## RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 8 Linear Equations Ex 8B.

**Other Exercises**

- RS Aggarwal Solutions Class 8 Chapter 8 Linear Equations Ex 8A
- RS Aggarwal Solutions Class 8 Chapter 8 Linear Equations Ex 8B
- RS Aggarwal Solutions Class 8 Chapter 8 Linear Equations Ex 8C

**Question 1.**

**Solution:**

The numbers be 8x and 3x (∵ Ratio is 8 : 3)

Sum = 143

According to the condition

8x + 3x = 143

=> 11x = 143

=> x = \(\\ \frac { 143 }{ 11 } \) = 13

First number = 8x = 8 x 13 = 104

and second number = 3x = 3 x 13 = 39 Ans.

**Question 2.**

**Solution:**

Let the number = x

According to the condition,

x – \(\frac { 2 }{ 3 } x\) = 20

=> \(\\ \frac { 3x-2x }{ 3 } \) = 20

=> \(\\ \frac { x }{ 3 } \) = 20

=> x = 20 x 3 = 60

Hence original number = 60 Ans.

You can also Download NCERT Solutions for Class 8 Maths to help you to revise complete Syllabus and score more marks in your examinations.

**Question 3.**

**Solution:**

Let the number = x

According to the condition

\(\frac { 4 }{ 5 } x-\frac { 2 }{ 3 } x\) = 10

=> \(\\ \frac { 12x-10x }{ 15 } \) = 10

=> \(\\ \frac { 2x }{ 15 } \) = 10

=> 2x = 10 x 15

x = \(\\ \frac { 10 x 15 }{ 2 } \) = 75

Hence number = 75 Ans.

**Question 4.**

**Solution:**

Let first part = x

then second part = 24 – x (∵ Sum = 24)

According to the condition,

7x + 5 (24 – x) = 146

=> 7x + 120 – 5x = 146

=> 2x = 146 – 120 = 26

=> x = \(\\ \frac { 26 }{ 2 } \) = 13

First part = 13

and second part = 24 – 13 = 11 Ans.

**Question 5.**

**Solution:**

Let number = x

According to the condition

\(\frac { x }{ 5 } +5=\frac { x }{ 4 } -5\)

=> \(\\ \frac { x }{ 5 } \) – \(\\ \frac { x }{ 4 } \)= – 5 – 5

=> \(\\ \frac { 4x-5x }{ 20 } \) = – 10

=> \(\\ \frac { -x }{ 20 } \) = – 10 = \(\\ \frac { x }{ 20 } \) = 10

=> x = 10 × 20 = 200

Hence number = 200 Ans.

**Question 6.**

**Solution:**

Ratio between three numbers = 4:5:6

Let the largest number = 6x

smallest number = 4x

and third number = 5x

According to the condition,

6x + 4x = 5x + 55

=> 10x = 5x + 55

=> 10x – 5x = 55

=> 5x = 55

=> x = \(\\ \frac { 55 }{ 5 } \) = 11

Numbers will be 4x = 4 × 11 = 44,

5x = 5 × 31 = 55 and 6x = 6 × 11 = 66

Hence numbers are 44, 55, 66 Ans.

**Question 7.**

**Solution:**

Let the number = x

According to the condition,

4x + 10 = 5x – 5

=> 4x – 5x = – 5 – 10

=> – x = – 15

x = 15

Hence number = 15 Ans.

**Question 8.**

**Solution:**

Ratio between two numbers 3: 5

Let first number = 3x

and second number = 5x

Now according to the condition.

3x + 10 : 5x + 10 = 5 : 7

**Question 9.**

**Solution:**

Let first odd number = 2x + 1

second number 2x + 3

and third number = 2 + 5

According to the condition.

2x + 1 + 2x + 3 + 2x + 5 = 147

=> 6x + 9 = 147

=> 6x = 147 – 9 = 138

=> x = \(\\ \frac { 138 }{ 6 } \) = 23

Hence first odd number 2x + 1

= 23 x 2 + 1 = 46 + 1 = 47

Second number 47 + 2 = 49

and third number = 49 + 2 = 51 Ans.

**Question 10.**

**Solution:**

Let first even number = 2x

second number = 2x + 2

and third number = 2x + 4

According to the condition,

2x + 2x + 2 + 2x + 4 = 234

=> 6x + 6 = 234

=> 6x = 234 – 6

=> 6x = 228

=> x = \(\\ \frac { 228 }{ 6 } \) = 38

First even number = 2x = 38 x 2 = 76

second number = 76 + 2 = 78

and third number 78 + 2 = 80 Ans.

**Question 11.**

**Solution:**

The sum of two digits = 12

Let the ones digit of the number = x

then tens digit = 12 – x

and number = x + 10 (12 – x)

= x + 120 – 10x = 120 – 9x

Reversing the digits,

ones digit of new number = 12 – x

and tens digit = x

the number = 12 – x + 10x = 12 + 9x

According to the condition,

12 + 9x = 120 – 9x + 54

=> 9x + 9x

=> 120 + 54 – 12

=> 174 – 12

=> 18x = 162

=> x = \(\\ \frac { 162 }{ 18 } \) = 9

Original number = 120 – 9x

= 120 – 9 x 9

= 120 – 81

= 39

Hence number 39 Ans.

Check :Original number= 39

Sum of digits = 3 + 9 = 12

Now reversing its digit the new number

will be = 93

and 93 – 39 = 54 which is given.

**Question 12.**

**Solution:**

Let units digit of the number = x

then tens digit = 3x

and number = x + 10

3x = x + 30x = 31x

on reversing the digits.

units digit = 3x

and tens digit = x

then number 3x + 10x = 13x

According to the condition,

31x – 36 = 13x

=> 31x – 13x = 36

=> 18 x 36

=> x = \(\\ \frac { 36 }{ 18 } \) = 2

The original number = 31x = 31 x 2 = 62

Hence number = 62 Ans.

Check : Number = 62

tens digit = 2 x 3 = 6

On reversing the digit, the new number will be = 26

62 – 26 = 36 which is given.

**Question 13.**

**Solution:**

Let numerator of a rational number = x

Then its denominator = x + 7

**Question 14.**

**Solution:**

Let numerator of a fraction = x

The denominator = 2x – 2

**Question 15.**

**Solution:**

Let breadth of the rectangle = x cm

then length = x + 7

Area = l × b = (x + 7) × x

In second case,

Length of the new rectangle = x + 7 – 4

= x + 3 cm

and breadth = x + 3

Area = (x + 3)(x + 3)

According to the condition,

(x + 3)(x + 3) = x(x + 7)

x2 + 3x + 3x + 9 = x2 + 7x

=> x2 + 6x – 7x – x2 = – 9

=> x = – 9

=> x = 9

Length of the original rectangle

= > x + 7 = 9 + 7 = 16 cm

and breadth = x = 9 cm. Ans.

**Question 16.**

**Solution:**

Let length of rectangle = x m

then width = \(\frac { 2 }{ 3 } x\)m

Perimeter = 2 (l + b) m

=> \(2\left( x+\frac { 2 }{ 3 } x \right) =180\)

**Question 17.**

**Solution:**

Let the length of the base of the triangle = x cm

then altitude = \(\frac { 5 }{ 3 } x\)cm

Area = \(\\ \frac { 1 }{ 2 } \) base x altitude

**Question 18.**

**Solution:**

Let ∠A, ∠B and ∠C are the three angles of a triangle and

Let ∠A + ∠B = ∠C

But ∠A + ∠B + ∠C = 180°

∠C + ∠C = 180°

=> 2∠C = 180°

∠C = 90°

and ∠A + ∠B = 90°

Let ∠A = 4x and ∠B = 5x

4x + 5x = 90°

=> 9x = 90°

x = \(\frac { { 90 }^{ o } }{ 9 } \) = 10°

∠A = 4x = 4 x 10° = 40°

and ∠B = 5x = 5 x 10° = 50°

and ∠C = 90° Ans.

**Question 19.**

**Solution:**

Time taken downstream = 9 hour

and time taken upstream = 10 hour

Speed of stream = 1 km/h

Let the speed of a steamer in still water

= x km/h.

Distance downstream = 9(x + 1) km.

and upstream = 10 (x – 1) km.

According to the condition,

10(x – 1) = 9(x + 1)

l0x – 10 = 9x + 9

=> 10x – 9x = 9 + 10

=> x = 19

Hence speed of steamer in still water = 19 km/h

and distance = 9(x + 1)

= 9(19 + 1)

= 9 x 20km

= 180km. Ans.

**Question 20.**

**Solution:**

The distance between two stations = 300 km

Let the speed of first motorcyclists = x km/h

and speed of second motorcyclists = (x + 7)km/h

Distance covered by the first = 2x km

and distance covered by the second = 2 (x + 7) km

= 2x + 14 km

Distance uncovered by them = 300 – (2x + 2x + 14)kms.

According to the condition,

300 – (4x + 4) = 34

=> 300 – 4x – 14 = 34

=> 300 – 14 – 34 = 4x

=> 4x = 300 – 48

=> 4x = 252

=> x = \(\\ \frac { 252 }{ 4 } \) = 63

Speed of the first motorcyclists = 63km/h

and speed of second = 63 + 7 = 70 km/h

Check. Distance covered by both of them

= 2 x 63 + 2 x 70 = 126 + 140 = 266

Total distance = 300 km.

Distance between them = 300 – 266 = 34 km.

which is given.

**Question 21.**

**Solution:**

Sum of three numbers = 150

Let first number = x

then second number = \(\frac { 5 }{ 6 } x\)

and third number = \(\\ \frac { 4 }{ 5 } \) of second

**Question 22.**

**Solution:**

Sum of two pans = 4500

Let first part = x

then second part = 4500 – x

According to the condition,

5% of x = 10% of (4500 – x)

**Question 23.**

**Solution:**

Let age of Rakhi = x years

then her mother’s age = 4x

After 5 years,

Rakhi’s age = x + 5

and her mother’s age = 4x + 5

According to the conditions,

4x + 5 = 3 (x + 5)

=> 4x + 5 = 3x + 15

=> 4x – 3x = 15 – 5

=> x = 10

Rakhi ‘s present age = 10 years

and her mother’s age = 4 x 10

= 40 years Ans.

**Question 24.**

**Solution:**

Let age of Monu = x year

His father’s age = x + 29

and his grandfather’s age = x + 29 + 26

= x + 55

and sum of their ages = 135 years

Now,

x + x + 29 + x + 55 = 135

=> 3x + 84 = 135

=> 3x = 135 – 84 = 51

=> x = \(\\ \frac { 51 }{ 3 } \) = 17 years

Monu’s age = 17 years

His father age = 17 + 29 = 46 years

and his grandfather’s age = 46 + 26 = 72 years

**Question 25.**

**Solution:**

Let age of grandson = x year

Then his age = 10x

But 10x = x + 54

=> 10x – x = 54

=> 9x = 54

=> x = \(\\ \frac { 54 }{ 9 } \) = 6

Grand’s son age = 6 years

and his age = 6 x 10 = 60 years

**Question 26.**

**Solution:**

Let age of elder cousin = x years

and age of younger = (x – 10) years.

15 years ago,

age of older, cousin = x – 15 years

and age of younger = x – 10 – 15

= (x – 25) years.

According to the condition,

x – 15 = 2(x – 25)

=> x – 15 = 2x – 50

=> 2x – x = 50 – 15

=> x = 35

Age of elder cousin = 35 years

and age of younger = 35 – 10

= 25 years Ans.

**Question 27.**

**Solution:**

Let number of deer = x

Hope given RS Aggarwal Solutions Class 8 Chapter 8 Linear Equations Ex 8B are helpful to complete your math homework.

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Aditya yadav says

Sir please add test paper (book)