These NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 Questions and Answers are prepared by our highly skilled subject experts.

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

Question 1.

Answer:

Question 2.

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

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

Answer:

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

L.C.M of 9, 3 and 21 = 63

∴ The descending order is \(\frac{2}{3}, \frac{8}{21}, \frac{2}{9}\)

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

L.C.M of 5, 7 and 10 = 70

∴ The descending order is \(\frac{7}{10}, \frac{3}{7}, \frac{1}{5}\)

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?

\(\frac{4}{11}\) | \(\frac{9}{11}\) | \(\frac{2}{11}\) |

\(\frac{3}{11}\) | \(\frac{5}{11}\) | \(\frac{7}{11}\) |

\(\frac{8}{11}\) | \(\frac{1}{11}\) | \(\frac{6}{11}\) |

(Along the first row \(\frac{4}{11}+\frac{9}{11}+\frac{2}{11}=\frac{15}{11}\))

Answer:

Sum of numbers along the

1^{st}row = \(\frac{4}{11}+\frac{9}{11}+\frac{2}{11}\) = \(\frac{4+9+2}{11}\)

\(\frac{15}{11}\)

Since, the sum of numbers in each case is the same therefore, it is a magic square.

Question 4.

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

Answer:

Length of a rectangle = 12\(\frac{1}{2}\) cm = \(\frac{25}{2}\) cm

Width of a rectangle = 10\(\frac{2}{3}\)cm = \(\frac{32}{3}\) cm

Perimeter of a rectangle = 2 (length + breadth)

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 + AE

(ii) Perimeter of rectangle BCDE

= 2 [length + breadth]

= 2 [CD + DE]

Thus, the perimeter of ΔABE is greater.

Question 6.

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

Answer:

Width of the picture = 7\(\frac { 3 }{ 5 }\)cm

\(\frac { 38 }{ 5 }\)cm

Required width of the picture

= 7\(\frac { 3 }{ 10 }\) cm = \(\frac { 73 }{ 10 }\) cm

The picture should be trimmed by

The picture should be trimmed by \(\frac { 3 }{ 10 }\) cm.

Question 7.

Ritu ate \(\frac { 3 }{ 5 }\) 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:

Ritu ate \(\frac { 3 }{ 5 }\) part of an apple, therefore part of the apple ate by Somu = 1 – \(\frac { 3 }{ 5 }\)

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

\(\frac{3}{5}>\frac{2}{5}\)

∴ Ritu had the larger share by = \(\frac{3}{5}-\frac{2}{5}\)

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

Thus, Ritu had the larger share by \(\frac{1}{5}\) part of an apple.

Question 8.

Michael finished colouring a picture in \(\frac{7}{12}\) hour. Vaibhav finished colouring the

same picture in \(\frac{3}{4}\) hour. Who worked longer? By what fraction was it longer?

Answer:

Time taken by Michael = \(\frac{7}{12}\) hour

Time taken by Vaibhav = \(\frac{3}{4}\) hour.

Thus, Vaibhav worked longer by \(\frac{1}{6}\) hour.