- E Since all radii of a triangle are equal, PQ= PR.
Since PQ= QRtoo, the triangle must be equilateral.
Since its area is , the lengths have the measures
shown in the diagram. The circle has a radius of 6.
The shaded region is equal to
the area of the sector minus
the area of the triangle.
Since the central angle is
60°, the sector has an area
that is^1 ⁄ 6 of the whole circle,
or (^1 ⁄ 6 )(π(6)^2 ) = 6π. Subtract-
ing the area of the triangle
gives.
(Chapter 10, Lesson 3: The Pythagorean Theorem)
(Chapter 10, Lesson 5: Areas and Perimeters)
(Chapter 10, Lesson 8: Circles) - C If the ratio of boys to girls in a class is 3 to 5,
then 3/(3 + 5) =^3 ⁄ 8 of the class is boys and 5/(3 + 5) =^5 ⁄ 8
of the class is girls. This means there are (^3 ⁄ 8 )(160) = 60
boys and (^5 ⁄ 8 )(160) = 100 girls in the senior class. Sim-
ilarly, the fraction of boys in the junior class is^3 ⁄ 5 and
the fraction of girls is^2 ⁄ 5. If there are xstudents in the
junior class, then there are (^3 ⁄ 5 )xboys and (^2 ⁄ 5 )xgirls in
the junior class. If the ratio of boys to girls is 1:1 when
the classes are combined, then
60 + (^3 ⁄ 5 )x= 100 + (^2 ⁄ 5 )x
Subtract 60 and (^2 ⁄ 5 )x:(^1 ⁄ 5 )x= 40
Multiply by 5: x= 200
(Chapter 8, Lesson 7: Word Problems)
(Chapter 7, Lesson 4: Ratios and Proportions)
Section 3- A A six-month hiatus(break) would cause her
skills to weaken,something she might fear. atrophy=
weaken from disuse; align=line up; disseminate=
spread like seed - C Domineeringopinions are overbearing and
preachy. vindictive=inspired by revenge; pedantic=
acting like a know-it-all; conciliatory=acting to bring
people together; treacherous=betraying someone’s
confidence; didactic=preachy; dogmatic=conde-
scendingly preachy; prosaic=ordinary - A The missing word must refer to Walter’s in-
ability to make up his mind. vacillation=inability to
make up one’s mind; solicitation=request for help;
rejuvenation=restoration of one’s youth; admonish-
ment=mild reproof
693 π−93
628 MCGRAW-HILL’S SAT
- B It is easier to pick a simple value for the “start-
ing” population in 1980, like 100. Since the popula-
tion increased by 10% from 1980 to 1990, the 1990
population must have been (100)(1.10) = 110. Since it
decreased by 10% from 1990 to 2000, the 2000 popu-
lation must have been (110)(0.90) = 99. From 1980 to
2000, then, the percent change was (99 – 100)/100 =
–1/100 = –1%.
(Chapter 7, Lesson 5: Percents) - E According to the definition of g, g(3) = 2f(3) – 1.
According to the table, f(3) = 11, so g(3) = 2f(3) – 1 =
2(11) – 1 = 22 – 1 = 21.
(Chapter 11, Lesson 2: Functions) - C Although you may substitute 5yfor xas a first
step, it’s probably easier to simplify the expression
first:
Factor:Simplify:Substitute:Simplify:
(Chapter 8, Lesson 4: Working with Roots)
(Chapter 8, Lesson 5: Factoring)
(Chapter 8, Lesson 6: Inequalities, Absolute Values,
and Plugging In)
- C Think of numbers that are larger than their
squares. This excludes negatives, because the squares
of negatives are always positive. It also excludes num-
bers greater than 1, because the squares of these are
bigger than the original numbers. Therefore, 0 < x< 1.
This means I and II are true, but not III.
(Chapter 9, Lesson 3: Numerical Reasoning Problems)
18.C Believe it or not, you don’t need to find the
two midpoints in order to answer this question. You
need to know only that the distance between the
two mid-points is half of the distance between the
two endpoints. The distance between the endpoints
is (3x+ 2) – (–x– 4) = 3x+ 2 + x+ 4 = 4x+ 6. Half of
this is 2x+ 3.
(Chapter 10, Lesson 4: Coordinate Geometry)
44 yy=5 yy−xy−()xy−2()xxyy^22 −+^2Q
R
P
33 √ (^33)
6
30 °
30 °