These Class 8 Maths Extra Questions Chapter 1 Rational Numbers will help students prepare well for the exams.
Class 8 Maths Chapter 1 Extra Questions Rational Numbers
Class 8 Maths Rational Numbers Extra Questions
Question 1.
The following relation is used to illustrate the property of multiplicative identity. A and B are numbers.
\(\frac{1}{4}\) × A = B.
What is the value of A and B ?
Solution:
Multiplicative identity property :
For a ∈ R, a × 1 = 1 × a = a
∴ \(\frac{1}{4}\) × 1 = \(\frac{1}{4}\)
⇒ A = 1 and B = \(\frac{1}{4}\)
Question 2.
Observe the number ine below. It is given that GH = HI = IJ = JK
What is the rational number represented by point J?
Solution:
Every unit divided into four parts.
∴ J = \(\frac{-7}{4}\) = -1\(\frac{3}{4}\)
Question 3.
Which of the two numbers is greater? \(\frac{-4}{-5}\) or \(\frac{-5}{8}\)
Solution:
\(\frac{-4}{-5}\) = \(\frac{4}{5}\)
∴ \(\frac{4}{5}\) > \(\frac{-5}{8}\)
so \(\frac{-4}{5}\) is greater.
Question 4.
Use suitable properties of operations and simplify :
\(\frac{2}{9}\) × (-\(\frac{5}{12}\)) – \(\frac{5}{12}\) × \(\frac{4}{9}\)
Solution:
Question 5.
Multiply \(\frac{7}{18}\) to the reciprocal of (\(\frac{-5}{13}\))
Solution:
The reciprocal of (\(\frac{-5}{13}\)) = \(\frac{-13}{5}\)
∴ \(\frac{7}{18}\) × (\(\frac{-13}{5}\)) = \(\frac{-91}{90}\)
Question 6.
Write the following rational numbers in ascending order.
\(\frac{-5}{8}\), \(\frac{7}{-10}\), \(\frac{-3}{5}\)
Solution:
Question 7.
Name the property under multiplication used in the following.
i) \(\frac{-2}{3}\) × 1 = 1 × \(\frac{-2}{3}\) = \(\frac{-2}{3}\)
ii) \(\frac{14}{3}\) × \(\frac{4}{7}\) = \(\frac{4}{7}\) × \(\frac{14}{3}\)
Solution:
i) \(\frac{-2}{3}\) × 1 = 1 × \(\frac{-2}{3}\) = \(\frac{-2}{3}\)
Multiplicative Identity
ii) \(\frac{14}{3}\) × \(\frac{4}{7}\) = \(\frac{4}{7}\) × \(\frac{14}{3}\)
Commutative
Rational Numbers Extra Questions Class 8
Question 1.
Find a rational number between \(\frac{3}{7}\) and \(\frac{4}{8}\).
Solution:
\(\frac{3}{7}\) = \(\frac{3 \times 8}{7 \times 8}\) = \(\frac{24}{56}\)
\(\frac{4}{8}\) = \(\frac{4 \times 7}{8 \times 7}\) = \(\frac{28}{56}\)
∴ The rational numbers between \(\frac{3}{7}\) and \(\frac{4}{8}\) = \(\frac{25}{56}\), \(\frac{26}{56}\), \(\frac{27}{56}\)
Extra Questions of Rational Numbers Class 8
Question 1.
Arrange the following rational numbers in the ascending order :
\(\frac{5}{4}\), \(\frac{2}{3}\), \(\frac{6}{8}\), \(\frac{8}{5}\), \(\frac{9}{2}\).
Solution:
Rational numbers = \(\frac{5}{4}\), \(\frac{2}{3}\), \(\frac{6}{8}\), \(\frac{8}{5}\), \(\frac{9}{2}\)
L.C.M of the denominators 4, 3, 8, 5, 2 is 120.
Question 2.
Using appropriate properties solve the given below.
\(\frac{2}{5}\) × \(\frac{-3}{7}\) – \(\frac{1}{14}\) – \(\frac{3}{7}\) × \(\frac{3}{5}\)
Solution:
\(\frac{2}{5}\) × \(\frac{-3}{7}\) – \(\frac{1}{14}\) – \(\frac{3}{7}\) × \(\frac{3}{5}\)
= \(\frac{2}{5}\) × \(\frac{-3}{7}\) – \(\frac{3}{7}\) × \(\frac{3}{5}\) – \(\frac{1}{14}\) (By commutativity)
= \(\frac{2}{5}\) × \(\frac{-3}{7}\) + (\(\frac{-3}{7}\)) × \(\frac{3}{5}\) – \(\frac{1}{14}\)
= \(\frac{-3}{7}\)(\(\frac{2}{5}\) + \(\frac{3}{5}\)) – \(\frac{1}{14}\) (By Distributivitity)
= \(\frac{-3}{7}\) × 1 – \(\frac{1}{14}\) (By Multiplicative Identity)
= \(\frac{- 6 – 1}{14}\) = \(\frac{-1}{2}\)
Rational Numbers Class 8 Extra Questions
Question 1.
Find \(\frac{3}{7}\) + (\(\frac{-6}{11}\)) + (\(\frac{-8}{21}\)) + (\(\frac{5}{22}\)).
Solution:
\(\frac{3}{7}\) + (\(\frac{-6}{11}\)) + (\(\frac{-8}{21}\)) + (\(\frac{5}{22}\)).
Do you think the properties of commu¬tativity and associativity made the calculations easier ?
Question 2.
Find \(\frac{-4}{5}\) × \(\frac{3}{7}\) \(\frac{15}{16}\) × (\(\frac{-14}{9}\)).
Solution:
We have
\(\frac{-4}{5}\) × \(\frac{3}{7}\) \(\frac{15}{16}\) × (\(\frac{-14}{9}\))
= (-\(\frac{4 \times 3}{5 \times 7}\)) × (\(\frac{15 \times(-14)}{16 \times 9}\))
ADDITIVE IDENTITY : Zero is called the identity of addition of Rational numbers / Integers / Whole numbers.
It means by zero to a certain number, we get the previous number is Result.
0 + (Previous) number = (Same) number (Result)
0 + 6 = 6, 0 + (-8) = -8
0 + \(\frac{2}{3}\) = \(\frac{2}{3}\) (Whole, Integer, Rational respectively)
a + 0 = a
Thus multiplying a number by 1 (one) we get same number as result.
(a × 1) = a where ‘a’ is Whole / Integer / Rational
for example :
0 × 1 = 0 (‘0’ is whole number)
– 7 × 1 = – 7 (-7 is Integer)
\(\frac{-2}{3}\) × 1 = \(\frac{-2}{3}\)
(\(\frac{-2}{3}\) is a Rational number)
Hence 1 is called Multiplicative Identity for Rational numbers.
Properties Check list for all numbers
Question 3.
Write the additive inverse of the following :
i) \(\frac{-7}{19}\)
Solution:
\(\frac{7}{19}\) is the additive inverse of \(\frac{-7}{19}\) because
\(\frac{-7}{19}\) + \(\frac{7}{19}\) = \(\frac{- 7 + 7}{19}\) = \(\frac{9}{19}\) = 0
i) \(\frac{21}{112}\)
Solution:
The additive inverse of \(\frac{21}{112}\) is \(\frac{-21}{112}\) (Check!)
Question 4.
Verify that – (- x) is the same as x for
i) x = \(\frac{13}{17}\)
The additive inverse of x = \(\frac{13}{17}\) is – x
= \(\frac{-13}{17}\) since \(\frac{13}{17}\) + (\(\frac{-13}{17}\)) = 0
The same equality \(\frac{13}{17}\) + (\(\frac{-13}{17}\)) = 0
shows that the additive inverse of \(\frac{-13}{17}\)
is \(\frac{13}{17}\) or – (\(\frac{-13}{17}\)) = \(\frac{13}{17}\) i.e., – (- x) = x.
ii) x = \(\frac{-21}{31}\)
Solution:
Additive inverse of x = \(\frac{-21}{31}\) is – x= \(\frac{21}{31}\)
since \(\frac{-21}{31}\) + \(\frac{21}{31}\) = 0
The same equality \(\frac{-21}{31}\) + \(\frac{21}{31}\) = 0,
shows that the additive inverse of \(\frac{21}{31}\) is \(\frac{-21}{31}\), i.e., – (- x) = x.
Question 5.
Find \(\frac{2}{5}\) × \(\frac{-3}{7}\) – \(\frac{1}{14}\) – \(\frac{3}{7}\) × \(\frac{3}{5}\)
Solution:
Question 6.
Write any 3 rational numbers between -2 and 0.
Solution:
-2 can be written as \(\frac{-20}{10}\) and 0 as \(\frac{0}{10}\).
Thus we have \(\frac{-19}{10}\), \(\frac{-18}{10}\), \(\frac{-17}{10}\), \(\frac{-16}{10}\), \(\frac{-15}{10}\), ….. , \(\frac{-1}{10}\) between -2 and 0
You can take any three of these.
Question 7.
Find any ten rational numbers between \(\frac{-5}{6}\) and \(\frac{5}{8}\).
Solution:
We first convert \(\frac{-5}{6}\) and \(\frac{5}{8}\) to rational numbers with the same denominators.
\(\frac{-5 \times 4}{6 \times 4}\) = \(\frac{-20}{24}\) and \(\frac{5 \times 3}{8 \times 3}\) = \(\frac{15}{24}\)
Thus we have \(\frac{-19}{24}\), \(\frac{-18}{24}\), \(\frac{-17}{24}\), …….., \(\frac{14}{24}\) as the rational numbers between \(\frac{-20}{24}\) and \(\frac{15}{24}\).
You can take any ten of these.
Question 8.
Find a rational number between \(\frac{1}{4}\) and \(\frac{1}{2}\).
Solution:
We find the mean of the given rational numbers.
(\(\frac{1}{4}\) + \(\frac{1}{2}\)) ÷ 2 = (\(\frac{1 + 2}{4}\)) ÷ 2 = \(\frac{3}{4}\) × \(\frac{1}{2}\) = \(\frac{3}{8}\)
\(\frac{3}{8}\) lies between \(\frac{1}{4}\) and \(\frac{1}{2}\).
This can be seen on the number line also.
We find the mid point of AB which is C, represented by (\(\frac{1}{4}\) + \(\frac{1}{2}\)) ÷ 2 = \(\frac{3}{8}\)
We find that \(\frac{1}{4}\) < \(\frac{3}{8}\) < \(\frac{1}{2}\).
If a and b are two rational numbers, then \(\frac{a+b}{2}\) is a rational number between a and
b such that a < \(\frac{a+b}{2}\) < b.
This again shows that there are countless number of rational numbers between any two given rational numbers.
Question 9.
find three rational numbers between \(\frac{1}{4}\) and \(\frac{1}{2}\).
Solution:
We find the mean of the given rational numbers.
As given in the Jabove example, the mean is \(\frac{3}{8}\) and \(\frac{1}{4}\) < \(\frac{3}{8}\) < \(\frac{1}{2}\).
We now find another rational number between \(\frac{1}{4}\) and \(\frac{3}{8}\).
For this, we again find the mean of \(\frac{1}{4}\) and \(\frac{3}{8}\). That is,
Thus, \(\frac{5}{16}\), \(\frac{3}{8}\), \(\frac{7}{16}\) are the three rational numbers between \(\frac{1}{4}\) and \(\frac{1}{2}\).
This can clearly be shown on the number line as follows :
