Tag: Maths RS Aggarwal Solutions Class 8

RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1E

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1E.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1A
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1B
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1C
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1D
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1E
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1F
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1G
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1H

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:
Product of two numbers = – 9
one number = – 12
Let second number = x

Question 5.
Solution:
Product of two rational numbers = \(\\ \frac { -16}{ 9 } \)
One number = \(\\ \frac { -4 }{ 3 } \)

Question 6.
Solution:
Let x be multiplied

Question 7.
Solution:
Let x be multiplied

Question 8.
Solution:
Let required number = x

Question 9.
Solution:
sum of \(\\ \frac { 13 }{ 5 } \) and \(\\ \frac { -12 }{ 7 } \)
= \(\\ \frac { 13 }{ 5 } \) + \(\\ \frac { -12 }{ 7 } \)

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:
(i) No, not always closed under division.
(ii) No, not always commutative.
(iii) No, not always associative.
(iv) No. It is not possible to divide any number by zero.

Hope given RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1E are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3E

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3E.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3A
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3B
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3C
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3D
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3E
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3F
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3G
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3H

Evaluate:

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:

Question 13.
Solution:
Finding the square root of 2509 by division we find that 9 is left as remainder

9 must be subtracted to get the perfect square 100.
Least number to be subtracted = 9

Question 14.
Solution:
Finding the square root of 7581 by division method, we find that 12 is left as remainder.
12 must be subtracted from 7581 to get a perfect square i.e., 7581 – 12 = 7569

(i) The least number to be subtracted = 12
(ii) Perfect square = 7569
(iii) and square root = 87 Ans.

Question 15.
Solution:
Finding the square root of 6203 by division method, we find that 38 is to be added to get a perfect square.
(i) Least number to be added = 38
(ii) Perfect square = 6241
(iii) Square root = 79 Ans.

Question 16.
Solution:
Finding the square root of 8400 by long division method, we find that 64 is to be added to 8400,
We, get 8400 + 64 = 8464

Least number to be added = 64
Perfect square = 8464
Square root = 92 Ans.

Question 17.
Solution:
Least four-digit number = 1000

Finding the square root of 1000 by the division method, we find that 24 must be added to get a perfect square of 4 digits.
Perfect square = 1000 + 24 = 1024 Ans.
square root of 1024 = 32

Question 18.
Solution:
Greatest number of five digits = 99999
Finding the square root of 99999
We get remainder = 143

Required perfect square = 99999 – 143 = 99856
and square root = 316 Ans

Question 19.
Solution:
Area of a square field = 60025 m²
Let its side = a

Hope given RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3E are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3B.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3A
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3B
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3C
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3D
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3E
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3F
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3G
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3H

Question 1.
Solution:
We know that a number ending in 2, 3, 7 or 8 is never a perfect square. So, (i) 5372, (ii) 5963, (iii) 8457, (iv) 9468 cannot be perfect square
Again number ending in an odd number of zeros is also never a perfect square.
Among (v) 360, (vi) 64000, (vii) 2500000 each one has odd zeros at its end. So, there cannot be a perfect square.

Question 2.
Solution:
We know that the square of an even number is also an even number.
(i) 196 (iii) 900, (v) 324 are the squares of even numbers.

Question 3.
Solution:
We know that square.of an odd number is alway is an odd number and square of an even number is always an even number.
Therefore the (ii) 961, (iv) 8649 (v) 4225 are squares of odd numbers.

Question 4.
Solution:
We know that sum of the first n odd natural numbers = n² Therefore.
(i) ∵ It ends with 13
and 1 + 3 + 5 + 7 + 9 + 11 + 13 is the sum of first 7 odd numbers
∵Its sum = (7)² = 49
(ii) 1 + 3 + 5 + 7 + 9+ 11 + 13 + 15 + 17 + 19
Here, n = 10
∵Sum = n² = (10)² = 100
(iii) (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23)
Here, n = 12
Sum = n² = (12)² = 144 Ans.

Question 5.
Solution:
(i) 81 = (9)² = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 (Sum of first 9 odd numbers)
(ii) 100 = (10)² =1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 (Sum of first 10 odd numbers)

Question 6.
Solution:
We know that 2m, m² – 1 and m² + 1 is a Pythagorean triplet where m > 1
(i) One number = 6
∵ 2m = 6 => m = 3
∵ Other members of triplet will be
m² – 1 = (3)² – 1 = 9 – 1 = 8
and m² + 1 = (3)² + 1 = 9 + 1 = 10
∵ Pythagorean triplet is 6, 8, 10
(ii) Let 2m = 14 = m = \(\\ \frac { 14 }{ 2 } \) = 7
m² – 1 = (7)² – 1 = 49 – 1 = 48
and m² + 1 = (7)² + 1 = 49 + 1 = 50
∵Pythagorean triplet = 14, 48, 50
(iii) Let 2m = 16 => m = 8
m² – 1 = (8)² – 1 = 64 – 1 = 63
and m² + 1 = (8)² + 1 = 64 + 1 = 65
∵Pythagorean triplet =16, 63, 65
(iv) Let 2m = 20 => m = 10
∵m² – 1 = (10)² – 1 = 100 – 1 = 99
and m² + 1 = (10)² + 1 = 100 + 1 = 101
∵Pythagorean triplet = 20, 99, 101 Ans.

Question 7.
Solution:
We know that :
(n + 1)² – n² = {(n + 1) + n}
Therefore :
(i) (38)² – (37)² = 38 + 37 = 75
(ii) (75)² – (74)² = 75 + 74 = 149
(iii) (92)² – (91)² = 92 + 91 = 183
(iv) (105)² – (104)² = 105 + 104 = 209
(v) (141)² – (140)² = 141 + 140 = 281
(vi) (218)² – (217)² = 218 + 217 = 435

Question 8.
Solution:
We know that (a + b)² = a² + 2ab + b²
(i) (310)² = (300 + 10)²
= (300)² + 2 x 300 x 10 + (10)²
= 90000 + 6000 + 100 = 96100
(ii) (508)² = (500 + 8)²
= (500)² + 2 x 500 x 8 + (8)²
= 250000 + 8000 + 64 = 258064
(iii) (630)² = (600 + 30)²
= (600)² + 2 x 600 x 30 + (30)²
= 360000 + 36000 + 900 = 396900

Question 9.
Solution:
We know that (a – b)² = a² – 2ab + b²
(i) (196)² = (200 – 4)²
= (200)² – 2 x 200 x 4 + (4)²
= 40000 – 1600 + 16
= 40016 – 1600 = 38416
(ii) (689)² = (700 – 11)²
= (700)² – 2 x 700 x 11 +(11)²
= 490000 – 15400 + 121
= 490121 – 15400 = 474721
(iii) (891)² = (900 – 9)²
= (900)² – 2 x 900 x 9 + (9)²
= 810000 – 16200 + 81
= 810081 – 16200 = 793881

Question 10.
Solution:
Using (a – b) (a + b) = a² – b²
(i) 69 x 71 = (70 – 1) (70 + 1)
= (70)² – (1)² = 4900 – 1
= 4899
(ii) 94 x 106 = (100 – 6) (100 + 6)
= (100)² – (6)²
= 10000 – 36 = 9964

Question 11.
Solution:
Using (a – b) (a + b) – a² – b²
(i) 88 x 92 = (90 – 2) (90 + 2)
= (90)² – (2)²
= 8100 – 4 = 8096
(ii) 78 x 82 = (80 – 2) (80 + 2)
= (80)² – (2)²
= 6400 – 4 = 6396

Question 12.
Solution:
(i) The square of an even number is even
(ii) The square of an odd number is odd
(iii) The square of a proper fraction is less than the given fraction.
(iv) n² = the sum of first n odd natural numbers. Ans.

Question 13.
Solution:
(i) False: No. of digits of a perfect square can be even or odd.
(ii) False: Square of a prime number is not a prime number.
(iii) False: It is not always possible.
(iv) False: It is not always possible.
(v) True: The product of two squares is always a perfect square.

Hope given RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3B are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1D

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1D.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1A
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1B
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1C
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1D
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1E
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1F
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1G
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1H

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:
(i) The product of a rational number and its reciprocal is 1.
(ii) Zero has no reciprocal.
(iii) The numbers 1 and -1 are their own reciprocal.
(iv) Zero is not the reciprocal of any number.
(v) The reciprocal of a, where a≠0, is \(\\ \frac { 1 }{ a } \)
(vi) The reciprocal of \(\\ \frac { 1 }{ a } \) where a≠0 is a
(vii) The reciprocal of a positive rational number is positive.
(viii) The reciprocal of a negative rational number is negative.

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RS Aggarwal Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3A

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3A.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3A
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3B
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3C
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3D
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3E
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3F
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3G
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3H

Question 1.
Solution:
(i) 441
= 3 x 3 x 7 x 7

Question 2.
Solution:
(i) 1225
= 5 x 5 x 7 x 7

Question 3.
Solution:
(i) Factors of 3675
3 x 5 x 5 x 7 x 7

Question 4.
Solution:
(i) 1575
= 3 x 3 x 5 x 5 x 7

Question 5.
Solution:
The largest two digit number = 99

Finding the square root of 99, we get remainder = 18
∴The greatest two digit number which is a perfect square will be = 99 – 18 = 81

Question 6.
Solution:
The largest 3 digit number = 999

Finding the square root of 999, we get remainder = 38
∴The greatest 3 digit number which is a perfect square = 999 – 38 = 961

Hope given RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3A are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 2 Exponents Ex 2C

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 2 Exponents Ex 2C.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 2 Exponents Ex 2A
• RS Aggarwal Solutions Class 8 Chapter 2 Exponents Ex 2B
• RS Aggarwal Solutions Class 8 Chapter 2 Exponents Ex 2C

Objective Questions :
Tick the correct answer in each of the following:

Question 1.
Solution:
Answer = (c)

Question 2.
Solution:
Answer = (d)

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:
Answer = (b)

Question 6.
Solution:

Question 7.
Solution:
Answer = (a)

Question 8.
Solution:
Answer = (a)

Question 9.
Solution:
Answer = (d)

Question 10.
Solution:
Answer = (d)

Question 11.
Solution:

Question 12.
Solution:

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:
3670000 = 3.670000 x 1000000
= 3.67 x 10⁶ (c)

Question 16.
Solution:

Question 17.
Solution:

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RS Aggarwal Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3C

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3C.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3A
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3B
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3C
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3D
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3E
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3F
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3G
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3H

Question 1.
Solution:
(23)² = 529

Question 2.
Solution:
(35)² = 1225

Question 3.
Solution:
(52)² = 2704

Question 4.
Solution:
(96)² = 9216

Find the value of each of the following using the diagonal method :

Question 5.
Solution:
(67)² = 4489 Ans.

Question 6.
Solution:
(86)² = 7396 Ans

Question 7.
Solution:
(137)² = 18769 Ans

Question 8.
Solution:
(256)² = 65536 Ans.

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RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1G

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1G.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1A
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1B
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1C
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1D
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1E
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1F
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1G
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1H

Question 1.
Solution:
Total length of rope = 11 m.
Sum of lengths of two parts

Question 2.
Solution:
Total weight of rice and drum
= \(40\frac { 1 }{ 6 } \) kg
= \(13\frac { 3 }{ 4 } \)

Question 3.
Solution:
Total weight of three types of fruits
= \(19\frac { 1 }{ 3 } \) kg
= \(\\ \frac { 58 }{ 3 } \) kg

Question 4.
Solution:
Total earnings = Rs 160

Question 5.
Solution:
Cost of 1m cloth = Rs \(63\frac { 3 }{ 4 } \)

Question 6.
Solution:
Distance covered in 1 hour
= \(60\frac { 2 }{ 5 } \)
= \(\\ \frac { 302 }{ 5 } \)

Question 7.
Solution:
Length of rectangular park
= \(36\frac { 3 }{ 5 } \) m
= \(\\ \frac { 183 }{ 5 } \)
and breadth = \(16\frac { 2 }{ 3 } \) = \(\\ \frac { 50 }{ 3 } \) m

Question 8.
Solution:
Side of a square plot = \(8\frac { 1 }{ 2 } \) m
= \(\\ \frac { 17 }{ 2 } \) m
Area = (Side)² = Side x Side
= \(\\ \frac { 17 }{ 2 } \) x \(\\ \frac { 17 }{ 2 } \) m²
= \(\\ \frac { 289 }{ 4 } \) m²
= \(72\frac { 1 }{ 4 } \) m²

Question 9.
Solution:
Cost of 1 litre petrol = Rs \(63\frac { 3 }{ 4 } \)
= Rs \(\\ \frac { 255 }{ 4 } \)
Cost of 34 litres of petrol
= \(\\ \frac { 255 }{ 4 } \) x 34
= \(\\ \frac { 255X17 }{ 2 } \)
= \(\\ \frac { 4335 }{ 2 } \)
= Rs \(2167\frac { 1 }{ 2 } \)

Question 10.
Solution:
Distance covered in 1 hour = 1020 km.
Distance covered in \(4\frac { 1 }{ 6 } \) hours
= 1020 x \(4\frac { 1 }{ 6 } \)
= 1020 x \(\\ \frac { 25 }{ 6 } \) km
= \(\\ \frac { 25500 }{ 6 } \)
= 4250 km. Ans.

Question 11.
Solution:
Cost of \(3\frac { 1 }{ 2 }\) metres cloth
= Rs \(166\frac { 1 }{ 4 }\)

Question 12.
Solution:
Total length of piece of chord
= \(71\frac { 1 }{ 2 }\)
No. of pieces = 26

Question 13.
Solution:
Area of a room = \(65\frac { 1 }{ 4 }\) m²
Breadth = \(5\frac { 7 }{ 16 }\) m

Question 14.
Solution:
Product of two fractions = \(9\frac { 3 }{ 5 }\)
= \(\\ \frac { 48 }{ 5 } \)
One fraction = \(9\frac { 3 }{ 7 }\) = \(\\ \frac { 66 }{ 7 } \)

Question 15.
Solution:
Let total number of students =1
and no.of boys = \(\\ \frac { 5 }{ 8 } \)

Question 16.
Solution:
Let no of pages = 1
Then no. of pages read = \(\\ \frac { 7 }{ 9 } \)

Question 17.
Solution:
Total amount, Rita has = Rs 300
Amount spent on notebooks = \(\\ \frac { 1 }{ 3 } \) of 300

Question 18.
Solution:
Total amount earned by Amit = Rs 32000
Amount spent on food = \(\\ \frac { 1 }{ 4 } \) of Rs 32000

Question 19.
Solution:
Let number = 1
Then difference between \(\\ \frac { 3 }{ 5 } \) and \(\\ \frac { 2 }{ 7 } \)

Question 20.
Solution:
Let total number of spectators = 1
No. of spectators in covered place = \(\\ \frac { 2 }{ 7 } \) of
1 = \(\\ \frac { 2 }{ 7 } \)

Hope given RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1G are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 2 Exponents Ex 2A3

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 2 Exponents Ex 2A.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 2 Exponents Ex 2A
• RS Aggarwal Solutions Class 8 Chapter 2 Exponents Ex 2B
• RS Aggarwal Solutions Class 8 Chapter 2 Exponents Ex 2C

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:
Let x be the required number which is multiplied

Question 12.
Solution:
Let x be the required number

Question 13.
Solution:
5^2x+1 ÷ 25 = 125

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RS Aggarwal Class 8 Solutions Chapter 1 Rational Numbers Ex 1F

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1F.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1A
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1B
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1C
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1D
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1E
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1F
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1G
• RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1H

Question 1.
Solution:
Rational number between \(\\ \frac { 1 }{ 4 } \) and \(\\ \frac { 1 }{ 3 } \)
= \(\\ \frac { 1 }{ 2 } \) \(\left[ \frac { 1 }{ 4 } +\frac { 1 }{ 3 } \right] \)
= \(\\ \frac { 1 }{ 2 } \) \(\left[ \frac { 3+4 }{ 12 } \right] \)
= \(\\ \frac { 1 }{ 2 } \) x \(\\ \frac { 7 }{ 12 } \)
= \(\\ \frac { 7 }{ 24 } \)

Question 2.
Solution:
Solution. Rational number between 2 and 3
= \(\\ \frac { 1 }{ 2 } \)(2 + 3)
= \(\\ \frac { 1 }{ 2 } \) x 5
= \(\\ \frac { 5 }{ 2 } \)

Question 3.
Solution:
Rational number between \(\\ \frac { -1 }{ 3 } \) and \(\\ \frac { 1 }{ 2 } \)
= \(\\ \frac { 1 }{ 2 } \) \(\left[ \frac { -1 }{ 3 } +\frac { 1 }{ 2 } \right] \)
= \(\\ \frac { 1 }{ 2 } \) \(\left[ \frac { -2+3 }{ 6 } \right] \)
= \(\\ \frac { 1 }{ 2 } \) x \(\\ \frac { 1 }{ 6 } \)
= \(\\ \frac { 1 }{ 12 } \)

Question 4.
Solution:
First rational number between -3 and -2

Question 5.
Solution:
First rational number between 4 and 5

Question 6.
Solution:
First rational numbers between \(\\ \frac { 2 }{ 3 } \) and \(\\ \frac { 3 }{ 4 } \)

Question 7.
Solution:

Question 8.
Solution:
– 1 = \(\\ \frac { -5 }{ 5 } \) and 2 = \(\\ \frac { 10 }{ 5 } \)
12 rational number between – 1 and 2 can be

Hope given RS Aggarwal Solutions Class 8 Chapter 1 Rational Numbers Ex 1F are helpful to complete your math homework.

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