NCERT Solutions for Class 6 Maths Chapter 7 Fractions

NCERT Solutions for Chapter 7 Class 6 Maths, "Fractions," covers the following exercises:

Exercise 7.1: Fractional Units and Equal Shares

This exercise introduces the concept of fractional units and how to divide objects into equal shares. It helps students understand the basic idea of fractions.

Exercise 7.2: Fractional Units as Parts of a Whole

Here, students learn to identify fractional units as parts of a whole. This exercise reinforces the connection between fractions and everyday objects.

Exercise 7.3: Measuring Using Fractional Units

This exercise focuses on measuring lengths and quantities using fractional units. Students practice applying fractions in practical measuring scenarios.

Exercise 7.4: Marking Fraction Lengths on the Number Line

Students learn to mark fractions on a number line in this exercise. It helps them visualize the position of fractions relative to whole numbers.

Exercise 7.5: Mixed Fractions

This exercise introduces mixed fractions and teaches students how to convert between mixed numbers and improper fractions.

Exercise 7.6: Equivalent Fractions

In this exercise, students explore the concept of equivalent fractions, learning how different fractions can represent the same value.

Exercise 7.7: Comparing Fractions

Students learn to compare fractions based on their sizes in this exercise. They practice determining which fractions are greater or smaller.

Exercise 7.8: Addition and Subtraction of Fractions

This exercise focuses on the addition and subtraction of fractions, providing students with step-by-step methods to solve these operations.

Exercise 7.9: A Pinch of History

This section gives a brief history of fractions, helping students understand the development and significance of fractions in mathematics.

Access NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Exercise 7.1

1. Three guavas together weigh 1 kg. If they are roughly of the same size, each guava will roughly weigh ____ kg.

Ans: Each guava will roughly weigh ⅓ kg.

2. A wholesale merchant packed 1 kg of rice in four packets of equal weight. The weight of each packet is ____ kg.

Ans: The weight of each packet is ¼ kg.

3. Four friends ordered 3 glasses of sugarcane juice and shared it equally among themselves. Each one drank ____ glass of sugarcane juice.

Ans: Each one drank ¾ glass of sugarcane juice.

4. The big fish weighs ½ kg. The small one weighs ¼ kg. Together they weigh ____ kg.

Ans: Together they weigh ¾ kg.

5. Find out and discuss the words for fractions that are used in the different languages spoken in your home, city, or state. Ask your grandparents, parents, teachers, and classmates what words they use for different fractions, such as for one and a half, three quarters, one and a quarter, half, quarter, and two and a half, and write them here:

Ans: 

To complete this task, you'll need to ask people around you (grandparents, parents, teachers, classmates) how they refer to different fractions in your local language or dialect. Here are some examples in Hindi for common fractions:

• Quarter (¼): Paauna or Chauthai

• Half (½): Aadhaa

• Three-quarters (¾): Teen Paav

• One and a quarter (1¼): Sawa ek

• One and a half (1½): Dedh

• Two and a half (2½): Dhaai

6. 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. 

To arrange the fractions in order from smallest to largest, we first convert them to numerical fractions:

• Quarter = $\dfrac{1}{4}$​

• Half = $\dfrac{1}{2}$

• Three quarters = $\dfrac{3}{4}$​

• One and a quarter = $\dfrac{1}{4}$ = $\dfrac{5}{4}$​

• One and a half = $\dfrac{1}{2}$ = $\dfrac{3}{2}$​

• Two and a half = $\dfrac{1}{2}$ = $\dfrac{5}{2}$​

Arranged in order from smallest to biggest:

1. Quarter $\dfrac{1}{4}$​

2. Half $\dfrac{1}{2}$

3. Three quarters $\dfrac{3}{4}$​

4. One and a quarter $\dfrac{1}{4}$​

5. One and a half $\dfrac{1}{2}$​

6. Two and a half $\dfrac{1}{2}$

Exercise 7.2

1. By dividing the whole chikki into 6 equal parts in different ways, we get $\dfrac{1}{6}$ chikki pieces of different shapes. Are they of the same size?

Ans:

Even though the pieces of chikki are divided into different shapes, they are still of the same size. Each piece represents $\dfrac{1}{6}$ of the whole chikki, meaning that all the pieces have the same area, even if their shapes are different. This illustrates that fractions represent a part of a whole, regardless of the shape of the part, as long as the total area is equally divided.

2. What is the fractional unit of chikki shown below?

Ans:

We get this piece by breaking the chikki into 3 equal pieces. So this is $\dfrac{1}{3}$ chikki.

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

Ans:

a.$\dfrac{1}{6}$

b. $\dfrac{1}{3}$​

c. $\dfrac{1}{3}$​

d. $\dfrac{1}{6}$​

e. $\dfrac{1}{6}$​

f. $\dfrac{1}{3}$​

g. $\dfrac{1}{6}$​

h. $\dfrac{1}{6}$​

Exercise 7.3

1. Continue this table of ½ for 2 more steps.

Ans: 

2. Can you create a similar table for ¼?

3. Make ⅓ using a paper strip. Can you use this to also make ⅙?

Ans: Yes, by folding the strip with ⅓ marked, you can divide it further to create ⅙.

4. Draw a picture and write an addition statement as above to show:

a. 5 times ¼ of a roti

Ans: ¼ + ¼ + ¼ + ¼ + ¼ = 5/4 or 1¼

b. 9 times ¼ of a roti

Ans: ¼ + ¼ + ¼ + ¼ + ¼ + ¼ + ¼ + ¼ + ¼ = 9/4 or 2¼

5. Match each fractional unit with the correct picture:

$\dfrac{1}{3}$ $\dfrac{1}{8}$ $\dfrac{1}{8}$ $\dfrac{1}{6}$

Ans:

⅓: Matches with the third picture (the one with three equal parts, one shaded).

⅕: Matches with the second picture (the one with five equal parts, one shaded).

⅛: Matches with the first picture (the one with eight equal parts, one shaded).

⅙: Matches with the fourth picture (the one with six equal parts, one shaded).

Exercise 7.4

Find the lengths of the various blue lines shown below. Fill in the boxes as well.

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 blue line in the box or in your notebook.

Ans:

2. Here, a unit is divided into 5 equal parts. Write the fraction that gives the length of the blue lines in the respective boxes or in your notebook.

Ans:

3. Now, a unit is divided into 8 equal parts. Write the appropriate fractions in your notebook​

Ans:

$\dfrac{2}{8}$

$\dfrac{3}{8}$​

$\dfrac{4}{8}$​

$\dfrac{5}{8}$

$\dfrac{6}{8}$​

$\dfrac{7}{8}$​

$\frac{8}{8}$​

Figure it Out

1. On a number line, draw lines of lengths $\dfrac{1}{10}$, $\dfrac{3}{10}$​, and $\dfrac{4}{5}$.

Ans: On the number line, draw lines corresponding to the lengths $\frac{1}{10}$​, $\dfrac{3}{10}$​, and $\dfrac{4}{5}$​. These can be marked by dividing the number line into 10 equal parts and placing the fractions accordingly.

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

Ans: Five more fractions you can mark on the number line could be $\dfrac{1}{2}$​, $\dfrac{3}{5}$​, $\dfrac{7}{10}$​, $\dfrac{9}{10}$​, and 1. Mark these fractions on the number line by dividing it further if necessary.

3. How many fractions lie between 0 and 1? Think, discuss with your classmates, and write your answer.

Ans: Between 0 and 1, there are an infinite number of fractions, as you can always divide any segment of the number line into smaller parts. Examples include $\dfrac{1}{2}$​, $\dfrac{1}{3}$​, $\dfrac{2}{3}$​, $\dfrac{4}{5}$​, and many more.

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 $\dfrac{1}{2}$​. So the blue line is $\dfrac{1}{2}$​ units long. Write the fraction that gives the length of the black line in the box.

Ans: The blue line is $\dfrac{1}{2}$​ units long.

The black line is 2 units long. The fraction for the length of the black line is $\dfrac{2}{1}$​.

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

Ans: The fractions for the lengths of the black lines in the respective boxes are:

Exercise 7.5

Questions:

1. How many whole units are there in $\dfrac{7}{2}$​?

Ans: 27​ has 3 whole units and a remaining fraction of $\dfrac{1}{2}$​.

2. How many whole units are there in $\dfrac{4}{3}$​ and $\dfrac{7}{3}$​?

Ans:

• 4​ has 1 whole unit and $\dfrac{1}{3}$​ remaining.

• $\dfrac{7}{3}$​ has 2 whole units and $\dfrac{1}{3}$​ Remaining

Figure it Out:

1. Figure out the number of whole units in each of the following fractions:

a. $\dfrac{8}{3}$​
b. $\dfrac{11}{5}$​
c. $\dfrac{9}{4}$​

Ans: a. $\dfrac{8}{3}$ has 2 whole units and $\dfrac{2}{3}$ remaining.
b. $\dfrac{11}{5}$​ has 2 whole units and $\dfrac{1}{5}$​ remaining.
c. $\dfrac{9}{4}$​ has 2 whole units and $\dfrac{1}{4}$​ remaining.

2. Can all fractions greater than 1 be written as such mixed numbers?

Ans: Yes, all fractions greater than 1 can be written as mixed numbers, which include a whole number part and a fractional part.

3. Write the following fractions as mixed fractions (e.g., $\dfrac{9}{2}$ = 4 $\dfrac{1}{2}$​):

$\dfrac{8}{3}$​

$\dfrac{11}{5}$​

$\dfrac{19}{4}$​

$\dfrac{47}{9}$​

$\dfrac{12}{5}$​

$\dfrac{19}{6}$​

Ans:

$\dfrac{8}{3}$​ = 2 $\dfrac{2}{3}$

$\dfrac{11}{5}$ =2​ $\dfrac{1}{5}$

$\dfrac{19}{4}$​ =4 $\dfrac{3}{4}$

$\dfrac{47}{9}$​ =5 $\dfrac{2}{9}$

$\dfrac{12}{5}$​ =2 $\dfrac{2}{5}$

$\dfrac{19}{6}$​ =3 $\dfrac{1}{6}$

Figure it Out

1. ​Write the following mixed numbers as fractions:

a. 3 $\dfrac{1}{4}$
b. 7$ \dfrac{2}{3}$​
c. 9$ \dfrac{4}{9}$​
d. 3$ \dfrac{1}{6}$​
e. 2$ \dfrac{3}{11}$​
f. $3 \dfrac{9}{10}$

Ans:

a. 3 $\dfrac{1}{4}$ = $\dfrac{13}{4}$
b. 7$ \dfrac{2}{3}$​ = $\dfrac{23}{3}$​
c.9$ \dfrac{4}{9}$​  = $\dfrac{85}{9}$​
d. 3$ \dfrac{1}{6}$​  = $\dfrac{19}{6}$​
e. 2$ \dfrac{3}{11}$ = $\dfrac{25}{11}$​
f. $3 \dfrac{9}{10}$ = $\dfrac{39}{10}$​

Exercise 7.6

Answer the following questions after looking at the fraction wall:

1. Are the lengths $\dfrac{1}{2}$​ and $\dfrac{3}{6}$​ equal?

The lengths $\dfrac{1}{2}$​ and $\dfrac{3}{6}$ are equal because $\dfrac{3}{6}$​ simplifies to $\dfrac{1}{2}$​.

2. Are $\dfrac{2}{3}$​ and $\dfrac{4}{6}$​ equivalent fractions? Why?

Yes, $\dfrac{2}{3}$​ and $\dfrac{4}{6}$​ are equivalent fractions because when you multiply both the numerator and denominator of $\dfrac{2}{3}$ by 2, you get $\dfrac{4}{6}$​.

3. How many pieces of length $\dfrac{1}{6}$​ will make a length of $\dfrac{1}{2}$​?

Three pieces of length $\dfrac{1}{6}$​ will make a length of $\dfrac{1}{2}$​, because $\dfrac{3}{6} = \dfrac{1}{2}$.

4. How many pieces of length $\dfrac{1}{6}$​ will make a length of $\dfrac{1}{3}$​?

Two pieces of length $\dfrac{1}{6}$​ will make a length of $\dfrac{1}{3}$​, because $\dfrac{2}{6} = \dfrac{1}{3}$.

Figure it out

1. Are $\dfrac{3}{6}$, $\dfrac{4}{8}$, and $\dfrac{5}{10}$​ equivalent fractions? Why?

Answer: Yes, $\dfrac{3}{6}$​, $\dfrac{4}{8}$​, and $\dfrac{5}{10}$​ are equivalent fractions because they all simplify to $\dfrac{1}{2}$.

2. Write two equivalent fractions for $\dfrac{2}{6}$.

Answer: Two equivalent fractions for $\dfrac{2}{6}$​ are:

• $\dfrac{1}{3}$ (simplified by dividing the numerator and denominator by 2)

• $\dfrac{4}{12}$​ (multiplied both the numerator and denominator by 2)

3. Write as many equivalent fractions as you can for $\dfrac{4}{6}$

Answer:​ Some equivalent fractions for $\dfrac{4}{6}$​ are:

• $\dfrac{2}{3}$ (simplified by dividing by 2)

• $\dfrac{8}{12}$​ (multiplied both the numerator and denominator by 2)

• $\dfrac{12}{18}$​ (multiplied both the numerator and denominator by 3)

• $\dfrac{16}{24}$​ (multiplied both the numerator and denominator by 4)

Figure it out:

1. Three rotis 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: ______.

Answer: The fraction of roti each child gets is: $\dfrac{3}{4}$​.

Division fact: $3 \div 4 = \frac{3}{4}$​.

Addition fact:$\dfrac{3}{4} + \dfrac{3}{4} + \dfrac{3}{4} + \dfrac{3}{4} = 3$

Multiplication fact: $4 \times \dfrac{3}{4} = 3$

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.

Answer:

• Each child gets $\dfrac{2}{4} = \dfrac{1}{2}$

• Division fact: $2 \div 4 = \frac{1}{2}$

• Addition fact:$\dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2} = 2.$

• Multiplication fact: $4 \times \dfrac{1}{2} = 2$.

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

Answer: Anil would get $\dfrac{2}{5}$​ of a cake.

a. 5 glasses of juice shared equally among 4 friends is the same as ____ glasses of juice shared equally among 8 friends.
So,$\frac{5}{4} = \frac{\_}{8}$​.

Answer:

5 glasses of juice shared equally among 4 friends is the same as 10 glasses of juice shared equally among 8 friends.

So, $\dfrac{5}{4} = \dfrac{10}{8}$​.

b. 4 kg of potatoes divided equally in 3 bags is the same as 12 kg of potatoes divided equally in ____ bags.
So, $\dfrac{4}{3} = \dfrac{12}{\_}$​.

Answer:

4 kg of potatoes divided equally in 3 bags is the same as 12 kg of potatoes divided equally in 9 bags.

So,$\dfrac{4}{3} = \dfrac{12}{9}$​.

c. 7 rotis divided among 5 children is the same as ____ rotis divided among ____ children.

So,$\dfrac{7}{5} = \dfrac{\_}{\_}$. 

Answer:

7 rotis divided among 5 children is the same as 14 rotis divided among 10 children.

So,$\dfrac{7}{5} = \frac{14}{10}$.

4. Find equivalent fractions for the given pairs of fractions such that the fractional units are the same.

a. $\dfrac{7}{2}$ $and $\dfrac{3}{5}$
b. $\dfrac{8}{3}$​ and $\dfrac{5}{6}$​
c. $\dfrac{3}{4}$ and $\dfrac{3}{5}$
d. $\dfrac{6}{7}$​ and $\dfrac{8}{5}$​
e. $\dfrac{9}{4}$​ and $\dfrac{5}{2}$
f. $\dfrac{1}{10}$​ and $\dfrac{2}{9}$​
g. $\dfrac{8}{3}$ and $\dfrac{11}{4}$
h. $\dfrac{13}{6}$ and $\dfrac{1}{9}$

Answer:

a. Equivalent fractions: $\dfrac{35}{10}$​ and $\dfrac{6}{10}$

b. Equivalent fractions: $\dfrac{16}{6}$​ and $\dfrac{5}{6}$

c. Equivalent fractions: $\dfrac{15}{20}$​ and $\dfrac{12}{20}$

d. Equivalent fractions: $\dfrac{30}{35}$​ and $\dfrac{56}{35}$​

e. Equivalent fractions: $\dfrac{18}{8}$​ and $\dfrac{20}{8}$

f. Equivalent fractions: $\dfrac{9}{90}$​ and $\dfrac{20}{90}$​

g. Equivalent fractions: $\dfrac{32}{12}$​ and $\dfrac{33}{12}$​

h. Equivalent fractions: $\dfrac{39}{18}$​ and $\dfrac{2}{18}$​

Figure it out:

Express the following fractions in lowest terms:

a. $\dfrac{17}{51}$

Find the GCD: The GCD of 17 and 51 is 17 (since 51 = 3 × 17).

Simplify:$\dfrac{17 \div 17}{51 \div 17} = \dfrac{1}{3}$​

Lowest term: $\dfrac{1}{3}$

B. $\dfrac{64}{144}$

Find the GCD: The GCD of 64 and 144 is 16.

Simplify:$\dfrac{64 \div 16}{144 \div 16} = \dfrac{4}{9}$

Lowest term: $\dfrac{4}{9}$

C. $\dfrac{126}{147}$

Find the GCD: The GCD of 126 and 147 is 21.

Simplify:$\dfrac{126 \div 21}{147 \div 21} = \dfrac{6}{7}$​

Lowest term: $\dfrac{6}{7}$​

D. $\dfrac{527}{112}$

1. Find the GCD: The GCD of 527 and 112 is 1 (no common factors).

2. Simplify: Since the GCD is 1, it is already in the lowest terms.$\dfrac{527 \div 1}{112 \div 1} = \frac{527}{112}$

Lowest term:$\dfrac{527}{112}$

Exercise 7.7

1. Compare the following fractions and justify your answers:

a.$\dfrac{8}{3}$ and $\dfrac{5}{2}$
b. $\dfrac{4}{9}$​ and $\dfrac{3}{7}$​
c. $\dfrac{7}{10}$​ and $\dfrac{9}{14}$​
d. $\dfrac{12}{5}$ and $\dfrac{8}{5}$​
e. $\dfrac{9}{4}$ and $\dfrac{5}{2}$

Answer:

a.$\dfrac{8}{3} $> $\dfrac{5}{2}$ ​because $8 \times 2 = 16$ and $5 \times 3 $= 15, and 16>15.

b. $\dfrac{4}{9} > \frac{3}{7}$​ because $4 \times 7 = 28$ and $3 \times 9 = 27$, and 28>27.

c.$\dfrac{7}{10} > \dfrac{9}{14}$​ because the equivalent fractions are $\dfrac{49}{70}$​ and $\dfrac{45}{70}$​, and 49>45.

d.$\dfrac{12}{5}$ > $\dfrac{8}{5}$​ because both fractions have the same denominator, and 12>8.

e. $\dfrac{9}{4} $> $\dfrac{5}{2}$​ because $\dfrac{9}{4}$ = 2.25 and $\dfrac{5}{2}$ = 2.5, and 2.25>2.5.

2. Write the following fractions in ascending order: 

a.$\dfrac{7}{10}$, $\dfrac{11}{15}$, $\dfrac{2}{5}$
b.$\dfrac{19}{24}$, $\dfrac{5}{6}$, $\dfrac{7}{12}$​

Answer:

Ascending order: 

a.$\dfrac{2}{5}$, $\dfrac{7}{10}$, $\dfrac{11}{15}$​

b.$\dfrac{7}{12}$, $\dfrac{19}{24}$, $\dfrac{5}{6}$​

3. Write the following fractions in descending order: 

a.$\dfrac{25}{16}$, $\dfrac{7}{8}$, $\dfrac{13}{4}$, $\dfrac{17}{32}$
b.$\dfrac{3}{4}$, $\dfrac{12}{5}$, $\dfrac{7}{12}$, $\dfrac{5}{4}$​

Answer:

Descending order:

a.$\dfrac{13}{4}$, $\dfrac{25}{16}$, $\dfrac{7}{8}$, $\dfrac{17}{32}$​

b.$\dfrac{12}{5}$, $\dfrac{5}{4}$, $\dfrac{3}{4}$, $\dfrac{7}{12}$​

Exercise 7.8

1. Add the following fractions using Brahmagupta’s method:

a.$\dfrac{2}{7} + \dfrac{5}{7} + \dfrac{6}{7}$

Answer:

Since the denominators are the same, add the numerators:
$\dfrac{2+5+6}{7} = \dfrac{13}{7}$

b.$\dfrac{3}{4} + \dfrac{1}{3}$

Answer:

Find the least common denominator (LCD) of 4 and 3, which is 12:
$\dfrac{3}{4}$ = $\dfrac{9}{12}$​ and$\dfrac{1}{3}$ = $\dfrac{4}{12}$
$\dfrac{9+4}{12}$ = $\dfrac{13}{12}$

c.$\dfrac{2}{3} + \dfrac{5}{6}$

Answer:

The LCD of 3 and 6 is 6:
$\dfrac{2}{3}$ = $\dfrac{4}{6}$​ and $\dfrac{5}{6}$​ stays the same.
$\dfrac{4+5}{6}$ = $\dfrac{9}{6}$= $\dfrac{3}{2}$​

d.$\dfrac{2}{7} + \dfrac{2}{7}$

Answer:

Since the denominators are the same:
$\dfrac{2+2}{7}$ = $\dfrac{4}{7}$

e.$\dfrac{3}{4} + \dfrac{1}{3} + \dfrac{1}{5}$

Answer:

The LCD of 4, 3, and 5 is 60:
$\dfrac{3}{4}$ = $\dfrac{45}{60}$, $\dfrac{1}{3}$ = $\dfrac{20}{60}$, $\dfrac{1}{5}$ = $\dfrac{12}{60}$​
$\dfrac{45+20+12}{60}$ = $\dfrac{77}{60}$

f. $\dfrac{2}{3} + \dfrac{4}{5}$

Answer:

The LCD of 3 and 5 is 15:
$\dfrac{2}{3} $= $\dfrac{10}{15}$ and $\dfrac{4}{5}$ = $\dfrac{12}{15}$
$\dfrac{10+12}{15} $= $\dfrac{22}{15}$​

g. $\dfrac{4}{5} + \dfrac{3}{4}$

Answer:

The LCD of 5 and 4 is 20:
$\dfrac{4}{5}$ = $\dfrac{16}{20}$ and $\dfrac{3}{4}$ = $\dfrac{15}{20}$

$\dfrac{16+15}{20}$ = $\dfrac{31}{20}$​

h. $\dfrac{3}{5} + \dfrac{5}{8}$

Answer:

The LCD of 5 and 8 is 40:
$\dfrac{3}{5}$ = $\dfrac{24}{40}$​ and $\dfrac{5}{8}$ = $\dfrac{25}{40}$

$\dfrac{24+25}{40}$ = $\dfrac{49}{40}$​

i. $\dfrac{9}{2} + \dfrac{5}{4}$

Answer:

The LCD of 2 and 4 is 4:
$\dfrac{9}{2}$ = $\dfrac{18}{4}$ and $\dfrac{5}{4}$​ stays the same.
$\dfrac{18+5}{4} = \dfrac{23}{4}$

j. $\dfrac{8}{3} + \dfrac{2}{7}$

Answer:

The LCD of 3 and 7 is 21:
$\dfrac{8}{3}$ = $\dfrac{56}{21}$ and $\dfrac{2}{7} = \dfrac{6}{21}$
$\dfrac{56+6}{21}$ = $\dfrac{62}{21}$​

k. $\dfrac{3}{4} + \dfrac{1}{3} + \dfrac{1}{5}$

Answer:

The LCD of 4, 3, and 5 is 60:
$\dfrac{3}{4} $= $\dfrac{45}{60}$, $\dfrac{1}{3}$ = $\dfrac{20}{60}$, $\dfrac{1}{5}$ = $\dfrac{12}{60}$​
$\dfrac{45+20+12}{60} $= $\dfrac{77}{60}$​

l. $\dfrac{2}{3} + \dfrac{4}{5} + \dfrac{7}{7}$

Answer:

The LCD of 3, 5, and 7 is 105:
$\dfrac{2}{3}$ = $\dfrac{70}{105}$, $\dfrac{4}{5}$ = $\dfrac{84}{105}$, $\dfrac{7}{7}$ = $\dfrac{105}{105}$​

$\dfrac{70+84+105}{105} $= $\dfrac{259}{105}$

M. $\dfrac{9}{2} + \dfrac{5}{4} + \dfrac{7}{6}$

Answer:

The LCD of 2, 4, and 6 is 12:
$\dfrac{9}{2}$ = $\dfrac{54}{12}$, $\dfrac{5}{4}$ = $\dfrac{15}{12}$, $\dfrac{7}{6}$ = $\dfrac{14}{12}$​
$\dfrac{54+15+14}{12} $= $\dfrac{83}{12}$​

2. Rahim mixes $\dfrac{2}{3}$​ liters of yellow paint with $\dfrac{3}{4}$​ liters of blue paint to make green paint. What is the volume of green paint he has made?

Answer:

The LCD of 3 and 4 is 12:
$\dfrac{2}{3}$ = $\dfrac{8}{12}$​ and $\dfrac{3}{4}$ = $\dfrac{9}{12}$

$\dfrac{8}{12}$ + $\dfrac{9}{12}$ = $\dfrac{17}{12}$ = 1 $\dfrac{5}{12}$​ liters of green paint.

3. Geeta bought $\dfrac{2}{5}$​ meter of lace and Shamim bought $\dfrac{3}{4}$​ meter of the same lace to put a complete border on a tablecloth 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?

Answer:

The LCD of 5 and 4 is 20:
$\dfrac{2}{5}$ = $\dfrac{8}{20}$​ and $\dfrac{3}{4} = \dfrac{15}{20}$​
$\dfrac{8}{20}$ + $\dfrac{15}{20}$ = $\dfrac{23}{20}$ = 1 $\dfrac{3}{20}$​ meters.
They have 1 meter and $\dfrac{3}{20}$​ meters of lace, which is more than enough to cover the 1-meter border.

Figure it out:

1. ​$\dfrac{5}{8}$  - $\dfrac{3}{8}$ 

Solution:

$\dfrac{5}{8}$ - $\dfrac{3}{8}$ = $\dfrac{5 - 3}{8}$ = $\dfrac{2}{8}$ = $\dfrac{1}{4}$​

2.$\dfrac{7}{9}$ - $\dfrac{5}{9}$​

$\dfrac{7}{9}$- $\dfrac{5}{9}$ = $\dfrac{7 - 5}{9}$ = $\dfrac{2}{9}$​

3. $\dfrac{10}{27}$ -$ \dfrac{1}{27}$​

$\dfrac{10}{27}$ - $\dfrac{1}{27}$ = $\dfrac{10 - 1}{27}$ = $\dfrac{9}{27}$ = $\dfrac{1}{3}$​

Figure it out:

1. Carry out the following subtractions using Brahmagupta’s method:

a.$\dfrac{8}{15}$ -$\dfrac{3}{15}$​

Solutions:

Since the denominators are the same:
$\dfrac{8 - 3}{15} $=$ \dfrac{5}{15}$ =$ \dfrac{1}{3}$​

b.$\dfrac{2}{5}$ - $\dfrac{4}{15}$​
Solutions:

The least common denominator (LCD) of 5 and 15 is 15:
$\dfrac{2}{5}$= $\dfrac{6}{15}$, so $\dfrac{6}{15} $- $\dfrac{4}{15}$ = $\dfrac{2}{15}$​

c. $\dfrac{5}{6} - \dfrac{4}{9}$
The LCD of 6 and 9 is 18:
$\dfrac{5}{6} $= $\dfrac{15}{18}$​ and$\dfrac{4}{9} $= $\dfrac{8}{18}$
So,$\dfrac{15}{18}$ - $\dfrac{8}{18}$ = $\dfrac{7}{18}$​

d. $\dfrac{2}{3} - \dfrac{1}{2}$
The LCD of 3 and 2 is 6:
$\dfrac{2}{3}$ = $\dfrac{4}{6}$​ and $\dfrac{1}{2}$ = $\dfrac{3}{6}$​
So,$\dfrac{4}{6}$ - $\dfrac{3}{6}$ =$\dfrac{1}{6}$​

2. Subtract as indicated:

a.$\dfrac{13}{4}$​ from $\dfrac{10}{3}$​

Solution:

The LCD of 4 and 3 is 12:
$\dfrac{13}{4}$ = $\dfrac{39}{12}$​ and $\dfrac{10}{3}$ = $\dfrac{40}{12}$​
So,$\dfrac{40}{12}$ - $\dfrac{39}{12}$ = $\dfrac{1}{12}$​

b. $\dfrac{18}{5}$​ from $\dfrac{23}{3}$
Solution:

The LCD of 5 and 3 is 15:
$\dfrac{18}{5}$ = $\dfrac{54}{15}$​ and $\dfrac{23}{3}$ = $\dfrac{115}{15}$​
So, $\dfrac{115}{15}$ - $\dfrac{54}{15}$ = $\dfrac{61}{15}$​

c. $\dfrac{29}{7}729​ from 457\dfrac{45}{7}$
Solution:

Since the denominators are the same:
$\dfrac{45}{7} $-$ \dfrac{29}{7}$ = $\dfrac{16}{7}$​

3. Solve the following problems:

a. Jaya’s school is $\dfrac{7}{10}$​ km from her home. She takes an auto for $\dfrac{1}{2}$​ km and then walks the remaining distance. How much does she walk daily?

The LCD of 10 and 2 is 10:
$\dfrac{7}{10} $- $\dfrac{1}{2}$ = $\dfrac{7}{10}$ -$ \dfrac{5}{10}$ =$ \dfrac{2}{10}$ = $\dfrac{1}{5}$​ km
So, Jaya walks $\dfrac{1}{5}$​ km daily.

b. Jeevika takes $\dfrac{10}{3}$ minutes to complete a round of the park and her friend Namit takes $\dfrac{13}{4}$​ minutes. Who takes less time and by how much?

The LCD of 3 and 4 is 12:
$\dfrac{10}{3}$ = $\dfrac{40}{12}$​ and $\dfrac{13}{4}$ = $\dfrac{39}{12}$​
Jeevika takes$\dfrac{40}{12}$​ minutes and Namit takes $\dfrac{39}{12}$​ minutes.
Jamit takes less time by$\dfrac{1}{12}$​ minute.

Exercise 7.9

1. Can you find three different fractional units that add up to 1? It turns out there is only one solution to this problem (up to changing the order of the 3 fractions)! Can you find it? Try to find it before reading further.

Solution:

Yes, one solution to the problem of finding three different fractional units that add up to 1 is:

$\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6} = 1$

This combination works because:

• $\dfrac{1}{2} = \dfrac{3}{6}$​

• $\dfrac{1}{3} = \dfrac{2}{6}$

• $\dfrac{1}{6}$ remains the same.

When you add them together:

$\dfrac{3}{6} + \dfrac{2}{6} + \dfrac{1}{6} = \dfrac{6}{6}$ = 1

So, $\dfrac{1}{2}$, $\dfrac{1}{3}$​, and $\dfrac{1}{6}$​ are the three different fractions that add up to 1.

2. Can you find four different fractional units that add up to 1? It turns out that this problem has six solutions! Can you find at least one of them? Can you find them all? You can try using similar reasoning as in the cases of two and three fractional units—or find your own method! 

Solution:

Here are four different fractional units that add up to 1, along with a few combinations:

One Solution:

1. $\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6} + \dfrac{1}{6} = 1$

Six Solutions:

1. $\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6} + \dfrac{1}{6} = 1$

2. $\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} = 1$

3. $\dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} = 1$

4. $\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} = 1$

5. $\dfrac{1}{3} + \dfrac{1}{3} + \dfrac{1}{3} + \dfrac{1}{3} = 1$

6. $\dfrac{1}{5} + \dfrac{1}{5} + \dfrac{1}{5} + \dfrac{1}{5} = 1$

Different Combinations of Four Different Fractions:

1. $\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{12} + \dfrac{1}{4} = 1$

2. $\dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} = 1$

3. $\dfrac{1}{5} + \dfrac{1}{10} + \dfrac{1}{10} + \dfrac{1}{10} = 1$

Benefits of NCERT Solutions for Class 6 Maths Chapter 7 Fractions

• Provides easy-to-follow explanations about fractions, helping students understand concepts like proper and improper fractions clearly.

• Offers clear steps and methods for adding, subtracting, multiplying, and dividing fractions, making it simpler for students to learn and apply these operations effectively.

• Helps students learn the basics of fractions, which is essential for understanding more advanced math topics in the future.

• Includes various practice problems that improve student’s ability to recognise and work with fractions, enhancing their problem-solving skills.

• The FREE PDF download allows students to study and practice at their own pace, making learning more convenient and adaptable.

Class 6 Maths Chapter 7: Exercises Breakdown

Important Study Material Links for Class 6 Maths Chapter 7 - Fractions

Conclusion 

NCERT Solutions for Class 6 Maths Chapter 7, "Fractions," helps students understand how to work with fractions, including operations like addition, subtraction, multiplication, and division. The clear explanations make it easier to learn these important concepts and their practical uses. Practising the exercises improves students' skills in handling fractions, which is essential for more advanced maths topics. With the solutions available as a FREE PDF download, students can study and review at their own pace, making learning both convenient and effective.

Chapter-wise NCERT Solutions Class 6 Maths 

The chapter-wise NCERT Solutions for Class 6 Maths are given below. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

Related Important Links for Maths Class 6

Along with this, students can also download additional study materials provided by Vedantu for Maths Class 6.