Math Changes to the "New" SAT

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 = .5x – 15, and then adding 17.50 to each side of this last equation gives us C = .5x + 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 (x1,y1) and (x2,y2) 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(x0,y0) is any point on the line.

In this problem m = .5 and (x0,y0) = (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 = .5x – 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 = .5x + 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.