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24 Oct 2023, 01:30
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B
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Difficulty:
45%
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63% (01:29) correct
37%
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If a, b and c are even integers greater than 10, is a+b divisible by 6?
(1) b=2c
(2) a=2b
(1) b=2c
(2) a=2b
24 Oct 2023, 04:26
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If a, b and c are even integers greater than 10, is a+b divisible by 6?
All we know is that all the three are multiple of 2.
Thus, a + b can be such that
1a. a and b are equal and each a multiple of 3, leaving us with a YES. E.g. 12 + 12 = 24
1b. Both are a non-multiple of 3, leaving us with with a NO. E.g. 14 + 14 = 28
2a. Both are not equal but sum gives us a multiple of 3, leaving us with a YES. E.g. 14 + 16 = 30
2b. Both are not equal but sum also is a non-multiple of 3, leaving us with a YES. E.g. 16 + 20
and So on..........
So many more possibilities exists.
(1) b=2c
Nothing about c so
1. a = 12 c = 14
b = 28
a + b = 40, NO
2. a = 12, c = 12
b = 24
a + b = 36, YES
INSUFFICIENT.
(2) a=2b
a + b = 2b + b = 3b
Since we already know that b is even i.e. 2x where x is any integer greater than 5,
a + b = 3*2x = 6x
Hence a multiple of 6.
SUFFICIENT.
Answer B.
All we know is that all the three are multiple of 2.
Thus, a + b can be such that
1a. a and b are equal and each a multiple of 3, leaving us with a YES. E.g. 12 + 12 = 24
1b. Both are a non-multiple of 3, leaving us with with a NO. E.g. 14 + 14 = 28
2a. Both are not equal but sum gives us a multiple of 3, leaving us with a YES. E.g. 14 + 16 = 30
2b. Both are not equal but sum also is a non-multiple of 3, leaving us with a YES. E.g. 16 + 20
and So on..........
So many more possibilities exists.
(1) b=2c
Nothing about c so
1. a = 12 c = 14
b = 28
a + b = 40, NO
2. a = 12, c = 12
b = 24
a + b = 36, YES
INSUFFICIENT.
(2) a=2b
a + b = 2b + b = 3b
Since we already know that b is even i.e. 2x where x is any integer greater than 5,
a + b = 3*2x = 6x
Hence a multiple of 6.
SUFFICIENT.
Answer B.
Aryaa03
Joined: 12 Jun 2023
Last visit: 19 Feb 2026
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GMAT Focus 1: 645 Q86 V81 DI78 
23 Dec 2023, 23:38
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Bunuel
Sol: Option-B
Let a=2m, b=2n and c=2p where, m, n and p are greater than 5
Is a+b divisible by 6 ?
a+b = 2m+2n = 2*(m+n) , therefore a+b will be divisible by 6 if (m+n) is a multiple of 3
Stat-1:
b=2c
There is no relation given between a and b, thus Insufficient
To cross-check:
Let, b= 28, c=14
If a=16 ; (a+b) = (16+28) = 44 , which is not divisible by 6
If a=14 ; (a+b) = (14+28) = 42 , which is divisible by 6
Stat-2:
a=2b , This implies m=2n
(a+b) = 2*(m+n) = 2*(2n+n) = 2*3*n , for n>5, this is divisible by 6. Sufficient
