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

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

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New post 04 Dec 2018, 00:58
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This one's all about pattern recognition.

Notice that the equation is giving us a weighted average. Because y is bigger than x, the average must be weighted towards y. Visually:

x---------avg---y
10--------avg---20

Thus, the answer must be between 15 and 20, exclusive.

Among the answer choices, only 18 fits the bill.

Choose D.
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Re: If x, y, and k are positive numbers such that 10*x/(x+y)+ 20  [#permalink]

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New post 04 Dec 2018, 00:59
2
pzazz12 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


I suspect that many test-takers would not recognize that this is a weighted average problem. For those test-takers, a straightforward and efficient approach would be to plug in the answer choices, which represent the value of k.

Answer choice C: k= 15
(10x + 20y)/(x+y) = 15
10x + 20y = 15x + 15y
5y = 5x
y = x
Doesn't work because the problem states that x<y.

We need y to be larger, so let's try answer choice D: k=18
10x + 20y = 18x + 18y
2y = 8x
y/x = 8/2
Success! x<y.

The correct answer is D.

The process would be very quick if you recognized right away that since k = one of the answer choices, you could simply plug them in for k.
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Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)  [#permalink]

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New post 27 May 2019, 13:24
Bunuel wrote:
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 \(18\).


Bunuel
Which values confirm the highlighted part?
Also, could you help me to figure out the highlighted part without putting the value?
Thanks__
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If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)  [#permalink]

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New post 27 May 2019, 13:33
GMATGuruNY wrote:
pzazz12 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


I suspect that many test-takers would not recognize that this is a weighted average problem. For those test-takers, a straightforward and efficient approach would be to plug in the answer choices, which represent the value of k.

Answer choice C: k= 15
(10x + 20y)/(x+y) = 15
10x + 20y = 15x + 15y
5y = 5x
y = x
Doesn't work because the problem states that x<y.

We need y to be larger, so let's try answer choice D: k=18
10x + 20y = 18x + 18y
2y = 8x
y/x = 8/2
Success! x<y.

The correct answer is D.

The process would be very quick if you recognized right away that since k = one of the answer choices, you could simply plug them in for k.

GMATGuruNY
From your explanation it seems that y=x in choice C. Next step, you choose choice D (where the value is more than 15). So, how did you confirm that the value which is more than 15 will make the condition legit (x<y)?
I'll appreciate your help.
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Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)  [#permalink]

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New post 27 May 2019, 14:07
EvaJager wrote:
ausadj18 wrote:
Can someone show me the shortcut to solving number 148 from the OG:

If x,y and k are positive numbers such that (x/x+y)(10)+(y/x+y)(20)=k and x <y, which of following could be value for k?

a) 10
b) 12
c)15
d)18
e) 30


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

\(\frac{10y}{x+y}>\frac{10y}{2y}=5\) and \(\,\,\frac{10y}{x+y}<\frac{10y}{y}=10\), therefore \(10+5<k<10+10\) or \(15<k<20\).

Answer D.

Hi Bunuel, IanStewart
Could you help me to comprehend the highlighted part?
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Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)  [#permalink]

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New post 27 May 2019, 21:29
Asad wrote:
Bunuel wrote:
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 \(18\).


Bunuel
Which values confirm the highlighted part?
Also, could you help me to figure out the highlighted part without putting the value?
Thanks__


Check here: https://gmatclub.com/forum/if-x-y-and-k ... 31-20.html
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Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)  [#permalink]

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New post 01 Sep 2019, 09:08
BANON wrote:
If x, y, and k are positive numbers such that \(\frac{x}{x + y}*10 + \frac{y}{x + y}*20 = k\) and if x < y, which of the following could be the value of k?

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


Given:
1. x, y, and k are positive numbers such that \(\frac{x}{x + y}*10 + \frac{y}{x + y}*20 = k\)
2. x < y

Asked: Which of the following could be the value of k?

\(\frac{x}{x + y}*10 + \frac{y}{x + y}*20 = k\)
\(\frac{x+y}{x + y}*10 + \frac{y}{x + y}*10 = k\)
\(10 + \frac{y}{x + y}*10 = k\)
SInce y>x
\(y> \frac{(x+y)}{2}\)
\(\frac{y}{x+y}>\frac{1}{2}\)
\(k > 10 + \frac{1}{2}*10\)
\(k > 15\)
\(10 + \frac{y}{x + y}*10 = k\)
Since \(\frac{y}{x+y}< 1\)
\(k< 10 + 1*10\)
k<20
15<k<20
Only option D 18 satisfies these conditions

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

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New post 01 Sep 2019, 09:11
Asad wrote:
EvaJager wrote:
ausadj18 wrote:
Can someone show me the shortcut to solving number 148 from the OG:

If x,y and k are positive numbers such that (x/x+y)(10)+(y/x+y)(20)=k and x <y, which of following could be value for k?

a) 10
b) 12
c)15
d)18
e) 30


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

\(\frac{10y}{x+y}>\frac{10y}{2y}=5\) and \(\,\,\frac{10y}{x+y}<\frac{10y}{y}=10\), therefore \(10+5<k<10+10\) or \(15<k<20\).

Answer D.

Hi Bunuel, IanStewart
Could you help me to comprehend the highlighted part?


Asad
Please see my solution for explanation of the highlighted part in Bunuel's solution.
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Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)  [#permalink]

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New post 01 Sep 2019, 21:58
10x/ (X+Y) + 20Y/(X+Y) = K

10*{ (X+2Y)/ (X+Y) } = K

(X+2Y) / (X+Y) = K/10 and we know that X<Y, so here we plug in values.

(X+2Y) / (X+Y) = 10/10......Can't be true as Y will have to be 0 for this.
(X+2Y)/ (X+Y) = 12/10.... Gives us 2 equations. X+ 2Y = 12 and X+Y = 10...solving this we get X= 8 and Y =2, so incorrect.
(X+2Y) / (X+Y) = 15/10...similar to above, we will get two equations and by solving them, we get X= 5 and Y= 5, so incorrect.
(X+2Y) / (X+Y) = 18/10.......X+ 2Y= 18 and X+Y = 10........solving them we get, X= 2 and Y = 8...This is the correct solution.
(X+2Y) / (X+Y) = 30/10......we get -ve value of X. Hence the answer is (D)
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If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)  [#permalink]

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New post 09 Sep 2019, 10:52
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 \(1\)



Hey Bunuel


Can you please calrify this Query? I would like an answer for this as well.
Thanks in advance.
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Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)  [#permalink]

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New post 09 Sep 2019, 10:57
Inten21 wrote:
uvee wrote:
Bunuel wrote:
\(10*\frac{x}{x+y}+20*\frac{y}{x+y}=k\)

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

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

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 \(1\)



Hey Bunuel


Can you please calrify this Query? I would like an answer for this as well.
Thanks in advance.


Check here: https://gmatclub.com/forum/if-x-y-and-k ... l#p1873381
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Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)  [#permalink]

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New post 16 Nov 2019, 00:24
I encountered this in a GMATFocus test.

Here's the GMATFocus solution.
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Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)   [#permalink] 16 Nov 2019, 00:24

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