Last week, guest author, **Dr. Steve Warner** previewed the changes coming to the re-designed SAT next March. In addition to a detailed explanation, he posed a sample math question and now provides an in-depth answer here. Dr. Warner’s latest book, **New SAT Math Problems**, is still available at a discount on Amazon.

Gary takes a New York City cab 6 miles to work and must pay $17.50 for the cab ride. After work, Gary takes another New York City cab 10 miles to visit his family and must pay $27.50. During both of these rides, Gary was charged a “drop fee” (an initial charge when the cab’s meter is activated) of *d* dollars, plus an additional *m* dollars for each 1/5 of a mile travelled. What is the value of* md *?

**Solution:** We can model the information here with a linear equation *C* = *mx* + *d*, where *m* is the cost of the cab ride per 1/5 mile, *d* is the drop fee, and *C* is the total cost for a cab that travels *x/*5 miles.

We are essentially given two points on the line, but we should convert miles to 1/5 miles as we write down the points. So 6 miles is equivalent to an *x*-value of 6 ⋅ 5 = 30 units of miles, and 10 miles is equivalent to an *x*-value of 10 ⋅ 5 = 50 units of 1/5 miles. It follows that the two points given are (30,17.50) and (50,27.50).

We can now find the slope of the line passing through these two points: m=(27.50 – 17.50 )/(50 – 30)= 10/20 = .50.

Using the point (30,17.50) and the slope *m* = .5 we can now write an equation of the line in point-slope form: *C *– 17.50 = .5(*x *– 30).

Distributing the .5 on the right hand side of the equation gives us C – 17.50 = .5*x* – 15, and then adding 17.50 to each side of this last equation gives us *C* = .5*x *+ 2.50.

It follows that *m* = .5 and *d* = 2.50. So *md* = **1.25** or 5/4.

**Notes: **(1) The slope of a line passing through the points (*x*1,*y*1) and (*x*2,*y*2) is

Slope = *m* = rise/run = *y2* –*y1 / **x2 –x1*

In this problem the two points are (30,17.50) and (50,27.50).

(2) Using the points (6,17.50) and (10,27.50) would have led to a wrong answer (unless a conversion was done at the end) because is being measured in 1/5 miles and not miles.

(3) The** slope-intercept form of an equation of a line **is** ***y*** = mx + b**

**where**

*m*is the slope of the line and b is the y-coordinate of the

*y*-intercept, i.e. the point (0,

*b*) is on the line. Note that this point lies on the

*y*-axis.

In this problem we used the letter *C* (for cost) instead of *y*, and *d* (for drop) instead of *b*.

It turned out that *m* = .5 and *d* = 2.50.

(4) The **point-slope form of an equation of a line** is

*y – y0 = m (x—x0)*

where ** m **is the slope of the line and

**(**) is any point on the line.

*x*0,*y*0In this problem *m* = .5 and (*x*0,*y*0) = (30,17.50).

(5) We could have used the point (50, 27.50) instead when writing an equation of the line in point-slope form. In this case we would get *C* – 27.50 = .5*x* – 50. I leave it to the reader to show that when you solve this equation for *C* you get the same slope-intercept form as before.

(6) As an alternative to using point-slope form, after finding *m*, we could plug one of the points into the slope-intercept form of the equation and solve for *d* as follows:

*C* = .5*x* + *d*

17.50 = .5(30) + *d*

17.50 = 15 + *d*

*d* = 17.50 – 15 = 2.5

We now have *m* = .5 and *d* = 2.5, so that *md* = 1.25 or 5/4 .

(7) If we were to take our points as (6,17.50) and (10, 27.50), we would get a slope of *m* = (27.50 – 17.50)/(10 – 6) = 10/4 = 2.50. This is the cost per mile. We can change this to the cost per 1/5 mile by dividing by 5 to get 2.50/5 = .5 as before.

(8) **Technical note: **In reality a NYC cab also charges *m* dollars for each 60 seconds of “wait time” when the cab is stuck in traffic. The question given here avoided this extra complication.