Class 7

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 4 Lines and Angles Ex 1 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 4th Lesson Lines and Angles Exercise 1

Question 1.
Name the figure drawn below.

Solution:
Line segment \(\overline{\mathrm{AB}}\)
Ray \(\overline{\mathrm{DC}}\)
Line \(\overline{\mathrm{SY}}\)
Point P

Question 2.
Draw the figures for the following.
(i) \(\overline{\mathrm{OP}}\)
(ii) Point X
(iii) \(\overline{\mathrm{RS}}\)
(iv) \(\overline{\mathrm{CD}}\)
Solution:

Question 3.
Name all the possible line segments in the figure.

Solution:
\(\overrightarrow{\mathrm{AB}}, \widehat{\mathrm{BC}}, \overline{\mathrm{CD}}, \overline{\mathrm{AC}}, \overline{\mathrm{AD}}, \overline{\mathrm{BD}}\)

Question 4.
Write any five examples of angles that you have observed arround.
Example: The angle formed when a scissor is opened.
Solution:
i) Angle formed when a door is opened.
ii Angle between two adjacent edges of a blackboard.
iii) Angle between two adjacent edges of a ruler.
iv) Angle between two adjacent edges of a set square.
v) Angle formed at elbow.

Question 5.
Identify the following given angles as acute, right or obtuse.

Solution:
i) acute
ii) obtuse.
iii) right
iv) acute
v) obtuse.

Question 6.
Name all the possible angles you can find in the following figure. Which are acute, right, obtuse and straight angles?

Solution:
Acute angles = ∠AOF,∠FOE, ∠EOD, ∠DOC, ∠COB, ∠EOC, ∠DOB, ∠DOF
Obtuse angles = ∠AOD, ∠AOC, ∠BOF
Right angle = ∠AOE,∠BOE

Question 7.
Which of the following pairs of lines are parallel? Why?

Solution:
Figure (i) and (iv) are pairs of parallel lines because however long they are produced, they
never intersect each other.

Question 8.
Which of the following lines are intersecting?

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 3 Simple Equations Ex 3 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 3rd Lesson Simple Equations Exercise 3

Question 1.
Write the information given in the picture in the form of an equation. Also, find ‘x’ in the following figure.

Solution:
From the figure x + 11 = 15
∴ x = 15 – 11 (transposing +11)
∴ x = 4cm

Question 2.
Write the information given in the picture in the form of an equation. Also, find ‘y in the following figure.

Solution:
From the figure y + 8 = 13
y = 13 – 8 (transposing 8)
∴ y = 5cm

Question 3.
If we add 7 to twice a number, we get 49. Find the number.
Solution:
Let the number be x
Then twice the number = 2x
Onaddingi = 2x + 7
By problem, 2x + 7 = 49
2x = 49 – 7 (transposing + 7)
2x = 42
x = \(\frac { 42 }{ 2 }\) (transposing x 2)
x = 21

Question 4.
If we subtract 22 from three times a number, we get 68. Find the number.
Solution:
Let the number be x
Then three times the number = 3x
On subtracting 22 ⇒ 3x – 22
By problem, 3x – 22 = 68
3x = 68 + 22 (transposing -22)
3x = 90
x = \(\frac { 90 }{ 3 }\)(transposingx3)
x = 30
∴ The required number 30

Question 5.
Find a number which when multiplied by 7 and then reduced by 3 is equal to 53.
Solution:
Let the number be x
Multiplied by 7 ⇒ 7x
Then reduced by 3 ⇒ 7x – 3
By problem, 7x – 3 = 53
7x = 53 + 3 (transposing – 3)
7x = 56
x = \(\frac { 56 }{ 7 }\) (transposing x 7)
x = 8 .
∴ The required number = 8

Question 6.
Sum of two numbers is 95. 1f one exceeds the other by 3, find the numbers.
Solution:
Let one number be = x.
Then the other number x – 3
Sumofthenumbers x + x – 3 = 2x – 3
By problem, 2x – 3 = 95
2x = 95 + 3 (transposing – 3)
2x = 98
x = \(\frac { 98 }{ 2 }\) (transposing x 2)
x = 49
∴if one number x 49 then the other number x – 3 = 49 – 3 = 46

Question 7.
Sum of three consecutive integers is 24. Find the integers.
Solution:
Let the three integers be = x, x + 1, x 2
Sumoftheintegers = x + x + 1 + x + 2 = 3x + 3
By problem, 3x + 3 = 24’
3x = 24 – 3 (Transposing + 3)
3x = 21
x = \(\frac { 21 }{ 3 }\) (transposing x 3)
x = 7
∴ The integers x = 7
x + 1 = 7 + 1 = 8
x + 2 = 7 + 2 = 9

Question 8.
Find the length and breadth of the rectangle given below if its perimeter is 72m.

Solution:
Length of the rectangle = 5x + 4
Breadth of the rectangLe = x – 4
Perimeter of the rectangle = 2 x (length + breadth)
= 2x[(5x + 4)+(x – 4)]
= 2[5x + 4 + x – 4]
= 2(6x)
= 12x
Byproblem, 12x = 72
x = \(\frac { 72 }{ 12 }\) (transposing x 12)
x = 6
∴ Length of the rectangle = 5x + 4 = 5 x 6.4 = 34cm
Breadth of the rectangÌe = x – 4 = 6 – 4 = 2cm

Question 9.
Length of a rectangle exceeds its breadth by 4 m. 1f the perimeter of the rectangle is 84 m, find its length and breadth.
Solution:
Let the breadth be x
Then its length = x + 4
Perimeter of the rectangle = 2 x (length + breadth)
=2[x + 4 + x]
= 2 (2x + 4)
= 4x + 8
By problem, 4x + 8 = 84
4x = 84 – 8 (transposing + 8)
4x = 76
x = \(\frac{76}{4}\) (transposing x 4)
x = 19
∴ Length of the rectangle =x+4= 19+4 = 23m
Breadth of the rectangle = 19m

Question 10.
After 15 years, Hema’s age will become four times that of her present age. Find her present age.
Solution:
Let the preser age of Hema be x years
After 15 years Hema Age = 4x
By problem, x + 15 = 4x
x + 15 – 4x = 4x – 4x (Subtracting 4x from both sides)
– 3x + 15 = 0
– 3x = – 15
x = \(\frac{-15}{-3}\) transposing x ( – 3)]
x = 5
∴ Her present age is 5 years.
∴ Her present age is 5 years.

Question 11.
A sum of ₹.3000 is to be given in the form of 63 prizes. Ifthe prize money is either ₹. 100 or.25. Find the number of prizes of each type.
Solution:
Let the number of ₹ 100 prizes be x
Then the number of ₹ 25 prizes be = 63 – x
Value of the prizes = 100x + (63 – x) x 25
= 100x+ 1575 – 25x
= 75x + 1575
By problem, 75x + 1575 = 3000
75x = 3000 – 1575
75x = 1425
x = \(\frac { 1425 }{ 75 }\)
x = 19
∴ ₹ 100 prizes = 19
₹25 prizes= 63 – x =63 – 19 = 44

Question 12.
A number is divided into two parts such that one part is 10 more than the other. Ifthe two parts are in the ratio 5:3, find the number and the two parts.
Solution:
Let one part be x
Then the other part = x + 10
Ratio of these two parts = x + 10 : x
Byproblem, x + 10 : x=5:3
∴ \(\frac{x+10}{x}=\frac{5}{3}\)
3(x+ 10)=5x
3x + 30 = 5x
30 = 5x – 3x
2x = 30
x = \(\frac{30}{2}\)
x = 15
If one part is 15 then the other part must be x + 10 = 15 + 10 = 25
∴ The number is 15 + 25 40

Question 13.
Suhana said, “multiplying my number by 5 and adding 8 to it gives the same answer as subtracting my number from 20”. Find Suhana’s numbers.
Solution:
Let Suhana’s number be x
Muhtplying by S and adding 8 to that number 5x + 8
Subtracting that number from 20 = 20 – x
By problem above two are equal.
i.e. 5x + 8 = 20 – x
5x + x = 20 – 8
6x = 12
x = \(\frac{12}{6}\) = 2
x = 2
∴ Suhanas number is 2

Question 14.
The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest mark is 87. What is the lowest mark?
Solution:
Let the Lowest mark of the class = x
Twice the least mark = 2x
On adding 7 = 2x + 7
By problem; 2x + 7 = 87
2x = 87 – 7
2x = 80
x = \(\frac{80}{2}\) = 40
x = 40
∴ The lowest mark = 40

Question 15.
In adjacent figure find the magnitude of each of the three angles formed?
(Hint: Sum of all angles at a point on a line is 180°)
Solution:
(Hint : Sum of all angles at a point on a line is 180°)

We know sum of angles at a point = 180°
∴ x° + 2x° + 3x° = 180
6x° = 180°
x = \(\frac{180}{6}\) = 30
∴ The angles are x = 30°
2x = 2 x 30 = 60°
3x° = 3 x 30° = 90°

Question 16.
Solve the following riddle:
I am a number
Tell my identity
Take me two times over
And add a thirty six.
To reach a century
Solution:
Let the number be x
Two times the number = 2x
On adding 36 = 2x 36
By problem, 2x + 36 = 100 – 4
2x 36 = 96
2x = 96 – 36
2x = \(\frac{60}{2}\)
x = 30

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 3 Simple Equations Ex 2 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 3rd Lesson Simple Equations Exercise 2

Question 1.
Solve the following equations without transposing and check your result.
(i) x + 5 = 9
(ii) y – 12 = -5
(iii) 3x + 4 = 19
(iv) 9z = 81
(v) 3x + 8 = 5x + 2
(vi) 5y + 10 = 4y – 10
Solution:
(i) x + 5 = 9
Solution:
i) x + 5 = 9
x + 5 – 5 9 – 5 (subtract 5 from both sides)
x = 4
Check
LHS = x + 5
(substituting x = 4)
= 4 + 5 = 9
RHS = 9
∴ L.H.S = R.H.S

ii) y – 12 = – 5
y – 12 = – 5
y – 12 + 12= – 5 + 12 (add l2onbothsides)
y = 7
Check
LHS = y – 12
= 7 – 12= – 5
RHS = -5
∴ L.H.S = R.H.S

iii) 3x+4= 19
3x + 4 = 19
3x + 4 – 4 = 19 – 4
(subtract 4 from both sides)
3x = 15
\(\frac{3 x}{3}=\frac{15}{3}\) (Divide both sides by3)
x = 5
Check
LHS = 3x + 4
= 3 x 5 + 4
= 15 + 4 = 19
RHS = 19
∴ L.,H.S = R.H.S

iv) 9z = 81
\(\frac{9 z}{9}=\frac{81}{9}\) (Divide both sides by 9)
z = 9
Check
LHS = 9z = 9 x 9 = 81
RHS = 81
∴ LHS = RHS

v) 3x + 8 = 5x + 2
3x + 8 = 5x + 2
3x + 8 – 8 = 5x + 2 – 8
(Adding -8 on both sides)
3x = 5x – 6
3x – 5x = 5x – 6 – 5x
(Subtract 5x from both sides)
-2x = -6
\(\frac{-2 x}{-2}=\frac{-6}{-2}\)(Divide both sides by -2)
x = 3
Check
LHS = 3x + 8 = 3(3) + 8 = 9 + 8 = 17
RHS = 5x + 2 = 5(3) + 2 = 15 + 2 = 17
∴ LHS = RHS

(vi) 5y + 10 = 4y – 10
5y + 10 = 4y – 10
5y + 10 – 1o = 4y – 10 – 10
(Subtract 10 from both sides)
5y =4y – 20
5y – 4y = 4y – 20 – 4y
(Substract ty from both sides)
y = – 20
Check
LHS = 5y + 10 = 5 x( – 20) + 10 = – 100+ 10= – 90
RHS = 4y – 10 = 4 x ( – 20) – 10 = – 80 – 10 = -90
∴ LHS = RHS

Question 2.
Solve the following equations by transposing the terms and check your result.
(i) 2 + y = 7
Solution:
y= 7 – 2 (transposlng+2)
y = 5
Check:
LHS = 2 + y = 2 + 5 = 7
RHS = 7
∴ L.H.S = R.H.S

(ii) 2a – 3 = 5
2a – 3 = 5
2a = 5 + 3 (transposing – 3)
2a = 8
(transposing x 2)
a = 4
Check
LHS = 2a – 3 = 2 x 4 – 3 = 8 – 3 = 5
RHS =5
∴ L.H.S = R.H.S

(iii) 10 – q = 6
10 – q = 6
– q = 6 – 10(transposing + 10)
– q – 4
q = \(\frac{-4}{-1}\) = 4 (transposing x ( – 1)
Check
LHS= 10 – q= 10 – 4= 6
RHS = 6
∴ L.H.S = R.H.S

(iv) 2t – 5 = 3
2t – 5 = 3
2t – 5 = 3 (transposing – 5)
2t = 3 + 5
2t = 8
t = \(\frac{8}{2}\) (transposing x (2))
Check
LHS=2t – 5= 2 x 4 – 5 = 8 – 5 = 3
RHS = 3
∴ L.H.S = R.H.S

(v) 14 = 27 – x
14 = 27 – x
0 = 27 – x – 14 (transposing + 14)
0 = 13 – x (transposIng – x)
x = 13
Check
LHS = 14
RHS = 27 – x = 27 – 13 = 14
∴ L.H.S = R.FIS

(vi) 5(x + 4) = 35
5(x + 4) = 35
x + 4 = \(\frac{35}{5}\) (lransposingx5)
x + 4 = 7
x = 7 – 4 (transposing + 4)
x = 3
Check
LHS = 5(3 + 4) = 5 x 7 = 35
RHS = 35
∴ L.H.S = R.H.S

(vii) -3x = 15
– 3x= 15
x = \(\frac{15}{-3}\) (transposingx( – 3))
x= – 5
Check
LHS = – 3x = -3x( – 5)= 15
RHS= 15
∴ L.H.S = R.H.S

(viii) 5x – 3 = 3x – 5
5x – 3 = 3x – 5
5x = 3x – 5 + 3(transposing – 3)
5x = 3x – 2
5x – 3x = – 2(trarisposing + 3x)
2x = – 2
x = \(\frac{-2}{2}\) (transposingx2)
x= – 1
C heck
LHS = 5x – 3 = 5x( – 1) – 3 = – 5 – 3 = – 8
RHS = 3x – 5 = 3x( – 1) – 5 = – 3 – 5 = – 8
∴ L.H.S = R.H.S

(ix) 3y + 4 = 5y – 4
3y + 4 = 5y – 4
3y = 5y – 4 – 4 (transposing + 4)
3y – 5y = – 8 (transposing + 5y)
– 2y = – 8
y = \(\frac{-8}{-2}\) =4 (transposingx( – 2)
y = 4
Check
LHS = 3y + 4 = 3 x (4) + 4 = 12 + 4 = 16
RHS = 5y – 4 = 5 x (4) – 4 = 20 – 4= 16
∴ L.H.S = R.H.S

(x) 3(x-3)=5(2x + 1)
3(x – 3)=5(2x+ 1)
3(x – 3)= 5(2x ÷ 1)
3x – 9= 10x+5
3x = 10x + 5 ÷ 9 (transposing – 9)
3x = 10x + 14
3x – 10x = 14(transposing+ lOx)
– 7x =14
x = \(\frac{14}{-7}\)(transposing x ( – 7))
x = – 2
Check
LHS = 3(x – 3) = 3[( – 2) – 3] = 3x( – 5) = – 15
RHS = 5(2x + 1) = 5 x [2( – 2) + 1] = 5 x [ – 4 + 1]
= 5 x ( – 3)= – 15
∴ LH.S = R.H.S

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 1 Integers Ex 7 Textbook Questions and Answers.

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

Question 1.
In a class test containing 15 questions, 4 marks are given for every correct answer and (-2) marks are given for every incorrect answer. (i) I3harathi attempts all questions but only 9 answers are correct. What is her total score’? (ii) One of her friends Hema answers only 5 questions correct. What will be her total score?
Solution:
i) Bharathi’s score = 4 × Number of correct answers + (-2) × Number of wrong answers
=4 × 9 + (-2) × 6 = 36 + (-12) = 24
ii) Hema’s score = 4 × (5) + (-2) × 10 = 20 + (-20) = 0

Question 2.
A cement company earns a profit of ₹ 9 per bag of white cement sold and a loss of ₹ 5 per bag of grey cement sold.
(i) The company sells 7000 bags of white cement and 6000 bags of grey cement in a month. What is its profit or loss?
(ii) What is the number of white cement bags it must sell to have neither profit nor loss, ifthe number of grey bags sold is 5400.
Solution:
i) Profit on 7000 white cement bags = 9 × 7000 = ₹ 63,000
Loss on 6000 grey cement bags = 5 × 6000 = ₹ 30,000
Net profit or loss = 63,000 – 30,000 = ₹ 33,000 profit
ii) When there is no loss then
Profit on white cement bags = loss on grey cement bags
Loss on grey cement bags = 5 × 5400 = ₹ 27,000
∴ Number of white cement bags = \(\frac{27,000}{9}\) = 3,000

Question 3.
The temperature at 12 noon was 10°C above zero. ¡fit 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 midnight’?
Solution:
The temperature at 12 noon = 10°C
Given temperature = 8°C below zero = – 8°C
Difference between the temperature = 10 – (- 8°) = 18°C
∴ Time lasted = 18 + 2 = 9 hours
i.e., at 9 o’ clock the temperature would be – 8C
At mid-night the temperature 10 + 12(-2) = 10 + (- 24) = – 14°C

Question 4.
In a class test (+3) marks are given for every correct answer and (-2) marks are given for every incorrect answer and no marks for not attempting any question.
(i) Radhika scored 20 marks. if she has got 12 correct answers, how many questions has she attempted incorrectly?
(ii) Mohini scores (-5) marks in this test, though she has got 7 correct answers.
How many questions has she attempted incorrectly?
Solution:
i) Let the no. of incorrect answers = x
So score = 3 × No. of correct answers + (-2) No. of incorrect answers
Radhika score = 20 marks
⇒ 20 = 3 × 12 + (-2) x ⇒ 20 = 36 – 2x ⇒ 2x = 36 – 20 ⇒ x = 16/2 = 8
∴ No. of incorrect questions = 8

ii) Mohini scores = -5
Marks for correct answers = 7 × 3 = 21
Marks for incorrect answers = -5-21 = -26
∴ Number of incorrect answers = -26 ÷ 2 = -13

Question 5.
An elevator descends into a mine shaft at the rate of 6 meters per minute. If the decent starts from 10 rn above the ground level, how long will it take to reach -350 m.
Solution:
Total length to be descend = -350 – 10 = -360 m
Rate of the speed of the elevator 6m per minute
Total time to descend = \(\frac{360}{6}\) = 60 minutes = 1 hour

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 1 Integers Ex 6 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 1st Lesson Integers Exercise 6

Question 1.
Fill the following blanks.
(i) -25 ÷ ……………… = 25
(ii) ………………. ÷ 1 = -49
(iii) 50 ÷ 0 =……………….
(iv) 0 ÷ 1 = …………………
Solution:
(i) -25 ÷ -1 = 25
(ii) -49 ÷ 1 = -49
(iii) 50 ÷ 0 = 50 cannot divide by 0
(iv) 0 ÷ 1 = 0

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 1 Integers Ex 5 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 1st Lesson Integers Exercise 5

Question 1.
Verify the following.
(i) 18 × [7 + (-3)] = [18 × 7] + [18 x (-3)]
(ii) (-21) × [(-4) + (-6)] = [(-21) × (-4) ] + [(-21) × (-6)]
Solution:
(i) 18 × [7 + (-3)] = [18 x 7] + [18 x (-3)]
LHS: 18 × (7 + (-3)] = 18 × (4) = 72
RHS:[18 × 7] + [18 × (-3)] = 126 + (-54) = 72
∴ LHS = RHS

(ii) (-21) × [(-4) + (-6)] = [(-21) × (-4) ] + [(-21) x (-6)]
LHS:(-21) × [(-4) + (-6)] = (-21) × (-10) = 210
RHS: [(-21) × (-4)] + [(-21) × (-6)] = (84) + (126)= 210
∴ LHS=RHS

Question 2.
(1) For any integer a, what is(-1) × a equal to9
(ii) Determine the integer whose product with (-1) is 5
Solution:
i) For any integer a
(- 1) × a = -a
ii) …………. × (-1) = 5
∴ (-5) × (-1) = 5

Question 3.
Find the product, using suitable properties.
(i) 26 × (-48)+(-48) × (-36)
(ii) 8 × 53 × (-125)
(iii) 15 × (-25) × (4) × (-10)
(iv) (-41) × 102
(v) 625 × (-35) +(625) × 65
(vi) 7 × (50 – 2)
(vii) (-17) × (-29)
(viii) (-57) × (-19) + 57
Solution:
(i) 26 × (-48) + (-48) × (-36)
= -48 × [26 + (-36)]
= -48 × (-10)
= 480

(ii) 8 × 53 × (- 125)
424 × (- 125)
= -53000

(iii) 15 × (-25) × (-4) × (-10)
= (-375) × (40)
= -15000

(iv) (-41) × 102
= -4182

(v) 625 × (-35) + (-625) × 65
= 625 × [(-35) + (-65)]
=625 × (-100)
= – 62500

(vi) 7 × (50 – 2)
= 7 × 48
= 336

(vii) (-17) × (-29)
= 493

(viii) (-57) × (-19) + 57
=1083 + 57
= 1140

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 1 Integers Ex 4 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 1st Lesson Integers Exercise 4

Question 1.
Fill in the blanks.
(i) ( – 100) × ( -6) =
(ii) ( – 3) × …………. = 3
(iii) 100 × ( – 6) = ………….
(iv) ( – 20) × (- 10) = ………….
(v) 15 × (-3) = ………….
Solution:
(i) ( – 100) × ( -6) =
(ii) ( – 3) × …………. = 3
(iii) 100 × ( – 6) = ………….
(iv) ( – 20) × (- 10) = ………….
(v) 15 × (-3) = ………….

Question 2.
Find each of the following products.
(i) 3 × ( – 1)
(ii) ( – 1) × 225
(iii) ( – 2 ) × ( – 30)
(iv) ( – 316) × ( – 1)
(v) (-15) × 0 × (-18)
(vi) (-12) × (-11) × (10)
(vii) 9 × ( – 3) × ( – 6)
(viii) ( – 18) × ( – 5) × ( – 4)
(ix) ( – 1) × ( – 2) × ( – 3) × 4
(x) ( – 3) × ( – 6) × ( – 2) × ( – 1)
Solution:
(i) 3 × ( – 1) = -3
(ii) ( – 1) × 225 = -225
(iii) ( – 2 ) × ( – 30) = 630
(iv) ( – 316) × ( – 1) = 316
(v) (-15) × 0 × (-18) = 0
(vi) (-12) × (-11) × (10) = 1320
(vii) 9 × ( – 3) × ( – 6) = 162
(viii) ( – 18) × ( – 5) × ( – 4) = -360
(ix) ( – 1) × ( – 2) × ( – 3) × 4 = -24
(x) ( – 3) × ( – 6) × ( – 2) × ( – 1) = 36

Question 3.
A certain freezing process requires that room temperature be lowered from 40°C at the rate of 5°C every hour. What will be the room temperature 10 hours after the process begins?
Solution:
At present, the room temperature 40°C
The total decrease in temperature after 10 hours at the rate of 5°C a hour = 10 × 5 = 50°
∴ Temperature after 10 hours = 40° – 50° = – 10°C.

Question 4.
In a class test containing 10 questions, ‘3’ marks are awarded for every correct answer and ( – 1) mark is for every incorrect answer and ‘0’ for questions not attempted.
(i) Gopi gets 5 correct and 5 incorrect answers. What is his score?
(ii) Reshma gets 7 correct answers and 3 incorrect answers. What is her score?
(iii) Rashmi gets 3 correct and 4 incorrect answers out of seven questions she attempts. What is her score?
Solution:
i) GopI’sscoe =5 × 3 + 5( – 1)= 15 + ( – 5) = 10
ii) Reshmasscore = 7 × 3 + 3 × ( – 1) = 21 +( – 3) = 18
iii) Rashmis score = 3 × 3 + 4( – 1) = 9 + ( – 4) = 5

Question 5.
A merchant on selling rice earns a profit of ₹ 10 per bag of basmati rice sold and a loss of ₹ 5 per bag of 1 non-basmati rice.

(i) He sells 3,000 bags of basmati rice and 5,000 bags ofnon-basmati rice in a month. What is his profit or loss in a month?
(ii) What is the number of basinati rice bags he must sell to have neither profit nor loss, if the number of bags of non-basmati rice sold is 6,400.
Solution:
i) Profit on 3,000 bags of basmati rice = 3,000 × 10 = ₹ 30,000
Loss on 5,000 bags of non-basmati rice = 5,000 × 5 = ₹ 25,000
Net profit or loss = 30,000 – 25,000 = ₹ 5000 profit
Another Method : Net profit or loss = 3,000 × 10 + 5,000(-5)
= 30,000 – 25,000 = ₹ 5,000

ii) As there is no loss or gain
Profit on basmati rice= Loss on non-basmati rice
But loss on nonbasrnati rice = 6,400 × 5 = ₹ 32,000
∴ Number of bags of basmati rice sold = \(\frac{32,000}{10}\) = 3200

Question 6.
Replace the blank with an integer to make it a true statement.
(i) (- 3) × ______________ = 27
(ii) 5 × ______________ = – 35
(iii) _______ × ( – 8) = – 56
(iv) _______ × ( – 12) = 132
Solution:
(i) (- 3) × -9 = 27
(ii) 5 × -7 = – 35
(iii) 7 × ( – 8) = – 56
(iv) 11 × ( – 12) = 132

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 1 Integers Ex 3 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 1st Lesson Integers Exercise 3

Question 1.
Represent the following subtractions on the number line.
(i) 7 – 2
(ii) 8 – ( – 7)
(iii) 3 – 7
(iv) 1 5 – 14
(v) 5 – ( – 8)
(vi) ( – 2) – ( – 1)
Solution:
(i) 7 – 2 = 5

(ii) 8 – ( – 7) = 15

(iii) 3 – 7 = -4

(iv) 1 5 – 14 = 1

(v) 5 – ( – 8) = 13

(vi) ( – 2) – ( – 1) = -2 + 1 = -1

Question 2.
Solve the following.
(i) 17 – ( – 14)
(ii) 13 – ( – 8)
(iii) 19 – ( – 5)
(iv) 15 – 28
(v) 25 – 33
(vi) 80 – ( – 50)
(vii) 150 – 75
(viii) 32 – ( – 18)
Solution:
(i) 17 – ( – 14) = 31
(ii) 13 – ( – 8) = 21
(iii) 19 – ( – 5) = 24
(iv) 15 – 28 = -13
(v) 25 – 33 = -8
(vi) 80 – ( – 50) = 130
(vii) 150 – 75 = 75
(viii) 32 – ( – 18) = 50

Question 3.
Express ‘ – 6’ as the difference between a negative integer and a whole number.
Solution:
– 6=( – 4) – (2) ( – 5) – (1)=( – 2) – (4)

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 1 Integers Ex 2 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 1st Lesson Integers Exercise 2

Question 1.
Represent the following additions on a number line.
(i) 5 + 7
(ii) 5 + 2
(iii) 5 + ( – 2)
Solution:

Question 2.
Compute the following.
(i) 7 + 4
(ii) 8+( – 3)
(iii) 11 + 3
(iv) 14 + ( – 6)
(v) 9 + ( – 7)
(vi) 14+( – 10)
(vii) 13 + ( – 15)
(viii) 4 + ( – 4)
(ix) 10 +( – 2)
(x) 100 +( – 80)
(xi) 225 +( – 145)
Solution:
(i) 7 + 4 = 11
(ii) 8 +( – 3) = 5
(iii) 11 + 3 = 14
(iv) 14 + ( – 6) = 8
(v) 9 + ( – 7) = 2
(vi) 14 +( – 10) = 4
(vii) 13 + ( – 15) = -2
(viii) 4 + ( – 4) = 0
(ix) 10 +( – 2) = 8
(x) 100 +( – 80) = 20
(xi) 225 +( – 145) = 80

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 3 Simple Equations Ex 1 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 3rd Lesson Simple Equations Exercise 1

Question 1.
Write L.H.S and R.H.S of the following simple equations.
(i) 2x = 10
(ii) 2x – 3 = 9
(iii) 4z + 1 = 8
(iv) 5p + 3 = 2p + 9
(v) 14 = 27 – y
(vi) 2a – 3 = 5
(vii) 7m = 14
(viii) 8 = q + 5
Solution:

Problem / Equation	L.H.S	R.H.S
(i) 2x= 10	2x	10
(ii) 2x – 3 = 9	2x – 3	9
(iii) 4z + 1 = 8	4z + 1	8
(iv) 5p + 3 = 2p + 9	5p + 3	2p + 9
(v) 14 = 27 – y	14	27 – y
(vi) 2a – 3 = 5	2a-3	5
(vii) 7m = 14	7m	14
(viii) 8 = q + 5	8	q + 5

Question 2.
Solve the following equations by trial and error method.
(i) 2 + y = 7
(ii) a – 2 = 6
(iii) 5m = 15
(iv) 2n = 14
Solution:
(i) 2 + y = 7
if y = 1; LHS = 2 + y = 2 + 1 = 3 ≠ 7
If y = 2; LHS = 2 + 2 = 4 ≠ 7
If y = 3; LHS = 2 + 3 = 5 ≠ 7
If y = 4; LHS = 2 + 4= 6 ≠ 7
If y = 5; LHS = 2 + 5 = 7 = 7 = RHS
∴ y = 5 is the solution of 2 + y = 7

(ii) a – 2 = 6
As a – 2 = 6 ; the value of ‘a’ must be greater than 6.
If a = 7; LHS = 7 – 2 = 5 ≠ 6
If a = 8; LHS = 8 – 2 = 6 = RHS
∴ a = 8 is the solution of a – 2 = 6

(iii) 5m = 15
If m = 1 then LHS = 5 × 1 = 5 ≠ b15
m = 2 then LHS = 5 × 2 = 10 ≠ 15
m = 3 then LHS = 5 × 3 = 15 = 15
∴ m = 3 is the solution of 5m = 15

(iv) 2n = 14
If n = 1 then LHS = 2 × 1 = 2 ≠ 14
n = 2 then LHS = 2 × 2 = 4 ≠ 14
n = 3 then LHS = 2 × 3 = 6 ≠ 14
n = 4 then LHS = 2 × 4 = 8 ≠ 14
n = 5 then LHS = 2 × 5 = 10 ≠ 14
n = 6 then LHS = 2 × 6 = 12 ≠ 14
n = 7 then LHS = 2 × 7 = 14 ≠ 14
∴ n = 7 is the solution of 2n = 14

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 2 Fractions, Decimals and Rational Numbers Ex 7 Textbook Questions and Answers.

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

Question 1.
Write any three equivalent rational numbers to each of the following
i) \(\frac { 2 }{ 3 }\)
ii) \(\frac { -3 }{ 8 }\)
Solution:

Question 2.
What is the equivalent rational number for \(\frac { -15 }{ 36 }\) with
(i) denominator 12
(ii) numerator 75?
Solution:
i) \(\frac{-15}{36}=\frac{-15 \div 3}{36 \div 3}=\frac{-5}{12}\)
ii) \(\frac{-15}{36}=\frac{-15 \times 5}{36 \times 5}=\frac{-75}{180}\)

Question 3.
Mark the following rational numbers on the number line.
(i) \(\frac { 1 }{ 2 }\)
(ii) \(\frac { 3 }{ 4 }\)
(iii) \(\frac { 3 }{ 2 }\)
(iv) \(\frac { 10 }{ 3 }\)
Solution:

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 2 Fractions, Decimals and Rational Numbers Ex 6 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 2nd Lesson Fractions, Decimals and Rational Numbers Exercise 6

Question 1.
Solve the following.
(i) 0.3 × 6
(ii) 7 × 2.7
(iii) 2.71 × 5
(iv) 19.7 × 4
(v) 0.05 × 7
(vi) 210.01 × 5
(vii) 2 × 0.86
Solution:
(i) 0.3 × 6 = 1.8
(ii) 7 × 2.7 = 18.9
(iii) 2.71 × 5 = 13.55
(iv) 19.7 × 4 = 78.8
(v) 0.05 × 7 = 0.35
(vi) 210.01 × 5 = 1050.05
(vii) 2 × 0.86 = 1.72

Question 2.
Find the area of a rectangle whose length is 6.2 cm and breadth is 4 cm.
Solution:
Length of the rectangle = 6.2 cm
Breadth of the rectangle = 4 cm
Area of the rectang’e = Length × Breadth
= 6.2 × 4 = 24.8cm²

Question 3.
Solve the following.
(i) 21.3 × 10
(ii) 36.8 × 10
(ii) 53.7 × 10
(iv) 168.07 × 10
(v) 131.1 × 100
(vi) 156.1 × 100
(vii) 3.62 × 100
(viii) 43.07 × 100
(ix) 0.5 × 10
(x) 0.08 × 10
(xi) 0.9 × 100
(xii) 0.03 × 1000
Solution:
(i) 21.3 × 10 = 213
(ii) 36.8 × 10 = 368
(ii) 53.7 × 10 = 537
(iv) 168.07 × 10 = 1680.7
(v) 131.1 × 100 = 13110
(vi) 156.1 × 100 = 15610
(vii) 3.62 × 100 = 362
(viii) 43.07 × 100 = 4307
(ix) 0.5 × 10 = 5
(x) 0.08 × 10 = 0.8
(xi) 0.9 × 100 = 90
(xii) 0.03 × 1000 = 30

Question 4.
A motor bike covers a distance of 62.5 km.consuming one litre of petrol. How much distance does it cover for 10 litres of petrol?
Solution:
Distance covered for 1 lit, of petrol = 62.5 km
∴ Distance covered for 10 lit, of petrol = 62.5 × 10 = 625 km

Question 5.
Solve the following.
(i) 1.5 × 0.3
(ii) 0.1 × 47.5
(iii) 0.2 × 210.8
(iv) 4.3 × 3.4
(v) 0.5 × 0.05
(vi) 11.2 × 0.10
(vii) 1.07 × 0.02
(viii) 10.05 × 1.05
(ix) 101.01 × 0.01
(x) 70.01 × 1.1
Solution:
(i) 1.5 × 0.3 = 0.45
(ii) 0.1 × 47.5 = 4.75
(iii) 0.2 × 210.8 = 42.16
(iv) 4.3 × 3.4 = 14.62
(v) 0.5 × 0.05 = 0.025
(vi) 11.2 × 0.10 = 1.12
(vii) 1.07 × 0.02 = 0.0214
(viii) 10.05 × 1.05 = 10.5525
(ix) 101.01 × 0.01 = 1.0101
(x) 70.01 × 1.1 = 77.011

Question 6.
Solve the following.
(i) 2.3 ÷ 100
(ii) 0.45 ÷ 5
(iii) 44.3 ÷ 10
(iv) 127.1 ÷ 1000
(v) 7 ÷ 35
(vi) 88.5 ÷ 0.15
(vii) 0.4 ÷ 20
Solution:
(i) 2.3 ÷ 100 = \(\frac{23}{10}\) ÷ 100 = \(\frac{23}{10} \times \frac{1}{100}=\frac{23}{1000}\) = 0.023
(ii) 0.45 ÷ 5 = \(\frac{45}{100}\) ÷ 5 = \(\frac{45}{100} \times \frac{1}{5}=\frac{9}{100}\) = 0.09
(iii) 44.3 ÷ 10 = 44.3 × \(\frac{1}{10}\) = 4.43
(iv) 127.1 ÷ 1000 = 127.1 × \(\frac{1}{1000}\) = 0.1271
(v) 7 ÷ 35 = 7 × \(\frac{1}{3.5}=\frac{7 \times 10}{3.5 \times 10}=\frac{70}{35}\) = 2
(vi) 88.5 ÷ 0.15 = \(\frac{885}{10} \div \frac{15}{100}\) = \(\frac{885}{10} \times \frac{100}{15}\) = 590
(vii) 0.4 ÷ 20 = \(\frac{4}{10}\) ÷ 20 = \(\frac{4}{10} \times \frac{1}{20}=\frac{1}{10 \times 5}=\frac{1}{50}\) = 0.02

Question 7.
A side of a regular polygon is 3.5 cm in length. The perimeter of the polygon is 17.5 cm.
How many sides does the polygon have?
Solution:
Side of each length of the polygon . = 3.5 cm
Total length of all sides = perimeter = 17.5 cm
Number of sides of the polygon = 17.5 + 3.5
= \(\frac{175}{10} \div \frac{35}{10}\) = \(\frac{175}{10} \times \frac{10}{35}\) = 5

Question 8.
A rain fall of 0.896 cm. was recorded in 7 hours, what was the average amount of rain per
hour?
Solution:
Total rainfall recorded in 7 hours = 0896 cm
∴ Average rainfall (for 1 hour) = 0.896 ÷ 7
= \(\frac{896}{1000} \div 7=\frac{896}{1000} \times \frac{1}{7}=\frac{128}{1000}\) = 0.128