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NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12

Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7

September 17, 2024 by LearnCBSE Online

Get the simplified Class 6 Maths NCERT Solutions of Ganita Prakash Chapter 7 Fractions textbook exercise questions with complete explanation.

Ganita Prakash Class 6 Maths Chapter 7 Solutions Fractions

NCERT Solutions for Class 6 Maths Ganita Prakash Chapter 7 Fractions

7.1 Fractional Units and Equal Shares Figure it Out (Page No. 152 – 153)

Fill in the blanks with fractions.

Question 1.
Three guavas together weigh 1 kg. If they are roughly of the same size, each guava will roughly weigh ______kg.
Solution:
Given, weigh of 3 guavas = 1 kg,
then weigh of each of the guava = \(\frac{\text { total weight }}{3}=\frac{1}{3}\)kg
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 1

Question 2.
A wholesale merchant packed 1 kg of rice in four packets of equal weight. The weight of each packet is __kg.
Solution:
Here, 1 kg of rise is packed in 4 packets of equal weight, then weight of each packet
= \(\frac{\text { Total weight }}{\text { No. of packets }}=\frac{1}{4}\)
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 2

Question 3.
Four friends ordered 3 glasses of sugarcane juice and shared it equally among themselves. Each one drank ____ glass of sugarcane juice.
Solution:
Amount of sugarcane consumed by four friends = 3 glasses .
Amount of sugarcane consumed by each friend = \(\frac{3}{4}\)glass
Hence, each one drank \(\frac{3}{4}\)glass of sugarcane juice.

Question 4.
The bis fish weighs \(\frac{1}{2}\) kg. The small one weighs \(\frac{1}{4}\) kg. Together they weigh ____ kg.
Solution:
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 3
Given the weighs of big fish = \(\frac{1}{2}\)kg and the weighs of small fish = \(\frac{1}{4}\)kg
Total weight of both fish = \(\frac{1}{2}+\frac{1}{4}=\frac{2+1}{4}\) kg
= \(\frac{3}{4}\) kg

Question 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:
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 4
Now, arranging the fraction words in order of size from the smallest to the biggest parts Quarter < Half < Three Quarters < One and a quarter < One and a half < Two and a half.

Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7

7.2 Fractional Units as Parts of a Whole Figure it Out (Page No. 155)

Question 1.
The figures below show different fractional units of a whole chikki. How much of a whole chikki is each piece? (See figure NCERT Textbook page 155)
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 6
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 5
Solution:
(a) We get this piece by breaking the chikki into 12 equal parts. Hence it is \(\frac{1}{2}\)
(b) We get this piece by breaking the chikki into 4 equal parts. Hence it is \(\frac{1}{4}\)
(c) We get this piece by breaking the chikki into 8 equal parts. Hence it is \(\frac{1}{8}\)
(d) We get this piece by breaking the chikki into 6 equal parts. Hence it is \(\frac{1}{6}\)
(e) We get this piece by breaking the chikki into 8 equal parts. Hence it is \(\frac{1}{8}\)
(f) We get this part by breaking the chikki into 6 equal pieces. Hence it is \(\frac{1}{6}\)
(g) We get this part by breaking the chikki into 24 equal pieces. Hence it is \(\frac{1}{24}\)
(h) We get this part by breaking the chikki into 24 equal pieces. Hence it is \(\frac{1}{24}\)

7.3 Measuring Using Fractional Units Figure it Out (Page No. 158)

Question 1.
Continue this table of \(\frac{1}{2}\) for 2 more steps.
Solution:
Here
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 7
represents a full roti (whole)

Step 1.
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 8
= \(\frac{1}{2}\) = 1 times half

Step 2.
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 9
= \(\frac{1}{2}\) + \(\frac{1}{2}\) = 2 times half

Step 3.
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 10
= \(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\) = 3 times half

Step 4.
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 11
= \(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\) = 4 times half

Step 5.
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 12
= \(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\) = 5 times half

Step 6.
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 13
= \(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\) = 6 times half
= 6 × \(\frac{1}{2}=\frac{6}{2}\) = 3

Step 7.
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 14
= \(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\) = 7 times half
= 7 × \(\frac{1}{2}=\frac{7}{2}\)

Question 2.
Can you create a similar table for \(\frac{1}{4}\)?
Solution:
Yes, we can create a similar table for \(\frac{1}{4}\). Here
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 15
represents a full roti (whole).
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 16

Question 3.
Make \(\frac{1}{3}\) using a paper strip. Can you use this to also make \(\frac{1}{6}\)?
Solution:
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 17
Yes, we can use this \(\frac{1}{3}\) to make \(\frac{1}{6}\).
When we divide each part of this \(\frac{1}{3}\) it is equal to Half of \(\frac{1}{3}=\frac{1}{3} \times \frac{1}{2}=\frac{1}{6}\)
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 18

Question 4.
Draw a picture and write an addition statement as above to show:
(a) 5 times of a roti
(b) 9 times of a roti
Solution:
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 19
Represents a full roti (whole)
(a)
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 20
5 times \(\frac{1}{4}\) of a roti
= \(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{5}{4}\)
1 full and \(\frac{1}{4}\) roti
(b)
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 21
9 times \(\frac{1}{4}\) of a roti
= \(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{9}{4}\)
2 full and \(\frac{1}{4}\) of roti

Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7

Question 5.
Match each fractional unit with the correct picture:
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 22
Solution:
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 23

7.4 Marking Fraction Lengths on the Number Line Figure it Out (Page No. 160)

Question 1.
On a number line, draw lines of lengths \(\frac{1}{10}, \frac{3}{10}\) and \(\frac{4}{5}\).
Solution:
Step 1. Draw a line l. Mark a point O on it.
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 24
Step 2. Mark point A at a distance of 1 unit from O.
Step 3. Divide OA into 10 equal parts.
Here OP represents \(\frac{1}{10}\), OQ represents \(\frac{3}{10}\), and OR represents \(\frac{8}{10}=\frac{4}{5}\)

Question 2.
Write five more fractions of your choice and mark them on the number line.
Solution:
Step 1. Let a number line OJ is divided into 10 equal parts.
Step 2. Now mark points A, B, C, D,…., J on it.
Step 3. Here
OF represents \(\frac{6}{10}=\frac{3}{5}\)
OG represents \(\frac{7}{10}\)
OB represents \(\frac{2}{10}=\frac{1}{5}\)
OH represents \(\frac{8}{10}=\frac{4}{5}\)
OI represents \(\frac{9}{10}\)
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 25

Question 3.
How many fractions lie between 0 and 1? Think, discuss with your classmates, and write your answer.
Solution:
There are an infinite number of fractions between 0 and 1.
Example: \(\frac{3}{5}, \frac{4}{5}, \frac{7}{10} \frac{1}{2}\) etc.

Question 4.
What is the length of the pink 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 pink line is y units long. Write the fraction that gives the length of the black line in the box.
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 26
Solution:
Length of black line is \(\frac{1}{2}\);
Length of black line is \(\frac{1}{2}\) + \(\frac{1}{2}\) + \(\frac{1}{2}\)
Fraction that gives length of black line = \(\frac{3}{2}\)

Question 5.
Write the fraction that gives the lengths of the black lines in the respective boxes.
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 27
Solution:
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 28

Intext Questions

Question 1.
Here, the fractional unit is dividing a length of 1 unit into three equal parts. Write the fraction that gives the length of the pink line in the box or in your notebook. (Page 159)
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 29
Solution:
Here number line OR is divided into three equal parts OP, PQ and QR.
Hence length of pink line = OP + PQ = \(\frac{1}{3}+\frac{1}{3}=\frac{2}{3}\)

Question 2.
Here, a unit is divided into 5 equal parts. Write the fraction that gives the length of the pink lines in the respective boxes or in your notebook.
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 30
Solution:
Here number line OT = 1 unit is divided into five equal parts OP, PQ, QR, RS and ST.
Hence length of pink line OQ = OP + PQ = \(\frac{1}{5}+\frac{1}{5}=\frac{2}{5}\)
Now, length of pink line OS = OP + PQ + QR + RS = \(\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=\frac{4}{5}\)
Hence, OQ = \(\frac{2}{5}\) OS = \(\frac{4}{5}\)

Question 3.
Now, a unit is divided into 8 equal parts. Write the appropriate fractions in your notebook Solution:Here number line OH is divided into 8 equal parts OA, AB, BC, CD, DE, EF, FG and GH.
Solution:
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 31
Also, OA = \(\frac{1}{8}\), OB = \(\frac{2}{8}\), OC = \(\frac{3}{8}\), OH = \(\frac{8}{8}\) = 1

7.5 Mixed Fractions Figure it Out (Page No. 162)

Question 1.
How many whole units are there in \(\frac{7}{2}\)?
Solution:
Here \(\frac{7}{2}\) = 7 times \(\frac{1}{2}\)
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 32
= 1 + 1 + 1 + \(\frac{1}{2}\)
= 3 + \(\frac{1}{2}\)
= 3\(\frac{1}{2}\)
Hence 3 whole units are there in \(\frac{7}{2}\).

Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7

Question 2.
How many whole units are there in \(\frac{4}{3}\) and in \(\frac{7}{3}\)?
Solution:
Here \(\frac{4}{3}\) = 4 times \(\frac{1}{3}\) = \(\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\)
= \(\frac{1+1+1}{3}\) + \(\frac{1}{3}\)
= 1 + \(\frac{1}{3}\) = 1\(\frac{1}{3}\)
Hence 1 whole unit are there in \(\frac{4}{3}\) and \(\frac{7}{3}\) = 7 times \(\frac{1}{3}\)
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 33
Hence 2 whole units are therein \(\frac{7}{3}\).

7.5 Mixed Fractions Figure it Out (Page No. 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
Fraction ← \(\frac{7}{3}\) = 2 + \(\frac{1}{3}\) → Mixed number
[This number is thus also called ‘two and one third’. We also write it as 2\(\frac{1}{3}\).]
Solution:
(a) \(\frac{8}{3}\)
Numerator 8 is split/divided into as many 3’s i.e. denominator no.: 8 = 3 + 3 + 2
= 1 + 1 + \(\frac{2}{3}\) = 2 \(\frac{2}{3}\)
∴ No. of whole units in \(\frac{8}{3}\) = 2 whole units.

(b) \(\frac{11}{5}=\frac{5}{5}+\frac{5}{5}+\frac{1}{5}\) [Numerator 11 = 5 + 5 + 1]
= 1 + 1 + \(\frac{1}{5}\)
= 2\(\frac{1}{5}\)

(c) \(\frac{9}{4}=\frac{4}{4}+\frac{4}{4}+\frac{1}{4}\)
= 1 + \(\frac{1}{4}\) [Numerator 9 = 4 + 4 + 1]
= 2\(\frac{1}{4}\)
∴ No. of whole units in \(\frac{9}{4}\) = 2 whole units.

Question 2.
Can all fractions greater than 1 be written as such mixed numbers?
A mixed number /mixedfraction contains a whole number (called the whole part) and a fraction that is less than 1 (called the fractional part).
Solution:
Yes, all fractions greater than 1 can be written as mixed fractions/numbers.

Question 3.
Write the following fractions as mixed fractions (e.g., \(\frac{9}{2}\) = 4\(\frac{1}{2}\))
(a) \(\frac{9}{2}\)
Solution:
\(\frac{9}{2}=\frac{2}{2}+\frac{2}{2}+\frac{2}{2}+\frac{2}{2}+\frac{1}{2}\)
= 1 + 1 + 1 + 1 + \(\frac{1}{2}\)
= 4\(\frac{1}{2}\)

(b) \(\frac{9}{5}\)
Solution:
\(\frac{9}{5}=\frac{5}{5}+\frac{4}{5}\)
= 1 + \(\frac{4}{5}\)
= 1\(\frac{4}{5}\)

(c) \(\frac{21}{19}\)
Solution:
\(\frac{21}{19}=\frac{19}{19}+\frac{2}{19}\)
= 1 + \(\frac{2}{19}\)
= 1\(\frac{2}{19}\)

(d) \(\frac{47}{9}\)
Solution:
\(\frac{47}{9}=\frac{9}{9}+\frac{9}{9}+\frac{9}{9}+\frac{9}{9}+\frac{9}{9}+\frac{2}{9}\)
= 1 + 1 + 1 + 1 + 1 + \(\frac{2}{9}\)
= 5 + \(\frac{2}{9}\)
= 5\(\frac{2}{9}\)

(e) \(\frac{12}{11}\)
Solution:
\(\frac{\mathbf{1 2}}{\mathbf{1 2}}=\frac{11}{11}+\frac{1}{11}\)
= 1 + \(\frac{1}{11}\)
= 1\(\frac{1}{11}\)

(f) \(\frac{19}{6}\)
Solution:
\(\frac{19}{6}=\frac{6}{6}+\frac{6}{6}+\frac{6}{6}+\frac{1}{6}\)
= 1 + 1 + 1 + \(\frac{1}{6}\)
= 3 + \(\frac{1}{6}\)
= 3 \(\frac{1}{6}\)

7.5 Mixed Fractions Figure it Out (Page No. 163)

Question 1.
Write the following mixed numbers as fractions:
(a) 3\(\frac{1}{4}\)
Solution:
Here 3\(\frac{1}{4}\) = 3 + \(\frac{1}{4}\)
[Splitting 3\(\frac{1}{4}\) into 3 and \(\frac{1}{4}\)]
= 1 + 1 + 1 + \(\frac{1}{4}\)
= \(\frac{4}{4}+\frac{4}{4}+\frac{4}{4}+\frac{1}{4}=\frac{13}{4}\)

(b) 7\(\frac{2}{3}\)
Solution:
Here 7\(\frac{2}{3}\) = 7 + \(\frac{2}{3}\)
[Splitting 7\(\frac{2}{3}\) into 7 and \(\frac{2}{3}\)]
= 1 + 1 + 1 + 1 + 1 + 1 + 1 + \(\frac{2}{3}\)
= \(\frac{3}{3}+\frac{3}{3}+\frac{3}{3}+\frac{3}{3}+\frac{3}{3}+\frac{3}{3}+\frac{3}{3}+\frac{2}{3}\)
= \(\frac{23}{3}\)

(c) 9\(\frac{4}{9}\)
Solution:
Here 9 \(\frac{4}{9}\) = 9 + \(\frac{4}{9}\)
[Splitting 9\(\frac{4}{9}\) into 9 and \(\frac{4}{9}\)]
= 9 times \(\frac{9}{9}+\frac{4}{9}\)
= \(\frac{81}{9}+\frac{4}{9}=\frac{85}{9}\)

(d) 3\(\frac{1}{6}\)
Solution:
Here 3\(\frac{1}{6}\) = 3 + \(\frac{1}{6}\)
[Splitting 3\(\frac{1}{6}\) into 3 and \(\frac{1}{6}\)]
= 3 times \(\frac{6}{6}\) + \(\frac{1}{6}\)
= \(\frac{18}{6}+\frac{1}{6}\)
= \(\frac{19}{6}\)

Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7

(e) 2\(\frac{3}{11}\)
Solution:
We have 2\(\frac{3}{11}\) = 2 + \(\frac{3}{11}\)
[Splitting 2\(\frac{3}{11}\) into 2 and \(\frac{3}{11}\)]
= 2 times \(\frac{11}{11}+\frac{3}{11}\)
= \(\frac{22}{11}+\frac{3}{11}\)
= \(\frac{25}{11}\)

(f) 3\(\frac{9}{10}\)
Solution:
3\(\frac{9}{10}\) = 3 + \(\frac{9}{10}\)
[Splitting 3\(\frac{9}{10}\) into 3 and \(\frac{9}{10}\)]
= 3 times \(\frac{10}{10}+\frac{9}{10}\)
= \(\frac{30}{10}+\frac{9}{10}=\frac{39}{10}\)

7.6 Equivalent Fractions Figure it Out 7.7 Simplest form of a Fraction Figure it Out (Page No. 165)

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

Question 2.
Write two equivalent fractions for \(\frac{2}{6}\).
Solution:
Here \(\frac{2}{6}\) → \(\frac{2 \times 2}{6 \times 2}, \frac{2 \times 3}{6 \times 3}, \frac{2 \times 4}{6 \times 4}\)
Hence \(\frac{4}{12}, \frac{6}{18}, \frac{8}{24}\) are equivalent fractions of \(\frac{2}{6}\).

Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7

Question 3.
\(\frac{4}{6}\) = ___________ = ___________ = ___________ = ___________________
(Write as many as you can)
Solution:
Here,
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 34

Intext Questions

Answer the following questions after looking at the fraction wall: [Page 164]
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 35

Question 1.
Are the lengths \(\frac{1}{2}\) and \(\frac{3}{6}\) equal?
Solution:
Yes, here lengths \(\frac{1}{2}\) and \(\frac{3}{6}\) = \(\frac{1}{2}\)
Lengths are equal.

Question 2.
Are \(\frac{2}{3}\) and \(\frac{4}{6}\) equivalent fractions? Why?
Solution:
Yes, lengths \(\frac{2}{3}\) and \(\frac{4}{6}\) = \(\frac{1}{3}\) are equivalent fraction, as they have same length.

Question 3.
How many pieces of length \(\frac{1}{6}\) will make a length of \(\frac{1}{2}\)?
Solution:
Total no.of pieces = \(\frac{\frac{1}{2}}{\frac{1}{6}}=\frac{1}{2} \times \frac{6}{1}=\frac{6}{2}\) = 3
Hence three pieces of length \(\frac{1}{6}\) will make a length of \(\frac{1}{2}\)

Question 4.
How many pieces of length \(\frac{1}{6}\) will make a length of \(\frac{1}{3}\)?
Solution:
Total no. of pieces = \(\frac{\frac{1}{3}}{\frac{1}{6}}=\frac{1}{3} \times \frac{6}{1}=\frac{6}{3}\) = 2
Hence two pieces of length \(\frac{1}{6}\) will make a length of \(\frac{1}{3}\).

7.7 Simplest form of a Fraction Figure it Out (Page No. 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 roti each child gets is _________________ .
Division fact:
Addition fact:
Multiplication fact:
Compare your picture and answers with your classmates!
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 36
Solution:
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 37
Division fact:
3 wholes divided in 4 parts
3 ÷ 4 = \(\frac{3}{4}\)

Addition fact:
Four times \(\frac{3}{4}\) added gives 3 wholes
= \(\frac{3}{4}+\frac{3}{4}+\frac{3}{4}+\frac{3}{4}\)
= \(\frac{12}{4}\) = 3

Multiplication fact:
4 parts of \(\frac{3}{4}\) make 3 wholes
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.
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 38
Solution:
As 2 rotis have to be shared equally by 4 children we divide each roti in 4 parts and give
(a) 1 part of each roti to each child as shown below:
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 39
(b) 2 parts to each child as shown below
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 40
Division fact:
2 wholes divide in 4 parts
2 ÷4 or \(\frac{2}{4}=\frac{1}{2}\)

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

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

Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7

Question 3.
Anil was in a group where 2 cakes were divided equally among 5 children. How much cake would Anil get?
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 41
Solution:
Anil is in a group where 2 cakes were divided equally among 5 children.
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 42
Each cake gets divided into 5 parts and Anil gets one part from each cake i.e. \(\frac{1}{5}+\frac{1}{5}=\frac{2}{5}\)

7.7 Simplest form of a Fraction Figure it Out (Page No. 168 – 169)

Question 1.
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, \(\frac{5}{4}\) = \(\frac{?}{8}\).
(b) 4 kg of potatoes divided equally in 3 bags is the same as 12 kgs of potatoes divided equally in ____________ bags. So, \(\frac{4}{3}\) = \(\frac{12}{?}\).
(c) 7 rods divided among 5 children is the same as rods divided among children. So, \(\frac{7}{5}\) = ____________
Solution:
(a) Here, the amount of juice each friend gets when 5 glasses are shared among 4 friends = \(\frac{\text { number of glasses }}{\text { number of friends }}=\frac{5}{4}\)
Now to determine how many glasses of juice would be needed to give each of the 8 friends the same amount = 8 × \(\frac{5}{4}\)
= 10 glasses
So, 10 glasses of juice shared equally among 8 friends is the same as 5 glasses of juice shared equally among 4 friends.
∴ \(\frac{5}{4}=\frac{10}{8}\)

(b) Here 4 kg of potatoes divided equally in 3 bags then amount of potatoes per bag = \(\frac{4 \mathrm{~kg}}{3 \text { bags }}=\frac{4}{3}\) 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
\(\frac{12 \mathrm{~kg}}{x \text { bags }}=\frac{4}{3}\) kg per bag
⇒ 12 × 3 = 4 × x
⇒ 36 = 4x
⇒ x = \(\frac{36}{4}\)
⇒ x = 9
∴ \(\frac{4}{3}=\frac{12}{9}\)

(c) Dividing 7 rotis among 4 children gives 7 each child = \(\frac{7}{5}\) of a roti. We can find an
equivalent fraction by multiplying both the numerator and the denominator by the same number. For example, multiplying both by 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
∴ \(\frac{7}{5}=\frac{14}{10}\)

Intext Questions

Question 1.
Find equivalent fractions for the given pairs of fractions such that the fractional units are the same. (Page 172)
(a) \(\frac{7}{2}\) and \(\frac{3}{5}\)
Solution:
Given fractions are \(\frac{7}{2}\) and \(\frac{3}{5}\)
Here, the denominators are 2 and 5.
And least common multiple of 2 and 5 is 10.
Hence for both fractions let’s have same denominator of 10.
Now for \(\frac{7}{2}\) multiply both the numerator and the denominator by 5.
\(\frac{7}{2}=\frac{7 \times 5}{2 \times 5}=\frac{35}{10}\)
And for \(\frac{3}{5}\) multiply both the numerator and the denominator by 2, we get,
\(\frac{3 \times 2}{5 \times 2}=\frac{6}{10}\)
Hence, the equivalent fractions with the same denominator are:
\(\frac{35}{10}\) and \(\frac{6}{10}\)

(b) \(\frac{8}{3}\) and \(\frac{5}{6}\)
Solution:
Given fractions are \(\frac{8}{3}\) and \(\frac{5}{6}\)
Here, the denominators are 3 and 6.
And least common multiple of 3 and 6 is 6.
Now for \(\frac{8}{3}\) multiply both the numerator and the denominator by 2.
\(\frac{8}{3}=\frac{8 \times 2}{3 \times 2}=\frac{16}{6}\)
\(\frac{5}{6}\) already have a denominator 6.
Hence, the equivalent fractions with the same denominator are:
\(\frac{16}{6}\) and \(\frac{5}{6}\)

(c) \(\frac{3}{4}\) and \(\frac{3}{5}\)
Solution:
Given fractions are \(\frac{3}{4}\) and \(\frac{3}{5}\)
Here, the denominators are 4 and 5.
And least common multiple of 4 and 5 is 20.
Now for \(\frac{3}{4}\) multiply both the numerator and the denominator by 5.
\(\frac{3}{4}=\frac{3 \times 5}{4 \times 5}=\frac{15}{20}\)
And for \(\frac{3}{5}\) multiply both the numerator and the denominator by 4, we get
\(\frac{3}{5}=\frac{3 \times 4}{5 \times 4}=\frac{12}{20}\)
So, the equivalent fractions with the same denominator are:
\(\frac{15}{20}\) and \(\frac{12}{20}\)

(d) \(\frac{6}{7}\) and \(\frac{8}{5}\)
Solution:
Given fractions are \(\frac{6}{7}\) and \(\frac{8}{5}\)
Here, the denominators are 7 and 5.
And least common multiple of 7 and 5 is 35.

Now for \(\frac{6}{7}\) multiply both the numerator and the denominator by 5.
\(\frac{6}{7}=\frac{6 \times 5}{7 \times 5}=\frac{30}{35}\)
And for \(\frac{8}{5}\) multiply both the numerator and the denominator by 7, we get
\(\frac{8}{5}=\frac{8 \times 7}{5 \times 7}=\frac{56}{35}\)
So, the equivalent fractions with the same denominator are:
\(\frac{30}{35}\) and \(\frac{56}{35}\)

(e) \(\frac{9}{4}\) and \(\frac{5}{2}\)
Solution:
Given fractions are \(\frac{9}{4}\) and \(\frac{5}{2}\)
Here, the denominators are 4 and 2.
And least common multiple of 4 and 2 is 4.
Now for \(\frac{5}{2}\) multiply both the numerator and the denominator by 2.
\(\frac{5}{2}=\frac{5 \times 2}{2 \times 2}=\frac{10}{4}\)
and \(\frac{9}{4}\) already have a denominator 4
So, the equivalent fractions with the same denominator are:
\(\frac{9}{4}\) and \(\frac{10}{4}\)

(f) \(\frac{1}{10}\) and \(\frac{2}{9}\)
Solution:
Given fractions are and \(\frac{1}{10}\) and \(\frac{2}{9}\)
Here, the denominators are 10 and 9.
And least common multiple of 10 and 9 is 90.
Now for \(\frac{1}{10}\) multiply both the numerator and the denominator by 9.
\(\frac{1}{10}=\frac{1 \times 9}{10 \times 9}=\frac{9}{90}\)
And for 2 multiply both the numerator and the denominator by 10, we get
\(\frac{2}{9}=\frac{2 \times 10}{9 \times 10}=\frac{20}{90}\)
So, the equivalent fractions with the same denominator are:’
\(\frac{9}{90}\) and \(\frac{20}{90}\)

(g) \(\frac{8}{3}\) and \(\frac{11}{4}\)
Solution:
Given fractions are \(\frac{8}{3}\) and \(\frac{11}{4}\)
Here, the denominators are 3 and 4.
And least common multiple of 3 and 4 is 12.
Now for \(\frac{8}{3}\) multiply both the numerator and the denominator by 4.
\(\frac{8}{3}=\frac{8 \times 4}{3 \times 4}=\frac{32}{12}\)
And for \(\frac{11}{4}\) multiply both the numerator and the denominator by 3, we get
\(\frac{11}{4}=\frac{11 \times 3}{4 \times 3}=\frac{33}{12}\)
So, the equivalent fractions with the same denominator are:
\(\frac{32}{12}\) and \(\frac{33}{12}\)

(h) \(\frac{13}{6}\) and \(\frac{1}{9}\)
Solution:
Given fractions are \(\frac{13}{6}\) and \(\frac{1}{9}\)
Here, the denominators are 6 and 9.
And least common multiple of 6 and 9 is 18.

Now for \(\frac{13}{6}\) multiply both the numerator and the denominator by 3.
\(\frac{13}{6}=\frac{13 \times 3}{6 \times 3}=\frac{39}{18}\)

And for \(\frac{1}{9}\) multiply both the numerator and the denominator by 2, we get
\(\frac{1}{9}=\frac{1 \times 2}{9 \times 2}=\frac{2}{18}\)

So, the equivalent fractions with the same denominator are:
\(\frac{39}{18}\) and \(\frac{2}{18}\)

Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7

7.7 Simplest form of a Fraction Figure it Out (Page No. 173)

Question 1.
Express the following fractions in lowest terms:
(a) \(\frac{17}{51}\)
Solution:
Here 51 is divisible by 17 then \(\frac{51}{17}\) = 3
∴ \(\frac{17}{51}=\frac{17}{17 \times 3}=\frac{1}{3}\)
which is the lowest term.

(b) \(\frac{64}{144}\)
Solution:
Here 64 and 144 are both multiples of 16. So we divide both by 16.
\(\frac{64}{144}=\frac{64 \div 16}{144 \div 16}\)
= \(\frac{4}{9}\)

(c) \(\frac{126}{147}\)
Solution:
Here, \(\frac{126}{147}\)
= \(\frac{126 \div 7}{147 \div 7}\)
[∵ HCF of 126 and 147 is 7 × 3 = 21]
= \(\frac{18 \div 3}{21 \div 3}\)
= \(\frac{6}{7}\)

(d) \(\frac{525}{112}\)
Solution:
Here 525 and 112 are both multiples of 7, we divide both by 7.
= \(\frac{525 \div 7}{112 \div 7}\)
= \(\frac{75}{16}\)

7.8 Comparing Fractions Figure it Out (Page No. 174)

Question 1.
Compare the following fractions and justify your answers:
(a) \(\frac{8}{3}, \frac{5}{2}\)
Solution:
(a) Given fractions are \(\frac{8}{3}\) and \(\frac{5}{2}\).
Here LCM of denominators 3 and 2 is 6
then multiplying and dividing \(\frac{8}{3}\) by 2 and \(\frac{5}{2}\) by 3 then
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 43

(b) \(\frac{4}{9}, \frac{3}{7}\)
Solution:
Given fractions are \(\frac{4}{9}\) and \(\frac{3}{7}\).
Here LCM of denominators 9 and 7 is 63.
then multiplying and dividing \(\frac{4}{9}\) by 7 and \(\frac{3}{7}\) by 9 then
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 44

(c) \(\frac{7}{10}, \frac{9}{14}\)
Solution:
Given fractions are \(\frac{7}{10}\) and \(\frac{9}{14}\).
Here LCM of denominators 10 and 14 is 70.
then multiplying and dividing \(\frac{7}{10}\) by 7 and \(\frac{9}{14}\) by 5 then
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 45

(d) \(\frac{12}{5}, \frac{8}{5}\)
Solution:
Given fractions are \(\frac{12}{5}, \frac{8}{5}\).
Here \(\frac{12}{5}\) > \(\frac{8}{5}\)

(e) \(\frac{9}{4}, \frac{5}{2}\)
Solution:
Given fractions are \(\frac{9}{4}, \frac{5}{2}\).
Here LCM of denominators 4 and 2 is 4.
then multiplying and dividing \(\frac{5}{2}\) by 2 and
Equivalent fractions with 4 are \(\frac{5 \times 2}{2 \times 2}\)
\(\frac{9}{4}>\frac{10}{4}\)
∴ \(\frac{9}{4}>\frac{5}{2}\)

Question 2.
Write following fractions ascending order.
(a) \(\frac{7}{10}, \frac{11}{15}, \frac{2}{5}\)
Solution:
The given fractions are \(\frac{7}{10}, \frac{11}{15}, \frac{2}{5}\)
Let us find LCM of denominator 10, 15, 5
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 46
∴ LCM of 10, 15 and 5 = 2 × 3 × 5 = 30
Now let us make denominator of each fractions as LCM
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 47
Hence given fractions in ascending order are: \(\frac{2}{5}, \frac{7}{10} \frac{11}{5}\)

(b) \(\frac{19}{24}, \frac{5}{6}, \frac{7}{12}\)
Solution:
The given fractions are \(\frac{19}{24}, \frac{5}{6}, \frac{7}{12}\)
Here LCM of 24, 6, 12 is 24.
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 48
On arranging in ascending Order, we get
\(\frac{14}{24}, \frac{19}{24}, \frac{20}{24}\)
⇒ \(\frac{7}{12}, \frac{19}{24}, \frac{5}{6}\)

Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7

Question 3.
Write the following fractions descending order.
(a) \(\frac{25}{16}, \frac{7}{8}, \frac{13}{4}, \frac{17}{32}\frac{3}{4}, \frac{12}{5}, \frac{7}{12}, \frac{5}{4}\)
Solution:
The given fractions are \(\frac{25}{16}, \frac{7}{8}, \frac{13}{4}, \frac{17}{32}\frac{3}{4}, \frac{12}{5}, \frac{7}{12}, \frac{5}{4}\)
Let us find LCM of denominator 16, 8,4, 32
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 49
LCM of 16, 8,4, 32 = 2 × 2 × 2 × 2 × 2 = 32
Now let us make denominator of each fractions as LCM thus
∴ \(\frac{25 \times 2}{16 \times 2}, \frac{7 \times 4}{8 \times 4}, \frac{13 \times 8}{4 \times 8}, \frac{17 \times 1}{32 \times 1}\)
\(\frac{50}{32}, \frac{28}{32}, \frac{104}{32}, \frac{17}{32}\)
On arranging in descending order, we get
\(\frac{104}{32}>\frac{50}{32}>\frac{28}{32}>\frac{17}{32}\)
∴ \(\frac{13}{4}>\frac{25}{16}>\frac{7}{8}>\frac{17}{32}\)

Hence given fractions in descending order are \(\frac{13}{4}, \frac{25}{16}, \frac{7}{8}\) and \(\frac{17}{32}\)

(b) \(\frac{3}{4}, \frac{12}{5}, \frac{7}{12}, \frac{5}{4}\)
Solution:
Given fractions are \(\frac{3}{4}, \frac{12}{5}, \frac{7}{12}, \frac{5}{4}\)
Here LCM of 4, 5, 12, 4 is 60
Now let us make denominator of each fraction as LCM then
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 50
Hence fractions in descending order are
\(\frac{12}{5}>\frac{5}{4}>\frac{3}{4}>\frac{7}{12}\)

7.9 Relation to Number Sequences Figure it Out (Page No. 179)

Question 1.
Add the following fractions using Brahmagupta’s method:
(a) \(\frac{2}{7}+\frac{5}{7}+\frac{6}{7}\)
Solution:
Here \(\frac{2}{7}+\frac{5}{7}+\frac{6}{7}\)
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 51

(b) \(\frac{3}{4}+\frac{1}{3}\)
Solution:
Here \(\frac{3}{4}+\frac{1}{3}\)
Here LCM of denominators 4 and 3 is 12
∴ Equivalent fraction of \(\frac{3}{4}\) with denominators 12 is \(\frac{9}{12}\) and equivalent fraction of \(\frac{1}{3}\) with denominators 12 is \(\frac{4}{12}\)
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 52

(c) \(\frac{2}{3}+\frac{5}{6}\)
Solution:
Given \(\frac{2}{3}+\frac{5}{6}\)
Now LCM of 3 and 6 is 6.
Expressing as equivalent fractions with denominators 6, we get
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 53

(d) \(\frac{2}{3}+\frac{2}{7}\)
Solution:
Here \(\frac{2}{3}+\frac{2}{7}\)
Now LCM of 3 and 7 is 21
Expressing as equivalent fractions with denominators 21, we get
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 54

(e) \(\frac{3}{4}+\frac{1}{3}+\frac{1}{5}\)
Solution:
Here \(\frac{3}{4}+\frac{1}{3}+\frac{1}{5}\)
Now LCM of 4, 3, 5 is 60.
Expressing as equivalent fractions with denominators 60, we get
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 55

Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7

(f) \(\frac{2}{3}+\frac{4}{5}\)
Solution:
Here \(\frac{2}{3}+\frac{4}{5}\)
Now LCM of 3 and 5 is 15
Expressing as equivalent fractions with denominators 15, we get
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 56

(g) \(\frac{4}{5}+\frac{2}{3}\)
Solution:
Here \(\frac{4}{5}+\frac{2}{3}\)
Now LCM of 5 and 3 is 15
Thus expressing as equivalent fractions with denominators 15, we get
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 57

(h) \(\frac{3}{3}+\frac{5}{8}\)
Solution:
Given \(\frac{3}{3}+\frac{5}{8}\)
Here LCM of 5 and 8 is 40
Expressing as equivalent fractions with denominators 40, we get
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 58

(i) \(\frac{9}{2}+\frac{5}{4}\)
Solution:
Here \(\frac{9}{2}+\frac{5}{4}\)
Now LCM of 2 and 4 is 4.
Expressing as equivalent fractions with denominators 4, we get
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 59

(j) \(\frac{8}{3}+\frac{2}{7}\)
Solution:
Given \(\frac{8}{3}+\frac{2}{7}\)
Here LCM of 3 and 7 is 21
Expressing as equivalent fractions with denominators 21, we get
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 60

(k) \(\frac{3}{4}+\frac{1}{3}+\frac{1}{5}\)
Solution:
Here \(\frac{3}{4}+\frac{1}{3}+\frac{1}{5}\)
Now LCM of 4, 3, 5 is 60
Expressing as equivalent fractions with denominators 60, we get
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 61

(l) \(\frac{2}{3}+\frac{4}{5}+\frac{3}{7}\)
Solution:
Here \(\frac{2}{3}+\frac{4}{5}+\frac{3}{7}\)
Now LCM of 3, 5 and 7 is 105.
Expressing as equivalent fractions with denominators 105, we get
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 62

(m) \(\frac{9}{2}+\frac{5}{4}+\frac{7}{6}\)
Solution:
Given \(\frac{9}{2}+\frac{5}{4}+\frac{7}{6}\)
Here LCM of 2, 4, 6 is 12.
Now expressing as equivalent fractions with denominators 12, we get
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 63

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:
Given quantity of yellow paint = \(\frac{2}{3}\) litres
and quality of blue paint = \(\frac{3}{4}\) litres
Volume of green paint made
= \(\frac{2}{3}\) litres + \(\frac{3}{4}\) litres
Here LCM of 3 and 4 is 12
Now expressing as equivalent fractions with denominator 12, we get
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 64

Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7

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:
Given length of lace bought by Geeta = \(\frac{2}{5}\) meter
and length of lace bought by Shamim = \(\frac{3}{4}\) meter
Total length of lace = \(\frac{2}{5}\) + \(\frac{3}{4}\)
LCM of 5 and 4 is 20
Now expressing as equivalent fractions with denominator 20, we get
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 65
Total lace required = Perimeter = 1 m
Hence total lace Geeta and Shamim have together is sufficient to cover the whole border.

7.9 Relation to Number Sequences Figure it Out (Page No. 181)

Question 1.
\(\frac{5}{8}-\frac{3}{8}\)
Solution:
Given \(\frac{5}{8}-\frac{3}{8}\)
As fractional unit is same i.e., \(\frac{1}{8}\) we shall simply subtract numerators keeping fractional unit as \(\frac{1}{8}\)
Then \(\frac{5}{8}-\frac{3}{8}=\frac{5-3}{8}\)
= \(\frac{2}{8}=\frac{1}{4}\) (representing in simplest form)

Question 2.
\(\frac{7}{9}-\frac{5}{9}\)
Solution:
Given \(\frac{7}{9}-\frac{5}{9}\)
As fractional unit is same i.e., \(\frac{1}{9}\) we shall simply subtract numerators keeping fractional unit as \(\frac{1}{9}\)
\(\frac{7}{9}-\frac{5}{9}\)
= \(\frac{7-5}{9}=\frac{2}{9}\)

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

7.9 Relation to Number Sequences Figure it Out (Page No. 182)

Question 1.
Carry out the following subtractions using Brahmagupta’s method:
(a) \(\frac{8}{15}-\frac{3}{15}\)
Solution:
Given \(\frac{8}{15}-\frac{3}{15}\)
Fractional unit for both fractions is \(\frac{1}{15}\) then
\(\frac{8}{15}-\frac{3}{15}=\frac{8-3}{15}\)
= \(\frac{5}{15}=\frac{1}{3}\)

(b) \(\frac{2}{5}-\frac{4}{15}\)
Solution:
Given \(\frac{2}{5}-\frac{4}{15}\)
Here LCM of 5 and 15 is 15. Fractional unit for both fractions should be \(\frac{1}{15}\)
then \(\frac{2 \times 3}{5 \times 3}-\frac{4 \times 1}{15 \times 1}\)
= \(\frac{6}{15}-\frac{4}{15}\)
= \(\frac{6-4}{15}\)
= \(\frac{2}{15}\)

(c) \(\frac{5}{6}-\frac{4}{9}\)
Solution:
Given \(\frac{5}{6}-\frac{4}{9}\)
Hence LCM of 6 and 9 is 18. Fractional unit for both fractions should be \(\frac{1}{18}\) then
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 66

(d) \(\frac{2}{3}-\frac{1}{2}\)
Solution:
Given \(\frac{2}{3}-\frac{1}{2}\)
Here LCM of 3 and 2 is 6. Fractional unit for both fractions should be \(\frac{1}{6}\)
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 67

Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7

Question 2.
Subtract as indicated:
(a) \(\frac{13}{4}\) from \(\frac{10}{3}\)
Solution:
Given y – y
Here, LCM of 3 and 4 is 12.
Fractional unit for both fractions should be \(\frac{1}{12}\)
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 68

(b) \(\frac{18}{5}\) from \(\frac{23}{3}\)
Solution:
Here, \(\frac{23}{3}-\frac{18}{5}\)
Now, LCM of 3 and 5 is 15.
Fractional unit = \(\frac{1}{15}\) for both fractions
Hence
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 69

(c) \(\frac{29}{7}\) from \(\frac{45}{7}\)
Solution:
Here fractional = \(\frac{1}{7}\)for both fractions.
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 70

Question 3.
Solve the following problems:
(a) Java’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?
Solution:
Given distance between Jaya’s school and home is \(\frac{7}{10}\) km and distance covered by Jaya in auto is \(\frac{1}{2}\) km.
∴ Distance Jaya covered by walking = \(\frac{7}{10}\)km – \(\frac{1}{2}\)km
LCM of 10 and 2 is 10.
Fractions Class 6 NCERT Solutions Ganita Prakash Maths Chapter 7 71
Hence Jaya walks \(\frac{1}{5}\) km or 200 meters to reach her school.

(b) Jeevika takes \(\frac{10}{3}\) minutes to take a complete round of the park and her 13 friend Namit takes \(\frac{13}{4}\) minutes to do the same. Who takes less time and by how much?
Solution:
Time taken by Jeevika to cover 1 round of park = \(\frac{10}{3}\)mm
Time taken by Namit to cover 1 round of 13 park= \(\frac{13}{4}\) min

To find who takes less time we need to compare \(\frac{10}{3}\) and \(\frac{13}{4}\)
Here, LCM of 3 and 4 is 12
\(\frac{10 \times 4}{3 \times 4}\) and \(\frac{13 \times 3}{4 \times 3}\)
[Expressing fractions in fractional unit \(\frac{1}{12}\)]
= \(\frac{40}{12}\) and \(\frac{39}{12}\)
\(\frac{40}{12}>\frac{39}{12}\)

Hence Namit takes less time
\(\frac{40}{12}-\frac{39}{12}\)
= \(\frac{40-39}{12}=\frac{1}{12}\)
Namit takes \(\frac{1}{2}\) of minutes less than Jeevika.

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