Class 6

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 TS 6th Class Maths Solutions Chapter 3 Playing with Numbers InText Questions to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 3 Playing with Numbers Exercise InText Questions

Do This

Question 1.
Are 953, 9534, 900, 452 divisible by 2? Also check by actual division.
Answer:
(i) The given number is 953.
This number is divisible by 2 because it has not any one of the digits 0, 2, 4, 6 or 8 in its units place.

(ii) The given number is 9534.
This number is divisible by 2 because it has 4 in its units place.

(iii) The given number is 900.
This number is divisible by 2 because it has ‘0’ in its units place.

(iv) The given number is 452.
This number is divisible by 2 because it has 2 in its units place.

Do This

Check whether the following numbers are divisible by 3 ?
(i) 45986
Answer:
The given number is 45986.
Sum of the digits = 4 + 5 + 9 + 8 + 6 = 32
32 is not a multiple of 3.
So the given number is not divisible by 3.

(ii) 36129
Answer:
The given number is 36129.
Sum of the digits = 3 + 6 + 1 + 2 + 9 = 21
21 is a multiple of 3.
So the given number is divisible by 3.

(iii) 7874
Answer:
The given number is 7874.
Sum of the digits = 7 + 8 + 7 + 4 = 26
26 is not a multiple of 3.
So the given number is not divisible by 3.

Try These

Question 1.
Is 7224 divisible by 6 ? Why ?
Answer:
The given number has 4 in its units place.
So it is divisible by 2.
The sum of the digits of the given number is 7 + 2 + 2 + 4 = 15.
It is a multiple of 3.
So the given number is divisible by 3.
If a number is divisible by both 2 and 3 then it is divisible by 6 also.

Question 2.
Give two examples of 4 digit numbers which are divisible by 6.
Answer:
9648 and 3756.

Question 3.
Can you give an example of a number which is divisible by 6 but not by 2 and 3, why ?
Answer:
A number is divisible by 6 only when it is divisible by 2 and 3.
So, it is not possible to give an example for such number.

Do This

Question 1.
Test whether 9846 is divisible by 9 ?
Answer:
Number = 9846
Sum of the digits = 9 + 8 + 4 + 6 = 27 27
\(\frac{27}{9}\) = 3 9
∴ 9846 is divisible 9.

Question 2.
Without acutal division, find whether 8998794 is divisible by 9 ?
Answer:
Number = 8998794
Sum of the digits = 8 + 9 + 9 + 8 + 7 + 9 + 4 = 54
\(\frac{54}{9}\) = 6
∴ 8998794 is divisible by 9.

Question 3.
Check whether 786 is divisible by both 3 and 9 ?
Answer:
Number = 786
Sum of the digits = 7 + 8 + 6 = 21
\(\frac{21}{3}\) = 7
So 786 is divisible by 9.
But 21 is not divisible by 9.
So 786 is not divisible by 9.

Do This

Question 1.
Find the factors of 80.
Answer:
Factors of 80 are 1, 2, 4, 5, 8, 10, 16, 20, 40, 80.

Question 2.
Do all the factors of a given number divide the number exactly ? Find the factors of 28 and verify by division.
Answer:
Yes, all the factors of a given number divide the number exactly.
Factors of 28 are 1, 2, 4, 7, 14, 28.

Verification:
\(\frac{28}{28}\) = 1, \(\frac{28}{14}\) = 2, \(\frac{28}{7}\) = 4, \(\frac{28}{4}\) = 7, \(\frac{28}{2}\) = 14

Question 3.
3 is a factor of 15 and 24. Is 3 a factor of their difference also ?
Answer:
Yes. (The difference between 15 and 24 is 9 and 9 is a multiple of 3.)

Try These

Question 1.
What is the smallest prime number?
Answer:
The smallest prime number is 2.

Question 2.
What is the smallest composite number?
Answer:
The smallest composite number is 4.

Question 3.
What is the smallest odd composite number?
Sol.
The smallest odd composite number is 6.

Question 4.
Give 5 odd and 5 even composite numbers.
Answer:
The odd composite numbers are
9, 15, 21, 25, 27 etc.
The even composite numbers are
4, 6, 8, 10, 12 etc.

Question 5.
Is 1 prime or composite and why?
Answer:
The number 1 has only one factor i.e. (itself). So, 1 is neither prime nor composite.

Question 6.
Can you guess a prime number which when on reversing Its digits, gives another prime number?
(Hint : Take a 2 digit prime number)
Answer:
13 is a prime number. On reversing its digits it becomes 31, which is also a prime number.

Question 7.
You know 311 is a prime number. Can you find the other two prime numbers just by rearranging the digits?
Answer:
Given prime number is 311.
The other two prime numbers just by rearranging the digits are 113 and 131.

Do This

From the following numbers identify different pairs of co-primes. 2, 3, 4, 5, 6, 7, 8, 9 and 10.
Answer:
(2, 3); (2, 5); (2, 7); (3, 5); (3, 7); (5, 7)

Do This

Question 1.
Write the prime factors of 28 and 36 through division method.
Answer:

∴ Prime factors of 36 is 2 × 2 × 3 × 3

Question 2.
Write the prime factors of 42 by factor tree method.
Answer:

∴ Prime factors of 42 is 2 × 3 × 7

Do This

Find the HCF of 12, 16 and 28 by prime factorization method.
Answer:

The common factors of 12, 16 and 28 is 2 × 2 = 4
Hence, HCF of 12, 16 and 28 is 4.

Do This

Find the HCF of 28, 35 and 49 by division method^
Answer:
First find the HCF of any two numbers.
Let us find the HCF of 28 and 35

Last divisor is 7
∴ HCF of 28 and 35 is 7.
Then find the HCF of the third numbern and the HCF of first two numbers.
Let us find the HCF of 49 and 7.

HCF of 49 and 7 is 7.
∴ The HCF of 28, 35 and 49 is 7.

Think, Discuss and Write

What is the HCF of any two
(i) Consecutive numbers ?
Answer:
The HCF of any two consecutive numbers is 1.

(ii) Consecutive even numbers ?
Answer:
2

(iii) Consecutive odd numbers ?
Answer:
1 (one)

What do you observe ? Discuss with your friends.
Answer:
It was observed that the HCF of two consecutive numbers and consecutive odd numbers is same, i.e., 1.

Try This

Question 1.
find the LCM of
(i) 3, 4
(ii) 10, 11
(iii) 5, 6, 7
(iv) 10, 30
(v) 4, 12, 24
(vi) 3, 12
What do you observe ?
Answer:
(i) LCM of 3 and 4 = 3 × 4 = 12
(ii) LCM of 10 and 11 = 10 × 11 = 110
(iii) LCM of 5, 6 and 7 = 5 × 6 × 7 = 210
(iv) LCM of 10 and 30 = 10 × 1 × 3 = 30
(v) LCM of 4, 12 and 24 = 4 × 3 × 2 = 24
(vi) LCM of 3 and 12 = 3 × 4 = 12
It is observed that the LCM of two numbers will be their product, if the given numbers have no common factor except 1.

Think, Discuss and Write

When will the LCM of two or more numbers be their own product ?
Answer:
If the numbers are co-primes or relatively prime numbers then the LCM of two or more numbers be their own product.

Think, Discuss and Write

Question 1.
What is the LCM and HCF of twin prime numbers?
Answer:
Let the twin primes may be (3, 5)
LCM of 3, 5 is their product 3 × 5 = 15
HCF of 3, 5 is 1.
(for any type of twin prime)

Question 2.
Interpret relationship between LCM and HCF of any two numbers?
Answer:
Consider the two numbers be 14 and 21.
Now find LCM of 14 and 21.

∴ LCM of 14 and 21 = 7 × 2 × 3 = 42
Now find HCF of 14 and 21.

∴ HCF of 14 and 21 is 7.
Relation between LCM and HCF of 14 and 21:
42 × 7 = 14 × 21 = 294
Product of LCM and HCF of two numbers = Product of two numbers.

Do This

Divisibility Rule for 4:

Question 1.
Is 100000 is divisible by 4? Why?
Answer:
100000 = 1000 × 100
The given number is a multiple of 100.
We know, 100 is divisible by 4.
∴ The given number (i.e., 100000) is divisible by 4.

Question 2.
Give an example of a 2 digit number that is divisible by 2 but not divisible by 4?
Answer:
22, 26, 30, 34, 38 98 .
All the above two digit numbers are divisible by 2 but not divisible by 4.

Do This

Question 1.
Is 76104 divisIble by 8?
Answer:
The number formed by the last three digits is 104. It is divisible by 8.
Hence, the given number is divisible by 8.

Question 2.
Write the numbers that are divisible by 8 and lie between 100 and 200.
Answer:
104, 112, 120, 128, 136, 144, 152, 160, 168, 176, 184, 192 are ail divisible by 8.
They lie between 100 and 200.

Page No. 92 (45)

Divisibility Rule for 11:

Using the division rule of ’11’. Fill the following table.

Answer:

1221 is a Palindrome number, which on reversing their digits gives the same number.
Thus, every Palindrome number with even number of digits, is always divisible by 11.

Write a Palindrome number of 6 digits and verify whether it is divisible by 11 or not.
Answer:
Palindrome number which on reversing their digits gives the same number.
Every Palindrome number with even number is always divisible by 11.

∴ The 6 digited Palindrome number is 123321. It is divisible by 11.

Students can practice Telangana 6th Class Maths Textbook Solutions Chapter 5 Measures of Lines and Angles Ex 5.3 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 5 Measures of Lines and Angles Exercise 5.3

Question 1.
Which of the following are models for parallel lines, perpendicular lines and which are neither of them.
(i) The vertical window bars.
(ii) Railway lines (Track)
(iii) The adjacent edges of door.
(iv) The letter ‘V’ in English alphabet.
(v) The opposite edges of Blackboard.
Answer:
(i) For parallel lines —— The vertical window bars.
(ii) For parallel lines —— Railway lines (Track)
(iii) For perpendicular lines —— The adjacent edges of door.
(iv) Neither parallel lines nor perpendicular lines —— The letter ‘V’ in English alphabet.
(v) For parallel lines The opposite edges of Blackboard.

Question 2.
Trace the copy of set squares (Geometry box) on a paper and mark the perpendicular edges.
Answer:

Question 3.
ABCD is a rectangle. \(\overline{\mathbf{A C}}\) and \(\overline{\mathbf{B D}}\) are diagonals. Write the pairs of parallel lines, perpendicular lines from the figure in symbolic form. Also write pairs of intersecting lines.

Answer:
(a) Parallel lines : AB || CD, AD || BC.
(b) Perpendicular lines: AD ⊥ DB, AB ⊥ BC, BC ⊥ CD, CD ⊥ DA
(c) Pair of intersecting lines: AC, BD.

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 Telangana 6th Class Maths Textbook Solutions Chapter 7 Fractions and Decimals Ex 7.3 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 7 Fractions and Decimals Exercise 7.3

Question 1.
Write shaded portion as fraction. Arrange them in ascending or descending order using sign ‘<‘, ‘=’, ‘>’ between the fractions.
(i)

Answer:
\(\frac{3}{8}, \frac{6}{8}, \frac{4}{8}, \frac{1}{8}\); Arranging them in ascending order, we get \(\frac{1}{8}<\frac{3}{8}<\frac{4}{8}<\frac{6}{8}\)
Descending order: \(\frac{6}{8}>\frac{4}{8}>\frac{3}{8}>\frac{1}{8}\)

(ii)

Answer:
\(\frac{8}{9}, \frac{4}{9}, \frac{3}{9}, \frac{6}{9}\); Arranging them in ascending order, we get \(\frac{3}{9}<\frac{4}{9}<\frac{6}{9}<\frac{8}{9}\)
Descending order: \(\frac{8}{9}>\frac{6}{9}>\frac{4}{9}>\frac{3}{9}\)

Question 2.
Show \(\frac{2}{6}, \frac{4}{6}, \frac{8}{6}, \frac{5}{6}\) and \(\frac{6}{6}\) on the number line. Also arrange them in ascending order.
Answer:

Question 3.
Look at the figures and write ‘<‘ or’>’, ‘=‘ between the given pairs of fractions:

(i) \(\frac{1}{6}\) ________ \(\frac{1}{3}\)
(ii) \(\frac{3}{4}\) ________ \(\frac{2}{6}\)
(iii) \(\frac{2}{3}\) ________ \(\frac{2}{4}\)
(iv) \(\frac{6}{6}\) ________ \(\frac{3}{3}\)
(v) \(\frac{5}{6}\) ________ \(\frac{5}{5}\)
Make five more such problems and ask your friends to solve them.
Answer:
(i) \(\frac{1}{6}\) < \(\frac{1}{3}\)
(ii) \(\frac{3}{4}\) > \(\frac{2}{6}\)
(iii) \(\frac{2}{3}\)> \(\frac{2}{4}\)
(iv) \(\frac{6}{6}\) = \(\frac{3}{3}\)
(v) \(\frac{5}{6}\) < \(\frac{5}{5}\)

Question 4.
Fill with the appropriate sign. (‘<‘, ‘=’, ‘>’)
(i) \(\frac{1}{2}\) _______ \(\frac{1}{5}\)
Answer:
\(\frac{1}{2}\) > \(\frac{1}{5}\)

(ii) \(\frac{2}{4}\) _______ \(\frac{3}{6}\)
Answer:
\(\frac{2}{4}\) = \(\frac{3}{6}\)

(iii) \(\frac{3}{5}\) _______ \(\frac{2}{3}\)
Answer:
\(\frac{3}{5}\) < \(\frac{2}{3}\)

(iv) \(\frac{3}{4}\) _______ \(\frac{2}{8}\)
Answer:
\(\frac{3}{4}\) > \(\frac{2}{8}\)

(v) \(\frac{3}{5}\) _______ \(\frac{6}{5}\)
Answer:
\(\frac{3}{5}\) < \(\frac{6}{5}\)

(vi) \(\frac{7}{9}\) _______ \(\frac{3}{9}\)
Answer:
\(\frac{7}{9}\) > \(\frac{3}{9}\)

Question 5.
Answer the following. Also write how you solved them.
(i) Is \(\frac{5}{9}\) equal to \(\frac{4}{5}\)?
Answer:
We write the fractions \(\frac{5}{9}\) and \(\frac{4}{5}\) having the same denominators

(ii) Is \(\frac{9}{16}\) equal to \(\frac{5}{9}\)?
Answer:
We write the fractions \(\frac{9}{16}\) and \(\frac{5}{9}\) having the same denominators

(iii) Is \(\frac{4}{5}\) equal to \(\frac{16}{20}\)?
Answer:
We write the fractions \(\frac{4}{5}\) and \(\frac{16}{20}\) having the same denominators

(iv) Is \(\frac{1}{15}\) equal to \(\frac{4}{30}\)?
Answer:
We write the fractions \(\frac{1}{15}\) and \(\frac{4}{30}\) having the same denominators

Question 6.
Varshith read 25 pages of a story book containing 100 pages. Lalitha read \(\frac{2}{5}\) of the same story book. Who read less? Give reason.
Answer:
Total number of pages in the story book = 100
Number of pages that Varshith read = 25
Number of pages that Lalitha read
= \(\frac{2}{5}\) of 100
= \(\frac{2}{5}\) × 100 = 40
So, Varshith read less pages.

Question 7.
Fill the appropriate (+ or -) sign in the blank space.
(i)

Answer:
+

(ii)

Answer:
–

(iii)

Answer:
+

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 – \(\frac{2}{3}\)
Answer:
1 – \(\frac{2}{3}=\frac{3}{3}-\frac{2}{3}=\frac{3-2}{3}=\frac{1}{3}\) [∵ 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 – \(\frac{2}{15}\)
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 \(\) 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 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
= \(\frac{11}{5}\)
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.
(∵ \(\frac{44}{20}-\frac{35}{20}=\frac{44-35}{20}=\frac{9}{20}\))

Students can practice Telangana 6th Class Maths Textbook Solutions Chapter 7 Fractions and Decimals Ex 7.2 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 7 Fractions and Decimals Exercise 7.2

Question 1.
Which group of fractions are like fractions among the following?
(i) \(\frac{2}{7}, \frac{3}{7}, \frac{4}{7}\)
Answer:
\(\frac{2}{7}, \frac{3}{7}, \frac{4}{7}\) are like fractions

(ii) \(\frac{1}{9}, \frac{2}{9}, \frac{4}{9}\)
Answer:
\(\frac{1}{9}, \frac{2}{9}, \frac{4}{9}\) are also like fractions

(iii) \(\frac{3}{7}, \frac{4}{9}, \frac{7}{11}\)
Answer:
\(\frac{3}{7}, \frac{4}{9}, \frac{7}{11}\) are unlike fractions

Question 2.
Write five groups of like fractions.
Answer:

Question 3.
From each of these, Identify like fractional numbers.
(i) \(\frac{2}{3}, \frac{5}{3}, \frac{1}{3}, \frac{4}{6}\)
Answer:
\(\frac{2}{3}, \frac{5}{3}, \frac{1}{3}, \frac{4}{6}\left(=\frac{2}{3}\right)\) are like fractional numbers

(ii) \(\frac{1}{7}, \frac{3}{5}, \frac{2}{5}, \frac{1}{9}\)
Answer:
\(\frac{3}{5}\) and \(\frac{2}{5}\) are like fractional numbers

(iii) \(\frac{7}{8}, \frac{8}{7}, \frac{2}{8}, \frac{7}{5}\)
Answer:
\(\frac{7}{8}\) and \(\frac{2}{8}\) are like fractional numbers

Students can practice Telangana 6th Class Maths Textbook Solutions Chapter 7 Fractions and Decimals Ex 7.1 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 7 Fractions and Decimals Exercise 7.1

Question 1.
Out of these which are proper fractional numbers?
(i) \(\frac{3}{2}\)
(ii) \(\frac{2}{5}\)
(iii) \(\frac{1}{7}\)
(iv) \(\frac{8}{3}\)
Answer:
\(\frac{2}{5}\) and \(\frac{1}{7}\) are proper fractional numbers.

Question 2.
Which of these are improper fractional numbers?
(i) \(\frac{2}{7}\)
(ii) \(\frac{7}{11}\)
(iii) \(\frac{9}{7}\)
(iv) \(\frac{13}{2}\)
(v) \(\frac{7}{3}\)
Write where each of the above improper fractional numbers would lie on the number line.
Answer:
\(\frac{7}{3}\) and \(\frac{13}{2}\) are improper fractional numbers.
\(\frac{7}{3}\) lies in between 2 and 3.

\(\frac{13}{2}\) lies in between 6 and 7.

Question 3.
Pick out the mixed fractions from these.
(i) \(\frac{3}{5}\)
(ii) 1\(\frac{2}{7}\)
(iii) \(\frac{7}{2}\)
(iv) 2\(\frac{3}{5}\)
Answer:
1\(\frac{2}{7}\) and 2\(\frac{3}{5}\) are mixed fractions.

Question 4.
Convert the following Improper fractions Into mixed fractions.
(i) \(\frac{7}{3}\)
(ii) \(\frac{11}{2}\)
(iii) \(\frac{9}{4}\)
(iv) \(\frac{27}{4}\)
Answer:

Question 5.
Convert the following mixed fractions into Improper fractions.
(i) 1\(\frac{2}{7}\)
(ii) 3\(\frac{2}{8}\)
(iii) 10\(\frac{2}{9}\)
(iv) 8\(\frac{7}{9}\)
Answer:

Students can practice Telangana 6th Class Maths Textbook Solutions Chapter 5 Measures of Lines and Angles Ex 5.2 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 5 Measures of Lines and Angles Exercise 5.2

Question 1.
Write ‘True’ or ‘False’. Correct all those that are false.
(i) An angle smaller than right angle is acute angle. ___________
Answer:
True

(ii) A right angle measures 180°. ___________
Answer:
False
A right angle measures 90°. (correct)

(iii) A straight angle measures 90°. ___________
Answer:
False
A straight angle measures 180°. (correct)

(iv) The measure greater than 180° is a reflex angle. ___________
Answer:
True

(v) A complete angle measures 360°. ___________
Answer:
True

Question 2.
Which angles in the adjacent figure are acute and which are obtuse ? Check your estimation by measuring them. Write their measures too.

Answer:
In the adjacent figure The measure of ∠1 is 80°
The measure of ∠2 is 100°
The measure of ∠3 is 80°
The measure of ∠4 is 100°
∠l and ∠3 are acute angles because they are less than 90°.
∠2 and ∠4 are obtuse angles because they are greater than 90° and less than 180°.

Question 3.
What is the measure of these angles ? Which is the largest angle ? Draw an angle larger than the largest angle.

Answer:
∠ABC = 70°
∠FED = 120°
∠RQP = 90°
∠FED is the largest angle.

An angle larger than the largest ∠DEF is ∠STU = 150°

Question 4.
Write the type of angle formed between the long hand and short hand of a clock at the given timings. (Take the small hand as the base)
(i) At 9’0 clock in the morning
(ii) At 6’0 clock in the evening
(iii) At 12 noon
(iv) At 4’0 clock in the afternoon
(v) At 8’0 clock in the night.
Answer:
The angle formed between the long hand and short hand of a clock.
(i) At 9’O clock in the morning is right angle.
(ii) At 6’O clock in the evening is straight angle.
(iii) At 12 noon no angle is formed because the two hands coincide.
(iv) At 4’O clock in the afternoon is obtuse angle.
(v) At 8’O clock in the night is reflex angle.

Question 5.
Match the angles by measure. Draw figures for these as well.

Answer:

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.