SAT Word Problems

 

Word Problems using the Point-Slope and Slope-Intercept Equations of a Line

 

In this section, you will learn how to:

 

  • Determine what word problems can be solved using a linear equation
  • Use the information in the problem to create a linear equation to describe the situation

 

When you are given word problems in which something happens at a constant rate—growing at the same rate, something being used at the same rate—and you have data that could be arranged in ordered pairs—see if you can solve the problem by creating a linear equation.

 

Example 1:     Charlie and his family are living through another Florida hurricane.  Their electricity has been off for 3 straight days.  When Charlie lights a candle, it is 6 inches high.  After 2 hours, it is 4 inches high.  If the candle burns at a constant rate, how tall will be after burning for 3 hours?

 

We can solve this problem by creating a linear equation. Charlie lights the candle at time t = 0.  We have the ordered pair (0, 6) which tells us that at time 0, the candle is 6 inches high.  We also have the ordered pair (2, 4) which tells us that after 2 hours, the candle is 4 inches high.

 

Because we have two ordered pairs, we know enough to find the equation of the line using the point-slope form y – y1 = m(x – x1).

 

First, find the slope by using the ordered pairs.  Let (x­1, y1) = (0, 6) and (x2, y2) = (2, 4).

 

The slope, me, = .

 

The slope is negative because the candle decreases in length as time goes on.

 

Now that we know the slope, we can substitute into the equation using either point.  I am using (0, 6).  y – 6 = –1(x – 0); y – 6 = –1x or y = – x + 6.

 

Since x is equal to the time and we want time = 3, substitute 3 into the equation and get

 y = –3 + 6 = 3.

 

The candle is 3 inches tall after it has burned for 3 hours.

 

Example 2:  Rhonda is saving money for a trip to a cheerleading competition.  Her grandmother gives her $100 and she gets a job babysitting for $6 an hour. Write an equation in slope-intercept form to describe how much money she has at any given time.

 

Slope represents the rate of change and Rhonda’s money increases by a constant rate of $6 per hour.   Before she has worked a single hour—in other words, when she has worked 0 hours, she has $100.  We can represent this as (0, 100) and recognize this as a y-intercept.  Putting the equation in slope-intercept form—y = mx + b—we have

 y = 6x + 100.

 

Try the following problems.

 

 

  1. When a tree is planted, it stands 7 inches tall.  After 3 years, it stands 40 inches tall.  How tall will the tree be after 10 years?

 

  1. Charlie’s young cousin, Joshua, is 50 inches tall when he is 8 years old.  He is 60 inches tall when he is 12 years old.  If he continues to grow at the same rate, how tall will he be when he is 14?

 

  1. When Joshua was 8, he weighed 68 pounds.  When he was 10, he weighed 80 pounds.  If he continues to grow at the same rate, how much will he weigh when he is 12 years old?

 

  1. Jessica is selling Girl Scout cookies.  After1 hour, she has sold 5 boxes and after 3 hours, she has sold 15 boxes.  Write a linear equation to describe this situation.

 

  1. In problem 4 above, if Jessica sells for 7 hours, how many boxes of cookies has she sold?

 

6.  Charlie is doing a study of Florida’s rainfall levels during hurricanes.    Fortunately, he picked a good year because that year 4 hurricanes hit Florida.  During Hurricane Jeanne, 1.5 inches of rain fell in the first ½ hour and 3 inches of rain fell during the first hour. If the rain continues to fall at a constant rate (after the first half hour), how much rain falls in 5 hours?

 

  1. When the stock exchange opens, a stock is valued at $8.  After two hours of trading, the stock is valued at $5.  Assuming the stock price changes at a constant rate, what is the price of the stock after 3.5 hours of trading?

 

  1. Rhonda’s cheerleading team holds a car wash to raise money for its trip to the state competition.  After 2 hours, the team has raised $50.  After 3 hours, the team has raised $75.  How much money will the team have raised after 7 hours, assuming they earn money at a constant rate?

 

  1. A student is saving money for college.  His grandmother gives him a $500 savings bond for a Christmas present.  Afterwards, he gets a job for $7 an hour at the car wash.  Write a linear equation describing how much money the student will have if he works for h hours.

 

 

  1. A kid is selling magazine subscriptions door-to-door.   He earns $5 for each subscription sold.  If he sells more than 10 subscriptions, he earns a bonus of $2 for each additional subscription.  Write an equation describing how much money he makes.  Let n = number of subscriptions and assume he sells more than 10 subscriptions.

 

 


Answers

 

  1. 117 inches.  Ordered pairs are (0, 7) and (3, 40).  The slope is 11. Equation is y – 7 = 11(x – 0) or y = 11x + 7.  Substituting, we get y = 11(10) =+ 7 = 110 + 7 = 117.

 

  1. He will be 65 inches tall.  Ordered pairs are (8, 50) and (12, 60).  The slope is 2.5 and the equation is  y – 50 = 2.5(x – 8) which becomes y = 2.5x + 30; y = 2.5(14) + 30 = 35 + 30 = 65.

 

  1. He will weigh 92 pounds.  The ordered pairs are (8, 68) and (10,80).  The slope is 6; equation is y – 80 = 6(x – 10 ), which becomes  y = 6x + 20.

 

  1. y = 5x.  The ordered pairs are (1, 5) and (3, 15).  The slope is 5.  y – 5 = 5(x – 1) ;

 

y – 5 = 5x – 5; y = 5x.

 

  1. 35 boxes. If y = 5x and x = 7, y = 5(7) = 35.

 

  1. 15 inches.  The equation is y = 3x 

 

  1. The stock will be worth $2. 75.  The ordered pairs are (0, 8) and (2, 5).  The slope is

 

– 1.5.  y – 8 = –1.5 (x – 0).  Therefore, y = – 1.5x + 8.  After 3.5 hours of trading, the stock is worth:

 

 y = (3.5) + 8.  This is equal to – 5.25 + 8 = 2.75.

 

8.  The group raises $175 after 7 hours.

 

(2, 50) (3, 75)

 

slope: 

y

y = 25x

 

After 7 hours, the group has raised y = 25(7) = 175

 

  1. y = 7h + 500

 

We are already given the slope: it is 7.  Remember that slope indicates a rate of increase and the wages increase by a constant $7 per hour. The y-intercept is (0, 500).  In slope-intercept form, this is y = 7h + 500.

 

10.             M = 50 + 7(n

 

    1. Money made by selling the first 10 subscriptions = 10(5) = 50.
    2. Money made by selling any number of subscriptions greater than 10:

7(n – 10).  Notice that we must subtract 10 from n because those first 10 subscriptions do not earn the seller $7 but only $5. 

    1. The total money earned, M, is the amount of money earned selling the first 10 subscriptions plus the total amount of money earned selling additional subscriptions after the first 10.
    2. M = 50 + 7(n

 

 

 

 

 

Finding Missing Constants in Functions

 

In this section, you will learn how to find the value of a missing constant in a linear or quadratic equation.

 

Example 1:     A line with equation y =  – x + k passes through the point (2, 5). Find the value of k.

 

Begin by replacing x in the equation with 2 and y by 5.

 

5 = – 2 + k.

 

Now, solve for k by adding 2 to both sides:

 

5 + 2 = – 2 + 2 + k

 

7 = 0 + k

 

7 = k

 

Hence, k = 7 and the equation is y = – x + 7.

 

Example 2:     The function f(x) = x2 + 2x + c.  The function passes through the point

(– 1, 1).  Find the value of c. Then, write the equation.

 

In the function, replace x with –1 and f(x), which is equivalent to y, with 1. 

 

1 = (–1)2 + 2(–1) + c

1 = 1 – 2 + c

1 = – 1 + c

2 = c.

 

The equation becomes f(x) = x2 + 2x + 2.

 

Example 3:     The functions f(x) = x2 and g(x) = a – x2 intersect at the point (3, 9).  Find the value of a.  Then, find the equation of g(x).

 

Because f(3) and g(3) intersect at the same point, 9, we can set f(3) and g(3) equal to each other: remember that f(3) = 32 = 9 and g(3) = a – 9

9 = a – 9.

18 = a. 

g(x) = 18 – x2

 

Example 4:     If y = 3x + k and the line passes through the point (s, s + 2), find an expression for k.

 

Replace x with s and y with s + 2: s + 2 = 3s + k. Solving, get k =  –2s + 2.


Try the following problems.

 

  1. If f(x) = x + k and the line passes through (0, 2), find k.

 

  1. If f(x) = and the line passes through (s, s – 2), find k.

 

  1. If f(x) = 2x2 + 3x + c and the graph passes through (1, 2), find c.

 

 

  1. If two functions f(x) = x2 + 2x + 1 and g(x) = a – x2 intersect at (0,1), find a.

 

 

 

  1. If f(x) = (x – a)2 and g(x) = x2 intersect at (1,1), find the value of a.

Answers

 

1.      k = 2

2.      k =

3.      c = – 3

 

4.      a = 1.

 

Explanation:  The two equations intersect at (0,1).  Thus: x2 +2x + 1 = a – x2 when x = 0.

Since f(0) = 1 and g(0) = a,

 

1 = a

 

 

5.      a = 2. 

 

When x = 1, f(1) = (1 – a)2 and g(1) = 1.  Because f(1) and g(1) are the same point, set these expression equal:  (1 – a)2 = 1. 

 

There are several ways to solve from here:

 

  1. Use the square root property and take the square root of both sides.  Remember that the square root of 1 is  and you must consider both values.

 

1 – a = 1. This means that a = 0.  This is not practical because it means f and g are the same function.  We are assuming that they are different functions.

 

Also consider 1 – a =  –1.  This gives a = 2, which works.

 

  1. You can also square the binomial (1 – a)2 = (1 – a)(1 – a) = 1 – 2a + a2

 

Therefore, 1 – 2a + a2 = 1. Subtracting 1 from both sides, we get – 2a + a2 = 0.

 

We can also write this as a2 – 2a = 0.

Factor: a (a – 2) = 0.

Set each factor equal to 0 and solve for a: a = 0 and a – 2 = 0.  Thus, a = 0 and a = 2.

We can disregard a = 0 because it would make f and g the same function.  Therefore, the answer is a = 2.

 

 

 

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