AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors

Andhra Pradesh AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors World Textbook Exercise Questions and Answers.

AP State Syllabus 5th Class Maths Solutions Chapter 5 Multiples and Factors

Question 1.
Observe which numbers are not divisible?

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 1

Answer: Which of the above numbers are exactly divisible by 2?
Answer:
2410, 1282, 3784, 6728, 5466

b. Observe the units place of the numbers which are divisible by 2.
Answer:
0, 2, 4, 6, 8.

c. Are all these numbers even numbers ? Yes / no. (Yes)
Answer:
So a number is divisible by 2, if the digit at its one place is either 0/2/4/6/8.

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors

Do these: (TextBook Page No.62)

Question 1.
Circle the following numbers which are divisible by 2.

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 2

Answer:

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 3

Question 2.
Write any 5, four-digit numbers which arc divisible by 2 ?
Answer:
25680, 45,622, 78,964, 87,766 and 97.678

Exercise 1:

Question 1.
Find the numbers which are divisible by 2. Write the reason for the numbers which are not divisible?
a) 3458
b)56745
c) 3850
d) 6736
e) 6733
f) 3394
Answer:
3458, 3850, 6736 and 3394 are divisible by 2. Remaining numbers are not divisible.
Reason : The digits of the numbers of ones places is 3 and 5.

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors

Question 2.
Find the numbers which are divisible by 5 and 19. Write the reason for the numbers which are not divisible ?
a) 3568
b) 3540
c) 6585
d) 7550
e) 4235
f) 7200
g) 7865
h) 5880
i) 4440
j) 8198
k) 8645
Answer:
b, d, f, h, j are divisible by both 5 and 10. Remaining numbers are not divisible by 5 and 10.
Reason : The numbers which have n’t zero at their ones place.

Question 3.
The numbers below are divisible by 5. Fill in the blanks with suitable digit.
a) 786_
Answer:
786_ (0/5)

b)560_
Answer:
560_(0/5)

c)785_
Answer:
785_(0/5)

d) 555_
Answer:
555_ (0/5)

e) 586_
Answer:
586(0/5)

f) 786_
Answer:
786_(0/5)

g) 584_
Answer:
584_(0/5)

h) 100_
Answer:
100_(0/5)

Question 4.
Write any 5 numbers which are exactly divisible by 2 and 5.
Answer:
2540, 62570, 250, 367280 and 764520.

Question 5.
Write any 5 numbers which are exactly divisible by 2,5 and 10 ?
Answer:
86540, 79980, 89960, 45570 and 76540.

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors

Do these: (TextBook Page No.65)

Question 1.
Circle the number which is exactly divisible by 3 and 9 and write correct reason.

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 4

Answer:

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 5

Reason : If the digital root of the given number is 9, then the number is exactly divisible by 3 and 9.

Question 2.
Write any 5 numbers which are exactly divisible by 3 and 9?
Answer:
1350, 1476, 0342, 1539 and 1629.

Do these: (TextBook Page No.66)

Question 1.
Circle the numbers which are divisible by 4. Give the reason, if is not divisible by 4.

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 47

Answer:

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 48

a, b, d, f and g are divisible by 4. Remaining numbers last two digits is not divisible by 4. So they are not divisible by 4.

Question 2.
Write the missing number in the blank to make the number exactly divisible by 4.
a) 323_
Answer:
323_ (2 / 6)

b) 304_
Answer:
304_ (0)

c) 58_6
Answer:
58_ 6 (1/3)

d) 53_ _
Answer:
53_ _ (04 / 0812 / 16 / 20 / 24 / 28 / 36/ 40)

e) 65_ _
Answer:
65_ _
(04 / 08 / 12 / 16 / 20 / 24 / 28/ 36 / 40).

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors

Do these (Textbook Page No. 67)

Question 1.
Check whether the following numbers are divisible by 6 or not.
1) 210
2) 162
3) 625
4) 120
5) 156
Answer:
210, 162, 120 and 156 are divisible by ’6’.

Question 2.
Change the digits of the following numbers to make them divisible by 6.
1) 543
2) 231
3) 5463
4) 1002
5) 4815
Answer:

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 6

Do these: (TextBook Page No.68)

Question 1.
Find the following numbers which are divisible by 8.
a) 2456
b) 3971
c) 824
d) 923
e) 2780
f) 93624
g) 76104
Answer:
a) In 2456, 456 is divisible by 8. So, 2456 is divisible by 8.
b) In 3971, 971 is not divisible by 8. So, 3971 is not divisible by 8.
c) 824 is divisible by 8.
d) 923 is not divisible by 8.
e) In 2780, 780 is not divisible by 8. So, 2780 is not divisible by 8.
f) In 93624, 624 is divisible by 8. So, 93624 is divisible by 8.
g) In 76104, 104 is divisible by 8. So, 76104 is divisible by 8.

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors

Exercise 2:

Question 1.
Circle the following numbers which are divisible by 2 (by using divisibility rule).
3624 3549 7864 8420 8500 8646 5007 7788
Answer:

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 49

Question 2.
Find out which of the following numbers are divisible by 6.
1276 43218 71218 71826 4734 3743
Answer:
(i) Given number = 1276
Ones place is 6 that is even.
1276 is divisible by 2.
Digital root of 1276 is 1 + 2 + 7 + 6 = 16
16 is not divisible by 3.
So, 1276 is not divisible by 6.

(ii) Given number = 43218
Ones place is 8 that is even.
43218 is divisible by 2.
Digital root of 43218 is 4 + 3 + 2 + 1 + 8 = 18
18 is divisible by 3.
So, 43218 is divisible by ‘6’.

(iii) Given number = 71218
Ones place is 8 that is even.
71218 is divisible by ‘2’
Digital root of 71218 is
7 + 1 + 2 + 1 + 8 = 19
19 is not divisible by ‘3’
So, 71218 is divisible by ‘6’.

(iv) Given number = 71826
Ones place is ‘6’ that is even.
71826 is divisible by ‘2’
Digital root of 71826 is 7 + 1 + 8 + 2 + 6 = 24
24 is divisible by ‘3’
So, 71826 is divisible by’6′.

(v) Given number = 4734
Ones place is ‘4’ that is even.
4734 is divisible by ‘2’
Digital root of 4734 is 4 + 7 + 3 + 4 = 18
18 is divisible by’3′
So, 4734 is divisible by ‘6’.

(vi) Given number = 3743
Ones place is ‘3’ that is odd.
3743 is not divisible by ‘2’
So, 3743 is not divisible by ‘6’.

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors

Question 3.
The number 50 AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 8 9 is exactly divisible by 9. Fill the AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 8 with the correct number.
Answer:
Given number 50 AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 8 19 is exactly divisible by 9.
So, digital root is 5 + 0 + ? + 1 + 9 = 15 + ?
= 15 + 3 = 18 = 1 + 8 = 9
∴ 9 is divisible by ‘9’.

Question 4.
The number 4 AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 8 468 is exactly divisible by 6. Fill the AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 8 with the correct number.
Answer:
Given number 4 AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 8 468 is exactly divisible by 6. So, digital roots is 4 + AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 8 + 4 + 6 + 8 = 22 + 2 = 24
24 is divisible by 3.
So, 4 2 468 is divisible by 6.

Question 5.
Fill the blanks with suitable digits. So that it can be divisible by 2 and 10.
678_ 588_ 388_ 222_
364_ 786_ 666_ 788_
Answer:
Given numbers are divisible by ’10’ they have ‘0’ in its units place and that number also divisible by ‘2’.
6780 5880 3880 2220
3640 7860 6660 7880

Question 6.
Find the numbers which are divisible by 4 and 8.
2104 726352 1800 32256 52248 25608
Answer:
Observe the last two digits of the given numbers 2104, 726352, 1800, 32256, 52248 and 25608.
There are 04, 52, 00, 56, 48 and 08 at the end of the numbers.
Hence all these numbers are multiples of 4.
So, 2104, 726352, 1800, 32256, 52248 and 25608 are divisible by 4.

Divisible by 8 :
i) Given number 2104
Last three digits of given 104 is divisible by 8.
So, 2104 is divisible by 8.

ii) Given number 726352
Last three digits of given 352 is di-visible by 8.
So, 726352 is divisible by 8.

iii) Given number 1800
Last three digits of given 800 is divisible by 8.
So, 1800 is divisible by 8.

iv) Given number 32256
Last three digits of given 256 is divisible by 8.
So, 32256 is divisible by 8.

v) Given number 52248
Last three digits of given 248 is divisible by 8.
So, 52248 is divisible by 8.

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors

Question 7.
Try whether the numbers are divisible by 2, 3, 4, 5, 6, 8, 9 and 10.
a) 333
b) 128
c) 225
d) 7535
e) 8289
f) 99483
g) 67704
h) 67713
i) 9410
j) 67722
k) 20704
l) 35932
m) 85446
n) 90990
o) 18540
Answer:
a) Given number is 333
Digital root of 333 is 3 + 3 + 3 = 9
Hence it is multiple of 3 and 9.
So, 333 is divisible by 3 and 9.

b) Given number is 128.
Ones place is 8 that is even.
128 is divisible by 2.
Last two digits of the given number is 28 is multiple of 4.
So, 128 is divisible by 4.
128 ÷ 8 = 16.
So, 128 is divisible by 8
Hence 128 is divisible by 2, 4 and 8.

c) Given number is 225
Ones place is ‘5’, so it is divisible by 5.

d) Given number is 7535
Ones place is 5, so, it is divisible by 5.

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors

e) Given number is 8289.
Digital root of 8289 is 8 + 2 + 8 + 9 = 27 = 2 + 7 = 9
9 is multiple of 3 and 9.
So, 8289 is divisible by 3 and 9.

f) Given number is 99483.
Digital root of 99483 is 9 + 9 + 4 + 8 + 3 = 33 = 3 + 3 = 6
6 is multiple of 3, so, 99483 is divisible by 3.

g) Given number is 67704
Ones place is 4 that is even.
67704 is divisible by 2.
Digital root of 67704 is 6 + 7 + 7 + 0 + 4 = 24 = 2 + 4 = 6.
6 is multiple of 3, so, 67704 is divisible by 3.
The number is divisible by 2 and 3, so, 67704 is divisible by 6.
Last two digits of the number is 04 is divisible by 4. So, 67704 is divisible by 4.
Last three digits of the number is 704 is divisible by 8.
So, 67704 is divisible by 8.
∴ 67704 is divisible 2, 3,4, 6 and 8.

h) Given number is 67713
Digital root of 67713 is 6 + 7 + 7 + 1 + 3 = 24 = 2 + 4 = 6
6 is multiple of 3.
So, 67713 is divisible of 3.

i) Given number is 9410
Last digit of given number is ‘0’ that is even.
So, 9410 is divisible by 2, 5 and 10.

j) Given number is 67722
Ones place is 2 that is even.
67722 is divisible by 2.
Digital root of 67722 is 6 + 7 + 7 + 2 + 2 = 24 = 2 + 4 = 6
‘6’ is divisible by 3.
So, 67722 is divisible by 6.

k) Given number is 20704.
Ones place is 4 that is even.
20704 is divisible by 2.
Last two digits of the number is 04 is divisible by 4.
So, 20704 is divisible by 4.
Last three digits of the number is 704 is divisible by 8. So, 20704 is divisible by 8.
∴ 20704 is divisible 2, 4 and 8

l) Given number is 35932
Ones place is 2 that is even.
35932 is divisible by 2.
Last two digits of the number is 32
32 is multiple of 4.
So, 35932 is divisible by 4.
∴ 35932 is divisible 2 and 4

m) Given number is 85446
Ones place is 6 that is even.
85446 is divisible by 2.
Digital root of 85446 is 8 + 5 + 4 + 4 + 6 = 27 = 2 + 7 = 9
∴ ‘9’ is multiple of 3 and 9.
So, 85446 is divisible by 3 and 9.
∴ 85446 is divisible by 2,3 and 9.

n) Given number is 90990
Last digit is ‘0’.
So, it is divisible by 2,5 and 10.

o) Given number is 18540 Last digit is ‘0’.
So it is divisible by 2, 5 and 10.

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors

Question 8.
Find the missing digit that would make each number divisible by the given number.
a) 395 _ by 10
Answer:
395 0

b)24305 _ by 9
Answer:
24305 4

c) 69839 _ by 3 and 9
Answer:
698391

d) 271 _ 8 by 6
Answer:
2710 8

e) 20710 _
Answer:
20710_

f) 5027 _ 5 by 3 and 5
Answer:
50273 5

g) 145 _ 2 by 8
Answer:
14512

h) 92048 _ by 2
Answer:
92048 _ (0/2/4/6/8)

i) 23405 _ by 5
Answer:
(0/5)

Answer: For eiven ‘e’ problem divisible by is not given. So, the problem is not solved.

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors

Question 9.
Find the smallest number that is to be added to 289279, so that it can be divisible by 8.
Answer:
Given number is 289279.
Last three digits of given number is 279.
279 is not divisible by 8, but 280 is divisible by 8.
289279 is added by 1 it is divisible by 8.

Do these: (TextBook Page No. 69 & 70)

Question 1.
Write first ten multiples of the following.
a) 3 b) 5 c) 8 d) 9 e) 10
Answer:
a) Ten multiples of 3 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30
b) Ten multiples of 5 = 5, 10, 15, 20, 25, 30, 35, 40, 45, 50
c) Ten multiples of 8 = 8, 16, 24, 32, 40, 48, 56, 64, 72, 80
d) Ten multiples of 9 = 9, 18, 27, 36, 45, 54, 63, 72, 81, 90
e) Ten multiples of 10 = 10, 20, 30, 40, 50, 60, 70, 80, 90, 100

Question 2.
Find out the multiples of 2, 3, 5 from 1 to 20. Write separately.
Answer:
First 1 to 20 multiples of 2 = 2, 4, 6, 8, 10, 12, 14, 16, 18, 20
First 1 to 20 multiples of 3 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30
First 1 to 20 multiples of 5 = 5, 10, 15,20

Question 3.
Write down the first 10 multiples of 7.
Answer:
First 10 multiples of 7 is = 7, 14, 21, 28, 35, 42, 49, 56, 63, 70.

Question 4.
Find out the multiples of 7, 8, 10 from the following numbers and write separately.
20, 14, 45, 24, 32, 35, 90, 8, 7, 10, 441, 385
Answer:
Multiples of 7 from given = 7, 14, 35, 385, 441
Multiples of 8 from given = 8, 24, 32
Multiples of 10 from given = 10, 20, 90

Question 5.
Find out the numbers which are not the multiples of 3.

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 9

Answer:
8 26 32 28 88 48

Question 6.
Write the odd multiples of 9 less than hundred.
Answer:
9, 27, 45, 63, 81, 99.

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors

Do these: (TextBook Page No.71)

Question 1.
Write first 10 multiples of the following numbers and list the com-mon multiples.
a) 2 and 4
b) 4 and 12
c) 6 and 8
d) 5 and 10
Answer:
a) Multiples of 2 : 2, 4, 6, 8, 10, 12, 14, 16, 18, 20.
Multiples of 4 : 4, 8, 12, 16, 20, 24, 28, 32, 36, 40
Common multiples of 2 and 4 : 4, 8, 12, 16, 20

b) Multiples of 4 : 4, 8, 12, 16, 20, 24, 28, 32, 36, 40
Multiples of 12 : 12, 24, 36, 48, 60, 72, 84, 96, 108, 120
Common multiples of 4 and 12 : 12, 24, 36

c) Multiples of 6 : 6, 12, 18, 24, 30, 36, 42, 48, 54, 60
Multiples of 8 : 8, 16, 24, 32, 40, 48, 56, 64, 72, 80
Common multiples of 6 and 8 : 24, 48

d) Multiples of 5 : 5, 10, 15, 20, 25, 30, 35, 40, 45, 50
Multiples of 10 : 10, 20, 30, 40, 50, 60, 70, 80, 90, 100
Common multiples of 5 and 10 : 10, 20, 30, 40, 50.

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors

Do this: (TextBook Page No. 72)

Find the LCM for the following sets of numbers.
Answer:
Multiples of 12 : 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, …………..
Multiples of 15: 15, 30, 45, 60, 75, 90, 105, 120, 135, 150, …………
Common multiples of 12 and 15 = 60, 120
Least common multiple of 12 and 15 = 60.

2) Multiples of 16: 16, 32, 48, 64, 80, 96, 112, 128, 144, 160
Multiples of 20 : 20, 40, 60, 80, 100, 120, 140, 160, 180, 200, ………..
Common multiples of 16 and 20 = 80, 160.
Least common multiple of 16 and 20 = 80

3) Multiples of 8:
8, 16, 24, 32, 40, 48, 56, 64, 72, 80, ……….., 120, ………..
Multiples of 12: 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, …………
Multiples of 20: 20, 40, 60, 80, 100, 120, 140, 160, 180, 200, …………
Least common multiple of 16 and 20 = 80
Common multiples of 8, 12 and 20 = 120
Least common multiple of 8, 12 and 20 = 120

4) Multiples of 15 : 15, 30, 45, 60, 75, 90, 105, 120, 135, 150 ……….
Multiples of 20 : 20, 40, 60, 80, 100, 120, 140, 160, 180, 200
Common multiples of 15 and 20 = 60, 120
Least common multiple of 15 and 20 = 60

5) Multiples of 6 : 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, …………
Multiples of 9: 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, ………..
Multiples of 12 : 12, 24, 36, 48, 60, 72, 84, 96, 108, 120
Common multiples of 6, 9 and 12 = 36
Least common multiple of 6, 9 and 12 = 36.

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors

Try this: (TextBook Page No. 72)

Find the LCM for the following pairs of numbers. What do you ob-serve.
1) 15, 30
2) 4, 16
3) 5, 15
4) 6, 18
Answer:
1) Multiples of 15 : 15, 30, 45.60. 75.90, 105, 120, 135,150
Multiples of 30 : 30,60, 90. 120, 150, 180, 210, 240, 270, 300, …………
Common multiples of 15 and 30
30, 60, 90, 120, 150, ……………..
Least common multiples of 15 and 30 = 30

2) Multiples of 4 :
4, 8, 12, 16, 20, 24, 28, 32, 36, 40, …………
Multiples of 16 : 16, 32, 48, 64, 80, 96, 112, 128, 144, 160, …………..
Common multiples of 4 and 16 = 16, 32, 48 …………
LCM of 4 and 16 = 16

3) Multiples of 5 : 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, …………..
Multiples of 15 : 15, 30, 45, 60, 75, 90, 105, 120, 135, 150, ………….
Common multiples of 5 and 15 = 15, 30, 45, 60
LCM of 5 and 15 = 15

4) Multiples of 6 :
6, 12, 18,24, 30, 36,42,48, 54, 60, ……………
Multiples of 18 :
18, 36, 54, 72,90, 108, 126, 144, 162, 180, ……….
Common multiples of 6 and 18 = 18, 36, 54
LCM of 6 and 18 = 18
Observation :
In a given pair of numbers, if one of them is multiple of other number than the biggest number is LCM of the number.

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors

Do these: (TextBook Page No.75)

Question 1.
Find all the factors of the following numbers.
a) 21
b) 38
c) 72
d) 96
Answer:
a) 21 = 1 × 21 = 3 × 7 = 7 × 3 = 21 × 1
Thus, all the factors of 21 are : 1, 3, 7 and 21.

b) 38 = 1 × 38
= 2 × 19
Thus, all the factors of 38 are : 1, 2, 19 and 38.

c) 72 = 1 × 72
= 2 × 36
= 3 × 24
= 4 × 18
= 6 × 12
= 8 × 9
Thus, all the factors of 72 are : 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 and 72.

d) 96 = 1 × 96
= 2 × 48
= 3 × 32
= 4 × 24
= 6 × 16
= 8 × 12
Thus, all the factors of 96 are : 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48 and 96.

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors

Question 2.
Find out whether the first number is a factor of the second number.
a) 14; 322
b) 26; 832
c) 35; 425
d) 56; 3500
e) 8; 48
f) 14; 37
g) 15; 75
h) 12; 72
Answer:
a) Yes, 14 is a factor of 322.
b) Yes, 26 is a factor of 832.
c) No, 35 is not a factor of 425.
d) No, 56 is not a factor of 3500.
e) Yes, 8 is a factor of 48.
0 No, 14 is not a factor of 37.
g) Yes, 15 is a factor of 75.
h) Yes, 12 is a factor of 75.

Question 3.
What are the factors of 66 ?
Answer:
Factors of 66 = 1 × 66
= 2 × 33
= 3 × 22
= 6 × 11
Factors are 1, 2, 3, 6, 11, 22, 33 and 66.

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors

Question 4.
Write all the even factors of 64.
Answer:
Factors of 64 = 1 × 64 = 2 × 32
= 4 × 16 = 8 × 8
Even Factors of 64 are : 2, 4, 8, 16, 32 and 64.

Question 5.
List out the numbers, which are prime / composite below 20.
Answer:

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 10

Fun activity:

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 11

Answer the following questions:

Question 1.
What are the prime numbers between 1 to 10 ?
Answer:
2, 3, 5, 7.

Question 2.
What are the prime numbers between 10 to 20 ?
Answer:
11, 13, 17, 19.

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors

Question 3.
What are the prime numbers between 20 to 50 ?
Answer:
23, 29, 31, 37, 41, 43, 47.

Question 4.
How many prime numbers are there between 1 to 50 ?
Answer:
15 prime numbers are there. (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 49).

Question 5.
What are the prime numbers between 50 to 100 ?
Answer:
53, 59, 61, 67, 71, 73, 79, 83, 87, 89, 97.

Question 6.
How many prime numbers are there between 50 to 100 ?
Answer:
10 prime numbers are there between 50 to 100.

Question 7.
Do you observe any speciality in these prime numbers ? What is it ?
Answer:
All numbers are odd, ‘2’ is only one even prime.

Question 8.
Are the all prime numbers even or odd ?
Answer:
All are odd except 2.

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors

Do these: (TextBook Page No.78)

a) 52
b) 100
c) 88
d) 96
e) 90
Answer:
a) AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 12

∴ Prime factorisation of 52 = 2 × 2 × 13

b) AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 13

∴ Prime factorisation of 100 = 2 × 2 × 5 × 5

c) AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 14

∴ Prime factorisation of 88 = 2 × 2 × 2 × 11

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors

d) AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 15

∴ Prime factorisation of 96 = 2 × 2 × 2 × 2 × 2 × 3

e) AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 16

∴ Prime factorisation of 90 = 2 × 3 × 3 × 5

Question 2.
The prime factorisation of 12 × 15 is
Answer:

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 17

∴ 12 × 15 = 2 × 2 × 3 × 3 × 5.

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors

Question 3.
Match the following:

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 18

Answer:

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 19

Question 4.
5 × 2 × 3 × 3 is the prime factorisation of _________
Answer:
90

Do this: (TextBook Page No.79)

Question 1.
Find the common factors of the following numbers and represent in the diagram.
a) 6 and 12
b) 12 and 20
c) 9 and 18
d) 11 and 22
Answer:
a) Factors of 6 = 1, 2, 3, 6
Factors of 12 = 1, 2, 3, 4, 6, 12

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 20

b) Factors of 12 = 1, 2, 3, 4, 6, 12
Factors of 20 = 1, 2, 4, 5, 10, 20

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 21

c) Factors of 9 = 1, 3, 9
Factors of 18 = 1, 2, 3, 6, 9, 18

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 22

d) Factors of 11 = 1, 11
Factors of 22 = 1,2, 11, 22

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 23

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors

Do this: (TextBook Page No.80)

Find the HCF of the following pairs of numbers by writing common factors.
1) 21 and 28
2) 34 and 20
3) 33 and 39
4) 16 and 36
5) 12 and 18
6) 80 and loo
Answer:
1) Factors of 21 = 1, 3, 7, 21
Factors of 28 = 1, 2, 4, 7, 14, 28
Common factors of 21 and 28 = 1, 7
HCF of 21 and 28 = 7.

2) Factors of 34 = 1, 2, 17, 34
Factors of 20 = 1, 2, 4, 5, 10, 20
Common factors of 34 and 20 = 1, 2
HCF of 34 and 20 = 1, 2.

3) Factors of 33 = 1, 3, 11, 33
Factors of 39 = 1, 3, 13, 39
Common factors of 33 and 39 = 1, 3
HCF of 33 and 39 = 3.

4) Factors of 16 = 1, 2, 4, 8, 16
Factors of 36 = 1, 2, 3, 4, 9, 12, 18, 36
Common factors of 16 and 36 = 1, 2, 4

5) Factors of 12 = 1, 2, 3, 4, 6, 12
Factors of 18 = 1, 2, 3, 6, 9, 18
Common factors of 12 and 18 = 1,2, 3, 6
HCF of 12 and 18 = 6

6) Factors of 80 = 1, 2, 4, 5, 8, 10, 16, 20, 40, 80
Factors of 100 = 1, 2, 4, 5, 10, 20, 25, 50, 100
Common factors of 80 and 100 = 1, 2, 4, 5, 10, 20
HCF of 80 and 100 = 20.

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors

Try this: (TextBook Page No.80)

Find the HCF for the following pair of numbers. What do you observe ?
1) 4, 16
2) 4, 12
3) 5, 15
4) 14, 42
Answer:
1) Factors of 4 = 1, 2, 4
Factors of 16 = 1,2, 4, 8, 16
Common factors of 4 and 16 = 1, 2, 4
HCF of 4 and 16 = 4.

2) Factors of 4 = 1, 2, 4
Factors of 12 = 1,2, 3, 4, 6, 12
Common factors of 4 and 12 = 1, 2, 4
HCF of 4 and 12 = 4.

3) Factors of 5 = 1, 5
Factors of 15 = 1, 3, 5, 15
Common factors of 5 and 15 = 1, 5
HCF of 5 and 15 = 5.

4) Factors of 14 = 1, 2, 7
Factors of 42 = 1, 2, 3, 6, 7, 14, 21, 22
Common factors of 14 and 42 = 1, 2, 7
HCF of 14 and 42 = 7

Observation :
In pair of numbers, if one of them is multiple of the other. The smallest number is the HCF of the pair of number.

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors

Do these: (TextBook Page 82)

Question 1.
Find LCM and HCF by prime factorisation method for the following.
a) 15, 48
b) 18, 42, 48
c) 15, 25, 30
d) 10, 15, 25
e) 15, 18, 36, 20
Answer:
a) Given numbers 15 and 48.

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 24

Prime factorisation of 15 = 1 × 3 × 5
Prime factorisation of 48 = 2 × 2 × 2 × 2 × 3
Common factors = 1 × 3
Other factors = 5 × 2 × 2 × 2 × 2
LCM = 1 × 3 × 5 × 2 × 2 × 2 × 2 = 240
HCF = 1 × 3 = 3.

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors

b) Given numbers 18, 42 and 48.

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 25

Prime factorisation of 18 = 1 × 2 × 3 × 3
Prime factorisation of 42 = 1 × 2 × 3 × 7
Common factors = 1 × 2 × 3
Other factors = 3 × 7 × 2 × 2 × 2
LCM = 1 × 2 × 3 × 3 × 7 × 2 × 2 × 2 = 1008
HCF = 1 × 2 × 3 = 6

c) Given numbers 15, 25 and 30

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 26

Prime factorisation of 15 = 1 × 3 × 5
Prime factorisation of 25 = 1 × 5 × 5
Prime factorisation of 30 = 1 × 2 × 3 × 5
Common factors = 1 × 5
Other factors = 3 × 3 × 5 × 2
LCM = 1 × 5 × 3 × 3 × 5 × 2 = 450
HCF = 1 × 5 = 5.

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors

d) Given numbers 10, 15 and 25

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 27

Prime factorisation of 10 = 1 × 2 × 5
Prime factorisation of 15 = 1 × 3 × 5
Prime factorisation of 25 = 1 × 5 × 5
Common factors = 1 × 5 = 5
Other factors = 2 × 3 × 5 = 30
LCM = 5 × 30 = 150
HCF = 5

e) Given numbers 15, 18, 36 and 20

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 28

Prime factorisation of 15 = 1 × 3 × 5
Prime factorisation of 18 = 1 × 2 × 3 × 3
Prime factorisation of 36 = 1 × 3 × 3 × 2 × 2
Prime factorisation of 20 = 1 × 2 × 2 × 5
Common factors = 1
Other factors = 3 × 5 × 2 × 3 × 3 × 3 × 2 × 2 × 2 × 5 = 194,400
LCM = 194, 400
HCF = 1.

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors

Question 2.
Find LCM and HCF by division method.
a) 16, 28, 36
b) 12, 18, 42
c) 30, 75, 90
d) 24, 32, 48
e) 12, 15, 18

Answer:
a) Given numbers 16, 28 and 36

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 29

LCM = 2 × 2 × 4 × 7 × 9 = 1008
HCF = 2 × 2 = 4

b) Given numbers 12, 18 and 42

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 30

LCM = 2 × 3 × 2 × 3 × 7 = 252
HCF = 2 × 3 = 6

c) Given numbers 30, 75, 90

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 31

LCM = 5 × 3 × 2 × 5 x 6
HCF = 5 × 3 = 15

d) Given numbers 24, 32, 48

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 32

LCM = 4 × 2 × 3 × 4 × 6 = 576
HCF = 4 × 2 = 8

e) Given numbers 12, 15 and 18

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 33

LCM = 3 × 2 × 2 × 5 × 6 = 360
HCF = 3

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors

Exercise 3:

Solve the following word-problems:

Question 1.
There are some fruits in a basket. If we arrange 4 or 6 or 8 or 10 fruits in a pile, no fruits are left in the bas¬ket. What is the minimum number of fruits in the basket ?

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 34

Answer:
Arrangement of fruits in a pile are 4 or 6 or 8 or 10
We need to find the LCM of 4, 6, 8, 10

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 35

LCM = 2 × 2 × 1 × 3 × 2 × 5 = 120
The minimum number of fruits in the basket = 120.

Question 2.
Ramu has 16 blue marbles and 12 white ones. If he wants to arrange them in identical groups without leaving any marbles, what is the maximum number in each group Ramu can make?

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 36

Answer:
Number of marbles Ramu has = 16 and 12
We need to find the HCF of 12 and 16

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 37

HCF = 4
Maximum number of marbles Ramu can make = 4.

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors

Question 3.
Two Neon lights are turned on at the same time. One blinks for every 4 seconds and other blinks for every 6 seconds. In 60 seconds how many times will they blink at a time ?

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 38

Answer:
Blink time of lights = 4 sec and 6 sec
LCM of 4 and 6 = AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 39
LCM = 2 × 2 × 3 = 12 sec
The lights blinks at a time for every 12 sec.
Given time = 60 sec
Number of times will the lights blink = 60 ÷ 12 = 5 times.

Question 4.
There are 40 girls and 32 boys, who want to participate in state-level games competition. If each team must have the same number of girls and the same number of boys.
i) What is the maximum number in each team that can participate in state-level games ?
ii) How many boys and girls will be on each team?

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 40

Answer:
Number of boys 32 and girls participate in the game is 32 and 40
i) To find maximum number in each team can participate in state-level game, we need to find
HCF of 32 and 40

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 41

∴ HCF = 2 × 2 × 2 = 8
The maximum number in each team = 8 numbers.

ii) To find out boys and girls will be an each team, we find LCM of 32 and 40.

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 42

∴ LCM of 32 and 40 = 2 × 2 × 2 × 4 × 5 = 160.

Question 5.
Find the least number of sheets of paper required to make notebooks containing 32 sheets or 40 sheets or 48 sheets without a single sheet leaving behind.

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 43

Answer:
To find the least number of sheets of paper required to make note books, we need to find LCM of 32, 40, 48.

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 44

LCM = 2 × 2 × 2 × 4 × 5 × 6 = 960
∴ 960 of sheets of paper required to make notebooks containing given sheets.

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors

Question 6.
What is the least number of chairs needed for an auditorium so that they can be arranged either 27 in a row or 33 in a row?

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 45

Answer:
To find least number of chairs needed for an auditorium are need to find LCM of 27 and 33.

AP Board 5th Class Maths Solutions 5th Lesson Multiples and Factors 46

LCM = 3 × 9 × 11 = 297.
∴ 297 chairs needed for an auditorium.

AP Board 10th Class Biology Notes Chapter 3 Transportation

Students can go through AP State Board 10th Class Biology Notes Chapter 3 Transportation to understand and remember the concept easily.

AP State Board Syllabus 10th Class Biology Notes Chapter 3 Transportation

→ All living organisms need nutrients, gases, liquids, etc. for the growth and maintenance of the body.

→ In lower organisms like Amoeba and Hydra, all the materials are transported through a simple process like diffusion, osmosis, etc.

→ In higher organisms developed a specialized system called circulatory system for the transport of materials in the body.

→ Our pulse rate is equal to our heartbeat rate.

→ In 1816 Rene Laennec discovered the stethoscope to hear a heartbeat.

→ The circulatory system consists of the heart, blood, and blood vessels.

→ The heart is the vital organ located in between the lungs and protected by the rib cage.

→ The size of our heart is approximately the size of our fist.

→ Pericardial membranes covering the heart protect it from mechanical shocks and injuries.

→ The heart is divided into four parts.

→ The upper two parts of the heart are called atria or auricles and the lower ones are called ventricles.

→ The coronary vessels supply blood to the muscles of the heart.

→ The walls of the ventricles are relatively thicker than atrial walls.

→ Arteries are the blood vessels that supply blood to various organs in the body.

→ The largest artery is the Aorta arises from the upper part of the left ventricle.

→ The pulmonary artery carries deoxygenated blood from the heart to the lungs. It arises from the right ventricle.

→ The superior vena cava collects blood from the anterior parts of the body and the inferior vena cava collects blood from the posterior parts of the body opens into the right atrium of the heart.

→ The two atria and the two ventricles are separated from each other by muscular partitions called septa.

AP Board Solutions AP Board 10th Class Biology Notes Chapter 3 Transportation

→ One-way walls that permit blood to flow in one direction were noticed in the veins of legs by Girolamo Fabrici (Ita.ly) in 1574.

→ William Harvey dissected the hearts of dead people and studied the valves between each atrium and its ventricle and noticed they were one-way valves.

→ Malpighi is called the smallest blood vessels which connect the smallest arteries and veins as capillaries.

→ Arteries carry oxygenated blood from the heart to body parts whereas veins carry deoxygenated blood from body parts to the heart.

→ Atrium and ventricle of the same side are connected by atrium ventricular aperture.

→ The human heart starts beating around the 21st day during embryonic development.

→ One contraction and one relaxation of atria and ventricles are called one cardiac cycle.

→ When the walls between the atria and ventricles are closed forcibly, we can listen to the first sharp sound of the heart ‘lubb’.

→ The valves which are present in the blood vessels are closed to prevent the backward flow of blood, we can listen to a dull sound of the heart ‘dubb’.

→ The cardiac cycle includes an active phase systole and a resting phase the diastole of atria and ventricles.

→ One cardiac cycle is completed in approximately 0.8 seconds.

→ The time needed for atrial contraction is 0.11 – 0.14 seconds.

→ The time needed for ventricular contraction is 0.27 – 0.35 seconds.

AP Board Solutions AP Board 10th Class Biology Notes Chapter 3 Transportation

→ If blood flows through the heart only once for completing one circulation it is called single circulation. eg: Fish.

→ If the blood flows through the heart twice for completing one circulation it is called double circulation. eg: Frog, Reptiles, Aves, and Mammals.

→ To supply nutrients to the cells, the liquid position of the blood with nutrients flows out of the capillaries. This is called tissue fluid.

→ The lymphatic system transports the tissue fluid into the main bloodstream.

→ Blood is a substance that contains solid and liquid particles whereas lymph is the substance that contains blood without solid particles.

→ In unicellular organisms like an amoeba, the protoplasm shows Brownian movements because the nutrients and oxygen are distributed throughout the protoplasm equally. Sponges use seawater for transportation by beating of flagella that are present in their body.

→ In cnidarians, saclike gastrovascular cavity digests the food and transports nutrients to each and every cell of the body. e.g: Hydra and Jellyfish.

→ In Platyhelminthes digestive system supplies food to all the cells directly and the excretory system collects wastes from each cell individually. e.g: Fasciola hepatica.

→ In nematyhelmenthes, the pseudocolor has taken up the function of collection and distribution of materials.

→ In Annelids the first acoelomate animals have developed a pulsative vessel to move the fluid and the transporting medium is blood.

→ The Arthropods have developed a pulsative organ heart to pump the blood.

→ The transport system that supplies nutrients to the tissues directly is an open type of circulatory system. e.g: Arthropods, many mollusks, and lower chordates.

→ In the closed type of circulatory system blood flows in blood vessels and then supplies nutrients to the body time. e.g: Cephalopod mollusks and octopus.

→ Doctors measure blood pressure with a device called a Sphygmomanometer. The normal blood pressure of a person is 120 / 80.

AP Board Solutions AP Board 10th Class Biology Notes Chapter 3 Transportation

→ People who have high B.P. during the resting period are said to have hypertension.

→ The straw yellowish-colored fluid portion after the formation of the blood clot is serum.

→ Vitamin K helps in the coagulation of blood.

→ Haemophilia is the result of a genetic disorder in which the blood may not coagulate. Thalassemia is an inherited disorder that is related to blood.

→ Osmosis plays a very important role in water absorption by root hairs.

→ Root pressure also plays a significant role in the absorption and movement of water in the xylem.

→ The evaporation of water through leaves is called transpiration.

→ Water also evaporates through the lenticels of the stem.

→ Fully grown maize plant transpires 15 liters per week.

→ The tissue that helps in the conduction of water is the xylem and the tissue that transports food in the phloem.

→ There is a relationship between transportation and transpiration in plants.

→ Mineral salts are necessary for plant nutrition and they are obtained from the soil solution through root hair.

→ Biologists studied food transportation in plants with the help of aphids.

→ Circulation: The movement of blood around the body.

→ Auricles: The upper chambers of the heart are called Auricles.

AP Board Solutions AP Board 10th Class Biology Notes Chapter 3 Transportation

→ Ventricles: The lower chambers of the heart are called Ventricles.

→ Pulse: The regular heartbeat while blood as it is sent around the body that can be felt in different places especially on the inner part of the wrist.

→ Artery: ‘the blood vessel that carries oxygenated blood to all body parts except pulmonary artery.

→ Vein: The blood vessel that collects deoxygenated blood except for the pulmonary vein.

→ Stethoscope: Instrument used to hear the sounds of heartbeat and sounds of Lungs.

→ Aorta: The main artery of the heart supplying oxygenated blood to all body parts. It arises from the left ventricle of the heart.

→ Capillary: It is a very fine blood vessel that connects the smallest arteries and veins.

→ Systole: The contraction phase of the heart is known as systole.

→ Diastole: The relaxation phase of the heart is called diastole.

→ Cardiac Cycle: One cardiac cycle includes an active phase systole and resting phase diastole of atria (auricles) and ventricles. The time required for completion of one cardiac cycle is approximately 0.8 seconds.

→ Blood Pressure (B.P.): The pressure with which the blood flows in the blood vessels is called blood pressure. The normal B.R of a person is 120/80.

→ Lymph: Lymph is the substance that contains blood without solid particles. A clear liquid containing white blood cells helps to clear the tissues of the body.

→ Single circulation: If the blood flows through the heart only once for completing one circulation, it is called single circulation. e.g: Fish.

→ Double circulation: If the blood flows through the heart twice for completing one circulation is called double circulation. e.g: Frog, Reptiles, Aves, mammals.

→ Coagulation of blood: It is clotting of blood when the injury occurs. Blood platelets in the blood play an important role in the coagulation of blood and prevent loss of blood.

→ Sphygmomaiìoineter: It is an instrument used by doctors to measure blood pressure.

AP Board Solutions AP Board 10th Class Biology Notes Chapter 3 Transportation

→ Prothrombin: A protein present in blood plasma that is converted into active thrombin during coagulation.

→ Thrombin: An enzyme in blood plasma that causes the clotting of blood by converting fibrinogen to fibrin.

→ Fibrinogen: A blood protein that helps in the clotting of blood when the vessels are injured.

→ Fibrin: It is an insoluble protein formed from fibrinogen during the clotting of blood. It forms a fibrous mesh that obstructs the flow of blood.

→ Root hair: A thin hair-like outgrowth of an epidermal cell f a plant root that absorbs water and minerals from the sou.

→ Radicle: it is the first part of a seedling to emerge from the seed during the process of germination. it develops into a root,

→ Root pressure: Root pressure is osmotic pressure within the cells of a root system that causes sap to rise through ‘ plant stem to the leaves.

→ Plant nutrIents: These are the elements required for the normal growth of the plants. Plants absorb most of these nutrients from the soil through roots.

→ Xylem: It Is the water-conducting tissue in plants comprising b0th living and nonliving tissue.

→ Phloem: It is the food transporting tissue in plants consisting only of living tissue.

→ Vascular bundles: It is part of the transport system in vascular plants. It is a strand of conducting vessels in the stem or leaves of a plant typically with phloem on the outside and xylem on the inside.

→ Right atrIum: The upper right chamber of the heart receives deoxygenated blood through cava veins.

→ Left atrIum: The upper left chamber of the heart receives oxygenated blood from the lungs through the pulmonary vein.

→ Right ventrIcle: The lower right chamber of the heart receives deoxygenated blood from the right atrium. Sends deoxygenated blood to lungs through the pulmonary artery.

→ Left ventrIcle: The lower left chamber of the heart from which the main Aorta that supplies oxygenated blood to aH body parts except for lungs.

AP Board Solutions AP Board 10th Class Biology Notes Chapter 3 Transportation

→ Butcher: A person whose trade is cutting up and selling meat ¡n a shop.

→ Stump: Apart, as of a branch, limb, or tooth. remaining after the main part has been cut away. broken off nr worn clown.

→ Pericardial membrane: The membrane that surrounds the heart and protects it from mechanical shocks.

→ Coronary vessels: The vessels that deliver oxygen-rich blood to the muscles of the heart.

→ Superior vena cava: Receives deoxygenated blood from the head and arms and chest and opens into the right atrium of the heart.

→ Inferior vena cava (or) Post caval vein: It Is the large vein that carries deoxygenated blood from the lower half of the body into the right atrium of the heart.

→ Dangling: Hanging or swinging loosely.

→ Spurts: A sudden strong flow of a liquid, a sudden increase in something.

→ Brownian movement: The erratic random movement of microscopic particles in a fluid, as a result of continuous bombardment from molecules of the surrounding medium.

→ Parazone: i) Any multicellular invertebrate of the group Parazoa. which consists of sponges.
ii) A primitive multicellular marine animal whose porous body is supported by a fibrous skeletal framework.

→ Cnidarians: An aquatic invertebrate animal of the phylum cnidaria, which comprises coelenterates.

→ Gastrovascular Cavity: Functions ¡n both digestion and distribution of nutrients and particles to all parts of the body.

→ Nematyhelmenthes: A phylum including the nematodes and worms and sometimes the acanthocephalans, rotifers, gastrotrichs and marine, organisms.

AP Board Solutions AP Board 10th Class Biology Notes Chapter 3 Transportation

→ Pulsative organ: A minute muscular organ functioning as an accessory heart In various insects.

→ Lymphatic system: The network of vessels through which lymph drains from the tissues into the blood.

→ Lymphatic circulation: it is the process by which the lymphatic system circulates a clear fluid called lymph throughout the body.

→ Open type of circulatory system: Blood vessels are absent. Blood supplies nutrients directly to the tissues. e.g: Arthropods, many mollusks.

→ Closed type of circulatory system: Blood supplies nutrients and flows in the blood vessels. e.g: Cephalopod mollusks, higher animals (e.g: Octopus).

→ Edema: A condition characterized by an excess of watery fluid collecting In the cavities or tissues of the body.

→ Systolic pressure: The blood pressure during the contraction of the left ventricle of the heart.

→ Diastolic pressure: The blood pressure after the contraction of the heart while the chambers of the heart that refill with blood.

→ Thromboklnase: An enzyme liberated from blood platelets that convert prothrombin into thrombin as blood starts to clot.

→ Entangle: Cause to become twisted together with or caught in.

→ Serum: A protein-rich liquid that separates out when blood coagulates.

→ Squash: Crush or squeeze with a force so that it becomes flat, soft, or out of shape.

→ Tensile strength: The resistance of a material to breaking under tension.

→ Oak: It ¡s a tree or shrub ¡n the genus Quercus belongs to the family Fagaceae. Oakwood has great strength and hardness and is very resistant to insect and fungal attacks because of its high tannin content.

→ Aphids: A small bug that feeds by sucking sap from plants.

→ Cambium: A cellular plant tissue from which phloem, xylem, or cork grows by division resulting in secondary thickening.

→ Hemophilia: It Is a disease condition in which coagulation of blood may not occur due to genetic defect.

→ Badger: A heavily built omnivorous nocturnal mammal of the weasel family, typically having a grey and black coat.

→ Sycamore: It ¡s a name that is applied at various times and places to several different types of trees, but with somewhat similar leaf forms.

→ Gnaw: Bite at or nibble something persistently.
AP Board 10th Class Biology Notes Chapter 3 Transportation 1
AP Board 10th Class Biology Notes Chapter 3 Transportation 2
→ William Harvey (1578-1657) – Father Of Cardio Vascular Medicine:

  • William Harvey was an English Physician born in 1578.
  • He was the first scientist to describe completely and ¡n detail the systematic circulation and properties of blood being pumped to the brain and body by the heart.
  • He wrote a book on the circulation of blood called “De Motu Cordis”.
  • He concluded that blood flows through the arteries and comes back to the heart by veins.
  • Harvey dissected the hearts of dead people and studied valves between each atrium and its ventricles.

AP Board 3rd Class EVS Solutions 3rd Lesson Animals Around Us

Andhra Pradesh AP Board 3rd Class EVS Solutions 3rd Lesson Animals Around Us Textbook Exercise Questions and Answers.

AP State Syllabus 3rd Class EVS Solutions Lesson 3 Animals Around Us

I. Conceptual Understanding:

Question 1.
Tell and write the names of five pet animals.
Answer:
Dog, Cow, Ox, Sheep, Goat, Buffalo, Horse, Cat etc., are pet animals.

Question 2.
Write any three differences between animals and birds ?
Answer:
Difference between animals and birds.

AnimalsBirds
1. Have four legs.1. Have two legs.
2. Don’t have wings.2. Have wings.
3. They have bones filled with bone marrow3. They have hollow bones

AP Board 3rd Class EVS Solutions 3rd Lesson Animals Around Us

Question 3.
How do animals move from one place to another ?
Answer:
Some animals walk, some crawl, some jump and some animals swim. Birds fly from one place to another place.

Question 4.
How can we help the animals and birds around us ?
Answer:

  1. Put bowls of water for stray animals, especially during summer.
  2. Avoid deforestration.
  3. Feed the birds with grains like Rice, Bajra, Channa etc.
  4. We should protect birds and animals.
  5. We should not disturb nests, eggs of birds.

AP Board 3rd Class EVS Solutions 3rd Lesson Animals Around Us

II. Questioning and Hypothesis:

Question 5.
Say who I am ?

A) I am long and shiny. I have no legs and no ears. I crawl and live in an ant hill. Who am I ? Who am I ?
Answer:
Snake

B) I live in water, I never sleep. I breathe with gills. Who am I ? Who am I ?
Answer:
Fish.

C) I have four legs. I give milk. I eat leaves. Who am I ? Who am I ?
Answer:
I give milk.

D) I have wings. I fly high in the sky. I can see smaller things on the ground. Who am I?
Answer:
Bird.

AP Board 3rd Class EVS Solutions 3rd Lesson Animals Around Us

III. Experiments & Field Observations:

Question 6.
Some birds can fly, some birds cannot. Observe birds in your surround-ings and categorize them under the proper headings ?

AP Board 3rd Class EVS Solutions 3rd Lesson Animals Around Us 1

Answer:

S.No.Can flyCan not fly
1.CrowHen
2.PigeonPeacock
3.SparrowPenguin
4.ParrotOstritch

AP Board 3rd Class EVS Solutions 3rd Lesson Animals Around Us

IV. Information Skills & Project Work:

Question 7.
Collect or draw the pictures of two animals of each category and paste them in the given space.

AP Board 3rd Class EVS Solutions 3rd Lesson Animals Around Us 2

Answer:
Student Activity.

V. Drawing Pictures and Model Making:

Question 8.
Colour the pictures given below.

AP Board 3rd Class EVS Solutions 3rd Lesson Animals Around Us 3

Answer:
Student Activity.

VI. Appreciation, values and creating awareness towards bio-diversity:

Question 9.
You know that mosquitoes harm us. Write three preventive measures to be taken to keep mosquitoes away.
Answer:
Preventive measure to keep away mosquitoes.

  1. Dump out any standing water near our home.
  2. Use mosquito nets.
  3. Wear light coloured clothing.
  4. Use mosquito repellent.

AP Board 3rd Class EVS Solutions 3rd Lesson Animals Around Us

Additional Questions:

I. Conceptual Understanding:

Question 1.
How do the domestic and pet animals help us ?
Answer:
Dogs, Cats, Cows, goats, Buffaloes, Ducks, Hen are same examples of domestic animals.

Uses of domestic animals :

  1. Dogs guard our house.
  2. Cats catch mice.
  3. Cows, goats give us milk.
  4. Oxen and buffaloes help farmers in farming.
  5. Hens and ducks lay eggs.
  6. Goats, sheep & hen give us meat.
  7. Horses and Donkey carry loads.

Question 2.
What are wild animals ? Give examples ?
Answer:
Some animals live only in forest without being introduced by humans. They are called wild animals.
Ex : Lion, Tiger, bear and elephant.

Question 3.
What are terrestrial animals ?
Answer:
The animals that live on land are called ‘Terrestrial animals’.
Ex : Cow, Dog, Cat, Hen etc.,

Question 4.
What are aquatic animals ?
Answer:
The animals that live in water are called “Aquatic animals.
Ex : Fishes, Octopus, Star fish, Dolphin etc.

AP Board 3rd Class EVS Solutions 3rd Lesson Animals Around Us

Question 5.
What are amphibians ? Which helps them ?
Answer:
The animals that live both on land and in water are called ‘Amphibians’.
Their moist skin, webbed feet, strong hind limbs help them to live on land and in water.
Ex: Frog, Salamander.

Question 6.
What are herbivores ? Give examples ?
Answer:
Some animals eat only grass and plant products. They are called herbivores.
Ex : Cows, Bullocks, Donkeys, Horses, Elephants, Deer, Goat etc.,

Question 7.
What are carnivores ? Give examples ?
Answer:
Flesh (animal) eating animals are called ‘Carnivores’. Ex : tiger, Lions, Foxes, Crocodiles etc.,

Question 8.
What are Omnivores ? Give examples ?
Answer:
Some animals eat both grass and flesh (animals.) They are Omnivores.
Ex : Human beings, Bears, Crows, Monkeys, Dogs etc.,

AP Board 3rd Class EVS Solutions 3rd Lesson Animals Around Us

II. Questioning and Hypothesis:

Question 9.
I suck animals blood. Who am I?
Answer:
Mosquito.

Question 10.
I suck nectar from flowers. Who am 1?
Answer:
Butterflies and honey bees.

Question 11.
I consume dead animals. I am called ‘Scavenger’, Who am I?
Answer:
Vultures, Crows and Foxes.

AP Board 3rd Class EVS Solutions 3rd Lesson Animals Around Us

III. Experiments & Field Observations:

Observe the picture of the agricultural farm.

AP Board 3rd Class EVS Solutions 3rd Lesson Animals Around Us 4

Question 1.
Name the animals you see in the picture ?
Answer:
Buffaloes, Monkey, Squirrel, Salamander, Crane, Crow, Parrot, Dog, Snake.

Question 2.
What animals you see in your surroundings.
Answer:
I can see all these animals in my surroundings.

AP Board 3rd Class EVS Solutions 3rd Lesson Animals Around Us

IV. Information Skills & Project Work:

Question 1.
Look at the picture. mark D for domestic animals and W for wild animals.

AP Board 3rd Class EVS Solutions 3rd Lesson Animals Around Us 5

Answer:
Student activity.

Question 2.
Animals move in different ways to go from one place to another observe the picture write how do they move about ?

AP Board 3rd Class EVS Solutions 3rd Lesson Animals Around Us 6

1. _______ and _______ can fly
2. _______ and _______ can crawl
3. _______ and _______ can walk
4. _______ and _______ can hop
5. _______ and _______ can swim
6. _______ and _______ can jump.
Answer:
1. Pigeon and crow can fly.
2. Lizard and Snake can crawl.
3. Cat and dog can walk.
4. Kangaroo and frog can hop.
5. Fish and Octopus can swim.
6. Rabbit and Monkey can jump.

AP Board 3rd Class EVS Solutions 3rd Lesson Animals Around Us

Activity 3:

Question 3.
Observe the picture, tell and write the names of the animals and their living places.

AP Board 3rd Class EVS Solutions 3rd Lesson Animals Around Us 7

Answer:
Student activity.

Activity 4:

Question 4.
Write the names of animals according to the places they live.

AP Board 3rd Class EVS Solutions 3rd Lesson Animals Around Us 8

Answer:
Student Activity.

AP Board 3rd Class EVS Solutions 3rd Lesson Animals Around Us

V. Matching :

Question 1.
Match the following animals with their living places.

AP Board 3rd Class EVS Solutions 3rd Lesson Animals Around Us 9

Answer:
1. i
2. c
3. d
4. e
5. g
6. f
7. a
8. b
9. h

Question 2.
Match the following animals with their pray.

AP Board 3rd Class EVS Solutions 3rd Lesson Animals Around Us 10

Answer:

AP Board 3rd Class EVS Solutions 3rd Lesson Animals Around Us 11

AP Board 3rd Class EVS Solutions 3rd Lesson Animals Around Us

Question 3.
Match the following animals with their sounds.
mooing, coos, squeak, chirp, cowl, meows, roars, Brays, neighs, grunts, barks.
Answer:

  1. Calf – mooing
  2. Cuckoo – coos
  3. Lizard – squeaking
  4. Cricket – chirping
  5. Dog – barks.
  6. Cat – meows
  7. Crow – caws
  8. tiger – roars
  9. Horse – neighs
  10. Pig – grunts
  11. Donkey – brays
  12. Goat – coos

Question 4.
If you find a baby bird fell down from the nest. What will you do?
Answer:
If I find a baby bird fell down from the nest, I will run and pick up the baby bird, and place it carefully in the nest.

Multiple Choice Questions:

Question 1.
Cow baby is _______
a) Cuckoo
b) lamb
c) calf
d) puppy
Answer:
c) calf

Question 2.
The sound of cricket is _______
a) squeaking
b) chirping
c) mooing
d) cool
Answer:
b) chirping

Question 3.
A hen is kept in _______
a) chicken coap
b) stable
c) sty
d) kennel
Answer:
a) chicken coap

AP Board 3rd Class EVS Solutions 3rd Lesson Animals Around Us

Question 4.
Vultures, crows and foxes consume dead animals. They are called _______
a) decomposers
b) scavengers
c) herbivores
d) Omnivores
Answer:
b) scavengers

Question 5.
Frog, Salamander comes under _______
a) aquatic animals
b) terrestrial animals
c) amphibians
d) None
Answer:
c) amphibians

Question 6.
Animals that live on land are called _______
a) Terrestrial Animals
b) Aquatic animals
c) Amphibians
d) None
Answer:
a) Terrestrial Animals

AP Board 3rd Class EVS Solutions 3rd Lesson Animals Around Us

Question 7.
We should be _______ of animals and birds.
a) sad
b) kind
c) bad
d) None
Answer:
b) kind

Question 8.
_______ suck nectar from flowers.
a) Butterflies
b) Mosquitoes
c) Ants
d) None
Answer:
a) Butterflies

Question 9.
_______ suck the blood from animals.
a) Flies
b) Mosquitoes
c) Ants
d) None
Answer:
b) Mosquitoes

AP Board 3rd Class EVS Solutions 3rd Lesson Animals Around Us

Question 10.
_______ is the dog’s shelter.
a) Stable
b) nest
c) kennel
d) None
Answer:
c) kennel

AP Board 9th Class Maths Solutions Chapter 7 Triangles Ex 7.2

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 7 Triangles Ex 7.2 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 7th Lesson Triangles Exercise 7.2

Question 1.
In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at ‘O’. Join A to O. Show that (i) OB = OC (ii) AO bisects ∠A.
AP Board 9th Class Maths Solutions Chapter 7 Triangles Ex 7.2 1
Solution:
Given that in ΔABC
AB = AC
Bisectors of ∠B and ∠C meet at ‘O’.
To prove
i) OB = OC
∠B = ∠C (Angles opposite to equal, sides)
\(\frac{1}{2} \angle \mathrm{B}=\frac{1}{2} \angle \mathrm{C}\) (Dividing both sides by 2)
∠OBC = ∠OCB
⇒ OB = OC (∵ Sides opposite to equal angles in ΔOBC)

ii) AO bisects ∠A.
In ΔAOB and ΔAOC
AB = AC (given)
BO = CO (already proved)
∠ABO = ∠ACO (∵ ∠B =∠C)
∴ ΔAOB ≅ ΔAOC
⇒ ∠BAO = ∠CAO [ ∵ CPCT of ΔAOB and ΔAOC]
∴ AO is bisector of ∠A.

AP Board 9th Class Maths Solutions Chapter 7 Triangles Ex 7.2

Question 2.
In ΔABC, AD is the perpendicular bisector of BC (see given figure). Show that ΔABC is an isosceles triangle in which AB = AC
AP Board 9th Class Maths Solutions Chapter 7 Triangles Ex 7.2 2
Solution:
Given that AD ⊥ BC; AD = DC
In ΔABD and ΔACD
AD = AD (common)
BD = DC (given)
∠ADB = ∠ADC (given)
∴ ΔABD ≅ ΔACD (∵ SAS congruence)
⇒ AB = AC (CPCT of ΔABD and ΔACD)

AP Board 9th Class Maths Solutions Chapter 7 Triangles Ex 7.2

Question 3.
ABC is an isosceles triangle in which altitudes BD and CE are drawn to equal sides AC and AB respectively (see figure). Show that these altitudes are equal.
AP Board 9th Class Maths Solutions Chapter 7 Triangles Ex 7.2 3
Solution:
Given that AC = AB; BD ⊥ AC; CE ⊥ AB
In ΔBCD and ΔCBE
∠BDC = ∠CEB (90° each)
∠BCD = ∠CBE (∵ angles opp. to equal sides of a triangle)
BC = BC
∴ ΔBCD ≅ ΔCBE (∵ AAScongruence)
⇒ BD = CE (CPCT)

AP Board 9th Class Maths Solutions Chapter 7 Triangles Ex 7.2

Question 4.
ABC is a triangle in which altitudes BD and CE to sides AC and AB are equal (see figure). Show that
i) ΔABD ≅ ΔACE
ii) AB = AC i.e., ABC is an isosceles triangle.
AP Board 9th Class Maths Solutions Chapter 7 Triangles Ex 7.2 4
Solution:
Given that BD ⊥ AC; CE ⊥ AC
BD = CE
Now in ΔABD and ΔACE
∠ADB = ∠AEC (∵ given 90°)
∠A = ∠A (commori angle)
BD = CE
∴ ΔABD = ΔACE (∵ AAS congruence)
⇒ AB = AC (∵ C.P.C.T)

AP Board 9th Class Maths Solutions Chapter 7 Triangles Ex 7.2

Question 5.
ΔABC and ΔDBC are two isosceles triangles on the same base BC (see figure). Show that ∠ABD = ∠ACD.
AP Board 9th Class Maths Solutions Chapter 7 Triangles Ex 7.2 5
Solution:
AP Board 9th Class Maths Solutions Chapter 7 Triangles Ex 7.2 6
Given that ΔABC and ΔDBC are isosceles.
To prove ∠ABD = ∠ACD
Join A and D.
Now in ΔABD and ACD
AB = AC (∵ equal sides of isosceles triangles)
BD = CD (∵ equal sides of isosceles triangles)
AD = AD (∵ common side)
∴ ΔABD ≅ ΔACD (∵ SSS congruence)
⇒ ∠ABD = ∠ACD (CPCT)

AP Board 5th Class EVS Solutions 3rd Lesson Clothes We Wear

Andhra Pradesh AP Board 5th Class EVS Solutions 3rd Lesson Clothes We Wear Textbook Exercise Questions and Answers.

AP State Syllabus 5th Class EVS Solutions Chapter 3 Lesson We Wear

I. Conceptual understanding:

Question 1.
Write a brief note on the uses of air?
Answer:
Uses of air :

  1. Air is the most primary need to live without air. We can’t live even for minutes of time.
  2. Electric power can be generated by air through wind mills.
  3. Through wind mills water can be lifted from wells for cultivation.
  4. Air is also used for drying clothes, smelling flowers, drink juice with a straw, to fly a kite, to ride a bicycle to play a flute etc.

Question 2.
Name the kind of clothes we wear in different seasons?
Answer:
We feel convenient with different kinds of clothes in different seasons.
Ex: During Summer we wear cotton clothes.
During winter season we wear woolen clothes.
During rainy season we wear rain coats which are made up of water proof material.

Question 3.
Write differences between natural and artificial fibres?
Answer:

Natural fibreArtificial fibre
I. It is obtained from nature.1. It is man made.
2. The fibre structures cannot be changed.2. Fibre structures can be changed.
3. These are comfortable3. Not comfortable.
4. No chemicals are used4. Chemical substances are needed for processing
5. Environmental friendly5. Some fabrics are harmful to the environment.
6. Their life time is less6. They last longer.

AP Board 5th Class EVS Solutions 3rd Lesson Clothes We Wear

II. Questioning and Hypothesis:

Question 4.
Name the reasons why people use woollen clothes in the winter season?
Answer:
We wear woollen clothes in winter because air is a poor conductor of heat and woollen clothes do not allow the transfer of body heat to the environment and keep us warm. The wool is structured so, there is air trapped inside.

III. Experiments and field observations:

Question 5.
Wash your dress with a detergent soap and write your experiences?
Answer:
My experiences when I wash my dress with a detergent soap:

  1. If my new dress is cotton, when I wash it with a detergent soap it looses its colour for one or two washes.
  2. If my dress is silk I will dry it in a shady place.Otherwise they loose their colour.

IV. Information Skills & Project Work:

Question 6.
Collect pieces of cloth and classify them into natural fabrics and artificial fabrics and paste them on the chart. Display the chart.
Answer:
Student activity.

AP Board 5th Class EVS Solutions 3rd Lesson Clothes We Wear

V. Drawing Pictures and Model Making:

Question 7.
Draw the picture of different types of dresses.
Answer:
Student activity.

VII. Appreciation:

Question 8.
Say what you like at tailor’s shop?
Answer:
At tailors shop I will find different types of dresses in different colours and different models which I like most.There I can also find work sarees with different colours and designs. Cotton shirts will also be colourful.

AP Board 5th Class EVS Solutions 3rd Lesson Clothes We Wear

Additional Questions:

Question 1.
What are natural fibres? Give examples?
Answer:
Fibres that are obtained from nature i.e either from plants or animlas are called ” natural fibres”.
Eg : Cotton, jute are plant fibres, silk, wool are animals fibres.

Question 2.
What are artificial fibres? Give examples?
Answer:
Man made fibres made in factories by using chemical process are called ” Synthetic fibres”.
Eg: Polyester, terylene, rayon, and nylon.

Question 3.
Why do people wear different types of clothes in different areas and ages?
Answer:
People wear different types of clothes in different areas and different ages. It depends upon the material, comfort climate etc., People in cold countries use warm clothes made of wool. In hot counties use clothes made of cotton. Small kids use soft textured clothes.

AP Board 5th Class EVS Solutions 3rd Lesson Clothes We Wear

Question 4.
How do clothes protect us?
Answer:
Clothes protect us from heat,cold,rain. They absorb sweat and keep us cool. They proctect us from germs, insect bites, dust and pollution. They help us to look smart.

Question 5.
Why should we wash our clothes?
Answer:
We should wash our clothes regularly. While playing, clothes may get dirty and wet with sweat. If we wear clothes without washing properly, we may suffer from skin diseases. Clothes should be washed with a detergent soap and dry under the sun.

Question 6.
What are the properties of air ?
Answer:

  1. Air has weight.
  2. Air occupies space.
  3. Air exerts pressure.
  4. Air produces and carries sound and smell.

AP Board 5th Class EVS Solutions 3rd Lesson Clothes We Wear

II. Questioning and Hypothesis:

Question 7.
Observe these figure and put (✓) mark where air occupies space.

AP Board 5th Class EVS Solutions 3rd Lesson Clothes We Wear 1

Answer:

AP Board 5th Class EVS Solutions 3rd Lesson Clothes We Wear 2

Question 8.
Write what you have observed from these experiments?
Answer:
(1) AP Board 5th Class EVS Solutions 3rd Lesson Clothes We Wear 3 Air has weight.

(2) AP Board 5th Class EVS Solutions 3rd Lesson Clothes We Wear 4 Air produces and carries sound.

(3) AP Board 5th Class EVS Solutions 3rd Lesson Clothes We Wear 5 Air occupies space.

(4) AP Board 5th Class EVS Solutions 3rd Lesson Clothes We Wear 6 Air exerts pressure.

AP Board 5th Class EVS Solutions 3rd Lesson Clothes We Wear

III. Information Skills & Project Work:

Question 9.
People wear uniform that suits their profession. Let us use the following pictures and identify their profession?

AP Board 5th Class EVS Solutions 3rd Lesson Clothes We Wear 7

Answer:

Student Activity.

Multiple Choice Questions:

Choose the correct answer:

Question 1.
Cotton clothes are obtained from ________ plant.
A) cotton
B) flax
C) coconut
D) jute
Answer:
A) cotton

Question 2.
Which of the following are natural fibres _______ .
A) cotton
B) wool
C) silk
D) jute
E) all the above
Answer:
E) all the above

Question 3.
Silk is obtained from _______ .
A) silk worm
B) sheep
C) cotton plant
D) none
Answer:
A) silk worm

AP Board 5th Class EVS Solutions 3rd Lesson Clothes We Wear

Question 4.
Silk worm feeds on the _______.
A) cotton leaves
B) mulberry leaves
C) both
D) none.
Answer:
B) mulberry leaves

Question 5.
Rain coats & umbrella material comes under ________ fibres.
A) natural
B) artificial
C) both
D) none
Answer:
B) artificial

Question 6.
Sweaters are made up of ________ .
A) silk
B) cotton
C) wool
D) none
Answer:
C) wool

Question 7.
We get wool from ________
A) silkworms
B) jute plant
C) sheep
D) none
Answer:
C) sheep

AP Board 5th Class EVS Solutions 3rd Lesson Clothes We Wear

Question 8.
We wear _______ clothes in winter.
A) woollen
B) cotton
C) silk
D) none
Answer:
A) woollen

Question 9.
We wear _______ clothes in summer.
A) woollen
B) cotton
C) silk
D) none
Answer:
B) cotton

Question 10.
_______ plant is used to make linen.
A) Jute
B) Cotton
C) Flax
D) None
Answer:
C) Flax

Question 11.
Gunny bags, ropes are made from _______
A) cotton
B) jute
C) wool
D)none
Answer:
B) jute

AP Board 5th Class EVS Solutions 3rd Lesson Clothes We Wear

Question 12.
We should not _______ air.
A) breath
B) use
C) pollute
D) none
Answer:
C) pollute

AP Board 10th Class Social Studies Notes Chapter 4 Climate of India

Students can go through AP State Board 10th Class Social Studies Notes Chapter 4 Climate of India to understand and remember the concept easily.

AP State Board Syllabus 10th Class Social Studies Notes Chapter 4 Climate of India

→ What have been the general conditions, year after year over thirty years or more gives us the climate,

→ The elements of weather and climate are temperature, atmospheric pressure, wind, humidity, and precipitation,

→ Climograph shows average monthly values of maximum temperature, minimum temperature, and rainfall for a given place.

→ The factors that affect climate are called climatic controls. These include

  • Latitude
  • Land-water relationship
  • Relief and
  • Upper air circulation.

→ The average temperatures for the year drop as you go further away from the Equator.

→ Darker areas such as heavily vegetated regions tend to be good absorbers; lighter areas, such as snow and ice-covered regions, tend to be good reflectors.

→ The difference between the temperature of day and night and that of summer and winter is not much. This is known as ‘equable climate’.

→ Hills and mountains will have lower temperatures than locations on the plain.

→ The climate of India is affected by the movement of upper air currents in the atmosphere above 12,000 m called ‘Jet Streams’.

→ The temperature in the Indian landmass considerably reduces from Mid-November and this cold season continues till February.

→ During the hot season, as we move from the southern to the northern part of the country, the average temperature increases.

AP Board Solutions AP Board 10th Class Social Studies Notes Chapter 4 Climate of India

→ The monsoon forms in the tropical area approximately between 20eN and 208S latitudes,

→ The bulk of annual rainfall in India is received from the southwest monsoon.

→ The retreat of the monsoon is marked by clear skies and a rise in temperature.

→ As the Earth started to take shape from a fire bail to a planet, many gases were released.

→ These gases did not escape into outer space because of the Earth’s gravitational pull, and it still holds them back.

→ The atmosphere traps a lot of solar energy that reaches the Earth by preventing it from totally escaping back into space. This is called the Greenhouse effect.

→ Much of the warming has been occurring since thelhdfiStrfal Revolution is because of human activities. Hence, the current global warming is Anthropogenic Globed Warming.

→ Methane is said to be even more powerful than carbon dioxide as a greenhouse gas. c& AGW (Anthropogenic Global Warming) is causing many changes in the distribution of heat in the Earth system.

→ One of the human activities that contribute to global warming is deforestation.

→ An international effort to form an agreement whereby all countries try to reduce their emission of greenhouse gases has so far not been achieved.

→ Climograph: Graphs that show average monthly values of maximum temperature, minimum temperature, and rainfall for a given place.

→ Weather: The state of the atmospheric conditions over an area at a particular time.

→ Monsoon: The seasonal reversal of wind system in India.

→ Insolation (IncomIng solar radiation): The heat coming from the Sun to the Earth’s surface In the form of rays.

→ Jet streams: The fast flowing upper air currents in a narrow belt In the upper atmosphere above 12,000 m.

→ Pressure zone: The exertion of force by one body on the surface of another.

→ Global Warming: Atmosphere traps a lot of solar energy that reaches Earth by preventing it from totally escaping into space This is called Global Warming.

AP Board Solutions AP Board 10th Class Social Studies Notes Chapter 4 Climate of India

→ Climate Weather conditions followed a similar general pattern over many years over a large area.

→ Climatic controls The factors that affect climates like latitude, relief, land-water relationship, and upper air circulation.

→ Equable climate: The climatic condition where the differences between the temperatures of day and night and summer and winter are not much

→ Trade winds: Winds blowing from sub4roplcal high-pressure belt to equatorial low-pressure belt by reflecting towards the west are called trade winds.

→ Western disturbances: Cyclone depressions coming from the mediterranean sea.

→ Loo: A dry hot wind that blows In the northern plain of India.

→ Bursting monsoon: The pre-monsoon showers towards the end of the summer season in the Deccan Plateau.

→ Mango showers: The pre-monsoon showers In Andhra Pradesh helping the early ripening of mangoes.

→ The onset of the monsoon: By the beginning of June, the arrival of the Arabian Sea branch and the Bay of Bengal branch in India is called the onset of the monsoon.

→ The onset of the monsoon: By the beginning of June, the arrival of the Arabian Sea branch and the Bay of Bengal branch in India is called the onset of the monsoon.

→ Anthropogenic Global Warming: Global warming that is caused by humans since the Industrial Revolution.

→ October heat: WIth high temperature and humidity, the weather becomes oppressive during the retreat of the monsoon is called October heat.
AP Board 10th Class Social Studies Notes Chapter 4 Climate of India 1
AP Board 10th Class Social Studies Notes Chapter 4 Climate of India 2

AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.2

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 12th Lesson Factorisation Exercise 12.2

AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.2

Question 1.
Factorise the following expression
i) a2 + 10a +25
ii) l2 – 16l + 64
iii) 36x2 + 96xy + 64y2
iv) 25x2 + 9y2 – 30xy
v) 25m2– 40mn + 1 6n2
vi) 81x2 – 198 xy + 12ly2
vii) (x+y)2 – 4xy
(Hint : first expand ( x + y)2 )
viii) l4 + 4l2m2 + 4m4
Solution:
i) a2 + 10a +25
= (a)2 + 2 × a × 5 + (5)2
It is in the form of a2 + 2ab + b2
a2 + 2ab + b2= (a + b)2
∴ a2 + 10a + 25 = (a + 5)2 = (a + 5) (a + 5)

AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.2

ii) l2 – 16l + 64
l2 – 16l + 64
= (l)2 – 2 × l × 8 + (8)2
It is in the form of a2 – 2ab + b2
a2 – 2ab + b2 = (a – b)2
∴ l2 – 16l + 64 = (l – 8)2 = (l – 8) (l – 8)

iii) 36x2 + 96xy + 64y2
36x2 + 96xy + 64y2
= (6x)2 + 2 × 6x × 8y + (8y)2
It is in the form of a2 + 2ab + b2
a2 + 2ab + b2 = (a + b)2
∴ 36x2 + 96xy + 64y2
= (6x + 8y)2 = (6x + 8y) (6x + 8y)

AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.2

iv) 25x2 + 9y2 – 30xy
25x2 + 9y2 – 30xy
= (5x)2 + (3y)2 – 2 × 5x × 3y
It is in the form of a2 + b2 – 2ab
a2 + b2 – 2ab = (a – b)2
∴ 25x2 + 9y2 – 30xy
= (5x – 3y)2 = (5x – 3y) (5x – 3y)

v) 25m2– 40mn + 1 6n2
25m2 – 40mn + 16n2
= (5m)2 – 2 × 5m × 4n + (4n)2
It is in the form of a2 – 2ab + b2
a2 – 2ab + b2 = (a – b)2
∴ 25m2 – 40mn + 16n2
= (5m – 4n)2
= (5m – 4n) (5m – 4n)

vi) 81x2 – 198 xy + 12ly2
81x2 – 198xy + 121y2
= (9x)2 – 2 × 9x × 11y + (11y)2
It is in the form of a2 – 2ab + b2
a2 – 2ab + b2 = (a – b)2
∴ 81x2 – 198xy + 121y2
= (9x – 11y)2 – (9x – 11y) (9x – 11y)

vii) (x+y)2 – 4xy
(Hint : first expand ( x + y)2 )
= (x + y)2 – 4xy
= x2 + y2 + 2xy – 4xy
= x2 + y2 – 2xy = (x – y)2 = (x – y)(x – y)

viii) l4 + 4l2m2 + 4m4
l4 + 4l2m2 + 4m4
= (l2)2 + 2 × l2 × 2m2 + (2m2)2
It is in the form of a2 + 2ab + b2
a2 + 2ab + b2 = (a – b)2
∴ l4 + 4l2m2 + 4m4
= (l2 + 2m2)2 = (l2 + 2m2) (l2 + 2m2)

AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.2

Question 2.
Factorise the following
i) x2 – 36
ii) 49x2 – 25y2
iii) m2 – 121
iv) 81 – 64x2
v) x2y2 – 64
vi) 6x2 – 54
vii) x2 – 81
viii) 2x -32 x5
ix) 81x4 – 121x2
x) (p2 – 2pq + q2)-r2
xi) (x+y)2 – (x-y)2
Solution:
i) x2 – 36
x2 – 36
⇒ (x)2 – (6)2 is in the form of a2 – b2
a2 – b2 = (a + b) (a – b)
∴ x2 – 36 = (x + 6) (x – 6)

AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.2

ii) 49x2 – 25y2
= (7x)2 – (5y)2
= (7x + 5y) (7x – 5y)

iii) m2 – 121
m2 -121
= (m)2 – (11)2
= (m + 11) (m – 11)

iv) 81 – 64x2
81 – 64x2
= (9)2 – (8x)2
= (9 + 8x) (9 – 8x)

v) x2y2 – 64
= (xy)2 – (8)2
= (xy + 8)(xy – 8)

vi) 6x2 – 54
6x2 – 54
= 6x2 – 6 x 9 ‘
= 6(x2 – 9)
= 6[(x)2 – (3)2]
= 6(x + 3) (x – 3)

AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.2

vii) x2 – 81
x2 – 81
= x2 – 92
= (x + 9 )(x – 9)

viii) 2x – 32 x5
2x – 32 x5
= 2x – 2x x 16x4
= 2 x (1 – 16x4)
= 2x [12) – (4x2)2]
= 2x (1 + 4x2) (1 – 4x2)
= 2x (1 + 4x2) [(15 – (2x)2]
= 2x (1 + 4x2) (1 + 2x) (1 – 2x)

ix) 81x4 – 121x2
81x4 – 121x2
– x2 (812 – 121)
= x2[(9x)2 – (11)2]
= x2 (9x + 11) (9x -11)

x) (p2 – 2pq + q2)-r2
(p2 – 2pq + q2) – r2
= (p – q)2 – (r)2 [∵ p2 – 2pq + q2 = (p – q)2]
= (p – q + r) (p – q – r)

xi) (x + y)2 – (x – y)2
(x + y)2 – (x – y)2
It is in the form of a2 – b2
a = x + y, b = x- y
∴ a2 – b2 =(a + b)(a-b)
= (x + y + x – y) [x + y- (x – y)]
= 2x [x + y-x + y]
= 2x x 2y = 4xy

AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.2

Question 3.
Factorise the expressions
(i) lx2 + mx
(ii) 7y2 + 35Z2
(iii) 3x4 + 6x3y + 9x2Z
(iv) x2 – ax – bx + ab
(v) 3ax – 6ay – 8by + 4bx
(vi) mn + m + n + 1
(vii) 6ab – b2 + 12ac – 2bc
(viii) p2q – pr2 – pq + r2
(ix) x (y + z) -5 (y + z)

(i) lx2 + mx
lx2 + mx
= l × x × x + m × x = x(lx + m)

(ii) 7y2 + 35z2
7y2+ 35z2
= 7 × y2 + 7 × 5 × z2
= 7(y2 + 5z2)

(iii) 3x4 + 6x3y + 9x2Z
3x4 + 6x3y + 9x2Z
= 3 × x2 × x2 + 3 × 2 × x × x2 × y + 3 × 3 × x2 × z
= 3x2 (x2 + 2xy + 3z)

AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.2

(iv) x2 – ax – bx + ab
x2 – ax – bx + ab
= (x2 – ax) – (bx – ab)
= x(x – a) – b(x – a)
= (x – a) (x – b)

(v) 3ax – 6ay – 8by + 4bx
3ax – 6ay – 8by + 4bx
= (3ax – 6ay) + (4bx – 8by)
= 3a (x – 2y) + 4b (x – 2y)
= (x – 2y) (3a + 4b)

(vi) mn + m + n + 1
mn + m + n + 1
= (mn + m) + (n + 1)
= m (n + 1) + (n + 1)
= (n + 1) (m + 1)

(vii) 6ab – b2 + 12ac – 2bc
6ab – b2 + 12ac – 2bc
= (6ab – b2) + (12ac – 2bc)
= (6 × a× b – b × b) + (6 × 2 × a × c – 2 × b × c)
= b [6a – b] + 2c [6a – b]
= (6a – b) (b + 2c)

(viii) p2q – pr2 – pq + r2
p2q – pr2 – pq + r2
= (p2q – pr2) – (pq – r2)
= (p × p × q – p × r × r) – (pq – r2)
= P(pq – r2) – (pq – r2) × 1
= (pq – r2)(p – 1)

(ix) x (y + z) -5 (y + z)
= x(y + z) – 5(y + z)
= (y + z) (x – 5)

AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.2

Question 4.
Factorise the following
(i) x4 – y4
(ii) a4 – (b + c)4
(iii) l2 – (m – n)2
(iv) 49x2 – \(\frac{16}{25}\)
(v) x4 – 2x2y2 + y4
(vi) 4 (a + b)2 – 9 (a – b)2
Solution:
= (x2)2 – (y2)2 is in the form of a2 – b2
a2 – b2 = (a + b) (a – b)
x4 – y4 = (x2 + y2)(x2 – y2)
= (x2 + y2)(x + y)(x – y)

(ii) a4 – (b + c)4
a4 – (b + c)4
= (a2)2 – [(b + c)2]2
= [a2 + (b + c)2] [a2 – (b + c)2] ,
= [a2 + (b + c)2] (a + b + c) [a – (b + c)]
= [a2 + (b + c)2] (a + b + c) (a – b – c)

(iii) l2 – (m – n)2
l2 – (m – n)2
= (l)2 – (m – n)2
= [l + m – n] [l – (m – n)]
= [l + m -n] [l – m + n]

AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.2

(iv) 49x2 – \(\frac{16}{25}\)
= (7x)2 – (\(\frac{4}{5}\))2
= (7x+ (\(\frac{4}{5}\)) (7x – (\(\frac{4}{5}\))

(v) x4 – 2x2 y2 + y4
= (x2 )2 – 2x2 y2 + (y2 )2
It is in the form of a2 – 2ab + b2
a2 – 2ab + b2 = (a – b)2
∴ x4 – 2x2 y2 + y4 = (x2 – y2 )2
= [(x)2 – (y)2 ]2
= [(x + y) (x – y)]2
= (x + y)2 (x – y)2
[∵ (ab)m = a m . bn ]

(vi) 4 (a + b)2 – 9 (a – b)2
4 (a + b)2 – 9 (a – b)2
= [2(a + b)]2 – [3(a – b)]2
= [2(a + b) + 3(a- b)] [2(a + b)-3(a- b)]
= (2a + 2b + 3a – 3b) (2a + 2b – 3a + 3b)
= (5a – b) (5b – a)

AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.2

Question 5.
Factorise the following expressions
(i) a2+ 10a + 24
(ii) x2 +9x + 18
(iii) p2 – 10q + 21
(iv) x2 – 4x – 32
Solution:
(i) a2+ 10a + 24
a2 + 10a + 24 .
= a2 + 6a + 4a + 24
= a x a + 6a + 4a + 6 × 4
= a(a + 6) + 4(a + 6)
= (a + 6) (a + 4) (or)
a2 + 10a + 24
AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.2 1
∴ a2 + 10a + 24 = (a + 6) (a + 4)

(ii) x2 + 9x + 18
x2 + 9x + 18
= (x + 3) (x + 6)
AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.2 2
∴ x2 + 9x + 18 = (x + 3) (x + 6)

(iii) p2 – 10q + 21
p2 – 10p + 21
= (P – 7) (p – 3)
AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.2 3
∴ p2 – 10p + 21 = (p – 7)(p – 3)

AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.2

(iv) x2 – 4x – 32
x2 – 4x – 32
= (x – 8) (x + 4)
AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.2 4
∴ x2 – 4x – 32 = (x – 8) (x + 4)

Question 6.
The lengths of the sides of a triangle are integrals, and its area is also integer. One side is 21 and the perimeter is 48. Find the shortest side.
Solution:
Perimeter of a triangle
= AB + BC + CA = 48
⇒ c + a + b = 48
AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.2 5
The solutions of Harmeet, Rosy are wrong.
AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.2 6
∴ Srikar had done it correctly.
⇒ 21 + a + b = 48
⇒ a + b = 48 – 21 = 27
∴ The lengths of a, b should be 10, 17
∴ a + b > c [the sum of any two sides of a triangle is greater than the 3rd side]
∴ 10 + 17 > 2
27 > 21 (T).
∴ The length of the shortest side is 10 cm.

AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.2

Question 7.
Find the values of ‘m’ for which x2 + 3xy + x + my – in has two linear factors in x and y, with integer coefficients.
Solution:
Given equation is x2 + 3xy + x + my – m ……….(1)
Let the two linear equations in x and y be (x + 3y + a) and (x + 0y + b).
Then (x + 3y + a) (x + 0y + b)
= x2 + 0xy + bx + 3xy + 0y2 + 3by + ax + 0y + ab
= x2 + bx + ax + 3xy + 3by + ab ………….. (2)
Comparing equation (2) with (1),
x2 + 3xy + x + my – m
= x2 + (a + b)x + 3xy + 3by + ab
Equating the like terms on both sides,
ab = – m ………….. (3)
(a + b)x = x ⇒ a + b = 1 ……………. (4)
3by = my ⇒ 3b = m ⇒ b = \(\frac{\mathrm{m}}{3}\)
Substitute ‘b’ value in equation (4),
a = \(1-\frac{m}{3}=\frac{3-m}{3}\)
ab = -m
[ ∵ from (3)]
put a & b value then ,
\(\left(\frac{3-m}{3}\right)\left(\frac{m}{3}\right)\) = -m
\(\frac{3 \mathrm{~m}-\mathrm{m}^{2}}{9}\)= -m
⇒ 3m – m2 = – 9m
⇒ m2 – 12m = 0
⇒ m(m – 12) = 0
⇒ m = 0 (or) m = 12
lf m = 12

∴ b = \(\frac{12}{3}\) = 4&a = \(\frac{3-\mathrm{m}}{3}=\frac{3-12}{3}\)
= \(\frac{-9}{3}\) = -3
∴ Linear factors are (x + 3y – 3), (x + 4) If m = 0
b = \(\frac{0}{3}\) = 0 & a = \(\frac{3-0}{3}=\frac{3}{3}\) = 1
∴ Linear factors are (x + 3y + 1), x.

AP Board 9th Class Maths Solutions Chapter 10 Surface Areas and Volumes Ex 10.4

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 10 Surface Areas and Volumes Ex 10.4 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 10th Lesson Surface Areas and Volumes Exercise 10.4

AP Board 9th Class Maths Solutions Chapter 10 Surface Areas and Volumes Ex 10.4

Question 1.
The radius of a sphere is 3.5 cm. Find its surface area and volume.
Solution:
Radius of the sphere, r = 3.5 cm
AP Board 9th Class Maths Solutions Chapter 10 Surface Areas and Volumes Ex 10.4 1

AP Board 9th Class Maths Solutions Chapter 10 Surface Areas and Volumes Ex 10.4

Question 2.
The surface area of a sphere is 1018\(\frac{2}{7}\) cm2 . What is its volume ?
Solution:
Surface area of sphere = 4πr2
= 1018\(\frac{2}{7}\) cm2
AP Board 9th Class Maths Solutions Chapter 10 Surface Areas and Volumes Ex 10.4 2
= 3054.857cm3
≅ 3054.86cm3

Question 3.
The length of equator of the globe is 44 cm. Find its surface area.
Solution:
Length of the equator of the globe 2πr = 44 cm.
2 × \(\frac{22}{7}\) × r = 44
∴ r = \(\frac{44 \times 7}{2 \times 22}\) = 7cm
∴ surface area = 4πr2
= 4 × \(\frac{22}{7}\) × 7 × 7
= 4 × 22 × 7
= 616cm2

AP Board 9th Class Maths Solutions Chapter 10 Surface Areas and Volumes Ex 10.4

Question 4.
The diameter of a spherical ball is 21 cm. How much leather is required to prepare 5 such balls?
Solution:
Diameter of the spherical ball d’ = 21 cm
Thus, its radius r = \(\frac{d}{2}=\frac{21}{2}\) = 10.5 cm
Surface area of one ball = 4πr2
= 4 × \(\frac{22}{7}\) × 10.5 × 10.5
= 88 × 1.5 × 10.5 = 1386 cm2
∴ Leather required for 5 such balls
= 5 × 1386 = 6930 cm2

Question 5.
The ratio of radii of two spheres is 2 : 3. Find the ratio of their surface areas and volumes.
Solution:
Ratio of radii r1 : r2 = 2 : 3
Ratio of surface area
= 4πr12 : 4πr22
= 22: 32 = 4 : 9
Ratio of volumes
= 4/3 πr13 : 4/3 πr23
= 23 : 33 = 8 : 27

Question 6.
Find the total surface area of hemisphere of radius 10 cm. (Use π = 3.14)
Solution:
Radius of the hemisphere = 10 cm
Total surface area of the hemisphere = 3πr2
= 3 × 3.14 × 10 × 10
= 9.42 × 100
= 942 cm2

AP Board 9th Class Maths Solutions Chapter 10 Surface Areas and Volumes Ex 10.4

Question 7.
The diameter of a spherical balloon increases from 14 cm. to 28 cm. as air is being pumped into it. Find the ratio of surface areas of the balloons in the
two cases.
Solution:
The diameter of the balloon, d = 14 cm
Thus, its radius, r = \(\frac{d}{2}=\frac{14}{2}\) = 7 cm
∴ Surface area = 4πr2 = 4 × \(\frac{22}{7}\) × 7 × 7
= 88 × 7 = 616cm2
When air is pumped, the diameter = 28 cm
thus its radius = \(\frac{d}{2}=\frac{28}{2}\) = 14 cm
Its surface area = 4πr2
= 4 × \(\frac{22}{7}\) × 14 × 14
= 88 × 28 = 2464 cm2
Ratio of areas = 616 : 2464
= 1 : 4

(OR)

Original radius = \(\frac{14}{2}\) = 7 cm
Increased radius = \(\frac{28}{2}\) = 14cm
Ratio of areas = r12 : r22
= 72 : 142
= 7 × 7 : 14 × 14
= 1:4

AP Board 9th Class Maths Solutions Chapter 10 Surface Areas and Volumes Ex 10.4

Question 8.
A hemispherical bowl is made of brass, 0.25 cm thickness. The inner radius of the bowl is 5 cm. Find the ratio of outer surface area to inner surface area.
Solution:
Inner radius of the hemisphere ‘r’ = 5 cm
Outer radius of the hemisphere ‘R’
= inner radius + thickness
= (5 + 0.25) cm = 5.25 cm
Ratio of areas = 3πR2: 3πr2
= R2 : r2
= (5.25)2: 52
= 27.5625 : 25
= 1.1025:1
= 11025 : 10000
= 441 : 400
[Note : If we read “radius as diameter” then we get the T.B. answer]

Question 9.
The diameter of a lead ball is 2.1 cm. The density of the lead used is 11.34 g/c3. What is the, weight of the ball ?
Solution:
The diameter of the ball = 2.1 cm
Thus, its radius, r = \(\frac{d}{2}=\frac{2.1}{2}\) = 1.05 cm
Volume of the ball V’ = \(\frac{4}{3}\)πr3
= \(\frac{4}{3} \times \frac{22}{7}\) x 1.053 = \(\frac{101.87}{21}\)
∴Weight of the ball = Volume × density
= 4.851 × 1.34
= 55.010

AP Board 9th Class Maths Solutions Chapter 10 Surface Areas and Volumes Ex 10.4

Question 10.
A metallic cylinder of diameter 5 cm 1 and height 3 \(\frac{1}{3}\) cm is melted and cast into a sphere. What is its diameter ?
Solution:
Diameter of the cylinder’d’ = 5 cm
Thus, its radius, r = \(\frac{d}{2}=\frac{5}{2}\) = 2.5 cm
Height of the cylinder,
AP Board 9th Class Maths Solutions Chapter 10 Surface Areas and Volumes Ex 10.4 3
Volume of the cylinder
AP Board 9th Class Maths Solutions Chapter 10 Surface Areas and Volumes Ex 10.4 5
Given that cylinder melted to form sphere
∴ Volume of the sphere = Volume of the cylinder
AP Board 9th Class Maths Solutions Chapter 10 Surface Areas and Volumes Ex 10.4 4
(Where r is the radius of the sphere)
r3 = \(\frac{3}{4}\) × 2.5 × 2.5 × \(\frac{10}{3}\)
r3 = 2.53
∴ r = 2.5 cm
Hence its diameter, d = 2r
= 2 × 2.5 = 5 cm

AP Board 9th Class Maths Solutions Chapter 10 Surface Areas and Volumes Ex 10.4

Question 11.
How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold ?
Solution:
Diameter of the hemispherical bowl = 10.5 cm
Thus its radius = \(\frac{d}{2}=\frac{10.5}{2}\) = 5.25cm
Quantity of milk, the bowl can hold = Volume of the bowl = \(\frac{2}{3}\)πr3
= \(\frac{2}{3} \times \frac{22}{7}\) × 5.25 × 5.25 × 5.25
= 303.1875 cm3
= \(\frac{303.1875}{1000}\) lit = 0.303 lit.

Question 12.
A hemispherical bowl has diameter 9 cm. The liquid is poured into cylindrical bottles of diameter 3 cm and height 3 cm. If a full bowl of liquid is Riled in the bottles, find how many
bottles are required ?
Solution:
Diameter of the hemispherical bowl ‘d’ = 9 cm
Its radius, r = \(\frac{d}{2}=\frac{9}{2}\) = 4.5cm
Volume of its liquid = Volume of the bowl = \(\frac{2}{3}\) πr3
= \(\frac{2}{3} \times \frac{22}{7}\) × 4.5 × 45 × 4.5
Diameter of the cylindrical bottle, d = 3 cm
Its radius, r = \(\frac{d}{2}\)
= \(\frac{3.0}{2}\)
= 1.5cm

Height of the bottle, h = 3 cm
Let the number of bottles required = n
Then total volumes of these n bottles = n πr2h
But this is equal to volume of the bowl
Hence n. \(\frac{22}{7}\) × 1.5 × 1.5 × 3
= \(\frac{2}{3} \times \frac{22}{7}\) × 4.5 × 4.5 × 4.5
∴ n = \(\frac{2}{3} \times \frac{20.25}{1.5}\) = 9
∴ Number of bottles required = 9

AP Board 3rd Class EVS Solutions 2nd Lesson Plants Around Us

Andhra Pradesh AP Board 3rd Class EVS Solutions 2nd Lesson Plants Around Us Textbook Exercise Questions and Answers.

AP State Syllabus 3rd Class EVS Solutions Lesson 2 Plants Around Us

I. Conceptual Understanding:

Question 1.
What are the parts of a plant? Label with a diagram?
Answer:

AP Board 3rd Class EVS Solutions 2nd Lesson Plants Around Us 1

Roots, stem, flower, fruit and leaves are the parts of a plant.

Question 2.
How do roots help the plant ?
Answer:

  1. Roots of the plant are below the soil and most important part of the plant.
  2. Roots fix the plant in the ground.
  3. They absorb water and nutrients from the soil and send them to the stems, leaves etc.

Question 3.
How does stem help the plant ?
Answer:

  1. Stem carries water and nutrients from the root to all the other parts of a plant.
  2. Stem supports the plant.

AP Board 3rd Class EVS Solutions 2nd Lesson Plants Around Us

II. Questioning and Hypothesis:

Question 4.
See the houses of Sita and Lakshmi. What questions would you ask them regarding plants?

AP Board 3rd Class EVS Solutions 2nd Lesson Plants Around Us 2

Answer:

Questions to SitaQuestions to Lakshmi
1. Do you like plants?1. Don’t you like plants?
2. What are the uses of plants?2. Did you get fresh air ?
3. What plants do you have in your garden?3. Is it hot or cool in your house?
4. Did you get your own vegetables?4. Did you buy all vegetables?

III. Experiments and Field Observations:

Question 5.
Observe trees, Shrubs, herbs climbers and creepers in your surroundings and name them.
Answer:

AP Board 3rd Class EVS Solutions 2nd Lesson Plants Around Us 3

AP Board 3rd Class EVS Solutions 2nd Lesson Plants Around Us

IV. Information Skills & Project Work:

Question 6.
Collect some aromatic leaves in your surroundings and name them by smelling only.
Answer:
Student activity.

V. Drawing Pictures and Model Making:

Question 7.
Draw a tree which you find in your surroundings. Colour it.
Answer:
Student activity.

Question 8.
Colour the leaves.

AP Board 3rd Class EVS Solutions 2nd Lesson Plants Around Us 4

Answer:
Student Activity.

VI. Appreciation, values and creating awareness towards bio-diversity.

Question 9.
How do you feel if you see some one cutting the branches of the trees around you. What will you do then ?
Answer:
If I find some one cutting the branches of trees around us. Then I will stop them by explaining them about the importance of growing plants and uses of trees.

Question 10.
What will you do if you see the fallen leaves on your school ground ?
Answer:
If I found fallen leaves in my school ground I will collect them and use them in making manure by the following way.
Dig a pit in the ground. Keep the fallen leaves and left over (kitchen scaps, egg shells etc) in it and cover the pit leave for a few days. They decompose and turn into manure. Use this manure for healthy growth of a plant in the garden.

AP Board 3rd Class EVS Solutions 2nd Lesson Plants Around Us

Additional Questions:

I. Conceptual Understanding:

Question 1.
What is a trunk?
Answer:

  1. As the plant grows bigger, the stem strengthens, these thick stems are called trunks. The trunks are covered by bark.
  2. Some examples of trees that have trunks are Banyan, Tamarind. Mango etc.

Question 2.
What are shrubs?
Answer:
Shrubs are small plants with hard stems. Ex : Rose, Hibiscus.

Question 3.
What are herbs?
Answer:
Herbs are very small plants with soft and green stems. Ex : Tulasi, Wheat.

Question 4.
What are climbers?
Answer:
Climbers are the plants that grow on support. Ex : Grapevine, Bitter gourd.

AP Board 3rd Class EVS Solutions 2nd Lesson Plants Around Us

Question 5.
What are creepers?
Answer:
Creepers are the plants that creep on the ground. Ex: Watermelon, Pumpkin.

Question 6.
What are the parts of a leaf? Label them with a diagram.

AP Board 3rd Class EVS Solutions 2nd Lesson Plants Around Us 6

Answer:

AP Board 3rd Class EVS Solutions 2nd Lesson Plants Around Us 5

Parts of a leaf are Apex, Leaf margin,, petiole, vein.

Question 7.
Why do we call leaves as food factories of a plant ?
Answer:
Leaves are called food factories of a plant, because plants prepare their food in the green leaves with the help of air, water and sunlight with the process of photosynthesis.

Question 8.
How can you make manure ?
Answer:
Dig a pit in the ground. Keep the fallen leaves and left over (kitchen scaps, egg shells etc) in it and cover the pit leave for a few days. They decompose and turn into manure. Use this manure for healthy growth of a plant in the garden..

AP Board 3rd Class EVS Solutions 2nd Lesson Plants Around Us

II. Experiments & Field Observations:

Question 1.
Have you Observe the leaves of different plants ? All the leaves same in size, shape, colour and smell ? What are your observations ?

AP Board 3rd Class EVS Solutions 2nd Lesson Plants Around Us 7

Answer:

LeavesObservation
1. Banana leafVery big.
2. HibiscusBroad and marins are like a saw.
3. Papaya LeavesLook like our palm.
4. Coconut LeavesHave long veins.
5. Tamarind LeavesSmall.
6. Pudina, Tulasi & CorianderHave different aroma.

Question 2.
Collect a few leaves of lemon, mango, neem, tulsi, pudina and coriander. Crush the leavs and smell them. Do they smell the same ? Do you know how different leaves are useful to us ?
Answer:
All the leaves do not have same smell. They differ in their smells.

Uses of Different Leaves:

  1. Coriander, Curry leaves, drumstick leaves – used to eat
  2. Tea leaves – use to make tea powder
  3. Neem and Tulasi leaves – used in medicines.
  4. Banana leaves, Banyan leaves, Sal tree leaves – used to make disposable plates and bowls to serve food.

AP Board 3rd Class EVS Solutions 2nd Lesson Plants Around Us

III. Drawing Pictures and Model Making:

Question 1.
Make a leaf album.
Answer:
Student activity.

Question 2.
Make the beautiful pictures given here which are made of leaves and paste them in your notes.

AP Board 3rd Class EVS Solutions 2nd Lesson Plants Around Us 8

Answer:
Student activity.

IV. Activity:

Question 1.
What are the uses of plants ?

AP Board 3rd Class EVS Solutions 2nd Lesson Plants Around Us 9

Answer:
Uses of Plants :

  1. Plants are gifts of nature.
  2. They give us food.
  3. We get fresh air from the plants.
  4. Plants absorb carbon-di-oxide and release oxygen which we breathe.
  5. The roots of big plants hold the soil and prevent soil erosion.

AP Board 3rd Class EVS Solutions 2nd Lesson Plants Around Us

Multiple Choice Questions:

Question 1.
Pick the odd one out from the following.
a) Water melon
b) Pumpkin
c) Coriander
d) Strawberry
Answer:
c) Coriander

Question 2.
_________ are the food factories of a plant.
a) Branches
b) Leaves
c) Stem
d) Roots
Answer:
b) Leaves

Question 3.
_________ are very small plants with soft and green stems.
a) Herbs
b) Creepers
c) Trees
d) Climbers
Answer:
a) Herbs

AP Board 3rd Class EVS Solutions 2nd Lesson Plants Around Us

Question 4.
_________ fix the plant in the ground.
a) Stem
b) Roots
c) Leaves
d) Flowers
Answer:
b) Roots

Question 5.
Thick stems are called _________
a) roots
b) trunks
c) nails
d) trees
Answer:
b) trunks

Question 6.
_________ absorbs nutrients and water from the soil.
a) Stems
b) Roots
c) Leaves
d) Branches
Answer:
b) Roots

AP Board 3rd Class EVS Solutions 2nd Lesson Plants Around Us

Question 7.
_________ carries water and nutrients to all the plants of a plant.
a) Roots
b) Stems
c) Leaves
d) Trees
Answer:
b) Stems

Question 8.
Plants absorb _________ and release _________.
a) oxygen, hydrogen
b) carbon-di-oxide, Oxygen
c) hydrogen, Oxygen
d) oxygen, carbon-di-oxide
Answer:
b) carbon-di-oxide, Oxygen

Question 9.
_________ are used to make manure.
a) Fallen leaves
b) Branches
c) Roots
d) Trunk
Answer:
a) Fallen leaves

AP Board 3rd Class EVS Solutions 2nd Lesson Plants Around Us

Question 10.
_________ of the following are used in medicines.
a) Coconut, sapota
b) Tulasi, neem
c) Mango, Tamarind
d) Banyan tree
Answer:
b) Tulasi, neem.

AP Board 9th Class Maths Solutions Chapter 9 Statistics Ex 9.2

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 9 Statistics Ex 9.2 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 9th Lesson Statistics Exercise 9.2

AP Board 9th Class Maths Solutions Chapter 9 Statistics Ex 9.2

Question 1.
Weights of parcels in a transport office are given below.
AP Board 9th Class Maths Solutions Chapter 9 Statistics Ex 9.2 1
Find the mean weight of the parcels.
Solution:

Weight in kg xiNo. of parcels fix1fi
50251250
65342210
75382850
90403600
110475170
120161920

Σfi = 200
Σfixi = 17000
\(\begin{array}{l}
\overline{\mathrm{x}}=\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}=\frac{17000}{200}=\frac{170}{2} \\
\overline{\mathrm{x}}=85
\end{array}\)
Mean = 85

AP Board 9th Class Maths Solutions Chapter 9 Statistics Ex 9.2

Question 2.
Number of familles In a village in correspondence with the number of children are given below.
AP Board 9th Class Maths Solutions Chapter 9 Statistics Ex 9.2 2
Find the mean number of children per family.
Solution:

No. of childrens xiNo. of families fix1fi
0110
12525
23264
31030
4520
615

Σfi = 84
Σfixi = 144
\(\overline{\mathrm{x}}=\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}=\frac{144}{84}\)
Mean = 1.714285

Question 3.
If the mean of the following frequency distribution is 7.2, find value of ‘k’.
AP Board 9th Class Maths Solutions Chapter 9 Statistics Ex 9.2 3
Solution:
AP Board 9th Class Maths Solutions Chapter 9 Statistics Ex 9.2 4
Σfi = 40 + k;
Σfixi = 260 + 10k
Given that \(\overline{\mathrm{x}}\) = 7.2
But \(\overline{\mathrm{x}}=\frac{\Sigma \mathrm{f}_{1} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
7.2 = \(\frac{260+10 k}{40+k}\)
288.0 + 7.2k = 260 + 10k
10k – 7.2k = 288 – 260
2.8k = 28
k = \(\frac{28}{2.8}\) = 10

AP Board 9th Class Maths Solutions Chapter 9 Statistics Ex 9.2

Question 4.
Number of villages with respect to their population as per India census 2011 are given below.
AP Board 9th Class Maths Solutions Chapter 9 Statistics Ex 9.2 5
Find the average population in each village.
Solution:

Population (in thousands xi)Villages fix1fi
1220240
51575
3032960
2035700
1536540
8756

Σfi = 145 Σfixi = 2571 thousands
\(\overline{\mathrm{x}}=\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
Mean = \(\frac{2571}{145}\) = 17.731 thousands

Question 5.
A FLATOUN social and financial educational programme initiated savings programme among the high school children in Hyderabad district. Mandal wise savings in a month are given in the following table.

MandalNo. of schoolsTotal amount saved (in rupees
Amberpet62154
Thirumalgiri62478
Saidabad5975
Khairathabad4912
Secunderabad3600
Bahadurpura97533

Find arithmetic mean of school wise savings in each mandal. Also find the arithmetic mean of saving of all schools.
Solution:
AP Board 9th Class Maths Solutions Chapter 9 Statistics Ex 9.2 6
Σfi = 33
Σfixi = 14652
Mean = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
\(\bar{x}=\frac{14652}{33}\) = ₹ 444 (Mean savings per school)

AP Board 9th Class Maths Solutions Chapter 9 Statistics Ex 9.2

Question 6.
The heights of boys and girls of IX class of a school are given below.
AP Board 9th Class Maths Solutions Chapter 9 Statistics Ex 9.2 7
Compare the heights of the boys and girls.
[Hint: Fliid median heights of boys and girls]
Solution:
AP Board 9th Class Maths Solutions Chapter 9 Statistics Ex 9.2 8
Boys median class =\(\frac{37+1}{2}=\frac{38}{2}\)= 19th observation
∴ Median height of boys = 147 cm
Girls median class = \(\frac{29+1}{2}=\frac{30}{2}\) = 15th observation
∴ Median height of girls = 152 cm

Question 7.
Centuries scored and number of cricketers in the world are given below.
AP Board 9th Class Maths Solutions Chapter 9 Statistics Ex 9.2 9
Find the mean, median and mode of the given data.
Solution:
AP Board 9th Class Maths Solutions Chapter 9 Statistics Ex 9.2 10
AP Board 9th Class Maths Solutions Chapter 9 Statistics Ex 9.2 12

Question 8.
On the occasion of New year’s day a sweet stall prepared sweet packets. Number of sweet packets and cost of each packet is given as follows
AP Board 9th Class Maths Solutions Chapter 9 Statistics Ex 9.2 11
Find the mean, median and mode of the given data.
Solution:
AP Board 9th Class Maths Solutions Chapter 9 Statistics Ex 9.2 13
N = Σfi = 150
Σfixi = 12000
Mean = \(\overline{\mathrm{x}}=\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}=\frac{12000}{150}=80\)
Median = average of (\(\frac{N}{2}+1\) and \(\frac{N}{2}\) terms = average of 75 and 76 observation = 75
Mode = 50

AP Board 9th Class Maths Solutions Chapter 9 Statistics Ex 9.2

Question 9.
The mean (average) weight of three students is 40 kg. One of the students Ranga weighs 46 kg. The other two students, Rahim and Reshma have the same weight.
Find Rahim’s weight. cgigB)
Solution:
Weight of Ranga = 46 kg
Weight of Reshma = Weight of Rahim = x kg say
Average = \(\frac{\text { Sum of the weights }}{\text { Number }}\) = 40kg
∴ 40 = \(\frac{46+x+x}{3}\)
3 x 40 = 46 + 2x
2x = 120 – 46 = 74
∴ x = \(\frac{74}{2}\) = 37 .
∴ Rahim’s weight = 37 kg.

AP Board 9th Class Maths Solutions Chapter 9 Statistics Ex 9.2

Question 10.
The donations given to an orphanage home by the students of different classes of a secondary school are given below.

ClassDonation by each student in (Rs)No. of students donated
VI515
VII715
VIII1020
IX1516
X2014

Find the mean, median and mode of the data.
Solution:
AP Board 9th Class Maths Solutions Chapter 9 Statistics Ex 9.2 14
Σfi = 80
Σfixi = 900
Mean \(\overline{\mathrm{x}}=\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}=\frac{900}{80}=11.25\)
Median = Average of \(\left(\frac{\mathrm{N}}{2}\right)\) and \(\left(\frac{\mathrm{N}}{2}+1\right)\) terms of \(\frac{80}{2},\left(\frac{80}{2}+1\right)\) terms
= average of 40 and 41 terms = ₹10
Mode = ₹ 10

Question 11.
There are four unknown numbers. The mean of the first two numbers is 4 and the mean of the first three is 9. The mean of all four numbers is 15; if one of the four numbers is 2 find the other numbers.
Solution:
We know that mean = \(\frac{\text { sum }}{\text { number }}\)
Given that, Mean of 4 numbers = 15
⇒ Sum of the 4 numbers = 4 x 15 = 60
Mean of the first 3 numbers = 9
⇒ Sum of the first 3 numbers = 3 x 9 = 27
Mean of the first 2 numbers = 4
⇒ Sum of the first 2 numbers = 2 x 4 = 8
Fourth number = sum of 4 numbers – sum of 3 numbers = 60 – 27 = 33
Third number = sum of 3 numbers – sum of 2 numbers = 27 – 8 = 19
Second number = Sum of 2 numbers – given number = 8-2 = 6
∴ The other three numbers are 6, 19, 33.

AP Board 4th Class English Solutions 1st Lesson Three Butterflies

Andhra Pradesh AP Board 4th Class English Solutions 1st Lesson Three Butterflies Textbook Exercise Questions and Answers.

AP State Syllabus 4th Class English Solutions Chapter 1 Three Butterflies

Textbook Page No. 1

Pre-Reading
AP Board 4th Class English Solutions 1st Lesson Three Butterflies 1

Activity 1

I. Look at the picture and answer the questions :

Question 1.
What do you see in the picture ?
Answer:
I see a girl with umbrella and other without an umbrella in the rain.

Question 2.
Do you like to play in the rain ?
Answer:
Yes, I do.

AP Board 4th Class English Solutions 1st Lesson Three Butterflies

Question 3.
What do you carry when you go out in rain ?
Answer:
I carry an umbrella or a rain coat in the rain.

Question 4.
If you were in the place of the girl holding an umbrella, what would you do ?
Answer:
I would ask the other girl to come and share my umbrella.

Textbook Page No. 4

Comprehension :

Activity-2

Fill in the blanks.

1. The butterflies lived in a _____
Answer: garden

2. Butterflies suck ____ from flowers.
Answer: nectar

3. The rain hit the _____ like pebbles.
Answer: butterflies

4. _____ was watching the butterflies from behind the clouds.
Answer: The sun

AP Board 4th Class English Solutions 1st Lesson Three Butterflies

Textbook Page No. 5

Vocabulary :

Activity-3

Find the plural forms of the singular nouns from the reading text. Write on the table.

AP Board 4th Class English Solutions 1st Lesson Three Butterflies 2
Answer:
AP Board 4th Class English Solutions 1st Lesson Three Butterflies 3

Singular nouns form their plurals by adding ‘s’, ‘es’, or ‘ies’ depending upon the word endings.

AP Board 4th Class English Solutions 1st Lesson Three Butterflies

Activity-4

Fill the given crossword puzzle with the plurals of the given singulars from the hint box.
AP Board 4th Class English Solutions 1st Lesson Three Butterflies 4
Answer:
AP Board 4th Class English Solutions 1st Lesson Three Butterflies 5

Textbook Page No. 6

Grammar

Activity-5

Underline the material nouns in the sentences given below.

1. _____ is used to make jewellery.
Answer: Silver

2. The hen lays ____.
Answer: eggs

3. I use _____ for my hair.
Answer: coconut oil

4. I get new _____ on my birthday.
Answer: clothes

5. Drink ____ for good health.
Answer: milk

6. My sister likes _____.
Answer: cheese

7. ____ is used in constructions.
Answer: Iron

8. My father has a ______
Answer: mobile phone

AP Board 4th Class English Solutions 1st Lesson Three Butterflies

Textbook Page No. 9

Writing

Activity-6

Read the following and punctuate the sentence correctly.

rani mary rajesh basha and rehana are playing cricket “don’t you go to school” asked an old man “today is Sunday Sunday is a holiday” they said

Rewrite the above passage using punctuation marks the full stop (.), comma (,), question mark (?) and capital letters, wherever necessary.
___________________________________
___________________________________
___________________________________
Answer:
Rani, Mary, Rajesh, Basha and Rehana are playing cricket. “Don’t you go to school?” asked an old man. “Today is Sunday, Sunday is a holiday” they said.

Activity-7

Complete the conversation using the hints given below.

Three butterflies wanted to leam dancing, singing and swimming. First, they went to a peacock and asked …
Butterflies : Dear Peacock, we know that you are a good dancer. We want to leam dancing. Would you teach us to dance ?
Peacock : With pleasure, Dear Butterflies, I can teach you to dance. Secondly, three butterflies went to a cuckoo and asked …
Butterflies : Dear Cuckoo, ___________________________________
___________________________________
___________________________________
Answer: we know that you are a good singer. We want to leam singing. Would you teach us to sing?
Cuckoo : ___________________________________
Answer: With pleasure, Dear Butterflies, I can teach you to sing.

Lastly, the three butterflies went to a fish and asked…

Butterflies : ___________________________________
___________________________________
___________________________________
Answer: Dear Fish, we know that you are a good swimmer. We want to leam swimming. Would you teach us to swim?
Fish : ___________________________________
_______________________________________
Answer: With pleasure, Dear Butterflies, I can teach you to swim.

AP Board 4th Class English Solutions 1st Lesson Three Butterflies

Textbook Page No. 11

Listening and Responding

Activity-8

Listen to the conversation between a butterfly and a caterpillar.

Caterpillar : Hi, Butterfly.
Butterfly : Hi dear, caterpillar !
Caterpillar : You are so beautiful!
Butterfly : You too.
Caterpillar : Really!? Am I beautiful?
Butterfly : Yes. We both are beautiful.
Caterpillar : Is it so ?
Butterfly : Yes, it is.
AP Board 4th Class English Solutions 1st Lesson Three Butterflies 6
Caterpillar : You have beautiful wings. But I don’t have.
Butterfly : You, see, I too was once just like you.
Caterpillar : Hard to believe!
Butterfly : Believe it., I still can remember crawling just like you.
Caterpillar : You are’ lucky! You can fly. But I can’t.
Butterfly ; Yes dear,, Caterpillar! Waiting for you to fly together.

Textbook Page No. 10

Answer the following questions.

Question 1.
“You too will become lik me”; who is ‘you’ and who is ‘me’ in the sentence ?
Answer:
‘You is Caterpillar and ‘me’ is butterfly.

Question 2.
“We both are beautiful”, said butterfly. What is your opinion on this statement ? Justify.
Answer:
This statement is true. The Caterpillar turns into a butterfly. So, both of them are beautiful.

Question 3.
“Difficult to believe”, what is that difficult to believe ?
Answer:
It is difficult to believe that one the butterfly was also a caterpillar.

Question 4.
Which one do you like, the caterpillar or the butterfly ? Why?
Answer:
I like the butterfly because it is colourful.

Question 5.
Which one do you usually chase, the butterfly or the caterpillar? Why?
Answer:
I usually chase a butterfly because it gives fun and joy to catch it.

Question 6.
What happens to the butterfly if you catch them ?
Answer:
The butterfly’s powdery material on its wings, sticks to our fingers.

Textbook Page No. 13

Activity-9

Roleplay

Now, convert the following instructions into questions for asking permission by using “can”, “may”. Role play the conversation given below.

Note to the teacher : Form pairs and ask them to play their roles.

1. Take your eraser.
2. Go to toilet.
3. Close the door.
4. Ask you a question.
5. Use your bicycle.
6. Open the window.
AP Board 4th Class English Solutions 1st Lesson Three Butterflies 7
Answer:
AP Board 4th Class English Solutions 1st Lesson Three Butterflies 8

Textbook Page No. 14

Activity-10

Dictation :

Listen to the teacher carefully and write down the paragraph dictated by your teacher.
______________________________________________________________________
______________________________________________________________________
______________________________________________________________________
Answer:
Suddenly it started raining and the three butterflies got wet in the rain. They found it difficult to fly. They looked for shelter. They saw a Sunflower nearby. It was watching them curiously.

Activity – 11

Language game: (Group activity)
Language game – Asking permission.
Material required : Dice and beads.

Procedure :

1. Make the students form a group of four at least.
2. Ask them to throw the dice, read the number and count the number of grids and move the bead on the board.
3. And ask them to ask permission using the hint given in the grid. The other student can give or deny permission.
4. The one who moves to the end is to be declared winner of the game.

Student Activity

AP Board 4th Class English Solutions 1st Lesson Three Butterflies 9

Three Butterflies

Summary :

Once there lived three butterflies who were friends. One was white, one red and other yellow. They aid everything together. One day as they were playing, Suddenly it grew dark and strong wind blew out. Seeing this the red butterfly said to stop playing as it will rain. It started to rain and the three butterflies had no shelter. Then, they saw a sunflower nearby. They asked the sunflower for shelter. But the sunflower wished to give shelter only to its favorite red and yellow butterflies. But, two of them refused as they don’t want to leave their friend. Then the butterflies saw a white lily. When they asked it for help, it wished to help only the white one. Then, the white butterfly refused as it don’t want to leave its friends. Three of them became totally wet and it was difficult for them to fly. The sun, being pleased seeing the love among butterflies, chased the clouds away. The butterflies dried their wings and were happy again.

సారాంశము

ఒక తోటలో మూడు సీతాకోక చిలుకలు ఉండేవి. అవి ఎంతో మంచి స్నేహితులు. అందులో ఒకటి ఎరుపుగా, మరొకటి పసుపుగా, వేరొకటి తెల్లగా ఉండేవి. ప్రతిరోజూ కలిసి ఆడుకుంటూ ఉండేవి. ఒకనాడు అవి ఆడుకుంటుండగా మబ్బులు పట్టి, బాగా గాలి వీచింది. కాసేపటికి పెద్దగా వర్షం కురవడం మొదలైంది. సీతాకోక చిలుకలు తడిచిపోతున్నాయి. ఇదంతా అక్కడ ప్రొద్దు తిరుగుడు పువ్వు గమనిస్తుంది. సీతాకోక చిలుకలు దానిని సహాయం కోరాయి. కానీ అది మాత్రం దానికి ఇష్టమైన ఎరుపు, పసుపు సీతాకోక చిలుకలకు మాత్రమే నివాసం ఇస్తానన్నది. తమలో ఒకరిని విడిచివేయుట ఇష్టం లేక అక్కడ నుండి వచ్చేశాయి. తర్వాత తెలుపు లిల్లీ కనిపించింది. లిల్లీని ఆశ్రయం కోసం అర్థించగా, లిల్లీ తనకు ఇష్టమైన తెలుపు రంగు సీతాకోక చిలుకకు మాత్రమే సహాయం చేస్తాను అంటుంది. తన స్నేహితులను అలా వదిలేయడం ఇష్టం లేక లిల్లీ సహాయాన్ని నిరాకరిస్తాయి. ఇదంతా సూర్యుడు గమనించాడు. వాటి మధ్య ప్రేమ, ఆప్యాయతలు చూసి మేఘాలను తరిమేస్తాడు. దీనితో ఆ మూడు సీతాకోక చిలుకలు తమ రెక్కలను ఆరబెట్టుకొని మళ్ళీ సంతోషంగా జీవించసాగాయి.

Glossary

nectar = a sweet liquid produced by plants and collected by bees; తేనె
shelter = a place that protects one from bad weather; ఆశ్రయం
pebbles = small stones; గులకరాళ్ళు
dismay = a feeling of unhappiness and disappointment; దిగులు
chased = ran after (someone or something) to catch them పరిగెత్తించు
sucked = to pull in liquid or air through you mouth wither using your teeth; పీల్చుట

AP Board 10th Class Biology Notes Chapter 2 Respiration

Students can go through AP State Board 10th Class Biology Notes Chapter 2 Respiration to understand and remember the concept easily.

AP State Board Syllabus 10th Class Biology Notes Chapter 2 Respiration

→ The respiratory system is the integrated system of organs involved in the intake and exchange of oxygen and carbon dioxide between an organism and the environment.

→ Respiration is the process by which food is broken down for the release of energy.

→ From his early experiment, Lavoisier thought that the gas liberated on heating charcoal was like fixed air (CO,).

→ Lavoisier also felt that whatever it was in the atmosphere helped in the burning of phosphorous.

→ Lavoisier noted that the air we breathe out precipitated lime water while that after heating metal did not.

→ Lavoisier also found that something beyond the lungs occurred to produce carbon dioxide and body heat.

→ The pathway of air in our body through the respiratory system – Nostrils → Nasal cavity → Pharynx → Larynx → Trachea → Bronchi → Bronchioles → Lungs → Alveoli.

→ By respiratory system, we usually mean the passage that transport air into the lungs and to the microscopic air sacs called alveoli and vice versa.

→ The inhaled air passes through the nasal cavity its temperature is brought close to that of the body, and it takes up water vapour from it. Air is also filtered in the nasal cavity.

→ Epiglottis, a flap-like muscular valve controls movements of air and food towards their respective passages.

→ When air passes over the vocal cords in the larynx, it causes them to vibrate and produce sounds.

→ The trachea or windpipe channels air to the lungs and it divides into two bronchi – one leading to each lung.

→ The bronchi further divide into small and smaller branches called bronchioles.

AP Board Solutions AP Board 10th Class Biology Notes Chapter 2 Respiration

→ Bronchioles finally terminate in clusters of air sacs called alveolus in the lungs which are very small and numerous.

→ The gaseous exchange takes place in alveoli as blood capillaries take up oxygen and expel CO2.

→ Breathing is the process of inhaling and exhaling.

→ When the volume of the chest cavity is increased, its internal pressure decreases and the air from the outside rushes into the lungs. This is known as inspiration.

→ During expiration, the chest wall is lowered and moves inward and the diaphragm relaxes and assumes its dome shape.

→ Two membranes called pleura covers the lungs.

→ Oxygen is carried in the blood by binding haemoglobin which is present in the red blood cells.

→ Aerobic respiration occurs in the presence of oxygen-producing a lot of energy, CO2 and water.

→ Anaerobic respiration and fermentation occur in the absence of oxygen-producing less amount of energy.

→ ATP – Adenosine Tri Phosphate.

→ The energy released due to the breakdown of sugar molecules is stored in ATP.

→ Each ATP molecule gives 7200 calories of energy. This energy is stored in the form of phosphate bonds.

AP Board Solutions AP Board 10th Class Biology Notes Chapter 2 Respiration

→ In all organisms, glucose is oxidized in two stages. In the first stage called glycolysis, it is converted into two molecules of pyruvic acid.

→ In the second stage, if oxygen is available, pyruvic acid is oxidized to CO2 and water releases a lot of energy.

→ If oxygen is inadequate or not available, pyruvic acid is converted into either ethanol or lactic acid and has very little amount of energy.

→ Accumulation of lactic acid results in muscular pain.

→ Yeast grows rapidly if it is supplied v^th glucose in solution.

→ The combustion of glucose gives us carbon dioxide, water and energy.

→ The reasons for the development of different types of respiratory organs are body size, availability of water and the type of respiratory system.

→ The tracheal respiratory system is seen in insects like cockroaches, grasshoppers, etc.

→ In bronchial respiration, organisms respire with the help of gills or bronchus, e.g: Fish.

→ Respiration through the skin is known as cutaneous respiration, e.g: Frog, Earthworm and Leeches.

→ The gaseous exchange takes place mainly in the stomata of leaves. The gaseous exchange also takes place on the plants like the surface of roots, lenticels on the stem.

→ Photosynthesis is a process of synthesis or an anabolic process that occurs in chloroplasts whereas respiration is a catabolic process that breakdown down complex food molecules in cells.

→ Respiration occurs in the cytoplasm and mitochondria where carbohydrates are burned to produce energy.

AP Board Solutions AP Board 10th Class Biology Notes Chapter 2 Respiration

→ For a plant’s metabolism, both photosynthesis and respiration are required.

→ During the daytime, the rate of photosynthesis is usually higher than that of respiration while at night it is just reverse in most plants.

→ Aerobic respiration: Respiration that is carried out in the presence of oxygen.

→ Anaerobic respiration: Respiration is carried out in the absence of oxygen.

→ Alveoli: The structural and functional unit of lungs. Very small and numerous chambers are present in the lungs.

→ Pharynx: It is the common passage for food and air.

→ Trachea: It Is also known as the windpipe. It connects the nose and mouth to the lungs.

→ Bronchi: It is the part of the trachea that enter the lungs. These conduct air into the lungs.

→ Bronchioles: The bronchus branches into smaller tubes which in turn become bronchioles.

→ Epiglottis: A flap-like muscular valve that controls movements of air and food towards their respective passages.

→ Anabolic process: it ¡s the phase of metabolism In which simple substances are synthesized into complex materials of living tissue.

→ Catabolic process: Metabolic breakdown of complex molecules into simple ones.

→ Aerial roots: Plants that have their roots in very wet places such as ponds or marshes are unable to obtain oxygen. These plants form roots above the soil surface and take in oxygen through these roots. These roots are called aerial roots. Ex: mangrove.

→ Lenticels: The exchange of gases takes place from the stem through the openings called lenticels.

AP Board Solutions AP Board 10th Class Biology Notes Chapter 2 Respiration

→ Fermentation: In the absence of oxygen, yeast cells convert glucose to ethanol (alcohol). This process is called Fermentation. For the production of alcohol, molasses, grape juice and germinating barley or wheat grains are used as raw materials.

→ Energy currency: ATP is called the energy currency of the cell. ATP (Adenosine Tri Phosphate) serves as the primary energy source in the cell. This complex molecule is critical for all life from the simplest to the most complex ATP. it is a complex nanomachine that serves as the primary energy currency of the cell.

→ Glycolysis: It is the first phase of respiration ¡n which glucose molecule is converted into two molecules of pyruvic acid.

→ Respiration: It is the process by which food is broken down to release energy in the cell.

→ Breath: The air is taken into or expelled from the lungs.

→ Combustion: A chemical process in which substances combine with the oxygen in the air to produce heat and light.

→ Vitiated air: Air from which oxygen has been removed. Air containing a reduced percentage of oxygen.

→ Chalky acid gas: Carbon dioxide.

→ Diaphragm: The layer of the muscle between the lungs and the stomach, used especially to control breathing.

→ Inspiration or Inhalation: it is the stage of external respiration in which air or water is taken into the respiratory organ.

→ Expiration or Exhalation: It is the stage of external respiration in which air or water is sent out of the respiratory organ.

→ Pleura: These are the membranes covering the lungs.

AP Board Solutions AP Board 10th Class Biology Notes Chapter 2 Respiration

→ Oxidation: It is the process of the addition of oxygen atoms from the substance or removal of hydrogen atoms from it.

→ Pant: To breath quickly with short noisy breaths because of running, climbing. etc.

→ Janus Green B solution: It is used to test the presence of oxygen in the medium.
(or)
Diazine solution It is blue ¡n colour and turns pink when oxygen is in short supply around it.

→ Dough: A mixture of flour and water ready to be baked into bread.

→ Fractional distillatIon: It is the separation of a mixture into its component parts, or fractions, such as in separating chemical compounds by their boiling point by heating them in to temperature at which one or more fractions of the compound will evaporate.

→ Tracheal respiration: It is the respiration that occurs through tube-like structures (trachea) present in insects.

→ Bronchial respiration: Respiration that occurs through gills is called bronchial respiration. Ex: Fish

→ Cutaneous respiratIon: If the respiration occurs through the skin that type of respiration is called cutaneous respiration. e.g: Frog, Earthworm and Leeches.

→ Orchid: A plant that has flowers with three parts, the middle one being like a lip.

→ Gradient: The rate at which oxygen changes, or increases and decreases between one region and another region.

AP Board Solutions AP Board 10th Class Biology Notes Chapter 2 Respiration

→ Marsh: An area of low flat land that is always wet and soft.
AP Board 10th Class Biology Notes Chapter 2 Respiration 1

→Antoine Lavoisier (26-08-1743 – 08-05-1794) – Father Of Modern Chemistry:

  • Antoine Lavoisier was a French nobleman, chemist and leading figure in the 1 8th century chemical revolution.
  • He has developed the experimentally based theory of the chemical reactivity of oxygen.
  • Lavoisier coauthored the modern system for naming chemical substances.
  • He also worked on combustion, respiration and oxidation of metals in 1772.