SOLVING WORD PROBLEMS USING LINEAR MODELS

Plan for Solving a Word Problem :

(i) Find out what numbers are asked for from the given information.

(ii) Choose a variable to represent the number(s) described in the problem. Sketch or a chart may be helpful.

(iii) Write an equation that represents relationships among the numbers in the problem.

(iv)  Solve the equation and find the required numbers.

(v) Answer the original question. Check that your answer is reasonable.

A linear function y = mx + b can be used as a model for many types of real life word problems which involve a constant rate of change.

Example 1 : 

A person travels home from work at a constant speed. Ten minutes after leaving work he is 20 miles from home, and 20 minutes after leaving work he is 12 miles from home. If he continues to travel at the same speed, how long will it take him to arrive home from work ?

Solution :

The problem asks for the number of minutes it takes to travel from work to home.

Start with the linear equation y = mx + b, in which y is the distance in miles from home, and x is the time in minutes.

When x = 10, y = 20

When x = 20, y = 12

First Equation : 

20 = 10m + b

Second Equation : 

20 = 10m + b

By subtracting the second equation from the first equation we get

8 = -10m

Divide each side by -10.

-⁴⁄₅ = m

Substitute m = -4/5 into the first equation. 

20 = 10(-⁴⁄₅) + b

20 = -8 + b

Add 8 to each side.

28 = b

In y = mx + b, replace m with -⁴⁄₅ and b with 28.

y = (-⁴⁄₅)x + 28

When he arrives home, y = 0. 

0 = (-⁴⁄₅)x + 28

Solve for x. 

(⁴⁄₅)x = 28

Multiply each side by 5.

4x = 140

Divide both sides by 4.

x = 35

It takes 35 minutes from work to home.

Example 2 :

At the beginning of a trip, the tank of Chloe’s car was filled with 12 gallons of gas. When she travels constantly on the highway 60 miles per hour, the car consumes 1 gallon of gas per 35 miles. If she traveled 5 hours and 15 minutes on the highway with a constant speed of 60 miles per hour, how many gallons of gas are left in the tank?

(A)  3

(B)  4

(C)  5

(D)  6

Solution :

Start with the linear equation y = mx + b, in which y is the number of gallons gas left in the tank after x miles of distance travelled.

It is given that the car consumes 1 gallon of gas for every 35 miles of distance. That is, amount of gas left in the tank is decreasing at the rate of 1 gallon per 35 miles.

m = -¹⁄₃₅

y = mx + b

y = (-¹⁄₃₅)x + b

Here, b is the amount of gas in the tank initially, that is, 12 gallons.

y = (-¹⁄₃₅)x + 12 ----(1)

Distance travelled in 5 hours 15 minutes :

= speed  time

= 60  5¼

= 60  ²¹⁄₄

= 15  21

= 315 miles

Substitute x = 315 in (1).

y = (-¹⁄₃₅)(315) + 12 

y = -9 + 12

y = 3

After Chloe traveled 5 hours and 15 minutes, amount of gas left in the tank is 3 gallons.

Example 3 :

A rock climber is climbing up a 450 feet high cliff. By 9:30 am. the climber reached 90 feet up the cliff and by 11:00 am, he has reached 210 feet up the cliff. If he climbs with a constant speed, by what time will he reach the top of the cliff?

(A)  1 : 45 pm

(B)  2 : 00 pm

(C)  2 : 15 pm

(D)  2 : 30 pm

Solution :

Start with the linear equation y = mx + b, in which y is the height climbed in x hours.

We can assume the following values corresponding to thew 9:30 am and 11.00 am.

9.30 am ----> 0 hours

11.00 am ----> 1.5 hours

Average rate of climbing :

= 80 feet per hour

y = mx + b

y = 80x + b

Here, b is the initial height. That is 90 feet at 9.30 am (or 0 hours).

y = 80x + 90

At the top of the cliff, y = 450.

450 = 80x + 90

360 = 80x

x = 4.5 hours

4.5 hours later from 9 : 30 am is 2 pm.

Therefore, the correct answer is option (B).

Example 4 :

In 2005 a house was purchased for $280,000 and in 2013 it was sold at $334,000. Assuming that the value of the house increased at a constant annual rate what would be the price of the house in the year 2018?

(A)  $354,250

(B)  $361,000

(C)  $367,750

(D)  $374,500

Solution :

Start with the linear equation y = mx + b, in which y is the height climbed in x hours.

We can assume the following values corresponding to thew 2005, 2013 and 2018.

2005 ----> 0

2013 ----> 8

2018 ----> 13

Average rate of change in the price of the house :

= $6,750 per year

y = mx + b

y = 6,750x + b

Here, b is the initial height. That is $280,000 in 2005.

y = 6,750x + 280,000

To find the price of the house in the year 2018, substitute x = 13.

y = 6,750(13) + 280,000

y = 87,750 + 280,000

y = 367,750

Therefore, the correct answer is option (C).

Example 5 :

To join Eastlake Country Club one must pay d dollars for a one time membership fee and pay w dollars for a monthly fee. If the first month is free for the club, what is the total amount, y, x months after a person joined the club, in terms of d, w, and x?

(A)  y = wx - 1 + d

(B)  y = w(x - 1) + d

(C)  y = d(x - 1) + w

(D)  y = dx - 1 + w

Solution :

Since the first month is free for the club, the amount of monthly fee x months after a person joined the club is w(x - 1), and the total amount including the one time membership is

y = w(x - 1) + d

Therefore, the corrrect answer is option (B).

Example 6 :

From 1990 to 2000 The population of city A rose from 12,000 to 28,000 and the population of city B rose from 18,000 to 24,000. If the population of the two cities increased at a constant rate, in what year was the population of both cities the same?

Solution :

Average rate of change in the population of city A :

= 1,600 per year

Average rate of change in the population of city B :

= 600 per year

Let x be the number of years from 1990 and y be the population after 1990.

The population of city A after 1990 would be

y = 1600x + 12000

The population of city B after 1990 would be

y = 600x + 18000 

To find the year the population of both cities were the same, let the two equations be equal.

y = y

1600x + 12000 = 600x + 18000

1000x = 6000

x = 6

The year which is 6 years from 1990 is 1996.

In the year 1996, the population of both cities was the same.

Example 7 :

An empty 1,200 gallon tank is filled with water at a rate of 6 gallons of water per minute. At the same time, another 1,200 gallon tank full of water is being drained at a rate of 9 gallons per minute. How many minutes will it take for the amount of water in both tanks to become the same?

Solution :

Let y be the amount of water in the both the tanks after x minutes.

The empty 1,200 gallon tank is filled with water at a rate of 6 gallons of water per minute.

Amount of water in the emty tank after x minutes :

y = 1200x

The empty 1,200 gallon tank is filled with water at a rate of 6 gallons of water per minute.

Amount of water in the emty tank after x minutes :

y = 6x

Another 1,200 gallon tank full of water is being drained at a rate of 9 gallons per minute.

Amount of water in the second tank full of water after x minutes :

y = -9x + 1200

When the amount of water in both tanks to become the same,

y = y

6x = -9x + 1200

15x = 1200

x = 80

It take 80 minutes for the amount of water in both tanks to become the same.

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