WORD PROBLEMS ON SOLVING LINEAR EQUATIONS

Problem 1 :

The sum of three consecutive even natural numbers is 48. Find the greatest of these numbers.

Solution :

Let the three consecutive even natural numbers be x, x + 2, and x + 4.

As per the question,

x + x + 2 + x + 4 = 48

3x + 6 = 48

3x = 48 - 6

3x = 42

x = 14

Now,

The three consecutive even natural numbers are

14, 14 + 2, 14 + 4

14, 16, 18

So, the greatest of these number is 18.

Problem 2 :

The sum of three consecutive odd natural numbers is 69. Find the prime number out of these numbers.

Solution :

Let the three consecutive odd natural numbers be x, x + 2, and x + 4.

As per the question,

x + x + 2 + x + 4 = 69

3x + 6 = 69

3x = 69 - 6

3x = 63

x = 21

Now,

The three consecutive odd natural numbers are

21, 21 + 2, 21 + 4

21, 23, 25

So, the prime number among these three is 23.

Problem 3 :

The sum of three consecutive numbers is 156. Find the number which is a multiple of 13 out  of these numbers.

Solution :

Let the three consecutive numbers be x, x + 1, and x + 2.

As per the question,

x + x + 1 + x + 2 = 156

3x + 3 = 156

3x = 156 - 3

3x = 153

x = 51

Now,

The three consecutive odd natural numbers are

51, 51 + 1, 51 + 2

51, 52, 53

52 is exactly divisible by 3.

So, 52 is the multiple of 13 among these three numbers.

Problem 4 :

Divide 54 into two parts such that one part is 2/7 of the other.

Solution :

Let the one part be x.

And the other part will be 2/7x.

By given question,

x + 2/7x = 54

(7x + 2x)/7 = 54

7x + 2x = 54(7)

9x = 378

x = 42

Now,

One part = x = 42

Other part = 2/7x = 2/7(42)

Other part = 12

So, 54 is divided into 42 and 12.

Problem 5 :

Sum of the digits of a two-digit number is 11. The given number is less than the number obtained by interchanging the digits by 9. Find the number.

Solution :

Let the digit at one's place = x

Then, the digit at ten's place = y

Given, the sum of the digits of a two-digit number is 11.

x + y = 11 ----(1)

The number with digits x and y represented as 10y + x.

Now, Interchange the number with digits x and y represented as 10x + y.

By given question,

(10x + y) - (10y + x) = 9

10x + y - 10y + x = 9

9x - 9y = 9

Dividing by 9 on both sides, 

x - y = 1 ----(2)

Add (1) and (2), we get

x + y + x - y = 11 + 1

2x = 12

x = 6

By applying x = 6 in equation (1), we get

x + y = 11

6 + y = 11

y = 11 - 6

y = 5

Now,

Digit at one's place = x = 6

Digit at ten's place = y = 5

So, the required number is 56.

Problem 6 :

Two equal sides of a triangle are each 4m less than three times the third side. Find the dimensions of the triangle, if its perimeter is 55m.

Solution :

Let the third side of triangle be x m.

Then, two equal sides of triangle = (3x - 4)m

Given, perimeter of a triangle is 55 m.

Formula for the perimeter of a triangle,

P = a + b + c

55 = (3x - 4) + (3x - 4) + x

55 = 7x - 8

55 + 8 = 7x

63 = 7x

x = 9

Now, 

Third side = x = 9 m

Two sides = (3x - 4) m

= (3(9) - 4) m

= 23 m

So, the dimensions of the triangle are 9m, 23m and 23m.

Problem 7 :

After 12 years, Kanwar shall be 3 times as old as he was 4 years ago. Find his present age.

Solution :

Let Kanwar present age be x year.

After 12 yrs, Kanwar age is (x + 12)yr

And 4 years ago,  Kanwar age is (x - 4)yr

As per the question,

x + 12 = 3(x - 4)

x + 12 = 3x - 12

x - 3x = -12 - 12

-2x = -24

x = 12

So, the Kanwar present age is 12 year.

Problem 8 :

If 1/2 is subtracted from a number and the difference is multiplied by 4, the result is 5. What is the number ?

Solution :

Let x be the number.

As per the question,

(x - 1/2)4 = 5

4x - 2 = 5

4x = 5 + 2

4x = 7

x = 7/4

So, the number is 7/4.

Problem 9 :

The sum of four consecutive integers is 266. What are the integers ?

Solution :

Let the four consecutive integers be x, x + 1, x + 2, x + 3

As per the question,

x + x + 1 + x + 2 + x + 3 = 266

4x + 6 = 266

4x = 266 - 6

4x = 260

x = 65

Now,

 x, x + 1, x + 2, x + 3

65, 65 +1, 65 + 2, 65 + 3

65, 66, 67, 68

So, the integers are 65, 66, 67, 68.

Problem 10 :

Find a number whose fifth part increased by 30 is equal to its fourth part decreased by 30.

Solution :

Let x be the number.

Assume one fifth part of the number is x/5.

And one fourth part of the number is x/4.

As per the question,

x/5 + 30 = x/4 - 30

(x + 150)/5 = (x - 120)/4

By cross multiplication,

4(x + 150) = 5(x - 120)

4x + 600 = 5x - 600

4x - 5x = -600 - 600

-x = -1200

x = 1200

So, the number is 1200.