Class 6

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

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 3 HCF and LCM Ex 3.2 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 3rd Lesson HCF and LCM Ex 3.2

Question 1.
Using divisibility rules, determine which of the following numbers are divisible by 11.
i) 6446
ii) 10934
iii) 7138965
iv) 726352
Answer:
i) 6446
If the difference between the sum of the digits at odd places and the sum of the digits at even places of a number is either 0 or a multiple of 11. Then the number is divisible by 11.
Sum of the digits at odd places = 6 + 4 = 10
Sum of the digits at even places = 4 + 6 = 10
Difference 10 – 10 = 0
So, 6446 is divisible by 11.

ii) 10934
Sum of the digits at odd places = 4 + 9 + 1 = 14
Sum of the digits at even places = 3 + 0 = 3
Difference = 14 – 3 = 11
is a 11 multiple by divisibility rule for 11.
So, 10934 is divisible by 11.

iii) 7138965
Sum of the digits at odd places = 5 + 9 + 3 + 7 = 24
Sum of the digits at even places = 6 + 8 + 1 = 15
Difference = 24 – 15 = 9
is not a multiple by divisibility rule for 11
So, 7138965 is not divisible by 11.

iv)726352
Sum of the digits at odd places = 2 + 3 + 2 = 7
Sum of the digits at even places = 5 + 6 + 7 = 18
Difference = 18 – 7 = 11
is a 11 multiple by divisibility rule for 11.
So, 726352 is divisible by 11.

Question 2.
Write all the possible numbers between 2000 and 2100, that are divisible by 11.
Answer:
Numbers between 2000 and 2100 are 2001, 2002, 2003, ……… , 2097, 2098, 2099.
If we divide 2000 by 11 we get remainder 9.
By adding 2 to the 2000, then we get 2002.
Check by 11 divisibility rule, 2002 is divisible by 11. (difference of sum of odd places digits and sum of even places digits is ‘0’.)
Then 11 multiples after 2002 are 2013, 2024, 2035, 2046, 2057, 2068, 2079, 2090, 2101, …… 2112.
Therefore, 2002, 2013, 2024, 2035, 2046, 2057, 2068, 2079 and 2090 are the numbers divisible by 11 in between 2000 and 2100.

Question 3.
Write the nearest number to 1234 which is divisible by 11.
Answer:
If we divide 1234 by 11 we get remainder 2.
So, 1234 – 2 = 1232 is divisible by 11
(Difference of sum of odd places digits and sum of even places digits of 1232 is ‘0’)
Therefore, the nearest number to 1234 which is divisible by 11 is 1232.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 3 HCF and LCM Ex 3.1 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 3rd Lesson HCF and LCM Ex 3.1

Question 1.
Which of the following numbers are divisible by 2, by 3 and by 6?
Answer:
i) 237192 has 2 in its one’s place.
The number which has 0, 2, 4, 6 and 8 in its ones place is divisible by 2.
237192 has 2 in its ones place. So, 237192 is divisible by 2.
(OR)
237192 is an even number and Hence divisible by 2.
Sum of the digits = 2 + 3 + 7 + 1 + 9 + 2 = 24 is a multiple of 3 of 237192.
If the sum of the digits of a number is a multiple of 3.
Then the number is divisible by 3.
So, 237192 is divisible by 3.
If a number is divisible by both 2 and 3, then it is also divisible by 6.
237192 is divisible by both 2 and 3.
Therefore 237192 is divisible by 6.

ii) 193272 has 2 in its ones place.
So, 193272 is divisible by 2.
[ Sum of the digits = 1 + 9 + 3 + 2 + 7 + 2 = 24 is a multiple of 3.
So, 193272 is divisible y 3.
193272 is divisible by both 2 and 3.
Therefore 193272 is divisible by 6.

iii) 972312 has 2 in its ones place.
So, 972312 is divisible by 2.
Sum of the digits = 9 + 7 + 2 + 3 + 1 + 2 = 24 is a multiple of 3
So, 972312 is divisible by 3.
972312 is divisible by both 2 and 3.
Therefore 972312 is divisible by 6.

iv) 1790184 has 4 in its ones place.
So, 1790184 is divisible by 2.
Sum of the digits = 1 + 7 + 9 + 0 + 1 + 8 + 4 = 30 is a multiple of 3.
So, 1790184 is divisible by 3.
1790184 is divisible by both 2 and 3.
Therefore 1790184 is divisible by 6.

v) 312792 has 2 in its ones place.
So, 312792 is divisible by 2.
Sum of the digits = 3 + 1 + 2 + 7 + 9 + 2 = 24 is a multiple of 3.
So, 312792 is divisible by 3.
312792 is divisible by both 2 and 3.
Therefore 312792 is divisible by 6.

vi) 800552 has 2 in its ones place.
So, 800552 is divisible by 2.
Sum of the digits = 8 + 0 + 0 + 5 + 5 + 2 = 20 is not a multiple of 3.
So, 800552 is not divisible by 3.
800552 is divisible by 2 but not by 3.
Therefore 800552 is not divisible by 6.

vii) 4335 has 5 in its ones place.
So, 4335 is not divisible by 2.
Sum of the digits = 4 + 3 + 3 + 5 = 15 is a multiple of 3.
So, 4335 is divisible by 3.
4335 is not divisible by 2 but it is divisible only by 3.
Therefore 4335 is not divisible by 6.

viii) 726352 has 2 in its ones place.
So, 726352 is divisible by 2.
Sum of the digits = 7 + 2 + 6 + 3 + 5 + 2 = 25 is not a multiple of 3.
So, 726352 is not divisible by 3.
726352 is divisible by 2 but not divisible by 3.
Therefore 726352 is not divisible by 6.

Question 2.
Determine which of the following numbers are divisible by 5 and by 10.
25, 125, 250, 1250, 10205, 70985, 45880.
Check whether the numbers that are divisible by 10 are divisible by 2 and 5.
Answer:
The numbers with zero or five at ones place are divisible by 5.
The numbers which have 0, 2, 4, 6 and 8 in its units place are divisible by 2.
The numbers with zero at ones place are divisible by 10.

Therefore the numbers that are divisible by 10 are also divisible by both 2 and 5.

Question 3.
Make 3 different 3 digit numbers using 2, 3, 4 where each digit can be used only once, Check which of these numbers is divisible by 9.
Answer:
3 different 3-digit numbers using 2, 3, 4 are 234, 342, 243
Given digits 2, 3 & 4; their sum = 2 + 3 + 4 = 9,
Any number formed by these digits is always divisible by 9.
a) Sum of the digits of 234 = 2 + 3 + 4 = 9 is divisible by 9.
If the sum of the digits of the number is divisible by 9.
So, 234 is divisible by 9.
b) Sum of the digits of 342 = 3 + 4 + 2 = 9 is divisible by 9.
So, 342 is divisible by 9.
c) Sum of the digits of 243 = 2 + 4 + 3 = 9 is divisible by 9.
So, 243 is divisible by 9.

Question 4.
Write different 2 digit numbers using digits 5, 6, 7. Check whether these numbers are divisible by 2, 3, 5, 6 and 9.
Answer:
Different 2 digit numbers using 5, 6, 7 are 56, 57, 65, 67, 75, 76
56, 76 are divisible by 2
57, 75 are divisible by 3
65, 75 are divisible by 5
There is no number which is divisible by 6. [both 2 and 3 and there by 6]
There is no number which is divisible by 9.

Question 5.
Find the smallest number that must be added to 128, so that it becomes exactly divisible by 5.
Answer:
Given number be 128.
A number to be divisible by 5, its unit digit must be ‘0’ or ‘5’.
So, 128 + 2 = 130 is divisible by 5
128 + 7 = 135 is divisible by 5
In these two numbers, 2 is the smallest number.
The smallest number to be added is 2.

Question 6.
Find the smallest number that has to be subtracted from 276 so that it becomes exactly divisible by 10.
Answer:
Given number be 276.
A number to be divisible by 10, its units digit must be ‘0’.
So, 276 – 6 = 270 is divisible by 10.
276 – 16 = 260 is divisible by 10.
In these numbers, 6 is the smallest number.
∴ The smelliest number to be subtracted is 6.

Question 7.
Write all the numbers between 100 and 200 which are divisible by 6.
Answer:
The numbers in between 100 and 200 are
101, 102, 103, 104 ……. 198, 199
102 is divisible by 2 and 3. So 102 is divisible by 6.
Adding 6 to it successively we get
102, 108, 114, 120 126, 132, 138, 144, 150, 156, 162, 168, 174, 180, 186, 192, 198 are multiples of 6 and are divisible by 6.

Question 8.
Write the greatest four digit number which is divisible by 9. Is it divisible by 3? What do you notice?
Answer:
The greatest four digit number is 9999.
Sum of the digits of 9999 = 9 + 9 + 9 + 9 = 36 is divisible by 9
So, 9999 is divisible by 9.
Sum of the digits of 9999 is also multiple of 3.
Therefore, 9999 is also divisible by 3.
We notice that the numbers which are divisible by 9 are always divisible by 3.

Question 9.
Which of the following are divisible by 8?
(i) 1238
(ii) 13576
(iii) 93624
(iv) 67104
Answer:
i) 1238
The number formed by the last 3-digits is 238.

If the number formed by the last 3-digits in the same order is divisible by 8, then the number is divisible by 8.
238 is not divisible by 8. So, 1238 is not divisible by 8.

ii) 13576

The number formed by the last 3-digits is 576. 576 is divisible by 8. So, 13576 is divisible by 8.

iii) 93624

The number formed by the last 3-digits is 624. 624 is divisible by 8. So, 93624 is divisible by 8.

iv) 67104

The number formed by the last 3-digits is 104. 104 is divisible by 8. So, 67104 is divisible by 8.

Question 10.
Write the nearest number to 12345 which is divisible by 4.
Answer:
Given number be 12345.
A number to be divisible by 4, the number formed by the last two digits is divisible by 4.
So, 12345 – 1 = 12344 is divisible by 4
12345 – 2 = 12343 is not divisible by 4
12345 – 3 = 12342 is not divisible by 4
12345 – 4 = 12341 is not divisible by 4
12345 – 5 = 12340 is divisible by 4
and
12345 + 1 = 12346 is not divisible by 4
12345 + 2 = 12347 is not divisible by 4
12345 + 3 = 12348 is divisible by 4
12340, 12344 and 12348 are divisible by 4.
∴ 12344 is the nearest number to 12345 which is divisible by 4.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 6 Basic Arithmetic Ex 6.2 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 6th Lesson Basic Arithmetic Ex 6.2

Question 1.
Check whether the following are in proportion? or not?
(a) 10, 12, 15, 18
(b) 11, 16, 16, 21
(c) 8, 13, 17, 19
(d) 30, 24, 20, 16
Answer:
a) 10, 12, 15, 18
Given numbers are 10, 12, 15, 18.
If a, b, c, d are in proportion i.e., a : b :: c : d,
then product of extremes = product of means (a.d = b.c)

a : b :: c : d
10 × 18 = 12 × 15
180 = 180
If the product of extremes is equal to the product of means, then the given numbers are in proportion.
So, 10, 12, 15 and 18 are in proportion,

b) 11, 16, 16, 21
Given numbers are 11, 16, 16, 21.
If a, b, c, d are in proportion i.e., a : b :: c : d, then product of extremes = product of means (a.d = b.c)

11 × 21 = 16 × 16
231 ≠ 256
Here the product of extremes is not equal to the product of means, then the given numbers are not in proportion.
So, 11,16, 16 and 21 are not in proportion.

c) 8, 13, 17, 19
Given numbers are 8, 13, 17,19.
If a, b, c, d are in proportion i.e., a : b :: c : d,
then the product of extremes = product of means (a.d = b.c)

8 × 19 = 13 × 17
152 ≠ 221
Here the product of extremes is not equal to the product of means, then the given numbers are not in proportion.
So, 8,13, 17 and 19 are not in proportion.

d) 30, 24, 20, 16
Given numbers are 30, 24, 20, 16.
If a, b, c, d are in proportion i.e., a : b :: c : d,
then the product of extremes = product of means (a.d = b.c)
30 : 24 :: 20 : 16

480 = 480
Here the product of extremes is equal to the product of means, then the given numbers are in proportion.
So, 30, 24, 20 and 16 are in proportion.

Question 2.
Write true or false for each of the following.
a) 4 : 2 :: 14 : 7
b) 21 : 7 :: 15 : 5
c) 13 : 12 :: 12 : 13
d) 5 : 6 :: 7 : 8
Answer:
a) 4 : 2 :: 14 : 7
Given proportion is 4 : 2 :: 14 : 7
If a, b, c, d are in proportion, then a.d = b.c

4 × 7 = 2 × 14
28 = 28
So, the given proportion is true.

b) 21 : 7 :: 15 : 5
Given proportion is 21 : 7 : : 15 : 5
If a, b, c, d are in proportion, then a.d = b.c

21 × 5 = 7 × 15
105 = 105
So, the given proportion is true.

c) 13 : 12 :: 12 : 13
Given proportion is 13 : 12 :: 12 : 13
If a, b, c, d are in proportion, then a.d = b.c

13 × 13 = 12 × 12
169 ≠ 144
So, the given proportion is false.

d) 5 : 6 :: 7 : 8
Given proportion is 5 : 6 :: 7 : 8
If a, b, c, d are in proportion, then a.d = b.c

5 × 8 = 6 × 7
40 ≠ 42
So, the given proportion is false.

Question 3.
Check whether the following form a proportion? Write middle terms and extremes where the ratios form a proportion.
a) 15 cm, 1 m and Rs. 45, Rs. 300
b) 20 ml, 2 l and Rs. 100, Rs. 10,000
Answer:
a) 15 cm, 1 m and Rs. 45, Rs. 300
Given values are 15 cm, 1 m and Rs. 45, Rs. 300
Ratio of lengths = 15 cm : 1 m (convert metres into cm)
= 15 cm : 100 cm = 3 : 20 (write in simplest form)
Ratio of amounts = Rs. 45 : Rs. 300 = 45 : 300 = 3 : 20
The two ratios are equal.
So, 3 : 20 :: 3 : 20 They are in proportion.
Middle terms are 20, 3 and extremes are 3, 20.

b) 20 ml, 2 l and Rs. 100, Rs. 10,000
Given values are 20 ml, 2 l and Rs. 100, Rs. 10,000
Ratio of quantities = 20 ml : 2 l (convert litres into ml)
= 20 : 2000 (write in simplest form)
= 1 : 100
Ratio of amounts = Rs. 100 : Rs. 10,000 (write in simplest form)
= 1 : 100
The two ratios are equal.
So, 1 : 100 = 1 : 100
∴ They are in proportion.
Middle terms are 100, 1 and extremes are 1, 100.

Question 4.
Find the missing numbers in the following proportions.
a) 8 : 12 :: : 48
b) 15 : :: 105 : 98
c) 34 : 102 :: 27 :
Answer:
a) 8 : 12 :: : 48
Given proportion is 8 : 12 :: : 48
Let the missing number = x
If the given numbers are in proportion, then
Product of extremes = Product of means

8 × 48 = 12 × x (or) 12x = 8 × 48

∴ Missing number in the proportion = 32

b) 15 : :: 105 : 98
Given proportion is 15 : : 105 : 98
Let the missing number = a
If the given numbers are in proportion, then Product of extremes = product of means

15 × 98 = 105 × a (or) 105 a = 15 × 98

∴ a = 14
Missing number in the proportion

c) 34 : 102 :: 27 :
Given proportion is 34 : 102 :: 27 :
Let the missing number = y
If the given numbers are in proportion,
then the product of extremes = product of means

Missing number in the proportion = 81

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 2 Whole Numbers Unit Exercise Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 2nd Lesson Whole Numbers Unit Exercise

Question 1.
Choose the appropriate symbol from < or > and place it in the blanks.
i) 8 ……. 7
ii) 5 ……. 2
iii) 0 ……. 1
iv) 10 ……. 5
Answer:
i) 8 …. > …. 7
ii) 5 …. > …. 2
iii) 0 …. < …. 1
iv) 10 …. > …. 5

Question 2.
Present the successor of 11 and predecessor of 5 on the number line.
Answer:
i)

Successor of 11 is 12.

ii)

Predecessor of 5 is 4.

Question 3.
Which of the statements are true ( T ) and which are false ( F ). Correct the false statements.
i) There is a natural number that has no predecessor. ( T )
ii) Zero is the smallest whole number. ( T )
iii) A whole number on the left of another number on the number line, is greater than that number. (F)
Answer:
i) There is a natural number that has no predecessor. ( T )
ii) Zero is the smallest whole number. ( T )
iii) A whole number on the left of another number on the number line, is greater than that number. ( F )

Question 4.
Give the results without actually performing the operations, using the given information.
i) 28 × 19 = 532 , then 19 × 28 =
ii) a × b = c , then b × a =
iii) 85 + 0 = 85 , then 0 + 85 =
Answer:
i) 28 × 19 = 532 , then 19 × 28 = 532
ii) a × b = c , then b × a = c
iii) 85 + 0 = 85 , then 0 + 85 = 85

Question 5.
Find the value of the following:
i) 368 × 12 + 18 × 368
ii) 79 × 4319 + 4319 × 11
Answer:
i) 368 × 12 + 18 × 368 – 368 × 12 + 368 × 18
Distributive property of multiplication over addition.
= 368 × (12 + 18)
= 368 × 30 = 11040

ii) 79 × 4319 + 4319 × 11
= 79 × 4319 + 11 × 4319
Distributive property of multiplication over addition.
= (79 + 11) × 4319
= 90 × 4319;
= 388710

Question 6.
Chandana and Venu purchased 12 note books and 10 note books respectively. The cost of each note book is Rs. 15. Then how much amount should they pay to the shopkeeper?
Answer:
Number of note books purchased by Chandana = 12
Number of note books purchased by Venu = 10
Total number of note books purchased together = 12 + 10
Cost of each note book = Rs. 15
Cost of (12 + 10) note books = (12 + 10) × 15
= 22 × 15
The amount paid to the shopkeeper = Rs. 330

Question 7.
Match the following.
i) 3 + 1991 + 7 = 3 + 7 + 1991                        [ ]                A) Additive identity
ii) 2 × 68 × 50 = 2 × 50 × 68                           [ ]                B) Multiplicative identity
iii) 1                                                                  [ ]                C) Commutative under addition
iv) 0                                                                  [ ]                D) Distributive property of multiplication over addition
v) 879 × (100 + 30) = 879 × 100 + 879 × 30  [ ]                E) Commutative under multiplication
Answer:
i) C
ii) E
iii) B
iv) A
v) D

Question 8.
Study the pattern:
91 × 11 × 1 = 1001
91 × 11 × 2 = 2002
91 × 11 × 3 = 3003
Write next seven steps. Check, whether the result is correct.
Answer:
91 × 11 × 1 = 1001
91 × 11 × 7 = 7007
91 × 11 × 8 = 8008
91 × 11 × 9 = 9009
91 × 11 × 10 = 10010

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 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

Students can go through AP Board 6th Class Maths Notes Chapter 12 Data Handling to understand and remember the concepts easily.

AP State Board Syllabus 6th Class Maths Notes Chapter 12 Data Handling

→ Data: Information which is in the shape of numbers or words or pictures which help us in taking decisions is called data.
If data is expressed in numbers it is called numerical data.
Eg: The marks obtained by five students at an examination is 25, 32, 28,14 & 24.
If data is expressed in words it is called data in words.
Eg : The colours liked by some students are Red, black, pink, white, etc.

→ Frequency: Number of times a particular observation occurs in a given data is called its frequency.

→ Frequency distribution table: A table showing the frequency or count of various items is called a frequency distribution table.
A data can be arranged in a tabular form using tally marks. The data arranged in a tally table is easy understood and interpret.

→ Pictograph: If a data is arranged using pictures, then it is called a pictograph.

→ Bar graph: If a data is arranged using rectangles, then it is called a bar graph.
The rectangles can be either vertical or horizontal in a bar graph.

→ Scale: Scale is a convenient way to represent the data. It quantifies what a single unit represents in a given bar graph or pictograph.

Students can go through AP Board 6th Class Maths Notes Chapter 11 Perimeter and Area to understand and remember the concepts easily.

AP State Board Syllabus 6th Class Maths Notes Chapter 11 Perimeter and Area

→ Perimeter: The perimeter of a polygon is sum of all its sides.
The perimeter of an equilateral triangle is P = 3 × side
The perimeter of a rectangle P = 2 (length + breadth)
And its area A = length × breadth A = l × b
The perimeter of a square is P = 4 × side And its area A = side × side (or)
A = s × s The circumference of a circle C = 2πr where r is the radius of the circle.

→ Area: The region occupied by a plane figure is called its area.
To find the area of a complex figure, we divide the given shape into the combination of rectangles, squares and triangles where ever necessary.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 2 Whole Numbers Ex 2.3 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 2nd Lesson Whole Numbers Exercise 2.3

Question 1.
Study the pattern.

1 × 8 + 1 = 9
12 × 8 + 2 – 98
123 × 8 + 3 = 987
1234 × 8 + 4 = 9876
12345 × 8 + 5 = 98765

Write the next four steps. Can you find out how the pattern works?
Answer:

12 × 8 + 2 = 98
123 × 8 + 3 = 987
1234 × 8 + 4 = 9876
12345 × 8 + 5 = 98765
123456 × 8 + 6 = 987654
1234567 × 8 + 7 = 9876543
12345678 × 8 + 8 = 98765432
123456789 × 8 + 9 = 987654321

The digits on the L.H.S. are in increasing order and the digits on the result (right side) are in decreasing order.

Question 2.
How would we multiply the numbers 13680347, 35702369 and 25692359 with 9 mentally? What is the pattern that emerges?
Answer:
i) 13680347 × 9 = 13680347 × (10 – 1)
Distributive property of multiplication over subtraction.
= 13680347 × 10 – 13680347 × 1
= 136803470 – 13680347
= 123123123

ii) 35702369 × 9 = 35702369 × (10 – 1)
Distributive property of multiplication over subtraction.
= 35702369 × 10 – 35702369 × 1
= 357023690 – 35702369
= 321321321

iii) 25692359 × 9 = 25692359 × (10 – 1)
Distributive property of multiplication over subtraction.
= 25692359 × 10 – 25692359 × 1
= 256923590 – 25692359
= 231231231