Test 7 Section 7 - #16 (page 788)

It's a lot easier to think about this question if you put it in an xy-coordinate plane.  Let's make Diane's house the origin:

What we want to do is compare the green dotted line to the distance Diane usually drives, and see how much shorter her drive would be if she could just go directly.  Easy part first: her usual commute is 16 mi + 15 mi + 4 mi = 35 mi.  How do we know how long the green line is?  Well, it looks suspiciously like a right triangle to me, so let's Pythagorize it!  (Note: don't say "Pythagorize" in public...people will look at you funny.)  We'll call our green line d:

15² + 20²  = d²

225 + 400 = d²

625 = d²

25 = d

As is very often the case on the SAT, since they prefer to work with integers, we're looking at a big ol' 3-4-5 triangle (15-20-25 in this case).  Get quick at recognizing those, and you'll save yourself a few seconds by avoiding the Pythagorean Theorem.

OK. So Diane usually drives 35 miles, but if she could drive directly, it would only be 25 miles.  Her drive would be 10 miles shorter.  That's choice (C).

Test 7 Section 7 - #15 (page 788)

It would take us way too long to list all the numbers from 1-100 and check which are perfect squares.  Luckily we don't have to.  All we need to do is count all the ones that ARE, and then we know all the rest are NOT.  To do so, let's just list squares of integers until they get bigger than 100:

1² = 1
2² = 4
3² = 9
4² = 16
5² = 25
6² = 36
7² = 49
8² = 64
9² = 81
10² = 100
11² = 121 (Oops, too big!)

There are 100 numbers between 1 and 100 (inclusive), and we just counted 10 of them that ARE perfect squares. That means 90 are not.  Choice (E) is our answer.

Test 7 Section 7 - #13 (page 787)

Remember our conditions:

• One number is odd.
• One number is a multiple of 5.
• One number is the day of the month of Kyle's birthday.

The key is the part where they say that each number satisfies exactly one of the conditions.  That means, for example, if the number 15 were in a combination, we'd have to eliminate that choice, because 15 is both odd and a multiple of 5.  It satisfies 2 conditions, but the question said that's not allowed.  We can actually eliminate 3 choices just because of this: (B), (C), and (D) all have odd multiples of 5 (25, 15, and 15, respectively).

(A) 14-20-13
(B) 14-25-13
(C) 15-18-16
(D) 20-15-20(E) 34-30-21

So how do we know whether to pick (A) or (E)?  Well, if it were (E), our odd number would be 21, our multiple of 5 would be 30, and that'd mean our day of the month would have to be 34.  But no month has 34 days.  So we have to pick (A): 13 is odd, 20 is a multiple of 5, and 14 is a day of the month.

Test 7 Section 7 - #12 (page 787)

If we're trying to identify the segment with a slope of -1, the first thing I want to do is eliminate the segments that have positive slopes.  We know any of these segments that are going up as we're going from left to right have a positive slope, so we're not even going to consider them.  That eliminates segments OC and OD.

OA, OB, and DC all have negative slopes.  If you've got a good eye for slopes,  you've got a good feeling about (E) right now, because OA looks way too steep, and OB looks way too gradual.  It's a good idea to confirm though, so let's put the points into the slope formula just to be sure.

DC goes between points (1, 3) and (3, 1).  So our slope is:

Yep, (E) is our answer.

Note that we can use this same formula to find the slopes of all the segments for practice, or just for fun (hah!):  Slope of OA = -3, slope of OB = -1/3, slope of OC = 1/3, slope of OD = 3.

Test 7 Section 7 - #15 (page 789)

They're asking for the number of integers from 1-100 that aren't the square of an integer, but it's going to be a lot easier and faster to count how many are the square of an integer*:

1² = 1
2² = 4
3² = 9
4² = 16
5² = 25
6² = 36
7² = 49
8² = 64
9² = 81
10² = 100 ...and we're done! (Any higher will be out of the range we care about.)

Count how many we've got. I'm counting 10.

There are 100 integers between 1 and 100 inclusive (inclusive just means "including the ends," so we are counting both 1 and 100), and only 10 of them are squares of integers. Which means 90 aren't. The answer is (E).

* This is kinda like a shaded region area problem in that way...it's easier to look for the opposite of what they ask for, and then subtract.

Test 7 Section 7 - #18 (page 789)

When they say they're looking for the "sum of consecutive integers from -22 to x, inclusive, is 72" they mean this:

-22 + -21 + -20 + ... + x = 72

That's the way your math teacher would write it, anyway. :)

What you need to recognize here is that for a while, you're going to be getting nothing but negative numbers for your sum.  -22 + -21 + -20 = -63, for example.  And it's only going to get worse when you add in -19.  You're going to have to go a long way before your numbers start to go positive again.  Think for a minute about when these numbers will first start getting positive.  Or, even better: ask yourself when your sum is going to be zero.

-22 + -21 + -20 + ... + 20 + 21 +22 = 0

You'll have zero when you add up every consecutive integer from -22 to 22; the positive and negative numbers will cancel each other out.  From there, we're cruising.  Watch the bolded bits.

-22 + -21 + -20 + ... + 20 + 21 +22 + 23 = 23
-22 + -21 + -20 + ... + 20 + 21 +22 + 23 + 24 = 47
-22 + -21 + -20 + ... + 20 + 21 +22 + 23 + 24 + 25 = 72

The last number we had to add was 25, so that's our x.  (B) is our answer.

Of course, the long way is just to get crackin' with the calculator (if your fingers are fast it won't take that long) and start adding things up.  When you get to 72, you're done.  No shame in doing it that way, especially if it's the first way you see, and you start on it right away.