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 :  

=  1/4 ⋅ y  

=  y / 4

Given : Total profit earned on these two products is $43.

So, we have

x/3  +  y/4  =  43

L.C.M of (3, 4) is 12.

Then, we have

(4x / 12)  +  (3y / 12)  =  43

Simplify.

(4x + 3y) / 12  =  516

Multiply both sides by 12.

4x + 3y  =  516 ------(2)

From (1), we can plug x  =  150 - y in (2). 

4(150 - y) + 3y  =  516 

Simplify.

600 - y  =  516

Subtract 600 from both sides.

- y  =  - 84

y  =  84

Plug y  =  84 in (1).

(1)------> x  =  150 - 84

x  =  66

Hence, the cost prices of two products are $66 and $84. 

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