Exercise 2.1
"Fractions & Decimals" Chapter 2 - Introduction - Class 7
Q 1, Ex 2.1 - Fractions and Decimals - Chapter 2 - Maths Class 7th - NCERT
Solve :
(i) $2-\dfrac{3}{5}$
Sol :
$\dfrac{2}{1}-\dfrac{3}{5}$ $=\dfrac{2\times 5}{1\times 5}-\dfrac{3}{5}$ $=\dfrac{10-3}{5}$ $=\dfrac{7}{5}$
(ii) $4+\dfrac{7}{8}$
Sol :
$\dfrac{4}{1}+\dfrac{7}{8}$ $=\dfrac{4\times 8}{8}+\dfrac{7}{8}$ $=\dfrac{(4\times 8)+7}{8}$ $=\dfrac{39}{8}$ $=4\dfrac{7}{8}$
(iii) $\dfrac{3}{5}+\dfrac{2}{7}$
Sol :
$\dfrac{3}{5}+\dfrac{2}{7}$ $= \dfrac{3\times 7}{5\times 7}+\dfrac{2\times 5}{7\times 5}$ $=\dfrac{21+10}{35}$ $=\dfrac{31}{35}$
(iv) $\dfrac{9}{11}-\dfrac{4}{15}$
Sol :
$\dfrac{9}{11}-\dfrac{4}{15}$ $\dfrac{}{}-\dfrac{}{}$
(v) $\dfrac{7}{10}+\dfrac{2}{5}+\dfrac{3}{2}$
Sol :
$\dfrac{7}{10}+\dfrac{2}{5}+\dfrac{3}{2}$ $\dfrac{}{}+\dfrac{}{}+\dfrac{}{}$ $=\dfrac{}{}$ $=\dfrac{26}{10}$ $=\dfrac{13}{5}$ $=2\dfrac{3}{5}$
(vi) $2\dfrac{2}{3} + 3 \dfrac{1}{2}$
Sol :
$2\dfrac{2}{3} + 3 \dfrac{1}{2}$
(vii) $8\dfrac{1}{2}-3\dfrac{5}{8}$
Sol :
$=8\dfrac{1}{2}-3\dfrac{5}{8}$
$=\dfrac{17\times 4}{2\times 4}-\dfrac{29}{8}$
$=\dfrac{68-29}{8}$
$=\dfrac{39}{8}=4\dfrac{7}{8}$
Q 2, Ex 2.1 - Fractions and Decimals - Chapter 2 - Maths Class 7th - NCERT
Question 2:
Arrange the following in descending order:
(i) 
(ii) 
(ii) Answer:
(i) 
Changing them to like fractions, we obtain
Since 42>24>14,
(ii) 
Changing them to like fractions, we obtain
As 49 > 30>14,
Question 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?
 |  |  | (Along the first row  ) |
 |  |  | |
 |  |  |
Answer:
Along the first row, sum = 
Along the second row, sum =
Along the third row, sum = 
Along the first column, sum = 
Along the second column, sum = 
Along the third column, sum =
Along the first diagonal, sum = 
Along the second diagonal, sum =
Since the sum of the numbers in each row, in each column, and along the diagonals is the same, it is a magic square.
Q 4, Ex 2.1 - Fractions and Decimals - Chapter 2 - Maths Class 7th - NCERT
Question 4:
A rectangular sheet of paper is 
cm long and 
cm wide.
cm long and cm wide.
Find its perimeter.
Answer:
Length =
Breadth = 
Perimeter = 2 × (Length + Breadth)
Q 5, Ex 2.1 - Fractions and Decimals - Chapter 2 - Maths Class 7th - NCERT
Question 5:
Find the perimeters of (i) ΔABE (ii) the rectangle BCDE in this figure. Whose perimeter is greater?
Answer:
(i) Perimeter of ΔABE = AB + BE + EA
(ii)
Perimeter of rectangle = 2 (Length + Breadth)
Perimeter of ΔABE =
Changing them to like fractions, we obtain
As 531 > 430,
Perimeter (ΔABE) > Perimeter (BCDE)
Q 6, Ex 2.1 - Fractions and Decimals - Chapter 2 - Maths Class 7th - NCERT
Question 6:
Salil wants to put a picture in a frame. The picture is 
cm wide.
cm wide.
To fit in the frame the picture cannot be more than
cm wide. How much should the picture be trimmed?
cm wide. How much should the picture be trimmed?Answer:
Width of picture =
Required width =
Page No 32:
Q 7, Ex 2.1 - Fractions and Decimals - Chapter 2 - Maths Class 7th - NCERT
Question 7:
Ritu ate 
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?
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:
Part of apple eaten by Ritu =
Part of apple eaten by Somu = 1 − Part of apple eaten by Ritu
= 
Therefore, Somu ate 
part of the apple.
part of the apple.
Since 3 > 2, Ritu had the larger share.
Difference between the 2 shares =
Therefore, Ritu’s share is larger than the share of Somu by
.
.Q 8, Ex 2.1 - Fractions and Decimals - Chapter 2 - Maths Class 7th - NCERT
Question 8:
Michael finished colouring a picture in 
hour. Vaibhav finished colouring the same picture in 
hour. Who worked longer? By what fraction was it longer?
hour. Vaibhav finished colouring the same picture in hour. Who worked longer? By what fraction was it longer?Answer:
Time taken by Michael =
Time taken by Vaibhav =
Converting these fractions into like fractions, we obtain
Since 9 > 7,
Vaibhav worked longer.
Difference =$\frac{9}{12}-\frac{7}{12}$
$=\frac{2}{12}$
$=\frac{1}{6}$ hour
$=\frac{2}{12}$
$=\frac{1}{6}$ hour
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