In this chapter, we provide KSEEB SSLC Class 8 Maths Chapter 8 Linear Equations in One Variable Ex 8.2 for English medium students, Which will very helpful for every student in their exams. Students can download the latest KSEEB SSLC Class 8 Maths Chapter 8 Linear Equations in One Variable Ex 8.2 pdf, free KSEEB SSLC Class 8 Maths Chapter 8 Linear Equations in One Variable Ex 8.2 pdf download. Now you will get step by step solution to each question.

**Karnataka State Syllabus Class 8 Maths Chapter 8 Linear Equations in One Variable Ex 8.2**

Question 1.

If 4 is added to a number and the sum is multiplied 3, the result is 30. Find the number.

Answer:

Let the number be ‘x’.

If 4 is added to it will be x + 4.

Sum is multiplied by 3 the result is 30.

∴ (x + 4)3 = 30

3x +12 = 30

3x = 30 – 12

3x = 18

x = 183

x = 6

∴The number is 6

Question 2.

Find three consecutive odd numbers whose sum is 219.

Answer:

Let the odd number be ‘x’.

The next two consecutive numbers are x + 2 and x + 4

x + (x + 2) + (x + 4) = 219

3x + 6 = 219

3x = 219 – 6

3x = 213

x = 2133

x = 71

x + 2 = 71 + 2 = 73

x + 4 = 71 + 4 = 75

Three consecutive odd numbers are 71, 73, 75

Question 3.

A number subtracted by 30 gives 14 subtracted by 3 times the number. Find the number.

Answer:

Let the number be x

Number subtracted by 30 = 30 – x

14 subtracted by 3 times the number 3x – 14

∴30 – x = 3x – 14

30+ 14 = 3x + x

44 = 4x

x = 444

x =11

∴ The number is 11

Question 4.

If 5 is subtracted from three times a number the result is 16. Find the number.

Answer:

Let the number be x, 5 is subtracted from 3 times the number the result is 16.

3x – 5 = 16

3x= 16 + 5

3x = 21

x = 213

x = 7

∴ The number is 7

Question 5.

Find two numbers such that one of them exceeds the other by 9 and their sum is 81.

Answer:

Let the number is x. The other number is x + 9.

Their sum is 81

∴ x + (x + 9) = 81

2x = 81 – 9

2x = 72

x = 722

x = 36

x + 9 = 36 + 9 = 45

∴ The number are 36 and 45

Question 6.

Prakruthi’s age is 6 time Sahil’s age. After 15 years prakruthi will be 3 times as old as Sahil. Find their age.

Answer:

Let Sahil’s present age be x. Prakruthi’s present age is 6x, 15 years later Sahil age will be (x + 15) years and Prakruthi age will be (6x + 15) years.

Given that the Prakruthi age will be 3 times as old as Sahil.

∴6x +15 = 3(x + 15)

6x + 15 = 3x + 45

6x – 3x = 45 – 15

3x = 30

x = 303 = 10

Sahils age = x = 10 years Prakruthis age = 6x = 6 x 10 = 60 years

Question 7.

Ahmed’s father is thrice as old as Ahmed. After 12 years, his age will be twice that of his son. Find their present age.

Answer:

Ahmed’s presents age be x

Fathers present age = 3x

12 years later Ahmed’s age = x + 12

and father’s age = 3x + 12

Given 3x + 12 = 2(x + 12)

3x+ 12 = 2x + 24

3x – 2x = 24 – 12

x = 12 years

∴ Ahmed’s age = 12 years

Fathers age = 3x = 3 × 12 = 36 years.

Question 8.

Sanju is 6 years older than his brother Nishu. If the sum of their ages is 28 years what are their present ages.

Answer:

Let Nishu’s age be ‘x’ Sanju’s age = x + 6

Sum of their ages = 28 x + (x + 6) = 28

2x + 6 = 28

2x = 28 – 6

2x = 22

x = 222

x = 11

Nishu’sage = x = 11 years

Sanju’ s age = x + 6 = 11 + 6 = 17 years

Question 9.

Viji is twice as old as his brother Deepu. If the difference of their ages is 11 years, find their present age.

Answer:

Let Deep’s age be ‘x’, Viji’s age is 2x

Difference of their age = 11

2x – x = 11

x = 11

∴ Deepu’s age = x = 11 years

Viji’s age = 2x = 2 x 11 = 22 years

Question 10.

Mrs. Joseph is 27 years older than her daughter Bindu. After 8 years she will be twice as old as Bindu. Find their present ages.

Answer:

Let Bindu’s present age be ‘x’ years.

Mrs. Joseph’s present age = x + 27 years.

After 8 years Bindu’s age = x + 8 and Mrs.

Josephs age = x + 27 + 8

= x + 35 years.

Given that x + 35 = 2(x + 8)

x + 35 = 2x + 16

35 – 16 = 2x – x

19 = x

x= 19

Bindu’s present age = 19 years

Mrs.Joseph’s age = x + 27 = 19 + 27 = 46 years

Question 11.

After 16 years Leena will be three times as old as she is now. Find her present age.

Answer:

Let Leena’s present age be ‘x’ years

16 years later she will be (x + 16) years

Given x + 16 = 3x

16 = 2x

x = 162

x = 8 years

Leena’ s present age = 8 years

Question 12.

A rectangle has a length which is 5 cm less than twice its breadth. If the length is decreased by 5 cm and breadth is increased by 2 cm the perimeter of the resulting rectangle will be 74 cm. Find the length and breadth of the original rectangle.

Answer:

Let the breadth of the original rectangle be ‘b’ twice the breadth is 2b.

The length of the rectangle is 5 cm less than twice the breadth.

∴ Length = 2b – 5

If the length is decreased by 5 then length

is 2b – 5 – 5 = 2b – 10

If the breadth is increased by 2cm then breadth is b + 2 cm.

Perimeter of new rectangle = 2(length + breadth)

74 = 2(2b – 10 + b + 2)

74 = 2(3b – 8)

74 + 16 = 6b 90 = 6b

b = 906

b = 15cm.

Breadth of the original rectangle = 15 cm

Length of the original rectangle = 2b – 5

= 2 × 15 – 5

= 30 – 5

= 25 cm

Question 13.

The length of a rectangular field is twice its breadth. If the perimeter of the field is 288m. Find the dimensions of the field.

Answer:

Let breadth of the rectangular field bee ‘x’m

∴ Its length = 2x

Its perimeter = 288

2(1 + b) = 288

2(2x + x) = 288

2(3x) = 288

6x = 288

x = 2886 = 48

Its length = 2x = 2 × 48 = 96m

breadth = x = 48m

Question 14.

Srishti’s salary is the same as 4 times Azar’s salary. If together they earn Rs 3750 a month find their individual salaries.

Answer:

Let Axar’s salary be x. Sristi’s salary is 4x together they earn Rs. 3750

∴ x + 4x = 3750

5x = 3750

x = 37505

x = 750

4x = 4 × 750 = 3000

∴ Azar’s salary is Rs.750 and Sristi’s salary is Rs.3000

**All Chapter KSEEB Solutions For Class 8 maths**

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