WORD PROBLEMS ON SIMULTANEOUS EQUATIONS

In this section, you will learn how to solve word problems on simultaneous equations.

The picture shown below tells us the trick to solve word problems on simultaneous equations.

Word Problems on Simultaneous Equations - Examples

Example 1 :

In a school, there are 880 students in total. If there is 20% more boys than girls, find the number of boys and girls in the school.

Solution :

Let "x" and "y" be the no. of boys and girls respectively.

Then, we have

x + y = 880 ------(1)

Given : There is 20% more boys than girls.

So, we have

x = 120% of y

x = 1.2y ------(2)

Substitute x = 1.2y in (1)

1.2y + y = 880

Simplify.

2.2y = 880

Divide by 2.2

y = 400

Substitute y = 400 in (2).

x = 1.2 ⋅ 400

x = 480

So, the number of boys in the school is 480 and girls is 400.

Example 2 :

Mr. Lenin invests some amount in deposit A and some amount in deposit B. The total money invested is $2500. He gets 10% income on deposit A and 20% income on deposit B. If the total income earned be $380, find the amount invested in A and B separately

Solution :

Let "x" and "y" be the amounts invested in A and B respectively.

Then, we have

x + y = 2500

Subtract y from both sides.

x = 2500 - y ------(1)

Given : Mr. Lenin gets 10% income on deposit A, 20% income on deposit B and the total income earned be $380.

So, we have

10% of x + 20% of y = 380

0.1x + 0.2y = 380

Multiply both sides by 10.

x + 2y = 3800 -------(2)

From (1), we can plug x = 2500 - y in (2).

(2)------> 2500 - y + 2y = 3800

Simplify.

2500 + y = 3800

Subtract 2500 from both sides.

y = 1300

Substitute y = 1300 in (1).

(1)------> x = 2500 - 1300

x = 1200

So, the amount invested in the deposit A is $1200 and B is $1300.

Example 3 :

The sum of two numbers is 209. If one number is 7 less than two times of the other, then find the two numbers.

Solution :

Let "x" and "y" be the two numbers.

Given : The sum of two numbers is 209.

x + y = 209 ------(1)

Given : One number is 7 less than two times of the other.

So, we have

x = 2y - 7 -------(2)

From (2), we can plug x = 2y - 7 in (1).

(1)------> 2y - 7 + y = 209

Simplify.

3y - 7 = 209

Add 7 to both sides.

3y = 216

Divide both sides by 3.

y = 72

Substitute y = 72 in (2).

x = 2 ⋅ 72 - 7

x = 144 - 7

x = 137

So, the two numbers are 137 and 72.

Example 4 :

The perimeter of the rectangle is 158 cm. If the length is 7 more than 3 times the width, find the area of the rectangle.

Solution :

Let "x" and "y" be the length and width of the rectangle respectively.

Given : Perimeter of the rectangle is 158 cm.

So, we have

2x + 2y = 158

Divide both sides by 2.

x + y = 79 ------(1)

Given : The length is 7 more than 3 times the width.

So, we have

x = 3y + 7 --------(2)

From (2), we can plug x = 3y + 7 in (1).

(1)------> 3y + 7 + y = 79

Simplify.

4y + 7 = 79

Subtract 7 from both sides.

4y = 72

Divide both sides by 4.

y = 18

Substitute y = 18 in (2).

x = 3 ⋅ 18 + 7

x = 61

Therefore, the length of the rectangle is 61 cm and width is 18 cm.

Then, area of the rectangle is

= length ⋅ width

= 61 ⋅ 18

= 1098

So, the area of the rectangle is 1098 sq.cm.

Example 5 :

A trader gains one third of the cost price as profit on a product and one fourth of the cost price as profit on other product. Total profit earned on these two products is $43. The sum of the cost prices of two products is $150. Find the cost price of each product.

Solution :

Let "x" and "y" be the cost prices of the two products.

Then, we have

x + y = 150

Subtract y from both sides.

x = 150 - y -------(1)

Let the trader gain one third of the cost price as profit on the product whose cost price is x.

Then, profit on the product whose cost price is x :

= 1/3 ⋅ x

= x / 3

Let the trader gain one fourth of the cost price as profit on the product whose cost price is y.

Then, profit on the product whose cost price is y :