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If x, y, and k are positive numbers such that (x/(x+y))(10)

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Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]

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New post 07 May 2014, 02:17
kanusha wrote:
Hi Bunnel ,
Can you pls Explain how these Steps Came in your Solution

10*\frac{x}{x+y}+20*\frac{y}{x+y}=k

10*\frac{x+2y}{x+y}=k

10*(\frac{x+y}{x+y}+\frac{y}{x+y})=k

Finally we get: 10*(1+\frac{y}{x+y})=k

I Didn't Understood how The Steps 3 and 4 Came? :(


\(\frac{x+2y}{x+y}=\frac{x+y}{x+y}+\frac{y}{x+y}=1+\frac{y}{x+y}\).

Hope it's clear.
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Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]

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New post 31 Jan 2015, 17:27
Hi,

I found plugging in the answer choices to be easier. It could take a few seconds more based on the speed but at least it made the problem tractable. A and C were easily eliminated. The later choice between B, D and E was also easy.

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Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]

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New post 31 Jan 2015, 18:34
2
Hi All,

You can take advantage of the answer choices and some math "logic" to get to the correct answer. Let's TEST THE ANSWERS....

We're told that X, Y and K are POSITIVE and that X < Y.

We're also told that 10X/(X+Y) + 20Y/(X+Y) = K. We're asked what COULD be the value of K, which means that there's more than one possible answer. Since the answers are NUMBERS, one of the them MUST be a possible answer.

We can manipulate the given equation into:

10X + 20Y = K(X+Y)

Since all of the variables are POSITIVE, K MUST be between 10 and 20. Here's the proof:

IF.....K=10, then the equation becomes...
10X + 20Y = 10X + 10Y
20Y = 10Y
Since Y is positive, 20Y = 10Y is NOT possible.
Eliminate Answer A

In that same way, K can't be 20 (or greater) because the end equation would be an impossibility.

With the remaining 3 answers, we can TEST the possibilities...

IF...K = 12, then the equation becomes...
10X + 20Y = 12X + 12Y
8Y = 2X
4Y = X
In this scenario, X > Y which is the OPPOSITE of what we were told. This is NOT the answer.
Eliminate B.

IF....K=15, then the equation becomes...
10X + 20Y = 15X + 15Y
5Y = 5X
Y = X
In this scenario, X = Y, which is NOT a match for what we were told. This is NOT the answer.
Eliminate C.

IF....K=18, then the equation becomes....
10X + 20Y = 18X + 18Y
2Y = 8X
Y = 4X
Here, X < Y, which IS a match for what we were told. This IS a POSSIBLE answer.

Final Answer:


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Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]

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New post 12 Sep 2015, 21:32
Bunuel wrote:
BANON wrote:
If x, y, and k are positive numbers such that x/(x+y)*10 + y/(x+y)*20 = k and if x < y, which of the following could be the value of k?
A. 10
B. 12
C. 15
D. 18
E. 30


\(10*\frac{x}{x+y}+20*\frac{y}{x+y}=k\)

\(10*\frac{x+2y}{x+y}=k\)

\(10*(\frac{x+y}{x+y}+\frac{y}{x+y})=k\)

Finally we get: \(10*(1+\frac{y}{x+y})=k\)


We know that \(x<y\)

Hence \(\frac{y}{x+y}\) is more than \(0.5\) and less than \(1\)

\(0.5<\frac{y}{x+y}<1\)

So, \(15<10*(1+\frac{y}{x+y})<20\)


Only answer between \(15\) and \(20\) is \(18\).

Answer: D (18)

There can be another approach:

We have: \(\frac{10x+20y}{x+y}=k\), if you look at this equation you'll notice that it's a weighted average.

There are \(x\) red boxes and \(y\) blue boxes. Red box weight is 10kg and blue box weight 20kg, what is the average weight of x red boxes and y blue boxes?

k represents the weighted average. As y>x, then the weighted average k, must be closer to 20 than to 10. 18 is the only choice satisfying this condition.

Answer: D (18)

what will happen when x>y in this question? since we dont know the value of x and y they can be 1:100 or something like that. what will be the ans then?
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Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]

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New post 13 Sep 2015, 12:14
Hi anik19890,

You ask a couple of different questions.

First, the prompt tells us that X < Y, so we CAN'T have a situation in which X > Y (or even X = Y). THAT specific inequality helps to define the correct answer.

Second, the prompt asks for which of the following COULD be the value of K, meaning that there is more than one possible answer (but only one of the 5 answer choices is correct). Based on the inequality we're given, we COULD have X = 1, Y = 100 (as you described), which would give us...

(1/101)(10) + (100/101)(20) = K

10/101 + 2000/101 = K
2010/101 = K

But that answer is NOT among the answer choices.

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If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]

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New post 25 Jan 2016, 07:30
great question! as all GMAC questions are. should the source be mgmat this would be an insane lengthy calculation.

i actually triaged this the following way: 10, 30 are out because we are given 10 and 20 in the prompt. for the same reason 15 is out (30/2). choosing between B and D and picking D because D gives a more favourable stats for correct answers.

back to why this q is great: you can easily solve this by picking numbers where x<y and quickly see a pattern that k is fluctuating between 15 and 20. u can also go a long way and do algebra and see that (1+ y/(x+y)) never exceeds 2 hence k would never be greater than 20. you can do the weighted average approach! wow. this is the true GMAT type of question that gives you a welath of tools to approach the answer in reasonable amount of time.
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Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]

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New post 25 Jan 2016, 08:25
To Find
Exact value of'K' to be an INTEGER
put (x=1 & y=9)
10*{1+y/(x+y)}=K
=> k=10(1+9/10)
=> k=19
SINCE K=19 IS NOT IN OPTION NEAREST ANSWER IS K=18 ; (15<K<20)
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Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]

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New post 25 Jan 2016, 08:33
vaibhavpanchal wrote:
To Find
Exact value of'K' to be an INTEGER
put (x=1 & y=9)
10*{1+y/(x+y)}=K
=> k=10(1+9/10)
=> k=19
SINCE K=19 IS NOT IN OPTION NEAREST ANSWER IS K=18 ; (15<K<20)


You can quickly see that as x,y,k are all positive integers with x<y and all the options are integers, try with a combination of x+y that will divide both 10 and 20. Try x=1 and y=4 (keep it simple for the calculations).

Once you substitute x=1 and y=4, you get k =18. No need to check what range does 'k' lie in.

Hence D is the correct answer.
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Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]

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New post 28 Jul 2016, 11:08
Aloha Everyone => This is such an awesome question .
Here is my approach
Simplifying the equation => 10/x+y[x+2y] =k
Now plugging in the values
Option A => K=10 => x+y=x+2y => y=0 => Rejected as x,y,k are all greater than 0.
Option B => K=12 => 5x+10y=6x+6y => x=4y => Impossible as x<y
Option C=> K=15 => 3x+3y=2x+4y => x=y => Rejected as it is given y>x
Option D => K=18 => 5x+10y =9x+9y => 4x=y => May be possible....
Option E => K =30 => 3x+3y = x+2y => 2x+y = 0 => Impossible as they are all positive numbers.

Hence Only D is possible .

Smash That D
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Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]

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New post 29 Jul 2016, 01:55
value will be always between 15<k<20 try any number it will fall in this range

only D satisfies that.
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Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]

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New post 28 May 2017, 06:39
(10x + 20y)/(x+y) = k  10x + 20y = kx + ky  20y - ky = kx - 10x  y(20 - k) = x(k - 10)  (20 - k)/(k - 10) = x/y  and since 0 < x < y, then 0 < x/y < 1, and it must be that 0 < (20 - k) / (k - 10) < 1. From the answer choices we can be sure k - 10 isn't negative, so we can multiply through this inequality by k-10 to find that 0 < 20 - k < k - 10, or that 15 < k < 20.
You can see this by separating the three-part inequality 0 < 20 - k < k - 10 into two inequalities:  0 < 20 - k, and adding k to both sides, k < 20  20 - k < k - 10, and adding k and 10 to both sides, we find that 30 < 2k, or 15 < k  Putting those two inequalities together we have 15 < k < 20. => K=18

Option D
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Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]

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New post 02 Jun 2017, 19:25
My logic behind solving this problem was that I was looking for sum of X and Y to be a multiple of 10 since I wanted to get an integer after the first and second parenthesis. So i picked X=1 and Y=4 and got the correct answer. I just tried to plug in 2 and 3 and it worked as well. hope this approach helps :)
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Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]

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New post 13 Jun 2017, 21:40
I used VeritasPrepKarishma's method. This is the basic formula for weighted average.

Since it is 10x + 20y/x+y then k has to fall between 10 and 20. And since x < y then its closer to 20. Hence 18.
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Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]

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New post 20 Jun 2017, 07:36
BANON wrote:
If x, y, and k are positive numbers such that (x/(x+y))(10) + (y/(x+y))(20) = k and if x < y, which of the following could be the value of k?

A. 10
B. 12
C. 15
D. 18
E. 30


We are given:

(x/(x+y))(10) + (y/(x+y))(20) = k

[10x/(x+y)] + [20y/(x+y)] = k

We can combine the two fractions on the left side of the equation because they have the same denominator, (x + y).

[(10x + 20y)/(x+y)] = k

We see that we have a weighted average equation in which x items have an average of 10, and another y items have an average of 20 and a weighted average of k. In this case, the value of k must be between 10 and 20. However, since x is less than y, the weighted average (or k) must be closer to 20 than to 10. Thus k must be 18.

Answer: D
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Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]

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New post 20 Jun 2017, 18:19
Bunuel wrote:
BANON wrote:
If x, y, and k are positive numbers such that x/(x+y)*10 + y/(x+y)*20 = k and if x < y, which of the following could be the value of k?
A. 10
B. 12
C. 15
D. 18
E. 30


\(10*\frac{x}{x+y}+20*\frac{y}{x+y}=k\)

\(10*\frac{x+2y}{x+y}=k\)

\(10*(\frac{x+y}{x+y}+\frac{y}{x+y})=k\)

Finally we get: \(10*(1+\frac{y}{x+y})=k\)


We know that \(x<y\)

Hence \(\frac{y}{x+y}\) is more than \(0.5\) and less than \(1\)

\(0.5<\frac{y}{x+y}<1\)

So, \(15<10*(1+\frac{y}{x+y})<20\)


Only answer between \(15\) and \(20\) is \(18\).

Answer: D (18)

There can be another approach:

We have: \(\frac{10x+20y}{x+y}=k\), if you look at this equation you'll notice that it's a weighted average.

There are \(x\) red boxes and \(y\) blue boxes. Red box weight is 10kg and blue box weight 20kg, what is the average weight of x red boxes and y blue boxes?

k represents the weighted average. As y>x, then the weighted average k, must be closer to 20 than to 10. 18 is the only choice satisfying this condition.

Answer: D (18)


Hi, Bunuel! I have a silly question. Can you please explain how you arrived at the following?
We know that \(x<y\)

Hence \(\frac{y}{x+y}\) is more than \(0.5\) and less than \(1\)
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Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]

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New post 20 Jun 2017, 22:01
uvee wrote:
Bunuel wrote:
BANON wrote:
If x, y, and k are positive numbers such that x/(x+y)*10 + y/(x+y)*20 = k and if x < y, which of the following could be the value of k?
A. 10
B. 12
C. 15
D. 18
E. 30


\(10*\frac{x}{x+y}+20*\frac{y}{x+y}=k\)

\(10*\frac{x+2y}{x+y}=k\)

\(10*(\frac{x+y}{x+y}+\frac{y}{x+y})=k\)

Finally we get: \(10*(1+\frac{y}{x+y})=k\)


We know that \(x<y\)

Hence \(\frac{y}{x+y}\) is more than \(0.5\) and less than \(1\)

\(0.5<\frac{y}{x+y}<1\)

So, \(15<10*(1+\frac{y}{x+y})<20\)


Only answer between \(15\) and \(20\) is \(18\).

Answer: D (18)

There can be another approach:

We have: \(\frac{10x+20y}{x+y}=k\), if you look at this equation you'll notice that it's a weighted average.

There are \(x\) red boxes and \(y\) blue boxes. Red box weight is 10kg and blue box weight 20kg, what is the average weight of x red boxes and y blue boxes?

k represents the weighted average. As y>x, then the weighted average k, must be closer to 20 than to 10. 18 is the only choice satisfying this condition.

Answer: D (18)


Hi, Bunuel! I have a silly question. Can you please explain how you arrived at the following?
We know that \(x<y\)

Hence \(\frac{y}{x+y}\) is more than \(0.5\) and less than \(1\)


\(\frac{y}{x+y}\)

Since both x and y are positive, then the denominator is greater than numerator. So, \(\frac{y}{x+y}<1\).

Next, if x were equal to y, then we'd have y/(2y) = 1/2 but x < y, so denominator is less than 2y, which makes the fraction greater than 1/2.

Hope it's clear.
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Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]

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New post 21 Jun 2017, 07:54
Thank you so much Bunuel!!
That seemed lot easier now
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Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]

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New post 08 Jul 2017, 05:58
BANON wrote:
If x, y, and k are positive numbers such that (x/(x+y))(10) + (y/(x+y))(20) = k and if x < y, which of the following could be the value of k?

A. 10
B. 12
C. 15
D. 18
E. 30


Here's my alternate approach,

10x+20y=kx+ky

Start plugging in values from the given options:
1. 10x+20y=10x+10y =>2y=y; absurd WRONG
2. 10x+20y=12x+12y => 8y=2x; since x<y this is false; WRONG
3. 10x+20y=15x+15y => x=y; since x<y this is false; WRONG
4. 10x+20y=18x+18y => 2y=8x; looks good; lets keep it till we check the next option
5. 10x+20y=30x+30y =>-10y=10x; this means one number out of x & y is -ve; this condition is wrong as per our question since both x and y are +ve

Thus "D" is the clear answer. We dont require any knowledge of weighted average nor any complex maths is involved.

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Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]

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New post 19 Jan 2018, 08:31
Could you guys please explain if the answer holds true when x<0 and y>0 ? Thanks !
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Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]

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New post 19 Jan 2018, 20:26
Hi Phlaryu,

The prompt tells us that X, Y and K are all POSITIVE, so you shouldn't be wasting time on the outcome if X < 0.

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Re: If x, y, and k are positive numbers such that (x/(x+y))(10)   [#permalink] 19 Jan 2018, 20:26

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