NCERT Solutions Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1

NCERT Solutions Class 7 Maths
Chapter – 2 (Fractions and Decimals)

The NCERT Solutions in English Language for Class 7 Mathematics Chapter – 2 Fractions and Decimals Exercise 2.1 has been provided here to help the students in solving the questions from this exercise.

Chapter : 2 Fractions and Decimals

Exercise – 2.1

1. Solve:
(i) 2 – \mathbf{\frac{3}{5}}

(ii) 4 + \mathbf{\frac{7}{8}}

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

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

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

(vi) \mathbf{2\frac{2}{3} +3\frac{1}{2}}

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

Solution –

(i) 2 – \mathbf{\frac{3}{5}}

= \frac{2}{1} - \frac{3}{5}

LCM of 1, 5 = 5
= \frac{2\times 5 - 3\times 1}{5}

= \frac{10 - 3}{5}

= \frac{7}{5}

(ii) 4 + \mathbf{\frac{7}{8}}

= \frac{4}{1} + \frac{7}{8}
LCM of 1, 8 = 8

= \frac{4 \times 8 + 7\times1}{8}

= \frac{32 +7}{8}

= \frac{39}{8} = 4\frac{7}{8}

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

LCM of 5, 7 = 35

= \frac{3\times 7 + 2\times 5}{5\times 7}

= \frac{21+10}{35}

= \frac{31}{35}

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

LCM of 11, 15 = 165

= \frac{9\times 15 - 4\times 11}{11\times 15}

= \frac{135 - 44}{165}

= \frac{91}{165}

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

LCM of 10, 5 and 2 = 10

= \frac{7\times 1 + 2\times2 + 3\times5}{10}

= \frac{7+4+15}{10}

= \frac{26}{10} = \frac{13}{5} = 2\frac{3}{5}

(vi) \mathbf{2\frac{2}{3} +3\frac{1}{2}}

= \frac{8}{3} + \frac{7}{2}

LCM of 3, 2 = 6

= \frac{8\times 2 + 7\times 3}{6}

= \frac{16+21}{6}

= \frac{37}{6} = 6\frac{1}{6}

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

= \frac{17}{2} - \frac{29}{8}
LCM of 2, 8 = 8

= \frac{17\times 4 - 29\times 1}{8}

= \frac{68-29}{8}

= \frac{39}{8} = 4\frac{7}{8}

2. Arrange the following in descending order:

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

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

Solution –

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

LCM of 9, 3, 21 = 63

\frac{2}{9} = \frac{2\times 7}{9\times 7} = \frac{14}{63}

\frac{2}{3} = \frac{2\times 21}{3\times 21} = \frac{42}{63}

\frac{8}{21} = \frac{8\times 3}{21\times 3} = \frac{24}{63}

Clearly,

\frac{42}{63} > \frac{24}{63} > \frac{14}{63}

Hence,

\frac{2}{3} > \frac{24}{63} > \frac{2}{9}

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

LCM of 5, 7, 10 = 70

\frac{1}{5} = \frac{1\times 14}{5\times14} = \frac{14}{70}

\frac{3}{7} = \frac{3\times 10}{7\times10} = \frac{30}{70}

\frac{7}{10} = \frac{7\times 7}{10\times7} = \frac{49}{70}

Clearly,

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

Hence,

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

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?

\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}

Solution –
Sum along the first row = \frac{4}{11}  + \frac{9}{11} + \frac{2}{11} = \frac{15}{11}

Sum along the second row = \frac{3}{11} + \frac{5}{11} + \frac{7}{11}  = \frac{15}{11}

Sum along the third row = \frac{8}{11} + \frac{1}{11} + \frac{6}{11} = \frac{15}{11}

Sum along the first column = \frac{4}{11} + \frac{3}{11} + \frac{8}{11} = \frac{15}{11}

Sum along the second column = \frac{9}{11}  + \frac{5}{11} + \frac{1}{11} = \frac{15}{11}

Sum along the third column = \frac{2}{11} + \frac{7}{11} + \frac{6}{11} = \frac{15}{11}

Sum along the first diagonal = \frac{4}{11} + \frac{5}{11} + \frac{6}{11} = \frac{15}{11}

Sum along the second diagonal = \frac{2}{11} + \frac{5}{11} + \frac{8}{11} = \frac{15}{11}

Since the sum of the numbers in each row, in each column, and along the diagonals is the same, it is a magic square.

4. A rectangular sheet of paper is \mathbf{12\frac{1}{2}} cm long and \mathbf{10\frac{2}{3}} cm wide. Find its perimeter.
Solution –
Length of sheet = 12\frac{1}{2} cm = \frac{25}{2} cm

Breadth of sheet = 10\frac{2}{3} cm = \frac{32}{3} cm
We know that,
Perimeter of the rectangle = 2 × (length + breadth)

= 2 × \left [ \frac{25}{2} + \frac{32}{3}\right ]
LCM of 2, 3 = 6

= 2 × \left ( \frac{25\times 3 + 32 \times 2}{6} \right )

= 2 × \left ( \frac{75+64}{6} \right )

= 2 × \left ( \frac{139}{6} \right )

= \frac{139}{3} cm = 46\frac{1}{3} cm

Hence, the perimeter of the sheet of paper is 46\frac{1}{3} cm.

5. Find the perimeters of (i) ΔABE (ii) the rectangle BCDE in this figure. Whose perimeter is greater?
NCERT Class 7 Maths Solution

Solution –
Given  AB = \frac{5}{2} cm, AE = 3\frac{3}{5} = \frac{18}{5} cm, BE = 2\frac{3}{4} = \frac{11}{4} cm, ED = \frac{7}{6} cm

(i) Perimeter of ∆ABE
Perimeter of the triangle = Sum of all sides
Perimeter of ∆ABE = AB + BE + EA

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

The LCM of 2, 4 and 5 = 20

= \frac{5\times10 + 11\times5 + 18\times4}{20}

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

= \frac{177}{20} = 8\frac{17}{20} cm

(ii) Perimeter of the rectangle BCDE

Perimeter of the rectangle = 2 × (length + breadth)
Perimeter of rectangle BCDE = 2 × (BE + ED)

= 2 × [\frac{11}{4} + \frac{7}{6}]

The LCM of 4, 6 = 12

= 2 × \frac{11 \times 3 +7\times2}{12}

= 2 × \frac{33 +14}{12}

= 2 × \frac{47}{12} = \frac{47}{6} = 7\frac{5}{6}

Perimeter of triangle ABE = \frac{177}{20}

Perimeter of rectangle BCDE = \frac{47}{6}

LCM of 20 and 6 = 60

\frac{177}{20} = \frac{177\times 3}{20\times 3} = \frac{531}{60}

\frac{47}{6} = \frac{47\times 10}{6\times 10} = \frac{470}{60}

Clearly, \frac{531}{60} > \frac{470}{60}

Hence, \frac{177}{20} > \frac{47}{6}

∴ Perimeter of ∆ABE > Perimeter of Rectangle BCDE

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

The width of Picture = 7\frac{3}{5} = \frac{38}{5} cm

Required width of a Frame = 7\frac{3}{10} = \frac{73}{10} cm

∴ The picture should be trimmed by = \frac{38}{5} – \frac{73}{10} 
The LCM of 5, 10 = 10

= \frac{38 \times 2 - 73\times 1}{10}

= \frac{76-73}{10}

= \frac{3}{10} cm

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

7. Ritu ate \mathbf{\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?
Solution –
Part of apple eaten by Ritu is = \frac{3}{5}

Part of apple eaten by Somu is = 1 – Part of apple eaten by Ritu

= 1 – \frac{3}{5}

The LCM of 1, 5 = 5

= \frac{1\times 5-3}{5}

= \frac{5-3}{5} = \frac{2}{5}

∴ Part of apple eaten by Somu is \frac{2}{5}

So, \frac{3}{5} > \frac{2}{5} hence, Ritu ate larger size of apple.

Now, the difference between the two shares = \frac{3}{5} – \frac{2}{5}

= \frac{3-2}{5} = \frac{1}{5}

Thus, Ritu’s share is larger than share of Somu by \frac{1}{5}

8. Michael finished colouring a picture in \mathbf{\frac{7}{12}} hour. Vaibhav finished colouring the same picture in \mathbf{\frac{3}{4}}hour. Who worked longer? By what fraction was it longer?
Solution –
Time taken by Michael to colour the picture is = \frac{7}{12} hr.

Time taken by the Vaibhav to colour the picture is = \frac{3}{4} hr.
The LCM of 12, 4 = 12

\frac{7}{12} = \frac{7\times 1}{12\times 1} = \frac{7}{12}

\frac{3}{4}  = \frac{3\times 3}{4\times 3} = \frac{9}{12}

Clearly, \frac{7}{12} < \frac{9}{12}

Hence, \frac{7}{12} < \frac{3}{4}

Thus, Vaibhav worked for longer time.
So, Vaibhav worked longer time by = \frac{3}{4} – \frac{7}{12}

= \frac{9}{12} – \frac{7}{12}

= \frac{9-7}{12}

= \frac{2}{12} 

= \frac{1}{6} hour longer.

 

 

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