5.5 Answer Explanations
legislature had at least 15 members (to
allow for the "5 additional members of the
legislature" that could have voted against the
measure, for a total of^1 —of the members
3
voting against it). However, beyond this we
know essentially nothing from statement 2.
In particular, depending on the number of
members of the legislature (which we have
not been given), any number of members
could have voted against the measure. For
example, exactly one member could have
voted against the measure, in which case the
legislature would have had (1 + 5) x 3 = 18
members. Exactly two members could have
voted against the measure, in which case the
legislature would have had (2 + 5) x 3 = 21
members, and so on for 3 members voting
against, 4 members voting against, etc.;
NOT sufficient.
Considering the statements 1 and 2 together,
the reasoning is similar to the reasoning for
statement 2, but with the further condition that
the total number of members of the legislature
is divisible by 12 (so as to allow that both exactly
1
4
—of the members did not vote on the measure
while exactly -^1 could have voted against the
3
measure). For example, it could have been the
case that the legislature had 24 members. In
this case, -^1 of the members would have been 8
3
members, and, consistent with statements 1 and
2, 3 of the members (8 - 5) could have voted
against the measure. Or the legislature could
have had 36 members, in which case, consistent
wit 'h statements 1 and 2,^1
—(36) 3 -^5 =^12 -^5 =^7
members could have voted against the measure.
Th e correct answer 1s E·
both statements together are still not sufficient.
DS05986
- If y * 0, is lxl = 1?
(1) X=—y
IYI
(2) !xi =-x
Algebra
Can we determine whether lxl = 1?
1'._
(1) Given that x = IYI, we consider two cases:
y
y > 0 and y < 0. If y > 0, then IYI = y and
—
=
y
IYI
—= 1. So if y > 0, then x = l and, of course,
y y
lxl =l.Ify<O,then IYI =(-l)yandx=—
= y = y 旧
(-1)—= (-1)(1) = -1. So lxl =
(-l)y y
1.If y < 0, then I xi = 1. In both of the two
cases, lxl = 1; SUFFICIENT.
(2) Given that lxl = -x, all we know is that xis
not positive. For example, both -4 and -5
satisfy this condition on x: l-41 = 4 = -(-4)
and l-5 1 = 5 = -(-5); NOT sufficient.
1h e correct answer 1s A· ，
statement 1 alone is sufficient.
0S08306
- If xis a positive integer, what is the value of x?
．｀
．．
.. l2
，'，
x三5
n
—X =nandn"i'O.
Algebra
(1) It is given that xis a positive integer. Then,
x三五
X^4 =x
x^4 —x=O
x(x—1)(正+X + 1) = 0
Thus, the positive integer value of x being sought
will be a solution of this equation. One solution
of this equation is x = 0, which is not a positive
integer. Another solution is x = 1, which is a
positive integer. Also, x 2 + x + 1 is a positive
integer for all positive integer values of x, and so x
2 + x + 1 = 0 has no positive integer solutions.
Thus, the only possible positive integer value of x
is 1; SUFFICIENT.
(2) It is given that n -:I-O. Then,
given
square both sides
subtract x from both sides
factor left side
given
multiply both sides by x
divide both sides by n, where n -::J:. 0
