WORD PROBLEMS ON LINEAR EQUATIONS

Problem 1 :

If the numerator of a fraction is increased by 2 and the denominator by 1, it becomes 1. In case, the numerator is decreased by 4 and the denominator by 2, it becomes 1/2. Find the fraction.

Solution :

Let x/y be the fraction.

Given : If the numerator is increased by 2 and the denominator by 1, the fraction becomes 1.

(x + 2)/(y + 1) = 1

(x + 2)/(y + 1) = 1

x + 2 = y + 1

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

Given : In case the numerator is decreased by 4 and the denominator by 2, the fraction becomes 1/2.

(x - 4)/(y - 2) = 1/2

2(x - 4) = 1(y - 2)

2x - 8 = y - 2

2x - y = 6 ----(2)

Solving (1) and (2), we get

x = 7 and y = 8

Hence, the fraction is 7/8.

Problem 2 :

A park charges $10 for adults and $5 for kids. How many many adults tickets and kids tickets were sold, if a total of 548 tickets were sold for a total of $3750 ?

Solution :

Let x be the no. of adult tickets and y be the no. of kids tickets sold.

Given :A total of 548 tickets were sold.

x + y = 548 ----(1)

Given : Cost of each adult ticket is $10 and kid ticket is $5 and tickets were sold for a total of $3750.

10x + 5y = 3750

2x + y = 750 ----(2)

Solving (1) & (2), we get

x = 202 and y = 346

Hence, the number of adults tickets sold is 202 and kids tickets is 346.

Problem 3 :

A number consists of three digits of which the middle one is zero  and the sum of the other digits is 9. The number formed by interchanging the first and third digits is more than the original number by 297.Find the number.

Solution :

Let x0y be the three digit number. (As per the given information, middle digit is zero).

Given : The sum of the other digits is 9.

x + y = 9 ----(1)

By interchanging the first and third digits, the number we get is

y0x

Given : The number formed by interchanging the first and third digits is more than the original number by 297.

y0x - x0y = 297

(100y + x) - (100x + y) = 297

100y + x - 100x - y = 297

-99x + 99y = 297

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

Solving (1) & (2), we get

x = 3 and y = 6

x0y = 306

Hence the three digit number is 306.

Problem 4 :

A manufacturer produces 80 units of a product at a cost of $22000 and 125 units at a cost of $28750. Assuming the cost curve to be linear, find the equation of the cost curve and then use it to estimate the cost of 95 units.

Solution :

Since the cost curve is linear, its equation will be

y = Ax + B.

(Here y = Total cost, x = no. of units)

Given : The total cost of 80 units of the product is $22000.

22000 = 80A + B

80A + B = 22000 ----(1)

Given : The total cost of 125 units of the product is $28750.

28750 = 125A + B

125A + B = 28750 ----(2)

Solving (1) and (2), we get

A = 150 and B = 10000

Then, the equation of the cost curve is

y = 150x + 10000 ----(3)

Estimate the cost of 95 units : 

Substitute x = 95 in (3).

y = 150(95) + 10000

= 14250 + 10000

= $24,250

Problem 5 :

Y is older than X by 7 years. 15 years back X's age was 3/4  of Y's age. Find the present their present ages.

Solution :

Let x be the present age of X and y be the present age of Y.

Given : Y is older than X by 7 years.

y = x + 7 ----(1)

15 years back :

X's age = x - 15

Y's age = y - 15

Given : 15 years back X's age was 3/4  of Y's age.

(x - 15) = 3/4 ⋅ (y - 15)

4(x - 15) = 3(y - 15)

4x - 60 = 3y - 45

4x = 3y + 15 ----(2)

Solving (1) and (2), we get

x = 36 years

y = 43 years

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