GMAT® Official Guide 2019 Quantitative Review
z be represented by 2n + 1 and 2n + 3,
where n can be any integer. Since 28 = w + z,
it follows that
28 = (2n + 1) +(2n + 3)
28 = 4n+ 4
24 = 4n6 = nsimplify
subtract 4 from
both sides
divide both sides
by4Thus, w = 2(6) + 1 = 13, z = 2(6) + 3 = 15, and
hence exactly one value can be determined
for wz; SUFFICIENT.
The correct answer is B;
statement 2 alone is sufficient.DS02474 !!_- If abc * 0, is Q = _!!__?
c b
(1) a = l c
(2) C = 1
Algebra Fractions
a
Since _k_ = ~ + e = ~ x .! = .!!.... and
e b b e be
a
b - = a+ -b = ax -eae = - .. 1t 1s to b e
e e b b 'determined whether .!!.... = ae.
be b
(1) Given that a = 1, the equation to be
1nvest1gate.. d , -a = -ae , 1s. -^1 = -e. Tu· 1s
be b be b
equation can be true for some nonzero
values of band e (for example, b = e = 1) and
false for other nonzero values of b and e (for
example, b = 1 and e = 2); NOT sufficient.
(2) Given that e = 1, the equation to be
1nvest1gate.. d , -a = -ae ,. 1s -a = -a. Tu1s ·
be b b b
equation is true for all nonzero values of a
and b; SUFFICIENT.
The correct answer is B;
statement 2 alone is sufficient.DS1447i- The arithmetic mean of a collection of 5 positive
integers, not necessarily distinct, is 9. One additional
positive integer is included in the collection and the
arithmetic mean of the 6 integers is computed. Is the
arithmetic mean of the 6 integers at least 10?
(1) The additional integer is at least 14.
(2) The additional integer is a multiple of 5.Arithmetic Statistics
Since the arithmetic mean of the 5 integers is 9,
the sum of the 5 integers divided by 5 is equal to
9, and hence the sum of the 5 integers is equal
to (5)(9) = 45. Let x be the additional positive
integer. Then the sum of the 6 integers is 45 + x,
and the arithmetic mean of the 6 integers is
--^45 + x. D eterm1ne. w h et h er --45+x ~ 10 , or
6 6
equivalently, whether 45 + x ~ 60, or equivalently,
whether:>.:~ 15.(1) Given that x ~ 14, then x could equal 14
and x ~ 15 is not true, or x could equal 15
and x ~ 15 is true; NOT sufficient.
(2) Given that xis a multiple of 5, then x could
equal 10 and x ~ 15 is not true, or x could
equal 15 and x ~ 15 is true; NOT sufficient.
Taking (l) and (2) together, then xis a multiple
of 5 that is greater than or equal to 14, and
so x could equal one of the numbers 15, 20, 25,
30,. ... Each of these numbers is greater than or
equal to 15.The corn!ct answer is C;
both statements together are sufficient.
DS11003- A certain list consists of 400 different numbers. Is the
average (arithmetic mean) of the numbers in the list
greater than the median of the numbers in the list?
(1) Of the numbers in the list, 280 are less than the
average.
(2) Of the numbers in the list, 30 percent are
greater than or equal to the average.Arithmetic Stati tics
In a list of 400 numbers, the median will be
halfway between the 200th and the 201st
numbers in the list when the numbers are ordered
from least to greatest.