SAT Math Solutions – Part 2 Two days ago I gave you two math problems for the revised SAT. To see that post click the following link: SAT Math Problems for the Revised SAT You can find a solution to the first problem here: SAT Math Problems and Solutions for the Revised SAT Today I will solve the second problem. Level 2 Data Analysis A puddle with an area of 500 cm2, is monitored by scientists for the number of paramecia present. The scientists are interested in two distinct species, let’s call them “species A” and “species B.” At time t = 0, the scientists measure and estimate the amount of species A and species B present in the puddle. They then proceed to measure and record the number of each species of paramecium present every hour for 12 days. The data for each species, were then fit by a smooth curve, as shown in the graph above. Which of the following is a correct statement about the data above? (A) At time t = 0, the number of species B present is 150% greater than the number of species A present. (B) At time t = 0, the number of species A present is 75% less than the number of species B present. (C) For the first 3 days, the average growth rate of species B is higher than the average growth rate of species A. (D) The growth rate of both species A and species B decreases for the last 8 days. Solution: The last 8 days correspond to times t = 4 through t = 12. During this time, the growth rate of both species is decreasing. So the answer is choice (D). Notes: (1) 300 = 200 + 2(100), and therefore 300 is 200% greater than 100. This eliminates choice (A). (2) We can also use the percent change formula Here the original value is 100 and the change is 300 – 100 = 200. It follows that Percent Change = (200/100)*100 = 200%. (3) To eliminate choice (B) we can use the percent change formula again with original value 300 and change 300 – 100 = 200: .Percent Change = (200/300)*100 = 200/3 = 66 2/3%. (4) We can compute the average growth rate over the interval from t = a to t = b, by computing the slope of the line passing through the points (a,f(a)) and (b,f(b)). That is, we would compute m =(f(b) – f(a))/(b – a). For example, over the first 3 days, the average growth rate of species A is approximately (400 – 100)/(3 – 0) = 300/3 = 100 paramecia per day. The two points we used here were (0,100) and (3,400). Similarly, over the first 3 days, the average growth rate of species B is approximately (350 – 300)/(3 – 0) = 50/3 = 16 2/3 paramecia per day. The two points we used here were (0,300) and (3,350). This eliminates choice (C). (5) It should be noted that we do not actually need to compute the growth rates to determine which growth rate is higher. We can simply look at the “steepness” of the two curves. An easy way to do this is to draw a “tangent line” to each curve at the point where we wishto examine the growth rate. Here is an example of such an analysis at t = 2: Notice that the tangent line for species A rises faster than the tangent line for species B. This shows that species A is growing faster than species B at time t = 2. (6) We can use a similar analysis as we did in note 5 to see that the growth rate is decreasing for each species between times t = 4 and t = 12. Here as an example of such an analysis for Species B: Notice that the tangent line at time t = 4 is steeper then the tangent line at time t = 8. This suggests that the growth rate of species B is decreasing from t = 4 to t = 12. If you liked this article, please share it with your Facebook friends: And don’t forget to check out my social media pages: Comments comments