MODELLING WITH LINEAR FUNCTIONS WORD PROBLEMS

Problem 1 :

Jimmy is having a birthday party at the zoo. The zoo has a fixed fee for birthday parties, plus a fee per person. Jimmy is told the total charge for 10 people, including himself, would be $97.50 and the total charge for 20 people, including himself, would be $175. Determine the:

a) independent and dependent variables

b) rate of change

c) initial value

d) the total charge for 17 people

e) the number of people who could come for $500

Solution :

Here two types of fees, they are collecting. One is fixed fee and other is fee per person.

y = mx + b ----(1)

Here b is the fixed charge and m is the charge per person.

(number of person, charge)  ==> (10, 97.50) and (20, 175)

a) independent variable = x (charge per person)

dependent variables = y (total charge)

b) Using formula, 

Slope = (175 - 97.50) / (20 - 10)

= 77.5/10

= 7.75

c) To find initial value, we will apply the slope 

y = 7.75x + b

Applying the point (10, 97.50), we get

97.50 = 7.75(10) + b

b = 97.50 - 77.5

b = 20

So, initial fee is $20.

d) the total charge for 17 people

Applying the value of m and b, we get

y = 7.75x + 20

Here number of people = 17

y = 7.75(17) + 20

y = 131.75 + 20

y = 151.75

e) To find the number of people. we should apply y = 500

500 = 7.75x + 20

500 - 20 = 7.75x

7.75x = 480

x = 480/7.75

x = 61.9

Approximately 62.

So, 62 is the number of people.

Problem 2 :

Jimmy is driving home from a vacation. His car is on cruise control so he maintains a constant speed. After 3 hours of driving, he is 740 km from home. After 6 hours, he is 461 km from home. Determine the:

a. independent and dependent variables

b. rate of change

c. initial value

d. distance after 8 h and 15 m.

e. time it will take him to get home?

Solution :

y = mx + b

Here b be the initial value and x be the number of hours travelling.

a) Independent variable = time

Dependent variable = Distance covered from home

b. Writing the relation as set of ordered pairs, we get

(3, 740) and (6, 461)

Slope or rate of change = (461 - 740) / (6 - 3)

= -279/3

= -93

-93 miles per hour

c. Applying the value of m,

y = -93x + b

Applying (3, 740), we get

740 = -93(3) + b

b = 740 + 279

b = 1019

y = -93x + 1019

d. 8 hours and 15 minutes = 8.25 hours

1 hour = 60 min

1 min = 1/60

15 min = 15/60 = 0.25 hours

y = -93(8.25) + 1019

y = 251.75 km

e. When y becomes 0, he has reached home.

0 = -93x + 1019

-1019 = -93x

x = 1019/93

x =10.95

So, he will take 10.95 hours to reach home.

Problem 3 :

Jimmy and Karen rented cars from the same company. The company charges an initial fee plus a charge per km. Jimmy drove 240 km and was charged $59.40. Karen drove 490 km and was charged $74.40.

Determine the:

a. rate of change

b. initial cost

c. the charge after 837km

d. the number of km you can drive for $200

Solution :

(240, 59.40) and (490, 74.40)

a) Rate of change = (74.40 - 59.40) / (490 - 240)

= 15/250

= 0.06

b) Applying the rate of change, we get

y = 0.06x + b

Applying the point (240, 59.40), we get

59.40 = 0.06(240) + b

59.40 - 14.4 = b

b = 45

y = 0.06x + 45

c) Charge after 837 km

y = 0.06(837) + 45

y = 50.22 + 45

y = 95.22

d)  y = 0.06x + 45

y = 200

200 = 0.06x + 45

200 - 45 = 0.06x

155 = 0.06x

x = 155/0.06

x = 2583.3

Problem 4 :

An insurance company has an initial charge to insure jewelry, plus a charge per dollar value of the jewelry. A ring with a value of $3500 costs $189.50 to insure. A ring with a value of $5900 costs $297.50 to insure.

Determine the:

a. Rate of Change

b. Initial charge

c. Cost to insure a $12000 ring

d. The value of a ring you could insure for $100

Solution :

Here two quantities are having relationships.

a)  Worth of ring and amount to insure. Based on worth of ring the amount to insure will differ.

(3500, 189.50) (5900, 297.50)

rate of change = (297.50 - 189.50) / (5900 - 3500)

= 108/2400

= 0.045

b) To find the initial value, let us apply m = 0.045

y = 0.045x + b

Applying the point (3500, 189.50), we get

189.50 = 0.045(3500) + b

b = 189.50 - 157.5

b = 32

y = 0.045x + 32

c) Cost of ring = 12000

y = 0.045x + 32

y = 0.045(12000) + 32

y = 540 + 32

y = 572

For the ring, we have to invest $572

d) When y = 100

100 = 0.045x + 32

100 - 32 = 0.045x

0.045x = 68

x = 68/0.045

x = 1511.1

Problem 5 :

A school decides to sell t-shirts to raise money. If they sell 20 shirts, they will lose $30. If they sell 100 shirts, they will make $650.

Determine the:

a. rate of change

b. initial value

c. number of shirts they need to sell to break even

Solution :

(20, -30) (100, 650)

a) Rate of change = (650 + 30) / (100 - 20)

= 680/80

= 8.5

Rate of change is $8.5.

b)

y = 8.5x + b

Applying the point (20, -30), we get

-30 = 8.5(20) + b

b = -30 - 170

b = -200

y = 8.5x - 200

c)  Break even point means there is no profit and loss

So, y = 0

8.5x - 200 = 0

8.5x = 200

x = 200/8.5

x = 23.5

x = 24

So, 24 shirts.

Problem 6 :

Lanny got a short term job selling computers. He is paid on commission. In order to impress customers, he bought a few nice suits. If he has $20000 in sales, he will lose $140. If he has $30000 in sales, he will make a $90 profit.

Determine the:

a. rate of change and initial value

b. the amount he needs to sell to break even

c. The amount he needs to sell in order to make $1000 profit

Solution :

(20000, -140) and (30000, 90)

a) Rate of change = (90 + 140) / (30000 - 20000)

= 230/10000

= 0.023

b) Break even point, when y = 0

y = 0.023x + b

Applying the point (20000, -140), we get

-140 =  0.023(20000) + b

-140 - 460 = b

b = -600

y = 0.023x - 600

c)

applying y = 0

0 = 0.023 x - 600

600 = 0.023x

x = 600/0.023

x = 26086

d) Profit = 1000

y = 1000

1000 = 0.023 x - 600

1000 + 600 = 0.023x

1600/0.023 = x

x = 69565