GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 31 Mar 2020, 02:35 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)

Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 62378
Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)  [#permalink]

### Show Tags

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.
_________________
Intern  B
Joined: 26 Aug 2017
Posts: 3
Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)  [#permalink]

### Show Tags

Could you guys please explain if the answer holds true when x<0 and y>0 ? Thanks !
EMPOWERgmat Instructor V
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 16322
Location: United States (CA)
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]

### Show Tags

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,
Rich
_________________
Kaplan GMAT Instructor Joined: 21 Jun 2010
Posts: 64
Location: Toronto
Re: If x, y, and k are positive numbers such that 10*x/(x+y)+ 20  [#permalink]

### Show Tags

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.
Director  G
Joined: 04 Aug 2010
Posts: 547
Schools: Dartmouth College
Re: If x, y, and k are positive numbers such that 10*x/(x+y)+ 20  [#permalink]

### Show Tags

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.
_________________
GMAT and GRE Tutor
New York, NY

Available for tutoring in NYC and long-distance.
SVP  V
Joined: 03 Jun 2019
Posts: 2438
Location: India
GMAT 1: 690 Q50 V34 WE: Engineering (Transportation)
Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)  [#permalink]

### Show Tags

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
Intern  S
Joined: 11 Jun 2019
Posts: 34
Location: India
Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)  [#permalink]

### Show Tags

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)
VP  D
Joined: 14 Feb 2017
Posts: 1377
Location: Australia
Concentration: Technology, Strategy
Schools: LBS '22 (I)
GMAT 1: 560 Q41 V26
GMAT 2: 550 Q43 V23 GMAT 3: 650 Q47 V33 GMAT 4: 650 Q44 V36 GMAT 5: 650 Q48 V31
GMAT 6: 600 Q38 V35
GMAT 7: 710 Q47 V41 GPA: 3
WE: Management Consulting (Consulting)
Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)  [#permalink]

### Show Tags

I encountered this in a GMATFocus test.

Here's the GMATFocus solution.
Attachments Capture.JPG [ 219.02 KiB | Viewed 127 times ]

_________________
Here's how I went from 430 to 710, and how you can do it yourself: Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)   [#permalink] 15 Nov 2019, 23:24

Go to page   Previous    1   2   [ 28 posts ]

Display posts from previous: Sort by

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