Category: Class 6

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 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 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

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 2 Whole Numbers Ex 2.2 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 2nd Lesson Whole Numbers Exercise 2.2

Question 1.
Find the sum by suitable rearrangement.
i) 238 + 695 + 162
ii) 154 + 197 + 46 + 203
Answer:
i) 238 + 695 + 162 = 238 + 162 + 695 (Commutative property)
= (238 + 162) + 695 (Associative property)
= 400 + 695
= 1095

ii) 154 + 197 + 46 + 203 = 154 + 46 + 197 + 203 (Commutative property)
= (154 + 46) + (197 + 203) (Associative property)
= 200 + 400
= 600

Question 2.
Find the product by suitable rearrangement.
i) 25 × 1963 × 4
ii) 20 × 255 × 50 × 6
Answer:
i) 25 × 1963 × 4 = 25 × (1963 × 4)
= 25 × (4 × 1963) (Commutative property)
= (25 × 4) × 1963 (Associative property)
= 100 × 1963 = 196300

ii) 20 × 255 × 50 × 6 = 20 × 50 × 255 × 6 (Commutative property)
= (20 × 50) × (255 × 6) (Associative property)
= 1000 × 1530
= 15,30,000

Question 3.
Find the product using suitable properties.
1)205 × 1989 ii) 1991 × 1005
Answer:
i) 205 × 1989 = (200 + 5) × 1989
(Distributive property of multiplication over addition)
= (200 × 1989) + (5 × 1989)
= 397800 + 9945
= 407745

ii) 1991 × 1005
= 1991 × (1000 + 5)
Distributive property of multiplication over addition.
= (1991 × 1000)+ (1991 × 5)
= 1991000 + 9955
= 2000955

Question 4.
A milk vendor supplies 56 liters of milk in the morning and 44 liters of milk in the evening to a hostel. If the milk costs Rs. 50 per liter, how much money he gets per day?
Answer:
Capacity of milk supplied in the morning = 56 l
Capacity of milk supplied in the evening = 44 l
Total capacity of milk supplied in one day = (56 + 44)l
Cost of one liter milk = Rs. 50
Cost of (56 + 44) liters milk =(56 + 44) × 50
= 100 × 50
= Rs. 5000
∴ Money got by vendor per day = Rs. 5000

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 2 Whole Numbers Ex 2.1 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 2nd Lesson Whole Numbers Exercise 2.1

Question 1.
How many whole numbers are there in between 27 and 46?
Answer:
Number of whole numbers upto 27 is 28 (from zero to 27)
Number of whole numbers upto 45 is 46 (excluding 46)
Number of whole numbers between 27 and 46 = 46 – 28 = 18
They are 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45.

Question 2.
Find the following using number line.
i) 6 + 7 + 7
ii) 18 – 9
iii) 5 × 3
Answer:
i) 6 + 7 + 7
Answer:
Draw the number line starting from zero (0).

Starting from 6, we make 7 jumps to the right of 6 on the number line. Then we reach 13. Again make 7 jumps to the right of 13. Then we reach 20.
So, 6 + 7 + 7 = 20

ii) 18 – 9
Answer:

Draw the number line starting from zero (0).
Start from 18. We make 9 jumps to the left of 18 on the number line.
Then we reach 9.
So, 18 – 9 = 9

iii) 5 × 3
Answer:

Draw the number line starting from zero (0).
Start from 0 and make 3 jumps to the right of the zero on the number line.
Now, treat 3 jumps as one step.
So, to make 5 steps (i.e., 3, 6, 9, 12 and 15) move on the right side, we read 15 on the number line.
So, 5 × 3 = 15

Question 3.
In each pair, state which whole number on the number line is on the right of the other number.
i) 895, 239
Answer:
As 239 < 895, we conclude that 895 is on the RHS of 239.

895 is right of 239 on the number line.

ii) 1001, 10001
Answer:
As 1001 < 10001, we conclude that 10001 is on the RHS of 1001.

10001 is right of 1001 on the number line.

iii) 15678, 4013
Answer:
As 4013 < 15678, we can say that 15678 is on the RHS of 4013.

15,678 is right of 4,013 on the number line.

Question 4.
Mark the smallest whole number on the number line.
Answer:
We know that zero is the smallest whole number mark it on the number line.

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

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

Question 1.
Write each of the following in numeral form.
i) Hundred crores hundred thousands and hundred.
Answer:
Indian system: 100,01,00,100.

ii) Twenty billion four hundred ninety seven million pinety six thousands four hundred seventy two.
Answer:
20,497,096,472

Question 2.
Write each of the following in words in both Hindu-Arabic and International system.
i) 8275678960
ii) 5724500327
iii) 1234567890
Answer:
i) 8275678960
Hindu – Arabic system: 827,56,78,960
Eight hundred twenty seven crores fifty six lakhs seventy eight thousand nine hundred and sixty.
International system: 8,275,678,960
Eight billion two hundred seventy five million six hundred seventy eight thousand nine hundred and sixty.

ii) Hindu-Arabic system: 572,45,00,327
Five hundred seventy two crores forty five lakhs three hundred and twenty seven. International system: 5,724,500,327
Five billion seven hundred twenty four million five hundred thousand three hundred and twenty seven,

iii) 1234567890
Hindu-Arabic system: 123,45,67,890
One hundred twenty three crores forty five lakhs sixty seven thousand eight hundred and ninety.
International system: 1,234,567,890
One billion two hundred thirty four million five hundred sixty seven thousand eight hundred and ninety.

Question 3.
Find the difference between the place values of the two eight’s in 98978056.
Answer:
Place values of 8 in Hindu-Arabic system of the given number are 8,000 and 80,00,000
Difference = 80,00,000 – 8,000 = 79,92,000
Place values of 8 in International system of the given number are 8000 and 8,000,000
Difference = 8,000,000 – 8,000 = 79,92,000

Question 4.
How many 6 digit numbers are there in all?
Answer:
Number of 6 digit numbers = greatest 6 digit number – greatest 5 digit number
= 9,99,999 – 99,999 = 9,00,000.

Question 5.
How many thousands make one million?
Answer:
1000 thousands can make one million.
1 Million = 1000 Thousands.

Question 6.
Collect ‘5’ mobile numbers and arrange them in ascending and descending order.
Answer:
Let the 5 mobile numbers are: 9247568320, 9849197602, 8125646682, 6305481954, 7702177046
Ascending order: 6305481954, 7702177046, 8125646682, 9247568320, 9849197602
Descending older: 9849197602, 9247568320, 8125646682, 7702177046, 6305481954

Question 7.
Pravali has one sister and one brother. Pravali’s father earned one million rupees and wanted to distribute the amount equally. Estimate approximate amount each will get in lakhs and verify with actual division.
Answer:
One million = 10,00,000 = 10 lakhs
Pravali’s father distributed 10 lakhs amount to his 3 children equally.
So, the share of each children = 10 lakhs ÷ 3 = Rs. 3,33,333
= Rs. 3,00,000 (approximately)

Question 8.
Government wants to distribute Rs. 13,500 to each farmer. There are 2,27,856 farmers in a district. Calculate the total amount needed for that district. (First estimate, then calculate)
Answer:

Question 9.
Explain terms Cusec, T.M.C, Metric tonne, Kilometer.
Answer:
a) Cusec: A unit of flow equal to 1 cubic foot per second.
1 Cusec = 0.028316 cubic feet per second = 28.316 litre per second.
Cusec is the unit to measure the liquids in large numbers quantity.

b) TMC: TMC is the unit to measure the water in large quantity.
TMC means Thousand Million Cubic feet.
1 TMC = 0.28316000000 litre
= 28.316 billion litre
= 2831.6 crores litre

c) Metric tonne: Metric tonne is the unit of weight.
Metric tonne = 1000 kg = 10 quintals
We should use this unit in measuring crops, paddy, dall, etc.

d) Kilometer: Kilometre is the unit of length.
1 kilometer = 1000 meters.
We should use this unit is measuring distance between villages, towns, cities,…. etc.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 1 Numbers All Around us Ex 1.4 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 1st Lesson Numbers All Around us Exercise 1.4

Question 1.
Write some daily life situation where we can use large numbers.
Answer:
We should use large number in our daily life in
a) Counting money at banks, etc.
b) Population of the city or state or country or world.
c) Austronautical distances, etc.

Question 2.
A box of medicine contains 3,00,000 tablets each weighing 15mg. What is the total weight of all the tablets in the box in grams and in kilograms?
Answer:
Weight of a tablet = 15 mg
Weight of 3,00,000 tablets = 300000 × 15
Weight of one box of tablets = 45,00,000 mg
we know 1000 mg = 1 gram
To convert mg into grams we have to divide grams by 1000 mg = 45,00,000 ÷ 1000
Weight of one box of tablets in grams = 4500 grams.
We know 1000g = 1kg
To convert ‘g’ into kilograms we have to divide kilograms by 1000g = \(\frac{4500}{1000}\) = 4.5 kg

Question 3.
Damodhar wants to buy onions in Kurnool market. Each onion bag weighs 45 kg. He loaded 326 onion bags with 45kg in a lorry. Find the total weight of onions in kilograms and quintals.
Answer:
Weight of one bag onions = 45 kg
Weight of 326 bag onions = 326 × 45 = 14,670 kg
We know, 100 kg = 1 quintal.
To convert kgs into quintal we have to divide kgs by 100
= 14670 ÷ 100 = 146.7 quintals.

Question 4.
The population of 4 South Indian States according to 2011 Census:
Andhra Pradesh: 8,46,65,533, Karnataka: 6,11,30,704, Tamil Nadu: 7,21,38,958 and Kerala: 3,33,87,677. What is the total population of South Indian States?
Answer:
Population of Andhra Pradesh = 8,46,65,533
Karnataka = 6,11,30,704
TamilNadu = 7,21,38,958
Kerala = 3,33,87,677
Total population of 4 states = 25,13,22,872

Question 5.
A famous cricket player has so far scored 28,754 runs in International matches. He wishes to complete 50,000 runs in his career. How many more runs does he need?
Answer:
Number of runs wishes to complete = 50,000
Number of runs scored by the player = 28,754
Number of runs needed = 50,000 – 28,754 = 21,246 runs

Question 6.
In an election, the successful candidate registered 1,32,356 votes and his nearest rival secured 42,246 votes. Find the majority of successful candidate.
Answer:
Number of votes secured by the winner = 1,32,356
Number of votes secured by the rival = 42,246
Number of more votes secured by the winner = 1,32,356 – 42,246 = 90,110
Majority of the winner = 90,110 votes.

Question 7.
Write the greatest and smallest six digit numbers formed by all the digits 6, 4, 0, 8, 7, 9 and find the sum and difference of those numbers.
Answer:
Given digits are 6, 4, 0, 8, 7, 9
The greatest number formed by the digits = 9,87,640
The smallest number formed by the digits = 4,06,789
Sum of the numbers = 9,87,640 + 4,06,789 = 13,94,429
Difference between numbers = 987640 – 406789 = 580851

Question 8.
Haritha has Rs. 1,00,000 with her. She placed an order for purchasing 124 ceiling fans at Rs. 726 each. How much money will remain with her after the purchase?
Answer:
Cost of each fan = Rs. 726
Cost of 124 fans = 726 × 124 = Rs. 90,024
Money with Haritha = Rs. 1,00,000
Cost of 124 fans = Rs. 90,024
Remaining money after purchasing = Rs. 9,976

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 1 Numbers All Around us Ex 1.3 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 1st Lesson Numbers All Around us Exercise 1.3

Question 1.
Write each of the following numbers in digits by using International place value chart. Also write them in expanded form.
i) Nine million seven hundred thousand and six hundred five.
Answer:
9,700,605: 9,000,000 + 700,000 + 600 + 5

ii) Seven hundred million eight hundred seventy two thousand and four hundred seven.
Answer:
700,872,407: 700,000,000 + 800, 000 + 70,000 + 2000 + 400 + 7

Question 2.
Rewrite each of the following numerals with proper commas in the International system of numeration and write the numbers in word form.
i) 717858
ii) 3250672
iii) 75623562
iv) 956237676
Answer:
i) 717,858: Seven hundred seventeen thousand eight hundred and fifty eight.
ii) 3,250,672: Three million two hundred fifty thousand six hundred and seventy two.
iii) 75,623,562: Seventy five million six hundred twenty three thousand five hundred and sixty two.
iv) 956,237,676: Nine hundred fifty six million two hundred thirty seven thousand six hundred and seventy six.

Question 3.
Write the following number names in both Indian and International systems.
i) 6756327
ii) 45607087
iii) 8560707236
Answer:
i) 6756327
Indian system:
67,56,327: Sixty seven lakh fifty six thousand three hundred and twenty seven.
International system:
6,756,327: Six million seven hundred fifty six thousand three hundred and twenty seven.

ii) 45607087 Indian system:
4,56,07,087: Four crores fifty six lakhs seven thousand eighty seven.
International system:
45,607,087: Forty five million six hundred seven thousand and eighty seven.

iii) 8560707236 Indian system:
856,07,07,236: Eight hundred fifty six crores seven lakh seven thousand two hundred and thirty six.
International system:
8,560,707,236: Eight billion five hundred sixty million seven hundred seven thousand two hundred and thirty six.

Question 4.
Express the following numbers in other system.

S.No.	Indian	International
1.	42,56,876
2.		800,000,000
3.	956,76,72,345
4.		6,303,448,433

Answer:

S.No.	Indian	International
1.	42,56,876	4,256,876
2.	80,00,00,000	800,000,000
3.	956,76,72,345	9,567,672,345
4.	630,34,48,433	6,303,448,433

Question 5.
Write the following numbers in International system (Word Form).
i) Twenty Nine crore thirty five lakh forty six thousand seven hundred and fifty three.
Answer:
293,546,753
Word form: Two hundred ninety three million five hundred forty six thousand seven hundred and fifty three.

ii) Thousand crore ninety nine lakh and forty three.
Answer:
10,009,900,043
Word form: Ten billions nine million nine hundred thousand and forty three.

Question 6.
Write following numbers in Indian system (Word Form).
i) Nine billion twenty four million fifty thousand and seventy two.
Answer:
902,40,50,072
Word form: Nine hundred two crores forty lakhs fifty thousand and seventy two.

ii) Seven hundred billions six millions four thousand seven hundred and five.
Answer:
70000,60,04,705
Word form: Seventy thousand crores sixty lakhs four thousand seven hundred and five.

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

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 1 Numbers All Around us Ex 1.1 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 1st Lesson Numbers All Around us Exercise 1.1

Question 1.
Identify the greatest and smallest among the following numbers.

S.No.	Numbers	Greatest	Smallest
1.,	67456, 76547, 15476, 75460
2.	64567, 66000, 78567, 274347
3.

Create Your Own Problem on Block No: 3 and fill the above table.
Answer:

S.No.	Numbers	Greatest	Smallest
1.	67456,76547, 15476, 75460	76547	15476
2.	64567, 66000, 78567, 274347	274347	64567
3.	95234, 572594, 82630, 830942	830942	82630

Question 2.
Write the given numbers in ascending and descending order.

S.No.	Numbers	Descending order
1.	75645, 77845, 24625, 85690
2.	6790, 27895, 16176, 50000

S.No.	Numbers	Ascending order
1.	75645, 77845, 24625, 85690
2.	6790, 27895, 16176, 50000

Answer:

S.No.	Numbers	Ascending order
1.	75645, 77845, 24625, 85690	24625, 75645, 77845, 85690
2.	6790, 27895, 16176, 50000	6790, 16176, 27895, 50000

S.No.	Numbers	Descending order
1.	75645, 77845, 24625, 85690	85690, 77845, 75645, 24625
2.	6790, 27895, 16176, 50000	50000, 27895, 16176, 6790

Question 3.
Write the numbers in word form.

S.No.	Number	Word Form
1.	73,062
‘2.	1,80,565
3.	25,45,505
4.

Create Your Own Problem on Block No:4 and fill the above table.
Answer:

S.No.	Number
1.	73,062	Seventy three thousand sixty two
2.	1,80,565	One lakh eighty thousand five hundred and sixty five
3.	25,45,505	Twenty five lakhs forty five thousand five hundred and five
4.	88,88,888	Eighty eight lakhs eighty eight thousand eight hundred and eighty eight

Question 4.
Write the numbers in figures.

S.No.	Word Form	Number
1.	Sixty thousand sixty six	60,066
2.	Seventy eight thousand four hundred and fourteen
3.	Nine lakhs ninety six thousand and ninety
4.

Create Your Own Problem on Block No:4 and fill the above table.
Answer:

S.No.	Word Form	Number
1.	Sixty thousand sixty six	60,066
2.	Seventy eight thousand four hundred and fourteen	78,414
3.	Nine lakhs ninety six thousand and ninety	9,96,090
4.	Fifty eight lakhs sixty seven thousand four hundred and thirty two	58,67,432

Question 5.
Write 4 digit numbers is many as possible with 6, 0, 5, 7 digits.
Answer:
5067, 5076, 5706, 5760, 5670, 5607
6057, 6075, 6705, 6750, 6570, 6507
7056, 7065, 7605, 7650, 7506, 7560

Question 6.
Form the greatest and smallest numbers with given digits and find the difference without repetition.

Answer:

Question 7.
Observe the table and fill the empty boxes.

Answer:

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.