Chapter 2 – Fractions
You will Learn
1. Multiplications of Fractions
2. Fractions as an Operator
3. Reciprocal of Fractions
4. Division of Fractions
Fractions
A fraction is a number which can be written in the form
of a , where a and b both are natural numbers.
b
a is numerator and b is the denominator of fraction
Examples of fractions :
8
2
5
, 3 , 13
7
a
.
b
Like Fractions : - The fractions which have the same denominator.
Examples -
1 4 7
, ,
9 9 9
Unlike Fractions : - The fractions with different denominators.
Examples - 2 , 5 , 7
3 7 9
Proper Fraction : - Fractions with numerator smaller than the denominator.
Examples -
3 8 2
, ,
4 9 7
Improper Fraction : - Fractions with numerator greater than the denominator.
Examples -
9 7 8
, ,
5 2 3
Mixed Fractions or Numbers : - Some fractions can be expressed as a
combination of whole number as well as proper fraction.
1
Example : 7 can be expressed as 2
3
3
Simplest or Lowest form of Fractions : - The fractions with no common
factors other than 1.
Examples :
2 5 7
, ,
3 7 9
Equivalent Fraction : - If
c
mxa
, where m is any integer, then a and
=
d
mxb
b
c are called equivalent fractions.
d
Examples :
6
2x3
=
10
2x5
Multiplication of Fractions
Multiplying a Whole Number by a Fraction : To multiply a fraction by a
whole number, we simply multiply the numerators together and the denominator
remains the same.
2
Examples : - 5 x
2 10
,
=
7
7
38 x
17
19
=
38 x 17
19
1
= 34
Fraction as an Operator: When a fraction is used with the word ‘of’, it acts as
an operator. It acts as operator ‘Multiplication’.
2
of 14
7
Example : -
2
=
2
x 14 = 2 x 2 = 4
7
1
Here ‘of’ represents ‘x’ (multiplication)
Multiplying a Fraction by a Fraction: During multiplication of fractions,
product of Numerators is the Numerator and the product of Denominators is
the Denominator.
Example : -
=
3
9
x
4
2
3
9
3x9
27
x
=
=
4
2
4x2
8
Multiplication of mixed fractions: During multiplication of fractions, product
of Numerators is the Numerator and the product of Denominators is the
Denominator.
3
9
x
4
2
Example : -
=
3
9
3x9
27
x
=
=
4
2
4x2
8
Some Solved Examples
Alternative Method
Example1 : - 4 x 4
1
2
=
x
4
½
4
16
2
=
18
= 16 + 2
Example2 : -
1
1
=
x 12
3
8
=
=
x
12
1
3
4
4
+
4
1
24
1
8
1
24
1
24
2
4 x 4
1
2
=4 x
9
2
=
4x9
2
= 2 x 9 = 18
1
1
3
x 12
1
8
=
1
3
x
97
8
=
97
24
= 4
1
24
Some More Solved Examples
Statement Examples
Example 3 : 2 of 20
3
9
2 27 2 x 9
=
x
=
= 18
3
1
1
9
3 36 3 x 9
x
=
= 27
4
1
1
3
1
of 15
5
=
1
1x3
15
x
=
= 3
5
1
1
Example 4 : 3 of 36
4
=
Problem 1 : - Rajesh has 15 chocolates. He gave 1/5th of the total
to his friend Ram. How many chocolates did Ram get?
Solution : - As ‘of’ represent multiplication so,
Problem 2 : - Disha and Shreya were given Rs. 156 as pocket money
by their mother. Disha spent 1/3rd of it and Shreya spent 1/4th of
the remaining. Find how much did each spend?
Solution:
Money spent by Disha
52
1
156 1 x 52
1
x
=
= 52
of 156 =
3
1
3
1
Remaining Money = 156 – 52 = 104
26
1 104 1 x 26
1
=
= 26
of 104 = x
Money spent by Shreya =
4
1
4
1
Exercise 2.1
Multiply 1.
9
(c)
1
9
27
20
27 20 9 x 1
x
=
=
x
=
40 21 2 x 7
14
40
21
2
9
7
1
9
81 25
81 25 9 x 1
(f)
x
=
=
x
=
16
100 36 100 36 4 x 4
4
Problem 3 : - In a class of 50 students, 1/5th of the total like
to play football, 3/5th like to play cricket, and the remaining
play basketball. Find the number of students who played :
Basketball, Football and Cricket.
Solution:
Total class strength = 50
10
Student like Football =
4
1
of 50
5
=
1
Multiply 2.
(g)
(i)
1
1
13
7
91
1
4 x 3 =
x =
= 15
3
2 3 2 6
6
1
1
50
x
= 10
5
10
Student like Cricket =
3
of 50
5
=
3
50
x
= 30
5
1
1
25 37 925
1
8 x9 =
x
=
= 77
3 4
3 4
12
12
Now, Remaining student like Basketball
= 50 – 10 – 30 = 10
Exercise 2.1
Problem 5 : Raj reads
1/4th
if a book in 1 hour. How much part of the book will he read in 3
1
5 hours?
Solution:
Raj reads 1/4th book in = 1 Hour
Therefore, time need to read full book = 4 Hours
4
The part of the book to be read in 3
4
1
1
1
16
1
hours =
=
x 3 =
x
5
4
5
4
5
5
1
More Examples
Find
=
= 9
2
3
1
2
3
1
of
x
3
4
1
3
4
2
1
of
18
9
x
18
2
Find
=
3
1
2
3
1
= 40
of
x
3
4
1
3
4
2
1
of 80
40
x
80
Reciprocal of a Fraction
The reciprocal (or the multiplicative inverse) of a fraction is a new fraction in which
the places of the numerator and denominator are interchanged. If ‘a/b’ is a fraction
then ‘b/a’ is reciprocal of fraction ‘a/b’.
Examples :
Reciprocal of 5 is
1
5
3
5
Reciprocal of
is
5
3
Division of Fractions
Dividing a Whole Number by a Fraction : To divide a whole number by a fraction,
we multiply the whole number by the reciprocal of the fraction.
1
6 ÷
2
= 6 x
2
= 12
1
Dividing a Fraction by a Whole Number : To divide a fraction by a whole number ,
we multiply the fraction by the reciprocal of the whole number.
1
3
÷ 9 =
4
3
1
1
1
x
=
=
9
4x3
12
4
3
Division of Fractions
Dividing a Fraction by a Fraction: To divide a fraction by a fraction , we multiply the
fraction by the reciprocal of the other fraction.
1
3
3
÷
=
4
20
5
3
20
1x5
5
x
=
=
4
3
1x1
1
1
1
=5
Exercise 2.2
Problem 3. : Solve
Problem 1. : Find the Reciprocal
3
5
(a). Reciprocal of
5 is
(d). Reciprocal of
is
e.
5
3
d.
5
1
÷
4
=
5
5
x
4
1
5
1
=
3x4
5x1
f.
2
2
3
÷
5
=
x
5
=
12
5
= 2
2
5
5
2
x
5
3
=
5x5
2x3
=
25
6
2
2
x
1
= 4
1
6
5
=
and
and
1
1x2
=
5x1
5
2
1
÷
2
1
3
1
1
Problem 2. : Find
3
1
2
5
1
x 2
5
1
5
x
and
2
1
1
5
x
2
1
=
1x2
5x1
Problem 5. : Identify pattern and write next three numbers in the series.
d.
1
, 1,
2
1
1 , 2,
2
1 , , ,
2
2
1
1
1
1
, 1, 1 , 2 2 , 3, 3 , 4
2
2
2
2
e.
1 1 3
1
1
, , , 1 , 1 , 1 , __ , __ , __
4 2 4
2
2
1 1 3
1
1
3
1
, , , 1 , 1 , 1 , 1 , 2 ,2
4 2 4
4
2
4
4
Problem 7. : Write = or = in the colored box without actually working out the answers.
a.
2
3 5
+1
+
3
4 6
3
5
2
+ +1
4
6
3
2
5
2
3
3 5
+1
=
+ +1
+
3
4
6
3
4 6
Associative property of Addition, (a + b) +c = a + (b + c)
For other problems, check for the properties of integers for Division, Multiplication and Subtraction
Problem 8. : A farmer had a square plot of land. He gave ¼ of it to his wife as shown in figure.
He asked his four sons to share the remaining portion of the land equally and he said that all of
them must get the land equally and in the same shape. Help the sons to divide the land.
Solution : ¼ of the land given to the wife.
For wife
1 4–1 3
=
Hence, remaining part of the plot = 1 – =
4
4
4
Now, remaining part of land to be divided equally among the 4 sons = (Remaining Land) ÷ (4)
=
3
4
÷ 4 =
3
4x4
=
3
16
Word Problems
Solved Examples
Problem 1 : The product of a fraction and the difference of 3
Solution : Missing fraction x 3
Missing fraction x
5
1
– 2
6
3
23
7
–
6
3
23
7x2
Missing fraction x
–
6
3x2
23
14
Missing fraction x
–
6
6
23 – 14
Missing fraction x
6
Missing fraction x
9
= 1
6
5
1
and 2
is 1. Find the fraction.
6
3
= 1
= 1
= 1
Missing fraction = 1 ÷
2
= 1
Missing fraction =
6
9
3
= 1
Missing fraction =
2
3
9
6
Exercise 2.3
Problem 1 : A 440m length of road is being repaired. After one day, the workers had repaired ¾ of
the road. What length of the road is left unrepaired?
Solution : Total length of road being repaired = 440 m
Length of road repaired in one day = ¾ of 440 m
110
3
3
440
= 330 m
of 440 =
x
4
1
4
1
Therefore, road remained unrepaired = 440 – 330 = 110 m
Problem 2 : 1/3 of the 6 dozen fruits in a fruit seller’s cart are apples. If ¼ of all the fruits are
oranges and the rest of the fruits are bananas, find how many dozen bananas are there in the fruit
seller’s cart?
Solution : Total fruits in fruit seller’s cart = 6 dozen
Apples = 1/3 of 6 dozen
1
1
1
6
= 2 dozen
of 6 =
x
3
1
3
1
Now, Oranges = 1/4 of 6 dozen =
3
3
1
1 6
=
dozen
of 6 =
x
4 1
2
4
2
3
12 4 3
6
–
2
–
=
– –
Therefore, left fruits are bananas =
2
2
2 2
=
12 –4 – 3
5
=
=
2
2
2
Multiply and divide 6 and 2 by 2 to
get equal denominator
1
dozen
2
Or 2.5 dozen
Problem 3 : 1/8th of the passengers of a train were children. If there were 40 children
travelling in the train on a certain day, how many adults were there in that train that day?
Solution : 1/8 of the total passengers = Children
1/8 x Total passengers = 40
Total Passengers = 40 ÷ 1/8
8
= 40 x
= 320
1
Therefore, Adults = Total passengers – children = 320 – 40 = 280
Problem 4 : In a class of 48 students, ¼ of them regularly watch a particular TV programme.
How many of the students do not regularly watch the programme?
Solution : Students watch a particular TV programme = ¼ of the 48
12
=
1
48
x
= 12
4
1
1
Therefore, students do not watch a particular TV programme regularly = 48 – 12 = 36
Problem 5 : The product of a fraction with the sum of 5
Solution : (Fraction) x 5 1 + 6 1 =3
3
3
(Fraction) x
16
19
+
3
3
(Fraction) x
16 + 19
3
(Fraction) x
35
3
(Fraction) = 3 ÷
=3
=3
=3
35
3x3
=
3
35
Therefore, Fraction =
9
35
1
1
and 6 is 3 . What is the fraction?
3
3
Problem 6 : A tree is standing by the side of a building. 2/5th of the height of the tree is above the
building level. The building is 9 meters high. What is the height of the tree?
Solution : Height of the tree = (Height of the building) + (Height of the tree above building level)
2
x (Height of Tree) = 9
5
(Height of Tree) –
Height of Tree –
2 Height of Tree
= 9
5
5 (Height of tree) – 2 (Height of tree)
=
5
3 (Height of tree)
=
5
Height of tree = 9 ÷
9
9
3
5
3
Height of tree = 9 x
5
3
1
Hence, Height of tree = 3 x 5 = 15 meters
Problem 7 : When a fraction is divided by the difference of 1/2 and 1/6, we get 2/3. What is the fraction?
Solution : As per statement, Fraction ÷ (1/2 – 1/6) = 2/3
(Fraction) ÷
1
1
–
2
6
(Fraction) ÷
3–1
6
2
3
=
=
2
3
1
2
6
Fraction ÷
=
2
3
3
Fraction x
3
2
=
1
3
Fraction =
2
3
÷
3
1
Fraction =
2
3
x
Hence, Fraction =
1
3
2x1
2
=
3x3
9
Problem 8 : Jaya had three ribbons with her. The length of the green ribbon was 12 meters 50 cm.
The red ribbon was 3/5 of the length of the green ribbon. The yellow ribbon was 2½ times the red
ribbon. What is the length of the yellow ribbon.?
Solution : As per statement, Length of Green ribbon = 12.50 meters
Therefore, length of Red ribbon = 3/5 of (length of Green ribbon)
5
25
3
15
3 x 1250
= x 12.50 =
=
= 7.50 meters
5 x 100
2
5
1 2
Length of Yellow ribbon = 2½ of (Length of Red ribbon)
1
15
5
15
= 2
x
=
x
2
2
2
2
=
75
= 18.75 meters
4
Or
= 2
1
5 7.50
x 7.50 =
x
2
2
1
3.75
5
7.50
= 5 x 3.75 = 18.75 meters
=
x
2
1
1
Problem 9 : In a flower bouquet, 13/30 of the flowers were roses, 2/5 of the flowers were orchids and the
remaining were gladioli. If the number of gladiola flowers were 7, how many flowers were there in the
bouquet?
Solution : As per statement,
Roses = 13/30 of Bouquet
Orchids = 2/5 of Bouquet
Gladiola = 7
Flowers in Bouquet = (Roses + Orchids + Gladiola)
13
2
of Flowers in Bouquet +
of Flowers in Bouquet + 7
30
5
Let, flowers in Bouquet = a
Therefore,
=
a =
13 a
2 a
+
+ 7
30
5
a
13a
2a
–
–
= 7
1
30
5
1
5a
30
=
7
6
30a 13a 12a
–
–
= 7
30
30
30
30a - 13a - 12a
= 7
30
a
6
=
7
Hence, Flowers in Bouquet is a = 7 x 6 = 42
Problem 10 : 1800 people attended a wedding reception. 7/18 of them were men and 11/24 of them were
ladies. The remaining were children. What fraction of them were children and how many children were
present?
Solution : As per statement,
Men =
Men = 7/18 of 1800
Ladies = 11/24 of 1800
Children = 1800 – Men - Ladies
7
7
of 1800 =
x 1800
18
18
100
Men =
7 1800
x
= 7 x 100 = 700
18
1
Now, Ladies
Ladies=
11
11
of 1800 =
x 1800
24
24
75
Ladies =
Hence, Children = 1800 – 700 – 825 = 1800 – 1525 = 275
11 1800
x
= 11 x 75 = 825
24
1
Fraction of Children attended the wedding =
11
55
11
275
=
1800
72
360
72
Problem 11 : Rohan has a collection of stamps. His collection is 3¾ times that of Pranav’s collection of stamps
which is 320 stamps. How many stamps does Rohan have?
Solution : As per statement,
Rohan’s collection = 3¾ of Pranav’s collection
= 3
3
4
of 320
80
=
15
4
x 320
1
Therefore Rohan’s Collection of stamps = 15 x 80 = 1200
Problem 12 : Ankitesh and his friend Jimmy together sold tickets for a charity show for Rs. 3800. If
3
Ankitesh sold 1
times as that by Jimmy, how much did each of them sell?
8
3
Solution : As per statement,
Ankitesh’s sold tickets = 1
of Jimmy’s sold tickets
8
Now, Jimmy’s sold tickets + Ankitesh’s sold tickets = 3800
Jimmy’s sold tickets + 11/8 Jimmy’s sold tickets = 3800
Jimmy’s sold tickets x 1 + 11
1
8
= 3800
Jimmy’s sold tickets x 8 + 11
8
8
= 3800
Jimmy’s sold tickets x 8 + 11
8
Jimmy’s sold tickets x
19
8
Jimmy’s sold tickets = 3800 ÷
200
= 3800
Jimmy’s sold tickets =
3800 x
8
19
1
= 3800
19
8
= 200 x 8 = 1600
Jimmy’s sold tickets = 1600
Ankitesh’s sold tickets = 3800 – 1600 = 2200
Some More Solved Problems
Problem : How many months are there in 2/5 of a century?
Solution : One century = 100 Years
One year = 12 Months
Now,
20
2
2 100
= 2 x 20 = 40 years
of Century = x
5
5
1
Hence, months in 2/5 of the century = 40 x 12 = 480 months
Problem : How many seconds are there in 2/3 of 2 hours?
Solution : One Hour = 3600 Seconds ------------(as One Hour has 60 minutes and 1 minute has 60 seconds)
So, 2 Hours = 2 x 3600 = 7200 Seconds
Now,
2400
2
2 7200
= 2 x 2400 = 4800 Seconds
of 2 hours = x
3
3
1
Hence, months in 2/5 of 2 hours = 4800 Seconds
Some More Solved Problems
Problem : A designer uses 2/3 meter of cloth to make a blouse for a dancer. If 150 students are
taking part in the dance and those many blouses are needed, then how many meters if cloth will
be designer be using?
Solution : One student require cloth for blouse = 2/3 of one meter
Total students taking part in dance = 150
Therefore, total cloth the designer using = 150 x (2/3 of meter)
30
=
2
150
x
=
3
2 x 30 = 60 meters
1
Problem : There are 120 seats in the balcony of a theatre. If this is 1/5th of the total seats, what is
the total number of seats in the theatre?
Solution : Seats in Balcony = 120
Therefore, 1/5 of Total seats = 150 (seats in balcony)
Now,
1
of Total seats = 120
5
Total seats = 120 ÷
1
x Total seats = 120
5
Total seats = 120 x
1
5
5
= 120 x 5 = 600 seats
1
Hence, total seats in Theatre = 600 Seats