Miguel is centimeters tall. At 2:00 p.m. one day, his shadow is centimeters long, and the shadow of a nearby fence post is centimeters long. In terms of , what is the height, in centimeters, of the fence post?
Note that the answer on the SAT website follows a correct calculation but marked the wrong answer as the correct one. Don’t be alarmed – this only shows there are humans on the other side of the exam !
Lets try and see what would be a good approach to this question. We should also analyze what makes it confusing and work on techniques to reduce the mess.
The questions starts with a statement:
How could we read this information? f() means that if you place a variable inside the function will spit out what ever is written on the other side. We also note that there are limitation on what we can place in the function (also known as the Domain of the function) because we can only place n that are integers (good ! No fractions !).
What does this function do? Lets take an integer n – how about 7? If we calculate f(2 times 7 = 14) we will get 2 times the value of f(7) – whatever that maybe.
OK – not too bad – this defines our function and we understand what it can do.
What other information are we given?
OK – so when we place into the function the value 4 we get back 4.
That means that f(2 times n)=2 times f(n) and if 2 times n = 4 then n must be 2. So f(2X2)=2f(2)=4 so f(2) should be 2.
Now we are given options to chose from what f() actually does.
Since we combined the two pieces of information and concluded f(2) must give us 2 and we know f(4)=4 we can check which functions actually spit out these values when we plug in 2 or 4.
By elimination only B satisfies both conditions.
In a class of seniors, there are boys for every girls. In the junior class, there are boys for every girls. If the two classes combined have an equal number of boys and girls, how many students are in the junior class?
We now have to decide what would be the best strategy to solve. Lets consider the information.
Class I – the senior class – has 80 students with a 3:5 boys to girls. This implies that out of every 8 students (3+5) 3 are boys and 5 are girls. This mean there are 30 boys and 50 girls in this class.
Class II- the junior class – has ? number of students with 2 girls for every 3 boys or 3:2 boys to girls. We don’t know how many students are in the junior class and actually this is the number we are looking for as our answer.
What else do we know? We know that when mixed together the resulting population has a 1:1 ratio of boys to girls or equal number of boys and girls.
Several possible ways to solve.
One – write an equation – if X is the number of students in the junior class then the number of boys in the class is 3/5 of X and the number of girls is 2/5 of X.
That means that in the combined class the number of all the boys is 30 + (3/5)*X and the number of girls is 50+ (2/5)*X. However these two expressions are equal !
We then write an equation: 30 + (3/5)*X = 50+ (2/5)*X or when simplified (1/5)*X=20 and then X=100
Two – plug in. We know that the junior class has a ratio of 3:2 – it will make sense then that the total number of students in the junior class will be a multiplicity of 5 (answers A and C can be ignored). If we plug in 80 we have 48 boys and 32 girls – but we need a difference of 20 to account for the excess girls in the senior class. If we check 120 we get 72 boys and 48 girls – too many. You can see the 100 is the correct answer – plug in is a legitimate option, wise plug in is even better.
Three – If you understand that the junior class should have 20 more boys than girls and that these 20 boys are 1/5 of the class because there are 3/5 boys and only 2/5 girls in the junior class – it is straight forward to find the answer – 100 students in the junior class.
Now – let see what this question is really about!
It is a “What can’t be” question because it asks
Which mean we need to know something about each option.
There are 3 points of a function that are given to us and from their coordinates we need to conclude something about the function.
What happens between the points given? who knows? depends on the function.
It is now wise to remember that in the SAT
we know (or I hope you know) that a line is defined by two points. This seems the most easy option to disprove. If the rise between X=0 to X=1 is 5 =[-4-(-9)] then if we move another two units to the right (to get to X=3 – our run) we should add another 10 units to the rise and get to [10+(-4)]=6 but when X=3 we have 0 – which means these three points are not on a line- QED
Now lets look on the other options – the idea is that a. we can learn something b. we are not always going to be smart and check the option that is the right answer first.
All we need to find or imagine is a case in which these points will fulfill the statement – remember the way this question is phrased means we are looking for an answer that is always NOT TRUE.
Could be – if it starts at -9 goes up to -4 and continue to 0 – it makes sense it has a maximum somewhere. Those of you who have taken Calculus can relate to some of the fundamental theorems of Calculus !
So we agree this is not always wrong (in fact it is always true)
Well lets see – we have three y values: -9, -4, and 0 all are equal or smaller than zero. It is possible the graph never crosses the X axes – Does the graph have to cross the X axes and have values greater than zero? NO – so this statement is not always false and can be at some instances true and thus we can cross it out.
WHAT? well lets think what this statement means. Draw a few points on the bottom half of a piece of paper and connect them. Now fold the paper in the middle and copy the points to the upper part of the paper. Connect the points on the top part. You just created a symmetric function with respect to the line created by you folding the paper.
Now can you imagine you can do the same with the 3 points given? probably yes – so this option goes away because it can be true and is not always false.
We discussed this is impossible. and this is the answer to choose.
Well, is the graph of a parabola (y=aX^2+bX+c) can be drawn to include these points? yes – we can imagine a part of a parabola passing through any three given points – just sketch them and see.
Notice we have all the types of answers here – the correct one – an answer that can never be
Answers that are always true and answers that can be true but are not necessarily so.
Be ready for this kind of a mix. Read the answers as part of your reading of the question and always start with the one that look the most easy to check.
In the mid of Hurricane Sandy it seems a little out of place to think of it in terms of math, however someone like me, watching nature’s powers unfold, thinks about the beautiful ways in which physical powers manifest themselves and do so according to laws we can describe using mathematics.
If you want, Mathematics is one language we humans use to describe Nature. One can see this use at work by going to the NOAA website – you can see how math can help predict, using sophisticated mathematical models, how this storm will unfold as well as advanced mathematical modeling for other projects http://www.nesdis.noaa.gov/Research.html .
Hope this storm will not cause too much damage. Keep safe.