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Around the World in 80 Questions (Day 2): If a, b, and c are positive

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Bunuel
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Around the World in 80 Questions (Day 2): If a, b, and c are positive [#permalink] Around the World in 80 Questions (Day 2): If a, b, and c are positive   18 Jul 2023, 08:00
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42% (01:59) correct 58% (02:22) wrong based on 313 sessions
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If a, b, and c are positive integers, is 2a + 121b - 5c divisible by 11?

(1) a + 3c is a multiple of 22

(2) a - 19c is a multiple of 33


 


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Re: Around the World in 80 Questions (Day 2): If a, b, and c are positive [#permalink] Around the World in 80 Questions (Day 2): If a, b, and c are positive   18 Jul 2023, 08:23
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Asked: If a, b, and c are positive integers, is 2a + 121b - 5c divisible by 11?

Is (2a + 11*11b - 5c) divisible by 11 ?
Is (2a-5c) divisible by 11 ?

(1) a + 3c is a multiple of 22
a + 3c = 22k ; where k is an integer.
2a + 6c = 44k
2a -5c + 11c = 44k
2a - 5c = 44k - 11c = 11(4k-c)
2a - 5c is a multiple of 11.
SUFFICIENT

(2) a - 19c is a multiple of 33
a - 19c = 33k ; where k is an integer
2a - 38c = 66k
2a - 5c - 33c = 66k
2a - 5c = 33c + 66k = 11(3c+6k)
2a - 5c is a multiple of 11.
SUFFICIENT

IMO D
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Re: Around the World in 80 Questions (Day 2): If a, b, and c are positive [#permalink] Around the World in 80 Questions (Day 2): If a, b, and c are positive   18 Jul 2023, 08:28
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since a, b and c are positive we can use the given statements as follows:

Stat..(1): a+3c=22k

if we change the given relation as a=22k-3c and substituting it in 2a+121b-5c
it will be like 2(22k-3c)+121b-5c, solving further it will be 44k+121b-11c
and if we change it like 11(4k+11b-c) then it is a multiple of 11
Therefore, (1) suff..

Stat..(2): a-19c=33k

if we change the given relation as a=33k+19c and substituting it in 2a+121b-5c
it will be like 2(33k+19c)+121b-5c, solving further it will be 66k+121b+33c
and if we change it like 11(6k+11b+3c) then it is a multiple of 11
Therefore, (2) suff..

Hence answer is D, both statements are sufficient alone.

Thank you,
Hope it helps... :)
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Re: Around the World in 80 Questions (Day 2): If a, b, and c are positive [#permalink] Around the World in 80 Questions (Day 2): If a, b, and c are positive   18 Jul 2023, 09:08
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Bunuel wrote:
If a, b, and c are positive integers, is 2a + 121b - 5c divisible by 11?

(1) a + 3c is a multiple of 22

(2) a - 19c is a multiple of 33


 


This question was provided by GMAT Club
for the Around the World in 80 Questions

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121b is divisible by 11. We need to find out if 2a - 5c is divisible by 11.

1)

a + 3c = 22*x

2a + 6c = 2*22*x

2a + 6c - 11c = 2 * 22 * x - 11c

2a - 5c = 11(2x-c)

Hence, 2a - 5c is a multiple of 11. Sufficient.

2)

a - 19c = 33*x

2a - 38c = 33*x

2a -38c + 33c = 33* x + 33c

2a - 5c = 33(3x+c)

Hence, 2a - 5c is a multiple of 11. Sufficient.

IMO D
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Re: Around the World in 80 Questions (Day 2): If a, b, and c are positive [#permalink] Around the World in 80 Questions (Day 2): If a, b, and c are positive   18 Jul 2023, 10:43
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Bunuel wrote:
If a, b, and c are positive integers, is 2a + 121b - 5c divisible by 11?

(1) a + 3c is a multiple of 22

(2) a - 19c is a multiple of 33


 


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for the Around the World in 80 Questions

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answer is D
IN STATEMENT 1 ;
a+3c is a multiple of 22
let us try to put the value of a & c
for c be 1 then for a be 19 if we put these value of a & c in equation 2a+121b-5c than we got the value which is divisible by 11 and similarly we try to put the value of c be 2 then a be 16 and similarly putting the value in equation we got the value which is divisible by11.
IN STATEMENT 2 we follow the similiar method and got the value divisible by 11.
THEREFORE THE RIGHT ANSWER IS "D"
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Re: Around the World in 80 Questions (Day 2): If a, b, and c are positive [#permalink] Around the World in 80 Questions (Day 2): If a, b, and c are positive   18 Jul 2023, 12:04
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Bunuel wrote:
If a, b, and c are positive integers, is 2a + 121b - 5c divisible by 11?

(1) a + 3c is a multiple of 22

(2) a - 19c is a multiple of 33



Using Statement 1:

2a + 6c = Multiple of 44 = 44*k; where k is a positive integer

then, 2a + 6c - 6c + 121b -5c = 44*k+ 121b - 11c = 44*k + 11(11b-c)
Since both 44 and 11 are divisible by 11, statement 1 is sufficient to provide an answer.

Using Statement 2:

2a-38c = Multiple of 66 = 66*K; where K is a positive integer

then, 2a - 38c + 38c + 121b - 5c = 66*K + 121b + 33c = 66*K + 11(11b- 3c)
Since both 66 and 11 are divisible by 11, statement 2 is sufficient to provide an answer.

IMO OA should be D.
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Re: Around the World in 80 Questions (Day 2): If a, b, and c are positive [#permalink] Around the World in 80 Questions (Day 2): If a, b, and c are positive   19 Jul 2023, 03:31
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\(121b\) will always be a value divisible by 11, so to determine if \(2a + 121b - 5c\) is divisible by 11, the relationship between \(2a - 5c\) will determine if the equation as a whole is divisible by 11. If the relation ship adds or subtracts a multiple of 11 to/from \(121b\) then the equation will be divisible by 11.

(1) a + 3c is a multiple of 22

Plugging in values to check divisibility:

when \(a = 1\) \(c = 7\) then \(2a - 5c = -33\) Divisible by 11
when \(a = 4\) \(c = 6\) then \(2a - 5c = -22\) Divisible by 11
when \(a = 19\) \(c = 1\) then \(2a - 5c = -33\) Divisible by 11
when \(a = 41\) \(c = 1\) then \(2a - 5c = -77\) Divisible by 11

Sufficient


(2) a - 19c is a multiple of 33

Plugging in values to check divisibility:

when \(a = 5\) \(c = 2\) then \(2a - 5c = 0\) Divisible by 11
when \(a = 24\) \(c = 3\) then \(2a - 5c = -33\) Divisible by 11
when \(a = 52\) \(c = 1\) then \(2a - 5c = 99\) Divisible by 11
when \(a = 71\) \(c = 2\) then \(2a - 5c = 132\) Divisible by 11


Sufficient

Answer D
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Re: Around the World in 80 Questions (Day 2): If a, b, and c are positive [#permalink]
19 Jul 2023, 03:31
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