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If 0<a<b and k= (2a+7b)/b , which of the following must be true?

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If 0<a<b and k= (2a+7b)/b , which of the following must be true?  [#permalink]

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New post Updated on: 23 Oct 2015, 03:27
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If 0 < a < b and k = (2a + 7b)/b , which of the following must be true?

A. k < 2
B. k < 7
C. k < 9
D. k > 9
E. k > 11

Originally posted by Michelle7194 on 23 Oct 2015, 00:38.
Last edited by Bunuel on 23 Oct 2015, 03:27, edited 2 times in total.
Renamed the topic and edited the question.
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Re: If 0<a<b and k= (2a+7b)/b , which of the following must be true?  [#permalink]

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New post 23 Oct 2015, 07:05
9
1
Michelle7194 wrote:
If 0 < a < b and k = (2a + 7b)/b , which of the following must be true?

A. k < 2
B. k < 7
C. k < 9
D. k > 9
E. k > 11


Here's another approach:

k = (2a + 7b)/b
= 2a/b + 7b/b
= 2(a/b) + 7
Since 0 < a < b, we know that a/b is less than 1, which means that 2(a/b) is some number less than 2.
So, we get k = (some number less than 2) + 7
From here, we can see that k must be less than 9
Answer: C

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Brent
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Re: If 0<a<b and k= (2a+7b)/b , which of the following must be true?  [#permalink]

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New post 23 Oct 2015, 03:25
2
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Michelle7194 wrote:
if 0<a<b and k= (2a+7b)/b , which of the following must be true?

a. k<2
b. k<7
c. k<9
d. k>9
e. k>11

I would appreciate some explanation with the answer. Thank you :)


Please follow posting guidelines (link in my signature) especially writing the correct topic title. Also do mention the source of the question if you select the tag: "Source-Other Please Specify".

As for your question, it is very simple and straightforward with number plug in.

Assume a=1 and b=2 as a<b, you get k = (2+14)/2 = 8. This eliminates all but option C , making it the correct answer.
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Re: If 0<a<b and k= (2a+7b)/b , which of the following must be true?  [#permalink]

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New post 29 Dec 2016, 09:17
Assume , a = 1 , b = 2
\(K = \frac{2+14}{2}\)
\(\frac{16}{2}\)
K = 8
K<9
C
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  [#permalink]

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New post 14 Jun 2017, 00:25
Who said that 'a' and 'b' are full integers ? So we have to assume that always ?

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If 0<a<b and k= (2a+7b)/b , which of the following must be true?  [#permalink]

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New post 14 Jun 2017, 00:47
1
k=\(\frac{2a+7b}{b}\)
k =2(\(\frac{a}{b})\)+7

given a>0;b>0 --> \(\frac{a}{b}\)= proper fraction
\(\frac{a}{b}\)<1
k<2(1)+7
k<9
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Re: If 0<a<b and k= (2a+7b)/b , which of the following must be true?  [#permalink]

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New post 14 Jun 2017, 01:32
1
DadhichD wrote:
Who said that 'a' and 'b' are full integers ? So we have to assume that always ?

Posted from my mobile device


No, you should not assume that a variable is necessarily an integer if you are not explicitly told so. Having said that, solutions above do not assume that a and b are integers. For example check solution here: https://gmatclub.com/forum/if-0-a-b-and ... l#p1591061
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If 0<a<b and k= (2a+7b)/b , which of the following must be true?  [#permalink]

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New post 17 Jun 2017, 16:54
Michelle7194 wrote:
If 0 < a < b and k = (2a + 7b)/b , which of the following must be true?

A. k < 2
B. k < 7
C. k < 9
D. k > 9
E. k > 11


Let a = 1, b = 2.

\(k = \frac{(2a + 7b)}{b}\)

\(k = \frac{2a}{b} + \frac{7b}{b}\)

\(k = \frac{2*1}{2} + 7\)

\(k = 1+ 7 = 8\)

\(k < 9\). Answer (C)..
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Re: If 0<a<b and k= (2a+7b)/b , which of the following must be true?  [#permalink]

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New post 20 Jun 2017, 06:34
Michelle7194 wrote:
If 0 < a < b and k = (2a + 7b)/b , which of the following must be true?

A. k < 2
B. k < 7
C. k < 9
D. k > 9
E. k > 11


We can simplify the given equation:

k = (2a + 7b)/b

k = 2a/b + 7b/b

k = 2a/b + 7

Since 0 < a < b, a/b must be less than 1, so 2a/b must be less than 2; thus, k < 2 + 7 or k < 9.

Answer: C
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Re: If 0<a<b and k= (2a+7b)/b , which of the following must be true?  [#permalink]

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New post 24 Jun 2017, 14:24
\(k = \frac{(2a + 7b)}{b}\)

\(k = 2\frac{a}{b} + 7\)

As we know that \(0 < a < b\) we know that value if \(2\frac{a}{b}\) should be\(<= 2\)

\(k < 2 + 7\)

\(k < 9\)

Hence, Answer is C
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Re: If 0<a<b and k= (2a+7b)/b , which of the following must be true?  [#permalink]

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