The DAV Class 7 Maths Book Solutions Pdf and DAV Class 7 Maths Chapter 2 Worksheet 6 Solutions of Operations on Rational Numbers offer comprehensive answers to textbook questions.
DAV Class 7 Maths Ch 2 WS 6 Solutions
Question 1.
Correct the following statements:
(i) Between two rational numbers, we can find only one rational number.
Answer:
Between two fractions, we can find infinitely many fractions.
(ii) Between two rational numbers, we can find as many integers as we like.
Answer:
Between two rational numbers, we can find as many rational numbers as we like.
(iii) Between two integers, we can find as many integers as we like.
Answer:
Between two integers, we can find as many rational numbers as we like.
Question 2.
Find a rational number between:
(i) 2 and 4
Answer:
Rational number between
2 and 4 = \(\frac{1}{2}\)(2 + 4) = \(\frac{1}{2}\) x 6 = 3
(ii) – 2 and – 6
Answer:
Rational number between
– 2 and – 6 = \(\frac{1}{2}\)[-2 + (-6)]
= \(\frac{1}{2}\)[-2 – 6]
= \(\frac{1}{2}\)[-8]
= -4
(iii) \(\frac{1}{4}\) and \(\frac{-3}{4}\)
Answer:
Rational number between
\(\frac{1}{4}\) and \(\frac{-3}{4}\)
= \(\frac{1}{2}\left[\frac{1}{4}+\left(\frac{-3}{4}\right)\right]\)
= \(\frac{1}{2}\left[\frac{1}{4}-\frac{3}{4}\right]\)
= \(\frac{1}{2}\left[\frac{1-3}{4}\right]\)
= \(\frac{1}{2}\left[\frac{-2}{4}\right]=\frac{-1}{4}\)
(iv) \(\frac{-2}{3}\) and \(\frac{-7}{3}\)
Answer:
Rational number between
\(\frac{-2}{3}\) and \(\frac{-7}{3}\)
= \(\frac{1}{2}\left[\frac{-2}{3}+\left(\frac{-7}{3}\right)\right]\)
= \(\frac{1}{2}\left[\frac{-2}{3}-\frac{7}{3}\right]\)
= \(\frac{1}{2}\left[\frac{-2-7}{3}\right]\)
= \(\frac{1}{2}\left[\frac{-9}{3}\right]\)
= \(\frac{-3}{2}\)
Question 3.
Insert three rational numbers between:
(i) \(\frac{4}{13}\) and \(\frac{1}{13}\)
Answer:
First rational number between \(\frac{4}{13}\) and \(\frac{1}{13}\)
= \(\frac{1}{2}\left(\frac{4}{13}+\frac{1}{13}\right)\)
= \(\frac{1}{2}\left(\frac{4+1}{13}\right)\)
= \(\frac{1}{2} \times \frac{5}{13}=\frac{5}{26}\)
Second rational number = \(\frac{1}{2}\left(\frac{4}{13}+\frac{5}{26}\right)\)
= \(\frac{1}{2}\left(\frac{4 \times 2+5 \times 1}{26}\right)\)
= \(\frac{1}{2}\left(\frac{8+5}{26}\right)\)
= \(\frac{1}{2} \times \frac{13}{26}=\frac{1}{4}\)
Third rational number = \(\frac{1}{2}\left(\frac{5}{26}+\frac{1}{13}\right)\)
= \(\frac{1}{2}\left(\frac{5 \times 1+1 \times 2}{26}\right)\)
= \(\frac{1}{2}\left(\frac{5+2}{26}\right)\)
= \(\frac{1}{2} \times \frac{7}{26}=\frac{7}{52}\)
Hence the required rational numbers are \(\frac{5}{26}, \frac{1}{4}, \frac{7}{52}\)
(ii) \(\frac{-7}{10}\) and \(\frac{11}{10}\)
Answer:
First rational number between \(\frac{-7}{10}\) and \(\frac{11}{10}\)
= \(\frac{1}{2}\left(\frac{-7}{10}+\frac{11}{10}\right)\)
= \(\frac{1}{2}\left(\frac{-7+11}{10}\right)\)
Second rational number = \(\frac{1}{2}\left(\frac{-7}{10}+\frac{1}{5}\right)\)
= \(\frac{1}{2}\left(\frac{-7+2}{10}\right)\)
Third rational number = \(\frac{1}{2}\left(\frac{1}{5}+\frac{11}{10}\right)\)
= \(\frac{1}{2}\left(\frac{-7+2}{10}\right)\)
= \(\frac{1}{2} \times \frac{13}{10}=\frac{13}{20}\)
Hence the required rational numbers are
(iii) \(\frac{-4}{3}\) and \(\frac{-19}{3}\)
Answer:
First rational number between \(\frac{-4}{3}\) and \(\frac{-19}{3}\)
= \(\frac{1}{2}\left(\frac{-4}{3}-\frac{19}{3}\right)\)
= \(\frac{1}{2} \times\left(\frac{-4-19}{3}\right)\)
= \(\frac{1}{2} \times\left(\frac{-23}{3}\right)=\frac{-23}{6}\)
Second rational number = \(\frac{1}{2}\left(\frac{-4}{3}-\frac{23}{6}\right)\)
= \(\frac{1}{2}\left(\frac{-8-23}{6}\right)\)
= \(\frac{1}{2} \times\left(-\frac{31}{6}\right)\)
= \(\frac{1}{2} \times \frac{-31}{6}=\frac{-31}{12}\)
Third rational number = \(\frac{1}{2}\left(\frac{-23}{6}-\frac{19}{3}\right)\)
= \(\frac{1}{2}\left(\frac{-23}{6}-\frac{19}{3}\right)\)
= \(\frac{1}{2} \times\left(\frac{-61}{6}\right)=\frac{-61}{12}\)
Hence the required rational number are \(\frac{-23}{6}, \frac{-31}{12}\) and \(\frac{-61}{12}\)
(iv) \(\frac{1}{8}\) and \(\frac{-15}{8}\)
Answer:
First rational number between \(\frac{1}{8}\) and \(\frac{-15}{8}\)
= \(\frac{1}{2}\left(\frac{1}{8}-\frac{15}{8}\right)\)
= \(\frac{1}{2} \times\left(\frac{1-15}{8}\right)\)
= \(\frac{1}{2} \times\left(\frac{-14}{8}\right)\)
Second rational number = \(\frac{1}{2}\left(\frac{1}{8}-\frac{7}{8}\right)\)
= \(\frac{1}{2} \times\left(\frac{1-7}{8}\right)\)
= \(\frac{1}{2} \times\left(\frac{-6}{8}\right)\)
Third rational number = \(\frac{1}{2}\left(\frac{-7}{8}-\frac{15}{8}\right)\)
= \(\frac{1}{2}\left(\frac{-7-15}{8}\right)\)
= \(\frac{1}{2} \times\left(\frac{-22}{8}\right)\)
Hence the required rational number are \(\frac{-7}{8}, \frac{-3}{8}\) and \(\frac{-11}{8}\)
Question 4.
Find five rational numbers between:
(i) \(\frac{-4}{7}\) and \(\left|\frac{-4}{7}\right|\)
Answer:
The given rational numbers are \(\frac{-4}{7}\) and \(\left|\frac{-4}{7}\right|\)
First rational number = \(\frac{1}{2}\left(\frac{-4}{7}+\frac{4}{7}\right)\)
= \(\frac{1}{2}\) × 0 = 0
Second rational number = \(\frac{1}{2}\) (\(\frac{-4}{7}\) + 0)
= \(\frac{1}{2} \times \frac{-4}{7}=\frac{-2}{7}\)
Third rational number = \(\frac{1}{2}\)(0 + \(\frac{4}{7}\))
= \(\frac{1}{2} \times \frac{4}{7}=\frac{2}{7}\)
Fourth rational number = \(\frac{1}{2}\)(0 + \(\frac{-2}{7}\))
= \(\frac{1}{2}\)(0 – \(\frac{2}{7}\))
= \(\frac{1}{2} \times \frac{-2}{7}=\frac{-1}{7}\)
Fifth rational number = \(\frac{1}{2}\)(0 + \(\frac{2}{7}\))
= \(\frac{1}{2} \times \frac{2}{7}=\frac{1}{7}\)
Hence the required rational numbers are 0, \(\frac{-2}{7}, \frac{2}{7}, \frac{-1}{7}\) and \(\frac{1}{7}\)
(ii) \(\frac{-8}{3}\) and \(\left|\frac{-8}{3}\right|\)
Answer:
= \(\frac{-8}{3}\) and \(\frac{8}{3}\)
First rational number = \(\frac{1}{2}\left(\frac{-8}{3}+\frac{8}{3}\right)\)
= \(\frac{1}{2}\) × 0 = 0
Second rational number = \(\frac{1}{2}\)(\(\frac{-8}{3}\) + 0)
= \(\frac{1}{2} \times \frac{-8}{3}=\frac{-4}{3}\)
Third rational number = \(\frac{1}{2}\)(0 + \(\frac{8}{3}\))
= \(\frac{8}{3}\)
Fourth rational number = \(\frac{1}{2}\)(0 + \(\frac{-4}{3}\))
= \(\frac{1}{2}\)(0 – \(\frac{4}{3}\))
= \(\frac{1}{2} \times \frac{-4}{3}=\frac{-2}{3}\)
Fifth rational number = \(\frac{1}{2}\)(0 + \(\frac{4}{3}\))
= \(\frac{1}{2} \times \frac{4}{3}=\frac{2}{3}\)
Hence the required rational numbers are 0, \(\frac{-4}{3}, \frac{4}{3}, \frac{-2}{3}\) and \(\frac{2}{3}\)
DAV Class 7 Maths Chapter 2 Value Based Questions
Question 1.
Rohit donated \(\frac{1}{5}\) of his monthly income to an Non-Government Organisation working for the education of the girl child, spend \(\frac{1}{4}\) of his salary on food, \(\frac{1}{3}\) on rent and \(\frac{1}{15}\) on other expenses. He is left with ₹ 9000.
(i) Find Rohit’s monthly salary.
(ii) What values of Rohit are depicted here?
(iii) Why is the education, specially for girls, important?
Answer:
(i) Let Rohit’s monthly salary be ₹ x.
Amount donated to NGO = \(\frac{1}{5}\) × x = \(\frac{x}{5}\)
Amount spent on food = \(\frac{1}{4}\) × x = \(\frac{x}{4}\)
Amount spent on rent = \(\frac{1}{3}\) × x = \(\frac{x}{3}\)
Amount spent on other expenses = \(\frac{1}{15}\) × x = \(\frac{x}{15}\)
According to the question, we have
x – \(\left(\frac{x}{5}+\frac{x}{4}+\frac{x}{3}+\frac{x}{15}\right)\) = 9000
⇒ x – \(\left(\frac{60 x+75 x+100 x+20 x}{300}\right)\) = 9000
⇒ x – \(\left(\frac{255 x}{300}\right)\) = 9000
⇒ \(\frac{300 x-255 x}{300}\) = 9000
⇒ \(\frac{45 x}{300}\) = 9000
⇒ x = \(\frac{300 \times 9000}{45}\) = 60,000
Therefore, Rohit’s monthly salary is ₹ 60,000.
(ii) Values: Sensitivity towards girls education, kindness, etc.
(iii) Education helps in overall development of girls themselves and society.