Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
07 Dec 2020, 06:45
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
73% (02:02) correct
27%
(02:05)
wrong
based on 216
sessions
History
Date
Time
Result
Not Attempted Yet
If a, b and c are positive integers such that a is a factor of b, and a is a multiple of c, which of the following is NOT necessarily an integer?
A. bc/a
B. (a + b)/c
C. (c - b)/a
D. (a + c)/c
E. 2bc/a
A. bc/a
B. (a + b)/c
C. (c - b)/a
D. (a + c)/c
E. 2bc/a
07 Dec 2020, 06:55
Kudos
Bookmarks
IMO C.
Here's how I would break it down:
"a is a factor of b" -> b = am (let m be a non zero integer)
"a is a multiple of c" -> a = ck ( k is again a non zero integer)
And combining both: b = ckm
Now we can simply plug in the values and see which ones will always be an integer.
Some are easy to detect for eg bc/a & 2bc/a because 'a' is a factor of 'b.'
C) (c-b)/a = (c-ckm)/ ck -> we will still have 'k' left in the denominator thus not necessarily an integer.
Posted from my mobile device
Here's how I would break it down:
"a is a factor of b" -> b = am (let m be a non zero integer)
"a is a multiple of c" -> a = ck ( k is again a non zero integer)
And combining both: b = ckm
Now we can simply plug in the values and see which ones will always be an integer.
Some are easy to detect for eg bc/a & 2bc/a because 'a' is a factor of 'b.'
C) (c-b)/a = (c-ckm)/ ck -> we will still have 'k' left in the denominator thus not necessarily an integer.
Posted from my mobile device
07 Dec 2020, 07:07
Kudos
Bookmarks
Bunuel
a is a factor of b
This means b/a = j (where j is some integer)
a is a multiple of c
This means a/c = k (where k is some integer)
This also means that (b/a)(a/c) = jk
Simplify to get: b/c = jk
If j and k are integers, we know that jk is an integer
A. bc/a
We already know that: b/a = j (where j is some integer)
If we multiply both sides of the equation by c we get: cb/a = jc
Since j and c are both integers, we know that jc is an integer, which means bc/a is ALWAYS an integer
B. (a + b)/c
(a + b)/c = a/c + b/c
= k + jk
= some integer (we already know k and jk are integers, it's also true that k + jk is an integer
C. (c - b)/a
(c - b)/a = c/a - b/a
We know that b/a is an integer, but we don't know anything about c/a
So let's test some values
From the given information, it COULD be the case that a = 4, b = 12 and c = 2
This means (c - b)/a = (2 - 12)/4 = -10/4 = -2.5, and -2.5 is NOT an integer
Answer: C
Cheers,
Brent
