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NCERT Solutions for Class 8 Maths Chapter 1- Rational Numbers is a resourceful article which was developed by GFG experts to aid students in answering questions they may have as they go through problems from the NCERT textbook.
This chapter contains the following topics:
Class 8 Maths NCERT Solutions Chapter 1 Exercises: |
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(i) -2/3 × 3/5 + 5/2 – 3/5 × 1/6
(ii) 2/5 × (- 3/7) – 1/6 × 3/2 + 1/14 × 2/5
Solution:
(i) -2/3 × 3/5 + 5/2 – 3/5 × 1/6
Given equation: -2/3 × 3/5 + 5/2 – 3/5 × 1/6
By regrouping we get,
= -2/3 × 3/5 - 3/5 × 1/6 + 5/2
= 3/5 (-2/3 - 1/6)+ 5/2 [taking 3/5 as common]
= 3/5 ((-2×2/3×2 -1×1/6×1 )+ 5/2 [by using distributive property]
= 3/5 ((-4-1)/6)+ 5/2
= 3/5 ((–5)/6)+ 5/2
= – 15/30 + 5/2 [Dividing -15 and 30 by 2 we get -1/2]
= – 1/2 + 5/2
= 4/2
= 2
Therefore,
-2/3 × 3/5 + 5/2 – 3/5 × 1/6 = 2
(ii) 2/5 × (- 3/7) – 1/6 × 3/2 + 1/14 × 2/5
Given equation: 2/5 × (- 3/7) – 1/6 × 3/2 + 1/14 × 2/5
By regrouping we get,
= 2/5 × (-3/7) + 1/14 × 2/5 – (1/6 × 3/2)
= 2/5 × (-3/7 + 1/14) – 3/12
= 2/5 × ((-6 + 1)/14) – 3/12 [by using distributive property]
= 2/5 × ((-5)/14)) – 1/4
= (-10/70) - 1/4 [Dividing -10 and 70 by 10 we get -1/7]
= -1/7 - 1/4
= (-4 -7)/28
= -11/28
Therefore,
2/5 × (- 3/7) – 1/6 × 3/2 + 1/14 × 2/5 = -11/28
(i) 2/8
(ii) -5/9
(iii) -6/-5
(iv) 2/-9
(v) 19/-16
Solution:
We know that the additive inverse of x will be -x,
(i) 2/8
Given: 2/8
Additive inverse of 2/8 will be -2/8
(ii) -5/9
Given: -5/9
Additive inverse of -5/9 will be 5/9
(iii) -6/-5
Given: -6/-5
-6/-5 = 6/5 [Dividing both by -1 ]
Additive inverse of 6/5 will be -6/5
(iv) 2/-9
Given: 2/-9
2/-9 = -2/9
Additive inverse of -2/9 will be 2/9
(v)19/-16
Given: 19/-16
19/-16 = -19/16
Additive inverse of -19/16 will be 19/16
(i) x = 11/15
(ii) x = -13/17
Solution:
(i) x = 11/15
Given, x = 11/15
Since, additive inverse of x will be -x
Therefore, the additive inverse of 11/15 will be -11/15 (as 11/15 + (-11/15) = 0)
We can also represent the following as 11/15 = -(-11/15)
Thus, -x = -11/15
-(-x) = -(-11/15) = (11/15) = x
Hence, verified: -(-x) = x
(ii) -13/17
Given, x = -13/17
Since, additive inverse of x will be -x as x + (-x) = 0
Therefore, the additive inverse of -13/17 will be 13/17 as 13/17 + (-13/17) = 0
We can also represent the following as 13/17 = -(-13/17)
Thus, -x = -13/17
-(-x) = -(-13/17) = (13/17) = x
Hence, verified: -(-x) = x
(i) -13
(ii) -13/19
(iii) 1/5
(iv) -5/8 × (-3/7)
(v) -1 × (-2/5)
(vi) -1
Solution:
We know that the multiplicative inverse of x will be 1/x as a × 1/a = 1
(i)-13
Given: -13
The multiplicative inverse of -13 will be -1/13
(ii) -13/19
Given: -13/19
The multiplicative inverse of -13/19 will be -19/13
(iii) 1/5
Given: 1/5
The multiplicative inverse of 1/5 will be 5
(iv)-5/8 × (-3/7)
Given: -5/8 × (-3/7)
-5/8 × (-3/7) = 15/56
The multiplicative inverse of 15/56 will be 56/15
(v)-1 × (-2/5)
Given: -1 × (-2/5)
-1 × (-2/5) = 2/5
The multiplicative inverse of 2/5 will be 5/2
(vi)-1
Given: -1
The multiplicative inverse of -1 will be -1
(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
Given: -4/5 × 1 = 1 × (-4/5) = -4/5
It is representing the property of multiplicative identity.
(ii) -13/17 × (-2/7) = -2/7 × (-13/17)
Given: -13/17 × (-2/7) = -2/7 × (-13/17)
It is representing the property of commutativity.
(iii)-19/29 × 29/-19 = 1
Given: -19/29 × 29/-19 = 1
It is representing the property of multiplicative inverse
Solution:
Given: 6/13 × (Reciprocal of -7/16)
Since, reciprocal of -7/16 = 16/-7 = -16/7
Therefore,
6/13 × (-16/7) = -96/91
Solution:
Given: 1/3 × (6 × 4/3) = (1/3 × 6) × 4/3
Here, the product of their multiplication does not change. Hence, Associativity Property is used in the given equation.
Solution:
Given: -1 1/8 which is equal to -9/8
Since it is the multiplication inverse, therefore the product should be 1.
8/9 × (-9/8) = -1 ≠ 1
Hence, 8/9 is not the multiplication inverse of -1 1/8
Solution:
Give: 3 1/3 = 10/3
Since it is the multiplication inverse, therefore the product should be 1.
0.3 × 10/3 = 3/3 = 1
Hence, 0.3 is the multiplicative inverse of 3 1/3.
(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.
Since, 0 = 0/1
Therefore, the reciprocal of 0 = 1/0, which is not defined.
Hence, the rational number that does not have a reciprocal is 0.
(ii) The rational numbers that are equal to their reciprocals.
Since, 1 = 1/1
Therefore, the reciprocal of 1 = 1/1 = 1
Similarly,
-1 = -1/1
Therefore, the reciprocal of -1 = -1/1 = -1
Hence, the rational numbers that are equal to their reciprocals are 1 and -1
(iii) The rational number that is equal to its negative.
Since negative of 0 = -0 = 0
Therefore, the rational number that is equal to its negative is 0.
(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 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 .
(i) 7/4 (ii) -5/6
Solution:
(i) In number line we have to cover zero to positive integer 1 which signifies the whole no 1, after that we have to divide 1 and 2 into 4 parts and we have to cover 3 places away from 0, which denotes 3/4. And the total of seven places away from 0 represents 7/4. P represents 7/4.
(ii) For representing - 5/6 we have to divide 0 to - 1 integer into 6 parts and we have to go 5 places away from 0 for - 5/6.
Solution:
We have to divide 0 to - 1 integer into 11 parts and the distance of 2, 5, 9 from 0 towards the left of it represents - 2/11, -5/11, -9/11 marked A, B, C, respectively.
Solution:
We can write the number 2 as 6 / 3
Hence, we can write, the five rational numbers which are smaller than 2 are:
1 / 3 , 2 / 3 , 3 / 3 , 4/ 3 , 5 / 3
Solution:
For finding rational numbers between fractions we have to take L. C. M. of their denominators or its multiples. Here L. C. M. Of 5 and 2 is 10 and for finding fractions between them we have to take multiple of 10. Let us take 20 as denominator.
So,
-2 / 5 = (- 2 / 5) × (4 / 4) = -8 / 20
Also,
1 / 2 = (1 / 2) × (10 / 10) = 10 / 20
Hence ten rational numbers between - 2 / 5 to 1 / 2 are same as rational numbers between - 8 / 20 and 10 / 20. And those are as follows
-7 / 20, -6 / 20, -5 / 20, -4 / 20, -3 / 20, -2 / 20, -1 / 20, 0, 1 / 20, 2 / 20
(i) 2/3 and 4/5 (ii) - 3/2 and 5/3 (iii)1/4 and 1/2
Solution:
(i) 2 / 3 and 4 / 5
For finding rational numbers between fractions we have to take L. C. M. of their denominators or its multiples.
Here L. C. M. Of 3 and 5 is 15
And we take the denominators as multiple of 15, as 60
Hence
2 / 3 = ( 2 / 3 ) × ( 20 / 20 ) = 40 / 60
4 / 5 = ( 4 / 5 ) × ( 12 / 12 ) = 48 / 60
Five rational numbers between 2 / 3 and 4 / 5 same as five rational numbers between
40 / 60 and 48 / 60
Therefore, Five rational numbers between 40 / 60 and 48 / 60 are as follows
41 / 60, 42 / 60, 43 / 60, 44 / 60, 45 / 60
(ii) -3 / 2 and 5 / 3
Similarly,
L. C. M. of 2 and 3 is 6.
Here we take denominators same as 6.
-3 / 2 = ( -3 / 2 ) × ( 3 / 3 ) = -9 / 6
5 / 3 = ( 5 / 3 ) × ( 2 / 2 ) = 10 / 6
Hence five rational numbers between -3 / 2 and 5 / 3 are same as five rational numbers between -9 / 6 and 10 / 6 and those are as follows
-8 / 6, -7 / 6, -1, -5 / 6, -4 / 6
(iii) 1 / 4 and 1 / 2
Here L. C. M. of 4 and 2 is 8.
Here we take denominator as multiple of 8 say 32.
Hence
1 / 4 = ( 1 / 4 ) × (8 / 8) = 8 / 32
1 / 2 = ( 1 / 2 ) × ( 16 / 16 ) = 16 / 32
Hence five rational numbers between 1 / 4 and 1 / 2 are same as five rational numbers between 8/32 and 16/32 and those are as follows
9 / 32, 10 / 32, 11 / 32, 12 / 32, 13 / 32
Solution:
We can write -2 as -10 / 5
Hence five rational numbers greater than -2 are as follows
-1 / 5, -2 / 5, -3 / 5, -4 / 5 ,-1
Solution:
L .C. M. of 4 and 5 is 20. For finding rational number between them we should make denominator same or multiple of L .C.M.
Here we take 80.
3 / 5 = ( 3 / 5) × ( 16 / 16 ) = 48 / 80
3 / 4 = ( 3 / 4 ) × ( 20 / 20 ) = 60 / 80
Ten rational numbers between 3 / 5 and 3 / 4 are same as ten rational numbers between 48 / 80 and 60 / 80
Ten rational numbers between 48 / 80 and 60 / 80 are as follows
49 / 80, 50 / 80, 51 / 80, 52 / 80, 54 / 80, 55 / 80, 56 / 80, 57 / 80, 58 / 80, 59 / 80
