# NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals (EX 2.1) Exercise 2.1

Fractions and Decimals are covered in the Class 7 Maths Chapter 2 Exercise 2.1 Solutions. This very basic but important chapter in Mathematics, presented in Class 7, has four main divisions or subjects in Class 7 Maths Exercise 2.1. The key points discussed in the NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1, are explained in a detailed manner on the Extramarks website. To properly comprehend the concepts associated with the chapter, Extramarks suggests students carefully read through each topic. This will undoubtedly enable students to fully make use of Extramarks’ NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1 for understanding the crucial chapter on Fractions and Decimals.

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Topics relating to the introduction of Fractions are covered in NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1. Through this assignment, students can recall the study of fractions, including their addition and subtraction, as well as Proper, Improper, and Mixed fractions. The NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1, is an essential resource for covering the complete Math curriculum because it offers a variety of questions that measure students’ conceptual comprehension.

## Class 7 Maths Chapter 2 Fractions and Decimals Exercise 2.1

The key concepts of Fractions and Decimals covered in the chapter are reviewed in the NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1. The fundamentals of fractions and decimals, as well as addition and subtraction of fractions and decimals, can all be learned by students using the NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1. There are eight questions in the NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1, four of which are short answers and four of which are longer in nature.

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### Important Topics under NCERT Solutions for Class 7 Maths Chapter 2 – Fractions and Decimals, Exercise 2.1

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## Access NCERT Solutions for Class 7 Maths Chapter 2 – Fractions and Decimals

Students should refer to the NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1

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### NCERT Solutions for Class 7 Maths Chapter 2 Exercise 2.1

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### NCERT Solutions for Class 7 Maths

The basis for higher courses is laid around Class 7. Therefore, it is a crucial class in students’ careers. Only studying for classes is insufficient in today’s fast-paced, fiercely competitive world. Students should conduct independent research, learn all the concepts and theories, and be able to apply the theories in real-world situations. Students who use Extramarks’ NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1 are better able to learn and achieve their goals. All of the solutions to the NCERT textbook are included on the Extramarks website and were created by Extramarks’ experts. These solutions have been written so that the students can study and comprehend all the concepts more effectively and efficiently. The language is basic and suitable for students.

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### Proper Fractions

A proper fraction is one in which the numerator (the higher portion of the fraction) is less than the denominator (the lower portion). The component of a whole is represented by a correct fraction.

For example- A person must take 2/3 of the cake when they divide a cake into 3 pieces and want 2 pieces from each of them. The cake is divided into three sections, and they are taking two of those parts.

### Improper Fraction

An improper fraction is one in which the numerator exceeds the denominator. A cake cannot be cut into 3 pieces and then cut into 4 pieces. The result would be – 4/3 in a mathematical sense.

A person needs to get a brand-new 3-slice cake in order to fix the issue of being unable to remove 4 pieces out of 3. They just need to take one slice from the original 3-piece cake because they already have 3 slices from the new entire. Mathematically, it would appear as

1 whole ⅓ or 1⅓.

A mixed fraction is an improper fraction with this altered appearance. 5/2, 11/9, 800/700, and it includes more improper fractions. Examples of mixed fractions are 1⅓, 100⅔, 11¾ etc.

The NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1 offer students the confidence they need to perform better on the CBSE examination along with a selection of activities and problem sets.

### Addition and subtraction of fractions

Students only need to add the numerators of fractions with the same denominator and can leave the denominator alone.

For instance, 5/2 + 6/2 = 12/2.

If the fractions are, for example, 5/2 and 4/6, they must first get the LCM of the denominators before dividing each one by the LCM. The appropriate numerator must now be multiplied by the quotient of the previous division. For each fraction involved, the procedure must be repeated. The sum of the outcomes of the multiplications would be the resulting numerator, and the resultant LCM would be the resulting denominator.

The same is true of subtraction.

Students can access the NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1 are recognised as being immensely beneficial for preparing for the CBSE Class 7 examination.

### To Find the Ascending Order or Descending Order

Students must remember this before they proceed: The fraction with the larger numerator is thought to be the greatest when two fractions have the same denominator. And the numerator fraction with the smallest value is regarded as the lowest. Now, if they want to determine the order in ascending –

5/2, ¾, ⅛, 8/4

First, they must equalise the denominator. These fractions can be multiplied with other fractions as long as the denominators are the same in each instance. These fractions can be multiplied by other fractions. As a result, 5/2 multiplied by 4/4 will equal 20/8 (which finally equals 5/2, keeping the number constant).

### NCERT Solutions for Class 7

Mathematical chapters cannot be fully grasped by only viewing the PDF files. Students will undoubtedly have questions about any chapter. The Extramarks website offers online lessons where students can ask questions right away in order to erase these doubts from their minds. The learning process can be made enjoyable and interesting by the experts. Students’ understanding of the chapters will improve with the help of this visual learning method, as Extramarks adds graphics and animations to its videos to enhance students’ learning. They can use live classes to get ready for their examinations. In the live classes, students can rediscover their concepts of Mathematics. The instructors in the live classes will help them develop their capacity for logical reasoning in addition to teaching students Mathematics, its numerous formulas, and procedures.

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Q.1

$\begin{array}{l}\text{Solve}:\\ \text{(i) 2-}\frac{3}{5}\text{(ii) 4+}\frac{7}{8}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{(iii)}\frac{3}{5}+\frac{2}{7}\text{(iv)}\frac{9}{11}-\frac{4}{15}\\ \text{(v)}\frac{7}{10}\text{+}\frac{2}{5}\text{+}\frac{3}{2}\text{(vi) 2}\frac{2}{3}+3\frac{1}{2}\text{(vii) 8}\frac{1}{2}-3\frac{5}{8}\end{array}$

Ans.

$\begin{array}{l}\left(i\right)\text{\hspace{0.17em}}2-\frac{3}{5}=\frac{10-3}{5}=\overline{)\frac{7}{5}}\\ \left(ii\right)\text{\hspace{0.17em}}4+\frac{7}{8}=\frac{32+7}{8}=\frac{39}{8}=\overline{)4\frac{7}{8}}\\ \left(iii\right)\text{\hspace{0.17em}}\frac{3}{5}+\frac{2}{7}=\frac{21+10}{35}=\overline{)\frac{31}{35}}\\ \left(iv\right)\text{\hspace{0.17em}}\frac{9}{11}-\frac{4}{15}=\frac{135+44}{165}=\frac{179}{165}=\overline{)1\frac{14}{165}}\\ \left(v\right)\text{\hspace{0.17em}}\frac{7}{10}+\frac{2}{5}+\frac{3}{2}=\frac{7+4+15}{10}=\frac{26}{10}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{13}{5}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2\frac{3}{5}\\ \left(vi\right)\text{\hspace{0.17em}}2\frac{2}{3}+3\frac{1}{2}=\frac{8}{3}+\frac{7}{2}=\frac{16+21}{6}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{37}{6}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=6\frac{1}{6}\\ \left(vii\right)\text{\hspace{0.17em}}8\frac{1}{2}-3\frac{5}{8}=\frac{17}{2}-\frac{29}{8}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{68-29}{8}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{39}{8}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=4\frac{7}{8}\end{array}$

Q.2

$\begin{array}{l}\mathrm{Arrange}\mathrm{the}\mathrm{following}\mathrm{in}\mathrm{descending}\mathrm{order}.\\ \left(\mathrm{i}\right)\frac{2}{9},\frac{2}{3},\frac{8}{21}\left(\mathrm{ii}\right)\frac{1}{5},\frac{3}{7},\frac{7}{10}\end{array}$

Ans.

$\begin{array}{l}\text{First},\text{we find the LCM of 9},\text{3 and 21}.\\ \text{LCM of 9},\text{3 and 21}=\text{63}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{2×7}{9×7},\text{\hspace{0.17em}}\frac{2×21}{3×21},\text{\hspace{0.17em}}\frac{8×3}{21×3}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{14}{63},\text{\hspace{0.17em}}\frac{42}{63},\frac{24}{63}\\ \text{Since},\text{42}>\text{24}>\text{14}.\\ \text{Therefore},\\ \frac{42}{63}>\text{\hspace{0.17em}}\frac{24}{63}>\frac{14}{63}\\ ⇒\overline{)\frac{2}{3}>\frac{8}{21}>\frac{2}{9}}\\ \left(ii\right)\text{\hspace{0.17em}}\frac{1}{5},\text{\hspace{0.17em}}\frac{3}{7},\text{\hspace{0.17em}}\frac{7}{10}\\ \text{First},\text{find the LCM of 5},\text{7 and 1}0\\ \text{LCM of denominators 5},\text{7 and 1}0\text{is 7}0.\\ \text{So},\text{\hspace{0.17em}}\\ \frac{1}{5},\text{\hspace{0.17em}}\frac{3}{7},\text{\hspace{0.17em}}\frac{7}{10}\\ ⇒\frac{1×14}{5×14},\frac{3×10}{7×10},\frac{7×7}{10×7}\\ ⇒\frac{14}{70},\frac{30}{70},\frac{49}{70}\\ \text{Since},\text{49}>\text{3}0\text{}>\text{14}.\\ \text{Therefore},\text{\hspace{0.17em}}\\ \frac{49}{70}>\text{\hspace{0.17em}}\frac{30}{70}>\text{\hspace{0.17em}}\frac{14}{70}\\ ⇒\overline{)\frac{7}{10}>\frac{3}{7}>\frac{1}{5}}\end{array}$

Q.3 In a magic square, the sum of the numbers in each row, in each column and along the diagonal is the same. Is this a magic square?

$\begin{array}{ccc}\frac{4}{11}& \frac{9}{11}& \frac{2}{11}\\ \frac{3}{11}& \frac{5}{11}& \frac{7}{11}\\ \frac{8}{11}& \frac{1}{11}& \frac{6}{11}\end{array}$

Ans.

$\begin{array}{l}\text{Finding the sum of first row}:\\ Sum\text{\hspace{0.17em}}of\text{\hspace{0.17em}}1st\text{\hspace{0.17em}}row=\frac{4}{11}+\frac{9}{11}+\frac{2}{11}\\ =\frac{4+9+2}{11}\\ =\frac{15}{11}\\ \text{Finding the sum of second row}:\\ \text{Sum of 2nd row}=\frac{3}{11}+\frac{5}{11}+\frac{7}{11}\\ =\frac{15}{11}\\ \text{Finding the sum of third row}:\\ \text{Sum of 3rd row}=\frac{8}{11}+\frac{1}{11}+\frac{6}{11}\\ =\frac{8+1+6}{11}\\ =\frac{15}{11}\\ \text{Finding the sum of first column}:\\ \text{Sum of 1st column}=\frac{4}{11}+\frac{3}{11}+\frac{8}{11}\\ =\frac{4+3+8}{11}\\ =\frac{15}{11}\\ \text{Finding the sum of second column}:\\ \text{Sum of 2nd column}=\frac{9}{11}+\frac{5}{11}+\frac{1}{11}\\ =\frac{9+5+1}{11}\\ =\frac{15}{11}\\ \text{Finding the sum of third column}:\\ \text{Sum of 3rd column}=\frac{2}{11}+\frac{7}{11}+\frac{6}{11}\\ =\frac{2+7+6}{11}\\ =\frac{15}{11}\\ \text{Now},\text{we must find diagonal sum}.\\ \text{Sum of}diagonal\text{\hspace{0.17em}}from\text{\hspace{0.17em}}left\text{\hspace{0.17em}}\text{​}bottom\text{\hspace{0.17em}}to\text{\hspace{0.17em}}right\text{\hspace{0.17em}}top\\ =\frac{8}{11}+\frac{5}{11}+\frac{2}{11}\\ =\frac{8+5+2}{11}\\ =\frac{15}{11}\\ \text{Sum of}diagonal\text{\hspace{0.17em}}from\text{\hspace{0.17em}}left\text{\hspace{0.17em}}\text{​}bottom\text{\hspace{0.17em}}to\text{\hspace{0.17em}}right\text{\hspace{0.17em}}top\\ =\frac{6}{11}+\frac{5}{11}+\frac{4}{11}\\ =\frac{6+5+4}{11}\\ =\frac{15}{11}\\ \text{Since the sum of the numbers in each row},\text{in each}\\ \text{column along the diagonal is the same},\text{which is equal}\\ \text{to}\text{\hspace{0.17em}}\frac{15}{11}.\\ \text{Therefore},\text{given square is magical}.\end{array}$

Q.4

$\begin{array}{l}\mathrm{A}\mathrm{rectangular}\mathrm{sheet}\mathrm{of}\mathrm{paper}\mathrm{is}\text{\hspace{0.17em}}12\frac{1}{2}\text{\hspace{0.17em}}\mathrm{cm}\text{\hspace{0.17em}}\mathrm{long}\mathrm{and}\\ 10\frac{2}{3}\text{\hspace{0.17em}}\mathrm{cm}\text{\hspace{0.17em}}\mathrm{wide}.\mathrm{Find}\mathrm{its}\mathrm{perimeter}.\end{array}$

Ans.

$\begin{array}{l}\text{We are given}:\\ \text{Length}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{rectangular}\text{\hspace{0.17em}}\text{sheet}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{paper}=12\frac{1}{2}\text{cm}=\frac{25}{2}\text{cm}\\ \text{Width}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{rectangular}\text{\hspace{0.17em}}\text{sheet}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{paper=10}\frac{2}{3}\text{cm}=\frac{32}{3}\text{cm}\\ \text{Since},\text{\hspace{0.17em}}\overline{)\text{Perimeter of a rectangle}=\text{2}×\left(\text{length}+\text{width}\right)}\\ \text{Therefore},\\ \text{Perimeter of the given rectangular sheet of paper}\\ =2×\left(\frac{25}{2}+\frac{32}{3}\right)\text{cm}\\ =2×\left(\frac{75+64}{6}\right)\text{cm}\\ =2×\frac{139}{6}\text{cm}\\ =\frac{139}{3}\text{cm}=\overline{)46\frac{1}{3}\text{cm}}\end{array}$

Q.5

$\begin{array}{l}\mathrm{Find}\mathrm{}\mathrm{the}\mathrm{}\mathrm{perimeters}\mathrm{}\mathrm{of}\mathrm{}\left(\mathrm{i}\right)\mathrm{\Delta ABE}\mathrm{}\left(\mathrm{ii}\right)\mathrm{}\mathrm{the}\mathrm{}\mathrm{rectangle}\\ \mathrm{BCDE}\mathrm{}\mathrm{in}\mathrm{}\mathrm{this}\mathrm{}\mathrm{figure}.\mathrm{}\mathrm{Whose}\mathrm{}\mathrm{perimeter}\mathrm{}\mathrm{is}\mathrm{}\mathrm{greater}?\end{array}$

Ans.

$\begin{array}{l}\text{Since},\text{the Perimeter of a triangle}=\text{side}+\text{side}+\text{side}\\ \text{Therefore},\text{Perimeter of triangle ABE}=\text{AB}+\text{BE}+\text{AE}\\ =\frac{5}{2}+2\frac{3}{4}+3\frac{3}{5}\text{cm}\\ =\frac{5}{2}+\frac{11}{4}+\frac{18}{5}\text{cm}\\ =\frac{50+55+72}{20}\text{cm}\\ =\frac{177}{20}\text{cm}\\ \text{In rectangle BCDE},\text{length BE}=2\frac{2}{4}\text{cm}=\frac{11}{4}\text{cm}\\ \text{Width ED}=\frac{7}{6}\text{cm}\\ \text{Since},\text{\hspace{0.17em}}\overline{)\text{Perimeter of a rectangle}=\text{2}×\left(\text{length}+\text{width}\right)}\\ \text{Therefore},\text{perimeter of rectangle BCDE}=2×\left(BE+ED\right)\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2×\left(\frac{11}{4}+\frac{7}{6}\right)\text{cm}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}=2×\left(\frac{33+14}{12}\right)\text{cm}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}=2×\frac{47}{12}\text{cm}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}=\frac{47}{6}\text{cm}\\ \text{Hence},\text{the perimeter of}\text{\hspace{0.17em}}\Delta ABC=\overline{)\frac{177}{20}\text{cm}}\\ \text{and}\\ \text{The perimeter of rectangle BCDE}=\overline{)\frac{47}{6}\text{cm}}\\ \text{Now},\text{comparing both the perimeter to find which is}\\ \text{greater}.\\ \text{Since},\\ \text{The LCM of denominators of both the perimeter}\left(\text{2}0\text{and 6}\right)\text{is}\\ \text{equal to 6}0.\\ \text{Therefore},\\ \frac{177}{20},\text{\hspace{0.17em}}\frac{47}{6}⇒\frac{177×3}{20×3},\frac{47×10}{6×10}⇒\frac{531}{60},\frac{470}{10}\\ \text{Since},\text{531 is greater than 47}0.\\ \text{Therefore},\\ \frac{531}{60}>\frac{470}{60}⇒\frac{177}{20}>\frac{47}{6}\\ \text{Thus},\text{the perimeter of triangle ABE is greater than}\\ \text{the perimeter of rectangle BCDE}.\end{array}$

Q.6

$\begin{array}{l}\mathrm{Sali}\mathrm{}\mathrm{wants}\mathrm{}\mathrm{to}\mathrm{}\mathrm{put}\mathrm{}\mathrm{a}\mathrm{}\mathrm{picture}\mathrm{}\mathrm{in}\mathrm{}\mathrm{a}\mathrm{}\mathrm{frame}.\mathrm{}\mathrm{The}\mathrm{}\mathrm{picture}\mathrm{}\mathrm{is}\mathrm{ }7\frac{3}{5}\mathrm{cm}\mathrm{ }\mathrm{wide}.\mathrm{}\\ \mathrm{To}\mathrm{}\mathrm{fit}\mathrm{}\mathrm{in}\mathrm{}\mathrm{the}\mathrm{}\mathrm{frame}\mathrm{}\mathrm{the}\mathrm{}\mathrm{picture}\mathrm{}\mathrm{cannot}\mathrm{}\mathrm{be}\mathrm{}\mathrm{more}\mathrm{}\mathrm{than}\mathrm{ }7\frac{3}{10}\mathrm{cm}\mathrm{wide}.\mathrm{}\\ \mathrm{How}\mathrm{}\mathrm{much}\mathrm{}\mathrm{should}\mathrm{}\mathrm{the}\mathrm{}\mathrm{picture}\mathrm{}\mathrm{be}\mathrm{}\mathrm{trimmed}?\end{array}$

Ans.

$\begin{array}{l}\text{We are given}:\\ \text{The}\text{\hspace{0.17em}}\text{width}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{picture=7}\frac{\text{3}}{\text{5}}\text{cm}\\ \text{Required}\text{\hspace{0.17em}}\text{width}\text{\hspace{0.17em}}\text{of picture to be fit in frame=7}\frac{3}{10}\text{cm}\\ \text{Therefore},\text{picture to be trim}=7\frac{3}{5}-7\frac{3}{10}\text{cm}\\ =\frac{38}{5}-\frac{73}{10}\text{cm}\\ =\frac{3}{10}\text{cm}\\ \text{Thus picture should be trimmed by}\text{\hspace{0.17em}}\overline{)\frac{3}{10}\text{cm}}\end{array}$

Q.7

$\begin{array}{l}\mathrm{Ritu}\mathrm{ate}\text{\hspace{0.17em}}\frac{3}{5}\text{\hspace{0.17em}}\mathrm{part}\mathrm{of}\mathrm{an}\mathrm{apple}\mathrm{and}\mathrm{the}\mathrm{remaining}\mathrm{apple}\mathrm{was}\\ \mathrm{eaten}\mathrm{by}\mathrm{her}\mathrm{brother}\mathrm{Somu}.\mathrm{How}\mathrm{much}\mathrm{part}\mathrm{of}\mathrm{the}\mathrm{apple}\\ \mathrm{did}\mathrm{Somu}\mathrm{eat}?\mathrm{Who}\mathrm{had}\mathrm{the}\mathrm{larger}\mathrm{share}?\mathrm{By}\mathrm{how}\mathrm{much}?\end{array}$

Ans.

$\begin{array}{l}\text{Since},\text{Ritu ate\hspace{0.17em}}\frac{3}{5}\text{\hspace{0.17em}part of an apple and her brother eaten the}\\ \text{rest part of the apple}.\\ \text{So},\text{the part of apple left after eaten by Ritu}=1-\frac{3}{5}\\ =\frac{5-3}{5}\\ =\frac{2}{5}\\ \text{Therefore},\text{Somu ate\hspace{0.17em}}\frac{2}{5}\text{\hspace{0.17em}parts of apple}.\\ \text{Now comparing the part eaten by them\hspace{0.17em}}\frac{3}{5},\text{\hspace{0.17em}​}\frac{2}{5}\\ \text{Since},\text{3}>\text{2}.\\ \text{Therefore}\\ \frac{3}{5}>\frac{2}{5}\\ \text{That means Ritu ate larger part}.\\ \text{The difference in both parts}=\frac{3}{5}-\frac{2}{5}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{3-2}{5}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{5}\\ \text{Thus,}\\ \text{Somu ate}\frac{2}{5}\text{\hspace{0.17em}}\text{part of apple}.\\ \text{Ritu ate the larger part of apple}\\ \text{Ritu ate}\text{part more apple than her brother Somu}.\end{array}$

Q.8

$\begin{array}{l}\mathrm{Michael}\mathrm{finished}\mathrm{colouring}\mathrm{a}\mathrm{picture}\mathrm{in}\text{\hspace{0.17em}}\frac{7}{12}\mathrm{hour}.\\ \mathrm{Vaibhav}\mathrm{finished}\mathrm{colouring}\mathrm{the}\mathrm{same}\mathrm{picture}\mathrm{in}\text{\hspace{0.17em}}\frac{3}{4}\text{\hspace{0.17em}}\mathrm{hour}.\\ \mathrm{Who}\mathrm{worked}\mathrm{longer}?\mathrm{By}\mathrm{what}\mathrm{fraction}\mathrm{was}\mathrm{it}\mathrm{longer}?\end{array}$

Ans.

$\begin{array}{l}\text{We are given}:\\ \text{Michael\hspace{0.17em}worked\hspace{0.17em}for\hspace{0.17em}}\frac{\text{7}}{\text{12}}\text{hour}\\ \mathrm{Vaibhav}\text{\hspace{0.17em}}\mathrm{worked}\text{\hspace{0.17em}}\mathrm{for}\text{\hspace{0.17em}}\frac{3}{4}\mathrm{hour}\phantom{\rule{0ex}{0ex}}\text{Therefore},\\ \text{In order to find the longer hour of work we have to compare}\\ \text{both the fractions\hspace{0.17em}}\frac{7}{12},\text{\hspace{0.17em}}\frac{3}{4}\\ \text{The LCM of 12 and 4}=\text{12}\\ \mathrm{So},\frac{7}{12},\frac{3}{4}=\frac{7}{12},\frac{9}{12}\\ \text{Therefore},\text{\hspace{0.17em}}\frac{9}{12}>\frac{7}{12}\\ \text{That means Vaibhav worked for longer hour}.\\ \text{Difference in their working hour\hspace{0.17em}\hspace{0.17em}}=\frac{9}{12}-\frac{7}{12}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{9-7}{12}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{2}{12}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\overline{)\frac{1}{6}}\\ \text{Thus},\text{Vaibhav worked for}\frac{1}{6}\text{\hspace{0.17em}hour more than Michael}.\end{array}$

## FAQs (Frequently Asked Questions)

### 1. Which chapters in Mathematics for Class 7 are crucial in terms of exams?

Every chapter is significant for achieving higher examination scores. However, students may choose which chapters are more significant, and they should concentrate on those according to the PDF. The chapters on Fractions and Decimals, Data Handling, Simple Equations, and Lines And Angles have the highest weightage and are therefore more significant, according to the marks distribution. Geometry, Mensuration, Quantity Comparison, and Integers are also crucial.

### 2. How do students get good grades in Mathematics in Class 7?

The subject that necessitates frequent practice is Mathematics. No matter how skilled students are, if they do not personally tackle the difficulties, the questions get hard and time-consuming. Every day, students must schedule at least an hour to spend answering the questions. They should complete all of the NCERT exercises and verify the solutions with Extramarks’ NCERT solutions.

### 3. Is it easy to find NCERT solutions on the Extramarks website?

The Extramarks website offers easy access to NCERT solutions. Students can quickly obtain the solutions on the Extramarks website. In addition, It offers its students live expert workshops, a self-evaluation centre, K12 study materials for their revision plans, and many other resources to help them succeed on their exams. One of the best resources for assisting students in their exam preparation is Extramarks, which provides NCERT solutions. This is accurate because every subject that might be covered in the yearly exams is covered in the NCERT textbook.

The solutions on the Extramarks website are recommended by subject-matter experts for use in students’ studies. In order to help students grasp the fundamental ideas of the topic, the NCERT solutions are suitable for students who find the chapter difficult. If students want to finish the chapter questions more quickly, which will help them do well on their exam, they must practice the NCERT solutions.

### 4. Are the NCERT solutions difficult for students to understand?

Students will not find the NCERT solutions to be difficult. They can simply study the solutions and excel in the key themes to perform well in their exams with consistent practice and the appropriate assistance from Extramarks. Extramarks provides thorough answers to the NCERT questions to help students comprehend the material. Students can also take part in doubt sessions offered on the Extramarks website if they are

reluctant to ask questions in front of their peers at school.