LINEAR EQUATIONS WITH ONE UNKNOWN

Problem 1 :

9c + 1 = 10

Solution  :

Given , 9c + 1 = 10

9c + 1 – 1 = 10 - 1

9c =  10 – 1

9c  =  9

c  =  9/9

c  =  1

Problem 2 :

6y – 5  =  7

Solution :

Given , 6y – 5  =  7

6y – 5 + 5 =  7 + 5

6y  =  7 + 5

6y  =  12

y  =  12/6

y   =  2

Problem 3 :

8  =  3a – 4

Solution :

Given , 8  =  3a – 4

8 + 4  =  3a – 4 + 4

8 + 4  =  3a

12  =  3a

12/3  =  a

4  =  a

Problem 4 :

m/5 + 9 = 11

Solution :

Given , m/5 + 9 = 11

m/5 + 9 – 9 =  11 - 9 

m/5  =  2

m  =  2 × 5 

m  =  10

Problem 5 :

13 + 7x =  27

Solution :

Given , 13 + 7x =  27

13 + 7x – 13 = 27 – 13

7x  =  27 – 13

7x  =  14

x =  14/7

x  =  2

Problem 6 :

17 – q = 6

Solution :

Given , 17 – q = 6

17 – q – 17 =  6 - 17

-q  =  -11

q  =  11 

Problem 7:

(n – 31)/4 =  2

Solution :

 Given , (n – 31)/4 =  2

n – 31  =  8

n  =  8 + 31

n  =  39 

Problem 8:

1 + 2r  =   35

Solution :

Given , 1 + 2r =   35

1 + 2r – 1  = 35 - 1

2r  =  35 – 1

2r  =  34

r  = 34 /2

r = 17

Problem 9:

42 + 5t =  8t

Solution :

Given , 42 + 5t =  8t

42 + 5t – 42 =  8t - 42

    5t = 8t – 42

5t – 8t = -42

-3t  =  - 42

t = 42/3

t = 14

Problem 10:

4p – 3 =  17

Solution :

Given ,4p – 3 =  17

4p – 3 + 3  =  17 + 3

4p = 17 + 3

4p  =  20

P = 20/4

P = 5

Problem 11 :

The length of a rectangular field is thrice its width. If the perimeter of this field is 800 m , find the length of the field

Solution :

Perimeter of the rectangle = 800 m

Let length be l and width be w.

l = 3w

Perimeter of the rectangle = 2(l + w)

= 2(3w + w)

= 2(4w)

= 8w

8w = 800

w = 800/8

w = 100

Length of = 3(100)

= 300 m

So, length and width of the rectangle are 300 m and 100 m respectively.

Problem 12 :

After 18 years, Swarnim will be 4 times as old as he is now. His present age is _____

Solution :

Let x be the present age of Swarnim.

After 18 years, the age of Swarnim = x + 18

x + 18 = 4x

Subtracting x on both sides

18 = 4x - x

3x = 18

x = 18/3

x = 6

So, his present age is 6 years.

Problem 13 :

The denominator of a rational number is greater than the numerator by 10. If the numerator is increased by 1 the and denominator is decreased by 1, then expression for new denominator is _________.

Solution :

Let the required fraction be x/y

Numerator = x, denominator = y

y = x + 10

Numerator increased by 1, then denominator is decreased by 1

x + 1 / (y - 1)

= (x + 1) / (x + 10 - 1)

= (x + 1) / (x + 9)

Problem 14 :

The area of a rectangular garden 50 m long is 300 sq. m.. Find the width of the garden?

Solution :

Area of garden = 300 sq.m

Length = 50 m

Let w be the width of the rectangle.

50w = 300

Dividing by 50 on both sides

w = 300/50

w = 6

So, width of the rectangle is 6 m.

Problem 15 :

The area of a rectangular garden is 2900 sq.m. If its width is 29m, find the length of the garden ?

Solution :

Area of garden = 300 sq.m

Length = 50 m

Let w be the width of the rectangle.

50w = 300

Dividing by 50 on both sides

w = 300/50

w = 6

So, width of the rectangle is 6 m.

Problem 16 :

Three fourth of a number exceeds its one third by 60.Find the number. 

Solution :

Let x be the number.

Three fourth of the number = 3x/4

One third of the number = 1x/3

3x/4 = 60 + x/3

3x/4 = (180 + x) / 3

9x = 4(180 + x)

9x = 720 + 4x

9x - 4x = 720

5x = 720

x = 720/5

x = 144

So, the required number is 144.

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