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SCERT AP 7th Class Maths Solutions Pdf Chapter 6 Data Handling Ex 6.4 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 6th Lesson Data Handling Ex 6.4

Question 1.
In the adjacent Double bar graph, the number of students of a class is shown according to the academic year.
Answer the following questions on the basis of this Double bar graph,

(i) In which academic year, number of girls wrere more than number of boys in the school ?
Answer:
In the year 2016-17 number of girls were more than number of boys in the school.

(ii) In which academic year, number of both girls and boys in the school were equal ?
Answer:
In the year 2015-16 number of bot.fi’ girls and boys in the school were equal.

(iii) What was the total number of students in the school in the academic year 2013-14 ?
Answer:
In the academic year 2013-14
Number of boys = 70
Number of girls = 50 .

Total = 120
Total number of students in the school in the academic year 2013-14 is 120.

Question 2.
The marks in Mathematics and Science of five students of class 7 are given in the table. Exhibit these by the vertical double bar graph by given information.

Answer:

Question 3.
The adjacent Pie chart gives the expenditure on various items during a month for a family. In the figure angles made by each sector at the center are given then answer the following : i) If the expenditure on rent is ?3000, then how much amount they saved ?

Answer:
Angle of sector on Rent = 90°
Angle of sector on Savings = 90°
Angle of sectors are equal. So, their amounts are also equal. Expenditure on rent is ₹3000.

(ii) On which item the expenditure is minimum ?
Answer:
Smallest angle of sector on Education is the item of expenditure is minimum.
So, expenditure on education is minimum.

(iii) On which item the expenditure is maximum ?
Answer:
Largest angle of sector on Food is the item of expenditure is maximum.
So, expenditure on food is maximum.

Question 4.
The following data shows the number of students opting different subjects in college.

Construct a Pie diagram to represent the above data.
Answer:
The angle of each sector will depend on the ratio between the number of students each subject and total number of students.
Angle of sector = \(\frac{\text { Value of the item }}{\text { Sum of the values of all items }}\) × 360°

Steps of construction:

1. Draw a circle with any convenient radius and mark it’s centre as ‘O’.
2. Mark a point A, somewhere on the circumference and join OA.
3. Construct ∠AOB = 90° to represent angle of the sector of Botany.
4. Construct ∠BOC = 120° to represent angle of the sector of Mathematics.
5. Construct ∠COD = 40° to represent angle of the sector of Physics.
6. Construct ∠DOE = 60° to represent angle of the sector of Chemistry.
7. Construct ∠EOF = 20° to represent angle of the sector of Economics.
8. Now ∠FOA = 30° to represent angle of the sector of Commerce.

SCERT AP 7th Class Maths Solutions Pdf Chapter 6 Data Handling Ex 6.3 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 6th Lesson Data Handling Ex 6.3

Question 1.
Find the Median of the following:
(i) 7, 3, 15, 0, 1, 71, 19, 4, 17.
Answer:
Given data: 7, 3, 15. 0, 1, 71, 19, 4, 17.
Arrange the given observations in ascending order.

In nine observations the fifth observation 7 is the middle most value.
∴ Median = 7

(ii) 12, 23, 11, 18, 15, 20, 86, 27.
Answer:
Given data : 12, 23, 11, 18, 15, 20, 86, 27.
Arrange the given observations in ascending order:

In eight observations the 4th and 5th observations are 18 and 20.
Here, we have two middle most values 18 and 20.
Median = Average of the two middle most values.
= \(\frac{18+20}{2}\) = \(\frac{38}{2}\) = 19
∴ Median of the data = 19.

Question 2.
The number of pages in text books of different subjects are 421, 175, 128, 117, 150, 145, 147 and 113 find median of given data.
Answer:
Given data : 421, 175, 128, 117, 150, 145, 147, 113
Arrange the given observations in ascending order

In eight observations the 4th and 5th observations are 145 and 147.
Here, we have two middle most values 145 and 147.
Median = Average of the two middle most values
= \(\frac{145+147}{2}\) = \(\frac{292}{2}\) = 146
∴ Median of the data = 146.

Question 3.
The weekly sales of motor bikes in a showroom for the past 14 weeks are 10, 6, 8, 3, 5, 6, 4, 7, 12, 13, 16, 10, 4 and 7. Find the Median of the data.
Answer:
Given data: 10, 6, 8, 3, 5, 6, 4, 7, 12, 13, 16, 10, 4, 7
Arrange the given observations in ascending order.

In fourteen observations the 7th and 8th observations are 7 and 7.
Here we have two middle most values 7 and 7.
Median = Average of the two middle most values
= \(\frac{7+7}{2}\) = \(\frac{14}{2}\) = 7
∴ Median of the data = 7

Question 4.
Find the Median of 0.3, 0.25, 0.32, 0.147, 0.19, 0.2 and 7.1.
Answer:
Given data: 0.3, 0.25, 0.32, 0.147, 0.19, 0.2, 7.1.
Arrange the given observations in ascending order.

In seven observations the 4th observation 0.25 is the middle most value.
∴ Median. = 0.25

Question 5.
If the Median of observations 2x, 3x, 4x, 5x, 6x, (x > 0) is 28, then find the value of ‘x’.
Answer:
Given data : 2x, 3x, 4x, 5x, 6x and median is 28.
Data is arranged in the ascending order

In five observations the third observation 4x is the middle most value. .
∴ Median – 4x = 28
⇒ \(\frac{4 x}{4}\) = \(\frac{28}{4}\) = 7
∴ x = 7

SCERT AP 7th Class Maths Solutions Pdf Chapter 5 Triangles InText Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 5th Lesson Triangles InText Questions

Check Your Progress [Page No. 92]

Question 1.
Classify the following triangles according to the length of their sides:

Answer:

[Page No. 93]

Question 2.
Classify the following triangles according to the measure of their angles.

Answer:

Let’s Think [Page No. 97]

Question 1.
Can you find a triangle in which each angle is less than 60°?
Answer:
Sum of angles in a triangle is 180°.
If each angle is less than 60°, then the sum of angles of a triangle is < 180°. So, triangle cannot form. So, we cannot find a triangle in which each angle is less than 60°.

Check Your Progress [Page No. 99]

Question 1.
Find the value of ‘x’ in the given figure.

Answer:
In the given figure,
∠HAC = ∠TAE = a (∵ Vertically opposite angles)
In ∆AHC,
we know, ∠H + ∠C + ∠HAC 180°
⇒ 60° + 80° + a = 180°
⇒ 140° + a – 140° = 180°- 140° … a = 40°.
∴ ∠HAC = ∠TAE = a = 40°

In ∆ATE, we know,
∠T + ∠TAE + ∠E = 180°
⇒ x + a + 70° = 180°
⇒ x + 40° + 70° = 180°
⇒ x + 110°- 110° = 180°- 110°
∴ x = 70°

Let’s Do Activity [Page No. 101]

Question 1.
Measure the length of the sides of ∆ABC and fill the following table.

Answer:

From the above table, we can conclude that the sum of the lengths of any two sides of a triangle is greater than the length of the third side.

Question 2.
Take the same measurements of the previous activity and note the results as given below.
Answer:

From the above table we can conclude that the difference of the lengths of any two sides of a triangle is less than the length of the third side.

Question 3.

Fill the following table as per measurements given in the above triangles:

What do you observe from the last four columns in above table ?

• In any triangle the opposite side to the biggest angle is bigger than the other two sides.
• In any triangle the opposite side to the smallest angle is smaller than the other two sides.

Answer:

Let’s Think [Page No. 102]

Question 1.
What are the measurements of angles of an equilateral triangle?
Answer:

In ∆ADI
AD = DI – AI = 6 cm (All angles are equal)
∠A = ∠D = ∠I = 60° (All angles are equal)
So, in an equilateral triangle,

All Sides are equal in length. Each angle is equal to 60°.

Puzzle Time [Page No. 104]

Question 1.
Find the interior angles of all triangles by using given dues.

Answer:

Hints:
Each angle in an equilateral triangle is 60°.
Straight angle = 180°
In an isosceles triangle, two angles are equal.
In a scalene triangle all angles are unequal.
Right angle = 90°.
Obtuse angle > 90°.

Examples:

Question 1.
In a triangle, two angles are 43° and 57°. Find the third angle.
Answer:
Given two angles of a triangle are 43° and 57°.
Sum of these two angles = 43° + 57° = 100°
In a triangle the sum of the interior angles is 180°.
∴ Third angle = 180° – 100° = 80°

Question 2.
Find the value of ‘x’ in the following triangles :

Answer:
From the figure
∠A + ∠B + ∠C = 180° (v Sum of three angles in triangle is 180°)
⇒ 65° + 60° + x = 180°
⇒ 125° + x = 180°
∴ x =180°- 125° = 55°

Answer:
From the figure
∠P + ∠Q + ∠R = 180°
x + x + 80° = 180°
⇒ 2x + 80° = 180°
⇒ 2x = 180°-80°
⇒ 2x = 100°
⇒ x = \(\frac{100^{\circ}}{2}\) = 50°
∴ x = 50°

Answer:
From the figure
∠X + ∠Y + ∠Z – 180°
⇒ x + 90° + 42° = 180°
⇒ x + 132° = 180°
⇒ x = 180°- 132°
∴ x = 48°

Question 3.
Find the values of x and y in the given triangle.

Answer:
In the ∆ABC
∠ACD + ∠ACB = 180° (linear pair of angles)
110° + y = 180°
⇒ y = 180°- 110°
∴ y = 70° ……………….(i)

∠BAC + ∠ACB + ∠CBA = 180°
⇒ x + y + 60° = 180°
⇒ x + 70° + 60° = 180° (From (i))
⇒ x + 130° = 180°
∴ x = 50°

Question 4.
In a right angled triangle one acute angle is 44°, then find the other acute angle.
Answer:
We know that, sum of the two acute angles in a right angled triangle is 90°.
Given that, in the right angled tri-angle one of the acute angles = 44°
Other acute angle in right angled triangle = 90° – 44° = 46°

Question 5.
The angles of a triangle are (x + 10)°, (x – 20)° and (x + 40)°. Find the value of x and the measure of the angles.
Answer:
Given that the angles of the triangle are
(x + 10)°, (x – 20)° and (x + 40)°
(x + 10)° + (x – 20)° + (x + 40)° = 180°
⇒ x + 10° + x – 20° + x + 40° = 180°
⇒ 3x + 30° = 180°
⇒ 3x – 180° – 30°
⇒ 3x = 150°
⇒ x = \(\frac{150^{\circ}}{3}\) = 50°

The angles are,
x + 10° = 50° + 10° = 60°
x – 20° = 50° – 20° = 30°
x + 40° = 50° + 40° = 90°
∴ Measure of the angles are 60°, 30° and 90°.

Question 6.
In the ∆ABC, exterior angle at ∠C = 105° and ∠A=65°. Find the other interior opposite angle.
Answer:
Given that, exterior angle at ∠C = 105°.
One of the interior opposite angle ∠A = 65°.

The other interior opposite angle is ∠B .
∠A + ∠B =105°(∵ Exterior angle property of a triangle)
65° + ∠B = 105°
⇒ ∠B = 105° – 65° = 40°

Question 7.
Find the exterior angle of the given triangle.
Answer:

From the figure, ∠P = 30°, ∠Q = 35°
Exterior angle at R = ∠P + ∠Q (∵ Exterior angle property of a triangle)
= 30° + 35° = 65°

Question 8.
Find the values of ‘x’ and ‘y’ in the following figure.

Answer:
∠LMN + 110° = 180° (Linear pair angles)
∠LNM = 180° -110° = 70°
∠LMN = ∠LNM (Angles opposite to equal sides)
y = 70°
⇒ x + y – 110° (Exterior angle property)
⇒ x + 70° = 110°
∴ x = 110°-70° – 40°

SCERT AP 7th Class Maths Solutions Pdf Chapter 5 Triangles Unit Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 5th Lesson Triangles Unit Exercise

Question 1.
How many right angles exist in a triangle?
Answer:
Only one right angle can exist in a triangle.

Question 2.
Which is the longest side in ∆XYZ having a right angle at ‘Z’?
Answer:

In ∆XYZ, z = 90°
In triangle, the side opposite to the largest angle is longest side. .
So, XY is the longest side in ∆XYZ.

Question 3.
Is the sum of any two angles of a tri-angle always greater than the third angle? Give examples to justify your answer.
Answer:

In ∆ADI, ∠A = 70°, ∠D = 50°, ∠I = 60°
∠A + ∠D = 70° + 50° =120° > 60° ∴ ∠A + ∠D > ∠I.
∠D + ∠I = 50° + 60° = 110° > 70° ∴ ∠D + ∠I > ∠A.
∠A + ∠I = 70° + 60° = 130° > 50° ∴ ∠A + ∠I > ∠D.
Yes, sum of any two angles of a triangle always greater than the third angle.

Question 4.
Choose any three measures from the following to make three different triangular wooden frames.
11 m, 9 m, 3 m, 7 m and 5 m.
Answer:
We know that in a triangle sum of any two sides is greater than the third side.
(i) 3m, 7m, 5m
(ii) 3m, 9m, 11m
(iii) 3m, 7m, 9m

Question 5.
Write any two possible measurements to be suitable for the following triangles.
(i) Right angled triangle.
Answer:

In ∆SRI,
∠S = 40°, ∠R = 90°, ∠I = 50°
SR = 4 cm, RI = 3 cm and SI = 5 cm

(ii) Obtuse angled triangle.
Answer:

In ∆MDI, .
∠A = 200, ∠D = 1200, ∠I = 40°
AD = 4 cm, DI = 5 cm and AI = 7cm

(iii) Acute angled triangle.
Answer:

In ∆KIS,
∠K = 50°, ∠I – 60°, ∠S = 70°
KI = 4 cm, IS = 6 cm, KS = 8 cm

Question 6.
Find the value of ‘x’ and ‘y’ in the adjacent figure.

Answer:
In ∆ACL, AC = CL
So, ∆ACL is an isosceles triangle.

In isosceles triangle, the angles which are opposite to equal sides are also equal.
So, ∠A = ∠L = x°
∠C = 56° (Vertically opposite angle)

In ∆ACL,
we know ∠A + ∠C + ∠L = 180°
⇒ x + 56° + x = 180°
⇒ 2x + 56° – 56° = 180° – 56°
⇒ 2x =124°
⇒ \(\frac{2 x}{2}=\frac{124^{\circ}}{2}\)
∴ x = 62°
Exterior angle at A = ∠C + ∠L
⇒ y = 56 + 62
⇒ y = 118°

Question 7.
In ∆ABC, ∠A is four times to ∠B and ∠C is five times to ∠B. Find the three angles.
Answer:

In ∆ABC, let ∠B = x°
then ∠A 4 times of ∠B = 4x°
∠C = 5 times of ∠B = 5x°

In ∆ABC, we know
∠A + ∠B + ∠C = 180°
⇒ 4x + x + 5x = 180°
⇒ 10x = 180°

x = 18°
∠B = 18°,
∠A = 4x° = 4 × 18° = 72°,
∠C = 5x° = 5 × 18° = 90°
∴ Three angles are 18°, 72° and 90°.

Question 8.
Ladder was faced to a wall, One end of the ladder was making 70° with the floor. Find the angle of the other end of the ladder with the wall.
Answer:

Let A be the one end of the ladder.
C is other end of the ladder.
B is the foot of the wall.

So, ∆ABC is formed.
In ∆ABC, ∠A = 70°, ∠B = 90°

We know, in ∆ABC,
∠A + ∠B + ∠C = 180° ‘
70° + 90° + ∠C = 180°
160° + ∠C = 180°
160° + ∠C – 160° = 180° – 160°
∠C = 20°
So, angle of the other end of the ladder is 20°.

Question 9.
Write the possible measurements of angles in the following table. One example is given for you.
Answer:

SCERT AP 7th Class Maths Solutions Pdf Chapter 5 Triangles Ex 5.4 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 5th Lesson Triangles Exercise 5.4

Question 1.
Two sides of a triangle are 5cm and 4cm respectively. Write any three possible measurements that suit the third side.
Answer:
Given two sides of a triangle are 5 cm and 4 cm.
We know that, in triangle sum of lengths of any two sides is always greater than the third side, and in the triangle difference of lengths of any two sides is always less than the third side.
So, 5 cm, 4 cm, 2 cm
5 cm, 4 cm, 3 cm
5 cm, 4 cm, 4 cm
5 cm, 4 cm, 5 cm
5 cm, 4 cm, 6 cm
5 cm, 4 cm, 7 cm
Measurements that suit for third side are 2 cm, 3 cm, 4 cm, 5 cm, 6 cm, 7 cm.

Question 2.
The lengths of line segments are 3 cm, 5 cm, 6 cm and 9 cm.
(i) From the above measurements which group of the line segments can form a triangle.
Answer:
3 cm, 5 cm, 6 cm and 5 cm, 6 cm, 9 cm groups form triangle.
Because, sum of any two sides is greater than the third side in this group.

(ii) Which group of line segments cannot form a triangle, give the reason?
Answer:
We know that, sum of the lengths of any two sides is always greater than the third side.
3 cm, 6 cm, 9 cm does not form a triangle.
Because sum of 3 cm and 6 cm is not greater than 9 cm.

Question 3.
Find the value of ‘x’ in the following triangles :

Answer:
Given, in ∆PEN, PE = 4 cm, PN = 4 cm ∠E = 65°, ∠N = x°
Two sides are equal in length.
So, ∆PEN is an isosceles triangle.

In isosceles triangle, the angles which are opposite to equal sides are also equal.
So, ∠N = ∠E
∴ x = 65°

Answer:
Given, in ∆BGA,
∠B = 56°, ∠G = x, ∠A = 56° and BG = x cm, GA = 4.3 cm
We know that, sum of angles of a triangle is 180°.
In ∆BGA, ∠B + ∠G + ∠A = 180°
⇒ 56° + x + 56° = 180°
⇒ 112° + x = 180° :
⇒ 112° + x – 112° =180°- 112°
∴ x = 68°

Two angles are equal in ∆BGA.
So, ∆BGA is an isosceles triangle.
In isosceles triangle the sides which are opposite to equal angles are also equal.
So, BG = GA
∴ x = 4.3 cm

Question 4.
∆ABC is an isosceles triangle in which AB = AC. If ∠A = 80°, then find ∠B and ∠C.
Answer:

In isosceles triangle the angles which are opposite to equal sides are also equal.
So, ∠B = ∠C = x°
In ∆ABC, we know
∠A + ∠B + ∠C — 180°
⇒ 80° + x + x = 180°
⇒ 80° + 2x – 80° = 180° – 80°
⇒ 2x = 100°
⇒ \(\frac{2 x}{2}=\frac{100^{\circ}}{2}\)
∴ x = 50°
∴ ∠B = ∠C = 50°

Question 5.
Find the values of ‘x’ in each of the below triangles.

Answer:
In ∆PQR, PQ = QR and ∠P = 53°
∆PQR is an isosceles triangle.

In isosceles triangle, the angles which are opposite to equal sides are also equal.
So, ∠P = ∠R = 53°
In ∆PQR, we know
∠P + ∠Q + ∠R = 180°
⇒ 53° + x + 53° = 180°
⇒ 106° + x = 180°
⇒ 106° + x- 106° = 180°- 106°
∴ x = 74°

Answer:
In ∆LMN, LM = LN and exterior angle at L = 110° .
∆LMN is an isosceles triangle.
In isosceles triangle, the angles which are opposite to equal sides are also equal.

So, ∠M = ∠N = x° .
We know that an exterior angle of a triangle is equal to the sum of its opposite interior angles.
Exterior angle at
L = ∠M + ∠N = 110°
⇒ x + x = 110°
⇒ 2x = 110°
⇒ \(\frac{2 x}{2}=\frac{110}{2}\)
∴ x = 55°

Question 6.
Which of the following statements are true in the following diagram.

(i) OY < OT
Answer:

In ∆TOY, ∠T + ∠O + ∠Y = 180°
⇒ ∠T + 90° + 50° = 180°
⇒ ∠T + 140° = 180°
⇒ ∠T + 140°- 140° = 180° – 140°
∴ ∠T – 40°
We know that side opposite to the largest angle is longest.
OY< OT – True

(ii) TY < TO
Answer:
TY < TO False

(iii) ∠Y < ∠T
Answer:
∠Y < ∠T
50° < 40° – False

(iv) TY < OY
Answer:
TY < OY – False

SCERT AP 7th Class Maths Solutions Pdf Chapter 5 Triangles Ex 5.3 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 5th Lesson Triangles Exercise 5.3

Question 1.
Write the exterior angles of ∆XYZ.

Answer:
Exterior angles of ∆XYZ are ∠PXY, ∠ZYQ and ∠XZR.
Exterior angle of X is ∠PXY. Exterior angle of Y is ∠QYZ.
Exterior angle of Z is ∠XZR.

Question 2.
Find the exterior angles in each of the following triangles :

Answer:
From the figure
∠A = 60°, ∠B = 73°
Exterior angle of a triangle is equal to the sum of its opposite interior angles.

Exterior angle at
C = ∠A + ∠B
= 600 + 730
= 133°

Answer:
From the figure,
∠D = 90°, ∠E = 30°
Exterior angle of a triangle is equal to the sum of its opposite interior angles.
Exterior angle at
F = ∠D + ∠E
= 90° +.30°
= 120°

Question 3.
Find the value of ‘x’ in the following figures.

Answer:
From the figure, ∠A = 35°, ∠B = x° and exterior angle at C = 70°
Exterior angle of a triangle is equal to the sum of its opposite interior angles. Exterior anjpNtt
C = ∠A + ∠B = 70°
⇒ 35° + x= 70°
⇒ 35° + x – 35° = 70° – 35°
∴ x = 35°

Answer:
From the figure, ∠P = 4x, ∠Q = 3x and exterior angle at R = 119°
Exterior angle of a triangle is equal to the sum of its opposite interior angles.
Exterior angle at R = ∠P + ∠Q = 119°
⇒ 4x + 3x = 119°
⇒ 7x = 119°
⇒ \(\frac{7 x}{7}=\frac{119^{\circ}}{7}\)
∴ x = 17°

Question 4.
If the exterior angle of a triangle is 110° and it’s interior opposite angles are x° and (x + 10)°, then find the value of ‘x’.
Answer:
Given interior opposite angles are x° and (x + 10)°.
Exterior angle =110° .
Exterior angle of a triangle is equal to the sum of its opposite interior angles.
x + x + 10° = 110°
⇒ 2x + 10°- 10° = 110°- 10°
⇒ 2x = 100°
⇒ \(\frac{2 x}{2}=\frac{100^{\circ}}{2}\)
∴ x = 50°

Question 5.
Find the values of V and ‘y’ in each of the following figures.

Answer:
From the figure, ∠A = 40°, ∠B = 60°, ∠D = 45°
Exterior angle at C = x°
Exterior angle at E = y°

Exterior angle of a triangle is equal to the sum of its opposite interior angles.
Exterior angle at C = ∠A + ∠B
x = 40° + 60°
∴ x = 100°

Exterior angle at E = ∠C + ∠D
y = x + 45°
y = 100° + 45°
∴ y = 145°

Answer:
From the figure, ∠M = y, ∠N = 70°
Exterior angle at M = x°
Exterior angle at L = 120°

Exterior angle of a triangle is equal to the sum of its opposite interior angles.
Exterior angle at L = ∠M + ∠N = 120°
⇒ y + 70° = 120°
⇒ y + 70 – 70 = 120 – 70
∴ y = 50°

∠QLN + ∠MLN = 180° (linear pair of angles)
120° + ∠MLN = 180°
120° + ∠MLN – 120° = 180°- 120°
∠MLN = 60°
∴ ∠L = 60°

Exterior angle at M = ∠L + ∠N
x = 60° + 70° = 130°
∴ x = 130°
∴ x = 130° and y = 50°

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 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 3 Simple Equations Unit Exercise Questions and Answers.

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

Question 1.
Choose the correct answer.
(i) Which of the following numbers satisfy the equation – 6 + m = – 10 ?
(a) 2
(b) 4
(c) – 4
(d) – 2
Answer:
(c) – 4

Explaination:
– 6 + m = -10
⇒ – 6 + m + 6 = – 10 + 6
(Add 6 on both sides)
⇒ m = – 4

(ii) The equation having – 2 as solution is
(a) x + 2 = 5
(b) 7 + 3x = 1
(c) 2x + 3 = 7
(d) 2(x + 1) = 4
Answer:
(b) 7 + 3x = 1

Explaination
(a) x + 2 = 5
⇒ x + 2 – 2 = 5 – 2 (Subtract 2 on both sides)
⇒ x = 3 it is not – 2

(b) 7 + 3x = 1
⇒ 7 + 3x – 7 = 1 – 7 (Subtract 7. on both sides)
⇒ 3x = – 6
⇒ \(\frac{3 x}{3}\) = \(\frac{-6}{3}\) (Divide by 3 on both sides)
∴ x = – 2
so, solution is – 2
Similarly, check (c) and (d) also.

(iii) If a and b are positive integers, then the solution of the equation ax = b will always be a ____________ .
(a) positive number
(b) negative number
(c) 1
(d) 0
Answer:
(a) positive number

Explaination:
If a and b are positive integers and b < a, then ax ± b also positive integer.

(iv) The equation which can’t be solved in integers is
(a) 2(x – 3) = 10
(b) \(\frac{x}{3}\) = 5
(c) 5 – 3m = 1
(d) 2k + 1 = 1
Answer:
(c) 5 – 3m = 1

Explaination:
(a) 2(x – 3) = 10
⇒ 2x – 6 = 10 (Distributive property)
⇒ 2x – 6 + 6 = 10 + 6 (Add 6 on both sides)
⇒ 2x = 16
⇒ \(\frac{2x}{2}\) = \(\frac{16}{2}\) (Divide by 2 on both sides)
⇒ x = 8 is an integer.

(b) \(\frac{x}{3}\) = 5
⇒ \(\frac{x}{3}\) × 3 = 5 × 3 (Multiply by 3 on both sides)
⇒ x = 15 is an integer.

(c) 5 – 3m = 1
⇒ 5 – 3m – 5 = 1 – 5 (Subtract 5 on both sides)
⇒ – 3m = – 4
⇒ \(\frac{-3 m}{-3}\) = \(\frac{-4}{-3}\)
⇒ m = \(\frac{4}{3}\) is not an integer.

(d) 2k + 1 = 1
⇒ 2k + 1 – 1 = 1 – 1 (Subtract 1 on both sides)
⇒ 2k = 0
⇒ \(\frac{2 k}{2}\) = \(\frac{0}{2}\) (Divide by 2 on both sides)
⇒ k = 0 is an integer
So, our answer is c.

(v) Which of the following is not allowed in a given equation?
(a) Adding the same number to both sides of the equation.
(b) Subtracting the same number from both sides of the equation.
(c) Multiplying both sides of the equation by the same non-zero number.
(d) Dividing both sides of the equation by the same number.
Ans.
(d) Dividing both sides of the equation by the same number.

Question 2.
Fill in the blanks.
(i) If 2y – 1 = 5, then value of 5y + 3 is ____________
Answer:
Given 2y – 1 = 5
⇒ 2y – 1 + 1 = 5 + 1 (Add 1 on both sides)
⇒ 2y = 6
⇒ \(\frac{2 y}{2}\) = \(\frac{6}{2}\) (Divide by 2 on both sides)
∴ y = 3
then 5y + 3 = 5(3) + 3
= 15 + 3 = 18

(ii) Changing of term from one side of equation to other side is called __________
Answer:
Transposition.

(iii) If the sum of two numbers is 60. One is thrice the other, then the equation formed is ___________.
Answer:
Let one number be x.
Other number = thrice of first number = 3(x) = 3x
Sum of two numbers = 60
⇒ x + 3x = 60
⇒ 4x = 60

(iv) If ‘x’ is a natural number, then the solution of x – 8 = – 8 is _________ .
Answer:
x – 8 = – 8
x – 8 + 8 = – 8 + 8 (Add 8 on both sides)
x = 0 is not a natural number.
So, given equation has no solution.

(v) 13 subtracted from twice of a number gives 3, then the number
Answer:
Let the number be x.
Twice the number = 2x
13 is subtracted from twice the number
⇒ 2x- 13 = 3
⇒ 2x – 13 + 13 = 3 + 13 (Add 13 on both sides)
⇒ 2x = 16
⇒ \(\frac{2 x}{2}\) = \(\frac{16}{2}\) (Divide 2 on both sides)
⇒ x = 8
∴ Number is 8.

Question 3.
Check whether the value given in the brackets is a solution to the given equation or not.
(a) 2n + 5 = 19 (n = 7)
Answer:
Given 2n + 5 = 19
Substitute n = 7 in the given equation
2(7) + 5 = 19
⇒ 14 + 5 = 19
⇒ 19 = 19
∴ LHS = RHS
So, n = 7 is the solution of the given equation.

(b) \(\frac{3 m}{5}\) – 7 = 1 (m = 10)
Answer:
Given \(\frac{3 m}{5}\) – 7 = 1
Substitute m = 10 in the given equation
\(\frac{3(10)}{5}\) – 7 = 1
⇒ 3 × 2 – 7 = 1
⇒ 6 – 7 = 1
∴ – 1 = 1
LHS ≠ RHS
So, m = 10 is not the solution of given equation.

Question 4.
Solve 5 – 2k = – 3 using trial and error method.
Answer:

For k = 4, LHS = RHS
So, k = 4 is the solution of given equation.

Question 5.
Write the following equations in mathe-matical statement form.
(a) 2m + 7 = 21
Answer:
7 more than twice the m is 21.

(b) \(\frac{n}{7}\) = 4
Answer:
One seventh of n is 4.

Question 6.
Give the steps you will use to separate the variables and then solve the equation.
(a) 7(x – 3) = 28
Answer:
Given 7(x – 3) = 28
⇒ 7x – 21 = 28 (Distributive property)
⇒ 7x – 21 + 21 = 28 + 21 (Add 21 on both sides)
⇒ 7x = 49
⇒ \(\frac{7 x}{7}\) = \(\frac{49}{7}\) (Divide by 7 on both sides)
⇒ x = 7

(b) 8y – 9 = 15
Answer:
Given 8y – 9 = 15
⇒ 8y – 9 + 9 = 15 + 9 (Add 9 on both sides)
⇒ 8y = 24
⇒ \(\frac{8 y}{8}\) = \(\frac{24}{8}\) (Divide by 8 on both sides)
⇒ y = 3

Question 7.
Solve the following equations and check the result (Method of Transposition)
(a) 9(a + 3) + 7 = 22
Given 9(a + 3) + 7 = 22
⇒ 9(a + 3) + 7 – 7 = 22 – 7 (Subtract 7 on both sides)
⇒ 9(a + 3) = 15
⇒ 9a + 27 = 15 (Distributive property)
⇒ 9a + 27 – 27 = 15 – 27 (Subtract 27 on both sides)
⇒ 9a = – 12
⇒ \(\frac{9 a}{9}\) = \(\frac{-12}{9}\) (Divide by 9 on both sides0
⇒ a = \(\frac{-4}{3}\)

Check:
Substitute a = \(\frac{-4}{3}\) in the given equation.
LHS = 9(a + 3) + 7
= \(9\left(\frac{-4}{3}+\frac{3}{1}\right)\) + 7
= \(9\left(\frac{-4+9}{3}\right)\) + 7

= 15 + 7 = 22 = RHS
Hence verified.

(b) 25 = 18 – 7(b – 6)
Answer:
Given 25 = 18 – 7(b – 6)
⇒ 25 – 18 = l8 – 7(b – 6) – 18 (Subtract 18 on both sides)
⇒ 7 = – 7(b – 6)
⇒ \(\frac{7}{-7}\) = \(\frac{-7(b-6)}{-7}\) (Divide by – 7 on both sides)
⇒ – 1 = b – 6
⇒ – 1 + 6 = b – 6 + 6 (Add 6 on both sides)
⇒ 5 = b
∴ b = 5 (By transposition)

Check: Substitute b = 5 in the
given equation.
RHS = 18- 7(b – 6) – = 18-7(5-6)
= 18 – 7(- 1)
= 18 + 7 = 25 = LHS
Hence verified.

Question 8.
Six times a number is 72. Find the number.
Answer:
Let the number be x.
Given, six times of number = 72
⇒ 6x = 72
⇒ \(\frac{6 x}{6}\) = \(\frac{72}{6}\)
⇒ x = 12
∴ The number is 12.

Question 9.
Three-fourth of a number is more than one-fourth of same number by 2. Find the number.
Answer:
Let the number be x.
Three fourth of number = \(\frac{3}{4}\) of x = \(\frac{3 x}{4}\)
One-fourth of number = \(\frac{1}{4}\) of x = \(\frac{x}{4}\)
Three-fourth of number
= One-fourth of number + 2
⇒ \(\frac{3 x}{4}=\frac{x}{4}+\frac{2}{1}\)
⇒ \(\frac{3 x}{4}=\frac{x+8}{4}\)
⇒ \(\frac{3 x}{4} \times 4=\frac{x+8}{4} \times 4\) (Multiply by 4 on both sides)
⇒ 3x = x + 8
⇒ 3x – x = x + 8 – x (Subtract x on both sides)
⇒ 2x = 8
⇒ \(\frac{2 x}{2}\) = \(\frac{8}{2}\) (Divide by 2 on both sides)
⇒ x = 4

Question 10.
In the given figure the perimeter of the square is 40m. Then find x.

Answer:
From the above figure,
Side of square = (3x – 5) m
Perimeter of square = 40 m
⇒4 × side = 40
⇒ 4 × (3x – 5) = 40
⇒ \(\frac{4(3 x-5)}{4}\) = \(\frac{40}{4}\) (Divide by 4 on both sides)
⇒ 3x – 5 = 10
⇒ 3x – 5 + 5 = 10 + 5 (Add 5 on both sides)
⇒ 3x = 15
⇒ \(\frac{3 x}{3}\) = \(\frac{15}{3}\) (Divide by 3 on both sides)
∴ x = 5 m

Question 11.
Jeevan is 3 years younger than his brother Sasi. If the sum of their present ages is 19 years. What are their present ages ?
Answer:
Let the present age of Sasi is x years.
The present age of Jeevan = (x – 3) years
Sum of their present ages = 19
⇒x + x – 3 = 19
⇒ 2x – 3 = 19
⇒ 2x – 3 + 3 = 19 + 3 (Add 3 on both sides)
⇒ 2x = 22
⇒ \(\frac{2 x}{2}\) = \(\frac{22}{2}\) (Divide by 2 on both sides)
⇒ x = 11 years
∴ Present age of Sasi x = 11 years
Present age of Jeevan = x – 3
= 11 – 3 = 8 years

Question 12.
The length of a rectangle is 20m more than its width. If the perimeter of the rectangle is 100m, then find the length and breadth of rectangle.
Answer:
Let the width of a rectangle (b) = x m
Then the length of a rectangle (l)
= 20 more than its width
= (x + 20) m
Perimeter of the rectangle = 100 m
⇒ 2(1 + b) = 100
⇒ 2(x + 20 + x) = 100
⇒ 2(2x + 20) = 100
⇒ 4x + 40 = 100
⇒ 4x + 40 – 40 = 100 – 40 (Subtract 40 on both sides)
⇒ 4x = 60 4x 60
⇒ \(\frac{4 x}{4}\) = \(\frac{60}{2}\) (Divide by 4 on both sides)
⇒ x = 15 m
∴ Width x = 15 m
Length = x + 20 = 15 + 20 = 35 m

Question 13.
In a family, the consumption of rice is 4 times that of wheat. The total consumption of the two cereals in a month is 30kg. Find the quantities of rice and wheat consumed in the family.
Answer:
Let the quantity of wheat consumed in the month x kg. .
Quantity of rice = 4 times of wheat = 4x kg
Quantity of rice + Quantity of wheat = 30 kg
⇒ 4x + x = 30
⇒ 5x = 30
⇒ \(\frac{5 x}{5}\) = \(\frac{30}{5}\) (Divide by 5 on both sides)
⇒ x = 6 kg
∴ Quantity of wheat = 6 kg
Quantity of rice = 4x = 4 × 6 = 24 kg

Question 14.
The teacher tells the students in the class that the highest marks obtained by a student in the class is 7 more than twice the lowest marks. If the highest mark is 93, then what is the lowest mark ?
Answer:
Let the lowest mark = x then the highest mark
twice the lowest mark + 7 = 93
⇒ 2x + 7 = 93
⇒ 2x + 7 – 7 = 93 – 7 (Subtract 7 on both side)
⇒ 2x = 86
⇒ \(\frac{2 x}{2}\) = \(\frac{86}{2}\) (Divide by 2 on both sides)
⇒ x = 43
∴ Lowest mark = 43

Question 15.
A man travelled \(\frac{4}{5}\) of his journey by train, \(\frac{1}{7}\) by bus and the remaining 16 km by auto. What is the length of his total journey ?
Answer:
Let the length of total journey = x km
Travelled by train = \(\frac{4}{5}\) of x = \(\frac{4x}{5}\) km
Travelled by bus = \(\frac{1}{7}\) of x = \(\frac{x}{7}\) km
Travelled by auto =16 km
Total length of journey = \(\frac{4x}{5}\) + \(\frac{x}{7}\) + 16
⇒ x = \(\frac{4 x}{5}+\frac{x}{7}+\frac{16}{1}\)
⇒ x = \(\frac{28 x+5 x+16 \times 35}{35}\)
⇒ x = \(\frac{33 x+560}{35}\)

(Multiply 35 on both sides)
⇒ 35x = 33x + 560
⇒ 35x – 33x = 33x + 560 – 33x (Subtract 33x on both sides)
⇒ 2x = 560
⇒ \(\frac{2 x}{2}\) = \(\frac{560}{2}\) (Divide by 2 on both sides)
⇒ x = 280 km
Therefore, length of total journey = 280 km

SCERT AP 7th Class Maths Solutions Pdf Chapter 3 Simple Equations Ex 3.3 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 3rd Lesson Simple Equations Exercise 3.3

Question 1.
Solve the following equations and check the result.
(i) 5x – 17 = 18
Answer:
Given 5x – 17 = 18
⇒ 5x – 17 + 17 = 18 + 17 (Add 17 on both sides)
⇒ 5x = 35
⇒ \(\frac{5 x}{5}\) = \(\frac{35}{5}\) (Divide by 5 on both sides)
⇒ x = 7

Check:
Substitute x = 7 in 5x – 17 = 18
LHS = 5x – 17
= 5(7) – 17
= 35 – 17 = 18 = RHS
Hence verified.

(ii) 29 – 7y = 1
Answer:
Given 29 – 7y = 1
⇒ 29 – 7y – 29 = 1 – 29 (Subtract 29 on both sides)
⇒ – 7y = -28 .
⇒ \(\frac{-7 y}{-7}\) = \(\frac{-28}{-7}\) (Divide by – 7 on both sides)
⇒ y = 4

Check:
Substitute y = 4 in
29 – 7y = 1
LHS = 29 – 7y
= 29 – 7(4)
= 29 – 28 = 1 = RHS
Hence verified.

(iii) a – 2.3 = 1.5
Answer:
Given a – 2.3 = 1.5
⇒ a – 2.3 + 2.3 = 1.5 + 2.3 (Add 2.3 ort both sides)
⇒ a = 3.8

Check:
Substitute a = 3.8 in a – 2.3 =1.5
LHS = a – 2.3
= 3.8 – 2.3
= 1.5 = RHS
Hence verified.

(iv) b + 3\(\frac{1}{2}\) = \(\frac{7}{4}\)
Answer:

(v) \(\frac{7 p}{10}\) + 9 = 15
Answer:

(vi) 6(q – 5) = 42.
Answer:
Given 6(q – 5) = 42
⇒ \(\frac{6(q-5)}{6}\) = \(\frac{42}{6}\) (Divide by 6 on both sides)
⇒ q – 5 = 7
⇒ q – 5 + 5 = 7 + 5 (Add 5 on both sides)
⇒ q = 12

Check:
Substitute q = 12 in
6(q – 5) = 42
LHS = 6(q – 5)
= 6(12 – 5)
= 6(7) = 42 = RHS
Hence verified.

(vii) – 3(m + 5) + 1 = 13
Answer:
Given – 3(m + 5)4-1 = 13
⇒ – 3(m + 5)+ 1 – 1 = 13 – 1 (Subtract 1 on both sides)
⇒ – 3(m + 5) = 12
⇒ \(\frac{-3(m+5)}{-3}\) = \(\frac{12}{-3}\) (Divide by – 3 on both sides)
⇒ m + 5 = – 4
⇒ m + 5 – 5 = – 4 – 5 (Subtract 5 on both sides)
⇒ m = – 9

Check:
Substitute m = – 9 in – 3(m + 5) + 1 = 13
LHS = – 3(m + 5) + 1
= – 3(- 9 + 5) + 1 = (- 3 × – 4) + 1
= + 12 + 1 = 13 = RHS
Hence verified.

(viii) \(\frac{n}{2}+\frac{n}{3}+\frac{n}{5}\) = 31
Answer:

Question 2.
Solve the following equations and check the result.

(i) 3(p – 7) – 4 = 5
Answer:
Given 3(p – 7) – 4 = 5
⇒ 3(p – 7) – 4 + 4 = 5 + 4 (Add 4 on both sides)
⇒ 3(p – 7) = 9
⇒ \(\frac{3(p-7)}{3}\) = \(\frac{9}{3}\) (Divide by 3 on both sides)
⇒ p – 7 = 3
⇒ p – 7 + 7 = 3 + 7 (Add 7 on both sides)
⇒ p = 10

Check:
Substitute p = 10 in
3(p – 7) – 4 = 5
LHS = 3(p – 7) – 4.
= 3(10 – 7) – 4.
= 3 × 3 – 4
= 9 – 4 = 5 = RHS
Hence verified.

(ii) 5(q – 3) – 3(q – 2) = 0
Answer:
Given 5(q – 3) – 3(q – 2) = 0
⇒ 5q – 15 – 3q + 6 = 0 (Distributive property)
⇒ 2q – 9 = 0 .
⇒ 2q – 9 + 9 = 0 + 9 (Add 9 on both sides)
⇒ 2q = 9
⇒ \(\frac{2 \mathrm{q}}{2}\) = \(\frac{9}{2}\) (Divide by 2 on both sides)
⇒ q = \(\frac{9}{2}\)

Check : Substitute q = \(\frac{9}{2}\) in
5(q – 3) – 3(q – 2) = 0
LHS = 5(q – 3) – 3(q – 2)
= 5\(\left(\frac{9}{2}-3\right)\) – 3\(\left(\frac{9}{2}-2\right)\)
= 5\(\left(\frac{9-6}{2}\right)\) – 3\(\left(\frac{9-4}{2}\right)\)
= 5 × \(\frac{3}{2}\) – 3 × \(\frac{5}{2}\)
= \(\frac{15}{2}-\frac{15}{2}\) = 0 = RHS
Hence verified.

(iii) 4x – 0.3x – 1.2 = 0.6
Answer:
Given 0.4x – 0.3x – 1.2 = 0.6
⇒ 0.1 x – 1.2 = 0.6
⇒ 0.1 x – 1.2 + 1.2 = 0.6 + 1.2 (Add 1.2 on both sides)
⇒ 0.1 x = 1.8
⇒ \(\frac{1 \mathrm{x}}{10}\) = \(\frac{18}{10}\)

(Multiply by 10 on both sides)
⇒ x = 18

Check:
Substitute x = 18 in
0.4x – 0.3x – 1.2 = 0.6
LHS = 0.4x – 0.3x – 1.2
= 0.4(18) – 0.3(18) – 1.2
= 7.2 – 5.4 – 1.2
= 7.2 – 6.6 = 0.6 = RHS
Hence verified.

(iv) 4(3y + 4) = 7.6
Answer:
Given 4(3y + 4) = 7.6.
⇒ \(\frac{4(3 y+4)}{4}\) = \(\frac{7.6}{4}\) (Divide by 4 on both sides)
⇒ 3y + 4 = 1.9
⇒3y + 4 – 4= 1.9 – 4 (Subtract 4 on both sides)
⇒ 3y = – 2.1
⇒ \(\frac{3 y}{3}\) = – \(\frac{2.1}{3}\) (Divide by 3 on both sides)
⇒ y = – 0.7

Check:
Substitute y = – 0.7 in
4(3y + 4) = 7.6
LHS = 4(3y + 4)
= 4[3 (- 0.7) + 4]
= 4[- 2.1 + 4]
= 4 × 1.9 = 7.6 = RHS
Hence verified.

(v) 20 – (2r – 5) = 25
Answer:
Given 20 – (2r – 5) = 25

(Subtract 20 on both sides)
⇒ – (2r – 5) = 5
⇒ – (2r -5) × – 1 = 5 × – 1
(Multiply by – 1 on both sides)
⇒ 2r – 5 = 5
⇒ 2r – 5 + 5= – 5 + 5 (Add 5 on both sides)
⇒ 2r = 0
⇒ \(\frac{2 \mathrm{r}}{2}\) = \(\frac{0}{2}\) (Divide by 2 on both sides)
⇒ r = 0

Check:
Substitute r = 0 in
20 – (2r – 5) = 25
LHS = 20 – (2r – 5)
= 20 – [2(0) – 5]
= 20 – [0 – 5]
= 20 – (- 5)
= 20 + 5 = 25 = RHS
Hence verified.

(vi) 3(5 – t) – 2(t – 2) = – 1
Answer:
Given 3(5 – t) – 2(t – 2) = – 1
⇒ 15 – 3t – 2t + 4 .= – 1 (Distributive property)
⇒ 19 – 5t = – 1

(Subtract 19 on both sides)
⇒ – 5t = – 20
⇒ \(\frac{-5 t}{-5}\) = \(\frac{-20}{-5}\) (Divide by – 5 on both sides)
⇒ t = 4

Check:
Substitute t = 4 in
3(5 -1) – 2(t – 2) = – 1
LHS = 3(5 – t) – 2(t – 2)
= 3(5 – 4) – 2(4 – 2)
= 3(1) – 2(2)
= 3 – 4 = – 1 = RHS
Hence verified.

(vii) 3(2k + 1) – 2(k – 5) – 5(5 – 2k) = 16
Answer:
Given 3(2k + 1) – 2(k T 5) – 5(5 – 2k) = 16
⇒ 6k + 3-2k + 10-25 + 10k = 16 (Distributive property)
⇒ 14k – 12 = 16
⇒ 14k – 12 + 12 – 16 + 12 (Add 12 on both sides)
⇒ 14k = 28
⇒ \(\frac{14 \mathrm{k}}{14}\) = \(\frac{28}{14}\) (Divide by 14 on both sides)
⇒ k = 2

Check:
Substitute k = 2 in
3(2k + 1) – 2(k – 5) – 5(5 – 2k) = 16
LHS = 3(2k + 1) – 2(k – 5) – 5(5 – 2k)
= 3[2 × 2 + 1] – 2[2 – 5] – 5[5 – 2 × 2]
= 3[4 + 1] – 2(- 3) – 5(5 – 4)
= 15 + 6 – 5 – 16 = RHS
Hence verified.

(viii) \(\frac{3 \mathrm{~m}}{4}\) – 5m – \(\frac{3}{4}\)
Answer:
⇒ – 17m – 3 = 48
⇒ – 17m – 3 + 3 =48 + 3 (Add 3 on both sides)
⇒ – 17m = 51
⇒ \(\frac{-17 \mathrm{~m}}{-17}\) = \(\frac{51}{-17}\) (Divide by – 17 on both sides)
⇒ m = – 3

(ix) \(\frac{4 n}{5}+\frac{n}{4}-\frac{n}{2}=\frac{11}{10}\)
Answer:

(x) \(\frac{x}{2}-\frac{4}{5}+\frac{x}{5}+\frac{3 x}{10}=\frac{1}{5}\)
Answer:

Question 3.
Write any three equivalent equations having the solution x = 2.
Answer:

Question 4.
Write any three equivalent equations having the solution a = – 5.
Answer:

SCERT AP 7th Class Maths Solutions Pdf Chapter 3 Simple Equations Ex 3.2 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 3rd Lesson Simple Equations Exercise 3.2

Question 1.
Give the steps you will use to separate the variables and then solve the equation.
(i) \(\frac{5 m}{3}\) = 10
Answer:
Given \(\frac{5 m}{3}\) = 10
⇒ \(\frac{5 m}{3}\) × 3 = 10 × 3 (Multiply with 3 on both sides)
⇒ 5m = 30
⇒ \(\frac{5 m}{5}\) = \(\frac{30}{5}\) (Divide both sides by 5)
⇒ m = 6

Check:
Substitute m = 6 in the given equation,
LHS = \(\frac{5 m}{3}\) = \(\frac{5(6)}{3}\) = \(\frac{30}{3}\) = 10 = RHS.
Hence verified.

(ii) 4M – 23 = 13
Answer:
Given 4n – 23 = 13

⇒ 4n = 36
⇒ \(\frac{4 n}{4}\) = \(\frac{36}{4}\) (Divide both sides by 4)
⇒ n = 9

Check:
Substitute n = 9 in the given equation.
LHS = 4n – 23
= 4(9) – 23
= 36 – 23 = 13 = RHS
Hence verified.

(iii) – 5 + 3x = 16
Answer:
Given – 5 + 3x = 16
⇒ – 5 + 3x + 5 = 16 + 5 (Add 5 on both sides)
⇒ 3x = 21
⇒ \(\frac{3 x}{3}\) = \(\frac{21}{3}\) (Divide by 3 on both sides)
⇒ x = 7

Check:
Substitute x = 7 in the given equation
LHS = – 5 + 3x
= – 5 + 3(7)
= – 5 + 21 = 16 = RHS
Hence verified.

(iv) 2(y – 1) =8
Answer:
Given 2(y – 1) = 8
⇒ 2y – 2 = 8 (Distributive property)
⇒ 2y – 2 + 2 = 8 + 2 (Add 2 on both sides)
⇒ 2y = 10
⇒ \(\frac{2 y}{2}\) = \(\frac{10}{2}\) (Divide by 2 on both sides)
⇒ y = 5

Check: Substitute y = 5 in the given equation
LHS = 2(y – 1)
= 2(5 – 1)
= 2 × 4 = 8 = RHS
Hence verified.

Question 2.
Solve the following simple equations and check the results.
(i) 3x = 18
Answer:
Given 3x = 18
⇒ \(\frac{3 x}{3}\) = \(\frac{18}{3}\) (Divide by 3 on both sides)
⇒ x = 6

Check: Substitute x = 6 in 3x = 18
LHS ⇒ 3x = 3(6) = 18 = RHS
Hence verified.

(ii) \(\frac{b}{7}\) = – 2
Sol.
Given \(\frac{b}{7}\) = – 2
⇒ \(\frac{b}{7}\) × 7 = – 2 × 7 (Multiply by 7 on both sides)
⇒ b = – 14

Check: Substitute b = – 14 in
\(\frac{b}{7}\) = – 2
LHS = \(\frac{b}{7}\) = \(\frac{- 14}{7}\) = – 2 = RHS
Hence verified

(iii) – 2x = – 10
Answer:
Given – 2x = – 10
⇒ \(\frac{-2 x}{-2}\) = \(\frac{-10}{-2}\) (Divide by – 2 on both sides)
⇒ x = 5

Check:
Substitute x = 5 in
– 2x = – 10
LHS = – 2x
= – 2(5) = – 10 = RHS
Hence verified.

(iv) 10 + 6a = 40
Answer:
Given 10 + 6a = 40
⇒ 10 + 6a – 10 = 40 – 10 (Subtract 10 on both sides)
⇒ 6a = 30
⇒ \(\frac{6a}{6}\) = \(\frac{30}{6}\) (Divide by 6 on both sides)
⇒ a = 5

Check: Substitute a = 5 in 10 + 6a = 40
LHS = 10 + 6a
= 10 + 6(5)
= 10 + 30
= 40 = RHS
Hence verified.

(v) – 7m = 21
Answer:
Given – 7m = 21
⇒ \(\frac{-7 m}{-7}\) = \(\frac{21}{-7}\) (Divide by – 7 on both sides)
⇒ m = – 3

Check:
Substitute m = in – 7m = 21
LHS = – 7m
= – 7 (- 3)
= 21 = RHS
Hence verified.

(iv) 4p + 7 = – 21
Answer:
Given 4p + 7 = – 21
⇒ 4p + 7 – 7 = – 21 – 7 (Subtract 7 on both sides)
⇒ 4p = – 28
⇒ \(\frac{4p}{4}\) = \(\frac{-28}{4}\) (Divide by 4 on both sides)
⇒ p = – 7

Check:
Substitute p = – 7 in 4p + 7 = – 21
LHS = 4p + 7
= 4(- 7) + 7
= – 28 + 7
= – 21 = RHS
Hence verified.

(vii) 3x – \(\frac{1}{3}\) = 5
Answer:

Check: Substitute x = \(\frac{16}{9}\) in 3x – \(\frac{1}{3}\) = 5
LHS = 3x – \(\frac{1}{3}\)
= 3\(\left(\frac{16}{9}\right)\) – \(\frac{1}{3}\)
= \(\frac{16}{3}-\frac{1}{3}\)
= \(\frac{16-1}{3}\) = \(\frac{15}{3}\)
= 5 = RHS
Hence verified.

(viii) 18 – 7n = – 3
Answer:
Given 18 – 7n = – 3
⇒ 18 – 7n – 18 = – 3 – 18 (Subtract 18 on both sides)
⇒ – 7n = – 21
⇒ \(\frac{-7 n}{-7 n}\) = \(\frac{-21}{-7}\) (Divide by – 7 on both sides)
⇒ n = 3

Check:
Substitute n = 3 in 18 – 7n = – 3
LHS = 18 – 7n
= 18 – 7(3)
= 18 – 21
= – 3 = RHS
Hence verified.

(ix) 3(k + 4) = 21
Sol.
Given 3(k + 4) = 21
⇒ 3k + 12 = 21 (Distributive property)
⇒ 3k + 12 – 12 = 21 – 12 (Subtract 12 on both sides)
⇒ 3k = 9
⇒ \(\frac{3 \mathrm{k}}{3}\) = \(\frac{9}{3}\) (Divide by 3 on both sides)
⇒ k = 3

Check:
Substitute k = 3 in 3(k + 4) = 21
LHS = 3(k + 4)
= 3(3 + 4)
= 3 × 7 = 21= RHS
Hence verified.

(x) 9 (a + 1) + 2 = 11
Answer:
Given 9(a + 1) + 2 = 11
⇒ 9(a + 1) + 2 – 2 = 11 – 2 (Subtract 2 on both sides)
⇒ 9(a + 1) = 9
⇒ \(\frac{9(a+1)}{9}\) = \(\frac{9}{9}\) (Divide by 9 on both sides)
⇒ a + 1 = 1
⇒ a + 1 – 1 = 1 – 1 (Subtract 1 on both sides)
⇒ a = 0

Check:
Substitute a = 0 in 9(a + 1) + 2 = 11
LHS = 9(a + 1) + 2
= 9(0 + 1) + 2
= 9(1) + 2
= 9 + 2 = 11 = RHS
Hence verified.

SCERT AP 7th Class Maths Solutions Pdf Chapter 3 Simple Equations Ex 3.1 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 3rd Lesson Simple Equations Exercise 3.1

Question 1.
Write the equations of the following mathematical statements.
(i) A number x decreased by 5 is 14.
Answer:
Given number = x
Number decreased by 5 = x – 5
∴ x – 15 = 14.

(ii) Eight times of y plus 3 is -5.
Answer:
Given number = y
Eight times of y = 8 ∙ y
Eight times of y plus 3 = 8y + 3
∴ 8y + 3 = – 5

(iii) If you add one fourth of z to 3 you get 7.
Answer:
Given number = z
One fourth of z = \(\frac{1}{4}\) ∙ z = \(\frac{z}{4}\)
One fourth of z is added to 3 = \(\frac{z}{4}\) + 3
Result is 7
∴ \(\frac{z}{4}\) + 3 = 7.

(iv) If you take away 5 from 3 times of m, you get 11.
Answer:
Given number = m
3 times of m = 3m
Take away 5 from 3 times of m = 3m – 5
Result is 11
∴ 3m – 5 = 11

(v) Sum of angles 2x, (x – 30) is a right angle.
Answer:
Given angles 2x, (x – 30)
Sum of angles 2x, (x – 30)
= 2x + x – 30 = 3x – 30
Sum of angles is right angle (90°).
∴ 3x – 30 = 90°

(vi) The perimeter of a square of side ‘a’ is 14 m.
Answer:
Given side of a square = a
Perimeter = 4 ∙ side = 4 ∙ a
Given perimeter = 14 m
∴ 4a = 14 m

Question 2.
Write the following equations in statement form.
(i) m – 5 = 12
Answer:
A number m is decreased by 5 is 12.

(ii) \(\frac{\mathbf{a}}{\mathbf{3}}\) = 4
Answer:
One third of a is 4.

(iii) 4x + 7 = 15
Answer:
Sum of 4 times of x and 7 is 15.
(or)
7 is added to 4 times of x is 15.

(iv) 2 – 3y = 11
Answer:
2 is decreased by 3 times of y is 11.
(or)
3 times of y is subtracted from 2 is 11.

Question 3.
Check whether the value given in the brackets is a solution to the given equation or not.
(i) 5n – 7 = 23 (n = 6)
Answer:
Given 5n – 7 = 23
When n = 6
L.H.S = 5n – 7
= 5(6) – 7
= 30 – 7
= 23
R.H.S =23
Here, L.H.S = R.H.S
So, n = 6 is a solution of the given equation.

(ii) \(\frac{p}{4}\) – 7 = 5 (p = 8)
Answer:
Given \(\frac{p}{4}\) – 7 = 5; When p = 8
LHS = \(\frac{p}{4}\) – 7

RHS = 5
– 5 ≠ 5
Here, LHS ≠ RHS
So, p = 8 is not a solution of the given equation.

(iii) 5 – 2x = 19 -7
Answer:
Given 5 – 2x = 19
When x = – 7
LHS = 5 – 2x
= 5 – 2(- 7) = 5 + 14 = 19
RHS = 19
19 = 19
Here, LHS = RHS
So, x = – 7 is a solution of the given equation.

(iv) 2 + 3(m – 1) = 5 (m = -2)
Answer:
Given 2 + 3(m – 1) = 5
When m = – 2
LHS = 2 + 3(m – 1)
= 2 + 3(- 2 – 1)
= 2 + 3(- 3) = 2 – 9 = – 7
RHS = 5
– 7 ≠ 5
Here, LHS ≠ RHS
So, m = – 2 is not a solution of given equation.

Question 4.
Solve the following equations using trial and error method,
(i) 3x – 7 = 5
Answer:

For x = 4, LHS = RHS
So, x = 4 is the solution of given equation.

(ii) 5 – y = – 1
Answer:

For y = 6, LHS = RHS.
So, y = 6 is the solution of given equation.