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If b, c, and n are positive numbers such that 30(2b/(c+2b))

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If b, c, and n are positive numbers such that 30(2b/(c+2b)) [#permalink] New post 30 Nov 2010, 20:22
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If b, c, and n are positive numbers such that 30(2b/(c+2b)) + 15(c/(c+2b)) = n, and c < b, which of the following could be the value of n?

A. 15
B. 25
C. 27
D. 35
E. 45
[Reveal] Spoiler: OA

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Last edited by Bunuel on 13 Dec 2012, 03:25, edited 3 times in total.
Renamed the topic and edited the question.
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Re: Tough Fraction Question - Find possible value for n [#permalink] New post 30 Nov 2010, 20:37
martie11 wrote:
Source: Knewton

Solution to follow. Good luck.


I'm thinking E.

45+60b/c+15C/2b=N

C<B, and we know that all numbers are positive, so you can set C=0 and N will be 45.
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Re: Tough Fraction Question - Find possible value for n [#permalink] New post 30 Nov 2010, 20:48
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martie11 wrote:
Source: Knewton

Solution to follow. Good luck.


I'm thinking E.

45+60b/c+15C/2b=N

C<B, and we know that all numbers are positive, so you can set C=0 and N will be 45.

Not quite.

I have updated the Q with OA. Solution to follow.
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Re: Tough Fraction/Algebra Question - Find possible value for n [#permalink] New post 30 Nov 2010, 22:34
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just cross multiply, we get
(60-2n)b = (n-15)c
now, as given, c<b ==> b/c > 1
==> b/c = (n-15) / (60-2n) > 1

case 1: 60-2n < 0 (negative) ==> 60<2n ==> n>30
==> from (n-15) / (60-2n) > 1 we can write (n-15) < (60-2n) as 60-2n is negative the sign changes
==> 3n<75 ==> n<25 and n>30 as derived above and this is not posible hence 60-2n is not negative

case2: 60-2n>0 positive ==> 60>2n ==> n< 30
==> from (n-15) / (60-2n) > 1 we can write (n-15) > (60-2n) as 60-2n is positive the sign DOES NOT change
==> 3n>75 ==> n>25 and n<30 as derived above
and this says n is between 25 and 30

25<n<30

the only possible answer in this range is 27.

answer "C"

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Re: Tough Fraction/Algebra Question - Find possible value for n [#permalink] New post 30 Nov 2010, 22:47
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martie11 wrote:
Source: Knewton

Solution to follow. Good luck.


You are basically given that:
15 (\frac{4b+c}{2b+c}) = 15 (1 + \frac{2b}{2b+c}) = n and c < b

Thus, the largest value of the denominator should be lesser than the case where c= b. Then \frac{2b}{2b+c} > \frac{2b}{3b} which is \frac{2}{3}

Thus, n > 15 (1 + \frac{2}{3}) which means n > 15 (\frac{5}{3} and hence n > 25.

The largest value to this occurs when c = 0, and if you substitute that in to the original equation, we get \frac{2b}{2b+c} < \frac{2b}{2b} which is 1. Thus n < (15)(2) which means n < 30.

The only value of n which falls in this range is 27 and hence the answer is C.
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Re: Tough Fraction/Algebra Question - Find possible value for n [#permalink] New post 30 Nov 2010, 23:05
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30(2b)/(c+2b) + 15(c)(c+2b) = n
15*((c+2b)/(c+2b)) + 60b/(c+2b)=n
n = 15 + 30b/(c+2b)

if c is much much less than b, the fractional part approaches 15
if c is almost equal to b, the fractional part approaches 10

These are the upper and lower bounds on the fractional part

15+10 < n < 15+15

Hence, n=27 is a possible solution
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Re: Tough Fraction/Algebra Question - Find possible value for n [#permalink] New post 30 Nov 2010, 23:35
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The equation can be written as:
(15/c+2b)(4b+c) = n
(15/c+2b)(c+2b +2b) = n
which gives us:
15 + 30b/(c+2b) = n

now we know that n is positive integer, so, 30b/(c+2b) has to be positive integer.

now, lets deal with 30*b/(c+2b), we know one more condition that c<b

lets take c=1, b=2; 30*2/5 = 12
lets take c=2,b=4; same result i.e 12

So, n=15+12=27

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Re: Tough Fraction/Algebra Question - Find possible value for n [#permalink] New post 01 Dec 2010, 19:15
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martie11 wrote:
Source: Knewton

Solution to follow. Good luck.


Exactly like question no 148 of OG12. Just values are different.

This question can be easily solved using weighted averages, if you are comfortable with the concept.

Note here that the question is
\frac{{15 *c + 30*2b }}{{c + 2b}} = n

The formula of weighted averages looks like this: \frac{{ax + by}}{{x + y}} =Weighted Average

So n will be more than 15 but less than 30 and will depend on the relation between c and b.
Since c < b, then c:2b is less than 1:2
Then weighted average (n) will lie between 25 and 30.
Only possible answer is 27.
Note: You cannot find the exact value of n here. Just the range in which it lies.
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Re: Tough Fraction/Algebra Question - Find possible value for n [#permalink] New post 02 Dec 2010, 22:25
Thank you for all your feedback.
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Re: Tough Fraction/Algebra Question - Find possible value for n [#permalink] New post 13 Dec 2012, 03:14
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The values in the answer choices are quite close so it's certain that if I plug in,
the resulting value will be very near one of the options. It's a rigid range.

Let c=1 and b=2

n=30(4/5) + 15(1/5)=27 I got lucky!

Try something else like c=2 and b=10

n=315/11 = 28 Like I said very rigid range.

Answer: 27
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Re: If b, c, and n are positive numbers such that 30(2b/(c+2b)) [#permalink] New post 13 Dec 2012, 03:25
Expert's post
martie11 wrote:
If b, c, and n are positive numbers such that 30(2b/(c+2b)) + 15(c/(c+2b)) = n, and c < b, which of the following could be the value of n?

A. 15
B. 25
C. 27
D. 35
E. 45


Very similar question to practice: if-x-y-and-k-are-positive-numbers-such-that-x-x-y-128231.html
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Re: If b, c, and n are positive numbers such that 30(2b/(c+2b)) [#permalink] New post 15 Dec 2012, 00:17
Bunuel wrote:
martie11 wrote:
If b, c, and n are positive numbers such that 30(2b/(c+2b)) + 15(c/(c+2b)) = n, and c < b, which of the following could be the value of n?

A. 15
B. 25
C. 27
D. 35
E. 45


Very similar question to practice: if-x-y-and-k-are-positive-numbers-such-that-x-x-y-128231.html


Thanks for pointing out that they are the same question. It helped me now really understand it. :)
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Re: Tough Fraction/Algebra Question - Find possible value for n [#permalink] New post 15 Dec 2012, 20:49
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martie11 wrote:
Source: Knewton

Solution to follow. Good luck.


Exactly like question no 148 of OG12. Just values are different.

This question can be easily solved using weighted averages, if you are comfortable with the concept.

Note here that the question is
\frac{{15 *c + 30*2b }}{{c + 2b}} = n

The formula of weighted averages looks like this: \frac{{ax + by}}{{x + y}} =Weighted Average

So n will be more than 15 but less than 30 and will depend on the relation between c and b.
Since c < b, then c:2b is less than 1:2
Then weighted average (n) will lie between 25 and 30.
Only possible answer is 27.
Note: You cannot find the exact value of n here. Just the range in which it lies.

Hey Karishma... I have a question.... I saw the explanation on your blog but still it did not clear... when taking weighted averages.... when a<b, the weighted average will be to the right of midpoint of the weights right..... in this case... we are taking weighted averages of c, 2b..... since c<b c is also less than 2b..... so the weighted average of c, 2b is closer to 2b.... weights are 15,30 and midpoint is 22.5... so it should be in the range of 22.5 to 30... then how did you conclude n to be between 25 and 30 instead of 22.5?
I agree with your answer and was able to work it out conventional way but looking for some time saving tricks.... Thanks
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Re: If b, c, and n are positive numbers such that 30(2b/(c+2b)) [#permalink] New post 16 Dec 2012, 03:02
Ans:
solving the equation we get n= 15(c+4b)/(c+2b) the max value will occur when c =0 which is 30 and the min value will occur when c=b which is >25 , therefore the answer is (C).
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Re: Tough Fraction/Algebra Question - Find possible value for n [#permalink] New post 16 Dec 2012, 03:17
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Amateur wrote:
Hey Karishma... I have a question.... I saw the explanation on your blog but still it did not clear... when taking weighted averages.... when a<b, the weighted average will be to the right of midpoint of the weights right..... in this case... we are taking weighted averages of c, 2b..... since c<b c is also less than 2b..... so the weighted average of c, 2b is closer to 2b.... weights are 15,30 and midpoint is 22.5... so it should be in the range of 22.5 to 30... then how did you conclude n to be between 25 and 30 instead of 22.5?
I agree with your answer and was able to work it out conventional way but looking for some time saving tricks.... Thanks



Average of two numbers lies at their mid point. Weighted average of two numbers lies closer to the number which has higher weight.
The average of 15 and 30 will be 22.5
The weighted average of 15 and 30 where w1 and w2 are the weights of 15 and 30 respectively will depend on the values of w1 and w2.

w1/w2 = c/2b
c/b < 1 (because c < b) so c/2b < 1/2

If c/2b were equal to 1/2, w1/w2 = 1/2 so the average would be at 25. (Plug it in the formula if you are not sure how we arrived at 25).
But since c/2b is actually less than 1/2, the weighted average will fall between 25 and 30.
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Re: Tough Fraction/Algebra Question - Find possible value for n [#permalink] New post 16 Dec 2012, 20:35
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Click here for detailed explanation on weighted averages: Weighted Averages on GMAT
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Re: If b, c, and n are positive numbers such that 30(2b/(c+2b)) [#permalink] New post 16 Dec 2012, 22:50
If b, c, and n are positive numbers such that 30(2b/(c+2b)) + 15(c/(c+2b)) = n, and c < b, which of the following could be the value of n?

A. 15
B. 25
C. 27
D. 35
E. 45

Now, 60b/(c+2b) + 15c/(c+2b) = n
=> 60b+15c/(c+2b)=n
=> 60b+15c=nc+2nb
=> b(60-2n)=c(n-15)

Option 1:
Put n = 15
b(30)=c(0)
Does not satisfy the equation.

Option 2:
Put n = 25
b(10)=c(10)
Since c<b, this is also not the answer

Option 3:
Put n = 27
b(8)=c(12)
Since c<b, this could be the answer.
Like, b=3 and c=2

Option 4:
Put n = 35
b(-ve)=c(20)
b,c and d cannot be -ve. Hence, not the answer.

Option 5:
Put n = 45
b(-ve) = c(20)
b,c and d cannot be -ve. Hence, not the answer.

Therefore, only possible soultion is Option c.
Re: If b, c, and n are positive numbers such that 30(2b/(c+2b))   [#permalink] 16 Dec 2012, 22:50
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