NCERT Solutions For Class 6 Maths Chapter 7 Fractions
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Students can access the NCERT Solutions For Class 6 Maths Chapter 7 Fractions. Curated by experts according to the CBSE syllabus for 2023–2024, these step-by-step solutions make Maths much easier to understand and learn for the students. These solutions can be used in practice by students to attain skills in solving problems, reinforce important learning objectives, and be well-prepared for tests.
Introduction
Arya, Abhimanyu, and Vivek shared lunch. Arya has brought two sandwiches, one made of vegetables and one of jam. The other two boys forgot to bring their lunch. Arya agreed to share his sandwiches so that each person would 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?
(a) Arya has divided the sandwich into 3 equal parts. So each person will get one part.
(b) Each boy receives 1 / 3 part
∴ The required fraction is 1 / 3
What fraction of a day is 8 hours?
There are 24 hours in a day
We have 8 hours
Hence, the required fraction is 8 / 24
Identify the error, if any.
(i) The shaded portion is not half
Hence, this is not 1 / 2
(ii) Since the parts are not equal
The shaded portion is not 1 / 4
(iii) Since the parts are not equal
∴ The shaded portion is not 3 / 4
Colour the part according to the given fraction.
Write the fraction representing the shaded portion.
(i) Number of parts = 4
Shaded portion = 2
∴ Fraction = 2 / 4
(ii) Number of parts = 9
Shaded portion = 8
∴ Fraction = 8 / 9
(iii) Number of parts = 8
Shaded portion = 4
∴ Fraction = 4 / 8
(iv) Number of parts = 4
Shaded portion = 1
∴ Fraction = 1 / 4
(v) Number of parts = 7
Shaded portion = 3
∴ Fraction = 3 / 7
(vi) Number of parts = 12
Shaded portion = 3
∴ Fraction = 3 / 12
(vii) Number of parts = 10
Shaded portion = 10
∴ Fraction = 10 / 10
(viii) Number of parts = 9
Shaded portion = 4
∴ Fraction = 4 / 9
(ix) Number of parts = 8
Shaded portion = 4
∴ Fraction = 4 / 8
(x) Number of parts = 2
Shaded portion = 1
∴ Fraction = 1 / 2
Kristin received a CD player for her birthday. She bought 3 CDs and received 5 others as gifts. What fraction of the total CDs did she buy, and what fraction did she receive as gifts?
Number of CDs Kristin bought from the market = 3
Number of CDs received as gifts = 5
Total number of CDs Kristin has = 3 + 5 = 8
∴ Fraction of CD she bought = 3 / 8
∴ The fraction of CDs received as gifts = 5 / 8
Write the natural numbers from 102 to 113. What fraction of them are prime numbers?
Natural numbers from 102 to 113 are
102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113
Total number of natural numbers given = 12
Number of prime numbers = 4 [103, 107, 109, 113]
∴ The required fraction = 4 / 12 = 1 / 3
Write the natural numbers from 2 to 12. What fraction of them are prime numbers?
Natural numbers from 2 to 12 are
2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
Total number of natural numbers given = 11
Number of prime numbers = 5
∴ The required fraction = 5 / 11
Kanchan dyes dresses. She had to dye 30 dresses. She has so far finished 20 dresses. What fraction of dresses has she finished?
Total number of dresses Kanchan has to dye = 30 dresses
Number of dresses she has finished = 20 dresses
∴ The required fraction = 20 / 30 = 2 / 3
What fraction of these circles has X’s in them?
Total number of circles in the figure = 8
Number of circles having Xs in them = 4
∴ The required fraction = 4 / 8 = 1 / 2
What fraction of an hour is 40 minutes?
There are 60 minutes in 1 hour
∴ 1 hour = 60 minutes
Hence, the required fraction = 40 / 60
A Fraction
Draw number lines and locate the points on them:
(a) 1 / 2, 1 / 4, 3 / 4, 4 / 4
(b) 1 / 8, 2 / 8, 3 / 8, 7 / 8
(c) 2 / 5, 3 / 5, 8 / 5, 4 / 5
(a) 1 / 2, 1 / 4, 3 / 4, 4 / 4
Here, divide the number line from 0 to 1 into four equal parts
C = 2 / 4 = 1 / 2
B = 1 / 4
D = 3 / 4 and
E = 4 / 4 = 1
(b) 1 / 8, 2 / 8, 3 / 8, 7 / 8
Divide the number line from 0 to 1 into eight equal parts
B = 1 / 8
C = 2 / 8
D = 3 / 8
H = 7 / 8
(c) 2 / 5, 3 / 5, 8 / 5, 4 / 5
From the given number line, we have
C = 2 / 5
D = 3 / 5
E = 4 / 5
I = 8 / 5
Express the following as improper fractions:
(a)
(b)
(c)
(d)
(e)
(f)
(a) (7 × 4 + 3) / 4 = 31 / 4
∴ The improper form is 31 / 4
(b) (5 × 7 + 6) / 7 = 41 / 7
∴ The improper form is 41 / 7
(c) (2 × 6 + 5) / 6 = 17 / 6
∴ The improper form is 17 / 6
(d) (10 × 5 + 3) / 5 = 53 / 5
∴ The improper form is 53 / 5
(e) (9 × 7 + 3) / 7 = 66 / 7
∴ The improper form is 66 / 7
(f) (8 × 9 + 4) / 9 = 76 / 9
∴ The improper form is 76 / 9
Express the following as mixed fractions:
(a) 20 / 3
(b) 11 / 5
(c) 17 / 7
(d) 28 / 5
(e) 19 / 6
(f) 35 / 9
(a) 20 / 3
∴ 20 / 3 =
(b) 11 / 5
∴ 11 / 5 =
(c) 17 / 7
∴ 17 / 7 =
(d) 28 / 5
∴ 28 / 5 =
(e) 19 / 6
∴ 19 / 6 =
(f) 35 / 9
∴ 35 / 9 =
Fraction on the Number Line
Write the fractions. Are all these fractions equivalent?
(a)
(i) The shaded portion is 1 / 2
(ii) The shaded portion is 2 / 4 = (2 / 2) / (4 / 2) = 1 / 2
(iii) The shaded portion is 3 / 6 = (3 / 3) / (6 / 3) = 1 / 2
(iv) The shaded portion is 4 / 8 = (4 / 4) / (8 / 4) = 1 / 2
Hence, all fractions are equivalent.
(b)
(i) The shaded portion is 4 / 12 = (4 / 4) / (12 / 4) = 1 / 3
(ii)The shaded portion is 3 / 9 = (3 / 3) / (9 / 3) = 1 / 3
(iii) The shaded portion is 2 / 6 = (2 / 2) / (6 / 2) = 1 / 3
(iv) The shaded portion is 1 / 3
(v) The shaded portion is 6 / 15 = (6 / 3) / (15 / 3) = 2 / 5
All the fractions in their simplest form are not equal
Hence, they are not equivalent fractions.
Reduce the following fractions to the simplest form:
(a) 48 / 60
(b) 150 / 60
(c) 84 / 98
(d) 12 / 52
(e) 7 / 28
Solutions:
(a) 48 / 60 = (12 × 4) / (12 × 5)
= 4 / 5
(b) 150 / 60 = (30 × 5) / (30 × 2)
= 5 / 2
(c) 84 / 98 = (14 × 6) / (14 × 7)
= 6 / 7
(d) 12 / 52 = (3 × 4) / (13 × 4)
= 3 / 13
(e) 7 / 28 = 7 / (7 × 4)
= 1 / 4
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.
Total number of pencils Ramesh had = 20
Number of pencils used by Ramesh = 10
∴ Fraction = 10 / 20 = 1 / 2
Total number of pencils Sheelu had = 50
Number of pencils used by Sheelu = 25
∴ Fraction = 25 / 50 = 1 / 2
Total number of pencils Jamaal had = 80
Number of pencils used by Jamaal = 40
∴ Fraction = 40 / 80 = 1 / 2
Therefore, each has used up an equal fraction of pencils, i.e. 1 / 2
Write the fractions and pair up the equivalent fractions from each row.
(a) 1 / 2
(b) 4 / 6 = (4 / 2) / (6 / 2)
= 2 / 3
(c) 3 / 9 = (3 / 3) / (9 / 3)
= 1 / 3
(d) 2 / 8 = (2 / 2) / (8 / 2)
= 1 / 4
(e) 3 / 4
(i) 6 / 18 = (6 / 6) / (18 / 6)
= 1 / 3
(ii) 4 / 8 = (4 / 4) / (8 / 4)
= 1 / 2
(iii) 12 / 16 = (12 / 4) / (16 / 4)
= 3 / 4
(iv) 8 / 12 = (8 / 4) / (12 / 4)
= 2 / 3
(v) 4 / 16 = (4 / 4) / (16 / 4)
= 1 / 4
The following are the equivalent fractions
(a) and (ii) = 1 / 2
(b) and (iv) = 2 / 3
(c) and (i) = 1 / 3
(d) and (v) = 1 / 4
(e) and (iii) = 3 / 4
Replace ☐ in each of the following with the correct number:
(a) 2 / 7 = 8 / ☐
(b) 5 / 8 = 10 / ☐
(c) 3 / 5 = ☐ / 20
(d) 45 / 60 = 15 / ☐
(e) 18 / 24 = ☐ / 4
(a) Given
2 / 7 = 8 / ☐
2 × ☐ = 7 × 8
☐ = (7 × 8) / 2
= 28
(b) Given
5 / 8 = 10 / ☐
☐ = (8 × 10) / 5
= 16
(c) Given
3 / 5 = ☐ / 20
☐ = (3 × 20) / 5
= 12
(d) Given
45 / 60 = 15 / ☐
☐ = (15 × 60) / 45
= 20
(e) Given
18 / 24 = ☐ / 4
☐ = (18 × 4) / 24
= 3
Find the equivalent fraction of 3 / 5 having
(a) denominator 20
(b) numerator 9
(c) denominator 30
(d) numerator 27
(a) We require denominator 20
Let M be the numerator of the fractions
∴ M / 20 = 3 / 5
5 × M = 20 × 3
M = (20 × 3) / 5
= 12
Therefore, the required fraction is 12 / 20
(b) We require numerator 9
Let N be the denominator of the fractions
∴ 9 / N = 3 / 5
3 × N = 9 × 5
N = (9 × 5) / 3
= 15
Therefore the required fraction is 9 / 15
(c) We require denominator 30
Let D be the numerator of the fraction
∴ D / 30 = 3 / 5
5 × D = 3 × 30
D = (3 × 30) / 5
= 18
Therefore the required fraction is 18 / 30
(d) We require numerator 27
Let N be the denominator of the fraction
∴ 27 / N = 3 / 5
3 × N = 5 × 27
N = (5 × 27) / 3
= 45
Therefore the required fraction is 27 / 45
Find the equivalent fraction of 36 / 48 with
(a) numerator 9
(b) denominator 4
(a) Given numerator = 9
∴ 9 / D = 36 / 48
D × 36 = 9 × 48
D = (9 × 48) / 36
D = 12
Hence, the equivalent fraction is 9 / 12
(b) Given, denominator = 4
∴ N / 4 = 36 / 48
N × 48 = 4 × 36
N = (4 × 36) / 48
= 3
Hence, the equivalent fraction is 3 / 4
Check whether the given fractions are equivalent:
(a) 5 / 9, 30 / 54
(b) 3 / 10, 12 / 50
(c) 7 / 13, 5 / 11
(a) Given 5 / 9 and 30 / 54
We have 5× 54 = 270
9 × 30 = 270
5 × 54 = 9 × 30
Hence, 5 / 9 and 30 / 54 are equivalent fractions
(b) Given 3 / 10 and 12 / 50
We have 3 × 50 = 150
10 × 12 = 120
3 × 50 ≠ 10 × 12
Hence, 3 / 10 and 12 / 50 are not equivalent fractions
(c) Given 7 / 13 and 5 / 11
We have 7 × 11 = 77
5 × 13 = 65
7 × 11 ≠ 5 × 13
Hence, 7 / 13 and 5 / 11 are not equivalent fractions
Match the equivalent fractions and write two more for each.
(i) 250 / 400 (a) 2 / 3
(ii) 180 / 200 (b) 2 / 5
(iii) 660 / 990 (c) 1 / 2
(iv) 180 / 360 (d) 5 / 8
(v) 220 / 550 (e) 9 / 10
(i) 250 / 400
= (5 × 50) / (8 × 50)
= 5 / 8
25 / 40 and 30 / 48 are two more fractions
(ii) 180 / 200
= (9 × 20) / (10 × 20)
= 9 / 10
18 / 20 and 27 / 30 are two more fractions
(iii) 660 / 990
= (2 × 330) / (3 × 330)
= 2 / 3
20 / 30 and 200 / 300 are two more fractions
(iv) 180 / 360
= (1 × 180) / (2 × 180)
= 1 / 2
20 / 40 and 30 / 60 are two more fractions
(v) 220 / 550
= (2 × 110) / (5 × 110)
= 2 / 5
20 / 50 and 40 / 100 are two more fractions
∴ The equivalent fractions are
(i) 250 / 100 = (d) 5 / 8
(ii) 180 / 200 = (e) 9 / 10
(iii) 660 / 990 = (a) 2 / 3
(iv) 180 / 360 = (c) 1 / 2
(v) 220 / 550 = (b) 2 / 5
Proper Fractions
Look at the figures and write ‘<’ or ‘>’, ‘=’ between the given pairs of fractions.
(a) 1 / 6 ☐ 1 / 3
(b) 3 / 4 ☐ 2 / 6
(c) 2 / 3 ☐ 2 / 4
(d) 6 / 6 ☐ 3 / 3
(e) 5 / 6 ☐ 5 / 5
(a) Here, the numerators are the same. So, the fraction having lesser denominator is greater
∴ 1 / 6 < 1 / 3
(b) 3 / 4 = (3 × 3) / (4 × 3)
= 9 / 12
2 / 6 = (2 × 2) / (6 × 2)
= 4 / 12
Between 4 / 12, 9 / 12
Both fractions have the same denominators. So, the fraction having greater numerator will be greater
∴ 9 / 12 > 4 / 12
3 / 4 > 2 / 6
(c) Here, the numerators are the same. So, the fraction having lesser denominator is greater
∴ 2 / 3 > 2 / 4
(d) We get 6 / 6 = 1 and 3 / 3 = 1
So, 6 / 6 = 3 / 3
(e) Here, the numerators are the same. So, the fraction having lesser denominator is greater
∴ 5 / 6 < 5 / 5
Find answers to the following. Write and indicate how you solved them.
(a) Is 5 / 9 equal to 4 / 5?
(b) Is 9 / 16 equal to 5 / 9?
(c) Is 4 /5 equal to 16 / 20?
(d) Is 1 / 15 equal to 4 / 30?
(a) 5 / 9, 4 / 5
Convert these fractions into like fractions
5 / 9 = (5 / 9) × (5 / 5)
= 25 / 45
4 / 5 = (4 / 5) × (9 / 9)
= 36 / 45
∴ 25 / 45 ≠ 36 / 45
Hence, 5 / 9 is not equal to 4 / 5
(b) 9 / 16, 5 / 9
Convert into like fractions
9 / 16 = (9 / 16) × (9 / 9)
= 81 / 144
5 / 9 = (5 / 9) × (16 / 16)
= 80 / 144
∴ 81 / 144 ≠ 80 / 144
Hence, 9 / 16 is not equal to 5 / 9
(c) 4 / 5, 16 / 20
16 / 20 = (4 × 4) / (5 × 4)
= 4 / 5
∴ 4 / 5 = 16 / 20
Hence, 4 / 5 is equal to 16 / 20
(d) 1 / 15, 4 / 30
4 / 30 = (2 × 2) / (15 × 2)
= 2 / 15
∴ 1 / 15 ≠ 4 / 30
Hence, 1 / 15 is not equal to 4 / 30
How quickly can you do this? Fill appropriate sign. ( ‘<’, ‘=’, ‘>’)
(a) 1 / 2 ☐ 1 / 5
(b) 2 / 4 ☐ 3 / 6
(c) 3 / 5 ☐ 2 / 3
(d) 3 / 4 ☐ 2 / 8
(e) 3 / 5 ☐ 6 / 5
(f) 7 / 9 ☐ 3 / 9
(g) 1 / 4 ☐ 2 / 8
(h) 6 / 10 ☐ 4 / 5
(i) 3 / 4 ☐ 7 / 8
(j) 6 / 10 ☐ 3 / 5
(k) 5 / 7 ☐ 15 / 21
(a) Here, the numerators are the same. So, the fraction having lesser denominator is greater
∴ 1 / 2 > 1 / 5
(b) 2 / 4 = 1 / 2 and 3 / 6 = 1 / 2
∴ 2 / 4 = 3 / 6
(c) 3 / 5 = (3 × 3) / (5 × 3)
= 9 / 15
2 / 3 = (2 × 5) / 3 × 5)
= 10 / 15
Here, between 9 / 15 and 10 / 15 both have the same denominators. Hence, the fraction having greater numerator will be greater.
∴ 3 / 5 < 2 / 3
(d) Here, 2 / 8 = 1 / 4
As, 3 / 4 and 1 / 4 have the same denominators. Hence, the fraction having greater numerator will be greater
∴ 3 / 4 > 2 / 8
(e) Here, the denominators are the same. So, the fraction having greater numerator will be greater
∴ 3 / 5 < 6 / 5
(f) Here, the denominators are the same. So, the fraction having greater numerator will be greater
∴ 7 / 9 > 3 / 9
(g) We know 2 / 8 = 1 / 4
Hence, 1 / 4 = 2 / 8
(h) 6 / 10 = (3 × 2) / (5 × 2)
= 3 / 5
Between 3 / 5 and 4 / 5
Both have the same denominators. So, the fraction having greater numerator will be greater
∴ 6 / 10 < 4 / 5
(i) 3 / 4 = (3 × 2) / (4 × 2)
= 6 / 8
Between 6 / 8 and 7 / 8
Both have same denominators. So, the fraction having greater numerator will be greater
∴ 3 / 4 < 7 / 8
(j) 6 / 10 = (3 × 2) / (5 × 2)
= 3 / 5
∴ 6 / 10 = 3 / 5
(k) 5 / 7 = (5 × 3) / (7 × 3)
= 15 / 21
∴ 5 / 7 = 15 / 21
Rafiq exercised for 3 / 6 of an hour, while Rohit exercised for 3 / 4 of an hour. Who exercised for a longer time?
Rafiq exercised = 3 / 6 of an hour
Rohit exercised = 3 / 4 of a hour
3 / 6, 3 / 4
Convert these into like fractions
3 / 6 = (3 × 2) / (6 × 2)
= 6 / 12
3 / 4 = (3 × 3) / (4 × 3)
= 9 / 12
Clearly, 9 / 12 > 6 / 12
∴ 3 / 4 > 3 / 6
Therefore, Rohit exercised for a longer time than Rafiq.
The following fractions represent just three different numbers. Separate them into three groups of equivalent fractions, by changing each one to its simplest form.
(a) 2 / 12
(b) 3 / 15
(c) 8 / 50
(d) 16 / 100
(e) 10 / 60
(f) 15 / 75
(g) 12 / 60
(h) 16 / 96
(i) 12 / 75
(j) 12 / 72
(k) 3 / 18
(l) 4 / 25
(a) 2 / 12 = (1 × 2) / (6 × 2)
= 1 / 6
(b) 3 / 15 = (1 × 3) / (5 × 3)
= 1 / 5
(c) 8 / 50 = (4 × 2) / (25 × 2)
= 4 / 25
(d) 16 / 100 = (4 × 4) / (25 × 4)
= 4 / 25
(e) 10 / 60 = (1 × 10) / (6 × 10)
= 1 / 6
(f) 15 / 75 = (1 × 15) / (5 × 15)
= 1 / 5
(g) 12 / 60 = (1 × 12) / (5 × 12)
= 1 / 5
(h) 16 / 96
= (1 × 16) / (6 × 16)
= 1 / 6
(i) 12 / 75 = (4 × 3) / (25 × 3)
= 4 / 25
(j) 12 / 72 = (1 × 12) / 6 × 12)
= 1 / 6
(k) 3 / 18 = (1 × 3) / (6 × 3)
= 1 / 6
(l) 4 / 25
Totally there are 3 groups of equivalent fractions.
1 / 6 = (a), (e), (h), (j), (k)
1 / 5 = (b), (f), (g)
4 / 25 = (c), (d), (i), (l)
In a class A of 25 students, 20 passed with 60% or more marks; in another class B of 30 students, 24 passed with 60% or more marks. In which class was a greater fraction of students getting 60% or more marks?
Total number of students in Class A = 25
Students passed in first class in Class A = 20
Hence, fraction = 20 / 25
= 4 / 5
Total number of students in Class B = 30
Students passed in first class in Class B = 24
Hence, fraction = 24 / 30
= 4 / 5
∴ An equal fraction of students passed in first class in both the classes
Ila read 25 pages of a book containing 100 pages. Lalita read 2 / 5 of the same book. Who read less?
Total number of pages a book has = 100 pages
Lalita read = 2 / 5 × 100 = 40 pages
Ila read = 25 pages
∴ Ila read less than Lalita.
Write the shaded portion as a fraction. Arrange them in ascending and descending order using correct sign ‘’ between the fractions:
(c) Show 2 / 6, 4 / 6, 8 / 6 and 6 / 6 on the number line. Put appropriate signs between the fractions given.
5 / 6 ☐ 2 / 6, 3 / 6 ☐ 0, 1 / 6 ☐ 6 / 6, 8 / 6 ☐ 5 / 6
(a) The first circle shows 3 shaded parts out of 8 equal parts. Hence, the fraction is 3 / 8
The second circle shows 6 shaded parts out of 8 equal parts. Hence, the fraction is 6 / 8
The third circle shows 4 shaded parts out of 8 equal parts. Hence, the fraction is 4 / 8
The fourth circle shows 1 shaded part out of 8 equal parts. Hence, the fraction is 1 / 8
The arranged fractions are:
1 / 8 < 3 / 8 < 4 / 8 < 6 / 8
(b) The first square shows 8 shaded parts out of 9 equal parts. Hence, the fraction is 8 / 9
The second square shows 4 shaded parts out of 9 equal parts. Hence, the fraction is 4 / 9
The third square shows 3 shaded parts out of 9 equal parts. Hence, the fraction is 3 / 9
The fourth square shows 6 shaded parts out of 9 equal parts. Hence, the fraction is 6 / 9
The arranged fractions are:
3 / 9 < 4 / 9 < 6 / 9 < 8 / 9
(c) Each unit length should be divided into 6 equal parts to represent the fractions 2 / 6, 4 / 6, 8 / 6 and
6 / 6 on number line. These fractions can be represented as follows:
5 / 6 > 2 / 6
3 / 6 > 0
1 / 6 < 6 / 6
8 / 6 > 5 / 6
Compare the fractions and put an appropriate sign.
(a) 3 / 6 ☐ 5 / 6
(b) 1 / 7 ☐ 1 / 4
(c) 4 / 5 ☐ 5 / 5
(d) 3 / 5 ☐ 3 / 7
(a) Here both fractions have the same denominators. So, the fraction with the greater numerator is the highest factor
∴ 3 / 6 < 5 / 6
(b) Multiply by 4
1 / 7 = (1 × 4) / (7 × 4)
= 4 / 28
Multiply by 7
1 / 4 = (1 × 7) / (4 × 7)
= 7 / 28
Here 4 < 7
∴ 1 / 7 < 1 / 4
(c) Here both fractions have the same denominators. So, the fraction with the greater numerator is the highest factor
∴ 4 / 5 < 5 / 5
(d) Here both numerators are the same. So, the fraction having less denominator will be the highest factor
∴ 3 / 7 < 3 / 5
Make five more such pairs and put appropriate signs.
(a) 5 / 8 < 6 / 8
Here, the denominators are the same. So, the fraction having greater numerator is the highest factor
(ii) 5 / 8 > 2 / 8
Here, the denominators are the same. So, the fraction having greater numerator is the highest factor
(iii) 6 / 13 > 6 / 18
Here, the numerators are the same. So, the fraction having lesser denominator will be the highest factor
(iv) 5 / 25 > 3 / 25
Here, the denominators are the same. So, the fraction having greater numerator is the highest factor
(v) 9 / 50 < 9 / 45
Here, the numerators are the same. So, the fraction having lesser denominator will be the highest factor
Improper and Mixed Fractions
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?
Wall space painted by Shubham in a room = 2 / 3
Wall space painted by Madhavi in a room = 1 / 3
Total space painted by both = (2 / 3 + 1 / 3)
= (2 + 1) / 3
= 3 / 3
= 1
∴ Shubham and Madhavi together painted 1 complete wall in a room.
Fill in the missing fractions.
(a) 7 / 10 – ▯ = 3 / 10
(b) ▯ – 3 / 21 = 5 / 21
(c) ▯ – 3 / 6 = 3 / 6
(d) ▯ + 5 / 27 = 12 / 27
(a) Given 7 / 10 – ▯ = 3 / 10
▯ = 7 / 10 – 3 / 10
▯ = (7 – 3) / 10
▯ = 4 / 10
▯ = 2 / 5
(b) Given ▯ – 3 / 21 = 5 / 21
▯ = 5 / 21 + 3 / 21
▯ = (5 + 3) / 21
▯ = 8 / 21
(c) Given ▯ – 3 / 6 = 3 / 6
▯ = 3 / 6 + 3 / 6
▯ = (3 + 3) / 6
▯ = 6 / 6
▯ = 1
(d) Given ▯ + 5 / 27 = 12 / 27
▯ = 12 / 27 – 5 /27
▯ = (12 – 5) / 27
▯ = 7 /27
Javed was given 5 / 7 of a basket of oranges. What fraction of oranges was left in the basket?
Fraction of oranges given to Javed = 5 / 7
Fraction of oranges left in the basket = 1 – 5 / 7
= 7 / 7 – 5 / 7
= (7 – 5) / 7
= 2 / 7
Solve:
(a) 1 / 18 + 1 / 18
(b) 8 / 15 + 3 / 15
(c) 7 / 7 – 5 / 7
(d) 1 / 22 + 21 / 22
(e) 12 / 15 – 7 / 15
(f) 5 / 8 + 3 / 8
(g) 1 – 2 / 3 (1 = 3 / 3)
(h) 1 / 4 + 0 / 4
(i) 3 – 12 / 5
(a) 1 / 18 + 1 / 18
= (1 + 1) / 18
= 2 / 18
= 1 / 9
(b) 8 / 15 + 3 / 15
= (8 + 3) / 15
= 11 / 15
(c) 7 / 7 – 5 / 7
= (7 – 5) / 7
= 2 / 7
(d) 1 / 22 + 21 / 22
= (1 + 21) / 22
= 22 / 22
= 1
(e) 12 /15 – 7 / 15
= (12 – 7) / 15
= 5 / 15
= 1 / 3
(f) 5 / 8 + 3 / 8
= (5 + 3) / 8
= 8 / 8
= 1
(g) 1 – 2 / 3
= 3 / 3 – 2 / 3
= (3 – 2) / 3
= 1 / 3
(h) 1 / 4 + 0
= 1/ 4
(i) 3 – 12 / 5
= 15 / 5 – 12/ 5
= (15 – 12) / 5
= 3 / 5
Write these fractions appropriately as additions or subtractions:
(a) Total number of parts each rectangle has = 5
No. of shaded parts in the first rectangle = 1, i.e., 1 / 5
No. of shaded parts in the second rectangle = 2, i.e., 2 / 5
No. of shaded parts in the third rectangle = 3, i.e., 3 / 5
Clearly, the fraction represented by the third rectangle = Sum of the fractions represented by the first and second rectangles
Hence, 1 / 5 + 2 / 5 = 3 / 5
(b) Total number of parts each circle has = 5
We may observe that the first, second and third circles represent 5, 3 and 2 shaded parts out of 5 equal parts, respectively. Clearly, the fraction represented by the third circle is the difference between the fractions represented by the first and second circles.
Hence, 5 / 5 – 3 / 5 = 2 / 5
(c) Here, we may observe that the first, second and third rectangles represent 2, 3 and 5 shaded parts out of 6 equal parts, respectively. Clearly, the fraction represented by the third rectangle is the sum of fractions represented by the first and second rectangles.
Hence, 2 / 6 + 3 / 6 = 5 / 6
Equivalent Fractions
Fill in the boxes.
(a) ▯ – 5 / 8 = 1 / 4
(b) ▯ – 1 / 5 = 1 / 2
(c) 1 / 2 – ▯ = 1 / 6
(a) ▯ – 5 / 8 = 1 / 4
▯ = 1 / 4 + 5 / 8
▯ = [(1 × 2 + 5)] / 8
▯ = 7 / 8
(b) ▯ – 1 / 5 = 1 / 2
▯ = 1 / 2 + 1 / 5
▯ = [(1 × 5) + (1 × 2)] / 10
▯ = (5 + 2) / 10
▯ = 7 / 10
(c) 1 / 2 – ▯ = 1 / 6
▯ = 1 / 2 – 1 / 6
▯ = [(1 × 3) – (1 × 1)] / 6
▯ = (3 – 1) / 6
▯ = 2 / 6
▯ 1 / 3
Jaidev takes 
Time taken by Jaidev to walk across the school ground =
Time taken by Rahul to walk across the school ground = 7 / 4 minutes
Convert these fractions into like fractions.
11 / 5 = 11 / 5 × 4 / 4
= (11 × 4) / (5 × 4)
= 44 / 20
7 / 4 = 7 / 4 × 5 / 5
= (7 × 5) / (4 × 5)
= 35 / 20
Clearly, 44 / 20 > 35 / 20
11 / 5 > 7 / 4
∴ Rahul takes less time than Jaidev to walk across the school ground.
Difference = 11 / 5 – 7 / 4
= 44 / 20 – 35 / 20
= 9 / 20
Hence, Rahul walks across the school ground by 9 / 20 minutes.
Asha and Samuel have bookshelves of the same size, partly filled with books. Asha’s shelf is 5 / 6 th full, and Samuel’s shelf is 2/ 5 th full. Whose bookshelf is more full? By what fraction?
Fraction of Asha’s bookshelf = 5 / 6
Fraction of Samuel’s bookshelf = 2 / 5
Convert these fractions into like fractions.
5 / 6 = 5 / 6 × 5 / 5
= (5 × 5) / (6 × 5)
= 25 / 30
2 / 5 = 2 / 5 × 6 / 6
= (2 × 6) / (5 × 6)
= 12 / 30
25 / 30 > 12 / 30
5 / 6 > 2 / 5
∴ Asha’s bookshelf is more full than Samuel’s bookshelf.
Difference = 5 / 6 – 2 / 5
= 25 / 30 – 12 / 30
= 13 / 30
Nandini’s house is 9 / 10 km from her school. She walked some distance and then took a bus for 1 / 2 km to reach the school. How far did she walk?
Distance of the school from house = 9 / 10 km
Distance she travelled by bus = 1 / 2 km
Distance walked by Nandini = Total distance of the school – Distance she travelled by bus
= 9 / 10 – 1 / 2
= [(9 × 1) – (1 × 5)] / 10
= (9 – 5) / 10
= 4 / 10
= 2 / 5 km
∴ Distance walked by Nandini is 2 / 5 km
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?
Total length of wire = 7 / 8 metre
Length of one piece of wire = 1 / 4 metre
Length of the other piece of wire = Length of the original wire and this one piece of wire
= 7 / 8 – 1 / 4
= [(7 × 1) – (1 × 2)] / 8
= (7 – 2) / 8
= 5 / 8
∴ Length of the other piece of wire = 5 / 8 metre
Complete the addition and subtraction box.
(a) 2 / 3 + 4 / 3
= (2 + 4) / 3
= 6 / 3
= 2
1 / 3 + 2 / 3
= (1 + 2) / 3
= 3 / 3
= 1
2 / 3 – 1 / 3
= (2 – 1) / 3
= 1 / 3
4 / 3 – 2 / 3
= (4 – 2) / 3
= 2 / 3
1 / 3 + 2 / 3
= (1 + 2) / 3
= 3 / 3
= 1
Hence, the complete given box is
(b) 1 / 2 + 1 / 3
= [(1 × 3) + (1 × 2)] / 6
= (3 + 2) / 6
= 5 / 6
1 / 3 + 1 / 4
= [(1 × 4) + (1 × 3)] / 12
= (4 + 3) / 12
= 7 / 12
1 / 2 – 1 / 3
= [(1 × 3) – (1 × 2)] / 6
= (3 – 2) / 6
= 1 / 6
1 / 3 – 1 / 4
= [(1 × 4) – (1 ×3)] / 12
= (4 – 3) / 12
= 1 / 12
1 / 6 + 1 / 12
= [(1 × 2) + 1] / 12
= (2 + 1) / 12
= 3 / 12
= 1 / 4
Hence, the complete given box is
Naina was given 
Fraction of cake Naina got =
Fraction of cake Najma got =
The total amount of cake given to both of them = 3 / 2 + 4 / 3
= [(3 × 3) + (4 × 2)] / 6
= (9 + 8) / 6
= 17 / 6
=
Sarita bought 2 / 5 metres of ribbon and Lalita 3 /4 metres of ribbon. What is the total length of the ribbon they bought?
Ribbon length bought by Sarita = 2 / 5 metre
Ribbon length bought by Lalita = 3 / 4 metre
The total length of the ribbon bought by both of them = 2 / 5 + 3 / 4
Taking LCM 20,
= [(2 × 4) + (3 × 5)] / 20
= (8 + 15) / 20
= 23 / 20 metre
∴ The total length of the ribbon bought by both Sarita and Lalita is 23 / 20 metre
Solve
(a) 2 / 3 + 1 / 7
(b) 3 / 10 + 7 / 15
(c) 4 / 9 + 2 / 7
(d) 5 / 7 + 1 / 3
(e) 2 / 5 + 1 / 6
(f) 4 / 5 + 2 / 3
(g) 3 / 4 – 1 / 3
(h) 5 / 6 – 1 / 3
(i) 2 / 3 + 3 / 4 + 1 / 2
(j) 1/ 2 + 1 / 3 + 1 / 6
(k)
(l)
(m) 16 / 5 – 7 / 5
(n) 4 / 3 – 1 / 2
(a) 2 / 3 + 1/ 7
Taking LCM
[(2 × 7) + (1 × 3)] / 21
= (14 + 3) / 21
= 17 / 21
(b) 3 / 10 + 7 / 15
Taking LCM 30,
= [(3 × 3) + (7 × 2)] / 30
= (9 + 14) / 30
= 23 / 30
(c) 4 / 9 + 2/ 7
Taking LCM 63,
= [(4 × 7) + (2 × 9)] / 63
= (28 + 18) / 63
= 46 / 63
(d) 5 / 7 + 1 / 3
Taking LCM 21,
= [(5 × 3) + (1 × 7)] / 21
= (15 + 7) / 21
= 22 / 21
(e) 2 / 5 + 1 / 6
Taking LCM 30,
= [(2 × 6) + (1 × 5)] / 30
= (12 + 5) / 30
= 17 / 30
(f) 4 / 5 + 2 / 3
Taking LCM 15,
= [(4 × 3) + (2 × 5)] / 15
= (12 + 10) / 15
= 22 / 15
(g) 3 / 4 – 1 / 3
Taking LCM 12,
= [(3 × 3) – (1 × 4)] / 12
= (9 – 4) / 12
= 5 / 12
(h) 5 / 6 – 1 / 3
Taking LCM 6,
= [(5 × 1) – (1 × 2)] / 6
= (5 – 2) / 6
= 3 / 6
= 1 / 2
(i) 2 / 3 + 3 / 4 + 1 / 2
Taking LCM 12,
= [(2 × 4) + (3 × 3) + (1 × 6)] / 12
= (8 + 9 + 6) / 12
= 23 / 12
(j) 1 / 2 + 1 / 3 + 1 / 6
Taking LCM 6,
= [(1 × 3) + (1 × 2) + (1 × 1)] / 6
= (3 + 2 + 1) / 6
= 6 / 6
= 1
(k)
= [(3 × 1) + 1] / 3 + [(3 × 3) + 2] / 3
= (3 + 1) / 3 + (9 + 2) / 3
= 4/ 3 + 11 / 3
= (4 + 11) / 3
= 15 / 3
= 5
(l)
= [(3 × 4) + 2] / 3 + [(3 × 4) + 1] / 4
= 14 / 3 + 13 / 4
= [(14 × 4) + (13 × 3)] / 12
= (56 + 39) / 12
= 95 / 12
(m) 16 / 5 – 7 / 5
= (16 – 7) / 5
= 9 / 5
(n) 4 /3 – 1 / 2
Taking LCM 6,
= [(4 × 2) – (1 × 3)] / 6
= (8 – 3) /6
= 5 / 6
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Frequently Asked Questions
The NCERT solution for Class 6 Chapter 7: Fractions is important as it provides a structured approach to learning, ensuring that students develop a strong understanding of foundational concepts early in their academic journey. By mastering these basics, students can build confidence and readiness for tackling more difficult concepts in their further education.
Yes, the NCERT solution for Class 6 Chapter 7: Fractions is quite useful for students in preparing for their exams. The solutions are simple, clear, and concise allowing students to understand them better. They can solve the practice questions and exercises that allow them to get exam-ready in no time.
You can get all the NCERT solutions for Class 6 Maths Chapter 7 from the official website of the Orchids International School. These solutions are tailored by subject matter experts and are very easy to understand.
Yes, students must practice all the questions provided in the NCERT solution for Class 6 Maths Chapter 7: Fractions as it will help them gain a comprehensive understanding of the concept, identify their weak areas, and strengthen their preparation.
Students can utilize the NCERT solution for Class 6 Maths Chapter 7 effectively by practicing the solutions regularly. Solve the exercises and practice questions given in the solution.