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If m is an integer greater than 9 but less than 20, is n greater than

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If m is an integer greater than 9 but less than 20, is n greater than [#permalink]

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If m is an integer greater than 9 but less than 20, is n greater than the average (arithmetic mean) of m and 20?

(1) n = 3m
(2) The distance on the number line between n and 20 is less than the distance on the number line between n and m.

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Re: If m is an integer greater than 9 but less than 20, is n greater than [#permalink]

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New post 26 Jan 2018, 00:01
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IMO D

Statement= n>(m+20)/2?
1) If m=10, n=30 & m+20/2=15, therefore n>average of m+20. Stands true for all values. Sufficient.
2) For all values of m, if n is closer to 20 than it is to m on the number line then n>average of m+20, holds true for all values of m from 9 to 19. Therefore Sufficient.
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Re: If m is an integer greater than 9 but less than 20, is n greater than [#permalink]

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New post 01 Feb 2018, 10:31
A is sufficient.

however B can as well be sufficient because distance between n and 20 is less than n and m.

for example m can assume values from 10 to 19. there fore middle value being 14.5. Hence as per option B n has to be greater than 14.5 and less than 20. it cannot exceed 20 or take higher values than 20 because at this point all value of n will not satisfy condition B. Hence 14.5<= n<=20

Answer D
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If m is an integer greater than 9 but less than 20, is n greater than [#permalink]

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Bunuel wrote:
If m is an integer greater than 9 but less than 20, is n greater than the average (arithmetic mean) of m and 20?

(1) n = 3m
(2) The distance on the number line between n and 20 is less than the distance on the number line between n and m.


It should be D

From the stem, we know the max and min for \(m\)

\(9 < m < 20\) - so \(m\) can range between \(10\) and \(19\). Arithmetic mean of m and 20 can be \(\frac{10+20}{2} = 15\) or \(\frac{19+20}{2} = 19.5\) at the extremes. Question asks if \(n\) is greater than these values.

Statement 1: Sufficient
\(n = 3m\) so minimum value of \(n = 45\) and maximum is \(57\), which makes it greater than the AM of \(m\) and \(20\) in all cases.

Statement 2: Sufficient
Since \(m\) is to the left of \(20\), and \(n\) is closer to \(20\) than it is to \(m\), this tells us \(n\) is always to the right of \(m\). The diagram attached shows both possible cases and arrows represent Arithmetic mean of \(m\) and \(20\) and in both cases, \(n\) shows up as greater than the Arithmetic mean.
Attachments

diagram.jpg
diagram.jpg [ 19.14 KiB | Viewed 679 times ]


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Re: If m is an integer greater than 9 but less than 20, is n greater than [#permalink]

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New post 03 Feb 2018, 06:22
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jedit wrote:
Bunuel wrote:
If m is an integer greater than 9 but less than 20, is n greater than the average (arithmetic mean) of m and 20?

(1) n = 3m
(2) The distance on the number line between n and 20 is less than the distance on the number line between n and m.


It should be D

From the stem, we know the max and min for \(m\)

\(9 < m < 20\) - so \(m\) can range between \(10\) and \(19\). Arithmetic mean of m and 20 can be \(\frac{10+20}{2} = 15\) or \(\frac{19+20}{2} = 19.5\) at the extremes. Question asks if \(n\) is greater than these values.

Statement 1: Sufficient
\(n = 3m\) so minimum value of \(n = 45\) and maximum is \(57\), which makes it greater than the AM of \(m\) and \(20\) in all cases.

Statement 2: Sufficient
Since \(m\) is to the left of \(20\), and \(n\) is closer to \(20\) than it is to \(m\), this tells us \(n\) is always to the right of \(m\). The diagram attached shows both possible cases and arrows represent Arithmetic mean of \(m\) and \(20\) and in both cases, \(n\) shows up as greater than the Arithmetic mean.


Good explanation for statement 2 man.Thanks.
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If m is an integer greater than 9 but less than 20, is n greater than [#permalink]

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New post Updated on: 03 Feb 2018, 11:14
jedit wrote:
Bunuel wrote:
If m is an integer greater than 9 but less than 20, is n greater than the average (arithmetic mean) of m and 20?

(1) n = 3m
(2) The distance on the number line between n and 20 is less than the distance on the number line between n and m.


It should be D

From the stem, we know the max and min for \(m\)

\(9 < m < 20\) - so \(m\) can range between \(10\) and \(19\). Arithmetic mean of m and 20 can be \(\frac{10+20}{2} = 15\) or \(\frac{19+20}{2} = 19.5\) at the extremes. Question asks if \(n\) is greater than these values.

Statement 1: Sufficient
\(n = 3m\) so minimum value of \(n = 45\) and maximum is \(57\), which makes it greater than the AM of \(m\) and \(20\) in all cases.

Statement 2: Sufficient
Since \(m\) is to the left of \(20\), and \(n\) is closer to \(20\) than it is to \(m\), this tells us \(n\) is always to the right of \(m\). The diagram attached shows both possible cases and arrows represent Arithmetic mean of \(m\) and \(20\) and in both cases, \(n\) shows up as greater than the Arithmetic mean.



Hi jedit

how did you define that \(m\) can range between \(10\) and \(19\). ? Did you break this \(9 < m < 20\) into two inequalities

\(9 < m\) --> \(1+9<m+1\) --> \(10<m+1\) like this ? :?

\(m < 20\) --> \(1+m<20+1\) --> \(1+m<21\) like this ? :?

thanks! :-)

Originally posted by dave13 on 03 Feb 2018, 10:41.
Last edited by dave13 on 03 Feb 2018, 11:14, edited 1 time in total.
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Re: If m is an integer greater than 9 but less than 20, is n greater than [#permalink]

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New post 03 Feb 2018, 10:49
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dave13 wrote:
jedit wrote:
Bunuel wrote:
If m is an integer greater than 9 but less than 20, is n greater than the average (arithmetic mean) of m and 20?

(1) n = 3m
(2) The distance on the number line between n and 20 is less than the distance on the number line between n and m.


It should be D

From the stem, we know the max and min for \(m\)

\(9 < m < 20\) - so \(m\) can range between \(10\) and \(19\). Arithmetic mean of m and 20 can be \(\frac{10+20}{2} = 15\) or \(\frac{19+20}{2} = 19.5\) at the extremes. Question asks if \(n\) is greater than these values.

Statement 1: Sufficient
\(n = 3m\) so minimum value of \(n = 45\) and maximum is \(57\), which makes it greater than the AM of \(m\) and \(20\) in all cases.

Statement 2: Sufficient
Since \(m\) is to the left of \(20\), and \(n\) is closer to \(20\) than it is to \(m\), this tells us \(n\) is always to the right of \(m\). The diagram attached shows both possible cases and arrows represent Arithmetic mean of \(m\) and \(20\) and in both cases, \(n\) shows up as greater than the Arithmetic mean.



Hi jedit

how did you define that \(m\) can range between \(10\) and \(19\). ?

thanks! :-)



Hey dave13

The answer to your question has been highlighted. It is clearly mentioned in the
question stem that m is an integer, greater than 9 and less than 20.

Hope that answers your question!
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If m is an integer greater than 9 but less than 20, is n greater than [#permalink]

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New post 03 Feb 2018, 11:20
Hi pushpitkc thanks for your answer :)



One question when do I need to break this \(9 < m < 20\) into two inequalities

\(9 < m\) --> \(1+9<m+1\) --> \(10<m+1\) like this ? :?

\(m < 20\) --> \(1+m<20+1\) --> \(1+m<21\) like this ? :?


i thought i was supposed to do something like this ....or even perhaps subtract +1 from both sides of inequility :? :-)
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Re: If m is an integer greater than 9 but less than 20, is n greater than [#permalink]

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dave13 wrote:
Hi pushpitkc thanks for your answer :)



One question when do I need to break this \(9 < m < 20\) into two inequalities

\(9 < m\) --> \(1+9 \(10

\(m < 20\) --> \(1+m<20+1\) --> \(1+m<21\) like this ? :?


i thought i was supposed to do something like this ....or even perhaps subtract +1 from both sides of inequility :? :-)


Hi dave13

Recall our earlier discussion on inequality and let's try to use the properties in this question. we are given

9
average of \(m\) & \(20\) will be a number that is mid-point of \(m\) & \(20\). Mathematically it will be \(\frac{m+20}{2}\). We need to find whether \(n>\frac{m+20}{2}\)

To know the range of the average add \(20\) to both sides of the inequality (1)

9+20
\(\frac{29}{2}<\frac{m+20}{2}<\frac{40}{2} =>14.5<\frac{m+20}{2}<20\) (dividing both sides of the inequality by a positive number will not change the sign)

So from the question stem we got the range of average of \(m\) & \(20\), which is between \(14.5\) & \(20\)

Statement 1: \(n=3m\). Now multiply inequality (1) by \(3\), we have

\(9*3<3*m<20*3=>27<3m<60=>27
so \(n\) is between \(27\) & \(60\) whereas average is between \(14.5\) & \(20\), clearly \(n>\frac{m+20}{2}\). Sufficient

Statement 2: Given \(n\) is closer to \(20\) than \(m\). we know the average of \(m\) & \(20\) will be mid point i.e equidistant from both \(m\) & \(20\) and as \(n\) is closer to \(20\) so \(n>average\). Sufficient

Option D
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Re: If m is an integer greater than 9 but less than 20, is n greater than [#permalink]

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New post 04 Feb 2018, 06:47
niks18 wrote:
dave13 wrote:
Hi pushpitkc thanks for your answer :)



One question when do I need to break this \(9 [m]1+9 [m]10 [m]1+m [m]1+m\frac{m+20}{2}\)

To know the range of the average add \(20\) to both sides of the inequality (1)

9+2014.52727\frac{m+20}{2}[/m]. Sufficient

Statement 2: Given \(n\) is closer to \(20\) than \(m\). we know the average of \(m\) & \(20\) will be mid point i.e equidistant from both \(m\) & \(20\) and as \(n\) is closer to \(20\) so \(n>average\). Sufficient

Option D




Hello niks18 :)

Thanks a lot for taking time to explain :) i just have one question :-)

Why do you add +20 to both sides ?

Just wanna give you example so as you understand which moment i dont understand

ok i googled and found this. please have a look at the question with explanation below (and see my highlighted comment )
----------------------------------------------------------------------------------------------------------------------------
Technique: Boundary Testing

If 2 < x - 6 < 10 and 25 < y + 10 < 45, what inequality represents the range of values of x + y?

1.) Solve each inequality separately.
2 < x - 6 < 10
2 + 6 < x - 6 + 6 < 10 + 6 (see -6 turns positive when adding 6 to both sides, whereas when you add 20 to both sides +20 is still positive and not negative -20, why :? Moreover 6 in the middle is canceled out :? can you explain the difference between your tecinique and this one
8 < x < 16

25 < y + 10 < 45
25 - 10 < y + 10 - 10 < 45 - 10
15 < y < 35

2.) Combine each inequality by using the boundary of each inequality to find the end of the combined (i.e., summed, x + y) inequality.

2a.) Find the smallest possible value of the inequality.
In the first inequality: x is 8.000...0001
In the second inequality: y is 15
23 < x + y

2b.) Find the largest possible value of the inequality.
In the first inequality: x is 16
In the second inequality: y is 34.9999...
x + y < 51
3.) Combine each value from step 2 to find the inequality that encapsulates x + y.

3a.) Find the smallest possible value of the combined inequality.
8.000...0001 + 15 produces x + y > 23

3b.) Find the largest possible value of the combined inequality.
16 + 34.9999 produces y < 51

Putting these together: 23 < x + y < 51
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If m is an integer greater than 9 but less than 20, is n greater than [#permalink]

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New post 04 Feb 2018, 07:02
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dave13 wrote:
niks18 wrote:
dave13 wrote:
Hi pushpitkc thanks for your answer :)



One question when do I need to break this \(9 [m]1+9 [m]10 [m]1+m [m]1+m\frac{m+20}{2}\)

To know the range of the average add \(20\) to both sides of the inequality (1)

9+2014.52727\frac{m+20}{2}[/m]. Sufficient

Statement 2: Given \(n\) is closer to \(20\) than \(m\). we know the average of \(m\) & \(20\) will be mid point i.e equidistant from both \(m\) & \(20\) and as \(n\) is closer to \(20\) so \(n>average\). Sufficient

Option D




Hello niks18 :)

Thanks a lot for taking time to explain :) i just have one question :-)

Why do you add +20 to both sides ?

Just wanna give you example so as you understand which moment i dont understand

ok i googled and found this. please have a look at the question with explanation below (and see my highlighted comment )
----------------------------------------------------------------------------------------------------------------------------
Technique: Boundary Testing

If 2 23

3b.) Find the largest possible value of the combined inequality.
16 + 34.9999 produces y < 51

Putting these together: 23 < x + y < 51


Hi dave13,

at first there seems to be some formatting issue in my post which somehow I am unable to rectify. Hope you can understand that.

Now coming to your question -
2<x-6<10, now you are adding 6 on both sides of the inequality. why we add 6? because I need range of x only and if I add 6 to x-6, then 6-6=0, leaving only x

=> 2+6<(x-6)+6<10+6

=>8<x<16-----------(1)

now where is -6 turning positive here? this is a simple addition. stick to the basic and do not complicate the question simply because it is an inequality.

now 25<y+10<45, again I need range of y only so to remove 10 i will now subtract 10 from both sides of the inequality

25-10<y+10-10<25-10

15<y<35-------(2)

now as I need range of x+y I will simply add inequality (1) & (2). Note Here I can directly add inequalities because ALL SIGNS ARE SAME

so on adding I will get 8+15<x+y<16+35

=> 23<x+y<51
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If m is an integer greater than 9 but less than 20, is n greater than [#permalink]

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New post 04 Feb 2018, 07:08
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Hi dave13

I am unable to understand why there are so many formatting issues in the forum occurring today, but I hope you got the solution to your queries
If m is an integer greater than 9 but less than 20, is n greater than   [#permalink] 04 Feb 2018, 07:08
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If m is an integer greater than 9 but less than 20, is n greater than

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