GMAT® Official Guide 2019 Quantitative Review
PS00ll l- For all positive integers m and v, the expression me v
represents the remainder when m is divided by v. What
is the value of ((98 e 33) e 17) -(98 e (33 e 17))?
(A) -10
(B) -2
(C) 8
(D) 13
(E) 17Arithmetic
First, for ((98 0 33) 017), determine 98 0 33,
which equals 32, since 32 is the remainder
when 98 is divided by 33 (98 = 2(33) + 32).
Then, determine 32 017, which equals 15,
since 15 is the remainder when 32 is divided by
17 (32 = 1(17) + 15). Thus, ((98 0 33) 0 17) = 15.Next, for (98 0 (33 017)), determine 33 017,
which equals 16, since 16 is the remainder
when 33 is divided by 17 (33 = 1(17) + 16).
Then, determine 98 0 16, which equals 2,
since 2 is the remainder when 98 is divided by
16 (98 = 6(16) + 2). Thus, (98 0 (33 0 17)) = 2.Finally, ((98 0 33) 0 17 - (98 0 (33 017)) =
15 - 2 = 13.The correct answer is D.PS13841Year-end Investment Value
18,000 -+-----------
16,000 -+-------
~ ~ 14,000
:g 12,000
u:i 10,000
2 8,000
~ 6,000 +--f-".LL<J-----l'/
~ 4,000
2,000
0 +--_._------L----'----'-------L--'-------'---'----
2000 2001 2002 2003
year- bonds ~ stocks D cash
84. The chart above shows year-end values for Darnella's
investments. For just the stocks, what was the increase
in value from year-end 2000 to year-end 2003?(A) $1,000
(B) $2,000
(C) $3,000
(D) $4,000
(E) $5,000Arithme1tic
From the graph, the year-end 2000 value for
stocks is 9,000 - 6,000 = 3,000 and the year-end
2003 value for stocks is 10,000 -5,000 = 5,000.
Therefore, for just the stocks, the increase in
value from year-end 2000 to year-end 2003 is
5,000-3,000 = 2,000.The correct answer is B.
PS05775- If the sum of the reciprocals of two consecutive odd
integers is ~!, then the greater of the two integers is
(Al 3
(B) 5
(C) 7
(D) 9
(E) 11Arithmetic
The sum of the reciprocals of 2 integ :rs, a and b,
1s. 1 1 - + - = --a+ b. Th ere1ore, r since. -^12. 1s t e h sum o f
a b ab 35
the reciprocals of 2 consecutive odd integers, the
integers must be such that their sum is a multiple
of 12 and their product is the same multiple of 35
so that the fraction reduces to^12. Considering
35
the simplest case where a+ b = 12 and ab= 35,
it is easy to see that the integers are 5 and 7
since 5 and 7 are the only factors of 35 that are
consecutive odd integers. The larger of these is 7.Algebraically, if a is the greater of the two
integers, then b = a - 2 anda+(a-2) 12
a(a-2) 352a-2 12
a(a-2) 35
35(2a -2) = l2a(a -2)
70a - 70 = l2a^2 - 24a
0 = l2a^2 - 94a + 70
0=2(6a-5)(a-7)Thus, 6a - 5 = 0, so a=-,^5 or a - 7 = 0, so a= 7.
6
Since a must be an integer, it follows that a= 7.The correct answer is C.