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305
PRACTICE TEST 2
ARGOPREP.COM/GRE
QUANTITATIVE REASONING
ANSWER KEY: SECTION 5
a
a
a a
a a
a a
he su o the arcs is the circle
9
a
i pli
a
This shows that the smallest measurement for arc
a
is 40°. This makes the greatest arc (2
a
) in this instance only 80° which is
smaller
than one-
third the circumference of the circle. Because you cannot determine whether the
largest arc is smaller or greater than one-third the circumference, the correct
answer choice is
D
.
3. B.
This problem tests your ability to use number theory and the property of reciprocals in order to solve an inequality.
You are given several pieces of information, all relating the variables
a, b,
and
c
to the number 1. To solve this problem start with the last two given
statements:
abc
Gi en
ac
1
Divide by
b
b
Next, relate
a
and
c
to
b:
ab
Gi en
a
1
Divide by
b
b
Note: because we are given that
b
is positive we know the inequality will not
change direction.
Now let
x
be a positive number such that: 1
a
. Return to the inequality:
x
1 < 1
Substitution
x
b
x
>
b
Reciprocal inequality
Using the same chain of reasoning for
c
:
cb
< 1
Given
Given
c
< 1
Divide by b
b
Let
y
be a positive number such that: 1
c
.
y
1 < 1
Substitution
y
b
y
>
b
Reciprocal inequality
If
y
>
b
and
x
>
b
then
xy
>
b
.
Now return to the given equation:
abc
Gi en
1
b
1
ubstitution
x
y
b xy
Multiply by
xy
306
ARGOPREP.COM/GRE
QUANTITATIVE REASONING
ANSWER KEY: SECTION 5
Because you have already demonstrated that
xy b
then
b xy
cannot be true. Therefore
a
and
c
cannot be
positive
numbers. This
only leaves negative numbers:
ab
< 1
Given
a
< 1
Divide by
b
b
Now let
x
be a
negative
number such that: 1
a
.
x
1 < 1
Substitution
x
b
x
<
b
Reciprocal inequality of a negative number
Using the same chain of reasoning for
c
:
cb
< 1
Given
c
< 1
Divide by
b
b
Let
y
be a
negative
number such that: 1
c
.
y
1 < 1
Substitution
y
b
y
<
b
Reciprocal inequality of a negative number
1
You now have a situation where it is possible that
xy b
. So looking at Quantity A:
a c
you can easily see that Quantity A will be a
negative value while Quantity B will be positive. The correct answer choice is
B
.
4. D.
his proble tests our abilit to si pli a s ste o e uations in order to find the range o ans ers. ou are
given two averages:
a b c
b c d
and
3
3
You are asked to compare Quantity A which is the average of
a
and
d
. Start
by simplifying the averages:
a b c
erage gi en
3
a b c
ultipl b call this e uation
b c d
erage gi en
3
b c d
ultipl b call this e uation
a
–
d
a d
d
dd
d
to both sides
a d
27
d
Divide by 2
2
2
307
PRACTICE TEST 2
ARGOPREP.COM/GRE
QUANTITATIVE REASONING
ANSWER KEY: SECTION 5
You now have an expression for the average of
a
and
d.
From here you see it is not possible to reduce the number of
variables further, so you will have to compare this expression against Quantity B to see if you can resolve the problem.
For Quantity B start with:
a b c
uation
a
b
–
c
Isolate the
a
variable
a b c d
96 –
b
–
c b c d
Substitute for
a
on the right hand side
4
4
d
=
Simplify
4
You now have an expression for Quantity B. Compare:
Quantity A ? Quantity B
27
d
2
d
?
Substitution
2
4
8
d
d
Multiply by 4
7
d
ubtract
d ro both sides
At this point you can see that the relationship cannot be resolved because you cannot determine the value of
d
.
You cannot determine the relationship from the information given. The correct
answer choice is
D
.
5. B.
This problem tests your ability to generate an algebraic equation from a geometric diagram. The arcs in the diagram
have equal radii and touch at exactly one point. Quantity A is the shaded region
and can be written:
Shaded Area = Area of the square – unshaded area
First
calculate the area of the square using
r
as the radius:
Area of the square
r
r
r
2
Now calculate the area enclosed by the arcs:
unshaded area
1 p
r
2
1 p
r
2
p
r
2
2
4
Now compare the two quantities:
Quantity A ? Quantity B
16
r
2
p
r
2 ? 3p
r
2
Comparison of the Quantities
16
r
2 ? 6p
r
2
Add 3p
r
2 to both sides
p
Divide by 2
r
2
.
ubstitute appro i ation or p
Quantity B is greater than Quantity A. The correct answer choice is
B
.
308
ARGOPREP.COM/GRE
QUANTITATIVE REASONING
ANSWER KEY: SECTION 5
6. B.
This problem tests your ability to solve a system of equations. Start by substituting the given values into the given
functions. For
f
(3):
f
(
x
Ax
2
Bx
Gi en
f
A
(3)2
B
ubstitution
A
B
ubstitute and si pli
A
B
Subtract 7 from both sides
A B
Divide by 3
A
B
Multiply by 4
Now do the same for
f
(4):
f
A
(4)2
B
ubstitution
A
B
ubstitute and si pli
A
B
Substitute and simplify
Now solve the system of equations:
A
B
From
f
(3)
A
B
From
f
(4)
A
f
(3) –
f
(4)
A
9
Solve for
A
4
Now use
f
to find
B
:
A
B
–9
B
Substitute
4
B
Simplify
65
B
Solve for
B
4
B
is greater than
A
, therefore Quantity B is greater than Quantity A. The correct
answer choice is
B
.
7. D.
This problem tests your ability to determine the range of a function. You are given the following equation and
asked to compare
x
and 0:
4
x
3 – 12
x
2 – 36
x
2
x
– 3
or this t pe o proble it is easiest to use a process o eli ination. he first thing to note is that because the
expression is equal to zero the only valid solutions of
x
will occur when: 4
x
3
x
2
x
309
PRACTICE TEST 2
ARGOPREP.COM/GRE
QUANTITATIVE REASONING
ANSWER KEY: SECTION 5
The best approach to solve this problem is the process of elimination. Because you are comparing solutions for
x
to the value zero you can start by checking if 0 is a valid solution: 4(0)3
2
ubstitute
Answer choice C can be eliminated.
Next, look to see if there is a solution for
x
that is
greater
than 0. For this you only have to prove that there is
a positive solution for
x.
here is no need to find the actual solution. tart b choosing a couple o con enient
values for
x.
Start with 1:
2
3
2
4 1
1
1
ubstitute
2
2
2
1
i pli the e ponents
2
From this conclusion you can reason that if there is some number
x
greater
than 1 which results in the expression 2
being
less
than zero, there must be some value between 1
and that number for which the expression
will be
zero 2
(i.e. a solution for
x
).
Now try the value 2 for
x
:
4(2)3
2
ubstitute
i pli the e ponents
You have shown there is a solution for
x
which is
greater
than 0.
Now, see if there is a solution which is
less
than 0. Remember you have already established that for
x
the
polynomial is 27 (and greater than 0), so you are looking to see if there is a negative number which will result in
the polynomial having a value less than zero. For convenience choose
the value –10: 3
2
ubstitute
i pli the e ponents
From this you have shown there is at least one value for
x
which is
less
than 0 that will be a solution for the polynomial.
Because there is both a positive and a negative solution the answer cannot be determined. The correct answer
choice is
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