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If 5 < x < 10 and y = x + 5, what is the greatest possible

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If 5 < x < 10 and y = x + 5, what is the greatest possible  [#permalink]

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New post Updated on: 13 Jun 2012, 03:04
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If 5 < x < 10 and y = x + 5, what is the greatest possible integer value of x + y ?

A. 18
B. 20
C. 23
D. 24
E. 25

Originally posted by shivanigs on 12 Jun 2012, 22:05.
Last edited by Bunuel on 13 Jun 2012, 03:04, edited 1 time in total.
Edited the question, added the answer choices and OA
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Re: Inequalities Question - Request help  [#permalink]

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New post 13 Jun 2012, 03:03
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shivanigs wrote:
Hi,

Request your help to understand the concept behind the following question :

If 5 < x < 10 and y = x + 5,what is the greatest possible integer value of x + y?


Please read and follow: 11-rules-for-posting-133935.html Pay attention to the points #3, 7 and 8.

Original question is below:

If 5 < x < 10 and y = x + 5, what is the greatest possible integer value of x + y ?

A. 18
B. 20
C. 23
D. 24
E. 25

Since \(y=x+5\) then \(x+y=x+(x+5)=2x+5\). So, we need to find the greatest possible integer value of \(2x+5\).

Multiply \(5 < x < 10\) by 2: \(10<2x<20\). Now add 5 to each part of the inequality: \(15<2x+5<25\). As you can see the greatest possible integer value of \(2x+5\) is 24.

Answer: D.

Hope it's clear.
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Re: Inequalities Question - Request help  [#permalink]

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New post 12 Jun 2012, 23:14
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shivanigs wrote:
Hi,

Request your help to understand the concept behind the following question :

If 5 < x < 10 and y = x + 5,what is the greatest possible integer value of x + y?

Hi,

To find the greastest possible integer value of x + y, it is not necessary that both x & y should be integers,
x, y should be chosen in such a way that their sum is an integral value.

so, to find, max value of (x + y)
x + y = 2x + 5
maximum value of x such that 2x is integer would be 9.5 (given, 5 < x < 10)
x + y (maximum) = 19 + 5 =24

Let me know if you need any more assistance on this topic.

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Re: Inequalities Question - Request help  [#permalink]

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New post 13 Jun 2012, 04:52
Bunuel wrote:
shivanigs wrote:
Hi,

Request your help to understand the concept behind the following question :

If 5 < x < 10 and y = x + 5,what is the greatest possible integer value of x + y?


Please read and follow: 11-rules-for-posting-133935.html Pay attention to the points #3, 7 and 8.

Original question is below:

If 5 < x < 10 and y = x + 5, what is the greatest possible integer value of x + y ?

A. 18
B. 20
C. 23
D. 24
E. 25

Since \(y=x+5\) then \(x+y=x+(x+5)=2x+5\). So, we need to find the greatest possible integer value of \(2x+5\).

Multiply \(5 < x < 10\) by 2: \(10<2x<20\). Now add 5 to each part of the inequality: \(15<2x+5<25\). As you can see the greatest possible integer value of \(2x+5\) is 24.

Answer: D.

Hope it's clear.


Dear Bunuel,
1.)can't understand what is the need to take 2x+5? wont it be easy to calculate with x less than 10.?
2.) when x is not integer but x+y to be integer - we can take x=9.5 and y = 14.5 - giving 24 - is this right?
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Re: Inequalities Question - Request help  [#permalink]

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New post 13 Jun 2012, 04:59
kashishh wrote:
Bunuel wrote:
shivanigs wrote:
Hi,

Request your help to understand the concept behind the following question :

If 5 < x < 10 and y = x + 5,what is the greatest possible integer value of x + y?


Please read and follow: 11-rules-for-posting-133935.html Pay attention to the points #3, 7 and 8.

Original question is below:

If 5 < x < 10 and y = x + 5, what is the greatest possible integer value of x + y ?

A. 18
B. 20
C. 23
D. 24
E. 25

Since \(y=x+5\) then \(x+y=x+(x+5)=2x+5\). So, we need to find the greatest possible integer value of \(2x+5\).

Multiply \(5 < x < 10\) by 2: \(10<2x<20\). Now add 5 to each part of the inequality: \(15<2x+5<25\). As you can see the greatest possible integer value of \(2x+5\) is 24.

Answer: D.

Hope it's clear.


Dear Bunuel,
1.)can't understand what is the need to take 2x+5? wont it be easy to calculate with x less than 10.?
2.) when x is not integer but x+y to be integer - we can take x=9.5 and y = 14.5 - giving 24 - is this right?


What do you mean by "need"? One can solve a question with different approaches and you can choose the approach you personally find easier.
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Re: If 5 < x < 10 and y = x + 5, what is the greatest possible  [#permalink]

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New post 13 Jun 2012, 05:09
Dear Bunuel,

Apologies for the wrongful post..did not know the rules.Have gone thru the same,will not happen again.Thanks for your help!

Regards
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Re: If 5 < x < 10 and y = x + 5, what is the greatest possible  [#permalink]

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New post 13 Jun 2012, 09:29
Dear Bunuel,
sorry for the word 'need'. i just only wanted to know whether the way i did is right.
thanx for the reply.
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Re: If 5 < x < 10 and y = x + 5, what is the greatest possible  [#permalink]

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New post 13 Jun 2012, 09:43
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Re: If 5 < x < 10 and y = x + 5, what is the greatest possible  [#permalink]

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New post 14 Oct 2013, 10:36
The question asks what is the greatest integer value of 2X+5, where 5<X<10, the greatest value of 2X which is integer will be achieved if X=9.5

Hence the greatest integer value of X+Y = 2x9.5+5=24, answer - D
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Re: If 5 < x < 10 and y = x + 5, what is the greatest possible  [#permalink]

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New post 03 Aug 2017, 23:59
I am getting it 23.

x = { 6,7,8,9} as 5<x<10

y=x+5 . Let's plug in values

x=6 y=11
x=7 y=12
x=8 y=13
x=9 y=14

Now as we have to find maximum value I chose x=9 and y=14

So x+y=9+14=23

Can someone let me know where I am making mistake here
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Re: If 5 < x < 10 and y = x + 5, what is the greatest possible  [#permalink]

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New post 05 Aug 2017, 03:37
Tango3 wrote:
I am getting it 23.

x = { 6,7,8,9} as 5<x<10

y=x+5 . Let's plug in values

x=6 y=11
x=7 y=12
x=8 y=13
x=9 y=14

Now as we have to find maximum value I chose x=9 and y=14

So x+y=9+14=23

Can someone let me know where I am making mistake here


Hi Tango3, the question here asks for the max integer value of x+y, but nowhere it mentions x and y are integers.

x+y = x+(x+5) = 2x + 5. So, to make the sum an integer, x should assume the value 9.5

So the max sum is 24.
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Re: If 5 < x < 10 and y = x + 5, what is the greatest possible   [#permalink] 05 Aug 2017, 03:37
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