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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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25 Jan 2018, 22:51
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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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25 Jan 2018, 23:01
1
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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01 Feb 2018, 09:31
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

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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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02 Feb 2018, 06:17
2
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 [ 19.14 KiB | Viewed 810 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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03 Feb 2018, 05:22
1
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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Posts: 1101
If m is an integer greater than 9 but less than 20, is n greater than  [#permalink]

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Updated on: 03 Feb 2018, 10:14
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, 09:41.
Last edited by dave13 on 03 Feb 2018, 10:14, 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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03 Feb 2018, 09:49
2
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.

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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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03 Feb 2018, 10:20

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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04 Feb 2018, 00:38
1
dave13 wrote:

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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04 Feb 2018, 05:47
niks18 wrote:
dave13 wrote:

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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04 Feb 2018, 06:02
1
dave13 wrote:
niks18 wrote:
dave13 wrote:

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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04 Feb 2018, 06:08
1
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 &nbs [#permalink] 04 Feb 2018, 06:08
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