CBSE Class 6 Maths Important Questions Chapter 7 - Fractions

Very Short Answer Questions                                                1 Mark

1. Write fraction representing the shaded portion

Ans: Given- A figure with some shaded squares in it.

We have to find fractions for the shaded portion.

Total number of squares $=12$

Number of shaded boxes $=6$

$\therefore $ the shaded portion $=\dfrac{6}{12}$

2. Shade in the given figure: $\dfrac{5}{9}$

Ans: Given: A figure having $9$ boxes.

We have to shade 5 boxes out of a total 9 boxes.

Therefore,

3. Write in fraction form of eight-ninths.

Ans: We are given eight-ninths

To find: the fraction form of eight-ninths

Eight - ninths means eight parts out of nine

So, the fraction will be $\dfrac{8}{9}$

4. Write down the fraction with numerator 3, denominator 9

Ans: Given, numerator =3

Denominator =9

We have to find the fraction.

We know that the numerator is the above part of a fraction and the denominator is the below part of the fraction.

$\therefore$ The fraction is $\dfrac{3}{9}$.

5. Fill up the blanks

1. $\mathbf{\dfrac{1}{12}\square 1}$

Ans: Given: $\dfrac{1}{12}\square 1$

We have to put a sign between the terms like $<,>,=$

Solve, $\dfrac{1}{12}$

$=0.83$

We can see that the value is less than one.

$\therefore \dfrac{1}{12}<1$

2. $\mathbf{\dfrac{6512}{6512}\square 1}$

Ans: Given: $\dfrac{6512}{6512}\square 1$

We have to put a sign between the terms like $<,>,=$

Solve, $\dfrac{6512}{6512}$

$=1$

Therefore,

$\dfrac{6512}{6512}=1$

6. Compare $\mathbf{\dfrac{4}{5}}$ and $\mathbf{\dfrac{3}{5}}$

Ans: Given: two terms $\dfrac{4}{5},\,\dfrac{3}{5}$

We have to compare the given terms.

We can see that the denominator of both the terms is the same. So we will compare the numerators only.

Here, 

4>3

 $\therefore \dfrac{4}{5}>\dfrac{3}{5}$

Short Answer Questions                                                           2 Marks

1. Find $\mathbf{\dfrac{3}{4}}$ of $\mathbf{12.}$

Ans: Given, two terms $\dfrac{3}{4},12$

We have to find $\dfrac{3}{4}$ of $12.$

We know that $x$ of $y$ means $x\times y$

$\therefore \dfrac{3}{4}$ of $12$\[\]

$ =\dfrac{3}{4}\times 12 $

$ =9 $

2. What fraction of an hour is 35 minutes?

Ans: Given, a time

We have to find $35$ minutes will be what fraction of an hour.

We know that $1$ hour $=60$ minutes

$\therefore $ fraction will be $\dfrac{35}{60}.$

3. The figure given can be written in the fraction form as $\mathbf{\dfrac{2}{3}.}$ Say true or false.

Ans: Given: figure

We have to find if the figure shows fraction $\dfrac{2}{3}.$

As we can see that all parts of the figure are not similar. Therefore, it cannot be represented as a fraction. So, the statement is False.

4. Name the numerator and denominator in the $\mathbf{\dfrac{16}{20}}$

Ans: Given: $\dfrac{16}{20}$

To find: numerator and denominator

We know that the numerator is the above part of the fraction and the denominator is the below part.

$\therefore $ Numerator $=16$

Denominator $=20$

5. Convert $\mathbf{\dfrac{30}{8}}$ into a mixed fraction

Ans: Given: $\dfrac{30}{8}$

To find: mixed fraction of the given expression.

We got mixed fraction by dividing the fraction

We know if  $\dfrac{x}{y}=z$ then mixed fraction is $z\dfrac{x}{y}$

$\therefore $ $\dfrac{30}{8}$

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

$ =3\dfrac{3}{4}$

6. Convert $\mathbf{6\dfrac{7}{9}}$ into improper fraction.

Ans: Given: $6\dfrac{7}{9}$

To find the improper form of the given expression

We know that \[z\dfrac{x}{y}=\dfrac{y\times z+x}{y}\]

So, $6\dfrac{7}{9}=9\times 6+7$ as numerator

Therefore, the improper fraction will be $\dfrac{61}{9}.$

7. Fill in the blanks $\mathbf{\dfrac{54}{63}=\dfrac{6}{\square }}$

Ans:  Given: $\dfrac{54}{63}=\dfrac{6}{\square }$

We need to fill the blank

On Left Hand Side, divide numerator and denominator y by $9$

Thus, $\dfrac{54}{63}\div \dfrac{9}{9}$

$=\dfrac{6}{7}$

Thus, the number which has to be filled at blank is $7.$

8. Simplify:

1. $\mathbf{\dfrac{3}{8}+\dfrac{4}{8}+\dfrac{2}{8}}$

Ans: Given: $\dfrac{3}{8}+\dfrac{4}{8}+\dfrac{2}{8}$

We have to simplify the fractions by adding them

As the denominator of each fraction is same then the $\text{L}\text{.C}\text{.M}$ will be $8.$

Now, simply add the numerator of each fraction, we get

$=\dfrac{3+4+2}{8}$
$=\dfrac{9}{8}.$

2. $\mathbf{\dfrac{8}{9}-\dfrac{6}{9}}$

Ans: We have to find the difference between the fractions.

We can see that the denominator of both fractions is same then the $\text{L}\text{.C}\text{.M}$will be $9.$

Now, simply subtract the numerators, we get

$ \dfrac{8-6}{9} $

$ =\dfrac{2}{9} $

Short Answer Questions                                                           3 Marks

1. Seema has 28 books. She gave $\mathbf{\dfrac{4}{10}}$ to Meera. How many books does Meena has? How much is left with Seema?

Ans: Given: Total books Seema has $=28$

Seema gave books to Meera $\dfrac{4}{10}$ of $28$

$\therefore $ Books with Meena $=\dfrac{4}{10}\times 28$

$=11$ Books

Now, books left with Seema $=28-11$

$=17$ Books

2. Represent $\mathbf{3\dfrac{2}{5}}$ on the number line.

Ans: Given: $3\dfrac{2}{5}$

We have to represent the fraction on the number line.

We can write the fraction as,

$3\dfrac{2}{5}=3+\dfrac{2}{5}$

Therefore, we represent it on the number line as 

3. Write a fraction equivalent to $\mathbf{\dfrac{4}{5}}$ with numerator 16.

Ans: Given: $\dfrac{4}{5}$

We have to find a fraction so that the numerator of fraction is $16.$

Multiply the numerator and denominator of the given fraction with $4$to get the required fraction.

$ \dfrac{4}{5}\times \dfrac{4}{4} $ 

$ =\dfrac{16}{20} $

4. Write a fraction equivalent to $\mathbf{\dfrac{42}{60}}$ with denominator 10.

Ans: Given: $\dfrac{42}{60}$

We have to find a fraction so that the denominator of fraction is $10.$

Divide the numerator and denominator of the given fraction with $6$ to get the required fraction.

$ \therefore \dfrac{42}{60}\div \dfrac{6}{6} $

$ =\dfrac{7}{10} $

5. Simplify $\mathbf{\dfrac{7}{10}}$ into the simplest form.

Ans: Given: $\dfrac{7}{10}$

We have to find the simplest form of the fraction.

The HCF of the terms \[7\text{ }\!\!\And\!\!\text{ }10\] is \[1.\]

Thus, the fraction $\dfrac{7}{10}$ is already in its simplest form.

6. Simplify \[\mathbf{2\dfrac{7}{9}+\dfrac{11}{15}+\dfrac{9}{24}-3\dfrac{1}{4}}\]

Ans: Given: \[2\dfrac{7}{9}+\dfrac{11}{15}+\dfrac{9}{24}-3\dfrac{1}{4}\]

We need to simplify the given expression. So we’ll find \[\text{LCM}\]and then simplify the expression.

\[\text{LCM}\]of numbers \[9,15,24,4=360\]

$ 2\dfrac{7}{9}+\dfrac{11}{15}+\dfrac{9}{24}-3\dfrac{1}{4} $ 

$ =\dfrac{25}{9}+\dfrac{11}{15}+\dfrac{9}{24}-\dfrac{13}{4} $

$ =\dfrac{1000+264+135-1170}{360} $

$ =\dfrac{1399-1170}{360} $

$ =\dfrac{229}{360} $

7. Subtract $\mathbf{3\dfrac{7}{8}-5\dfrac{1}{6}}$

Ans: Given: $3\dfrac{7}{8}-5\dfrac{1}{6}$

We need to simplify the given expression. So we’ll find \[\text{LCM}\]and then simplify the expression.

\[\text{LCM}\] of the numbers $6,8=24$

$ 3\dfrac{7}{8}-5\dfrac{1}{6} $

$ =\dfrac{31}{6}-\dfrac{31}{8}$ 

$ =\dfrac{31\times 4-31\times 3}{24} $

$ =\dfrac{124-93}{24} $

$ =\dfrac{31}{24} $

Long  Answer Questions                                                           4 Marks

1. Write four equivalent fraction for each of the following:

1. $\mathbf{\dfrac{3}{7}}$

 Ans: Given: $\dfrac{3}{7}$

To find: four equivalent fraction

Multiply and divide numerator and denominator with four different numbers

$ \dfrac{3}{7}\times \dfrac{2}{2}=\dfrac{6}{14} $

$ \dfrac{3}{7}\times \dfrac{3}{3}=\dfrac{9}{21} $

$ \dfrac{3}{7}\times \dfrac{4}{4}=\dfrac{12}{28} $

$ \dfrac{3}{7}\times \dfrac{5}{5}=\dfrac{15}{35} $

2. $\mathbf{\dfrac{300}{900}}$

Ans: To find four equivalent fraction

Multiply and divide numerator and denominator with four different numbers

$ \dfrac{300}{900}\div \dfrac{2}{2}=\dfrac{150}{450}$ 

$ \dfrac{300}{900}\div \dfrac{3}{3}=\dfrac{100}{300} $

$ \dfrac{300}{900}\div \dfrac{5}{5}=\dfrac{60}{180} $

$ \dfrac{300}{900}\div \dfrac{10}{10}=\dfrac{30}{90} $ 

2. Show that $\mathbf{\dfrac{6}{7}}$ and $\mathbf{\dfrac{42}{49}}$ are equivalent fractions.

Ans: Given: Fractions, $\dfrac{6}{7}$, $\dfrac{42}{49}$

We need to show that both the fractions are equivalent.

Thus, $\dfrac{6}{7}=\dfrac{42}{49}$

Cross multiply, we get

$ 6\times 49=249........(1) $

$ 7\times 42=294........(2) $

$ \Rightarrow (1)=(2) $

Therefore, we can say that the given fractions are equivalent.

3. Reduce into simplest form: $\mathbf{\dfrac{225}{500}}$

Ans: Given: $\dfrac{225}{500}$

We need to find the simplest form of the given fraction.

Divide by \[5,\] we get

$\dfrac{225}{500}\div \dfrac{5}{5}=\dfrac{45}{100}$

Again, divide by \[5,\] we get

$\dfrac{45}{100}\div \dfrac{5}{5}=\dfrac{9}{20}$

We can see that the HCF of the terms of the fraction is $1$

$\therefore \dfrac{9}{20}$ is the simplest form of the given fraction.

4. Convert \[\mathbf{\dfrac{1}{4},\dfrac{5}{8},\dfrac{13}{24}}\] into like fractions.

Ans: Given: \[\dfrac{1}{4},\dfrac{5}{8},\dfrac{13}{24}\]

We need to convert the given fractions into like fractions.

The LCM of the denominators will be

\[\begin{align} & 4\left| \!{\underline {\, 4,8,24 \,}} \right.  \\ & 2\left| \!{\underline {\, 1,2,6 \,}} \right.  \\ & \left| \!{\underline {\, 1,1,3 \,}} \right.  \\ \end{align}\]

Therefore, LCM = $4\times 2\times 3$ 

$\quad\quad\quad\quad\quad\quad=24$ 

$=\dfrac{\left( 6\times 1 \right),\left( 5\times 3 \right),\left( 13 \right)}{24}$ 

$=\dfrac{6,15,13}{24}$

So, the required like fractions are $\dfrac{6}{24},\dfrac{15}{24},\dfrac{13}{24}$

5. Compare $\mathbf{\dfrac{8}{13}\text{and}\dfrac{8}{7}}$

Ans: Given: fractions $\dfrac{8}{13}\text{and}\dfrac{8}{7}$

We need to compare the fractions.

To compare the fractions the denominator of the fractions must be the same.

To convert into like terms, take LCM then we get

$ =13\times 7 $

$ =91 $

$ \therefore \dfrac{8\times 7,8\times 13}{91} $

$ =\dfrac{56,104}{91} $

$ \therefore \dfrac{56}{91}<\dfrac{104}{91} $

6. Roshni bought a material of length $\mathbf{3\dfrac{2}{5}\text{m}}$ and one more piece of length $\mathbf{2\dfrac{7}{10}\text{m}\text{.}}$ How much material did she purchase in all?

Ans: Given: Length of first material bought by Roshni $=3\dfrac{2}{5}\text{m}$

Length of second material bought by Roshni $=2\dfrac{7}{10}\text{m}$

Total length of material purchased by Roshni will be

$ =3\dfrac{2}{5}+2\dfrac{7}{10} $

$ =3+\dfrac{2}{5}+2+\dfrac{7}{10} $

$ =5+\dfrac{2\times 2+7}{10} $

$ =5+\dfrac{4+7}{10} $

$ =5+\dfrac{11}{10} $

$ =\dfrac{61}{10} $

$ =6\dfrac{1}{10}\text{m} $

7. Ram bought $\mathbf{6\dfrac{1}{2}}$ litres of milk. Out of this $\mathbf{5\dfrac{1}{4}}$ litres was used. How much is the remaining milk?

Ans: Total milk bought by Ram $=6\dfrac{1}{2}\text{litres}$

Milk Used $=5\dfrac{1}{4}\text{litres}$

Remaining milk with Ram will be

$ =6\dfrac{1}{2}-5\dfrac{1}{4} $

$ =6+\dfrac{1}{2}-\left[ 5+\dfrac{1}{4} \right] $

$ =6-5+\left[ \dfrac{1}{2}-\dfrac{1}{4} \right] $

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

$ =1\dfrac{1}{4}\text{litres} $

Very Long  Answer Questions                                             6 Marks

1.  Classify each of the following into proper, improper and mixed fractions.

1. $\mathbf{\dfrac{1}{5}}$

 Ans: Proper fraction

2. $\mathbf{12}$

Ans: Improper fraction

3. $\mathbf{3\dfrac{1}{5}}$

Ans: Proper fraction

4. $\mathbf{\dfrac{15}{6}}$

Ans: Improper fraction

5. $\mathbf{\dfrac{15}{20}}$

Ans: Proper fraction

2. Compare $\mathbf{\dfrac{5}{8}}$ and $\mathbf{\dfrac{4}{9}}$

Ans: Given: $\dfrac{5}{8}$ and $\dfrac{4}{9}$

We need to compare the given fractions so we will convert both fractions into like fractions by taking LCM.

$ \text{LCM = 72} $

$ \text{=}\dfrac{5\times 9,4\times 8}{72} $

$ =\dfrac{45,32}{72} $

$ =\dfrac{5}{8}>\dfrac{4}{9} $

3. Arrange the following in ascending and descending order $\mathbf{\dfrac{2}{3},\dfrac{3}{4},\dfrac{7}{10},\dfrac{8}{15},\dfrac{5}{8}}$

Ans: Given: fractions $\dfrac{2}{3},\dfrac{3}{4},\dfrac{7}{10},\dfrac{8}{15},\dfrac{5}{8}$

We need to find the ascending and descending order of the fractions.

We will first convert the fractions in like terms and then find the order.

To convert into like terms. LCM will be

$\begin{align} & 4\left| \!{\underline {\, 3,4,10,15,8 \,}} \right.  \\ & 5\left| \!{\underline {\, 3,1,10,15,2 \,}} \right.  \\ & 3\left| \!{\underline {\, 3,1,2,3,2 \,}} \right.  \\ & 2\left| \!{\underline {\, 1,1,2,1,2 \,}} \right.  \\ & \left| \!{\underline {\, 1,1,1,1,1 \,}} \right.  \\ & \text{LCM = 4}\times \text{5}\times \text{3}\times \text{2} \\ & \text{=120} \\ \end{align}$

Then the fractions will be

$ \dfrac{2\times 20,3\times 30,7\times 12,8\times 8,5\times 15}{120} $

$ =\dfrac{80,90,84,64,75}{120}$

$=\dfrac{80}{120},\dfrac{90}{120},\dfrac{84}{120},\dfrac{64}{120},\dfrac{75}{120} $

Ascending order will be

$=\dfrac{64}{120},\dfrac{75}{120},\dfrac{80}{120},\dfrac{84}{120},\dfrac{90}{120} $

$ =\dfrac{8}{15},\dfrac{5}{8},\dfrac{2}{5},\dfrac{7}{10},\dfrac{3}{4} $

Descending order will be

$=\dfrac{90}{120},\dfrac{84}{120},\dfrac{80}{120},\dfrac{75}{120},\dfrac{64}{120} $

$ =\dfrac{3}{4},\dfrac{7}{10},\dfrac{2}{3},\dfrac{5}{8},\dfrac{8}{15} $

5 Important Formulas of Class 6 Chapter 7 You Shouldn’t Miss!

• Addition of Fractions: $\dfrac{a}{b} + \dfrac{c}{d} = \dfrac{ad + bc}{bd}$

• Subtraction of Fractions: $\dfrac{a}{b} - \dfrac{c}{d} = \dfrac{ad - bc}{bd}$

• Multiplication of Fractions: $\dfrac{a}{b} \times \dfrac{c}{d} = \dfrac{a \times c}{b \times d}$

• Division of Fractions: $\dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a}{b} \times \dfrac{d}{c}$

• Equivalent Fractions: $\dfrac{a}{b} = \dfrac{ka}{kb}$

Benefits of Class 6 Chapter 7 Maths Important Questions

• Better Understanding: Important questions help clarify key concepts in fractions, making it easier for students to see how they apply in real life.

• Focused Practice: Students can practice specific areas within the chapter, helping them strengthen their knowledge of fractions.

• Improved Problem-Solving Skills: Regular practice helps students become better at solving different types of fraction problems.

• Exam Preparation: Practising important questions helps students get familiar with what to expect in exams, making them feel more confident.

• Self-Assessment: These questions allow students to check their understanding and find areas where they need more practice.

• Homework Help: Important questions provide useful examples, supporting students with their homework and learning outside the classroom.

• Foundation for Future Topics: Understanding fractions well helps students learn more complex math topics in higher grades.

Conclusion 

The Important Questions for Class 6 Maths Chapter 7 - Fractions from Vedantu is a helpful resource for students. It provides a straightforward way to practise and understand fractions, laying a solid foundation for future math topics. By using these questions, students can prepare for their exams effectively and gain confidence in solving problems. Don't miss this chance to improve your learning experience. Access this resource today and take a step toward achieving better results in your studies!

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