GMAT® Official Guide 2019 Quantitative Review
(i) The measure of LCAB is 180 - 2y since the
sum of the measures of the angles of b. CAB
is 180.(ii) LDAB is a right angle (because DB is a
diameter of the circle) and so
z + x + (180 -2y) = 90, or, equivalently,
2y-X - Z = 90.(iii) z + 90 + y = 180 since the sum of the
measures of the angles of right triangle
b.DAB is 180, or, equivalently, z = 90 - y.(iv) x = 2z because the measure of exterior angle
LAOC to MOD is the sum of the measures
of the two opposite interior angles, LODA
and LOAD.(v) y = 2x because the measure of exterior angle
LACE to b.OCA is the sum of the measures
of the two opposite interior angles, LCOA
andLCAO.Multiplying the .final equation in (iii) by 2 gives
2z = 180 - 2y. But, x = 2z in (iv), so x = 180 - 2y.
Finally, the sum of the measures of the angles of
b.CAB is 180 and soy+ y + x = 180. Then from
(v), 2x + 2x + x = 180, 5x = 180, and x = 36.The correct answer is B.
PS16967- An airline passenger is planning a trip that involves
three connecting flights that leave from Airports A, B,
and C, respectively. The first flight leaves Airport A
every hour, beginning at 8:00 a.m., and arrives at
Airport B 2 ½ hours later. The second flight leaves
Airport B every 20 minutes, beginning at 8:00 a.m.,
and arrives at Airport C 1 ½ hours later. The third flight
leaves Airport C every hour, beginning at 8:45 a.m.
What is the least total amount of time the passenger
must spend between flights if all flights keep to their
schedules?
(Al 25 min
(Bl 1 hr 5 min
(Cl 1 hr 15 min
(Dl 2 hr 20 min
(El 3 hr 40 minArithme1tic
Since the flight schedules at each of Airports A,
B, and C are the same hour after hour, assume
that the passenger leaves Airport A at 8:00 and
arrives at Airport B at 10:30. Since flights from
Airport B leave at 20-minute intervals beginning
on the hour, the passenger must wait 10 minutes
at Airport B for the flight that leaves at 10:40
and arrives at Airport C 11 hours or 1 hour
6
10 minutes later. Thus, the passenger arrives at
Airport Cat 11 :50. Having arrived too late for
the 11:45 flight from Airport C, the passenger
must wait 55 minutes for the 12:45 flight. Thus,
the least total amount of time the passenger
must spend waiting between flights is 10 + 55 =
65 minutes, or 1 hour 5 minutes.The correct answer is B.
PS07426- If n is a positive integer and n^2 is divisible by 72, then
the largest positive integer that must divi Je n is
(Al 6
(Bl 12
(Cl 24
(Dl 36
(El 48Arithmetic
Since n^2 is divisible by 72, n^2 = 72k for some
positive integer k. Since n^2 = 72k, then 72k must
be a perfoct square. Since 72k = (2^3 )(3^2 )k, then
k = 2m^2 for some positive integer min order for
72k to be a perfect square. Then, n^2 = 72k =
(2^3 )(3^2 )(2m^2 ) = (2^4 )(3^2 )m^2 = [(2^2 )(3)(m)]2, and
n = (2^2 )(3)(m). The positive integers that MUST
divide n are 1, 2, 3, 4, 6, and 12. Therefore, the
largest positive integer that must divide n is 12.The corriect answer is B.