AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.3

AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.3

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.3 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 10th Lesson Practical Geometry Ex 10.3

AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.3

Question 1.
Construct the following angles with the help of a protractor.
i) ∠ABC = 65°
ii) ∠PQR = 136°
iii) ∠Y = 45°
iv) ∠O = 172°
Solution:
i) ∠ABC = 65°
AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.3 1
Steps of construction :

  1. Draw a ray \(\overrightarrow{\mathrm{BC}}\) of any length.
  2. Place the centre point of the protractor at B and the line be aligned with \(\overrightarrow{\mathrm{BC}}\).
  3. Mark a point A at 65°.
  4. Join BA. ∠ABC = 65° is formed.
    Hence the required angle ∠ABC = 65° is constructed.

ii) ∠PQR = 136°
Steps of construction :

  1. Draw a ray \(\overrightarrow{\mathrm{QR}}\) of any length,
  2. Place the centre point of the protractor at Q and the line be aligned with \(\overrightarrow{\mathrm{QR}}\).
  3. Mark a point P at 136°. ‘
  4. Join PQ. ∠PQR = 136° is formed.
    Hence the required angle ∠PQR = 136° is constructed.

AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.3 2

AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.3

iii) ∠Y = 45°
Steps of construction:

  1. Draw a ray \(\overrightarrow{\mathrm{YZ}}\) of any length.
  2. Place the centre point of the protractor at Y
    and the line be aligned with \(\overrightarrow{\mathrm{YZ}}\).
  3. Mark a point X at 45°.
  4. Join YX. ∠XYZ = 45° is formed.
    Hence the required angle ∠XYZ = 45° is constructed.

AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.3 3

iv) ∠O = 172°
AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.3 4
Steps of construction:

  1. Draw a ray \(\overrightarrow{\mathrm{OT}}\) °f any length.
  2. Place the centre point of the protractor at O and the line be aligned with \(\overrightarrow{\mathrm{OT}}\)
  3. Mark a point D at 172°.
  4. Join OD. ∠DOT = 172° is formed.
    Hence the required angle ∠DOT = 172° is constructed.

Question 2.
Copy the following angles in your note book and find their bisectors:
AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.3 5
Solution:
AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.3 6

AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.2

AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.2

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.2 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 10th Lesson Practical Geometry Ex 10.2

AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.2

Question 1.
Draw a line segment PQ = 5.8 cm and construct its perpendicular bisector using ruler and compasses.
Solution:
AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.2 1
Steps of construction :

  1. Draw a line segment \(\overline{\mathrm{PQ}}\) = 5.8 cm.
  2. Set the compasses as radius with more than half of the length of \(\overline{\mathrm{PQ}}\).
  3. With P as centre, draw arcs below and above the line segment.
  4. With the same radius and Q as centre draw two arcs above and below the line segment to cut the previous arcs. Name the intersecting points of arcs as X and Y.
  5. Join the points X and Y. So, the line I is the perpendicular bisector of PQ.
    Hence l is required perpendicular bisector of PQ which meets at A.

Question 2.
Ravi made a line segment of length 8.6 cm. He constructed a bisector of AB on C. Find the length of AC and BC.
AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.2 2
Solution:
AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.2 3
As it is a bisector, it divides the line segment into two equal parts.
Each equal part is half of AB (8.6 cm) = \(\frac{\mathrm{AB}}{2}=\frac{8.6}{2}\) = 4.3 cm
∴ AC = BC = 4.3 cm

AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.2

Question 3.
Using ruler and compasses, draw AB = 6.4 cm. Locate its mid point by geometric construction.
Solution:
AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.2 4

  1. Draw a line segment \(\overline{\mathrm{PQ}}\) = 6.4 cm
  2. Set the compasses as radius with more than half of the length of \(\overline{\mathrm{PQ}}\).
  3. With A as centre, draw arcs below and above the line segment.
  4. With the same radius and B as centre draw two arcs above and below the line segment to cut the previous arcs. Name the intersecting points of arcs as X and Y.
  5. Join the points X and Y. So, the line l is the perpendicular bisector of AB.
    Hence l is the required perpendicular bisector of AB which meets at M.

AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1

AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 10th Lesson Practical Geometry Ex 10.1

AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1

Question 1.
Construct a line segment of length 6.9 cms using ruler and compass.
AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1 1
Solution:
Steps of construction:

  1. Draw a line l. Mark a point A on the line l.
  2. Place the metal pointer of the compass on the zero mark of the ruler. Open the compass, so that pencil point touches the 6.9 cm mark on the ruler.
  3. Place the pointer at A on the line l and draw an arc to cut the line. Mark the point where the arc cuts the line as B
  4. On the line l, we got the line segment AB of required length.
    AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1 2

Question 2.
Construct a line segment of length 4.3 cms using ruler.
Solution:
AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1 3
Steps of construction:

  1. Place the ruler on paper and hold it firmly.
  2. Mark a point with a sharp edged pencil against 0 cm mark on the ruler. Name the point as P.
  3. Mark another point against 3 small divisions just after the 4 cm mark. Name this point as Q.
  4. Join points P and Q along the edge of the ruler.
    Therefore, PQ is the required line segment of length 4.3 cm.

AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1

Question 3.
Construct a circle with centre M and radius 4 cm.
Solution:
AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1 4
Steps of construction:

  1. Mark a point M on the paper.
  2. By using compasses take 4 cm as the radius on with the scale.
  3. Place the metal point on M and draw a circle from M.
    Hence required circle is constructed with center M and radius 4 cm.

Question 4.
Draw any circle and mark three points A, B and C such that
(i) A is on the circle
(ii) B is in the interior of the circle
(iii) C is in
Solution:
AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1 5

AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry InText Questions

AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry InText Questions

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry InText Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 10th Lesson Practical Geometry InText Questions

AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry InText Questions

Check Your Progress (Page No. 139)

Question 1.
Measure the lengths of \(\overline{\mathrm{AP}}\) and \(\overline{\mathrm{PB}}\) in both the constructions. What do you observe?
Solution:
In the construction we observe that \(\overline{\mathrm{AP}}=\overline{\mathrm{PB}}\)
That is P can divide AB into two equal parts.
AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry InText Questions 1

Let’s Think (Page No. 140)

Question 1.
In the construction of perpendicular bisector in step 2, what would happen if we take the length of radius to be smaller than half the length of \(\overline{\mathrm{AB}}\) ?
Solution:
If we take the length of radius to be smaller than half the length of \(\overline{\mathrm{AB}}\)
Arcs cannot intersect each other. So, we can’t construct the perpendicular bisector to the given line \(\overline{\mathrm{AB}}\)

AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry InText Questions

(Page No. 144)

Question 1.
Construct angles of 180°, 240°, 300°.
Solution:
AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry InText Questions 2
i) ∠AOB = 180°
Steps of construction:

  • Draw any ray \(\overline{\mathrm{OA}}\) of any length.
  • Place the pointer of the compasses at ‘O’ with ‘O’ as centre any convinient radius draw an arc cutting OA at M.
  • With M as centre and without altering radius draw an arc which cuts the previous arc at P.
  • Draw an arc with the same radius from P which cuts the previous arc at Q and from Q draw another arc which meets at R.
  • Join OR. ( \(\overline{\mathrm{OB}}\) ). Then ∠AOB is the required angle.
    Hence the required angle ∠AOB = 180° is constructed.

ii) ∠PQR = 240°
Steps of construction:

  • Draw any ray \(\overline{\mathrm{QP}}\) of any length.
  • Place the pointer of the compasses at O with ‘O’ as centre any convinient radius draw an arc cutting QP at A.
  • With A as centre and without altering radius draw an arc which cuts the previous arc at B.
  • Draw an arc with the same radius from B which cuts the previous arc at C and from C draw another arc which meets at D.
  • Draw an arc with the same radius from D draw an arc which cuts the previous arc at E.
  • Join \(\overrightarrow{\mathrm{QE}}(\overrightarrow{\mathrm{QR}})\). Then ∠PQR is the required angle.
    Hence the required angle ∠PQR = 240° is constructed.

AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry InText Questions 3

AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry InText Questions

iii) ∠XYZ – 300°
Steps of construction :

  • Draw any ray \(\overline{\mathrm{YZ}}\) of any length.
  • Place the pointer of the compasses at Y with Y as centre any convinient radius draw an arc cutting YZ at P.
  • With P as centre and without altering radius draw arc which quts the previous arc at Q and from Q draw another arc which cuts the previous arc at R.
  • Draw an arc with the same radius from R which cuts the previous arc at S and from S draw another arc which meets at T.
  • Draw an arc with the same radius from T which cuts the previous arc at U.
  • Join \(\overrightarrow{\mathrm{YU}}(\overrightarrow{\mathrm{YX}})\). Then ∠XYZ is required angle.
    Hence the required angle ∠XYZ = 300° is constructed.
    AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry InText Questions 4

(Page No. 145)

Question 1.
Construct an angle of 45° by using compasses.
Solution:
45°
Steps of construction :
i) Draw any ray \(\overline{\mathrm{OP}}\) of any length.
ii) Draw arcs with the same radius from A to B and from B to C which cuts the previous arc at B and C respectively.
iii) Draw arcs from B and from C with same radius which can intersect at X.
iv) Join \(\overrightarrow{\mathrm{OX}}(\overrightarrow{\mathrm{OD}})\) i.e., ∠POD = 90°.
v) Draw the bisector to ZPOD which is \(\overrightarrow{\mathrm{OQ}}\)
vi) Now, ∠POQ = ∠QOD = \(\frac{\angle \mathrm{POD}}{2}=\frac{90}{2}\) = 45°
∴ ∠POQ = 45° (Q.E.D)
AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry InText Questions 5

AP Board 6th Class Maths Solutions Chapter 9 2D-3D Shapes InText Questions

AP Board 6th Class Maths Solutions Chapter 9 2D-3D Shapes InText Questions

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 9 2D-3D Shapes InText Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 9th Lesson 2D-3D Shapes InText Questions

AP Board 6th Class Maths Solutions Chapter 9 2D-3D Shapes InText Questions

(Page No. 125)

Question 1.
Draw six different types of rough sketches of polygons in your notebook. In which case, it is not possible to form a polygon ?
Hence, what is the least number of sides needed to form a polygon ? Obviously three.
Solution:
AP Board 6th Class Maths Solutions Chapter 9 2D-3D Shapes InText Questions 1
From the above figure we conclude that with one side and two sides cannot form a polygon. So, atleast three sides needed to form a polygon.

AP Board 6th Class Maths Solutions Chapter 9 2D-3D Shapes InText Questions

Check Your Progress (Page No. 127)

Look at the adjacent figure.
AP Board 6th Class Maths Solutions Chapter 9 2D-3D Shapes InText Questions 2

Question 1.
Which points are marked in the interior of A GHI ?
Solution:
The points marked in the interior of AGHI are A, B and 0.

Question 2.
Which points sure marked on the triangle?
Solution:
The points marked on the triangle are G, P, H, I and Y.

Question 3.
Which points are’marked in the exterior of A GHI ?
Solution:
The points marked in the exterior of AGHI are M, R, S, X and Z.

(Page No. 130)

Question 1.
Take a rectangular sheet (like a post-card). Fold it along its length so that one half fits exactly over the other half. Is this fold a line of symmetry ? Why ?
Solution:
Yes. Because the two parts of the rectangular sheet (post-card) coincide exactly with each other.
So, the folded line is the line of symmetry.

AP Board 6th Class Maths Solutions Chapter 9 2D-3D Shapes InText Questions

Question 2.
Open it up now and again fold along its width in the same way. Is this second fold also a line of symmetry ? Why ?
Solution:
Yes. Because the two parts of the rectangular sheet (post-card) coincide exactly with each other.
So, the second folded line also a line of symmetry.

Question 3.
Do you find that these two lines are the lines of symmetry ?
Solution:
Yes. These two lines are the lines of symmetry. One is line of symmetry along length and the other is line of symmetry along width.

Project (Page No. 130)

Question 1.
Collect symmetrical figures from your surroundings and prepare a scrap book.
Solution:
AP Board 6th Class Maths Solutions Chapter 9 2D-3D Shapes InText Questions 3

AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts InText Questions

AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts InText Questions

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts InText Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 8th Lesson Basic Geometric Concepts InText Questions

AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts InText Questions

(Page No. 111)

i) How many rays are there ?
Solution:
Four
AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts InText Questions 1

ii) How many coins are close to player Q ?
Solution:
B, C and D coins are close to player Q.

iii) While striking with striker there is a possibility of a coin touches with any other. Draw all such possibilities in the given picture by means of the line segments.
Solution:
\(\overline{\mathrm{CB}}\) and \(\overline{\mathrm{DE}}\) .

iv) How many such line segments can be drawn in the picture ?
Solution:
\(\overline{\mathrm{AB}}, \overline{\mathrm{BC}}, \overline{\mathrm{ED}}, \overline{\mathrm{AC}}, \overline{\mathrm{AE}}, \overline{\mathrm{CD}}, \overline{\mathrm{CE}}, \overline{\mathrm{AD}}, \overline{\mathrm{BD}} \text { and } \overline{\mathrm{BE}}\)

AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts InText Questions

Let’s Do (Page No. 112)

Question 1.
Observe the table and their notations and fill the gaps.
AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts InText Questions 2
Solution:
AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts InText Questions 3

Lets Explore (Page No. 114)

Question 1.
Measure the lengths of all line segments in the given figures by using divider and scale. Then compare the sides of the given figures.
AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts InText Questions 4
Solution:
i) In ΔABC; \(\overline{\mathrm{AB}}\) = 2.2 cm; \(\overline{\mathrm{BC}}\) = 2 cm and \(\overline{\mathrm{AC}}\) = 2.2 cm ‘
2.2 cm = 2.2 cm > 2 cm i.e., \(\overline{\mathrm{AB}}=\overline{\mathrm{AC}}>\overline{\mathrm{BC}}\)
Two sides are equal and one side is different in length.

ii) In PQRS rectangle \(\overline{\mathrm{PS}}=\overline{\mathrm{QR}}\) = 2.7 cm; \(\overline{\mathrm{P Q}}=\overline{\mathrm{RS}}\) = 1.8 cm and \(\overline{\mathrm{PR}}=\overline{\mathrm{QS}}\) = 3.2 cm.
Opposite sides are equal and diagonals are equal in length,

iii) In KLMN square \(\overline{\mathrm{KL}}=\overline{\mathrm{LM}}=\overline{\mathrm{MN}}=\overline{\mathrm{KN}}\) = 1.8 cm
All sides are equal in length.

AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts InText Questions

Let’s Do (Page No. 115)

Question 1.
(i) Find the parallel lines in the below figure. Name, write and read them.
AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts InText Questions 5
Solution:
a) l, m are parallel lines.
We denote this by writing l || m and can be read as l is parallel to m.
b) n, o are parallel lines.
We denote this by writing n||o and can be read as n is parallel to o.
c) p, q are parallel lines.
We denote this by writing p||q and can be read as p is parallel to q.

ii) Find the intersecting lines in the above figure. Name, write and read them.
Solution:
Intersecting lines are (l, q); (m,q); (m, r); (n,q); (p, r); (o, r); (o, q); (q, r);

iii) Find the concurrent lines in the above figure. Name, write and read them.
Solution:
Three or more lines passing through the same point are called concurrent lines. Concurrent lines are (l, o, p) & (m, n, p).

iv) Find the perpendicular lines in the above figure. Name, write and read them.
Solution:
a) p, o are perpendicular lines.
We denote this by writing p ⊥ o and can be read as p is perpendicular to o.
b) p, n are perpendicular lines.
We denote this by writing p ⊥ n and can be read as p is perpendicular to n.
c) n, q are perpendicular lines.
We denote this by writing n ⊥ q and can be read as n is perpendicular to q.
d) q, o are perpendicular lines.
We denote this by writing q⊥ o and can be read as q is perpendicular to o.

AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts InText Questions

(Page No. 119)

Question 1.
Measure the angles at the vertices.
AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts InText Questions 6
Solution:
AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts InText Questions 7
In triangle ABC,
m∠BAC = 60°
m∠ABC = 60°
m∠ACB = 60°

AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts InText Questions 8
In triangle XYZ,
m∠YXZ = 40°
m∠XYZ = 70°
m∠XZY = 70°

AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts InText Questions

AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts InText Questions 9
In triangle PQR,
m∠QPR = 35°
m∠PQR = 38°
m∠PRQ = 107°

AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts Ex 8.1

AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts Ex 8.1

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts Ex 8.1 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 8th Lesson Basic Geometric Concepts Ex 8.1

AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts Ex 8.1

Question 1.
In the given figure there are some points marked. Name them.
AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts Ex 8.1 1
Solution:
AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts Ex 8.1 2

Question 2.
Join the points given below. Name the line segments so formed in the figure.
AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts Ex 8.1 3
Solution:
AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts Ex 8.1 4

\(\overline{\mathrm{AD}}\) read as segment AD
\(\overline{\mathrm{BC}}\) read as segment BC
AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts Ex 8.1 5

\(\overline{\mathrm{AB}}\) read as segment AB

\(\overline{\mathrm{DC}}\) read as segment DC

AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts Ex 8.1 6

\(\overline{\mathrm{AB}}\) read as segment AB
\(\overline{\mathrm{BC}}\) read as segment BC
\(\overline{\mathrm{CD}}\) read as segment CD
\(\overline{\mathrm{DE}}\) read as segment DE
\(\overline{\mathrm{EF}}\) read as segment EF
\(\overline{\mathrm{FA}}\) read as segment FA

AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts Ex 8.1

Question 3.
Identify the following from the given figure.
(i) Any six points.
(ii) Any six line segments. (Starts with G)
(iii) Any six rays. (With initial point I)
(iv) Any three lines.
AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts Ex 8.1 7
Solution:
Points are : A, B, C, D, E, F
G, H, I, and J
ii) Line segments are : \(\overline{\mathrm{GF}}, \overline{\mathrm{GE}}, \overline{\mathrm{GD}}, \overline{\mathrm{GH}}, \overline{\mathrm{GC}} \text { and } \overline{\mathrm{GA}}\)
iii) Rays are: \(\overrightarrow{\mathrm{IA}}, \overrightarrow{\mathrm{IC}}, \overrightarrow{\mathrm{IB}}, \overrightarrow{\mathrm{ID}}, \overrightarrow{\mathrm{JA}}, \overrightarrow{\mathrm{JC}}, \overrightarrow{\mathrm{JE}}\)
iv)Lines are : AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts Ex 8.1 8

Question 4.
Write ‘True’ or ‘False’.
(i) A line has two end points.
(ii) Ray is a part of line.
(iii) A line segment has two end points.
(iv) We can draw many lines through two points.
Solution:
i) A line has two end points. (False)
ii) Ray is a part of line.(True)
iii) A line segment has two end points.(True)
iv) We can draw many lines through two points.(False)

AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts Ex 8.1

Question 5.
Draw and Name :
(T) Line containing point K.
QS) Draw a circle and a line such that the line intersects the circle.
a) at no points b) at one point c) at two points
(iii) Can we draw a line and a circle with 3 intersecting points ?
Solution:
AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts Ex 8.1 9
iii) No, we can’t draw a line and a circle with 3 intersecting points. As a line can intersect the circle at two points only.

Question 6.
List out all capital letters in English alphabet, that you can write with 3 line segments by using all.
Solution:
AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts Ex 8.1 10

AP Board 6th Class Maths Solutions Chapter 12 Data Handling InText Questions

AP Board 6th Class Maths Solutions Chapter 12 Data Handling InText Questions

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 12 Data Handling InText Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 11th Lesson Data Handling InText Questions

AP Board 6th Class Maths Solutions Chapter 12 Data Handling InText Questions

Let’s Do (Page No. 159)

Question 1.
Take a die. Throw it and record the number. Repeat the activity 40 times and record the numbers. Represent the data in a frequency distribution table using tally marks.
Solution:
AP Board 6th Class Maths Solutions Chapter 12 Data Handling InText Questions 1
AP Board 6th Class Maths Solutions Chapter 12 Data Handling InText Questions 2

AP Board 6th Class Maths Solutions Chapter 12 Data Handling InText Questions

(Page No. 159)

Question 1.
In what way is the bar graph better than the pictograph ?
Solution:
Bar graphs are better indicative as they show exact numerical value. Also, to indicate negative values and positive values, it looks easier in a bar graph.

AP Board 6th Class Maths Solutions Chapter 11 Perimeter and Area InText Questions

AP Board 6th Class Maths Solutions Chapter 11 Perimeter and Area InText Questions

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 11 Perimeter and Area InText Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 11th Lesson Perimeter and Area InText Questions

AP Board 6th Class Maths Solutions Chapter 11 Perimeter and Area InText Questions

Check Your Progress (Pg.No. 147)

Question 1.
Find the perimeters of the given figures.
AP Board 6th Class Maths Solutions Chapter 11 Perimeter and Area InText Questions 1
In the above figures (I) and (ii), find the perimeter of ΔKLM, ΔKMN and □ KLMN.
(a) Compare the perimeters of ΔKLM and □ KLMN
Compare the perimeters of ΔKMN and □ KLMN.
What can you say? –
Solution:
i) In ΔKLM, given KL = 2cm; LM = 2.6 cm; MK = 3.8 cm
Perimeter of ΔKLM = KL + LM + MK
= 2 + 2.6 + 3.8
= 8.4 cm

AP Board 6th Class Maths Solutions Chapter 11 Perimeter and Area InText Questions

In ΔKMN, given KM = 3.8 cm; MN = 2cm; KN = 2.6 cm
Perimeter of ΔKMN = KM + MN + KN
= 3.8 + 2 + 2.6 = 8.4 cm

ii) In □ KLMN, given KL = 2cm, LM = 2.6cm, MN = 2cm and KN = 2.6 cm
Perimeter of □ KLMN = KL + LM + MN + KN = 2 + 2.6 + 2 + 2.6 = 9.2 cm

a) By comparing the perimeter of ΔKLM and □ KLMN
8.4 cm < 9.2 cm
∴ Perimeter of ΔKLM < Perimeter of □ KLMN
By comparing the perimeter of ΔKMN and □ KLMN
8.4 cm < 9.2 cm
∴ Perimeter of ΔKMN < Perimeter of □ KLMN.
By comparing the perimeters of triangle with quadrilateral ‘
Perimeter of the triangle is always less than the perimeter of quadrilateral.

Let’s Explore (Pg.No. 150)

Question 1.
If the radius of a circle is doubled, then, what is the change in its circumference?
Solution:
If the radius is r, then its circumference of the circle is 2πr.
If radius is doubled means r = 2r (new)
Write r = 2r in 2πr.
Then, the circumference = 2π x 2r
= 4πr = 2.27πr
Circumference is doubled.
So, if the radius of a circle is doubled, then the circumference is also doubled.

Question 2.
If the radius is halved, then, what is the change in its circumference?
Solution:
If the radius is r, then its circumference of the circle is 2πr.
If radius is halved means r = \(\frac{\mathbf{r}}{2}\) (new)
Write r = \(\frac{\mathbf{r}}{2}\) in 2πr.
Then, the circumference = 2π x \(\frac{r}{2}=\frac{1}{2}\). 2πr
Circumference is halved.
So if the radius of a circle is halved, then the circumference is also halved.

AP Board 6th Class Maths Solutions Chapter 11 Perimeter and Area InText Questions

Check Your Progress (Page No. 151)

Question 1.
Find the area of a square with side 16 cm.
Solution:
Given side of the square (s) = 16 cm
Area of the square A = s x s = 16 x 16 = 256 sq.cm

Question 2.
Length and breadth of a rectangle are 16 cm and 12 cm respectively. Find its area.
Solution:
Given length of the rectangle l = 16 cm
breadth of the rectangle b = 12 cm
Area of the rectangle A = l x b = 16 x 12 = 192 sq.cm

Let’s Think (Page No. 151)

Question 1.
Find the perimeter and area of a square of side 4 cm. Are these same? Give some examples to support your answer.
Solution:
AP Board 6th Class Maths Solutions Chapter 11 Perimeter and Area InText Questions 2

AP Board 6th Class Maths Solutions Chapter 11 Perimeter and Area InText Questions

From the above table,
If side is 4 cm, then only its perimeter and areas are equal.
In other cases perimeter and areas are not same.

AP Board 6th Class Maths Solutions Chapter 2 Whole Numbers InText Questions

AP Board 6th Class Maths Solutions Chapter 2 Whole Numbers InText Questions

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 2 Whole Numbers InText Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 2nd Lesson Whole Numbers InText Questions

AP Board 6th Class Maths Solutions Chapter 2 Whole Numbers InText Questions

Question 1.
Fill the following table with the successor and predecessor of the numbers provided. (Page No. 15)

S.No. Natural number Predecessor Successor
1. 135
2. 237
3. 999

Solution:

S.No. Natural number Predecessor

Successor

1. 135 134 136
2. 237 236 238
3. 999 998 1000

Discuss

Question 1.
Which number has no successor ? (Page No. 15)
Answer:
Each and every number has a successor.

Question 2.
Which number has no predecessor? (Page No. 15)
Answer:
Zero (0) has no predecessor in the set of whole numbers.

Check Your Progress (Page No. 16)

Question 1.
Which is the smallest whole number?
Answer:
Zero(0) is the smallest whole number.

AP Board 6th Class Maths Solutions Chapter 2 Whole Numbers InText Questions

Let’s Think (Page No. 16)

Question 1.
Are all natural numbers whole numbers? .
Solution:
Yes. All the natural numbers are whole numbers.

Question 2.
Are all whole numbers natural numbers?
Solution:
No. All the whole numbers are not natural numbers.

Let’s Do (Page No. 17)

Show these on number line:

i) 5 + 3
Solution:
AP Board 6th Class Maths Solutions Chapter 2 Whole Numbers InText Questions 1
Draw the number line starting with ‘O’.
Start from 5, make 3 jumps to the right of 5 on the number line. We reach 8.
So, 5 + 3 = 8

ii) 5 – 3
Solution:
AP Board 6th Class Maths Solutions Chapter 2 Whole Numbers InText Questions 2
Draw the number line starting with zero (0).
Start from 5, make 3 jumps to the left of 5 on the number line, we reach 2.
So, 5 – 3 = 2

AP Board 6th Class Maths Solutions Chapter 2 Whole Numbers InText Questions

iii) 3 + 5
Solution:
AP Board 6th Class Maths Solutions Chapter 2 Whole Numbers InText Questions 3
Draw the number line starting with zero (0).
Start from 3, we make 5 jumps to the right of 3 on the number line. We reach 8. So, 3 + 5 = 8

iv) 10 + 1
Solution:
AP Board 6th Class Maths Solutions Chapter 2 Whole Numbers InText Questions 4
Draw the number line which starts with zero (0).
Start from 10, make 1 jump to the right of 10 on the number line. We reach’ll. So, 10 + 1 = 11

v) 8 – 5
Solution:
AP Board 6th Class Maths Solutions Chapter 2 Whole Numbers InText Questions 5
Draw the number line which starts with zero (0).
Start from 8, we make 5 jumps to the left of 8 on the number line. We reach 3. So, 8-5 = 3

Let’s Explore (Page No. 17)

Find the following by using number line:

Question 1.
Which number should be deducted from 8 to get 5?
Solution:
AP Board 6th Class Maths Solutions Chapter 2 Whole Numbers InText Questions 6
Draw the number line, which starts with zero (0).
To get 5 from 8. We have to start from 8. We make 3 jumps to the left of 8 on the number line, we reach 5. As we are moving on left side we have minus sign.
So, 8 – 3 = 5
Therefore to get 5 we deduct 3 from 8.

AP Board 6th Class Maths Solutions Chapter 2 Whole Numbers InText Questions

Question 2.
Which number should be deducted from 6 to get 1?
Solution:
AP Board 6th Class Maths Solutions Chapter 2 Whole Numbers InText Questions 7
Draw the number line starting with zero (0).
To get 1 from 6 we have to start from 6.
We make 5 jumps to the left of 6 on the number line. We reach 1. As we have moved to left side, we have 6-5 = 1 Threfore to get 1 we deduct 5 from 6.

Question 3.
Which number should be added to 6 to get 8?
Solution:
AP Board 6th Class Maths Solutions Chapter 2 Whole Numbers InText Questions 8
Draw the number line starts with zero (0).
To get 8 from 6, we have to start from 6.
We make 2 jumps to the right of 6 on the number line to reach 8; So, 6 + 2 = 8 ‘
Therefore to get 8 we add 2 to 6.

Question 4.
How many 6 are needed to get 30?
Solution:
AP Board 6th Class Maths Solutions Chapter 2 Whole Numbers InText Questions 9
Draw the number line starts with zero (0).
Start from 0 and make 6 jumps to the right of the zero as the number line. Now, treat 6 jumps as one step. So, to reach 30, we make 5 steps.
So, 5 × 6 = 30

Question
Raju and Gayatri together made a number line and played a game on it.
AP Board 6th Class Maths Solutions Chapter 2 Whole Numbers InText Questions 10
Raju asked “Gayatri, where will you reach if you jump thrice, taking leaps of 3, 8 and 5”? Gayatri said the first leap will take me to 3 and then from there I will reach 11 in the second step and another five steps from there to 16′.
Draw Gayatri’s steps and verify her answers.
Play this game using addition and subtraction on this number line with your friend.
Solution:
AP Board 6th Class Maths Solutions Chapter 2 Whole Numbers InText Questions 11

AP Board 6th Class Maths Solutions Chapter 2 Whole Numbers InText Questions

Lets Think (Page No. 19)

Question 1.
Are the whole numbers closed under subtraction?
Solution:
8 – 5 = 3, a whole number
5 – 8 = -3 is not a whole number
Therefore whole numbers are not closed under subtraction.

Question 2.
Are the whole numbers closed under division?
Solution:
6 ÷ 3 = 2, a whole number
3 ÷ 6 = \(\frac{3}{6}\), not a whole number 6
Therefore, whole numbers are not closed under division.

Check Your Progress (Page No. 19)

Question 1.
Find out 12 ÷ 3 and 4 ÷ 7.
Solution:
12 ÷ 3
12 is divided by 3 means, we subtract 3 from 12 repeatedly till, we get zero i.e., we subtract 3 from 12 again and again till, we get zero.
12 – 3 = 9 once
9-3 = 6 twice
6-3 = 3 thrice
3-3 = 0 four times
So, 12-3 = 4

42 ÷ 7
42 is divided by 7 means, we subtract 7 from 42 repeatedly, i.e., we subtract 7 from 42 again and again till, we get zero a number less than 7.
42 – 7 = 35 once
35 – 7 = 28 twice
28 – 7 = 21 thrice
21-7 = 14 four times
14-7 = 7 five times
7 – 7 = 0 six times
i.e., we can subtract 7 from 42 for 6 times successively. ,
So, 42 ÷ 7 = 6.

AP Board 6th Class Maths Solutions Chapter 2 Whole Numbers InText Questions

Question 2.
What would 6 4-0 and 9 4- 0 be equal to?
Solution:
6 ÷ 0
6 is divided by 0 means, we subtract 0 from 6 repeatedly i.e., we subtract 0 from 6 again and agian from 6.
6 – 0 = 6 once
6 – 0 = 6 twice
6 – 0 = 6 thrice and ……………….
If we subtract zero from 6 successively we can’t get zero at any end.
So, 6 ÷ 0 is not a number that we can reach.
So, division of a whole number by 0 does not give a known number as answer, so it is not defined.
Similarly 9 ÷ 0 is not defined.
So, we can’t say whether they are equal or not.

Let’s Explore (Page No. 20)

Take few examples and check whether

a) Subtraction is commutative over whole numbers or not?
Solution:
Let’s take two whole numbers 4 and 6
6 – 4 = 2 and (4 – 6) = – 2 is not a whole number.
So, 6 – 4 ≠ 4 – 6.
Therefore we say that subtraction is not commutative over the whole numbers.

b) Division is commutative over whole numbers or not ?
Solution:
Let’s take two whole numbers 8 and 2
8 ÷ 2 = 4 and (2 ÷ 8) = \(\frac{1}{4}\) is not a whole number.
So, 8 ÷ 2 ≠ 2 ÷ 8
Therefore, we say that division is not commutative over the whole numbers.

AP Board 6th Class Maths Solutions Chapter 2 Whole Numbers InText Questions

Check Your Progress (Page No. 22)

Verify the following:
i) (5 × 6) × 2 = 5 × (6 × 2)
Solution:
L.H.S : (5 × 6) × 2 = 30 × 2 = 60
R.H.S : 5 × (6 × 2) = 5 × 12 = 60
∴ L.H.S = R.H.S
So (5 × 6) × 2 = 5 × (6 × 2)
∴ Multiplication of whole numbers is associative.

ii) (3 × 7) × 5 = 3 × ( 7 × 5 )
Solution:
L.H.S : (3 × 7) × 5 = 21 × 5 = 105
R.H.S : 3 × (7 × 5) = 3 × 35 = 105
∴ (3 × 7) × 5 = 3 × (7 × 5)
∴ Multiplication of whole numbers is associative.

Check Your Progress (Page No. 22)

Use the commutative and associative properties to simplify the following:

i) 319 + 69 + 81
Solution:
319 + 69 + 81 = 319 +(81 + 69) (Commutative property)
= (319 + 81) + 69 (Associative property)
= 400 + 69 = 469

ii) 431 + 37 + 69 + 63
Solution:
431 + 37 + 69 + 63
= 431 + (37 + 69) + 63
= 431 + (69 + 37) + 63 (Commutative property)
= (431 + 69) + (37 + 63) (Associative property)
=(431 + 69) + 100
= 500+ 100 = 600

AP Board 6th Class Maths Solutions Chapter 2 Whole Numbers InText Questions

iii) 2 × (71 × 5)
Solution:
2 × (71 × 5) = 2 × (5 × 71) (Commutative property)
= (2 × 5) × 71 (Associative property)
= 10 × 71
= 710

iv) 50 × 17 × 2
Solution:
50 × (17 × 2) = 50 × (2 × 17) (Commutative property)
= (50 × 2) × 17 (Associative property)
= 100 × 17 = 1700

Let’s Think (Page No. 22)

a) Is(8 ÷ 2) ÷ 4 = 8 ÷ (2 ÷ 4)?
Is there any associative property for division ?
Check if this property holds for subtraction of whole numbers too.
Solution:
a) (8 ÷ 2) ÷ 4 = (8 ÷ 2) ÷ 4
= 4 ÷ 4 = 1
8 ÷ (2 ÷4) = 8 ÷ \(\left(\frac{2}{4}\right)\)
= 8 – \(\frac{4}{2}\) = 8 × 2 = 16
So, (8 ÷ 2) ÷ 4 ≠ 8 ÷ (2 ÷ 4) .
Therefore, associative property does not holds in division.

b) Is (8 – 2) – 4 = 8 – (2 – 4) ?
Solution:
(8 – 2) – 4 = 6 – 4 .
= 2
8 – ( 2 – 4 ) = 8 – ( – 2)
= 8 + 2 = 10
So, (8 – 2) – 4 ≠ 8 – (2 – 4)
Therefore, associative property does not holds in subtraction,
i. e., whole numbers are not associative w.r.t. subtraction.
They are not equal.
So whole numbers do not satisfy Associative property w.r.t. Subtraction & Division.

AP Board 6th Class Maths Solutions Chapter 2 Whole Numbers InText Questions

Find using distributive property : (Pg. No. 22)
i) 2 × (5 + 6)
ii) 5 × (7 + 8)
Solution:
i) 2 × (5 + 6)
Given, 2 × (5 + 6) = (2 × 5) + (2 × 6)
By using distributive property of multiplication over addition.
2 × 11 = 10 + 12
22 = 22
L.H.S. = R.H.S

ii) 5 × (7 + 8)
Given, 5 × (7 + 8) = (5 × 7) + (5 × 8)
By using distributive property of multiplication over addition.
5 × 15 = 35 + 40
75 = 75
L.H.S = R.HS

iii) 19 × 7 + 19 × 3
Given, (19 × 7) + (19 × 3) = 19 × (7 + 3)
By using distributive property of multiplication over addition.
133 + 57 = 19 × 10
190 = 190
L.H.S = R.H.S

Find : i) 25 × 78 ii) 17 × 26 iii) 49 × 68 + 32 × 49 using distributive property. (Page. No. 22)
Solution:
i) 25 × 78
Given, 25 × 78 = 25 × (80 – 2)
By using distributive property of multiplication over subtraction.
= (25 × 80) – (25 × 2)
= 2000 – 50 = 1950
∴ 25 × 78 = 1950

ii) 17 × 26
Given, 17 × 26 = (10 + 7) × 26
By using distributive property of multiplication over addition.
= (10 × 26) + (7 × 26)
= 260 + 182 = 442
7.17 × 26 = 442

(OR)

= 17 × (30 – 4)
By using distributive property of multiplication over subtraction.
= (17 × 30) – (17 × 4)
= 510 – 68 = 442

AP Board 6th Class Maths Solutions Chapter 2 Whole Numbers InText Questions

iii) 49 × 68 + 32 × 49
Given, 49 × 68 + 32 × 49 = (49 × 68) + (49 × 32)
By using distributive property of multiplication over addition.
= 49 × (68 + 32) .
= 49 × 100
∴ (49 × 68) + (32 × 49) = 4900

Let’s Explore (Page. No. 25)

Question 1.
Which numbers can be shown as a line only?
Solution:
Two or more than two numbers can be shown as a line.
i.e., 2, 3, 4, 5, 6,

Question 2.
Which numbers can be shown as rectangles?
Solution:
6, 8, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 26, 27,.

Question 3.
Which numbers can be shown as squares?
Solution:
4, 9, 16, 25.

AP Board 6th Class Maths Solutions Chapter 2 Whole Numbers InText Questions

Question 4.
Which numbers can be shown as triangles?
Solution:
3, 6, 10, 15, 21, :
Note : Starting from 3; +3, +4, +5, +6, …………………. are all triangular numbers.

AP Board 6th Class Maths Solutions Chapter 1 Numbers All Around us InText Questions

AP Board 6th Class Maths Solutions Chapter 1 Numbers All Around us InText Questions

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 1 Numbers All Around us InText Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 1st Lesson Numbers All Around us InText Questions

AP Board 6th Class Maths Solutions Chapter 1 Numbers All Around us InText Questions

Write the numbers in expanded form. (Page No. 5)

Question 1.
96,08,54,039
Solution:
96,08,54,039 = 9 × 10,00,00,000 + 6 × 1,00,00,000 + 8 × 10,00,000 + 5 × 10,000 + 4 × 1000 + 3 × 10 + 9 × 1
Ninety six crores eight lakhs fifty four thousand and thirty nine.

Question 2.
857,90,00,756
Solution:
857,90,00,756 = 8 × 100,00,00,000 + 5 × 10,00,00,000 + 7 × 1,00,00,000 + 9 × 10,00,000 + 7 × 100 + 5 × 10 + 6 × 1
Eight hundred fifty seven crores ninety lakhs seven hundred and fifty six.

1 Crore = 10 Ten Lakhs
= 100 Lakhs
= 1000 Ten Thousands
= 10,000 Thousands
= 1,00,000 Hundreds
= 10,00,000 Tens
= 1,00,00,000 Unit’s

Check Your Progress (Page No. 6)

Question 1.
Write 10 crores and 100 crores as in the above table.
Solution:
Ten crores = 10 One crores
= 100 Ten lakhs
= 1000 Lakhs
= ,10,000 Ten thousands
= 1,00,000 Thousands
= 10,00,000 Hundreds
= 1,00,00,000 Tens
= 10,00,00,000 Units

AP Board 6th Class Maths Solutions Chapter 1 Numbers All Around us InText Questions

Hundred crores = 100 One crores
= 10 Ten crores
= 10,000 Lakhs
= 1.0. 000 Ten thousands
= 10.0. 000 Thousands
= 1.0. 00.000 Hundreds
= 10.0. 00.000 Tens
= 100.0. 00.000 Units

Check Your Progress (Page No. 8)

Question 1.
Write remaining numbers of the above table in the International System.
Solution:
AP Board 6th Class Maths Solutions Chapter 1 Numbers All Around us InText Questions 9

Question 2.
Fill the boxes in the table with your own numbers and write in words in the International system.
Solution:
a) 896800705

Put comma for each period 896,800,705 in International System.
In expanded form :
= 8 ×x 1,000,000,000 + 9 × 10,000,000 + 6 × 1,000,000 + 8 × 100,000 + 7 × 100 + 5 × 1

In word form :
Eight hundred ninety six millions eight hundred thousand seven hundred and five.

AP Board 6th Class Maths Solutions Chapter 1 Numbers All Around us InText Questions

b) 239176507857
Put comma for each period 239,176,507,857 in International System.
In expanded form :
= 2 × 100,000,000,000 + 3 × 10,000,000,000 + 9 × 1,000,000,000 + 1 × 100,000,000 + 7 × 10,000,000 + 6 × 1,000,000 + 5 × 100,000 + 7 × 1,000 + 8 × 100 + 5 × 10 + 7 × 1
In word form :
Two hundred thirty nine billion one seventy six million five hundred seven thousand eight hundred and fifty seven.

c) 452069258932
Put comma for each period 452,069,258,932
In expanded form :
= 4 × 100,000,000,000 + 5 × 10,000,000,000 + 2 × 1,000,000,000 + 6 × 10,000,000 + 9 × 1,000,000 + 2 × 100,000 + 5 × 10,000 + 8 × 1,000 + 9 × 100 + 3 × 10 + 2 × 1
In word form :
Four hundred fifty two billion sixty nine million two hundred fifty eight thousand nine hundred and thirty two.

d) 839241367054
Put comma for each period 839,241,367,054
In expanded form :
8 × 100,000,000,000 + 3 × 10,000,000,000 + 9 × 1,000,000,000 + 2 × 100,000,000 + 4 × 10,000,000 + 1 × 1,000,000 + 3 × 100,000 + 6 × 10,000 + 7 × 1.000 + 5 × 10 + 4 × 1
In word form :
Eight hundred thirty nine billion two hundred forty one million three hundred sixty seven thousand and fifty four.

e) 342056743298
Put comma for each period 342,056,743,298
In expanded form :
3 × 100,000,000,000 + 4 × 10,000,000,000 + 2 × 1,000,000,000 + 5 × 10,000,000 + 6 × 1,000,000 + 7 × 100,000 + 4 × 10,000 + 3 × 1,000 + 2 × 100 + 9 × 10 + 8 × 1
In word form :
Three hundred forty two billion fifty six million seven hundred forty three thousand two hundred and ninety eight.

Check Your Progress (Page No.12)

Question 1.
Round off each to the nearest ten, hundred and thousands.
(1) 56,789 (2) 86,289 (3) 4,56,726 (4) 5,62,724
Solution:

S.No. Number Nearest ten Nearest hundred | Nearest thousand
1. 56,789 56,790 56,800 57,000
2. 86,289 86,290 86,300 86,000
3. 4,56,726 4,56,730 4,56,700         ’ 4,57,000
4. 5,62,724 5,62,720 5,62,700 5,63,000

Let’s Explore (Page No.12)

Question 1.
Discuss with your friends about rounding off numbers. Consider the population of A.P., Telangana and India in 2011. Round off the numbers to the nearest lakhs.
Solution:

State Population in 2011 Round off the nearest lakhs
Andhra Pradesh 4,92,94,020 4,93,00,000
Telangana 3,52,86,757 3,53,00,000
India 1,21,08,54,977 1,21,09,00,000

AP Board 6th Class Maths Solutions Chapter 1 Numbers All Around us InText Questions

Estimate the sum by rounding and verify the result. (Page No.12)

Question 1. 8756 + 723
Solution:
Given 8756 + 723
First estimate by rounding = 8800 + 700 = 9500
AP Board 6th Class Maths Solutions Chapter 1 Numbers All Around us InText Questions 1
Thus sum is 9,479.
Think
9479 is close to the estimate of 9500.

Question 2.
56723 + 4567 + 72 + 5
Solution:
Given 56723 + 4567 + 72 + 5
First estimate by rounding = 56720 + 4570 + 70 + 10 = 61370
AP Board 6th Class Maths Solutions Chapter 1 Numbers All Around us InText Questions 2

The sum is 61,367.
Think
61367 is close to the estimate of 61370.

Question 3.
656724 + 8567
Solution:
Given 656724 + 8567
First estimate by rounding = 657000 + 9000 = 666000
AP Board 6th Class Maths Solutions Chapter 1 Numbers All Around us InText Questions 3
The sum is 6,65,291.

Think
665291 is close to the estimate of 666000.

Question 4.
60756 + 2562 + 72
Solution:
Given 60756 + 2562 + 72
First estimate by rounding = 60760 + 2560 + 70 = 63390
AP Board 6th Class Maths Solutions Chapter 1 Numbers All Around us InText Questions 4
The sum is 63,390.
Think
63390 is equal to the estimate of 63390.

AP Board 6th Class Maths Solutions Chapter 1 Numbers All Around us InText Questions

Estimate the difference by rounding and verify the result.Pg. No. 13)

Question 1.
7023 – 856
Solution:
Given, 7023 – 856
First estimate by rounding = 7000 – 900 = 6100
AP Board 6th Class Maths Solutions Chapter 1 Numbers All Around us InText Questions 5

Think
6167 is close to the estimate of 6100

Question 2.
9563 – 2847
Solution:
Given, 9563 – 2847
First estimate by rounding = 10000 – 3000 = 7000
AP Board 6th Class Maths Solutions Chapter 1 Numbers All Around us InText Questions 6

Think
6716 is close to the estimate of 7000

Question 3.
52007 – 6756
Solution:
Given, 52007 – 6756
First estimate by rounding = 52000 – 7000 = 45000
AP Board 6th Class Maths Solutions Chapter 1 Numbers All Around us InText Questions 7
Think
45251 is close to the estimate of 45000

AP Board 6th Class Maths Solutions Chapter 1 Numbers All Around us InText Questions

Question 4.
95625 – 4235
Solution:
Given, 95625 – 4235
First estimate by rounding = 95600 – 4200 = 91400 .
AP Board 6th Class Maths Solutions Chapter 1 Numbers All Around us InText Questions 8
Think
91390 is close to the estimate of 91400.

Estimate the product by rounding and verify the result.

Question 1.
63 × 85
Solution:
Given, 63 × 85
First estimate by rounding = 60 × 90 = 5400,
Rounding the result to hundreds = 5400

AP Board 6th Class Maths Solutions Chapter 1 Numbers All Around us InText Questions 10

Think
5355 is close to the estimate of 5400.

Question 2.
636 × 78
Solution:
Given, 636 × 78
First estimate by rounding = 640 × 80 = 51200
Rounding the result to hundreds = 51200
AP Board 6th Class Maths Solutions Chapter 1 Numbers All Around us InText Questions 11
Think
49608 is close to the estimate of 51200.

Question 3.
506 × 85
Solution:
Given, 506 × 85
First estimate by rounding = 500 × 90 = 45000
AP Board 6th Class Maths Solutions Chapter 1 Numbers All Around us InText Questions 12
Think
43010 is close to the estimate of 45000.

Question 4.
709 × 98
Solution:
Given, 709 × 98
First estimate by rounding = 700 × 100 = 70000

AP Board 6th Class Maths Solutions Chapter 1 Numbers All Around us InText Questions 13
Think
69482 is close to the estimate of 70000.

AP Board 6th Class Maths Solutions Chapter 1 Numbers All Around us InText Questions

Estimate the quotient by rounding and verify the result.

Question 1.
936 ÷ 7
Solution:
Given, 936 ÷ 7
Divide 936 ÷ 7
First estimate by rounding 1000 ÷ 10 = 100

AP Board 6th Class Maths Solutions Chapter 1 Numbers All Around us InText Questions 14
Think
133 is close to the estimate of 100.

Question 2.
956 ÷ 17
Solution:
Given, 956 ÷ 17
Divide 956 ÷ 17
First estimate by rounding 1000 – 20 = 50
AP Board 6th Class Maths Solutions Chapter 1 Numbers All Around us InText Questions 15
Think
56 is close to the estimate of 50.

Question 3.
859 ÷ 23
Given, 859 ÷ 23
Divide 859 ÷ 23
First estimate by rounding 860 ÷ 20 = 43
AP Board 6th Class Maths Solutions Chapter 1 Numbers All Around us InText Questions 16
Think
37 is close to the estimate of 43.

AP Board 6th Class Maths Solutions Chapter 1 Numbers All Around us InText Questions

Question 4.
708 ÷ 32
Given, 708 ÷ 32
Divide 708 ÷ 32
First estimate by rounding 710 ÷ 30 = 23
AP Board 6th Class Maths Solutions Chapter 1 Numbers All Around us InText Questions 17
Think
22 is close to the estimate of 23.

AP Board 6th Class Maths Solutions Chapter 7 Introduction to Algebra Unit Exercise

AP Board 6th Class Maths Solutions Chapter 7 Introduction to Algebra Unit Exercise

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 7 Introduction to Algebra Unit Exercise Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 7th Lesson Introduction to Algebra Unit Exercise

Question 1.
The cost of one fan is Rs. 1500. Then what is the cost of ‘n’ fans?
Answer:
AP Board 6th Class Maths Solutions Chapter 7 Introduction to Algebra Unit Exercise 1
Given cost of one fan = Rs. 1500
Number of fans = n
Cost of n fans = cost of one fan × no. of fans = 1500 × n
∴ Cost of n fans = 1500n

Question 2.
Srinu has number of pencils. Raheem has 4 times the pencils as of Srinu. How many pencils does Rahim has? Write an expression.
Answer:
Let number of pencils Srinu has = x
Number of pencils Raheem has = 4 times of Srinu
= 4 × x
∴ Number of pencils Raheem has = 4x

AP Board 6th Class Maths Solutions Chapter 7 Introduction to Algebra Unit Exercise

Question 3.
Parvathi has 5 more books than Sofia. How many books are with Parvathi? Write an expression choosing any variable for number of books.
Answer:
Let number of books Sofia has = y
Given Parvathi has 5 more books than Sofia
Number of books Parvathi has = 5 books more than Sofia
= y + 5
∴ Number of books Parvathi has = y + 5

Question 4.
Which of the following are equations?
i) 10 – 4p = 2
ii) 10 + 8x < – 22
iii) x + 5 = 8
iv) m + 6 = 2
v) 22x – 5 = 8
vi) 4k + 5 > – 100
vii) 4p + 7 = 23
viii) y < – 4
Answer:
i) 10 – 4p = 2
We know that, a mathematical statement involving equality symbol is called an equation.
10 – 4p = 2 has equality symbol.
So, it is an equation.

ii) 10 + 8x < – 22
We know that, a mathematical statement involving equality symbol is called an equation.
10 + 8x < – 22 has no equality symbol.
So, it is not an equation [so it is an inequation]

iii) x + 5 = 8 We know that, a mathematical statement involving equality symbol is called an equation.
x + 5 = 8 has equality symbol.
So, it is an equation.

AP Board 6th Class Maths Solutions Chapter 7 Introduction to Algebra Unit Exercise

iv) m + 6 = 2 We know that, a mathematical statement involving equality symbol is called an equation.
m + 6 = 2 has equality symbol.
So, it is an equation.

v) 22x – 5 = 8 We know that, a mathematical statement involving equality symbol is called an equation.
22x – 5 = 8 has equality symbol.
So, it is an equation.

vi) 4k + 5 > – 100
We know that, a mathematical statement involving equality symbol is called an equation.
4k + 5 > -100 has no equality symbol.
So, it is not an equation.
It is an inequation.

vii) 4p + 7 = 23
We know that, a mathematical statement involving equality symbol is ailed an equation.
4p + 7 = 23 has equality symbol.
So, it is an equation.

viii) y < – 4
We know that, a mathematical statement involving equality symbol is called an equation.
y < – 4 has no equality symbol.
So, it is not an equation.
It is an inequation.

Question 5.
Write L.H.S and R.H.S of the following equations:
i) 7x + 8 = 22
ii) 9y – 3 = 6
iii) 3k – 10 = 2
iv) 3p – 4q = -19
Answer:
i) 7x + 8 = 22
Given equation is 7x + 8 = 22
LHS = 7x + 8
RHS = 22

ii) 9y – 3 = 6
Given equation is 9y – 3 = 6
LHS = 9y – 3
RHS = 6

iii) 3k – 10 = 2
Given equation is 3k – 10 = 2
LHS = 3k – 10
RHS = 2

AP Board 6th Class Maths Solutions Chapter 7 Introduction to Algebra Unit Exercise

iv) 3p – 4q = -19
Given equation is 3p – 4q = -19
LHS = 3p – 4q
RHS = -19

Question 6.
Solve the following equations by trial and error method.
i) x – 3 = 5
ii) y + 6 = 15
iii) y = -1
iv) 2k – 1 = 3
Answer:
i) x – 3 = 5
Given equation is x – 3 = 5
If x = 1, then the value of x – 3 = 1 – 3 = -2 ≠ 5
If x = 2, then the value of x – 3 = 2 – 3 = -l ≠ 5
If x = 3, then the value of x – 3 = 3 – 3 = 0 ≠ 5
If x = 4, then the value of x – 3 = 4 – 3 = l ≠ 5
If x = 5, then the value of x – 3 = 5 – 3 = 2 ≠ 5
If x = 6, then the value of x – 3 = 6 – 3 = 3 ≠ 5
If x = 7, then the value of x – 3 = 7 – 3 = 4 ≠ 5
If x = 8, then the value of x – 3 = 8 – 3 = 5 = 5
From the above when x = 8, the both LHS and RHS are equal.
∴ Solution of the equation x – 3 = 5 is x = 8

ii) y + 6 = 15
Given equation is y + 6 = 15
If y = 1, then the value of y + 6 = 1 + 6 = 7 ≠ 15
If y = 2, then the value of y + 6 = 2 + 6 = 8 ≠ 15
If y = 3, then the value of y +6 = 3 + 6 = 9 ≠ 15
If y = 4, then the value of y + 6 = 4 + 6 = 10 ≠ 15
If y = 5, then the value of y + 6 = 5 + 6 = 11 ≠ 15
If y = 6, then the value of y + 6 = 6 + 6 = 12 ≠ 15
If y = 7, then the value hf y + 6 = 7 + 6 = 13 ≠ 15
If y = 8, then the value of y + 6 = 8 + 6 = 14 ≠ 15
If y = 9, then the value of y + 6 = 9 + 6 = 15 = 15
From the above when y = 9, the both LHS and RHS are equal.
∴ Solution of the equation y + 6 = 15 is y = 9

AP Board 6th Class Maths Solutions Chapter 7 Introduction to Algebra Unit Exercise

iii) \(\frac{m}{2}\) = -1
Given equation is \(\frac{m}{2}\) = -1
If m = 1, then the value of \(\frac{m}{2}\) = \(\frac{1}{2}\) ≠ -1
If m = 2, then the value of \(\frac{m}{2}\) = \(\frac{2}{2}\) = 1 ≠ -1
If m = 3, then the value of \(\frac{m}{2}\) = \(\frac{3}{2}\) ≠ -1
Here, we are not getting negative values.
If we take (substitute) m as a negative number we will get negative value.
If m = -1, then the value of \(\frac{m}{2}\) = \(\frac{-1}{2}\) ≠ -1
If m = -2, then the value of \(\frac{m}{2}\) = \(\frac{-2}{2}\) = -1 = -1
From the above when m = -2, the both LHS and RHS are equal.
∴ Solution of the equation \(\frac{m}{2}\) = -1 is m = -2.

iv) 2k – 1 = 3
Given equation is 2k – 1 = 3
If k = 1, then the value of 2k – 1 = 2(1) – 1 = 2 – 1 = 1 ≠ 3
If k = 2, then the value of 2k – 1 = 2(2) – 1 = 4 – 1 = 3 = 3
From the above when k – 2, the both LHS and RHS are equal.
Solution of the equation 2k – 1 = 3 is k = 2.