Class 7

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

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

Question 1.
Multiply the following. Write the answer as a mixed fraction.
(i) \(\frac { 3 }{ 6 }\) x 10
(ii) \(\frac { 1 }{ 3 }\) x 4
(iii) \(\frac { 6 }{ 7 }\) x 2
(iv) \(\frac { 2 }{ 9 }\) x 5
(v) 15 x \(\frac { 2 }{ 5 }\)
Solution:
(i) \(\frac { 3 }{ 6 }\) x 10 = \(\frac { 30 }{ 6 }\) = 5 i.e., integer part 5; fraclon part = 0
(ii) \(\frac { 1 }{ 3 }\) x 4 = \(\frac { 4 }{ 3 }\) = 1\(\frac { 1 }{ 3 }\)
(iii) \(\frac { 6 }{ 7 }\) x 2 = \(\frac { 12 }{ 7 }\) = 1\(\frac {5 }{ 7 }\)
(iv) \(\frac { 2 }{ 9 }\) x 5 = \(\frac { 10 }{ 9 }\) = 1\(\frac { 1 }{ 9 }\)
(v) 15 x \(\frac { 2 }{ 5 }\) = \(\frac { 30 }{ 5 }\) = 6 i.e., integer part = 6; fraction part = o

Question 2.
Shade: (i) \(\frac { 1 }{ 2 }\) of the circles in box (a)
(ii) \(\frac { 2 }{ 3 }\) of the triangles in box (b)
(iii) \(\frac { 3 }{ 5 }\)of the rectangles in box (c)
(iv) \(\frac { 3 }{ 4 }\) of the circles in box (d)


Solution:

Question 3.
Find (i) \(\frac { 1 }{ 2 }\) of 12 (ii) \(\frac { 2 }{ 5 }\) of 15
Solution:
(i) \(\frac { 1 }{ 2 }\) of 12 = \(\frac { 1 }{ 2 }\) x 12 = \(\frac { 12 }{ 3 }\) = 4

(ii) \(\frac { 2 }{ 5 }\) of 15 = \(\frac { 2 }{ 5 }\) x 15 = \(\frac { 30 }{ 5 }\) = 6

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

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

Question 1.
Fill up the blanks
(i) The line which intersects two or more lines at distinct points is called __________
(ii) If the pair of alternate interior angles are equal then the lines are ______________
(iii) The sum of interior angles on the same side of the transversal are supplementary then the lines are ____________
(iv) If two lines intersect each other then the number of common points they have
Solution:
i) The line which intersects two or more lines at distinct points is called transversal.
ii) If the pairs of alternate interior angles are equal then the lines are parallel.
ii) The sum of interior angles on the same side of the transversal are supplementary then the lines are parallel.
iv) If two lines intersect each other then the number of common points they have Only one.

Question 2.
In the adjacent figure, the lines ‘P and ‘m’ are parallel and ‘n’ is a transversal. Fill in the blanks for all the situations given below.

(i) If ∠1 = 80° then∠2 =
(ii) If ∠3 = 45° then ∠7 =
(iii) If ∠2 = 90° then ∠8 =
(iv) If ∠4 = 100° then ∠8 =
Solution:
(i) If ∠1 = 80° then∠2 = 100
(ii) If ∠3 = 45° then ∠7 = 45
(iii) If ∠2 = 90° then ∠8 = 90
(iv) If ∠4 = 100° then ∠8 = 100

Question 3.
Find the measures of x,y and z in the figure, where l || BC

Solution:
∠y = 75° (alt. tnt. angles)
∠z = 45° (ait. tnt. angles)
∠x + ∠y + ∠z = 180° (∵ int. angles of a triangle = 1800)
75° + 45° + ∠x = 180°
∠x = 180°- 120°
= 60°

Question 4.
ABCD is a quadrilateral in which AB || DC and AD || BC. Find ∠b, ∠c and ∠d.
Solution:
∠b + ∠50° = 180° (int. angles on the same side of AB)
∠b = 180° – 50° = 130°
Similarly ∠b + ∠c = ∠c + ∠d = ∠d + ∠50 = 180°
∠130° + ∠c = 180°
∠c = 180° – 130° = 50°
Also ∠c + ∠d = 50° +∠d = 180°
∠d = 180° – 50° = 130°

Question 5.
In a given figure, ‘l’ and m’ are intersected by a transversal ‘n’. Is 1 || m?

Solution:
∠AQP and ∠CRS are exterior angles on the same side of the transversal n and their sum
100° + 80° = 180°
Hence l//m.

Question 6.
Find ∠a, ∠b, ∠c, ∠d and ∠e in the figure? Give reasons.

Note: Two anow marks pointing in the same direction represent parallel lines.
Solution:
∠a = ∠50° (∵ alt. int, angle between two parallel lines)
∠b = 50° (∵ alt. int, angle between two parallel lines)
∠b = ∠c = 50° (∵ alt. int. angles between two parallel lines)
∠c = ∠d = 50° ( ∵ alt. int angles between two parallel lines)
∠e = ∠d = 50°(∵ alt. int. angles between two parallel lines)

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 4 Lines and Angles Ex 6 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 4th Lesson Lines and Angles Exercise 6

Question 1.
Name two pairs of vertically opposite angles in the figure.

Solution:
Vertically opposite angles are
∠AOC, ∠BOD and ∠BOC, ∠AOD

Question 2.
Find the measure of x, y and z without actually measuring them.

Solution:
From the figure
∠y and 160° are vertically opposite angles and hence equal.
∴ ∠y=1600
Also ∠x = ∠z and (. vertically opposite angles)
x + 160° = 180° (Linear pair of angles)
∴ ∠x = 180° – 160° = 20°
∠z = 20° (∵ ∠x, ∠z are vertically opposite)
∴ x = 20°
y = 160°
z = 20°

Question 3.
Give some examples of vertically opposite angles in your surroundings.
Solution:
i) Angles between legs of a folding cot/scamp cot.
ii) Angles between legs of a folding chair.
iii) Angles between plates of a scissors.

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 4 Lines and Angles Ex 5 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 4th Lesson Lines and Angles Exercise 5

Question 1.
Draw the following pairs of angles as adjacent angles. Check whether they fonn linear pair.

Solution:
Linear pair of angles

Linear pair of angles

Do not form linear pair.

Question 2.
Niharika took two angles- 1300 and 500 and tried to check whether they form a linear pair. She made the following picture.
Solution:
Yes. These two angles do not form a pair of linear angles.
Because she made angle more than 180°.

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

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

Question 1.
Which of the following are adjacent angles?

Solution:
(i)
i) a and b are adjacent angles.

ii) c and d are adjacent angles.

iii) e, f are not adjacent angles.

Question 2.
Name all pairs of adjacent angles in the figure. How many pairs of adjacent angles are formed? Why these angles are called adjacent angles?
Solution:
(∠AOL). ∠AOC). (∠AOC. ∠BOC)
(∠BOC. ∠BOD), (∠BOD. ∠AOD) are the four pairs of adjacent angles.
These are adjacent since they have a common vertex, common arm and on either sides of the common arm.

Question 3.
Can two adjacent angles be supplementary? Draw figure.
Solution:
Yes. Two adjacent angles can be supplementary.

Question 4.
Can two adjacent angles be complementary? Draw figure.
Solution:

Yes. Two adjacent angles can be complementary.

Question 5.
Give four examples of adj acent angles in daily life.
Example: Angles between the spokes at the centre ofa cycle wheel.
(i) ____________________________
(ii) ____________________________
(iii) ____________________________
(iv) ____________________________
Solution:
i) Angles between (Door and Wall). (Door and frame)
ii) Angles between the 3 hands of a clock at a point of time.
iii) Adjacent angles in the Rangoli.
iv) Angles between spokes of the window design.

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 4 Lines and Angles Ex 3 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 4th Lesson Lines and Angles Exercise 3

Question 1.
Which of the following pairs of angles are supplementary?

Solution:
i) Sum of the angles = 110° + 70° =180°
Hence they are supplementary.

ii) Sum of the angles = 90° + 90° =180°
hence they are supplementary.

iii) Sum of the angles = 50°+ 140° =190° ≠ 180°
Hence they are not supplementary.

Question 2.
Find the supplementary angles of the given angles.
(i) 105° (ii) 95° (iii) 150° (iv) 20°
Solution:
(i) 105°
Supplementary angle of 105° = 180° – 105° = 75°
(ii) 95°
Supplementary angle of 95° = 180° – 95° = 85
(iii) 150°
Supplementary angle of 150° = 180° – 150° = 30°
(iv) 20°
Supplementary angle of 20° = 20° = 180° – 20° = 60°

Question 3.
Two acute angles cannot form a pair of supplementary angles. Justify.
Solution:
As the measure of an acute angle is less than 90°, sum of two acute angles < 90° + 90° = 180°
Hence two acute angles cant brin a pair of supplementary angles.

Question 4.
Two angles are equal and supplementary to each other. Find them.
Solution:
Let the angle be x°
then its supplementary 180° – x°
By problem, x° =180° – x°
x° + x° = 180°
x° = \(\frac{180^{\circ}}{2}\)
x° = 90°

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 4 Lines and Angles Ex 2 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 4th Lesson Lines and Angles Exercise 2

Question 1.
Which of the following pairs of angles are complementary?

Solution:
In fig (i) 110° – 30° = 140° ≠ 90. Hence they are not complementary.
In fig (ii) 90° + 90° = 180 ≠ 90°. Hence they are not complementary.
In fig (iii) 80° + 10° = 90° = 90. Hence they are complementary.

Question 2.
Find the complements of the given angles. Find the complementary angles of the following.
(i) 25°
(ii) 40°
(iii) 89°
(iv) 55°
Solution:
i) Complementary angle of 25° = 90° – 25° = 65°
ii) Complementary angle of 40° = 90° – 40° = 50°
ill) Complementary angle of 89° = 90° – 89° = 1°
iv) Complementary angle of 55° = 90° – 55 = 35°

Question 3.
Two angles are complement to each other and are also equal. Find them.
Solution:
Let the angle be x°
Then Its complement = (900_ x°)
By problem, x° = 90° – x°
x° + x°= 90°
2x = 90°
x = \(\frac{90^{\circ}}{2}\)
x = 45°

Question 4.
Manasa says, “Each angle in any pair of complementary angles is always acute”. Do
Solution:
Yes. Each angle ¡n any pair of complementary angles is always acute.
As sum of two angles is 90°, each of the angle must be less than 90°.

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