RBSE Class 6 Maths Chapter 7 Fractions Solutions

Practicing RBSE Class 6 Maths Solutions and Class 6 Maths Chapter 7 Fractions Solutions Question Answer helps develop logical thinking and accuracy.

Fractions Class 6 Solutions

Ganita Prakash Class 6 Chapter 7 Solutions Fractions

Figure it Out (Page 152)

Fill in the blanks with fractions.
1. Three guavas together weigh 1 kg. If they are roughly of the same size, each guava will roughly weigh ___________ kg.
2. A wholesale merchant packed 1 kg of rice in four packets of equal weight. The weight of each packet is ___________ kg.
3. Four friends ordered 3 glasses of sugarcane juice and shared it equally among themselves. Each one drank ___________ glass of sugarcane juice.
4. The big fish weighs \(\frac{1}{2}\) kg. The small one weighs \(\frac{1}{4}\) kg. Together they weigh ___________ kg.
5. Arrange these fraction words in order of size from the smallest to the biggest in the empty box below:
One and a half, three quarters, one and a quarter, half, quarter, two and a half.

Solution:
1. \(\frac{1}{3}\)
2. \(\frac{1}{4}\)
3. \(\frac{3}{4}\)
4. \(\frac{3}{4}\)
5. Quarter < half < three quarter < one and a quarter < one and a half < two and a half

Figure it Out (Page 155)

Question 1.
The figures below show different fractional units of a whole chikki. How much of a whole chikki is each piece?

Solution:
A whole chikki is as follows :

On this basis, we can measure the fractional unit of the given chikki pieces.
(a) We get this piece by breaking the whole chikki into 12 equal pieces. So this is \(\frac{1}{12}\) chikki.

(b) We get this piece by dividing the whole chikki into 4 equal parts. So this is \(\frac{1}{4}\) chikki.

(c) We get this piece by dividing the whole chikki into 8 equal parts. So this is \(\frac{1}{8}\) chikki.

(d) We get this piece by dividing the whole chikki into 6 equal parts. So this is \(\frac{1}{6}\) chikki.

(e) We get this piece by dividing the whole chikki into 8 equal parts. So this is \(\frac{1}{8}\) chikki.

(f) We get this piece by dividing the whole chikki into 6 equal parts. So this is \(\frac{1}{6}\) chikki.

(g) We get this piece by breaking the whole chikki into 24 equal parts. So this is \(\frac{1}{24}\) chikki.

(h) We get this piece by breaking the whole chikki into 24 equal parts. So this is \(\frac{1}{24}\) chikki.

Figure it Out (Page 158)

Question 1.
Continue this table of \(\frac{1}{2}\) for 2 more steps.
Solution:

Two more steps :

Question 2.
Can you create a similar table for \(\frac{1}{4}\) ?
Solution:
Yes, we can create a similar table for \(\frac{1}{4}\).

Question 3.
Make \(\frac{1}{3}\) using a paper strip. Can you use this to also make \(\frac{1}{6}\) ?
Solution:

Yes, we can use it to also make \(\frac{1}{6}\). We can get \(\frac{1}{6}\)
\(\frac{1}{6}\) by dividing \(\frac{1}{3}\) strip into two equal parts.

\(\frac{1}{3}\) × \(\frac{1}{2}\) = \(\frac{1}{6}\)

Question 4.
Draw a picture and write an addition statement as above to show:
(a) 5 times \(\frac{1}{4}\) of a roti
(b) 9 times \(\frac{1}{4}\) of a roti
Solution:

Question 5.
Match each fractional unit with the correct picture:

Solution:

Figure it Out (Page 160)

Question 1.
On a number line, draw lines of lengths \(\frac{1}{10}\), \(\frac{3}{10}\) and \(\frac{4}{5}\).
Solution:

(i) OA = \(\frac{1}{10}\)
(ii) OC = \(\frac{3}{10}\)
(iii) OH = \(\frac{8}{10}\) = \(\frac{4}{5}\)

Question 2.
Write five more fractions of your choice and mark them on the number line.
Solution:

(1) OA = \(\frac{1}{7}\)
(2) OB = \(\frac{2}{7}\)
(3) OC = \(\frac{3}{7}\)
(4) OE = \(\frac{5}{7}\)
(5) OF = \(\frac{6}{7}\)

Question 3.
How many fractions lie between 0 and 1 ? Think, discuss with your classmates, and write your answer.
Solution:
There are infinite fractions between 0 and
1, e.g. \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{8}\), \(\frac{1}{16}\), \(\frac{1}{32}\), \(\frac{1}{64}\) ………………..

Question 4.
What is the length of the blue line and black line shown below? The distance between 0 and 1 is 1 unit long, and it is divided into two equal parts. The length of each part is \(\frac{1}{2}\). So the blue line is \(\frac{1}{2}\) unit long. Write the fraction that gives the length of the black line in the box.

Solution:
Length of the blue line = \(\frac{1}{2}\) unit
Length of the black line = \(\frac{1}{2}\) + \(\frac{1}{2}\) + \(\frac{1}{2}\)
= 3 × \(\frac{1}{2}\) = \(\frac{3}{2}\) units
∴ Required fraction =

Question 5.
Write the fraction that gives the lengths of the black lines in the respective boxes.

Solution:

Figure it Out (Page 162)

Question 1.
How many whole units are there in \(\frac{7}{2}\)?
Solution:

Hence, \(\frac{7}{2}\) contains 3 whole units.

Question 2.
How many whole units are there in \(\frac{4}{3}\) and in \(\frac{7}{3}\)?
Solution:
\(\frac{4}{3}\) = 4 × \(\frac{1}{3}\) = 4 times \(\frac{1}{3}\)
= \(\frac{1}{3}\) + \(\frac{1}{3}\) + \(\frac{1}{3}\) + \(\frac{1}{3}\)

So, there are 2 whole units in \(\frac{7}{3}\).

Figure it Out (Page 162)

Question 1.
Figure out the number of whole units in each of the following fractions:
(a) \(\frac{8}{3}\)
(b) \(\frac{11}{5}\)
(c) \(\frac{9}{4}\)
We saw that

This number is thus also called ‘two and two thirds’. We also write it as 2\(\frac{2}{3}\).
Solution:

Question 2.
Can all fractions greater than 1 be written as such mixed numbers?
Solution:
Yes, all fractions greater than 1 can be written as mixed numbers.

Question 3.
Write the following fractions as mixed fractions (e.g., \(\frac{9}{2}\) = 4\(\frac{1}{2}\))
a) \(\frac{9}{2}\)
b) \(\frac{9}{5}\)
c) \(\frac{21}{19}\)
d) \(\frac{47}{9}\)
e) \(\frac{12}{11}\)
f) \(\frac{19}{6}\)
Solution:

Figure it Out (Page 163)

Question 1.
Write the following mixed numbers as fractions:
(a) 3\(\frac{1}{4}\)
(b) 7\(\frac{2}{3}\)
(c) 9\(\frac{4}{9}\)
(d) 3\(\frac{1}{6}\)
(e) 2\(\frac{3}{11}\)
(f) 3\(\frac{9}{10}\)
Solution:

Figure it Out (Page 165)

Question 1.
Are \(\frac{3}{6}\), \(\frac{4}{8}\), \(\frac{5}{10}\)equivalent fractions? Why?
Solution:
Here, the simplest form of \(\frac{3}{6}\) = \(\frac{3 \div 3}{6 \div 3}\) = \(\frac{1}{2}\)
[∵ HCF of 3 and 6 = 3]
\(\frac{4}{8}\) = \(\frac{4 \div 4}{8 \div 4}\) = \(\frac{1}{2}\)
[∵ HCF (4, 8) = 4]
The simplest form of \(\frac{5}{10}\) = \(\frac{5 \div 5}{10 \div 5}\) = \(\frac{1}{2}\)
[∵ HCF (5, 10) = 5]

So, \(\frac{3}{6}\), \(\frac{4}{8}\), \(\frac{5}{10}\) are equivalent fractions, as the simplest form all of these fractions is \(\frac{1}{2}\).

Question 2.
Write two equivalent fractions for \(\frac{2}{6}\).
Solution:
Two equivalent fractions for
\(\frac{2}{6}\) = \(\frac{2 \times 2}{6 \times 2}\), \(\frac{2 \times 3}{6 \times 3}\)
= \(\frac{4}{12}\), \(\frac{6}{18}\)

Question 3.
\(\frac{4}{6}\) = …………………… (write as many as you can)
Solution:

Figure it Out (Page 166)

Question 1.
Three rods are shared equally by four children. Show the division in the picture and write a fraction for how much each child gets. Also, write the corresponding division facts, addition facts, and, multiplication facts. Fraction of rod each child gets is _____________ .
Division fact:
Addition fact:
Multiplication fact:
Compare your picture and answers with your classmates!

Solution:
Here, 3 rotis are shared equally by four children.

(a) Division Facts: Dividing three whole units in 4 equal parts = 3 ÷ 4 = \(\frac{3}{4}\)

(b) Addition Facts: Adding \(\frac{3}{4}\) four times gives
whole 3, i.e. \(\frac{3}{4}\) + \(\frac{3}{4}\) + \(\frac{3}{4}\) + \(\frac{3}{4}\) = \(\frac{12}{4}\) = 3

(c) Multiplication Facts: Multiplying \(\frac{3}{4}\) by 4 we get whole 3, i.e., 4 × \(\frac{3}{4}\) = 3

Question 2.
Draw a picture to show how much each child gets when 2 rotis are shared equally by 4 children. Also, write the corresponding division facts, addition facts, and multiplication facts.
Solution:
Here, 2 rotis are shared equally by 4 children, so we divide each roti in 4 equal parts. The share of each child is as follows—

(a) Division facts : Dividing 2 whole units in 4 equal parts = 2 ÷ 4 = \(\frac{2}{4}\) = \(\frac{1}{2}\)

(b) Addition facts : \(\frac{2}{4}\) + \(\frac{2}{4}\) + \(\frac{2}{4}\) + \(\frac{2}{4}\) + \(\frac{8}{4}\) = 2

(c) Multiplication facts : 4 × \(\frac{2}{4}\) = 2

Question 3.
Anil was in a group where 2 cakes were divided equally among 5 children. How much cake would Anil get?


Solution:
2 cakes were divided equally among 5 children. So, Anil would get = 2 ÷ 5 = \(\frac{2}{5}\) cake

Figure it Out (Page 168)

Find the missing numbers :
(a) 5 glasses of juice shared equally among 4 friends is the same as ___________ glasses of juice shared equally among 8 friends.
So,

(b) 4 kg of potatoes divided equally in 3 bags is the same as 12 kgs of potatoes divided equally in ___________ bags.
So,

(c) 7 rotis divided among 5 children is the same as ________ _________ rotis divided among ________ _________ children.
So,
Solution:
(a) Here, 5 glasses of juice are shared equally among 4 friends, so share of each friend = \(\frac{5}{4}\).
Now to determine how many glasses ofjuice would be needed to each of the 8 friends the same amount
= 8 × \(\frac{5}{4}\) = \(\frac{40}{4}\) = 10 glasses

So, 10 glasses of juice shared equally among 4 friends is the same as 5 glass of juice shared equally among 4 friends.
∴

(b) Here, 4 kg of potatoes divided equally in 3 bags then amount of potatoes per bag
= \(\frac{4 \mathrm{~kg}}{3 \text { bags }}\) = \(\frac{3}{4}\)kg per bag
Let x is the number of bags for 12 kg of potatoes. Where each bag has the same amount of potatoes then,

(c) Dividing 7 rotis among 5 children, share of each child = \(\frac{7}{5}\) of a roti.
We can find an equivalent fraction by multiplying both the numerator and denominator by the same number i.e. 2.
\(\frac{7 \times 2}{5 \times 2}\) = \(\frac{14}{10}\)
So, 7 rotis divided among 5 children is the same as 14 rotis divided among 10 children.
∴

Figure it Out (Page 173)

Question 1.
Express the following fractions in lowest terms:
a) \(\frac{17}{51}\)
b) \(\frac{64}{144}\)
c) \(\frac{126}{147}\)
d) \(\frac{525}{112}\)
Solution:

Figure it Out (Page 174)

Question 1.
Compare the following fractions and justify your answers :
a) \(\frac{8}{3}\), \(\frac{5}{2}\)
b) \(\frac{4}{9}\), \(\frac{3}{7}\)
c) \(\frac{7}{10}\), \(\frac{9}{14}\)
d) \(\frac{12}{5}\), \(\frac{8}{5}\)
e) \(\frac{9}{4}\), \(\frac{5}{2}\)
Solution:


(d) The fractions are : \(\frac{12}{5}\), \(\frac{8}{5}\)
Here, the denominators of both fractions are same.
∴ \(\frac{12}{5}\) > \(\frac{8}{5}\)

(e) The given fractions are: \(\frac{9}{4}\), \(\frac{5}{2}\)
LCM of 4 and 2 is 4, so \(\frac{5}{2}\) = \(\frac{5 \times 2}{2 \times 2}\) = \(\frac{10}{4}\)
Here, \(\frac{9}{4}\) < \(\frac{10}{4}\)
So, \(\frac{9}{4}\) < \(\frac{5}{2}\)

Question 2.
Write the following fractions in ascending order.
a) \(\frac{7}{10}\), \(\frac{11}{15}\), \(\frac{2}{5}\)
b) \(\frac{19}{24}\), \(\frac{5}{6}\), \(\frac{7}{12}\)
Solution:
(a) The given fractions are : a) \(\frac{7}{10}\), \(\frac{11}{15}\), \(\frac{2}{5}\)
Here LCM of denominators 10, 15, 5


So, the ascending order of fractions : a) \(\frac{7}{12}\), \(\frac{19}{24}\), \(\frac{5}{6}\)

Question 3.
Write the following fractions in descending order.
(a) \(\frac{25}{16}\), \(\frac{7}{8}\), \(\frac{13}{4}\), \(\frac{17}{32}\)
(b) \(\frac{3}{4}\), \(\frac{12}{5}\), \(\frac{7}{12}\), \(\frac{5}{4}\)
Solution:
(a) The given fractions are : (a) \(\frac{25}{16}\), \(\frac{7}{8}\), \(\frac{13}{4}\), \(\frac{17}{32}\)
Finding LCM of 16, 8, 4, 32


(b) The given fractions are : \(\frac{3}{4}\), \(\frac{12}{5}\), \(\frac{7}{12}\), \(\frac{5}{4}\)
LCM of 4, 5, 12 and 4

Figure it Out (Page 179)

Question 1.
Add the following fractions using Brahmagupta’s method:
a) \(\frac{2}{7}\) + \(\frac{5}{7}\) + \(\frac{6}{7}\)
b) \(\frac{3}{4}\) + \(\frac{1}{3}\)
c) \(\frac{2}{3}\) + \(\frac{5}{6}\)
d) \(\frac{2}{3}\) + \(\frac{2}{7}\)
e) \(\frac{3}{4}\) + \(\frac{1}{3}\) + \(\frac{1}{5}\)
f) \(\frac{2}{3}\) + \(\frac{4}{5}\)
g) \(\frac{4}{5}\) + \(\frac{2}{3}\)
h) \(\frac{3}{5}\) + \(\frac{5}{8}\)
i) \(\frac{9}{2}\) + \(\frac{5}{4}\)
j) \(\frac{8}{3}\) + \(\frac{2}{7}\)
k) \(\frac{3}{4}\) + \(\frac{1}{3}\) + \(\frac{1}{5}\)
l) \(\frac{2}{3}\) + \(\frac{4}{5}\) + \(\frac{3}{7}\)
m) \(\frac{9}{2}\) + \(\frac{5}{4}\) + \(\frac{7}{6}\)
Solution:

Question 2.
Rahim mixes \(\frac{2}{3}\) litres of yellow paint with \(\frac{3}{4}\) litres of blue paint to make green paint. What is the volume of green paint he has made?
Solution:
Volume of yellow paint = \(\frac{2}{3}\) litres
Volume of blue paint = \(\frac{3}{4}\) litres
By mixing these two paints—
Volume of green paint = \(\frac{2}{3}\) litres + \(\frac{3}{4}\) litres
= (\(\frac{2}{3}\) + \(\frac{3}{4}\)) litres
= (\(\frac{2 \times 4}{3 \times 4}\) + \(\frac{3 \times 3}{4 \times 3}\)) litres [∵ LCM of 3 and 4 = 12]

Question 3.
Geeta bought \(\frac{2}{5}\) meter of lace and Shamim bought \(\frac{3}{4}\) meter of the same lace to put a complete border on a table cloth whose perimeter is 1 meter long. Find the total length of the lace they both have bought. Will the lace be sufficient to cover the whole border?
Solution:
Length of lace bought by Geeta = \(\frac{2}{5}\) meter
Length of lace bought by Shamim = \(\frac{3}{4}\) meter
∴ Total length of the lace = (\(\frac{2}{5}\) + \(\frac{3}{4}\)) meter
= (\(\frac{2 \times 4}{5 \times 4}\) + \(\frac{3 \times 5}{4 \times 5}\)) meter [∵ LCM 5, 4 = 20]
= (\(\frac{8+15}{20}\)) = \(\frac{23}{20}\) meter
(\(\frac{20}{20}\) + \(\frac{3}{20}\))
(1 + \(\frac{3}{20}\))meter
= 1 \(\frac{3}{20}\) meter
Length of the whole border = Perimeter = 1 meter.
Since 1\(\frac{3}{20}\) m > 1 m, so lace will be sufficient to cover the whole border.

Figure it Out (Page 181)

Question 1.
\(\frac{5}{8}\) – \(\frac{3}{8}\)
Solution:
\(\frac{5}{8}\) – \(\frac{3}{8}\) = \(\frac{5 – 3}{8}\) = \(\frac{2}{8}\) = \(\frac{1}{4}\)

Question 2.
\(\frac{7}{9}\) – \(\frac{5}{9}\)
Solution:
\(\frac{7}{9}\) – \(\frac{5}{9}\) = \(\frac{7 – 5}{9}\) = \(\frac{2}{9}\)

Question 3.
\(\frac{10}{27}\) – \(\frac{1}{27}\)
Solution:
\(\frac{10}{27}\) – \(\frac{1}{27}\) = \(\frac{10 – 1}{27}\) – \(\frac{9}{27}\) = \(\frac{1}{3}\)

Figure it Out (Page 182)

Question 1.
Carry out the following subtractions using Brahmagupta’s method:
a) \(\frac{8}{15}\) – \(\frac{3}{15}\)
b) \(\frac{2}{5}\) – \(\frac{4}{15}\)
c) \(\frac{5}{6}\) – \(\frac{4}{9}\)
d) \(\frac{2}{3}\) – \(\frac{1}{2}\)
Solution:
a) \(\frac{8}{15}\) – \(\frac{3}{15}\)
= a) \(\frac{8 – 3}{15}\) – \(\frac{5}{15}\) = \(\frac{1}{3}\)

b) \(\frac{2}{5}\) – \(\frac{4}{15}\)

Question 2.
Subtract as indicated :
a) \(\frac{13}{4}\) – \(\frac{10}{3}\)
b) \(\frac{18}{5}\) – \(\frac{23}{3}\)
c) \(\frac{29}{7}\) – \(\frac{45}{7}\)
Solution:

Question 3.
Solve the following problems :
(a) Jaya’s school is \(\frac{7}{10}\) km from her home. She takes an auto for \(\frac{1}{2}\) km from her home daily, and then walks the remaining distance to reach her school. How much does she walk daily to reach the school?
(b) Jeevika takes \(\frac{10}{3}\) minutes to take a complete round of the park and her friend
Namit takes \(\frac{13}{4}\) minutes to do the same.
Who takes less time and by how much?
Solution:
(a) Distance between Jaya’s home and school = \(\frac{7}{10}\) km
Distance covered by Jaya by auto = \(\frac{1}{2}\) km
Distance covered by Jaya by walking
= \(\frac{7}{10}\) km – \(\frac{1}{2}\) km
= (\(\frac{7}{10}\) – \(\frac{1}{2}\)) km
= (\(\frac{7}{10}\) – \(\frac{1 \times 5}{2 \times 5}\)) km [∵ LCM of 10 and 2 = 10]
= (\(\frac{7}{10}\) – \(\frac{5}{10}\)) km
= \(\frac{7 – 5}{10}\) = \(\frac{2}{10}\) km = \(\frac{1}{5}\) km
= \(\frac{1}{5}\) × 1000 m = 200 m
Hence, Jaya walks daily \(\frac{1}{5}\) km or 200 m to reach the school.

(b) Time taken by Jeevika to take a complete round = \(\frac{10}{3}\) minutes
Time taken by Namit to take a complete round = \(\frac{10}{3}\) minutes
Comparing these fractions—
\(\frac{10}{3}\) = \(\frac{10 \times 4}{3 \times 4}\) = \(\frac{40}{12}\)
and \(\frac{13 \times 3}{4 \times 3}\) = \(\frac{39}{12}\) [∵ LCM of 3 and 4 = 12]
∵ \(\frac{39}{12}\) < \(\frac{40}{2}\), so Namit takes less time.
Again, \(\frac{40}{12}\) – \(\frac{39}{12}\) = \(\frac{1}{12}\) minutes
Hence, Namit takes \(\frac{1}{12}\) minutes less time.

Fractions Class 6 Question Answer

Fractions Class 6 Extra Questions

Multiple Choice Questions—

Question 1.
The fraction representing the shaded portion is:

(a) \(\frac{2}{3}\)
(b) \(\frac{1}{5}\)
(c) \(\frac{4}{5}\)
(d) \(\frac{5}{5}\)
Answer:
(b) \(\frac{1}{5}\)

Question 2.
Which of the following is an improper fraction?
(a) \(\frac{3}{2}\)
(b) \(\frac{2}{3}\)
(c) \(\frac{0}{2}\)
(d) \(\frac{1 + 2}{3}\)
Answer:
(a) \(\frac{3}{2}\)

Question 3.
The equivalent fraction of \(\frac{2}{5}\) having numerator 6 is :
(a) \(\frac{6}{5}\)
(b) \(\frac{6}{10}\)
(c) \(\frac{6}{15}\)
(d) 7
Answer:
(c) \(\frac{6}{15}\)

Question 4.
The simplest form of \(\frac{36}{54}\) will be :
(a) \(\frac{18}{27}\)
(b) \(\frac{3}{2}\)
(c) \(\frac{2}{3}\)
(d) \(\frac{72}{108}\)
Answer:
(c) \(\frac{2}{3}\)

Question 5.
The appropriate sign between the fractions
\(\frac{1}{3}\) □ \(\frac{1}{5}\) .
(a) <
(b) =
(c) >
(d) None of these
Answer:
(c) >

Question 6.
The sum of \(\frac{1}{2}\) and \(\frac{1}{5}\) will be :
(a) \(\frac{1}{7}\)
(b) \(\frac{2}{7}\)
(c) \(\frac{2}{10}\)
(d) \(\frac{7}{10}\)
Answer:
(d) \(\frac{7}{10}\)

Question 7.
The value of \(\frac{5}{6}\) – \(\frac{3}{4}\) will be :
(a) \(\frac{2}{2}\)
(b) \(\frac{8}{10}\)
(c) \(\frac{15}{24}\)
(d) \(\frac{1}{12}\)
Answer:
(d) \(\frac{1}{12}\)

Question 8.
The equivalent fraction of \(\frac{2}{3}\) is :
(a) \(\frac{10}{15}\)
(b) \(\frac{3}{2}\)
(c) \(\frac{5}{6}\)
(d) \(\frac{15}{10}\)
Answer:
(a) \(\frac{10}{15}\)

Question 9.
The sum of 3\(\frac{1}{5}\) + \(\frac{3}{5}\) will be :
(a) \(\frac{16}{5}\)
(b) \(\frac{19}{5}\)
(c) \(\frac{1}{5}\)
(d) \(\frac{18}{5}\)
Answer:
(b) \(\frac{19}{5}\)

Question 10.
Which of the following fractions is the greatest of all?
\(\frac{1}{5}\), \(\frac{1}{3}\), \(\frac{1}{7}\)
(a) \(\frac{1}{7}\)
(b) \(\frac{1}{5}\)
(c) \(\frac{1}{3}\)
(d) None of these
Answer:
(c) \(\frac{1}{3}\)

Question 11.
The simplest form of \(\frac{48}{72}\) is :
(a) \(\frac{2}{3}\)
(b) \(\frac{4}{6}\)
(c) \(\frac{24}{36}\)
(d) None of these
Answer:
(a) \(\frac{2}{3}\)

Question 12.
The sum of \(\frac{1}{9}\), \(\frac{2}{9}\), and \(\frac{6}{9}\), will be :
(a) \(\frac{12}{9}\)
(b) \(\frac{2}{9}\)
(c) 1
(d) None of these
Answer:
(c) 1

Fill in the blanks—

1. In a proper fraction, …………………….. is always smaller than …………………….. .
Answer:
numerator, denominator

2. In a …………………….. fraction, one part is whole and one part is fraction.
Answer:
mixed

3. The equivalent fraction of \(\frac{1}{9}\) =
Answer:
15

4. \(\frac{5}{28}\) + \(\frac{4}{28}\) = …………………….. .
Answer:
\(\frac{9}{28}\)

5. Denominator of fraction \(\frac{7}{12}\) is …………………….. .
Answer:
12

6. If – \(\frac{1}{3}\) = \(\frac{1}{3}\) then correct fraction in the  will be …………………. .
Answer:
\(\frac{2}{3}\)

7. The largest of the fractions \(\frac{1}{5}\), \(\frac{1}{7}\), \(\frac{1}{11}\) is …………………. .
Answer:
\(\frac{1}{5}\)

8. Converting the mix fraction 4\(\frac{2}{5}\) as an improper fraction, we get …………………….. .
Answer:
\(\frac{22}{5}\)

Write True/False for the following statements—

1. In a proper fraction denominator is always smaller than numerator. (True/False)
2. Improper fraction can be converted into mixed fraction. (True/False)
3. Two fractions are called equivalent fractions if both of these represent same quantity. (True/Falsc)
4. Fractions whose numerators are not equal are not equivalent fractions. (True/False)
Answer:
1. False
2. True
3. True
4. False

Make the right match—

Question 1.


Answer:
(i) – (b), (ii) – (a). (iii) – (d). (iv) – (c).

Question 2.

part (A)	Part (B)
(i) \(\frac{48}{60}\)	(a) \(\frac{3}{13}\)
(ii) \(\frac{150}{60}\)	(b) \(\frac{1}{4}\)
(iii) \(\frac{84}{98}\)	(c) \(\frac{4}{5}\)
(iv) \(\frac{12}{52}\)	(d) \(\frac{6}{7}\)
(v) \(\frac{7}{28}\)	(e) \(\frac{5}{2}\)

Answer:
(i) – (c), (ii) – (e). (iii) – (d). (iv) – (a), (v) – (b).

part (A)	Part (B)
(i) \(\frac{48}{60}\)	(c) \(\frac{4}{5}\)
(ii) \(\frac{150}{60}\)	(e) \(\frac{5}{2}\)
(iii) \(\frac{84}{98}\)	(d) \(\frac{6}{7}\)
(iv) \(\frac{12}{52}\)	(a) \(\frac{3}{13}\)
(v) \(\frac{7}{28}\)	(b) \(\frac{1}{4}\)

Very Short Answer Type Questions—

Question 1.
Show \(\frac{1}{3}\) on a number line.
Solution:

Question 2.
Write two improper fractions with denominator 7.
Solution:
\(\frac{8}{7}\), \(\frac{12}{7}\)

Question 3.
Express \(\frac{11}{3}\) as mixed fraction.
Solution:
\(\frac{11}{3}\) = \(\frac{9 + 2}{3}\) = \(\frac{9}{3}\) + \(\frac{2}{3}\) = 3 + \(\frac{2}{3}\) = 3\(\frac{2}{3}\)

Question 4.
Find the difference between \(\frac{5}{6}\) and \(\frac{2}{6}\).
Solution:
\(\frac{5}{6}\) – \(\frac{2}{6}\) = \(\frac{3}{6}\) i.e. \(\frac{1}{2}\)

Question 5.
Find two equivalent fractions of \(\frac{3}{5}\).
Solution:
\(\frac{3 \times 2}{5 \times 2}\) = \(\frac{6}{10}\), \(\frac{3 \times 3}{5 \times 3}\) = \(\frac{9}{15}\)

Question 6.
Write three improper fractions.
Solution:
\(\frac{5}{4}\), \(\frac{6}{5}\), \(\frac{7}{2}\)

Question 7.
Find the equivalent fraction of \(\frac{2}{9}\) having denominator 63.
Solution:

Question 8.
Express \(\frac{17}{4}\) as mixed fraction.
Solution:

Question 9.
Kanchan dyes dresses. She had to dye 30 dresses. She has dyed 20 dresses so far. What fraction of dresses has she finished?
Solution:
Kanchan had to dye dresses = 30
She has dyed dresses = 20
∴ Required fraction = \(\frac{20}{30}\) = \(\frac{20 \div 10}{30 \div 10}\) = \(\frac{2}{3}\)

Question 10.
Write the natural numbers from 2 to 12. What fraction of them are prime numbers?
Solution:
Total natural numbers from 2 to 12
= 11{2, 3, 4, 5, 6, 7, 8,9, 10, 11, 12}
Total prime numbers from 2 to 12 = 5(2, 3, 5, 7, 11)
∴ Required fraction = \(\frac{5}{11}\)

Question 11.
Write the natural numbers from 102 to 113. What fraction of them are prime numbers?
Solution:
Total natural numbers from 102 to 113 = 12{102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113}
Total prime numbers from 102 to 113 = 4{103, 107, 109. 113}
∴ Required fraction = \(\frac{4}{12}\) = \(\frac{4 \div 4}{12 \div 4}\) = \(\frac{1}{3}\)

Question 12.
What fraction of an hour is 40 minutes?
Solution:
Number of minutes in an hour = 60
Given minutes = 40
∴ Required fraction: \(\frac{40}{60}\) = \(\frac{40 \div 20}{60 \div 20}\) = \(\frac{2}{3}\)

Short Answer Type Questions—

Question 1.
Find the equivalent fraction of \(\frac{36}{48}\) having :
(a) numerator 9
(b) denominator 4
Solution:
(a) ∵ 36 ÷ 9 = 4, so we divide the numerator and denominator by 4 :
\(\frac{36}{48}\) = \(\frac{36 \div 4}{48 \div 4}\) = \(\frac{9}{12}\)

(b) ∵ 48 ÷ 4 = 12, so we divide the numerator and denominator by 12 :
\(\frac{36}{48}\) = \(\frac{36 \div 12}{48 \div 12}\) = \(\frac{3}{4}\)

Question 2.
Add \(\frac{2}{5}\) and \(\frac{3}{7}\).
Solution:
\(\frac{2}{5}\) and \(\frac{3}{7}\)
LCM of 5 and 7 = 5 × 7 = 35, then we have :

Question 3.
Subtract \(\frac{2}{5}\) from \(\frac{5}{7}\).
Solution:
\(\frac{5}{7}\) – \(\frac{2}{5}\)
LCM of 7 and 5 = 7 × 5 = 35, then we get
\(\frac{5}{7}\) – \(\frac{2}{5}\) = \(\frac{5 \times 5}{7 \times 5}\) – \(\frac{2 \times 7}{5 \times 7}\) = \(\frac{25}{35}\) – \(\frac{14}{35}\)
= \(\frac{25 – 14}{35}\) – \(\frac{11}{35}\)

Question 4.
Ila read 25 pages of a book containing 100 pages. Lalita read \(\frac{1}{2}\) of the same book. Who read less?
Fraction of pages Ila read = \(\frac{25}{100}\)
Fraction of pages Lalita read
= \(\frac{1}{2}\) = \(\frac{1 \times 50}{2 \times 50}\) = \(\frac{50}{100}\)
Clearly, \(\frac{25}{100}\) < \(\frac{50}{100}\), thus Ila read less.

Question 5.
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 \(\frac{3}{6}\) of an hour.
Rohit exercised \(\frac{3}{4}\) of an hour.
Now, \(\frac{3}{6}\) = \(\frac{3 \times 2}{6 \times 2}\) = \(\frac{6}{12}\)
and, \(\frac{3}{4}\) = \(\frac{3 \times 3}{4 \times 3}\) = \(\frac{9}{12}\)
∴ \(\frac{9}{12}\) > \(\frac{6}{12}\) i.e., \(\frac{3}{4}\) > \(\frac{3}{6}\)
Therefore, Rohit exercised for a longer time.

Question 6.
Yasha bought \(\frac{2}{5}\) meter of ribbon and
Lalita bought \(\frac{1}{4}\) meter of ribbon.
What is the total length of ribbon they bought?
Solution:
Length of the ribbon bought by Yasha = \(\frac{2}{5}\)m
Length of the ribbon bought by Lalita = \(\frac{1}{4}\) m
Total length of the ribbon bought (\(\frac{2}{5}\) + \(\frac{1}{4}\)) m
= (\(\frac{2 \times 4}{5 \times 4}\) + \(\frac{1 \times 5}{4 \times 5}\)) m = (\(\frac{8}{20}\) + \(\frac{5}{20}\)) m
\(\frac{8 + 5}{20}\) = \(\frac{13}{20}\) m

Essay Type Questions-—

Question 1.
Asha and Samuel have bookshelves of the same size. Asha’s bookshelf is \(\frac{5}{6}\) th full and Samuel’s bookshelf is \(\frac{2}{5}\) th full. Whose bookshelf is more full and by what fraction?
Solution:

∴ Asha’s bookshelf is more full.
Fraction by which Asha’s bookshelf is more full
= \(\frac{5}{6}\) – \(\frac{2}{5}\) = \(\frac{25}{30}\) – \(\frac{12}{30}\)
= \(\frac{25 – 12}{30}\) = \(\frac{13}{30}\)

Question 2.
Jaidev takes a round of school ground in 2\(\frac{1}{5}\) minutes. Rahul takes \(\frac{7}{4}\) minutes to do the same work. Who takes less time and by how much?
Solution:
We know that, \(\frac{7}{4}\) = 1\(\frac{3}{4}\)
Clearly, 1\(\frac{3}{4}\) < 2\(\frac{1}{5}\) [∵ 1 < 2]
∴ Rahul takes less time to take a round of school ground.
Now, 2\(\frac{1}{5}\) – 1\(\frac{3}{4}\) = (\(\frac{11}{5}\) – \(\frac{7}{4}\))
= (\(\frac{11 \times 4}{5 \times 4}\) – \(\frac{7 \times 5}{5 \times 4}\)) = (\(\frac{44}{20}\) – \(\frac{35}{20}\)) = \(\frac{9}{20}\)
Therefore, Rahul takes \(\frac{9}{20}\) minutes lesser time.

Question 3.
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 her/his pencils.
Solution:
Fractions of pencils used up by Ramesh, Sheelu and Jamaal are \(\frac{10}{20}\), \(\frac{25}{50}\) and \(\frac{40}{80}\) respectively.

Therefore, the fraction of pencils used by each of them is equal.