Fractions Class 6 Ex 7.1
Question 1.
Write the fraction representing the shaded portion.
Solution:
(i) Total number of parts = 4
Number of shaded parts = 2
∴ Fraction =
(ii) Total number of parts = 9
Number of shaded parts = 8
∴ Fraction =
(iii) Total number of parts = 8
Number of shaded parts = 4
∴ Fraction =
(iv) Total number of parts = 4
Number of shaded parts = 1
∴ Fraction =
(v) Total number of parts = 7
Number of shaded parts = 3
∴ Fraction =
(vi) Total number of parts = 12
Number of shaded parts = 3
∴ Fraction =
(vii) Total number of parts = 10
Number of shaded parts = 10
∴ Fraction =
(viii) Total number of parts = 9
Number of shaded parts = 4
∴ Fraction =
(ix) Total number of parts = 8
Number of shaded parts = 4
∴ Fraction =
(x) Total number of parts = 2
Number of shaded part = 1
∴ Fraction =
Question 2.
Colour the part according to the given fraction.
Solution:
Question 3.
Identify the error, if any.
Solution:
(a) Since the shaded part is not half.
∴ This is not
(b) Since, the parts are not equal.
∴ Shaded part is not
(c) Since, the part are not equal.
∴ Shaded part is not
Question 4.
What fraction of a day is 8 hours?
Solution:
Since, a day has 24 hours and we have 8 hours,
∴ Required fraction =
Question 5.
What fraction of a hour is 40 minutes?
Solution:
Since I hours = 60 minutes
∴ Fraction of 40 minutes =
Question 6.
Arya, Abhimanyu and Vivek shared lunch. Arya has brought two sandwiches, one made of vegetable and one of Jam. The other two boys forgot to bring their lunch. Arya agreed to share his sandwiches so that each person will have an equal share 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 has divided his sandwich into three equal parts.
So, each of them will get one part.
(b) Each one of them will receive
∴ Required fraction =
Question 7.
Kanchan dyes dresses. She had to dye 30 dresses. She has so far finished 20 dresses. What fraction of dresses has she finished?
Solution:
Total number of dresses to be dyed = 30
Number of dresses finished = 20
∴ Required fraction =
Question 8.
Write the natural numbers from 2 to 12. What fraction of them are prime numbers?
Solution:
Natural numbers between 2 and 12 are;
2,3,4, 5, 6, 7, 8, 9, 10,11, 12
Number of given natural numbers = 11
Number of prime numbers = 5
∴ Required fraction =
Question 9.
Write the natural numbers from 102 to 113. What fraction of them are prime numbers?
Solution:
Natural numbers from 102 to 113 are;
102,103,104,105,106, 107,108, 109,110, 111, 112,113
Total number of given natural numbers = 12
Prime numbers are 103, 107, 109, 113
∴ Number of prime numbers = 4
∴ Required fraction =
Question 10.
What fraction of these circles have X’s in them?
Solution:
Total number’of circles = 8
Number of circles having X’s in them = 4
Required fraction =
Question 11.
Kristin received a CD player for her birthday. She bought 3 CDs and received 5 others as gifts. What fraction of her total CDs did she buy and what fraction did she receive as gifts?
Solution:
Number of CDs bought by her from the market = 3
Number of CD’s received as gifts = 5
∴ Total number of CDs = 3 + 5 = 8
∴ Fraction of CD (bought) =
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Fractions Class 6 Ex 7.2
Question 1.
Draw number lines and locate the points on them.
Solution:
We have divided the number line from 0 to 1 into four equal parts.
C represents
B represents
D represents
and E represents
We have divided the number line from 0 to 1 into eight equal parts.
B represents
C represents
D represents
and H represents
From the above number line, we have
C represents
D represents
E represents
and I represents
Question 2.
Express the following as mixed fractions:
Solution:
Question 3.
Express the following as improper fractions:
Solution:
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Fractions Class 6 Ex 7.3
Question 1.
Write the fractions. Are all these fractions equivalent?
Solution:
Since all the fractions in their simplest form are not equal.
∴ They are not equivalent fractions.
Question 2.
Write the fractions and pair up the equivalent fractions from each row.
Solution:
The following pairs fractions:represent the equivalent fractions.
(a) and (ii) =
(b) and (iv) =
(c) and (i) =
(d) and (v) =
(e) and (iii) =
Question 3.
Replace
Solution:
Question 4.
Find the equivalent fraction of
(a) denominator 20
(b) numerator 9
(c) denominator 30
(d) numerator 27
Solution:
(a) Here, we require denominator 20.
Let N be the numerator of the fractions.
∴ The required fraction is
(b) Here, we required numerator 9.
Let D be the denominator of the fraction.
∴ The required fraction is
(c) Here, we required denominator 30.
Let N be the numerator of the fraction.
∴ The required fraction is
(d) Here, we required numerator 27.
Let D be the denominator of the fraction.
∴ The required fraction is
Question 5.
Find the equivalent fraction of
(a) numerator 9
(b) denominator 4
Solution:
(a) Given that numerator = 9
So, the equivalent fraction is
(b) Given that denominator = 4
∴
⇒ N =
So, the equivalent fraction is
Question 6.
Check whether the given fractions are equivalent:
Solution:
(a)
We have 5 x 54 = 270
and 9 x 30 = 270
Here 5 x 54 = 9 x 30
∴
(b)
We have 3 x 50 = 150
and 10 x 12 = 120
Here 3 x 50 ≠ 10 x 12
∴
(c)
We have 7 x 11 = 77 and 5 x 13 = 65
Here 7 x 11 ≠ 5 x 13
∴
Question 7.
Reduce the following fractions to simplest form:
Solution:
Question 8.
Ramesh had 28 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 her/his pencils.
Solution:
Ramesh used up 10 pencils out of 20 pencils.
Sheelu used up 25 pencils out of 50 pencils.
Jamaal used up 40 pencils out of 80 pencils.
Yes, each has used up an equal fractions, i.e.,
Question 9.
Match the equivalent fractions and write two more for each.
Solution:
Two additional examples of equivalent fractions are
Two additional examples of equivalent fractions are
Two additional examples of equivalent fractions are
Two additional examples of equivalent fractions are
Two additional examples of equivalent fractions are
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Fractions Class 6 Ex 7.4
Question 1.
Write shaded portion as fraction. Arrange them in ascending and descending order using correct sign ‘<‘, ‘=’, ‘>’ between the fractions.
(c) Show
Solution:
(a) Total number of divisions = 8
(i) Number of shaded parts = 3
∴ Fraction =
(ii) Total number of divisions = 8
Number of shaded parts = 6
∴ Fraction =
(iii) Total number of divisions = 8
Number of shaded parts = 4
∴ Fraction =
(iv) Total number of divisions = 8
Number of shaded part = 1
∴ Fraction =
Now the fractions are:
(b)(i) Total number of divisions = 9
Number of shaded parts = 8
∴ Fraction =
(ii) Total number of divisions = 9
Number of shaded parts = 4
∴ Fraction =
(iii) Total number of divisions = 9
Number of shaded parts = 3
∴ Fraction =
(iv) Total number of divisions = 9
Number of shaded parts = 6
∴ Fraction =
∴ Fractions are
Question 2.
Compare the fractions and put an appropriate sign.
Solution:
Here, denominators of the two fractions are same and 3 < 5.
Here, numerators of the fractions are same and 7 > 4.
Here, denominators of the two fractions are same and 4 < 5.
Here, numerators of the two fractions are same and 5 < 7.
Question 3.
Make five more such pairs and put appropriate signs.
Solution:
Question 4.
Look at the figures and write ’<’, or ’>’ ’=’ between the given pairs of fractions.
Make five more such problems and solve them with your friends
Solution:
Make five more such problems yourself and solve them with your friends.
Question 5.
How quickly can you do this? Fill appropriate sign. ‘<‘, ‘=’, ‘>’.
Solution:
Question 6.
The following fractions represent just three different numbers. Separate them into three groups of equivalent fractions, by changing each one to its simplest form.
Solution:
Now grouping the above fractions into equivalent fractions, we have
Question 7.
Find answers to the following. Write and indicate how you solved them.
Solution:
By cross-multiplying, we get
5 x 5 = 25 and 4 x 9 = 36
Since 25 ≠ 36
By cross-multiplying, we get
9 x 9 = 81 and 16 x 5 =80
Since 81 ≠ 80
By cross-multiplying, we get
4 x 20 = 80 and 5 x 16 = 80
Since 80 = 80
By cross-multiplying, we get
1 x 30 = 30 and 4 x 15 = 60
Question 8.
Ila read 25 pages of a book containing 100 pages.
Lalita read
Solution:
Ila reads 25 pages out of 100 pages.
Lalita reads
Comparing
1 x 5 = 5 and 2 x 4 = 8
Since 5 < 8
∴
Hence Ila reads less pages.
Question 9.
Rafiq exercised for
Solution:
Rafiq exercised for
Rohit exercised for
Comparing
3 x 4 = 12 and 3 x 6 = 18
Since 12 < 18
∴
Hence Rohit exercised for longer time.
Question 10.
In a class A of 25 students, 20 passed in first class, in another class B of 30 students, 24 passed in first class. In which class was a greater fraction of students getting first class?
Solution:
In class A, 20 students passed in first class out of 25 students.
∴ Fraction of students getting first class
In class B, 24 students passed in first class out of 30 students.
∴ Fraction of students getting first class
Comparing the two fractions, we get
Hence, both the class A and B have the same fractions.
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Fractions Class 6 Ex 7.5
Question 1.
Write these fractions appropriately as additions or subtractions.
Solution:
(a) The given figure represents the addition of
Thus the given diagrams can be represented as
(b) The given figure represents the difference between 1 and
Thus, the given diagrams can be represented as
(c) The given figure represents addition of
Thus, the given diagrams can be represented as
Question 2.
Solve:
Solution:
Question 3.
Shubham painted
Solution:
Fraction of wall painted by Shubham =
Fraction of wall painted by Madhavi =
Fraction of wall painted by Shubham and Madhavi
Thus the fraction of wall painted by both = 1
Question 4.
Fill in the missing fractions.
Solution:
Question 5.
Javed was given
Solution:
Fraction of basket of oranges =
Fraction of basket as a whole can be taken as 1.
∴ Fraction of basket of oranges left
Thus, the required fraction =
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Fractions Class 6 Ex 7.6
Question 1.
Solve
Solution:
Sarita bought
Solution:
Length of ribbon bought by Sarita =
Length of ribbon bought by Lalita =
∴ Length of ribbon bought by Sarita and Lalita
Hence, the required length =
Question 3.
Naina was given 1
Solution:
Piece of cake given to Naina = 1
Piece of cake given to Najma = 1
Piece of cake given to Naina and Najma
Hence the total amount of piece given to both = 2
Question 4.
Fill in the boxes:
Solution:
Here, missing number is
Here, missing number is
Here, missing number is
Question 5.
Complete the addition-subtraction box.
Solution:
Thus the box may be completed as follows:
Question 6.
A piece of wire
Solution:
Total length of the wire =
Length of one piece of wire =
∴ Length of the other piece =
LCM of 8 and 4 = 8
Hence, the length of the other piece =
Question 7.
Nandini’s house is
Solution:
Total distance from house to school =
Distance travelled by Nandini by bus =
∴ Distance travelled by her on foot
Hence, the distance travelled by her on foot =
Question 8.
Asha and Samuel have bookshelves of the same size partly filled with books. Asha’s shelf is
Solution:
Asha’s shelf is
and Samuel’s shelf is
Comparing
LCM of 6 and 5 = 30
Hence, Asha’s shelf is full more than Samuel’s shelf.
Hence,
Question 9.
Jaidev takes 2
Solution:
Jaidev takes 2
Rahul takes 2
Comparing 2
So, the time take to cover the same distance by Rahul is less than that of Jaidev.
Hence, Rahul takes