Category: Class 6

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 4 Integers Unit Exercise Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 4th Lesson Integers Unit Exercise

Question 1.
Write the integers for the following situations.
i) A kite is flying at a height of 225 m in the sky. ( )
ii) A whale is at a depth of 1250 m in the ocean. ( )
iii) The temperature in Sahara desert is 12°C below freezing temperature. ( )
iv) Ravi withdrawn Rs. 3800 from ATM using his debit card. ( )
Answer:
i) 225 m
ii) – 1250 m
iii) – 12°C
iv) -3800

Question 2.
Justify the following statements with an example.
i) A positive integer is always greater than a negative integer.
ii) All positive integers are natural numbers.
iii) Zero is greater than a negative integer.
iv) There exist infinite integers in the number system.
v) All whole numbers are integers.
Answer:
i) A positive integer is always greater than a negative integer:
Consider
a) 3, -5 are two integers
3 > -5
b) – 10, 1 are two integers
1 > -10
∴ A positive integer is always greater than a negative integer.

ii) All positive integers are natural numbers:
Positive integers = 1, 2, 3, …..
Natural numbers = 1, 2, 3, …..
So, all positive integers are natural numbers.

iii) Zero is greater than a negative integer:
On a number line, for a given pair of numbers the number on R.H.S is always greater than the number on L.H.S.
All negative integers lie on the L.H.S. of zero, on a number line.
As such all negative numbers are less than zero or zero is greater than all negative integers.

iv) There exist infinite integers in the number system:
If we write integer on a number line, as the line extends on both sides endlessly so as the integers.
(Or)
For every integer there exists another integer which is 1 more than the given integer. Hence the integers are infinite.

v) All whole numbers are integers:
Whole numbers : 0, 1, 2, 3,……
Integers: ……, -3, -2, -1, 0, 1, 2, 3,…..
So, whole number are part of integers.
∴ All whole numbers are integers.

Question 3.
Represent
i) 3+4
ii) 8+(-3)
iii) -7-2
iv) 6-(5)
v)-5-(-4)
on number line.
Answer:
i) 3 + 4 = +7.

On the number line we first move 3 steps to the right of 0 to reach +3.
Then, we move 4 steps to the right of +3 and reach + 7.

ii) 8 + (-3) = +5.

On the number line we first move 8 steps to the right of 0 to reach +8.
Then, we move 3 steps to the left of +8 and reach +5.

iii) -7 -2 = -9

On the number line we first move 7 steps to the left of 0 to reach -7.
Then we move 2 steps to the left of -7 and reach -9.

iv) 6 – (5) = + 1

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

v) -5 – (-4) = -1

-5 – (-4) = -5 + 4 = -1 (∵ -(-a) = a)
On the number line, we first move 5 steps to the left of 0 to reach -5.
Then, we move 4 steps to the right of -5 and reach -1.

Question 4.
Write all the integers lying between the given numbers.
i) 7 and 12
ii) -5 and -1
iii) -3 and 3
iv) -6 and 0
Answer:
i) 7 and 12
Integers lying between 7 and 12 are 8, 9, 10, ll.

ii) -5 and -1
Integers lying between -5 and -1 are -2, -3 and -4.

iii) -3 and 3
Integers lying between -3 and 3 are -2, -1, 0, 1, 2.

iv) -6 and 0
Integers lying between -6 and 0 are -5, -4, -3, -2, -1.

Question 5.
Arrange the following integers in ascending order and descending order.
-1000, 10, -1, -100, 0, 1000, 1, -10
Answer:
Given numbers -1000, 10, -1, -100, 0, 1000, 1, -10
Ascending order: – 1000, -100, -10, -1, 0, 1, 10, 1000
Descending order: 1000, 10, 1, 0, -1, -10, -100, -1000

Question 6.
Write a real life situation for each of the following integers.
i) -200 m
ii) +42°C
iii) -4800(cr)
iv) -3.0 kg
Answer:
i) -200 m
In the Singareni coal mines workers will go 200 m below the ground level.
ii) +42°C
In summer average temperature of May month is 42°C.
iii) Rs. 4800 (crores)
Central Government sanctioned Rs. 4800(crores) for education in the Annual Budget.
iv) – 3.0 kg
In a Primary School the ground balance of rice = -3 kg

Question 7.
Find:
i) (-603) + (603)
ii) (-5281) +(1825)
iii) (-32) + (-2) + (-20) + (-6)
Answer:
i) (-603) + (603)
Sum of a number and its additive number is 0.
Additive inverse of – 603 is 603.
So, -603 + 603 = 0

ii) (-5281) + (1825)
= – 5281 + 1825 = – 3456

iii) (-32) + (-2) + (-20) + (-6)
= -32 – 2 – 20 – 6 = -60

Question 8.
Find:
i)(-2) – (+1)
ii) (-270) – (-270)
iii) (1000) – (-1000)
Answer:
i) -2 – (+1)
= -2 -1 = -3

ii) – 270 – (-270)
= -270 + 270 [∵ -(-a) = a]
= 0 [-a + a = 0]

iii) 1000 – (-1000)
= 1000 + 1000 [∵ -(-a) = a]
= 2000

Question 9.
In a quiz competition, where negative score for wrong answer is taken,Team A scored +10, -10, 0, -10, 10, -10 and Team B scored 10,10, -10,0,0,10 in 6 rounds successively .Which team wins the competition? How?
Answer:
Score of Team A = (+10), (-10), 0, (-10), 10, (-10)
Total score of Team A = (+10) + (-10) + (0) + (-10) + (10) + (-10)
= +10 – 10 + 0 – 10 + 10 – 10
= +20 – 30 = -10
Score of Team B = 10, 10, -10, 0, 0, 10
Total Score of Team B = (+10) + (+10) + (-10) + 0 + 0 + (+10)
= +10 + 10 – 10 + 10
= + 30 – 10
= + 20
-10 < + 20
Team A < Team B
So, Team B is the winner. Because, Team B got more score than Team A.

Question 10.
An apartment has 10 floors and two cellars for car parking under the basement. A lift is now, at the ground floor. Ravi goes 5 floors up and then 3 floors up, 2 floors down and then 6 floors down and come to lower cellar for taking his car. Count how many floors does Ravi travel all together? Represent the result on a vertical number line.
Answer:

First move:
Ground to 5th floor up = 5
(Can reach 5th floor)
Second move:
5th floor to 3 floors up = 3
(Can reach 8)
Third move:
8th floor to 2 floors down = 2
(Can reach 6th floor)
Fourth move:
6th floor to 6 floors down = 6
(Can reach ground floor)
Final move:
Ground to lower cellar = 2
(Can reach cellar -2)
No. of floors Ravi travelled = 18

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 4 Integers Ex 4.4 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 4th Lesson Integers Ex 4.4

Question 1.
Find:
i) 40 – (22)
ii) 84 – (98)
iii) (-16) + (-17)
iv) (-20) – (13)
v) (38) – (-6)
vi) (-17) – (-36)
Answer:
i) 40 – (22) = +40 – 22
= +(18 + 22)-22
= 18 +(22-22)
= 18 + 0
∴ 40-(22) = + 18

ii) 84 – (98) = 84 – 98
= + 84 – 84 – 14
= (84 – 84) – 14
= 0 – 14
∴ 84 – (98) = – 14

iii) (-16) + (-17) = -16 – 17 = – 33
∴ (-16) + (-17) = -33

iv) (-20) – (13) = – 20 – 13
= -33
∴ (-20) – (13) = – 33

v) 38 – (-6) = 38 + 6
We know – (-a) = a = + 44
∴ 38 – (-6) = + 44

vi) (-17) – (-36) = -17 + 36
We know – (- a) = a
= -17 + 17 + 19
= (-17+17) + 19
= + 19
∴ (-17) – (-36) = + 19

Question 2.
Fill in the boxes with >, < or = sign:
(i) (-4) + (-5) (-5) – (-4)
(ii) (-16) – (-23) (-6) + (-12)
(iii) 44 – (-10) 47 +(-3)
Answer:
(i) (-4) + (-5) (-5)- (-4)
-4-5  -5 + 4
-9 -1

(ii) (-16) – (-23) (-6) + (-12)
-16 + 23 -6-12  [∵ -(-a) = a]
-16 + 16 + 7 -18
+7 -18

(iii) 44 – (-10) 47 + (-3)
44+10 47-3.
54 44 + 3 – 3
54 44

Question 3.
Fill in the blanks:
i) (-13) + ———– = 0
ii) (-16) + 16 = ———–
iii) (-5) + ———– = -14
iv) ———– + (2 – 16) = – 22
Answer:
i) (-13) + ———– = 0
We know, additive inverse of a is -a (or) – a is a.
Additive inverse of -13 is 13. So, -13 + 13 = 0

ii) (-16) + 16 = ———–
We know, sum of a number and its additive inverse is 0.
i.e., (-a) + a = 0
So, (-16) + (16) = 0

iii) (-5) + ———– = -14
-5 + (-9) = -14

iv) ———– + (2 – 16) = – 22
———– + (-14) = -22
– 8 + (-14) = – 22

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 4 Integers Ex 4.3 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 4th Lesson Integers Ex 4.3

Question 1.
Add the following integers using number line.
i) 7 + (-6)
ii) (-8) + (-2)
iii) (-6) + (-5) + (+2)
Answer:
i) 7 + (-6)

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

ii) (-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 left of -8 and reach -10.
So,(-8) + (-2) = -10

iii) (-6) + (-5) + (+2)

On the number line, we first move 6 steps to the left of 0 to reach -6.
We move 5 steps to the left of -6 and reach -11.
Then, we move 2 steps to the right of -11 and reach -9.
So, (-6) + (-5) + (+2) = -9

Question 2.
Add without using number line.
(i) 10 + (-3)
(ii) 10 + (+16)
(iii) (-8) + (+8)
Answer:
i) 10 + (-3)
10 + (-3) = 7 + 3 + (-3) = 7 + (3 + (-3))
= 7 + 3 – 3 = 7 + 0
∴ 10 + (-3) = +7

ii) -10 + (+16)
-10 + (+16)
= -10 + 10 + 6
= (-10 + 10) + 6
= 0 + 6
∴ -10 + (+16) = + 6

iii) (-8) + (+ 8)
-8 + (+8) = -8 + 8 = 0
∴ -8 + (+8) = 0

Question 3.
Find the sum of i) 120 and -274 ii) -68 and 28
Answer:
i) 120 and-274
Method -1: Sum = + 120 + (-274)
= + 120 + (- 120 – 154)
= + 120 – 120 – 154
∴ 120 + (-274) = -154

Method – II:
As the given numbers have opposite sign, we subtract one from other

As 274 is having negative (-) sign, the answer is – 154.

ii) -68 and 28
Sum = – 68 + (28)
= – 40 – 28 + 28
∴ – 68 + 28 = – 40

Question 4.
Simplify:
i) (-6) + (-10) + 5 + 17
ii) 30 + (-30) + (-60) + (-18)
Answer:
i) (-6) + (-10) + 5 + 17
– 6 – 10 + 5 + 17 = -16 + 22
= -16 + 16 + 6
= + 6
∴ (-6) + (-10) + 5 + 17 = + 6

ii) 30 + (-30) + (-60) + (-18)
= 30 + (-30 -60 -18)
= 30 + (-108)
= 30 – 108 = – 78
∴ 30 + (-30) + (-60) + (-18) = -78

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 4 Integers Ex 4.2 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 4th Lesson Integers Ex 4.2

Question 1.
Put appropriate symbol > or < in the boxes given.
i) -1 0
ii) -3 -7
iii) -10  +10
Answer:
i) -1 < 0
ii) -3 > -7
iii) -10 < +10

Question 2.
Write the following integers in increasing and decreasing order:
i) -7, 5, -3
ii) -1, 3, 0
iii) 1, 3, -6
iv) -5, – 3, -1
Answer:

S.No.	Numbers	Increasing order	Decreasing order
i)	-7, 5, -3	-7 < 5 < -3	5 > -3 > -7
ii)	-1, 3, 0	-1 < 0 < 3	3 > 0 > -1
iii)	1, 3, -6	-6 < 1 < 3	3 > 1 > -6
iv)	-5, -3, -1	-5 < -3 < -1	-1 > -3 > -5

Question 3.
Write True or False.
i) Zero is on the right of -3 ( )
ii) -12 and +12 represent on the number line the same integer ( )
iii) Every positive integer is greater than zero ( )
iv) (-100) > (+100) ( )
Answer:
i) True
ii) False
iii) True
iv) False

Question 4.
Find all integers which lie between the given two integers. Represent them on number line:
i) -1 and 1
ii) -5 and 0
iii) -6 and -8
iv) 0 and -3
Answer:
i) Given integers -1 and +1

Integers between -1 and 1 is 0.

ii) Given integers -5 and 0

Integers between -5 and 0 are -4, -3, -2, -1.

iii) Given, integers -6 and -8

Integers between -6 and -8 is -7.

iv) Given integers 0 and -3

Integers between 0 and -3 are -1, -2.

Question 5.
The temperature recorded in Shimla is -4°C and in Kufri is -6°C on the same day. Which place is colder on that day? Why?
Answer:
Given temperatures of Shimla and Kufri are -4°C and -6°C.
We know that -6°C < – 4°C
Therefore, temperature in Kufri is colder than Shimla.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 4 Integers Ex 4.1 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 4th Lesson Integers Ex 4.1

Question 1.
Write True or False against each of the following statements.
i) -7 is on the right side of -6 on the number line.
ii) Zero is a positive number.
iii) 29 is on the right side of zero on the number line.
iv) -1 lies between the integers -2 and 1.
v) There are nine integers between -5 and +5.
Answer:
i) False
ii) False
iii) True
iv) True
v) True

Question 2.
Observe the following number line and answer the following questions.

i) Which is the nearest positive integer to -1?
ii) How many negative numbers you will find on the left side of Zero?
iii) How many integers are there in between -3 and 7?
iv) Write 3 integers lesser than -2.
v) Write 3 integers more than -2.
Answer:
i) +1 is the nearest positive integer to -1.
ii) On the given number line negative numbers to left of zero are -1, -2, -3, -4, -5.
So, there are 5 in number.
iii) On the number line integers between -3 and 7 are -2, -1, 0, 1, 2, 3, 4, 5, 6.
So, there are 9 in number.
iv) Integers less than -2 means numbers left side of -2.
They are -3, -4, -5, -6, -7, ………
So, 3 integers lesser than -2 are -3, -4, -5.
v) Integers more than -2 means numbers right side of -2.
They are -1, 0, 1, 2, 3, 4, ……..
So, 3 integers more than -2 are -1, 0, 1.

Question 3.
Represent the integers on a number line as given below.
i) Integers lies between -7 and -2.
ii) Integers lies Between -2 and 5.
Answer:

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

AP State Syllabus 6th Class Maths Solutions 3rd Lesson HCF and LCM Unit Exercise

Question 1.
Classify the given numbers according to their divisibility.

Answer:

(OR)

Question 2.
Write the divisibility rule by 11 with one example.
Answer:
Divisibility by 11:
A given number is divisible by 11, if the difference between the sum of the digits at odd places and the sum of the digits at even places (from the right) is either ‘O’ or a multiple of 11.
Ex: 123321
Sum of the digits at odd places = 1 + 3 + 2 = 6.
Sum of the digits at even places = 2 + 3 + 1 = 6
Their difference = 6 – 6 = 0
So, 123321 is divisible by 11.

Question 3.
Fill the table with correct answer.

Answer:

Question 4.
Find HCF of 70, 105 and 175 by prime factorization method.
Answer:
Given numbers are 70, 105 and 175.
The HCF of 70, 105 and 175 can be formed by prime factorization as follows:

Thus, 70 = 2 × 5 × 7
105 = 3 × 5 × 7
175 = 5 × 5 × 7
Common factors of 70, 105 and 175 are: 5, 7
Their product: 5 × 7 = 35
Hence, HCF of 70, 105 and 175 is 35.

Question 5.
Find HCF of 18, 54, 81 by continued division method.
Answer:
Given numbers are 18, 54 and 81 To find the HCF of 18, 54 and 81
First find the HCF of 18 and 54

HCF of 18 and 54 is 18.
Then find the HCF of the third number and the HCF of first two numbers.
i.e., let us find the HCF of 81 and 18

Last divisor is 9 when remainder is zero.
∴ HCF of 18, 54 and 81 is 9.

Question 6.
Find LCM of 4, 12, 24 by two methods.
Answer:
Given numbers are 4, 12 and 24
a) LCM by prime factors method:
Factors of 4 = 2 × 2
Factors of 12 = 2 × 2 × 3
Factors of 24 = 2 × 2 × 2 × 3
LCM of 4, 12 and 24= 2 × 2 × 2 × 3 = 24

b) Division method:

Thus, the LCM of 4, 12 and 24 is 2 × 2 × 2 × 3 = 24

Question 7.
What is the capacity of the largest vessel which can empty the oil from three vessels containing 32 liters, 24 liters and 48 liters respectively, an exact number of times?
Answer:
Given capacity of three vessels are 32 liters, 24 liters and 48 liters.
To find the capacity of the largest vessel.
We have to find the HCF of 32, 24 and 48.
First find the HCF of 32 and 24.
HCF of 32 and 24 is 8.

Then, find the HCF of third number and the HCF of first two numbers.
i.e., let us find the HCF of 48 and 8.

HCF of 32, 24 and 48 is 8.
Hence, the capacity of the largest vessel which can empty the oil, then the three vessels containing 32 l, 24 l and 48 l of exact number of times is 8 liters.

Question 8.
HCF, LCM of two numbers are 9 and 54 respectively. If one of those two numbers is 18, find the other number.
Answer:
Given, HCF of two numbers = 9
LCM of two numbers = 54
One of the two numbers a = 18
Then, the other number b = ?
We know, that the product of two numbers = LCM × HCF
a × b = LCM × HCF
18 × b = 54 × 9
b = \(\frac{54 × 9}{18}\) = 27
∴ The other number = 27

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 3 HCF and LCM Ex 3.7 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 3rd Lesson HCF and LCM Ex 3.7

Question 1.
Find the LCM and HCF of the following numbers and check their relationship.
i) 15, 24
ii) 8, 25
iii) 12, 48
iv) 30, 48
Answer:
i) 15,24
The given numbers are 15 and 24.

LCM of 15 and 24 = 2 × 2 × 2 × 3 × 5 = 120
HCF of 15 and 24 = 3
Now, LCM × HCF = 120 × 3 = 360
Product of two numbers = 15 × 24 = 360,
∴ LCM × HCF = Product of the numbers.

ii) 8,25
The given numbers are 8 and 25.

LCM of 8 and 25 = 2 × 2 × 2 × 5 × 5 = 200
HCF of 8 and 25 = 1
Now LCM × HCF = 200 × 1 = 200
Product of two numbers = 8 × 25 = 200
∴ LCM × HCF = Product of the numbers

iii) 12, 48
The given numbers are 12 and 48.

LCM of 12 and 48 = 2 × 2 × 2 × 2 × 3 = 48
HCF of 12 and 48 = 12
Now LCM × HCF = 48 × 12 = 576
Product of two numbers = 12 × 48 = 576
∴ LCM × HCF = Product of the numbers

iv) 30, 48
Given numbers are 30 and 48.

LCM of 30 and 48 = 2 × 2 × 2 × 2 × 3 × 5 = 240
HCF of 30 and 48 = 6
LCM × HCF = 240 × 6 = 1440
Product of the numbers = 30 × 48 = 1440
∴ LCM × HCF = Product of the numbers.

Question 2.
If the LCM of two numbers is 290 and their product is 7250, what will be its HCF?
Answer:
Given LCM of two numbers = 290
Product of two numbers = 7250
HCF of two numbers = ?
We know that, LCM × HCF = Product of two numbers
290 × HCF = 7250
HCF = \(\frac{7250}{290}\) = 25
∴ HCF of two numbers = 25

Question 3.
The product of two numbers is 3276. If their HCF is 6, find their LCM.
Answer:
Given HCF of two numbers = 6
Product of two numbers = 3276
LCM of two numbers = ?
We know that, LCM × HCF = Product of two numbers
LCM × 6 = 3276
LCM = \(\frac{3276}{6}\) = 546
∴ LCM of two numbers = 546

Question 4.
The HCF of two numbers is 6 and their LCM is 36. If one of the numbers is 12, find the other.
Answer:
Given LCM of two numbers = 36
HCF of two numbers = 6
One of the numbers a = 12
Let second number b = ?
We know that, product of two numbers = LCM × HCF
a × b = LCM × HCF
12 × b = 36 × 6
b = \(\frac{36 × 6}{12}\) = 18
∴ Second number b = 18

Question 5.
Can two numbers have 16 as their HCF and 384 as their LCM? Give reason.
Answer:
Here, if we divide 384 by 16
\(\frac{384}{16}\) = 24 (No remainder)
If HCF is the factor LCM of two numbers, then it is possible.

Question 6.
Can two numbers have 14 as their HCF and 204 as their LCM? Give reasons in support of your answer.
Answer:
Here, if we divide 204 by 14
\(\frac{204}{14}\) = 14 and the remainder is 8.
HCF is not the factor of LCM to two numbers. So, it is not possible.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 3 HCF and LCM Ex 3.6 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 3rd Lesson HCF and LCM Ex 3.6

Question 1.
Find the LCM of the following numbers by prime factorization method.
i) 12 and 15
ii) 15 and 25
iii) 14 and 21
Answer:
i) 12 and 15
To find the LCM of 12 and 15.
First find the factors of 12,15 and then find the common factors of 12 and 15.
12 = 2 x 2 x 3; 15 = 3 x 5

Prime factorization of 12 = 2 x 2 x 3
Prime factorization of 15 = 3 x 5
Common Factors of 12, 15 = 3
Extra factors of 12 and 15 = 2 x 2 x 5
∴ LCM of 12 and 15 = Common factor x extra factors
= 3 x 2 x 2 x 5 = 60
Hence the LCM of 12 and 15 is 60.

ii) 15 and 25
To find the LCM of 15 and 25.
First find the factors of 15, 25 and then find the common factors of 15 and 25.

Prime factorization of 15 = 3 x 5
Prime factorization of 25 = 5 x 5
Common factors of 15, 25 = 5
Extra factors of 15 and 25 = 3 x 5
∴ LCM of 15 and 25 = Common factor x extra factors
= 5 x 3 x 5 = 75
Hence the LCM of 15 and 25 is 75.

iii) 14 and 21
To find the LCM of 14 and 21.
First find the factors of 14, 21 and then find the common factors of 14 and 21.

Prime factorization of 14 = 2 x 7
Prime factorization of 21 = 3 x 7
Common factors of 14, 21 = 7
Extra factors of 14 and 21 = 2 x 3
LCM of 14 and 21 = Common factor x extra factors
= 7 x 2 x 3 = 42
Hence the LCM of 14 and 21 is 42.

Question 2.
Find the LCM of the following numbers by division method.
i) 84, 112, 196
ii) 102, 119, 153
iii) 45, 99, 132, 165
Answer:
i) 84,112,196

∴ Thus, the LCM of 84, 112 and 196 is
2 x 2 x 2 x 2 x 3 x 7 x 7 = 2352

ii) 102, 119, 153

Thus, the LCM of 102,119 and 153 is
2 x 3 x 3 x 7 x 17 = 2142

iii) 45, 99, 132, 165

Thus, the LCM of 45, 99, 132 and 165 is
2 x 2 x 3 x 3 x 5 x 11 = 1980

Question 3.
Find the smallest number which when added to 5 is exactly divisible by 12,14 and 18.
Answer:
To find the number exactly divisible by 12, 14 and 18.
We have to find the LCM of 12, 14 and 18.

LCM of 12, 14 and 18 is 2 x 2 x 3 x 3 x 7 = 252
To get the smallest number we have to subtract 5 to the LCM of 12, 14 and 18 i.e., 252 – 5 = 247
Therefore, the smallest number which when added to 5 is exactly divisible by 12, 14 and 18 is 247.

Question 4.
Find the greatest 3 digit number which when divided by 75, 45 and 60 leaves
i) No remainder
ii) The remainder 4 in each case.
Answer:
i) To find the number exactly divisible (No remainder) by 75, 45 and 60.

We have to find the LCM of 75, 45 and 60.
LCM of 75, 45 and 60 is 900.
So, 990 is the greatest 3 – digit number which is exactly i divisible by the 75, 45 and 60.

ii) To get the remainder 4 when we divide the greatest 3 – digit number by 75, 45 and 60.
We have to add 4 to the greatest 3 – digit number, which is exactly divisible by 75, 45 and 60.
By adding 4 to 900 we get 904 (900 + 4)
Therefore, the greatest 3 – digit number divisible by 75, 45 and 60 by leaving remainder 4 is 994.

Question 5.
Two bells ring together. If the bells ring at every 3 minutes and 4 minutes respec-tively, at what interval of time, will they ring together again?
Answer:
Time interval for the first bell = 3 minutes
Time interval for the second bell = 4 minutes
Time interval for the bells together = LCM of 3 and 4
= 2 x 2 x 3 = 12 minutes

After 12 minutes the two bells again ring together.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 3 HCF and LCM Ex 3.5 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 3rd Lesson HCF and LCM Ex 3.5

Question 1.
Find the HCF of the following by prime – factorization and by continued division method.
i) 48, 64
ii) 126, 216
iii) 40, 60, 56
iv) 10, 35, 40
Answer:
i) 48,64
Answer:
Factorization method:
48 = 2 x 2 x 2 x 2 x 3
64 = 2 x 2 x 2 x 2 x 2 x 2
HCF = 2 x 2 x 2 x 2 = 16
∴ HCF of 48 and 64 is 16.
Continued division method:

HCF = 16

ii) 126, 216
Factorization method:
Answer:
126 = 2 x 3 x 3 x 7
216 = 2 x 2 x 2 x 3 x 3 x 3
HCF = 2 x 3 x 3 = 18
∴ HCF of 126 .and 216 is 18.
Continued division method:

HCF = 18

iii) 40, 60, 56
Answer:
Factorization method:
40 = 2 x 2 x 2 x 5
60 = 2 x 2 x 3 x 5
∴ HCF of 40, 56 and 60 is 4.
Continued division method:

First find the HCF of the smallest and the biggest of given numbers i.e., 40 and 60
Now, find the HCF of 20 and the remaining number 56.

HCF = 4

iv) 10, 35,40
Answer:
Factorization method
10 = 2 x 5
35 = 5 x 7 40 = 2 x 2 x 2 x 5
HCF = 5
Continued division method:

First find the HCF of the smallest and biggest numbers of the given numbers, i.e., 10 and 40.
HCF of 10 and 40 is 10.
10 is a factor of 40.
So, HCF (10, 40) is 10.
Now, find HCF of 10 and the remaining number i.e., 10 and 35.

HCF of 10 and 35 is 5.
∴ HCF of 10, 35 and 40 is 5.

Question 2.
Two milk cans have 60 and 165 liters of milk. Find a can of maximum capacity which can exactly measure the milk in two cans.
Answer:
Capacity of first can = 60 l
Capacity of second can = 165 l
Capacity of required can = HCF (60,165)

HCF of 60 and 165 is 15.
Therefore maximum capacity of the required can is 15 liters.

Question 3.
Three different containers contain different quantities of milk whose measures are 403 lit, 465 lit, 527 liters. What biggest measure must be’ there to measure all different quantities in exact number of times?
Answer:
Quantity of first milk container = 403 l
Quantity of second milk container = 465 l
Quantity of third milk container = 527 l
Quantity of required milk container = HCF of (403, 465,527)
First find the HCF of 403 and 527.

HCF of 403 and 527 is 31.
Now, find HCF of 31 and the remaining number i.e., 31 and 465
The HCF of 31 and 465 is 31.

Therefore, the HCF of 403, 465 and 527 is 31.
Maximum quantity of the required container is 31 litres.

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

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 3 HCF and LCM Ex 3.3 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 3rd Lesson HCF and LCM Ex 3.3

Question 1.
Write all the factors for the following numbers.
(i) 24   (ii) 56   (iii) 80   (iv) 98
Answer:
i) 24
24 = 1 × 24
24 = 3 × 8
24 = 2 × 12
24 = 4 × 6
Factors of 24 are 1,2, 3, 4, 6, 8, 12 and 24.

ii) 56
56 = 1 × 56
56 = 7 × 8
56 = 2 × 28
56 = 4 × 14
Factors of 56 are 1, 2, 4, 7, 8, 14, 28 and 56.

iii)80
80 = 5 × 16
80 = 2 × 40
80 = 8 × 10
80 = 4 × 20
Factors of 80 are 1, 2, 4, 5, 8, 10, 16, 20, 40 and 80.

iv) 98
98 = 7 × 14
98 = 2 × 49
Factors of 98 are 1, 2, 7, 14, 49, 98.

Question 2.
What is the greatest prime number between 50 and 100
Answer:
The greatest prime number between 50 and 100 is 97.

Question 3.
The numbers 13 and 31 are prime numbers. Both these numbers have same digits 1 and 3. Find 2 more such pairs of prime numbers below 100.
Answer:
17 and 71; 37 and 73; 79 and 97.

Question 4.
Express the following numbers as the sum of two odd primes.
(i) 18   (ii) 24    (iii) 36   (iv) 44
Answer:
i) 18 = 7 + 11 = 5 + 13
ii) 24 = 5 + 19 = 7 + 17 = 11 + 13
iii) 36 = 5 + 31 = 7 + 29 = 13 + 23 = 47 + 19
iv) 44 = 3 + 41 = 7 + 37 = 13 + 31

Question 5.
Write seven consecutive composite numbers less than 100.
Answer:
Consecutive composite numbers are 8, 9; 9, 10; 14, 15; 15, 16; 20, 21; 21, 22; 24, 25; 25, 26; 26, 27; 27, 28; 32, 33; 33, 34; 34, 35; 35, 36;…….

Question 6.
Write two prime numbers whose difference is 10.
Answer:
Two prime numbers whose difference is 10 are (7, 17); (13, 23); (31, 41); (43, 53); (61, 71); (79, 89);…….

Question 7.
Write three pairs of prime numbers less than 20, whose sum is divisible by 5.
Answer:

Pair of primes	Sum	Is divisible by 5 or not ?
2,3	2 + 3 = 5	Yes
3,7	3 + 7 = 10	Yes
7, 13	7 + 13 = 20	Yes