Test 4 Section 9 - #16 (page 613)

Plug in. Say x = 2. Then y = 2 + 1/2 = 2.5. Now try your conditions:

I. y ≠ x
Yep, that's true.

II. y is an integer.
Nope, not true. 2.5 is not an integer.

III. xy > x²
This is the tricky one, but it is true with our plug-in -- (2.5)(2) > 2². It will ALWAYS be true because y will always be slightly bigger than x.

So the answer is I and III are true. That's (D).

Test 4 Section 9 - #15 (page 613)

Plug in to solve this one. (Note, it doesn't matter if the equation doesn't work with the numbers you pick, since you're not asked to solve. This is just to help you see how the equation should go.) Say the shortest side, x, is 6. Then the other two sides are 8 and 10. So the equation you'd set up is:

6² + 8² = 10²

Or, in terms of x:

x² + (x + 2)² = (x + 4)²

That's choice (C).

Test 4 Section 6 - #15 (page 597)

Even though this is a grid-in question, we can actually plug-in to solve. That's because they don't ask us for the areas themselves, they ask us for the RATIO of the areas.

Say QS = 1, QV = 3, PT = 3, and PR = 4. That's going to mean that SV = 2, since it's the difference between QV and QS.

Our areas:

▵PST = ½(SV)(PT) = ½(2)(3) = 3
▵PQR = ½(QV)(PR) = ½(3)(4) = 6

So the ratio of ▵PST to ▵PQR = 3/6, or 1/2. You could also grid in .5.

Test 4 Section 6 - #6 (page 594)

This one works kinda like a ratio problem. The key is to recognize that if the ratio of white eggs to brown eggs is 2 to 3, then the ratio of, say, white eggs to TOTAL eggs is 2 to 5. (You could also, of course, do brown eggs to TOTAL eggs 3 to 5. Potayto potahto.)

The total number of eggs in the basket has to be a multiple of 5. The only answer choice that isn't is (B) 12, so that's the answer.

Test 4 Section 3 - #18 (page 585)

Big time plug-in here. Say x = 10, and z = 2. So the first shirt will cost $10, and each shirt after that will cost $10 - $2 = $8. Now say the customer buys n = 5 shirts. Those shirts would cost

$10 for the first one

+ $8 for the second

+ $8 for the third

+ $8 for the fourth

+ $8 for the fifth

= $42 

So we're looking for the answer that gives us 42 when we plug in the number we chose above.

(A) x + (n - 1)(x - z)


We can translate that to
10 + (5 - 1)(10 - 2) = 10 + (4)(8) = 42.

That's our answer! Make sure, when you plug in, that you try the other answers too to make sure they don't work. In this case they don't, so we're done.

Test 5 Section 2 - #14 (page 641)

This one looks intimidating, but if you start plugging in you'll see the way pretty quickly.

Start with the lowest ordered pair you can use: (1,1). Yep, that works; it gives you 2(1) + 3(1) < 6, or 5 < 6. Note, though, that there aren't any others: (2,1) gives you 7, and (1,2) gives you 8. Both of those are bigger than 6, so the only ordered pair of positive integers that works is (1,1).

Test 3 Section 8 - #12 (page 547)

The question doesn't say anything about these numbers other than that they're integers, so why not assume that 10 of them are:

-1, -2, -3, -4, -5, -6, -7, -8, -9, -10

Then you just need the 11th one to cancel those out. All the numbers above add up to -55, so make the last number +55. Now your sum is zero!

If you're going to have a sum equal to zero, you're going to need at least one positive number, but you don't necessarily need any more than that.

Test 1 Section 3 - #20 (page 401)

This #20 is a...drumroll please...PLUG-IN! Say k = 10. So the salesperson's commission is 10% of his sales.

He sells 2 cars, for $14,000 each, so his commission will be 10% of $28,000 in sales:

That's $2,800 in commission for our pal the salesman.

Just drop 10 in for k in all the answers, and see which ones gives you $2,800:

(A) 280(10) = 2,800

(B) 7,000(10) = 70,000

(C) 28,000(10) = 280,000

(D) 14,000/[100 + 2(10)] =116.666... 

(E) [28,000 + 10]/100 = 280.1  

Only (A) works!

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!

2010 PSAT - Saturday - Section 2 (Math)

The following are TestTakers' suggestions for solving the problems on the October 16th, 2010 PSAT, Section 2 (a math section). Click "Read More" to view!

Test 6 Section 4- #4 (page 713)

Do some plug-in to figure out what they want:  

2^2 +1 = 5

3^2 + 1 = 10

4^2 +1 = 17

5^2 + 1 = 26

...and so on.

We're looking for a number that's 1 more than a perfect square.  The one that works for us is 7^2 + 1 = 49+1 = 50.  Choice E.

Test 7 Section 9 - #10 (page 797)

This is a PLUG-IN question. They're saying he put up n posters, but let's just say he wanted to put up 5.

 The question tells us he uses 4 6-inch pieces of tape to put up each poster (seems wasteful to me, but whatever). That's 24 inches of tape per poster, or 2 feet of tape per poster.  Remember, the question asks for answers in feet. So if he puts up 5, as we're plugging in for n, he's using 5x2 feet of tape, or 10 feet. If he started with 300 feet, he'll be left with 290 feet.

Now, go to the answer choices, and see which one gives you 290 when you plug 5 in for n:

(A) 300 - 6(5) = 270
(B) 300 - 2(5) = 290
(C) 300 - 5 = 295
(D) 300 - (1/2)(5) = 297.5
(E) 300 - (1/4)(5) = 298.75

Test 7 Section 3 - #17 (page 773)

See variables, think PLUG-IN, sometimes even in the grid-in section!  Pick 3 prime numbers greater than 2:

3, 5, 7

Now just do what the problem asks: 3 x 5 x 7 = 105.  How many factors does 105 have (including 1 and 105)?

1 x 105

3 x 35

5 x 21

7 x 15

Any more?  Nope.  That's 8 factors.

Test 7 Section 3 - #7 (page 770)

Variables in the question and in the answer?  I'm thinking PLUG-IN.  But we're not going to plug in directly for t here, we're going to plug in for the two numbers that differ by 1. Let's say our numbers are 2 and 3. So since t is the sum of those numbers, t = 5.

The question asks us for the greater of our two numbers, so we're looking for the answer that gives us 3, when we plug in t = 5.  Always try them all to be sure when you're plugging in, but the only one that works is going to be (C): plug in 5 for t and you end up with 6/2 = 3.

Test 2 Section 5 - #14 (page 468)

Tough one (grid-ins usually are). The solution to this one, though, can be found by PLUGGING IN (with a little BACKSOLVE flavor added in for good measure)! Just start by trying any number for x. How about x = 3? Just to be clear, we're looking for f(3) to be LESS THAN 3 (that's what they say in the question: f(a) < a).

f(x) = |3x - 17|

f(3) = |3(3) - 17|

f(3) = |9-17|

f(3) = |-8|

f(3) = 8

Ok, so that didn't work; 8 is not less than 3. But here's where that BACKSOLVE flavor comes in: now we have a better idea of how the pieces fit together. We want x to get a little bigger, so that 3x gets closer to 17.  That way, f(x) will be small.  Why not try x = 5 next?

f(5) = |3(5) - 17|

f(5) = |15-17|

f(5) = |-2|

f(5) = 2

GREAT! 2 < 5, so x = 5 works! (A few other numbers will work too: anything between 4.25 and 8.5...though you'd probably be best to stick with integers and just try 5, 6, or 7.)

Test 2 Section 5 - #15 (page 468)

This one has hints of PLUGiIN and BACKSOLVE, even though it's a grid-in! We don't need any of that stuff about how many candies are in the jar...that's just extraneous information. All we need to know is where Ari's starting from (3 red, 4 green) and how many more he takes. We want him to have more red pieces than green pieces by the end, but just barely. Here's how I set it up:

RED	GREEN	Comments
3	4	This is our starting point.
+7	+6	Just plugging in 2 numbers that are close together and add up to 13.
=10	=10	OOPS!  That's not enough red...Ari has the same amount of each color now.

Now it's kinda clear where to go: add one more red (and one less green):

RED	GREEN	Comments
3	4	This is our starting point (again).
+8	+5	Plug in new numbers.
=11	=9	That'll do it!  Any less red and he'd have the same number of each (as we saw above).

Test 5 Section 4 - #16 (page 655)

We can kinda PLUG-IN here, even though there aren't answer choices.  It's a percent question at the heart of it, so we just need to find a way to be dealing with 100, to make our lives easier.

What we want is for 4b to be equal to 100.  That way, it's easy for us to figure out what 125% of 4b is. So let's say b = 25, which makes 4b = 100.

125% of 100 is 125, so we need to solve:

a +2b = 125 

Plug-in b = 25:

a + 2(25) = 125 

a + 50 = 125

a = 75

Then just keep on plugging in:

a/b = 75/25

a/b = 3

Our answer is 3.

Test 4 Section 6 - #8 (page 595)

I see variables, so I'm thinking PLUG-IN (yes, even in a triangle problem)!  Plug-in works really well in triangles, provided you follow 2 very important rules:

1. Don't break a triangle!  A triangle's angles must add up to 180°.
2. Don't break a straight line!  A straight line ALSO must be 180°.

So we can't plug-in willy-nilly, but with practice we can get very good and very quick at using plug-in, even in geometry questions.  I'm choosing to make a = 80, and b = 60.  Let's leave c alone for now.

It's super important from here to be careful filling in the rest of the angles.  (This is why I haven't plugged-in for c yet.)  I'll use a black pen to fill in what I can from here, based on the numbers I chose.  I've gotta make sure my triangles that I've already given two angles to add up to 180°, and that the straight lines on the left and bottom of the big triangle also end up being 180°.  Here goes:

Now it's clear what c has to be: c = 20.  Let's fill that in (in red again -- I ran out of colored pens), and fill in the rest of the angles.

Ok.  Do a quick check and make sure all the triangles (including the big one, four in all) add up to 180°, and make sure all the lines do as well.  We good?  Good.  That was the hard part, but with practice you'll find that you can do this very quickly.  And as always, the beauty of plug-in is that once you've done the work, you can be super sure of your answer.

From here, all we do is check the answer choices with our numbers plugged in.  Remember, we said a = 80 and b = 60.  We're looking for the answer that gives us c = 20.

(A) 80 + 3(60) - 180 = 80
(B) 2(80) + 2(60) - 180 = 100
(C) 180 - 80 - 60 = 40
(D) 360 - 80 - 60 = 220
(E) 360 - 2(80) - 3(60) = 20

Bingo.

There are a few other ways to do this problem, too.  Read on if you're interested.

Notice that the labeled angles actually outline a quadrilateral.  The angles in a quadrilateral always add up to 360°, just like a triangle's angles always total 180°.  So c is going to be what's left over when we start at 360° and take away everything else.

c = 360 - 2a - 3b.

You might also have noticed that there's only one answer choice that actually acknowledges the fact that there are 3 b's and 2 a's in the diagram.  Sometimes the SAT will do that, and it's often a giveaway.  On the hardest multiple choice question in the section I don't know if I'd feel comfortable just picking (E) and moving on based on that, but there you go.

Test 2 Section 8 - #16 (page 486)

The key here is to plug in a few different numbers and look for a pattern. Start with n=2. If you have a 2x2 square, all 4 of the squares would lie along the boundary, so k=4. Now try a new number. If n=3, you'd have a 3x3 gameboard and only 8 of the squares would be along the boundary (the other square is smack in the middle), so k=8. Try one more number to detect a pattern. When n=4, you have a 4x4 gameboard, and 12 squares make up the border (the other 4 are in the middle), so k=12. So far we have k=4, k=8, and k=12... see the pattern? Multiples of 4. The only answer that's a multiple of 4 is (E).

Test 2 Section 5 - #7 (page 465)

If you have a graphing calculator, this one is probably easiest to do using Plug-In. First, plug in anything you want for a... let's go with a=3. Now graph that equation: y=3x²+2. Next graph the SECOND equation (y=x²+2) and compare the two graphs. You'll see that the second graph is wider than the first, so the answer is (B).