Author name: Prasanna

Haryana Board HBSE 8th Class Sanskrit Solutions रुचिरा भाग 3

• Chapter 1 सुभाषितानि
• Chapter 2 बिलस्य वाणी न कदापि में श्रुता
• Chapter 3 डिजीभारतम्
• Chapter 4 सदैव पुरतो निधेहि चरणम्
• Chapter 5 कण्टकेनैव कण्टकम्
• Chapter 6 गृहं शून्यं सुतां विना
• Chapter 7 भारतजनताऽहम्
• Chapter 8 संसारसागरस्य नायकाः
• Chapter 9 सप्तभगिन्यः
• Chapter 10 नीतिनवनीतम्
• Chapter 11 सावित्री बाई फुले
• Chapter 12 कः रक्षति कः रक्षितः
• Chapter 13 क्षितौ राजते भारतस्वर्णभूमिः
• Chapter 14 आर्यभटः
• Chapter 15 प्रहेलिकाः

Haryana State Board HBSE 6th Class Maths Solutions Chapter 8 Decimals Ex 8.2 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 8 Decimals Exercise 8.2

Question 1.
Complete the table with the help of these boxes and use decimals to write the number :

Solution:

Question 2.
Write the numbers given in the following place value table in decimal form:

Solution:
(i) The number is 3 x 1 + 2 x \(\frac{1}{10}\) + 5 x \(\frac{1}{100}\) = 3 + 0.2 + 0.05 = 3.25.
(ii) Thenumberis 1 x 100 + 0 x 10 + 2 x 1 + 6 x \(\frac{1}{10}\) + 3 x \(\frac{1}{100}\)
= 100 + O + 2 + 0.6 + 0.03 = 102.63.
(iii) The number is 3 x 10 + 0 x 1 + 0 x \(\frac{1}{10}\) + 2 x \(\frac{1}{100}\) + 5 x \(\frac{1}{1000}\)
= 30 + 0 + 0 + 0.02 + 0.005 = 30.025.
(iv) The numberis 2 x 100+ 1 x 10 + 1 x 1 + 9 x \(\frac{1}{10}\) + 0 x \(\frac{1}{100}\) + 2 x \(\frac{1}{1000}\)
= 200 + 10 + 1 + 0.9 + 0.00 + .002 = 211.902.
(v) The number is 1 x 10 + 2 x 1 + 2 x \(\frac{1}{10}\) + 4 x \(\frac{1}{100}\) + 1 x \(\frac{1}{1000}\)
= 10 + 2 + 0.2 + 0.04 + 0.001 = 12.241.

Question 3.
Write the following decimals in the place value table :
(a) 0.29, (b) 2.08, (c) 19.60, (d) 148.32, (e) 200.812.
Solution:

Question 4.
Write each of the following as decimals :

Solution:

Question 5.
Write each of the following decimals in words :
(a) 0.03
(b) 1.20
(c) 17.38
(d) 108.56
(e) 10.07
(f) 210.109
(g) 0.032
(h) 5.008.
Solution:
(a) 0.03 = Zero point zero three or three hundredths.
(b) 1.20 = One point two zero.
(c) 17.38 = Seventeen point three eight.
(d) 108.56 = One hundred eight point five six.
(e) 10.07 = Ten point zero seven.
(f) 210.109 = Two hundred ten point one zero nine.
(g) 0.032 = Zero point zero three two.
(h) 5.008 = Five point zero zero eight.

Question 6.
Between which two points numbers on the number line does each of the given number lie ?
(a) 0.06
(b) 0.45
(c) 0.19
(d) 0.66
(e) 0.92
(f) 0.57
(g) 0.03
(h) 0.20.
Solution:
All the given numbers lie between whole number 0 and 1.
(a) 0 and 0.1, nearer to the number 0.1.
(b) 0.4 and 0.5, nearer to the number 0.5.
(c) 0.1 and 0.2, nearer to the number 0.2.
(d) 0.6 and 0.7, nearer to the number 0.7.
(e) 0.9 and 1.0, nearer to the number 0.9.
(f) 0.5 and 0.6, nearer to the number 0.6.
(g) 0.0 and 0.1, nearer to the number 0.0.
(h) 0.1 and 0.3, nearer to the number 0.2.

Question 7.
Write as fractions in the lowest terms :

(a) 0.60
(b) 0.05
(c) 0.75
(d) 0.18
(f) 0.82
(g) 0.004
(h) 0.125.
(i) 0.066.
Solution:

Haryana State Board HBSE 6th Class Maths Solutions Chapter 8 Decimals Ex 8.1 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 8 Decimals Exercise 8.1

Question 1.

Solution:

Question 2.
Write the following decimals in the (a) 19.4 (6) 0.3 (c) 10.6 (d) 205.9
Solution:

Question 3.
Write each of the following as decimals:
(a) 7 tenths
(b) Two tens, 9 tenths
(c) Fourteen point six
(d) One Hundred and 2 ones
(e) Six hundred point eight.
Solution:
(a) 7 tenths = \(\frac{7}{10}\) = 0.7
(b) Two tens, 9 tenths = 20 + \(\frac{9}{10}\) = 20.9.
(c) Fourteen point six = 14.6.
(d) Hundred and two ones = 102.0.
(e) Six hundred point eight = 600.8.

Question 4.
Write each of the following as decimals:

Solution:

Question 5.
Write the following decimals as fractions. Reduce the fractions to their lowest forms : (a) 0.6, (b) 2.5, (c) 1.0, (d) 3.8, (e) 13.7, (f) 21.2, (g) 6.4.
Solution:
(a) 0.6 = \(\frac{6}{10}=\frac{3}{5}\)
(b) 2.5 = \(\frac{25}{10}=\frac{5}{2}\)
(c) 1.0 = 1.
(d) 3.8 = \(\frac{38}{10}=\frac{19}{5}\)
(e) 13.7 = \(\frac{137}{10}\)
(f) 21.2 = \(\frac{212}{10}=\frac{106}{5}\)
(g) 6.4 = \(\frac{64}{10}=\frac{32}{5}\)

Question 6.
Express the following as cm using decimals :
(a) 2 mm
(b) 30 mm
(c) 116 mm
(d) 4 cm 2 mm
(e) 162 mm
(f) 83 mm.
Solution:
(a) 2 mm = \(\frac{2}{10}\) = 0.2 cm.
(b) 30 mm = \(\frac{30}{10}\) = 3.0 cm
(c) 116 mm = \(\frac{116}{10}\) = 11.6 cm
(d) 4 cm 2 mm = 4 +\(\frac{2}{10}\) = 4.2 cm
(e) 162mm = 11 + \(\frac{52}{10}\) = 11 + 5.2 = 16.2 cm.
(f) 83 mm =\(\frac{83}{10}\) = 8.3 cm.

Question 7.
Between which two whole numbers on the number line are the given numbers ? Which one is nearer the number ?
(a) 0.8, (b) 5.1, (c) 2.6, (d) 6.4, (e) 9.0, (f) 4.9.

Solution:
(a) 0-1, and nearer to the number 1.
(b) 5-6, and nearer to the number 5.
(c) 2-3, and nearer to the number 3.
(d) 6-7, and nearer to the number 6.
(e) 8-10, and nearer to the number 9.
(f) 4-5, and nearer to the number 5.

Question 8.
Show the following decimal numbers on the number line :
(a) 0.2,
(b) 1.9,
(c) 1.1,
(d) 2.5.
Solution:

Question 9.
Write the decimal number represented by the points A, B, C and D on the given number line :

Solution:
(i) Point A represents the number 0.8.
(ii) Point B represents the number 1.3
(iii) Point C represents the number 2.2
(iv) Point D represents the number 2.9.

Question 10.
(a) The length of Ramesh’s notebook is 9 cm 5 mm. What will be its length in cm ?
(b) The length of a young gram plant is 65 mm. Express its length in cm.
Solution:
(a) 9 cm 5 mm = 9 + \(\frac{5}{10}\) = 9.5 cm (b) 65 mm = \(\frac{65}{10}\) = 6.5 cm.

Haryana State Board HBSE 6th Class Maths Solutions Chapter 7 Fractions Ex 7.4 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 7 Fractions Exercise 7.4

Question 1.
Write each fraction. Arrange them in ascending and descending order using correct sign ‘<‘, ‘=’,‘>‘ between the fractions:

Solution:

Question 2.
Compare the fractions and put an appropriate sign.

Solution:

Question 3.
Make 5 more such pairs and make appropriate signs.
Solution:

Question 4.
Name the fractions and arrange them in ascending order:

Solution:

Question 5.
Look at the figures and write ‘<‘ or ‘>’, ‘=’ between the pairs of fractions.

Make 5 more such problems and solve them with your friends.
Solution:

Question 6.
How quickly can you compare these? Fill appropriate signs.

Solution:

Question 7.
The following fractions represent just three different numbers. Separate them into three grups of equal fractions by changing each one to its simplest fraction.

Solution:

Question 8.
Find answers to the following. Write and indicate how you solved them:
(a) Is \(\frac{5}{9}\) equal to\(\frac{4}{5}\) ?
(b) Is \(\frac{9}{16}\) equal to \(\frac{5}{9}\) ?
(e) Is \(\frac{4}{5}\) equal to \(\frac{16}{20}\)?
(d) Is \(\frac{1}{15}\) equal to \(\frac{4}{30}\)?
Solution:
(a) \(\frac{5}{9} \neq \frac{4}{5}\) because 5 x 5 ≠ 9 x 4 i.e., 25 ≠ 36
(b) \(\frac{9}{16} \neq \frac{5}{9}\) because 9 x 9 ≠ 16 x 5 i.e., 81 ≠ 80
(c) \(\frac{4}{5}=\frac{16}{20}\) because 4 x 20 = 5 x 16 i.e., 80 = 80
(d) \(\frac{1}{15} \neq \frac{4}{30}\) because 1 x 30 ≠ 15 x 4 i.e., 30 ≠ 60.

Question 9.
Ila read 25 pages of a book containing 100 pages. Lalita read \(\frac{1}{2}\) of the same book. Who read less ?
Solution:
Pages read by Ila = 25
Pages read by Lalita = \(\frac{1}{2}\) x 100 = 50
∵ 25 < 50
∴ Ila read less pages.

Question 10.
Rafiq exercised for \(\frac{3}{6}\) of an hour, while Rohit exercised for \(\frac{3}{4}\) of an hour.
Who exercised for a longer time ?
Solution:
Rafiq exercised for \(\frac{3}{6}=\frac{1}{2}\) of an hour = \(\frac{1}{2}\) x 60 = 30 min.
Rohit exercised for \(\frac{3}{4}\) of an hour = \(\frac{3}{4}\) x 60 = 45 min.
∴ Rohit exercised for a longer time.

Question 11.
In a class A of 25 students 20 passed in first class; in another class B of 30 students, 24 passd in first class. In which class were there more number of students getting first class ?
Solution:
Fraction of the students passed in first class in class A = \(\frac{20}{25}=\frac{4}{5}\)
Fraction of the students passed in first class in class B = \(\frac{24}{30}=\frac{4}{5}\)
∴ Equal number of students got first class in both the classes.

Haryana State Board HBSE 6th Class Maths Solutions Chapter 7 Fractions Ex 7.5 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 7 Fractions Exercise 7.5

Question 1.
Write these fractions appropriately as additions or subractions:

Solution:

Question 2.
Complete the statements given with each figure.

Solution:

Question 3.

Solution:

Question 4.
Shubham painted 2/3 of the wall space in his room. His sister Madhavi helped and painted 1/3 of the wall space. How much did they paint together ?
Solution:
Shubham painted 2/3 of the wall space in his room.
Madhavi painted 1/3 of the wall space.
∴ The part they painted together = \(\frac{2}{3}+\frac{1}{3}=\frac{2+1}{3}=\frac{3}{3}\) = 1
Hence, they painted the complete wall.

Question 5.
Kamlesh bought 3\(\frac{1}{2}\) kg of sugar whereas Anwar bought 2\(\frac{1}{2}\) kg of sugar. Find the total amount of sugar bought by both of them.
Answer:
Amount of sugar bought by Kamlesh
= 3\(\frac{1}{2}\)kg
Amount of sugar bought by Anwar
= 2\(\frac{1}{2}\)kg
Total amount of sugar bought by both of them
= \(3 \frac{1}{2}+2 \frac{1}{2}=\frac{7}{2}+\frac{5}{2}[/atex]
= [latex]\frac{7+5}{2}=\frac{12}{2}\) = 6 kg

Question 6.
Fill in the missing fractions :


Solution:

Question 7.
The teacher taught 3/5 of the book. Mahesh revised 1/5 more on his own. How much does he still have to revise ?
Solution:
Teacher taught \(\frac{3}{5}\) of the book.
Mahesh revised \(\frac{1}{5}\) more on his own.
∴ Total part of the book revised by Mahesh
\(\frac{3}{5}+\frac{1}{5}=\frac{3+1}{5}=\frac{4}{5}\)
∴ He still have to revise
= \(1-\frac{4}{5}=\frac{5-4}{5}=\frac{1}{5}\)

Question 8.
Javed was given 5/7 of a bucket of oranges. What fraction of oranges was left in the basket ?
Solution:
∴ Javed was given \(\frac{5}{7}\)of oranges.
∵ Fraction of oranges left in the basket
= \(1-\frac{5}{7}=\frac{7-5}{7}=\frac{2}{7}\)

Haryana State Board HBSE 6th Class Maths Solutions Chapter 7 Fractions Ex 7.6 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 7 Fractions Exercise 7.6

Question 1.
(a) Solve :

Solution:

Question 2.
Sarita bought \(\frac{2}{5}\) m of ribbon and Lalita \(\frac{3}{4}\) m of ribbon. What was the total length of the ribbon they bought ?
Solution:
Ribbon bought by Sarita = \(\frac{2}{5}\) m
Ribbon bought by Lalita = \(\frac{3}{4}\) m
Total length of ribbon = \(\frac{2}{5}+\frac{3}{4}=\frac{2 \times 4+3 \times 5}{20}\)
= \(\frac{8+15}{20}=\frac{23}{20}=1 \frac{3}{20} \mathrm{~m}\)

Question 3.
Naina was given 1\(\frac{1}{2}\) piece of cake and Najma was given 1\(\frac{1}{3}\) piece of cake.
Find the total amount of cake given to both of them.
Solution:
Cake given to Naina = 1\(\frac{1}{2}\) = \(\frac{3}{2}\)
Cake giveen to Najma = 1\(\frac{1}{3}\) = \(\frac{4}{3}\)
∴ Total amount of cake given to both of them
= \(\frac{3}{2}+\frac{4}{3}=\frac{3 \times 3+2 \times 4}{6}\)
= \(\frac{9+8}{6}=\frac{17}{6}=2 \frac{5}{6}\)

Question 4.
Fill in the boxes:

Solution:

Question 5.
Complete the addition, subtraction box.

Solution:

Question 6.
A piece of wire 7/8 metre long broke into two pieces. One piece was 1/4 metre long. How long is the other piece ?
Solution:
Total length of a piece of wire = \(\frac{7}{8}\) m
Length of one piece = \(\frac{1}{4}\) m
∴ Length of the other piece = \(\frac{7}{8}-\frac{1}{4}=\frac{7-2}{8}=\frac{5}{8}\)m

Question 7.
Nandni’s house is \(\frac{9}{10}\) km. from her school. She walked some distance and then took a bus for \(\frac{1}{2}\) km upto the school. How far did she walk ?
Solution:
Total distance between house and school = \(\frac{9}{10}\) km
Distance covered by bus = \(\frac{1}{2}\) km
∴ Distance walked by her = \(\frac{9}{10}-\frac{1}{2}=\frac{9-5}{10}\)
= \(\frac{4}{10}=\frac{2}{5}\) km.

Question 8.
Asha and Samuel have bookshelves of the same size. Asha’s shelf is \(\frac{5}{6}\) full of books and Samuel’s shelf is \(\frac{2}{5}\) full. Whose bookshelf is more full ? By what fraction?
Answer:
Asha’s shelf is \(\frac{5}{6}\) full of books,
Samuel’s shelf is \(\frac{2}{5}\) full of books
Now, \(\frac{5}{6}-\frac{2}{5}=\frac{5 \times 5-2 \times 6}{30}=\frac{25-12}{30}=\frac{13}{30}\)
∴ Asha’s bookshelf is more full by \(\frac{13}{30}\) fraction.

Question 9.
Jaydev takes 2\(\frac{1}{5}\) minutes to walk across the school ground. Rahul takes \(\frac{7}{4}\) minutes to do the same. Who takes less time and by what fraction ?
Solution:
Time taken by Jaydev to walk across the school ground = \(2 \frac{1}{5}=\frac{11}{5}\) minutes
Time taken by Rahul to do the same = \(\frac{7}{4}\) minutes
Now, \(\frac{11}{5}-\frac{7}{4}=\frac{11 \times 4-5 \times 7}{20}=\frac{44-35}{20}=\frac{9}{20}\)
∴ Rahul takes less time by \(\frac{9}{20}\) fraction.

Haryana State Board HBSE 6th Class Maths Solutions Chapter 7 Fractions Ex 7.3 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 7 Fractions Exercise 7.3

Question 1.
Write the fractions. Are all these fractions equivalent ?

Solution:
(a) \(\frac{1}{2}, \frac{2}{4}, \frac{3}{6}, \frac{4}{8}\). All these fractions are equivalent.
(b) ,\(\frac{4}{12}, \frac{3}{9}, \frac{2}{6}, \frac{1}{3}, \frac{6}{15}\). First four fractions are equivalent. All fractions are not equivalent. 14 a bo 15

Question 2.
Write the fractions and pair up the equivalent fraction from each row A and B.

Solution:

Question 3.
Replace in each of the following by the correct number:

Solution:

Question 4.
Find the equivalent fraction of \(\frac{3}{5}\) having
(a) denominator 20
(b) numarator 9
(c) denominator 30
(d) numarator 27
Solution:

Question 5.
Find the equivalent fraction of \(\frac{36}{48}\) with
(a) numarator 9
(b) denominator 4
Solution:

Question 6.
Check whether the given fractions are equivalent :
(a) \(\frac{5}{9}, \frac{30}{54}\)
(b) \(\frac{3}{10}, \frac{12}{50}\)
(c) \(\frac{7}{13}, \frac{5}{11}\)
Solution:

Question 7.
Reduce the following fractions to their simplest forms :
(a) \(\frac{48}{60}\)
(b) \(\frac{150}{60}\)
(c) \(\frac{84}{98}\)
(d) \(\frac{12}{52}\)
(e) \(\frac{7}{28}\)
Solution:

Question 8.
Ramesh had 20 pencils, Sheelu had 50 pencils and Jamaal had 80 pencils. After 4 months, Ramesh used up 10 pencils, Sheelu used up 25 pencils, and Jamaal used up 40 pencils. What fraction did each use up ? Check if each has used up an equal fraction of their pencils ?
Solution:
∵ Ramesh had 20 pencils and he used up 10 pencils.
Fraction of pencils used by Ramesh = \(\frac{10}{20}=\frac{1}{2}\)
∵ Sheelu had 50 pencils and she used up 25 pencils.
∴ Fraction of pencils used by Sheelu = \(\frac{25}{50}=\frac{1}{2}\)
∵ Jamaal had 80 pencils and he used up 40 pencils.
∴ Fraction of pencils used up by Jamaal = \(\frac{40}{80}=\frac{1}{2}\)
Yes, each had used up an equal fraction of their pencils.

Question 9.
Match the equivalent fractions and write other 2 for each :

Solution:
Solution:
To reduce \(\frac{250}{400}\) we find the H .C.F. of 250 and 400.
We have

∴ 250 = 2 x 5 x 5 x 5
and 400 = 2 x 2 x 2 x 2 x 5 x 5
∴ H.C.F. of 250 and 400
= 2 x 5 x 5 = 50
\(\frac{250}{400}=\frac{250 \div 50}{400 \div 50}=\frac{5}{8}\)
Thus (i) is matched to (d).

(ii) To reduce \(\frac{180}{200}\) we find the HCF of 180 and 200.
We have

∴ 180 = 2 x 2 x 3 x 3 x 5
and 200 = 2 x 2 x 2 x 5 x 5
∴ HCF of 180 and 200
= 2 x 2 x 5 = 20
\(\frac{180}{200}=\frac{180 \div 20}{200 \div 20}=\frac{9}{10}\)
Thus, (ii) is matched to (e).

(iii) To reduce \(\frac{660}{990}\), we find the HCF of
660 and 990.
We have

∴ 660 = 2 x 2 x 3 x 5 x 11
and 990 = 2 x 3 x 3 x 5 x 11
∴ HCF of 660 and 990
= 2 x 3 x 5 x 11 = 330
\(\frac{660}{990}=\frac{660 \div 330}{990 \div 330}=\frac{2}{3}\)
Thus, (iii) is matched to (a).

(iv) To reduce \(\frac{180}{360}\), we find the HCF of 180 and 360.
∴ 180 = 180
360 = 180 x 2
∴ HCF of 180 and 360 = 180
\(\)
Thus, (iv) is matched to (c).

(v) To reduce \(\frac{220}{550}\), we find the HCF of 220
and 550 We have
220 = 550
and 220 = 2 x 110
HCF of 220 and 550 = 110
\(\)
Thus, (v) is matched to (b).

Haryana State Board HBSE 6th Class Maths Solutions Chapter 7 Fractions Ex 7.2 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 7 Fractions Exercise 7.2

Question 1.
Draw number lines and place the points on them.

Solution:

Question 2.
Express the following as mixed fractions :

Solution:

Question 3.
Express the following as improper fractions:

Solution:

Haryana State Board HBSE 6th Class Maths Solutions Chapter 7 Fractions Ex 7.1 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 7 Fractions Exercise 7.1

Question 1.
Give the fraction representing the shaded partion?

Solution:

Question 2.
Colour the part according to the fraction given

Solution:

Question 3.
Identify the error, if any ?

Solution:
(i) This is not \(\frac{1}{2}\), because the two parts are not equal.
(ii) This is not \(\frac{1}{4}\) , because the four parts are not equal.
(iii) This is not \(\frac{3}{4}\), because the four parts are not equal.

Question 4.
What fraction of a day is 8 hours ?
Solution:
∵ A day has 24 hours
∴ 8 hours = \(\frac{8}{24}=\frac{1}{3}\)

Question 5.
What fraction of an hour is 40 minutes ?
Solution:
∵ An hour has 60 minutes
∴ 40 minutes = \(\frac{40}{60}=\frac{2}{3}\)

Question 6.
Arya, Abhimanyu and Vivek share for lunch. Arya brought two sandwiches, one made of vegetable and one of jam. The other two boys forget theirs lunch. Arya agree to share his sandwiches so that each person will have an equal shares of each sandwich.
(a) How can Arya divide his sandwiches so that each person has an equal share ?
(b) What part of a sandwich will each boy receive ?
Solution:
(a) Arya can divide each sandwich into three equal parts, total 6 parts.
(b) Each boy will receive l/3rd part of each sandwich i.e. each gets 2 parts.

Question 7.
Kanchan has three frocks that she wears while playing. The material is good, but the colours are faded. Her mother buys same blue dye and uses it on two of the frocks. What fraction of Kanchan’s playing frocks did her mother dye ?
Solution:
Her mother dyed 2/3rd of Kanchan’s playing frocks.

Question 8.
Write the natural numbers from numbers ?
Solution:

Prime numbers are : 2, 3, 5, 7, 11 i.e., 5.
Hence, \(\frac{5}{11}\) fraction are prime numbers.

Question 9.
Write the natural numbers from 102 to 103 numbers ? What fraction of them are prime numbers?
Solution:

Prime numbers are : 103, 107, 109, 113 i.e., 4
Total natural numbers are 12.
Hence, \(\frac{4}{12}=\frac{1}{3}\) fraction are prime numbers.

Question 10.
What fraction of these circles have X’s in them ?

Solution:
(a) \(\frac{5}{8}\)
(b) \(\frac{3}{8}\)
(c) \(\frac{4}{8}\)

Question 11.
Dinesh, Sumit, Ram, Joy, Marshal, Imran, Jayant, Babu, Kabir and Rohan decide to play basketball. The first five boys are of one team and the rest are of the other team.
(a) What fraction of the boys is of the first team ?
(b) What, does this fraction mean ? Are equal parts involved ? If so, equal in what sense?
Solution:
(a) ∵ There are ten boys in all. and no. of boys in the first team = 5
∴ Fraction of the boys in the first team
= \(\frac{5}{10}=\frac{1}{2}\)

(b) Fraction \(\frac{1}{2}\) means that half the boys are in the first team. Yes, equal parts are involved.
This shows that five boys are in the first team and five boys are in the second team.

Question 12.
Kristin receive a C.D. Player for her birthday. She has been collecting C.D’s since then. She bought 3 C.D’s and received 5 others as gifts. What fraction of her total C.D’s did she buy ?
Solution:
Total number of C.D’s she has = 3 + 5 = 8
Number of C.D’s bought by her = 3
Hence, the required fraction of C.D’s she bought
= \(\frac{3}{8}\)

Haryana State Board HBSE 6th Class Maths Solutions Chapter 6 Integers InText Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 6 Integers InText Questions

Try These (Page 148)

Question 1.
Write the following numbers with appropriate signs :
(a) 100 m below sea level (b) 25°C above 0° temperature
(c) 15°C below 0° temperature (d) 5 number less than 0.
Solution:
(a) +100 m, (b) +25°C, (c) -15°C, (d) -5.

Try These (Page 151)

Question 1.
Mark -3, 7, -4, -8,1 and -5 on the number line. Can you tell which is the largest integer amongst these integers ?
Solution:

7 is the largest integer amongst these integers.

Try These (Page 152)

Question 1.
Compare the following pairs of numbers using > or < :

From the above exercise, we arrived at the following conclusions, give examples of each :
(a) Every positive integer is larger than every negative integer.
e.g. 3 > – 1, 5 > -2, 6 > -3 etc.
(b) Zero is less than every positive integer.
e.g. 0 < 2, 0 < 5, 0 < 10 etc.
(c) Zero is larger than every negative integer.
e.g. 0 > -3, 0 > -7, 0 > -15 etc.
(d) Zero is neither a negative integer nor a positive integer.
(e) Farther a number from zero on the right, larger is its value.
e.g. 7 > 5, 8 > 3, 10 > 2 etc.
(f) Farther a number from zero on the left, smaller is its value.
e.g. -5 < -3, -7. < -4, -10 < -6 etc.

Try These (Page 159)

Find the answers of the following additions:

Question 1.
(a) (-11)+ (-12)
(b) (+10) + (+4)
(c) (-32)+ (-25)
(d) (+23) + (+40).
Solution:
(a) (-11) + (-12) = -23
(b) (+10) + (+4) =+14
(c) (-32) + (-25) = -57
(d) (+23) + (+40) = +63.

Try These (Page 160)

Question 1.
Find the solution of the following :
(a) (-7) + (+8)
(b) (-9) + (+13)
(c) (+7) + (-10)
(d) (+12) + (-7).
Solution:
(a) (-7) + (+8) = +1
(b) (-9) + (+13) = +4
(c) (+7) + (-10) = -3
(d) (+12) + (-7) = +5.

Try These (Page 162)

Question 1.
Find the solution of the following additions using a number line :
(a) (-2) + 6
(b) (-6) + 2
Make two such questions and solve them using the number line.
Solution:
(a)

First we move 2 steps to the left of 0 reaching -2 and then from this point we move 6 steps to the right. We reach the point 4. Thus, (-2) + (+6) = 4.

(b)

First we move 6 steps to the left of 0 reaching -6 and then from this point we move 2 steps to the right. We reach the point -4. Thus, (-6) + (+2) = -4.

Question 2.
Find the solution of the following additions using a number line :
(i) (-4) + (+3)
(ii) (+4) + (-3)
Solution:
(i)

First we move 4 steps to the right of 0 reaching 4 and then from this point we move 3 steps to the right. We reach the point -1. Thus, (-4) + (+3) = -1.

(ii)

First we move 4 steps to the right of 0 reaching 4 and then from this point we move 3 steps to the left. We reach the point 1. Thus, (+4) + (-3) = 1.

Question 3.
Find the solution of the following without using a number line :
(a) (+7) + (-11), (b) (-13) + (+10), (c) (-7) + (+9), (d) (+10) + (-5).
Make five such questions and solve them.
Solution:
(a) (+7) + (-11) = -4,
(b) (-13) + (+10) = -3,
(c) (-7) + (+9) = +2,
(d) (+10) + (-5) = 5.

Question 4.
Find the solution of the following without using a number line :
(i) (+5) + (+3),
(ii) (-6) +(-2),
(iii) (-5)+(+15),
(iv) (-8) + (+5),
(v) (+7) + (+10).
Solution:
(i) (+5) + (+3) = +8,
(ii) (-6) + (-2) = -8,
(iii) (-5) + (+15) = + 10
(iv) (-8) + (+5) = -3,
(v) (+7) + (+10) = 17.