Srinivas

Students can practice Telangana SCERT Class 6 Maths Solutions Chapter 14 Understanding 3D and 2D Shapes Ex 14.2 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 14 Understanding 3D and 2D Shapes Exercise 14.2

Question 1.
Examine whether the following are polygons if not why ?

Answer:
i) Figure is kept open. So it is not a polygon.
ii) Figure is a closed one made up of 4 line segments. So it is a polygon.
iii) Figure is a circle. It is curved. It is not made up of line segments. So it is not a polygon.

Question 2.
Count the number of sides of the polygons given below and name them.

Answer:
Figure (i) has 5 sides. It is called a pentagon.
Figure (ii) has 8 sides. It is called an octagon.
Figure (iii) has 6 sides. It is called a hexagon.
Figure (iv) has 3 sides. It is called a triangle.

Question 3.
Identify the regular polygons among the figures given below :

Answer:
Figure (i) is a square. Its sides and angles are all equal. So it is a regular polygon.
Figure (iv) is a regular hexagon. It has equal sides and its angles are equal.
Figure (vi) is an equilateral triangle. Its sides are equal. Its angles are equal.

Students can practice Telangana SCERT Class 6 Maths Solutions Chapter 14 Understanding 3D and 2D Shapes Ex 14.1 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 14 Understanding 3D and 2D Shapes Exercise 14.1

Question 1.
A triangular pyramid has a triangle at its base. It is also known as a tetrahedron. Find the number of

i) No. of Faces :
ii) No. of Edges :
iii) No. of Vertices :
Answer:
A triangular pyramid (tetrahedron) has
i) No. of Faces : 4
ii) No. of Edges : 6
iii) No. of Vertices : 4

Question 2.
A square pyramid has a square at its base. Find the number of

i) No. of Faces :
ii) No. of Edges :
iii) No. of Vertices :
Answer:
A square pyramid has
i) No. of Faces 5
ii) No. of Edges : 8
iii) No. of Vertices : 5

Question 3.
Fill the table.

Answer:

Question 4.
A triapgular prism is often in the shape of a kaleidoscope. It has triangular faces.

i) No. of triangular Faces :
ii) No. of rectangular Faces :
iii) No. of Edges :
iv) No. of Vertices :
Answer:
i) No. of triangular Faces : 2
ii) No. of rectangular Faces : 3
iii) No. of Edges : 9
iv) No. of Vertices : 6

Students can practice Telangana SCERT Class 6 Maths Solutions Chapter 12 Symmetry InText Questions to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 12 Symmetry InText Questions

Do This

Question 1.
Match each letter with its mirror image. The dotted line with every letter shows in the mirror.

Can you think of more such alphabets and words which will remain the same in their mirror image ?
Answer:

Yes. They are 0, X,H, I, which are same as mirror images.

Try This

Question 1.
Place a mirror along the dotted lines and draw their mirror images.

Do you observe any change ? Are angles in the images equal to the angles in the given figures ?
Answer:
We can observe the angles formed in the original figures and its mirror images are same.

Do This

Question 1.
In the figures given below find which are symmetric figures.

Can we find line of symmetry for every figure ?
Answer:
(i), (ii) figures are symmetric figures.
(iii) & (iv) are not symmetric figures.

Try This

Write the letters of English alphabet A to Z and find out which have
i) Vertical lines of symmetry
ii) Horizontal lines of symmetry
iii) No lines of symmetry
Answer:
i) Vertical lines of symmetry :

ii) Horizontal lines of symmetric

iii) No lines of symmetric

Try These

Question 1.
Draw any five objects which have a line of symmetry.
Answer:
The following figure having the line of symmetry.

Question 2.
Draw any five objects which are not symmetric.
Answer:

Think, Discuss And Write

Question 1.
If the paper is folded four times how many lines of symmetry can be formed with paper cutting ?
Answer:
If a paper is folded 4 times then 8 symmetric lines are formed.

Question 2.
To cut four similar figures side by side by folding the paper, how many folds are needed ?
Answer:
Two paper foldings are to be needed to form 4 similar figures.

Students can practice Telangana SCERT Class 6 Maths Solutions Chapter 13 Practical Geometry InText Questions to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 13 Practical Geometry InText Questions

Try These

Question 1.
Construct two circles with same radii in such a way that

i) the circles intersect at two points.
ii) touch each other at one point only.
Answer:
i) Circles intersect at two points :
ii) Touch each other at one point only :

Think, Discuss And Write

Question 1.
How would you check whether It Is perpendicular or not ? Note that it passes through P as required.
Answer:

The perpendicularity can be measured by protractor.
So, the perpendicular line always passes through P’.

Do This

Question 1.
Measure the lengths of \(\overline{\mathrm{A P}}\) and \(\overline{\mathrm{B P}}\) in both the constructions. Are they equal?
Answer:

\(\overline{\mathrm{A P}}\) = 1.7 cm
\(\overline{\mathrm{B P}}\) = 1.7 cm
yes. They are equal

Think, Discuss And Write

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 \(\overline{\mathrm{A B}}\)?
Answer:
Explanation:
1) Construct a line segment \(\overline{\mathrm{A B}}\) with a suitable radius.
2) To construct a perpendicular line to the given line, take the radius more than half of a given line segment \(\overline{\mathrm{A B}}\).
3) With the centres A, B draw two arcs as mentioned above.
4) The two arcs are intersected at a point. Name it P, Q.
5) Join these two points by a straight line. Name it as l. ∴ l ⊥ \(\overline{\mathrm{A B}}\).

If we does not taken the radius half of the given line segment, then the perpendicular bisector doesn’t formed. Since the arcs doesn’t interest each other.

Do This

Question 1.
Construct angles of 180°, 240°, 300°.
Answer:

Question 2.
Construct an angle of 45° by using compasses.
Answer:
Steps of construction:

1) Construct \(\overline{\mathrm{OA}}\) line segment with suitable radius.
2) Draw an arc on \(\overline{\mathrm{OA}}\) from the centre ‘O’, its cuts \(\overline{\mathrm{OA}}\) at ‘C’.
3) Draw two arcs with the centres O, C with equal radii the two arcs meet at ‘D’.
4) Draw another two arcs with the centres D, C they meet at point E. Join O, E.
5) With the centres D, G draw another two arcs, they meet at point F. Join O, F.
∴ ∠AOB = 45°
So we get the required angle.

Students can practice Telangana SCERT Class 6 Maths Solutions Chapter 14 Understanding 3D and 2D Shapes InText Questions to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 14 Understanding 3D and 2D Shapes InText Questions

Try These

Question 1.
(i) What is the shape of the face of a cube ?
(ii) What is the shape of the face of a cuboid ?
Answer:
i) Square shape.
ii) Rectangular shape.

Question 2.
Ramesh has collected some boxes in his room. Pictures of these are given here. How many are cubes and how many are cuboids?

Answer:
Number of cubes = 3
Number of cuboids = 4

Question 3.
Ajith has made a cuboid by arranging cubes of 2 cms each. What is the length, breadth and height of the cuboid so formed ?

Answer:
Length of the first cuboid = 2 + 2 = 4 cm
breadth = 2 cm height = 2cm
Length of the second cuboid =2 + 2 + 2 = 6 cm
breadth = 2 cm height = 2 cm

Think, Discuss And Write

Question 1.
What is the difference between a cylinder and a cone with respect to the number of faces, vertices and edges? Discuss with your friends.
Answer:

Do This

Question 1.
Fill the table accordingly :

The cylinder, the cone and the sphere have no straight edges. What is the base of a cone ? Is it a circle ? The cylinder has two bases. What shape is the base ? Of course, a sphere has no face ! Think about it.
Answer:

The base of a cone is circle.
The shape of the base of a cylinder is circle.

Do This

Question 1.
Draw ten polygons with different shapes in your notebook.
Answer:

Question 2.
Use match-sticks or broom-sticks and form closed figures using :
i) Six sticks
ii) Five sticks
iii) Four sticks
iv) Three sticks
v) Two sticks
In which case was it not possible to form a polygon ? Why ?
Answer:

Observation : In the fifth case it is not possible to form a polygon.
Conclusion : We find that we could not form a polygon using two sticks.
Reason : A polygon must have at least three sides.

Try This

Question 1.
Find out the differences :

Measure the lengths of the sides and angles of (i) and (ii). What did you find ?
Answer:
i) Length of sides
AB = 3.2 cm
BC = 1. 2 cm
CD = 2 cm
ED – 2 6 cm
AE = 1. 6 cm
angles ∠A = 125°
∠B = 70°
∠C = 160°
∠D = 105°
∠E = 80°
∴ sides and angles are not equal.
The given figure is an irregular polygon (pentagon).
∴ Sides, angles are not equal.

ii) In the given figure AB = BC = CD = DE
= AE = 1.9 cm
Each angle = 108°
angles ∠A = ∠B = ∠C = ∠D = ∠E = 108°
∴ Sides, angles are equal.
The given figure is a regular pentagon.

Students can practice Telangana SCERT Class 6 Maths Solutions Chapter 12 Symmetry Ex 12.1 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 12 Symmetry Exercise 12.1

Question 1.
Check whether the given figures are symmetric or not ? Draw the line of symmetry as well.

Answer:

The dotted lines represents a line of symmetry.

Question 2.
Draw a line of symmetry for each of the figures, wherever possible.

Answer:

Question 3.
In the figure, l is the line of symmetry.
Complete the diagram to make it symmetric.

Answer:

Question 4.
Complete the figures such that the dotted line is the line of symmetry.

Answer:

Students can practice Telangana 6th Class Maths Textbook Solutions Chapter 7 Fractions and Decimals InText Questions to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 7 Fractions and Decimals InText Questions

Try These

Question 1.
How will you represent the following pictorially ?
(i) \(\frac{3}{4}\)
Answer:
\(\frac{3}{4}\)

(ii) \(\frac{2}{8}\)
Answer:
\(\frac{2}{8}\)

(iii) \(\frac{1}{3}\)
Answer:
\(\frac{1}{3}\)

(iv) \(\frac{5}{8}\)
Answer:
\(\frac{5}{8}\)

Question 2.
Write the fraction representing the shaded portion.

Answer:
(i) \(\frac{1}{3}\)
(ii) \(\frac{1}{4}\)
(iii) \(\frac{2}{6}\)

Do This

Question 1.
Write 5 proper fractions and draw them pictorially.
Answer:

Question 2.
Rani says that shaded portion in the given figure represents \(\frac{1}{4}\). Do you agree with her? Give reason to support your answer.

Answer:
Yes. The reason is out of fourportions only one portion is shaded.

Do This

Question 1.
Write improper fractions represented by the following pictures.

Answer:

Answer:

Answer:

Question 2.
Represent the following fractions pictorially :
\(\frac{7}{4}, \frac{5}{3}, \frac{7}{6}\)
Answer:

Do This

Question 1.
Write the following as mixed fractions
\(\frac{7}{2}, \frac{8}{5}, \frac{9}{4}, \frac{13}{5}, \frac{17}{3}\)
Answer:

Try These

Question 1.
Write the numerator and denominators of the following fractional numbers:
\(\frac{1}{3}, \frac{2}{5}, \frac{7}{2}, \frac{19}{3}, \frac{7}{29}, \frac{11}{13}, \frac{1}{7}, \frac{8}{3}\)
Answer:

Question 2.
Sort the following fractions into the category of proper and improper fractions. Also write improper fraction as mixed fractions:
\(\frac{1}{3}, \frac{2}{5}, \frac{7}{2}, \frac{19}{3}, \frac{7}{29}, \frac{11}{13}, \frac{1}{7}, \frac{8}{3}\)
Answer:
Improper fractions:

Proper fractions:
\(\frac{1}{3}, \frac{2}{7}, \frac{3}{5}, \frac{1}{9}\)

Do This

Question 1.
Show that following on number lines.
(i) \(\frac{7}{6}\)
Answer:

(ii) \(\frac{5}{2}\)
Answer:

(iii) \(\frac{7}{5}\)
Answer:

(iv) \(\frac{9}{6}\)
Answer:

Question 2.
Consider the following numbers. Which of these would lie on the number line
i) before 1
Answer:
Fractions before 1 are \(\frac{1}{3}, \frac{7}{9}, \frac{6}{11}\)

ii) between 1 and 2
Answer:
Fractions between 1 and 2 are \(\frac{7}{5}, \frac{9}{5}\)

Try These

Question 1.
Write 5 fractional numbers that are in the standard form.
Answer:
Standard form of 5 fractional numbers
= \(\frac{1}{2}, \frac{2}{3}, \frac{4}{5}, \frac{5}{6}, \frac{8}{9}\)

Question 2.
Write 5 fractional numbers that are not in standard form.
Answer:
\(\frac{10}{6}, \frac{144}{100}, \frac{51}{17}, \frac{48}{16}, \frac{36}{44}\)

Question 3.
Convert the following fractions into their standard form.
i) \(\frac{7}{28}\)
ii) \(\frac{15}{90}\)
iii) \(\frac{11}{33}\)
iv) \(\frac{39}{13}\)
Answer:

Think, Discuss And Write

Question 1.
Rafi says “there can be no equivalent fractions that are also like fractions”. Do you agree with him ? Explain your answer and justify.
Answer:
Equivalent fractions = \(\frac{1}{2}, \frac{2}{4}, \frac{8}{16}, \frac{32}{64}\)
Like fraction = \(\frac{3}{2}, \frac{5}{2}, \frac{7}{2}, \frac{8}{2}, \frac{9}{2}\)
Equivalent fractions are like fractions also. So, I agree with him.

Do This

Question 1.
Identify the biggest and the smallest in these group of fractional numbers.
(i) \(\frac{1}{7}, \frac{3}{7}, \frac{2}{7}, \frac{5}{7}\)
Answer:
Biggest fraction = \(\frac{5}{7}\)
Smallest fraction = \(\frac{1}{7}\)

(ii) \(\frac{1}{9}, \frac{13}{9}, \frac{11}{9}, \frac{5}{9}\)
Answer:
Biggest fraction = \(\frac{13}{9}\)
Smallest fraction = \(\frac{1}{9}\)

(iii) \(\frac{1}{3}, \frac{5}{3}, \frac{17}{3}, \frac{9}{3}\)
Answer:
Biggest fraction = \(\frac{17}{3}\)
Smallest fraction = \(\frac{1}{3}\)

Question 2.
Which of these is the smaller fraction ?
(i) \(\frac{2}{5}, \frac{3}{7}\)
Answer:

(ii) \(\frac{7}{8}, \frac{5}{4}\)
Answer:

(iii) \(\frac{3}{11}, \frac{1}{2}\)
Answer:

(iv) \(\frac{5}{6}, \frac{2}{3}\)
Answer:

Question 3.
Write the following fractional number in ascending order.
(i) \(\frac{1}{7}, \frac{13}{7}, \frac{11}{7}, \frac{5}{7}, \frac{15}{7}\)
Answer:
Ascending order:
\(\frac{1}{7}<\frac{5}{7}<\frac{11}{7}<\frac{13}{7}<\frac{15}{7}\)

(ii) \(\frac{2}{3}, \frac{5}{6}, \frac{3}{9}, \frac{24}{18}\)
Answer:

(iii) \(\frac{2}{3}, \frac{1}{2}, \frac{5}{6}, \frac{7}{12}\)
Answer:

(iv) \(\frac{1}{5}, \frac{1}{2}, \frac{1}{8}, \frac{1}{3}, \frac{1}{12}\)
Answer:

Do this

Question 1.
Write the following in descending order.
i) \(\frac{1}{9}, \frac{13}{9}, \frac{11}{9}, \frac{15}{9}, \frac{3}{9}\)
Answer:

ii) \(\frac{1}{6}, \frac{2}{3}, \frac{3}{9}, \frac{5}{6}\)
Answer:

iii) \(\frac{1}{5}, \frac{9}{5}, \frac{3}{5}, \frac{6}{5}\)
Answer:
Descending Order = \(\frac{9}{5}>\frac{6}{5}>\frac{3}{5}>\frac{1}{5}\)

iv) \(\frac{1}{4}, \frac{1}{2}, \frac{1}{8}, \frac{3}{4}\)
Answer:
Descending Order = \(\frac{3}{4}>\frac{1}{2}>\frac{1}{4}>\frac{1}{8}\)

Question 8.
Simplify:
(i) \(\frac{1}{18}+\frac{1}{18}\)
Answer:
\(\frac{1}{18}+\frac{1}{18}=\frac{1+1}{18}=\frac{2}{18}=\frac{1}{9}\)

(ii) \(\frac{8}{15}+\frac{3}{15}\)
Answer:
\(\frac{8}{15}+\frac{3}{15}=\frac{8+3}{15}=\frac{11}{15}\)

(iii) \(\frac{7}{7}-\frac{5}{7}\)
Answer:
\(\frac{7}{7}-\frac{5}{7}=\frac{7-5}{7}=\frac{2}{7}\)

(iv) \(\frac{1}{22}+\frac{21}{22}\)
Answer:
\(\frac{1}{22}+\frac{21}{22}=\frac{1+21}{22}=\frac{22}{22}\) = 1

(v) \(\frac{12}{15}-\frac{7}{15}\)
Answer:
\(\frac{12}{15}-\frac{7}{15}=\frac{12-7}{15}=\frac{5}{15}=\frac{1}{3}\)

(vi) \(\frac{5}{8}+\frac{3}{8}\)
Answer:
\(\frac{5}{8}+\frac{3}{8}=\frac{5+3}{8}=\frac{8}{8}\) = 1

(vii) 1 – \(\frac{2}{3}\left(1=\frac{3}{3}\right)\)
Answer:
1 – \(\frac{5}{8}+\frac{3}{8}=\frac{5+3}{8}=\frac{8}{8}\) [∵ 1 = \(\frac{3}{3}\)]

(viii) \(\frac{1}{4}+\frac{0}{4}\)
Answer:
\(\frac{1}{4}+\frac{0}{4}=\frac{1+0}{4}=\frac{1}{4}\)

(ix) 3 – \(\frac{12}{5}\)
Answer:
3 – \(\frac{12}{5}=\frac{3 \times 5}{1 \times 5}-\frac{12}{5}=\frac{15}{5}-\frac{12}{5}=\frac{15-12}{5}=\frac{3}{5}\)

Question 9.
Fill in the missing fractions.
i) \(\frac{7}{10}\) – ___ = \(\frac{3}{10}\)
Answer:

ii) __ – \(\frac{3}{21}=\frac{5}{21}\)
Answer:

iii) ___ – \(\frac{3}{3}=\frac{3}{6}\)
Answer:

iv) ___ – \(\frac{5}{27}=\frac{12}{27}\)
Answer:

Question 10.
Narendra painted \(\frac{2}{3}\) area of the wall in his room. His brother Ritesh helped and painted \(\frac{1}{3}\) area of the wall. How much did they paint together ?
Answer:
Area of the wall painted by Narendra = \(\frac{2}{3}\)
Area of the wall painted by Ritesh = \(\frac{1}{3}\)
Area of the wall painted by both
Narendra and Ritesh = \(\frac{2}{3}+\frac{1}{3}\)
= \(\frac{2+1}{3}=\frac{3}{3}\) = 1
Narendra and his brother Ritesh painted the complete wall.

Question 11.
Neha was given \(\frac{5}{7}\) of a basket of bananas. What fraction of bananas was left in the basket?
Answer:
The part of a basket of bananas given to Neha = \(\frac{5}{7}\)
The part of bananas left in the basket
= 1 – \(\frac{5}{7}=\frac{1 \times 7}{1 \times 7}-\frac{5}{7}=\frac{7}{7}-\frac{5}{7}=\frac{7-5}{7}=\frac{2}{7}\)

Question 12.
A piece of rod \(\frac{7}{8}\) metre long is broken into two pieces. One piece was \(\frac{1}{4}\) metre long. How long is the other piece?
Answer:
Length of a piece of rod = \(\frac{7}{8}\) metre
Length of one broken piece of rod = \(\frac{1}{4}\) metre.
Length of the other piece = \(\frac{7}{8}-\frac{1}{4}\)
= \(\frac{7}{8}-\frac{1 \times 2}{4 \times 2}=\frac{7}{8}-\frac{2}{8}=\frac{7-2}{8}=\frac{5}{8}\)m
∴ \(\frac{5}{8}\)m long is the other piece.

Question 13.
Renu takes 2\(\frac{1}{5}\) minutes to walk 5 around the school ground. Snigdha takes \(\frac{7}{4}\) minutes to do the same. Who takes less time and by what fraction?
Answer:
Time taken by Renu to walk around the school ground = 2\(\frac{1}{5}\) minutes
= \(\frac{11}{5}\) minutes
Time taken by Snigdha to walk around the school ground = \(\frac{7}{4}\) minutes
To find the person who takes less time to do the same, we write the fractions \(\frac{11}{5}\) and \(\frac{7}{4}\) having the same denominators.
\(\frac{11}{5} \times \frac{4}{4}=\frac{44}{20} ; \frac{7}{4} \times \frac{5}{5}=\frac{35}{20}\)
We know that \(\frac{35}{20}<\frac{44}{20}\)
Therefore, Snigdha takes \(\frac{9}{20}\) minutes less time to walk around the school ground.
(∵ \(\frac{44}{20}-\frac{35}{20}=\frac{44-35}{20}=\frac{9}{20}\))

Try These

Question 1.
(i) Write fractions for the following decimal and also find how many tenth parts are there in each :
0.4, 0.2, 0.8, 1.6, 5.4, 555.3, 0.9
Answer:
0.4 = \(\frac{4}{10}\)
0.2 = \(\frac{2}{10}\)
0.8 = \(\frac{8}{10}\)
4 tenth parts 2 tenth parts 8 tenth parts

1.6 = 1\(\frac{6}{10}\) = 1 + \(\frac{6}{10}\) 6 tenth parts
5.4 = 5\(\frac{4}{10}\) = 5 + \(\frac{4}{10}\) 4 tenth parts
555.3 = 553 + \(\frac{3}{10}\) = 3 tenth parts
0.9 = \(\frac{9}{10}\) = 9 tenth parts

(ii) Complete the following table.

Answer:

(iii) Complete the following table.

Answer:

(iv) Measure the length of these line segments and fill it in the table given below.

Answer:

Do This

Question 1.
Fill in the blanks.
(i) 325 paise = _________ rupees _________ paise = ₹ _________
(ii) 570 paise = _________ rupees _________ paise = ₹ _________
(iii) 2050 paise = _________ rupees _________ paise = ₹ _________
Answer:
(i) 325 paise = 3 rupees 25 paise = ₹ 3.25
(ii) 570 paise = 5 rupees 70 paise = ₹ 5.70
(iii) 2050 paise = 20 rupees 50 paise = ₹ 20.50

Do This

Question 1.
Find: i) 0.39 + 0.26
ii) 0.8 + 0.07
iii) 1.45 + 1.90
iv) 3.44 + 1.58
Answer:

Students can practice Telangana 6th Class Maths Textbook Solutions Chapter 7 Fractions and Decimals Ex 7.5 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 7 Fractions and Decimals Exercise 7.5

Question 1.
Sonu went to a shop. He wanted to buy a chiki and a toffee. One chiki costs ₹ 0.75 and a toffee costs ₹ 0.50. If he buys one each of them how much he has to pay to the shopkeeper. Sonu’s mother gave him ₹ 2. How much he will get in return? Suppose if his mother gave him ₹ 5, then how much will the shopkeeper return ?
Answer:
Cost of one chiki is ₹ 0.75
Cost of one toffee is₹ 0.50
Cost of one chiki and one toffee
= ₹ 0.75 + ₹ 0.50
= \(\frac{75}{100}+\frac{50}{100}=\frac{125}{100}\) = ₹ 1.25
The amount to be paid to the shop-keeper by Sonu = ₹ 1.25
Amount given to Sonu by his mother
= ₹ 2.00
The amount Sonu will get in return
= ₹ 2.00 – ₹ 1.25 = ₹ 0.75
If Rs. 5 is given by his mother, the amount returned by shopkeeper.
= ₹ 5.00 – ₹ 1.25 = ₹ 3.75
₹ 3.75 will the shopkeeper return.

Question 2.
Add the following decimal fractions,
i) 25.11 + 3.80
ii) 14.01 + 1.1 + 1.98
iii) 9.85 + 0.61
iv) 2.3 + 18.94
v) 2.57+ 3.75
Answer:
i) 25.11 + 3.80

∴ 25.11 + 3.80 = 28.91

ii) 14.01 + 1.1 + 1.98

∴ 14.01 + 1.1 + 1.98 = 17.09

iii) 9.85 + 0.61

∴ 9.85 + 0.61 = 10.46

iv) 2.3 + 18.94

∴ 2.3 + 18.94 = 21.24

v) 2.57 + 3.75

∴ 2.57 + 3.75 = 6.32

Question 3.
Abhishek travelled 5 km. 28 m. by bus, 2 km. 265 m. by car and the rest 1 km. 30 m. on foot. How much distance did he travel in all ?
Answer:
Abhishek travelled a distance by bus-= 5 km. 28m
Abhishek travelled a distance by car = 2 km. 265 m
Abhishek travelled a distance by foot = 1 km. 30 m
∴ Total distance did he travel in all = 5 km.
28 m + 2 km. 265 m + 1 km.30 m = 8 km. 323 m.

Question 4.
Mrs.Vykuntam bought 6.25 m of dress material for her elder daughter and 5.75 m for the younger one. How much dress material did she buy for her daughters ?
Answer:
Dress material for elder daughter
= 6.25 m
Dress material for younger one
= 5.75 m
∴ Total dress material bought for two daughters
= 6.25 m + 5.75 m
= 12 m.

Students can practice Telangana 6th Class Maths Textbook Solutions Chapter 7 Fractions and Decimals Ex 7.4 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 7 Fractions and Decimals Exercise 7.4

Question 1.
Fill in the blanks.
i) The fractional form of 0.8 is
ii) The integral part of 15.9 is
iii) The digit in the tenths place of 171.9 is
iv) The place value of 8 in 9.8 is
v) The point between the integral part and the decimal part of the decimal number is called
Answer:
i) \(\frac{8}{10}\)
ii) 15
iii) 9
iv) tenth (or) \(\frac{8}{10}\)
v) decimal point

Question 2.
Write the decimal for each of the following.
i) One hundred twenty five and four tenths
ii) Twenty and two tenths
iii) Eight and six tenths.
Answer:
i) 125.4
ii) 20.2
iii) 8.6

Question 3.
Write the following fractions in the decimal form using the decimal point.
(i) \(\frac{16}{100}\)
Answer:
\(\frac{16}{100}\) = 0.16

(ii) \(\frac{278}{1000}\)
Answer:
\(\frac{278}{1000}\) = 0.278

(iii) \(\frac{6}{100}\)
Answer:
\(\frac{6}{100}\) = 0.06

(iv) \(\frac{369}{100}\)
Answer:
\(\frac{369}{100}\) = 3.69

(v) \(\frac{16}{1000}\)
Answer:
\(\frac{16}{1000}\) = 0.016

(vi) \(\frac{345}{10}\)
Answer:
\(\frac{345}{10}\) = 34.5

(vii) \(\frac{907}{100}\)
Answer:
\(\frac{907}{100}\) = 9.07

Question 4.
Write the place value of each underlined digit. ‘
i) 34.26
ii) 8.88
iii) 0.91
iv) 0.50
v) 3.03
vi) 6.74
Answer:
i) The underlined digit is in ones place. So its place value is 4 x 1 = 4
ii) The underlined digit is in hundredths place. So its place value is \(\frac{8}{100}\)
iii) The underlined digit is in tenths place. So its place value is \(\frac{9}{10}\).
iv) The underlined digit is in tenths place. So its place value is \(\frac{5}{10}\).
v) The underlined digit is in hundredths place. So its place value is \(\frac{3}{100}\).
vi) The underlined digit is in tenths place. So its place value is \(\frac{7}{10}\).

Question 5.
Find greater in the following.
(i) 0 0.2 (or) 0.4
(ii) 70.08 (or) 70.7
(iii) 6.6 (or) 6.58
(iv) 7.4 (or) 7.35
(v) 0.76 (or) 0.8
Answer:
(i) 0.2 or 0.4 2
0.2 = \(\frac{2}{10}\) and 0.4 = \(\frac{4}{10}\)
\(\frac{2}{10}\) means 2 parts out of 10 parts.
\(\frac{4}{10}\) means 4 parts out of 10 parts.
∴ 4 > 2
(i.e) 0.4 > 0.2

(ii) 70.08 (or) 70.7
Whole part of both is 70.
0.08 = \(\frac{8}{100}\); 0.7 = \(\frac{7}{10}=\frac{7 \times 10}{10 \times 10}=\frac{70}{100}\)
0.08 means 8 parts out of 100 parts
0.7 means 70 parts out of 100 parts
∴ 70 > 8
(i.e.,) 70.7 > 70.08

(iii) 6.6 or 6.58
Whole part of both is 6.
0.6 = \(\frac{6}{10}=\frac{6 \times 10}{10 \times 10}=\frac{60}{100}\); 0.58 = \(\frac{58}{100}\)
0.6 means 60 parts out of 100 parts.
0.58 means 58 parts out of 100 parts.
∴ 60 > 58
(i.e.,) 6.6 > 6.58

(iv) 7.4 or 7.35
Whole part of both is 7.
0.4 = \(\frac{4}{10}=\frac{4 \times 10}{10 \times 10}=\frac{40}{100}\); 0.35 = \(\frac{35}{100}\)
0.4 means 40 parts out of 100 parts.
0.35 means 35 parts out of 100 parts.
∴ 40 > 35
(i.e.,) 7.4 > 7.35

(v) 0.76 or 0.8
Whole part of both is zero.
0.76 = \(\frac{76}{100}\); 0.8 = \(\frac{8}{10}=\frac{8 \times 10}{10 \times 10}=\frac{80}{100}\)
0.76 means 76 parts out of 100 parts.
0.8 means 80 parts out of 100 parts.
∴ 80 > 76
(i.e.,) 0.8 > 0.76

Question 6.
Rewrite in ascending order.
i) 0.04, 1.04, 0.14, 1.14
ii) 9.09, 0.99, 1.1, 7
Answer:
i) 0.04,1.04,0.14,1.14
By observation, we can say that
0.04 < 0.14 < 1.04 < 1.14 (ascending order)

ii) 9.09, 0.99, 1.1, 7
By observation, we can say that
0.99 < 1.1 < 7 < 9.09 (ascending order)

Question 7.
Rewrite in descending order
i) 8.6, 8.59, 8.09, 8.8
ii) 6.8, 8.66, 8.06, 8.68
Answer:
i) 8.6,8.59,8.09, 8.8
By observation, we can say that
8.8 > 8.6 > 8.59 > 8.09 (descending order)

ii) 6.8, 8.66, 8.06, 8.68
By observation, we can say that
8.68 > 8.66 > 8.06 > 6.8 (descending order)

Students can practice TS SCERT Class 6 Maths Solutions Chapter 10 Perimeter and Area Ex 10.1 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 10 Perimeter and Area Exercise 10.1

Question 1.
Find the perimeter of each of the following shapes :

Answer:
(i) Shape – 1 :
Given that AB = 40 cm ; BC = 50 cm
CD = 35 cm ; DE = 60 cm
and EA = 45 cm
Perimeter of shape – 1 = AB + BC + CD + DE + EA
= (40 + 50 + 35 + 60 + 45) cm
= 230 cm

(ii) Shape – 2 : Given that
AB = 8 cm; i.e., AB = HG = 8 cm;
GF = 2 cm; i.e., GF = HI = 2 cm;
ED = 5 cm; i.e., ED = JK = 5 cm;
IJ = 3 cm; i.e., IJ = LK = 3 cm;
AL = 3 cm; i.e., AL = BC = 3 cm
CD = EF = 3 cm
Perimeter of shape – 2 = AB + BC + CD + DE + EF + FG + GH + HI + IJ + JK + KL + LA
= (8 + 3 + 3 + 5 + 3 + 2 + 8 + 2 + 3 + 5 + 3+ 3)cm
= 48 cm

(iii) Shape – 3 : Given that
AB = 6 cm ; BC = 2 cm ; CD = 2 cm ;
DE = 2 cm ; EF = 2 cm ; FG = 2 cm ;
GH = 2 cm ; and HA = 6 cm
Perimeter of shape – 3
= AB + BC + CD + DE + EF + FG + GH + HA
= (6 + 2 + 2 + 2 + 2 + 2 + 2 + 6) cm
= 24 cm

(iv) Shape – 4 : Given that
AL = 4 cm (i.e.) AL = BC = 4 cm
AB = 2 cm (i.e.) AB = HG = 2 cm
JK = 2 cm (i.e.) JK = ED = 2 cm
GF = 4 cm (i.e.) GF = HI = 4 cm
KL = 4 cm (i.e.) KL = IJ = 4 cm
EF = 4 cm (i.e.) EF = CD = 4 cm
Perimeter of shape – 4
= AB + BC + CD + DE + EF + FG + GH + HI + IJ + JK + KL + LA
= 2 + 4 + 4 + 2 + 4 + 4 + 2 + 4 + 4 + 2 + 4 + 4
= 40 cm

Question 2.
Find the perimeter of each of the following figures.

What would be cost of putting a wire around each of these shapes given that 1 cm wire costs ₹ 15 ?
Answer:
(i)

Perimeter = AB + BC + CA
= (30 + 55 + 35) cm
= 120 cm
Cost of putting a wire around the given shape = 120 × Rs.15 = Rs. 1800

(ii)

ABCD is a rectangle so opposite sides are equal.
AB = CD = 40 cm
BC = AD = 20 cm
Perimeter = AB + BC + CD + DA
= 40 + 20 + 40 + 20
= 120 cm
Cost of putting a wire around the given shape at Rs.15 per 1 cm = 120 × Rs.15
= Rs. 1800

(iii)

ABCD is a square, so all the sides are equal.
∴ AB = BC = CD = DA = 30 cm
Perimeter = AB + BC + CD + DA
= 30 + 30 + 30 + 30
= 120 cm
Cost of putting a wire around the given shape = 120 × Rs.15
= Rs. 1800

(iv)

Given that the length of one side = 24 cm
ABCDEF is a regular hexagon (i.e.) it contains 6 equal sides.
Perimeter = 6 × side
= 6 × 24cm = 144 cm
Cost of putting a wire around the given shape = 144 × Rs.15 = Rs. 2160

Question 3.
How many different rectangles can you make with a 24 cm long string with integral sides and what are the sides of those rectangles in cm ?
Answer:
The length of string = 24 cm
Perimeter of the rectangle = 2(l + b) =24 cm
l + b = \(\frac{24}{2}\)cm = 12 cm
We can make different rectangles with 24 cm. long string as follows.

Question 4.
A flower bed is in the shape of a square with a side 3.5 m. Each side is to be fenced with 4 rows of ropes. Find the cost of rope required at ₹ 15 per meter.
Answer:
The shape of a flower bed is a square.
The length of the side of the square
= 3.5 m
Length of 4 rows of ropes on each side = 4 × 3.5 m = 14 m
Length of the rope on 4 sides
= 14 m × 4 = 56 m
Cost of rope required at Rs. 15 per meter = Rs. 56 × 15 = Rs. 840

Question 5.
A piece of wire is 60 cm long. What will be the length of each side if the string is used to form :
(i) an equilateral triangle
(ii) a square
(iii) a regular hexagon
(iv) a regular pentagon
Answer:
(i) Length of the wire = 60 cm
The perimeter of an equilateral triangle with side x cm = 3 × x = 3x
By problem,
3x = 60 cm
∴ x = \(\frac{60}{3}\) cm = 20 cm
Length of each side of the equilateral triangle = 20 cm

(ii) Length of the wire = 60 cm
The perimeter of the square with side ‘x’ cm = 4 × x = 4x
By problem,
4x = 60 cm
x= \(\frac{60}{4}\) cm = 15 cm
Length of each side of the square =15 cm

(iii) Length of the wire = 60 cm
The perimeter of the regular hexagon with side x’ cm = 6 × x = 6x
By problem,
6x = 60 cm
∴ x = \(\frac{60}{5}\) cm = 10 cm 6
Length of each side of the regular hexagon = 10 cm

(iv) Length of the wire = 60 cm
The perimeter of the regular pentagon with side ‘x’ cm = 5 × x = 5x
By problem,
5x = 60 cm
x = \(\frac{60}{5}\) cm = 12 cm
Length of each side of the regular pentagon = 12 cm.

Question 6.
Bunty and Bubly go for jogging every morning. Bunty goes around a square park of side 80 m. Bubly goes around a rectangular park with length 00 m and breadth 60 m. If they both fake 3 rounds, who covers more distance and by how much ?
Answer:
Bunty goes around a square park.
The length of the side of square park
= 80 m.
The perimeter of the square park
= 4 × 80 = 320 m.
Distance covered by Bunty in 3 rounds = 320m × 3 = 960 m.
Bubly goes around a rectangular park.
The length and breadth of the park are 90 m and 60 m respectively.
The perimeter of the rectangular park = 2 (length + breadth)
= 2[90 + 60]
= 2 × 150
= 300 m
Distance covered by Bubly in 3 rounds = 300m × 3 = 900 m.
Bunty covers greater distance by 60 m. (∵ 960 – 900 = 60 m)

Question 7.
The length of a rectangle is twice of its breadth. If its perimeter is 48 cm, find the dimensions of the rectangle.
Answer:
The perimeter of the rectangle = 48 cm Let the breadth of the rectangle be x cm The length of the rectangle .
= 2 × x = 2x cm Perimeter of the rectangle
= 2 (length + breadth)
= 2 (2x + x)cm
= 2 × 3x = 6x
By problem,
6x = 48
∴ x = \(\frac{48}{6}\) = 8
Breadth of the rectangle = 8 cm
Length of the rectangle = 2 × 8 = 16 cm

Question 8.
Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is the length of third side ?
Answer:
Two sides of a triangle are 12 cm and 14 cm.
Length of third side = x cm .
Perimeter of the triangle
= sum of the three sides = 12 + 14 + x = (26 + x) cm
By problem,
(26 + x) cm = 36 cm
∴ x = 36 – 26 = 10 cm
Length of third side of the triangle
= 10 cm.

Question 9.
Find the perimeter of each of the following shapes:
(i) A triangle of sides 3 cm., 4 cm. and 5 cm.
(ii) An equilateral triangle of side 9 cm.
(iii) An isosceles triangle with equal sides 8 cm each and third side of 6 cm.
Answer:
(i) The sides of a triangle are 3 cm, 4 cm, 5 cm
Perimeter of the given triangle
= sum of the three sides = (3 + 4 + 5)cm
= 12 cm

(ii) Side of the equilateral triangle = 9 cm
Perimeter of the equilateral triangle = (9 + 9 + 9)cm
= 27 cm (∵ All the 3 sides are equal)

(iii) The length of one of the equal sides of the isosceles triangle is 8 cm.
Length of the third side is 6 cm.
Perimeter of the given triangle
= (8 + 8 + 6) cm
= 22 cm

Students can practice Telangana 6th Class Maths Textbook Solutions Chapter 8 Data Handling InText Questions to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 8 Data Handling InText Questions

Try This

Question 1.
Give two examples of data in numerical figures.
Answer:
In a class of 36 students 15 like mangoes, 12 like apples, 9 like bananas.

Total number of students = 36

Question 2.
Give two examples of data in words.
Answer:
In a village of 100 families 45 read Eenadu, 35 read Sakshi, 20 read Andra Jyothi news papers.

Think, Discuss And Write

Question 1.
In what way Is the bar graph better than the pictograph?
Answer:

• Pictographs are difficult and time consuming to construct. But bargraphs easy to draw.
• In pictographs, each pictorial symbol represents a fixcd number of units. So, it becomes a problem to represent and read any fraction of that unit. For example suppose each pictorial symbol represents 50 units. Then 231 units are represented by 4 full pictures and a proportional fraction of the fifth. This proportional representation introduces error and is quite different to guess the correct value.
• But each bar represents only one value. The length or height of the bar indicates the value of the item. So, we represent correct value with bargraphs.
• Thats why hargraph is better than pictograph.