WORD PROBLEMS ON SIMULTANEOUS EQUATIONS

About "Word Problems on Simultaneous Equations"

Word Problems on Simultaneous Equations :

Solving word problems on simultaneous equations is sometimes a difficult job for some students.

Actually it is not. There is a simple trick behind it.

The picture given below tells us the trick.

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)

Plugging x = 1.2y in (2)

1.2y + y = 880

Simplify.

2.2y = 880

Divide by 2.2

y = 400

Plug y = 400 in (2).

x = 1.2 ⋅ 400

x = 480

Hence, 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

Plug y = 1300 in (1).

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

x = 1200

Hence, 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

Plug y = 72 in (2).

x = 2 ⋅ 72 - 7

x = 144 - 7

x = 137

Hence, 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

Plug 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

Hence, 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 :