Category: Class 7

SCERT AP 7th Class Maths Solutions Pdf Chapter 1 Integers InText Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 1st Lesson Integers InText Questions

Fill in the blanks: (Page No: 3)

Question 1.
7 × (-4) = – (7 × 4) = ___________
Answer:
– 28

Question 2.
2 × (-6) = – (2 × 6) = ___________
Answer:
– 12

Check Your Progress (Page No: 4)

Find the values of

(i) 4×(-8)
Answer:
4 × (-8) = -(4×8)
= – 32

(ii) 5 × (-20)
Answer:
5 × (-20) = – (5 × 20)
= – 100

(iii) 7 × (-8)
Answer:
7 × (-8) = – (7 × 8)
= – 56

(iv) 10 × (-9)
Answer:
10 × (- 9) = -(10 × 9)
= – 90

Fill in the blanks: (Page No: 4)

Question 1.
– 3 × 4 = ________ = 3 × (-4)
Answer:
– 12

Question 2.
– 4 × 4 = ________ = 4 × (-4)
Answer:
– 16

Let’s Explore: (Page No: 4)

Question 1.
Prepare a pattern to find (- 3) × 5 starting from 4 × 5.
Answer:
4 × 5 = 20
3 × 5 = 15
2 × 5 = 10
1 × 5 = 5
0 × 5 = 0

Question 2.
Prepare a pattern to find (- 7) × 3 starting from 5 × 3.
Answer:
5 × 3 = 15
4 × 3 = 12
3 × 3 = 9
2 × 3 = 6
1 × 3 = 3
0 × 3 = 0

Find the values of (Page No: 4)

Question 1.
(-6) × 7 = __________ = -(6 × 7) = – 42
Answer:
6 × (-7)

Question 2.
(-2) × 5 = __________ = -(2 × 5) = – 10
Answer:
2 × (-5)

Question 3.
(-3) × 6 = __________ = -(3 × 6) = – 18
Answer:
3 × (-6)

Question 4.
(-4) × 5 = __________ = -(4 × 5) = – 20
Answer:
4 × (-5)

Check Your Progress (Page No. 4)

Find the values of

(i) (-6) × 5
Answer:
(- 6) × 5 = -(6 × 5) = – 30

(ii) (- 15) × 2
Answer:
(- 15) × 5 = – (15 × 2) = – 30

(iii) (- 12) × 8
Answer:
(- 12) × 8 = – (12 × 8) = – 96

(iv) (- 10) × 6
Answer:
(- 10) × 6 = – (10 × 6) = – 60

Find the values of (Page No: 4)

Question 1.
(- 2) × (- 4) = _________
Answer:
8

Question 2.
– 2 × (- 5) = _________
Answer:
10

Check Your Progress (Page No: 5)

Question 1.
Prepare a pattern to find (-5) × (-4) starting from (-5) × 3.
Answer:
-5 × 3 = – 15
-5 × 2 = – 10
-5 × 1 = – 5
-5 × 0 = 0
-5 × (-1) = + 5
-5 × (-2) = + 10
-5 × (-3) = + 15
-5 × (-4) = + 20

Question 2.
Prepare a pattern to find (- 7) × (- 2) starting from (- 7) × 5.
Answer:
– 7 × 5 = – 35
– 7 × 4 = – 28
– 7 × 3 = – 21
– 7 × 2 = – 14
– 7 × 1 = – 7
– 7 × 0 = 0
– 7 × (- 1) = + 7
– 7 × (- 2) = + 14

Let’s Do Activity (Page No: 6)

Fill the grid by multiplying each number in the first column with each number in the hirst row and answer the following questions.

Answer:

Question 1.
Write your observations from the table.
Answer:
Product of two integers with same sign is always a positive integer and the product of two integers with opposite signs is always negative integer.

Question 2.
What happens when an integer multiplied with (-1) ?
Answer:
a × (-1) = – 1 × a = – a.
When an integer is multiplied by – 1, the sign of the integer changes from + ve to – ve or – ve to + ve.

Question 3.
When will we get product of two integers is zero ?
Answer:
The product of two integers is zero only when one of them is zero.

Division of Integers (Page No: 8)

Observe the following and fill the blanks

Answer:

Check Your Progress (Page No: 9)

Fill the following table.

Answer:

Let’s Do Activity (Page No: 10)

The fish in the pond below, carry some numbers. Choose any 4 pairs and carry out four multiplications with those numbers. Now, choose four other pairs and carry out divisions with those numbers.

Answer:
Multiplications:
(i) (- 10) × 6 = – 60
(ii) (- 4) × (- 9) = 36
(iii) 12 × 6 = 72
(iv) 6 × (-4) = – 24
(v) 8 × 6 = 48
(vi) 12 × 8 = 96

Divisions:
(i) (-36) ÷ 6 = – 6
(ii) (-100) ÷ (- 10) = 10
(iii) 45 ÷ (-9) = – 5
(iv) 72 ÷ 12 = 6
(v) (-56) ÷ 8 = – 7
(vi) (-28) ÷ (-4) = 7

Puzzle time (Page No: 11)

Jasvi says her favorite number through puzzle. Find it.

Answer:

(i) Closure property: Observe the following tables and complete them.

∴ Integers are closed under addition and multiplication.
Answer:

∴ Integers are closed under addition and multiplication.

Let’s Think (Page No: 11)

Can you find at least one pair of integers whose sum or product is not an integer ?
Sol.
No, we can’t find at least one pair of integers whose sum of product is not an integer.

Page No. 12

(i) Closure property:

∴ Integers are closed under subtraction, but need not closed under division.
Answer:

∴ Integers are closed under subtraction, but need not closed under division.

(ii) Commutative Law: Observe the following tables and complete them.

∴ Integers are commutative under addition and multiplication.
Answer:

∴ Integers are commutative under addition and multiplication.

∴ Integers are not commutative under subtraction and division.
Answer:

∴ Integers are not commutative under subtraction and division.

Page No. 13

(iii) Associative Law: Observe the following tables and complete them.

∴ Integers are associative under addition and multiplication.
Answer:

∴ Integers are associative under addition and multiplication.

∴ Integers are associative under subtraction and division.

iv) Identity property: Observe the following tables and complete them.

Answer:

Page No: 14

v. What should be added to – 3 to get additive identity 0 ?

A. We add + 3 to get additive identity 0. Observe the following.
(i) 4 + (- 4) = _________
Answer:
0

(ii) (-5) + 5 = _________
Answer:
0

(iii) (- 6) + _________ = 0
Answer:
6

Check Your Progress (Page No: 14)

Question 1.
Write the additive inverses of 5, -8, 1 and 0.
Answer:

Lets Think (Page No: 14)

Question 1.
What should be multiplied by 6 to get multiplicative identity 1 ? Is it exist in integers ?
Answer:
6 × \(\frac{1}{6}\) = 1 (OR) 6 ÷ 6
But \(\frac{1}{6}\) is not a integer.
Multiplicative identity of 6 is \(\frac{1}{6}\).
∴ \(\frac{1}{6}\) does not exist in integers.

Question 2.
Verify – 3 × [(-4) – 2] = [(- 3) × (- 4)] – [(- 3) × 2]. Is-multiplication distribute over subtraction of integers ? Write your observations.
Answer:
– 3 × [(- 4) – 2] = [(- 3) × (- 4)] – [(-3) × 2]
– 3 × [- 4 – 2] = [3 × 4] – [- (3 × 2)]
– 3 × (-6) = 12 – (-6)
3 × 6 = 12 + 6
18 = 18
L.H.S. = R.H.S.
Yes it is multiplication distribute over subtraction of integers.
That is a × (b – c) = (a × b) – (a × c)
Yes, it exists in integers.

Check Your Progress (Page No: 17)

Simplify the following.

(i) 5 × 6 – 6
Answer:
5 × 6 – 6
= 30 – 6 (Multiplication)
= 24 (Subtraction)

(ii) 24 ÷ 3 × 3 – 30
Answer:
24 ÷ 3 × 3 – 30 (Division)
= 8 × 3 – 30 (Multiplication)
= 24 – 30 (Subtraction)
= – 6

(iii) 5 × 5 – 5 ÷ 5 + 5
Answer:
5 × 5 – 5 ÷ 5 + 5 (Division)
= 5 × 5 – 1 + 5 (Multiplication)
= 25 – 1 + 5 (Addition)
= 25 + 5 – 1
= 30 – 1 (Subtraction)
= 29

Let’s Explore (Page No: 19)

Question 1.
If |x| = 15, then what will be the value of x ? Discuss.
Answer:
Given |x| = 15
If x > 0, then |x| = + 15
If x < 0, then |x| = – 15
So, – 15 < x < 15

Examples

Question 1.
A sump is full of water, when the motor started pumping the level of water decreasing 2 inches per minute then what is the level of water from the ground level after 20 minutes ?
Answer:
The change in level of water per minute = – 2 inches (decreasing 2 inches)
The level of water after 20 minutes = 20 × (-2) – 40 inches
So, the level of water in sump is 40 inches depth from the ground level.

Question 2.
An elevator begins from 20 m above the ground. It descends into a mine shaft at the rate of 6 m per minute. What will be its position after 15 minutes ?
Answer:
Since the elevator is going down, so the distance covered by it will be represented by a negative integer. Change in position of the elevator in one minute = – 6 m.
Change in position of the elevator in 15 minutes = 15 × (-6) = – 90 m.
So, the final position of the elevator – 20 + (-90) = – 70 m.
The elevator is at 70 m below the ground level.

Question 3.
In a test, ( + 5) marks are given for every correct answer and (-3) marks are given for every incorrect answer. Lakshmi gets 45 correct and 15 in-correct answers. What is her score ?
Answer:
Marks given for one correct answer = 5
Marks obtained for 45 correct answers = 45 × 5 = 225
Marks given for one incorrect answer = – 3
Marks obtained for 15 incorrect answers = 15 × (- 3) = – 45
∴ Lakshmi’s score = 225 + (- 45)
= 180.

Question 4.
A borewell machine drills down 72 feet per hour from surface of the earth. If the water is at 360 feet down from surface of earth, after how many hours it will touch the water layer ?

Answer:
Depth of drilling in one hour = – 72 feet
Depth of water layer from surface of earth = – 360 feet
Number of hours required
= – 360 ÷ (-72)
= 5
Hence, the borewell machine will touch water layer at 5 hours of drilling.

Question 5.
In a test, (+ 4) marks are given for every correct answer and (- 2) marks are given for every incorrect answer. Sasi answered all the questions and scored 26 marks from 8 correct answers. How many incorrect answers had Sasi attempted?
Answer:
Marks given for one correct answer = 4
So, marks given for 8 correct answers
= 4 × 8
= 32
Sasi score = 26
Marks obtained for incorrect answers = 26 – 32 = -6
Marks given for one incorrect answer = – 2
∴ Number of incorrect answers
= (- 6) ÷ (- 2)
= 3

Question 6.
Shop keeper Yaseen earns a profit of ₹ 20 per bag of Sonamasoori rice sold and loss of ₹ 12 per bag of Hamsa rice. In one week he gets neither profit nor loss, if he solds 1440 Sonamasoori rice bags. How many Hamsa rice bags did he sell ?
Answer:
In the given problem there is neither profit nor loss.
So, profit earned + loss incurred = 0
Profit earned = – loss incurred
Profit earned by selling one Sonamasoori rice bag = ₹ 20
Profit earned by selling 1440 Sonamasoori rice bags
= 1440 × 20
= ₹ 28800
Loss incurred by selling one Hamsa rice bag = ₹ 12,
which we denoted by – 12 Loss incurred by selling one Hamsa rice bags = ₹ – 28800
Total number of Hamsa rice bags sold
= (- 28800) ÷ (-12)
= ₹ 2400 bags.

Question 7.
Find the additive inverses of (+ 2) and (- 3).
Answer:
Additive inverse of + 2 = – (+2) = – 2
Additive inverse of – 3 = – (-3) = + 3.

Question 8.
Multiply the following using associative law.
(i) – 25 × (- 4) × 2 × (- 8)
Answer:
– 25 × (- 4) × 2 × (- 8)
= [- 25 × (- 4)] × 2 × (- 8)
= [100 × 2] × (-8)
= 200 × (-8) = – 1600

(ii) (- 20) × (- 2) × (- 5) × 7
Answer:
(- 20) × (-2) × (-5) × 7
= (- 20) × [(-2) × (-5)] × 7
= [(- 20) × 10] × 7
= – 200 × 7
= – 1400

Question 9.
Are (-42) × (-7) and (-7) × (-42) equal ? Which is this law ?
Answer:
(-42) × (-7) = + 294
(-7) × (-42) = + 294
∴ (- 42) × (- 7) = (- 7) × (- 42)
It is multiplicative commutative law.

Question 10.
Simplify 26 × (- 48) + (- 48) × (- 36) using suitable laws.
Answer:
26 × (- 48) + (- 48) × (- 36)
= (- 48) × 26 + (- 48) × (- 36) (Commutative law)
= (- 48) × [26 + (- 36)] (Distributive law)
= (- 48) × (- 10) = 480

Question 11.
Simplify: 3 × 2 + 8 ÷ 4
Answer
3 × 2 + 8 ÷ 4 (Division)
= 3 × 2 ÷ 2 (Multiplication)
= 6 + 2 (Addition)
= 8

Question 12.
Simplify : 7 × 6 – \(\overline{8-4}\).
Answer:
7 × 6 – \(\overline{8-4}\) (Vinculum)
= 7 × 6 – 4 (Multiplication)
= 42 – 4 (Subtraction)
= 38

Question 13.
Simplify: 18 + 64 ÷ 4 {26 – (14 – \(\overline{7-3}\))}
Answer:
18 + 64 ÷ 4 {26 – (14 – \(\overline{7-3}\))}
= 18 + 64 ÷ 4 {26 – (14 – 4)} (Vinculum)
= 18 + 64 ÷ 4 {26 – (14 – 4)} (Simple bracket)
= 18 + 64 ÷ 4{26 – 10} (Curly bracket)
= 18 + 64 ÷ 4 {16} (Of)
= 18 + 64 ÷ 64 (Division)
= 18 + 1 (Addition)
= 19

Practice Questions (Page No: 21)

Question 1.
12, 19, 26, 33, 40, 47,
(a) 57
(b) 54
(c) 52
(d) 50
Answer:
(b) 54

Explaination:
Given 12, 19, 26, 33, 40, 47, __________
Each number in the series obtained by adding 7 to the previous number.
So, next number is 47 + 7 = 54.

Question 2.
2, 13, 24, 35, 46, 57, __________
(a) 65
(b) 67
(c) 68
(d) 72
Answer:
(c) 68

Explaination:
Given 2, 13, 24, 35, 46, 57, _________
Each number in the series obtained by adding 11 to the previous number. So, next number is 57 + 11 = 68.

Question 3.
61,67.71,73,79.
(a) 89
(b) 87
(c) 85
(d) 83
Answer:
(d) 83

Explaination:
Given 61, 67, 71, 73, 79, _________

So, answer is 79 + 4 = 83

Question 4.
3, 7, 13, 21, 31,
(a) 43
(b) 48
(c) 51
(d) 53
Answer:
(a) 43

Explaination:
Given 3, 7, 13, 21, 31, ___________

So, answer is 31 + 12 = 43.

Question 5.
8, 12, 20, 32, 52, 84,
(a) 111
(b) 126
(c) 136
(d) 174
Answer:
(c) 136

Explaination:
Given 8, 12, 20, 32, 52, 84, ___________

So, answer is 84 + 52 = 136

Question 6.
23,28,38,53,73,98,
(a) 121
(b) 128
(c) 135
(d) 146
Answer:
(b) 128

Explaination:
Given 23, 28, 38, 53, 73, 98, _______

So, answer is 98 + 30 = 128

Question 7.
101, 97, 89, 83, 79, 73, 71, ___________
(a) 61
(b) 65
(c) 66
(d) 67
Answer:
(d) 67

Explaination:
Given 101, 97, 89, 83, 79, 73, 71, ________

So, answer is 73 – 6 = 67

Question 8.
4, 7, 11, 18, 29, 47, _______
(a) 67
(b) 76
(c) 84
(d) 92
Answer:
(b) 76

Explaination:
Given 4, 7, 11, 18, 29, 47, _______

So, answer is 29 + 47 = 76

Question 9.
76, 187, 298, 409, 520,
(a) 631
(b) 656
(c) 701
(d) 724
Answer:
(a) 631

Explaination:
Given 76, 187, 298, 409, 520, _______

So, answer is 520 + 111 = 631

Question 10.
0, 2, 5, 10, 17, 28, 41, ___________
(a) 50
(b) 53
(c) 57
(d) 58
Answer:
(d) 58

Given 0, 2, 5, 10, 17, 28, 41, ______

2, 3, 5, 7, 11, 13, 17 all are prime numbers.
So, answer is 41 + 17 = 58

Question 11.
36, 45, 53, 60, 66, 71, ________
(a) 84
(b) 78
(c) 75
(d) 73
Answer:
(c) 75

Explaination:
Given 36, 45, 53, 60, 66, 71, _______

So, the answer is 71 + 4 = 75.

Question 12.
0, 15, 45, 90, 150, 225, _________
(a) 295
(b) 300
(c) 315
(d) 360
Answer:
(c) 315

Explaination:
Given 0, 15, 45, 90, 150, 225, ___________
0, 0 + (15× 1), 15 + (15 × 2), 45 + (15 × 3), 90 + (15 × 4), 150 + (15 × 5), 225 + (15 × 6), 0, 0 + 15, 15 + 30, 45 + 45, 90 + 60, 150 + 75, 225 + 90
0, 15, 45, 90, 150, 225, 315
So, the answer is 225 + (15 × 6) = 315

Question 13.
18, 23, 25, 30, 32, 37, _______
(a) 43
(b) 41
(c) 39
(d) 38
Answer:
(c) 39

Explaination:
Given 18, 23, 25, 30, 32, 37, ________

So, the answer is 32 + 7 = 39

Question 14.
4, 7, 11, 18, 29, 47, _________
(a) 71
(b) 76
(c) 77
(d) 82
Answer:
(b) 76

Explaination:
Given 4, 7, 11, 18, 29, 47, _______

So, the answer is 29 + 47 = 76

Question 15.
12, 18, 21, 27, 30, 36, 39, ______
(a) 43
(b) 45
(c) 49
(d) 52
Answer:
(b) 45

Explaination:
Given 12, 18, 2f, 27, 30, 36, 39, ______
3 × 4, 3 × 6, 3 × 7, 3 × 9, 3 × 10, 3 × 12, 3 × 13, 3 × 15
So, the answer is 3 × 15 = 45

SCERT AP 7th Class Maths Solutions Pdf Chapter 4 Lines and Angles Review Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 4th Lesson Lines and Angles Review Exercise

Question 1.
Observe the figure and name the points, line segments, rays and lines from the figure.

Answer:
Points: A, B, C, D, E, G
Line segments : \(\overline{\mathrm{AB}}, \overline{\mathrm{AD}}, \overline{\mathrm{AE}}, \overline{\mathrm{BD}}\overline{\mathrm{DE}}, \overline{\mathrm{BC}}, \overline{\mathrm{CD}}, \overline{\mathrm{BE}}\)
Rays : \(\overrightarrow{\mathrm{BA}}, \overrightarrow{\mathrm{DA}}, \overrightarrow{\mathrm{BE}}, \overrightarrow{\mathrm{AE}}, \overrightarrow{\mathrm{EA}}, \overrightarrow{\mathrm{DE}}\)
Lines : \(\overrightarrow{\mathrm{AE}}\)

Question 2.
Observe the figure and write intersect-ing lines and concurrent lines.

Answer:
Intersecting lines : l, p
Concurrent lines : l, m, n

Question 3.
Draw a line segment PQ = 6.3 cm.
Answer:

Question 4.
Name any three possible angles in the adjacent figure.

Answer:
Angles: ∠POQ, ∠POR, ∠POS, ∠OOS, ∠ROS, ∠QOR.

Question 5.
Write the type of angles you observed in the given clock.

Answer:
Angles (i) Acute angle
(ii) Right angle
(iii) Obtuse angle
(iv) Straight angle
(v) Reflex angle.

Question 6.
One right angle is equal to __________ degrees.
Answer:
90.

Question 7.
Write any two acute angles and any two obtuse angles.
Answer:
Acute angles :

Obtuse Angles:

Question 8.
Observe the parallel and perpendicular lines in the given figure. Write them using symbols ∥, ⊥
Answer:

Answer:
From the given figure.
Parallel lines : l ∥ m.
Perpendicular lines : l ⊥ n and m ⊥ n.

Question 9.
Measure and write angle ∠AOB with the help of protractor.

Answer:
∠AOB = 40°.

SCERT AP 7th Class Maths Solutions Pdf Chapter 4 Lines and Angles Unit Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 4th Lesson Lines and Angles Unit Exercise

Question 1.
Find the complementary, supplementary and conjugate angle of 36°.
Answer:
Complementary angle of 36° is 90° – 36° – 54°
Supplementary angle of 36° is 180° – 36° = 144°
Conjugate angle of 36° is 360° – 36° = 324°

Question 2.
Observe the figure and write any 4 pairs of adjacent angles.

Answer:
Adjacent angle of ∠AOB is ∠BOC.
Adjacent angle of ∠BOC is ∠COD.
Adjacent angle of ∠COD is ∠DOE.
Adjacent angle of ∠DOE is ∠EOF.

Question 3.
In the given figure the lines l and m intersect at O. Find x.

Answer:
Given l and m intersecting at O.
x° + 40° = 120° (vertically opposite angles)
x + 40° – 40°= 120° – 40°
∴ x = 80°

Question 4.
In the given figure \(\overline{\mathbf{A E}}\) is a straight line. If the ratio of angles ∠1, ∠2, ∠3, ∠4 in the given figure is 1:2 :3 : 4, then find the angles.

Answer:
Given \(\overline{\mathbf{A E}}\) is a straight line.
Ratio of angles ∠1, ∠2, ∠3, ∠4 is 1 : 2 : 3 : 4 that is 1x: 2x: 3x: 4x

The sum of the angles at a point on the same side of the line is 180°
∠1 + ∠2 + ∠3 + ∠4 = 180°
⇒ 1x + 2x + 3x + 4x = 180°
⇒ 10x = 180
⇒ \(\frac{10 x}{10}=\frac{180}{10}\)
∴ x = 18°
2x = 2 × 18° = 36°
3x = 3 × 18° = 54°
4x = 4 × 18° = 72°
Therefore the angles are 18°, 36°, 54°, 72°.

Question 5.
Write any two examples for linear pair of angles in your surroundings.
Answer:
Electric pole, Tree/Pen stand, etc.

Question 6.
Mani said, “Two obtuse angles can form a pair of conjugate angles.” Do you agree? Justify your answer.
Answer:
Obtuse angle is always less than 180°. Sum of two obtuse angles is less than 360°.
So, I do not agree, that two obtuse angles cannot form a pair of conjugate angles.

Question 7.
Draw a pair of adjacent angles which are not supplementary to each other.
Answer:
(i)

∠AOB and ∠BOC are adjacent angles.
∠AOB + ∠BOC = 50° + 60°
= 110° ≠ 180°

∠AOB and ∠BOC are not supplementary.
(ii)

∠POQ and ∠QOR are adjacent angles.
∠POQ + ∠QOR = 70° + 80°
= 150° ≠ 180°
∠POQ and ∠QOR are not supplementary.

Question 8.
In the figure, if l ∥ m, t is a transversal. Find ∠1 and ∠2.

Answer:
Given l ∥ m and t is a transversal.
∠1 = 110° (vertically opposite angles)
∠1 + ∠2 = 180° (co-interior angles are supplementary)
110° + ∠2 = 180°
110° + ∠2 – 110° = 180° – 110°
∠2 – 70°
∠1 = 110° and ∠2 = 70°

Question 9.
A line p intersects two lines l and m at two distinct points. Observe the figure and fill in the blanks :

(i) The line ‘p’ is known as ________, ________
(ii) ∠1 and ∠5 is a pair of ________ angles.
(iii) ∠4 and ∠6 is a pair of ________ angles.
(iv) ∠3 and ∠6 is a pair of ________ angles.
Answer:
(i) transversal line,
(ii) corresponding
(iii) alternate interior
(iv) co-interior

Question 10.
In the given figure \(\overrightarrow{\mathbf{C F}} \| \overrightarrow{\mathbf{B D}}, \overrightarrow{\mathbf{B E}}\) is transversal. ∠CAE = 135°, then find ∠ABD

Answer:
Given \(\overrightarrow{\mathrm{CF}} \| \overrightarrow{\mathrm{BD}}\) and \(\overrightarrow{\mathrm{BE}}\) is transversal, ∠CAE — 135°
∠BAF = ∠CAE = 135° (vertically opposite angles)
∴ ∠BAF = 135°
∠BAF + ∠ABD = 180° (co-interior angles are supplementary)
135° + ∠ABD = 180°
135° + ∠ABD – 135° =180°- 135°
∴ ∠ABD = 45°

SCERT AP 7th Class Maths Solutions Pdf Chapter 4 Lines and Angles Ex 4.4 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 4th Lesson Lines and Angles Exercise 4.4

Question 1.
In the given figure, two lines p ∥ q arid r is transversal, If ∠3 = 135°, then find the remaining angles.

Answer:
Given ∠3 = 135
∠1 = ∠3 = 135° (vertically opposite angles)
∠1 = ∠5 = 135° (corresponding angles)

∠3 = ∠8 = 135° (corresponding angles)
∠3 = ∠5 = 135° (Alternate interior angles)

∠1 + ∠7 = 180° (co-exterior angles are supplementary)
135° + ∠7 = 180°
∠7 = 180°- 135° = 45°
∠6 = ∠7 = 45° (vertically opposite angles)
∠6 = ∠2 = 45° (corresponding angles)
∠4 = ∠7 = 45° (corresponding angles)
∴ ∠1 = 135°, ∠2 = 45°, ∠3 = 135°, ∠4 = 45°, ∠5 = 135°, ∠6 = 45°, ∠7 = 45°, ∠8 = 135°

Question 2.
In the given figure, \(\overleftrightarrow{\mathbf{AB}}\) || \(\overleftrightarrow{\mathbf{CD}}\) and \(\overleftrightarrow{\mathbf{DE}}\) is a transversal. Find x.

Answer:
Given \(\overleftrightarrow{\mathbf{AB}}\) || \(\overleftrightarrow{\mathbf{CD}}\) and \(\overleftrightarrow{\mathbf{DE}}\) is a transversal line.
∠ABE = ∠CDB (corresponding angles)
∠ABF + ∠FBE = ∠CDB (we know ∠ABE = ∠ABF + ∠FBE)
From the figure ∠ABF = 35°,
∠FBE = x° and ∠CDB =100°
35° + x = 100°
⇒ 35° + x – 35° = 100° – 35°
∴ x = 65°

Question 3.
In the given figure, m || n and p is transversal. Find x and y.

Answer:
Given m || n and p is transversal.
So, ∠x = 120° (vertically opposite angles)
45 + y = x (corresponding angles)
⇒ 45 + y = 120° (we know ∠x = 120°)
⇒ 45 + y – 45 = 120-45°
⇒ y = 75°
x = 120° and y = 75°

Question 4.
In the given figure, \(\overrightarrow{\mathbf{A B}}\|\overrightarrow{\mathbf{C D}}\| \overrightarrow{\mathbf{F E}}\). Find x, y and ∠AEC.

Answer:
Given \(\overrightarrow{\mathbf{A B}}\|\overrightarrow{\mathbf{C D}}\| \overrightarrow{\mathbf{F E}}\)
∠x = 20° (Alternate interior angles)
∠y = 33° (Alternate interior angles)
∠AEC = ∠AEF + ∠FEC
∠AEC = ∠x + ∠y
∠AEC = 20° + 33° = 53°
∴ x = 20°, y = 33° and ∠AEC = 53°

Question 5.
In the given figure, a transversal t intersects two lines p and q. Check whether p ∥q or not.

Answer:
If co-interior angles are supplementary, then the lines are parallel.
100° + 80° = 180° (co-interior angles supplementary)
So, p and q are parallel to each other.

Question 6.
In the given figure if l ∥m, then find x, y and z.

Answer:
Given l ∥m and AC is transversal.
z = 40° (Alternate interior angles)
l ∥m and AB is transversal. , and x + (y + z) = 180° (co-interior angles are supplementary)
x + y + z = 180°
(x + y) + 40° = 180° (co-interior angles are supplementary) (∵ z = 40°)
x + y + 40° = 180°
x + y + 40°-40° = 180° – 40°
∴ x + y = 140°
But x = y (given)
2x = 140°
∴ x = \(\frac{140^{\circ}}{2}\) = 70° = y
So, x = 70°, y = 70° and z = 40°

Question 7.
In the given figure p, q, r and s are parallel lines and t is a transversal. Find x, y and z.

Answer:
p ∥ q
So, 80° + x = 180° (co-interior angles)
80° + x-80° = 180°-80°
x = 100°

q ∥ r
x + y = 180° (co-interior angles)
100° + y = 180° (we know x – 100°)
100° + y – 100° = 180°- 100°
y = 80°

r∥s
y = z (Alternate exterior angles)
y = z = 80° (we know y = 80°)
x = 100°, y = 80° and z = 80°

Question 8.
In the given figure \(\overrightarrow{\mathbf{A B}} \| \overrightarrow{\mathbf{C D}}\) and E is a point in between them. Find x + y + z. (Hint : Draw a parallel line to \(\overrightarrow{\mathbf{A B}}\) through E)

Answer:
Given \(\overrightarrow{\mathbf{A B}} \| \overrightarrow{\mathbf{C D}}\)

Draw \(\overrightarrow{\mathrm{EF}} \| \overrightarrow{\mathrm{AB}}\)
∠BAE = x,
∠AEF = p
∠FEC = q,
∠ECD = z
∠AEC = ∠AEF + ∠FEC
y = p + q

AB ∥ EF and AE is a transversal.
∠BAE + ∠AEF = 180° (co-interior angles are supplementary)
x + p = 180° ……………….(1)
CD ∥ EF and EC is a transversal.
∠FEC + ∠ECD = 180° (co-interior angles are supplementary’)
q + z = 180° …………….(2)

By adding (1) and (2)

x + y. + z = 360° (we know y = p + q)

Question 9.
Identify the pair of parallel lines in the given figure and write them.

Answer:
∠A = ∠E = 60° So, AL ∥ EH
∠B = ∠D = 90° So, BK ∥ DL
∠C = ∠F = 50° So, CJ ∥FG

SCERT AP 7th Class Maths Solutions Pdf Chapter 4 Lines and Angles Ex 4.3 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 4th Lesson Lines and Angles Exercise 4.3

Question 1.
Name three pairs of vertically opposite angles in the figure. If ∠AOB = 45°, then find ∠DOE.

Answer:
Vertically opposite angles :
∠AOB, ∠DOE; ∠BOC, ∠EOF; ∠COD, ∠AOF
Given ∠AOB = 45°
In the given figure, ∠AOB is the vertically opposite angle to ∠DOE.
So, ∠DOE = ∠AOB = 45°
∴ ∠DOE = 45°

Question 2.
In the given figure \(\overrightarrow{\mathbf{PQ}}\) is a straight line. Check whether x and y are vertically opposite angles or not. Give reason.

Answer:
\(\overrightarrow{\mathbf{PQ}}\) is a straight line. But \(\overrightarrow{\mathbf{SR}}\) is not a straight line.
If \(\overrightarrow{\mathbf{PQ}}\) and \(\overrightarrow{\mathbf{SR}}\) are intersecting lines the x and y become vertically opposite angles.
So, x and y are not vertically opposite angles.

Question 3.
Write any three examples for vertically opposite angles in your surroundings.
Answer:
Scissors, window grills, cross roads, rail cross junctions, etc.

Question 4.
In the given figure, the lines l and m intersect at point P. Observe the figure and find the values of x, y and z.

Answer:
Given l and m are intersecting lines at P.
∠y = 20° (vertically opposite angles)
∠x = ∠z (vertically opposite angles)
∠y + ∠x = 180° (linear pair)
⇒ 20° + ∠x = 180°
⇒ 20° + ∠x – 20° = 180°- 20
⇒ ∠x = 160°
∴ ∠x = ∠z = 160°
∠x = 160°, ∠y = 20° and ∠z = 160°.

Question 5.
In the given figure, two lines \(\overleftrightarrow{\mathbf{A D}}\) and \(\overleftrightarrow{\mathbf{E C}}\) intersects at O. Name two pairs of vertically opposite angles in the given figure.

Answer:
In the given figure \(\overleftrightarrow{\mathbf{A D}}\) and \(\overleftrightarrow{\mathbf{E C}}\) are intersecting at O.
∠AOE = ∠COD (Vertically opposite angles)
∠1 = ∠3
∠EOD = ∠AOC (Vertically opposite angles)
∠EOD = ∠AOB + ∠BOC (we know ∠AOC = ∠AOB + ∠BOC)
∠2 = ∠5 + ∠4

Question 6.
Two lines \(\overleftrightarrow{\mathbf{P S}}\) and \(\overleftrightarrow{\mathbf{Q T}}\) intersect at M. Observe the figure and find x.

Answer:
In the given figure \(\overleftrightarrow{\mathbf{P S}}\) and \(\overleftrightarrow{\mathbf{Q T}}\) are intersecting at M.
∠PMQ = ∠TMS (Vertically opposite angles)
∠QMS = ∠PMT (Vertically opposite angles)
∠QMR + ∠RMS = ∠PMT (we know ∠QMS = ∠QMR + ∠RMS)

But, given, ∠QMR = 40°, ∠RMS = x° and ∠PMT = 105°
⇒ 40° + x° – 40° = 105° – 40°
∴ x° = 65°

SCERT AP 7th Class Maths Solutions Pdf Chapter 4 Lines and Angles Ex 4.2 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 4th Lesson Lines and Angles Exercise 4.2

Question 1.
Observe the given figure and write any 2 linear pairs of angles.

Answer:
Linear pair of angles :
∠KOQ, ∠QOP
∠KOQ, ∠KOL
∠KOL, ∠LOP
∠QOP, ∠POL
∠KOM, ∠MOP
∠QOM, ∠MOL

Question 2.
Draw a pair of adjacent angles which are complementary to each other.
Answer:

In the figure ∠AOB, ∠BOC are adjacent angles and their sum is complementary.
∴ ∠AOB + ∠BOC = 50° + 40° = 90°

Question 3.
Draw a pair of adjacent angles which are supplementary to each other.
Answer:

In the figure ∠POQ, ∠OOR are linear pair and adjacent angles and their sum is supplementary.
∴ ∠POQ + ∠OOR = 110° + 70° = 180°

Question 4.
Give any two examples of adjacent angles in your surroundings.
Answer:
Fan, Bullock cart wheel, Clock, etc.

Question 5.
Observe the figure and write the adjacent angles.

Answer:
In the given figure, adjacent angles are
∠AOC, ∠COB
∠AOC, ∠COD
∠COD, ∠DOB
∠AOD, ∠DOB

Question 6.
Is it possible for the following pair of angles to form a linear pair? If yes, draw them. If not, give reason.
(i) 120°, 60°
Answer:
Yes.

120° + 60° = 180°
120°, 60° (Linear pair)

(ii) 98°, 102°
Answer:
No, their sum is not equal to 180°
98° + 102° = 200° > 180°
So, they are not 1irear pair.

Question 7.
Draw the following angles as linear pair. Write the straight line and common arm.

Answer:
∠BOA + ∠AOC = 180°

∴ ∠BOA and ∠AOC are linear pair.
BC is a straight line and OA is an arm.

Question 8.
In the given figure, AB is a straight line. O is a point on AB . Find the value of x.

Answer:
In the given figure \(\overrightarrow{\mathrm{AB}}\) is a straight line.
∠AOC + ∠COB = 180°

⇒ 2x + 30° + x = 180°
⇒ 3x + 30°= 180°
⇒ 3x + 30° – 30°= 180°
⇒ 3x = 150°
⇒ \(\frac{3 x}{3}=\frac{150^{\circ}}{3}\)
⇒ x = 50

Question 9.
Teacher asked students to check whether 40° and 140° form a linear pair or not by drawing angles. The students have drawn as follows. Who will get correct answer?

Answer:
Raheem: ∠AOD = 140°, ∠BOC = 40° they are not adjacent angles and not linear pair.
So, Raheem is not correct.

Mary: ∠COD = 140°, ∠BOC = 40° they are adjacent angles, but they are not linear pair.
So, Mary answer is not correct.

Roshitha: ∠XOZ = 140°, ∠ZOY = 40° they are adjacent angles and linear pair.
So, Roshitha answer is correct.

SCERT AP 7th Class Maths Solutions Pdf Chapter 4 Lines and Angles Ex 4.1 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 4th Lesson Lines and Angles Exercise 4.1

Question 1.
Find which of the following pair of angles are complementary and which are supplementary?

Answer:
90° + 20° = 1 10° ≠ 90° (or) ≠ 180°
So, they are not complementary or not supplementary.

Answer:
6o° + 30° = 90°
So, they are complementary angles.

Answer:
11°+ 79° = 90°
So, they arc complementary angles.

Answer:
125° + 55° = 180°
So, they are supplementary angles.

Answer:
50 + 130° = 180°
So, they are supplementary angles.

Answer:
40° + 120° = 160° ≠ 90° (or) 180°
So, they are not complementary angles or not supplementary angles.

Question 2.
If the ratio of two complementary angles is 2:3, then find the two angles.
Answer:
Given the ratio of two complementary angles = 2 : 3 = 2x : 3x
Sum of the complementary angles = 90° So, 2x + 3x = 90°
⇒ 5x = 90°
⇒ \(\frac{5 x}{5}=\frac{90^{\circ}}{5}\)
∴ x = 18°
Then the angles are
2x, 3x = 2(18°), 3(18°) = 36°, 54°

Question 3.
In the figure, ∠A and ∠Q are complementary angles. Find the value of x.

Answer:
Given ∠A = 9x°, ∠Q = 36° and
∠A and ∠Q are complementary angles.
So, ∠A + ∠Q = 90°
⇒ 9x + 36° = 90°
⇒ 9x + 36° – 36° = 90° – 36°
⇒ 9x = 54°
⇒ \(\frac{9 \mathrm{x}}{9}=\frac{54^{\circ}}{9}\)
⇒ x = 6° .

Question 4.
If ∠A and ∠B are conjugate angles and ∠A = ∠B. Find the two angles.
Answer:
Given ∠A and ∠B are conjugate angles and
∠A = ∠B = x°
So, ∠A + ∠B = 360°
⇒ x° + x° – 360°
⇒ 2x° = 360°
⇒ \(\frac{2 x}{2}=\frac{360^{\circ}}{2}\)
⇒ x = 180°
∴ ∠A = ∠B = x° = 180°

Question 5.
Draw a pair of complementary angles and a pair of supplementary angles.
Answer:

Question 6.
Teacher asked students to draw the complementary angle to the given angle. The students have drawn as follows. Who is correct?

Answer:
Sum of the two angles is 90°, then they are called complementary angles.
Complementary angle of 30° is 60°. So, Sujatha is correct.

Question 7.
In the given figure, ∠B and ∠E are supplementary angles. Find x.

Answer:
Given ∠B and ∠E are supplementary angles.
∠B + ∠E = 180°
⇒ 30° + 5x = 180°
⇒ 30° + 5x – 30° = 180°- 30°
⇒ 5x = 150°
⇒ \(\frac{5 x}{5}=\frac{150^{\circ}}{5}\)
∴ x = 30°

Question 8.
Ashritha said, “In the pair of supplementary angles, one angle must be obtuse angle.” Do you agree? Give reason.
Answer:
No, two acute angles cannot make supplementary angles. Two right angles (90°) can make supplementary angles. One acute and one obtuse angles can make supplementary angles.

Question 9.
Find the angle which is 40° more than its supplementary angle?
Answer:
Let one angle is x°.
Other angle is (x + 40)°.
Given two angles are supplementary.
So, x + x + 40° = 180°
⇒ 2x + 40° = 180°
⇒ 2x + 40° – 40° = 180° – 40°
⇒ 2x = 140°
⇒ \(\frac{2 x}{2}=\frac{140^{\circ}}{2}\)
Angle x = 70°
Other angle is x + 40° = 70° + 40° = 110°
∴ Angles are 70° and 110°.

Question 10.
Srinu said, “Two obtuse angles cannot be supplementary.” Do you agree? Justify your answer.
Answer:
Yes, sum of two obtuse angles is greater than 180°.

SCERT AP 7th Class Maths Solutions Pdf Chapter 2 Fractions, Decimals and Rational Numbers InText Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 2nd Lesson Fractions, Decimals and Rational Numbers InText Questions

Let’s Explore [Page No. 24]

Question 1.
Here are equivalent fractions containing 1 to 9 digits only once.
Eg: \(\frac{2}{6}=\frac{3}{9}=\frac{58}{174}\) or
\(\frac{2}{4}=\frac{3}{6}=\frac{79}{158}\)
Can you write some more?
Answer:
\(\frac{3}{21}=\frac{8}{56}=\frac{7}{49}\)
\(\frac{3}{27}=\frac{6}{54}=\frac{9}{81}\)
and so on.

Let’s Think [Page No. 24]

We know that different operations with the same pair of fractions usually give different answers. Observe the following calculations which are some interesting exceptions in fractions.
(1) 11/6 + 11/5 = 11/6 × 11/5
(2) 169/30 + 13/15 = 169/30 × 13/15
Answer:

Lets Do Activity [Page No. 241]

Question 1.
Make two dice with card board or wood. Paste colour chart papel- to the all faces of the each dice. Write any three like fractions and three unlike fractions on faces of each dice. Now in a group each time two students throw the dice who are sitting opposite to each other. Write the come up fraction on the dice In the given table and do the four fundamental operations with those two fractions. Submit the Ailed table to your teacher.
Answer:
Let the fractions be \(\frac{1}{3}, \frac{2}{3}, \frac{5}{3}\) and \(\frac{1}{2}, 1 \frac{3}{4}, \frac{5}{6}\)

[Page No. 27]

Observe the following table and fill in the blanks.

Check Your Progress [Page No. 28]

Find the product:
Question 1.
32.5 × 8
Answer:
32.5 × 8
= \(\frac{325}{10}\) × 8
= \(\frac{2600}{10}\)
∴ 32.5 × 8 = 260

Question 2.
94.62 × 7
Answer:
94.62 × 7
= \(\frac{9462}{100} \times \frac{7}{1}\)
= \(\frac{66234}{100}\)
∴ 94.62 × 7 = 662.34

Question 3.
109.761 × 31
Answer:
109.761 × 31
= \(\frac{109761}{1000}\) × 31
= \(\frac{3402591}{1000}\)
∴ 109.761 × 31 = 3402.591

Question 4.
61 × 2.39
Answer:
61 × 2.39
= 61 × \(\frac{239}{100}\)
= \(\frac{14579}{100}\)
∴ 61 × 2.39 = 145.79

Check Your Progress [Page No. 29]

Find the values of the following,
(i) 26.59 × 10
Answer:
26.59 × 10
Number of zeroes in 10 is one. So, decimal point has to shift to one place right in the product.
∴ 26.59 × 10 = 265.9

(ii) 206.5 × 100
Answer:
206.5 × 100
Number of zeroes in 100 is two. So, decimal point has to shift to two places right in the product.
∴ 206.5 × 100 = 20650

(iii) 206.5 × 1000
Answer:
206.5 × 1000
Number of zeroes in 1000 is three. So, decimal point has to shift to three places right in the product.
∴ 206.5 × 100 = 206500

(iv) 10.001 × 1000
Answer:
10.001 × 1000
Number of zeroes in 1000 is three. So, decimal point has to shift to three places right in the product.
∴ 10.001 × 1000 = 10001

Check Your Progress [Page No. 30]

Question 1.
Find the product of the following,
(i) 69.2 × 2.5
Answer:
69.2 × 2.5
= \(\frac{692}{10} \times \frac{25}{10}\)
= \(\frac{692 \times 25}{10 \times 10}\)
= \(\frac{17300}{100}\)
∴ 69.2 × 2.5 = 173

(ii) 20.61 × 3.09
Answer:
20.61 × 3.09
\(\frac{2061}{100} \times \frac{309}{100}=\frac{636849}{10000}\)
∴ 20.61 × 3.09 = 63.6849

(iii) 658.321 × 43.2
Answer:
658.321 × 43.2
= \(\frac{658321}{1000} \times \frac{432}{10}\)
= \(\frac{658321 \times 432}{1000 \times 10}\)
= \(\frac{284394672}{10000}\)
∴ 658.321 × 43.2 = 28439.4672

(iv) 206.005 × 0.07
Answer:
206.005 × 0.07
= \(\frac{206005}{1000} \times \frac{7}{100}\)
= \(\frac{206005 \times 7}{1000 \times 100}\)
= \(\frac{1442035}{100000}\)
∴ 206.005 × 0.07 = 14.42035

Let’s Explore [Page No. 30]

Question 1.
Observe the figure. Fill the blue boxes with suitable decimal numbers.
Answer:

→ I am a decimal number, who is half of one fourth of 100. Who am I?
Answer:
12.5

[Page No. 32]

Fill the following blanks:

Answer:

Check Your Progress [Page No. 32]

Question 1.
Find the following.
(i) 81.5 ÷ 10
Answer:
81.5 ÷ 10

∴ 81.5 ÷ 10 = 8.15

(ii) 4901.2 ÷ 100
Answer:
4901.2 ÷ 100

∴ 4901.2 ÷ 100 = 49.012

(iii) 7301.3 ÷ 1000
Answer:
7301.3 ÷ 1000

∴ 7301.3 ÷ 1000 = 7.3013

(iv) 1.2 ÷ 100
Answer:
1.2 ÷ 100

∴ 1.2 ÷ 100 = 0.012

[Page No. 33]

Question 2.
Find the following :
(i) 69.4 ÷ 2
Answer:
69.4 ÷ 2

∴ 69.4 ÷ 2 = 34.7

(ii) 56.32 ÷ 8
Answer:
56.32 ÷ 8

∴ 56.32 ÷ 8 = 7.04

(iii) -6.5 ÷ 4
Answer:
6.5 ÷ 4

∴ 6.5 ÷ 4 = 1.625

(iv) 108.7 ÷ 5
Answer:
108.7 ÷ 5

∴ 108.7 ÷ 5 = 21.74

Check Your Progress [Page No. 35]

Solve the following :
(i) 0.45 ÷ 0.9
Answer:
0.45 ÷ 0.9
∴ 0.45 ÷ 0.9 = 0.5

(ii) 2.125 ÷ 0.05
Answer:
2.125 ÷ 0.05

∴ 2.125 ÷ 0.05 = 42.5

(iii) 94.3 ÷ 0.004
Answer:
94.3 ÷ 0.004

= 943 × 25
∴ 94.3 ÷ 0.004 – 23575

iv) 10.25 ÷ 0.2
Answer:
10.25 ÷ 0.2

∴ 10.25 ÷ 0.2 = 51.25

Examples:

Question 1.
In the school out of 180 students, of the students are boys. Find the number of girls in the school?
Answer:
Number of students in the school = 180
Part oL the boys in the school = \(\frac{4}{9}\)
Number of boys = \(\frac{4}{9}\) of 180
= \(\frac{4}{9}\) × 180 = 80
∴ Number of girls = 180 – 80 = 100

Question 2.
if the cost of 22\(\frac{1}{2}\) kg. of apples (a box) in a whole sale market is ₹ 1170, then find the cost of 5 kg. of apples.

Answer:
Cost 0f 22\(\frac{1}{2}\)kg. of apples = ₹ 1170
Cost of 1 kg. ofapple = ₹ 1170 ÷ 22\(\frac{1}{2}\)
= ₹ 1170 × \(\frac{45}{2}\)
= ₹1170 × \(\frac{45}{2}\) = ₹ 52
= 1170 × \(\frac{2}{45}\)

∴ Cost of 5kg. óf apples = 5 × ₹ 52
= ₹ 260

Question 3.
If one side of a square is 3.8 cm, then find its perimeter.
Answer:

Number of sides to a square = 4
Side of a square = 3.8 cm
Each side of a square is equal.

Perime ret of a square = 4 × side
= 4 × 3.8
= 15.2 cm

Question 4.
Find :
(i) 239.27 × 10
Answer:
239.27 × 10 (number of zeroes in 10 is one. So, decimal point has to shift to one place right in the product).
∴ 239.27 × 10 = 2392.7

(ii) 5.305 × 100
Answer:
5.305 × 100 = 530.5

(iii) 23.1 × 1000
Answer:
23.1 × 1000 = 23100.0 = 23100

Question 5.
Bindu went to vegetable market with her mother to buy 3.5 kg Onions. If the cost of Onions is ₹ 18.50 per kg., then find the cost of 3.5 kg of Onions.
Answer:
Cost of 1 kg. Onions = ₹ 18.50
Cost of 3.5 kg. of Onions = ₹ 18.50 × 3.5 = 64.750
Cost of 3.5 kg. of Onions = ₹ 64.75

Step 1 : Multiply whole numbers ignoring decimals
35 × 1850 = 64750.

Step 2: As there are total 2 + 1 = 3 decimals, put decimal point after three digits from the right most to the product.
So, 3.5 × 18.50 = 64.750.

Question 6.
Madhuri is studying 7th class in Visakhapatnam. Her school teachers organised a tour to Araku valley by bus. Bus covered a distance of 98.5 km. in 2.5 hours. If the bus is travelled at the same speed in the journey, then find the distance travelled in 1 hour.
Answer:
Distance travelled by bus = 98.5 km. Time taken to travel this distance = 2.5 hours.
∴ Distance travelled by bus in 1 hour = 98.5 ÷ 2.5
= \(\frac{985}{25}\) = 39.4 km.
∴ Bus travelled in 1 hour = 39.4 km.

Practice Questions: [Page No. 39]

Question 1.
15, 27, 39, 51, 63,……………
(a) 85
(b) 75
(c) 65
(d) 73
Answer:
(b) 75

Explanation:
15, 27, 39, 51, 63,…………..
(13 × 1 + 2), (13 × 2 + 1), (13 × 3 + 0), (13 × 4 – 1), (13 × 5 – 2), ………………….
So, next number is (13 × 6 – 3)
= 78 – 3 = 75

Question 2.
2, 5, 10, 17, 26, 37, …………….
(a) 48
(b) 75
(c) 50
(d) 73
Answer:
(c) 50

Explanation:
2, 5, 10, 17, 26, 37, ……….
(1 × 1 + 1), (2 × 2 + 1), (3 × 3 + 1), (4 × 4 + 1), (5 × 5 + 1), (6 × 6 + 1), ………….
So, next number is (7 × 7 + 1) = 50

Question 3.
1, 6, 16, 31, 51, 76 ……………..
(a) 95
(b) 86
(c) 91
(d) 96
Answer:
No option.

Explanation:
1, 6, 16, 31, 51, 76, …………..
1, (1 + 5), (6 + 10), (16 + 15), (31 + 20), (51 + 25), (76 + 30), ……………
∴ So, next number is (76 + 30) = 106.

Question 4.
13, 14, 16, 20, 28, 44, …………..
(a) 76
(b) 75
(c) 87
(d) 73
Answer:
(a) 76

Explanation:
13, 14, 16, 20, 28, 44, …………
(12 + 2⁰), (12 + 2¹), (12 + 2²), (12 + 2³), (12 + 2⁴), (12 + 2⁵),
So, next number is 12 + 2⁶ = 12 + 64 = 76

Question 5.
28, 25, 30, 27, 32, 29, ………….
(a) 26
(b) 24
(c) 34
(d) 32
Answer:
(c) 34

Explanation:
28, 25, 30, 27, 32, 29, …………

So, next number is 32 + 2 = 34

Question 6.
3, -6, 12, -24, 48, -96, ………..
(a) 192
(b)- 102
(c)- 192
(d) 106
Answer:
(a) 192

Explanation:
3, -6, 12, -24, 48, -96, …………
3, (3 × -2), (-6 × -2), (12 × -2), (-24 × -2), (48 × -2)…………..
So, next number is (-96 × – 2) = 192.

Question 7.
1, 2, 6, 24, 120, 720, ……………….
(a) 920
(b)5040
(c) 1040
(d)4320
Answer:
(b)5040

Explanation:
1, 2, 6, 24, 120, 720, …………………
1, (1 × 2), (2 × 3), (6 × 4), (24 × 5), (120 × 6), ……………..
So, next number is 720 × 7 = 5040

Question 8.
63, 64, 67, 72, 79,……………
(a) 88
(b) 86
(c) 87
(d) 98
Answer:
(a) 88

Explanation:
63, 64, 67, 72, 79, ………..
63, (63 + 1), (64 + 3), (67 + 5), (72 + 7),….
So, next number is (79 + 9) = 88

Question 9.
9, 10, 22, 69, 280, …………..
(a) 1205
(b) 1425
(c) 1400
(d)1405
Answer:
(d)1405

Explanation:
9, 10, 22, 69, 280, ………….
9, (9 × 1 + 1), (10 × 2 + 2), (22 × 3 + 3), (69 × 4 + 4),………………
So, next number is (280 × 5 + 5) = 1405

Question 10.
729, 243, 81, 27, …………….
(a) 65
(b) 18
(c) 9
(d) 73
Answer:
(c) 9

Explanation:
729, 243, 81, 27 ………….
(729 ÷ 3 = 243), (243 ÷ 3 = 81), (81 ÷ 3 = 27), ……………….
So, next number is 27 ÷ 3 = 9

Question 11.
5, 15, 35, 75, 155,
(a) 215
(b) 305
(c) 315
(d) 265
Answer:
(c) 315

Explanation:
5, 15, 35, 75, 155, ………..
5 × 1, 5 × 3, 5 × 7, 5 × 15, 5 × 31, …………….
5(2¹ – 1), 5(2² – 1), 5(2³ – 1), 5(2⁴ – 1), 5(2⁵ – 1), ………..
So, next number is 5 x (2⁶ – 1) = 5 × 63 = 315

Question 12.
240, 240, 120, 40, ……………., 2.
(a) 10
(b) 20
(c) 18
(d) 35
Answer:
(a) 10

Explanation:
240, 240, 120, 40, ……………….,2.
240, 240 ÷ 1, 240 ÷ 2, 120 ÷ 3, 40 ÷ 4, 10 ÷ 5
So, next number is 40 ÷ 4 = \(\frac{40}{4}\) =10

Question 13.
20, 10, 10, 20, 80, ……….
(a) 320
(b) 640
(c) 400
(d) 80
Answer:
(b) 640

Explanation:
20, 10, 10, 20, 80, ……………
20, 20 × 2^-1, 10 × 2⁰, 10 × 2¹, 20 × 2²,
So, next number is 80 × 2³
= 80 × 8 = 640

Question 14.
7, 10, 8, 11, 9, 12, ………….
(a) 8
(b) 14
(c) 15
(d) 10
Answer:
(d) 10

Explanation:

So, next number is 9 + 1 = 10.

Question 15.
34, 30, 28, 24, 22, 18, ………….
(a) 16
(b) 14
(c) 20
(d) 15
Answer:
(a) 16

Explanation:
34, 30, 28, 24, 22, 18, ……

So, next number is (22 – 6) = 16

Number Series – 2

Question 1.
Adding or subtract of natural numbers:
Eg: 6, 7, 9, 12, 16, 21,…
(a) 21
(b) 25
(c) 27
(d) 28
Answer:
(c) 27

Explanation:
(6 + 1), (7 + 2,), (9 + 3), (12 + 4), (16 + 5)
so, next number is (21 + 6) = 27

Question 2.
Add the pattern:
Eg: 10, 20, 40, 70, 110, …
(a) 160
(b) 180
(c) 150
(d) 210
Answer:
(a) 160

Explanation:
(10 + 10,), (20 + 20), (40 + 30), (70 + 40)
so, next number is (110 + 50) = 160

Question 3.
Subtracting or adding odd numbers:
Eg: 27, 26, 23, 18, 11, …
(a) 4
(b) 2
(c) 9
(d) 5
Answer:
(b) 2

Explanation:
(27 – 1), (26 – 3), (23 – 5) (18 – 7)
so, next number is (11 – 9) = 2

Question 4.
Multiply with a fixed number
Eg: 5, 15, 45, 135, 405, ……..
(a) 1200
(b) 1215
(c) 850
(d) 925
Answer:
(b) 1215

Explanation:
(5 × 3), (15 × 3), (45 × 3), (135 × 3),
so, next number is (405 × 3) = 1215

Question 5.
Multiply and add with same natural numbers:
Eg: 5, 6, 14, 45 ………….
(a) 184
(b) 180
(c) 176
(d) 225
Answer:
(a) 184

Explanation: (5 × 1) + 1, (6 × 2) + 2, (14 × 3) +3,
so, next number is (45 × 4) + 4 = 184

Question 6.
Multiply and add with a different fixed numbers
Eg: 3, 9, 21, 45, 93 …
(a) 184
(b) 187
(c) 186
(d) 189
Answer:
(b) 187

Explanation:
(3 × 2) + 3, (9 × 2) + 3, (21 × 2) + 3, (45 × 2) + 3,
so, next number is (92 x 2) +3 = 187

Question 7.
Multiply with fixed number and add different numbers:
Eg: 12, 25, 52, 107.
(a) 196
(b) 207
(c) 214
(d) 218
Answer:
(d) 218

Explanation: (12 × 2) + 1, (25 × 2) + 2, (52 × 2) + 3,
so, next number is (107 × 2) + 4 = 218

Question 8.
Multiply the sequence number :
Eg: 7, 14, 42, 168, 840.
(a) 1680
(b) 5040
(c) 760
(d) 4200
Answer:
(b) 5040

Explanation:
(7 × 2), (14 × 3), (42 × 4), (168 × 5)
so, next number is (840 × 6) = 5040

Question 9.
Dividing with a fixed number
Eg: 256, 128, 64, 32, 16, …
(a) 8
(b) 4
(c) 16
(d) 10
Answer:
(a) 8

Explanation: (256/2), (128/2), (64/2),(32/2), ……….
so, next number is (16/2) = 8

Question 10.
Multiply and divide with a different fixed number.
Eg: 12, 60, 30, 150, 75,
(a) 325
(b) 150
(c) 375
(d) 300
Answer:
(c) 375

Explanation: (12× 5), (60/2), (30 × 5),(150/2), …
so, next number is (75 × 5) = 375

SCERT AP 7th Class Maths Solutions Pdf Chapter 2 Fractions, Decimals and Rational Numbers Review Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 2nd Lesson Fractions, Decimals and Rational Numbers Review Exercise

Question 1.
Write the following fractions in ascending order.
(i) \(\frac{3}{2}, \frac{5}{2}, \frac{1}{2}, \frac{17}{2}, \frac{9}{2}\)
Answer:
Given \(\frac{3}{2}, \frac{5}{2}, \frac{1}{2}, \frac{17}{2}, \frac{9}{2}\)
Given fractions are like fractions.

So, we should arrange the numerators in ascending order.
\(\frac{1}{2}<\frac{3}{2}<\frac{5}{2}<\frac{9}{2}<\frac{17}{2}\)

Ascending order: \(\frac{1}{2}<\frac{3}{2}<\frac{5}{2}<\frac{9}{2}<\frac{17}{2}\)

(ii) \(\frac{6}{5}, \frac{11}{10}, \frac{19}{5}, \frac{7}{10}, \frac{5}{10}\)
Answer:
Given \(\frac{6}{5}, \frac{11}{10}, \frac{19}{5}, \frac{7}{10}, \frac{5}{10}\)
Given fractions are not like fractions. So, first we have to convert them into like fractions.

Now arrange them in ascending order according to their numerators.
\(\frac{5}{10}<\frac{7}{10}<\frac{11}{10}<\frac{12}{10}<\frac{38}{10}\)
Ascending order: \(\frac{5}{10}<\frac{7}{10}<\frac{11}{10}<\frac{12}{10}<\frac{38}{10}\)

(iii) \(\frac{8}{3}, \frac{7}{6}, 3 \frac{1}{4}, \frac{5}{3}, \frac{11}{4}\)
Answer:
Given \(\frac{8}{3}, \frac{7}{6}, 3 \frac{1}{4}, \frac{5}{3}, \frac{11}{4}\)
\(\frac{8}{3}, \frac{7}{6}, \frac{13}{4}, \frac{5}{3}, \frac{11}{4}\)
We have to find the LCM of denominators to convert them into equal fraction.

LCM of numerators is 12

Now arrange the like fractions in ascending order according to their numerators.
\(\frac{14}{12}<\frac{20}{12}<\frac{32}{12}<\frac{33}{12}<\frac{39}{12}\)
Ascending order :
\(\frac{14}{12}<\frac{20}{12}<\frac{32}{12}<\frac{33}{12}<\frac{39}{12}\)

Question 2.
Calculate the following.
(i) \(\frac{3}{5}+\frac{7}{4}\)
Answer:
Given \(\frac{3}{5}+\frac{7}{4}=\frac{12+35}{20}=\frac{47}{20}\)

(ii) \(\frac{5}{6}+\frac{7}{12}\)
Answer:
Given \(\frac{5}{6}+\frac{7}{12}\)
= \(\frac{5 \times 2}{6 \times 2}+\frac{7}{12}\)
= \(\frac{10}{12}+\frac{7}{12}\)
= \(\frac{10+7}{12}=\frac{17}{12}=1 \frac{5}{12}\)

(iii) 1\(\frac{7}{8}-\frac{1}{5}\)
Answer:
Given 1\(\frac{7}{8}-\frac{1}{5}\)

(iv) 4\(\frac{1}{2}\) + 3\(\frac{1}{3}\)
Answer:
Given 4\(\frac{1}{2}\) + 3\(\frac{1}{3}\)

Question 3.
Simplify the following.
(i) \(\frac{1}{4}\) of 3
Answer:
Given \(\frac{1}{4}\) of 3
= \(\frac{1}{4} \times \frac{3}{1}=\frac{3}{4}\)

(ii) \(\frac{5}{8}\) of \(\frac{2}{3}\)
Answer:
Given \(\frac{5}{8}\) of \(\frac{2}{3}\)

(iii) \(\frac{15}{4} \times 2 \frac{1}{7}\)
Answer:

(iv) 3\(\frac{1}{3}\) × 2\(\frac{2}{5}\)
Answer:
Given 3\(\frac{1}{3}\) × 2\(\frac{2}{5}\)
= \(\frac{10}{3} \times \frac{12}{5}\)
= \(\frac{120}{15}=\frac{8}{1}\) = 8

Question 4.
Calculate the following.
(i) \(\frac{3}{4}\) ÷ 3
Answer:
Given \(\frac{3}{4}\) ÷ 3
= \(\frac{3}{4}\) ÷ \(\frac{3}{1}\)

We know reciprocial of \(\frac{3}{1}\) is \(\frac{1}{3}\)
= \(\frac{3}{4} \times \frac{1}{3}\)
= \(\frac{3}{12}\) (or) \(\frac{1}{4}\)

(ii) 8 ÷ 2\(\frac{1}{7}\)
Answer:
Given 8 ÷ 2\(\frac{1}{7}\)
= 8 ÷ 2\(\frac{15}{7}\)

We know reciprocial of \(\frac{15}{7}\) is \(\frac{7}{15}\)
= 8 × \(\frac{7}{15}\)
= \(\frac{8 \times 7}{15}\)
= \(\frac{56}{15}\) (OR) 3\(\frac{11}{15}\)

(iii) \(\frac{12}{7} \div \frac{2}{7}\)
Answer:
Given \(\frac{12}{7} \div \frac{2}{7}\)
We know reciprocial of \(\frac{12}{7} \div \frac{2}{7}\)
= \(\frac{12}{7} \times \frac{7}{2}\)
= \(\frac{12 \times 7}{7 \times 2}=\frac{84}{14}=\frac{6}{1}\) = 6

(iv) 5\(\frac{1}{2}\) ÷ 2\(\frac{9}{11}\)
Answer:
Given 5\(\frac{1}{2}\) ÷ 2\(\frac{9}{11}\)
= \(\frac{11}{2} \div \frac{31}{11}\)

We know reciprocial of \(\frac{31}{11}\) is \(\frac{11}{31}\)
= \(\frac{11}{2} \times \frac{11}{31}\)
= \(\frac{121}{62}\) (OR) 1\(\frac{59}{62}\)

SCERT AP 7th Class Maths Solutions Pdf Chapter 1 Integers Ex 1.2 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 1st Lesson Integers Exercise 1.2

Question 1.
Calculate the following,
(i) (- 96) ÷ 16
Answer:
(- a) ÷ b = – \(\frac{a}{b}\)
= \(\frac{-96}{16}\) = – 6

(ii) 98 ÷ (- 49)
Answer:
a ÷ (- b) = – \(\frac{a}{b}\)
= \(\frac{98}{-49}\) = – 2

(iii) (- 51) ÷ 17
Answer:
(- a) ÷ b = – \(\frac{a}{b}\)
= \(\frac{-51}{17}\) = – 3

(iv) 38 ÷ (- 19)
Answer:
a ÷ (- b) = – \(\frac{a}{b}\)
= \(\frac{38}{-19}\) = – 2

(v) (- 80) ÷ 20
Answer:
(- a) ÷ b = – \(\frac{a}{b}\)
= \(\frac{-80}{20}\) = – 4

(vi) (- 150) ÷ (- 25)
Answer:
(- a) ÷ (- b) = – \(\frac{-a}{-b}\) = \(\frac{a}{b}\)
= \(\frac{-150}{-25}\) = \(\frac{150}{25}\) = – 6

(vii) (- 600) ÷ 60
Answer:
(- a) ÷ b = – \(\frac{a}{b}\)
= \(\frac{-600}{60}\)
= – 10

(viii) (-54) ÷ 9
Answer:
(- a) ÷ b = – \(\frac{a}{b}\)
= \(\frac{- 54}{9}\)
= – 6

(ix) 130 ÷ 65
Answer:
a ÷ b = \(\frac{a}{b}\)
= \(\frac{130}{65}\)
= 2

(x) (- 315) ÷ ( -315)
Answer:
(- a) ÷ (- b) = \(\frac{-a}{-b}\) = \(\frac{a}{b}\)
= \(\frac{-315}{-315}\)
= \(\frac{315}{315}\)
= 1

Question 2.
The product of two integers is – 165. If one number is – 15, find the other integer.
Answer:
Given one number a = – 15
Other number b = ?
Product of two numbers = (- 165)
a × b = – 165
(- 15) × b = – 165
b = (- 165) × (- 15)
Other number b = 11

Question 3.
Because of Covid-19 a company lockdown for 6 months and got loss of ₹1,32,000 in the year 2020. Find the average loss of each month.

Answer:
Given loss of 6 months = ₹ 1,32,000
Average loss of each month = 1,32,000 ÷ 6
Average loss of each month = ₹ 22,000

Question 4.
The temperature at 12 noon was 10°C above zero. If it decreases at the rate of 2°C per hour until midnight, at what time would the temperature be 8°C below zero ? What would be the temperature at mid-night ?
Answer:
The temperature at 12 noon = 10°C above zero.
The temperature is decreasing at 2°C per hour 8°C below zero = – 8°C
So, total temperature falls 2°C in every one hour.
Therefore, to decrease 18°C, time taken
= 18/2 = 9 hrs.
Present time is 12 noon, so, the time when the temperature at 12 midnight will be
= (- 8° C) + (- 2° C × 3 hrs.)
= (- 8° C) + (- 6° C)
= 160
So, at 12 midnight the temperature will be – 14°C

Question 5.
A green grocer earns a profit of ₹ 7 per kg of tomato and got loss of ₹ 4 per kg of brinjal by selling. On Monday he gets neither profit nor loss,.if he sold 68 kgs of tomato. How many kgs of brinjal did he sell ?

Answer:
Given profit per kg of tomato = ₹ 7
Loss per kg of brinjal = ₹ 4
Weight of tomatoes sold = 68 kgs
Let number of kgs of tomato be x kg.
Number of kgs of brinjals be y kg.
7x – 4y = 0
7 × 68 – 4y = 0
– 4y = – 7 × 68
y = \(\frac{-7 \times 68}{-4}\)
∴ Weight of brinjals sold = + 7 × 17
= + 119 kgs.

Question 6.
In a test, each correct answer carry + 3 marks and incorrect answer – 1 mark, Sona attempted all the questions and scored + 20 marks though she got 10 correct answers.
(i) How many incorrect answers did she attempt ?
(ii) How many questions were given in the test ?
Answer:
Given marks for every correct answer = + 3
Given marks for every incorrect answer = – 1
No. of correct answer questions = 10
Total marks gained on correct answer questions = 10 × (+3) = + 30

(i) No. of incorrect answer questions = x
Total marks gained on incorrect answer questions = y × (-1) = – y.
Marks scored by Sona
⇒ + 30 + (- y) = + 20
⇒ – y = + 20 – 30
⇒ – y = – 10
⇒ y = 10
∴ No. of incorrect answers Sona attempts =10

(ii) Number of questions given in the test = No. of correct answer questions + No. of incorrect answer questions.
= 10 + 10 = 20 questions.

Question 7.
Write 5 pairs of integers (a, b) such that a ÷ b = -4.
(Ex : (12, – 3) because 12 + (-3) = -4)

Answer:
(i) (-16, +4) because (- 16) ÷ 4 = -4
(ii) (20, -5) because 20 ÷ (-5) = – 4
(iii) (—8, 2) because (-8) ÷ 2 =- 4
(iv) (24, -6) because 24 ÷ (-6) = – 4
(v) (-36, 9) because (-36) ÷ 9 = – 4

(OR)

SCERT AP 7th Class Maths Solutions Pdf Chapter 2 Fractions, Decimals and Rational Numbers Unit Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 2nd Lesson Fractions, Decimals and Rational Numbers Unit Exercise

Question 1.
Choose the correct answer,
(i) The set of integers are denoted by [ ]
(A) N
(B) W
(C) Z
(D) Q
Answer:
(C) Z

(ii) Number of decimal places in the product of 48.23 × 0.2 [ ]
(A) 2
(B) 3
(C) 1
(D)5
Answer:
(B) 3

(iii) Number of decimal places to the quotient of 537.1 + 10 [ ]
(A) 1
(B) 2
(C) 4
(D) 3
Answer:
(B) 2

(iv) An integer can be ____________
(A) negative
(B) positive
(C) zero
(D) all the above
Answer:
(D) all the above

Question 2.
Fill in the blanks:
(i) The numbers written in the form of \(\frac{p}{q}\), where p. q are integers and q ≠ 0 are __________ numbers
Answer:

(ii) 0.11 × 0.11 = __________
Answer:
0.11 × 0.11
= \(\frac{11}{100} \times \frac{11}{100}\)
= \(\frac{121}{10000}\) = 0.0121

(iii) Standard form of – \(\frac{15}{6}\) = ______
Answer:
Standard form of – \(\frac{15}{6}\)

(iv) Equivalent fraction to –\(\frac{2}{3}\) = __________
Answer:
Equivalent fraction to –\(\frac{2}{3}\)

Question 3.
Find the product:
(i) 2.1 × 6.3
Answer:
2.1 × 6.3
= \(\frac{21}{10} \times \frac{63}{10}\)
= \(\frac{1323}{100}\) = 13.23
∴ 2.1 × 6.3 = 13.23

(ii) 43.205 × 1.27
Answer:
43.205 × 1.27
= \(\frac{43205}{1000} \times \frac{127}{100}\)
= \(\frac{43205 \times 127}{1000 \times 100}\)
= \(\frac{5487035}{100000}\)
∴ 43.205 × 1.27 = 54.87035

(iii) 7.641 × 3.5
Answer:
7.64 1 × 3.5
= \(\frac{7641}{1000} \times \frac{35}{10}\)
= \(\frac{7641 \times 35}{1000 \times 10}\)
= \(\frac{267435}{10000}\)

∴ 7.641 × 3.5 = 26.7435.

(iv) 5.24 × 0.99
Answer:
5.24 × 0.99
= \(\frac{524}{100} \times \frac{99}{100}\)
= \(\frac{524 \times 99}{100 \times 100}\)
= \(\frac{51876}{10000}\)

∴ 5.24 × 0.99 = 5.1876

Question 4.
Solve the following :
(i) 61.24 ÷ 0.4
Answer:
61.24 ÷ 0.4

∴ 61.24 ÷ 0.4 = 153.1

(ii) 23.45 ÷ 1.5
Answer:
23.45 ÷ 1.5

∴ 23.45 ÷ 1.5 = 15.633….

(iii) 0.312 ÷ – 0.6
Answer:
0.312 ÷ – 0.6

∴ 0.312 ÷ – 0.6 = -0.52

(iv) 32.2 ÷ 2.2
Answer:
32,2 ÷ 2.2

∴ 32.2 ÷ 2.2 = 14.6363….,

Question 5.
Multiply 0.04 by – \(\frac{1}{2}\)
Answer:
Given multiply 0.04 by –\(\frac{1}{2}\)

Question 6.
Find standard form of – \(\frac{15}{35}\)
Answer:

Question 7.
A bus travelled 300 km In 7 \(\frac{1}{2}\)hours with uniform speed. Find how many km, it travelled in 1 hour.
Answer:
Given distance travelled by bus in 7\(\frac{1}{2}\)
hours = 300 km

Distance travelled by bus in 1 hour
= 300 ÷ 7\(\frac{1}{2}\)

Question 8.
Suvarna had 300. She spent \(\frac{1}{3}\) of her money on notebooks and \(\frac{1}{4}\) of the remaining on stationery items. How much money is left with ber?
Answer:
Total money Suvarna had = ₹ 300
Money spent on notebooks
= \(\frac{1}{3}\) of ₹ 300

Remaining amount
= Total money – spent on books
= 300 – 100 = ₹ 200

Money spent on stationery
= \(\frac{1}{4}\) of remaining amount

Money left with Suvarna
= Total money – spent on books – spent on stationery
= 300 – 100 – 50
= 300 – 150
∴ Money left with Suvarna = ₹ 150

Question 9.
One litre of diesel costs ₹ 84.65. What is the cost of 12.5 liters of diesel?
Answer:
Given cost of one litre diesel = ₹ 84.65
Cost of 12.5 liters diesel = 84.65 × 12.5
= \(\frac{8465}{100} \times \frac{125}{10}\)
= \(\frac{1058125}{1000}\) = ₹ 1058.125
∴ Cost of 12.5 litres diesel = ₹ 1058.125

Question 10.
Represent \(\frac{-2}{5}, \frac{-3}{5}, \frac{-1}{5}, \frac{3}{5}\) on same number line.
Answer:
Given fractions are \(\frac{-2}{5}, \frac{-3}{5}, \frac{-1}{5}, \frac{3}{5}\)
Ascending order: \(\frac{-3}{5}, \frac{-2}{5}, \frac{-1}{5}, \frac{3}{5}\)

SCERT AP 7th Class Maths Solutions Pdf Chapter 2 Fractions, Decimals and Rational Numbers Ex 2.3 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 2nd Lesson Fractions, Decimals and Rational Numbers Exercise 2.3

Question 1.
Fill ¡n the blanks in the table. One is done for you.
Answer:

Question 2.
Solve the following,
(i) 5.51 ÷ 2
Answer:
5.51 ÷ 2
= \(\frac{551}{100} \div \frac{2}{1}\)
= \(\frac{551}{100} \times \frac{1}{2}\)
= \(\frac{551 \times 1}{100 \times 2}\)
= \(\frac{551}{200}\)
∴ 5.51 ÷ 2 = 2.755

(ii) 38.4 ÷ 3
Answer:
38.4 ÷ 3
= \(\frac{384}{10} \div \frac{3}{1}\)
= \(\frac{384}{10} \times \frac{1}{3}\)
= \(\frac{384 \times 1}{10 \times 3}\)
= \(\frac{384}{30}\)
∴ 38.4 ÷ 3 = 12.8

(iii) 57.39 ÷ 6
Answer:
57.39 ÷ 6
= \(\frac{5739}{100} \div \frac{6}{1}\)
= \(\frac{5739}{100} \times \frac{1}{6}\)
= \(\frac{5739 \times 1}{100 \times 6}\)
= \(\frac{5739}{600}\)
∴ 57.39 ÷ 6 = 9.565

(iv) 562.1 ÷ 11
Answer:
562.1 ÷ 11
= \(\frac{5621}{10} \div \frac{11}{1}\)

= \(\frac{511}{10}\)
∴ 562.1 ÷ 11 = 51.1

(v) 0.7005 ÷ 5
Answer:
0.7005 ÷ 5
= \(\frac{7005}{10000} \div \frac{5}{1}\)

= \(\frac{1401}{10000}\)
∴ 0.7005 ÷ 5 = 0.1401

(vi) 9.99 ÷ 3
Answer:
9.99 ÷ 3
= \(\frac{999}{100} \div \frac{3}{1}\)

= \(\frac{333}{100}\)
∴ 9.99 ÷ 3 = 3.33

(vii) 13 ÷ 6.5
Answer:
13 ÷ 6.5
= \(\frac{13}{1} \div \frac{65}{10}\)

= \(\frac{2}{1}\)
∴ 13 ÷ 6.5 = 2

(viii) 10.01 ÷ 11
Answer:
10.01 ÷ 11
= \(\frac{1001}{100} \div \frac{11}{1}\)

= \(\frac{91}{100}\)
∴ 10.01 ÷ 11 = 0.91

(ix) 8 ÷ 0.32
Answer:
8 ÷ 0.32

∴ 8 ÷ 0.32 = 25

(x) 320.1 ÷ 33
Answer:
320.1 ÷ 33

∴ 320.1 ÷ 33 = 9.7

Question 3.
Solve the following divisions,
(i) 78.24 ÷ 0.2
Answer:
78.24 ÷ 0.2

∴ 78.24 ÷ 0.2 = 391.2

(ii) 4.845 ÷ 1.5
Answer:
4.845 ÷ 1.5
= \(\frac{4845}{1000} \div \frac{15}{10}\)

(iii) 0.246 ÷ 0.6
Answer:

(iv) 563.2 ÷ 2.2
Answer:
563.2 ÷ 2.2

∴ 563.2 ÷ 2.2 = 256

(v) 0.026 ÷ 0.13
Answer:
0.026 ÷ 0.13

∴ 0.026 ÷ 0.13 = 0.2

(vi) 4.347 ÷ 0.09
Answer:
4.347 ÷ 0.09
\(\frac{4347}{1000} \div \frac{9}{100}\)

∴ 4.347 ÷ 0.09 = 48.3

(vii) 3.9 ÷ 0.13
Answer:
3.9 ÷ 0.13

∴ 3.9 ÷ 0.13 = 30

(viii) 2032 ÷ 0.8
Answer:
20.32 ÷ 0.8

∴ 20.32 ÷ 0.8 = 25.4

(ix) 24.4 ÷ 6.1
Answer:
24.4 ÷ 6.1

∴ 24.4 ÷ 6.1 = 4

(x) 2.164 ÷ 0.008
Answer:
2.164 ÷ 0.008
\(\frac{2164}{1000} \div \frac{8}{1000}\)

∴ 2.164 ÷ 0.008 = 270.5

Question 4.
Solve the following.
(i) Divide 39.54 by 6
Answer:
39.54 by 6
= 39.54 ÷ 6

∴ 39.54 ÷ 6 = 6.59

(ii) Divide 72 by 10
Answer:
7.2 by 10
= 7.2 ÷ 10
= \(\frac{72}{10} \div \frac{10}{1}\)
= \(\frac{72}{10} \times \frac{1}{10}\)
= \(\frac{72 \times 1}{10 \times 10}=\frac{72}{100}\)
∴ 7.2 ÷ 10 = 0.72

(iii) Divide 5.2 by 1.3
Answer:
5.2 by 1.3
= 5.2 ÷ 1.3

∴ 5.2 ÷ 1.3 = 4

Question 5.
‘Sekhar travelled 154.5 km. in 5 hours with uniform speed on his bike. How much distance does he travel in one hour ?
Answer:
Given, distance travelled in 5 hours = 154.5 km

∴ Distance travelled by Sekhar in 1 hour = 30.9 km

Question 6.
If a Mason worked 100 hours in 12.5 days to construct a wall, then how many hours he totally worked in a day?

Answer:
Given number of hours worked in 12.5 days to.construct a wall = 100 hours.
Number of hours worked in one day
= 100 ÷ 12.5
= \(\frac{100}{1} \div \frac{125}{10}\)
= \(\frac{100}{1} \times \frac{10}{125}\)
= \(\frac{1000}{125}\)
∴ Number of hours worked by Mason in one day = 8 hours.

Question 7.
If the cost of dozen eggs is ₹ 61.80, then find the cost of an egg.
Answer:
Given, cost of dozen eggs = ₹ 61.80
We know 1 dozen =12
Cost of one egg = 61.80 ÷ 12

= \(\frac{515}{100}\)
∴ Cost of one egg = ₹ 5.15

Question 8.
If the price of a tablet strip containing 10 tablets is ₹ 26.5, then find the price of each tablet.
Answer:
Given, cost of 10 tablets = ₹ 26.5
Cost of each tablet = 26.5 ÷ 10
= \(\frac{265}{10} \div \frac{10}{1}\)
= \(\frac{265}{10} \times \frac{1}{10}\)
= \(\frac{265}{100}\)

∴ Cost of each tablet = ₹ 2.65.