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

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Intern
Joined: 23 Oct 2015
Posts: 1
If 0<a<b and k= (2a+7b)/b , which of the following must be true?  [#permalink]

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Updated on: 23 Oct 2015, 03:27
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15% (low)

Question Stats:

80% (01:43) correct 20% (01:51) wrong based on 340 sessions

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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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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

Cheers,
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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23 Oct 2015, 03:25
2
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

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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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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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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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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14 Jun 2017, 01:32
1
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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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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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.

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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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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$$<= 2$$

$$k < 2 + 7$$

$$k < 9$$

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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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03 Jul 2018, 22:05
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Re: If 0<a<b and k= (2a+7b)/b , which of the following must be true? &nbs [#permalink] 03 Jul 2018, 22:05
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