows how many sticks are needed to make a line of such shapes ?
ii) How many matchsticks are needed to form shape -12 ?
Solution:

Number of matchsticks 3 5 7 9
i) Let us know the pattern
S₁ = 3 = 2 + 1 = (1 × 2) + 1
S₂ = 5 = 4 + 1 = (2 × 2) + 1
S₃ = 7 = 6 + 1 = (3 × 2) + 1
S₄ – 9 = 8 + 1 = (4 × 2) + 1
Now the rule for this pattern is number of matchsticks.

ii) Used to make ‘n’ number of shapes is Sn = (n × 2) + 1 = 2n + 1
Number of matchsticks needed to form shape – 12 is
S₁₂ = 2(12) + 1 = 24 + 1 = 25 sticks.

Check Your Progress (Page No. 105)

Question 1.
Fill the following table as instructed. One is shown for you.

S.No.	Expression	Verbal Form
1.	y + 3	Three more than y
2.	2x – 1
3.	5z
4.	\(<table border=”2″>\)

Solution:

S.No.	Expression	Verbal Form
1.	y + 3	Three more than y
2.	2x – 1	One less than the double of x
3.	5z	5 times of z
4.	\(<table border=”2″>\)	Half of the m

Let’s Explore ? (Page No. 106)

Question 1.
Find the general rule for the perimeter of a rectangle. Use variables T and ‘b’ for length and breadth of the rectangle respectively.
Solution:
Given length of rectangle = l
breadth of rectangle = b
We know that the perimeter of rectangle is twice the sum of its length and breadth.
Sum of length and breadth = l + b
Twice the sum of length and breadth = 2 × (l + b)
Rule for the perimeter of a rectangle = 2(l + b)

Question 2.
Find the general rule for the area of a square by using the variable ‘s’ for the side of a square.
Answer:
Given side of a square = s
We know that the area of a square is the product of side and side.
Area of a square = side × side
Rule for the area of a square = s.s

Side	Area
1	1 × 1
2	2 × 2
3	3 × 3
4	4 × 4
………..	…………
s	S × s

(Page No. 107)

Question 1.
Find the nth term in the following sequences.
0 3, 6, 9, 12, ii) 2, 5, 8, 11, iii) 1, 4, 9, 16,
Solution:
i) Given number pattern is 3, 6, 9, 12,……………..
To find the nth term in the given pattern, we put the sequence in a table.

First number = 3 × 1
Second number = 3 × 2
nth number = 3 × n = 3n
So, the nth term of the pattern 3, 6, 9, 12, is 3n.

ii) Given number pattern is 2, 5, 8, 11,
To find the nth term in the given pattern, we put the sequence in a table.

First number = 2 = 3 × 1 – 1
Second number = 5 = 3 × 2 – 1
Third number = 8 = 3 × 3 – 1
nth number = 3 × n – 1 = 3n – 1
So, the nth term of the pattern 2, 5, 8, 11 is 3n – 1.

iii) Given number pattern is 1, 4, 9, 16,
To find the nth term in the given pattern, we put the sequence in a table.

First number =1 = 1 × 1
Second number = 4 = 2 × 2
Third number =9 = 3 × 3
nth number = n × n = n²
So, the nth term of the pattern 1, 4, 9, 16 is n².

Check Your Progress (Page No. 108)

Question 1.
Complete the table and find the value of ‘p’ for the equation \(\frac{\mathbf{p}}{\mathbf{3}}\) = 4

p	\(\frac{\mathbf{p}}{3}\) = 4	Condition satisfied ? Yes/ No
3
6
9
12

Solution:

p	\(\frac{\mathbf{p}}{3}\) = 4	Condition satisfied ? Yes/ No
3	\(\frac{3}{3}\) ≠1 ≠ 4	No
6	\(\frac{6}{3}\) ≠2 ≠ 4	No
9	\(\frac{9}{3}\) ≠ 3 ≠ 4	No
12	\(\frac{12}{3}\) ≠ 4 ≠ 4	Yes

Question 2.
Write LHS and RHS of following simple equations.
i) 2x + 1 = 10
ii) 9 = y – 2
iii) 3p + 5 = 2p + 10
Solution;
i) 2x+ 1 = 10
Given equation is 2x + 1 = 10
L.H.S = 2x + 1
R.H.S = 10

ii) 9 = y – 2
Given equation is 9 = y – 2
LHS = 9
RHS = y – 2

iii) 3p + 5 = 2p + 10
Given equation is 3p + 5 = 2p + 10
LHS = 3p + 5
RHS = 2p + 10

Question 3.
Write any two simple equations and write their LHS and RHS.
Solution:
i) Consider 8x + 3 = 4 is a simple equation.
L.H.S = 8x + 3
RHS = 4

ii) Consider 5a + 6 = 8a – 3 is a simple equation.
LHS = 5a + 6
RHS = 8a – 3

Let’s Explore (Page No. 109)

Observe for what value of m, the equation 3m = 15 has both LHS and RHS become equal.
Solution:
Given equation is 3m = 15
If m = 1, then the value of 3m = 3(1) = 3≠15 ∴ LHS ≠RHS
If m = 2, then the value of 3m = 3(2) = 6 ≠ 15 ∴ LHS ≠ RHS
If m = 3, then the value of 3m = 3(3) = 9≠15 ∴ LHS ≠ RHS
If m = 4, then the value of 3m = 3(4) = 12 ≠ 15 ∴ LHS ≠ RHS
If m = 5, then the value of 3m = 3(5) = 15 = 15 ∴ LHS = RHS
From the above when m = 5 the both LHS and RHS are equal

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 6 Basic Arithmetic InText Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 6th Lesson Basic Arithmetic InText Questions

Check Your Progress (Page No. 84)

Question 1.
Express the terms 45 and 70 by using ratio symbol.
Solution:
Given terms are 45 and 70
Ratio = 45 : 70
It can be read as 45 is to 70.

Question 2.
Write antecedent in the ratio 7:15.
Solution:
Given ratio 7 : 15
In the ratio first term is called antecedent.
In 7 : 15 antecedent is 7.

Question 3.
Write the consequent in the ratio 8 : 13.
Solution:
Given ratio 8 : 13
In the ratio second term is called consequent.
In 8 : 13 consequent is 13.

Question 4.
Express the ratio 35 : 55 in the simplest form.
Solution:
Given ratio 35 : 55 (or)
To write the ratio in the simplest form we have to divide by the common factor of two terms 35 and 55.
Common factor is 5.
Now divide by 5,
\(\frac{35}{55}=\frac{35 \div 5}{55 \div 5}=\frac{7}{11}\)
Simplest form of \(\) is \(\frac{7}{11}\)

Question 5.
In the given figure, find the ratio of
i) Shaded part to unshaded parts.
ii) Shaded part to total parts,
iii) Unshaded parts to total parts.

Solution:
i) In the given figur.e,
Number of shaded parts = 1
Number of unshaded parts = 3
Ratio = shaded parts : unshaded parts = 1:3

ii) Number of shaded parts = 1
Total parts = 4
Ratio = shaded parts : total parts = 1:4

iii) Number of unshaded parts = 3
Total parts = 4
Ratio = unshaded parts : total parts = 3:4

Question 6.
Express the following in the form of ratio.
a) The length of a rectangle is triple its breadth. ‘
b) In a school, the workload of teaching 19 sections has been assigned to 38 teachers.
Solution:
a) Let breadth of rectangle = x or one part = 1 part
length of rectangle = triple the breadth
= 3 x x = 3x = 3 parts
Ratio = l : b = x : 3x =\(\frac{1 x}{3 x}=\frac{1}{3}\) = 1:3

b) Given number of sections = 19
Number of teachers = 38
Ratio = 19 : 38 = \(\frac{19}{38}=\frac{1}{2}\) = 1 : 2

(Page No. 88)

Question 1.
Which ratio is larger in the following pairs ?
(a) 5 : 4 or 9 : 8
(b) 12 : 14 or 16 : 18
(c) 8: 20 or 12: 15
(d)4:7 or 7:11
Solution:
a) 5 : 4 or 9 : 8
Write the given ratios as fractions, we have 5 : 4 = \(\frac{5}{4}\) and 9 : 8 = \(\frac{9}{8}\)
Now find the LCM of the denominators of 4 and 8 is 8.
Make the denominator of the each fraction equal to 8.
We have \(\frac{5}{4} \times \frac{2}{2}=\frac{10}{8}\) and \(\frac{9}{8} \times \frac{1}{1}=\frac{9}{8}\)
Clearly we know that 10 > 9
∴ \(\frac{10}{8}>\frac{9}{8}\) (or) 10 : 8 > 9 : 8
10 : 8 is equal to 5 : 4
Therefore the larger ratio is 5 : 4.

b) 12 : 14 or 16:18
12 : 14 = \(\frac{12}{14}=\frac{6}{7}\) and 16 : 18 = \(\frac{16}{18}=\frac{8}{9}\)
Now find the LCM of the denominators of 7 and 9 is 63.
Make the denominator of the each fraction equal to 63.
we have \(\frac{6}{7} \times \frac{9}{9}=\frac{54}{63}\) and \(\frac{8}{9} \times \frac{7}{7}=\frac{56}{63}\)
Clearly, we know that 54 < 56
∴ \(\frac{54}{63}<\frac{56}{63}\) (or) 54:63 < 56:63
56 : 63 is equal to 16 : 18 (or) 8 : 9
∴ The larger ratio is 16 : 18.

c) 8 : 20 or 12 : 15
Write the given ratios as fractions we have
8:20 = \(\frac{8}{20}=\frac{2}{5}\) and 12:15 = \(\frac{12}{15}=\frac{4}{5}\)
\(\frac{2}{5}\) and \(\frac{4}{5}\)

Clearly \(\frac{2}{5}\) < \(\frac{4}{5}\)
i.e., 2:5 < 4 : 5 (or) 8: 20 < 12: 15
Therefore the larger ratio is 12 : 15.

d) 4: 7 or 7: 11
Write the given ratios as fractions, we have 4 7
4 : 7 = \(\frac{4}{7}\) and 7:11 = \(\frac{7}{11}\) .
Now find the LCM of the denominators of 7 and 11 is 77.
Make the denominators of the each fraction equal to 77.
We have \(\frac{4}{7} \times \frac{11}{11}=\frac{44}{77}\) and \(\frac{7}{11} \times \frac{7}{7}=\frac{49}{77}\)
\(\frac{44}{7}\) and \(\frac{49}{77}\)
Clearly we know that 44 < 49
∴ \(\frac{44}{77}<\frac{49}{77}\) (or) 44 : 77 < 49 : 77
i.e.,4: 7 < 7 : 11
Therefore the larger ratio is 7 : 11

Question 2.
Find three equivalent ratios of 12 : 16.
Solution:
Given ratio is 12 : 16
Write the given ratio as fraction we have 12:16= \(\frac{12}{16}=\frac{3}{4}\)
Now, write equivalent fractions of \(\frac{3}{4}\)

i. e., 6 : 8 = 9 : 12 = 12 : 16 = 15 : 20 = 18 : 24
∴ Equivalent ratios of 12 : 16 are 6 : 8, 9 : 12, 12 : 16, 15 : 20 and 18 : 24.

(Page No. 90)

Question 1.
Check whether the following terms are in proportion ?
1) 5,6,7,8
2) 3,5,6,10
3) 4,8,7,14
4) 2,12,3,18
Solution:
1) Given, 5, 6, 7, 8
If a, b, c, d are in proportion i.e., a : b :: c : d
If 5, 6, 7, 8 are in proportion i.e., 5 : 6 : : 7 : 8
We know that, product of extremes = Product of means [a x d : b x c]
5 x 8 = 6 x 7
40 ≠ 42
So, 5, 6, 7, 8 are not in proportion.

2) Given, 3, 5, 6, 10
If a, b, c, d are in proportion i.e., a : b :: c : d
If 3, 5, 6, 10 are in proportion i.e., 3 : 5 :: 6 : 10
We know that, product of extremes = Product of means a x d = b x c
3 x 10 = 5 x 6
30 = 30
So, 3, 5, 6, 10 are in proportion.

3) Given, 4, 8, 7, 14.
If a, b, c, d are in proportion i.e., a : b : : c : d
If 4, 8, 7, 14 are in proportion i.e., 4 : 8 : : 7 : 14
We know that, product of extremes = Product of means a x d = b x c
4 x 14 = 8 x 7
56 = 56
So, 4, 8, 7, 14 are in proportion.

4) Given, 2, 12, 3, 18
If a, b, c, d are in proportion i.e., a : b :: c : d
If 2, 12, 3, 18 are in proportion i.e., 2 : 12 : : 3 : 18
We know that, product of extremes = Product of means [ a x d = b x c ]
2 x 18 = 12 x 3
36 = 36
So, 2, 12, 3, 18 are in proportion.

Let’s Explore (Page No. 92)

Question 1.
Read the table and fill in the boxes.

Prepare two similar problems and ask your friend to solve them
Solution:

Weight	Cost of Tomato	Cost of Potato
5 kg	₹ 75	₹ 60
1 kg	₹15	₹ 12
3 kg	₹ 45	₹ 36

(Page. No. 94)

Question 1.
Represent the following in other forms.

Solution:

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts Ex 8.2 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 8th Lesson Basic Geometric Concepts Ex 8.2

Question 1.
Measure the lengths of the given line segments.

Solution:
i) \(\overline{\mathrm{AB}}\) = 2.4 cm
ii) \(\overline{\mathrm{PQ}}\) = 1.5 cm
iii) \(\overline{\mathrm{KL}}\) = 1cm, \(\overline{\mathrm{LM}}\) = 1 cm
\(\overline{\mathrm{KM}}=\overline{\mathrm{KL}}+\overline{\mathrm{LM}}\) =1 + 1 = 2 cm

Question 2.
Draw the following line segments.
i) AB = 6.3 centimeters ii) MN = 3.6 centimeters

Question 3.
Draw PQ = 4.6 cm and extend upto R such that PR = 6 cm.
Solution:

\(\overline{\mathrm{PR}}=\overline{\mathrm{PQ}}+\overline{\mathrm{QR}}\) = 4.6 + 1.4 = 6 cm

Question 4.
Draw a line segment \(\overline{\mathrm{OP}}\) with certain length and mark a point Q on it.
Check whether \(\overline{\mathrm{OP}}-\overline{\mathrm{PQ}}=\overline{\mathrm{OQ}}\)
Solution:

Given, \(\overline{\mathrm{OP}}\) = 8 cm; \(\overline{\mathrm{PQ}}\) = 3 cm
\(\overline{\mathrm{OP}}-\overline{\mathrm{PQ}}\) = 8cm – 3cm = 5 cm = \(\overline{\mathrm{OQ}}\)
∴ \(\overline{\mathrm{OP}}-\overline{\mathrm{PQ}}=\overline{\mathrm{OQ}}\) = 5cm

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 5 Fractions and Decimals InText Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 5th Lesson Fractions and Decimals InText Questions

(Page No. 63)

Question 1.
Is it true to say that 3 × \(\frac{1}{5}=\frac{1}{5}\) x 3?
Solution:
3 × \(\frac{1}{5}=\frac{1}{5}\) × 3. Yes, it is true.
By using commutative property over multiplication a × b = b × a
3 × \(\frac{1}{5}=\frac{1}{5}\) × 3 = \(\frac{3}{5}\)

Check Your Progress (Page No. 63)

Find :
i) \(5 \times 3 \frac{2}{7}\)
ii) \(2 \frac{5}{9} \times 3\)
iii) \(2 \frac{4}{7} \times 3\)
iv) \(3 \times 1 \frac{3}{4}\)
Solution:

Let’s Explore (Page No. 64)

Question 1.
Observe the products of fractions.
Have you observed the products of any two fractions is always lesser or greater than each of its fraction, write conclusion.
\(\frac{1}{5} \times \frac{2}{3}=\frac{2}{15}\) (Product of two proper fractions)
Solution:
Product of any two proper fractions is always less than each of its fraction.
i. e., \(\frac{2}{15}<\frac{1}{5} \text { and } \frac{2}{15}<\frac{2}{3}\)

ii) \(\frac{3}{2} \times \frac{5}{4}=\frac{15}{8}\) (Product of two improper fractions) •
Solution:
The product of any two improper fractions is always greater than each of its fraction.
i.e, \(\frac{3}{2}<\frac{15}{8} \text { and } \frac{5}{4}<\frac{15}{8}\)

iii) \(\frac{2}{3} \times \frac{5}{3}=\frac{10}{9}\) (Product of proper and improper fractions)
Solution:
The product of a proper fraction and an improper fraction is always greater than its proper fraction and less than its improper fraction.
i.e., \(\frac{2}{3}<\frac{10}{9} \text { and } \frac{5}{3}>\frac{10}{9}\)

(Pg. No. 66)

Question 1.
i) 4 ÷ \(\frac{1}{8}\)
ii) 9 ÷ \(\frac{3}{4}\)
iii) 7 ÷ \(\frac{2}{3}\)
iv) 35 ÷ \(\frac{7}{3}\)
v) 4 ÷ \(\frac{15}{8}\)
Solution:

(Pg. No. 67)

Question 1.
Observation these products and fill in the blanks.

Solution:

Check Your Progress (Page No. 68)

Question 1.
Write the reciprocal of fractions in the given table.

Solution:

(Reciprocal of a fraction \(\frac{\mathrm{a}}{\mathrm{b}}\) is \(\frac{\mathrm{b}}{\mathrm{a}}\))

(Page No. 69)

Question 1.
Find
i) \(\frac{7}{9}\) ÷ 4
ii) \(\frac{3}{4}\) ÷ 9
iii) 4\(\frac{1}{2}\) ÷ 6
iv) \(\frac{1}{5}\) ÷ 3
Solution:

Check Your Progress (Page No. 73, 74 & 75)

Question 1.
Fill in the blanks.

Fraction	Decimal Number	” Read as
\(\frac{6}{10}\)	0.6	Zero point six
\(\frac{37}{100}\)	0.37	Zero point three seven
	0.721	Zero point seven two one
		Seventeen point two

Solution:

Fraction	Decimal Number	Read as
\(\frac{6}{10}\)	0.6	Zero point six
\(\frac{37}{100}\)	0.37	Zero point three seven
\(\frac{721}{1000}\)	0.721	Zero point seven two one
\(\frac{172}{10}\)	17.2	Seventeen point two

Question 2.
Write the place value of the circled digits.

Solution:

Question 3.
a) 700 + 40 + 2 + \(\frac{1}{10}+\frac{3}{100}+\frac{6}{1000}\)
Solution:
700 + 40 + 2 + 0.1 + 0.03 + 0.006 = 742.136

b) 9000 + 800 + 3 + 0.2 + 0.05 + 0.007
Solution:
9000 + 800 + 3 + 0.2 + 0.05 + 0.007 = 983.257

c) 6000 + 400 + 20 + 1 + \(\frac{2}{10}+\frac{5}{100}+\frac{9}{1000}\)
Solution:
6000 + 400 + 20 + 1 + 0.2 + 0.05 + 0.009 = 6421. 259

d) 400 + 5+ \(\frac{1}{10}+\frac{8}{100}\)
Solution:
400 + 5 + 0.1 +0.08 = 405.18

Question 4.
Expand the following into decimals and fractional forms,
a) 164.238
b) 968.054
Solution:
a) 164.238 = 100 + 60 + 4 + 0.2 + 0.03 + 0.008
= 100 + 60 + 4 + \(\frac{2}{10}+\frac{3}{100}+\frac{8}{1000}\)

b) 968.054
Solution:
= 900 + 60 + 8 + 0.0 + 0.05 + 0.004
= 900 + 60 + 8 + 0 + \(\frac{5}{100}+\frac{4}{1000}\)
= 900 + 60 + 8 + \(\frac{5}{100}+\frac{4}{1000}\)

Question 5.
Write fractions as decimals.
1. \(\frac{23}{10}\) = ………..
2. \(\frac{6}{100}\) = ………..
3. \(\frac{3}{8}\) = ………..
4. \(\frac{2}{25}\) = ………..
Solution:
1. \(\frac{23}{10}\) = 2.3
2. \(\frac{6}{100}\) = 0.06
3. \(\frac{3}{8}\) = 0.375
4. \(\frac{2}{25}\) = 0.08

Question 6.
Write decimals as fractions in simplest form.
1. 0.2 = ……………
2. 0.38 = ……………
3. 1.62 = ……………
4. 8.1 = ……………
Solution:
1. 0.2 = \(\frac{2}{10}\)
2. 0.38 = \(\frac{38}{100}\)
3. 1.62 = \(\frac{162}{100}\)
4. 8.1 = \(\frac{81}{10}\)

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 4 Integers InText Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 4th Lesson Integers InText Questions

Check Your Progress?(Page No. 47)

Question 1.
Write any five positive integers.
Solution:
1, 2, 3, 4, 5, 6, 7,

Question 2.
Write any five negative integers.
Solution:
-1, -2, -3, -4, -5, -6,

Question 3.
Which number is neither positive nor negative?
Solution:
0 (zero)

Question 4.
Represent the following situations with integers,
(i) A gain of ₹ 500 ( )
(ii) Temperature is below 5°C ( )
Solution:
i) + 7 500
ii) – 5° C

Question 5.
Represent the following using either positive or negative numbers.
a) A bird is flying at a height of 25 meters above the sea level and a fish at a depth of 2 meters.
b) A helicopter is flying at a height of 60m above the sea level and a submarine is at 400m below sea level.
Solution:
a) Height of the flying bird 25 meters from th$ sea level = + 25 meters
Depth of the fish 2 meters from the sea level = – 2 meters
b) Height of the flying helicopter 60 meters from the sea level = + 60 m
Depth of the submarine 400 m from the sea level = – 400 m

Check Your Progress (Page No. 49)

Question 1.
Draw a vertical number line and represent -5,4,0,-6, 2 and 1 on it.
Solution:

Question 2.
Represent opposite integers of – 200 and + 400 on integer number line.
Solution:
Opposite integers means, additive inverse.
∴ Opposite integer (additive inverse) of – 200 is 200.
Opposite integer (additive inverse) of +400 is – 400.

Let’s Think (Page No. 50)

Question 1.
For any two integers, say 3 and 4, we know that 3 < 4.
Is it true to say -3 < -4? Give reason.
Solution:
On the number line, the value of a number increases as we move to right and decreases as we move to the left. As -3 lies right to -4 on the number line.
So, -3 < -4 is not true.

(Pg. No. 52)

Question 1.
What is additive inverse of 7 ?
Solution:
Additive inverse of 7 is -7.

Question 2.
What is additive inverse of -8 ?
Solution:
Additive inverse of -8 is 8.

Let’s Explore (Page.No. 52)

Question 1.
Find the value of the following using a number line.
i) (-3) + 5 ii) (-5) + 3
Make two questions on your own and solve them using the number line.
Solution:
i) (-3) + 5

On the number line, we first move 3 steps to the left of 0 to reach -3.
Then, we move 5 steps to the right of -3 and reach +2. So, (-3) + 5 = 2

ii) (-5) + 3

On the number line, we first move 5 steps to the left of 0 to reach -5. Then, we move 3 steps to the right of -5 and reach -2. So, (-5) + 3 = – 2

iii)(+6) + (-3)

On the number line, we first move 6 steps to the right of 0 to reach +6.
Then, we move 3 steps to the left of 6 and reach +3. So, (+6) + (-3) = 3

iv) (-4) + (-3)

On the number line, we first move 4 steps to the left of 0 to reach -4. Then, we move 3 steps to the left of -4 and reach -7.
So (-4) + (-3) = -7.

Question 2.
Find the solution of the following:
i) (+5) + (-5) (ii) (+6) + (-7) (iii) (-8) + (+2)
Ask your friend five such questions and solve them.
Solution:
i) (+5) + (-5) = 0

On the number line, we first move 5 steps to the right of 0 to reach +5.
Then, we move 5 steps to the left of +5 and reach 0.

ii) (+6) + (-7) = -1

On the number line, we first move 6 steps to the right of 0 to reach +6.
Then, we move 7 steps to the left of +6 and reach -1.

(iii) (-8) + (+2)

On the number line, we first move 8 steps to the left of 0 to reach -8. Then, we move 2 steps to the right of -8 and reach -6.

iv)(-4) + (+8) = +4

On the number line, we first move 4 steps to the left of 0 to reach -4. Then, we move 8 steps to the right of -4 and reach +4.

v) (+3) + (-4) = -1

On the number line, we first move 3 steps to the right of 0 to reach +3. Then, we move 4 steps to the left of +3 and reach -1.

vi) (+5) + (-6) = -1

On the number line, we first move 5 steps to the right of 0 to reach +5. Then, we move 6 steps to the left of +5 and reach -1.

vii) (+4) + (-4) = 0

On the number line, we first move 4 steps to the right of 0 to reach +4. Then, we move 4 steps to the left of +4 and reach 0.

viii) (-6) +(+4) =-2

On the number line, we first move 6 steps to the left of 0 to reach -6. Then, we move 4 steps to the right of -6 and reach -2.
So, (-6) + (+4) = -2

Let’s Explore (Page No. 55)

Question 1.
Take any two integers a and b. Check whether a+b is also an integer.
Case (i) : Consider two integers 3 and -2 (Positive and negative)
Sum = 3 + (-2) = +1 + 2 – 2 = +1 + 0 = +1 is also an integer.
Case (ii) : Consider two ihtegers 5 and 6 (Both are positive)
Sum = 5 +6 = + 11 is also an integer
Case (iii) : Consider two integers -4 and -6 (Both are negative)
Sum = -4 + (-6) = -4 -6 = -10 is also an integer
Case (iv) : Consider two integers -5 and 4 (Negative and positive)
Sum = -5 + 4 = -1 -4 + 4 = -1 + 0 = -1 is also an integer.
So, if a and b are integers, then their sum a + b is also an integer. Integers are closed under addition.

Question 2.
Check the following Properties on integers, a, b, c are any integers.
i) Closure Property under subtraction ‘
ii) Commutative Property under addition and subtraction (a + b = b + a ?, a – b = b – a?)
iii) Associative Property under addition and subtractioji.
(a + b) + c = a + (b + c) ? (a – b).- c = a – (b – c)?
Solution:
i) Closure Property under subtraction :
Case (i) : Consider two integers 4, -5 (positive and negative)
Then, difference a – b = 4 – (-5) = 4 + 5 = + 9is also an integer.
Case (ii) : Consider two integers 3, 8 (Both are positive)
Then, difference a – b = 3 – (+8) = 3-8
= +3 – 3 – 5 = 0 – 5 = -5is also an integer.
Case (iii) : Consider two integers -2, -6 (Both are negative) .
Then, difference a – b = -2 – (-6) = -2 + 6
= -2 + 2 + 4 = 0 + 4 = +4 is also an integer.

Case (iv) : Consider two integers -3, 2 (Negative and positive)
Then, difference a – b = -3 – (+2) = -3 -2 = -5 is also an integer.
So, if a and b are any two integers, then their difference a – b is also an integer. Integers are closed under subtraction.

ii) Commutative Property under addition and subtraction :
(a + b = b + a, a-b = b-a)
(A) Case (Q : Consider two integers-3 and 5 (Negative and positive)
Then, a + b = -3 + (+5)
= -3 + 5 = – 3 + 3 + 2 = 0 + 2 = + 2
b + a = +5 + (-3) = +2 + 3- 3 = +2 + 0 = + 2
∴ a + b = b + a

Case (ii) : Consider two integers +4 and +2 (Both are positive)
Then, a + b = +4 + (+2) = +4 + 2 = + 6 ‘b + a = +2 + (+4) = +2 + 4 = + 6
∴ a + b = b + a

Case (iii) Consider two integers -5 and -3 (Both are negative) Then, a + b = -5 + (-3) = -5 – 3 = -8 b + a = -3 + (-5) = – 3 – 5 = -8
∴ a + b = b + a

Case (iv) : Consider two integers +4 and -1 (Positive and negative)
Then, a + b = +4 + (*1) = +4 -1 = +3 + 1 -1 = +3 + 0 = +3 b + a = -1 + (+4) =-l + 4 = -l + l+ 3 = 0 + 3 = + 3
∴ a + b = b + a
So, integers are commutative under addition.

(B) Consider two integers -4 and +6 (Negative and positive)
Then, a – b = -4 – (+6) = -4 – 6 = -10 b-a = + 6-(-4) = +6 +4 = +10 -10*10
a – b ≠ b – a
So, integers are not commutative under subtraction.

iiO Associative property under addition and subtraction :
(a + b) + c = a + (b + c) ; (a – b) – c = a – (b – c)
(A) Case (i) : Consider any three integers 2, 4, -5
(a + b) + c = (2 + 4) + (-5) = 6 – 5 = +1 + 5- 5 = +1 + 0 = +1
a + (b + c) = 2 + (4 + (-5)) = 2 + (4 – 5) = 2 + (4 – 4 -1)
= 2 + (0-1) = + 2 – 1 = + 1 + 1 – 1 = +1 + 0 = +1
∴ (a + b) + c = a + (b + c)

Case (ii) : Consider any three integers +2, -5, +3
Then, (a + b) + c = [2 + (-5)] + 3 = [+2 -5] + 3 = +2 -2 -3 + 3 = 0 + 0 = 0
a + (b + c) = +2 + [( – 5)+3] = +2 + [-2 – 3 + 3] = +2 + (-2 + 0) = +2 -2 = 0
∴ (a + b) + c = a + (b + c)

Case (iii) : Consider any three integers 3, 4, 6
Then, (a + b) + c = [2 + (-5)] + 3 [+2-5] + 3
a + (b + c) = 3 + (4 + 6) = 3 + 10 = + 13
∴ (a + b) + c = a + (b + c)

Case (iv) : Consider any three integers -4, -2, +5
Then, (a + b) + c = [-4 + (-2)] + 5 = [-4 -2] + 5 = -6 + 5
= -1-5 + 5 = – 1 + 0 = – 1
a + (b + c) = -4 + [(-2) + 5] = -4 + [-2 + 5]
= _4 + [_2 + 2 +3] = -4 + 0 + 3 =-1-3+ 3
= -1 + 0 = -1
∴ (a + b) + c = a + (b + c)

Case (v) : Consider any three integers-3, 4, 1
Then, (a + b) + c = (-3 + 4) + 1 = (-3 + 3 + 1) + 1 = 0 + 1 + 1 = + 2
a + (b + c) = -3 + (4 + 1) = -3 + 5 = -3 + 3 + 2 = 0 + 2 = +2
∴ (a + b) + c = a + (b + c)

Case (vi) : Consider any three integers -2, 6, -7
Then, (a + b) + c = (-2 + 6) + (-7) = (-2 + 2 + 4) + (-7) = 0 + 4 – 7
= + 4 – 4 – 3 = 0 – 3 = -3
a + (b + c) = -2 + [6 + (-7)] = -2 + (6 – 7) = -2 + [+6 – 6 -1]
= -2 + (-1) = -2 -1 = -3
∴ (a + b) + c = a + (b + c)

Case (vii) : Consider any three integers +6, -3, -1
Then, (a + b) +c = [+6 + (-3)] + (-1) = (6 – 3) – 1 = (+3 +3 -3) -1
=+3-1 =+2 + 1 – 1 =+2 + 0 = + 2
a + (b + c) = +6 + [-3 + (-1)] = 6 + [-3 – 1] = 6 + (-4) .
= +2 + 4 – 4 = 2 + 0 = + 2
∴ (a + b) + c = a + (b + c)

Case(viii)
Consider any three integers -4, -1, -7
Then, (a + b) + c = [-4 + (-1)] + (-7) = (-4 -1) – 7 = -5 -7 = -12
a + (b + c) = -4 + [(-1) + (-7)] = -4 + [-1 -7] = -4 + (-8)
= -4 – 8 = -12
∴ (a + b) + c = a + (b + c)
From all the above cases we conclude that, integers are associative under addition.

(B) Consider any three integers +5, -4, 1
Then, (a – b) – c = (+5 – (-4)) – (+1) = (5 + 4) – 1 .
= + 9 – 1 = + 8 + 1 – 1 = + 8 + 0 = + 8
a – (b – c) = +5 – [- 4 – (+1)] = + 5 – [-4 – 1]
= + 5 – [ -5] =+ 5 + 5 = + 10
+ 8 ≠ +10
∴ (a – b) – c ≠ a – (b – c)
So, integers are not associative under subtraction.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 1 Numbers All Around us Ex 1.2 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 1st Lesson Numbers All Around us Exercise 1.2

Question 1.
Write each of the following in numeral form.
i) Sixty crores seventy five lakhs ninety two thousands five hundred and two.
Answer:
60, 75, 92, 502

ii) Nine hundred forty four crores six lakhs fifty five thousand four hundred and eighty six.
Answer:
944, 06, 55, 486

iii) Ten crores ten thousand and ten.
Answer:
10,00,10,010

Question 2.
Insert commas in the correct positions to separate periods and write the following numbers in words.
i) 57657560
ii) 70560762
iii) 97256775613
Answer:
i) 5,76,57,560: Five crores seventy six lakhs fifty seven thousands five hundred and sixty.
ii) 7,05,60,762: Seven crores five lakhs sixty thousands seven hundred and sixty two.
iii) 9725,67,75,613: Nine thousand seven hundred and twenty five crores sixty seven lakhs seventy five thousand six hundred and thirteen.

Question 3.
Write the following in expanded form.
i) 756723
ii) 60567234
iii) 8500756762
Answer:
i) 756723
Expanded form: 7 × 1,00,000 + 5 × 10,000 + 6 × 1,000 + 7 × 100 + 2 × 10 + 3 × 1
: 7 lakhs + 5 ten thousands + 6 thousands + 7 hundreds + 2 tens + 3 ones
Word form: Seven lakh fifty six thousand seven hundred and twenty three.

ii) 60567234
Expanded form: 6 × 1,00,00,000 + 5 × 1,00,000 + 6 × 10,000 + 7 × 1,000 + 2 × 100 + 3 × 10 + 4 × 1
: 6 crores + 5 lakhs + 6 ten thousands + 7 thousands + 2 hundreds + 3 tens + 4 ones
Word form: Six crore five lakh sixty seven thousand two hundred and thirty four.

iii) 8500756762
Expanded form: 8 × 1,00,00,00,000 + 5 × 10,00,00,000 + 7 × 1,00,000 + 5 × 10,000 + 6 × 1000 + 7 × 100 + 6 × 10 + 2 × 1
: 8 hundred crores + 5 ten crores + 7 lakhs & 5 ten thousands + 6 thousands + 7 hundreds + 6 tens + 2 ones
Word form: Eight hundred and fifty crore seven lakh fifty six thousand seven hundred and sixty two.

Question 4.
Determine the difference between the place value and the face value of 6 in 86456792.
Answer:
Given number is 86456792. By putting commas to separate periods the given number can be written as 8,64,56,792.
i) Place value of ‘6’ in thousand place = 6 x 1000 = 6,000
Face value of 6 = 6
Difference = 6,000 – 6 = 5,994
ii) Place value of ‘6’ in ten lakhs palce = 6 x 10,00,000 = 60,00,000
Face value of 6 = 6
Difference = 60,00,000 – 6 = 59,99,994