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Math Expert V
Joined: 02 Sep 2009
Posts: 59020
Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)  [#permalink]

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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 $$18$$.

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.

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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Could you guys please explain if the answer holds true when x<0 and y>0 ? Thanks !
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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)  [#permalink]

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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.

GMAT assassins aren't born, they're made,
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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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1
1
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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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.

(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 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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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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VP  V
Joined: 23 Feb 2015
Posts: 1297
If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)  [#permalink]

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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.

(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 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)?
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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EvaJager 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 $$15<k<20$$.

Hi Bunuel, IanStewart
Could you help me to comprehend the highlighted part?
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Math Expert V
Joined: 02 Sep 2009
Posts: 59020
Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)  [#permalink]

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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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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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EvaJager 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 $$15<k<20$$.

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

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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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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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 $$18$$.

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.

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.
Math Expert V
Joined: 02 Sep 2009
Posts: 59020
Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)  [#permalink]

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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 $$18$$.

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.

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.

Check here: https://gmatclub.com/forum/if-x-y-and-k ... l#p1873381
_________________ Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)   [#permalink] 09 Sep 2019, 10:57

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