rs aggarwal solutions for class 8 maths chapter 1
TRANSCRIPT
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
Exercise 1A Q1.
Q2.
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
Q3.
Q4
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
Q5.
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
Q6.
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
Q7.
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
Q8.
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
Q9.
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
Q10.
1. Every whole number is a rational number.Answer: TrueExplanation: By the definition of rational number, p/q where, q ≠ 0.We know that every whole number can be represented as a/1.From the above two statements we can conclude that every whole number is a rational number.
2. Every integer is a rational number.Answer: TrueExplanation: By the definition of rational number, p/q where, q ≠ 0.We know that every integer can be represented as a/1From the above two statements we can conclude that every integer is a rational number.
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
3. 0 is a whole number but it is not a rational number.Answer: FalseExplanation: By the definition of rational number, p/q where, q ≠ 0.We know that 0 can be represented as 0/1.From the above two statements we can conclude that 0 is a whole number and a rational number.
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
Exercise 1B Q1.
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
Q2.
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
Q3.
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
Exercise 1C Q1.
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
Q2.
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
Q3.
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
Q4.
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
Q5.
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
Q6.
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
Q7.
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
Q8.
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
Q9.
Q10.
Q11.
Q12.
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
Q13.
Q14.
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
1. Solutions:
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
Exercise 1D Q1.
Solution:
(3/5) × (-7/8)
Here by multiplying numerators and denominators
We get (3× -7) / (5 × 8) => -21/40
(ii)
(-9/2) × (5/4)
Here by multiplying numerators and denominators
We get (-9×5) / (2×4) => -45/8
Solution:
(-6/11) × (-5/3)
Here by multiplying numerators and denominators.
We get (-6×-5) / (11×3) = 30/33
further dividing to lowest terms by common divisor i.e. 3 = 30/33 => 10/11
(iv) Solution:
(-2/3) × (6/7)
Here by multiplying numerators and denominators.
We get (-2×6) / (3×7) = -12/21.
Further dividing to lowest terms by common divisor i.e. 3 = -12/21 => -4/7
(v) Solution:
(-12/5) × (10/-3)
Here by multiplying numerators and denominators.
We get (-12×10) / (5×-3) = -120/-15.
Further dividing to lowest terms by common divisor i.e. 3 = -120/-15 => 40/5
Further dividing to lowest terms by common divisor i.e. 5 = 40/5 => 8
(vi) Solution:
(25/-9) × (3/-10)
Here by multiplying numerators and denominators.
We get (25×3) / (-9×-10) = 75/90.
Further dividing to lowest terms by common divisor i.e. 15 = 75/90 => 5/6
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
(vii) Solution:
(5/-18) × (-9/20)
Here by multiplying numerators and denominators.
We get (5×-9) / (-18×20) = -45/-360 = 45/360.
Further dividing to lowest terms by common divisor i.e. 45 = 45/360 => 1/8
(viii) Solution:
(-13/15) × (-25/26)
Here by multiplying numerators and denominators.
We get (-13×-25) / (15×26) = 325/390.
Further dividing to lowest terms by common divisor i.e. 5 = 325/390 = 65/78
Further dividing to lowest terms by common divisor i.e. 13 = 65/78 => 5/6
(ix) Solution:
(16/-21) × (14/5)
Here by multiplying numerators and denominators.
We get (16×14) / (-21×5) = (224 ×-1) / (-105 ×-1) = -224/105.
Further dividing to lowest terms by common divisor i.e. 7 = -224 /105 = -32/15
(x) Solution:
Here by multiplying numerators and denominators.
We get (-7×24) / (6 ×1) = -168/6.
Further dividing to lowest terms by common divisor i.e. 2 = -168/6 = -84/3.
Further dividing to lowest terms by common divisor i.e. 3 = -84/3 => -28
(xi) Solution:
Here by multiplying numerators and denominators.
We get (7×-48) / (24 ×1) = -336/24 Further dividing to lowest terms by common divisor i.e. 6 = -336/24 = -56/4.
Further dividing to lowest terms by common divisor i.e. 6 = -84/6 => -14
(xii) Solution:
Here by multiplying numerators and denominators.
We get (-13×-10) / 5 = 130/5.
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
Further dividing to lowest terms by common divisor i.e. 5 = 130/5 = 26
1. Q2. Solution:
Firstly consider LHS
(3/7) × (-5/9) = (3×-5) / (7×9) = -15/63
Further by dividing -15/63 by using common divisor 3 we get, -5/21
RHS: (-5/9) × (3/7)(-5/9) × (3/7) = (-5×3) / (9×7) = -15/63
Further by dividing -15/63 by using common divisor 3 we get, -5/21
∴ LHS = RHS is verified.
(ii) Solution:
Firstly consider LHS
(-8/7) × (13/9) = (-8×13) / (7×9) = -104/63
RHS: (13/9) × (-8/7)(13/9) × (-8/7) = (13×-8) / (9×7) = -104/63
∴ LHS = RHS is verified.
(iii) Solution:
Firstly consider LHS
(-12/5) × (7/-36) = (-12×7) / (5×-36) = -84/-180 = 84/180
RHS: (7/-36) × (-12/5)(7/-36) × (-12/5) = (7×-12) / (-36×5) = -84/-180 = 84/180
∴ LHS = RHS is verified.
(iv) Solution:
Firstly consider LHS
-8 × (-13/12) = (-8×-13) / 12 = 104/12.
Further by dividing 104/12 by using common divisor 4 we get, 26/3
RHS: (-13/12) × -8(-13/12) × -8 = (-13×-8) / 12 = 104/12
Further by dividing 104/12 by using common divisor 4 we get, 26/3
∴ LHS = RHS is verified.
1. Q3. Solution:
Firstly consider LHS
((5/7) × (12/13)) × (7/18)
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
((5×12) / (7×13)) × (7/18)
(60/91) × (7/18)
((60×7) / (91×18)) = 420/1638
Further by dividing 420/1638 by using common divisor 42 we get, 10/39
RHS: (5/7) × ((12/13) × (7/18))
(5/7) × ((12×7) / (13×18))
(5/7) × (84/234)
(5×84) / (7×234) = 420/1638
Further by dividing 420/1638 by using common divisor 42 we get, 10/39
∴ LHS = RHS is verified.
(ii) Solution:
Firstly consider LHS
(-13/24) × ((-12/5) × (35/36))
(-13/24) × ((-12×35) / (5×36))
(-13/24) × (-420/180)
(-13×-420) / (24×180) = 5460/4320
RHS: ((-13/24) × (-12/5)) × (35/36)
((-13×-12) / (24×5)) × (35/36)
(156/120) × (35/36)
(156×35) / (120×36) = 5460/4320
Further by dividing 5460/4320 by using common divisor 10 we get, 546/432
Further by dividing 546/432 by using common divisor 6 we get, 91/72
∴ LHS = RHS is verified.
(iii) Solution:
Firstly consider LHS
((-9/5) × (-10/3)) × (21/-4)
((-9×-10) / (5×3)) × (21/-4)
(90/15) × (21/-4)
((90×21) / (15×-4)) = (1890/-60) × -1 = -1890/60
Further by dividing -1890/60 by using common divisor 10 we get, -189/6
Further by dividing -189/6 by using common divisor 3 we get, -63/2
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
RHS: (-9/5) × ((-10/3) × (21/-4))
(-9/5) × ((-10×21) / (3×-4))
(-9/5) × (-210/-12)
(-9×-210) / (5×-12) = (1890/-60) × -1 = -1890/60
Further by dividing -1890/60 by using common divisor 10 we get, 189/6
Further by dividing -189/6 by using common divisor 3 we get, -63/2
∴ LHS = RHS is verified.
1. Q4. Solution:
we know the commutative law states that, a × b = b × a
By using the above law
(-23/17) × (18/35) = (18/35) × (-23/17)
(ii) Solution:
we know the commutative law states that, a × b = b × a
By using the above law
-38 × (-7/19) = (-7/19) × (-38)
(iii) Solution:
we know the associative law states that, (a × b) × c = a × (b × c)
By using the above law
((15/7) × (-21/10)) × (-5/6) = (15/7) × ((-21/10) × (-5/6))
(iv) Solution:
we know the associative law states that, (a × b) × c = a × (b × c)
By using the above law
(-12/5) × ((4/15) × (25/-16)) = ((-12/5) × (4/5)) × (25/-16)
1. Q5.We know that the multiplicative inverse of a number (a/b) is (b/a)
(i) Solution:
The multiplicative inverse of (13/25) is (25/13)
(ii) Solution:
The multiplicative inverse of (-17/12) is (12/-17)
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
We put it in the standard form i.e. (12/-17) × -1 we get, -12/17
(iii) Solution:
The multiplicative inverse of (-7/24) is (24/-7)
We put it in the standard form i.e. (24/-7) × -1 we get, -24/7
(iv) Solution:
The multiplicative inverse of 18 is (1/18)
(v) Solution:
The multiplicative inverse of -16 is (1/-16)
We put it in the standard form i.e. (1/-16) × -1 we get, -1/16
(vi) Solution:
The multiplicative inverse of (-3/-5) is (-5/-3)
We put it in the standard form i.e. (-5/-3) × -1 we get, 5/3
(vii) Solution:
The multiplicative inverse of -1 is -1
(viii) Solution:
The multiplicative inverse of 0/2 is 2/0
Since 2/0 is undefined.
The multiplicative inverse of 0/2 is undefined
(ix) Solution:
The multiplicative inverse of (2/-5) is (-5/2)
(x) Solution:
The multiplicative inverse of (-1/8) is (8/-1)
We put it in the standard form i.e. (8/-1) × -1 we get, -8/1 = -8
1.
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
Exercise 1E 1.
i. Solution: Firstly convert ÷ to ×4/9 ÷ -5/12 = 4/9 × 12/-54×12 / 9×-5 = 48/-45By multiplying -1 on both numerator and denominator we get,48/-45 = 48×-1/-45×-1 = -48/45By dividing further by 3 into lowest terms we get,-48/45 = -16/15
ii. Solution: Firstly convert ÷ to ×-8/1 ÷ -7/16 = -8/1 × 16/-7-8×16 / 1×-7 = -128/-7By multiplying -1 on both numerator and denominator we get,-128/-7 = -128×-1/-7×-1 = 128/7
iii. Solution: Firstly convert ÷ to ×-12/7 ÷ (-18/1) = -12/7 × 1/-18-12×1 / 7×-18 = -12/-126By multiplying -1 on both numerator and denominatorwe get,-12/-126 = -12×-1 / -126×-1 = 12/126By dividing further by 6 into lowest terms we get,12/126 = 2/21
iv. Solution: Firstly convert ÷ to ×-1/10 ÷ -8/5 = -1/10 × 5/-8-1×5 / 10×-8 = -5/-80By multiplying -1 on both numerator and denominatorwe get,-5/-80 = -5×-1/-80×-1 = 5/80By dividing further by 5 into lowest terms we get,5/80 = 1/16
v. Solution: Firstly convert ÷ to ×-16/35 ÷ -15/14 = -16/35 × 14/-15-16×14 / 35×-15 = -224/-525By multiplying -1 on both numerator and denominatorwe get,-224/-525 = -224×-1/-525×-1 = 224/525By dividing further by 7 into lowest terms we get,224/525 = 32/75
vi. Solution: Firstly convert ÷ to ×-65/14 ÷ -13/7 = -65/14 × 7/13-65×7 / 14×13 = -455/182By dividing further by 7 into lowest termswe get,-455/182 = -65/26Dividing further by 13 into lowest terms we get,-65/26 = -5/2
2.
vii. Solution: Firstly consider LHS(13/5) ÷ (26/10)Convert ÷ to ×13/5 ÷ 26/10 = 13/5 × 10/2613×10 / 5×26 = 130/130 = 1RHS: (26/10) ÷ (13/5)Convert ÷ to ×26/10 ÷ 13/5 = 26/10 × 5/1326×5 / 10×13 = 130/130 = 1∴ LHS = RHS, The given statement is True.
viii. Solution: Firstly consider LHS-9/1 ÷ 3/4Convert ÷ to ×-9/1 ÷ 3/4 = -9/1 × 4/3-9×4 / 1×3 = -36/3 = -12RHS: 3/4 ÷ (-9)Convert ÷ to ×3/4 ÷ (-9/1) = 3/4 × 1/-93×1 / 4×(-9) = 3/-36 = 1/-12∴ LHS ≠ RHS, The given statement is false.
ix. Solution: Firstly consider LHS-8/9 ÷ -4/3Convert ÷ to ×-8/9 ÷ -4/3 = -8/9 × 3/-4-8×3 / 9×(-4) = -24/-36 = 2/3RHS: -4/3 ÷ -8/9Convert ÷ to ×-4/3 ÷ -8/9 = -4/3 × 9/-8-4×9 / 3×(-8) = -36/-24 = 3/2∴ LHS ≠ RHS, The given statement is false.
x. Solution: Firstly consider LHS-7/24 ÷ 3/-16Convert ÷ to ×-7/24 ÷ 3/-16 = -7/24 × -16/3-7×(-16) / 24×3 = 112/72 = 14/9RHS: 3/-16 ÷ -7/24Convert ÷ to ×3/-16 ÷ -7/24 = 3/-16 × 24/-73×24 / -16×-7 = 72/112 = 9/14∴ LHS ≠ RHS, The given statement is false.
3.
xi. Solution: Firstly consider LHS(5/9 ÷ 1/3) ÷ 5/2Convert ÷ to ×(5/9 × 3/1) ÷ 5/2(5×3 / 9×1) ÷ 5/215/9 ÷ 5/2Convert ÷ to ×15/9 × 2/515×2 / 9×5 = 30/45By dividing further by 15 into lowest terms we get,30/45 = 2/3RHS: 5/9 ÷ (1/3 ÷ 5/2)Convert ÷ to ×5/9 ÷ (1/3 × 2/5)5/9 ÷ (1×2 / 3×5)5/9 ÷ 2/15Convert ÷ to ×5/9 × 15/25×15 / 9×2 = 75/18By dividing further by 3 into lowest terms we get,75/18 = 25/6∴ LHS ≠ RHS, The given statement is false.
xii. Solution: Firstly consider LHS ((-16) ÷ 6/5) ÷ -9/10Convert ÷ to ×(-16/1 × 5/6) ÷ -9/10(-16×5 / 1×6) ÷ -9/10-80/6 ÷ -9/10Convert ÷ to ×-80/6 × 10/-9-80×10 / 6×-9 = -800/-54 = 800/54By dividing further by 2 into lowest terms we get,800/54 = 400/27RHS: (-16) ÷ (6/5 ÷ -9/10)Convert ÷ to ×-16 ÷ (6/5 × 10/-9)-16 ÷ (6×10 / 5×-9)-16 ÷ 60/-45Convert ÷ to ×-16/1 × -45/60-16×-45 / 1×60 = 720/60By
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
dividing further by 60 into lowest terms we get,720/60 = 12∴ LHS ≠ RHS, The given statement is false.
xiii. Solution: Firstly consider LHS(-3/5 ÷ -12/35) ÷ 1/14Convert ÷ to ×(-3/5 × 35/-12) ÷ 1/14(-3×35 / 5×-12) ÷ 1/14-105/-60 ÷ 1/14 = 105/60 ÷ 1/14Convert ÷ to ×105/60 × 14/1105×14 / 60×1 = 1470/60By dividing further by 30 into lowest terms we get,1470/60 = 49/2RHS: -3/5 ÷ (-12/35 ÷ 1/14)Convert ÷ to ×-3/5 ÷ (-12/35 × 14/1)-3/5 ÷ (-12×14 / 35×1)-3/5 ÷ -168/35Convert ÷ to ×-3/5 × 35/-168-3×35 / 5×-168 = -105/-840 = 105/840By dividing further by 5 into lowest terms we get,105/840 = 21/168
dividing further by 3 into lowest terms we get,
21/168 = 7/56 = 1/8
∴ LHS ≠ RHS, The given statement is false.
4.
Solution: We know that the given details,
The product of two rational numbers = -9
One rational number = -12
Let us consider the other number as x
Now, solving
-12 × x = -9
-12x = -9
x = -9/-12 = ¾
∴ the other rational number is ¾.
5.
Solution: We know that the given details,
The product of two rational numbers = -16/9
One rational number = -4/3
Let us consider the other number as x
Now, solving
-4/3 × x = -16/9
x = -16/9 ÷ -4/3
Convert ÷ to ×
x = -16/9 × 3/-4
x = -16×3 / 9×-4
x = -48/-36
By dividing further by 12 into lowest terms we get,
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
48/36 = 4/3
∴ the other rational number is 4/3.
6.
Solution: let us consider x to be multiplied to the number given i.e.
-15/56 × x = -5/7
x = -5/7 ÷ -15/56
Convert ÷ to ×
x = -5/7 × 56/-15
x = -5×56 / 7×-15
x = -280/-105
By dividing further by 35 into lowest terms we get,
280/105 = 8/3
∴ 8/3 has to be multiplied to -15/56 to get -5/7.
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
Exercise 1F 1.
Solution: Let us consider the rational number as x
So to find the rational number between ¼ and 1/3
By using the formula x= ½ (a/b + c/d)
x = ½(1/4 + 1/3)
by taking the LCM for 4 and 3 we get,
x = ½((1×3 + 1×4)/12)
= ½((3+4)/12)
= ½(7/12)
= ½ × 7/12
= 7/24
∴ the rational number between ¼ and 1/3 is 7/24.
2.
Solution: Let us consider the rational number as x
So to find the rational number between 2 and 3
By using the formula x= ½ (a/b + c/d)
x = ½(2 + 3)
x = ½(5)
= 5/2
∴ the rational number between 2 and 3 is 5/2.
3.
Solution: Let us consider the rational number as x
So to find the rational number between -1/3 and 1/2
By using the formula x= ½ (a/b + c/d)
x = ½(-1/3 + 1/2)
by taking the LCM for 3 and 2 we get,
x = ½((-1×2 + 1×3)/6)
= ½((-2+3)/6)
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
= ½(1/6)
= ½ × 1/6
= 1/12
∴ the rational number between -1/3 and 1/2 is 1/12.
4.
Solution: Let us consider the rational number as x
So to find the rational number between -3 and -2
By using the formula x= ½ (a/b + c/d)
x = ½(-3 + (-2))
= ½(-3-2)
= ½(-5) = -5/2
Now we find the rational number between -5/2 and -2. Since -5/2 also lies between -3 and -2.
Let us consider another rational number as y
So let us use the formula again
y= ½(-5/2 + (-2))
= ½ ((-5/2)-2)
= ½ ((-5-4)/2)
= ½ (-9/2)
= ½ × -9/2
= -9/4
∴ Two rational numbers between -3 and 2 is -5/2 and -9/4.
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
Exercise 1G 1.
Solution:
2. we know that the given details areThe length of the rope =11mLength of the 1st piece = 235mLength
of the 2nd piece = 3310mTotal length of the pieces = length of 1st piece + length of
2nd piece= 235 + 3310=(5×2+3)/5 + (10×3+3)/10=13/5 + 33/10
By taking the LCM for 5 and 10 is 10
= (13×2 + 33×1)/10
= (26+33)/10
=59/10m
Length of the remaining rope = length of the rope – total length of the pieces
= 11m – 59/10m
By taking the LCM
= (11×10 -59×1)/10
= (110-59)/10
=51/10
= 5110m
∴ The length of the remaining rope is 5110m.
2.
3. we know that the given details areA drum full of rice weighs = 4016kgEmpty drum weighs
= 1334kgWeight of rice in the drum = drum full of rice – weight of empty drum= 4016 – 13 ¾= 241/6 –
55/4By taking the LCM for 6 and 4 is 12= (241×2 – 55×3)/12
= (482 – 165)/12
= 317/12
= 26512kg
∴ The weight of rice in the drum =26512kg.
3.
Solution:
4. we know that the given details areWeight of three types of fruits = 1913kgWeight of apples
= 819kgWeight of oranges = 316kgWeight of pears = Weight of three types of fruits – (Weight of
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
apples + Weight of oranges)= 1913 –
(819 + 316)= 58/3 – (73/9 + 19/6)=
58/3 – (73×2 + 19×3)/18 (by taking LCM for 9 and 6 is 18)
= 58/3 – (146 + 57)/18
= 58/3 – 203/18
By taking LCM for 3 and 18 is 18
= (58×6 – 203×1)/18
= (348 – 203)/18
= 145/18
= 8118kg
∴ The weight of pears is 8118kg.
4.
Solution:
5. we know that the given details areTotal earnings = ₹160Earnings spent on tea and snacks =
₹2635Earnings spent on food = ₹5012Earnings spent on repairs = ₹1625∴ Total expenditure =
Earnings spent on tea and snacks + Earnings spent on food + Earnings spent on repairs= ₹2635 +
₹50 ½ + ₹1625= 133/5 + 101/2 + 82/5
By taking LCM for 5 and 2 is 10
= (133×2 + 101×5 + 82×2)/10
= (266 + 505 +164)/10
= 935/10
Total Savings = Total Earnings – Total Expenditure
= ₹160 – ₹935/10
By taking 10 as the LCM
= (160×10 – 935×1)/10
= (1600 – 935)/10
= 665/10 = 66 ½
∴ Total Savings on that day is ₹66 ½.
5.
Solution:
6. we know that the given details areCost of the cloth per meter = ₹66 ¾Total meters = 325metersTotal
cost of cloth = Cost of the cloth per meter × Total meters= ₹66 ¾ × 325= 267/4 × 17/5=
4335/20Further we can divide by 5 we get,
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
4335/20 = 867/4 = 216 ¾
∴ Cost of the cloth is ₹216 ¾.
6.
Solution:
7. we know that the given details areSpeed of the car = 6025 km/hr.Total hours = 6 ¼ hoursBy using
the formula, speed = distance/timeDistance of the car = speed of the car × time taken= 6025× 6
¼=302/5 × 25/4= 7550/20
= 755/2 = 377 ½
∴ Distance covered = 377 ½ km.
7.
Solution:
8. we know that the given details areLength of the park = 3635mBreadth of the park = 1623mBy using
the formula, area of rectangle = length × breadthArea of rectangular park = 3635 × 1623= 183/5m ×
50/3m= 9150/15m2 = 610m2∴ Area of the park = 610m2
8.
Solution:
9. we know that the given details areOne side measures = 812mBy using the formula,Area of square =
side × sideArea of square plot = 8 ½m × 8 ½m= 17/2m × 17/2m= 289/4m2 = 72 ¼ m2∴ Area of square plot is 72 ¼ m2
9.
Solution:
10. we know that the given details areCost of 1 liter petrol = ₹63 ¾Cost of 34 liters petrol = 34 × cost of 1 liter petrol= 34 × 63 ¾= 34 × 255/4= 8670/4We can further divide by 2 we get,= 4335/2
= 2167 ½
∴ Cost of 34 liters petrol is ₹2167 ½
10.
Solution:
11. we know that the given details areDistance covered in 1 hour = 1020kmDistance covered in 416 hour
= 416 × distance covered in 1 hour= 4 1/6 × 1020= 25/6 × 1020=25500/6= 4250km∴ Distance
covered in 416hours = 4250km.
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
Exercise 1H 1.
Solution:
By solving (-5/16 + 7/12) we get,
(-5/16 + 7/12)
By taking LCM for 16 and 12 is 48
(-5/16 + 7/12) = (-5×3 + 7×4)/48= (-15 + 28)/48= 13/48
1. 2.
Solution:
By solving (-5/16 + 7/12) we get,(8/-15 + 4/-3)
Firstly we have multiply both numerator and denominator by -1
(8×-1/-15×-1 + 4×-1/-3×-1) = (-8/15 + -4/3)
By taking LCM for 15 and 3 is 15
(-8/15 + -4/3) = (-8×1 + -4×5)/15= (-8 + (-20))/15= (-8-20)/15= -28/15
1. 3.
Solution:
By solving (7/-26 + 16/39) we get,(7/-26 + 16/39)
Firstly we have multiply both numerator and denominator by -1
(7×-1/-26×-1 + 16/39) = (-7/26 + 16/39)
By taking LCM for 26 and 39 is 78
(-7/26 + 16/39) = (-7×3 + 16×2)/78= (-21 + 32)/78= 11/78
1. 4.
Solution:
Firstly we have multiply both numerator and denominator by -1
(3/1 + 5/-7) = (3/1 + 5×-1/-7×-1)= (3/1+ (-5/7))
By taking LCM for 1 and 7 is 7
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
(3/1+ (-5/7)) = (3×7 + (-5×1))/7= (21 + (-5))/7= (21 – 5)/7= 16/7
1. 5.
Solution:
By solving (31/-4 + -5/8) we get,
Firstly we have multiply both numerator and denominator by -1
(31/-4 + -5/8) = (31×-1/-4×-1 + -5/8)= (-31/4 + -5/8)
By taking LCM for 4 and 8 is 8
(-31/4 + -5/8) = (-31×2 + -5×1)/8= (-62 + (-5))/8= (-62-5)/8= -67/8
1. 6.
Solution:
Let us consider x as a number to be added
7/12 + x = -4/15x = -4/15 – 7/12
By taking LCM for 15 and 12 is 60
x = (-4×4 – 7× 5)/60= (-16 – 35)/60= -51/60
We can further divide by 3 we get,-51/60 = -17/20
1. 7.
Solution:
we solve for,(2/3 + (-4/5) + 7/15 + (-11/20))
By taking LCM for 3, 5, 15 and 20 is 60
(2×20 + (-4×12) + 7×4 + (-11×3))/60 = (40 + (-48) + 28 + (-33))/60= (40 – 48 +28 – 33)/60= (68-81)/60= -13/60
1. 8.
Solution:
let us consider one of the number as x-5 + x = -4/3x = -4/3 + 5/1
By taking LCM for 3 and 1 is 3 we get,
x = (-4×1 + 5×3)/3= (-4+15)/3= 11/3
1. 9.
Solution:
let us consider one of the number as
RS Aggarwal Solutions for Class 8 Maths Chapter 1-
Rational Numbers
x-5/7 + x = -2/3x = -2/3 – (-5/7)= -2/3 + 5/7
By taking LCM for 3 and 7 is 21 we get,
x = (-2×7 + 5×3)/21= (-14+15)/21= 1/21
1. 10.
Solution:
let us consider one of the number as x-5/3 –x = 5/6-5/3 – 5/6 = xx = -5/3 – 5/6
By taking LCM for 3 and 6 is 6 we get,x = (-5×2 – 5×1)/6= (-10-5)/6= -15/6
Further we can divide by 3 we get,-15/6 = -5/2
1. 11.
Solution:
We know that for any real number a≠0 then, a-1= 1/a
∴ By using the formula, (-3/7)-1 = 1/ (-3/7)= 7/-3
By multiplying both numerator and denominator by -1 we get,
7/-3 = (7×-1/-3×-1) = -7/3