GMAT® Official Guide 2019 Quantitative Review
DSl7110- Is the perimeter of square S greater than the perimeter
of equilateral triangle T?
(1) The ratio of the length of a side of S to the
length of a side of Tis 4:5.
(2) The sum of the lengths of a side of S and a side
of Tis 18.Geometry
Letting s and t be the side lengths of square S and
triangle T, respectively, the task is to determine if
4s > 3t, which is equivalent ( divide both sides by 4t)
to d etermmmg.. "f 1 - s > -.^3
t 4I t 1s. given. t at h -s = -^4. s· mce -4 3 > - ,. 1t 10 r ll ows
t 5 5 4(1)that i > l; SUFFICIENT.
t 4
(2) Many possible pairs of numbers have the
sum of 18. For some of these (s,t) pairs it is
the case that i > l (for example, s = t = 9),
t 4
and for others of these pairs it is not the
case that i > l (for example, s = l and
t 4
t= 17); NOT sufficient.
The correct answer is A;
statement 1 alone is sufficient.
DSI 7136- If x + y + z > 0, is z > 1?
(1) Z>X+y+l
(2) X + y + 1 < 0Algebra Ir ~,
(1) The inequality x + y + z > 0 gives z > -x -y.
Adding this last inequality to the given
inequality, z > x + y + l, gives 2z > 1, or
z > .l, which suggests that (1) is not
2
sufficient. Indeed, z could be 2 (x = y = 0
and z = 2 satisfy both x + y + z > 0 and
z > x + y + l), which is greater than 1, and
z could be 1 (x = y = _ _land z = l satisfy
4 4 4
both x + y + z > 0 and z > x + y + l), which
is not greater than l; NOT sufficient.(2) It follows from the inequality x + y + z > 0
that: z > -(x + y). It is given that x + y + l < 0,
or (x + y) < - l, or -(x + y) > l. Therefore,
z > - (x + y) and -(x + y) > l, from which it
follows that: z > 1; SUFFICIENT.
The correct answer is B;
statement 2 alone is sufficient.
DS 07832- For all z, 1zl denotes the least integer greater than or
equal to z. Is ,xl = 0?
(1) - 1 < X < -0.1
(2) IX+ 0 .5 7 = 1Algebra
Determining if Ix l = 0 is equivalent to
determining if - 1 < x s 0. This can be inferred by
examinirng a few representative examples, such as
r -1.11 = -l,1-ll = -l,1 -0.9 7 = o,r-0.17 = 0,
10 7 = 0, and 10.1 7 = 1.(1) Given -1 < x < -0.l, it follows that
-1 < x s 0, since - 1 < x s 0 represents all
numbers x that satisfy- 1 < x < -0.l along
with all numbers x that satisfy- 0.1 s x SO;
SUFFICIENT.
(2) Given IX+ 0.5 7 = 1, it follows from the same
reasoning used just before (1) above that this
equality is equivalent to 0 < x + 0.5 s 1, which
in turn is equivalent to - 0.5 < x S 0.5. Since
from among these values of x it is possible for- 1 < x s Oto be true (for example, x = -0.l)
and it is possible for -1 < x s 0 to be false (for
example, x = 0.l), it cannot be determined if
Ix-] = 0; NOT sufficient.
The correct answer is A;
statement 1 alone is sufficient.