Class 6

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

AP State Syllabus 6th Class Maths Solutions 7th Lesson Introduction to Algebra Unit Exercise

Question 1.
The cost of one fan is Rs. 1500. Then what is the cost of ‘n’ fans?
Answer:

Given cost of one fan = Rs. 1500
Number of fans = n
Cost of n fans = cost of one fan × no. of fans = 1500 × n
∴ Cost of n fans = 1500n

Question 2.
Srinu has number of pencils. Raheem has 4 times the pencils as of Srinu. How many pencils does Rahim has? Write an expression.
Answer:
Let number of pencils Srinu has = x
Number of pencils Raheem has = 4 times of Srinu
= 4 × x
∴ Number of pencils Raheem has = 4x

Question 3.
Parvathi has 5 more books than Sofia. How many books are with Parvathi? Write an expression choosing any variable for number of books.
Answer:
Let number of books Sofia has = y
Given Parvathi has 5 more books than Sofia
Number of books Parvathi has = 5 books more than Sofia
= y + 5
∴ Number of books Parvathi has = y + 5

Question 4.
Which of the following are equations?
i) 10 – 4p = 2
ii) 10 + 8x < – 22
iii) x + 5 = 8
iv) m + 6 = 2
v) 22x – 5 = 8
vi) 4k + 5 > – 100
vii) 4p + 7 = 23
viii) y < – 4
Answer:
i) 10 – 4p = 2
We know that, a mathematical statement involving equality symbol is called an equation.
10 – 4p = 2 has equality symbol.
So, it is an equation.

ii) 10 + 8x < – 22
We know that, a mathematical statement involving equality symbol is called an equation.
10 + 8x < – 22 has no equality symbol.
So, it is not an equation [so it is an inequation]

iii) x + 5 = 8 We know that, a mathematical statement involving equality symbol is called an equation.
x + 5 = 8 has equality symbol.
So, it is an equation.

iv) m + 6 = 2 We know that, a mathematical statement involving equality symbol is called an equation.
m + 6 = 2 has equality symbol.
So, it is an equation.

v) 22x – 5 = 8 We know that, a mathematical statement involving equality symbol is called an equation.
22x – 5 = 8 has equality symbol.
So, it is an equation.

vi) 4k + 5 > – 100
We know that, a mathematical statement involving equality symbol is called an equation.
4k + 5 > -100 has no equality symbol.
So, it is not an equation.
It is an inequation.

vii) 4p + 7 = 23
We know that, a mathematical statement involving equality symbol is ailed an equation.
4p + 7 = 23 has equality symbol.
So, it is an equation.

viii) y < – 4
We know that, a mathematical statement involving equality symbol is called an equation.
y < – 4 has no equality symbol.
So, it is not an equation.
It is an inequation.

Question 5.
Write L.H.S and R.H.S of the following equations:
i) 7x + 8 = 22
ii) 9y – 3 = 6
iii) 3k – 10 = 2
iv) 3p – 4q = -19
Answer:
i) 7x + 8 = 22
Given equation is 7x + 8 = 22
LHS = 7x + 8
RHS = 22

ii) 9y – 3 = 6
Given equation is 9y – 3 = 6
LHS = 9y – 3
RHS = 6

iii) 3k – 10 = 2
Given equation is 3k – 10 = 2
LHS = 3k – 10
RHS = 2

iv) 3p – 4q = -19
Given equation is 3p – 4q = -19
LHS = 3p – 4q
RHS = -19

Question 6.
Solve the following equations by trial and error method.
i) x – 3 = 5
ii) y + 6 = 15
iii) y = -1
iv) 2k – 1 = 3
Answer:
i) x – 3 = 5
Given equation is x – 3 = 5
If x = 1, then the value of x – 3 = 1 – 3 = -2 ≠ 5
If x = 2, then the value of x – 3 = 2 – 3 = -l ≠ 5
If x = 3, then the value of x – 3 = 3 – 3 = 0 ≠ 5
If x = 4, then the value of x – 3 = 4 – 3 = l ≠ 5
If x = 5, then the value of x – 3 = 5 – 3 = 2 ≠ 5
If x = 6, then the value of x – 3 = 6 – 3 = 3 ≠ 5
If x = 7, then the value of x – 3 = 7 – 3 = 4 ≠ 5
If x = 8, then the value of x – 3 = 8 – 3 = 5 = 5
From the above when x = 8, the both LHS and RHS are equal.
∴ Solution of the equation x – 3 = 5 is x = 8

ii) y + 6 = 15
Given equation is y + 6 = 15
If y = 1, then the value of y + 6 = 1 + 6 = 7 ≠ 15
If y = 2, then the value of y + 6 = 2 + 6 = 8 ≠ 15
If y = 3, then the value of y +6 = 3 + 6 = 9 ≠ 15
If y = 4, then the value of y + 6 = 4 + 6 = 10 ≠ 15
If y = 5, then the value of y + 6 = 5 + 6 = 11 ≠ 15
If y = 6, then the value of y + 6 = 6 + 6 = 12 ≠ 15
If y = 7, then the value hf y + 6 = 7 + 6 = 13 ≠ 15
If y = 8, then the value of y + 6 = 8 + 6 = 14 ≠ 15
If y = 9, then the value of y + 6 = 9 + 6 = 15 = 15
From the above when y = 9, the both LHS and RHS are equal.
∴ Solution of the equation y + 6 = 15 is y = 9

iii) \(\frac{m}{2}\) = -1
Given equation is \(\frac{m}{2}\) = -1
If m = 1, then the value of \(\frac{m}{2}\) = \(\frac{1}{2}\) ≠ -1
If m = 2, then the value of \(\frac{m}{2}\) = \(\frac{2}{2}\) = 1 ≠ -1
If m = 3, then the value of \(\frac{m}{2}\) = \(\frac{3}{2}\) ≠ -1
Here, we are not getting negative values.
If we take (substitute) m as a negative number we will get negative value.
If m = -1, then the value of \(\frac{m}{2}\) = \(\frac{-1}{2}\) ≠ -1
If m = -2, then the value of \(\frac{m}{2}\) = \(\frac{-2}{2}\) = -1 = -1
From the above when m = -2, the both LHS and RHS are equal.
∴ Solution of the equation \(\frac{m}{2}\) = -1 is m = -2.

iv) 2k – 1 = 3
Given equation is 2k – 1 = 3
If k = 1, then the value of 2k – 1 = 2(1) – 1 = 2 – 1 = 1 ≠ 3
If k = 2, then the value of 2k – 1 = 2(2) – 1 = 4 – 1 = 3 = 3
From the above when k – 2, the both LHS and RHS are equal.
Solution of the equation 2k – 1 = 3 is k = 2.

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

AP State Syllabus 6th Class Maths Solutions 5th Lesson Fractions and Decimals Unit Exercise

Question 1.
The sum of two fractions is 5\(\frac{3}{9}\). If one fraction is 2\(\frac{3}{4}\), then find the other fraction.
Answer:
Given one fraction = 2\(\frac{3}{4}\) = \(\frac{11}{4}\)
Sum of two fractions = 5\(\frac{3}{9}\) = \(\frac{48}{9}\)
Now, \(\frac{11}{4}\) + second fraction = \(\frac{48}{9}\)
Second fraction = \(\frac{48}{9}\) – \(\frac{11}{4}\)
LCM of 9 and 4 is 36

Question 2.
A rectangle sheet of paper is of length 12\(\frac{1}{2}\) and breadth 10\(\frac{2}{3}\). Find its perimeter.
Answer:
Length of rectangular sheet = l = 12\(\frac{1}{2}\) = \(\frac{25}{2}\) m
Breadth of rectangular sheet = b = 10\(\frac{2}{3}\) = \(\frac{32}{3}\) m
Perimeter of rectangular sheet = 2(1 + b)

Question 3.
Simplify: \(\left(3 \frac{1}{6}-1 \frac{1}{3}\right)+\left(4 \frac{1}{6}-2 \frac{1}{3}\right)\)
Answer:
\(\left(3 \frac{1}{6}-1 \frac{1}{3}\right)+\left(4 \frac{1}{6}-2 \frac{1}{3}\right)\)
First convert the given mixed fractions into improper fractions.
\(\left(\frac{19}{6}-\frac{4}{3}\right)\) + \(\left(\frac{25}{6}-\frac{7}{3}\right)\)
Then subtract the fractions which are in brackets by making them like fractions.

Now add the fractions = \(\frac{11+11}{6}\) = \(\frac{22}{6}\) = 3\(\frac{4}{6}\)

Question 4.
By what number should 3\(\frac{1}{16}\) be multiplied to get 9\(\frac{3}{16}\)?
Answer:
Given fraction = 3\(\frac{1}{16}\) = \(\frac{49}{16}\)
Product = 9\(\frac{3}{16}\) = \(\frac{147}{16}\)
Fraction × fraction to be multiplied = \(\frac{147}{16}\)
\(\frac{49}{16}\) × other number = \(\frac{147}{16}\)
other number = \(\frac{147}{16}\) ÷ \(\frac{49}{16}\) = \(\frac{147}{16}\) × \(\frac{16}{49}\)
Other number = 3
∴ Number to be multiplied = 3.

Question 5.
The length of the staircase is 5\(\frac{1}{2}\) m. If one step is set at \(\frac{1}{4}\) m. then, how many steps will be there in the staircase?
Answer:
Given length of staircase = 5\(\frac{1}{2}\) m = \(\frac{11}{2}\) m
Length of each step = \(\frac{1}{4}\) m
Length of each step × number of steps = length of staircase
\(\frac{1}{4}\) × Number of steps = \(\frac{11}{2}\)
Number of steps = \(\frac{11}{2}\) ÷ \(\frac{1}{4}\) = \(\frac{11}{2}\) × \(\frac{4}{1}\) = 11 × 2 = 22
∴ Number of steps in the staircase = 22

Question 6.
Simplify: 23.5 – 27 + 35.4 – 17
Answer:
Given 23.5 – 27 + 35.4 – 17
First make them like decimals.
= 23.5 – 27.0 + 35.4 – 17.0
Add the positive numbers
= (+23.5 + 35.4) + (-27.0 – 17.0)
= 58.9 + (-44.0)
= 58.9 – 44.0
= 14.9

Question 7.
Sailaja bought 3.350kg of potatoes, 2.250kg of tomatoes and some onions. If the weight of the total items are 10.250 kg. Then, find the weight of onions.
Answer:
Weight of potatoes = 3.350 kg
Weight of tomatoes = 2.250 kg
Weight of total items = 10.250 kg
Weight of onions = ?
Weight of (potatoes + tomatoes + onions) = 10.250 kg
(3.350 + 2.250) + onions weight = 10.250
5.600 + onions weight = 10.250
Onions weight = 10.250 – 5.600
∴ Weight of onions = 4.650 kg.

Question 8.
What should be subtracted from 7.1 to get 0.713?
Answer:
Given number = 7.1
Difference = 0.713
Number to be subtracted = Number – difference
= 7.1 – 0.713
= 7.100 – 0.713
∴ Number to be subtracted = 6.387.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 7 Introduction to Algebra Ex 7.3 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 7th Lesson Introduction to Algebra Ex 7.3

Question 1.
Identify which of the following are equations.
i) x – 3 = 7
ii) l + 5 > 9
iii) p – 4 < 10
iv) 5 + m = -6
v) 2s – 2 = 12
vi) 3x + 5
Answer:
i) x – 3 = 7
We know that, a mathematical statement involving equality symbol is called an equation.
x – 3 = 7 has equality symbol. So, it is an equation.

ii) l + 5 > 9
We know that, a mathematical statement involving equality symbol is called an equation.
l + 5 > 9 has no equality symbol.
So, it is not an equation. [It is an inequation]

iii) p – 4 < 10
We know that, a mathematical statement involving equality symbol is called an equation.
p – 4 < 10 has no equality symbol.
So, it is not an equation. [It is an inequation]

iv) 5 + m = – 6
We know that, a mathematical statement involving equality symbol is called an equation.
5 + m = – 6 has equality symbol.
So, it is an equation.

v) 2s – 2 = 12
We know that, a mathematical statement involving equality symbol is called an equation.
2s – 2 = 12 has equality symbol.
So, it is an equation.

vi) 3x + 5
It is only an expression. It’s not an equation.

Question 2.
Write LHS and RHS of the following equations.
i) x – 5 = 6
ii) 4y = 12
iii) 2z + 3 = 7
Answer:
i) x – 5 = 6
Given equation is x – 5 = 6
LHS = x – 5
RHS = 6

ii) 4y = 12
Given equation is 4y = 12
LHS = 4y
RHS = 12

iii) 2z + 3 = 7
Given equation is 2z + 3 = 7
LHS = 2z + 3
RHS = 7

Question 3.
Solve the following equation by Trial & Error Method.
i) x + 3 = 5
ii) y – 2 = 7
iii) a + 4 = 9
Answer:
i) x + 3 = 5
Given equation is x + 3 = 5
If x = 1, then the value of x + 3 = 1 + 3 = 4 ≠ 5
If x = 2, then the value of x + 3 = 2 + 3 = 5 = 5
From the above when x = 2, then both LHS and RHS are equal.
∴ Solution of the equation x + 3 = 5 is x = 2

ii) y – 2 = 7
Given equation is y – 2 = 7
If y = 1, then the value of y – 2 = 1 – 2 = -1 ≠ 7
If y = 2, then the value of y – 2 = 2 – 2 = 0 ≠ 7
If y = 3, then the value of y – 2 = 3 – 2 = 1 ≠ 7
If y = 4, then the value of y – 2 = 4 – 2 = 2 ≠ 7
If y = 5, then the value of y – 2 = 5 – 2 = 3 ≠ 7
If y = 6, then the value of y – 2 = 6 – 2 = 4 ≠ 7
If y = 7, then the value of y – 2 = 7 – 2 = 5 ≠ 7
If y = 8, then the value of y – 2 = 8 – 2 = 6 ≠ 7
If y = 9, then the value of y – 2 = 9 – 2 = 7 = 7
From the above when y = 9, the both LHS and RHS are equal.
Solution of the equation y – 2 = 7 is y = 9

ii) a + 4 = 9
Given equation is a + 4 = 9
If a = 1, then the value of a + 4 = 1 + 4 = 5 ≠ 9
If a = 2, then the value of a + 4 = 2 + 4 = 6 ≠ 9
If a = 3, then the value of a + 4 = 3 + 4 = 7 ≠ 9
If a = 4, then the value of a + 4 = 4 + 4 = 8 ≠ 9
If a = 5, then the value of a + 4 = 5 + 4 = 9 = 9
From the above when a = 5, the both LHS and RHS are equal.
∴ Solution of the equation a + 4 = 9 is a = 5

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 7 Introduction to Algebra Ex 7.2 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 7th Lesson Introduction to Algebra Ex 7.2

Question 1.
Write the expressions for the following statements.
(i) 5 is added to three times z
(ii) 9 times ‘n’ is added to ’10’
(iii) 16 is subtracted from two times ‘y’
(iv) ‘y’ is multiplied by 10 and then x is added to the product
Answer:
i) Given 5 is added to three times z
three times z = 3z 5 is added to three times z = 3z + 5

ii) Given 9 times n is added to 10
9 times n = 9.n
9 times ‘n’ is added to 10 = 9n + 10

iii) Given 16 is subtracted from two times y
two times y = 2.y
16 is subtracted from two times y = 2.y – 16

iv) Given y is multiplied by 10 and then x is added to the product, y is multiplied by 10 = 10.y
x is added to the product = 10.y + x

Question 2.
Peter has ‘p’ number of balls. Number of balls with David is 3 times the balls with Peter. Write this as an expression.
Answer:
Given, number of balls with Peter = p
Number of balls with David = 3 times with the Peter
= 3.p
Expression is 3p.

Question 3.
Sita has 3 more note books than Geetha. Find the number of books that Sita has. Use any letter for the number of books that Geetha has.
Answer:
Given Sita has 3 more note books than Geetha.
Let the number of note books Geetha has = x
Number of note books Sita has = 3 more note books than Geetha
= x + 3

Question 4.
Cadets are marching in a parade. There are 5 cadets in each row. What is the rule for the number of cadets, for a given number of rows? Use ‘n’ for the number of rows.
Answer:
Given 5 cadets in each row.
Number of rows = n
Number of cadets in each row = 5
Number of cadets in n rows = 5 × n = 5n

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 5 Fractions and Decimals Ex 5.5 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 5th Lesson Fractions and Decimals Ex 5.5

Question 1.
Add the following:
i) 5.702, 5.2, 6.04 and 2.30
ii) 40.004; 44.444; 40.404 and 4.444
Answer:
i) 5.702, 5.2, 6.04 and 2.30
Given decimals fractions are 5.702; 5.2; 6.04 and 2.30
Convert the given decimals into like decimals.
5.702; 5.200; 6.040; 2.300
Then write the decimals in column with the decimal points directly below each other.
So, that tenth’s come under tenths, hundredths come under hundredths and so on, then add.

ii) 40.004; 44.444; 40.404 and 4.444
Given decimals are 40.004; 44.444; 40.404 and 4.444
They are like decimal fractions.
So, write the decimals in column with the decimal points directly below each other.
So, that tenths come under tenths, hundredths come under hundredths and so on, then add.

Question 2.
Do the following:
i) 426.326 – 284.482
ii) 5 – 3.009
iii) 2.107 – 0.31
Answer:
i) 426.326 – 284.482
Given decimals are 426.326 and 284.482.
They are like fractions.
So, write the decimals in column with the decimal points directly below each other.
So, that tenths come under tenths, hundredths come under hundredths and so on, then subtract.

ii) 5 – 3.009
Given decimals are 5 and 3.009 Make them as like fractions.
5.000 – 3.009
So, write the decimals in column with the decimal points directly below each other.
So, that tenths come under tenths, hundredths come under hundredths and so on. Then subtract.

iii) 2.107 – 0.31
Given decimals are 2.107 and 0.31, make them as like fractions.
2.107 – 0.310
So, write the decimals in column with the decimal points directly below each other.
So, that tenths come under tenths, hundredths come under hundredths and so on. Then subtract.

Question 3.
Akshara bought 3 m 40 cm cloth for her shirt and 1 m 10 cm cloth for skirt. Find the total cloth bought by her.
Answer:
Length of cloth bought for shirt = 3.40
Length of cloth bought for skirt = 1.10
Total length of cloth bought by Akshara = 3.40 + 1.10 = 4.50 cm

Question 4.
Write in decimals using the units written in brackets.
i) 90 rupees 75 paisa (Rs.)
ii) 49 m 20 cm (m)
iii) 12 kg 450 g (kg)
iv) 50 l 500 ml (l)
Answer:
i) 90 rupees 75 paisa = Rs. 90.75
ii) 49 m 20 cm = 49.20 m
iii) 12 kg 450 g = 12.450 kg
iv) 50 l 500 ml = 50.500 l

Question 5.
Convert into decimals and add.
i) 58 kg 100 g; 60 kg 350 g
ii) 80 m 15 cm; 72 m 30 cm
Answer:
i) 58 kg 100 g; 60 kg 350 g
Given values are 58 kg 100 g and 60 kg 350 g
First convert the given values into like decimal fractions.
58.100 kg and 60.350 kg
Then, write the decimals in column with the decimal points directly below each other.
So, that tenths come under tenths, hundredths come under hundredths and so on.
Then add:

ii) 80 m 15 cm; 72 m 30 cm
Given values are 80 m 15 cm and 72 m 30 cm.
First convert the given values into decimal fractions.
80.15 m and 72.30 m
Then, write the decimals in column with the decimal points directly below each other.
So, that tenths come under tenths, hundredths come under hundredths and so on. Then add:

Question 6.
Convert into decimals and subtract.
i) 14 kg 720 g from 16 kg 744 g
ii) 1 l 12 ml from 2 l 20 ml
Answer:
i) 14 kg 720 g from 16 kg 744 g
Given values are 14 kg 720 g and 16 kg 744 g.
First convert the given values into decimal fractions.
14.720 kg and 16.744 kg
Then, write the decimals in column with the decimal points directly below each other. So, that tenths come under tenths, hundredths come under hundredths and so on.
Then subtract:

ii) 1 l 12 ml from 2 l 20 ml
Given values are 1 l 12 ml and 2 l 20 ml
First convert the given values into decimal fractions.
1.012 l and 2.020 l
Then write the decimals in column with the decimal points directly below each other. So, that tenths come under tenths, hundredths come under hundredths and so on. Then subtract:

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 7 Introduction to Algebra Ex 7.1 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 7th Lesson Introduction to Algebra Ex 7.1

Question 1.
Find the rule which gives the number of matchsticks required to make the following matchstick patterns.
(i) A pattern of letter ‘T’
(ii) A pattern of letter ‘E’
(iii) A pattern of letter ‘Z’
Answer:

Letter	Matchsticks required for
	1	2	3	4	5	n
i) T	2	2 × 2	3 × 2	4 × 2	5 × 2	2n
ii) E	4	2 × 4	3 × 4	4 × 4	5 × 4	4n
iii) Z	3	2 × 3	3 × 3	4 × 3	5 × 3	3n

Question 2.
Make a rule between the number of blades required and the number of fans (say n) in a hall.
Answer:
Number of blades for one fan = 3 = 1 × 3
Number of blades for two fans = 3 + 3 = 2 × 3
Number of blades for three fans = 3 + 3 + 3 = 3 × 3
Number of blades for four fans = 3 + 3 + 3 + 3 = 4 × 3
………………………………………
Number of blades for n fans = 3 + 3 + …. (n times) = 3 × n
Rule of number of blades required for n fans = 3.n

Question 3.
The cost of one pen is Rs. 7, then what is the rule for the cost of ‘n’ pens ?
Answer:
Cost of one pen = Rs. 7 = 1 × 7
Cost of two pens = Rs. 7 + Rs. 7 = 2 × 7
Cost of three pens = Rs. 7 + Rs. 7 + Rs. 7 = 3 × 7
……………………………………..
Cost of n pens = Rs. 7 + Rs. 7 + Rs. 7 ….. n times = 7 × n
Rule for cost n pens = 7.n.

Question 4.
The rule for purchase of books is that the cost of q books is Rs. 25q, then find the price of one book.
Answer:
Given, rule for cost of q books = Rs. 25q
In this rule q is variable i.e., q = 1, 2, 3,….
To get the cost of one book, put q = 1 in rule 25q
∴ Cost of one book = 25 (q) = 25(1) = Rs. 25

Question 5.
Harshini says that she has 5 biscuits more than Padma has. How can you express the relationship using the variable ‘y’?
Answer:
Given, Harshini has 5 biscuits more than Padma.
Let number of biscuits Padma has = y
Number of biscuits Harshini has = 5 more than Padma
∴ Number of biscuits Harshini has = y + 5

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

AP State Syllabus 6th Class Maths Solutions 6th Lesson Basic Arithmetic Unit Exercise

Question 1.
Draw a four sided closed figure and divide it into some number of equal parts. Shade the figure with any colour so that the ratio of shaded parts to unshaded parts is 1 : 3.
Answer:

Given ratio of shaded parts: unshaded parts = 1 : 3
Total parts of a closed figure = 1 + 3 = 4
Consider a four sided closed figure square.
Divide it into 4 parts and shade one part.
Now, from the figure, shaded parts : unshaded parts = 1 : 3

Question 2.
Ramu spent \(\frac{2}{5}\)th of his money on a story book. Find the ratio of money spent to the money with him at present.
Answer:
Given money spent on story book = \(\frac{2}{5}\) th of his money.
That means from total 5 parts of money he spent 2 parts of money on story book. Then, out of 5 parts remaining parts of money with him is 3 parts.
So, ratio of money spent to the money with him at present
= parts of money spent : parts of money at him = 2 : 3

Question 3.
Divide Rs. 72,000 between Kesav and David in the ratio of 5 : 4.
Answer:
Amount to be divided = Rs. 72,000
Ratio of Kesav and David’s Shares = 5 : 4
Total shares = 5 + 4 = 9
Kesav’s share = \(\frac{5}{9}\) of 72000
\(\frac{5}{9}\) × 72,000 = 40,000/-
David’s share = \(\frac{4}{9}\) of 72,000
\(\frac{4}{9}\) × 72,000 = 32,000/-

Question 4.
The income of Kumar for 3 months is Rs. 15,000. If he earns the same amount for a month,
a) How much will he earn in 5 months?
b) In how many months, will he earn Rs. 95,000?
Answer:
Given Kumar’s 3 months income = Rs. 15,000
Kumar’s one month income = 15,000 ÷ 3
= \(\frac{15000}{3}\) = Rs. 5000
a) Kumar’s one month income = Rs. 5000
Kumar’s 5 months income = 5 × 5000 = Rs. 25,000
b) Time taken by Kumar to earn Rs. 5000 = 1 month
Time taken by Kumar to earn Rs. 95,000 = 95,000 ÷ 5000
= \(\frac{15000}{3}\) = 19 months = 1 year 7 months.

Question 5.
The cost of 16 chairs is Rs. 4,800. Find the number of chairs that can be purchased for Rs. 6,600.
Answer:
Cost of 16 chairs = Rs. 4800
Number of chairs that can be pruchased for Rs. 6600 = x (say)
∴ 16 : 4800 : : x : 6600
So, 16 x 6600 = 4800 x x
x = \(\frac{16 \times 6600}{4800}\) = 22 chairs
(OR)
Given cost of 16 chairs = Rs. 4800
Cost of one chair = 4800 ÷ 16
= \(\frac{4800}{3}\) = Rs. 300
Number of chairs can purchased for Rs. 6600 = 6600 ÷ cost of one chair
= 6600 ÷ 300
= \(\frac{6600}{300}\) = 22 chairs

Question 6.
What percentage of numbers from 1 to 30 has 1 or 9 in the unit’s digit?
Answer:
Total number from 1 to 30 = 30
Numbers ending with 1 or 9 are 1, 9,11, 19, 21, 29 = 6 numbers
Ratio = 6 : 30 = \(\frac{6}{30}\) = \(\frac{1}{5}\)
Percentage = \(\frac{1}{5}\) × 100% = 20%

Question 7.
In a college, 63% of students are less than 20 years of age. The number of students more than 20 years of age is — of number of students of 20 years of age which is 42. What is the total number of students in the college?
Answer:
Let the total number of students be x.
Then, number of students more than 20 years of age = (100 – 63)% of x
= 37% of x
Then, 37% of x = 42 + \(\frac{2}{3}\) of 48
\(\frac{37}{100}\) × x = 74
∴ x = 74 × \(\frac{100}{37}\) = 200

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 5 Fractions and Decimals Ex 5.4 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 5th Lesson Fractions and Decimals Ex 5.4

Question 1.
Which of the following are unlike decimal fractions ?
i) 5.03, 6.185
ii) 42.7, 7.42
iii) 16.003, 5.301
iv)15.81, 1.36
Answer:
i) 5.03, 6.185
5.3 has 2 decimal places
6.185 has 3 decimal places
Their decimal places are not equal.
So, these decimal fraction are unlike decimal fractions.

ii) 42.7, 7.42
42.7 has one decimal place
7.42 has two decimal places
Their decimal places are not equal.
So, these are unlike decimal fractions.

iii) 16.003, 5.301
16.3 has three decimal places
5.301 has three decimal places
Their decimal places are equal.
So, these are like decimal fractions.

iv) 15.81, 1.36
15.81 has two decimal places
1.36 has two decimal places
Their decimal places are equal.
So, these are like decimal fractions.

Question 2.
Change the following into like decimal fractions.
i) 0.802, 54.32, 873.274
ii) 4.78, 9.193, 11.3
iii) 16.003, 16.9, 16.19
Answer:
i) 0.802, 54.32, 873.274
Given decimal fractions are 0.802, 54.32, 873.274
The greatest number of decimal places is 3.
So, we convert all of them to equivalent decimals with 3 decimal places by placing sufficient zeroes.
0.802 = 0.802
54.32 = 54.320
873.274 = 873.274
Thus, 0.802, 54.32, 873.274 when converted to like decimals becomes 0.802, 54.320, 873.274.

ii) 4.78, 9.193, 11.3
Given decimal fractions are 4.78, 9.193, 11.3.
So, we convert all of them to equivalent decimals with 3 decimal places.
4.78 = 4.780
9.193 = 9.193
11.3 = 11.300
Thus, 4.78, 9.193, 11.3 when converted to like decimals becomes 4.780, 9.193, 11.300.

iii) 16.003, 16.9, 16.19
Given decimal fractions are 16.003, 16.9, 16.19
So, we convert all of them to equivalent decimals with three decimal places by providing sufficient zeroes.
16.3 = 16.003
16.9 = 16.900
16.19 = 16.190
Thus, 16.003, 16.9, 16.19 when converted to like decimals becomes 16.003, 16.900, 16.190.

Question 3.
Write the following in ascending order.
a) 7.26, 7.62, 7.2
b) 0.464, 0.644, 0.446, 0.664
c) 30.000, 30.060, 30.30
Answer:
a) 7.26, 7.62, 7.2
Given decimal fractions are 7.26, 7.62, 7.20
First compare the whole numbers.
Here whole numbers are equal. So, compare tenth’s place digits of the decimal fractions.
In 7.26 and 7.20 Tenths places are same.
But 7.62 Tenth place is more than the remaining numbers.
So,7.62 > 7.26 and 7.62 > 7.20
To compare 7.26 and 7.20 we should compare hundredth’s places. Then 7.26 > 7.20
∴ 7.62 > 7.26 > 7.20
(or) 7.20 < 7.26 < 7.62
Ascending order: 7.20; 7.26; 7.62

b) 0.464, 0.644, 0.446, 0.664
Given decimal fractions are 0.464, 0.644, 0.446, 0.664
Then, compare whole numbers.
Here whole numbers are same. Then, compare tenth’s place digits of the decimal fractions.
0.464, 0.446 are less than 0.644 and 0.664.
First make or check the given fraction are all like fractions or not?
Then, compare hundredth’s place digits
∴ 0.464 > 0.446, 0.664 > 0.644
and 0.664 > 0.644 > 0.464 > 0.446
(or) 0.446 < 0.464 < 0.644 < 0.664
Ascending order: 0.446; 0.464; 0.644; 0.664

c) 30.000, 30.060, 30.30
Given decimal fractions are 30.000, 30.060, 30.30
First make or check the given fractions are all like fractions or not?
So, 30.000; 30.060; 30.300 are like fractions.
Then, compare the whole numbers. They are same. So, compare, Tenth’s places of the decimal fractions.
30.300 > 30.060 > 30.000
(or) 30.000 < 30.060 < 30.300
Ascending order: 30.000; 30.060; 30.300

Question 4.
Arrange these numbers in descending order:
16.96; 16.42; 16.3; 16.03; 16.1; 16.99; 16.01
Answer:
Given decimal fractions are
16.96; 16.42; 16.3; 16.03; 16.1; 16.99; 16.01
First make or check the given fractions are all like fractions or not?
So, 16.96; 16.42; 16.30; 16.03; 16.10; 16.99; 16.01 are like fractions.
Then, compare whole numbers. They are same.
So, compare Tenth’s place digits of the decimal fractions.
16.96 < 16.99; 16.42 > 16.30 > 16.10 > 16.03 > 16.01
Then compare hundredth’s place digits.
So, 16.99 > 16.96 > 16.42 > 16.30 > 16.10 > 16.03 > 16.01 (or)
16.01 < 16.03 < 16.10 < 16.30 < 16.42 < 16.96 < 16.99
Descending order: 16.99; 16.96; 16.42; 16.30; 16.10; 16.03; 16.01.

Question 5.
Fill in the blanks by using appropriate symbols >, < or =.
i) 0.005 …… 0.0005
ii) 4.353 …… 4.2
iii) 58.3 …… 58.30
Answer:
i) 0.005 …… 0.0005
First make like decimal fractions
0.0050 …… 0.0005
By comparing Thousandth’s place digits.
So, 0.0050 > 0.0005

ii) 4.353 …… 4.2
First make like decimal fractions.
4.353 …… 4.200
Compare Tenth’s place digits, if whole numbers are same.
So, 4.353 > 4.200

iii) 58.3 …… 58.30
First make like decimal fractions.
58.30 …… 58.30
In these whole numbers and decimal place digits are same.
So, 58.30 = 58.30

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 5 Fractions and Decimals Ex 5.3 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 5th Lesson Fractions and Decimals Ex 5.3

Question 1.
Find the reciprocal of each of the following fractions.
i) \(\frac{5}{9}\)
ii) \(\frac{12}{7}\)
iii) 2\(\frac{1}{5}\)
iv) \(\frac{1}{8}\)
v) \(\frac{13}{11}\)
vi) \(\frac{8}{3}\)
Answer:

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

Question 2.
Simplify:
i) 15 ÷ \(\frac{3}{4}\)
ii) 6 ÷ 1\(\frac{4}{7}\)
iii) 3 ÷ 2\(\frac{1}{3}\)
iv) \(\frac{4}{9}\) ÷ 15
v) 4\(\frac{3}{7}\) ÷ 14
Answer:
i) 15 ÷ \(\frac{3}{4}\)
While dividing a whole number by a fraction, multiply the whole number with the reciprocal of the fraction.
Thus,

ii) 6 ÷ 1\(\frac{4}{7}\)
While dividing a whole number by a mixed fraction, first convert the mixed fraction into improper fraction and then multiply the whole number with the recipro¬cal of the improper fraction.
Thus,

iii) 3 ÷ 2\(\frac{1}{3}\)
While dividing a whole number by a mixed fraction, first convert the mixed fraction into improper fraction and then multiply the whole number with the reciprocal of the improper fraction.
Thus,

iv) \(\frac{4}{9}\) ÷ 15
While dividing a proper fraction by a whole number, multiply the proper fraction with the reciprocal of a whole number.
Thus,

v) 4\(\frac{3}{7}\) ÷ 14
While dividing mixed fraction by a whole number first convert the mixed fraction into improper fraction and then multiply the improper fraction with the reciprocal of a whole number.
Thus,

Question 3.
Find:
i) \(\frac{4}{9}\) ÷ \(\frac{2}{3}\)
ii) \(\frac{4}{11}\) ÷ \(\frac{8}{11}\)
iii) 2\(\frac{1}{3}\) ÷ \(\frac{3}{5}\)
iv) 5\(\frac{4}{7}\) ÷ 1\(\frac{3}{10}\)
Answer:
i) \(\frac{4}{9}\) ÷ \(\frac{2}{3}\)
While dividing a proper fraction by a proper fraction, multiply the proper fraction with the reciprocal of the fraction.
Thus,

ii) \(\frac{4}{11}\) ÷ \(\frac{8}{11}\)
While dividing a proper fraction by a proper fraction, multiply the proper fraction with the reciprocal of the fraction.
Thus,

iii) 2\(\frac{1}{3}\) ÷ \(\frac{3}{5}\)
While dividing a mixed fraction by a proper fraction, first convert the mixed fraction into improper fraction, then multiply it with the reciprocal of the proper fraction.
Thus,

iv) 5\(\frac{4}{7}\) ÷ 1\(\frac{3}{10}\)
While dividing a mixed fraction by a mixed fraction first convert the two mixed fraction into improper fraction, then multiply the improper fraction with the reciprocal of the second improper fraction.
Thus,

Question 4.
The product of two numbers is 25\(\frac{5}{6}\). If one of the number is 6\(\frac{2}{3}\). Find the other.
Answer:

Question 5.
By what number should 9\(\frac{3}{4}\) be multiplied to get 5\(\frac{2}{3}\)?
Answer:
Given one number = 9\(\frac{3}{4}\) = \(\frac{39}{4}\)
Product = 5\(\frac{2}{3}\) = \(\frac{17}{3}\)
\(\frac{39}{4}\) × Required Number = \(\frac{17}{3}\)
Required Number = \(\frac{17}{3}\) ÷ \(\frac{39}{4}\)
= \(\frac{17}{3}\) × \(\frac{4}{39}\)
= \(\frac{17 \times 4}{3 \times 39}\)
= \(\frac{68}{117}\)
Number to be multiplied = \(\frac{68}{117}\)

Question 6.
A bucket contains 34\(\frac{1}{2}\) litres of water. How many times do you get 1\(\frac{1}{2}\) litres of water?
Answer:
Capacity of Bucket = 34\(\frac{1}{2}\) litres = \(\frac{69}{2}\) litres
Capacity required = 1\(\frac{1}{2}\) = \(\frac{2 \times 1+1}{2}\) =\(\frac{3}{2}\)
Number of times we get = \(\frac{69}{2}\) ÷ \(\frac{3}{2}\) = \(\frac{69}{2}\) × \(\frac{2}{3}\) = 23 times.

Question 7.
The cost of 3\(\frac{3}{4}\) kg. of sugar is Rs. 121\(\frac{1}{2}\). Find its cost per 1 kg.
Answer:
Cost of 3\(\frac{3}{4}\) kg of sugar = Rs. 121\(\frac{1}{2}\)
Cost of 1kg of sugar

Cost of 1kg of sugar = Rs. 32.40.

Question 8.
The length of a rectangular field is 12\(\frac{1}{4}\) m. and its area is 65\(\frac{1}{3}\)m². Find its breadth.
Answer:
Length of a rectangular field = 12\(\frac{1}{4}\) = \(\frac{49}{4}\) m
Breadth of a rectangular field = ?
Area of a rectangular field = 65\(\frac{1}{3}\) = \(\frac{196}{3}\) sq.m
length × breadth = Area of a rectangular

∴ Breadth of a rectangular field = 5\(\frac{1}{3}\) m.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 5 Fractions and Decimals Ex 5.1 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 5th Lesson Fractions and Decimals Ex 5.1

Question 1.
Classify the fractions as proper, improper and mixed.
\(\frac{3}{4}\), \(\frac{6}{5}\), \(\frac{3}{2}\),\(\frac{4}{1}\), \(\frac{2}{3}\), \(\frac{1}{4}\), \(\frac{18}{13}\), 1\(\frac{5}{7}\),\(\frac{1}{3}\), 11\(\frac{1}{2}\)
Answer:
If in a fraction numerator is less than the denominator, then it is called a proper fraction.
Proper fractions are \(\frac{3}{4}\), \(\frac{2}{3}\), \(\frac{1}{4}\), \(\frac{1}{3}\)
If in a fraction numerator is greater than the denominator, then it is called an improper fraction.
Improper fractions are \(\frac{6}{5}\), \(\frac{3}{2}\), \(\frac{4}{1}\), \(\frac{18}{13}\)
A combination of a whole number and a proper fraction is called a mixed fraction.
Mixed fractions are 1\(\frac{5}{7}\), 11\(\frac{1}{2}\)

Question 2.
Write the following fractions in an ascending order.
i) \(\frac{3}{4}\), \(\frac{3}{2}\), \(\frac{2}{3}\), \(\frac{1}{5}\), \(\frac{18}{7}\)
ii) \(\frac{2}{7}\), \(\frac{3}{8}\), \(\frac{3}{4}\), \(\frac{5}{7}\), \(\frac{4}{9}\)
Answer:
Given fractions are \(\frac{3}{4}\), \(\frac{3}{2}\), \(\frac{2}{3}\), \(\frac{1}{5}\), \(\frac{18}{7}\)
These are unlike fractions.
To arrange the unlike fractions in ascending order / descending order first we have to convert them into equivalent fractions with LCM of their denominators, then compare the like fractions (i.e.,) we convert them to like fractions.
LCM of Denominators = 2 × 2 × 3 × 5 × 7 = 420

\(\frac{315}{420}\), \(\frac{630}{420}\), \(\frac{280}{420}\), \(\frac{84}{420}\), \(\frac{1080}{420}\)
These are like fractions. Now, we can compare them.
\(\frac{84}{420}\)< \(\frac{280}{420}\) < \(\frac{315}{420}\) < \(\frac{630}{420}\) < \(\frac{1080}{420}\)
i.e, \(\frac{1}{5}\) < \(\frac{2}{3}\) < \(\frac{3}{4}\) < \(\frac{3}{2}\) < \(\frac{18}{7}\)
∴ Ascending order: \(\frac{1}{5}\), \(\frac{2}{3}\), \(\frac{3}{4}\), \(\frac{3}{2}\), \(\frac{18}{7}\)
Descending order: \(\frac{18}{7}\), \(\frac{3}{2}\), \(\frac{3}{4}\), \(\frac{2}{3}\), \(\frac{1}{5}\)

ii) \(\frac{2}{7}\), \(\frac{3}{8}\), \(\frac{3}{4}\), \(\frac{5}{7}\), \(\frac{4}{9}\)
These are unlike fractions.
To arrange the unlike fractions in ascending order / descending order first we have to convert them into equivalent fractions with LCM of their denominators. Then compare the like fractions (i.e.,) we convert them to like fractions.
LCM of denominators = 2 × 2 × 2 × 3 × 3 × 7 = 504

\(\frac{144}{504}\), \(\frac{189}{504}\), \(\frac{378}{504}\), \(\frac{360}{504}\), \(\frac{224}{504}\)
These are like fractions. Now, we can compare them.
\(\frac{144}{504}\) < \(\frac{189}{504}\) < \(\frac{224}{504}\) < \(\frac{360}{504}\) < \(\frac{1080}{420}\)
i.e, \(\frac{2}{7}\) < \(\frac{3}{8}\) < \(\frac{4}{9}\) < \(\frac{5}{7}\) < \(\frac{3}{4}\)
∴ Ascending order: \(\frac{2}{7}\), \(\frac{3}{8}\), \(\frac{4}{9}\), \(\frac{5}{7}\), \(\frac{3}{4}\)
Descending order: \(\frac{3}{4}\), \(\frac{5}{7}\), \(\frac{4}{9}\), \(\frac{3}{8}\), \(\frac{2}{7}\)

Question 3.
Without doing calculation, find the result \(\frac{2}{3}\) + 1\(\frac{3}{4}\) + \(\frac{1}{3}\) – \(\frac{1}{4}\)
Answer:
Given \(\frac{2}{3}\) + 1\(\frac{3}{4}\) + \(\frac{1}{3}\) – \(\frac{1}{4}\)

Question 4.
Neha bought a cake. She ate \(\frac{7}{15}\) th of the cake immediately add in the afternoon she ate the remaining part. How much part die ate in the afternoon ?
Answer:
Whole cake = 1 = \(\frac{15}{15}\)
Neha divided the cake into 15 parts.
Part of a cake eaten by Neha = \(\frac{7}{15}\)
Remaining part of cake = Whole – eaten part immediately
= \(\frac{1}{1}\) – \(\frac{7}{15}\)
= \(\frac{15}{15}\) – \(\frac{7}{15}\)
= \(\frac{15-7}{15}\)
= \(\frac{8}{15}\)
∴ Part of a cake eaten by Neha in the afternoon = \(\frac{8}{15}\)

Question 5.
Simplify:
i) \(\frac{2}{5}\) + \(\frac{1}{3}\)
ii) \(\frac{5}{7}\) + \(\frac{2}{3}\)
iii) \(\frac{3}{5}\) – \(\frac{7}{20}\)
iv) \(\frac{17}{20}\) – \(\frac{13}{25}\)
Answer:
i) \(\frac{2}{5}\) + \(\frac{1}{3}\)
LCM of denominators = 3 × 5 = 15

ii) \(\frac{5}{7}\) + \(\frac{2}{3}\)
LCM of denominators = 7 × 3 = 21

iii) \(\frac{3}{5}\) – \(\frac{7}{20}\)
LCM of denominators = 2 × 2 × 5 = 20

iv) \(\frac{17}{20}\) – \(\frac{13}{25}\)
LCM of denominators = 2 × 2 × 5 × 5 = 100

Question 6.
Represent \(\frac{16}{5}\) pictorially
Answer:
Given fraction is \(\frac{16}{5}\) (Improper fraction)
\(\frac{16}{5}\) = Mixed fraction is 3\(\frac{1}{5}\)

3 + \(\frac{1}{5}\) = 3\(\frac{1}{5}\)

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 5 Fractions and Decimals Ex 5.2 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 5th Lesson Fractions and Decimals Ex 5.2

Question 1.
Find the product of the following.
i) 3 × \(\frac{5}{12}\)
ii) \(\frac{15}{8}\) × 12
iii) 1\(\frac{3}{4}\) × \(\frac{12}{21}\)
iv) \(\frac{4}{5}\) × \(\frac{12}{7}\)
Answer:

Question 2.
Which is greater?
i) \(\frac{1}{2}\) of \(\frac{6}{7}\) or \(\frac{2}{3}\) of \(\frac{3}{7}\)
ii) \(\frac{2}{7}\) of \(\frac{3}{4}\) or \(\frac{3}{5}\) of \(\frac{5}{8}\)
Answer:
i) \(\frac{1}{2}\) of \(\frac{6}{7}\) or \(\frac{2}{3}\) of \(\frac{3}{7}\)

\(\frac{3}{7}\) or \(\frac{2}{7}\)
So, \(\frac{3}{7}\) > \(\frac{2}{7}\)
∴ \(\frac{1}{2}\) of \(\frac{6}{7}\) > \(\frac{2}{3}\) of \(\frac{3}{7}\)

ii) \(\frac{2}{7}\) of \(\frac{3}{4}\) or \(\frac{3}{5}\) of \(\frac{5}{8}\)

Convert them into like fractions
LCM of denominators = 2 × 2 × 2 × 7 = 56
\(\frac{3 \times 4}{14 \times 4}\) or \(\frac{3 \times 7}{8 \times 7}\)
\(\frac{12}{56}\) or \(\frac{21}{56}\)
So, \(\frac{12}{56}\) < \(\frac{21}{56}\)
∴ \(\frac{2}{7}\) of \(\frac{3}{4}\) < \(\frac{3}{5}\) of \(\frac{5}{8}\)

Question 3.
Find:
i) \(\frac{7}{11}\) of 330
ii) \(\frac{5}{9}\) of 108
iii) \(\frac{2}{7}\) of 16
iv) \(\frac{1}{7}\) of \(\frac{3}{10}\)
Answer:
i) \(\frac{7}{11}\) of 330
= \(\frac{7}{11}\) × 330
= \(\frac{7}{11}\) × 11 × 30
= 7 × 30 = 210

ii) \(\frac{5}{9}\) of 108
= \(\frac{5}{9}\) × 108
= \(\frac{5}{9}\) × 9 × 12
= 5 × 12 = 60

iii) \(\frac{2}{7}\) of 16
= \(\frac{2}{7}\) × 16
= \(\frac{2×16}{7}\)
= \(\frac{32}{7}\) or 4\(\frac{4}{7}\)

iv) \(\frac{1}{7}\) of \(\frac{3}{10}\)
= \(\frac{1}{7}\) × \(\frac{3}{10}\)
= \(\frac{1 \times 3}{7 \times 10}\)
= \(\frac{3}{70}\)

Question 4.
If the cost of a notebook is Rs. 10\(\frac{3}{4}\). Then, find the cost of 36 books.
Answer:
Cost of each notebook = Rs. 10\(\frac{3}{4}\) (or) \(\frac{43}{4}\)
Cost of 36 notebooks = 36 × 10\(\frac{3}{4}\) = 36 × \(\frac{43}{4}\) = \(\frac{9 \times 4 \times 43}{4}\) = 9 × 43
Cost of 36 notebooks = Rs. 387

Question 5.
A motor bike runs 52\(\frac{1}{2}\) km using 1 litre of petrol. How much distance will it cover for 2\(\frac{3}{4}\) litres of petrol ?
Answer:
Distance covered by the motor bike 1 litre petrol = 52\(\frac{1}{2}\) km (or) \(\frac{105}{2}\) km
Distance covered by the motor bike for 2\(\frac{3}{4}\) litres of petrol = 2\(\frac{3}{4}\) of 52\(\frac{1}{2}\)
= \(\frac{11}{4}\) × \(\frac{105}{2}\)
= \(\frac{11 \times 105}{4 \times 2}\)
∴ Distance covered by the motor bike for 2\(\frac{3}{4}\) litres of petrol = \(\frac{1155}{8}\) = 144\(\frac{3}{8}\) km.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 3 HCF and LCM Ex 3.4 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 3rd Lesson HCF and LCM Ex 3.4

Question 1.
Prepare a factor tree for 90.
Answer:

Prime factorization of 90 is 2 × 3 × 3 × 5 = 90

Question 2.
Factorise 84 by division method.
Answer:

Prime factorization of 84 is 2 × 2 × 3 × 7 = 84

Question 3.
Write the greatest 4 digit number and express it in the form of its prime factors.
Answer:
The greatest 4 – digit number is 9999.

Prime factorization of 9999 is 3 × 3 × 11 × 101 = 9999

Question 4.
Write the prime factorization of 96 by factor tree method.
Answer:

Prime factorization of 96 is 2 × 2 × 3 × 2 × 2 × 2 = 96

Question 5.
I am the smallest number, having four different prime factors. Can you find me?
Answer:
The first four primes are 2, 3, 5, 7.
Their product = 2 × 3 × 5 × 7 = 210
So, the smallest number, having four different prime factors is 210.

Question 6.
Write the prime factorization of 28 and 36 through division method. Write the prime factorization of 42 by factor tree method.
Answer:

Prime factorization of 28 is 2 × 2 × 7 = 28

Prime factorization of 36 is 2 × 2 × 3 × 3 = 36

Prime factorization of 42 is 2 × 3 × 7 = 42