# WORD PROBLEMS ON SIMULTANEOUS EQUATIONS

Problem 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.

Problem 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.

Problem 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

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.

Problem 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.

Problem 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

Substitute y  =  84 in (1).

(1)------> x  =  150 - 84

x  =  66

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

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