AP State Syllabus AP Board 7th Class Maths Solutions Chapter 2 Fractions, Decimals and Rational Numbers Ex 1 Textbook Questions and Answers.

## AP State Syllabus 7th Class Maths Solutions 2nd Lesson Fractions, Decimals and Rational Numbers Exercise 1

Question 1.

Solve the following.

(i) 2 + \(\frac { 3 }{ 4 }\)

(ii) \(\frac{7}{9}+\frac{1}{3}\)

(iii) 1 – \(\frac{4}{7}\)

(iv) \(2 \frac{2}{3}+\frac{1}{2}\)

(v) \(\frac{5}{8}-\frac{1}{6}\)

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

Solution:

Question 2.

Arrange the following in ascending order.

(i) \(\frac{5}{8}, \frac{5}{6}, \frac{1}{2}\)

(ii) \(\frac{2}{5}, \frac{1}{3}, \frac{3}{10}\)

Solution:

Question 3.

Check whether in this square the sum of the numbers in each row and in each column and along the diagonals is the same.

Solution:

Question 4.

A rectangular sheet of paper is 5\(\frac{2}{3}\) cm long and 3\(\frac{1}{5}\) cm wide. Find its perimeter.

Solution:

Length of the rectangular sheet = 5\(\frac{2}{3}\) cm

Breadth/width of the rectangular sheet = 5\(\frac{2}{3}\) cm

Perimeter = 2 x (length + breadth)

Question 5.

The recipe requires 3\(\frac{1}{4}\) cups of flour. Radha has 1\( \frac{3}{8}\) cups of flour. How many more cups of flour does she need?

Solution:

Flour required for the recipe = 3\(\frac{1}{4}\) cups

Flour with Radha = 1\(\frac{3}{8}\) cups

More cups of flour required = \(3 \frac{1}{4}-1 \frac{3}{8}\)

= \(\frac{3 \times 4+1}{4}-\frac{1 \times 8+3}{8}\)

Question 6.

Abdul is preparing for his final exam. He has completed \(\frac{5}{12}\) part of his course content. Find out how much course content is left?

Solution:

Take content as 1 (i.e., full) Course completed = \(\frac{5}{12}\)

Course yet to be completed = 1 – \(\frac{5}{12}\)

= \(\frac{12 \times 1-5}{12}\)

\(\frac{12-5}{12}=\frac{7}{12}\)

Question 7.

Find the perimeters of(i) ΔABE (ii) the rectangle BCDE in this figure. Which figure has greater perimetre and by how much?

Solution:

i) Perimeter of ΔABE = AB + BE + AE

ii) Perimeter of BCDE = 2(BE + BC)

As \(\frac{116}{15}<\frac{153}{15}\), we conclude that the perimetre of ΔABE > Perimeter of BCDE