RBSE Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1

Rajasthan Board RBSE Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 Textbook Exercise Questions and Answers.

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

Question 1.
(i) 2 - \(\frac{3}{5}\)
Answer:

(ii) 4 + \(\frac{7}{8}\)
Answer:

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

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

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

(vi) 2 \(\frac{2}{3}+3 \frac{1}{2}\)
Answer:

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

Question 2.
Arrange the following in descending order:
(i) \(\frac{2}{9}, \frac{2}{3}, \frac{8}{21}\)
Answer:

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

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 \(\frac{4}{11}+\frac{9}{11}+\frac{2}{11}=\frac{15}{11}\).)
Answer:
Sum of the numbers along:

Since, sum in each case is equal, 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 the paper = 12\(\frac{1}{2}\) = \(\frac{25}{2}\)
Width of the paper = 10\(\frac{2}{3}\) = \(\frac{32}{3} \)cm
Perimeter of the paper = 2 × (length + width)

Question 5.
Find the perimeters of (i) ∆ABE
(ii) the rectangle BCDE in this figure. Whose perimeter is greater?

Answer:
(i) Perimeter of ∆ABE
= Sum of sides of ∆ABE
= AB + BE + AE

(ii) Perimeter of rectangle BCDE
= 2 × (length + width)
= 2 × (BE + ED)
= 2 × \(\left(2 \frac{3}{4}+\frac{7}{6}\right)\)

Again 8\(\frac{17}{20}\) > 7\(\frac{5}{6}\)
∴ Perimeter of ∆ABE is greater than perimeter of the rectangle BCDE.

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
Width of the frame = 7\(\frac{3}{10}\)cm = \(\frac{73}{10}\) cm
The picture is to be trimmed = \(\frac{38}{5}-\frac{73}{10}\)
= \(\frac{38 \times 2}{5 \times 2}-\frac{73}{10}=\frac{76}{10}-\frac{73}{10}=\frac{76-7.3}{10}=\frac{3}{10}\) cm

Question 7.
Ritu ate g 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:
Remaining part = \(\frac{1}{1}-\frac{3}{5}\)
= \(\frac{1 \times 5}{1 \times 5}-\frac{3 \times 1}{5 \times 1}=\frac{5}{5}-\frac{3}{5}=\frac{5-3}{5}=\frac{2}{5}\)
∴ Somu eat \(\frac{2}{5}\) part of the apple.
Again \(\frac{3}{5}\) > \(\frac{2}{5}\) and \(\frac{3}{5}\) - \(\frac{2}{5}\) = \(\frac{1}{5}\)
So, Ritu had the larger share and by \(\frac{1}{5}\) part.

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:
Michael finished the picture = \(\frac{7}{12}\) hour
Vaibhav finished the picuture = \(\frac{3}{4}\) hour
= \(\frac{3 \times 3}{4 \times 3}\) = \(\frac{9}{12}\) hour
∴ \(\frac{9}{12}\) > \(\frac{7}{12}\)
∴ Vaibhav worked longer and \(\frac{9}{12}-\frac{7}{12}\)
= \(\frac{2}{12}\) = \(\frac{1}{6}\)
i.e. Vaibhav worked longer by \(\frac{1}{6}\) hour.