NCERT Solutions for Class 6 Maths Chapter 7: Fractions - Exercise 7.4

Exercise 7.4

1. Write shaded portion as a fraction. Arrange them in ascending and descending order using correct sign ‘<’, ‘>’, ‘=’ between the fractions:

Ans: Observe the first circle. It is divided into eight parts out of which three parts are shaded. So, the fraction is $\dfrac{3}{8}$.

Observe the second circle. It is divided into eight parts out of which six parts are shaded. So, the fraction is $\dfrac{6}{8}$.

Observe the third circle. It is divided into eight parts out of which three parts are shaded. So, the fraction is $\dfrac{4}{8}$.

Observe the fourth circle. It is divided into eight parts out of which one part are shaded. So, the fraction is $\dfrac{1}{8}$.

Now, we will arrange the fraction in ascending order. If the denominator is the same, the smaller the numerator, the smaller is the fraction.

Ascending order: $\dfrac{1}{8} < \dfrac{3}{8} < \dfrac{4}{8} < \dfrac{6}{8}$

Now, we will arrange the fraction in descending order. If the denominator is the same, the greater the numerator, the larger is the fraction.

Descending order: $\dfrac{6}{8} > \dfrac{4}{8} > \dfrac{3}{8} > \dfrac{1}{8}$

b.

Ans: Observe the first square. It is divided into nine parts out of which eight parts are shaded. So, the fraction is $\dfrac{8}{9}$.

Observe the second square. It is divided into nine parts out of which four parts are shaded. So, the fraction is $\dfrac{4}{9}$.

Observe the third square. It is divided into nine parts out of which three parts are shaded. So, the fraction is \[\dfrac{3}{9}\].

Observe the fourth square. It is divided into nine parts out of which six parts are shaded. So, the fraction is $\dfrac{6}{9}$.

Now, we will arrange the fraction in ascending order. If the denominator is the same, the smaller the numerator, the smaller is the fraction.

Ascending order: $\dfrac{3}{9} < \dfrac{4}{9} < \dfrac{6}{9} < \dfrac{8}{9}$

Now, we will arrange the fraction in descending order. If the denominator is the same, the greater the numerator, the larger is the fraction.

Descending order: $\dfrac{8}{9} > \dfrac{6}{9} > \dfrac{4}{9} > \dfrac{3}{9}$

c. Show $\dfrac{2}{4}$,$\dfrac{4}{6}$, $\dfrac{8}{6}$ and $\dfrac{6}{6}$ on the number line. Put appropriate signs between the fractions given: 

$\dfrac{5}{6}\boxed{}\dfrac{2}{6}$, $\dfrac{3}{6}\boxed{}0$, $\dfrac{1}{6}\boxed{}\dfrac{6}{6}$, $\dfrac{8}{6}\boxed{}\dfrac{5}{6}$

Ans: Divide the number line between 0 and 1 into 6 parts and mark the points as follows.

Now, apply the sign by seeing the position of the fraction on the number line.

Therefore,

$\dfrac{5}{6}\boxed < \dfrac{2}{6}$,

$\dfrac{3}{6}\boxed < 0$,

$\dfrac{1}{6}\boxed > \dfrac{6}{6}$,

$\dfrac{8}{6}\boxed > \dfrac{5}{6}$

2. Compare the fractions and put an inappropriate sign.

a. $\dfrac{3}{6}\boxed{}\dfrac{5}{6}$

Ans: The fraction, $\dfrac{3}{6}$ shows that three parts are selected out of six parts and the fraction, $\dfrac{5}{6}$ shows that five parts are selected out of six parts. Thus, we know that $5 > 3$. Therefore, $\dfrac{3}{6}\boxed < \dfrac{5}{6}$.

b. $\dfrac{1}{7}\boxed{}\dfrac{1}{4}$

Ans: To solve the given question, make the denominator equal. To do so, multiply the numerator and denominator of $\dfrac{1}{7}$ and $\dfrac{1}{4}$ by 4 and 7 respectively.

$\dfrac{{1 \times 4}}{{7 \times 4}} = \dfrac{4}{{28}}$

$\dfrac{{1 \times 7}}{{4 \times 7}} = \dfrac{7}{{28}}$

The fraction, $\dfrac{4}{{28}}$ shows that four parts are selected out of twenty-eight parts and the fraction, $\dfrac{7}{{28}}$ shows that seven parts are selected out of twenty-eight parts. Thus, $\dfrac{4}{{28}}\boxed < \dfrac{7}{{28}}$.

Therefore, $\dfrac{1}{7}\boxed < \dfrac{1}{4}$.

c. $\dfrac{4}{5}\boxed{}\dfrac{5}{5}$

Ans: The fraction, $\dfrac{4}{5}$ shows that four parts are selected out of five parts and the fraction, $\dfrac{5}{5}$ shows that five parts are selected out of five parts. Thus, we know that $4 < 5$. Therefore, $\dfrac{4}{5}\boxed < \dfrac{5}{5}$.

d. $\dfrac{3}{5}\boxed{}\dfrac{3}{7}$

Ans: To solve the given question, make the denominator equal. To do so, multiply the numerator and denominator of $\dfrac{3}{5}$ and $\dfrac{3}{7}$ by 7 and 5 respectively.

$\dfrac{{3 \times 7}}{{5 \times 7}} = \dfrac{{21}}{{35}}$

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

The fraction, $\dfrac{{21}}{{35}}$ shows that twenty one parts are selected out of thirty five parts and the fraction, $\dfrac{{15}}{{35}}$ shows that fifteen parts are selected out of thirty five parts. Thus, $\dfrac{{21}}{{35}}\boxed > \dfrac{{15}}{{35}}$.

Therefore, $\dfrac{3}{5}\boxed > \dfrac{3}{7}$.

3. Make five more each pair and put appropriate signs.

a. $\dfrac{2}{6}\boxed{}\dfrac{4}{6}$

Ans: The fraction, $\dfrac{2}{6}$ shows that two parts are selected out of six parts and the fraction, $\dfrac{4}{6}$ shows that four parts are selected out of six parts. Thus, we know that $4 > 2$. Therefore, $\dfrac{2}{6}\boxed < \dfrac{4}{6}$.

b. $\dfrac{2}{5}\boxed{}\dfrac{3}{5}$

Ans: The fraction, $\dfrac{2}{5}$ shows that two parts are selected out of five parts and the fraction, $\dfrac{3}{5}$ shows that three parts are selected out of five parts. Thus, we know that $2 < 3$. Therefore, $\dfrac{2}{5}\boxed < \dfrac{3}{5}$.

c. $\dfrac{3}{9}\boxed{}\dfrac{5}{9}$

Ans: The fraction, $\dfrac{3}{9}$ shows that three parts are selected out of nine parts and the fraction, $\dfrac{5}{9}$ shows that five parts are selected out of nine parts. Thus, we know that $5 > 3$. Therefore, $\dfrac{3}{9}\boxed < \dfrac{5}{9}$.

d. $\dfrac{{13}}{{24}}\boxed{}\dfrac{{15}}{{24}}$ 

Ans: The fraction, $\dfrac{{13}}{{24}}$ shows that thirteen parts are selected out of twenty four parts and the fraction, $\dfrac{{15}}{{24}}$ shows that fifteen parts are selected out of twenty four parts. Thus, we know that $15 > 13$. Therefore, $\dfrac{{13}}{{24}}\boxed < \dfrac{{15}}{{24}}$.

e. $\dfrac{2}{5}\boxed{}\dfrac{1}{7}$

Ans: To solve the given question, make the denominator equal. To do so, multiply the numerator and denominator of $\dfrac{2}{5}$ and $\dfrac{1}{7}$ by 7 and 5 respectively.

$\dfrac{{2 \times 7}}{{5 \times 7}} = \dfrac{{14}}{{35}}$

$\dfrac{{1 \times 5}}{{7 \times 5}} = \dfrac{5}{{35}}$

The fraction, $\dfrac{{14}}{{35}}$ shows that fourteen parts are selected out of thirty five parts and the fraction, $\dfrac{5}{{35}}$ shows that five parts are selected out of thirty five parts. Thus, $\dfrac{{14}}{{35}}\boxed > \dfrac{5}{{35}}$.

Therefore, $\dfrac{2}{5}\boxed > \dfrac{1}{7}$.

4. Look at the figures and write ‘<’ or ‘>’, ‘=’ between the given pairs of fractions.

a. $\dfrac{1}{6}\boxed{}\dfrac{1}{3}$

Ans: Here, the numerators are the same. Thus, the fraction that consists of a lesser denominator is the greater.

$\dfrac{1}{6}\boxed{<}\dfrac{1}{3}$

b. $\dfrac{3}{4}\boxed{}\dfrac{2}{6}$

Ans: Let us solve by multiplying ‘3’ in numerator and denominator to the LHS,

$\dfrac{3}{4} = \dfrac{3\times3}{4\times3}$

$= \dfrac{9}{12}$

Now let us multiply ‘2’ in the numerator and denominator to the RHS

$\dfrac{2}{6} = \dfrac{2\times2}{6\times2}$

$= \dfrac{4}{12}$

Both the fractions have the same denominators. So, the fraction having a greater numerator will be the greater

$\dfrac{9}{12}\boxed{>}\dfrac{4}{12}$

$\therefore \dfrac{3}{4}\boxed{>}\dfrac{2}{6}$

c. $\dfrac{2}{3}\boxed{}\dfrac{2}{4}$

Ans: Here, the numerators are the same. So, the fraction having a lesser denominator is said to be greater.

$\dfrac{2}{3}\boxed{>}\dfrac{2}{4}$

d. $\dfrac{6}{6}\boxed{}\dfrac{3}{3}$

Ans: We get $\dfrac{6}{6} = 1 \text{and} \dfrac{3}{3} = 1$ thus $\dfrac{6}{6}\boxed{=}\dfrac{3}{3}$

e. $\dfrac{5}{6}\boxed{}\dfrac{5}{5}$

Ans: Here, the numerators are the same. So, the fraction having a lesser denominator is the greater.

$\dfrac{5}{6}\boxed{<}\dfrac{5}{5}$

5. How quickly can you do this? Fill appropriate sign (<,=,>):

a. $\dfrac{1}{2}\boxed{}\dfrac{1}{5}$

Ans: To solve the given question, make the denominator equal. To do so, multiply the numerator and denominator of $\dfrac{1}{2}$ and $\dfrac{1}{5}$ by 5 and 2 respectively.

$\dfrac{{1 \times 5}}{{2 \times 5}} = \dfrac{5}{{10}}$

$\dfrac{{1 \times 2}}{{5 \times 2}} = \dfrac{2}{{10}}$

The fraction, $\dfrac{5}{{10}}$ shows that five parts are selected out of ten parts and the fraction, $\dfrac{2}{{10}}$ shows that two parts are selected out of ten parts. We know that $5 > 2$.  Thus, $\dfrac{5}{{10}}\boxed > \dfrac{2}{{10}}$.

Therefore, $\dfrac{1}{2}\boxed > \dfrac{1}{5}$.

b. $\dfrac{2}{4}\boxed{}\dfrac{3}{6}$

Ans: To solve the given question, make the denominator equal. To do so, multiply the numerator and denominator of $\dfrac{2}{4}$ and $\dfrac{3}{6}$ by 6 and 4 respectively.

$\dfrac{{2 \times 6}}{{4 \times 6}} = \dfrac{{12}}{{24}}$

$\dfrac{{3 \times 4}}{{6 \times 4}} = \dfrac{{12}}{{24}}$

Both the fractions obtained are same. Thus, $\dfrac{{12}}{{24}}\boxed = \dfrac{{12}}{{24}}$.Therefore, $\dfrac{2}{4}\boxed = \dfrac{3}{6}$.

c. $\dfrac{3}{5}\boxed{}\dfrac{2}{3}$

Ans: To solve the given question, make the denominator equal. To do so, multiply the numerator and denominator of $\dfrac{3}{5}$ and $\dfrac{2}{3}$ by 3 and 5 respectively.

$\dfrac{{3 \times 3}}{{5 \times 3}} = \dfrac{9}{{15}}$

$\dfrac{{2 \times 5}}{{3 \times 5}} = \dfrac{{10}}{{15}}$

The fraction, $\dfrac{9}{{15}}$ shows that nine parts are selected out of fifteen parts and the fraction, $\dfrac{{10}}{{15}}$ shows that ten parts are selected out of fifteenparts. We know that $10 > 9$.  Thus, $\dfrac{9}{{15}}\boxed < \dfrac{{10}}{{15}}$.

Therefore, $\dfrac{3}{5}\boxed < \dfrac{2}{3}$.

d. $\dfrac{3}{4}\boxed{}\dfrac{2}{8}$

Ans: To solve the given question, make the denominator equal. To do so, multiply the numerator and denominator of $\dfrac{3}{4}$ by 2 .

$\dfrac{{3 \times 2}}{{4 \times 2}} = \dfrac{6}{8}$

Thus, $6 < 8$. Therefore, $\dfrac{3}{4}\boxed > \dfrac{2}{8}$.

e. $\dfrac{3}{5}\boxed{}\dfrac{6}{5}$

Ans: The fraction, $\dfrac{3}{5}$ shows that three parts are selected out of five parts and the fraction, $\dfrac{6}{5}$ shows that six parts are selected out of five parts. Thus, we know that $6 > 3$. Therefore, $\dfrac{3}{5}\boxed < \dfrac{6}{5}$.

f. $\dfrac{7}{9}\boxed{}\dfrac{3}{9}$

Ans: The fraction, $\dfrac{7}{9}$ shows that three parts are selected out of five parts and the fraction, $\dfrac{3}{9}$ shows that six parts are selected out of five parts. Thus, we know that $7 > 3$. Therefore, $\dfrac{7}{9}\boxed > \dfrac{3}{9}$.

g. $\dfrac{1}{4}\boxed{}\dfrac{2}{8}$

Ans: To solve the given question, make the denominator equal. To do so, multiply the numerator and denominator of $\dfrac{1}{4}$ by 2.

$\dfrac{{1 \times 2}}{{4 \times 2}} = \dfrac{2}{8}$

Therefore, $\dfrac{1}{4}\boxed = \dfrac{2}{8}$.

h. $\dfrac{6}{{10}}\boxed{}\dfrac{4}{5}$

Ans: Simplify the fraction, $\dfrac{6}{{10}}$ by dividing the numerator and denominator by 2.

$\dfrac{{{6}}}{{{{10}}}} = \dfrac{2}{5}$

Thus, $\dfrac{4}{5} > \dfrac{2}{5}$.

i. $\dfrac{3}{4}\boxed{}\dfrac{7}{8}$

Ans: To solve the given question, make the denominator equal. To do so, multiply the numerator and denominator of $\dfrac{3}{4}$ by 2.

$\dfrac{{3 \times 2}}{{4 \times 2}} = \dfrac{6}{8}$

Thus, $\dfrac{6}{8}\boxed < \dfrac{7}{8}$. Hence, $\dfrac{3}{4}\boxed < \dfrac{7}{8}$.

j. $\dfrac{6}{{10}}\boxed{}\dfrac{4}{5}$

Ans: To solve the given question, make the denominator equal. To do so, multiply the numerator and denominator of $\dfrac{4}{5}$ by 2.

$\dfrac{{4 \times 2}}{{5 \times 2}} = \dfrac{8}{{10}}$

Thus, $\dfrac{6}{{10}}\boxed < \dfrac{8}{{10}}$. Hence, $\dfrac{6}{{10}}\boxed < \dfrac{4}{5}$.

k. $\dfrac{5}{7}\boxed{}\dfrac{{15}}{{21}}$

Ans: To solve the given question, make the denominator equal. To do so, multiply the numerator and denominator of $\dfrac{5}{7}$ by 3.

$\dfrac{{5 \times 3}}{{7 \times 3}} = \dfrac{{15}}{{21}}$

Thus, $\dfrac{{15}}{{21}}\boxed = \dfrac{{15}}{{21}}$. Hence, $\dfrac{5}{7}\boxed = \dfrac{{15}}{{21}}$.

6. The following fractions represent just three different numbers. Separate them into three groups of equivalent fractions, by change each one to its simplest form:

a. $\dfrac{2}{{12}}$

Ans: We can see that 2 is a factor of 12. So, to convert it into the simplest form divide the numerator and denominator by 2.

$\dfrac{{{2}}}{{{{12}}}} = \dfrac{1}{6}$

b. $\dfrac{3}{{15}}$

Ans: We can see that 3 is a factor of 15. So, to convert it into the simplest form divide the numerator and denominator by 3.

$\dfrac{{{3}}}{{{{15}}}} = \dfrac{1}{6}$

c. $\dfrac{8}{{50}}$

Ans: The numbers 8, and 50 have 2 as their HCF. So, to convert it into the simplest form divide the numerator and denominator by 2.

$\dfrac{{{8}}}{{{{50}}}} = \dfrac{4}{{25}}$

d. $\dfrac{{16}}{{100}}$

Ans: The numbers 16, and 100 have 4 as their HCF. To convert it into the simplest form divide the numerator and denominator by 4.

$\dfrac{{{{16}}}}{{{{100}}}} = \dfrac{4}{{25}}$

e. $\dfrac{{10}}{{60}}$

Ans: We can see that 10 is a factor of 60. To convert it into the simplest form divide the numerator and denominator by 10.

$\dfrac{{{{16}}}}{{{{100}}}} = \dfrac{4}{{25}}$

f. $\dfrac{{15}}{{75}}$

Ans: To convert it into the simplest form divide the numerator and denominator by 10.

$\dfrac{{{{15}}}}{{{{75}}}} = \dfrac{1}{3}$

g. $\dfrac{{12}}{{60}}$

Ans: To convert it into the simplest form divide the numerator and denominator by 12.

$\dfrac{{{{12}}}}{{{{60}}}} = \dfrac{1}{5}$

h. $\dfrac{{16}}{{96}}$

Ans: Here, 16 is a factor of 96. So, to convert it into the simplest form divide the numerator and denominator by 16.

$\dfrac{{{{16}}}}{{{{96}}}} = \dfrac{1}{6}$

i. $\dfrac{{12}}{{75}}$

Ans: The highest common factor of 12 and 75 is 3. So, convert it into the simplest form divide the numerator and denominator by 3.

$\dfrac{{{{12}}}}{{{{75}}}} = \dfrac{4}{{15}}$

j. $\dfrac{{12}}{{72}}$

Ans: Here, 12 is a factor of 72. So, to convert it into the simplest form divide the numerator and denominator by 12.

$\dfrac{{{{12}}}}{{{{72}}}} = \dfrac{1}{6}$

k. $\dfrac{3}{{18}}$

Ans: Here, 3 is a factor of 18. So, to convert it into the simplest form divide the numerator and denominator by 3.

$\dfrac{{{3}}}{{{{18}}}} = \dfrac{1}{6}$

l. $\dfrac{4}{{25}}$

Ans: The given number is in simplest form as the highest common factor of 4 and 25 is 1.

Therefore, from the above simplified forms, part (b), (f), (g) are equivalent, parts (a), (e), (h), (j), (k) are equivalent and parts (c), (d), (i), (l) are equivalent.

7. Find answers to the following. Write and indicate how you solved them:

a. Is $\dfrac{5}{9}$ equal to $\dfrac{4}{5}$?

Ans: To solve the question, take the LCM of the denominators of 9 and 5.

The LCM of the denominator of 9 and 5 is 45.

Multiply numerator and denominator of $\dfrac{5}{9}$ by 4.

$\dfrac{{5 \times 5}}{{9 \times 5}} = \dfrac{{25}}{{45}}$

Multiply numerator and denominator of $\dfrac{4}{5}$ by 4.

$\dfrac{{4 \times 9}}{{5 \times 9}} = \dfrac{{36}}{{45}}$

Since $\dfrac{{25}}{{45}} \ne \dfrac{{36}}{{45}}$ then, $\dfrac{5}{9} \ne \dfrac{4}{5}$.

b. Is $\dfrac{9}{{16}}$ equal to $\dfrac{5}{9}$?

Ans: To solve the question, take the LCM of the denominators of 9 and 16.

The LCM of the denominator of 9 and 16 is 144.

Multiply numerator and denominator of $\dfrac{9}{{16}}$ by 16.

$\dfrac{{9 \times 144}}{{16 \times 144}} = \dfrac{{81}}{{144}}$

Multiply numerator and denominator of $\dfrac{5}{9}$ by 9.

$\dfrac{{5 \times 16}}{{9 \times 16}} = \dfrac{{80}}{{144}}$

Since $\dfrac{{81}}{{144}} \ne \dfrac{{80}}{{144}}$ then, $\dfrac{9}{{16}} \ne \dfrac{5}{9}$.

c. Is $\dfrac{4}{5}$ equal to $\dfrac{{16}}{{20}}$?

Ans: To solve the question, take the LCM of the denominators of 5 and 20.

The LCM of the denominator of 5 and 20 is 20.

Multiply numerator and denominator of $\dfrac{4}{5}$ by 4.

$\dfrac{{4 \times 4}}{{5 \times 4}} = \dfrac{{16}}{{20}}$

Since $\dfrac{{16}}{{20}} = \dfrac{{16}}{{20}}$ then, $\dfrac{4}{5} = \dfrac{{16}}{{20}}$.

Yes, $\dfrac{4}{5}$ is  equal to $\dfrac{{16}}{{20}}$.

d. Is $\dfrac{1}{{15}}$ equal to $\dfrac{4}{{30}}$?

Ans: To solve the question, take the LCM of the denominators of 15 and 30.

The LCM of the denominator of 15 and 30 is 30.

Multiply numerator and denominator of $\dfrac{1}{{15}}$ by 2.

$\dfrac{{1 \times 2}}{{15 \times 2}} = \dfrac{2}{{30}}$

Since $\dfrac{2}{{30}} \ne \dfrac{4}{{30}}$ then, $\dfrac{1}{{15}} \ne \dfrac{4}{{30}}$.

8. Ila reads 25 pages of a book containing 100 pages. Lalita read $\dfrac{2}{5}$ of the same book. Who reads less?

Ans: Ila read 25 pages out of 100 pages, that is, $\dfrac{{25}}{{100}}$.

Lalita read $\dfrac{2}{5}$ of the book.

Multiply the numerator and denominator of the fraction $\dfrac{2}{5}$ by 20 to make the denominator equal to $\dfrac{{25}}{{100}}$.

$\dfrac{{2 \times 20}}{{5 \times 20}} = \dfrac{{40}}{{100}}$

Since $40 > 25$ then, $\dfrac{{40}}{{100}} > \dfrac{{25}}{{100}}$.

Hence, Ila read less.

9. Rafiq exercised for $\dfrac{3}{6}$ of an hour, while Rohit exercised for $\dfrac{3}{4}$for an hour. Who exercised for a longer time?

Ans: To solve the given question, make the denominator equal. To do so, multiply the numerator and denominator of $\dfrac{3}{6}$ and $\dfrac{3}{4}$ by 4 and 6 respectively.

$\dfrac{{3 \times 4}}{{6 \times 4}} = \dfrac{{12}}{{24}}$

$\dfrac{{3 \times 6}}{{4 \times 6}} = \dfrac{{18}}{{24}}$

The fraction, $\dfrac{{12}}{{24}}$ shows that twelve parts are selected out of twenty four parts and the fraction, $\dfrac{{18}}{{24}}$ shows that eighteen parts are selected out of twenty four parts. Thus, $\dfrac{{12}}{{24}} < \dfrac{{18}}{{24}}$, that is,$\dfrac{3}{6} < \dfrac{3}{4}$. Hence, Rohit exercised for a longer time.

10. In a class A of 25 students, 20 passed in first class; in another class B of 30 students, 24 students in first class. In which class was a greater fraction of students getting first class?

Ans: The fraction for the given problem will be $\dfrac{{{\text{Students passed}}}}{{{\text{Total students}}}}$.

In class A, 20 students passed out of 25 students, that is, $\dfrac{{20}}{{25}}$. Therefore,

$\dfrac{{{{20}}}}{{{{25}}}} = \dfrac{4}{5}$

In class B, 24 students passed out of 30 students, that is, $\dfrac{{24}}{{30}}$. Therefore,

$\dfrac{{{{24}}}}{{{{30}}}} = \dfrac{4}{5}$

Hence, each class has the same fraction of students getting a first class.

NCERT Solutions for Class 6 Maths Chapter 7 Fractions Exercise 7.4

Opting for the NCERT solutions for Ex 7.4 Class 6 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 7.4 Class 6 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 6 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 6 Maths Chapter 7 Exercise 7.4 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

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