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Difficulty:
65%
(hard)
Question Stats:
55% (01:59) correct
45%
(01:56)
wrong
based on 142
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If a, b and c are integers, is (a·b) a multiple of 18?
(1) 2a = 3b
(2) 2b = 3c
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(1) 2a = 3b
(2) 2b = 3c
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
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Kyala1Jameson
Joined: 09 Mar 2021
Last visit: 11 Jul 2021
Posts: 28
Location: South Africa
Concentration: Leadership, Marketing
Schools: LBS MAM "23
GPA: 3.01
WE:Design (Computer Software)
Schools: LBS MAM "23
Posts: 28
imo C
If a, b and c are integers, is (a·b) a multiple of 18?
(1) 2a = 3b
b= 2a/3
if a = 3 , b = 2 => ab = 6
if a= 18 , b= 12 => ab = 18 (12) .. Multiple of 18 ...
Not sufficient
(2) 2b = 3c
No info about a so
not sufficient ..
no combining 1 and 2 ..
b= 2a/3
c = 2b/3
since minimum of b has to be 6 (in form of 6n , n 1,2,3 ...) to satisfy a b c are integer ..
if b = 6 , c = 4 , a = 9 ... ab = 18 (something) .. multiple of 18 ...
.. Sufficient
If a, b and c are integers, is (a·b) a multiple of 18?
(1) 2a = 3b
b= 2a/3
if a = 3 , b = 2 => ab = 6
if a= 18 , b= 12 => ab = 18 (12) .. Multiple of 18 ...
Not sufficient
(2) 2b = 3c
No info about a so
not sufficient ..
no combining 1 and 2 ..
b= 2a/3
c = 2b/3
since minimum of b has to be 6 (in form of 6n , n 1,2,3 ...) to satisfy a b c are integer ..
if b = 6 , c = 4 , a = 9 ... ab = 18 (something) .. multiple of 18 ...
.. Sufficient
Posts: 1,101
If a, b and c are integers, is (a·b) a multiple of 18?
Stat 1: 2a = 3b
It means, a = multiple of 3 and b= multiple of 2. Now, (a·b) can be multiple of 6, which means either multiple of 18 or not. Not sufficient
Stat 2: 2b = 3c
It means, b = multiple of 3 and c= multiple of 2. But we don't know anything about a. Not sufficient
Combining both, a = multiple of 3 and b= multiple of 2 & 3. Now, (a·b) can be multiple of 18. Sufficient
So, I think C.
Stat 1: 2a = 3b
It means, a = multiple of 3 and b= multiple of 2. Now, (a·b) can be multiple of 6, which means either multiple of 18 or not. Not sufficient
Stat 2: 2b = 3c
It means, b = multiple of 3 and c= multiple of 2. But we don't know anything about a. Not sufficient
Combining both, a = multiple of 3 and b= multiple of 2 & 3. Now, (a·b) can be multiple of 18. Sufficient
So, I think C.
If a, b and c are integers, is (a·b) a multiple of 18?
(1) 2a = 3b
a = 3b/2
Put b = 2, a = 3 and a.b = 6, which is not a multiple of 18
Put b = 6, a = 9 and a.b = 54, multiple of 18
Not sufficient
(2) 2b = 3c
b = 3c/2
Put c = 2, then b = 3 and put a = 1, a.b = 3
Put c = 2, then b = 3 and put a = 6, a.b = 18
Not sufficient
(1) + (2)
2b = 3c
b = 3c/2
2a = 3b
2a = 3(3c/2)
a = 9c/4
Since a, b, and C are integers. Minimum value of c = 4
For c = 4, a = 9, b = 6
a.b = 54
So, For any value of c, a is always a multiple of 9 and b is always a multiple if 6. Thus, a.b is always a multiple of 18
Sufficient
Option C
Posted from my mobile device
(1) 2a = 3b
a = 3b/2
Put b = 2, a = 3 and a.b = 6, which is not a multiple of 18
Put b = 6, a = 9 and a.b = 54, multiple of 18
Not sufficient
(2) 2b = 3c
b = 3c/2
Put c = 2, then b = 3 and put a = 1, a.b = 3
Put c = 2, then b = 3 and put a = 6, a.b = 18
Not sufficient
(1) + (2)
2b = 3c
b = 3c/2
2a = 3b
2a = 3(3c/2)
a = 9c/4
Since a, b, and C are integers. Minimum value of c = 4
For c = 4, a = 9, b = 6
a.b = 54
So, For any value of c, a is always a multiple of 9 and b is always a multiple if 6. Thus, a.b is always a multiple of 18
Sufficient
Option C
Posted from my mobile device
Bunuel
If a, b and c are integers, is (a·b) a multiple of 18?
Notice that we are told that all variables are integers.
(1) 2a = 3b --> a/b = 3/2. a is a multiple of 3 and b is a multiple of 2. ab must be a multiple of 6 but may or may not be a multiple of 18 (for example consider (a,b) = (3,2) and (a,b) = 9,6). Not sufficient.
(2) 2b = 3c --> b/c = 3/2. b is a multiple of 3. Not sufficient.
(1)+(2) b is a multiple of both 2 and 3, so it must be a multiple of 6 and since a is a multiple of 3, then ab must be a multiple of 6*3 = 18. Sufficient.
Answer: C.
Hope it's clear.
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