class 7 maths chapter 2 exercise 2.1

class 7 maths chapter 2 exercise 2.1 in simple solutions are given here. class 7 maths chapter 2 exercise 2.1 of NCERT Solutions for class 7 maths chapter 2 exercise 2.1 contains all the topics containing the basic information of Fractions. class 7 maths chapter 2 exercise 2.1 provide students the study of fractions including proper, improper and mixed fractions as well as their addition and subtraction. class 7 maths chapter 2 exercise 2.1 provides necessary material for solving different range of questions that test the students capability of understanding the concepts. It also helps the students of CBSE Class 7 Maths students.

Question 1:

Solve:

(i) $2-{\displaystyle \frac{3}{5}}$

(ii) $4+{\displaystyle \frac{7}{8}}$

(iii) ${\displaystyle \frac{3}{5}}+{\displaystyle \frac{2}{7}}$

(iv) ${\displaystyle \frac{9}{11}}-{\displaystyle \frac{4}{15}}$

(v) ${\displaystyle \frac{7}{10}}+{\displaystyle \frac{2}{5}}+{\displaystyle \frac{3}{2}}$

(vi) $2{\displaystyle \frac{2}{3}}+3{\displaystyle \frac{1}{2}}$

(vii) $8{\displaystyle \frac{1}{2}}-3{\displaystyle \frac{5}{8}}$

Solution:

(i) $2-{\displaystyle \frac{3}{5}}={\displaystyle \frac{10-3}{5}}={\displaystyle \frac{7}{5}}=1{\displaystyle \frac{2}{5}}$

(ii) $4+{\displaystyle \frac{7}{8}}={\displaystyle \frac{32+7}{8}}={\displaystyle \frac{39}{8}}=4{\displaystyle \frac{7}{8}}$

(iii) ${\displaystyle \frac{3}{5}}+{\displaystyle \frac{2}{7}}={\displaystyle \frac{3\times 7+5\times 2}{5\times 7}}={\displaystyle \frac{21+10}{35}}={\displaystyle \frac{31}{35}}$

(iv) ${\displaystyle \frac{9}{11}}-{\displaystyle \frac{4}{15}}={\displaystyle \frac{9\times 15-4\times 11}{15\times 11}}={\displaystyle \frac{135-44}{165}}={\displaystyle \frac{91}{165}}$

(v) ${\displaystyle \frac{7}{10}}+{\displaystyle \frac{2}{5}}+{\displaystyle \frac{3}{2}}={\displaystyle \frac{7\times 1+2\times 2+3\times 5}{10}}={\displaystyle \frac{7+4+15}{10}}$$={\displaystyle \frac{26}{10}}={\displaystyle \frac{13}{5}}=2{\displaystyle \frac{3}{5}}$

(vi) $2{\displaystyle \frac{2}{3}}+3{\displaystyle \frac{1}{2}}={\displaystyle \frac{8}{3}}+{\displaystyle \frac{7}{2}}$$={\displaystyle \frac{8\times 2+3\times 7}{3\times 2}}={\displaystyle \frac{16+21}{6}}={\displaystyle \frac{37}{6}}=6{\displaystyle \frac{1}{6}}$

(vii) $8{\displaystyle \frac{1}{2}}-3{\displaystyle \frac{5}{8}}={\displaystyle \frac{17}{2}}-{\displaystyle \frac{29}{8}}={\displaystyle \frac{17\times 4-29\times 1}{8}}=$${\displaystyle \frac{39}{8}}=4{\displaystyle \frac{7}{8}}$

Question 2. 

Arrange the following in descending order:

(i)${\displaystyle \frac{2}{9}},{\displaystyle \frac{2}{3}},{\displaystyle \frac{8}{21}}$

(ii)${\displaystyle \frac{1}{5}},{\displaystyle \frac{3}{7}},{\displaystyle \frac{7}{10}}$

Solution:

(i)${\displaystyle \frac{2}{9}},{\displaystyle \frac{2}{3}},{\displaystyle \frac{8}{21}}$

Converting into like fractions

${\displaystyle \frac{42}{63}},{\displaystyle \frac{24}{63}},{\displaystyle \frac{24}{63}}$

Arranging in descending orders

${\displaystyle \frac{42}{63}}>{\displaystyle \frac{24}{63}}>{\displaystyle \frac{24}{63}}$

Therefore,

${\displaystyle \frac{2}{3}}>{\displaystyle \frac{8}{21}}>{\displaystyle \frac{2}{9}}$

(ii)${\displaystyle \frac{1}{5}},{\displaystyle \frac{3}{7}},{\displaystyle \frac{7}{10}}$

Converting into like fractions

${\displaystyle \frac{14}{70}},{\displaystyle \frac{30}{70}},{\displaystyle \frac{49}{70}}$

Arranging in descending orders

${\displaystyle \frac{49}{70}}>{\displaystyle \frac{30}{70}}>{\displaystyle \frac{14}{70}}$

Therefore,

${\displaystyle \frac{7}{10}}>{\displaystyle \frac{3}{7}}>{\displaystyle \frac{1}{5}}$

Question 3. 

In a “magic square”, the sum of the numbers in each row, in each column and along

the diagonals is the same. Is this a magic square?

${\displaystyle \frac{4}{11}}$	${\displaystyle \frac{9}{11}}$	${\displaystyle \frac{2}{11}}$
${\displaystyle \frac{3}{11}}$	${\displaystyle \frac{5}{11}}$	${\displaystyle \frac{7}{11}}$
${\displaystyle \frac{8}{11}}$	${\displaystyle \frac{1}{11}}$	${\displaystyle \frac{6}{11}}$

Solution:

Along the first row ${\displaystyle \frac{4}{11}}+{\displaystyle \frac{9}{11}}+{\displaystyle \frac{2}{11}}={\displaystyle \frac{15}{11}}$

Along the Second row ${\displaystyle \frac{3}{11}}+{\displaystyle \frac{5}{11}}+{\displaystyle \frac{7}{11}}={\displaystyle \frac{15}{11}}$

Along the third row ${\displaystyle \frac{8}{11}}+{\displaystyle \frac{1}{11}}+{\displaystyle \frac{6}{11}}={\displaystyle \frac{15}{11}}$

Along the First column ${\displaystyle \frac{4}{11}}+{\displaystyle \frac{3}{11}}+{\displaystyle \frac{8}{11}}={\displaystyle \frac{15}{11}}$

Along the Second column ${\displaystyle \frac{9}{11}}+{\displaystyle \frac{5}{11}}+{\displaystyle \frac{1}{11}}={\displaystyle \frac{15}{11}}$

Along the third column ${\displaystyle \frac{2}{11}}+{\displaystyle \frac{7}{11}}+{\displaystyle \frac{6}{11}}={\displaystyle \frac{15}{11}}$

Along diagonal-1 ${\displaystyle \frac{4}{11}}+{\displaystyle \frac{5}{11}}+{\displaystyle \frac{6}{11}}={\displaystyle \frac{15}{11}}$

Along diagonal-2 ${\displaystyle \frac{8}{11}}+{\displaystyle \frac{5}{11}}+{\displaystyle \frac{2}{11}}={\displaystyle \frac{15}{11}}$

Yes this is a magic square.

Question 4. 

A rectangular sheet of paper is $12{\displaystyle \frac{1}{2}}$ cm long and $10{\displaystyle \frac{2}{3}}$ cm wide. Find its perimeter.

Solution:

Length of the paper=$12{\displaystyle \frac{1}{2}}$ cm

Breadth of the paper=$10{\displaystyle \frac{2}{3}}$

Perimeter of the paper= 2(Length of the paper + Breadth of the paper)

=$2(12{\displaystyle \frac{1}{2}}+10{\displaystyle \frac{2}{3}})$

=$2({\displaystyle \frac{25}{2}}+{\displaystyle \frac{32}{3}})$

=$2({\displaystyle \frac{25\times 3+32\times 2}{2\times 3}})$

=$2({\displaystyle \frac{75+64}{6}})$

=$2\times ({\displaystyle \frac{139}{6}})$

=$({\displaystyle \frac{139}{3}})$

=$(46{\displaystyle \frac{1}{3}})$

Perimeter of the paper= $(46{\displaystyle \frac{1}{3}})$ cm

Question 5. 

Find the perimeters of (i) $\Delta ABE$ (ii) the rectangle $BCDE$ in this figure. Whose perimeter is greater?

Answer: (i) In $AB={\displaystyle \frac{5}{2}}$ cm, $BE=2{\displaystyle \frac{3}{4}}$, $AE=3{\displaystyle \frac{3}{5}}$ and $DE={\displaystyle \frac{7}{6}}$,

The perimeter of  $\Delta ABE=AB+BE+AE$

$={\displaystyle \frac{5}{2}}+2{\displaystyle \frac{3}{4}}+3{\displaystyle \frac{3}{5}}$  

$={\displaystyle \frac{5}{2}}+{\displaystyle \frac{11}{4}}+{\displaystyle \frac{18}{5}}$

$={\displaystyle \frac{50+55+72}{20}}$

$={\displaystyle \frac{177}{20}}$

The perimeter of  $\Delta ABE=8{\displaystyle \frac{17}{20}}$ cm

(ii) In rectangle $BCDE$, 

Perimeter of rectangle = 2 (length + breadth)

=$2(2{\displaystyle \frac{3}{4}}+{\displaystyle \frac{7}{6}}$

=$2({\displaystyle \frac{11}{4}}+{\displaystyle \frac{7}{6}}$

=$2\times ({\displaystyle \frac{33+14}{12}}$

=${\displaystyle \frac{47}{6}}$ 

=$7{\displaystyle \frac{5}{6}}$ cm

Thus, the perimeter of rectangle BCDE is =$7{\displaystyle \frac{5}{6}}$ cm.

Comparing the perimeter of triangle and that of rectangle,

${\displaystyle \frac{177}{20}}$  cm$>7{\displaystyle \frac{5}{6}}$ cm

Therefore, the perimeter of triangle $ABE$ is greater than that of rectangle $BCDE$ .

Question 6. 

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

Answer: 

Given: The width of the picture = $7{\displaystyle \frac{3}{5}}$ cm and the width of picture frame =$7{\displaystyle \frac{3}{10}}$ cm

Therefore, the picture should be trimmed =$7{\displaystyle \frac{3}{10}}-7{\displaystyle \frac{3}{5}}$ =

=${\displaystyle \frac{73}{10}}-{\displaystyle \frac{38}{5}}$  =${\displaystyle \frac{76-73}{10}}$ cm

=${\displaystyle \frac{3}{10}}$ cm

Thus, the picture should be trimmed by ${\displaystyle \frac{3}{10}}$ cm.

Question 7.  

Ritu ate ${\displaystyle \frac{3}{5}}$ part of an apple and the remaining apple was eaten by her brother Somu.

How much part of the apple did Somu eat? Who had the larger share? By how much?

Answer: 

The part of an apple eaten by Ritu =${\displaystyle \frac{3}{5}}$

The part of an apple eaten by Somu =$1-{\displaystyle \frac{3}{5}}={\displaystyle \frac{2}{5}}$

Comparing the parts of apple eaten by both Ritu and Somu

Larger share will be more by part ${\displaystyle \frac{3}{5}}>{\displaystyle \frac{2}{5}}$.

Thus, Ritu’s part is more than Somu’s part by ${\displaystyle \frac{1}{5}}$.

Question 8. 

Michael finished colouring a picture in ${\displaystyle \frac{7}{12}}$ hour. Vaibhav finished colouring the same picture in ${\displaystyle \frac{3}{4}}$ hour. Who worked longer? By what fraction was it longer?

Answer: 

Time taken by Michael to colour the picture =${\displaystyle \frac{7}{12}}$ hour

Time taken by Vaibhav to colour the picture =${\displaystyle \frac{3}{4}}$ hour

Converting both fractions in like fractions, ${\displaystyle \frac{7}{12}}$ and ${\displaystyle \frac{3\times 3}{4\times 3}}$

Here, ${\displaystyle \frac{7}{12}}<{\displaystyle \frac{9}{12}}$

Thus, Vaibhav worked longer time.

Vaibhav worked longer time by ${\displaystyle \frac{3}{4}}-{\displaystyle \frac{7}{12}}={\displaystyle \frac{9-7}{12}}={\displaystyle \frac{2}{12}}={\displaystyle \frac{1}{6}}$ hour.

Thus, Vaibhav took ${\displaystyle \frac{1}{6}}$ hour more than Michael.