2010 PSAT - Saturday - Section 4 (Math)

The following are TestTakers' suggestions for solving the problems on the October 16th, 2010 PSAT, Section 4 (a math section). Click "Read More" to view!

21. Do yourself a favor and save 20 seconds by solving for the expression instead of doing the algebra here. It's good practice for doing so on harder questions! They give us 3x + 5 = 9, and they want 6x + 10.  How do you get from 3x + 5 to 6x + 10?  You double it!  So you also double the other side of the equation.
2(3x + 5) = (9)2
6x + 10 = 18
(B) is the answer.

22. When given fractions and asked to compare them, CONVERT THEM TO DECIMALS! You're looking for the fraction that lies between 1/2 (or 0.5) and 3/4 (or 0.75) on a number line. Only choice (C) works; 5/8 = 0.625.

23. This is a perfect example of a question that looks much harder than it is. All they're asking you to do is see which category increased the most in percent of total expenses from 1929 to 1997. So all you need to do is subtract each category in 1929 from the same category in 1997. You'll see Medical Care saw the greatest increase, from 4% to 17%.  Answer: (A).

24. A few ways to think about this one. First, you could just count the sides of each figure, since they tell you all the sides are of equal length. An octagon has 8 sides, a pentagon has 5. The ratio of their sides, if they have sides of the same length, has to be 8 to 5.

If you want to be sure, plug-in! Say each side has a length of 2. Now the perimeter of the 5-sided figure is 10, and the perimeter of the 8-sided figure is 16. The ratio of the 8-sided figure to the 5-sided one is 16 to 10, which still simplifies to 8 to 5. Answer choice (A).

25. Backsolve! They tell us the length of the rod is 16 feet, so a and b must add up to 16. Use that to see which answer works:

Answer choiceab (=16 - a)30a = 50b ?Comment
(C)88240 = 400NO!  b too big; make a bigger.
(D)106300 = 300YES!

26. Circle questions can usually be solved with ratios, and this one is no exception. We're going to solve this using "part to whole" ratios:

What we have here is a part of the area of the circle, corresponding to a 240° angle (not marked, but we know a circle has 360° and 120° makes up the unshaded region, so we subtract).

Let's find the area of the whole circle. The radius is 6, so our area (using πr2) is 36π.

Now let's put everything we know into our ratio to solve:
Solve that, and you'll get the area of the shaded region is 24π. Answer choice (D).

(You could also speed this up a bit by recognizing that you have 2/3 of the circle shaded, and just taking 2/3 of the area of the whole circle. Note that really, this is exactly the same thing as doing "part to whole.")

27. LOOK OUT! This is one of the most brutal RTFQ traps I've ever seen. Read the graph carefully and you'll notice that the grid lines are in units of two. If you made a mistake on this one, I bet it's because you picked (B) because you didn't read carefully. The real answer, of course, is (E).

28. PLUG-IN! Say x = 2. Now Cylinder 1 has a radius of 1 and a height of 8. Cylinder 2: radius = 4, height = 2. Cylinder 3: radius = 2, height = 4.

We can calculate the volume of a cylinder using the equation provided for us at the beginning of the section:
V = πr2h
For each of our 3 cylinders using our plugged-in values:
V1 = π(1)2(8) = 8π
V2 = π(4)2(2) = 32π
V3 = π(2)2(4) = 16π
Now just put them in order:
V1 < V3 < V2
That's answer choice (E).

29. Don't be thrown by the notation, just follow the instructions! 17 divided by 2.5 gives us 6.8.  Round that up to the nearest whole number like the question says, and you get 7.  Done.

30. Guess and check this one using factors of 24 for x and y.  The bigger factor has to be x, since x - y = 10.

x - y
So x has to be 12!

31. This one is tricky. A good way to think about it is to imagine the diagram to have started like this:
And then imagine someone having LIFTED the angle on the left from point H:

The angle to which you lift H will be the same as angle JPK...so x° is 180° - 166°.  The answer is 14.

32. This is another tricky question because of the order the information is given, but if you carefully organize the information, there's really no difficult math. Here's what the question says, and the answer, in diagram form:

33. Just be careful here and you'll be fine. All they want is the area of the triangle, and they give you both the base and the height. Just remember -- your base is AC, so you need to add AD and DC to get it. So our base is 1, and our height is 2/3, and once we remember that the formula for the area of a triangle is given to us in the beginning of the section, we have everything we need to calculate the answer:

34. Plug-in! They tell us all three variables are POSITIVE integers, which means we need to pick numbers for x and y that will make 45x - 60y a POSITIVE number. Since they want the LEAST POSSIBLE value for x, let's look at what happens when we make x and y both equal to 1:
 z = 45x - 60y
z = 45(1) - 60(1)
z = -15
No good, but that should give us a good idea where to go next: just make x = 2 (and leave y alone):
 = 45x - 60y
z = 45(2) - 60(1)
z = 30
We've shown x CAN'T be 1 and it CAN be 2, so 2 is the least possible value for x.

35. Plug-in AGAIN! This time, we just need to pick a number for n. Let's say it's 2. that means we have 12 women and 16 men in our group. What's the probability of randomly picking a woman out of the group? To answer this we just need to add 12 and 16 to get our total number of people (28), and then divide to get the probability. There's a 12/28 chance of picking a woman, which simplifies to 3/7. You could also grid in the decimal: .428 or .429 (rounding is optional; both answers would have been accepted).

36. "Directly proportional" means we're going to be doing ratios. Again, as long as you're careful here and don't get intimidated, you'll be fine:
Substitute and simplify:

Cross-multiply and solve: 

37. Ok, there are two ways to think about this one. First, pick a slope between 0.5 and 0.6 (I choose 0.55) and solve the slope formula:
 Substitute our coordinates and slope:

The range of the answers is: 5 < b < 6

You can also solve this question by plotting one of the given boundaries (it's much easier to use the slope of 0.5). Start at point (1, 0), and count. A slope of 0.5 is a slope of 1/2, which means the graph will rise (travel up the y-axis) 1 for every 2 it runs (or travels down the x-axis). So if it goes through (1, 0), it also goes through (3, 1), (5, 2), (7, 3), (9, 4), and (11, 5). Remember, we're looking for the b in (11, b). If the slope we're looking for is higher than 0.5, then b should be slightly higher than 5. 5.1 should be fine. Obviously, this method is slightly less exact, but it'll work in a pinch, like if you can't remember the slope formula on the test.

38. You really just need to know the math here (of course it's hard; it's the last question!). Ray has 7 pictures, 6 of which can go anywhere, and one of which (the one of his parents) needs to go in a certain spot. When you're trying to figure out how many possible arrangements there can be, you multiply possibilities*. Let's make a list of spaces for pictures, and count the possibilities for pictures that could go there:

1. 6 possibilities (any picture but the one of the parents)
2. 5 possibilities (any but the one we used in the first spot, and the parents)
3. 4 possibilities (follow the pattern...)
4. 1 possibility (this is where we put the parents)
5. 3 possibilities (we've already placed the other 4)
6. 2 possibilities (again...the rest have already been placed)
7. 1 possibility (the rest are all placed)

Then we multiply all the possibilities: 
And that's our answer: 720.

*Technically this is a permutation without replacement...more info on how to solve this kind of question here although I cannot stress enough how rare this kind of question is on the PSAT and SAT, especially when it's not the hardest question on the test. If you're only worried about your SAT score, you're really safe to ignore this.