CHAPTER 6 / WHAT THE SAT MATH IS REALLYTESTING 247
- 50 Move the shaded regions around, as shown
above, to see that they are really half of the square.
Since the area of the square is (10)(10) =100, the
area of the shaded region must be half of that, or 50. - A The number 3^40 is so big that your calculator is
useless for telling you what the last digit is. Instead,
think of 3^40 as being an element in the sequence
31 , 3^2 , 3^3 , 3^4 ,.... If you write out the first six terms
or so, you will see that there is a clear pattern to the
units digits: 3, 9, 27, 81, 243, 729,.... So the pattern
in the units digits is 3, 9, 7, 1, 3, 9, 7, 1,.... The
sequence repeats every four terms. Since 40 is a mul-
tiple of 4, the 40th term is the same as the 4th and
the 8th and the 12th terms, so the 40th term is 1.
Concept Review 3
- Don’t worry about the percent or about finding x.
Translate: 5 less than 28% of x^2 is 10 means
.28(x^2 ) − 5 = 10
Subtract 10: .28(x^2 ) − 15 = 0
So 15 less than 28% of x^2 is 0. - m= 3 + 6 + 9 +... + 93 + 96 + 99
n= 6 + 9 +... + 93
When you subtract nfrom m,all the terms cancel
except 3 + 96 + 99 =198.
extra bases. Each base has an area of π(2)^2 = 4 π, so
the surface area of the smaller cylinders is 2(4π) =
8 πgreater than that of the larger cylinder.- Your calculator is no help on this one because 4^134
is so huge. Instead, think of 4^134 as a term in the
sequence 4^1 , 4^2 , 4^3 , 4^4 ,.... What is the units digit
of 4^134? If you write out the first few terms, you
will see a clear pattern to the units digits: 4, 16,
64, 256,.... Clearly, every odd term ends in a 4
and every even term ends in a 6. So 4^134 must end
in a 6. - The first few terms are 1, 2, 3, 5, 8, 13, 21,.... Since
we are concerned only about the “evenness” and
“oddness” of the numbers, think of the sequence as
odd, even, odd, odd, even, odd, odd, even,.... No-
tice that the sequence repeats every three terms:
(odd, even, odd), (odd, even, odd), (odd, even,
odd),.... In the first 100 terms, this pattern re-
peats100/3 = 331 ⁄ 3 times. Since each pattern con-
tains 2 odd numbers, the first 33 repetitions
contain 66 odd numbers and account for the first
99 terms. The last term must also be odd because
each pattern starts with an odd number. There-
fore, the total number of odds is 66 + 1 =67.
Answer Key 3: Finding Patterns
SAT Practice 3
1.
- A If every term is 6 less than the square of the
previous term, then the second term must be
(3)^2 − 6 = 9 − 6 =3. The third term, then, is also
(3)^2 − 6 =3, and so on. Every term, then, must be
3, including the fifth. - C The sequence repeats every three terms:
(−4, 0, 4), (−4, 0, 4), (−4, 0, 4),.... Each one of the
groups has a sum of 0. Since 200/3 = 672 ⁄ 3 , the first
200 terms contain 67 repetitions of this pattern,
plus two extra terms. The 67 repetitions will have a
sum of 67(0) =0, but the last two terms must be − 4
and 0, giving a total sum of −4.
C
6
y
x
yx
+= +⎛
⎝⎜
⎞
⎠⎟
2 = ()=
3
2
210 20
- Don’t calculate the total surface area. Instead, just
notice that the two small cylinders, stacked to-
gether, are the same size as the large cylinder. But
remember that you are comparing surface areas,
not volumes.The surface areas are almost the
same, except that the smaller cylinders have two
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