Find the solution for any typical problem of linear equations by practicing our Worksheet on Word Problems on Linear Equation. You can find out different types of equations from simple to complex ones. We are offering solutions for every equation along with an explanation. Practice using the below Linear Equations Word Problems Worksheet and learn how to find a solution for a given linear equation problem quickly.

## Linear Equations Problems with Solutions

1. Convert the following statements into equations?

(a) 5 added to a number is 9.

(b) 3 subtracted from a number is equal to 12.

(c) 5 times a number decreased by 2 is 4.

(d) 2 times the sum of the numbers K and 7 is 13.

## Solution:

(a) The given statement is 5 added to a number is 9.

Let’s consider a number as ‘K’.

Then, adding 5 with K should be 9. That is 5 + K = 9.

By converting the given statement into an equation we will get the equation like 5 + K = 9.

(b) The given statement is 3 subtracted from a number is equal to 12.

Let’s consider a number as ‘K’.

Then, subtracting 3 from K should be 12. That is K – 3 = 12.

By converting the given statement into an equation we will get the equation as K – 3 = 12.

(c) The given statement is5 times a number decreased by 2 is 4.

Let’s consider a number as ‘K’.

Five times a number is equal to 5K.

5Kis decreased by 2 and that is 5K – 2.

Then, by decreasing 2 from 5K should be 4. That is5K – 2 = 4.

By converting the given statement into an equation we will get the equation as 5K – 2 = 4.

(d)The given statement is 2 times the sum of the number K and 7 is 13.

Let the sum of the number K and 7 is K + 7.

Now, 2 times the sum of K and 7 is 2 ( K + 7 ) which is equal to 13.

By converting the given statement into an equation we will get the equation as

2 ( K + 7 ) = 13.

2. A number is 12 more than the other. Find the numbers if their sum is 48?

## Solution:

The given statement is a number is 12 more than the other and if their sum is 48.

Let’s consider the first number as ‘K’.

If another number is 12 more than K and that is K + 12.

So, the sum of two numbers is

K + K + 12 = 48.

2K + 12 = 48.

2K = 48 – 12 = 36.

K = 36 ÷ 2 = 18

So first number K is 18 then other number is 12 + K = 12 + 18 = 30.

The sum of the two numbers 18 + 30 = 48.

The two numbers are 18 and 30.

3. Twice the number decreased by 22 is 48. Find the number?

## Solution:

The given statement is twice the number decreased by 22 is 48.

Let’s consider the number as ‘K’.

Twice the number is 2K.

2K is decreased by 22, that is 2K – 22.

So 2K – 22 = 48.

2k = 48 + 22 = 70.

K = 70 ÷ 2 = 35.

That is the number is K = 35.

4. Seven times the number is 36 less than 10 times the number. Find the number?

## Solution:

The given statement is Seven times the number is 36 less than 10 times the number.

Let’s consider a number as ‘K’.

Seven times the number is 7K.

Ten times the number is 10K.

36 less than ten times a number is 10K – 36.

7K should be equal to 10K – 36.

That is 7K = 10K – 36.

36 = 10K – 7K.

36 = 3K.

K = 36 ÷ 3 = 12.

The number is 12.

5. 4/5 of a number is more than 3/4 of the number by 5. Find the number?

## Solution:

The given statement is 4/5 of a number is more than 3/4 of the number by 5.

Let’s consider a number as ‘K’.

4 / 5 of a number is 4K / 5.

3 / 4 of a number is 3K / 4.

4K / 5 is more than 3K / 4 by 5.

That is 3K / 4 = 4K / 5 + 5.

By using LCM on RHS.

3K / 4 = ( 4K + 25 ) / 5.

By using cross multiplication on L.H.S and R.H.S for the above equation.

( 3K ) X 5 = ( 4K + 25 ) X 4

15K = 16K + 100.

K = 100.

The number is 100.

6. The sum of two consecutive even numbers is 38. Find the numbers?

## Solution:

The given statement is the sum of two consecutive even numbers is 38.

Let’s consider the consecutive even numbers are K + 2 and K + 4.

Sum of two consecutive even numbers is ( K + 2 ) + ( K + 4 ) = 38.

2K + 6 = 38.

2K = 38 – 6 = 32.

K = 32 ÷ 2 = 16.

K + 2 = 16 + 2 = 18.

K + 4 = 16 + 4 = 20

So the two consecutive even numbers are 18 and 20.

7. The sum of three consecutive odd numbers is 51. Find the numbers?

## Solution:

The given statement is the sum of three consecutive odd numbers is 51.

Let’s consider the consecutive odd numbers as K + 1, K + 3 and K + 5.

The Sum of three consecutive odd numbers is

( K + 1 ) + ( K + 3 ) + ( K + 5 ) = 51.

3K + 9 = 51.

3K = 51 – 9 = 42.

K = 42 ÷ 3 = 14.

K + 1 = 15.

K + 3 = 17.

K + 5 = 19.

The consecutive odd numbers are 15, 17, and 19.

8. Rene is 6 years older than her younger sister. After 10 years, the sum of their ages will be 50 years. Find their present ages?

## Solution:

The given statement Rene is 6 years older than her younger sister. After 1 0 years, the sum of their ages will be 50 years.

Let’s consider younger sister age as ‘K’.

Rene is 6 years older than her younger sister, that is 6 + K.

After 10 years the sum of Rene and Younger sister age will be equal to 50 can be written as

( K + 10 ) + ( K + 6 + 10 ) = 50.

2K + 26 = 50.

2K = 50 – 26 = 24.

K = 24 ÷ 2 = 12.

Younger sister age is 12.

Rene is 6 years older than her younger sister that is 12 + 6 = 18.

9. The length of a rectangle is 10 m more than its breadth. If the perimeter of a rectangle is 80 m, find the dimensions of the rectangle?

## Solution:

The given statement is the length of a rectangle is 10m more than its breadth and the perimeter of a rectangle is 80m.

Let’s consider the breadth of a rectangle as ‘K’.

b = K.

The length of a rectangle is 10m more than the breadth that is K + 10.

l = K + 10.

The perimeter of rectangle is 2 ( l + b ) = 80m.

2 ( K + 10 + K ) = 80m.

2K + 10 = 80 ÷ 2 = 40m.

2K = 40 – 10 = 30m

K = 30 ÷ 2 = 15m.

The breadth of a rectangle K = 15m.

Length of a rectangle K + 10 = 15 + 10 = 25m.

10. A 300 m long wire is used to fence a rectangular plot whose length is twice its width. Find the length and breadth of the plot?

## Solution:

The given statement is a 300 m long wire is used to fence a rectangular plot whose length is twice its width.

Let’s consider the width of the rectangle as ‘K’.

b = K.

The length of the rectangle is twice its width that is 2K.

l = 2K.

Perimeter of a rectangle is 2 ( l + b ) = 300.

2 ( K + 2K ) = 300m.

3K = 300 ÷ 2 = 150m.

K = 150 ÷ 3 = 50m.

The width of the rectangle is K = 50m.

The length of rectangle is 2K = 2 X 50 = 100m.

11. The denominator of a fraction is greater than the numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is 3/2, find the fraction?

## Solution:

The given statement is the denominator of a fraction is greater than the numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is 3/2.

Let’s consider the numerator as ‘K’.

The denominator of the fraction is greater than the numerator by 8 that is K + 8.

So fraction is K / K + 8.

The numerator is increased by 17 and the denominator is decreased by 1 which is

( K + 17 ) / ( K + 8 – 1 ) = 3 / 2.

By applying cross multiplication for above equation we get

2 ( K + 17 ) = 3 ( K + 7 ).

2K + 34 = 3K + 21.

34 – 21 = 3K – 2K.

K = 13.

The numerator of the fraction is K = 13.

Denominator of fraction is K + 8 = 13 + 8 = 21.

So fraction is 13 / 21.

12. A sum of $2700 is to be given in the form of 63 prizes. If the prize is either $100 or $25, find the number of prizes of each type?

## Solution:

The given statement is a sum of $2700 is to be given in the form of 63 prizes. If the prize is either $100 or $25.

Let’s consider ‘K’ as the number of the $100 prizes and ‘L’ as the number of the $25 prize.

100K + 25L = $2700 ——– (1).

Total number of prizes is 63 that is K + L = 63 ——- (2).

Multiply the equation (2) with 100 on both sides. That is

100K + 100L = $6300 ——– (3).

Subtract the equation (1) and (3).

100K + 25L = $2700.

100K + 100L = $6300.

(-)—–(-)——-(-)——————-

– 75L = – $3600.

L = 3600 ÷ 75 = 48.

Substitute the L value in equation number (2).

That is K + 48 = 63.

K = 63 – 48 = 15.

Finally, the number of prizes are 15 and 48.

13. In a class of 42 students, the number of boys is 2/5 of the girls. Find the number of boys and girls in the class?

## Solution:

The given statement isa class of 42 students, the number of boys is 2/5 of the girls.

Let’s consider the number of girls in the class is ‘K’.

So, the number of boys is 2 / 5 of the girls. That is 2K/ 5.

The total number of students in the class is 42. That is

2K / 5 + K = 42.

2K + 5K = 42 X 5 = 210.

7K = 210.

K = 210 ÷ 7 = 30.

The total number of girls in the class is K = 30

The total number of boys in the class is 2K / 5 = 2 X 30 / 5 = 2 X 6 = 12.

14. Among the two supplementary angles, the measure of the larger angle is 36° more than the measure of smaller. Find their measures?

## Solution:

The given statement is the two supplementary angles, the measure of the larger angle is 36° more than the measure of smaller.

Let’s consider the small-angle as ‘K’ and the larger angle as ‘L’.

‘K’ and ‘L’ are supplementary angles. That is K + L = 180° ———(1).

The larger angle is 36° more than the smaller angle. That is L = K + 36°.

Substitute the larger angle value in equation (1).

Then, K + K + 36° = 180°.

2K = 180° – 36° = 144°.

K = 144° ÷ 2 = 72°.

L = K + 36° = 72° + 36° = 108°.

The measures of the angles are 72° and 108°.

15. My mother is 12 years more than twice my age. After 8 years, my mother’s age will be 20 years less than three times my age. Find my age and my mother’s age?

## Solution:

The given statement is mother’s age is 12 years more than twice my age. After 8 years, my mother’s age will be 20 years less than three times my age.

Let’s consider my age as ‘K’.

My mother’s age is 12 years more than twice my age and that is 2K + 12.

After 8 years.

My age is K + 8.

My mother’s age is 2K + 12 + 8.

My mother’s age will be 20 years less than three times my age.

3(K + 8) – 20 = 2K + 20.

3K + 24 – 20 = 2K + 20.

3K + 4 = 2K + 20.

3K – 2K = 20 – 4 = 16.

My age is K = 16 years.

My mother’s age is 2K + 12 = 2(16) + 12 = 44 years.

16. In an isosceles triangle, the base angles are equal and the vertex angle is 80°. Find the measure of the base angles?

## Solution:

The given statement is in an isosceles triangle, the base angles are equal and the vertex angle is 80°.

As per the given information, the vertex angle is 80°.

The base angles are equal.

Triangle is K + K + 80° = 180°.

2K = 180° – 80° = 100°.

K = 100 ÷ 2 = 50°.

So, the measures of the base angles are 50°.

17. Adman’s father is 49 years old. He is 5 years older than four times Adman’s age. What is Adman’s age?

## Solution:

The given statement is an Adman’s father is 49 years old. He is 5 years older than four times Adman’s age.

Let’s consider an Adman’s age as ‘K’

Adman’s father is 49 years old.

Adman’s father is 5 years older than four times Adman’s age. That is

4K + 5 = 49.

4K = 49 – 5 = 44.

K = 44 ÷ 4 = 11 years

So, an Adman’s age is 11 years.

18. The cost of a pencil is 25 cents more than the cost of an eraser. If the cost of 8 pencils and 10 erasers is $12.80, find the cost of each?

## Solution:

The given statement is the cost of a pencil is 25 cents more than the cost of an eraser. If the cost of 8 pencils and 10 erasers is $12.80.

Let’s consider the cost of an eraser as ‘K’.

The cost of a pencil is 25 cents more than the cost of an eraser. That is K + 25.

The cost of 8 pencils and 10 erasers is $12.80.

1 dollar = 100 cents. So, $12.80 is 1280 cents.

8(K + 25) + 10K = 1280 cents.

8K + 200 + 10K = 1280 cents.

18K = 1280 – 200 = 1080 cents.

K = 1080 ÷ 18 = 60 cents.

The cost of an eraser is 60 cents.

Cost of pencils is K + 25 = 60 + 25 = 85 cents.

19. Divide 36 into two parts in such a way that 1/5 of one part is equal to 1/7 of the other?

## Solution:

The given statement is Divide 36 into two parts in such a way that 1/5 of one part is equal to 1/7 of the other.

By dividing the 36 into two parts

K / 5 = (36 – K) / 7.

By cross multiplying the above terms, we will get

7K = (36 – K)5.

7K = 36 X 5 – 5K.

7K + 5K = 180.

12K = 180.

K = 180 ÷ 12 = 15.

The first part is 15 and the second part is 36 – 15 = 21.

20. The length of the rectangle exceeds its breadth by 3 cm. If the length and breadth are each increased by 2 cm, then the area of the new rectangle will be 70 sq. cm more than that of the given rectangle. Find the length and breadth of the given rectangle?

## Solution:

The given statement is the length of the rectangle exceeds its breadth by 3 cm. If the length and breadth are each increased by 2 cm, then the area of the new rectangle will be 70 sq. cm more than that of the given rectangle.

Let’s consider the breadth of the rectangle as ‘K’.

b = K.

Then, the length of the rectangle exceeds its breadth by 3 cm. That is K + 3

l = K + 3.

If the length and breadth are each increased by 2cm is K + 2 and K + 3 + 2.

The area of a rectangle is 70sq. cm

Area of the rectangle is (l X b) = 70 sq. cm.

(K + 2) X (K + 5) = K(K + 3) + 70.

K^2 + 5K + 2K + 10 = K^2 + 3K + 70.

7K + 10 = 3K + 70.

4K = 60.

K = 15.

L = K + 3 = 15 + 3 = 18.

The length of the rectangle is 18 and the breadth of the rectangle is 15.