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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.

Haryana State Board HBSE 6th Class Maths Solutions Chapter 6 Integers Ex 6.3 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 6 Integers Exercise 6.3

Question 1.
Subtract:
(a) 35 – (20)
(b) 72 – (90)
(c) (-15) – (-18)
(d) (-20)-(13)
(e) 23 – (-12)
(f) (-32) – (-40)
Solution:
(a) 35 – (20) = 15.
(b) 72-(90) = -18.
(c) (-15) – (-18) = (-15) + 18 = 3.
(d) (-20) – 13 = -33.
(e) 23-(-12) = 23 + 12 = 35.
(f) (-32) – (-40) = (-32) + 40 = 8.

Question 2.
Fill in the blanks with >, < or = sign :
(a) (-3) + (-6) ……………. (-3) – (-6)
(b) (-21) – (-10) ……………. (-31) + (-11)
(c) 45-(-11) ……………. 57 +(-4)
(d.) (-25) -(-42) ……………. (-42) – (-25).
Solution:
(a) (-3) + (-6) = -9
and (-3)-(-6) = (-3) + 6 = 3
∵ (-9) < 3
∴ (-3) + (-6) < (-3) – (-6)

(b) (-21) – (-10) = -21 + 10 = -11
and (-31) + (-11) = – 42
∵ (-11) > (-42)
∴ (-21) – (-10) > (-31) + (-11)

(c) 45 – (-11) = 45+ 11 = 56
and 57 + (-4) = 53
56 > 53
∴ 45 – (-11) > 57 + (-4)

(d) (-25) – (-42) = (-25) + 42 = 17
and (-42) – (-25) = -42 + 25 = -17,
∵ 17 > -17
∴ (-25) – (-42) > (-42) – (-25).

Question 3.
Fill in the blanks :
(a) (-8) + = 0
(b) 13 + = 0
(c) 12 +(-12) =
(d) (-4) + =-12
(e) – 15 = – 10.
Solution:
(a) (-8) + 8 = 0
(b) 13 + (-13) = 0
(c) 12 + (-12) = 0
(d) (-4) + (-8) = -12
(e) 5 + (-15) = -10.

Question 4.
Find the value of:
(a) (-7)-8-(-25)
(b) (-13) + 32-8-1
(c) (-7) + (-8) + (-90)
(d) 50 – (-40) – (-2).
Solution:
(a) (-7) – 8 – (-25) = -7-8 + 25 = -15 + 25 = 10
(b) (-13) + 32 – 8 – 1 = 32 – 13 – 8 – 1
= 32 – 22 = 10
(c) (-7) + (-8) + (-90) = -7 – 8 – 90 = -105
(d) 50 – (-40) – (-2) = 50 + 40 + 2 = 92.

Haryana State Board HBSE 6th Class Maths Solutions Chapter 6 Integers Ex 6.2 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 6 Integers Exercise 6.2

Question 1.
Using the number line, write the integer which is :
(a) 3 more than 5
(b) 5 more than -5
(c) 6 less than 2
(d) 3 less than -2.
Solution:
(a) We want to find the integer which is 3 more than 5. So we start from 5 and proce 3 steps to the right of 5 to reach 8 as shown below. (Fig.)
∴ 3 more than 5 is 8.

(b) We want to find an integer which is 5 more than -5. So we start from -5 and proi 5 steps to the right of-5 to reach 0 as shown below (Fig.).

∴ 5 more than -5 is 0.

(c) We want to find an integer which is 6 less than 2. So we start from 2 and move 6 steps to the left of 2 to reach -4 as shown in Fig. , ∴ 6 less than 2 is -4.

(d) We want to find an integer is 3 less than -2. So we start from -2 and move 3 steps to the left to reach-5 as shown in Fig. 6.13.
∴ 3 less than -2 is -5.

Question 2.
Use a number line and add the following integers :
(a) 9 + (-6)
(b) 5 + (-11)
(c) (-1) + (-7)
(d) (-5) + 10
(e) (-1) + (-2) + (-3)
(f) (-2) + 8 + (-4).
Solution:
(a) 9 + (—6) = 3

(b) 5 + (-11) = -6

(c) (-1) + (-7) = -8

(d) (-5) + 10 = 5

(e) (-1) + (-2) + (-3) = -6

(f) (-2) + 8 + (-4). or 8 + (-2) + (-4) = 8 + (-6) = 2

Question 3.
Add without using a number line:
(a) 11 + (-7)
(b) (-13) + 18
(c) (-10) + (+19)
(d) (-250) + (+150)
(e) (-380) + (-270)
(f) (-217) + (-100).
Solution:
(a) 11 + (-7) = 4
(b) (-13) + 18 = 5
(c) (-10) + (+19) = 9
(d) (-250) + (+150) = -100
(e) (-380) + (-270) = -650
(f) (-217) + (-100) = -317.

Question 4.
Find the sum of:
(a) 137 and -354
(b) -52 and 52
(c) -312 , 39 and 192
(d) -50, -200 and 300.
Solution:
(a) 137 + (-354) = -217
(b) (-52) + (52) = 0,
(c) (-312) + 39 + 192 = (-312) + 231 = -81,
(d) (-50) + (-200) + 300 = (-250) + 300 = 50.

Question 5.
Find the value of:
(а) (-7) + (-9) + 4 + 16
(b) (37) + (-2) + (-65) + (-8).
Solution:
(a) (-7) + (-9) + 4 + 16 = (-16) + 20 = 4.
(b) (37) + (-2) + (-65) + (-8) =
37 + (-2 – 65 – 8) = 37 + (-75) = -38.
Note : Additive Inverse : Numbers such as 3 and -3, 2 and -2, added to each other give the sum zero. They are called additive inverse of each other.

Haryana State Board HBSE 6th Class Maths Solutions Chapter 6 Integers Ex 6.1 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 6 Integers Exercise 6.1

Question 1.
Write opposites of the following :
(a) Increase in weight
(b) 30 km north
(c) 326 BC
(d) Loss of Rs. 700
(e) 100 m above sea level.
Solution:
(a) Decrease in weight,
(b) 30 km south,
(c) 326 AD,
(d) Gain of Rs. 700,
(e) 100 m below sea level.

Question 2.
Represent the following numbers as integers with appropriate signs :
(a) An aeroplane is flying at a height of two thousand metres above the ground.
(b) A sub-marine is moving at a depth of eight hundred, metres below the sea level.
(c) A deposit of Rs. two hundred.
(d) Withdrawal of Rs. seven hundred.
Solution:
(a) +2000,
(b) -800,
(c) +200,
(d) -700.

Question 3.
Represent the following numbers on a number line :
(a) +5 (b) -10 (c) +8 (d) -1 (e) -6.
Solution:

Question 4.
Adjacent is a vertical line, representing integers. Observe it and locate the following points :
(a) If point D is +8, then which point, is -8 ?
(b) Is point. G a negative integer or a positive integer ?
(c) Which is the point, opposite to H ?
(d) Write integers for points B and E.
(e) Which point, marked on this number line has the least, value, if) Arrange all the points in decreasing order of value.
Solution:
(a) F,
(b) negative integer,
(c) A,
(d) +4, -10,
(e) E,
(f) D, C, B, A, O, H, G, F, E.

Question 5.
Following is the list of temperatures of five places in India, on a particular day of the year.

(a) Write the temperatures of these places in the form of integers in the blank column.
(b) Following is the number line representing the temperature in degree Celsius.

Plot the name of the city against its temperature.
(c) Which is the coolest place ?
(d) Write the names of the places whose temperatures are above 10°C.
Solution:
(a) -10°C, -2°C, +30°C, + 20°C, -5°C
(b)

(c) Siachin is the coolest place.
(d) Delhi and Ahmedabad.

Question 6.
In each of the following pairs, which number is to the right of the other on the number line ?
(a) 2, 9,
(b) -3, -8,
(c) 0, -1,
(d) -11, 10,
(e) -6, 6,
(f) 1, -100.
Solution:
(a) 9, (b) -3, (c) 0, (d) 10, (e) 6, (f) 1.

Question 7.
Write all the integers between the given pairs (write them in the increasing order).
(a) 0 and -7
(b) -4 and 4-
(c) -S and -15
(d) -30 and -23.
Solution:
(a) -6,-5,-4,-3,-2,-1.
(b) -3, -2, -1, 0, 1, 2, 3.
(c) -14, -13, -12, -11, -10, -9.
(d) -29, -28, -27, -26, -25, -24.

Question 8.
(a) Write four negative integers greater than -20.
(b) Write four negative integers less than -10.
Solution:
(a) -19, -18, -17, -16, (b) —11, -12, -13, -14.

Question 9.
For the following statements, write true (T) or false (F). If the statement is false, correct the statement.
(a) -8 is to the right of-10 on a number line.
(b) -100 is to the right of -50 on a number line.
(c) Smallest negative number is -1.
(d) -26 is larger than -25.
Solution:
(a) T,
(b) F, -100 is to the left of -50 on the number line,
(c) F, there is no smallest negative number,
(d) F, -26 is smaller than -25.

Question 10.
Draw a number line and answer the following :
(a) Which number will we reach if we move 4 numbers to the right of -2 ?
(b) Which number will we reach if w,e move 5 numbers to the left of 1 ?
(c) If we are at -8 on the number line, in which direction should we move to reach -13 ?
(d) If we are at -6 on the number dine, in which direction should we move to reach -1 ?

Solution:
(a) 2,
(b) -4,
(c) left,
(d) right.