How many even integers N exist such that x

Question

How many even integers N exist such that x < N < x + 10?

(1) x is not odd

(2) x is not an integer

B

rohit8865 · Accepted Answer

(1) if x= 0.1 then total no. of even integers (0.1< N < 10.1) = 5
if x=2 then total number of even integers (2<N<12) = 4
Not suff

(2) if x= non integer
then N= 5 always
sufficient

Ans B

musunna · Answer

This question test the concept of number line. 'Not odd' means any other integers, fraction etc except ODD number. Now just plug the given equation those 'not odd' values.

Ans: B

Bunuel · Answer

Check similar questions to practice: how-many-integers-n-are-there-such-that-r-n-s-166396.html how-many-integers-are-there-such-that-v-n-w-129065.html Hope it helps.

HKD1710 · Answer

x < N < X+10

Stmt-1: 
if x is not odd then what else can it be? it can be even. but do you see the second stmt as well, oh yes! x can be fraction too

So you have two cases:
when x is even say x=0 then you have 0 < 2,4,6,8 < 10 - in total 4 even numbers
when x is fraction say x=0.5 then you have 0.5 < 2,4,6,8,10 < 10.5 - in total 5 even numbers

hence statement-1 is  insufficient .

stmt-2: 
when x is fraction say x=0.5 then you have 0.5 < 2,4,6,8,10 < 10.5 - in total 5 even numbers
when x is fraction say x=1.5 then you have 1.5 < 2,4,6,8,10 < 11.5 - in total 5 even numbers
when x is fraction say x=2.5 then you have 2.5 < 4,6,8,10,12 < 12.5 - in total 5 even numbers

Sufficient .

devanshu92 · Answer

Easy deal by number picking. Questions asks the number of even integers between x and (x+10)

1. If x is an even integer
Ex: x = 0 ----> 0< 2,4,6,8 < 10 = 4

If x is not an integer but is even
Ex: x = 0.5 ----> 0.5 < 2,4,6,8,10 < 10.5 = 5

Clearly not sufficient.

2. If x is not an integer
Ex: x = 0.5 -----> 0.5 < 2,4,6,8,10 < 10.5 = 5

Sufficient.

Option B

Kudos if it helps

nightvision · Answer

why x can't be negative? 
not odd doesn't mean x can't be negative.

fskilnik · Answer

x\,\,\, < \,\,\,N\,\,{	ext{even}}\,\,\, < \,\,\,x + 10

? = \# N

\left( 1 ight)\,\,x 
e {	ext{odd}}\,\,\,\left\{ \begin{gathered}
 \,{	ext{Take}}\,\,x{	ext{ = 0}}\,\,\,\, \Rightarrow \,\,\,\,? = 4\,\,\,\,\,\,\,\left  \hfill \
 \,{	ext{Take}}\,\,x = 0.1\,\,\,\, \Rightarrow \,\,\,\,? = 5\,\,\,\,\,\,\,\left  \hfill \ 
\end{gathered} ight.

\left( 2 ight)\,\,x 
e \operatorname{int} \,\,\,\mathop \Rightarrow \limits^{\left( * ight)} \,\,\,x < \left\langle x ightangle \leqslant N \leqslant \left\langle {x + 9} ightangle < x + 10

\Rightarrow \,\,\,\,\left\{ \begin{gathered}
 \,\left\langle x ightangle \,\,{	ext{odd}}\,\,\,\, \Rightarrow \,\,\,\,\left\langle {x + 9} ightangle \,\,{	ext{even}}\,\,\,\, \Rightarrow \,\,\,\,{	ext{?}} = {	ext{5}}\,\,\,\left  \hfill \
 \,\left\langle x ightangle \,\,{	ext{even}}\,\,\,\, \Rightarrow \,\,\,\,\left\langle {x + 9} ightangle \,\,{	ext{odd}}\,\,\,\, \Rightarrow \,\,\,\,{	ext{?}} = {	ext{5}}\,\,\,\left  \hfill \ 
\end{gathered} ight.\,\,\,\, \Rightarrow \,\,\,\,\,? = 5

\left( * ight)\,\,\left\langle r ightangle \,\, = \,\,{	ext{smallest}}\,\,{	ext{integer}}\,\,{	ext{greater}}\,\,{	ext{than}}\,\,r

The correct answer is therefore (B).

This solution follows the notations and rationale taught in the GMATH method.

Regards, 
Fabio.

fskilnik · Answer

Hi, nightvision ! When statement (1) is considered, a BIFURCATION is all you need to prove insufficiency. (You COULD bifurcate (1) using negative numbers, of course.) Statement (2) is a lot more interesting (and its corresponding rigorous treatment much harder). My solution (posted above) deals with all possible scenarios. If you cannot feel comfortable with it, let me give you a suggestion: Substitute x by 0.1 (so that will be 1) and do all the reasoning. Afterwards, do exactly the same with 1.1 (in this case will be 2). If you prefer negative numbers, LoL, try x = -1.1 (in which case equals -1). Afterwards, do exactly the same with x = -0.1 (in this case equals 0). I believe after (at most) all 4 substitutions mentioned, you will REALLY understand all my arguments! Regards and success in your studies, Fabio.

itisSheldon · Answer

Veritas Prep  OFFICIAL EXPLANATION

There are various ways to approach a problem like this one.

Students who are familiar with the principles of evenly spaced sets (so-called “Fence Post Problems”) can immediately infer that the number of even integers in this set will depend on whether the excluded endpoints x and x+10 are themselves even or not.

For these students, it is important to be careful when interpreting statement (1). When reading that x is not odd, recognize that two possibilities still remain – even, of course, but also non-integer. One case excludes the endpoints, but the other does not, so the number of elements in the set will change by 1 between these two cases (it will be either 4 or 5, respectively). Statement (2), however, guarantees that the endpoints are not excluded, so the number of even integers will simply be Range/Spacing = 10/2 = 5.

Other students may wish to pick a few possible values for x to reach the same conclusion. In that case, it’s important to consider different types of numbers. What types? Since the additional statements reference whether x is odd and whether it’s an integer, it would be wise to choose at least one even integer, one odd integer, and one non-integer.

For x is even, we will always find four even integers N satisfying x < N < x+10. E.g. x=10, 10 < N < 20, N could be 12, 14, 16, or 18.

For x is odd, we will always find five even integers N satisfying x < N < x+10. E.g. x=1. 1 < N < 11, N could be 2, 4, 6, 8, or 10.

For x is a non-integer, we will once again always find five even integers N satisfying x < N < x+10. E.g. x=-0.5. -0.5 < N < 9.5, N could be 0, 2, 4, 6, or 8.

Using either approach, statement (1) is not sufficient to answer the question, but statement (2) is.  The answer is B.

wjMBA · Answer

Hi,

How many even integers N exist such that x < N < x + 10?

(1) x is not odd

(2) x is not an integer

I have another question, so we ignore statement (1) and acknowledge that it is insufficient, but where does statement (2) or the question tell us that x can't be odd?

For example it doesn't say that 3 < 4, 6, 8, 10, 12 < 13 does it?

Or does the question imply that the numbers are sequential?

Basshead · Answer

x < N < x + 10

Scenarios:  
 1.1 < N < 10.1  gives us 5 even integers
 2 < N < 12  gives us 4 even integers
 1 < N < 11  gives us 5 integers

(1) If x = not odd, we can still get four or five even integers, depending on if x = non-integer or x = even. INSUFFICIENT.

(2) If x = not an integer, we will have 5 even integers. SUFFICIENT.

Answer is B.

bumpbot · Answer

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How many even integers N exist such that x < N < x + 10?

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