PSEB 8th Class Maths Solutions Chapter 1 Rational Numbers InText Questions

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 1 Rational Numbers InText Questions and Answers.

PSEB 8th Class Maths Solutions Chapter 1 Rational Numbers InText Questions

Try These : (Textbook Page No.4)

Question 1.
Fill in the blanks in the following table :

Answer:

[Note : Rational numbers are not closed under division.]
e.g., $\frac {2}{3}$ ÷ 0 = ? This is not defined. That’s why our answer in the table is ‘No’.

Try These : (Textbook Page No.6)

Question 1.
Complete the following table:

Answer:

Try These : (Textbook Page No.9)

Question 1.
Complete the following table:

Answer:

Think, Discuss and Write : (Textbook Page No.11)

1. If a property holds for rational numbers, will it also hold for integers ? For whole numbers ? Which will ? Which will not ?
Answer:
( i ) Any property which is true for rational numbers is also true for integers except for any integers ‘a’ and ‘b’ (a ÷ b) is not necessarily an integer.
(ii) All properties which are true for rational numbers are also true for whole numbers also except:

• For ‘a’ and ‘b’ being whole numbers (a – b) may not be a whole number.
• For ‘a’ and ‘b’ being whole numbers (b ≠ 0), a ÷ b may not be a whole number.

Try These : (Textbook Page No.13)

1. Find using distributivity :

Question (i).
$\left\{\frac{7}{5} \times\left(\frac{-3}{12}\right)\right\}+\left\{\frac{7}{5} \times \frac{5}{12}\right\}$
Answer:
$\left\{\frac{7}{5} \times\left(\frac{-3}{12}\right)\right\}+\left\{\frac{7}{5} \times \frac{5}{12}\right\}$
= $=\frac{7}{5} \times\left[\frac{-3}{12}+\frac{5}{12}\right]$
= $\frac{7}{5} \times\left[\frac{-3+5}{12}\right]$
= $\frac{7}{5} \times \frac{2}{12}$
= $\frac{7}{5} \times \frac{1}{6}$
= $\frac {7}{30}$

Question (ii).
$\left\{\frac{9}{16} \times \frac{4}{12}\right\}+\left\{\frac{9}{16} \times \frac{-3}{9}\right\}$
Answer:
$\left\{\frac{9}{16} \times \frac{4}{12}\right\}+\left\{\frac{9}{16} \times \frac{-3}{9}\right\}$
= $\frac{9}{16} \times\left[\frac{4}{12}+\left(\frac{-3}{9}\right)\right]$
= $\frac{9}{16} \times\left[\frac{12+(-12)}{36}\right]$ …..(LCM = 36)
= $\frac{9}{16} \times\left[\frac{0}{36}\right]$
= $\frac{9}{16} \times 0$
= 0.

Try These : (Textbook Page No.17)

1. Write the rational number for each point labelled with a letter:

Question (i).

Answer:

Here, the rational number for-
the point A is $\frac {1}{5}$
the point B is $\frac {4}{5}$
the point C is $\frac {5}{5}$ or 1.
the point D is $\frac {8}{5}$
the point E is $\frac {9}{5}$

Question (ii).

Answer:

Here, the rational number for-
the point F is $\frac {-2}{6}$ or $\frac {-1}{3}$.
the point G is $\frac {-5}{6}$.
the point H is $\frac {-7}{6}$.
the point I is $\frac {-8}{6}$ or $\frac {-4}{3}$.
the point J is $\frac {-11}{6}$.