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If x is the average of three integers a, b, and c, is x a multiple of

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If x is the average of three integers a, b, and c, is x a multiple of  [#permalink]

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New post 06 Jan 2018, 05:30
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If x is the average of three integers a, b, and c, is x a multiple of 4?

(1) a, b, and c are consecutive even integers.
(2) a<b<c and a is multiple of 4.


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Re: If x is the average of three integers a, b, and c, is x a multiple of  [#permalink]

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New post 06 Jan 2018, 06:06
3
If x is the average of three integers a,b, and c, is x a multiple of 4?

(1) a, b, and c are consecutive even integers.
(2) a is multiple of 4.

Hi,
(a+b+c)/3=x

1.
Let a be the first even integer.Then b=a+2 and c= a+4
therefore (3a+6)/3=x
or a+2=x

Now if a= 2 then x is 4which is a multiple of 4
However if a=4 then x is 6.. not a multiple of 4
BCE

2. Insufficient on it's own

CE
Now after combining both 1 & 2 we get

a+2=x and a=4k

therefore a+2=4k+2 which will never be a multiple of 4.
Hence C is sufficient.

Let me know if there's a flaw
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Re: If x is the average of three integers a, b, and c, is x a multiple of  [#permalink]

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New post 06 Jan 2018, 09:44
1
chetan2u wrote:
If x is the average of three integers a,b, and c, is x a multiple of 4?

(1) a, b, and c are consecutive even integers.
(2) a is multiple of 4.


New question on average


Is (a+b+c)/3 = 4k, where k is a non negative integer. So the question is basically asking whether a+b+c = 12k, where k is a non negative integer.

(1) Lets look at various cases of consecutive even integers. If we take a set of 0, 2, 4 then the sum = 0+2+4 = 6, is NOT a multiple of 12. But if we take a set of 2, 4, 6 then the sum = 2+4+6 = 12, which IS a multiple of 12. So cant say. Not Sufficient.

(2) a is a multiple of 4, but if (b+c) gives us an odd number, then the sum will not be a multiple of 12. If b & c are both multiples of 4, then the average can be a multiple of 4. Not Sufficient.

Combining the two statements, a has to be a multiple of 4 and a,b,c are consecutive even integers. If we take a,b,c as (4,6,8) respectively then the sum = 4+6+8 = 18, is NOT a multiple of 12. If we take a,b,c as (4,2,6) respectively then the sum = 4+2+6 = 12, which IS a multiple of 12. So both cases are possible (we are given that a,b,c are consecutive even integers but we are not given that they are so in ascending order).
So Insufficient.

Hence E answer
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Re: If x is the average of three integers a, b, and c, is x a multiple of  [#permalink]

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New post 06 Jan 2018, 09:52
amanvermagmat wrote:
chetan2u wrote:
If x is the average of three integers a,b, and c, is x a multiple of 4?

(1) a, b, and c are consecutive even integers.
(2) a is multiple of 4.


New question on average


Is (a+b+c)/3 = 4k, where k is a non negative integer. So the question is basically asking whether a+b+c = 12k, where k is a non negative integer.

(1) Lets look at various cases of consecutive even integers. If we take a set of 0, 2, 4 then the sum = 0+2+4 = 6, is NOT a multiple of 12. But if we take a set of 2, 4, 6 then the sum = 2+4+6 = 12, which IS a multiple of 12. So cant say. Not Sufficient.

(2) a is a multiple of 4, but if (b+c) gives us an odd number, then the sum will not be a multiple of 12. If b & c are both multiples of 4, then the average can be a multiple of 4. Not Sufficient.

Combining the two statements, a has to be a multiple of 4 and a,b,c are consecutive even integers. If we take a,b,c as (4,6,8) respectively then the sum = 4+6+8 = 18, is NOT a multiple of 12. If we take a,b,c as (4,2,6) respectively then the sum = 4+2+6 = 12, which IS a multiple of 12. So both cases are possible (we are given that a,b,c are consecutive even integers but we are not given that they are so in ascending order).
So Insufficient.

Hence E answer

Answer should be C.

You did a mistake! On combining you took a case where you got X as a multiple when you considered numbers as
4+2+6 = 12
You should check again that the *statement1* says that they are even consecutive integers.

And statement 2 says a is multiple of 4

So numbers can be 4,6,8 or 8,10,12 or, 12,14,16 and so on.. The sum of all of them in the form of 3X doesn't make X as multiple of 4

So OA should be C

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Re: If x is the average of three integers a, b, and c, is x a multiple of  [#permalink]

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New post 06 Jan 2018, 09:56
chetan2u wrote:
If x is the average of three integers a,b, and c, is x a multiple of 4?

(1) a, b, and c are consecutive even integers.
(2) a is multiple of 4.


New question on average


Statement 1: as \(a\), \(b\) & \(c\) are consecutive even integers so we we have \(b=a+2\) & \(c=a+4\)

So average \(=\frac{a+a+2+a+4}{3}=a+2=x\)

if \(a=2\), then \(x=4\), a multiple of \(4\) but if \(a=4\), then \(x=6\), not a multiple of \(4\). Insufficient

Statement 2: \(a=4k\). But we do not know about other integers. Insufficient

Combining 1 & 2: we have \(x=a+2=4k+2\). Hence \(x\) will not be a multiple of \(4\). Sufficient

Option C
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Re: If x is the average of three integers a, b, and c, is x a multiple of  [#permalink]

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New post 07 Jan 2018, 07:36
stmt 1: let 3 consecutive integers be 2n-2, 2n , 2n+2. So (avg) x = [(2n-2) + (2n) + (2n+2)]/3 => x = 2n. x is even for sure but not definitely a multiple of 4 (put x = 1,3,5 etc). stmt 1 alone not enough

stmt 2: it says a is a multiple of 4, let's wrint a as 4k. Now the avg is (4k + b + c)/3. Does not give anything. stmt 2 not enough

now combine stmt 2 and 1.

From stmt 1 -> x = 2n
From stmt 2 -> 2n-2 multiple of 4 => 2(n-1) = 4k => n-1 = 2k => n = 2k - 1.
put the value of n in stmt 1, we get
x = 2(2k -1)
x = 4k - 2

Now we can definitely say x is not a multiple of 4. Hence C
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Re: If x is the average of three integers a, b, and c, is x a multiple of  [#permalink]

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New post 07 Jan 2018, 09:01
Hi,

Can someone verify if the above answers are correct?

Since X is the average of a+b+c, then a+b+c= 3x (since the average is calculated as a+b+c/3 = x).

Now, if you use both statements and plug in for example 10+12+14, you have 3x = 36, thus x=12, which is a multiple of 4.

Then there are plenty examples where x is not a multiple of 4, therefore using this logic E should be the answer.

Regards.
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Re: If x is the average of three integers a, b, and c, is x a multiple of  [#permalink]

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New post 07 Jan 2018, 09:35
krikre wrote:
Hi,

Can someone verify if the above answers are correct?

Since X is the average of a+b+c, then a+b+c= 3x (since the average is calculated as a+b+c/3 = x).

Now, if you use both statements and plug in for example 10+12+14, you have 3x = 36, thus x=12, which is a multiple of 4.

Then there are plenty examples where x is not a multiple of 4, therefore using this logic E should be the answer.

Regards.


Hi krikre

if you are using both the statements, then the highlighted portion is incorrect. it is mentioned that "a" is a multiple of 4 and 10 is not a multiple of 4. try using any multiple of 4 and then see if you are getting C or E
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Re: If x is the average of three integers a, b, and c, is x a multiple of  [#permalink]

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New post 17 Jan 2018, 11:46
Hi everyone,

I have a doubt about statement 1.

When it says "a, b, and c are consecutive even integers.", can I assume that a < b < c or this relation has to be explicit in the statement?

Could anyone clarify this issue?

Thanks!
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Re: If x is the average of three integers a, b, and c, is x a multiple of  [#permalink]

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New post 17 Jan 2018, 12:10
kdgodoy wrote:
Hi everyone,

I have a doubt about statement 1.

When it says "a, b, and c are consecutive even integers.", can I assume that a < b < c or this relation has to be explicit in the statement?

Could anyone clarify this issue?

Thanks!


Hi

For this question you can assume any order because ultimately you are taking summation of the 3 variables. But for other question types you must not assume the order unless stated even if the variables are consecutive.


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Re: If x is the average of three integers a, b, and c, is x a multiple of  [#permalink]

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New post 22 Jan 2018, 19:59
niks18 wrote:
kdgodoy wrote:
Hi everyone,

I have a doubt about statement 1.

When it says "a, b, and c are consecutive even integers.", can I assume that a < b < c or this relation has to be explicit in the statement?

Could anyone clarify this issue?

Thanks!


Hi

For this question you can assume any order because ultimately you are taking summation of the 3 variables. But for other question types you must not assume the order unless stated even if the variables are consecutive.


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Hi niks18

If we assume any order
Say b < a < c
Eg 10(b), 12(a), 14(c) average is divisible by 4

If we assume order,
say a<b<c

12, 14, 16 - average is not divisible by 4.

So for this question, we have to assume, a < b < c.

Please correct if I am wrong

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Re: If x is the average of three integers a, b, and c, is x a multiple of  [#permalink]

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New post 22 Jan 2018, 21:27
hellosanthosh2k2 wrote:
niks18 wrote:
kdgodoy wrote:
Hi everyone,

I have a doubt about statement 1.

When it says "a, b, and c are consecutive even integers.", can I assume that a < b < c or this relation has to be explicit in the statement?

Could anyone clarify this issue?

Thanks!


Hi

For this question you can assume any order because ultimately you are taking summation of the 3 variables. But for other question types you must not assume the order unless stated even if the variables are consecutive.


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Hi niks18

If we assume any order
Say b < a < c
Eg 10(b), 12(a), 14(c) average is divisible by 4

If we assume order,
say a
12, 14, 16 - average is not divisible by 4.

So for this question, we have to assume, a < b < c.

Please correct if I am wrong

Posted from my mobile device


Hi hellosanthosh2k2

This is the precise reason why statement 1 is not sufficient here. It has nothing to do with order but you cannot definitely say that average of 3 consecutive even integer will be divisible by 4.

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Re: If x is the average of three integers a, b, and c, is x a multiple of  [#permalink]

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New post 22 Jan 2018, 21:50
1
Hi niks18

my confusion is,

if a, b and c are consecutive integers in that order,
then say a = 4m (multiple of 4)
b = 4m + 2, c = 4m + 4

average = (12m + 6)/3 = 4m + 2 , which is not divisible by 4

but if b < a < c
b = 4m -2, a = 4m, c = 4m+2

average = (4m - 2 + 4m + 4m +2) = 12m/3 = 4m (which is divisible by 4)

so i think order does matter

Maybe i am wrong
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Re: If x is the average of three integers a, b, and c, is x a multiple of  [#permalink]

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New post 22 Jan 2018, 22:00
hellosanthosh2k2 wrote:
Hi niks18

my confusion is,

if a, b and c are consecutive integers in that order,
then say a = 4m (multiple of 4)
b = 4m + 2, c = 4m + 4

average = (12m + 6)/3 = 4m + 2 , which is not divisible by 4

but if b < a < c
b = 4m -2, a = 4m, c = 4m+2

average = (4m - 2 + 4m + 4m +2) = 12m/3 = 4m (which is divisible by 4)

so i think order does matter

Maybe i am wrong


Hi hellosanthosh2k2,

yes agreed when we are combining both the sentences then order does matter but for statement 1 order does not matter.

hi chetan2u

to remove the ambiguity around the question it would be better to mention the order of a, b, & c although as they are written alphabetically implying that a, b, c are in ascending order
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Re: If x is the average of three integers a, b, and c, is x a multiple of  [#permalink]

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New post 23 Jan 2018, 01:07
So niks18, can I assume that the order a < b < c when it is written alphabetically?

Because it was my question.

I approached the question as hellosanthosh2k2 posted.

When we assume any order because ultimately you are taking summation of the 3 variables, as you said, the two statements together aren't sufficient to answer the question.

Hence, the answer would be E.
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Re: If x is the average of three integers a, b, and c, is x a multiple of  [#permalink]

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New post 23 Jan 2018, 01:09
1
kdgodoy wrote:
So niks18, can I assume that the order a < b < c when it is written alphabetically?

Because it was my question.

I approached the question as hellosanthosh2k2 posted.

When we assume any order because ultimately you are taking summation of the 3 variables, as you said, the two statements together aren't sufficient to answer the question.

Hence, the answer would be E.


a, b, and c are consecutive even integers does NOT necessarily mean that a < b < c.
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Re: If x is the average of three integers a, b, and c, is x a multiple of &nbs [#permalink] 23 Jan 2018, 01:09
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