# WORD PROBLEMS ON SIMULTANEOUS LINEAR EQUATIONS

Problem 1 :

One number is greater than thrice the other number by 2. If 4 times the smaller number exceeds the greater by 5, find the numbers.

Solution :

Let x and y be the two numbers such that

x  >  y

Given : One number is greater than thrice the other number by 2.

Then, we have

x  =  3y + 2 -----(1)

Given : 4 times the smaller number exceeds the greater by 5.

Then, we have

4y - x  =  5 -----(2)

From (1), substitute (3y + 2) in (2).

(2)-----> 4y - (3y + 2)  =  5

Simplify.

4y - 3y - 2  =  5

y - 2  =  5

Add 2 to each side.

y  =  7

Substitute 7 for y in (1).

x  =  3(7) + 2

x  =  21 + 2

x  =  23

So, the two numbers are 23 and 7.

Problem 2 :

A two digit number is seven times the sum of its digits. The number formed by reversing the digits is 18 less than the given number. Find the given number.

Solution :

Let xy be the two digit number.

Given :  The two digit number is seven times the sum of its digits.

xy  =  7(x + y) -----(1)

In the two digit number xy, x is in tens place and y is in ones place.

Then, we have

xy  =  10 ⋅ x + 1 ⋅ y

xy  =  10x + y

Substitute (10x + y) for xy in (1).

(1)-----> 10x + y  =  7(x + y)

10x + y  =  7x + 7y

Simplify.

3x - 6y  =  0

Divide each side by 3.

x - 2y  =  0 -----(2)

Given :  The number formed by reversing the digits is 18 less than the given number.

The number formed by reversing the digits is yx.

Then, we have

xy - yx  =  18

[10 ⋅ x + 1 ⋅ y]  -  [10 ⋅ y + 1 ⋅ x]  =  18

(10x + y)  -  (10y + x)  =  18

Simplify.

10x + y - 10y - x  =  18

9x - 9y  =  18

Divide each side by 9.

x - y  =  2 -----(3)

Solving (2) and (3), we get

x  =  4 and y  =  2

So, the required two digit number is 42.

Problem 3 :

In a rectangle, if the length is increased and the width is reduced each by 2 cm then the area is reduced by 28 sq.cm . If the length is reduced by 1 cm and the width increased by 2 cm , then the area increases by 33 sq.cm . Find the area of the rectangle.

Solution :

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

Then, area of the rectangle is

=  x ⋅ y

Given : If the length is increased and the width is reduced each by 2 cm then the area is reduced by 28 sq.cm

Then, we have

(x + 2) ⋅ (y - 2)  =  x ⋅ y - 28

Simplify.

xy - 2x + 2y - 4  =  xy - 28

- 2x + 2y  =  - 24

Divide each side by 2.

- x + y  =  - 12 -----(1)

Given : If the length is reduced by 1 cm and the width increased by 2 cm , then the area increases by 33 sq.cm

Then, we have

(x - 1) ⋅ (y + 2)  =  x ⋅ y + 33

Simplify.

xy + 2x - y - 2  =  xy + 33

2x - y  =  35 -----(2)

Solving (1) and (2), we get

x  =  23 and y  =  11

Area of the original rectangle is

=  x ⋅ y

=  23 ⋅ 11

=  253 sq.cm

So, the area of the rectangle is 253 sq.cm.

Problem 4 :

Sum of incomes of A and B is \$2640. If B's income is 20% more than A, find the income of A and B.

Solution :

Let x and y be the incomes of A and B respectively.

Then,  we have

x + y  =  2640 -----(1)

Given : B's income is 20% more than A.

Then, we have

y  =  120% of x

y  =  1.2x -----(2)

Substitute 1.2x for y in (1).

x + 1.2x  =  2640

Simplify.

2.2x  =  2640

Divide each side by 2.2.

x  =  1200

Substitute 1200 for x in (2)

y  =  1.2 ⋅ 1200

y  =  1440

So, the income of A is \$1200 and B is \$1440.

Problem 5 :

Sum of the cost price of two products is \$50. Sum of the selling price of the same two products is \$52. If one is sold at 20% profit and other one is sold at 20% loss, find the cost price of each product.

Solution :

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

Given : Sum of the cost prices of the two products is \$50.

Then, we have

x + y  =  50 ------(1)

Given : One is sold at 20% profit and other one is sold at 20% loss.

Let us assume that x is sold at 20% profit.

Then, the selling price of x is

=  120% of x

=  1.2x

Let us assume that y is sold at 20% loss.

Then, the selling price of y is

=  80% of y

=  0.8y

Given : Sum of the selling price of the two products is \$52.

Then, we have

1.2x + 0.8y  =  52

Multiply both sides by 10.

12x + 8y  =  520

Divide each side by 4.

3x + 2y  =  130 -----(2)

Solving (1) and (2), we get

x  =  30 and y  =  20

So, the cost prices of two products are \$30 and \$20.

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