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If a is divisible by 2, is b + 5 an integer?  [#permalink]

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Question Stats: 68% (02:14) correct 32% (02:10) wrong based on 137 sessions

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If a is divisible by 2, is b + 5 an integer?

(1) The median of a and b is not an integer.
(2) The average (arithmetic mean) of 3a, b, and b + 10 is an even integer.

Something that occasionally throws me off and I often forget. When we get a variable on the gmat, what can we assume about it? Just that it is a real, rational number? We CANNOT automatically assume it's an integer correct?

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Originally posted by manimgoindowndown on 28 Feb 2013, 14:11.
Last edited by Bunuel on 01 Mar 2013, 01:52, edited 3 times in total.
Moved to DS forum and edited the question. Edited OA.
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Re: If a is divisible by 2, is b + 5 an integer?  [#permalink]

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Right. We only know that b is real and rational. If we could assume variables were integers, we wouldn't need any additional statements at all to answer the question, "Is b + 5 an integer?"
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Re: If a is divisible by 2, is b + 5 an integer?  [#permalink]

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manimgoindowndown wrote:
If a is divisible by 2, is b + 5 an integer?

(1) The median of a and b is not an integer.
(2) The average (arithmetic mean) of 3a, b, and b + 10 is an even integer.

Something that occasionally throws me off and I often forget. When we get a variable on the gmat, what can we assume about it? Just that it is a real, rational number? We CANNOT automatically assume it's an integer correct?

answer should be Option(b) as

from (1) :- either b is an odd integer or a real number of type x.y ...........hence not sufficient.
from (2) : (3a+b+b+10)/3 = 2k (let say)
so, 3a + 2b = 6k - 10
b = (3(2k-a) - 10)/2
b= 3(2k - a)/2 -5
as k is even, so 2k is also even also even - even is an even hence (2k-a) is divisible by 2
so, b = 3*some_integer - 5
hence b = integer...............hence condition 2 is sufficient

hence answer should be option(b)
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Re: If a is divisible by 2, is b + 5 an integer?  [#permalink]

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jbisht wrote:
manimgoindowndown wrote:
If a is divisible by 2, is b + 5 an integer?

(1) The median of a and b is not an integer.
(2) The average (arithmetic mean) of 3a, b, and b + 10 is an even integer.

Something that occasionally throws me off and I often forget. When we get a variable on the gmat, what can we assume about it? Just that it is a real, rational number? We CANNOT automatically assume it's an integer correct?

answer should be Option(b) as

from (1) :- either b is an odd integer or a real number of type x.y ...........hence not sufficient.
from (2) : (3a+b+b+10)/3 = 2k (let say)
so, 3a + 2b = 6k - 10
b = (3(2k-a) - 10)/2
b= 3(2k - a)/2 -5
as k is even, so 2k is also even also even - even is an even hence (2k-a) is divisible by 2
so, b = 3*some_integer - 5
hence b = integer...............hence condition 2 is sufficient

hence answer should be option(b)

I bolded two parts of your explanation. You originally indicated that 2k was even. k itself may be even or odd. You later asserted that k was even.
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Re: If a is divisible by 2, is b + 5 an integer?  [#permalink]

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manimgoindowndown wrote:
If a is divisible by 2, is b + 5 an integer?

(1) The median of a and b is not an integer.
(2) The average (arithmetic mean) of 3a, b, and b + 10 is an even integer.

I don't think the given OA is correct. Anyways, let a = 2p,where p is any integer.

From F.S 1, we have that$$\frac{(a+b)}{2}$$ = Not an integer

or$$\frac{2p}{2}+\frac{b}{2} = p+\frac{b}{2}$$= Not an integer. Assuming, b = 3 , we satisfy the fact statement and get a YES for the question stem. Again assuming b=1.2, we satisfy the fact statement but get a NO for the question stem.Insufficient.

From F.S 2, we have $$3a+b+(b+10)/3$$ = 2k, where k is again an integer.

Thus,$$3a/3 + (b+5)/3+(b+5)/3$$ = 2k

or 2/3*(b+5) = 2k-a = 2k-2p = 2(k-p)[an integer]

or (b+5) = An integer*3 = integer.

Sufficient.

B.
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Re: If a is divisible by 2, is b + 5 an integer?  [#permalink]

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2
NonYankee wrote:
manimgoindowndown wrote:
If a is divisible by 2, is b + 5 an integer?

(1) The median of a and b is not an integer.
(2) The average (arithmetic mean) of 3a, b, and b + 10 is an even integer.

Something that occasionally throws me off and I often forget. When we get a variable on the gmat, what can we assume about it? Just that it is a real, rational number? We CANNOT automatically assume it's an integer correct?

Right. We only know that b is real and rational. If we could assume variables were integers, we wouldn't need any additional statements at all to answer the question, "Is b + 5 an integer?"

From the stem we cannot assume that b is a rational number, it could be an irrational as well, for example $$\sqrt{2}$$.

If no other constraint is given about a variable, then all we can say that it's a real number.

If a is divisible by 2, is b + 5 an integer?

Notice that:
a is divisible by 2 implies that a is an even number.
The question basically asks whether b is an integer.

(1) The median of a and b is not an integer. b may or may not be an integer. Consider: a=2 and b=3 for an YES answer and a=2 and b=1/2 for a NO answer. Not sufficient.

(2) The average (arithmetic mean) of 3a, b, and b + 10 is an even integer --> $$3a+b+(b+10)=3*even$$ --> $$2b+10=3*even-3a$$ --> since a is even, then $$2b+10=3*even-3*even$$ --> $$2(b+5)=3(even-even)=3*even$$ --> $$b+5=3*\frac{even}{2}=3*integer=integer$$. Sufficient.

Hope it's clear.
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Re: If a is divisible by 2, is b + 5 an integer?  [#permalink]

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Bunuel wrote:
From the stem we cannot assume that b is a rational number, it could be an irrational as well, for example $$\sqrt{2}$$.

You make a good point. In any case, it's not an integer.
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Re: If a is divisible by 2, is b + 5 an integer?  [#permalink]

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manimgoindowndown wrote:
If a is divisible by 2, is b + 5 an integer?

(1) The median of a and b is not an integer.
(2) The average (arithmetic mean) of 3a, b, and b + 10 is an even integer.

Something that occasionally throws me off and I often forget. When we get a variable on the gmat, what can we assume about it? Just that it is a real, rational number? We CANNOT automatically assume it's an integer correct?

a = even

b+5 is an integer?

(1) The median of a and b is not an integer.

a+b/2 is not an integer.

Its given that a is even. The above expression can not be an integer if b is odd or fraction.

(2) The average (arithmetic mean) of 3a, b, and b + 10 is an even integer

3a+b+b+10/2= integer

3a+2b+10/2= integer

3a is even (a is even), 2 b is even and 10 is even. To have above equation as an integer b must be an integer.

Hence b+5 will be an integer.

B is the answer
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Re: If a is divisible by 2, is b + 5 an integer?  [#permalink]

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manimgoindowndown wrote:
If a is divisible by 2, is b + 5 an integer?

(1) The median of a and b is not an integer.
(2) The average (arithmetic mean) of 3a, b, and b + 10 is an even integer.

Something that occasionally throws me off and I often forget. When we get a variable on the gmat, what can we assume about it? Just that it is a real, rational number? We CANNOT automatically assume it's an integer correct?

From the question stem it is clear that $$a$$ is $$Even$$ and we need to know whether $$b$$ is integer

Statement 1: implies $$\frac{a+b}{2} =$$ Non Integer (NI). if $$a=2$$ & $$b=1.5$$, then median is non integer and $$b+5$$ is also a non integer

but if $$a=2$$ & $$b=3$$, then median is non integer but $$b+5$$ is an integer. Hence we have both a Yes & a NO. Insufficient

Statement 2: implies $$\frac{3a+b+b+10}{3}=Even => 3a+2b+10=3*Even$$

$$=>2b=Even-10-3a$$. Now as $$a$$ is even so $$3a$$ is even and $$10$$ is also even

so $$2b=Even-Even-Even=Even => b=\frac{Even}{2}=Integer$$. Hence $$b+5$$ will be an integer. Sufficient

Option B Re: If a is divisible by 2, is b + 5 an integer?   [#permalink] 23 Feb 2018, 21:15
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