By moiramuldoon
Use all the information in a problem.
That’s it — that’s the whole tip. In the SAT and the PSAT, there is no extra information in math problems. So if you solve the problem without using the fact that AB is congruent to BC or the fact that ABCD is a square, then you’ve got the wrong answer. You need to use all the information.
One way to be sure that you are using all the information is to cross out pieces as you use them. As you use each piece of information in your calculations, go back to the problem and softly cross out the information. (You don’t want the word to be unreadable, in case you come back to the problem….)
By moiramuldoon
There are several tips/strategies that everyone agrees help with the math section — picking real numbers and using them to replace variables in abstract problems, for example.
One strategy that I haven’t seen much talk about is using formulas.
It’s so easy and basic — and it’s helped all of my students. Essentially, if you are stuck on a problem, you write down the formula and then just substitute information in.
Here’s an example. (This is a very hard SAT math problem…)
The average of six numbers is 210. If all the numbers are distinct positive integers and one of the numbers is 24, what is the greatest possible value of one of the remaining five numbers?
This kind of question has freaked out a lot of students, but it’s really, really easy to set up — if you use the formula.
average = sum of terms divided by the number of terms
In this case, that would be
75 = [24+x+y+z+a+b]/6
Cross multiply and you get
450 = x+y+z+a+b+24
OK, this next part requires a little thinking:
In order to simplify, first we subtract 24 from both sides and get
426 =x+y+z+a+b
(This is a step that most students miss if they haven’t used the formula).
In order to figure out what the GREATEST possible value of one of those variables is, we have to think for a second. For one number to be really big, the rest would have to be small. And if all the numbers are distinct, positive integers, then the smallest they could be is 1, 2, 3, and 4. (Remember each number has to be distinct — otherwise, we could just use 1,1,1,1).
So if x =1, y=2, z=3, and a=4,
then
426 =1+2+3+4+b
and b = 416.
There’s your answer — that’s the biggest possible number that could make the equation true…