These Class 7 Maths Extra Questions Chapter 8 Rational Numbers will help students prepare well for the exams.
Class 7 Maths Chapter 8 Extra Questions Rational Numbers
Class 7 Maths Rational Numbers Extra Questions – 2 Marks
Question 1.
Which of the following rational number are positive?
i) \(\frac{-8}{-17}\)
Solution:
\(\frac{-8}{-17}\) = \(\frac{8}{17}\) a positive rational number.
ii) \(\frac{-9}{14}\)
Solution:
\(\frac{-9}{14}\) a Negative rational number
iii) \(\frac{-6}{-11}\)
Solution:
\(\frac{-6}{-11}\) = \(\frac{6}{11}\) a positive rational number.
iv) \(\frac{7}{-2}\)
Solution:
\(\frac{7}{-2}\) = \(\frac{-7}{2}\) a Negative rational number.
Question 2.
Express \(\frac{-2}{7}\) as a rational number with denominator
i) 14
ii) 42
Solution:
\(\frac{-2}{7}\)
i) Denominator 14; \(\frac{-2}{7}\) × \(\frac{2}{2}\) = \(\frac{-4}{14}\)
ii) Denominator 42; \(\frac{-2}{7}\) × \(\frac{6}{6}\) = \(\frac{-12}{42}\)
Question 3.
Express the following in simplest form
i) \(\frac{48}{60}\)
Solution:
\(\frac{48}{60}\) = \(\frac{12 \times 4}{12 \times 5}\) = \(\frac{4}{5}\)
ii) \(\frac{-25}{125}\)
Solution:
\(\frac{-25}{125}\) = \(\frac{-25}{125}=\frac{-5 \times 5}{5 \times 5 \times 5}\) = \(\frac{-1}{5}\)
Question 4.
Express the following rational number to the lowest form.
i) \(\frac{-60}{68}\)
Solution:
\(\frac{-60}{68}\) = \(\frac{-2 \times 2 \times 3 \times 5}{2 \times 2 \times 17}\) = \(\frac{-15}{17}\)
ii) \(\frac{11}{-1331}\)
Solution:
\(\frac{11}{-1331}\) = \(\frac{11}{-11 \times 11 \times 11}\) = \(\frac{1}{-11 \times 11}\) = \(\frac{1}{-121}\) = \(\frac{-1}{121}\)
Question 5.
The product of two rational number is \(\frac{1}{2}\). If one of them is \(\frac{7}{4}\), find the other.
Solution:
Product of two rational numebr = \(\frac{1}{2}\)
One rational number = \(\frac{7}{4}\)
Other rational number = \(\frac{1}{2}\) ÷ \(\frac{7}{4}\)
= \(\frac{1}{2}\) × \(\frac{4}{7}\) = \(\frac{2}{7}\)
Question 6.
Write True (T) or False (F) of the following statements.
i) All rational number are natural numbers.
ii) Multiplicative inverse of \(\frac{7}{4}\) is \(\frac{-7}{4}\)
Solution:
i) False
ii) False
Question 7.
Represent i) \(\frac{3}{4}\)
ii) \(\frac{7}{2}\) on the number line.
Solution:
Question 8.
Which of the following rational number is greater \(\frac{5}{3}\) or \(\frac{-2}{3}\)?
Solution:
\(\frac{5}{3}\) is a positive rational number.
\(\frac{-2}{3}\) is a negative rational number.
∴ \(\frac{5}{3}\) > \(\frac{-2}{3}\)
Question 9.
Add \(\frac{-7}{9}\) and \(\frac{2}{6}\)
Solution:
\(\frac{-7}{9}\) + \(\frac{2}{6}\) = \(\frac{-7}{9}\) + \(\frac{1}{3}\) = \(\frac{-7}{9}\) + \(\frac{1}{3}\) × \(\frac{3}{3}\)
= \(\frac{-7}{9}\) + \(\frac{3}{9}\) = \(\frac{-7+3}{9}\) = \(\frac{-4}{9}\)
Question 10.
Simplify : \(\frac{-7}{26}\) + \(\frac{16}{39}\)
Solution:
\(\frac{-7}{26}\) + \(\frac{16}{39}\)
= \(\frac{-7}{26}\) × \(\frac{3}{3}\) + \(\frac{16}{39}\) × \(\frac{2}{2}\)
= \(\frac{-21}{78}\) + \(\frac{32}{78}\)
= \(\frac{-21+32}{78}\) = \(\frac{9}{98}\)
Question 11.
Subtract \(\frac{-3}{7}\) from \(\frac{8}{9}\).
Solution:
\(\frac{8}{9}\) – \(\left(\frac{-3}{7}\right)\) = \(\frac{8}{9}\) + \(\frac{3}{7}\) = \(\frac{8}{9}\) × \(\frac{7}{7}\) + \(\frac{3}{7}\) × \(\frac{9}{9}\)
= \(\frac{56}{63}\) + \(\frac{27}{63}\) = \(\frac{56+27}{63}\) = \(\frac{83}{63}\)
Question 12.
What should be added to \(\frac{-7}{2}\) to get \(\frac{2}{7}\)?
Solution:
Let x is to be added.
x + \(\left(\frac{-7}{2}\right)\) = \(\frac{2}{7}\)
x – \(\frac{7}{2}\) = \(\frac{2}{7}\)
adding \(\frac{7}{2}\) on both sides
x – \(\frac{7}{2}\) + \(\frac{7}{2}\) = \(\frac{2}{7}\) + \(\frac{7}{2}\)
x = \(\frac{2}{7}\) × \(\frac{2}{2}\) + \(\frac{7}{2}\) × \(\frac{7}{7}\)
x = \(\frac{4}{14}\) + \(\frac{49}{14}\)
x = \(\frac{4+49}{14}\) ⇒ x = \(\frac{53}{14}\)
∴ \(\frac{53}{14}\) is to be added.
Question 13.
The sum of two rational number is 1 . One of the number is \(\frac{-3}{2}\) find the other rational number.
Solution:
Given sum of two rational number = 1
One of the rational number = \(\frac{-3}{2}\)
Other rational number = 1 – \(\left(\frac{-3}{2}\right)\)
= \(\frac{1}{1}\) + \(\frac{3}{2}\) = \(\frac{2}{2}\) + \(\frac{3}{2}\) = \(\frac{2+3}{2}\) = \(\frac{5}{2}\)
Question 14.
By what number should be multiply \(\frac{-3}{7}\), so that the product is \(\frac{7}{2}\)?
Solution:
Let the number to be multiplied = x.
According to the problem
x × \(\frac{-3}{7}\) = \(\frac{7}{2}\)
x = \(\frac{7}{2}\) × \(\frac{-7}{3}\) ⇒ x = \(\frac{-49}{6}\)
∴ \(\frac{-49}{6}\) is to be multiplied.
Question 15.
The cost of \(7 \frac{1}{2}\)m of cloth is ₹ \(12 \frac{3}{4}\) find its cost per metre.
Solution:
Let of \(7 \frac{1}{2}\)m of cloth = ₹ \(12 \frac{3}{4}\)
\(\frac{15}{2}\)m cloth cost = ₹\(\frac{51}{4}\)
Cost of 1m cloth = \(\frac{51}{4}\) ÷ \(\frac{15}{2}\)
= \(\frac{51}{4}\) × \(\frac{2}{15}\) = \(\frac{51}{2}\) × \(\frac{1}{15}\)
= \(\frac{17}{2}\) × \(\frac{1}{5}\) = \(\frac{17}{10}\) = ₹1.7
cloth of 1m cloth = ₹1.7
Rational Numbers Extra Questions Class 7 – 3 Marks
Question 16.
If x = \(\frac{7}{2}\), y = \(\frac{-2}{3}\) find \(\frac{x+y}{x-y}\)
Solution:
Question 17.
By what number should \(\frac{-33}{16}\) be divided get \(\frac{-11}{2}\)?
Solution:
Let ‘x’ is be divided
According to the problem
\(\frac{-33}{16}\) ÷ x = \(\frac{-11}{2}\)
\(\frac{-33}{16}\) × \(\frac{1}{x}\) = \(\frac{-11}{2}\)
\(\frac{1}{x}\) = \(\frac{11}{2}\) × \(\frac{16}{33}\)
\(\frac{1}{x}\) = \(\frac{8}{3}\) ⇒ x = \(\frac{3}{8}\)
By \(\frac{3}{8}\) is to be divided.
Question 18.
Simplify
i) 0 ÷ \(\frac{3}{5}\)
Solution:
0 ÷ \(\frac{3}{5}\) = 0
ii) \(\frac{8}{13}\) ÷ \(\frac{-8}{13}\)
Solution:
\(\frac{8}{13}\) ÷ \(\frac{-8}{13}\) = \(\frac{8}{13}\) × \(\frac{13}{-8}\) = \(\frac{1}{-1}\) = -1
iii) -4 ÷ \(\frac{1}{4}\)
Solution:
-4 ÷ \(\frac{1}{4}\) = -4 × \(\frac{4}{1}\) = -16
Question 19.
Simplify :
i) 16 ÷ \(\frac{1}{4}\)
Solution:
16 ÷ \(\frac{1}{4}\) = 16 × 4 = 64
ii) \(\frac{2}{3}\) ÷ \(\frac{3}{2}\)
Solution:
\(\frac{2}{3}\) ÷ \(\frac{3}{2}\) = \(\frac{2}{3}\) × \(\frac{2}{3}\) = \(\frac{4}{9}\)
iii) \(\frac{-17}{6}\) ÷ 0
Solution:
\(\frac{-17}{6}\) ÷ 0 is not defined.
Question 20.
Find the multiplicative inverse of
i) \(\frac{3}{2}\)
ii) 0
iii) \(\frac{-7}{4}\)
Solution:
i) Multiplicative inverse of \(\frac{3}{2}\) = \(\frac{2}{3}\)
ii) ‘0’ has no multiplicative inverse.
iii) Multiplicative inverse of \(\frac{-7}{4}\) is \(\frac{4}{-7}\)
Question 21.
Insert 3 rational number between \(\frac{7}{2}\) and \(\frac{2}{7}\).
Solution:
\(\frac{7}{2}\) and \(\frac{2}{7}\)
\(\frac{7}{2}\) × \(\frac{7}{7}\) and \(\frac{2}{7}\) × \(\frac{2}{2}\)
\(\frac{49}{14}\) and \(\frac{4}{14}\)
\(\frac{5}{14}\), \(\frac{6}{14}\), \(\frac{7}{14}\) are the three rational number between \(\frac{7}{2}\) and \(\frac{2}{7}\).
Question 22.
Simplify :
i) \(\frac{2}{3}\) – \(\frac{3}{2}\) + 1
Solution:
\(\frac{2}{3}\) – \(\frac{3}{2}\) + 1 = \(\frac{2}{3}\) × \(\frac{2}{2}\) – \(\frac{3}{2}\) × \(\frac{3}{3}\) + \(\frac{1}{1}\) × \(\frac{6}{6}\)
= \(\frac{4}{6}\) – \(\frac{9}{6}\) + \(\frac{6}{6}\) = \(\frac{4-9+6}{6}\) = \(\frac{10-9}{6}\) = \(\frac{1}{6}\)
ii) 1 – \(\frac{1}{2}\) – \(\frac{1}{2}\) – \(\frac{1}{2}\) – \(\frac{1}{2}\)
Solution:
1 – \(\frac{1}{2}\) – \(\frac{1}{2}\) – \(\frac{1}{2}\) – \(\frac{1}{2}\) = 1 – 4 × \(\frac{1}{2}\) = 1 – 2 = – 1
Question 23.
If a = \(\frac{2}{3}\), b = \(\frac{-3}{2}\), find \(\frac{a-2 b}{a+2 b}\)
Solution:
Question 24.
What should be subtracted from \(\left(\frac{4}{3}-\frac{2}{7}\right)\) to get \(\frac{-1}{2}\)
Solution:
\(\frac{4}{3}\) – \(\frac{2}{7}\) = \(\frac{4}{3}\) × \(\frac{7}{7}\) – \(\frac{2}{7}\) × \(\frac{3}{3}\) = \(\frac{28}{21}\) – \(\frac{6}{21}\) = \(\frac{28-6}{21}\) = \(\frac{22}{21}\)
According to the problem
\(\frac{22}{21}\) – x = \(\frac{-1}{2}\)
\(\frac{22}{21}\) + \(\frac{1}{2}\) = x
\(\frac{22}{21}\) × \(\frac{2}{2}\) + \(\frac{1}{2}\) × \(\frac{21}{21}\) = x
\(\frac{44}{42}\) + \(\frac{21}{45}\) = x ⇒ x = \(\frac{65}{42}\)
∴ \(\frac{65}{42}\) is to be subtracted.
Question 25.
Simplify: \(\frac{-2}{5}\) \(\frac{-3}{10}\) + \(\frac{4}{7}\)
Solution:
\(\frac{-2}{5}\)\(\frac{-3}{10}\) + \(\frac{4}{7}\) = \(\frac{-2}{5}\) × \(\frac{2}{2}\)\(\frac{-3}{10}\) + \(\frac{4}{7}\)
= \(\frac{-4}{10}\)\(\frac{-3}{10}\) + \(\frac{4}{7}\) = \(\frac{-7}{10}\) + \(\frac{4}{7}\)
= \(\frac{-7}{10}\) × \(\frac{7}{7}\) + \(\frac{4}{7}\) × \(\frac{10}{10}\) = \(\frac{-49}{70}\) + \(\frac{40}{70}\) = \(\frac{-49+40}{70}\) = \(\frac{-9}{70}\)
Extra Questions of Rational Numbers Class 7 – 5 Marks
Question 26.
Insert 5 rational number between \(\frac{1}{2}\) and \(\frac{7}{4}\).
Solution:
\(\frac{1}{2}\) and \(\frac{7}{4}\)
\(\frac{1}{2}\) × \(\frac{2}{2}\) and \(\frac{7}{4}\)
\(\frac{2}{4 }\) and \(\frac{7}{4}\)
\(\frac{2}{4}\) × \(\frac{2}{2}\) and \(\frac{7}{4}\) × \(\frac{2}{2}\)
\(\frac{4}{8}\) × \(\frac{14}{8}\)
\(\frac{5}{8}\), \(\frac{6}{8}\), \(\frac{7}{8}\), \(\frac{8}{8}\), \(\frac{9}{8}\) are 5 rational number between \(\frac{1}{2}\) and \(\frac{7}{4}\)
Question 27.
What should be added to \(\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{5}\right)\) to get \(\frac{-3}{4}\)?
Solution:
\(\frac{1}{2}\) + \(\frac{1}{3}\) – \(\frac{1}{5}\) = \(\frac{1}{2}\) × \(\frac{15}{15}\) + \(\frac{1}{3}\) × \(\frac{10}{10}\) – \(\frac{1}{5}\) × \(\frac{6}{6}\) = \(\frac{15}{30}\) + \(\frac{10}{30}\) – \(\frac{6}{30}\) = \(\frac{15+10-6}{30}\) = \(\frac{19}{30}\)
According to the problem
\(\frac{19}{30}\) + x = \(\frac{-3}{4}\)
x = \(\frac{19}{30}\) + \(\frac{3}{4}\)
x = \(\frac{19}{30}\) × \(\frac{4}{4}\) + \(\frac{3}{4}\) × \(\frac{30}{30}\)
x = \(\frac{76}{120}\) × \(\frac{90}{120}\)
x = \(\frac{76+90}{120}\) ⇒ x = \(\frac{166}{120}\) × \(\frac{83}{40}\)
∴ \(\frac{83}{40}\) should be added.
Question 28.
Add the sum of \(\left(\frac{1}{2}+\frac{3}{5}\right)\) and the multiplicative inverse \(\left(\frac{7}{4}-\frac{1}{3}\right)\)
Solution:
\(\frac{1}{2}\) + \(\frac{3}{5}\) = \(\frac{1}{2}\) × \(\frac{5}{5}\) + \(\frac{3}{5}\) × \(\frac{2}{2}\)
= \(\frac{5}{10}\) + \(\frac{6}{10}\) = \(\frac{5+6}{10}\) × \(\frac{11}{10}\)
\(\frac{7}{4}\) – \(\frac{1}{3}\) = \(\frac{7}{4}\) × \(\frac{3}{3}\) – \(\frac{1}{3}\) × \(\frac{4}{4}\)
= \(\frac{21}{12}\) – \(\frac{4}{12}\) = \(\frac{21-4}{12}\) × \(\frac{17}{12}\)
Multiplicative inverse of \(\frac{17}{12}\) = \(\frac{12}{17}\)
According to the problem
\(\frac{11}{10}\) + \(\frac{12}{17}\) = \(\frac{187+120}{170}\) = \(\frac{307}{170}\)
Question 29.
Simplify.
i) \(\frac{13}{6}\) × \(\frac{-24}{91}\)
ii) \(\frac{-5}{9}\) × \(\frac{63}{-625}\)
Solution:
Question 30.
The length and breadth of a rectangular garden model all given below in metres. Rakul wants to fense It 3 times. Find the length of wire required in metres.
Solution:
l = 1\(\frac{1}{2}\)m = \(\frac{3}{2}\)m, b = \(\frac{3}{4}\)m
Perimeter = 2(l+b) = \(2\left(\frac{3}{2}+\frac{3}{4}\right)\)
= \(2\left(\frac{6}{4}+\frac{3}{4}\right)\) = 2 × \(\frac{9}{4}\) = \(\frac{9}{2}\)m
Number of times fenced = 3.
∴ Total lenght of wire = 3 × \(\frac{9}{2}\)m
= \(\frac{27}{2}\)m = 13.5m