RS Aggarwal Solutions for Class 8 Maths Chapter 1 ...

RS Aggarwal Solutions for Class 8 Maths Chapter 1-

Rational Numbers

Exercise 1A Q1.

Q2.

https://byjus.com/?utm_source=pdf-click



Rational Numbers

Q3.

Q4




Rational Numbers

Q5.




Rational Numbers




Rational Numbers

Q6.




Rational Numbers




Rational Numbers

Q7.




Rational Numbers




Rational Numbers

Q8.




Rational Numbers




Rational Numbers

Q9.




Rational Numbers




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.




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.




Rational Numbers

Exercise 1B Q1.




Rational Numbers




Rational Numbers

Q2.




Rational Numbers




Rational Numbers

Q3.




Rational Numbers




Rational Numbers

Exercise 1C Q1.




Rational Numbers




Rational Numbers

Q2.




Rational Numbers




Rational Numbers

Q3.




Rational Numbers




Rational Numbers

Q4.




Rational Numbers




Rational Numbers

Q5.




Rational Numbers

Q6.




Rational Numbers




Rational Numbers

Q7.




Rational Numbers




Rational Numbers

Q8.




Rational Numbers




Rational Numbers

Q9.

Q10.

Q11.

Q12.




Rational Numbers

Q13.

Q14.




Rational Numbers

1. Solutions:




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)


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)


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)


We get (25×3) / (-9×-10) = 75/90.

Further dividing to lowest terms by common divisor i.e. 15 = 75/90 => 5/6




Rational Numbers

(vii) Solution:

(5/-18) × (-9/20)


We get (5×-9) / (-18×20) = -45/-360 = 45/360.


(viii) Solution:

(-13/15) × (-25/26)


We get (-13×-25) / (15×26) = 325/390.

Further dividing to lowest terms by common divisor i.e. 5 = 325/390 = 65/78


(ix) Solution:

(16/-21) × (14/5)


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:


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:


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:


We get (-13×-10) / 5 = 130/5.




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


∴ LHS = RHS is verified.

(ii) Solution:


(-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


(iii) Solution:


(-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


(iv) Solution:


-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



1. Q3. Solution:


((5/7) × (12/13)) × (7/18)




Rational Numbers

((5×12) / (7×13)) × (7/18)

(60/91) × (7/18)

((60×7) / (91×18)) = 420/1638


RHS: (5/7) × ((12/13) × (7/18))

(5/7) × ((12×7) / (13×18))

(5/7) × (84/234)

(5×84) / (7×234) = 420/1638



(ii) Solution:


(-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




(iii) Solution:


((-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






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



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


-38 × (-7/19) = (-7/19) × (-38)

(iii) Solution:

we know the associative law states that, (a × b) × c = a × (b × c)


((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)


(-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)




Rational Numbers

We put it in the standard form i.e. (12/-17) × -1 we get, -12/17

(iii) Solution:



(iv) Solution:

The multiplicative inverse of 18 is (1/18)

(v) Solution:

The multiplicative inverse of -16 is (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:


We put it in the standard form i.e. (8/-1) × -1 we get, -8/1 = -8

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




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,




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.




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.


So to find the rational number between 2 and 3


x = ½(2 + 3)

x = ½(5)

= 5/2

∴ the rational number between 2 and 3 is 5/2.

3.


So to find the rational number between -1/3 and 1/2


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)




Rational Numbers

= ½(1/6)

= ½ × 1/6

= 1/12

∴ the rational number between -1/3 and 1/2 is 1/12.

4.


So to find the rational number between -3 and -2


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.




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




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


= (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,




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.




Rational Numbers

Exercise 1H 1.

Solution:

By solving (-5/16 + 7/12) we get,

(-5/16 + 7/12)


(-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)


(-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)


(7×-1/-26×-1 + 16/39) = (-7/26 + 16/39)


(-7/26 + 16/39) = (-7×3 + 16×2)/78= (-21 + 32)/78= 11/78

1. 4.

Solution:


(3/1 + 5/-7) = (3/1 + 5×-1/-7×-1)= (3/1+ (-5/7))





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,


(31/-4 + -5/8) = (31×-1/-4×-1 + -5/8)= (-31/4 + -5/8)


(-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


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




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



RS Aggarwal Solutions for Class 8 Maths Chapter 1 ...

Documents

Transcript of RS Aggarwal Solutions for Class 8 Maths Chapter 1 ...