Class 8 Maths Chapter 1 Rational Numbers
1. Using appropriate properties find.
(i) -2/3 × 3/5 + 5/2 – 3/5 × 1/6
Solution:
-2/3 × 3/5 + 5/2 – 3/5 × 1/6
= -2/3 × 3/5– 3/5 × 1/6+ 5/2 (by commutativity)
= 3/5 (-2/3 – 1/6)+ 5/2
= 3/5 ((- 4 – 1)/6)+ 5/2
= 3/5 ((–5)/6)+ 5/2 (by distributivity)
= – 15 /30 + 5/2
= – 1 /2 + 5/2
= 4/2
= 2
(ii) 2/5 × (- 3/7) – 1/6 × 3/2 + 1/14 × 2/5
Solution:
2/5 × (- 3/7) – 1/6 × 3/2 + 1/14 × 2/5
= 2/5 × (- 3/7) + 1/14 × 2/5 – (1/6 × 3/2) (by commutativity)
= 2/5 × (- 3/7 + 1/14) – 3/12
= 2/5 × ((- 6 + 1)/14) – 3/12
= 2/5 × ((- 5)/14)) – 1/4
= (-10/70) – 1/4
= – 1/7 – 1/4
= (– 4– 7)/28
= – 11/28
2. Write the additive inverse of each of the following
Solution:
(i) 2/8
Additive inverse of 2/8 is – 2/8
(ii) -5/9
Additive inverse of -5/9 is 5/9
(iii) -6/-5 = 6/5
Additive inverse of 6/5 is -6/5
(iv) 2/-9 = -2/9
Additive inverse of -2/9 is 2/9
(v) 19/-16 = -19/16
Additive inverse of -19/16 is 19/16
3. Verify that: -(-x) = x for.
(i) x = 11/15
(ii) x = -13/17
Solution:
(i) x = 11/15
We have, x = 11/15
The additive inverse of x is – x (as x + (-x) = 0)
Then, the additive inverse of 11/15 is – 11/15 (as 11/15 + (-11/15) = 0)
The same equality 11/15 + (-11/15) = 0, shows that the additive inverse of -11/15 is 11/15.
Or, – (-11/15) = 11/15
i.e., -(-x) = x
(ii) -13/17
We have, x = -13/17
The additive inverse of x is – x (as x + (-x) = 0)
Then, the additive inverse of -13/17 is 13/17 (as 13/17 + (-13/17) = 0)
The same equality (-13/17 + 13/17) = 0, shows that the additive inverse of 13/17 is -13/17.
Or, – (13/17) = -13/17,
i.e., -(-x) = x
4. Find the multiplicative inverse of the
(i) -13 (ii) -13/19 (iii) 1/5 (iv) -5/8 × (-3/7) (v) -1 × (-2/5) (vi) -1
Solution:
(i) -13
Multiplicative inverse of -13 is -1/13
(ii) -13/19
Multiplicative inverse of -13/19 is -19/13
(iii) 1/5
Multiplicative inverse of 1/5 is 5
(iv) -5/8 × (-3/7) = 15/56
Multiplicative inverse of 15/56 is 56/15
(v) -1 × (-2/5) = 2/5
Multiplicative inverse of 2/5 is 5/2
(vi) -1
Multiplicative inverse of -1 is -1
5. Name the property under multiplication used in each of the following.
(i) -4/5 × 1 = 1 × (-4/5) = -4/5
(ii) -13/17 × (-2/7) = -2/7 × (-13/17)
(iii) -19/29 × 29/-19 = 1
Solution:
(i) -4/5 × 1 = 1 × (-4/5) = -4/5
Here 1 is the multiplicative identity.
(ii) -13/17 × (-2/7) = -2/7 × (-13/17)
The property of commutativity is used in the equation
(iii) -19/29 × 29/-19 = 1
Multiplicative inverse is the property used in this equation.
6. Multiply 6/13 by the reciprocal of -7/16
Solution:
Reciprocal of -7/16 = 16/-7 = -16/7
According to the question,
6/13 × (Reciprocal of -7/16)
6/13 × (-16/7) = -96/91
7. Tell what property allows you to compute 1/3 × (6 × 4/3) as (1/3 × 6) × 4/3
Solution:
1/3 × (6 × 4/3) = (1/3 × 6) × 4/3
Here, the way in which factors are grouped in a multiplication problem, supposedly, does not change the product. Hence, the Associativity Property is used here.
10. Write
(i) The rational number that does not have a reciprocal.
(ii) The rational numbers that are equal to their reciprocals.
(iii) The rational number that is equal to its negative.
Solution:
(i)The rational number that does not have a reciprocal is 0. Reason:
0 = 0/1
Reciprocal of 0 = 1/0, which is not defined.
(ii) The rational numbers that are equal to their reciprocals are 1 and -1.
Reason:
1 = 1/1
Reciprocal of 1 = 1/1 = 1 Similarly, Reciprocal of -1 = – 1
(iii) The rational number that is equal to its negative is 0.
Reason:
Negative of 0=-0=0
11. Fill in the blanks.
(i) Zero has _______reciprocal.
(ii) The numbers ______and _______are their own reciprocals
(iii) The reciprocal of – 5 is ________.
(iv) Reciprocal of 1/x, where x ≠ 0 is _________.
(v) The product of two rational numbers is always a ________.
(vi) The reciprocal of a positive rational number is _________.
Solution:
(i) Zero has no reciprocal.
(ii) The numbers -1 and 1 are their own reciprocals
(iii) The reciprocal of – 5 is -1/5.
(iv) Reciprocal of 1/x, where x ≠ 0 is x.
(v) The product of two rational numbers is always a rational number.
(vi) The reciprocal of a positive rational number is positive.